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OF THE

UNIVERSITY OF CALIFORNIA.

Class

STRUCTURAL ENGINEERING

BOOKS FOR STUDENTS

THE TESTING OF MATERIALS OF CON-
STRUCTION. A Text-Book for the Engineering
Laboratory and a Collection of the Results of Experi-
ment. By W. CAWTHORNE UNWIN, F.R.S., LL.D.
With 5 Plates and 206 Diagrams. 8vo, i8j. net.

ENGINEERING CONSTRUCTION IN IRON,
STEEL, AND TIMBER. By WILLIAM HENRY
WARREN. With 12 Folding Plates and 443 Diagrams.
Medium 8vo, 185. net.

APPLIED MECHANICS: Embracing Strength
and Elasticity of Materials, Theory and Design of
Structures, Theory of Machines and Hydraulics. A Text-
Book for Engineering Students. By DAVID ALLAN
Low (Whitworth Scholar), M.I.Mech.E., Professor of
Engineering, East London College (University of
London). With 850 Illustrations and 780 Exercises.
8vo, 7s. 6d. net.

ELEMENTARY APPLIED MECHANICS. By
ARTHUR MORLEY, M.Sc., M.I.Mech.E., Professor of
Mechanical Engineering in University College, Notting-
ham, and WILLIAM INCHLEY, B.Sc., Lecturer and
Demonstrator in Engineering in University College,
Nottingham. With 285 Diagrams and numerous
Examples. Crown 8vo, y. net.

MECHANICS FOR ENGINEERS: A Text-
Book of Intermediate Standard. By ARTHUR MORLEY,
M. Sc. , M. I. Mech. E. With 200 Diagrams and numerous
Examples. Crown 8vo, 4^. net.

STRENGTH OF MATERIALS. By ARTHUR
MORLEY, M.Sc., M.I.Mech.E. With 248 Diagrams
and numerous Examples. 8vo, js. 6d. net.

ELEMENTS OF MECHANICS. With Nu-
merous Examples, for the use of Schools and Colleges.
By GEORGE W. PARKER, M.A., of Trinity College,
Dublin. With 116 Diagrams and Answers to Examples.
8vo, 4*. 6d.

MECHANICS APPLIED TO ENGINEER-
ING. By JOHN GOODMAN, Wh.Sch., A.M.I.C.E.,
M.I.M.E., Professor of Engineering in the Leeds
University. With 714 Illustrations and numerous
Examples. Crown 8vo, 9^. net.

LONGMANS, GREEN, AND CO.

LONDON, NEW YORK, BOMBAY, AND CALCUTTA

STRUCTURAL ENGINEERING

JOSEPH HUSBAND

ASSOCIATE MEMBER AND WATT MEDALLIST OF THE INSTITUTION OF CIVIL ENGINEERS, ASSOCIATE

AND MEDALLIST OF THE ROYAL COLLEGE OF SCIENCE FOR IRELAND, MEMBER

OF THE FRENCH SOCIETY OF CIVIL ENGINEERS, HEAD OF

THE CIVIL ENGINEERING DEPARTMENT OF

THE UNIVERSITY OF SHEFFIELD,

AND

WILLIAM HARBY

SOMETIME ASSISTANT IN THE CIVIL ENGINEERING DEPARTMENT OF THE
UNIVERSITY OF SHEFFIELD

WITH 337 DIAGRAMS

LONGMANS, GREEN, AND CO.

39, PATERNOSTER ROW, LONDON

NEW YORK, BOMBAY, AND CALCUTTA

1911

PEEFACE.

DURING the last few years many excellent treatises have been published
on Structural Engineering, the majority of which are principally
devoted to a consideration of the subject from the purely theoretical
point of view, whilst a few others treat the subject mainly with regard
to the practical arrangement of structural details. The subject is one
of so wide a range that it is manifestly impossible to attempt anything
like a comprehensive survey of structures generally in a single volume.
Between the theoretical computation of the loads and stresses in a
structure and the evolution of a satisfactory practical design which
shall have due regard to the exigencies of practical construction, there
exists a gap which may only be bridged by considerable practical
experience and knowledge of shop methods.

The Authors have endeavoured in the present volume to deal with
the design of the more ordinary and commonly occurring structures
from both points of view, and whilst necessarily stating the main outlines
of theory involved, have extended the application of such theory to the
practical design of a considerable variety of structures and structural
details of everyday occurrence. The consideration of higher structures,
such as rigid and two-hinged metal arches, suspension bridges, and
structures of particular or unique character, has been intentionally
omitted in order to make room for such examples as will be of interest
to the majority of readers. It has been considered desirable to include
a short summary of the properties of structural materials and weights
of details in order that these may be readily available for reference in
one volume, and the chapter on materials is not intended to supply
other than a very brief compendium of the properties of materials. An
extended acquaintance with these is very necessary to the structural
designer, and such information in detail is readily accessible in treatises
on materials.

Wherever possible, numerical data and arithmetical, in preference
to analytical, methods have been adopted, and the use of mathematical
formulas has been avoided where not absolutely necessary. Although
this treatment may occasionally result in slightly more protracted
methods of calculation, the Authors are convinced, after many years of
both practical and teaching experience, that it will render the subject
matter more accessible to the greatest number of readers. The majority
of practical designers have neither the time nor opportunity for
acquiring an advanced knowledge of mathematics, and whilst not
decrying its desirability, an acquaintance with higher mathematics is
not necessary to the design of most ordinary structures. A thorough

a 2

235505

vi PREFACE

understanding of the elastic beam theory is necessary, and this, it is
hoped, has been stated in the fullest and simplest manner. The
consideration of column strength has been treated a little more fully
than is customary in works on Structural Engineering, and the Authors
are particularly indebted to Mr. J. M. Moncrieff, M.Inst.C.E., for
permission to make free use of the matter contained in his exhaustive
investigation on this subject. Points relative to the design of bridges
have been frequently alluded to and several bridge details illustrated ;
but as bridge design constitutes an extensive subject in itself, those
portions relating to it in the present volume are only intended to be
introductory. A brief section is devoted to tall building construction,
and for more detailed information readers may be referred to the many
excellent American treatises on this subject. It had been intended to
include a short notice of simple reinforced concrete structures, but
as these are already fully dealt with in specialized works and the
subject is rapidly attaining such large proportions, such notice has
been omitted through lack of space for adequate treatment. Masonry
structures and types of engineering masonry, which are inseparably
associated, have been given particular attention and several applied
designs carefully worked out.

It is hoped the information contained in the work may prove of
general utility to both draughtsmen and students, and although not
exclusively compiled for the use of students, it will be found practically
to cover the present syllabus of both the Ordinary and Honours Grade
Examinations in Structural Engineering held by the City and Guilds
of London Institute, whilst it will also be of valuable assistance to
those preparing for the Associate Membership Examination of the
Institution of Civil Engineers, and the B.Eng. degree of London
University.

The Authors desire to acknowledge their indebtedness to Messrs.
R. A. Skelton & Co. and Messrs. Mellowes & Co., Ltd., for information
kindly supplied respecting broad-flanged beams and glazing ; to Messrs.
J. M. Moncrieff and E.^Sandeman, MM.Inst.C.E., Mr. E. H. Stone,
M.Am.Soc.C.E. ; to the Engineering Standards Committee ; and to the
Minutes of Proceedings of the Inst. C.E., the Transactions of the
American Soc. O.E., the Memoir es of the French Soc. C.E., The Engineer,
Engineering^ Annales des Fonts et Chaussees, La Revue Technique, and
other journals, frequent references to which have been annotated.

J. H.
W. H.

SHEFFIELD,
August, 1911.

CONTENTS.

CHAPTER I

MATERIALS

Properties of Stone, Brick, Limes, and Cement — Natural and Artificial
Cements — Portland Cement — Manufacture — Composition — Influence
of Fine Grinding — Weight, Strength, and Elasticity — Tests — Aeration
of Cement — Aerating Sheds — Concrete — Aggregates — Method of
calculating bulk of Ingredients — Gauging and Mixing Plant — Concrete
in Mass — Concrete Blocks — Asphalte — Timber — Properties and
Strength — Seasoning — Preservation of Timber — Properties and
Strength of Cast Iron, Wrought Iron, and Steel

CHAPTER II

LOADS AND WORKING STRESSES

General Considerations regarding Dead and Live Loads — Weight of
Various Materials — Detailed Weights of Component Parts of Bridges
— Dead Load on Roofs — Weight of Principals, Coverings, and Minor
Details — Dead Load on Floors — Live Load on Floors and Bridges —
Weight of Locomotives — Equivalent Distributed Load on Main
Girders for Various Spans — Live Load on Cross-girders, Troughing,
and Rail-bearers — Live Load on Highway Bridges — Wind Load —
Intensity of Wind Pressure — Working Stresses — General Considera-
tions— Factor of Safety — Old Method of deciding Factor of Safety —
Wohler's Experiments — Launhardt — Weyrauch, Fidler's and Stone's
Formulae for determining Working Stresses — Applications .... 25

CHAPTER III

BENDING MOMENT AND SHEARING FORCE

Bending Moment— Shearing Force — Relation between Bending Moment
and Shearing Force — Diagrams of Bending Moment and Shearing
Force for simply supported Beams and Cantilevers under various
Conditions of Loading — Balanced and Unbalanced Cantilevers —
Bending Moment and Shearing Force due to Rolling Loads — Applica-
tion to Traction Engine, Locomotive, and Locomotive followed by
Uniform Rolling Load — Determination of Maximum Moments and
Equivalent Distributed Load — Bending Moment and Shearing Force
on Continuous and Fixed Beams — Characteristic Points — Determina-
tion of Pressures on Supports . . . . . * . . 50

viii CONTENTS

CHAPTER IV

BEAMS

PAGE

Moment of Resistance— Relation between Bending Moment and Moment
of Resistance — Position of Neutral Axis— Moment of Inertia— Modulus
of Section — Numerical Examples — Shear — Intensities of Shear on
Vertical and Horizontal Planes — Shear Intensity Diagrams — Applica-
tion to calculation of Rivet Pitch in Girders— General Considerations
of Design of Beams — Maximum permissible Deflection— Practical
Ratio of Depth to Span — Uses of Section Books — Details of Connec-
tions— Numerical Examples— Riveted Joints — Bracket Connections
— Detailed Design for Warehouse Floor 89

CHAPTER V

COLUMNS AND STRUTS

Relation between Ideal and Practical Columns — Method of Application of
Load— Fixation of Ends— Radius of Gyration— Calculation of—
Tabulation of Types of Columns and their Least Radius of Gyration
Euler's, Rankine's, and Straight-Line Formulae — Comparison of—
MoncriefFs Formula — Diagrams of Safe Working Stresses— Eccentric
Loading — Practical Considerations in selecting Type of Column —
Practical Details of Connections, Caps, and Bases — Column Founda-
tions— Design of Grillage Foundation— Practical Application of
Formulae to Design of Columns and Struts — Laterally Loaded Columns
—Foundation Blocks and Bolts— Live Loadtm Columns— Rivet Pitch
in Built-up Columns 132

CHAPTER VI

PLATE GIRDERS

Economic limiting Span and Depth — Sectional Area of Flanges — Thick-
ness of Web Plate — Stiffeners — Pitch of Rivets — Lateral Bracing-
Detailed Design of Crane Girders— Detailed Design of Bridge Girders 187

CHAPTER VII

LATTICE GIRDERS

Types and Spans for which Suitable — Stresses in Parallel Girders under
Symmetrical and Unsymmetrical Loading — Stresses in Girders of
Varying Depth — Maximum Stresses due to Rolling Loads — Detailed
Design of Lattice Crane Girders — Practical Detailed Construction of
Lattice Girders — Cross-Section of Booms — Main Bridge Girders —
End Posts — Ties — Struts — Expansion and Pin Bearings — Bowstring
Girders — Pin-connected Trusses— Parapet and Footbridge Girders . 213

CHAPTER VIII

DEFLECTION

Deflection — Deflection produced by one or more Loads on a Braced Girder
— Examples of Deflection of Parallel and Cantilever Girders — Hori-
zontal Displacement due to Deflection — Beams of Constant Cross-
Section — Deflection of Beams of Variable Cross-Section — Plate Girders 253

CONTENTS ix

CHAPTER IX

ROOFS

PAGE

Types — Proportions — Spacing of Principals — Loads — Reactions — Stress
Diagrams— Stresses by Method of Sections — Stresses in Three-Hinged
Arch Roof — Ends of Roofs — Wind Bracing — Wind Screen — Detailed
Design of Members — Numerical Examples — Detailed Design for Roof
Truss— Details of Covering 261

CHAPTER X

MISCELLANEOUS APPLICATIONS AND TALL BUILDINGS

Detailed Design for Lattice Roof Girder— Detailed Design for Crane Jib-
Detailed Design for Riveted Steel Tank— Tall Buildings— Types-
Live Loads on Floors and Roofs — Details of Construction — Floors —
Partitions — Roofs — Columns — Wind Bracing — Various Systems —
Stresses in Wind Bracing — Anchorage of Columns subject to Uplift . 296

CHAPTER XI

MASONRY AND MASONRY STRUCTURES

Masonry — Different Classes of Masonry employed for Engineering
Structures — Specifications — Rubble Concrete — Bond in Arched
Masonry — Masonry Piers — Footings— Pressure on Foundations —
Distribution of Pressure on Foundations and Joints — Retaining Walls
— Earth Pressure — Rankine's Formula — Rebhann's Method — Example
of Design — Surcharged Retaining Walls — Types of Walls — Design of
Panelled Retaining Walls — Methods of relieving Pressure against
Walls — Walls of Variable Section — Masonry Dams — Pressure of
Water — Lines of Resistance— Theoretical Profile and Necessity for
Inside Batter — Design of Section of Dam — Trapezium Law — High
and Low Dams — Overflow Dam — Actual Maximum Compressive
Stress by Various Methods — Profiles of Existing Dams — Shear Stress
in Dams — Effect of Foundation Bond on Shear Stress Distribution —
Present State of Opinion on Design of Dams — Masonry and Concrete
Arches — Disposition of Load — Luxembourg Arch — Equivalent Load
of Uniform Density — Line of Resultant Pressures — Three-Hinged
Masonry and Concrete Arches — Hinges for Masonry and Concrete
Arches — Line of Least Resistance in Rigid Masonry Arch — Design
for Three Hinged Concrete Arched Bridge — Abutments — Skew
Arches — Exact and Approximate Methods of Determining Joints in
Skew Arches — Tall Chimneys — Wind Pressure on — Design of Tall
Chimney . . 318

INDEX 391

LIST OF TABLES

TABLE PAGE

1. Properties of Stones . 3

2. Influence of Fineness of Grinding of Cement on the Strength of

Mortar 6

3. Influence of Fine Grinding on Weight of Cement 6

4. Tensile Strength of Cement 7

5. Elasticity of Cement and Cement Mortar 8

6. Percentage Voids in Aggregates 13

7. Weight and Strength of Timber 16

8. Analyses of Iron and Steel 19

9. Physical Properties of Iron and Steel 19

10. Tensile Strength of Wrought Iron 21

11. Bending Tests for Wrought Iron 21

12. Dimensions of Corrugated Sheets 22

13. Tensile Strength of Steel 23

14. Weight of Various Materials 26

15. Weight of Troughing and Floor Plating for Bridges 28

16. Weight of Rolled Steel Section Bars 31

17. Weight of Bolts, Nuts, and Rivet Heads 32

18. Sizes and Weights of Slates 33

19. Weight of " Eclipse " Glazing Bars . . . 33

20. Weight of Corrugated Sheeting 33

21. Weight of Asphalte on Floors and Roofs 34

22. Equivalent Distributed Loads on Railway Bridges 35

23. Traffic Loads on Highway Bridges 37

24. Working Stresses for Mild Steel by " Range " Formula 46

25. Moments of Inertia and Moduli of Sections 102

26. Ratios of Depth to Span of Mild Steel Beams 112

27. Deflection Formulae for Beams of Constant Cross -Section .... 259

28. Roof Slopes 263

29. Weights of Hollow Tile Floor Arches 312

30. Safe Pressure on Foundations and Masonry 330

31. Natural Slope and Weight of Earth 336

32. Coefficients of Effective Wind Pressure on Solids of Various Shapes . 386

CHAPTEE I.

MATERIALS.

Stone. — The principal engineering structures for which stone is
employed are masonry dams, piers, abutments and wing-walls for
bridges, arches and retaining walls. The essential characteristics of
stone for such works are durability, weight, and resistance to com-
pression. In selecting suitable stone, the following general properties
should receive consideration.

Durability. — Durability depends principally upon hardness, chemical
composition and relative fineness of grain, and is very essential in
masonry structures, since they are exposed to more or less severe
atmospheric influences, or are permanently or intermittently submerged
in water.

Weight. — The stability of structures such as dams, retaining walls,
and lofty piers, is dependent on the weight of the material used, whilst
in the case of ordinary walls and arches weight is not of such para-
mount importance.

Resistance to Compression. — All masonry structures being subject to
compressive stress, it is necessary that the material should possess an
adequate resistance to compression, especially in the case of massive
and lofty structures. Masonry is not employed under conditions
where any considerable tensile stress will be developed. Hardness
frequently decides the adoption or rejection of a stone which might
otherwise be employed, on account of excessive expense in working.

Porosity. — Porosity is undesirable, since it contributes to rapid
weathering and reduced resistance to compression. Very porous stones
are especially liable to be affected by frost.

The following aids to judgment with regard to the above properties
will be found useful. The best method of ascertaining the durability
of a proposed stone is by an inspection of existing structures built of
the same stone. Failing this, a careful examination should be made
of the weather-resisting qualities of an exposed face of the quarry
which has been undisturbed for a considerable length of time. The
weight and resistance to compression are obtained by applying suitable
tests, and figures relating to these will be found below. In all
important works, special tests should be made of the actual stone
employed. A suitable stone should exhibit on a clean fracture, a dense
and finely grained structure free from earthy matter. The porosity
of stone may be tested by immersing a dry sample in water and noting
the volume of water absorbed, at the same time observing the effect on
the clearness of the water, since this affords an indication of the

1 B

a c , ^ f STRUCTURAL ENGINEERING

amount of soft earthy matter contained in the stone. The principal
classes of stone employed in structural work are Granite, Limestone
and Sandstone.

Granite. — Granite is the most durable stone employed in con-
struction, but its hardness and expense of working restrict its use to
very exposed structures and those subject to the heaviest stresses.
Ordinary granite is composed of quartz, felspar and mica. Quartz is
silicon oxide, and practically indestructible. Felspar is a mixture of
silicates of alumina, soda, potash, or lime. It is much less durable
than quartz. Its colour varying from grey to red gives the distinctive
colour to the different varieties of granite. Mica is formed of sili-
cates of alumina and other earths crystallized in thin laminae which
readily split on weathering. It is the least durable constituent of the
granite. Syenitic granites contain hornblende in addition to the above
constituents. Hornblende is a silicate of lime and magnesia, is very
hard and durable, and imparts a dark green colour to the granite.
Syenifcic granites are tougher and more compact than ordinary granite,
but their scarcity makes the cost too great for general use. It is
largely employed for ornamental purposes on account of the high
polish it will take. An excess of iron in granite decreases its dura-
bility, and care should be exercised to select material without this
impurity. The iron is easily detected when the granite is exposed to
air, dark uneven patches of discoloration due to iron oxide forming
on the surface. Its weight varies from 160 to 190 pounds per cubic
foot, and the crushing load from 600 to 1200 tons per square foot. It
absorbs a very small percentage of water.

Limestone. — Limestone is very varied in composition and physical
characteristics, and includes all stones having carbonate of lime for
the principal constituent. Some varieties have a crystalline structure,
whilst others are composed entirely of shells and fossils cemented
together. Marble is the hardest and most compact limestone, but is
only used for decorative purposes. Other limestones make good build-
ing material, but care must be taken to select fine even-grained stone
containing no sand-cracks or vents. It is easily worked, and forms a
good evenly coloured surface. Its weight varies from 120 to 170
pounds per cubic foot, and the crushing resistance from 90 to 500 tons
per square foot. It is less durable than granite or sandstone, and
absorbs a relatively large percentage of water. It should always be
laid on its natural bed when building, and it is preferable to allow it to
season before using to free it from quarry sap or moisture it contains
when quarried.

Sandstone. — As its name implies, sandstone has for its chief con-
stituent sand, cemented together by various substances, and it is
largely upon the nature of the cementing material that the quality of
the stone depends, since the silica forming the sand is extremely
durable. The best cementing material is probably silicic acid, but
most sandstones have carbonate of lime, iron, alumina, or magnesia as
part of the cementing medium. In selecting sandstone, a recent fracture,
if examined through a lens, should be sharp and clean with grains of
a uniform size, well cemented together. Its weight should be at least
130 pounds per cubic foot, and it should not absorb more than 5 per

MATERIALS

cent, of its own weight of water when immersed for 24 hours. Sand-
stone is used for all the best ashlar work. The most important
variety for heavy engineering works is termed grit, from the formation
in which it occurs. It has a coarse-grained structure, is very strong,
hard, and durable, and is obtainable in very large blocks. For lighter
work freestone is used, being more easily worked.

TABLE 1. — PROPERTIES OP STONES.

Weight per
cubic foot.

Crushing weight
per square foot.

Percentage
absorption of
its own weiglt.

Granites —
Peterhead
Cornish.

Ibs.
165
162

tons.
800 to 1200

per cent.
0-25

Guernsey . .

187

950

Killiney (Dublin)
Sandstones—
Craigleith
Bramley Fall

171

140
132

690

860
240

3-6
3-7

Darley Top . ....

139

520

3-4

Grinshill Freestone . .
Bed Mansfield . . .
White Mansfield . .
Limestones —
Bath
Portland Whitbed . .

122
143
140

128
132

210
600
460

97
205

7-8
4-5
5-0

7-5
7-5

Bricks. — Bricks are manufactured from clay, sometimes mixed with
other earths, by moulding the clay to the required size and shape and
burning it in kilns. The quality of bricks depends upon the nature
and proportions of the constituents of the clay, and the heat to which
they are subjected in the kilns. A small quantity of lime, very finely
disseminated through the clay, is advantageous, as it assists vitrification
when burning ; but if in lumps, on being burnt, it is converted into
quicklime, which, on exposure to damp, sets up a slaking action with
consequent expansion and liability to split the bricks. The proportion
of iron contained in the clay influences the colour of the brick, which
may vary between yellow, red, blue, or black. Bricks should be burnt
until vitrification is just commencing. The characteristics of good
bricks are, freedom from flaws, cracks, quicklime, and salt ; they
should be of uniform colour, shape, and dimensions, and when struck
should give a clear ringing sound. The absorption of water is a good
indication of the quality of the brick. No brick should absorb more
than one-sixth its own weight of water.

A test often included in specifications is to place a saturated brick in
a temperature of 20° F., and subject it to a load of 836 Ibs. per square
inch. If there are any signs of injury, such bricks are liable to be
rapidly disintegrated by frost, and should be rejected.

There are many qualities of bricks, but the kinds most generally
employed in first-class engineering structures, are Staffordshire blue
bricks and the finer qualities of red bricks known as stock bricks.
Blue brick has a glazed surface, making it almost impervious to water,

4 STRUCTURAL ENGINEERING

is very durable, and has a high compressive resistance. It is used for
the face of works in contact with water, for piers and abutments of
bridges, retaining walls, and tunnels. Stock bricks are usually employed
for the interior portions of works. The crushing resistance of Stafford-
shire blue bricks varies between 0*9 and 2*8 tons per square inch, that
of stock bricks about 1 ton per square inch. The compressive stress
on bricks should never approach the crushing resistance, as the masonry,
of which they form part, is, as a structure, much weaker than the
individual bricks.

The weight and size of bricks vary according to the locality of
manufacture, a size frequently adopted being 9 ins. x 4J ins. x 3 ins.,
and weight about 9 Ibs.

When used in curved structures such as small arches, bricks should
always be tapered to the radius of the curve, to allow a uniform thick-
ness of mortar in the joints, and ensure an even bearing.

Specification for Brickwork. — The bricks to be laid in the bond
specified with joints not exceeding J inch in thickness. Every course
to be thoroughly flushed up with mortar, and the bricks well wetted
before laying. The work to be substantial, with neat and workman-
like appearance on the face. Heading joints to be truly vertically over
each other, and the horizontal joints perfectly straight and regular.
Any cutting to be neatly done. Arches to be turned in half-brick rings
with bond courses every five feet through the arch. No batts to be
used, except where absolutely necessary. All joints to be neatly struck
and drawn, and arch rings to be carefully kept in true curves. Special
care to be taken in laying the different rings in arched work exceeding
9 inches in thickness, so that on easing the centering, the separate rings
shall fully bear on each other throughout the whole length of the arch.
If any lower ring settles away from an upper ring, so as to cause a ring
joint to open, such joint shall not be pointed or stopped, excepting with
permission of the engineer or the clerk-of- works, and any such defective
rings are, if desired, to be entirely rebuilt. No bricks to be laid in
frosty weather, and newly executed work to be adequately covered over
at night. Circular work of less than three feet radius to be executed
with special radius bricks.

Lime. — Lime is obtained by heating limestone to redness in kilns.
This process, termed calcination, converts the limestone into quicklime,
which on being slaked with water, either by sprinkling or immersion
(the former is better), forms calcium hydrate. This substance, when
mixed with sand and water, and left to dry, changes into a solid mass
which is practically again limestone. The chemical action which takes
place is as follows : —

Limestone = CaCo3

(Carbonate of lime)

by calcining = CaO + C02->

(Quicklime) (Carbon dioxide)

on slaking, CaO -f H20 = CaH202

(Water j (Calcium hydrate)

on setting, CaH202 + C02 = OaCO, + H20->

Where the limestone is composed almost wholly of carbonate of
lime, the resulting quicklime, called fat or air-lime, will only harden

MATERIALS

in air, and unless used with a suitable sand, only the exposed por-
tions will set in reasonable time. Limestones containing clay, when
burnt form hydraulic lime, which has the property of setting more or
less rapidly under water or in damp situations. Such limes are said to
be feebly or strongly hydraulic according as the mortar of which they
form the active ingredient, sets slowly or rapidly under water.

CLASSIFICATION OF HYDEAULIC LIMES.

Class.

Per cent,
of clay.

Setting under water.

Feebly hydraulic
Ordinarily hydraulic
Strongly hydraulic .

5 to 12
15 to 20

20 to 30

Firm in 15 to 20 days. In 12 months hard as
soap. Dissolves with great difficulty.
Resists pressure of finger in 6 to 8 days. In
12 months hard as soft stone.
Firm in 20 hours. Hard in 2 to 4 days. In
6 months may be worked like a hard lime-
stone.

Artificial hydraulic lime can be made by calcining a mixture of fat
lime and clay, so as to have, as nearly as possible, the same percentage
composition as a natural hydraulic lime.

In the preparation of mortar, sand is added to the lime to prevent
excessive shrinkage and to save cost, but in no way affects the mixture
chemically. Sharp, clean, gritty sand should be used. The proportions
of lime and sand may be varied, but the following give good results
for ordinary purposes : 2*4 parts of sand to 1 part pure slaked lime,
and 1-8 parts of sand to 1 part of hydraulic lime.

Cement. — Cement is similar to the best hydraulic lime, but possesses
stronger hydraulic properties. There are two classes of cements — natural
and artificial. Natural cements are obtained by calcining naturally
occurring limestones which are found to produce cement. Roman
cement manufactured from nodules found in the London Clay is perhaps
the best known natural cement. Having no great ultimate strength, it is
unsuitable for use in heavy structures, and its use is confined to temporary
work where quick setting is of great importance. Artificial cements are
manufactured by calcining a mixture of clay with a suitable limestone.

Portland Cement. — Portland cement is the most frequently used
artificial cement. It is manufactured by mixing chalk and clay, or
limestone and clay, in such proportions that, after calcining, the mixture
forms cement having the following composition : Lime 58 per cent,
to 63 per cent., soluble silica 22 per cent., alumina 12 per cent., and
small percentages of oxide of iron and magnesia. There are two
methods of manufacture, termed the wet and dry processes. The wet
process consists of thoroughly mixing together chalk and clay in water,
allowing the mixture to dry, and burning it in a kiln at a very high
temperature. In the dry process, harder limestones and clay are used.
The limestone is crushed by machinery, and the clay roughly burnt
before mixing. They are then mixed together in the required propor-
tions, ground to a powder, and formed into bricks by adding sufficient

STRUCTURAL ENGINEERING

water to permit of moulding. These bricks are dried and burnt in
kilns, as in the former process. In both the foregoing processes the
effect of calcining the mixtures is to form clinker, which is subsequently
ground to the required degree of fineness. Thorough burning and
grinding are essential to good cement. To test the fineness of grinding,
samples are passed through a sieve having a specified number of meshes
per square inch, and the residue measured. The meshes per square inch
vary between 2500 and 32,000, and the permissible residue varies from
3 per cent, to 18 per cent, respectively. Table 2 shows the influence
of fine grinding on the tensile strength.

TABLE 2. — INFLUENCE OF FINENESS OF CEMENT ON STRENGTH
OF MORTAR.

Size of sand used

Fineness of cement used.

As received

Screened through.

2500 meshes
per sq. in.

3600

5800

Breaking stress. Lbs. per sq. in.

Screened through 2500 meshes per
sc[. in. ...

122-22

133-70

203-25

240-36

Extreme fine grinding of cement decreases its cohesive power, but
greatly increases its adhesive power and shortens the time of setting.
Thorough aeration before mixing, or the addition of a small percentage
of gypsum to the clinker before grinding, will reduce the rate of setting.
Gypsum should not be added where the cement is for use in sea-water,
and 1 per cent, should not, in any case, be exceeded. It is important
that the weight should be specified in connection with the degree of
fineness of grinding, and the following table indicates the relation
between weight and fineness of grinding.

TABLE 3. — INFLUENCE OF FINE GRINDING ON WEIGHT OF
CEMENT.

Weight of cement

Meshes per square
inch of sieve.

Percentage
residue.

per bushel.

per cubic foot.

per cent.

IDS.

2500

115

3600

112

5500

109

32000

The specific gravity varies between 3'1 and 3*15,

MATERIALS

Tests. — The strength of cement is ascertained by carrying out tensile
tests on briquettes made of the cement. The
briquettes are moulded to the standard form
shown in Fig. 1 by adding sufficient fresh
water to the cement to form an easily worked
paste. When set sufficiently to enable the
mould to be removed, the briquette is either
placed under water immediately or left in the
air for the first day, and then immersed until
the time for testing. The tests are generally
performed either 7 or 28 days after the cement
is gauged. The strength increases with the
age of the specimen. The 28 days' test should
show an increase of strength of 10 per cent,
to 25 per cent., as compared with the 7 days'
test. The following are the tensile strengths
of cement as required by various government
departments : —

Cross Section I"* I"
FlG. 1.

TABLE 4. — TENSILE STRENGTH OF CEMENT.

Tensile strength of
neat cement on
briquette of area of

Strength per
square inch.

Age of specimen when
tested.

2^ square inches.

India Office .

Ibs.
800

Ibs.
355

jl day in air
\6 days in water

Admiralty .

787

350

7 ,

War Office . . .

900

400

7 ,

350

155

2 ,

Trinity House . .

500

222

4 ,

700

333

The average tensile strength required in the British Standard
Specification l for Portland Cement is as follows : —

7 days from gauging, not less than 400 Ibs. per square inch

The average breaking stress of the briquettes 28 days after gauging
must show an increase on the breaking stress at 7 days after gauging
not less than : —

25% when the 7 day test is above 400 Ibs. and not above 450 Ibs.

20% „ „ „ 450 Ibs. „ ., 500 Ibs.

15% „ „ „ 500 Ibs. „ „ 550 Ibs.

10% „ „ „ 550 Ibs. „ „ 600 Ibs.

5% „ „ „ 600 Ibs.

the briquettes to remain in a damp atmosphere for the first day from
gauging, and to be afterwards immersed in fresh water until the time
for testing.

The above specification also requires that briquettes made from a

1 Reproduced from Report No. 12, revised August, 1910. By permission of
the Engineering Standards Committee.

STRUCTURAL ENGINEERING

mixture of one part of cement to three parts by weight of dry standard
sand, shall have the following average tensile strengths : —

7 days from gauging, not less than 150 Ibs. per square inch

The average breaking stress of the briquettes 28 days after gauging
must be not less than 250 Ibs. per square inch of section and the increase
in the breaking stress from 7 to 28 days must be not less than : —

20% when the 7 day test is above 150 Ibs. and not above 200 Ibs.
15% „ „ „ 200 Ibs. „ „ 300 Ibs.

10% „ „ „ 300 Ibs. „ „ 350 Ibs.

6% „ „ „ 350 Ibs.

The following table, deduced from experiments on beams of neat
cement and cement mortars, indicates the comparative elasticity of
cement and cement mortars at different ages.

TABLE 5. — ELASTICITY OF CEMENT AND CEMENT MORTAK.

Age of specimen.

1 week.

1 mouth.

3 months.

6 mouths.

Values of
E
in pounds
per sq. in.

Neat cement
Mortar 1 to 1
„ 2 to 1
„ 3 to 1

1,850,000
1,450,000
1,010,000

2,050,000
1,860,000
1,620,000
1,310,000

2,530,000
2,060,000
2,050,000
1,570,000

2,800,000
2,980,000
2,210,000
1,710,000

Le Chatelier Test. — To test the soundness of cement, the Le Chatelier
test is usually applied.1 The apparatus, illustrated in Fig. 2, consists of

a small split cylinder of spring brass, to which
are attached two indicators. *" The cylinder is
filled with cement gauged under the con-
ditions of the British Standard Specification,
and placed under water for 24 hours. It is
then removed from the water, and the dis-
tance between the ends of the indicators
accurately measured. The mould is again
immersed in cold water, which is brought to
boiling-point, and kept boiling for 6 hours.
After allowing the mould to cool, the distance
between the indicators is again measured.
The expansion should not exceed 10 milli-
metres after 24 hours' previous aeration of
the cement, nor 5 millimetres after seven
days' aeration.

It is usual to specify the time of setting,
a slow-setting cement taking between 2 and
7 hours to finally set, while a quick-setting
cement will set in 10 to 30 minutes.

Effects of Frost. — Cement, immediately
after frost, may appear considerably damaged, but will probably improve
with time, although it will not regain its original strength. Frost only

1 Reproduced from Report No. 12, revised August, 1910. British Standard
Specification for Portland Cement. By permission of the Engineering Standards
Committee.

_J_~±±TL

" FIG. 2.

MATERIALS 9

partially suspends the chemical action during the setting stage. Liabilty
to damage by frost decreases with the amount of water remaining in
the cement. In American practice, 1 Ib. of salt is dissolved in each
18 gallons of water used, if the temperature be 32° F., and 3 ounces
are added for every 3° of frost.

Aeration of Cement in Bulk. — On extensive works where concrete is
used in large quantity, special care is taken thoroughly to aerate the

10 STRUCTURAL ENGINEERING

cement before mixing. Fig. 3 is a cross-section of a suitable type of
cement aerating shed. The cement is run in at A, and spread on the
upper floors Yl in a layer 3 ins. to 6 ins. thick, where it remains for one
or two days. It is then dropped to the floors F2 by a suitable arrange-
ment of sliding grids or tilting boards, and left for a similar period.
After further exposure on the lower floors, it is collected in hoppers H,
from which it is drawn off into trucks at L as required. A fan at R
maintains a gentle current of fresh air, which enters by the ventilators
V, Y, and passes over the cement. Aeration or air-slaking proceeds more
efficiently under a slightly damp condition of the atmosphere than in
dry weather. Fig. 4 shows an enlarged detail of two of the floors with
sliding grids operated from the gangway G.

Concrete. — Concrete, one of the most important materials of con-
struction on account of its cheapness, ease of manufacture, strength and
durability, is formed by mixing together cement, sand, water, and some
other material, such as broken stones or bricks, slag, gravel, etc. The
mortar formed by the cement and sand is known as the matrix. The
proportion of sand to cement varies according to the quality of concrete
required. The sand should be sharp, clean, and free from salt. When
containing stones or lumps it should be screened, and washed if con-
taining clay or loam. The aggregate, i.e. the broken stone, etc., is
composed of the most suitable material to be found near the site. It
should be broken to pass through a 1 J or 2 inch ring. Hard, strong,
and heavy materials, with rough surfaces, make the best aggregates.

Mixing. — The mixing is done either by hand or special concrete
mixing machines, depending on the amount of concrete to be used.
When performed by hand, the materials of which the concrete is formed
are measured whilst dry, in boxes of sizes to suit the relative propor-
tions of the ingredients. They are then thoroughly mixed together on
a timber platform by turning the heap over at least twice, preferably
three times, after which water is sprinkled on the mixture, and the heap
again turned twice or thrice. Only sufficient water to make a plastic
mortar should be used, otherwise it is liable to wash away the
cement.

Fig. 5 illustrates the Gilbreth system of measuring the ingredients
for and mixing concrete, in great bulk. The sand, stone, and cement
are delivered by conveyers on to the platform A, and shovelled, as
required, into the hoppers B. These hoppers, as seen by the detail at
W, are three in number, placed side by side. The bottoms are open,
but rest almost in contact with the rim of a timber drum, D, which is
divided into three broad grooves. On the face of each hopper at C is
an adjustable hinged door with balance weight, such that when the
drum is at rest, the pressure of the door prevents any flow of material
from the hoppers. By means of power or the hand wheel at E and
chain gearing, the drum may be rotated in the direction of the arrow,
when the adjustable doors open and allow suitable volumes of the three
materials to flow over the drum into the hopper F of the mixer.
This is an inclined steel shoot, with frequent transverse rods, R, and
baffle plates, G, G, for turning over and thoroughly mixing the materials
during their descent. From a convenient supply at L, water is led
through the hose pipe H to a system of sprinklers, P, P, by means of

MATERIALS

FIG. 5.

12 STRUCTURAL ENGINEERING

which the concrete is thoroughly but not excessively watered during
mixing. The water supply is controlled by levers at K. The shoot
delivers into iron skips at T, which may be readily transferred by
cranes or aerial ropeways on to the site of the work. At S is a vertical
transverse section of the mixer showing the positions of the water-jets.
The lower end of the shoot is closed by a door, V. This arrangement
is largely employed on extensive works, such as reservoir dams, harbours,
and breakwaters, and is very efficient.

Bulk of Ingredients. — To determine the volume of the component
parts of the concrete, it is necessary (1) to decide on the proportion of
sand to cement to be used, (2) to ascertain the quantity of mortar the
above proportions will make, (3) to know as nearly as possible the per-
centage of voids in the broken stone or other aggregate to be used,
(4) to fix on a certain percentage of mortar in excess of the voids for com-
pletely isolating and surrounding the stones. Mortar equal to 10 per cent,
of the stone (including voids) may be considered a sufficient allowance for
the fourth requirement. The percentage of voids in the broken stone
may be ascertained by filling a tank of known capacity with the broken
stone, which has been previously thoroughly soaked, and measuring the
quantity of water required to fill up the tank. The bulk of mortal-
obtained from two parts of sand to one part of cement, varies between
three-quarters and five-sixths of the total bulk of the ingredients, and
four-fifths may be taken as a fair average.

Example 1. — To ascertain if the following proportions of constituents
will be a suitable mixture for a sound concrete.

1 part cement, 2 parts sand, 4 parts broken stone having 45 per
cent, of voids.

Let x = volume of cement before mixing.

Then 2x — volume of sand before mixing,

and 4# = volume of broken stone before mixing.
The volume of the mortar after mixing = |(# + 2#)

= 2-4z.

The volume of the voids = 4z x T4^ = l'8z.

The volume of mortar in excess of the voids = 2-4z — 1 Sx

= 0-6&.

The percentage excess of mortar com- ) _ 0-6
pared with the total volume of stone ] =' T X 10° = lo per Cent'

or a little more than would be required to ensure a thoroughly sound
concrete.

The volume of the broken stone (including voids) plus the volume
of the mortar in excess of the voids will be equal to the total volume of
the resulting concrete ;

i.e. 4# -f O'Qx = 100 parts of finished concrete,

from which x = 21*74 parts.

Therefore to make 100 parts of finished concrete the following
parts of constituents are required : —

MATERIALS 13

Cement = x = 21-74 parts.

Sand = 2x = 43'48 „
Broken stone = 4«c = 86-96 „

Total . . = 152-18 „

If the percentage excess of mortar be decided upon beforehand, the
calculation will be as follows : Required the ingredients to make 100
parts of finished concrete if the ratios are 1 part cement, 2^ parts sand,
and for the aggregate a shingle be used having 30 per cent, of voids,
the mortar in excess of the voids to be 12 per cent, of the aggregate.

Let x = volume of shingle required.

Then x -f- -^x =100 parts of finished concrete,
from which x = 89-3 parts.

Mortar required to fill the voids = 89'3 x T3& = 26*79 parts.
Mortar in excess of voids = 89'3 x ^ =10'71 parts.
Total mortar required is, therefore, equal to 37 '50 parts. Hence the
volume of mortar ingredients before mixing

= f X 37'5 = 46'9 parts,

and since the mortar is composed of 1 part of cement to 2| parts of
sand, the original volume of the cement required

= f X 46-9
= 13-4 parts.

Therefore the original volumes will be —

cement =13-4 parts

sand= 33-5 „
shingle = 89'3 „

The following table gives the approximate percentage of voids for
a number of suitable aggregates : —

TABLE 6. — PERCENTAGE VOIDS IN AGGREGATES.

Aggregate.

Percentage of voids.

Stone broken to 2£" gauge . . .
2" „ . . .

ir „ . . .

per cent.
37
40
42
33

Sand

Thames ballast (which contains
the necessary sand) ....

In positions where an inferior quality of concrete may be employed,
a larger proportion of sand is used in the mortar.

There are two general methods of building concrete structures :
(1) By depositing the concrete in position, in layers, immediately it

14 STRUCTURAL ENGINEERING

has been mixed, each layer being laid before the previous one has set,
so that the whole structure is one solid mass. (2) For breakwaters
and similar works where it is inconvenient to lay the concrete in mass,
blocks varying from 2 to 200 tons are made near the site, and after
setting are deposited in place by cranes. When structures are formed
in mass, special care should be taken to prevent weakness at the hori-
zontal planes between the layers. When using a slow-setting cement,
each layer should be rammed immediately after it has been deposited,
and the succeeding layer placed upon it as quickly as possible. For
quick-setting cements where the above procedure is impossible, the
upper surface of each layer, after setting, should be prepared by picking
to make it rough, washed and covered with a thin coating of neat
cement to ensure adhesion between the layers. The concrete should
never be disturbed whilst setting. In thick walls large, heavy stones,
called displacers or plums, are often used to reduce the quantity of
concrete. These plums should be placed no nearer together nor to
the face of the wall thati will allow a thickness of concrete of at least
6 inches to separate them from each other or from the face of the wall.

Asphalte. — Asphalte is used to a great extent in engineering works
for damp-proof courses, and layers in masonry and metal bridges, roads,
roofs, etc. It is a combination of bitumen and calcareous matter,
naturally or artificially combined. The natural asphaltes are usually
found as limestones saturated with 8 to 12 per cent, of bitumen. In
preparing such asphalte for use, the rock is ground to a powder, mixed
with sand or grit and heated with mineral tar. (Coal-tar should not
be used, being brittle, easily crushed, and readily softened under heat.)
The mastic, as the mixture is then called, is laid m situ as a thick
liquid. If the natural rock contains a large percentage of bitumen it
may be laid, after grinding and heating, as a powder, in which case it
must be thoroughly rammed, whilst hot, so as to form one solid mass.

The proportion of grit in the mastic will vary according to the
purpose for which the asphalte is to be used.

For roofs, lining of tanks, etc. . . 2 of grit to 18 of asphalte.
„ flooring, footways, etc. ... 2 „ 16 „

Many artificial asphaltes are manufactured, but they are generally
inferior to the natural varieties.

Timber. — The uses of timber in engineering structures may be
classified as follows : —

1. For marine works.

2. „ exposed structures other than marine works.

3. „ parts of structures under cover.

4. „ paving.

For the first three classes strength and durability are essential, and
for the last hardness is the chief quality required. All timber structures
situated in sea-water are subject to the attacks of sea-worms, which
bore into most varieties of timber, and reduce or entirely destroy its
strength. For such situations the most suitable timbers are greenheart
or jarrah, as these timbers contain an oil which renders them to some
extent immune from the attacks of sea-worms. For the second class
the timber is usually preserved from weathering by creosoting or other
means, which is unnecessary for the third class.

MATERIALS 15

Selection of Timber. — The selection of timber should be entrusted
only to a thoroughly experienced person, as the quality can only be
judged after much personal experience. For all the main members of
structures the heartwood only should be used, the outer portions, or
sapwood, being inferior in strength and durability. The annual rings
should be regular, close, and narrow. Darkness of colour is generally
a good indication of strength. When freshly cut, the timber should
have a sweet smell ; a disagreeable smell usually betokens decay. The
surface should be firm and bright when planed, and when sawn the
teeth of the saw should not be clogged. Knots should not be large,
numerous, or loose.

Classification of Timber according to Size. — Timber is converted from
the logs or balks to commercial sizes by different methods of sawing,
depending upon the uses to which the timber is to be put.

The usual sizes of timber employed in engineering construction are
the following : —

Whole or square timber 9 in. x 9 in. to 18 in. x 18 in.
Half timber . . . . 9 in. X 4J in. to 18 in. x 9 in.

Planks 11 in. to 18 in. wide by 3 in, to 6 in. thick

Deals 9 in. wide by 2 in. to 4 in. thick

Battens 4 J in. to 7 in. wide by 1 in. to 3 in. thick.

Varieties of Timber. — For engineering purposes the following are the
most generally used timbers : —

Pine, Baltic, sometimes known as red or yellow fir, consists of
alternate hard and soft rings. It has a strong resinous odour, is easily
worked, but is not nearly so durable as some of the harder timbers.
The best varieties are those in which the annual rings do not exceed
•^ in. in thickness, and contain little resinous matter. In seasoning
the maximum shrinkage is ^th part of its original width.

Sizes obtainable. — Balks 10 to 16 in. square, 18 to 45 ft. long.

Deals 2 to 5 in. thick, 9 in. wide, 18 to 50 ft. long.

Uses. — Decking, temporary work, or in positions where there is
little or no stress.

American Red or Yellow Pine, is clean, free from defects, and easily
worked, but is weaker and less durable than Baltic pine.

Sizes.— Balks 10 to 18 in. square, 16 to 50 ft. long.

Deals 2 to 5 in. thick, 9 in. wide, 16 to 50 ft. long.

Uses. — Similar to Baltic pine.

Pitch Pine, obtained from the southern states of America, is a
timber very largely employed in engineering works on account of its
strength and durability. It is very hard and heavy, and contains a
large proportion of sapwood full of resin. It is liable to cupshakes,
will not take paint, and is very hard to work.

Sizes.— Balks 10 to 18 in. square, 20 ft. to nearly 80 ft. long.

Deals 3 to 5 in. thick, 10 to 15 in. wide, 20 to 45 ft. long.

Uses. — For the heaviest timber structures, where strength and
durability are essential. The long lengths in which it is obtainable
make it very suitable for piles.

Oak, English, is the most durable of northern latitude timber. It

STRUCTURAL ENGINEERING

is very strong, hard, tough, and elastic. Contains gallic acid, which
increases its durability, but corrodes any iron penetrating the timber.

Uses. — It is too expensive for use in general engineering works,
and is only employed where extreme strength and durability are
required.

Oak, American, is similar in many respects to English, but is inferior
in strength and durability. It is sound, hard, tough, elastic, and
shrinks little in seasoning.

Sizes. — Balks 12 to 28 in. square, 25 to 40 ft. long.

Uses. — Similar to English oak.

Oreenheart is probably the strongest timber in use. It is dark
green to black in colour, and obtainable only from the northern part
of South America. It has a fine straight grain, is very hard and
heavy, has an, enormous crushing resistance, and contains an oil which
resists, to a great extent, the attacks of sea-worms. It is very apt to
split and splinter, and care must be taken when working it.

Sizes. — It is imported rough in logs 12 to 24 in. square, and up to
70 ft. in length.

Uses. — For piles, dock gates, jetties, and all marine structures.

Beech is black, brown, or white in colour, is light, hard, compact,
not hard to work, and is durable if kept constantly either wet or dry,
but if alternately wet and dry it quickly rots.

Use. — For piles and sleepers.

Elm possesses great strength and toughness. It has a close fibrous
grain, warps, and is difficult to work. It should be used when freshly
cut, and kept continually under water.

Uses. — For piles and fenders.

TABLE 7. — WEIGHT AND STRENGTH OF TIMBER.

Timber (seasoned).

Weight per
cubic foot.

Resistance to crush-
ing in direction of
grain in tons per
square inch.

Breaking load at
centre for square
beam 1" x 1" X 1
ft. span.

Pine Dantzi0"

Ibs.
36
34
30
42 to 48
48 to 60
54
60 to 70
43 to 53
36
41 to 55
60 to 63

3-1
2-1
1-8
3-0
2-9
3-1
5-8
3-4
2-6
2-3
3-2

4-5

6-8
4-2
4-6
5-4

Ibs.
400 to 700
390 to 570
470
350 to 500
500 750
500 700
900 1500
550 700
350 450
600 700
500 650

„ American Red ....
Yellow . . .
„ Pitch . . .

Oak, English
,, American .

Beech ....

Elm

Teak . .

Jarrah

Teak has a dark brown colour, is light, but very strong and stiff.
It contains a resinous oil, which makes it very durable, and able to
resist the attacks of ants. It is easily worked, but splinters.

Sizes.— Balks 12 to 15 in. square, 23 to 40 ft. long.

Uses. — Used to a great extent in shipbuilding for backing armour,

MATERIALS 17

as the oil it contains does not corrode iron. It is also used for decking
and structures liable to the attacks of ants.

Jarrah. — A red Australian timber, with a close wavy grain. It is
very durable, and contains an acid which partially resists sea-worms
and ants. It is full of resin, brittle, liable to cupshakes, and shrinks
and warps in the sun.

Sizes. — Balks 11 to 24 in. square, 20 to 40 ft. long.

Uses. — Similar to greenheart. It is also used to a large extent for
wood pavement.

Where two values are given in the above table for the crushing
resistance, the first is for timber moderately dry, and the second for
thoroughly seasoned timber. Timber when wet does not possess nearly
the same strength as when thoroughly dried.

Shearing Strength. — The shearing resistance of timber is very
variable, and few reliable experiments have been made to ascertain such
resistance. Rankine says the shearing strength along the grain is
practically the same as the tensile strength across the grain. The
following are approximate shearing values along the grain : —

Oak, elm, ash, birch . . . . . over 1000 Ibs. per sq. in.

Sycamore, Cuban pine „ 600 „ „

Norway pine, white pine, spruce . „ 400 „ „
English oak trenails, across the grain „ 4000 „ „

Decay of Timber. — Dry rot is caused by the confinement of gases,
produced by warmth and stagnant air, around timber. The fungus
thus produced feeds upon the wood, and reduces it to a powder. The
rot will spread to any wood in the vicinity. Unseasoned timber is
more liable to dry rot than seasoned. Thorough seasoning, ventilation
and protection from dampness are the best precautions against dry rot.
Wet rot is the decomposition of wood by chemical action through being
kept in a wet state. It is not infectious excepting through actual
contact. Thorough seasoning and preserving by painting, creosoting,
etc., will prevent wet rot.

Destruction of Timber. — All marine timber structures are liable to be
destroyed by sea-worms, chief amongst which is the Teredo. By
employing greenheart or jarrah the action of the sea-worms is to some
extent prevented. Other precautions adopted are, covering, after
tarring, of all timber below high-water level, with sheet zinc or copper,
or studding the whole surface with scupper nails. Where timber is
liable to be attacked by white ants, teak, jarrah, or other ant-resisting
timber should be selected for use.

Seasoning. — The object of seasoning is to expel any sap there may
be remaining in the timber. Natural seasoning is performed by
stacking balks in layers, under cover, and allowing a free circulation of
air to pass around each balk. This process requires a considerable
time, taking between three and twenty-six months, according to the
description of the timber and the sizes under treatment. Water
seasoning reduces the time occupied. By this method the timber is
placed under water, preferably in a stream, for a fortnight, after which
it is stacked under cover and allowed to thoroughly dry. Whilst under

18 STRUCTURAL ENGINEERING

water the sap is diluted and carried away, thus reducing the time
required to season when stacked.

Dessicating is seasoning the timber by enclosing it in a chamber and
circulating hot air around it. This method reduces the time required
for seasoning, but care must be taken that the heat is not sufficient to
cause the timber to split. This method of seasoning is unsuitable for
large balks.

Seasoning reduces the weight of timber by 20 per cent, to 33
per cent.

Preservation of Timber. — The most effective means of preserving
timber is, after thoroughly seasoning, to fill up the pores so as to
exclude all moisture from penetrating below the surface. Many
processes for accomplishing this have been adopted, but the most
successful is that of creosoting.

Creosoting. — By this process the timber is thoroughly impregnated
with dead oil of coal tar. Two methods of accomplishing this are in
use. The older method consists of placing the timber, after seasoning,
in an airtight cylinder, exhausting the air and then admitting the
creosoting oil, at a temperature of 120° F. and 170 Ibs. per square
inch pressure. After maintaining the pressure for a short time, the
oil is removed from the cylinder and the timber allowed to dry. By
this treatment up to 20 Ibs. of creosote per cubic foot of timber can
be injected.

A second method, known as the Rueping process, has latterly been
extensively adopted in America. After placing the seasoned timber in
the cylinder, the air pressure is increased to 75 Ibs. per square inch,
and whilst at that pressure the cylinder is filled with creosote. The
pressure is then gradually increased to 150 Ibs. per square inch, and
allowed to remain at such pressure for 15 minutes, after which the
pressure is reduced to that of the atmosphere and the creosote drained
from the cylinder. The air in the cells of the timber, having been
compressed to 150 Ibs. per square inch, expands and forces any surplus
creosote from the cells. To remove any oil from around the outside of
the pores a vacuum of 22 inches is created. The total operation
requires about 4J^ hours. By this treatment about 5 Ibs. of creosote
per cubic foot of timber is sufficient to preserve the timber against
decay in ordinary situations.

The amount of creosote required per cubic foot of timber will vary
according to the nature of the timber and the proportion of sapwood
in the piece.

Boucheries Process consists of forcing copper sulphate along the
fibres until the timber is thoroughly impregnated, and is very successful
in preventing dry rot ; but if the timber be exposed to the action of
water the salt is dissolved and carried away. It has not the power to
repel the attacks of white ants or sea-worms.

Kyan's Process. — Bichloride of mercury is, by this process, injected
into the pores of the timber. To some extent it prevents destruction
by ants and sea-worms, and is effective against dry rot, but is now
seldom employed.

Metals. — The metals of greatest use in construction are cast iron,
wrought iron, and steel. All being derived from iron ore, iron forms

MATERIALS

the chief constituent of each. The iron in the ore is in chemical
combination with other materials, which are partially removed by
smelting. The iron after smelting is known as pig iron. Cast iron is
manufactured directly from suitable pig iron by remelting and running
the molten metal into moulds. Wrought iron is made from pig iron
by processes termed refining, puddling, and rolling. There are a
number of methods of converting iron into steel, the chief being the
Bessemer, Siemens, and Siemens-Martin processes. The description
of the processes of manufacture cannot be entered into here, but may
be found in works on metallurgy.

The chief difference in the chemical composition of the three metals
is the proportion of carbon contained by each. The following table
gives typical percentages of carbon and other impurities in the metals.

TABLE 8. — ANALYSES OF IRON AND STEEL.

Cast iron.

Wrought iron.

Steel.

Carbon

Fer cent.
2-0 to 6-0

Per cent.
0 to 0-25

Per cent.
0'15 to 1-8

Silicon

0-2 to 2-0

0-032

0-015

Phosphorus
Manganese
Sulphur

0-038
0-013
0-014

0-004
trace
0-114

0-041
0-683
0-035

The carbon in cast iron exists partly in the state of a mechanical
mixture, when it is visible as black specks in the mass and gives the
iron a dark grey colour, and partly as an element in the chemical
compound. Free carbon makes the metal softer and more adaptable
for casting. When chemically combined, carbon imparts strength and
brittleness to the iron.

The chief differences of physical properties are shown in the
following table.

TABLE 9. — PHYSICAL PROPERTIES OF IRON AND STEEL.

Cast iron.

Wrought iron.

Steel.

Hard.

Soft.

Medium to hard.

Brittle.

Fusible.

Fusible when con-

taining a high per-

centage of carbon.

Malleable.

Ductile.

Forgeable.
Tenacious.

Highly elastic.
Temperable.

Weldable.

Cast Iron. — The use of cast iron for constructional purposes has
diminished to a very great extent in recent years owing to the improved

20 STRUCTURAL ENGINEERING

methods of manufacturing wrought iron and steel. Its use is now
restricted to columns, bed plates, cylinders, and similar members that
are subject to purely compressive stress.
Ultimate strength of cast iron —

Tension . . . . 7 to 11 tons per square inch.
Compression . . . 35 to 60 „ „

Shear . « . . . 8 to 13

The transverse strength is often specified in preference to the com-
pressive and tensional strengths. Test pieces, cast on the main casting
and afterwards removed, are tested by supporting them as beams and
applying central loads. A usual specification for such tests is as
follows : Test pieces to be 2 inches deep, 1 inch wide and 3 feet
G inches long, and placed on supports 3 feet apart. The central
breaking load to be not less than 28 cwts., and the deflection to be at
least YQ inch.

The very marked difference between the strength in tension and
compression makes it unsuitable for girders, and although formerly
used for such, the practice is now discontinued. The cost of production
and the high compressive resistance of cast iron render it an economical
material for use in compression members subject to steady dead loads,
but it is liable to fracture under sudden severe shocks. Grey iron, in
which there is a large percentage of free carbon, should be used for
casting. Columns should always be cast in a vertical position to ensure
a uniform density of metal, and to allow air bubbles and scorise to rise
to the head and be removed. In casting hollow columns the core is
more easily adjusted and kept in position by the above method than if
the column were cast in a horizontal position. Castings should be
clean, sound, and free from cinders, air holes, and blisters. Wavy
surfaces on castings indicate unequal shrinkage and want of uniformity
in the texture of the iron. Filled-up flaws can be detected by tapping
on the faces ; a dull sound is given out by the filling.

The uncertainty of obtaining perfectly sound castings necessitates a
high margin of safety being adopted.

Wrought Iron. — Wrought iron has a fibrous structure, and, compared
with cast iron, is soft and ductile. Its quality depends, in a great
measure, upon the rolling process it has undergone. The best quality
is produced by repeating the process of piling, welding, and rolling some
three or four times.

/Strength. — To ascertain the strength of wrought iron, samples
should be cut from each rolling and tested in tension, noting the
ultimate stress per square inch, the elastic limit, the percentage elonga-
tion, and the percentage reduction of area at fracture. The usual
requirements for strength are given in the following table : —

MATERIALS
TABLE 10. — TENSILE STRENGTH OF WROUGHT IRON.

Ultimate tensile stress
•per square inch.

Elongation per cent.

tons.

per cent.

Round or square bars

23 to 24

20 to 50

Flat bars

22 Qs

25 ^

Angle iron ....

T or H iron . . . .

p, , (grain lengthways .
b \ ,, cross ways

18
17

22
19

10
5

gheet (grain lengthways
' \ ,, cross ways .

20
18

22
19

10
5

The percentage elongation is a measure of the ductility of the
material. It is usual to specify bending tests in addition to the tensile
tests, and Table 11 shows the bending tests required by the Admiralty
for two qualities of wrought-iron plates.

TABLE 11. — BENDING TESTS FOR WROUGHT IRON.

Angle through which plates must bend without cracking.

Plates.

Hot.

Cold.

Thickness

1 inch thick
and under.

1 inch.

t inch.

* inch.

i inch.

B.B. grain lengthways

degrees.
125

degrees.
15

degrees.
25

degrees.
35

degrees.
70

B.B. grain cross ways .

B. grain lengthways .

B. grain cross ways

—

Rivets should bend double when cold without showing any signs
of fracture. Angles, tees, and channels should bend, whilst hot, to the
following shapes without fracture, Fig. 6.

A/Hf/es.

lees.

Channel.

FIG. 6.
Market Forms. — Sheet iron, so called when the thickness does not

STRUCTURAL ENGINEERING

exceed No. 4, B.W.G. (0-239 inch), is little used in engineering
construction.

Corrugated sheets are made by passing the sheets between grooved
rollers, after which* they are usually galvanized, i.e. coated with zinc,
to preserve them from rusting. The widths of the corrugations or
flutes are made 3, 4, or 5 inches. The width of sheet is specified in
terms of the number and width of the corrugations ; thus a 10/3 inch
sheet would cover a net width of 2 feet 6 inches when laid, the actual
width of the sheet being a little more to allow for side lap. The usual
sizes employed range from 18 B.W.G. to 22 B.W.G. in thickness. The
maximum ordinary length of sheet is 10 feet.

TABLE 12. — DIMENSIONS OF CORRUGATED SHEETS.

Thickness B.W.G.

Thickness in inches.

Widths of sheets.

No. 16

0-065

5/5",

6/5"

0-058

5/5",

6/5"

0-049

\ 5/5",

6/5", 6/4", 7/4",

0-042

/ 8/3",

10/3"

20
21

0-035
0-032

} 6/4",

7/4", 8/3", 10/3"

0-028

0-025
0-022

8/3",

10/3"

0-018

Plates. — The ordinary sizes are as follows : Thickness, J inch
to 1 inch ; width, 1 foot to 4 feet ; length, up to 15 feet. The
superficial area of a sheet must not exceed 30 square feet, nor the weight
be more than 4 cwts. Larger sheets are obtainable, but are charged
extra. Plates of less width than 12 inches are known as flats. Flats
can be readily obtained up to 25 feet in length.

Round and square bars are rolled in sizes from 3 inches to 6 inches
in diameter or side.

Wrought iron L, T, and H sections are obtainable, but are now
very little used in engineering structures. The sizes are similar to
those specified for mild steel.

Steel. — For all important structural works steel is the metal almost
exclusively used. The two chief varieties for such work are cast and
mild steel. The former is used for all important castings, and although
possessing many of the characteristics of cast iron, it is much superior
in strength, uniformity of texture, and as a material from which sound
castings may be produced.

Mild Steel. — For structural purposes the sizes of rolled sections,
strength and methods of testing have been standardized by the British
Engineering Standards Committee and published in specification form,
which is generally adopted for structural works.

MATERIALS

Strength. —

TABLE 13.— TENSILE STRENGTH OF STEEL.

Ultimate tensile
strength per
square inch.

Elastic limit.

Elongation.

Mild steel . . /
Cast steel (annealed)
Eivet steel . . .

tons.
28 to 32

30 to 40
26 to 30

tons. *
17 to 18

15 to 17
15 to 17

20 per cent, in 8 in.
10 to 20 per cent, in
25 per cent.

10 in.

Bending tests for mild steel. Test pieces, not less than 1J inches
wide, should withstand without facture, being doubled over until the
internal radius is not greater than 1J times the thickness of the test
piece. Rivet shanks should withstand without fracture, being bent
over, when cold, until the two parts of the shank touch.

Rolled Sections.— Flats. — Sections obtainable :—

Width, f inch to 12 inches ; thickness, ^ inch to 2 inches.
Maximum ordinary length, 40 feet.

Plates are rolled to maximum width, length, or area. The following
table shows the maximum dimensions for a few thicknesses of plates.

Thickness.

Length.

Width.

Area.

inches.

feet

inches.

sq. ft.

100

200

220

175

135

115

As the width multiplied by the length must not exceed the maximum
area, plates cannot possess both the maximum length and maximum
width.

Chequered Plates. — The maximum dimensions for chequered plates
will be found in the following table.

Thickness on plain.

Length.

Width.

Area.

inches.

feet.

inches.

sq. ft.

24 STRUCTURAL ENGINEERING

BmTded plates are made in sizes 3 feet to 5 feet square, the camber
varying between 2 and 3 inches. The thicknesses are ^ inch to J inch,
rising by y^ths.

Stamped Steel Troughing —The sizes of troughing will be found in
Table 15. The maximum dimensions of troughing stamped from
single plates are £ inch thick, 18 inches deep, and 36 feet long.

Round bars may be obtained up to 6 inches in diameter, the
maximum length up to If inches diameter being 60 feet. The maximum
length rapidly decreases for larger sections, 15 feet being the maximum
for 3 inches diameter and over. The sections of angles, tees, joists,
channels, zeds, and rails have been standardized, and the full lists,
published by the Engineering Standards Committee, should be con-
sulted. The maximum ordinary length for all the above sections is
40 feet. Some of the standard sections are used to a greater extent
than others, and consequently such sections are rolled more frequently,
and are easier to obtain at short notice. Rolling lists specifying these
sections are to be obtained from the makers. The weight per foot
length cannot be exactly adhered to by the rollers, and they reserve
the right to supply material within the limits of 2J per cent, under or
over the specified weights. Lengths beyond the ordinary maximum
may be obtained on payment of special "extras." Extras are also
charged for any workmanship, such as cutting to dead lengths,
bevelling, etc.

CHAPTEE II.

LOADS AND WORKING STRESSES.

Dead and Live Load. — General Considerations. — The load on a struc-
ture may be divided broadly into two classes — the dead load, and what
is usually termed live load. The dead load comprises the weight of the
structure itself, which is constantly imposed. The live load would
perhaps better be defined as incidental or intermittent load. It com-
prises all load which is applied to and removed from the structure at
intervals. The live load may be of very varied character. In the case
of bridges it consists of the weight of rolling traffic, such as trains or
road vehicles, together with pedestrian traffic and wind pressure. On
roofs the principal live load is that due to wind pressure ; on crane
girders, the weight of the traveller, together with the load lifted
increased by accelerating force ; on columns, wind pressure combined
with intermittent loads in cases where the columns support girders
subject to live loads. It will be noted that all structures in the open
are subject to wind pressure. Whilst the character of the dead load is
that of an unchangeable static load, that of the live load is very varied
on different structures. Thus the live load imposed on a short bridge
girder during the passage of a train at high speed differs greatly from
the very gradually applied pr essure as the tide rises against the gates of
a dock entrance, or the still more slowly increasing pressure behind a
dam as the reservoir gradually fills with water. Both these latter are,
however, examples of live load, although their effect on the structure is
much less intense than in the former case.

The dead load may be conveniently divided into two portions, one
comprising the weight of the main girders or frames of the structure,
and the other including the accessory parts of the structure necessary
for giving it the desired utility. The weight of the main girders of
a bridge, the principals of a roof, and the main columns and girders of
a framed building, are examples of the first subdivision, whilst the weight
of the flooring, permanent way or pavement, roof covering, etc., repre-
sents the second subdivision. In commencing to design any structure,
it is necessary to make as careful an estimate as possible of the dead
weight of the structure itself, and it is obvious the exact weight of a
structure is indeterminable until the structure has been completely
designed. The nominal amount of live load for which a structure is
to be designed is generally fairly accurately known, although owing to
the varied character of different live loads, their effect as regards the
stress brought into action is often a debatable point. Of the dead load,
the weight of the accessory parts of a structure is very closely estimable,

STRUCTURAL ENGINEERING

but the weight of such portions as main girders, roof principals, etc.,
may only be approximately estimated, and the closeness of agreement
between such estimated weight and the final weight of the structure as
designed, will determine whether the design is suitable or otherwise.
Frequently a re-calculation of the structure becomes necessary, owing to
the final weight considerably exceeding the estimated weight. This
matter constitutes one of the principal difficulties in design, and the
difficulty increases with the magnitude of structure and lack of precedent.
In a small structure exposed to the action of a considerable live load,
a very close estimate of the dead weight of the structure is of relatively
little importance, since by far the greater proportion of the stress will
be caused by the live load. In a very large structure, the bulk of the
stress will be caused by the dead load, and it is important to estimate
more accurately the probable weight of such a structure before com-
mencing its design. Unfortunately, the probable weight of large
structures is always more difficult to forecast, since fewer precedents
exist for purposes of comparison, and much greater judgment and
experience are essential, on account of the greater complexity of the
structure.

Innumerable formulae have been devised, which aim at giving the
probable weight of main girders, roof principals, piers, etc., but such
formulas can only furnish a general guide to be supplemented by careful
judgment, since the assumptions on which they are based are seldom
realized in the particular structure under consideration. Many such
formulae are framed on records of the weight of existing structures
similar to the type under consideration, and where ample precedent
exists the formulae will naturally be more reliable. Formulas, however,
are less reliable for the larger and more uncommon types of structures,
and it will be realized from the above remarks that considerable experi-
ence and judgment are absolutely necessary to undertaking the design
of large and important structures.

Dead Load. — Table 14 gives the weight of the various kinds of
masonry and materials employed in structural work.

TABLE 14. — WEIGHT OF VARIOUS MATERIALS.

Material.

Masonry.

Granite ashlar masonry in cement mortar

Freestone ashlar

Limestone rubble

Freestone rubble

Blue brickwork

Bed brickwork

Concrete.

Bubble concrete in masonry dams

Coke breeze concrete (1 to 6)

Ballast concrete

Clinker concrete (1, 2, 4)

Cement concrete (1 to 5)

Granolithic concrete (1 to 2)

Beinforced concrete, including average reinforcement

Weight in Ibs.
per cubic foot.

165
145
158

122 to 138
147
122

140 to 162
95
140
112
130
138
156

LOADS AND WORKING STRESSES 27

TABLE 14. — WEIGHT OF VARIOUS MATERIALS — continued.

Materials.

Weight in Ibs.
per cubic foot.

Ballast.

Limestone, 35 per cent, voids 110

40 , , 102

45 , , 93

Sandstone, 35 , , 94

45 , , 79-

Broken slag 2£ inches, containing about 35 per cent, voids . . 95 to 100

Gravel 90

Miscellaneous.

Slag, solid 150

Sand, damp 118

„ dry 90

York paving flags 154

Granite paving (Penrnaenmawr) 172

Asphalte 150

Timber.

Elm 36

Red pine and spruce fir 30 to 44

American yellow pine 30

Larch '. 31 to 35

Oak (English) 48 to 60

„ (American) " 54

Teak 41 to 55

Greenheart 60 to 70

Pitch pine 42 to 48

Jarrah (wood pavement) 60 to 63

Iron and Steel.

Cast iron 448

Wrought iron 480

Mild steel 490

Cast steel 492

Glass.

Flint 187

Plate and sheet 169

Fresh water 62-5

Sea water . . 64-125

Dead Load on Bridges. — In estimating the dead load on railway
bridges, the permanent way may be conveniently bulked as so much
per foot run. The following detail weights may be taken as repre-
sentative for the standard gauge of 4 feet 8j inches.

Sleepers, 9 ft. X 10 in. x 5 in. at 40 Ibs. per cub. ft. = 125 Ibs.
each. Chairs, 52 Ibs. Rails, 86 or 100 Ibs. per yard. Fishplates, 32
Ibs. per pair. Fishplate bolts, 5 Ibs. per set of four. Spikes, f Ib.
each.

STRUCTURAL ENGINEERING

Weight of 30 feet of single track — lbg

20 yards of rail at 86 Ibs 1720

11 sleepers (pine) at 125 Ibs 1375

22 chairs at 52 Ibs 1144

22 keys at 50 Ibs. per cub. ft. . . .... 68

2 pairs fishplates at 32 Ibs 64

2 sets fishplate bolts at 5 Ibs . 10

22 sets spikes and trenails at 4 Ibs 88

4469

Say 2 tons per 30 feet of single track, or ^ ton per foot run, = say 150
Ibs. per foot run. "With 100-lb. rails and suitably heavier chairs and
fastenings = 166 Ibs. per ft. run.

Rails. — Railway rails weighing 85, 86, and 100 Ibs. per yard are
in general use for main-line traffic, usually in 30 or 36 feet lengths.

Tramway Rails. — The following are in use. South London, 102
Ibs. per yard. Newcastle, 101. Leeds, 100. Birkenhead, 100. For
purposes of preliminary estimate, 105 Ibs. per yard may be taken.

Ballast and Pavement. — These weights are stated in Table 14.
The mean width of ballast may be taken as 12 feet for a single track,
23 feet for double track, and 45 feet for quadruple track, for standard
gauge. The average depth of ballast used is 18 to 19 inches, the lower
9 inches, called pavement, consisting of larger stones roughly hand
packed, leaving 9 to 10 inches of gravel or broken stone ballast above
to the upper level of sleepers. On the Great Central Cp.'s main line,
the quantity laid per mile of double track is 7500 cubic yards. The
lower layer of pavement is not laid on bridges.

Flooring. — The flooring of railway bridges consists of flat or buckled
plates on cross-girders and rail-bearers, or one of the many types of
troughing. For highway bridges, troughing levelled up with concrete,
or jack arches turned between cross-girders, are most suitable. Table 15
gives the weight of various floor details.

TABLE 15. — WEIGHT OF BRIDGE FLOOR DETAILS.

Detail.

Weight.

Lbs. per

sq. ft.

Flat Steel Plates.

Per thickness of \ inch .
Buckled Plates. Rise 2 to 3 inches.

\ inch thick

Troughing. Lap-jointed as Fig. 7, including rivets.

5-1

10-3
15-4
20-5

/T
FIG. 7.

LOADS AND WORKING STRESSES
TABLE 15. — WEIGHT OF BRIDGE FLOOR

Detail.

Weight.
Lbs. per
sq. ft.

1 foot 8 inch pitch, 6 inches deepj!
2 „ 0 „ „ 7J „ „ {j

2 „ 6 „ „ 10 „ „ {:

- inch thick •

19-71
23-45
23-56
31-15
23-69
31-39
24-61
32-68
25-30
33-61

26-08
34-27
25-58
33-70
26-49
34-94
27-07
35-73
29-03
38-29
37-97
47-18

34-00
34-63
34-98
35-06
35-18
45-47

55-48
68-61

2 „ 8 „ „ 12 „ „ 4]

2 „ 10 „ „ 14 „ „ {i

Butt-jointed as Fig. 8

If D

- P

/ : i ' \

i \

Fie
2 feet 0 inch pitch, 7g inches deep<
2 „ 6 „ „ 10 „ „ -
2 „ 8 „ „ 12 „ „ -

2 „ 10 „ „ 14 „ „ ,
2 „ G „ „ 14 „ „ i

2 „ 8 „ „ 15 „ „ «

Butt-jointed with reinforced boti

U P

/V^

£/._.i

,. 8.
•* inch thick .

. -j > »

i ' "

' 5

' 3 ' "

8 ' "

:2 > 5'
•

atn flange as Fig. 9.

/l\

^=^

FIG.

2 feet 0 inch pitch, 7i inches deep,
2 6 , „ 10
2 8 , , 12
2 10 , ,14
3 0 , , 16
3 0 , , 16
4 6 , , 18

" » » 5>

i/..l

3 inch plate, | inch straps
)> »
» »

5» »>

» >» *

J inch plate, jj inch straps
with double straps top and
bottom, £ inch thick .
with double straps top and
bottom, f inch thick .

30 STRUCTURAL ENGINEERING

TABLE 15. — WEIGHT OF BRIDGE FLOOR DETAILS — continued.

Detail.

Weight
Lbs. per
sq. ft.

Arched troughing as Fig. 10.

L_ . o . -J

FIG. 10.

20-60

" '

25-67

29-53

' "

24-48

33-00

20-31

Weight of Cross-girders. — In railway bridges with framed floors,
consisting of cross-girders, rail-bearers, and plate flooring, the most
economical spacing of cross-girders is from 7 ft. to 8 ft. The growing
practice of building locomotives with small wheels close together, tends
to a closer spacing in the near future. With these spacings, cross-
girders for double-track railway bridges are usually 26 fb. to 27 ft.
long, and 2 ft. 3 in. to 2 ft. 6 in. deep. Adopting these proportions,
the weight of one cross-girder may be taken as 2 '2 5 tons. For single-
track bridges with cross-girders 14 ft. to 15 ft. long, and 15 in. to
18 in. deep, the weight of one cross-girder may be taken as 0*5 ton.
Cross-girders between main braced girders with wide bays will be much
heavier, and an independent estimate is necessary. Cross-girders for
highway bridges will be appreciably lighter for the same spans, but
owing to the variable character of the traffic and width of highway
bridges, a preliminary estimate is necessary in each case.

Rail-learers. — Rail-bearers 7 ft. to 8 ft. long may be estimated at
700 Ibs. to 900 Ibs. each.

Jack arching is usually 9 inches to 13 J inches thick of brickwork,
and its weight is readily estimated.

Weight of Main Girders. — Plate Girders. The probable weight of
main plate girders, provided the depth is about one-tenth the span,
may be fairly closely obtained from the following formulae. Whenever
reliable information may be obtained as to the weight of well-designed
existing girders of the same span, similarly loaded, such records are
preferable to results calculated from formulae.

W = Total distributed or equivalent distributed load carried by
one girder, exclusive of its own weight. L = Span in feet.

Probable weight of plate girders. Depth about ^ span.

LOADS AND WORKING STRESSES

W v T

From 20 ft. to 40 ft. span = - * ' tons.

ooO

„ 60 ft. „ 90 ft. ,, =

Plate girders are seldom employed beyond 90 ft. span.
Probable weight of lattice girders. Depth about ~ span.

From 20 ft. to 50 ft. span = * tons.

„ 50ft. „ 90ft. „ =~ „

W y L

„ 90 ft. „ 120 ft. „ = -jgg- „

The weights of lattice girders of larger span should be carefully
estimated in detail, having regard to the probable stresses coming upon
the various members. The weight of main lattice girders is more
influenced by variations in arrangement of detail than is that of plate
girders.

Weight of Rolled Steel Section Bars. — Table 16 gives the weight per
foot run of the most commonly used rolled sections in mild steel. Full
lists of these are published in most section books.

TABLE 16. — WEIGHT OF ROLLED STEEL SECTION BARS.

Equal angles.

Unequal angles.

Size.

Weight IDS. per foot.

Size.

Weight Ibs. per foot.

in. in. in.

6 x 6 x |
5x5x5
4J X 4£ X £
4 x 4 x £
31 X 3J X *
3 x 3 x i
3 x 3 x i
2* x 2£ x i
2* X 2| X I

28-70
19-92
14-46
12-75
11-05
9-36
7-18
5-89
5-26

in. in. in.
6 X 4 X |
6 X 4 X i
6 x 3| X *
5 x 3" x I
4x3x|

19-92
16-15
15-31
12-75
11-05

CHANNELS.

in. in.
12 X 3*
10 X 3|
8x3

32-88
28-21
19-30

TEES.

ZED BARS.

in. in. in.
6 x 4 x i
6x3x|
5 x 4 x $
5x3x1

16-22
14-53
14-51
12-79

in. in.
10 x 3£
8 x 3£
5x3
4x3

28-16
22-68
14-17
12-26

STRUCTURAL ENGINEERING

The weights of several sizes of rolled joists are given on
146, together with other properties of those sections.

US-

TABLE 17. — WEIGHT OF BOLTS, NUTS, AND RIVET-HEADS.

Weight in Ibs. of

Diameter.

1 inch length of bolt.

One hexagonal nut
and bolt-head.

One square nut and
bolt-head.

100 rivet-heads.

in.

0-0313

0-057

0-071

0-0556

0-135

0-169

4-17

0-0869

0-261

0-330

8-15

—

10-85

0-1252

0-450

0-570

14-08

—

17-90

0-1703

0-720

0-90

22-36

—

27-50

0-2225

1-07

1-35

33-38

—

47-53

0-3476

2-09

2-63

65-19

0-5066

3-61

4-55

0-6815

5-70

7-20

0-8901

8-56

10-80

—

Iii estimating the weight of riveted work, an allowance of from
2 per cent, to 5 per cent, of the weight of the structure is usually made,
depending on the class of work, to cover the weight of rivet-heads.
Such allowance is usually excessive, and is really intended to cover the
wastage of drilling and punching the holes.

Dead Load on Roofs.— Principals. The probable weight of roof
principals of ordinary V-types may be fairly estimated from the follow-
ing formulas : —

W = Weight of one principal in Ibs. L = Span in feet. D = Dis-
tance apart of principals in feet.

For roofs with heavy covering, W = £ DL (1 + ^)
„ medium „ ' W= J DL (1 + &)
„ light „ W = ^DL(1+^)

Purlins and Common Rafters consist of angle, zed, or joist section if
of steel, or rectangular sections if timber. The weights of the former
are given above. The weight of timber members may be estimated on
a basis of 35 Ibs. to 40 Ibs. per cubic foot.

Roof Coverings. — Boarding 1 inch thick, 3 Ibs. to 3^ Ibs. per square
foot.

Slating. — The usual sizes and weights of slates used for roofing are
given in Table 18, for slates laid with 3-inch lap and nailed near the
centre. A square of slating is 100 square feet, and a nominal 1000 of
slates contains 1200.

LOADS AND WORKING STRESSES
TABLE 18. — SIZES AND WEIGHT OF SLATES.

Size.

Area covered by
1200 slates.

Weight per sq. foot
of roof surface.

in. in.

squares.

Ibs.

Doubles

13 X 6

2-5

8-25

Ladies

16 X 8

4-75

8-0

Countesses ....

20 X 10

7-0

8-0

Duchesses ....

24 X 12

10-0

8-5

Glazing. — The weight of glass such as used for roof covering is
14 ozs. per square foot for each •— inch in thickness. The sheets are
usually 2 feet wide and \ inch thick, and weigh 3j Ibs. per square foot.

Glazing Bars. — A large number of patent bars are in use. The
following table of the weights of Messrs. Mellowes and Co.'s " Eclipse "
glazing bars will afford a guide in estimating the weight of this detail.

TABLE 19. — WEIGHT OF GLAZING BARS.

" Eclipse " bar.

Bearing, centres of
purlins.

Weight in Ibs. per foot
run.

ft. in.

No. 7

6 0

2-796

„ 8

7 6

3-202

„ 9

8 6

3-578

„ 9A

9 3

3-906

„ 10

10 0

4-906

The. standard spacing of these bars is 24J inches.

Corrugated Sheeting. — These sheets are rolled of such sizes that the
side lap is about 1 inch, whilst the end lap is generally made 6 inches.
Allowing for these laps, Table 20 gives the weight per square foot for
various gauges.

TABLE 20. — WEIGHT OF CORRUGATED SHEETING.

Thickness B.W.G.

Weight in Ibs. per square foot,
including laps

2-78

2-20

1-80

1-49

1-12

Galvanized Fittings for Corrugated Sheets.— The weight of fittings
per 100 square feet of roof surface may be taken as 12- 6 Ibs. if the
sheets are attached to steel purlins by hook-bolts, and 7'6 Ibs. if screwed
to timber purlins.

Galvanized Ridging and Louvre Blades. — Ridging weighs from 1 Ib.
to 8 Ibs. per foot run for girths of from 10 inches to 36 inches, and
thicknesses of from No. 26 to No. 16 B.W.Gr. ; Louvre Blades, from
2-5 to 5 Ibs. per foot run for each blade of 11 inches girth, and thick-
nesses of from ~ inch to \ inch.

Stamped Steel Gutters usually vary from ~ inch to J inch thick, and
weigh from 9 Ibs. to 25 Ibs. per foot run according to section.

34 STRUCTURAL ENGINEERING

Lead for flashing and covering flats usually weighs from 5 to 7 Ibs.
per square foot, and allowing for laps, rolls, and nails, may be taken
at 5-8 to 8' 5 Ibs. per square foot.

Zinc. — The thickness of zinc sheeting follows the zinc gauge,
ranging from Nos. 9 to 16, and corresponding closely with Nos. 27, 25,
etc., to 19 of the B.W.G. respectively. Nos. 15 and 16 are generally
used for roofing. The sheets are 7 feet to 8 feet long by 2 feet
8 inches to 3 feet wide. Allowing for laps, the weight is l£ to If Ibs.
per square foot.

Asphalte.— The thickness and weight of asphalte laid on floors and
roof flats is given in Table 21.

TABLE 21. — WEIGHT OF ASPHALTE ON FLOORS AND ROOFS.

Thickness.

Wt. Ibs. per sq. it.

Roof flats and bridge floors

inch.
3

Goods warehouse floors

15§

Railway platforms

124

Waterproofing backs of arches

In laying asphalte and waterproof coating, care should be taken to
give the finished surface a good fall to ensure proper drainage.

Snow. — The snow load accumulates slowly, and may be treated as
dead load. 5 Ibs. per square foot of covered area is usually assumed in
England.

Dead Load on Floors. — Floors being of very varied construction, no
general weight per square foot may be cited. The type of floor being
decided, its weight per square foot may be readily estimated in detail
by reference to the preceding tables.

Live Load. — Live Load on Floors. The live load on floors of
different classes may be assumed as equivalent to the following dead
loads : —

For dwelling-houses, hotels, and hos-
pitals 60 to 70 Ibs. per square foot.

For schools, assembly halls, offices, and

retail shops 110 to 130 Ibs. „

For warehouse buildings 200 to 250 Ibs. „ „

Machine shop floors are examples where especially heavy loading
may occur. Usually the lighter classes of machine tools only are placed
on upper floors, heavier machinery requiring special foundations being
necessarily placed in the basement. The type of machines to be em-
ployed will determine the weight for which the floor is to be designed.
The load on such floors is less uniformly distributed than on ordinary
floors, and solid and rigid types should be adopted to ensure the best
distribution of the locally concentrated loads to the girders beneath.

Live Load on Bridges. — The intensity of the live load on railway
bridges is governed by the type of locomotives in use. These actually
impose heavy concentrated rolling loads upon the various members,
such loads depending on the weight of engine and tender and spacing

LOADS AND WORKING STRESSES

of axles. Two methods of treating the rolling load are in general use.
1. The concentrated load method, in which the bending moments and
shearing forces due to the actual concentrated loads at known spacings
are considered. 2. The equivalent distributed load method, in which
the system of actual concentrated loads is replaced by a distributed load
of uniform intensity which will create the same maximum bending
moment at any point on the span, as that caused by any position of
the concentrated axle loads. The former method is generally adopted
in American, and the latter in English practice.

Equivalent Distributed Load on Main Girders. — Fig. 11 shows the
loads and axle spacings of the types of locomotives which produce the

Tender 41 tons.

G. N. R.

Engine 66 fans.

! 610 \ 53 \

i I i i i i i i

Tender 36 fans. G-W-R- Engine 72 fans.

J6 /6 /6

& hns.

heaviest equivalent distributed load at present (1011) in use on
English railways ; and in Table 22 are given the equivalent uniformly
distributed loads due to these two types, for spans ranging from 10 to
200 feet.

TABLE 22. — EQUIVALENT DISTRIBUTED LOADS ON RAILWAY BRIDGES.

Equivalent distributed load in tons per foot run for single line of way.

Span in feet.

G.N.R.

G.W.R.

Span in feet.

G.N.R.

G.W.R.

3-69

3-46

1-96

3-11

3-35

100

1-93

1-92

2-88

3-18

110

1-92

2-61

2-95

120

1-92

2-45

2-77

130

1-92

2-36

2-61

140

1-92

1-91

2-22

2-47

150

1-92

1-91

2-12

2-34

160

1-92

1-91

2-10

2-24

170

1-92

1-90

2-08

2-10

180

1-92

1-90

2-04

2-01

190

1-92

1-90

2-00

1-99

200

1-92

1-89

36 STRUCTURAL ENGINEERING

The figures in Table 22 include an allowance of 2J per cent, over the
calculated values for possible future increase in the weight of locomotives,
and although these are outside values, a steady increase in the weight
of locomotives has taken place during the last few years.1

For larger spans, the live load is generally taken as that due to
the heaviest type of mineral or freight train, preceded by two of the
heaviest type of locomotives. The weight of a train made up of high
capacity wagons, or wagons loaded with boilers, girders, machinery,
etc., falls not far short of the weight of a train of engines. An
equivalent load of 1-5 tons per foot run fora single line of way probably
covers the effect of such a train for spans exceeding 200 feet. Spans
up to about 180 feet would be practically covered by three large
engines and tenders.

Live Load on Cross- Girders. — Cross-girders, however closely spaced,
have each in turn to carry the heaviest axle load, which is usually that
on the driving axle, or 19 tons per axle in present English practice.
As the distance between the two most heavily loaded axles is usually
from 5 to 8 feet, any closer spacing of the cross-girders is wasteful.
Cross-girders between lattice main girders are usually at wider spacing
than 8 feet, and in such cases the maximum loads will be equal to the
maximum reactions due to the rail-bearers or stringers when these are
most heavily loaded.

Live Load on Troughing. — With continuous trough flooring, it is
impossible for a single trough to carry the full load of the heaviest
axle, since the continuity of the floor causes the load to be distributed
over several trough sections to right and left of the load. The extent
of this distribution depends on the relative rigidity of rails and trough-
ing, but the maximum load coming upon any one trough will not
exceed one-half the actual axle load and may be as low as one-third.

Live Load on Rail-Bearers. — On short rail-bearers from 5 to 8 feet
span, the maximum live load will be the load on the heaviest axle when
at the centre of span. Longer rail-bearers will accommodate two,
three, or more axles simultaneously, and the worst position of such
axle loads for creating bending moment will determine the maximum
loading.

Live Load on Highway Bridges. — This, whilst not so intense as on
railway bridges, is very variable. Table 23 gives particulars of special
loads on highway bridges. For ordinary wheeled traffic over bridges
exempt from the special loads in Table 23, an equivalent distributed
load of 130 Ibs. per square foot of roadway should cover the require-
ments. For pedestrian traffic on footpaths, 120 Ibs. per square foot is
a sufficient allowance.

Wind Load. — The question as to the magnitude of wind pressure
and its mode of action on structures is one on which great uncertainty
still exists. The following facts, however, appear to be satisfactorily
established.

1. That exceptionally high intensities of pressure only prevail over
local and relatively small areas, and that the greater the area exposed
to the wind, the less will be the average pressure on such area. From

1 For further information on rolling loads, see Mins. Proceedings Inst. C.E.,
vol. clviii. p. 323 ; vol. cxli. p. 2 ; vol. clviii. p. 380.

LOADS AND WORKING STRESSES

W
§

7*1

s is

JB
1

§rH ^H

od 0-1-1 rH rH iH iH

O «5 O O O to 0^ O b- O O O <£>

44 i2 O

il §

^-> s-4 *r3

° fl ~ >

II 1 i

n o »o ^ 5 •*=

H §!nH PQ cfl

38 STRUCTURAL ENGINEERING

extended records of wind pressure obtained at the site of the Forth
Bridge, the average pressure upon a board 20 feet by 15 feet was gene-
rally only two-thirds of the pressure upon a small wind-gauge of l£
square feet area, both in moderate and high winds.

2. The total or effective pressure against thin flat plates is consider-
ably greater than the pressure calculated by multiplying the exposed
windward area by the wind pressure per square foot acting on the
windward face. This is due to the creation of a partial vacuum or
negative pressure behind the plate, caused by the suction effect of the
eddies set up by the resistance of the sharp edges to the free passage of
the wind current. Thus, taking the effective wind pressure on a thin
square plate as 100, the total pressure upon a cube presenting the same
area to the wind is represented by 80, and on a prism having a length
double that of the cube, by 72. The suction effect, therefore, rapidly
diminishes as the length of the obstacle in the direction of the wind
current increases. Roughly speaking, from 5 to | of the total effective
pressure on an exposed area of thin plating is due to the negative
pressure on the leeward side, and f to f to the positive pressure on the
windward side. The exact ratio in any particular case varies, however,
with the shape of the exposed surface.1 This has a direct bearing on
the effective pressure on the windward side of a V roof. The sharper
the angle at the ridge, the greater will be the negative pressure on the
leeward side, with consequent greater effective pressure than on roofs of
flatter slope.

3. The extent to which one plate surface will shelter another similar
one placed parallel with, and to leeward of the first, is actually very
small, unless the two surfaces are relatively close together. Thus the
windward girder of a bridge or the windward face of a lattice pier
offers little protection to the leeward girder or face. In the case of a
bridge with main plate girders and a continuous floor, the pressure
against the leeward girder will no doubt be less than on the windward
girder, but such reduction is less than is frequently assumed. Experi-
ments on square and rectangular plates show that the maximum shield-
ing effect occurs when the plates are separated by a distance of about
IK times their least cross dimension, and that the shielding effect of
long rectangular plates is considerably less than that of circular plates.

4. Results deduced from the overturning of railway rolling stock
show the mean side pressure to have varied from 26 to 34 Ibs. per
square foot, and, excepting in the case of structures in abnormally
exposed situations, it is unlikely that any considerable area is ever
exposed to a much higher positive mean pressure than 35 Ibs. per square
foot. Much higher pressures have been recorded by wind-gauges in
which the pressure is registered by the deflection of a spring, but such
pressures may be, and probably are, due to the dynamic effect of a
sudden gust of wind blowing for a very short period. At the same time
it should be remembered that a wind pressure of 30 Ibs. per square
foot applied suddenly will deflect the spring of a wind-gauge to the
same extent as a steadily applied pressure of GO Ibs. per square foot,
provided the sudden gust last long enough to produce a complete vibra-
tion of the spring. In a similar manner, a bridge or tall building will

1 Mins. Proceedings lust. C.E., vol. clvi. p. 78.

LOADS AND WORKING STRESSES 39

possess a time period of vibration depending upon its mass, which will,
however, be much longer than that of a light spring. If, then, a
sudden gust of wind prevail long enough to deflect a bridge or building
laterally through the full extent of its vibration, the stress produced
will be that due to the higher pressure as recorded by the spring-
loaded indicator of the wind-gauge.

5. The pressure of wind increases with the height above the ground,
so that the upper portions of lofty buildings, chimneys, piers, and
bridge girders are exposed to higher wind pressures than the lower
portions, which are further often sheltered by neighbouring buildings.

6. The pressure on convex surfaces is much less, and on concave
surfaces greater than on flat surfaces presenting the same projected
area to the direction of the wind current. Thus, on the area of a
circular chimney as seen in elevation, the effective wind pressure is only-
about half the actual pressure which would act on a thin flat surface of
similar outline and area.

Intensity of Wind-Pressure on Structures. — In the case of bridges, it
is generally assumed that no train will be traversing a bridge when the
wind pressure exceeds 30 to 35 Ibs. per square foot, whilst a maximum
pressure of 45 to 50 Ibs. may act on the structure when unloaded. The
area of the bridge exposed to the higher pressure will be from once to
about three times the area as seen in elevation, depending on the type
of construction. Thus a tubular girder with continuous roof and floor
will expose the elevational area only, a plate girder bridge from once to
twice that area, the degree of shelter afforded by the leeward girder
depending largely on the width of the bridge. A large bridge with
double lattice girders may expose an effective area equal to three times
the elevational area, or even more if the main girders be very broad.
The lower pressure of 30 Ibs. per square foot will act on the effective
area of the windward girder, and on the train considered as a continuous
screen, the upper edge being 12 feet 6 inches and the lower edge 2 feet
6 inches above the rails. In the latter case the train will largely shelter
the leeward girder, and the wind pressure will only act to any consider-
able extent on that portion of it (if any) projecting above the train.
The wind pressure on a moving train is, of course, to be considered as
a rolling load, since it travels with the train.

The suction effect previously referred to is most active in the case
of lattice-work structures which present a series of thin plate surfaces
to the wind. On tall buildings which have considerable depth in the
direction of the wind, the maximum effective pressure will be from § to
f of that assumed for open-work structures, or about 35 to 38 Ibs. per
square foot, whilst the larger area makes it still less probable that the
mean pressure will actually reach this figure. The usual allowance for
tall buildings and roof structures will be stated subsequently.

Further, it is obvious that judgment must be exercised in adopting
a suitable maximum wind pressure for a given structure. The lower
portions of many buildings, low roofs, etc., are often almost completely
sheltered by neighbouring and higher buildings, whilst for others in
lofty and more exposed situations the effect of the maximum wind
pressure must necessarily be considered.

40 • STRUCTURAL ENGINEERING

WORKING STRESSES AND SECTIONAL AREA OF MEMBERS.

General Considerations. — After making a careful estimate of the
load to be borne by a structure, the stresses in the various members
due to such load are next calculated. The methods of deducing the
stresses in the various members of structures due to specified loads
are considered in subsequent chapters. The sectional area of each
member then depends primarily on the magnitude of the stress
occurring in it, having due regard to a suitable disposition of the
material in the section according as the member is subject to tension,
compression, bending, etc., or to more than one of these actions com-
bined. In arranging the disposition of material in the cross-section,
further regard must be paid to convenience of making joints and con-
nections with neighbouring members and deduction of area occupied by
rivet-holes in the case of tension members. These features are fully
considered later, and at present attention is directed to the determi-
nation of the actual sectional area.

The working stress in any member is understood to be the actual
maximum stress in tons per square inch created in the member under
the most unfavourable conditions of loading. The principal difficulty
lies in determining at all accurately the actual maximum stress which
takes place in any member. In cases where the stress is entirely due
to a perfectly dead load, no doubt exists as to the value of the maximum
stress, and a relatively high working stress may safely be employed in
proportioning the sectional area. Examples of purely dead load
stresses in actual practice are, however, rare, and the maximum stresses
in the members of practical structures are generally due to the com-
bined effect of both dead and live loading. That portion of the stress
due to the dead load may be very closely estimated, whilst that caused
by the live load is in many cases a very indefinite quantity, depending
on the variable character and mode of application of the live load,
examples of which have already been mentioned. It is a well-
established fact that the actual momentary stress due to a suddenly
applied live load may be double that due to the same amount of static
or dead load. But between a suddenly applied and a very gradually
applied live load, there exists a series of infinite gradations, each of
which must be considered on its own particular merits. The additional
stress caused in such cases, over and above that due to the nominal
amount of the live load, is commonly referred to as dynamic stress or
stress due to impact, and there is little doubt that the time period of
application of the live load is the principal -factor in determining the
magnitude of the dynamic stress created.

Further, it appears from the results of numerous experiments that
a great number of repetitions of fluctuating stress in a member exer-
cises a more deteriorating effect on the material than a static stress of
equal maximum intensity. These facts obviously have a specially
important bearing on the design of members of bridges which are
most subject to frequent and rapidly applied fluctuations of load. It
is evident, therefore, in fixing the working stress, that regard should
be paid to the range of stress, or difference between maximum and

LOADS AND WORKING STRESSES 41

minimum stress in a member as well as the time period of application
of the live load. The character of live loads and degree of suddenness
of their application are so varied that a proper estimate of the pro-
bable dynamic effect, and therefore the actual maximum stress caused,
becomes a more or less complex matter. Many rules have been pro-
posed for making allowance for the above-mentioned influences in
designing the sectional area of members, and some of the principal of
these will now be stated. It may be mentioned, however, that the
present state of exact knowledge on this question is far from complete,
whilst useful investigation is much retarded by the difficulty of making
reliable experiments.

Factor of Safety. — Formerly the working stress was deduced by
dividing the ultimate or breaking strength of the material by 4, 5, 6,
etc., according to the reduction considered suitable for the character
of the load, and in cases where a fairly reliable estimate of the dynamic
effect of the live load is possible, this method used by persons pos-
sessing sound judgment and experience may produce satisfactory
results. Its abuse lies in its indiscriminate application to all or any
cases of live loading. It is necessary that the working stress be kept
safely within the minimum elastic limit of the material in order to avoid
permanent strain or set. For average mild steel, such as employed in
general structural work, the ultimate strength varies from 27 to 32
tons per square inch, and the elastic limit from about 14 tons per
square inch upwards, the latter being a more variable quantity than the
ultimate strength. The actual working stress should therefore not exceed
9, or at most 10 tons per square inch, in order to leave a reasonably safe
margin as regards the elastic limit. These figures correspond to a
factor of safety of 3 with respect to the ultimate strength.

For purely dead loads, that is where the actual maximum stress is
very closely estimable, a working stress of 9 tons per square inch for
mild steel may be safely adopted for tension numbers, and 8 tons per
square inch for compression numbers, excepting where the length of
the latter necessitates a reduction in accordance with the liability to
fail by buckling.

From the results of experiments on members subjected to fluctu-
ating stresses the following general deductions are made.

1. That a factor of safety of 4J should be employed when the
stress fluctuates between the maximum value and zero.

2. That a factor of safety of 9 should be employed when the stress
fluctuates between a certain amount of compression and a similar
amount of tension. This case would be met by trebling the nominal
live load stress and using a factor of safety of 3, as for a dead
load.

Case 2 probably represents the extreme effect of a variable load,
and between it and the case of a purely dead load there are numberless
others in which the ratio of dead to live load stress may have any
value. Taking these deductions as a basis, therefore, the factor of
safety for members subject to maximum and minimum stresses of the
same kind will vary between 3 and 4*5, and for members subject to
maximum and minimum stresses of opposite kinds the factor of safety
will vary between 4'5 and 9. The value chosen in any particular case

42 STRUCTURAL ENGINEERING

will depend principally on the extent to which the live or dead load
stress preponderates.

Most of the formulas devised for determining the sectional area or
working stress of a member, or what is practically the same thing —
the factor of safety to be employed, provide in effect, a sliding factor
of safety dependent upon the relative values of the nominal or apparent
maximum and stresses, whilst some take into account the degree of
suddenness of application of the live load.

Wohler's Experiments. — The following results were deduced from
a number of experiments made by Herr Wo'hler with a view to ascer-
taining the effect on the strength of materials when subjected to
known alternations of stress repeated until fracture ensued. The
material was found to break under a stress considerably less than that
which it would withstand when the load was applied steadily under
normal conditions of testing. The apparent loss of ultimate strength
also varied according to the difference between the maximum and
minimum stresses applied.1 Broadly stated the actual results were as
follows : — If t = breaking stress due to a steadily applied load with no
variation (usually called the static breaking stress), then under a large
number of applications of load varying between zero and u, the
material eventually broke under the load u when the apparent stress

caused was equal to - , or the strength of the material was apparently

reduced by one-half. Under the repeated application of a load varying
between -\-v and — v, that is between a certain compression and the
same amount of tension, the material eventually broke under the

load v when the apparent stress caused was equal to r, corresponding

with an apparent reduction in strength of two-thirds.

Launhardt-Weyrauch Formula. — This formula was devised to
express the reduced ultimate strength of a member when subject to
alternating nominal or apparent maximum and minimum stresses, as
indicated by the results of Wohler's experiments.

Let Max. S and Min. S = the higher and lower apparent stresses to
which the member is subjected ; t = static breaking stress of the
material in tons per square inch ; P = the reduced breaking stress in
tons per square inch due to the fluctuating load. Then —

To apply this to purposes of design, the suitable working stress p
tons per square inch is obtained by dividing the breaking stress P by
a factor of safety of 3, whence —

Taking / for mild steel = 27 tons, and for wrought iron = 21 tons
per square inch —

1 For a detailed discussion of the results of these experiments, see A Practical
Treatise on Bridge Construction, by T. Claxton Fidler.

LOADS AND WORKING STRESSES 43

and p = 4-6c(l + J • jj^-|) for wrought iron.

If Max. S and Min. S be of opposite kinds, Min. S will be denoted
as a negative stress, and the formula becomes —

Min.

EXAMPLE 2. — The maximum and minimum stresses in a mild steel
member are respectively 47 and 22 tons of tension. Determine the net
sectional area.

Working stress = 6(1 + \ • ff) = 7'4 tons per square inch, and net

4:7

sectional area = ^77 = 6*35 square inches.

EXAMPLE 3. — The maximum and minimum stresses leing respectively
47 tons tension and 10 tons compression, to determine the net sectional
area.

Here the minimum stress of 10 tons is negative. Working
stress = 6(1 — i ' 4?) = ^'35 tons per square inch, and net sectional

area = ^-^ = 8'8 square inches,
o oo

Relatively few members are actually allowed to suffer reversals of
stress, a counter-brace being usually provided to take up the reversed
stress, when the principal member is then subject to a stress fluctuating
between Max. S and zero. The working stress then equals 6(1 + 0)
= 6 tons per square inch, and the factor of safety = -/ = 4'5.

Where Max. S = Min. S, but is of opposite kind, the working
stress = 6(1— J) = 3 tons per square inch, and the factor of
safety = ~ = 9, as previously mentioned.

The apparently reduced breaking strength of the material when
subjected to alternating stresses is considered by some authorities to be
due to a deteriorated condition of the material to which the term
fatigue has been applied, and the results of the Wohler tests are
generally ascribed to some such condition of the material. It will be
noticed that the Launhardt-Weyrauch formula, being based on the
results of Wohler's tests, makes allowance for the so-called fatigue of
the material, but does not allow for the dynamic or impact effect of the
load, since the suddenness of application of the load does not enter into
the results deduced from the tests.

Modified Launhardt Formula. — The following modified form of
this formula is in use in America : —

p = 3'34(l + jjjj^-g) for wrought iron

the lower resulting value for p being assumed to cover the effects of
both " fatigue " and impact.

Whether it is necessary to make a separate allowance for the effects

44 STRUCTURAL ENGINEERING

of "fatigue" and dynamic stress is a much-debated question. Prof.
T. Claxton Fidler advances cogent reasons for supposing that the
effects of " fatigue " and dynamic action are one and the same.1

Fidler' s Dynamic Fo'rmula. — The following formula has been
proposed by Prof. T. Claxton Fidler for determining the sectional area
of members. It makes allowance for the dynamic effect of the
fluctuating load in a manner stated briefly as follows. If a member
be subject to an initial stress of, say, 2 tons per square inch due to the
dead load, the sudden application of a further load such as would
create an additional 3 tons of stress if applied very gradually, may
cause a momentary or dynamic increase of stress of 2 x 3 = 6 tons per
square inch, thus creating an actual maximum stress of 2 + 6 = 8 tons
per square inch. Similarly, the sudden application of a stress equal to
one-half the breaking strength of the material may readily account for
the failure of a bar by the creation of a dynamic stress equal to the
breaking stress. This is, broadly speaking, the line of argument by
which the results of Wohler's tests may be explained by reference to the
dynamic theory of stress, and, assuming this action to take place, the
material does not undergo any actual reduction of breaking strength,
but fails at its normal static breaking stress, simply because that stress
is momentarily created by the suddenness of application of the load.

If Max. S and Min. S represent as before, the maximum and
minimum stresses to which the member is subjected, the dynamic
increment of stress = w = Max. S — Min. S.

Hence the actual maximum stress to be provided for in designing
the sectional area = nominal maximum stress -f increment = Max. S
4- w ; and for mild steel tension members —

Max. S + w . ,

Net sectional area = — — ^— square inches

For compressive members not liable to buckle —

~ , . , Max. S + w . ,

Gross sectional area = - — = — — square inches.

For members subject to alternating stresses of opposite kinds,
Min. S will be negative, and iv = Max. S — (- Min. S) = Max. S
+ Min. S.

For wrought-iron tension members —

Max. S + w
Net sectional area = — x —

and for wrought-iron compression members —

Max. S + w

Gross sectional area =

Prof. Fidler recommends that in all cases where the increment w is
applied instantaneously, or practically so, its full value = Max. S
— Min. S, is to be added to the nominal Max. S. This will be the case in

1 Bridge Construction, T. C. Fidler, pp. 248 et seq.

LOADS AND WORKING STRESSES 45

web members of girders, cross-girders and longitudinals, whilst in the
case of the flanges of main girders exceeding 100 feet span, i.e. in which
an appreciable period of time elapses between minimum and maximum

loading, w is to be taken

Max. S - Min. S

This is, of course, a purely arbitrary allowance, but at the same time
reasonable. The formula used in this manner becomes very elastic,
and may obviously be applied generally, by varying w in accordance
with the time period of loading, but such modification must depend
entirely on the judgment of the designer.

EXAMPLE 4. — Required the sectional area for a mild steel member in
ivhich the stress alternates rapidly from 50 tons of tension to IS tons of
tension.

w = 50 - 18 = 32 tons

50 + 32
and sectional area = — ^ — = 9'1 square inches.

EXAMPLE 5. — The maximum stress in a mild steel strut is 57 tons of
compression and the minimum stress 12 tons of tension. Required the
sectional area exclusive of consideration of buckling tendency.

Here, w = 57 + 12 = 69 tons, and if the change from minimum to
maximum loading take place instantaneously, the actual maximum
stress to be designed for = 57 + 69 = 126 tons, whence-
sectional area = ~- = 18 square inches.

Stone's "Range" Formula.1 — The following formula has been
proposed by Mr. E. H. Stone, MJnst.C.E. Briefly, this formula
takes into account, separately, what is defined as the immediate effect
and the cumulative effect of the moving load. Quoting from the
author's original paper, the immediate effect is " observable every time
a train traverses a bridge, due to the sudden and violent manner in
which the load is applied — irregularities in the track, peculiarities of the
engine, or other causes." The cumulative effect is " an effect produced
in course of time by repeated loading and unloading." The object of
the inquiry was to establish a co-efficient to be applied to the nominal
moving load, which should express its total effect on the members of
the structure in terms of effective or equivalent fixed load.

1. The "immediate" effect of the moving load, as compared with
that due to the same amount of static or fixed load, is deduced from
experiments, information regarding which is given in the paper above
referred to.

2. The " cumulative " effect of the moving load, as compared with
that due to the same amount of fixed load, is deduced from Wohler's
experiments. The "immediate" and "cumulative" effects are then
combined in a rational manner to give the total effect of the moving
load, as compared with that of an equivalent amount of fixed load.
The results for mild steel are given in Table 24.

1 Transactions Am. Soc. C.E., vol. 41, pp. 467 to 553.

STRUCTURAL ENGINEERING

TABLE 24. — WORKING STRESSES FOR MILD STEEL BY RANGE
FORMULA — Safe Working Stress = 9 — (5 x R2).

Nominal load.

Effective load.

Working results.
Factor of safety =3

Co-efficient

to obtain

Permissible stress.

Total

Composition of
nominal load.

equivalent
of moving
load In

TVktol

Composition of
effective load.

Tons per sq. in.

nominal

terms of

xotai
effective

load.

fixed load.

i/,aj

Due to

Fixed

Moving

iouu.

Fixed

Moving

nominal

effective

load.

tons.

100

2-2500

225-OOOO

225-0000

4-00

9-00

100

2-5

97-5

2-1478

211-9105

2-5

209-4105

4-25

9-00

100
100

5
10

95
90

2-0585
1-9091

200-5575
181-8190

5
10

195-5575
171-8190

4-49
4'95

9-00
9-00

100

1-7889

167-0565

152-0565

5-39

9-00

100

1-6897

155-1760

135-1760

5-80

9-00

100

1-6061

i45'4575

120-4575

6-19

9-00

100

1-5344

137-4080

107-4080

6-55

9-00

100

33-3

66-6

1-4918

1327866

3-33

99-4533

6*78

9-00

100

1-4719

130-6735

95-6735

6*89

9-00

100

1-4167

125-0020

85-0020

7-20

9-00

100

1-3673

1202015

75-2015

?'49

9-00

100

1-3226

116-1300

66-1300

7-75

9-00

100

1-2817

112-6765

57-6765

7-99

9-00

100

1-2439

109-7560

49-7560

8-20

9-00

100

•2086

107-3010

42-3010

8'39

9-00

100

66-6

33-3

•1974

106-5800

66-6

39-9133

8-44

9-00

100

•1754

105-2620

35-2620

8-55

9-00

100

•1439

I03-5975

28-5975

8*69

9-00

100

•1136

102-2720

22-3720

8-80

9-00

100

•0844

101-2660

16-2660

8-89

9-00

100

•0559

100-5500

10-5590

8-95

9-00

100

1-0278

100-1300

5-1390

8-09

9-00

100

1-0000

lOO'OOOO

100

o-oooo

900

9-00

" It is found by experiment, as might have been expected, that the
co-efficient representing the extra effect of the moving load on a girder
(or on a member of a bridge truss) is a variable quantity depending on
the relative amount of fixed load and moving load in the total load,
being greatest where the proportion of moving load in the total load is
comparatively high."

The first column of Table 24 represents the total nominal load or
stress in the member, due to both fixed and moving load, expressed as
100 per cent. The second and third columns give the percentage of
nominal stress due to fixed and moving load respectively. The fourth
column contains the deduced co-efficients by which the nominal stress
due to moving load is to be multiplied in order to give effective or
equivalent fixed load stress. The values in the fifth column are obtained
by multiplying the percentages in column 3 by the co-efficients in
column 4, and adding the percentages in column 2, and express the
equivalent stress due to the total nominal load in terms of fixed or
dead load stress. Columns 6 and 7 show the composition of this
equivalent stress, that is, the proportion of it due to the effect of the

LOADS AND WORKING STRESSES 47

fixed load, and that due to the effect of the moving load. Since the
values in column 5 represent equivalent fixed load stress, a uniform
working stress of 9 tons per square inch may be adopted in every
instance, and the sectional area may be computed on the total effective
load at 9 tons per square inch, or on the total nominal load at some
reduced number of tons per square inch. The reduced working stress
to be adopted in any particular case

total nominal load
= \) tons x

total effective load

Thus, for a member having a total nominal stress of 100 tons, 30 tons
being due to dead load, and 70 tons to moving load, the equivalent
dead load stress from column 5 = 137*408, and sectional area

= • — ~ — = 15*27 square inches. If, however, it is desired to calculate
the sectional area with respect to the total nominal stress of 100 tons,
the reduced working stress to be employed = of 9 = 6*55 tons

per square inch, which is found in column 8, and sectional area
= g^r = 15*27 square inches as before. It may be noted that for

the same example, the sectional area by the Launhardt formula would
be 14*5 square inches, and by Claxton Fidler's formula, 18*9 square
inches.

The other values in column 8 are deduced similarly, and the table
exhibits at a glance the unit working stress to be adopted for a given
nominal total stress, of which the percentage composition is known.
A formula giving the working stress is readily obtained by plotting
the curve showing the relation between the working stresses of column 8
and the corresponding varying percentages of fixed and moving load
constituting the total nominal load. This curve being slightly modified
to allow for a reasonable amount of shock and jarring on lighter
members subject to considerable variation of stress, the formula
eventually deduced is as follows —

Let R = proportionate range of stress

Stress due to moving load Range of stress

~ Fixed load stress -|- moving load stress ~~ Total stress

Then safe working stress per square inch

= 9 - 5R2 for mild steel
and = 7 — 4R2 for wrought iron.

EXAMPLE 6. — A mild steel member is subject to a tensile stress of
57 tons, due to moving and fixed load combined, and 18 tons of tension,
due to fixed load alone. Required the sectional area by the Rang e formula.

Stress due to moving load = 57 - 18 = 39 tons. Hence, ratio
R = ff = 0-684.

Working stress = 9 - 5 X (0'684)2 = G*66 tons per square inch,

48 STRUCTURAL ENGINEERING

,. , total nominal load 57

and net sectional area = working stress = ^j. = 8-55 square

inches.

The working stress may also be taken from Table 24, thus—

Percentage of fixed load stress = £f x 100 = 31'6
From Table 24-

Working stress for 30 per cent, fixed load = 6 -55
>' ,, 33^ ,, ,, =• 6*78

„ „ 31-6 „ „ is practically the mean of

these two values, or
6* 66 tons per square
inch.

EXAMPLE 7. — A mild steel member is subject to a fixed load stress of
40 tons of tension, and alternations of 30 tons of tension and 30 tons of
compression caused by a moving load. To deduce the sectional area.

Here maximum tensile stress = 40 -f 30 = 70 tons. When the
moving load stress of 30 tons compression comes into operation, 30 of
the 40 tons tension due to the fixed load is neutralized, leaving a
"residual fixed load stress" of 10 tons of tension. The range of
stress is therefore 60 tons, whilst the " total stress " = stress due to
moving load -f residual fixed load stress at the instant the maximum
tensile stress of 70 tons is in operation, = 60 + 10 = 70 tons. It is
to be noticed that 60 of the 70 tons of maximum stress is really due to
the action of the moving load, and that for the moment, only 10 of the
40 tons tension originally due to the fixed load is in operation as fixed
load stress.

Hence, R = f§ = 0-86
and working stress = 9 — 5 x (0'86)2 = 5'3 tons per square inch

whence sectional area = ^ = 13 -2 square inches.

The above treatment of cases of fluctuating stresses is recommended
by Mr. Stone as being the most reasonable, and employed in this
manner, the "range" formula becomes applicable to all cases of
loading.

EXAMPLE 8. — A mild steel member is subject to a fixed load stress of
20 tons compression, and a moving load stress which alternates between
80 tons of compression and 40 tons of tension.

In this case the maximum compression = 20 + 80 = 100 tons.
At the instant the 40 tons of tension is created by the action of the
moving load, the whole 20 tons compression originally due to the
fixed load, is more than neutralized, and the effect due to the weight
of the structure itself here constitutes moving load effect, since it is
applied and removed each time the moving load comes into action.
The "residual fixed load stress" is therefore zero, and "total stress"
= range of stress + residual fixed load stress = 120 -f 0 = 120 tons.

LOADS AND WORKING STRESSES 49

., ,. -r, range of stress , ,

Hence the ratio R = — . . . . — — becomes = 1, for all cases

where actual reversal of stress takes place.

.*. Working stress = 9 — 5xl2 = 4 tons per square inch, and
sectional area = -^- = 30 square inches.

The " range " formula was primarily designed for use in connexion
with the determination of sectional areas for railway bridge members,
and consequently takes into account such influences of moving load as
pitching and bad balancing of locomotives, irregularities in track and
other causes of shock. If employed for designing sectional areas of
members of structures other than bridge girders — as, for example, roofs,
crane girders, etc. — the formula may be expected to err on the safe
side, since the effect of the moving or live load on these structures
is not nearly so severe as that on railway bridges. Further, the moving
load in such cases is not applied so rapidly as on bridges, and the
dynamic stress is therefore proportionately reduced.

The three formulae cited are representative of those in general use,
which in one form or another may be relegated to one of the above
types. It should be noticed that in every case a factor of safety of 3
is uniformly employed on the " equivalent dead load " stress, whilst on
the " nominal maximum stress " a variable factor of safety results from
using one or other of the formulas.

CHAPTEE III

FIG. 12.

BENDING MOMENT AND SHEARING FORCE.

Bending Moment.— Suppose the beam in Fig. 12 to be fixed at A and
loaded with 2 tons at the free end E, then the load multiplied by its

distance from A is called the moment

,2'. O**M of the load about A. Since the effect of

this moment is to cause bending of the
beam, the moment is further called the
bending moment at A. Its value = 2
tons x 12 feet =24 foot- tons.

Since the bending moment equals

L\^ load x distance, the bending moment

^^ at B distant 9 feet from E = 2 x 9 = 18

foot-tons. Similarly at C, the bending
moment = 2 x 6 = 12 foot-tons, at
D = 2 x 3 = 6 foot-tons, and at E
= 2x0 = 0. Plotting these values
below a horizontal line, ae, the points
so obtained lie on the straight line ef,
which is such that the vertical depths

between it and ae represent the bending moments at corresponding
points along the beam. The figure aef is called the bending moment
diagram for the beam. In this case, having obtained point /, the
diagram might have been completed by joining / to e, without calcu-
lating the bending moment for any intermediate points. It will be
seen that ae represents the length of the beam to any convenient scale,
which however has no connection with the scale to which the bending
moments are plotted. For instance, if ae be made 3 inches and af
2 inches, the scale for distance along the beam would be 4 feet to
1 inch, and for the bending moments, 12 foot-tons to 1 inch. The
practical value of a bending-moment diagram is to enable the bending
moment to be scaled off for any point along the beam, the diagram
being usually readily drawn after calculating the moment at a few
points only.

Shearing Force. — Apart from the bending action, the load has
another effect on the beam as indicated at A, Fig. 13. This is the
tendency to shear the beam vertically at any section such as A or B.
Such an effect is due to vertical shearing force. In this case it is equal
in amount to 2 tons at every vertical section along the beam. The
shearing force is represented diagrammatically below the bending

BENDING MOMENT AND SHEARING FORCE 51

FIG, 13.

moment diagram in Fig. 12, where GH again represents the span and
GK is made equal to 2 tons to scale. The depth of the rectangle
GHLK is a measure of the shearing force at any
section of the beam.

Relation between Bending Moment and Shear-
ing Force. — The following relation always exists
between the bending moment and sheariug force
at any section of a beam. The area of the
shearing-force diagram, between the free end of
the beam and any vertical section, equals the
bending moment at that section. Thus in Fig. 12 the area of the
rectangle GHLK, between the free end H and the vertical section
GK = GH x GK = 12 feet x 2 tons = 24 foot-tons, which was the
value obtained for the bending moment of. Again, considering
section 0, the area of the shearing-force diagram between H and M is
the rectangle MHLN, which = 6 feet x 2 tons = 12 foot-tons, the
bending moment previously obtained for section 0.

A beam such as the above, having one end fixed and the other free,
is called a cantilever. The bending moment in this case, if expressed
in general terms = W7 foot-tons, where W = the load and I — span.

Cantilever carrying any number of Concentrated Loads. — Suppose
the cantilever in Fig. 14 to be 30 feet long and to carry loads as indi-
cated. The bending moment at D
is nothing. At C the bending
moment = 5 tons x 10 feet = 50
foot-tons, which is plotted on the
bending-moment diagram at cc'. At
B there is the combined bending
moment, due to 5 tons acting at a
leverage of 22 feet, and G tons at a
leverage of 12 feet. Hence —

Bending moment at B = 5
X 22 + 6 x 12 = 182 foot-tons,
which is plotted at IV.

The bending moment at A
= 5 X 30 + 6x20 + 8x8 = 334 foot-
tons, which is plotted at aa'. The
diagram is completed by joining
points «', V, c', d by straight lines.
Note that the inclination of the
boundary line a'b'c'd changes under each point of application of the
loads. If the loads were very numerous and close together, the broken
line a'b'c'd would approximate to a curve.

The shearing force between D and C is 5 tons. Between C and B
it is augmented by the additional load of G tons, giving 5 + 6 = 11 tons,
and between B and A this 1 1 tons is further augmented by the load
of 8 tons, giving 5 + 6 + 8 = 19 tons. The construction of the
shearing-force diagram is indicated in the lower figure.

Cantilever carrying a Uniformly Distributed Load.— A distributed
load is any load which is applied continuously along the whole length of
a beam. In practice the majority of distributed loads to be considered

FIG. 14.

STRUCTURAL ENGINEERING

are uniformly distributed loads ; that is, each foot length of the beam
carries an equal amount of load, such, for example, as a beam supporting
a wall of uniform height. The dead weight of the beam itself, if of
uniform section, is also a uniform load.

Suppose the cantilever in Fig. 15 to be 16 feet long, and to carry
a load of 2 tons per foot run. The bending moment at A will be the

same as if the whole load were con-
centrated at its centre of gravity. The
total load = 16 x 2 = 32 tons, and its
centre of gravity is situated at the

-r-, 1 f sc^= — '" middle of its length ; that is, 8 feet

\.S^* from A. The bending moment at A

5 Jfa therefore = 32 x 8 = 256 foot- tons.

' / At the section B, the portion of the

load creating bending moment is that
distributed over the 12 feet length BE,
having its centre of gravity distant
0 feet from B.

/. bending moment at B

= (12 x 2) tons x 6 feet
= 144 foot-tons.

PIG. 15. Similarly, at section C, the bending

moment = (8x2) tons x 4 feet = 64

foot-tons, and at section D the bending moment = (4x2) tons x 2 feet
= 16 foot-tons. At E the moment is 0. These values are plotted as
before at aa\ ~bV, etc.

If, instead of considering only four sections of the beam, a very
great number were taken and the resulting bending moments plotted,
the points so obtained would be found to lie on an even curve passing
through a'b'c'd'e, which in this case is a semi-parabola tangent to the
horizontal line ae at e. Knowing this to be so, it is only necessary to
calculate the bending moment at A and to fit in the curve by the
geometrical contraction shown in Fig. 16. Set
out aa' = 256 ft.-tons to scale and ae — 16 ft.
to scale. Divide ae into any number of equal
parts and act' into the same number of equal
parts. Draw verticals through #, 5, c, and d,
and join e to points 1, 2, 3 on aa'. The inter-
sections of el with bV, e'2 with cc', and e3 with
FIG. 16. dd', give points #', c', a", through which a free

curve is drawn. More frequent points on the

curve may be obtained by dividing ae and aa' into a greater number of
parts.

The shearing force is nothing at the free end E, and increases by 2
tons for each foot length of the cantilever from E to A, so that at D it
is4:X 2 = 8 tons, at (J, 8 X 2 = 16 tons, at B, 24 tons, and at A, 32
tons. The S.F. diagram is therefore the triangle in the lower figure.
Note the relation between the two diagrams. The area of the S.F.

diagram between the free end G and centre of beam = — ^ — = 64

BENDING MOMENT AND SHEARING FORCE

FIG. 17.

ft.-tons, or the value of the B.M. obtained for section C. Expressed in

tvP
general terms, the B.M. at the fixed end = -^-, where w = load per

foot run and I = span in feet.

Beam supported at both Ends and carrying a Concentrated Load
at the Centre. — Suppose the beam in Fig. 17 to be 50 feet span, and
loaded at the centre with 8 tons.
Obviously half the load will be carried
by each support, giving rise to equal
upward reactions of 4 tons. If the
beam were inverted it would be similar
to a balanced cantilever supported at
the centre and having each end loaded
with 4 tons. The B.M. at the centre
therefore = 4 tons X 25 ft. = 100 ft.-
tons, or ^ load x i span. In this case
the beam bends with its concave side
uppermost, whereas in the cantilevers
previously considered the bending took
place with the concave side downwards.
It is customary to distinguish between
these two kinds of bending action by

designating the bending in Fig. 18, A as positive, and Fig. 18, B as
negative. Diagrams of positive bending moments are plotted above the
horizontal line representing the span, and negative
moments Mow.

The shearing force from the left-hand abut-
ment to the centre of the beam is 4 tons acting
upwards on the left of any section such as A,
Fig. 17, and downwards on the right of such
section. At the centre of the beam the upward
shear of 4 tons is more than neutralized by the downward acting load
of 8 tons, so that the shear now acts downwards on the left of any
section B, to the right of the centre and upwards on the right of such
section. As the relative direction of the shear reverses at the centre, it
is customary to designate the shearing force from the left abutment
to the centre of the beam as positive, and from the centre to the right
abutment as negative. From the centre to the right abutment the
vertical shear of 4 tons acts downwards on the left of every section,
until at the abutment it is neutralized by the upward reaction of 4
tons and becomes reduced to zero. This is shown on the S.F. diagram
by plotting CD = 4 tons above the horizontal line CE, and EF = 4
tons beloiv CE, and completing the diagram as shown. It should
be noted that, immediately under the load (in this case at the centre
of the beam), the diagram exhibits a positive shear of 4 tons and
a negative shear of 4 tons, so the actual resultant shear is + 4 - 4
= 0. This zero shear only exists at the central vertical section of the
beam, and, according to the diagram, it suddenly increases to the full
value of 4 tons on sections immediately to the right or left of the centre.
This state of shear only exists on the assumption that the load is applied
at a single point on the beam, a condition impossible of realization in

FIG. 18.

STRUCTURAL ENGINEERING

actual practice, since all loads must have an appreciable area of bearing
on the beams supporting them. As this point frequently gives rise to
some difficulty in correctly interpreting the meaning of shear force
diagrams, a reference to the following practical example will render it
more clear. Suppose the load of 8 tons in Fig. 19 to be applied by a
rolled joist having flanges 8 inches wide. The bearing area on the
beam AB is now 8 inches wide instead of being a single point, and the
" concentrated " load of 8 tons is actually a distributed load of 1 ton
per inch run over the central 8 inches of the length of AB. From
abutment A to the left-hand edge of the lower flange of the joist,
the shear is -f 4 tons as before. At 1 inch in from the edge of the
flange, a downward load of 1 ton has been applied, reducing the shear
to 4- o tons. At 2 inches in, the shear is further reduced by another
ton, and equals -f 2 tons ; at 3 inches it is reduced to + 1 ton, and
at the centre to zero. At 5 inches from the left-hand edge of
flange, 5 tons of downward acting load having been applied, the
shear is + 4 — 5 = — 1 ton ; at 6 inches it is + 4 — 6 = — 2 tons, etc.

6 ton*

8 fans.

FIG. 19.

FIG. 20.

Thus the shear is really gradually decreased from + 4 tons at the
left-hand edge of bearing to — 4 at the right-hand edge, and cannot
change abruptly as implied by the vertical steps on S.F. diagrams for
concentrated loads as usually drawn. The same is true for the bearing
surfaces on the supports, the shear increasing gradually from C to D
over the horizontal bearing of 8".

Beam supported at both Ends and carrying a Concentrated Load
at any Point.— Suppose the beam AB in Fig. 20 to be 40 ft. span, and
to carry a concentrated load of 10 tons at 24 feet from A. The reaction
at B (usually denoted RB) is found by taking moments about the
opposite end A of the beam. Thus—

RB x 40' = 10 tons X 24'
/. RB = 10 X ff = 6 tons.

The reaction RA at A must be the difference between the total load and
RB, or 10 - 6 = 4 tons. The B.M. at C = 6 tons x 16 ft. = 96 ft.-
tons, which enables the B.M. diagram to be plotted. The B.M. at C

BENDING MOMENT AND SHEARING FORCE

may also be obtained by working from the reaction at A. Thus 4 tons
X 24 ft. = 96 ft.-tons as before. The S.F. is + 4 tons from A to C,
and — 6 tons from C to B, and the diagram is plotted as shown. The
reactions at A and B are always inversely proportional to the segments
into which the span is divided by the load point C. Thus |§ of 10 tons
gives RA, the reaction at the opposite end to the 16 ft. segment, and fj
of 10 tons gives RB. By remembering this rule, the reactions may
readily be written down from a simple inspection of the position of the
load. Note. — The areas of the positive and negative portions of the
S.F. diagram for any beam are always equal, since each is a measure of
the B.M. at the same point.

Beam supported at both Ends and carrying any number of Con-
centrated Loads.— Suppose the beam in Fig. 21 to be 80 ft. span and
to carry loads as indicated.

Reaction RA = •— of 10 tons
+ fg of 16 tons + |§ of 6 tons
= 2 + 10 + 4J = 164 tons.
RB = 6 + 16 + 10 -161 = 15i
tons.

B.M. at E = RB x 16 ft.
= 154 X 16 = 248 ft.-tons.

Considering section D, the
reaction RB tends to bend the
length DB upwards, whilst the
intermediate load of 10 tons
acting at a leverage of 50 ft.
- 16 ft. = 34 ft., tends to
hold down the length DB, and
the B.M. at D will = RB x 50
ft. - 10 tons X 34 ft. = 154
X 50 - 340 = 435 ft.-tons.

The Bending Moment at C

is most readily obtained from FlG 2i

RA x 20ft. = 164 X 20 = 330

ft.-tons. The same result will be obtained if calculated from the oppo-
site end B of the beam. Thus B.M. at C = RB x 60 - 10 X 44 - 16
X 10 = 330 ft.-tons as before. Plotting 330, 435 and 248 ft.-tons
beneath C, D and E respectively, the B.M. diagram is obtained.

For the S.F. diagram, the shear from A to C = +164 tons- At C
it is reduced by 6 tons, giving +164 — 6= +104 tons> tne shear from
C to D. From D to E, the S.F. = + 104 - 16 = - 5j tons» and
from E to B, —54 — 10 = —154 fcons> which checks with the reaction
of 154 tons at B.

Beam supported at both Ends and carrying a uniformly distri-
buted Load over the Whole Span.— Suppose the beam AB in Fig. 22
to be 60 ft. span, and to carry a load of 14 tons per foot run. RA and
RB each equal half the total load = J x 60 x 14 = 45 tons. The
upward B.M. at C = 45 tons x 30 ft. = 1350 ft.-tons. The downicard
B.M. at C is due to the load of 45 tons, extending over CB, which may
be supposed concentrated at its centre of gravity distant 15 ft. from C.

.'. B.M. at C = 45 x 30 - 45 x 15 = 675 ft.-tons.

STRUCTURAL ENGINEERING

If the B.M. be calculated for several intermediate points between
B and C in a similar manner to that adopted for the cantilever in

YS^S/,//////^/S,,,,^S,SS,,/S,~,S,S,,,,,,S,A ^>* 15' fc^e Pl°^e(l values will be
^m Y//mm^/////^^^^^ t found to range themselves on a

parabola passing through abc.
Hence make cm = 675 ft.-tons, and
draw in each half of the curve by
the geometrical method previously
given. (It may be mentioned liere
that all B.M. diagrams for uniformly
distributed loads are bounded by
parabolic curves, and that diagrams
for concentrated loads are bounded
by straight lines.) The shearing
force at each abutment is equal to
the reaction of 45 tons, and de-
creases uniformly to zero at the
middle of the span. Expressed in
general terms the B.M. at the centre
w1
= -g- > w and I having the same significance as before.

Beam supported at both Ends, and carrying a uniformly distributed
Load over a Portion of the Span.— Suppose the beam in Fig. 23 to

FIG. 22.

FIG. 23.

be 100 ft. span, and to carry a load of 2 tons per foot run over a
length of 40 ft. as indicated. The reactions at A and B are the
same as if the total load were concentrated at the centre of gravity
of the distributed load on CD. Total load = 40x2 = 80 tons.
.*. RA = ^ of 80 = 48 tons, and RB = 32 tons. The B.M. at M,
supposing the load to be concentrated at M, would be 32 tons
X 60 ft. = 1920 ft.-tons, and triangle aeb> obtained by making

BENDING MOMENT AND SHEARING FORCE

me = 1920 ft.-tons, would be the B.M. diagram. Project C and D to
c and d respectively. Join c to d, cutting me in h. Bisect eh in Jc. cf
and dg are the bending moments at C and D, whether the load be all
concentrated at M, or distributed between C and D. Since the load is
actually distributed over CD and not concentrated at M, as above
supposed, the boundary of the required diagram from c to d will be
a parabolic curve passing through cM. The curve is drawn by the
same method as in Fig. 16, but is oblique, instead of rectangular, as
shown in the inset. The boundary of the diagram for the whole span
is then ackdb. This may be verified by calculating the moments at C,
M, and D. Thus—

B.M. at C = 48 x 20 = 960 ft.-tons.
D = 32 x 40 = 1280 „
M = 32 x 60 - (20 ft. x 2 tons) x 10 = 1520 ft.-tons.

These values will be found to agree with the scaled moments cf, dff, and
km. The shearing force from A to C equals the reaction at A, 48 tons.
From C towards D it diminishes at the rate of 2 tons per foot, so that
at 24 ft. from C the S.F. is zero. It further diminishes to —32 tons at
D, and remains constant from D to B. The maximum B.M. occurs at
P, 44 ft. from A, and vertically above p, where the S.F. diagram cuts
the horizontal base line, and is given by the area of the shear force
diagram between either end of the beam and pointy.

.*. Max. B.M. = area rect. qrst -f area triangle pqt

= 48 X 20 + i x 48 X 24 = 1536 ft.-tons,

which should correspond with the maximum scaled moment xy.

Note. — The following is a

special case of the above when 70' *j

the load extends from one abut-
ment partly over the span,
Fig. 24.

Taking the same span with
a load of 2 tons per ft. run, ex-
tending 70 ft. from A, the
same construction applies.
EA = ^> of (70x2) =91 tons,
and RB = 140 - 91 = 49 tons.

B.M. at M, supposing the
load concentrated = 49 X 65
= 3185 ft.-tons. Make me
— 3185 to scale. Join ae and
be. Project D to d. Join ad,
bisect eh in k and draw the
parabola akd.

The shearing force is + 91
tons at A, diminishing to zero
at P, distant 45-5 ft. from A,

and further diminishing to - 49 tons at D, then remaining constant
from D to B. The maximum B.M. occurs at P, and is given by area

FIG. 24.

STRUCTURAL ENGINEERING

12 forts.

of triangle prs = J x 91 X 45-5 = 2070J ft.-tons, which should corre-
spond with the scaled moment xy.

The bending moment on girders carrying concentrated loads consists

of two portions — that due to
the dead weight of the girder,
usually considered as a distri-
buted load, and that due to the
concentrated loads. In drawing
the diagram of moments, it is
convenient first to construct
separate diagrams for the dis-
tributed and concentrated
loads, and then to combine
them in order to obtain the
total B.M. diagram. The
following example illustrates
this method. Suppose the
girder in Fig. 25 weighs 0'25
ton per foot-run, and carries
concentrated loads as indi-
cated.

The B.M. at the middle of
the span due to the weight
of the girder

FIG. 25.

0-25 X 60 X 60

= 112-5 ft.-tons.

Make me = 112-5 ft.-tons to scale, and draw the parabola acb. The
reaction at A due to the concentrated loads at D and E = f J x 6
+ if X 12 = 5-7 tons. RB = 18 - 5'7 = 12-3 tons. B.M. at D due
to the concentrated loads alone = 5*7 X 33 = 188 '1 ft.-tons. B.M. at
E = 12-3 X 15 = 184-5 ft. -tons. Make df= 188-1, andcgr = 184*5 ft.-
tons to the same scale as used for me. Join afgl.

The total B.M. at E = ek + eg = ep.
-

M = me + mn = mr.

Other points between a and r and b and p may be found similarly, and
will lie on the curved boundary lines aq, qp, and pb, which complete the
total B.M. diagram.

For the S.F. diagram, the total reaction at A = 5-7 tons -f-
K60 X 1) = 13-2 tons, which equals the S.F. at A. From A to D the
S.F. diminishes at the rate of 0%25 ton per foot, and therefore
immediately to the left of D equals +13'2 - (33 X 0'25) = -J-4'95
tons. The load at D further reduces it by 6 tons, so that immediately
to the right of D the S.F. = -f 4-95 - 6'0 = -1'05 tons. From D
to E it is reduced by a 12-ft. length of the distributed load of J ton per
ft., and immediately to left of E the S.F. = -1-05 - (12 x 0'25)
= -4-05 tons. Immediately to right of E the S.F. = -4*05 - 12'0
= —16*05 tons, which is still further reduced between E and B by

BENDING MOMENT AND SHEARING FORGE

(15 X 0-25) = 3-75 tons, giving -16'05 -3'75 = -19-8 tons, which
checks with the reaction at B.

Bending Moment and Shear Force Diagrams for Balanced Canti-
levers.— In applying the cantilever principle to bridges, one or other
of the two arrangements in Fig. 26 is adopted.

Fig. 26, A, shows the elevation of a cantilever bridge in which a
single cantilever girder, GH, rests on two piers at P and Q, whilst the
outer ends GP and HQ of the girder project beyond the piers. The
two remaining openings KG and HL are each bridged by an indepen-
dent girder, one end of which rests on an abutment and the other on one
of the projecting arms of the cantilever. The condition of loading in
this case is as shown in the lower diagram, where the ends g and h
of the projecting arms each carry a concentrated load equal to one-half

y yr/r/nrs^^^^^

y/////////////^ *

iw! \fl 1 7

liw

that on the independent girders, whilst the cantilever girder GH further
supports its own weight together with the live load placed upon it.

In Fig. 26, B, two cantilever girders CD and EF are employed,
each supported at the centre by piers P and Q. The central opening
DE is bridged over by an independent girder resting on the cantilever
arms at D and E. The tendency of the weight of this central girder to
overbalance the cantilevers is counteracted by suitable balance weights
or " kentledge " applied at the ends 0 and F. The condition of loading
of the cantilevers CD and EF is then as shown in the lower diagram,
where the arms d and e each carry one-half of the weight W of the
central girder DE, and the opposite arms c and / each carry a similar
amount of balance weight.

The method of drawing the B.M. and S.F. diagrams for the
cantilever in case A is as follows. In this example the dead load
alone will be considered. Suppose the bridge to have the dimensions
indicated in Fig. 26, A, and the dead load to be 2 tons per foot run.

STRUCTURAL ENGINEERING

The weight of one of the detached spans KG or HL is 300 x 2 = 600
tons. One half this weight is concentrated on each end G and H of
the cantilever, whilst the whole length of 820 ft. is further loaded with
2 tons per foot run.

In Pig. 27 the B.M. at P and Q due to the concentrated loads
alone = 300 X 160 = 48,000 ft.-tons. Set off pr and qs each = 48,000
ft.-tons to scale. The B.M. at P and Q due to the distributed load on
the overhanging arms GP and QH = (160 X 2) X 80 ft. = 25,600 ft.-
tons. Set off pa and qb each = 25,600 ft.-tons to the same scale as pr
and qs. Since both these moments are acting simultaneously, the two
diagrams gpa and gpr require combining in the same manner as in Fig.
25, by adding together their vertical depths at several points. The

-500

-620

FIG. 27.

curve gt is the result, and the curve hv will be similar. The total
moment at P or Q = -48,000 -25,600 = -73,600 ft.-tons = pt.
The total load on the girder GH = 300 + 300 + (820 x 2) = 2240
tons. Since this load is symmetrically disposed, one-half or 1120 tons
will be borne by each of the piers P and Q. Considering the right-
hand half MH, the upward reaction at Q = 1120 tons. To obtain the
B.M. at M the middle of the span, take moments about M. The
downward acting moments = 300 tons x 410 ft. = 123,000 ft.-tons due
to the concentrated load at H + (410 x 2) tons x 205 ft. (the distance
of the e.g. of the distributed load on MH, from M) = 291,100 ft.-tons.
The upward acting moment = 1120 tons x 250 ft. = 280,000 ft.-tons.
The downward acting moment being the greater, the resultant moment
at M = - 291,100 + 280,000 = - 11,100 ft.-tons. Set off mn =

BENDING MOMENT AND SHEARING FOECE

11,100 ft.-tons, below the horizontal (since negative), and draw a parabola
through points /, M, and v. The complete B.M. diagram for the
cantilever is then the figure ytnvh, the moments being measured
vertically below gli.

The S.F. immediately to the right of G is - 300 tons. From G to
P a further downward load of (160 x 2) tons is to be subtracted,
giving - 300 - 320 = - 620 tons immediately to the left of pier P.
At P an upward force of + 1120 tons is applied, giving — 620 -f 1120
= + 500 tons immediately to the right of P. Between P and M this
is gradually diminished by the downward acting load of (250 x 2) tons,
giving -f- 500 — 500 = 0 at M. Similarly the shearing force falls to
— 500 tons immediately to the left of Q, becomes — 500 + 1120

400

-400

200

:>S$^^v$$S^$$§sS^^^

E |<? F

-400

+4CO

-2000

FIG. 28.

= + 620 tons to the right of Q, and falls to + 620 - 320 = + 300
tons immediately to the left of H, finally becoming = 4- 300 - 300 = 0,
at H, the point of application of the concentrated load of 300 tons.
Note that the area cdkf of the shear-force diagram between the free end
of the girder and the section over the pier P, equals the B.M. at P.

Thus area cdef = 300 x 160 = 48,000 ft.-tons,

area efk = 320 X — = 25,600 „
fi

giving area cdJcf = 73,600 ft.-tons, the value previously deduced for the
B.M. at P.

Considering next the cantilevers in Fig. 26, B, their outline is
reproduced in Fig. 28. Suppose the bridge to carry a dead load of

62 STRUCTURAL ENGINEERING

4 tons per foot run. The weight of the detached central girder = 200 x 4
= 800 tons. One-half this weight, or 400 tons, is applied as a con-
centrated load at D and E, and will require a balancing weight or
downward pull applied by means of anchor ties at each of the points
CandF.

The B.M. at P due to this concentrated load of 400 tons = 400 x
400' = 160,000 ft.-tons. Set off pr = 160,000 ft.-tons and join cr and
dr. The distributed load between 0 and P = 400 x 4 = 1600 tons,
and its moment about P considering it concentrated at its e.g., 200 ft.
from P = 1600 x 200 =320,000 ft.-tons. Set oft ps = 320,000 ft.-tons,
and draw the semi-parabolas cs and ds. Combining the two diagrams
as before, the resulting curves are cm and dm. The maximum B.M. at
P =pr + ps = 160,000 -f- 320,000 = 480,000 ft.-tons of negative moment.
A similar diagram efn will, of course, obtain for the cantilever EF. For
the shearing force, the total load on one pier = 400 (balance weight)
-f (800 X 4) distributed weight -f 400 (half-weight of detached span)
= 4000 tons = the upward reaction at P. Below C, set off — 400
tons. Between C and P a further downward acting load of (400 x 4)
tons gives a total negative shear of -400 - 1600 = - 2000 tons. The
upward reaction of -f 4000 tons converts this to + 2000 tons, which
again diminishes to + 400 tons at D.

Cantilever Girder with Unsymmetrical Load. — Taking the dimen-
sions of the cantilever bridge in Fig. 26, A, suppose the right-hand
half of the bridge to carry an additional uniformly distributed load of
2 tons per foot run. This would correspond fairly closely with the case
of two trains extending from the middle point M to the right-hand
abutment. The left-hand detached span KG weighs 600 tons as before,
so that the end G of the arm PG carries a concentrated load of 300
tons. The right-hand detached span HL, together with its additional
load of 2 tons per ft. run, now weighs 1200 tons, so that the end H of
the arm QH carries a concentrated load of 600 tons. The cantilever
girder GH also carries a load of 2 tons per foot run from G to M, and a
load of 4 tons per foot run from M to H. These loads are indicated in
Fig. 29. The loading being unsymmetrical, the reactions at P and Q
will no longer be equal. To obtain the reaction at Q, take moments
about the point P.

RQ X 500' + 300 X 160' +320 X 80' = 500 X 125' + 600 X 660' + 1640 x 455',

from which RQ = 2262-2 tons. The total load on the cantilever = 300
4. 000 + (410 X 2) + (410 X 4) = 3360 tons.

/. Reaction RP = 3360 - 2262'2 = 1097'8 tons.

The B.M. at P due to the concentrated load of 300 tons at G
= 300 x 160' = 48,000 ft.-tons. Set off pr = 48,000 ft.-tons and
join gr. The B.M. at P due to the distributed load of 2 tons per foot
extending over the arm GP = 320 x 80' = 25,600 ft.-tons. Set off
pa = 25,600 ft.-tons and draw the semi-parabola ga. Combining ga
and gr the resultant curve gt is obtained, the negative moment at the
pier P being = pa + pr = pt, or 25,600 -f 48,000 = 73,600 ft.-tons.
This portion of the diagram is identical with that in Fig. 27, since
the loading on the arm PG has not been altered.

BENDING MOMENT AND SHEARING FORCE

Proceeding similarly for the arm QH, the B.M. at Q due to the
concentrated load of 600 tons at H = 600 X 160' = 96,000 ft.-tons.
This is set off to scale at qs and hs joined. The B.M. at Q due
to the distributed load of 4 tons per foot run extending from Q to
H = 640 x 80' = 51,200 ft.-tons. This is set off to scale at qb, and
the semi-parabola drawn from I to h. Combining Jib and hs, the
curve hv is obtained, the negative moment at the pier Q being
= qb + qs = qv, or 96,000 -f 51,200 = 147,200 ft. -tons. The moments
between P and Q will be represented by a parabolic curve from t to v.

-300

-/d22-2

FIG. 29.

If the position of the point n on this curve, midway between P and
Q, be determined, the parabola may be drawn geometrically through
points t, n, and v. The B.M. at M, calculating from the left-hand side,
= - (300 X 410' + 820 X 205') acting downwards -f 1097'8 x 250'
(the moment of RP acting upwards) = - 16,650 ft.-tons negative
moment. Set off mn = 16,650 ft.-tons and draw the parabola tnv.
The complete diagram of moments is then bounded by gtnvh. Note
that if the moment at M had been positive, mn would have been set
off above the horizontal gh, and the parabola tnv would cut gh and
project above it.

64 STRUCTURAL ENGINEERING

The shearing force at G = — 300 tons

immediately to left of P = - 300 - (160 X 2) = - 620 tons
right of P = - 620 + 1097'8 = + 477'8

at M = + 477-8 - (250 X 2) = - 22'2 „

left of Q = - 22'2 - (250 X 4) = - 1022'2 „

„ right of Q = - 1022-2 + 2262-2 = + 1240 „

and at H = -f 1240 - (160 x 4) = + 600 „

which agrees with the concentrated load at the end of the arm QH.
The maximum negative moments occur at the piers where the shear
force diagram crosses the horizontal line. The minimum negative
moment between P and Q occurs at X, vertically over the point where
the shear force diagram crosses the horizontal a little to the left of M,
and equals xy to scale. Its value may also be obtained from the area
of the shear force diagram between G and X. Thus PX scaled or
calculated from the S.F. diagram = 238*9 feet. Then-
Negative area from G to P (dotted shading) = — ~ X 160

= - 73,600 ft.-tons.
Positive area from P to X (full shading) = + i x 477'8 x 238'9

= + 57,073-2 ft.-tons.

.*. Minimum negative moment at X = - 73,600 4- 57,073'2

= - 16,526-8 ft.-tons.

Bending Moment and Shearing Force Diagrams for Rolling Loads.
— The following simple cases of rolling loads will illustrate the methods
of drawing the B.M. and S.F. diagrams for concentrated loads which,
rolling over a span, successively occupy a number of different positions
upon it. Such cases occur when a locomotive crosses a bridge, or a
crane traveller moves from one position to another on the crane girders.
The concentrated loads are the weights supported by the various axles
and wheels. The intervals between the loads remain constant, dependent
upon the design of the locomotive or crane, etc. The treatment of
rolling loads will be best understood by considering first the case of a
single concentrated load rolling over a span.

Beam supported at both Ends and carrying a Single Concentrated
Rolling Load. — Suppose the beam in Fig. 30 to be 60 feet span, and
the load 6 tons. Divide the span into, say, six equal intervals of
10 feet, by lines I, II, III, IV, Y. These indicate five successive
positions of the load, and by calculating the B.M. for each position,
five values will be obtained, which being plotted and connected by a
curve, will give a diagram showing the B.M. for all intermediate
positions of the load. Thus -

Load at I, RA = 5 tons, and B.M. at I = 5 x 10 = 50 ft.-tons.

This is plotted to scale at 1 — 1' on the B.M. diagram, and al'b is the
B.M. diagram for the load standing at position I on the beam. Load
at II, R^ = 4 tons, and B.M. at II = 4 x 20 = 80 ft.-tons. Plot
2-2' ='80 ft.-tonp. Load at III, RA = 3 tons, and B.M. at III
= 3 x 30 = 90 ft.-tons, plotted at 3 - 3'. Load at IV, RA = 2 tons.

BENDING MOMENT AND SHEARING FORCE

65
Y,

B.M. at IV = 2 x 40 = 80 ft. -tons, plotted at 4 - 4'. Load at
RA = 1 ton. B.M. at Y = 1 x 50 = 50 ft.-tons, plotted at 5 - 5'.

The curve connecting these five points and the ends of the span
includes all the possible B.M. diagrams for the load occupying any
position on the span, and its height above the base line ab at any point
therefore gives the B.M. at the instant the rolling load passes that
particular point. The curve is a parabola, so that it is only necessary
to calculate the B.M. 3 — 3' when the load is at the centre, and then
draw in a parabola through the points a3'b. The triangular diagram

PIG. 30.

is obviously the same as that for a load at rest on the centre of a
beam, as in Fig. 17, and the height 3 — 3' of the parabola in this case

is therefore equal to — '. The parabola here is not to be confused with

the parabola showing the B.M. for a distributed load at rest. Here, the
parabola is the enclosing curve of a large number of individual triangular
diagrams, and has no connexion whatever with the diagram for a
distributed load.

The shearing force diagram is shown in the lower figure. "With
the load immediately to the right of point A, the whole 6 tons is
carried by the abutment A, giving a shear of -f- 6 tons at A, and nothing

66 STRUCTURAL ENGINEERING

at B. With the load at I, the shear at A equals the reaction of 5 tons,
and at B it equals — 1 ton, the diagram for this position being indicated
by the thin full line. With the load at II, the shear at A = 4- 4 tons,
and at B = — 2 tons, indicated by the long-dotted line. At position III,
the shear at A = + 3 tons, and at B, — 3 tons, the diagram being
shown by the heavy full line. With the load at IV the shear diagram
is as shown by the chain-dotted line, and at V by the short-dotted
line. The lines CD and EF, enclosing all the possible shear diagrams
as the load changes its position, give the maximum positive and
negative shearing forces for every position of the load. The shearing
forces are scaled off above or Mow the horizontal base line ED according
as the shear to left or right of the load is required, and not by scaling
the distance between CD and EF, which is of course everywhere equal
to 6 tons.

The following distinction between B.M. and S.F. diagrams for
stationary and rolling loads should be carefully noted. For a stationary
load the diagrams indicate the B.M. and S.F. existing simultaneously
at every point along the beam due to the particular position of the
load, which position is permanent. The rolling load diagrams are in
every case enveloping diagrams, such that the particular B.M. or S.F.
diagram consequent upon any specified position of the rolling load may
be readily obtained from them by projection. Thus in Fig. 30, the
parabola fll'2'3'4'5'# does not mean that moments = 1 — 1', 2 — 2',
3 — 3', etc., exist simultaneously, but that when the load is in position
I, the bending moment at section 1 = 1 — 1', at section II = r2, at
section III = s3, and at section IV = w4. Similarly when the load is
in position IV, the B.M. at section IV = 4 — 4', but at section II the
B.M. = r2, not 2 - 2'. With load at V, the B.M. at section V = 5 - 5',
and at section III = s3. With regard to the shearing force, when the
load is in position IV, the shearing force diagram is as shown by the
chain-dotted line, obtained by projecting point IV on to the lines CD
and EF, and ruling in the horizontals through + 2 and — 4. Thus
for this position of the load the shear from A to IV is constant and
= + 2 tons, whilst from IV to B it is constant and = — 4 tons.

In order to ascertain the B.M. at any point on a beam due to a
concentrated rolling load at any other point, the following simple
method obtains. Suppose, in Fig. 30, the load to be at position IV,
and the B.M. is required at II and III. Project point IV on to the
enveloping parabola, and join 4' to «, cutting verticals through II and
and III in r and c. The vertical heights 2r and 3c give the required
moments at sections II and III respectively. The application of this
method will be necessary in constructing the diagrams for two or more
rolling loads.

Beam supported at both Ends and carrying Two Unequal Con-
centrated Rolling Loads at a Constant Distance apart. — Suppose the
beam in Fig. 31 to be 50 feet span, the loads 5 tons and 10 tons
separated by a distance of 10 feet, and that they roll over the span
from left to right with the 5 ton load leading. Such a case represents,
for instance, the passage of a 15-ton traction engine over a 50-foot span.
Consider the loads as occupying a number of successive positions. In
the figure, ten positions distant 5 feet apart have been taken.

BENDING MOMENT AND SHEARING FORCE

The corresponding positions of the two loads for each 10 ft. of the
span are indicated by circles of distinctive lining. When two concen-
trated loads roll over a span, the maximum B.M. is found to occur
under the heavier load for certain positions of the loads, but under the
lighter load for certain other positions. It is consequently necessary

; j

fa \

*-M . 'f *

i i

V \

r v

1 V

II V

Hi 1

))

(

$ss

^ i

k._

FIG. 31.

to trace out the bending moments which occur beneath each load, two
distinct curves being obtained, an inspection of which at once indicates
the positions of the loads which give rise to the maximum moments,
as also the value of the moments. First draw the two parabolas acb
and ff56, representing the moments due to each load rolling over the

10 x 50
span separately. For the 10 tons load, me = - —— 125ft.-tons,

5 X 50
and for the 5 tons load w5 = - - = 62 J ft.-tons. Next consider the

68 STRUCTURAL ENGINEERING

loads approaching the span from the left hand. The 5 tons load rolls
over the first interval of 10 ft. before the 10 tons load reaches the
span. The arc a2 therefore constitutes the curve of moments under
the smaller axle for the interval A II. As the smaller load passes
beyond section I the larger load also rolls on to the span and causes
additional B.M. at the section occupied by the smaller load. When
the smaller load is at section III, the B.M. at III due to its own
weight = e3. The larger load is then at section 1, 10 ft. behind. Here,
it alone creates a moment =fg. Joining g to #, the height eh cut off
on the vertical through III (the position of the leading load) gives the
additional moment at section III due to the 10 tons load at section I.
The total moment at section III due to both loads then = e3 + eh.
Making 3k = eh, ek represents this total moment, and the curve is
sketched from 2 to k. When the smaller load arrives at section IV,
the moment due to its own weight alone = 04. The moment at
section II, ten feet behind, due to the larger load = In. Joining n to
#, the height 04 cut off on the vertical through IV gives the additional
moment at IV due to the 10 tons load at II. Making ±p = o±,p gives
another point on the curve. Repeating the construction for the
remaining sections, m5 = moment due to smaller load at V, and mq the
additional moment at V due to larger load at III. Adding m5 and mq,
point r is obtained, and so on. The curve a2Jcprb shown by the heavy
dotted line indicates the bending moments which occur under the
smaller load during the complete transit of both the loads across the span.

A similar construction is now made for ascertaining the moments
which occur under the larger load. If the loads be assumed to run
back over the span from right to left, the larger one leading, the
construction is identical, but is worked in the reverse direction. Thus
for the first 10 ft. from B to VIII, only the 10 tons load is on the
span, and the curve from b to s gives its moments. When the larger
load arrives at VII, its own moment = tu and the moment of the
smaller load at IX = z;9. Joining 9 to fl, tw is the additional moment
at VII due to the smaller load at IX, which being added to tu at ux
gives another point along the curve from s. Repeating the construc-
tion, the curve bsxyza, shown by the heavy full line, indicates the
bending moments under the heavier load during the transit of the two
loads. The two curves intersect at E, and it is at once seen that the
maximum moment occurs under the 10 tons load from a to E ; but from
E to b, the greater moments occur under the 5 tons load as shown by
the dotted curve from E to b falling outside the full curve Es#. The
complete curve of maximum moments is therefore az&b. If the loads
are free to pass over the span in the reverse direction, that is from left
to right with the larger load leading, the curve az&b would be reversed
left for right, when the branch to the right of the centre line mM.
would be similar to the left-hand branch azM. This will generally
be the case in practice.

If the two loads be added together and considered as a single
concentrated load of 15 tons, the outermost curve aLb would represent
the bending moments due to such a load rolling over the span. Its

central height mL= 15 * 5°= 187| ft.-tons. The height of this

BENDING MOMENT AND SHEARING FORCE 69

curve will not greatly exceed the height Pz of the actual curve of
moments, provided the distance between the two loads is small compared
with the span. In that case the curve aLJ) is sometimes substituted
for the real curve of moments having two halves each similar
to azM.

The S.F. diagrams are drawn in the lower figure corresponding to
the positions II, IV, VI, VIII and B of the smaller load, the same
type of line being employed as used for the circles denoting the
positions of the loads in the upper figure. With the larger load at A
and the smaller at II, the reaction and therefore the shear at A = 10
tons x f§ of 5 tons = + 14 tons. The S.F. diagram for this position
is indicated by the chain-dotted lines. HF = + 14 tons, which
immediately to the right of A is reduced by 10 tons, giving -f 4 tons,
which remains constant between A and II. At II the shear is further
reduced by 5 tons, giving -f 4 — 5 = — 1 ton, which remains constant
from II to B. The other diagrams will be readily traced, remembering
that the shear at A is, in every case, equal to the reaction at the
support due to any position of the loads. Thus taking smaller load at
VIII and larger at VI, indicated by the heavy line, the shear at A
= f§ of 10 tons + ~ of 5 tons = + 5 tons, which is constant from A to
VI, then reduced by 10 tons, giving — 5 tons up to VIII, where it is
again reduced by 5 tons, giving — 10 tons from VIII to B. The lines
FRG and HSK enclosing all the possible shear diagrams complete the
figure. It should be noticed that the boundary lines are broken at R
and S distant 10 ffc. or the load interval from the ends of the span.
Also that if KS and FR be produced to meet the horizontal base line
HG in points C and D respectively, the points C and D fall at distances
of 6f and 3^ ft. respectively from H and G. These distances are the
segments into which the load interval of 10 ft. is divided by the centre
of gravity of the two loads, and the loads are indicated by circles in
the lower figure, in positions for projecting the points C and D on to
the base line. It will be seen the loads are in the reverse position to
that in which they were supposed to cross the span. The shear force
diagram may be expeditiously drawn by setting off the reactions HF
and GK, placing the loads in the positions indicated and projecting the
centres of gravity X and Y to C and D respectively. By joining FD
and RG, KG and SH, the complete diagram is obtained without
drawing the individual shear diagrams. The end shears are unequal,
being + 14 and —13 tons, since the loads are unequal. If, however,
the loads may cross the span in the reverse order, the shear force
diagram would be inverted. In such a case, which commonly occurs in
practice, the larger portion of the shear diagram — here, the upper
portion — would be employed on each side of the horizontal base line HG.

If X and Y be projected to T and W on the base line ab of the
B.M. diagram, these points coincide with the ends of the parabolas
«EW and TE#. The vertical centre lines of these parabolas, on which
the maximum moments under each load are measured, are therefore
displaced If ft. and 3^ ft. respectively to left and right of the centre
line mM of the span. The maximum moment under the 10 tons load
= Pz = 163-3 ft.-tons and under the 5 tons load = QJ = 140-8
ft.-tons.

STRUCTURAL ENGINEERING

Beam supported at Both Ends and carrying Two Equal Con-
centrated Rolling Loads at a Constant Distance Apart. — This is a
special case of the last example. In Fig. 32 a span of 50 feet is taken
with two equal loads of 10 tons each, separated by a distance of 20 feet.
The construction is similar to that in Fig. 81. The parabola acb is

first drawn having me = -. — = 125 ft.-tons. This serves for both

+16

FIG. 32.

loads, since they are equal, and takes the place of curves abb and acb
in Fig. 81. Assuming the loads to roll over from left to right, the
curve «4 gives the moments under the leading load before the following
load comes on to the span. With leading load at m and follow-
ing load at d, the additional moment = me = cf, giving the total
moment = mf. Other points #, h, and Jc are obtained similarly, and
the curve a±ghd gives the maximum moments under the leading load.
The curve alng, giving the moments under the following load, will be

BENDING MOMENT AND SHEARING FORCE 71

similar and equal since the loads are equal. The total moments for
either direction of transit are given by the curve alnfgkl. The outer-
most curve aLb, having mL = — ^ — = 250 ft.-tons, indicates the

moments which would result from adding the two loads, and treating
them as a single concentrated load of 20 tons, which in this case
would be inadmissible, since m~L greatly exceeds the actual maximum
bending moment, rn or sg = 150 ft.-tons. This is due to the load
interval of 20 feet being so large relatively to the span. For the
shearing force diagram, the maximum end shears will occur when one
load is just over the abutment, and the other slightly more than 20 feet
along the span, so that FH and GK each =10 tons + f§ of 10 tons
= 16 tons. Placing the loads in the extreme positions, with centres of
gravity at X and Y, points C and D are obtained by projection and the
boundaries FRG and HSK of the shear force diagram determined.
Note that if GR and HS be produced to the opposite abutments, they
cut off heights representing +10 and —10 tons respectively.

The preceding method becomes very tedious if the number of axle
loads exceeds four or five, and the following construction is applicable
to any number of concentrated loads, or to a system of concentrated
and distributed loads combined. Fig. 33 shows the application of the
method to the case of the Atlantic engine and tender, indicated by the
axle loads on p. 35, rolling over a span of 60 feet. AB represents
the span of 60 feet, and wl9 w^ w3 . . . w8 the positions of the axle
loads when the engine is standing with the leading bogie axle over the
right-hand abutment B. Multiplying each load by its distance from A,
the following moments are obtained : —

iv1 X 60 feet 0 inches = 8'5 X 60 = 510 ft.-tons = Aa
w2 x 53 „ 9 , = 8'5 X 53-75 = 456'8 „ = ab

w3 X 48
«>4 X 41
W5 x 33
u>i X 24
tc? X 18
it's X 11

= 18 x 48*5 = 873 „ = be

= 18 X 41-66 = 750 „ = cd

= 13 X 33*66 = 437-6 „ = de

= 11-5 X 24-58 = 282-7 „ = ef

= 14-5 x 18-08 = 262-2 „ =fg

= 15 X 11-58 = 173-7 „ =gh

These values are set off to any convenient scale on the vertical line
Ah and hB joined. The total length Ah represents the upward
moment about A, of the reaction at B, due to all the loads in the
position indicated. Consequently, any intermediate ordinate as M
represents the moment of the reaction at B about the point Jc. Next,
join «B, project w2 to m on aB and join bm. Project w3 to n on ~bm
and join cw, and «/;4 to o on en and join do. Repeating this construction
until the last load w8 has been utilized, the broken line Bmnopqrsh is
obtained, which is such that the vertical ordinates intercepted between
it and the base line hB give the bending moments at any point of the
span for the axle loads in the given position. This may be proved
as follows : —

A/* = total reaction at B X AB, /. xt - RB X Bz, since triangles
BAh and Bxt are similar.

STRUCTURAL ENGINEERING

BENDING MOMENT AND SHEARING FORCE 73

A.a = n\ x AB, .*. xm = u\ x Bic, since triangles BA# and Exm
are similar.

But mt = xt — xm = RB X Bz (upward moment) — t^ x B# (down-
ward moment) = bending moment at section x.

Similarly at any section k, kl = upward moment RB x B&, and
Jcq = sum of downward moments of wl9 w2, w3 . . . w5 about k, whence
kl — Tcq = ql = bending moment at Je.

The moments under each axle for this position of the loads are
transferred to the horizontal base line XY in Fig. 34, being indicated
by the full line. (The vertical scale of moments for Fig. 34 is twice
that of Fig. 33.) If now the loads be supposed to roll backward from
B towards A, or what amounts to the same thing, the span AB be
supposed to move forward towards the right, the loads meantime
remaining stationary, a new position A^ of the span may be assumed
10 feet (or other convenient distance) in advance of AB. The vertical
ordinates between AjtrgponmBB] and the new base line A^ now give
the bending moments under the various axles for this new position of
the loads. These are also transferred to the common base line XY,
being indicated by the single chain-dotted line corresponding with
A^j. Other positions of the span, A2B2, A3B3 . . . A5B5, each 10 feet
in advance of the preceding one, are taken, and the bending moments
transferred to XY, Fig. 34, the resulting moment diagrams being
indicated by similar lining to that adopted for the corresponding base
lines in Fig. 33. By moving the span to the left of the loads it may
readily be ascertained if any greater moments are created than those
already plotted in Fig. 34. In this case no greater moments occur
than those plotted, and the maximum moments are those indicated by
the outer limits of the overlapping diagrams of Fig. 34. If an envelop-
ing parabola XLY be drawn just including these diagrams, such a
parabola will constitute the B.M. diagram due to a certain distributed
load of w tons per foot run which may be substituted for the actual axle
loads considered. The central height LM of this parabola scales 900
ft. -tons.

Hence ^ = * x 60 x 60 = 900, :. 10 = 2 tons.

8 8

That is, the equivalent distributed load for this type of locomotive for
a span of 60 feet is 2 tons per foot run.

The equivalent distributed load which will create at least the same
shearing force as the concentrated axle loads when placed in any position
on the span will be somewhat higher than that deduced from the
bending moments. Since, however, the shear force diagram for any
position of the axles is simply constructed, the maximum shear at any
point of the span is easily obtained after one or two trials.

When the load consists of a number of concentrated axle loads
followed by a distributed load of given intensity, the following
modification of the construction in Fig. 33 is to be observed.

In Fig. 35 four axle loads of 16 tons each are followed, after a
10 feet interval, by a distributed load of 1*6 tons per foot run, the span
being 50 feet. Placing the leading axle over B, the preceding con-
struction is repeated, but the distributed load is first treated as a load

STRUCTURAL ENGINEERING

of 25 x 1-6 = 40 tons, concentrated at its centre of gravity, 12'5 feet
from A. The resulting B.M. diagram is then abed . . . B. Project
C to c on bd, and join ac. Since the load between A and C is actually
distributed, bisect be in m, and draw in a parabolic arc through arm.
The required B.M. diagram is then amcd . . . B, which is utilized
after the manner of Fig. 33, for determining the varying moments
as the loads roll back from B towards A. The remainder of the
construction being similar to that of the last example is here omitted.

FIG. 35.

Beam supported at both Ends, and carrying a uniformly dis-
tributed Rolling Load. — This case is one which occurs very frequently
in practice, corresponding closely with that of the passage of a train
over a bridge span. A widely followed mode of procedure in such
cases is to substitute for the actual loads on the different axles of the
train, a uniform load per foot run sufficiently large to cause at least the
same bending moments at every point of the span as would be caused
by the concentrated axle loads. This materially reduces the labour
involved in drawing out the curves of moments by the method just
described, by substituting one parabola enveloping the several curves
obtained by the former construction.

In Fig. 36, a span of 50 feet is taken with a load of one ton per
foot run advancing over the span from left to right. When the
load covers the length A-I, the B.M. diagram is alb. As the load
successively covers the lengths A-II, A-III, and A-IV, the corre-
sponding moment diagrams are a2b, a3b, and a^b. These diagrams
are obtained by the method shown in Fig. 24. "When the load
covers the whole span from A to B, the B.M. diagram becomes
the parabola acb, having a central height me = £ x 1 X 50 x 50

BENDING MOMENT AND SHEARING FORCE

= 312-5 ft. -tons. The bending moment therefore gradually increases
as the load advances and the maximum moments occur when the span
is fully loaded, their value then being the same as for a stationary load
of the same intensity covering the whole span. The B.M. remains
constant so long as the moving load covers the whole span, but
gradually diminishes as the tail end of the load rolls off towards B.
With AI only covered by the load (10 tons), the reactions at A

-25

FIG. 36.

and B are respectively 9 tons and 1 ton, and the shear force diagram
is obtained by drawing a horizontal through — 1 as far as vertical
section I, and then an "inclined line upwards to -f 9. (See Fig. 24.)
In the upper figure the sectional shading showing the positions to
which the head of the load has advanced is drawn in distinctive lines
corresponding with those representing the B.M. and S.F. for those
positions. With load from A to II (20 tons), the reactions at A and B

76 STRUCTURAL ENGINEERING

are respectively 16 and 4 tons, and the corresponding S.F. diagram is
shown by the horizontal through - 4, and inclined line to +16.
The three remaining diagrams from — 9 to + 21, — 16 to + 24, and
- 25 to + 25 will readily be traced from the reactions in a similar
manner. The curve PS enveloping these five diagrams will include all
other possible shear diagrams for any intermediate positions of the load.
Its outline is a semi-parabola touching the base line PR at P, since the
depths 1, 4, 9, 16, and 25 at equal horizontal intervals equal the
squares of the numbers 1, 2, 3, 4, 5. A similar curve, QR, indicates
the shearing force as the tail end of the load rolls off the span from A
to B, or as the load rolls on to the span from B towards A. The four
intermediate diagrams in this case are omitted. It will be seen that
the shearing force diagram may be rapidly drawn by setting off PQ
and RS, each equal to half the total load required to cover the whole
span, and inserting the semi-parabolas PS and QR by the geometrical
method.

Continuous Beams. — A beam or girder is said to be continuous
when it bridges over more than one span, and rests on one or more
intermediate supports in addition to the two end supports. A con-
tinuous beam bends in the manner shown in Fig. 37. From A to a

A P _c

^a jb I q 6*

FIG. 37.

certain point P, the upper surface is concave and in compression whilst
the lower surface is convex and in tension. From P to Q the cur-
vature and bending are reversed, the upper portion being in tension
and the lower in compression. At Q another reversal of bending takes
place, with a corresponding change in the character of the stresses.
The points P and Q are known as points of contra-flexure, and at these
points the lending moment is nothing. The position of these points
depends on the character of the loading and section of beam. The
continuous beam ACB is equivalent to a system of two simple beams
ap and qb, and a cantilever pq, as indicated in the lower figure, the
points P and Q being projected to p and q, fixing the lengths of the
equivalent beams and cantilever. The stresses in the lower combi-
nation of simple beams are then identical with those in the continuous
girder alone. It is apparent from the .lower figure that the pressures
on the various supports differ considerably from those which would
obtain if the two openings were spanned by separate beams with an
interval at C. Thus, if the load be supposed uniformly distributed
from A to B, the support at A will carry half the load on the length
AP, and the support at B will carry half the load on the length
QB. The central support will carry the whole of the load on PQ,
together with the remaining halves of the loads on AP and QB. In
the case of two independent spans AC and BC, the central support

BENDING MOMENT AND SHEARING FORCE 77

would carry just half the total load from A to B, and each end support
one quarter of the total load. The central support beneath the con-
tinuous beam thus carries a greater proportion of the total load than
if supporting two simple beams only. When the positions of the
points of contra-flexure are determined, the B.M. at any section of the
beam, and the pressures on the supports, may readily be deduced by
reference to the equivalent system of simple beams and canti-
levers.

The following solutions, which illustrate typical cases, are based
on the assumptions that the supports are all at the same level and
the beam of uniform cross- section. They are therefore strictly appli-
cable to rectangular timber beams, rolled sections used as beams, and
plate girders of uniform section, whilst the results will be approxi-
mately correct for the generality of plate girders. The exact solution
for beams of very variable section is considerably involved, and beyond
the scope of this work. It should be distinctly remembered that a slight
subsidence in one or another of the supports of a continuous girder may
considerably modify the lending moment^ and consequently the stress, at
any section. Thus in Fig. 37, lowering of the central support C
would throw more of the load on the end supports A and B and
shorten the length of the convex portion PQ. The lengths ap and bq
of the equivalent simple beams would be thereby increased with a
corresponding increase in the bending moment upon them. Conversely,
subsidence of A or B or both, would increase the pressure on C and
cause the points P and Q to move outwards from the centre, with a
corresponding increase in the bending moment at the centre of the
cantilever pq. Variation in the loading further causes alteration in the
positions of the points P and Q, and therefore in the bending moment
also.

In Fig. 38, if the span BC carry a much heavier load than the span
AC, the effect is to cause relatively large deflection of BC and to spring

up the length AC. The points of contraflexure P and Q would then
move towards the left and the equivalent system of simple beams and
cantilever be as shown at apqb, with correspondingly modified bending
moments. Such action takes place in the case of a continuous girder
bridge with the live or rolling load advancing from one end support.
If P' denote the previous position of P when the girder was symmetri-
cally loaded, it will be seen that the upper and lower flanges of the
girder between P and P' undergo reversal of stress. In Fig. 37 the
upper flange from A to P is in compression under symmetrical loading,
whilst in Fig. 38, a portion of this flange, PP', is now in tension under

STRUCTURAL ENGINEERING

the unsymmetrical load. The stress in the lower flange between P and
P is also reversed from tension to compression.

For the above reasons, continuous girders are not very frequently
adopted, especially for moving loads. The reversal of stress over a
certain length of the girder may be suitably provided for, but a
relatively slight alteration in the level of the supports, after erection,
gives rise to unknown and possibly dangerous stresses. Several bridges,
in fact, originally erected as continuous girders have, on account of
unequal subsidence in the piers, been cut through in the neighbourhood
of the points of contra-flexure and so converted into actual simple
beams and cantilevers, the stresses in which are independent of slight
differences in level of the supports.1 As will be seen, however, a
continuous girder is more economical of material than several indepen-
dent spans together aggregating the same length, since the continuous
girder under similar loading is subject to less bending moment and may
therefore be designed of lighter section. It further possesses certain
advantages relative to ease of erection in lofty situations, which, however,
cannot be here considered in detail.

Characteristic Points of Bending Moment Diagrams.- — The bending
moments on continuous beams of uniform section may be readily

determined after finding what are
called the characteristic points of
the simple bending moment dia-
grams for each span considered in-
dependently. These characteristic
points are obtained as follows. In
Fig. 39 let AB represent the span,
and the parabola ACB the B.M.
diagram due to a uniformly dis-
tributed load. Divide the span AB

into three equal parts in points E and F, and at these points erect
perpendiculars EP and FQ. The height of points P and Q above the
base line AB is fixed by the condition that the moment of the area
ACB about one end of the span shall equal the moment of the rectangle

AKHB about the same point, or area ACB x --»- = area AKHB x -«-•

£- being a common factor, may, in this case, be eliminated, leaving

area ACB = area AKHB, and this condition is sufficient for fixing the
heights of P and Q in all cases where the B.M. diagram is symmetrical
about the vertical centre line, since G and G', the centres of gravity of
areas ACB and AKHB will both fall on CD, and the moment arm for

each area will equal -g-, or the half span. In the case of the parabola —

Area ACB = f CD x AB, and area AKHB = EP x AB
.'. EP x AB = f CD x AB, or EP = FQ = f CD

The characteristic points for a parabolic diagram are therefore
1 Mins. Proceedings Inst. C. E., vol. clxii. p. 245.

EOF

FIG. 39.

BENDING MOMENT AND SHEARING FORCE

obtained by erecting perpendiculars at each | of the span, and catting
off a height equal to f the central height of the parabola.

In Fig. 40, ACB is the B.M. diagram for a concentrated load at the
centre of span AB. The triangular area ACB, and the rectangle AKHB,
will obviously be equalized by drawing KH at half the height CD.
The characteristic points P and Q are therefore here situated at one-
half the central height CD above the base. Fig. 41 illustrates another

EOF

FIG. 40.

FIG. 41.

symmetrical case. The span AB is 20 feet, and equal loads of 6 tons
are carried at 4-feet distances from each end. Reactions at A and B
each equal 6 tons, and moments CK and DH == 6 x 4 = 24 ft.-tons,
AKHB being the B.M. diagram. Its area, measured by the scales used
for distance and bending moments, = AD X DH = 16 X 24 = 384.
The height of the rectangle of equal area on base AB = &£ = 19 '2
units on the B.M. scale. Marking E and F at the one-third points of
the span, the characteristic points are located at P and Q.

Characteristic Points for TInsymmetrical Bending Moment
Diagrams.— In Fig. 42, AB = 30 feet, and a concentrated load of
15 tons is applied at D, distant 6
feet from B.

RA = 3% of 15 = 3 tons, the
bending moment at D = 3 x 24
= 72 ft.-tons, and the B.M. dia-
gram is the triangle ACB, having
its centre of gravity at G, such that
MG = | MC. The moment of
area ABC will now be greater
about the end A of the span than
about B, and the characteristic
points will no longer be at the
same height above AB. Taking moments about A, and calling H
the height of the rectangle, having the same moment about A as the
triangle ACB- A_ AW1 x lg, = H x 8Q, x 15,

X 18 = II X 450
whence H = 43'2

which fixes the height of the characteristic point Q. Taking moments
about B, and calling h the height of the rectangle of equal moment —

FIG. 42.

whence h = 28 '8

STRUCTURAL ENGINEERING

which fixes the height of the characteristic point P. Any unsymmetrical
B.M. diagram may be treated in a similar manner.

Method of using the Characteristic Points for determining the
Bending Moment on Continuous Girders. — General Example. — A girder
is continuous over three spans of 30, 40, and 30 feet, and carries a load
of 2 tons per foot run over the first span, and 1*5 tons per foot run
over the second and third spans. To draw the B.M. and S.F. diagrams
and determine the points of contra-flexure and pressures on the
supports.

In Fig. 43 set out the spans AB, BC, and CD to scale, and upon
them draw the B.M. diagrams for the stated loads, assuming the spans

to be bridged by three independent girders instead of one continuous
girder. The central bending moments for the three spans are
respectively —

= 225 ft,tons,

X 40

= 300 ft,tons

and

1-5 x 30 X 30

= 168'75 ft.-tons. These values are set off to a

convenient scale at ab, cd, and ef, and the parabolas A#B, B^C, C/D,
drawn through them. Next mark the characteristic points of each
parabola by dividing each span into three equal parts, and making the
heights of 1 and 2 = f ab, 3 and 4 = f cd, and 5 and 6 = f ef. If the
outer ends of the girder rest freely on the supports at A and B, no
further use is made of the extreme characteristic points 1 and 6, which
may be disregarded. Commencing at A, a series of straight lines, AB',
B'C', C'D, require to be drawn, such that any two lines as AB', B'C',

BENDING MOMENT AND SHEARING FORCE 81

meeting over the intermediate support B, of the girder, will pass at
equal vertical distances above and below the characteristic points 2 and 3
on opposite sides of that support. Similarly the lines B'C', C'D, meeting
over the support C, must pass at equal vertical intervals below and
above the characteristic points 4 and 5. Also, whatever the number
of spans, the closing line C'D must terminate at the point D. Obviously
only one possible system of lines will fulfil these conditions, and their
directions are easily located after one or two trials. Thus, if the dotted
line Ag be taken as the first trial, it passes a little distance above 2,
and its direction, after reaching g, must be such that it passes the
same vertical distance beloiv 3 as it previously passed above 2. This
causes gh to pass considerably below 4, and hk must be continued to
pass an equal interval above 5. As this direction terminates at k
instead of at D, the series of dotted lines is not the one required, and
by tentatively lowering the first line A#, the correct directions AB'C'D
are ultimately obtained. It should be noted that any two lines meeting
over an intermediate support must pass on opposite sides of the two
characteristic points adjacent to that support, but that it is immaterial
whether they pass above or below either the right- or left-hand point.
A line is occasionally found to pass through one of the characteristic
points, in which case, the vertical interval being nothing, the succeeding
line beyond the next support must pass through the corresponding
adjacent characteristic point.

The broken line AB'C'D so found, constitutes a new base line from
which to measure the bending moments which actually obtain for the
continuous girder. The points pl9 p2, p^ p±, where this new base line
intersects the parabolic diagrams, determine the positions of the points
of contra-flexure, and projecting them to P15 P2, P3 and P4 on AD, their
horizontal distances apart may be scaled off. These are indicated in
the lower figure, which also shows the manner in which the continuous
girder may be divided into an equivalent system of simple beams and
cantilevers. The vertically shaded portions of the upper figure indicate
the bending moments on the continuous girder, the full lines denoting
positive moments, and the dotted, negative moments. At the points of
contra-flexure, the B.M. is of course zero, which necessitates these points
being made the points of junction between the simple beams and canti-
levers in the lower figure. The pressures on the supports A, B, 0, and D
are readily deduced from the lengths of the cantilever and simple girder
spans. Thus —

Pressure on A = f load on APX = £ X 22'7 X 2 = 22'7 tons.
„ B = £ load on APi + load on PjP2 -f | load on P2P3
= 2-2-7 + (7'3 X 2) + (9 X 1'6) + (| X 23 X T5)
= 68-05 tons.

C = I load on P2P3 4- load on P3P4 + \ load on P4D
= (\ X 23 X 1'5) + (16 X 1-5) + (i X 22 X 1'5)
= 57-75 tons.
D = i load on P4D = \ x 22 X To = 16'5 tons.

The sum of these pressures, 165 tons, of course equals the total load
on the beam. The continuous girder deflects in the manner shown by

STRUCTURAL ENGINEERING

the curved line VW. The shearing force diagram readily follows from
a consideration of the loads and pressures on supports. Note that the
inclined lines indicating the shear force cut the base line XY beneath
the central points of the equivalent girders AP1} P2P3, and P4D, where
the shear is zero, and the positive B.M. a maximum.

A few special cases may be noticed. In Fig. 44 a beam is con-
tinuous over two equal spans, AB and BC, and carries a uniform load

K._jj/ - «±« _ ' / ^ _ s-i _.*j

4i it /~ ~ / * i

b \e R i a

FIG. 44.

of w tons per foot run throughout. The parabolas A£B and EdC
having db = cd = -g-, represent the moments for the spans considered

independently. The characteristic points adjacent to the support B are
2 and 3. AB' and B'C are the only possible lines fulfilling the conditions
above mentioned, and they must obviously pass through the points 2
and 3. But since 22' = f db, and A2' = f AB, 22' also = f BB', whence
BB' = ab, or the negative B.M. over the pier B is equal in amount to

ivffi
the positive moment of -g- at the centre of either span considered

independently. The points of contra-flexure, p,p, evidently occur at
J I from A and C, since pP will then = f BB', and ep = J BB', which

FIG. 45.

in a parabola is the condition for e to be situated halfway between
I and B'. Hence the pressure on each end support A and C = \ of
f wl = f wl, and on B = f ui + J wl + f wl = Ij wl, or the central

BENDING MOMENT AND SHEARING FORCE

support carries 3J times the load on each end support. The equivalent
system of two simple beams and a cantilever is shown below.

Fig. 45 shows the moment diagram for three equal spans of I feet,
bridged by a continuous girder carrying a uniform load of w tons per
foot, from which the indicated pressures on the supports may be
deduced as shown.

A useful practical application occurs in the case of an open trough-
shaped conduit for carrying a canal, or a continuous pipe line conveying
water supply over several spans. Fig. 46 shows the character of the
moment diagram where six equal spans are involved.

FIG. 46.

EXAMPLE 9. — A girder is 42 feet long and is supported on walls at
either end and by a column at the centre. At 6 ft. intervals it carries
rolled joists, each of which imposes a floor load of 7 tons on the girder.
Required the B.M. diagram for the girder and the pressures on the
supports. Fig. 47.

FIG. 47.

Regarding the span AB as independent,
RA = 7 (if + £ 4- o3T) = 9 tons.
B.M. at D = 9 x « = 54 ft.-tons.
B.M. at E = 9 x 12 - 7 x 6 = 66 ft.-tons.
B.M. atF = 9xl8-7xl2-7x6 = 36 ft.-tons.

These are plotted at D^7, Ee, and F/, giving MefE as the moment
diagram for an independent girder between A and B. Since the ends

84 STRUCTURAL ENGINEERING

A and C are free, only the characteristic points adjacent to B are
required, and the loading being symmetrical, these will be situated at
the same height above AC. Set off BP = 1 AB. The diagram AdtfB
is divided up into constituent rectangles and triangles by the full lines,
and the sum of the moments of these areas about A will equal the
moment of the whole diagram about A. The individual centres of
gravity are located and marked by the small circles.

Moment of A AM about A = ~~- X 4 = 648
A deg „ = ^*i? x 10 = 360

A/FB

rect. J)^jgd
„ EF/A

.-. Mt.

5»

|
»»

of A^/B

X
X

J.<±

9
15

units.

:6 X 54

6 X 36
about A

= 2916
= 3240

= 9450

For the height H of the rectangle on AB having the same moment
about A, 21 X H x ^ = 9450, whence H = 42'85. Cut off Pp = 42-85
units on the B.M. scale employed when p is the characteristic point
required. Since the diagram is symmetrical, the new base line will be
obtained by joining A to p and producing to B'. The maximum B.M.
is obviously BB' = l£ x Pp = 1£ X 42-85 = 64-3 ft. -tons, and is nega-
tive, that is, the upper flange will be in tension and the lower in com-
pression over the support B. The points of contra-flexure are at S, S,
distant 15 ft. 8 in. from A or C. The lower figure shows the equivalent
system of simple beams and cantilever, and from the positions of the

3' 8" 9' 8"

two 7 ton loads which rest on sc, the reaction at C = TRT zp of 7 + J^r^r

of 7 = 5-95 tons. A similar reaction exists at A, whence pressure on
central column = total load — pressures on A and C = 42 — 11*9 = 30'1
tons.

The- S.F. diagram readily follows from the pressures, each step
scaling 7 tons.

Fixed Beams. — A beam or girder is said to be fixed at the ends
when it is so firmly built in or anchored down that a tangent, AB,
Fig. 48, to the curve of the bent beam at A is horizontal. In order to
realize this condition, it is evident there must be a sufficiently large
downward pressure or pull P applied to the portion AC of the beam,
as will create a reversed bending moment capable of balancing that
caused at A by the loads on the beam. The holding down force P may
be applied by the weight of masonry in the w.all above AC, or by
anchor rods taken down to a suitable depth. If, in the lower figure,
P' be not sufficient to create the same amount of moment as would
exist at A' if the beam were actually fixed, the end of the beam will tilt
up to some extent and bend as shown at A'C', when the tangent A'B' to

BENDING MOMENT AND SHEARING FORCE 85

the curve at A' will no longer be horizontal, and the beam will not
fulfil the condition of fixity of ends. In this case the point of contra-
flexure, which previously was
located at p, will move to some
point p' nearer to A', and the
B.M. at A' will be reduced to
that which P' is capable of pro-
ducing when acting at the

leverage P'A'. This will be FIG 43.

accompanied by a correspond-
ingly increased B.M. at the centre of the beam. A little consideration
will show the fallacy of assuming a beam to be fixed at the ends, simply
because it is apparently firmly built into a wall at either end.

In Fig. 48 suppose the beam to carry a central load of 2 tons over a
span of 20 feet. The B.M. at A, if the beam be actually fixed, will be

Wl 2 x 20

~o~= o — = 5 ft.-tons or 60 inch-tons. If the beam project, say,

18 inches into the wall and be fixed by the weight of brickwork resting

i ft"
on AC, then P x -IT must = 60 inch-tons, or P = 6§ tons. Assuming

a breadth of flange of 12 inches, the bearing area from A to 0 = \\
square feet, and the height h of the column of brickwork resting on
this area and weighing 6f tons, will be given by h x 1*5 X oVft = 6f
tons, whence h = 89 feet. This is supposing the column of brickwork
to actually rest on the end of the beam, whereas a portion of it would
probably be supported by the bond in the wall. Assuming a reasonable
height of wall above AC, say 30 feet, the beam would require to be
firmly built in for a minimum distance of 2 feet 7 inches at each end,
in order to realize fixed conditions, still supposing the weight of the
30 feet of brickwork to be wholly resting on AC. Probably the
majority of so-called fixed beams fall far short of the required degree of
fixation, with the result that if calculated as fixed beams they may be
stressed to nearly double the intensity intended in their design. The
moment of the holding-down force P about A, or P X | CA, in Fig. 48,
is called the moment of fixation, and the B.M. on the beam section at
A cannot exceed this moment of fixation. Consequently no beam or
girder should be assumed as having fixed ends, unless the actual
pressure upon the built-in or anchored-down ends is definitely known
to be equal to that required to produce the necessary moment of fixation
for balancing the B.M. due to the loading under consideration. Rela-
tively few girders in practice are intentionally designed as fixed beams.
Where it is necessary to fix the end of a girder, the necessary fixing
moment is provided by properly loading the end of the girder with a
definite balance weight, or by attaching to it anchor ties capable of
exerting a predetermined downward pull.

The bending moments on fixed beams of uniform section are readily
determined after locating the positions of the characteristic points of
the B.M. diagram for the beam considered as simply supported. The
straight line drawn through the two characteristic points constitutes the
new base line above and below which to measure the positive and
negative moments on the fixed beam.

STRUCTURAL ENGINEERING

Fig. 49 illustrates the case of a fixed beam of span / feet with a central

W/
concentrated load W. The triangle AOB of height = -7- is the moment

diagram for the simply supported beam. The height of the character-
istic points p, p, = \ CD, and joining these by EF the moment at the

Wl Wl

centre is i CD= + -0 , whilst EA = FB = - -"- is the moment at the

fixed ends. The points of contra-flexure are S, S, distant from each

end of the span. The fixed beam AB is equivalent to two fixed
cantilevers as and sb with a simply supported span ss carried between
them, as shown in the lower figure.

FIG. 49.

Fig. 50 gives the diagram of moments for a fixed beam of span /
feet carrying a uniform load of w tons per foot run. The parabola ACB

having CD = ^L ft.-tons, has the characteristic points p,p, at a height
above AB = f CD. The central B.M. on the fixed beam = £ CD
, and the end moments AE and BF are each

The length as may be found as

follows : —

The bending moment at
the centre of the independent
F w x ss2 wl* ,

span ss = — 3-- = ^77 > whence

x ~~ = ~~ Ik-tons.

ss = —r=. = 0'578Z. .*. as and

sb together = 0*422/ and as
= sb = 0-211/.

For any case of imsyin-

FlG 51 metrical loading the same con-

struction holds. Thus in Fig.

51, the fixed beam AB of 20 feet span carries a concentrated load
of 5 tons at C, distant 8 ft. from B. RA = £ of 5 = 2 tons, and the
moment CD for the beam simply supported = 2x12 =24 ft.-tons.
For the characteristic points —

BENDING MOMENT AND SHEARING FORCE

Moment of A ADO about A = 12 * 2 x 8 = 1152

„ A BDC
AADB

X 14| = 1408
= 2560

For the height H of rectangle on AB having the same moment
about A, H X 20 x f = 2560, /. H = 12'8. Mark P and Q at one-
third the span and make Pp = 12*8 units on the B.M. scale.

1 9 y

Similarly, moment of A ADC about B = _
A BDC „ = ?-*_
A ADB

X 12 = 1728

X 5± = 512

= 2240

For the height h of rectangle on AB having the same moment about
B, h X 20 X y- = 2240. .'. h = 11'2 units. Make Q# = 11-2 and
join the characteristic points j? and q. E^F is the new base line for
moments giving a maximum positive moment DM beneath the load and
negative moments EA and FB at A and B respectively. The equi-
valent system of two cantilevers and a simple beam is shown at

Beams fixed at One End and supported at the Other.— These are
not of much practical interest. Fig. 52 shows the B.M. diagram for a
beam fixed at A and supported at B carrying a distributed load of w
tons per foot run, and Fig. 53, the same beam with a concentrated load

FIG. 52.

of W tons at the centre. In these cases, the characteristic point
adjacent to the freely supported end B is neglected, since the B.M. at
this end is zero. The new base line BO is therefore drawn from B
through the characteristic pointy adjacent to the fixed end. The beam
AB is, in each case, equivalent to a cantilever as and supported beam sb
with one point of contra-flexure at S.

Beam fixed at Both Ends and continuous over Intermediate
Supports. — Fig. 54 represents a beam 50 ft. long fixed at each end and
supported at 20 feet from one end, carrying a load of 2 tons per
foot run.

STRUCTURAL ENGINEERING

The central heights of the parabolas ADB and BEG are given by

^ x 302 2 x 202

-= 225 ft.-tons and -^— = 100 ft. -tons respectively.

o o

The position of the new base line FGH is obtained by drawing two
lines through the characteristic points p and q adjacent to the fixed ends

FIG. 54.

A and 0, meeting at G vertically over B and passing equally below and
above the characteristic points p' and </ adjacent to the intermediate
support. Projecting down the four points of contra-flexure, the equiva-
lent system consists of three cantilevers asl9 s^ and s4c and two supported
beams s^ and s3s4. The negative moment AF is the maximum, scaling
169 ft.-tons.

CHAPTER IV.

BEAMS.

Moment of Resistance. — Vertical forces acting on a horizontal beam
produce a bending action in the beam. At any cross-section the bend-
ing action is proportional to the bending moment.

Suppose a loaded beam, Fig. 55, to be hinged at the centre. The
bending action would tend to close the portion between P and C and

FIG. 55.

open the lower portion P to T. If a block of material be placed at X
and a tie at Y to prevent movement about the hinge, it is evident that
the block at X would be compressed and the tie at Y stretched. Equi-
librium having been established, the bending moment at the section
must be equal to the moment of the forces in the block and tie about
the hinge at P.

Let C = compression in the block,
T = tension in the tie.

Then the moment of these forces about P

= T x 2/+C x x

= the bending moment on the section.

If the beam be made continuous, the material at the vertical section
through P would be subject to stresses similar to those in the block and
tie. All the material above some horizontal
plane, such as that passing through P, would
be in compression and all below that plane
in tension. Let the small arrows in Fig. 56
represent the stresses in the material at the
vertical section OPT, and Rc and R,, the
resultants of the compressive and tensile
stresses. Since all the forces causing bending
must act normally to the horizontal plane FIG. 56.

through P, the only forces acting parallel

to that plane are the forces Rc and RT, which must therefore be equal

P-- y

STRUCTURAL ENGINEERING

to produce equilibrium. The bending moment at the section OPT will
therefore be equal to the moment of the couple Rc x y or RT X y.
The moment of this couple is a measure of the strength of the beam
at the section and is known as the moment of resistance.

Suppose the block abdc, Fig. 57, be compressed to efhg. The upper
edge ef is decreased in length to a greater extent than the lower edge
gh, and as the stress must be proportional to the decrease of length, the
stress at ef is greater than the stress at gh. The decrease in length is
proportional to the distances of the edges ab and cd from the point P.
Therefore the stresses must also be proportional to the distances from
P. If the block had extended from P to C no alteration of length

N-Hr •=•*

FIG. 58.

would have occurred at P, demonstrating that the material at P is not
subject to any bending stress, whilst the maximum change in length
and therefore the greatest stress occurs at C. In a similar manner it
may be shown that the tensile stress below P varies from nothing at P
to a maximum at T, and at any point in the section is proportional to
the distance of that point from P.

At every vertical section of the beam there is some point P where
there is no direct stress. The plane containing all such points is known
as the neutral plane.

Position of the Neutral Axis. — The intersection of the neutral plane
with any cross -section of a beam is termed the neutral axis of the cross-
section. Let Fig. 58 represent the cross-section of a rectangular beam
divided into horizontal layers. If the intensity of compressive stress at
the upper surface (usually called the skin stress) =/c, and the intensity
of tensile stress at the lower surface =/«, then the average intensities
of stress in the layers above the neutral axis NA will be

Let a = sectional area of each layer.

Then the total stresses in the separate layers

= of & of & . . . of £

y< #c J yc

and the total compression above the neutral axis
= -fe/8 + ay? + • • • 4-

BEAMS

Similarly, the total tension below the neutral axis

Since total tension must equal total compression

7(^8

But

ye yt

4- at/i = ay's 4-

or the sum of the moments of the areas in tension is equal to the sum
of the moments of the areas in compression. This is the condition for
the neutral axis passing through the centre of gravity of the cross-
section. So long as the moduli of elasticity of the material in tension
and compression be the same, the neutral axis must always pass through
the centre of gravity of the cross-section whatever be its shape.

Moment of Resistance. — The stress in the top layer, Fig. 58, was

shown to be = afj^.

Its moment about the neutral axis

The total moment of the stresses above the neutral axis will therefore

ay?

Moment of the stresses below the neutral axis

(1)

(2)

The moment of resistance of the section is equal to the sum of the
expressions (1) and (2) when the layers are taken infinitely thin. It
will be seen that the portions of the expressions in the brackets are the
sums of all the small areas multiplied by the square of their distances
from the neutral axis. The moment of resistance may therefore be
written —

M.R. =

Sum of all the small
areas multiplied by
the square of their
distances from the
neutral axis

skin stress

distance of skin from N.A.

(8)

In sections symmetrical about the neutral axis the skin stress in
tension will be equal to the skin stress in compression, but for

STRUCTURAL ENGINEERING

unsymmetrical sections these stresses will not be equal. In unsym-
metrical sections, the skin stress, in the above expression, is that of
tension or compression, according as the denominator is the distance
from the neutral axis, of the skin subject to the stress adopted.

Moment of Inertia. — For any section, the sum of all the small areas
into which the section may be divided, multiplied by the square of
their distances from the neutral axis, is termed the moment of inertia
of the section, and is usually denoted by the letter I. The expression
(3) may then be written —

M R =

momenfc °f inertia x skin stress
distance of skin from N.A.

yc yt

The bending moment being equal to the moment of resistance,

The value of I is dependent on the distribution of the material
about the axis considered. The calculation of the moment of inertia
involves the use of the calculus, and it is not pro-
posed to give here the mathematical proof. The
formulae for a number of simple cases will be found
in Table 25, and from them I, for most ordinary
sections, may be calculated.

The following graphical method of obtaining
I will prove the accuracy of the formulas given for
rectangles.

To find the moment of inertia of the rectangle
ABCD, Fig. 59, about the side AB. Consider a
very thin horizontal strip ae of the rectangle at a
distance y from AB and of area /.

FIG. 59.

I of the strip = ly\

I for a similar strip at CD = Id2.

If the area I be reduced to I', so that

then I'd2 = If

If each horizontal strip of the rectangle be reduced in the same

, • • the square of its distance from AB ,1 c ,,

ratio, i.e. - —, the sum of all such reduced

areas multiplied by d2 will be the moment of inertia about AB.

Since the strips are very thin the length may be taken to represent
the area. The reduced length of ae will

7/2

= ae x u-r
a

BEAMS

The reduced lenths

f<J = b X

Taking a large number of strips and joining the extremities &,#, c,
etc., the points ~k,y, c, will be found to lie on a parabola passing through
A and D. The moment of inertia will then be equal to the area
multiplied by d2. The area of the parabolic segment D^AB

The " inertia area " CDcgkA will therefore

= IM
and I = ±bd x d2

sffcf8

Let the dimensions of the rectangle be —

b = 6", d = 12"
Then I about the side AB

6 X 123

= 3456 in.4

To obtain the moment of inertia about the neutral axis N.A.
Treating each half of the rectangle by the above graphic method,
two inertia areas, shown shaded, Fig. 60, are

obtained, the area of each being ^b-:- T

I for each half = ±b x ( )

For the whole rectangle

FIG. 60.

Again, let & = 6" and d = 12".

Then the moment of inertia of the rectangle about N.A.

EXAMPLE 10. — To find the moment of inertia of a rolled beam-
sect ion.

Let the section be 12" x 6" X J" metal with parallel sides, Fig. 61.

STRUCTURAL ENGINEERING

Since the section is symmetrical the neutral axis will be situated 6 in.
from the top and bottom.

The moment of inertia may be obtained by either of the following
methods : —

(1) Calculate the moment of inertia for the rectangle 12" x 6" and
subtract the moments of inertia of the two rectangles \ x d\.

(2) Calculate separately and add together the moments of inertia of
the two flanges and the web.

By the first method —

M.I. = I of 12" x 6" rectangle - 2 (I of 11" x 2f rectangle)

" 12 12
= 6 x 123 _ 2 X 2f X II3

12 ~~12T~

= 253-96 in.4

To find the moment of inertia by the second method it will be
necessary to consider the moment of inertia of a section about an axis
other than that passing through its centre of gravity.

j_i

VVVVVVV^ J

t-J."

. „

k -

1
1

M,-*

\^ i

jfx-

I
J-_

FIG. 61.

FIG. 62.

Moment of inertia of a section about any axis XX parallel to the axis
through its centre of gravity G, Fig. 62.

It has already been proved that for a rectangle : —

bd3
I about axis through the centre of gravity = -^ •

M3
I about one side = --.

In Fig. 62 let the axis XX be parallel to and distant R from the
neutral axis.

Treating the rectangle as composed of two rectangles, one above
and one below the axis XX, the sum of the moments of inertia of such
rectangles about XX will be the moment of inertia of the whole
rectangle about XX.

BEAMS 95

For the rectangle above the axis XX —

For the lower rectangle

IYY =

3
For the whole rectangle

Ixx-

But b x d = area of whole rectangle

£^3

and — = I of rectangle about the axis through its

centre of gravity.

Therefore the moment of inertia of the rectangle about the axis
XX is equal to its moment of inertia about the axis through its centre
of gravity plus the area of the rectangle multiplied by the distance
between the axes squared —

Ixx = ICG + AR2

This is true for all sections whatever may be the shape.

Returning to Example 10, second method. I of section = I of
web + I of two flanges. As the neutral axis passes through the centre
of gravity of the web, the moment of inertia of the web

= Of _ jxll3
~ 12 ~ 12

= 55-46 in.4

From the above proof the moment of inertia of each flange

6" x (iV

= ^-^L + 6"xf X(5f)2

= 99-25 in.4
The total moment of inertia for the section

= 55'4G + (2 x 99-25)
= 253-96 in.4

96 STRUCTURAL ENGINEERING

This result agrees with that of the preceding method.

The above method demonstrates the small resistance which the web
offers to the bending action. It was shown on p. 92 that the
moment of resistance was proportional to the moment of inertia,
therefore the proportion of the resistance to bending exerted by the

web will be a5°'*6, or with an area = — = 0'91 of that of the flanges,

£j t) O " J I) O

its resistance as compared with that of the flanges is only IQO.K = 0*28.

Hence it is desirable that the material which has to resist the bending
action be placed, within practical limits, as far from the neutral axis as
possible.

Modulus of Section. — It has already been proved that

MJLifl

/, the skin stress, is dependent only on the material of which the beam is
composed, but 1 and y are wholly dependent on the shape of the cross-
section of the beam. The quantity - is known as the modulus of the

section, and is a relative measure of the strength of a section. The
moment of resistance may then be written —

M.R. = skin stress x modulus of section
= /xZ

For sections symmetrical about the neutral axis the modulus of
section is equal to the moment of inertia divided by half the depth of
the section. Thus for a rectangular section

If'

—

The modulus of section may be found graphically by the following
method.

Graphical Method of obtaining the Modulus of Section. — Consider
a very thin layer, AB, in the flange of the beam section, Fig. 63, at a
distance y^ from the neutral axis. If the intensity of skin stress be

equal to/, the intensity of stress on the layer AB = f x'~
If the area of the strip = I

total stress on the layer = If—

BEAMS

If the area I be reduced in the ratio of — to ?, and the total stress

on the layer be considered to be distributed over the area I', the
intensity of stress on I' will

Reducing the area of all horizontal layers of the section in the ratio
of their distances from the neutral axis divided by y, an area for the
whole section will be obtained on which the
intensity of stress is equal to /. The modulus
of section will then be equal to the moment
of that area about the neutral axis.

Draw a base line parallel to the neutral axis
and at a distance y from it. Project the ex-
tremities of each layer on to the base line and
join the points thus obtained to any point (say
the centre of gravity of the section) on the
neutral axis. Then the area of the layer
between such lines will be the reduced area
required. The projections of the ends of the
layer AB on the base line are the points C
and D. Join C and D to E. The area ab
between the lines CE and DE is the area required.

In the triangles ODE and

But CD = AB

FIG. 63.

ab : CD : : yl : y

:. ab = CD x ^
y

/. ab = AB x ^

The area of equal stress intensity, called the modulus figure, for the
upper half will be C/cEd^D.

Let its area = Aj and its centre of gravity be distant <h from the
neutral axis. Then the total stress on the portion above the neutral
axis —

Moment about the neutral axis—

Since the section is symmetrical the moment of the stress in the
lower portion is equal to the moment of the stress in the upper
portion.

98 STRUCTURAL ENGINEERING

Therefore the total moment of resistance of the whole section

But

If D be the distance between the centres of gravity of the upper and
lower modulus figures

Z = DA1

For sections symmetrical about the neutral

M.R. =/Z
/. Z =

i axis

D =

FIG.

Since the total stresses above and below
the neutral axis are equal, the area of the
modulus figure below the neutral axis must
be equal to the area of the modulus figure
above the neutral axis.

Modulus Figure for Sections unsym-
metrical about the Neutral Axis. — In sections
such as the tee, Fig. 64, the centre of
gravity falls nearer to the lower surface
than the upper, and consequently the in-
tensity of skin stress at the lower surface
will be less than at the upper.

Let/ = intensity of skin stress at the lower surface.
y — distance of lower surface from the neutral axis.
fc = intensity of skin stress at the upper surface.
2A = distance of upper surface from neutral axis.

Two modulus figures may be drawn for the section, one having an
intensity equal to/c and the other an intensity equal to/.

The construction of the modulus figure for the upper skin stress, i.e.
/c, is shown in Fig. 64.

The base line for the upper portion is in the plane of the upper skin
where the stress = /c, but for the lower portion the base line being set
out at a distance yl below the neutral axis (i.e. where the stress would
equal /c), falls below the section. All layers below the neutral axis
must be projected on to the lower base line and joined to the point
selected on the neutral axis. The shaded area is the modulus figure for
the section.

Let the shaded area above the neutral axis = A.

Then total stress above neutral axis =/cA

Let d^ = distance of centre of gravity of upper shaded area from NA
dz= „ „ „ lower „

D = 4 + tk

BEAMS

Then the moment of resistance of the section

d2)

=/cAD

Construction for Modulus Figure having an Intensity of Stress equal
to ft, Fig. 65. — The intensity of stress ft occurs at a distance y below
the neutral axis, therefore at a distance
y above the neutral axis the intensity
will also be ft. The base line for the
upper portion is therefore drawn through
the plane ef. As the intensity of stress
in the material above this base line is
greater than /, the area aefb must be
increased. For any layer above the base
line, say ab, project on to the base line
in e and/; join e and /to a point on the
neutral axis and produce these lines to cut

the horizontal through ab in c and d. Then the length cd will be
the increased length of ab required. For all layers between the base
lines proceed as in the former construction.

Let A! = area of each portion of the modulus figure

Dj = distance between centres of gravity of shaded areas
= d, + 4

Then M.R. = AJV,

Knowing the bending moment at a vertical section of a beam, the
suitability of the cross-section for resisting it may be determined.

Bending moment = moment of resistance

FIG. 65.

or =/A1D1

If either quantity /CAD or/A^ be less than the bending moment
(after inserting a suitable value for ft or/c) the cross-section is not
strong enough to safely support the load on the beam and must be
increased. For beams composed of mild steel which has an equal
strength in tension and compression, it is only necessary to construct
the modulus figure for the larger intensity /c, as failure must occur
where the material is the more highly stressed. For cast iron and
other materials where the strength in tension does not equal the
strength in compression, both the maximum intensities fc and ft
produced by the bending moment, must be calculated and compared
with the allowable safe intensities for the material employed.

EXAMPLE 11. — To find the moment of resistance of a 4" x 6" x £" T
with parallel sides, Fig. 64.

The distance of the centre of gravity of the section from the lower

_ moment of all layers about lower edge

total area of section
_ 4" x 4" X y + 5£" X ^" X 8j"

4" x 4" + 54" X 4"
= 1-987"
Therefore y (Fig. 64) = 1-987"

«/i = 6 - 1-987 = 4-013"

100

STRUCTURAL ENGINEERING

Construct the modulus figure as in Fig. 64. Then the area of the
figure above the neutral axis

= area of shaded triangle.

_ i" v 4*013
-2 X.-JT-

= 1'003 sq. in.
Distance d^ of centre of gravity above the neutral axis NA

= | X 4-013

= 2-675"

Moment of triangle about neutral axis

= 1'003 X 2-675
= 2-68 in.3

The modulus area below the neutral axis must equal the area above
= 1-003 sq. in.

The centre of gravity of the lower area may be found by calculation
or by cutting out the figure in cardboard and suspending from two
points.

In the enlarged Fig. 66,

--T-

1 — -fA

/ 1 \

• 1-437' ;

1-987' , /

III

\ A,*'

! /

I i r

ftc

— °-\-

' I B\

i! \*

7/5*

•'11

PIG. 66.

Length CD = 4 x ^^ = 1-48 in.

Area ABDC = 1>9S + 1>48 x i
2

= 0-865 sq. in.

Length EF = J" x j~ = 0-185 in.
Area of triangle

OEF = 0-185 x 1-487 X \

= 0-138 sq. in.
Total area below the neutral axis

= 0-865 + 0-138 = 1-003 sq. in.

which corresponds with the area obtained for the upper portion.
Distance of centre of gravity of triangle OEF below neutral axis

= f X 1-487 = 0-991 in.
Moment of triangle OEF about neutral axis

= 0-138 X 0-991 = 0-137 in.3
Distance of centre of gravity of ABDC below CD

J(l-48 + 2 X 1*98)
= --~-/r-~ -.-J^r^—- = 0-27 in.

Distance of centre of gravity of ABDC from neutral axis
= 0-27 + 1*487 = 1-757 in.

BEAMS

Moment of ABDC about neutral axis

= 1-757 X 0-865 = 1'52 in.3

and distance dz of centre of gravity of the lower modulus figure from
the neutral axis

= ™ + °'137 = 1-65 in.
1-003

Moment of lower modulus figure about neutral axis

= 1-52 + 0-137 = 1-657 in.3
Then moment of resistance of the section

= /c(l-657 + 2-68)
= 4-337 /c

or =/c{l-003 x (1-65 + 2'675)}
= 4-337 /c

EXAMPLE 12.— What distributed load will a T 4" x 6" X i" support
over a span of 6 feet, the working stress (maximum skin stress) not to
exceed 7 tons per square inch ?

(1) When the 4 in. leg is horizontal.

From the previous example

M.B. = 4-337 fc

= 4-337 X 7

= 30-359 inch-tons.

Let w = tons per foot run supported by the beam.
Maximum bending moment

8
w X 62

ft. -tons

» v^ 1 O

inch-tons.

Note. — The moment of resistance being expressed in inches and
tons, the bending moment must also be expressed in those terms.

Then B.M. = M.R.

30-359
w = 0'551 ton per foot run.

2) "When the 4 in. leg is vertical.

he modulus of section may be readily calculated since the neutral
axis passes through the centres of gravity of both rectangles forming
theT.

(2)
Th

12 12

= 2-72

Modulus of section = Z = ^ = 1'86

M.R. = 1-36 X 7 = 9-52 inch-tons.

.-'*&.'-

102; ; ; V : J : ST^ClUR AL ENGINEERING
The bending moment will be the same as above

B.M. = M.R.

54w = 9-52

w = 0'176 ton per foot run.

Massing up of Sections. — The resistance to bending of a section
depends only on the disposition of the material normally to the neutral
axis and not to its relative position along the
axis. Sections of inconvenient shape, such as the
channel of Fig. 67, are massed together along
the neutral axis before constructing the modulus
figure.

Construction of Modulus Figure for a Rail
Section.— The centre of gravity of the section,
Fig. 68, is most readily found by cutting out the
section in good quality cardboard, suspending it
FIG. 67. from two points and finding where the verticals

through those points intersect. The centres of
gravity for the modulus figures may also be found in this way.

For any horizontal layer, say 41-41, proiect the extremities on to the
base line in IV, IV. Join IV, IV to a point in the neutral axis, such
lines cutting the layer ^-^ in 4, 4. Then the points 4, 4 will be on
the boundary of the modulus figure. The areas of the modulus figures
are obtained by the aid of a planimeter or calculated by the aid of
squared tracing paper. The rail section in the figure is a 100-lbs. rail
drawn full size.

TABLE 25. — MOMENTS OF INERTIA AND MODULI OF SECTIONS.

Section.

6
x-f

•-B-- •

Moment of inertia about
axis XX.

ED3
12

BD3
3

Modulus of section
about XX.

BD2

BEAMS

103

«« iy Y

V IV

III IV V

FIG

104

STRUCTURAL ENGINEERING

Section.

Moment of inertia about
axis XX.

Modulus of section
about XX.

- B—

x—

..... D .....

BD3 - bd3
12

BD3 - bd3
6D

7rD«

-CT = 0-0491D4

?rD3

s- = O0982D3

*154-_

32D

- B -i

BD3
~36~

BD2
^4~

BD3
12

BD3
4

BEAMS

105

Shear. — When any system of forces acts on a beam it produces a
vertical shearing action which tends to shear the beam in vertical
planes, as in Fig. 69, A. The bending
action creates differences of stress in the
horizontal layers of the beam and
thereby produces a horizontal shearing
action between the layers. If the
beam were composed of a number of
separate plates, they would slide upon
each other as in Fig. 69, B. In solid
beams the tendency for the layers to
slide upon each other is resisted by the shear stress in the material.

The method of calculating the vertical shearing force at any vertical
section of a beam has been explained in Chapter III.

In Fig. 70 consider the equilibrium of a portion of a beam acdb
lying between two vertical sections very close together. The horizontal
forces acting on it are, the horizontal stress on ac caused by the bend-
ing, the horizontal stress on bd acting in the opposite direction, and the

FIG. 69.

! =7

FIG. 70.

shear stress on cd, which is equal to the difference of the horizontal
stresses on ac and bd. The horizontal stress above c'c' at the section aa! is
equal to the area of the modulus figure above c'c' multiplied by the skin
stress /at that section.

But

B.M.ax|

Similarly at the section W the skin stress /i will be
/! = B.M, x |

Let Aj be the area of the modulus figure above c'c'.
difference of stress at the sections aa' and bb'

Then the

« A^B JL. - B.M.»)

(1)

Let a± = area of a thin horizontal strip distant y^ from the neutral

axis.
The area of the modulus figure for this strip

106 STRUCTURAL ENGINEERING

Let Aj be the sum of all such areas between a and c ;

then

The moment of the area aL about the neutral axis = a^

The total area of the section between a and c = A = 2^
and its moment about the neutral axis = 20^

Let Y be the distance of the centre of gravity of this area from the
neutral axis.

Then A x Y =

Also

from which

Ax Y

The total shear along cd will therefore

= ^L*J[(B.M.B - B.M.6) from (1)

Let the width of section c'c' = w \

Then the intensity of shear on the plane cd

f -

Ax Y
It* '

In the limit B.M.rt - B.M.6 = d (B.M.).

But the total vertical shear S on any section

Therefore

Consider a small rectangular prism of material in a loaded beam
(Fig. 71). The load and reaction produce shear stresses, acting in
w opposite directions on two vertical sides of
the prism.

To establish equilibrium there must be
another couple acting on the horizontal faces.
Let the total vertical stress = S
„ „ horizontal „ = Sx
Let the intensity of vertical stress = s
„ „ horizontal „ = s1

Then SZ> = Sirf

swdb =

FIG. 71.

:. 8=5

That is, the intensity of horizontal shear on the material must be
equal to the intensity of vertical shear.

BEAMS

107

Therefore the intensity of vertical shear at any point in the section
of a beam must

_ /> AYS

Intensity of Vertical Shear Stress on a Rectangular Beam Section. — To
find the intensity at the neutral axis.

A = area of section above NA = -g-
Y = distance of centre of gravity of area above NA
from the NA = 7

1 12

T =
= 12
bd d

^ is the mean intensity of shear on the section ; therefore the

intensity at the neutral axis is one and a half times the mean intensity.
Let the section of the beam be 10" x 6" and the vertical shear be 10
tons.

Then the mean shear intensity

= io~x~6 = °'166 ton Per gq- in-

Intensity at the neutral axis

•

= f X 0-166 ton per sq. in.
= 0'25 ton per sq. in.

By calculating the values of /, for
a number of other planes and plotting
them to a vertical line as in Fig. 72,
a diagram of shear intensity for the
section is obtained. The bounding
curve will be a parabola.

Distribution of Shear Stress in a Beam Section (Fig. 73). — The
stresses at a number of horizontal planes may be calculated by the
above formula and the values plotted to a vertical line, or the values
may be found from the modulus of section.

It has already been proved that

FIG. 72.

x = , whence A = Ax

Substituting this in the expression

108

STRUCTURAL ENGINEERING

The quantity ^r is constant for any particular vertical section.

Therefore the intensity of shear stress on any horizontal plane is pro-
portional to the area Ax of the portion of the modulus figure above
that plane divided by the width of the section at the plane. At any
plane cd, Fig. 73, the intensity will be equal to the shaded modulus

area divided by the width cd and multiplied by the constant '4- . The

intensity diagram may be constructed by calculating a series of values
of/, by the above method and plotting them to the line xx.

The diagram demonstrates (1) the small intensity of shear stress in
the flanges (section lined on the diagram), and (2) the almost even
distribution of stress over the web area. The maximum intensity = GK ;
the mean intensity over the whole section = GH. "When designing
beams with deep webs, the resistance to shear offered by the flanges is
usually neglected, and the web designed to resist the whole shearing
action.

BEAMS

109

EXAMPLE 13. — To find the pitch of rivets in the flanges of a plated
girder (Fig. 74).

Let the section of the girder be, one 18" X 7" rolled beam with one
12" x |" plate riveted to each
flange. Span of girder = 24 feet.
Load = 48 tons distributed.

The maximum vertical shear
will occur at the supports, and
be equal to 24 tons.

Shear intensity at the hori-
zontal plane between the plates
and rolled beam

= f =

AYS
ivl

FIG. 74.

where A = sectional area of plate.

Y = distance of centre of gravity of plate from N.A.
S = total vertical shear on section.
I = moment of inertia of section.

w has two values

= width of plate when calculating the intensity in the plate.
„ flange of beam „ „ „ joist.

Suppose the shear intensity along the plane under consideration to
remain constant for a horizontal length of 12 inches. The total shear
for this length would then be equal to the shear intensity multiplied by
the area of the plane.

Area for 12 inches length = w x 12.

Total shear stress for 12 inches length

= , X

12AYS

I
12 X

12 X | X 9T5s X 24

2225
= 9-04 tons.

Let the rivets be | inch diameter.

Resistance to single shear of one rivet = 3 tons.

Number of rivets required per foot length

9-04

= 3-01

Four rivets would therefore ba used, and being in pairs the pitch
would be G inches.

The shear decreases to nothing at the centre of span, and therefore
the pitch required would increase to a maximum at the centre of the
span. It is not advisable, however, to make the pitch of rivets in

110 STRUCTURAL ENGINEERING

such girders more than 6 inches, to avoid local buckling of plate, so
a uniform pitch would be kept throughout the full length of the
girder.

NOTE. — If the bearing resistance of the rivets be less than the
shearing resistance, the bearing resistance must be used in the above
calculation in place of the shearing resistance.

General Considerations of Design. — Type of Structure. — When
entering upon the design of any constructional work, the first conside-
ration is the type of structure to be employed. In nearly all cases
there will be a variety of types from which to select, and judgment —
to be obtained only from experience— will determine the most satis-
factory solution. The nature of the loads to be supported, character
of site, facilities for erection, bye-laws, etc., will influence the selection ;
but no general rules can be laid down, since the considerations vary in
every case. Upon the discretion of the designer, the success or failure
to provide an economical and satisfactory structure will depend.

Loads on Structures. — The loads a structure will be called upon to
bear have been fully discussed in Chapter II.

Arrangement of Members. — The type of structure having been
decided upon and the loads which it must support determined, it is
then necessary to arrange the positions of the different members of
the structure. The object of the structure as a whole is to transmit
forces from certain positions to a foundation, and the members of the
structure must be arranged in the most satisfactory manner for accom-
plishing this object. There are two general types of structures:
(1) those known as framed structures, in which the members are only
called upon to resist either direct tensile or compressive stresses — for
example, bridge girders, roofs, etc. ; and (2) structures composed partly
of members subject to bending action. For framed structures there
are recognized arrangements of members, but precautions must be
taken to ensure all the forces acting at the junctions of the members,
otherwise secondary stresses will be produced in the members and
connections. The second class includes a very wide range of structures,
which admits of no general rules applicable to the whole class. The

a'ble arrangements will depend on the relative positions of the
i, the foundations to which the loads may be transmitted, and any
obstacles between those positions. The final arrangement will be
selected after due consideration of the relative economy of the optional
systems. The arrangement should ensure a stiff frame against wind or
other lateral force which may tend to produce a racking action.

Design of Members. — Each member of a structure has a certain
function to perform, and the form of the member will depend upon the
nature of the stress it has to resist. Members subject to bending only
will at present be considered. It was shown in Chapter III. how to
obtain the bending moments and shear forces on such members, and in
the present chapter, having assumed the form of the member, the
stresses have been determined.

When selecting the form of a girder section due consideration must
be paid to the depth, as it is upon this property of the section that the
deflection of the beam depends. The allowable deflection varies with
the class of work, and is measured in terms of the span. The ratio of

BEAMS 111

deflection to span for first-class bridge work is as low as 1 to 2000,
whilst for small girders and rolled steel joists in ordinary buildings the
ratio may be as high as 1 in 400. It may be shown that the maximum

fl?
deflection for any beam of uniform section = a -TJT,

where a = a constant varying with the methods of loading and

end fixings.
/=the intensity of working (skin) stress in the material,

in tons per sq. in.
y = distance in inches from the neutral axis of the material

stressed to /tons per square inch.
E = modulus of elasticity of the material in tons.
L = span of beam in inches.
For beams symmetrical about the neutral axis —

y = 2", where D = depth of section in inches.
For such beams the above formula may be written —

ED
where A = deflection in inches.

For cantilevers having one concentrated end load a = J
„ „ a uniformly distributed „ a = J

For beams supported at each end, and central „ a = ±
„ „ „ distributed „ 0 = ^

To find the depth of a mild steel beam centrally loaded, if the ratio
of deflection to span must not exceed 1 to 1000, the working stress to
be 7 tons per square inch, and E to be 13,500 tons.

2/L2

A = ctt-^—-
ED

_L- = JL 2X7 L
1000 12 • 13500 'D
. D_ l_
' L~ il-57

i.e. the depth must be equal to TTTFy °f tne span, or if the span be 100
feet the depth must be —

100 =8-65 feet.

11-57

The following table gives the ratios of D to L for mild steel beams
for various ratios of deflection to span —

/ has been assumed = 7 tons per square inch
E „ = 13,500 tons.

112

STRUCTURAL ENGINEERING

TABLE 26. — RATIOS OF DEPTH TO SPAN OF MILD STEEL BEAMS.

Deflection
formula.

Ratios of deflection to span.

I to 400

1 to 600

1 to 1000

1 to 1500

Ratios of £

Ij L f

A-' /^

'\ '

1 JS

4-8

2-9

1-2
1

. 1 J

A-J/L2

; „, ' ' -H

1 By

9-6

6-4

3-9

2-6

»w

i /L2

it L — J-

A T5'E2/

11-6

7'7

A * /^

j, L-_-=^

*'I5

15-4

9-2

6-2

The values of ^ for beams irregularly loaded will lie between those

given for a single concentrated load at the centre, and a uniformly dis-
tributed load.

The values of j-for other values of /and E, say/ and E15 may be

^•Tjl

obtained by multiplying the ratios by Jj^.

The amount of lateral deflection on a beam will vary according to
the breadth of the beam. The forces producing lateral deflection in
a beam are usually of such a character that any reliable estimate
of them is impossible, and therefore no calculation for the breadth
based on such forces can be made. A practical rule is to make
the flanges at least ^ of the span where the beam is unsupported
laterally, or if such breadth be excessive the beam should be tied to

BEAMS

113

other members at distances of not more than thirty times the flange
width. In most cases beams are supported laterally by cross girders,
floors, etc., in which case the flange width is not of such importance.
Broad flanges make better bearings for cross girders, and provide more
space for connecting bolts and rivets.

Assuming the dimensions for the section by the above rules the
maximum stress in the member must be calculated by the method
previously described in this chapter. If such stress does not satisfy
the working stress of the material the section must be modified in
thickness, breadth, or increased depth until the stress is approximately
equal to the working stress.

It is usual to take the working stress some fraction of the ultimate
strength of the material. This allows a factor of safety for flaws in the
material and inability to accurately estimate the
loads on the members. If a factor of safety of 4
be used for steel having an ultimate strength of
28 tons per square inch, that is a working stress
of 7 tons per square inch, it does not imply that
the material may be stressed four times that
amount before failure occurs. The member
would fail to act in accordance with the accepted x*
laws of a beam immediately the elastic limit of
the material had been exceeded. If the member
be subjected to dead load only, or if the live load
be relatively small, a small factor of safety may
be adopted ; but if the live load greatly predomi-
nates, and is of a very varying character, a larger
factor of safety is necessary. The character of
the material employed also affects the factor of safety. For cast iron,
which may contain hidden flaws or initial stresses due to casting, a
higher factor of safety must be used than for mild steel, whose homo-
geneity is very reliable. The determination of suitable factors of
safety for various conditions of dead and live loading has already been
discussed in Chapter II.

Beams are often composed of one or more rolled sections, and
labour is saved in calculating the moments of resistance and inertia by
the use of tables given in section books issued by most constructional
firms, or in the lists of " Properties of British Standard Sections,"
compiled by the Engineering Standards Committee. The modulus of
section required for a beam is equal to the bending moment divided by

the working stress of the material ( -V-'

FIG. 75.

Having calculated this,

a section having such a modulus may be selected from the tables, due
regard being paid to the depth and shearing stress. The following is
an extract from the " Properties of British Standard Sections."

114

STRUCTURAL ENGINEERING

Reference No.
and code woi'd.

Size.

Standard thickness.

Radii.

Weight
per foot,
w

Sectional
area, a.

Centre of gravity.

AxB

B.S.B. 1.
Abscession

in.

in.
0-160

in.
0-248

in.
0-260

in.
0-130

Ibs.
4-00

in. 3
1-176

in.
0

Moments of inertia.

Radii of gyration.

Moments of resistance.

B.S.B.

No.

l'.v

in.*
1-657

in."
0-124

in.

1-187

in.
0-325

in.3
1-105

in.3
0-165

In column 1 the code word and reference number for use when
ordering the section are given. Columns 2 to 10 contain the physical
properties of the section. In columns 11 and 12 are given the moments
of inertia about the axes X-X and Y-Y. In ascertaining the strength
of a beam section to resist certain forces, the moment of inertia used
will be that about the axis normal to the forces. The least moment of
inertia of a section is also required in column calculations. Columns
15 and 16 are here called moments of resistance. This term must not
be confused with the moment of resistance denned in this chapter as
being the modulus of section multiplied by the skin stress. The
tabular value headed Moment of Resistance is actually the modulus of

the section, or -. The values are again given about both axes.

To use the above table to find what uniformly distributed load the
3" x 1J" beam would support over a span of L feet, the web of the
section to be vertical.

B.M. =M.R. x/

Let/= 7 tons per square inch,

Then

= M05X7

.-. w = ~j~ tons per foot run.

T 2

Note.— When using the formula -g- care must be taken to

express L in the correct units. If w be given, as is usual, in tons per
foot run, the total load on the beam will be equal to wL tons where L

is in feet. The bending moment in inch-tons will be ^— TT — 1 where L

BEAMS

115

Or taking L throughout in feet the formula

is the span in inches,
must be written - Q *"•

Columns 13 and 14 contain the radii of gyration,
and its use will be explained in Chapter Y.

EXAMPLE 14. — To find the modulus of section
for a compound beam.

Let the beam be composed of one 20" x 7j"
rolled beam and two 12" X f" plates riveted to
the rolled beam by £ in. diameter rivets (Fig. x
76).

Neglecting for the present the effect of the
rivets.

The moment of inertia of the rolled beam
about the horizontal axis, from the tables

Ix = 1671-291

Moment of inertia for the plates —
Ix = 2(1. + AR2)

This property

76.

5 x
8 *

= 1595-7 in.4

Total Ix for the section

= 1671-291 + 1595-7
= 3266-991 in.4

The rivets in the flanges are staggered, so that not more than two
rivets appear at any cross-section. If the rivet in the compression
flange completely fills up the hole, the total area of the compression
flange is not affected, but the liability of having rivets imperfectly
fitted makes it advisable, to ensure safety, to deduct the area of the
holes from the flange area when calculating the strength of the section.
It is apparent that the holes through the tension flange will reduce the
strength of that flange and must be taken into consideration. The
moment of inertia for the section will therefore be 3266-991 — Ix of
two rivet holes.

For £ in. rivets the holes are drilled jf in. diameter.

The mean thickness of the flanges, from the tables, = 1*01 in., and
may be taken to represent the mean length of the rivet in the beam.
The total length of the rivet will be I'Ol + 0'625 = 1-635 in.

Area of cross-section of hole = T635 x yf = 1*53 sq. in.
Ix of 2 rivet holes = 2J— — ^ — +1-53 x (9*8)2j

= 294-567 in.4

Ix of section = 3266-991 - 294-567
= 2972-424

™ i i f f
Modulus of section = - =

2972-424

279-75 in.

116

STRUCTURAL ENGINEERING

Note. — Rivet holes, if not symmetrically placed, will change the
position of the centre of gravity of the section, and therefore the
position of the neutral axis. In such cases the tables are of little
benefit.

Connections. — In all framed structures the different members are
fastened together by means of angles, plates, etc., and connecting
rivets or bolts. The available methods of connection will in some
cases determine the best cross-section of members to be employed.
The duty of a connection is to transmit forces from one member of a
structure to another, and all connections must receive the same care in
design as the members themselves.

The simplest method of connecting beams is to allow one beam to
rest on the flange of another and bolt them securely through the
flanges. Such a connection is shown in Fig. 77, A. If there are two
beams in line resting on the main girder, it is usual to fasten them
together by means of fish plates in the webs. This increases the lateral
stiffness of the beams and reduces the tendency to twist. Tapered washers
should be placed under the heads and nuts of all bolts having a bearing

FIG. 77.

on the inside faces of beam flanges, otherwise only a very small part of
the head or nut will be bearing on the flange. It is not always con-
venient to allow the secondary beams to rest on the top flange of the
main beam, and in such cases web connections have to be employed.
Fig 77, B, shows the connection of a comparatively small beam to a
larger beam. An angle or tee riveted to the web of the main beam
forms a bracket on which the small beam rests, and to increase the
stiffness of the connection cleats are bolted to the webs. Figs. 77,
C, D, show other connections where the beams are equal or nearly
so in depth. Although unusual, the lower flange of a beam is some-
times joggled, so as to rest upon the tapered portion of the main beam
flange (Fig. 77, D). This method is costly, and does not greatly increase
the efficiency of the connection. The angle connections between the
webs of rolled beams, and the number and spacing of rivets in them
have been standardized, and may be found in any maker's section book.
When using such standard connections, the strength of the rivets
or bolts should be checked to ensure that the strength is at least equal
to the shearing force at the connection. Bolts or rivets in such con-
nections may fail either by shearing or crushing. The shearing
strength of a rivet is equal to the area of its cross-section multiplied

BEAMS 117

by the shearing strength of the material. For rivet steel the shearing

strength is 21 to 22 tons per square inch, and factors of safety of from

4 to 6 are usually adopted, giving a working stress of 4 to 5 tons per

square inch. A rivet is said to be in single shear when for failure of

the connection to occur, it is only necessary to shear the rivet at one

section, as at a (Fig. 78). For the joint to fail in Fig.

78, #, the rivet must be sheared along two planes, or is

said to be in double shear. Although the area of material

sheared at b is twice that sheared at a, the strength of

a rivet in double shear is found, in practice, to be less

than twice the strength of a rivet in single shear. The

strength of a rivet in double shear is from 1J to Ij times

the strength of the rivet in single shear, and in the

following calculations will be assumed as 1^ times the

strength in single shear.

Let d = diameter of rivet in inches.

/, = safe shearing stress of material in tons per square inch.
S = vertical shear at connection in tons.

Then the strength of a rivet in single shear = - &/,

„ double „ = f.f^

Let n = number of rivets in single shear required to transmit the
shearing force S.

Then S = w

If n' = number of rivets required in double shear

The resistance to crushing offered by a rivet is equal to the
crushing (or bearing) resistance of the material multiplied by the area
of the rivet normal to the force. The safe bearing resistance of rivet
steel is from 7 to 10 tons, say 8 tons per square inch.

Let t = thickness of plate bearing on rivet,
fb = bearing stress of material.

Then the bearing resistance of one rivet = dtfb

The number of rivets required to transmit the shearing force S

The number of rivets required at a connection will be the larger
value of n in the expressions —

STRUCTURAL ENGINEERING

4S .

if the rivets are in single shear
double

>A "holes

118

and n" = ^

In practice rivet holes are punched or drilled ~ in. larger in diameter
than the rivets, and on closing the rivets should till up the holes, thus

increasing the sectional area of
the rivets, but calculations for
shear should be based on the
original diameter of the rivets.
Bolt holes are drilled similarly,
unless specified to be a driving
fit, and there is always a little
uncertainty as to the number
really in action. For this reason
more than are theoretically
necessary are usually employed
at connections.

EXAMPLE 15.— ^0 find the
number of f in. bolts and rivets

FIG. 79. required at the connection of a

24" X 7y rolled sfeel beam sup-
porting a uniformly distributed load of 50 tons, to another beam of
similar dimensions.

The connecting angles to be riveted to the beam a, Fig. 79, and
bolted to the beam b.

The vertical shear at the connection is equal to the reaction, i.e.
25 tons.

The rivets being in double shear, the shearing resistance of one rivet

1*-IX /71//CO

<H *+\r*+

' or rive

or rivek.
> »-7f

Let/

Then

5 tons per square inch,

n =

The bearing resistance of one rivet =

where t = thickness of web = 0-6 inch.
Let/& = 8 tons per square inch,

Then

n^xO-6x8=(Say)6

The bolts through the web of beam b being in single shear, the
shearing resistance of one bolt

Therefore

n =

(say) 9

X 5

BEAMS

The bearing resistance of one bolt = dtfb
where t = thickness of angle = J inch,

119

Therefore

n =

|X*X 8

= (say) 8

The theoretical number of bolts required is therefore 9, but as 12
may be conveniently employed this number will be adopted to allow for
a number being out of action.

The shearing resistance of the angles may be taken to be equal to
the minimum area, i.e. along a vertical section through the rivets or
bolts, multiplied by the resistance to shear of the material, although
this value will be somewhat small on account of the resistance offered
by the rivets or bolts to fracture at such a section.

In the above example the sectional area of the angles along the
vertical section through the rivets or bolts

= 2 (19J - 6 X Jf) X 4 = 13'875 sq. in.
The shearing resistance will therefore = 13-875 x 5 = 69'375 tons.

This is greatly in excess of the vertical shear, but the thickness of
angle cannot be much reduced, as such reduction would mean an in-
creased number of rivets required in bearing.

Joints in Tension Members. — Structural members subject to a purely
tensile stress are usually butt-jointed, with single or double covers at
the joints, as B and C, Fig. 80.
When only one cover is employed,
there is a tendency for the member
to bend near the joint, as at E. The
better construction is to employ two
covers at all such joints.

Failure of the joint may occur —

(1) By shearing all the rivets to
either side of the joint.

(2) By crushing the rivets.

(3) By pulling apart the cover
plate or plates along the weakest B
section.

(4) By tearing of the main plate.

Consider the joint Fig. 80; A.
Let t = thickness of member.

ti = „ cover plate or plates.

d = diameter of rivets.

d,= „ „ holes.

/ = safe shearing intensity on rivet.

/B = „ bearing „ „

w = width of member.

ft = safe tensile intensity on plates.

T = tension in member.

Then the shearing resistance of the rivets to either side of the
joint

120 STRUCTURAL ENGINEERING

= 2(W2-/) for single cover

£) for double covers.
The bearing resistance of the rivets to either side of the joint

= 2(dtfb) in member.

= 2(dtjfb) in single cover.

= 2(2^i/6) in double covers.

The resistance of the cover plate or plates to tension along the
section a-a

= (w - 2d1)t1ft for single cover.
= "2(w — 2^)^ ft for double covers.

The resistance of the member to tension alon the section a-a

The resistance in each of the above cases must be at least equal to
the tension in the member.

EXAMPLE 16. — A mild steel tension member is subject to a pull of 70
tons. Design a suitable section for the member and also a butt joint with
double cover plates.

The intensity of tensile stress not to exceed 7 tons per sq. in.
„ „ bearing „ „ 8 „

55 55 snear ,, ,, o 5) ,,

Double shear to be taken equal to If times single shear.

Adopting a rivet diameter of | inch.

The shearing resistance of one rivet = 5 tons.

The number of rivets required to either side of the joint

Let t — thickness of member.

Then the bearing resistance of one rivet = | x t x 8

and for the bearing resistance of the rivets to be equal to the pull in
the member

(| X t X 8)14 = 70 tons

from which t = f inch, say f inch.

•

Arranging the rivets as in Fig. 81, the weakest section at which
the member might fail would occur at either a-a or b-b. If the section
at a-a be assumed for the present as the weakest, let w = width of the
member,

Then (w - if )| x 7 = 70 tons,

from which w == 14'27 inches, say 14 1 inches.

For failure to occur along the section b-b, the plate must be torn
along that section and the leading rivet sheared. The strength along
b-b will therefore

= (14-5-2 x ii)|X 7 + 5

= 71*3 tons.

BEAMS

121

enough

along this section to resist

The member is therefore strong
the pull.

The strength along the section c-c is greater than along the section
&-&, since the reduction of plate area for the extra rivet in the section

FIG. 81.

c-c does not weaken that section to the same extent as the shearing
of the two extra rivets along b-b increases the resistance to failure
along c-c. For the same reasons failure would not take place by tearing
of the member along the sections d-d or e-e.

The joint may fail by te'aring the cover plates along e-e, or by the
crushing of the rivets in the covers.

Let tj, = thickness of each cover.

The resistance to tearing of the covers along e-e must be equal to
the pull on the member, or

(14J - 4 X 11)2^ X 7 = 70 tons

from which ^ = 0'46, say 0'5 inch.

Since the combined thickness of the covers is greater than the
thickness of the member, the bearing resistance of the rivets in the
covers will be greater than the bearing resistance in the member, and
failure would not occur by crushing of the rivets in the covers.

Examples of such joints applied to the ties of lattice girders will be
found in Fig. 193.

Riveted Connections subject to Bending Stresses. — Suppose, in Fig. 82 ,

FIG. 82.

a cantilever L inches long to be loaded at its outward end with W tons.

122 STRUCTURAL ENGINEERING

Then the bending moment at the connection of the cantilever with its
support = WL inch-tons.

This bending moment tends to produce a rotation about a hori-
zontal axis perpendicular to X-X, and subjects the rivets along the
section Y-Y to a horizontal shearing stress, in addition to the vertical
stress due to the vertical shear at the section. The moment of the
horizontal shearing stresses in the rivets along Y-Y about the hori-
zontal axis through X-X, must equal the bending moment at the
section. The moment of inertia of rivet c about the axis through
X-X = I0 + ay\

where Ib = moment of inertia of the rivet section about its longi-
tudinal axis.

a = area of cross-section of rivet in shear.

y = distance of centre of cross-section of rivet from the axis.

Since I0 is very small compared with ay2, the moment of inertia
may be considered equal to ay2.

The sum of the moments of inertia of the system of rivets will
then —

= Say2 = a (y2 -f y? + y22, etc. )
= 1

Let/s = maximum horizontal shear stress in outermost rivet.

y = distance of centre of that rivet from the axis.
Then the moment of resistance of the system of rivets

Therefore the bending moment at the section must
= B.M. = +

from which the maximum horizontal shearing stress in the rivets is
obtainable. The vertical shear will be equally distributed amongst the
rivets.

Let/fc = the maximum intensity of horizontal stress.
/„ = the intensity of vertical stress.

Then the maximum stress on the rivets will be the resultant of fh,

and/. = //?T/;2.

The bending moment also produces tension in the rivets above
X-X, in the other plane of the connection.

Then B.M. = M.R. =fj(y* + 2A2 + yas, etc.).

EXAMPLE 17. — Let ttie span of the cantilever be 5 feet, and the load
2 tons.

The bending moment at the connection

= 5x 12x2 = 1 20 inch-tons.
Assume a depth of 21 inches for the beam, and let it be connected

BEAMS

123

by two angle irons to the support. The moment of resistance of the
seven rivets of £ in. diameter along the section Y-Y, Fig. 82,

Since the rivets are in double shear, the area a will be equal to
twice the area of the cross-section of a f-in. rivet = 2 X 0*6 = 1*2
sq. in.

Then B.M. = M.R. to shearing.

/. f, = 3 '57 tons per sq. in.

Let/6 = maximum intensity of bearing stress on the rivets.
«! = the bearing area of one rivet.
t = thickness of web plate.

Then a^ = t X f in.
The moment of resistance to bearing of the line of rivets will

Let the thickness of the web plate = f in.

Then fli = f X | = 0'33 sq. in.
Again B.M. = M.R. to bearing

120 =/, X 0-33 X

/. fb= 12-9 tons per sq. in.

This is in excess of the safe bearing stress, and either more rivets
must be used or the web plate thickened to give a greater bearing area.
Suppose a second line of rivets be used, Fig. 83,

the,

and

M.E. =, x 0-33 X

/. 120 =fb X 14-48

120
fi = TITTQ = 8 '28 tons per sq. in

This stress would only occur on the two
extreme rivets, and may be safely adopted.

The addition of the second row of rivets
will decrease the horizontal shear stress in the
outermost rivets to 2 '3 6 tons per square inch.
The total maximum shear in the rivets

FIG. 83.

2'362 tons per sq. in.

124

STRUCTURAL ENGINEERING

The moment of resistance of the rivets along the lines Y-Y

'32 + 62 4- &

=f*xax 4(
B.M. = M.R.
120 =/ X 0-6 X

2\
/

/. ft = 3-57 tons per sq. in.

i.e. the maximum tension on the rivets is much below the safe working
stress, and a reduced number might be safely adopted. Only the
rivets in the upper half of the connection are stressed under the bend-
ing action, the bearing of the back of the bracket against the column
relieving the rivets in the lower half of the longitudinal stress. All

the fourteen rivets are subject to a slight vertical shear stress of mean

2
intensity = 1 . n.ft = 0'24 ton per square inch, so that the actual

JL~t /\ \) O

maximum intensity of stress will be slightly in excess of 3-57 tons per
square inch.

EXAMPLE 18. — Design of Floor for Warehouse.— Suppose the out-
line in Fig. 84 be the plan of a floor of a warehouse for which a design

is required.

The first consideration is the
kind of floor to be adopted ;
whether fireproof or not. It will
be supposed that a fireproof floor
is not required, and an ordinary
timber floor supported on steel
beams is decided upon.

Let the estimated live load on
the floor be equivalent to a dead
load of 1J cwts. per square foot
of floor area. The dead load may be
determined as the design proceeds.
A suitable covering for the floor
would be 1^ in. flooring secured to
11" x 3" timber joists spaced at
1' 6" centres. The span of the floor
joists will be determined by their
depth. For stiffness the span should
not exceed 10 times the depth ;
therefore the maximum span will be
10 x 11" = 9' 2". The arrangement shown in the figure makes the
maximum span 8' 0", or 8-6 times the depth of the joists.

Stress in Timber Floor Joists. — Each floor joist supports an area of
floor = !£' x 8' = 12 sq. feet. Load on each joist —

g \

^»-fl'-.

-fl'-

-e'-«

- a'-*

•-76'-*

-7

s"->

-76'-^

9 \

*»

,.,,..,

FIG. 84.

Weight of flooring = 12 x

x 85 =52 -5 Ibs.

one joist = 12 x 12 x 8 X 35 = 62-5
Live load = 12 X l X 112 = 2016

Total distributed load =2131 „
say = 19 cwts.

BEAMS 125

Weight of timber has been taken as 35 Ibs. per cubic foot.

Max. B.M.= ^-2

- 19 X 112 X 8 X 12

___

= 25,536 in.-lbs.

, . . 3 X 11 X 11

Modulus of section of joist = - — ^ —

= 60-5

25 536
Maximum stress in the timber = 'Q.5 = 422 Ibs. per sq. in.,

which gives a factor of safety of about 10. The timber joists in the
offset bay having less span would be stressed to a less extent, but for
uniformity 11" x 3" joists would be adopted throughout.
Let the working stresses for the steel beams be —

ft (tension or compression) = 7 tons per sq. in.
ft (shear intensity) not to exceed 3 tons per sq. in.

Primary Beams, 0, 13' 6" long. — The maximum loading for these
beams would be as shown in Fig. 85.

Reaction from each timber joist = 9*5 cwts.
Load at each bearing =19 „
Reactions = 85*5 „
Maximum B.M. (at centre)

= 85*5 X 6-75 - 19(1-5 + 3 + 4-5 + 6)
= 292-125 ft.-cwts.
= 175-275 in.-tons.

,, , , . .. . , 175-275

Modulus of section required = — - —

= 25-04

The depth of the section is controlled by the allowable deflection,

which may be taken in this case mm»m»»»m »<»*

as jJs of the span. The depth 9'\ ,v\ ,'6'\ ,'e'\ /y| /v| /v'| / V) ,'«'\9'

from Table 26 would require to J I

be about ~ of the span, since the (« ,jV «•!

loading approximates very nearly FIG g5
to a distributed load.

.-. Min. depth = 13'5o^ 12 = 7 in.

Referring to a list of properties of beam sections, it is found that an
8" X 6" x 35 Ibs. section has a modulus of section 27*649, also a
10" X 5" x 30 Ibs. section has a modulus of 29-137. The 10" x 5"
beam would be stiffer than the 8" x 6" beam and there would be a
savins' in weight by adopting that section.

The dead load of the beam = 30 x 13-5 = 405 Ibs.

Total reactions would then = 85'5 + 1'7 = 87'2 cwts.

126

STRUCTURAL ENGINEERING

Allowing for this extra dead load the modulus of section required
would be increased to 25*51, but still remain below that of the beam.

Area of the web of beam = 0*36 x 8'8
= 3-168 sq. in.

Average shear on web = ^g ^ 2Q

= T37 tons per sq. in.

The web is therefore strong enough to resist the shear.
Primary Beams, I, 12' 0" long (Fig. 86).— The load at each bear-
ing of joists will be as in the previous case = 19 cwts.
Reactions = 76 cwts.

Max. B.M. (at centre) = 76 x 6 -19 (0-75 + 2'25 -f 3'75 + 5-25)
= 328 ft. -cwts.
= 136-8 in.-tons.

Modulus of section required = — = —

= 19-54

A beam 8" X 5" X 28-02 Ibs. section having a modulus equal to
22 '339 would be strong enough, but the difference in weight of an

-i-

19 19 19 19 19 19 19 I9c»fs.

\ f6'[ / 6* I /tf'l' /V'l/ 6*1 /'tf*|/V]p*
* t T «» y * ^ -i-

174-4

174-4 cwfe.

„-

FIG. 86.

FIG. 87.

8" x 5" and a 10" x 5" section is only 1*97 Ibs. per foot, and so for
uniformity the 10" x 5" x 29*99 Ibs. section would be adopted.
Weight of beam = 3*2 cwts.
Total reactions = 77*6 cwts.

Shear on web = 1*2 tons per sq. in.

Beams, d (Fig. 87). — Reaction from each primary beam a = 87'2
cwts.

Load at each bearing = 174*4 cwts.
Reactions = 261'6 cwts.

Max. B.M. (at centre) = 261-6 x 16 -174-4 x 8
= 2790-4 ft.-cwts.
= 1674-24 in.-tons.

Modulus of section required = — - — = 239-18

32 x 12
Min. depth = — ^— = 16*7 in., say 17 in.

No standard beam section has the required modulus. A broad
flanged beam 20" x 12" (nominal size) x 138 Ibs. having a modulus of
section of 272 might be adopted.

Weight of beam =1*97 tons.

BEAMS 127

Bending moment at centre due to weight of beam

_ 1-97 x 32 x 12
= _. 94.55 m.-tons.

Total B.M. at centre = 1674-24 + 94-56
= 1768*8 in.-tons.

Modulus of section required = - — ^ — =252'7

The 20" x 12" beam is therefore strong enough to resist the bending.
Area of web = 16'5 X 0'76 = 12'54 sq. ins.
Total shear = 14*065 tons.
Average shear on web = 11 tons per sq. in.
Seams, f (Fig. 88). — Reaction from each primary beam = 77*6 cwts.
Load at each connection = 155-2 ISS.Z ISS.Z I55.2 clfA
cwts. e \ s' \ a \
Reaction = 232-8 cwts. I . i ' i . . i
Max. B.M. (at centre) "1, 32'

= 232-8 X 16 -155-2 X 8 FlG 88

= 2483-2 ft.-cwts.

= 1489-92 in.-tons.

1489'92
Modulus of section required = =— = 212*86

Min. depth of beam = 17 in.

A 24" x 7i" X 100 Ibs. beam has a modulus = 221-231
Weight of beam = 1-43 tons.

Max. B.M. due to weight of beam = -

= 68-64 in.-tons.

Total max. B.M. = 1489-92 + 68' 64
= 1558*56 in.-tons.

Modulus of section required = -^— - — = 222-65

This is slightly in excess of the modulus of a 24" X 7J" beam, but as
the maximum stress would only be 991.031 =7*04 tons per square inch,

this section may be adopted. ,648 «** K4* £^3.

Total reactions = 12-355 tons. e i »' \ *' \ * -,

Area of web = 9-6 sq. ins. C^

Average shear on web = 1*28 tons ^ 32' -1

per sq. in. FlG. 89>

Beam, e (Fig. 89). —

Reaction from each primary beam I = 77*6 cwts.

a= 87-2 „

Load at each connection =164-8 „

128 STRUCTURAL ENGINEERING

Reactions = 247*2 cwts.

Max. B.M. (at centre) = 247'2 X 1C -164'8 X 8

= 2636-8 ft.-cwts.
= 1582-08 in.-tons.

Modulus of section required = — ^ — = 226*01
Min. depth = 17 in.

A broad flanged beam will again be suitable ; 19" x 12" x 128 Ibs.
has a modulus of 244.

Weight of beam = 1*83 tons.

1-83 x 32 x 12
Max. B.M. due to weight of beam = - — - —

= 87-84 in.-tons.
Total B.M. = 1582-08 + 87*84
= 1669*92 in.-tons.

1669*92
Modulus of section required = — ^ — = 238-56

use ^6 c«&. which is less than the modulus of the beam.
rV \ 76 \ 76 Total reactions = 13-275 tons.

, ,* =^L Area of web = 10-35 sq.-in.

--226 - Average shear on web = 1-3 tons.

FIG. 90. Seams, g (Fig. 90).—

Load on floor joists in offset = 19 X 7'5 = 17'8 cwts.

8
Total load on each primary beam = 17*8 X 8= 142-4 cwts.

12 x 30
Weight of each primary beam = — —5 — = 3-2 cwts.

11Z

142*4 -I- 3*2
Reaction from each primary beam = - '—^ — - = 72*8 cwts.

Load at each connection = 145*6 cwts.
Reactions of beam# = 145-6 cwts.
Max. B.M. due to loads (between loads) = 145*6 x 7*5

= 1091-6 ft.-cwts.
= 654-96 in.-tons.

Modulus of section required = — = — = 93*56

Say 18" x 7" X 75 Ibs. with modulus of 127*7.
Weight of beam = 0-75 ton.

n ,, , . , . . , 0*75 X 22-5 x 12

B.M. due to weight of beam = —

= 25-31 in.-tons.
Total B.M. = 25-31 + 654'96
= 680-27 in.-tons.

Modulus of section required = — — — = 97*18

Total reactions = 7*655 tons.

Area of web = 8-8 sq. in.
Average shear on web = 0'87 ton per sq. in.

BEAMS 129

Beam h (Fig. 91).—

Reaction from beam/ = 12'355 tons. 20-41 20-41 has.

Reaction from beam g = 7*655 „ '*' \_ '*' \ '*'

Total load at connection = 20*01 „

Reactions of beam h = 20-01 „
Max. B.M. due to loads (between loads) = 20'01 x 12

= 240-12 ft.-tons
= 2881-44 in.-tons.

Modulus of section required = 288*'44 = 411-64

36 x 1 2
Minimum depth = — ^ — = say 19 in.

A compound girder, composed of two 20" x 7£" rolled joists, with
one 16" x |" plate riveted on each flange, has a modulus of 460.
Weight per foot length of girder = 252 Ibs.
Total weight of beam = 4'05 tons.

B.M. due to weight of beam = 4>Q5 x 36 x 12

= 218'7 in.-tons.

Total max. B.M = 2881-44 -f 218'7
= 3100-14 in.-tons.

Modulus of section required = 3100'14 = 442-9

4*05
Total reactions = 20'0l 4- —^- = 22'035 tons.

Area of webs = 21-6 sq. in.
Average shear on webs = 1-02 tons per sq. in.

Pitch of rivets in flanges.

Number of rivets required per foot (see Example 13)

12 AYS
= -j— 4-R(R = 3 tons for f rivets)

_ 12 x (16 X §) X 10-ft" X 22-035

4888 X $
= 1-9 rivets £' diar.

As the pitch should preferably not exceed 6 inches, a 6 -inch pitch
will be adopted.

Connections. — To avoid having too great a depth of floor the beams
must be fastened together by web connections. A suitable arrange-
ment is shown in Figs. 92 and 93.

The theoretical requirements for resisting the shear only have been
exceeded to add lateral stability to the connections. For example, at
thjB connections of the beams / to the girder A, the theoretical number
of 3 in. rivets in double shear required through the web of beams/

130

STRUCTURAL ENGINEERING

_ vertical shear
resistance of one rivet

12-355 ,

= (say) 4 for shear.

3-3

For bearing =

12-355

8 X 0-6 X 0-75

= (say) 4

FIG. 92.

Number of rivets required through the web of girder h in single
shear

= for shear

= (say) 6

FIG. 93.

Bearing on Walls. — The length of bearing on walls should not be
less than the depth of the beam, and a minimum length of 8 inches
should be allowed for beams of less than 8 inches in depth. Suppose
stone templates be used under the ends of all beams resting on walls,
and the safe-bearing pressure on such templates be 1 5 tons per square
foot. Dividing the reactions of the beams by 15 will give the bearing

BEAMS 131

area required, and again dividing by the flange width the required
length of bearing is obtained. In each case of the present example,
the length of bearing required by such calculation will be less than the
depth of the beam, and therefore the bearing lengths will be made
equal to the depths of the beams.

Angle runners are riveted to the girder h for supporting the ends
of the timber joists.

CHAPTER V.
COLUMNS AND STRUTS.

STRUCTURAL members which are exposed to compressive stress in the
direction of their length are classed generally as columns or struts,
the term column, pillar, or stanchion being more usually applied to the
main uprights of framed buildings, whilst compression members of
girders and trusses are referred to as struts. The practical design of
compression members, especially those in which the length is great
compared with the cross-sectional dimensions, is relatively a more
difficult problem than is the case with the majority of structural
members, since theory does not furnish so reliable a guide and con-
siderable judgment and experience are very essential.

The mathematical theory regarding the strength of columns has
been ably and thoroughly developed by numerous investigators. It is
based, however, on various assumptions which are never realized in
practice, and the absence of one or more of these assumptions
materially affects the capability of resistance of the column. The
assumptions made in regard to the ideal or theoretical column are as
follow : —

1. Perfect straightness of the physical* axis.

2. The load is considered as a purely compressive stress acting along
the axis of the column, or, in other words, centrally applied.

3. Uniformity of cross-section.

4. Uniform modulus of elasticity of the material of which the
column is constructed, throughout the whole length of the column and
over every part of any cross-section.

It is sufficiently difficult sensibly to realize these conditions in a
carefully prepared and mounted test column, whilst it is certain that
all practical columns and struts fail to comply with at least one and
usually more than one of these conditions. Considerable discrepancy,
therefore, exists between the theoretical strength of a column as
deduced from mathematical considerations and the practical strength
obtained by methods of testing, or from the observation of columns
which have failed in situ. It is important, however, to bear in mind
the above conditions, since the degree in which they are realized in any
particular column furnishes a valuable aid to judgment in deciding to
what extent the mathematical theory may be relied on.

The material employed and method of manufacture largely influence
the degree in which a practical column will tend to realize the above
conditions. No material, however carefully manufactured, is perfectly

132

COLUMNS AND STRUTS 133

uniform in structure and elasticity, although a very high degree of
uniformity is realized by modern methods of manufacture in the case
of mild steel, and too much emphasis has frequently been laid upon
the variable nature of the material in accounting for the discrepancies
between theory and practice. In the case of wrought iron and mild
steel the effect of cold straightening is to locally strain the material
beyond the limit of elasticity, with the result that the stiffness of the
fibres overstrained in tension is considerably lowered as regards resist-
ance to compressive stress, and the fibres overstrained in compression
are affected similarly as regards their resistance to tensile stress. Per-
manent internal tensile and compressive stresses are thus set up in the
material, the effect of which is by no means insignificant. Conclusions
regarding the resistance of columns, deduced from experimental tests,
may further be materially affected by the previous history of the
material. The following instances of the influence of history of
material have been given by the late Sir B. Baker in the case of
experiments carried out on solid mild steel columns, thirty diameters
in length, showing that the resistance varied according to previous
treatment, as follows : —

Tons per sq. in.

Annealed 14'5

Previously stretched 10 per cent. . . 12' 6

„ compressed 8 „ . . 22'1

9 „ . . 28-9

Straightened cold 11'8

The general adoption of machine riveting in the case of built-up
members is another frequent cause of local initial stress, as well as of
initial curvature. The effect of riveting up an assemblage of plates
and bar sections is to cause the various bars to stretch and creep past
each other in different degrees. This is most marked where light and
heavy sections are adjacent to each other, the lighter sections being
more severely stretched during riveting than the heavy. In symmetrical
sections, the camber caused by riveting down one side of a long member
will be sensibly neutralized by riveting along the opposite side, so that
the finished member may be apparently straight, but the ultimate effect
will be the creation of initial local stress in the material. In the case
of unsymmetrical sections, permanent curvature or waviness in the
direction of length with unavoidable occasional twisting results ^ from
the process of riveting, and these effects are difficult to minimize,
however carefully the work may be executed. Columns of cast iron or
cast steel are not subject to defects caused by riveting, but are influenced
by the usual hidden defects inherent to all castings, as well as by initial
stresses set up by unequal contraction. Hollow cast columns are espe-
cially liable to irregularity of cross-section due to the core getting
slightly out of centre during casting, which defect will be more marked
if the column be cast in a horizontal or inclined position.

In Fig. 94, let AB represent a hollow circular column having
irregular horizontal cross-sections as indicated. The geometrical axis
is the straight line AB, which, in the absence of defects of material
and irregularity of cross-section, would be assumed to be the true or

134

STRUCTURAL ENGINEERING

physical axis of the column. In such a case the line of application AB
of the load would coincide everywhere with the physical axis of the
column, and the resulting stress on every cross-
section would be purely compressive, with no
tendency to bend the column. Further, if the
physical axis of a long column were a perfectly
straight line, it would be possible to apply a
steadily increasing central load until the column
failed by direct crushing of the material. In
Fig. 94 the centres of gravity of the irregular
cross-sections occur at the points a, #, c, etc., and
these being transferred to their corresponding
positions #', #', c', on the elevation, the physical
axis of the column becomes the curved line
A a' b' c' . . . B. This alteration in outline of the
physical axis is due only to the considered defects
of cross-section, but it will be borne in mind that
non-uniformity of elasticity of the material and
initial camber may still further modify the posi-
tions of points a', b'9 c', etc. The above defects
are, in any practical column, quite unknown
quantities, so that it is impossible to state with
accuracy to what extent the physical axis does or
does not coincide with the geometrical axis.
The resulting effect of this deviation of the
physical axis from the geometrical axis is that at
cross-section No. 1 there is acting a direct com-
pression = P, and also a bending moment = P x a'\. At cross-section
No. 2, a direct compression = P and a B.M. = P X b'2, and so on. It
is the presence of this bending moment which determines the tendency
of the practical column to yield towards one side or the other, depending
on the outline of the physical axis. The points «, b, c, etc., may be
termed the centres of resistance of the various sections, being understood
to represent the points through which the resultant compression should
act in order to create uniform intensity of compression over the whole
cross-section, after making allowance for defects of form of cross-section
and variable elasticity of material. It follows that simple compression
on every cross-section of a column might only be ensured if the line of
action of the load coincided with the physical axis. Since, however,
the line of action of the load is a straight line and the physical axis
a curved or wavy line, it is practically impossible for any column not
to be subject to more or less bending moment at several sections
throughout its length.

Method of Application of Load. — In practice, relatively few
columns have the load centrally applied. In Fig. 95, A, a girder
carrying similar loads over two adjacent equal spans will impose a
resultant central load on the column. At B, two unequally loaded
girders connected to opposite sides of the same column will impose a
resultant load on the column, the line of action of which may be con-
siderably out of coincidence with the axis of the column. At C, the
load on a girder attached to one face of the column will impose a still

FIG. 94.

COLUMNS AND STRUTS

135

more eccentric load. This may be modified, as at D, by employing
two girders, G, G, instead of one, and carrying them on brackets
placed centrally on opposite faces of the column. This arrangement,
although more satisfactory theoretically, is often inconvenient in

FIG. 95.

practice, since it multiplies and complicates the connections, and inter-
feres with the arrangement of other members meeting on the same
column. In the case of compound columns built up of two or three
girder sections with tie-plates or lattice bracing as at E, the line of
application of the load becomes more difficult to define. Such columns
generally carry vertical loads W1? W2, and W3, due respectively to roof
weight and the loads handled by travelling cranes, whilst they are
further subject to bending moment caused by the horizontal wind
pressure P acting on the roof slope. The manner in which the
resultant load is shared by the three columns at any horizontal sec-
tion ss, will depend largely on the strength and rigidity of the bracing.
A further small amount of additional bending moment is caused by
the deflection of the column itself. In Fig. 96, the straight line AB
represents the original axis of the
column before the imposition of the
load. If a load P be applied at a
small eccentricity e, the resulting B.M.
will be P x 0, which will cause a
small deflection of the column indi-
cated by d. The ultimate B.M. at
the central section of the column,
after it has reached a state of equi-
librium, will then be P x (e + d). In
most practical cases, the deflection
being very small, the additional
moment P X d, due to that deflection, FIG. 96.
may be neglected. The jibs of cranes

and long horizontal or inclined struts in large bridge girders are
subject to additional deflection caused by their own weight, which still
further increases the B.M. upon them. Thus in Fig. 97 the straight
line AB represents the axis of a crane jib when in an unstrained
condition. An appreciable amount of sag will be caused by the dead

AJ"

FIG. 97.

136

STRUCTURAL ENGINEERING

weight of the jib, due to its inclined position, which will cause it to
assume some curved outline, as shown by the full line curve. When
lifting a weight W, the pressure on the jib due to the tensions W, W,
in the chain will augment the deflection, so that the axis of the jib
takes up some new position indicated by the dotted curve. A similar
action takes place in all horizontal and inclined structural members
subject to end thrust.

From the above remarks it will be apparent that the manner in
which the compressive load affects a column or strut is more complex
than is usually the case with tension members, and cannot be dismissed
by the assumption of a simple compressive stress acting at every
cross-section.

Methods of supporting or fixing the Ends of Columns.— Whilst
the above remarks apply to any column irrespective of the way in
which its ends are supported, the manner of support or attachment of
the ends of a column to adjacent members of a structure, greatly
influences the load that the column will safely carry. Four well-
defined methods of end support are easily recognized. These are
indicated diagrammatically in Fig. 98.

The column at A is said to be hinged or pin-ended, and under the

load deflects in a single curve. B illustrates a fixed-ended column which

under the load is constrained to deflect in a

A B c D treble curve with points of contra-flexure at

P and Q. In column C the lower end is
fixed and the upper end hinged or rounded,
the deflection causing a double curvature
with one point of contra-flexure at P. In
cases A, B, and C the upper end of the
column is supposed to be situated vertically
over the lower end, and to be incapable of
lateral movement, so that on deflection the
upper end becomes slightly depressed along
\i the vertical line. Fig. 98, D, represents a
column fixed at the lower end, but free to
FIG. 98. move laterally at the upper end when de-

flection under the load takes place. Such a

column will obviously bend in a single curve, and its behaviour will be
sensibly similar to one-half of the round-ended column at A, as may
be indicated by drawing in a similarly deflected lower half shown by
the dotted line.

In practice, so-called round-ended columns are constructed by
forming the ends to fit on round bearing pins or in hollow curved
sockets. Examples of these occur in crane jibs and in the struts of
pin-connected girders — a type almost universally adopted in American
practice. With regard to " pin-ended " columns, Mr. J. M. Moncrieff,
M.Inst.C.E., remarks, " It is quite useless to theorize with the view
of showing their superiority to round or pivot ends, owing to the fact
that their behaviour under load, even in a testing machine, depends
very largely on the closeness of the fit between pin and hole, upon the
smoothness or otherwise of the bearing surfaces, upon the diameter of
the pin in relation to the radius of gyration (of the column section),

COLUMNS AND STRUTS

137

and upon the presence, either accidental or premeditated, of a lubri-
cating medium." 1

In actual practice, a truly fixed-ended column seldom, if ever,
exists. Fixity of ends implies that the ends are so firmly held that, on
bending, the portions EP and FQ of the column in Fig. 98, B, remain
strictly tangent to the straight line EF. This is only possible where
the head and foot of the column are so rigidly held, or attached to
adjoining portions of a structure, as to be absolutely immovable laterally
— a condition difficult to realize in experiments with a testing machine,
and still more difficult of realization in practical structures. The
nearest approach to a fixed-ended column in practice is probably
exemplified in the case of the lowermost portion of a heavy column in
a framed building. The foot is secured to a heavy foundation block
of concrete sunk a considerable distance into the earth, and the head
secured to relatively heavy and rigid girders supporting the first floor
of the building. Even this, however, is not a truly fixed-ended
column, since the girders, no matter how rigid they may be, must
deflect to some extent under their load, and so permit of slight move-
ment of the head of the column, whilst the whole building is subject
to lateral movement due to wind pressure. The appearance of a
column, either on a working drawing or in situ frequently gives a very
false impression of its fixity. It is a commonly claimed advantage for
riveted connections in structural work that the compression members
are constrained to act as fixed-ended columns. This is in many cases
a quite erroneous assumption, since the degree of approximation to fixity
of ends depends entirely upon the relative stiffness of the column and
the other members of the structure attached to it, and the estimation
of this degree of fixity demands very careful consideration on the part
of the designer. The following: two cases cited by Mr. Moncrieff are
instructive and suggestive. In Fig. 99, assume a series of stiff gantry

U XU

FIG. 99.

vsw xvv^XVNNXVXVV

FIG. 100.

girders 2 ft. deep by 10 ft. span, riveted securely to the heads of
columns 30 ft. high, firmly braced together to preserve their vertically.
Assume also that the foundation blocks on which the columns rest are
very rigid, that the columns have large well-bolted bases, and that the
ratio of length to radius of gyration of these columns is very large,
and the columns, therefore, slender in proportion. Then the impo-
sition of load on any span will cause deflection in the girder, and the
ends of the girder will deviate from the vertical to a slight degree, but
the relative stiffness of the girders themselves, as compared with the

1 Transactions Am. Soc. C.E., vol. xlv. p. 358.

138

STRUCTURAL ENGINEERING

column, being high, the approximation to ideal fixity of ends would,
practically speaking, be of a high degree.

In Fig. 100, let the columns be spaced at 30 ft. oentres, retaining
the same depth of girder, 2 ft., and merely increasing the girder
sections to obtain the same value of working unit stress, while
increasing the radius of gyration of the columns to provide much
greater stiffness of column. Under these conditions the deflection of
the girder under load, and consequently the slope of the ends of the
girders where they are securely riveted to the column heads, would be
increased largely, and the columns would be subjected to heavy bending
stresses in addition to their direct load. These columns would be
much less heavily stressed if they had pin-joint connections to the
girders, and the apparent fixity of end given by a secure riveted
connection would actually be accompanied by severely prejudicial
secondary stresses.

A distinction requires to be drawn between " fixed-ended " and
" flat-ended " columns. The assumption has generally been made that

these two types of columns act in an
identical manner, and formulae giving the
permissible loads for both in one expres-
sion are frequently quoted. This is quite
erroneous, both theoretically and from
the evidence of practical tests. In the
case of a column with flat ends, that is,
in which the ends are merely kept in
contact with their bearing surfaces by
pressure, no tensile stress can be developed
at the ends, whilst with fixed ends a con-
siderable amount of tension may be
safely resisted. The two cases are illus-
trated in Fig. 101.

Fig. 101, A, represents a flat-ended
column in a deflected condition, such
that compressive stresses are set up at the
three sections, 1-1, 2-2, 3-3, of inten-
sities shown diagrammatically by the
shaded areas. So long as the stress on
the two end sections is entirely compres-
sive, the ends of the column will remain
in close contact with the bearing surfaces
applying the load, and the column will
behave in exactly the same manner as a
fixed-ended column of similar dimensions.

In such a case any bolts or other fastenings employed with a view to
fixing the ends of the column will be quite inoperative. Fig. 101, B,
represents a fixed-ended column such that compressive stresses of intensity
ab are set up on the left-hand side of sections 1-1 and 3-3, and the right-
hand side of section 2-2, and tensile stresses of intensity cd on the right-
hand side of sections 1-1 and 3-3, and on the left-hand side of section
2-2. In this case the tension cd at section 2-2 will be resisted by the
material of the column, whilst the tensile stresses cd at sections 1-1

FIG. 101.

COLUMNS AND STRUTS

139

and 3-3 will have to be resisted by the bolts or rivets which connect
the column with neighbouring parts of the structure. If the con-
necting bolts on the right-hand side of sections 1-1 and 3-3 of
column B were cut through so as to transform the column into a flat-
ended column whilst under load, the deflection would immediately
increase and the column alter its curvature, as indicated by the dotted
lines. Any increase of the load in column A would attempt to set up a
state of stress similar to that in column B, but column A being flat-ended,
and therefore incapable of resisting tension at its ends, would deflect
or spring further towards the left hand in a similar manner to the
supposed case of column B with its end connections severed. A
flat-ended column, although apparently as strong as a fixed-ended
column, may actually be on the verge of failure by excessive deflec-
tion, if the load be of such a nature as to set up incipient tension
along one edge of the bearing surfaces. This point is clearly
evidenced by the results of tests of flat-ended columns made by Mr.
Christie.

In actual structures it is not customary to employ purely flat-ended
columns, bolts or rivets being invariably inserted to make connections
with the foundation and upper members of the structure. These,
however, are often employed more with a view to convenience in erec-
tion and to prevent lateral movement of the column, than to specifically
resist tensile stresses which, under certain conditions of loading, may
become very severe. It is important, therefore, in designing columns
on the assumption of fixed ends, to ensure the bolted or riveted con-
nections being sufficiently strong or numerous to resist the above-
mentioned tensile stress. Fig. 98, D, represents a type of column
which most commonly occurs in connection with roof designs.

Fig. 102, A, illustrates the case of a detached roof carried by
columns fixed to a substantial foundation at F, F. The columns carry
a vertical load, W, due
to the weight of the
roof, whilst their upper -
ends are subject to ap-
preciable horizontal
movement due to the
wind pressure P. Fig.
102, B, is another ex-
ample of this type of
column, supporting an
" umbrella " or island
platform roof. In both
cases the columns deflect in a single curve.

From the preceding remarks it will be apparent that no column
in actual practice is ever subject to a uniform compressive stress per
square inch, since the bending action in combination with the direct
loading results in increasing the intensity of compression on the con-
cave or hollow side of the column and in decreasing the intensity of
compression on the convex side. Further, in cases where the bending
moment is large compared with the direct compression, the stress on
the convex side may be tensile instead of compressive. The maximum

FIG. 102.

140

STRUCTURAL ENGINEERING

stress per square inch on the section of a column under given condi-
tions of loading may only be arrived at by a calculation of the bending
moment as well as the direct compression per square inch. Most of
the column formulae in general use aim at giving the safe uniform com-
pression per square inch for columns of given dimensions and material.
Whilst this is the most convenient form in which to use such formula,
it should be remembered that the maximum stress on the material of
the column is usually considerably in excess of the uniform or average
stress exhibited by the formulae. Consideration will now be given to
some of the many formulas in common use.

Radius of Gyration. — Before proceeding to these, it is necessary
to comprehend what is implied by the term Radius of Gyration of a
column section. If, for any column section, the moment of inertia
about any axis be divided by the sectional area, the square root of the
resulting quotient gives the radius of gyration about that axis. Or, if

and

I = moment of inertia
r = radius of gyration

A = sectional area

As an example consider the two sections shown in Fig. 103, A and
B. A is a girder section of 30 sq. in. area. Section B is a box-section
having the same over-all dimensions and also the same sectional area.

For section A, moment of inertia about axis Y — Y = 167*5 in. units.

Sectional area = 30 sq. in. .*. r =

= 2*36 in.

For section B, moment of inertia about axis Y — Y=272'5 in. units.

Sectional area = 30 sq. in.

= 3'01 in.

h- — (S i

••i

Whilst possessing the same sectional area, section B has a decidedly
larger radius of gyration than section A. The increase in the radius

of gyration of section B is evi-
dently caused by the altered dis-
tribution of the web section, since
the flange section is the same for
both A and B. The radius of
gyration is thus seen to be de-
pendent on the shape of section or
distribution of material about the
axis in question. It will be
obvious merely from an inspec-
tion of the two sections that B
would form a stiffer column than
A, provided equal lengths were taken, and it may be stated generally
that the radius of gyration of a section affords a relative measure of
its stiffness to resist a bending or " buckling " action such as occurs
in columns. In order to form an estimate of the actual stiffness of
a column , the length as well as the radius of gyration of the cross-

I A i

*--d

>--*

\ ,2"--

-X

•f1"

r".

\ V (

FIG. 103.

COLUMNS AND STRUTS 141

section must be taken into account. Thus, if a column of section
B be twice the length of a column of section A, the former will be
less capable of supporting a given load than the latter, notwith-
standing its greater radius of gyration. In other words, the greater
length of column B rendering it more slender than column A, will
more than neutralize the advantage it possesses by reason of its
greater radius of gyration. If the length of a column be divided by

the radius of gyration of its cross-section, the resulting ratio - may be

regarded as a measure of its slenderness. Thus, if a column of section A
be 15 ft. long, and one of section B be 30 ft. long, then

for column A, 1 = H * 12" = 76

and for column B, 1 = ^WvT~ = 119>

or column B is considerably more slender than A, and consequently
less capable of carrying so great a load. It will be seen presently that

the ratio - constitutes an important term in all formulae which aim at

giving the safe or breaking loads for columns.

In the above example the radii of gyration were calculated about
the axis Y-Y. They may also be calculated about the axis X-X, or, if
desired, about any other axis passing through the section. Considering
the axis X-X of section A, the distribution of material is plainly
different from that with regard to Y-Y. The moments of inertia
about X-X and Y-Y will therefore have different values, and since the
sectional area remains the same whatever axis be considered, the radius
of gyration will necessarily vary with the moment of inertia. Thus

Moment of inertia of section A about X — X = 690 in. units.
Sectional area = 30 sq. in. /. r about X-X = \/ -^ = 4'8 in.

The previously calculated radius of gyration about Y-Y was 2-36",
and the significance of the two figures is that a column of section A is a
little more than four times as stiff to resist bending about the axis X-X
(or in the plane Y-Y), than about the axis Y-Y (or in the plane X-X).
If a column is equally free to spring or buckle towards one side or
another, it will naturally yield in that direction in which it is least stiff,
or, in other words, it will bend in the plane at right angles to the axis
about which its radius of gyration is least. Thus, columns A and B
would both more readily bend in the plane XX than in the plane YY,
since both have a smaller radius of gyration about the axis Y-Y than
about X-X. The least value of the radius of gyration for any given
section is commonly referred to as the Least Radius simply. In many
column sections the axis about which the radius of gyration is least is
easily recognized at sight. Such sections as a circle, hollow circle,
square, etc., have the same radius of gyration about any axis passing
through the centre. In some built-up sections, notably box sections,

142 STRUCTURAL ENGINEERING

as in Types 8 and 9, the radius of gyration may differ appreciably about
the axes X-X and Y-Y, and it is advisable to calculate both rather than
hastily assume the axis of least radius from inspection only. It will be
obvious that the most economical forms of column sections, so far as
load-bearing capacity is concerned, will be those having equal, or prac-
tically equal, radii of gyration abaut both principal axes. A built-up
section may generally be arranged to give this result, although practical
considerations, with regard to convenience of connections, sometimes
preclude the employment of such sections. The radii of gyration of
the elementary rolled sections will be found given in the list of Proper-
ties of British Standard Sections, as well as in most firms' section books.
It should be noticed that equal angle sections have the least radius
about an axis X-X (Fig. 104) passing through the centre of gravity of
the cross-section, and that they will consequently bend most readily
cornerwise, or in the plane Y-Y.

The radius of gyration of compound or built-up sections is readily
calculated from the given properties of the elementary sections of which
they are composed.

EXAMPLE 19. — To calculate the radii of gyration of the section in
Fig. 105, about the axes X-X and Y-Y.

1. About X-X. Obtain from the section book the following

FIG. 104. FIG. 105.

properties for a 10" X 6" x 42 Ibs. joist. Sectional area = 12'4 sq.
in. Moment of inertia about X-X = 212. Then
Mt. of inertia of two joists about X-X = 212 x 2 = 424

two plates „ „ = -jL X 14 (II3 - 103) = 386

Total I of section about X-X = 810
Total sectional area = 2xl4xJ + 2x 12-4 = 38'8 sq.-in.

/810

and r about X - X = A/ ^r8 = 4'57 m.

2. About Y-Y. From section book,

Mt. of inertia of one joist about y-y = 22'9

„ one „ ., Y-Y = 22'9 + 12'4 x (3J)2 = 174-8

two joists „ „ = 174-8 X 2 = 849'6
„ two plates „ „ = -^ X 1 X 14s = 228-G

Total I about Y-Y = 578^
and r about Y-Y = X/^rrr = 3'86 in.

COLUMNS AND STRUTS

143

The least radius is therefore about Y-Y and = 3*86 in. If desired
the 7 in. spacing between the joists might be increased in order to make
the radus of gyration about Y-Y equal to that about X-X. The
necessary spacing to effect this will be readily found after one or two
trials.

The following tables give the sectional area and least radius of
gyration for the most useful types of built-up columns, the figures
applying in every case to British Standard Sections.

Type 1 . — One joist with two or four plates.

Two plates.

Four plates.

Size of joist.

Size of plates.

Sectional

?• about

Sectional

r about

area.

YY.

area.

YY.

in.
9 X 7 X 58 Ibs.

in.
10 X J

27-06

2-188

37-06

2-397

10 X 8 X 70 ,

12 X i

32-60

2-571

44-60

2-839

12 X 6 X 54 ,

12 X J

27-88

2-490

39-88

2-810

14 X 6 X 57 ,

10 x £

26-76

2-039

36-76

2-309

16 X 6 X 62 ,

10 x |

30-73

2-066

43-23

2-333

18 X 7 X 75 ,

12 x f

37-06

2-473

52-06

2-794

Type 2, — Two joists with two or four plates.

Two plates.

Four plates.

Joist.

Plates.

Area.

r about
YY.

Area.

r about
YY.

in.

10 X 6 X 42 Ibs.

in.

in.
14 X *

38-70

3-86

52-70

3-91

12 X 6 X 54 ,

14 XJ

45-76

3-83

59-76

3-88

14 X 6 X 57 ,

16 X i

49-52

4-49

65-52

4-52

16 X 6 X 62 ,

16 X |

56-46

4-49

76-46

4-52

18 X 7 X 75 ,

18 X |

66-62

5-04

89-12

5-08

20 X 7* X 89 ,

9£

18 x f

74-84

5-05

97-34

5-08

Type 3. — Two joists with tie-plates or lattice bracing.

Joist.

Area.

r about YY.

in.
10 X 6 X 42 Ibs.

in.
1%

24-71

3-99

12 X 6 X 54 ,

31-76

3-98

14 X 6 X 57 ,

33-54

3-96

16 X 6 X 62 ,

36-45

3-94

18 X 7 X 74 ,

8£

44-13

4-49

20 X 7J X 89 ,

52-33

4-76

144 STRUCTURAL ENGINEERING

Type 4. — Three joists.

Two joists.

One joist.

Area.

r about YY.

in.

14X6 X 57 Ibs.
15X6 X 59 ,
16x6 X 62 ,
18x7 X 75 ,
20X7JX 89 ,
24X7^X100 ,

in.
12 X 6 X 54 Ibs.
12x6x54 „
12x6x54 „
14x6x57 „
14x6x57 „
16x6x62 „

49-42
50-57
52-33
60-90
69-10
77-01

4-70
5-04
5-32
6-18
6-99
7-99

Type 5.— Three joists.

Two joists.

One joist.

Area.

r about YY.

in.
8X6X35 Ibs.
8x6x35 ,,
9X7X58 „
9x7x58 ,,
10x8x70 „

in.

14x6 x 57 Ibs.
15X6 X 59 „
16x6 x 62 „
18x7 X 75 „
20X7JX 89 „

37-35
37-94
52-35
56-18
67-33

3-90
4-05
3-96
4-70
5-19

12x6x54 „

24x7^x100 „

61-15

5-83 . X-X

Type 6. — One joist with two channels.

Channels.

Joist.

Area.

r about YY.

in.
7X31
8X3£
9x3^x25-39 Ibs.
10x3^x28-21 „
12 X3|x 32-88 „
15x4

in.
6x5
6X5
6x5
8X6
8x6
8x6

19-25
20-72
22-29
26-88
29-63
34-95

2-25

2-57
2-88
3-08
3-67
4-60

Type 7. — Two channels with lattice bracing.

Channels.

I).

Area.

r, XX.

r, YY.

in.

7X3

in.

10-33

2-7

8x3£

13-36

2-95

9 X 3 J x 25-39 Ibs.

14-94

—

3-37

10X3JX28-21 „

4£

16-59

—

3-33

12x3^x32-88 „

6£

19-34

—

4-22

15x4

24-66

5-52

—

COLUMNS AND STRUTS

145

Type 8. — Two channels with two or four plates. Channel box-
section.

Two plates.

Four plates.

r*v* i

Area.

r, YY.

Area.

r, YY.

in.

7 X 3 X 17-66 Ibs.

10 X £

20"33

2-82

30-33

2-84

8X3JX22-72

12 X £

25-36

3-44

37-36

3-45

9 x 3.V X 25-39

12 X &

26-94

3-41

38-94

3-43

10x3^x28-21

12 X £

28-59

3-39

40-59

3-41

11 X 3i X 29-82

14 X }

31-54

4-16

45-54

4-12

12 X3ix 32-88

14 X |

6£

36-84

4-14

54-34

4-10

15x4 X 41-94

18 x |

47-17

5-51

69-67

5-41

Type 9.— Built-up box section.

Angles. Web plates.

Flange.

Area.

r, YY.

84X8JXJ

in.
24XJ

in.

14 XJ

51-00

3-71

3i X 3^ X J

24xi

16 X £

53-00

4-57

4 X4 x*

27 X J

18x|

60-00

5-13

4 X4 Xi

27 X J

18 x|

64-50

5-14

4;j x 4:7 X J

30 X ^

21 x*

68-00

5-97

4ix4jxi

30xi

24 x|

83-00

7-20

Type 10. — Four angles back to back with lacing.

Angles.

Area.

r, YY.

in.
3ix3 xj

12-00

1-67

4 X3 X|

13-00

1-99

5 X3 X*

15-00

2-53

5ix3ix£

17-00

2-72

6 X4 xj

19-00

2-91

6^ x 4* X 0-55

23-00

3-12

Type 11.-
side.

—Lattice-box section. Four an

Angles.

Area.

r, YY.

en *> tf». co co to

liH LCH I*(P-

XXXXXX

en *- Hi. co co to 5'

KH U|M U»- •

XXXXXX

tcHK|i-'K|Mtw|i-«u«u

in.

8
10
12
14
16
18

6-93
8-44
13-00
15-00
17-00
19-00

3-32
4-21
5-05
5-95
6-83
7-72

Four angles with lacing on each

146 STRUCTURAL ENGINEERING

Type 12.—" Gray " column. Eight angles with tie- plates.

Angles.

Area.

r, YY.

in.
3 X2JX|
3JX2JX*
4 X3 X£
4 X4 X£
41 X 41 X £
5 X5 XJ

in.
12
14
15
16
18
20

20-00
22-00
26-00
30-00
34-00
38-00

3-66
4-23
4-54
5-04
5-68
6-33

Type 13. — Broad-flanged beams used alone or with one plate on
each flange.

With two plates.

ArpA.

•>• w

olZG Oi D6&IH.

Area,

7 y IX.

Plates.

Area.

r, YY.

in.
8X 8X 371bs.

10-9

1-86

in.

9XJ

19-9

2-22

10X10X 55

16-3

2-30

12 Xi

28-3

2-85

12X12X 80

23-6

2-76

14X|

41-1

3-37

14X12X 96

28-1

2-74

16 Xf

48-1

3-64

15 X 12 X 101

29-6

2-73

18x1

52-1

3-98

16x12x107

31-6

2-72

18X|

54-1

3-94

18 X 12 X 121

35-6

2-68

18 xf

62-6

3-96

20x12x138

40-6

2-63

18 Xf

67-6

3-87

Type 14. — Solid circular section of diameter D.

Type 15. — Hollow circular section.

External diameter = D.
Internal „ = d.

r =

Type 16. — Solid rectangular or square section.
r = 0-289D.

Euler's Formula. — Assuming the conditions for an ideal round -
ended column as enumerated at the commencement of this chapter,
the following formula may be established by mathematical reasoning.

If P = ultimate or crippling load in tons ; E = Modulus of
elasticity of the material in tons per square inch ; / = length of
column in inches between centres of end bearings ; and I = moment
of inertia of cross-section in inch units,

COLUMNS AND STRUTS 147

9-87EI

Substituting Ar2 for I, where A = sectional area of column in square
inches and r = least radius of gyration in inches,

9-87EA

The ratio -, previously noted, occurs in the denominator, and

consequently the ultimate load P becomes smaller as -, or the slender-
ness of the column, increases. For any given material the value of E
is sensibly constant, and the formula may be written

P = constant X

This formula forms the basis of most of the "practical" formulae
intended to give the ultimate load on columns. Being based on the
assumption of an ideal column, it is not of much practical value, the
ultimate load as given by it representing the extreme outside limit of
load which a theoretically perfect column might withstand. By divid-
ing the ultimate load P by any desired factor of safety, as 3, 4, 5, etc.,
the corresponding safe load would be obtained. It is, further, more
convenient to express the safe load for a column in tons or pounds per
square inch of sectional area, so that including the factor of safety
and dividing P by the sectional area A, Euler's formula becomes
P 2F
^rr - P ~ 7T?>, where p — safe load in tons per square inch and

FxQ

F = factor of safety. Taking E as 13,400 tons per square inch for mild
steel and a factor of safety F = 4, the formula reduces to

= 33,064 -Hi
\r

It will be noticed that for low values of -, the formula would give

abnormally high values for p. Thus for a column for which - = say

40, p would = 20' G tons per square inch, which is of course quite outside
tt& practical range of load, since the elastic limit of the material would
be exceeded. The formula may therefore only be used practically up to

that value of - for which the safe working load p does not exceed the

safe crushing resistance of the material. If 6J tons per square inch be
fixed as the highest permissible value for p for mild steel columns, the

148

STRUCTURAL ENGINEERING

corresponding value of - is 70. The formula ceases then to have any
practical significance for columns in which the ratio of length to least
radius of gyration is less than 70. If increasing values of - beyond 70
be inserted and the corresponding values of p be worked out, the results
may be conveniently shown as in Fig. 106, by plotting values of -

0 20 40 6O BO 100 I2O I4O I6O ISO 20O 22O 24O 26O
RATIO £

FIG. 106.

horizontally and the corresponding values of p vertically. The result-
ing curve is marked No. 1.

Rankine's Formula may be taken as typical of a large class of
" practical " formulae, many of which are based on the results of actual
tests of columns. It may be expressed as

constant x ( -

where p = safe load in tons per square inch,/ = crushing resistance in
tons per square inch divided by the factor of safety, I and r being as
before. The constant in the denominator is variously modified accord-
ing to the conditions of end support. In this type of formula the safe
compression / tons per square inch which may be put on a very short
column is gradually reduced to p tons per square inch for longer

columns as the ratio - increases. Assuming 27 tons per square inch as

the ultimate crushing resistance of very short columns of mild steel, and
adopting a factor of safety of 4 as before, the formula becomes

p =

COLUMNS AND STRUTS 149

being the value of the constant for round-ended columns of mild
steel. The resulting values of p obtained from this formula are shown
by curve No. 2 in Fig. 106. The formulae of Gordon, Claudel, Ritter,
and Christie belong to this class.

Straight-line Formulae. — These are a class of formulae which have
been devised to give approximate results to those above mentioned.

They include the first power instead of the square of the term -. As
an example of these, the following may be taken

p = 6f(l - 0-00475-)

where p = safe load in tons per square inch for round-ended columns,
the factor of safety being 4 on the assumed ultimate strength of 27
tons for mild steel. The results of this formula are shown in Fig. 106,
by the straight line No 3. The object of the straight-line formula is to
obtain greater simplicity than is afforded by the various " curved "
formulse by avoiding the quadratic solutions which they necessitate.
It should be noticed, however, that the safe loads for columns of varying

- cannot be correctly given by such formulse. For the one in question,
the load becomes zero for a column having the ratio - = 210, which is

obviously incorrect. Fig. 106 shows that for ratios of - between 100

and 180, which include a fairly large proportion of practical columns,
the safe loads per square inch, whether calculated by Euler's, Rankine's, •
or the straight-line formula, do not greatly differ. In selecting a
suitable column section, the length, total load and character of end
supports are known beforehand. The type of cross-section desirable
will depend mainly upon total load and connections to be made with
the column. The method of selecting a suitable section of column by
the aid of the above formulse will then be as follows.

EXAMPLE 20. — Required a suitable box section in mild steel, formed of
two channels and two plates, to carry a, central load of 100 tons, the length
being 25ft. and the ends considered rounded.

1. Using Euler's Formula. — Try No. 2 section of Type 8.

r = 3-44 . 1 = 25 X 12 = 800" J = —^ = 87

A = 25-36 sq, in.

I
From curve No. 1, the safe load per square inch for - = 87 is 4' 3

tons. Hence total safe load = 25'36 x 4-3 = 109 tons. No. 1 section
will be found too small, and No. 2 would be adopted.

2. Using Rankine's Formula.—The safe load per square inch for -

= 87 from No. 2 curve is 3'4 tons, and total safe load = 25'36 X 3'4
= 86-2 tons. This section is therefore too small. Try No. 4 section of

150

STRUCTURAL ENGINEERING

Type 8. r = 3*39.

.*. - = = 88, and safe load per square inch is

E ,A

again practically 3*4 tons. A = 28*59 square inches, and total safe load
= 28-59 x 3*4 = 97*2 tons. As section No. 5 would be considerably
too large, No. 4 would be adopted. Note that this section has about 12
per cent, greater area than the one deduced by Euler's formula.

3. Using the Straight-line Formula. — Try section No. 2, Type 8.

- = 87 as before, and safe load per square inch from No. 3 line, Fig. 106,

= 3-9 tons. Hence total safe load = 25'36 x 3'9 = 98'9 tons. That
is, the same section as indicated by Euler's formula would be adopted
but with an apparently less margin of safety.

The above-mentioned formulae are open to the following objections.
Euler's formula is based on the assumption of a theoretically perfect
column, a case never realized in practice. The Rankine
type of formulas are based generally on the results of
limited numbers of practical tests. Any formula to be of
general practical value should show agreement, not only
with a very large number of reliable tests of varying
sections, but also tests covering a large range of ratios of
I to r, in addition to several tests at each ratio of I to r.
Few, if any, of the formulae in general use satisfactorily
approach these requirements. The straight-line formulae
can only be regarded as rough approximations, especially
when employed over a wide range of ratios of / to r.

Fixed-ended Columns. — In Fig. 107, if ACDB represent
the axis of a perfectly fixed-ended column of uniform
section in a bent or deflected state, the portions AC and
BD, after bending, remain tangent to the vertical line
AB. The central portion CD is also tangent to a vertical
line at its middle point M. C and D are points of contra-
flexure, and at these points, therefore, no bending moment
exists. In order to constrain the column to conform to
these conditions there must be the same bending moment
existing at A, M, and B. The points C and D must
therefore be situated midway along AM and BM re-
spectively, or, in other words, the central portion CD
which bends in a single curve will be equal to half the
FiG]~107. total length AB of the fixed-ended column, and will
behave similarly to a round-ended column. Further,
the deflections AE, BF, and MN will all be equal. Stated generally, a
fixed-ended column will carry the same load and be subject to the
same bending moment, and consequently the same stress, as a round-
ended column of half the length and the same sectional area. The
total central deflection MP, of the fixed-ended column AB, will, however,
be twice the deflection MN of the equivalent round-ended column CD,
under the same load.

The above statement must not be confused with the idea that a fixed-
ended column will carry twice the load of a round-ended column of equal
length, which is, of course, quite erroneous.

Hence, in selecting a section for a fixed-ended column of given

COLUMNS AND STRUTS 151

length, that section is 'employed which would be suitable for a round-
ended column of half the length, after making reasonable allowance for
the extent to which the presumably fixed-ended column fails to realize
the conditions of absolute fixity.

Moncrieff's Formula. — The most complete and reliable investigation
of the strength of columns is contained in a paper by Mr. J. M. Mon-
crieff, M.Inst.C.E., presented before the American Society of Civil
Engineers in 1900.1 The reader is referred to the original work for a
complete account of the reasoning and deductions. The general outline
of the method of inquiry followed is indicated in the following pages.
Limited space prohibits giving more than a brief outline, but the
reader is strongly recommended to study the original communi-
cation. The underlying principles upon which the reasoning is based
are —

1. That a perfectly centred column of perfect material and straight-
ness is probably never realized in practice ; and —

2. That the various disturbing influences preventing such
realization, each of which conduces to initial bending, are all
capable, as regards their ultimate effect, of being represented
or covered by an equivalent small eccentricity of loading.

In order to apply this principle to the case of practical
columns under intentional and apparently central loads, the
suitable value to be assigned to the " equivalent eccentricity "
required to be carefully determined. This value was finally
fixed after a careful analysis of the records of practically all
the really reliable tests of columns made in Europe and
America from 1840 down to date. In the course of this
analysis, the results of 1789 tests of ultimate strength of
columns of cast iron, wrought iron, steel, and timber were
carefully considered, so that the formulae deduced have the
merit of being far more representative of practical strength
than any previously proposed.

In Fig. 108, AB represents the axis of a round-ended
column under the action of a load P applied at a small FIG. 108.
eccentricity e. A = The resulting central deflection of the
column from the vertical AB. I = Length of column. It is easily
established that

where E = Modulus of elasticity of material and I = Moment of inertia
of cross-section.

The bending moment at the centre of the column then

y v

where fb is the stress per square inch caused by bending alone at a
distance y from the neutral axis of the cross-section. A = Area of

1 " The Practical Column under Central or Eccentric Loads." J. M. Moncrieff,
Transactions Am. Soc. C.E., vol. xlv. 1901.

152 STRUCTURAL ENGINEERING

cross-section, and r = Radius of gyration in the direction in which the
column bends.

Rearranging fb = ± -^a~> the positive sign denoting compres-

sion at the concave side of the column, and the negative sign tension at
the convex side. The direct compression on the column section, inde-

pendent of the eccentricity of loading, =fd= + T- = the average load

per square inch on the sectional area of the column. Hence, the total
stress per square inch at opposite edges of the section, due to both
bending and direct compression,

Substituting the value of A in equation (1)

48E

Using the + sign to determine the maximum compressive stress Fc
and transposing equation (2),

48E rF

rc _ _ y_e-\
-L/*

Similarly the — sign will determine the minimum stress F, at the
opposite edge of the section, which may or may not be tensile, according
as the tension due to bending exceeds, or does not exceed, the direct
compression. The suffix t is only used conveniently, and does not
necessarily imply that Ft is a tensile stress. Hence using the — sign
in (2),

_ ,

ye~\ d\

?j • • «

The formulse (1), (2), (3), and (4) are all general expressions ap-
plicable to round-ended columns of any given material and form of
section, and with any given value of eccentricity of loading probable

in practical work. For fixed-ended columns the value of - as given by

I
formulse (3) or (4) wil\ be doubled. For flat-ended columns, - will be

the same as given for fixed-ends up to the point where tension begins
to be developed at one edge of the end bearings. Since flat-ended
columns are incapable of resisting tensile stress, Ft in equation (4) must

COLUMNS AND STRUTS 153

be made = 0. For the ratio - at which tension begins to be developed
the formula then becomes —

Formulae (3) and (4) give the relation between -, the elasticity E of

the material, the maximum fibre stress Fc or F,, and the average load
per square inch, /d, on the column section which will produce that

maximum fibre stress. Formula (5) gives the relation between - and

the average load per square inch,/d, consistent with no tension being
set up in the material.

The use of equations (3) and (4) is as follows. Suppose that in a
given column section it is decided that a certain value of maximum
compressive stress Fc, or a certain value of minimum stress F,, is not to
be exceeded ; these values being inserted in the formulse together with

7/P

the values of E and — 2, corresponding with the material and the section
of column and eccentricity of loading adopted, the result gives at once
the value of - corresponding to /,, the average load per square inch.

*ip

It will be seen the formulse each contain the term --2. y and r depend

only on the form of section, e is the " equivalent eccentricity " to be
allowed for covering the defects of material, straightness of axis, etc.,
in the actual practical column. By careful comparison of results given
by the formula with those of the numerous tests previously mentioned,

?/P

Mr. Moncrieff found that by making p = 0*6, the formula expressed
very closely the strength of the weaker columns in the various series of

A I/)

tests, whilst by making —^ = 0*15, the strength of the stronger columns

was fairly represented. Since the practical strength is actually repre-
sented by the weaker experimental results, it appeared advisable not to

1JP

employ a lower value than 0*6 for the term '*-$. Adopting this value

and inserting suitable values for E and Fc, and employing a factor of
safety of 3 for dead loads, the results of the formulse when applied to
various materials are exhibited by the curves in Figs. 109, 110, 111,
and 112.

The average dead loads indicated by these curves correspond with
the following maximum stresses per square inch and Moduli of Elas-
ticity for the various materials.

154

STRUCTURAL ENGINEERING

U3d saNnod oven BJVS

COLUMNS AND STRUTS

155

O 83d SGNOOd QVOT 3JVS

156

STRUCTURAL ENGINEERING

Material.

FC) Ibs. per sq. in.

12,000

14,000,000

Wrought iron, of tensile strength, 45,000 to
50,000 Ibs. per square inch
Mild steel, of tensile strength, 60,000 to
70,000 Ibs. per square inch

18,000
24,000

28,000,000
30,000,000

Hard steel, of tensile strength of about
100,000 Ibs. per square inch

36,000

30,000,000

Yellow pine or pitch pine

2000

2 200000

1,300

1,400,000

French oak or Dantzic oak

2000

1,200000

The curves, Fig. 110, for the loads on flat-ended columns are
identical with those for fixed-ended columns up to the point where

1400
1300
1200
1100
1000
900
800
700
600
500
400
300
200
100
0

Safe Dead Loads frer square inch on

— =

=—

Cent-rally loaded Timber Co/umns.

Round and Fixed Ends.

- 5

1
c
>

Vr^

xl t px

_r-r--

*. —

^ — '

^-^-

(i""

„ — "

_-4-— • -f"""" ~~~ ~"

&**

'-^

I^-s^ss^f:^

=— *•

SO *O I3O '20 110 tOO 90 80 70 <

230 260 i40 220

ENDS.
ENDS.

FlG. 111.

-5<ffe Dead Loads per square inch
on Centrally loaded
Timber Columns
Flat Ends

tension begins to be developed, which occurs in the neighbourhood of
the ratio - = 110. Beyond this point the resistance of the flat-ended

COLUMNS AND STRUTS 157

column rapidly diminishes, as indicated by the sudden drop in the
curves.

In the original paper the factor of safety of 3 is applied to the
modulus of elasticity for reasons there fully explained. The use of the
factor of safety in this manner has an inappreciable influence on the
results of the formulae when applied to short columns, while its effect
gradually increases with the length, ultimately affording a factor of
safety of 3 against failure by instability or buckling, in the case of
very long columns, and the factor of safety against ultimate strength is
fairly even between these extremes.

In making use of the curves plotted from the above formula?, the
particular amount of " equivalent eccentricity " for which they provide
may be readily computed for any proposed column section. Thus
taking section No. 3, Type 3, the least radius r = 3 '9 6" and y = 6f.

Since -^ = 0-6, 6^^ = 0-6, and e = 1-4". The sectional area = 33'54
r2 ' 15*68

sq. in. Taking a mild steel column of say 30 ft. = 360 in. length

7 ^ P A
with fixed ends, - = ^^ = 91. From the curve for mild steel in Fig.

109, for fixed-ended columns, the safe dead load for this ratio = 12,800

• i_ TT -i, . i r j , i -, 33-54 X 12800
Ibs. per square inch. Hence the total safe dead lead =

= 191 tons. That is to say, in imposing an intended central dead load
of 191 tons on this column, provision is made for a possible eccentricity
of 1*4 in. to cover defects of elasticity, cross-section, initial curvature of
axis and slight inaccuracies in erection.

Columns under intentionally Eccentric Loads. — In adapting the
formula? to cases where the load is intentionally applied at a known

tie
eccentricity, it is necessary to add the value •-» = 0'6 as determined for

IIP

presumably centrally loaded columns, to the known value of % as

obtained from the intended eccentricity e and the dimensions and form
of sections which fix the values of y and r. Formulas (3) and (4) then
become —

and 4*E

where Fc and F, are, as before, the maximum permissible working
stresses in compression and tension respectively. Inserting the previous
tabular values for E and Fc and employing a factor of safety of 3, the

curves in Fig. 113 exhibit the relation between - and the average

working stress per square inch permissible for various degrees of eccen-
tricity of loading, in the case of mild steel columns with round ends.

158

STRUCTURAL ENGINEERING

Since the maximum compressive stress developed in a column of sym-
metrical section always ex-
ceeds the maximum tensile
stress, it is not necessary, in
the case of mild steel, to
use the formula containing
F,, as Ft may be taken, for
this material, under work-
ing loads, as equal to Fc, the
permissible compressive
stress. In the case of cast
iron it will, of course, be
necessary to use both ex-
pressions, and to adopt
whichever gives the lower

value to the ratio - for any

given load/d.

Practical Considera-
tions in selecting a Type
of Column. — Generally
speaking, the simplest forms
of cross-section are most
desirable. The principal
practical consideration is
usually the number and
kind of connections to be
made with horizontal girders
or joists, and whether such
are to be attached on one or
two only, or on all four
faces of the column. In
this respect Types 1, 2, 6, 8,
9, 12, and 13 are convenient
forms. Those columns with
the least amount of riveting
are cheaper, and less likely
to suffer distortion during
construction.

Single joist sections are
uneconomical as columns on
account of their small
radius of gyration as com-
pared with sectional area,
and consequently weight,
but are widely used by
reason of their cheapness
and ease of making con-
nections. Broad flange

beams are preferable as columns to the British Standard beams, the
additional width of flange giving them a considerably larger minimum

COLUMNS AND STRUTS 159

radius of gyration, weight for weight, and more latitude for riveted con-
nections with girders. They are rolled in 23 sizes from 7^" X 7|" X 31^
Ibs. per foot to 29J" x 11 if" X 177 Ibs. per foot.1 As a comparison,
the B.S.B., 10" x 8" x 70 Ibs. has a minimum radius of gyration of
1-865". The B.F.B., 11" x 11" X 70 Ibs. has a least radius of 2'58".
Both have practically the same weight and sectional area = 20*5 square
inches. Used as a fixed-ended column 25 ft. high, the

B.S.B. has - = pggT = 161, and safe load = 8850 Ibs. per sq. in.

7 ^00
theB.F.B.has- = ^ =116, „ =11,700

and the total safe loads would be 81 tons for the B.S.B. and 104-7
tons for the B.F.B., or 29 per cent, greater carrying power in favour of
the B.F.B.

In section Types 2, 3, 5, 6, 7, 8, and 9, the dimensions may readily
be modified, if desired, to give practically the same radius of gyration
about either axis. This is desirable in cases where a column is not
stiffened laterally in either direction by intermediate attachments.

The use of long columns in which the ratio - is very high, should

be avoided, since the working stress is very low, and the stiffness rela-
tively small. It is often advisable to employ a heavier section than
indicated in the case of very long columns in order to obtain necessary
stiffness, although at some sacrifice of economy on the score of
weight. Types 11, 12, 14, and 15 have the same radius of gyration in
any direction.

Types 3, 7, 10, 11, and 12, formed by bracing together two joists or
channels or four OT eight angles by means of tie-plates at intervals, are
inferior to columns with continuous webs or close lattice bracing,
especially when required to carry eccentric loads or to withstand heavy
lateral wind pressures as in the case of tall buildings. Type 10 is very
generally used for struts of lattice bridge girders. Type 11 forms a
light and economical section for very long struts in horizontal or
inclined positions. These are subject to deflection due to their own
weight, which unavoidably sets up the initial bending so detrimental
to columns. It is therefore important to keep down the dead weight
of such members. Crane jibs and very long bridge struts are usually
of this type. Type 12, known as the " Gray " column, is much favoured
in American practice. It is an economical section, easy to construct
and equally convenient for making connections with girders on all four
sides. The tie-plates should be closely spaced. This type is also used
with continuous plates between the pairs of angles, the column being
then much stronger.

Connections. — The usual connections required in the case of
columns are —

1. Columns to built-up plate girders.

2. Columns to joists or beam sections.

1 Handbook No. 14, Structural Steel, B. A. Skelton & Co.

160

STRUCTURAL ENGINEERING

3. Junctions in long columns where the cross-section is gradually
diminished from bottom to top.

4. Bases and caps.

Very great variety of arrangement of these details naturally exists
on account of the large number of column sections available. In
designing connections the following general principles should be
attended to.

1. Simplicity of detail in order to minimize expense in shop work,
and admit of rapid and easy assemblage of parts at site.

2. Riveted or bolted connections of beams or girders to columns
must in every case provide adequate shearing and bearing resistance to
the vertical load to be carried.

3. The joints should be arranged with a view to providing the
greatest possible lateral rigidity for the columns as well as stiff end
connections for the girdere, since the degree of lateral stability obtained
for intermediate points and heads of columns frequently decides
whether they may be considered as approximating to the strength of
fixed or round-ended columns. The capability of a structure to resist
horizontal wind pressure or racking action under live loads depends
entirely on the rigidity of the connections of vertical with horizontal
members in cases where the character of the structure does not permit
of efficient diagonal bracing.

4. Horizontal members should be attached to columns with the
least possible amount of eccentricity in order to minimize the bending
action on the columns. Simply supporting the load on wide brackets
without at the same time firmly riveting or bolting to the face of the
column is to be avoided, although a well-fitted bracket, where space
allows, will largely contribute to lateral rigidity.

The following figures illustrate suitable types of connections between
joists or girders and columns, such as commonly occur in practice.

Bolts are indicated by black
circles and rivets by open
circles. On extensive works
plant for putting in rivets in
situ is frequently installed,
in which case bolts will only
be required in such positions
as cannot be negotiated by
the riveter. Generally,
however, bolts are largely
used to save the expense of
riveting at site.

Fig. 114 shows the
usual connection of hori-
zontal joists with a

column of a single beam section. Top and bottom cleats A and B
connect the flanges of the joists J with the faces of the column C, and
a pair of web cleats W connect the web of the joist with the flange of
the column section. The top cleat A is often omitted, but its use greatly
stiffens the end of the joist.

Fig. 115 indicates a suitable arrangement for four broad flanged

FIG. 114.

COLUMNS AND STRUTS

161

joists meeting on a single B.F. joist column, together with a junction
between a lower heavier column section A, and an upper lighter one,

FIG. 115.

B. This type of joint is of common occurrence in high framed
buildings where the sections of the columns are diminished every two
or three storeys.

Fig. 116 illustrates a connection between heavy 3-beam compound
girders and a column consisting of two beam sections and two plates.
The width of the beam exceeds
that of the column, and pro-
hibits the use of web cleats.
Brackets or "stools" 0- are
riveted to the column faces,
having sufficient rivets to
resist the heavy end shear, and
the girders are bolted down to
the horizontal angles of the
stools, and through to the
flanges of the column by bolts
passing through the upper flange
cleats, which are previously
riveted to the upper flanges of
the girders. The vertical angles
which hold up the lower flange
cleats require packing plates at
P,P.

In the case of columns running through several storeys of a building,
it is desirable to preserve the continuity of the columns whenever
possible, and to connect all girders or joists to the faces of the columns.
Occasionally, however, this entails complexity of design, or one of the
girders may require to be continuous. Figs. 117 and 118 show the
detailed arrangement where a continuous plate girder A is carried over

FIG. 116.

162

STRUCTURAL ENGINEERING

the head of a lower column D, and two other slightly shallower girders
B and C meet from opposite sides in a direction at right angles with

_L__j riii

FIG. 117.

that of girder A. Fig. 117 shows a cross-section through girder A,
and Fig. 118 a cross-section of B or C. The lower flanges of the
girders are bolted down to the cap plate of column D, packings being
inserted as indicated. The ends of girders B and C are left just clear
of the web of girder A, but are bolted through it by four bolts, F.
The continuation column E, of lighter box section, is planted on the
upper flange of girder A, the under edges of the flange beneath the
column foot being packed up on channels or bent plates P. Substantial
gusset stiff eners S strengthen the web of girder A, and T-stiffeners T
the webs of B and C, and assist in transmitting the load from the upper
to the lower column. The timber floor joists J are carried on the upper
flanges of girders B and C. This is, generally speaking, an inferior
arrangement, since continuity of the columns, especially in tall buildings,

COLUMNS AND STRUTS

163

conduces to greater stiffness under the vertical loads, and more efficient
resistance to oscillation or racking due to lateral wind pressure.

iw^H-^H&fH
m

Fig. 119 illustrates a common case of column construction, where
two rail girders G carry the ends of travelling crane girders in adjacent
bays of a shop. The general arrangement is shown at A. The lower
part of the column, up to the level of the seatings of girders G, consists
of three joist sections with two plates each, laced together with diagonal
angle braces. Brackets B are built out to support the girder bearing
plates shown in the plan. Immediately beneath the girders the three
joists are tied together by deep gusset plates P, forming the back plate
for the brackets B. Above the level of the girders, the central joist,
reinforced by two channels, is carried up to support the adjacent spans
of the roof. In the side elevation E, the rail girders are removed.
Double and treble columns of this type frequently have the component
joists tied together with short rectangular plates. Diagonal bracing is,

1G4

STRUCTURAL ENGINEERING

Half Sect-ion on X-X. Plan wtth Girder G removed.

FIG. 119.

COLUMNS AND STRUTS

165

however, more satisfactory, as ensuring a more uniform distribution of
the total load between the individual joists.

Caps and Bases. — Columns are fitted with cap and base plates
attached to the faces bj angles. Before riveting on the cap and base
plates, the ends of the column, in all good-class work, are planed up
square with the axis to ensure a fair bearing on the end plates. The
plates and bar sections composing the column are, previously to riveting,
sawn to dead lengths, but the distortion consequent on riveting usually
leaves the ends uneven, thus necessitating the planing. The cap and

FIG. 120.

base plates are usually riveted with rivets having countersunk heads to
the outside, in order to leave flush surfaces for bearing on the foundation
block at the lower end, and against the lower flanges of the girders
supported at the upper end. Fig. 120 shows a typical cap and base
for a 20" x 12" broad flange column. Base plates vary in thickness
from f in. to 1^ in., and for columns carrying exceptional loads are
sometimes 1J in. to 2 in. Four, six, or more holes are left in the base
plate, at accessible points, through which pass the foundation bolts.
These latter are embedded in the concrete foundation block with the

166

STRUCTURAL ENGINEERING

screwed ends projecting ready for lowering the column into place. The
bolts are carefully placed in the correct positions by the aid of a wooden
template having holes drilled at the requisite spacings.

Columns are frequently mounted on steel or iron castings interposed
between the base plate and foundation block. This construction rnay

T ~
"t

r P

;

< >!

''•"'

Jll

FIG. 121.

FIQ. 122.

be adopted in cases where it is inconvenient to use a large base plate
with gussets or brackets for distributing the load from the column to
the foundation. Such built-up brackets frequently entail difficult and
expensive riveting, and the use of a casting effects the necessary dis-

COLUMNS AND STRUTS

1G7

tribution of load over a larger area in a cheaper manner. Fig. 121
shows a 15" x 15" Gray column mounted on a steel casting having a
36 inch square base. The bearing area of the casting on the foundation

FIG. 123.

block is thus 9 square feet, whilst the bearing area of the concrete
block on the actual ground would be still further increased. Fig. 122
shows the detail of a steel box-section column embedded at its lower

168

STRUCTURAL ENGINEERING

end in concrete filled into a hollow casting, which is bolted down to a
suitable foundation. This is a type of base frequently employed for
columns supporting warehouses where the basement is open to heavy
vehicular traffic. The base castings project 3 or 4 feet above the
pavement, and protect the columns from damage. In this example,
the drainage from the roof is led through down-pipes screwed to the
faces of the column, Fig. 123 is an example of a column of channel

FIG. 124.

box section reinforced with four angles. The column is 18 in. square,
and is provided with a 3 ft. 6 in. square base plate, 1 in. thick. Four
deep brackets, consisting of £ in. gusset plates, and 3^" X 3J" X J"
angles, assist in distributing the load to the base plate In arranging
the brackets at the base of a column, no useful end is served by
theoretical calculations as to the probable distribution of pressure
effected by any proposed system of bracketing. The essential points

COLUMNS AND STRUTS

169

to bear in mind are, to provide a reasonably thick base plate in pro-
portion to the load to be carried ; to place brackets in well-distributed
positions around the base, and to increase their number in accordance
with the size of column, having regard to convenience in riveting.
Small columns will generally be provided with two gusset plates, as in
Fig. 120. Box sections generally admit of four brackets, one on each
face, whilst double and treble compound columns may conveniently be

2 Feet

FIG. 125.

provided with eight, ten, or twelve brackets. Figs. 124 and 125 give
an example of a built-up base for a treble-joist column. Here, two
deep gusset plates are riveted to the flanges of the joists and bevelled
off at the ends to form four brackets, whilst other three brackets
are placed at right angles to the main gusset plates on either side.
The pressure transmitted to the base plate will naturally be more
intense immediately beneath the brackets, but by suitably increasing
their number and adopting a rational thickness of base plate, the
eventual distribution of pressure on the foundation will be sensibly
uniform. It is desirable in all heavy column construction that the
foundation bolts should pass through the horizontal tables of the
angles connecting the brackets with the base plates, so that the nuts
may bear on a considerable thickness of material. Where this cannot
be conveniently arranged, as in Fig. 124, wrought-iron or steel bridge
blocks B should grip the horizontal angles attached to the brackets,
and the foundation bolts be passed through the blocks. Brackets
should have a good depth in proportion to their offset from the column
face. It is a common practice to provide as many rivets through the
column faces and vertical angles of the brackets as will give a shearing
or bearing resistance at least equal to the total load on the column.
This rule provides for the transmission of the load through the brackets
to the base plate, independently of the bearing of the column end on
the base plate, which in rough work is often far from satisfactory.
Thus, in a column carrying 120 tons, assuming the resistance of £ inch
rivets in single shear as 3 tons, the minimum number through the

170

STRUCTURAL ENGINEERING

vertical bracket angles would be 40. If the rivets bear on | inch thick
angles, the bearing resistance of eacli rivet at 8 tons per square inch
= 8x^x| = 3^ tons, and the minimum number for bearing = 120
-^ 31 =-35.

Column Foundations. — Small columns carrying light loads are still
occasionally attached to stone foundation blocks, but the majority are
bolted down to concrete blocks having a basal area sufficiently large
to suitably distribute the pressure on the soil. The base plate should
rest on a sheet of 6 or 8 Ibs. lead laid on the top of the foundation
block. The heads of the foundation bolts are embedded near the
bottom of the block, and bear against flat or angle bars threaded on
to the bolts and laid longitudinally and transversely in the concrete.
The proportions of such foundation blocks are determined from the
character of the loads on the column and the safe pressure which may
be put on the soil, and detailed examples will be considered later on.

Grillage Foundations. — Foundations for columns on poor bearing
ground require the pressure distributing over a large area. In such
cases, a " spread " or " grillage " foundation is most suitable. As seen
in Fig. 127, the base casting rests on a layer of parallel beams or girders
projecting considerably beyond the edge of the casting. These rest
again on a second layer or "grill" of beams projecting beyond the
limits of the first grill, and these again may rest upon a third and
fourth layer if necessary. With each successive grill, the bearing area
is largely increased, until the intensity of pressure on the soil is reduced
to the desired value. The beams are tied together with bolts and
separators, the latter acting as stiffeners for the webs, and giving them
the necessary rigidity for resisting the vertical shearing force. The
projection of the beams takes the place of several footings in an

ordinary foundation, and the ad-
vantages of grillage foundations
are that the successive offsets may
be much longer than in the case
of masonry or concrete footings,
and the necessary bearing area is
obtained at a considerably less
depth than for a foundation block
of the ordinary type. The stack
of beams is eventually rammed up
with and encased in concrete,
which serves the double purpose
of aiding uniform distribution of
pressure amongst the separate
beams in each grill, and protecting
the steelwork from corrosion. In
very large grillage foundations,
the uppermost grill frequently
consists of built-up plate or box

girders, whilst the size of the beams diminishes in the various grills
from top to bottom. In Fig. 126 AB represents a single layer of
beams beneath a casting C carrying a loaded column. Let p tons per
foot run = the upward reaction of the ground against the under side of

FIG. 126.

COLUMNS AND STRUTS 171

each beam. The portion AD obviously acts as a cantilever, subject to
a uniformly distributed upward pressure of p tons per foot run, and
tends to bend as shown by the dotted lines on the right-hand side. There-

fore bending moment at D = *5- foot-tons = DE, and the semi-

parabola AE forms the B.M. diagram from A to D.

It is frequently stated in text-books and hand-books that DE is the
maximum bending moment on the grillage beams. Mr. Max am Ende,
M.Inst.C.E., has shown, however, that the maximum bending moment

o j

occurs at the centre of the beam, and = ~ -f ^5- foot-tons, where b is

'— —

the distance from the centre of the beam to the edge of the base
casting, or base plate, if no casting be employed.1 This may be
demonstrated as follows : —

Total upward pressure against AB = p x 2(a + b) tons = total
downward pressure exerted over the length 2b feet of the casting per
pitch width of beams. The central portion DK of the beam is,
therefore, subject to a downward pressure

_ 2p(a + b) p(a + b)

— — — 7 - tons per foot run,

and an upward pressure of p tons per foot run, or a resultant downward
pressure

p(a 4- b~} pa

— T — — P = ~r tons per foot run.

The span DK = 2b, hence B.M. due to the resultant downward load

pa (2b)2 pab ,
= V x -TT^ = ^r foot-tons.

0 O A

The parabola EHL having a central height GH = ^ rePresenfcs tne
moments over the span DK, due to the unbalanced load of y tons per

9%/M

foot, and since there is already a B.M. at D and K = ^-, the total

moment at F = DE + GH = FH = P~ + Pf.

It is sometimes urged that this moment will not be developed unless
the casting bends appreciably, and, no doubt, the extent to which the
moment FH may be approached in an actual foundation will depend
on the relative rigidity of the casting and the girders beneath it. With
a heavy steel casting, the yielding will be relatively small, but since the
beams in the first grill are often much deeper than the casting, their
rigidity is also considerable, and it is only reasonable, in designing the
beams, to provide for the maximum moment FH, even although the
actual moment may be somewhat less. Moreover, in the absence of a
rigid casting, the relatively flexible base-plate of the column will readily
1 Mins. Proceedings lust. C.E., vol. cxxviii. p. 36.

172 STRUCTURAL ENGINEERING

follow the deflection of the stiffer beams beneath it. The shearing
force diagram is shown below, the shear increasing from zero at the
ends of the beam to pa tons below D and K, and decreasing again to
zero at the centre.

The following example illustrates the method of designing a grillage
foundation for a column.

EXAMPLE 21. — A column carries a total load of 240 tons, and is
mounted on a casting having a base 4 feet square. The pressure on the
ground is not to exceed Ij tons per square foot. Required a suitable
grillage foundation.

240
Minimum ground area required = -yy~ =160 square feet. This may

be obtained by a rectangle 13 feet x 12 feet 6 inches. The heavier beams
will be in the upper grill and may be 13 feet long. Five beams may
be conveniently arranged beneath the four feet casting, in order to leave
suitable room between the flanges for packing in concrete. Hence
upward pressure per foot run against each beam of the upper grill = £
of ^ = |f tons ; a = 4 feet 6 inches, I = 2 feet.

,s/4-5 X 4-5 4-5 X 2\
/. B.M. at centre of each beam = ffl ~ -j ^ — I = 54

foot-tons = 54 x 12 = 648 inch-tons.

Employing a working stress of 9 tons per square inch, the required
modulus of section = £p = 72.

The B.S.B., 14" x 6" x 57 Ibs., has a modulus of 76-2. Hence the
first grill may consist of five 14" x 6" x 57 Ibs. beams spaced at
10 in. centres (Fig. 127).

Assuming 16 beams in the second grill, the upward pressure per

240
foot run against each beam = ^ of J^K = § tons ; a — 4 feet 3 inches,

b = 2 feet.

v> ,T r/4-25 X 4'25 4'25 X 2\
B.M. at centre of each beam = |H - ~^T~ "" + o / = 16

foot-tons = ^ x 12 = 191*25 inch-tons, and section modulus required

= ™™ = 21-25.
y

The B.S.B., 8" X 5" X 28 Ibs., has a modulus of 22-3. Hence the
second grill may consist of sixteen 8" x 5" x 28 Ibs. beams spaced at
10 in. centres. Four rows of plate separators may be placed between
the upper beams, and tube separators between the lower beams as
indicated in Fig. 127. A depth of 16 inches of concrete is shown
below the lower grill, the beams and casting being embedded as in the
figure.

Maximum shear at edge of casting for each beam in the upper
grill = f| x 4-5 = 16-6 tons. Sectional area of web = 12" x i" = fi

sq. inches. Mean shear stress on web = -y- = 2'8 tons per sq. inch.

Maximum shear, 4 feet 3 inches from end of each beam in the
lower grill = § x 4-25 = 5'1 tons. Sectional area of web = 6'5 X 0'35

= 2-27 sq. inches. Mean shear stress on web = - = 2*25 tons per

COLUMNS AND STRUTS

173

sq. inch. These stresses are well within the safe limit of working shear
if the beams are well supported by the separators and concrete.

•~n~-

nn
. ii

i
i1

'i i

1. ......'......... .......I

PIG. 127.

Practical Applications.— Some applications of the calculations to
practical examples of columns and struts will now be given. It must
be emphasized at the outset that the principal difitaftlto in designing

174

STRUCTURAL ENGINEERING

columns is to decide whether they may be treated as fixed-ended or
round-ended, or to what extent they may be considered as approximating
to one or other of these conditions of end support. In many cases this
is a matter for personal judgment only, and probably no two persons
would arrive at the same conclusion in a debatable case. As previously
noted, no column is ever as stiff as the theoretically fixed-ended column,
and in assuming any practical column to be actually fixed-ended, its
strength must thereby be somewhat over-estimated.

In many cases, notably those in which pin-ended connections are
employed, no doubt exists as to the columns being round-ended, whilst
in cases where considerable doubt exists as to the character of the end
support, the safest procedure is to treat the column as round-ended.

EXAMPLE 22. — A mild-steel bridge-strut, 18 ft. long, consisting of
four angles with lattice 'bracing similar to Type 10, is required to resist a
maximum compression of 55 tons, the ends being fitted with pin bearings.
Deduce a suitable section.

Taking section No. 3, Type 10, with four 5" x 3" x £" angles.
r = 2-53". / = 18 x 12 = 216". A = 15 square inches.

216
2-53

= 85

From the curve in Fig. 109, safe load for ratio - = 85, for round-ended
struts = 8400 Ibs. per square inch.

/. Total safe load =

15 °°

= 56J tons.

Fig. 128 indicates the general design. The width W will depend
on the interior breadth of the boom to which the strut is attached, and

FIG. 128.

does not affect its resistance to bending in the plane X-X. The con-
necting plates P, P, must be attached to the angles by a sufficient
number of rivets to provide the necessary resistance to shearing and
bearing, and will further be reinforced by additional plates to give the
necessary bearing area on the pins. The lacing bars usually consist of

COLUMNS AND STRUTS 175

flat bars from If in. to 2^ in. wide by ^ in. or f in. thick, and must
be sufficiently close to prevent local failure of the angles between two
junctions as J, J. This is very unlikely to occur with the usual pro-
portions of lacing in general use. The method of calculation for local
failure will be given in subsequent examples.

If the ends of the strut were securely riveted between very stiff
booms the strut would approximate to the condition of having fixed
ends. Section No. 1, Type 10, would then be suitable.

r = 1-67", - = p^ = 130, and safe load = 10,650 Ibs. per sq. in.

10650 x 12
A = 12 sq. in. and total safe load = — w±r\ = ^ tons.

EXAMPLE 23. — A strut of the type shown in Fig. 129 is 1 ft. 6 in.
long, and is required to resist a compression of 7 tons, with pin-connected
ends. Make a suitable design for the strut in mild steel.

"."_._" . j4y rr* /f\ ~~— fly

f- -f- — - - -y f

r--'''0*1 — i i i

--+ — i i r

K4'<

i~#-

FIG. 129.

Struts of this type, consisting of two flats with distance ferules at
F, F, are frequently used in roof trusses. By varying the length of
the ferules, the two bars may be separated by any desired distance to
give a suitable radius of gyration at the central section. They must,
however, be tied together at sufficiently frequent intervals to prevent
risk of local failure of one bar over the length FF. The dimensions
adopted are shown on the enlarged cross-section at S.

1. Considering local failure of one bar for the unsupported length
FF. Compression on one bar = \ - 3*5 tons. Sectional area = 2J"

3*5

X f" = fts(l- iQ- Pressure per square inch = — — = 2-24 tons, = 5017

i*0t)

Ibs. r = f x 0-289 (for rectangular section) = 0'18". The length
FF must be treated as round-ended, since the ferules are incapable of
fixing the ends in a strut of this type.

The ratio - for a safe stress of 5000 Ibs. per square inch = 125,
Fig. 109, for mild steel with round ends.

•*• .-T-^TT = 125, and I = 125 X 0'18 = 22'5" = 1' 10J"

U'lo

This determines the required distance between the ferules.

2. Considering failure of the strut as a whole with regard to
bending in the plane y-y.

176 STRUCTURAL ENGINEERING

7 QA

r = 2J" X 0-289 = 0'72", I = 90", and - = ^7 = 125
Safe load per square inch for - = 125 = 5000 Ibs.

Total safe load = :- - * $ * ° = C'97 tons> or PracticallJ

7 tons.

3. Considering failure as a whole with regard to bending in the
plane

Moment of inertia about y-y = ± x f(2'53 - 1'253) = 2-85.

Sectional area = 3| sq. in. /. r = \/o7j^? = 0*95"

I 90

- = /TT^ = 95, and safe stress per sq. in. = 7300 Ibs.

T yyo

.-. Total safe load = 3J X JiJg = 10'2 tons.

The strut is therefore slightly weakest as regards its resistance to

bending in the plane y-y.

EXAMPLE 24. — A mild-steel gantry
column for supporting the rail girder of
an overhead travelling crane is 24 ft.
high. The maximum load on it is 40
tons, and the columns are efficiently braced
together lengthwise of the gantry.

In Fig. 130, as regards bending
transversely, the columns will be equi-
valent to a round-ended column EF of
48 ft. length, or twice the length of the
Y~hj— I— Y actual column EG. For best resisting
i bending in the transverse plane, the
columns will be placed as at L in plan,
FIG. 130. so that the greatest radius of gyra-

tion comes into play. Taking the
broad-flanged beam section 11" x 11" X 70 Ibs., the greatest radius

I 48 x 12
of gyration = 4'73". Sectional area = 20-4 sq. in., and - =

= 127.

The safe load for mild-steel round-ended columns for this ratio
= 4800 Ibs. per square inch.

4800 x 20*4
/. Total safe load = - = 43-8 tons, or a little in excess

of the load to be carried. With regard to bending in the plane X-X,
the columns will realize a high degree of end fixation, and the section
adopted will have a large excess of strength considered as a fixed-ended
column 24 ft. long.

It should be noticed that a complete examination of this case would

COLUMNS AND STRUTS 177

necessitate a consideration of the wind pressure P acting on the head
and face of the column, which will create additional bending moment
at the basal section. The area exposed to the wind by the ends of
crane girders and face of rail girders would need to be fairly accu-
rately known in order to take account of this action. As this case is
similar to a subsequent example, it will be treated more fully later on.
For the present, assuming 80 square feet as the effective area exposed
to the wind and a maximum horizontal wind pressure during the
working of the crane of 25 Ibs. per square foot, the total horizontal
pressure at heads of windward and leeward columns = 80 x 25 = 2000
Ibs. This is resisted by two columns, giving 1000 Ibs. acting hori-
zontally at the head of each column, resulting in a bending moment at

the foot of the column = - "2240 — - = 129 inch-tons. To this

must be added the B.M. due to wind pressure on face of column.
Area of face = say 24 square feet. Total pressure on column = 24 x 25
= 600 Ibs., acting halfway up the column.

... B.M. = SQQX 12X12 = 39 inch-tons.
2240

Total B.M. due to wind pressure = 129 -f 39 = 168 inch-tons.
The section modulus of the 11" x 11" beam about X-X = 83'1.

i fiR
Hence bending stress due to wind pressure alone = OQ-.J =2 tons per

square inch. This calculation will give an indication of the probable
amount of stress likely to be caused by wind pressure under ordinary
working conditions, and the 11" x 11" beam section, as deduced
when the wind pressure was neglected, will evidently be too light.
Further disturbing action comes on the columns when the load is
being cross- traversed. The additional B.M. resulting therefrom cannot
be more than roughly approximated, but making a j^L

rational allowance for this, together with the wind pres- 2 tons. \ f
sure, a beam section 14" x 12" x 96 Ibs. would be a
reasonable section to employ.

EXAMPLE 25. — Deduce a suitable section for a mild-
steel column 20 ft. high, supporting a vertical load of 5
tons and a horizontal pressure of 2 tons due to wind,
acting at its upper end.

These conditions of loading occur in the case of
columns supporting roofs. The action of the loads is
similar to that in the last example, excepting that the
vertical load in the case of a roof is relatively small,
whilst the bending action due to the wind is consider-
able, and usually constitutes the principal cause of stress
in the column. This case will be considered more fully FIG>
than the last.

1. Calculate the horizontal deflection d, Fig. 131, due to the
pressure of two tons applied at the head of the column. If the lower
end be firmly fixed to an adequate foundation, the column is acting
under similar conditions to those obtaining in a cantilever fixed at one,
end and loaded at the other,

178 STRUCTURAL ENGINEERING

Deflection d = i . gj- , where I = 240", W = 2 tons, and E = 13,400.

The deflection cannot be calculated until I is known, and since
this depends on the cross-section adopted for the column, a prelimi-
nary value must be assumed for I, which will probably require
correction before finally deciding on the cross- section. Taking the
British Standard beam section 14" x 6" X 57 Ibs., from the section
book, A = 10*77 square inches. I = 533, and the section modulus
= 76-15.

B.M. at foot of column, due to horl. pressure = 2x240 = 480 in.-tons.

vertl. =5xl'29= 6'45

Total B.M. = 486-45 „
and stress due to bending at foot of column = _r , = ± 6-39 tons

/ D* J.O

per square inch, that is, 6 '39 tons per square inch compression on
concave or leeward face of section and 6*39 tons per square inch tension
on convex or windward face of section.

Direct compression = yrp^ =0*3 ton per square inch.

.*. Maximum compressive stress at concave side = 6'39 + 0'3 = 6*69
and „ tensile „ convex „ = ~6'39 + 0-3= -6'09

tons per square inch. As these stresses are well within the safe limit
for mild steel, implying a factor of safety of just over 4, the 14" x 6"
section may be adopted.

Note. — The above calculation neglects the small additional deflec-
tion caused by the vertical load of 5 tons, once the column is deflected
away from the vertical. In the great majority of cases, the additional
B.M. due to this deflection is very small compared with that due to the
horizontal pressure, and may be safely neglected. It should be taken
into account if the vertical load is very heavy, which is seldom the case
in roof construction. The B.M. of 6 '45 inch-tons, due to the deflection
of 1*29 in. is also very small compared with 480 inch-tons, and might in
this case also have been neglected without appreciably affecting the
result. The calculation further assumes exact central loading. If a
reasonable amount of " accidental " or " equivalent " eccentricity of
loading be assumed in accordance with the theory previously stated, it
would, in this case

_ „ _ 0-6 X r2 0'6 X (5'6)2 _ 2.fiq,,

-jT = - V- 9 '

and this again would have little influence on the total bending moment,
on account of the small vertical load. The inference to be drawn from
examples such as this, is that slight eccentricity of loading and deflec-
tion within reasonable limits have little bearing on the practical design
pf columns under light vertical loading, whilst the relatively large

COLUMNS AND STRUTS 179

horizontal load acting at the great leverage equal to the length of the
column practically converts the case into one of beam design. It is
advisable to calculate the deflection for any proposed section, in order
to make sure it does not exceed a reasonable limit, since a relatively
large deflection alternately to right and left would exercise a deteriorat-
ing effect on the stability of the foundation and on the connections at
the head of the column.

EXAMPLE 26. — Required a strut section consisting of two angles back
to bach, J in. apart, to resist a compression of IS tons. Length Wft. Ends
considered fixed. *

This type of strut, Fig. 132, is largely used in roof trusses and light
lattice girders. It is not an economical form since the load is necessarily
applied eccentrically. The rivets connect-
ing the strut with a junction plate or neigh-
bouring member, pass through the centre

Y////////////.

CO of the vertical legs V, V, of the angles, X- ir —

whilst the axis XX passing through the c * —

centre of gravity of the cross-section is dis-
tant e' from CC.

Assuming two angles 4" x 4" x i", the VYV

least radius is about the axis XX, which is FIG. 132.

situated T18 in. from the upper edge of the

section, and bending takes place most easily in the plane YY. The
centre line of rivets CC is 2'25 in. from upper edge and the eccentricity
e' is therefore = 2'25" -1-18" = 1'07".

Least radius of gyration about X-X =1-227 in.

The maximum intensity of compression will occur at the edges V, Y,
of the vertical legs. These are distant 2-82 in. from the axis XX. Hence,

ye' _ 2-82 x 1-07 _ 3'01 = 2
r2 ~ 1-227 X 1-227 ~~ 1'5 ~

and /. yi + 0-G = 2-6

The strut being considered fixed-ended and 10 ft. long, the equivalent

/ rn"
round-ended strut will be 5 ft. long and - = p^y = 49, say 50.

Referring to the curves of safe loads for eccentrically loaded mild

n *//

steel struts, Fig. 113, for various values of ^ + 0-6, the safe load for

ratio- = 50 and-^- -f O'G = 2-6, is 6000 Ibs. per square inch. (This

value has been interpolated between curves Nos. 2 and 3.)
Sectional area of two 4" x 4" x i" angles = 7'5 sq. in.

.'. Safe load = — gas = 20'09 tons,

or 2 tons more than the specified load. This section will therefore be
satisfactory.

180

STRUCTURAL ENGINEERING

x-ft i-'

Note — Care should be taken to ascertain about which axis the
radius of gyration is least. For struts formed of two equal angles, the
least radius is always about X-X. If two unequal angles be used with
the longer legs back to back the least radius may be about X-X or
Y-Y, depending on the size of angles used. The angles are riveted
together with f in. or J in. thick buttons or washers between, at sufficiently
close intervals to prevent local failure of one angle.

EXAMPLE 27. — A lattice box strut of Type 11, is required for a
horizontal wind brace betiveen the main girders of a bridge. The length

j - i*8 30 feet and maximum compression 15 tons.

Obtain a suitable section, treating the ends as
rounded.

In this example, the strut being in a hori-
x zontal position, three actions are to be considered .

1. The bending stress due to the dead
weight of the strut.

2. The bending stress due to the deflection
and end thrust.

3. The direct compression caused by the
end thrust of 15 tons. Assuming the section in Fig. 133, consisting
of four 2J" X 2J" X f" angles, laced on all four sides with 1J" x &"
lacing bars, the approximate weight of the strut for a length of 30 ft. is
about 1050 Ibs. This, acting as a distributed load, will create a bending
moment

5-23'

12"

FIG. 133.

1050 X 30 x 12
2240 X 8

= 21'1 inch-tons.

From the section book, the moment of inertia of one angle about
x-x = 0-989, and sectional area = T733 sq. in.

=> °'989 + 1<783 X

= 48'4

and Ix for four angles = 48*4 x 4 = 193-6, say 194. Total sectional
area = 1-733 x 4 = 6'93 sq. in.

.'. (Radius of gyration)2 =

194

= 27 '9

Making allowance for an " equivalent eccentricity " of loading

0'6r2 0'6 x 27'9 9.70,,
— -^- 2 79 ,

the deflection caused by the thrust of 15 tons acting at this eccentricity

15 X 3602 X 2-79

8EI - fP/2 " 8 X 13400 X 194 - | X 15 X 360';

= 0-29"

5 W/3
The deflection caused by the dead weight of the strut = ^^ x -jjrp

which, substituting the known values for W, ?, E and I, = 0-12 in. If
this be added to the equivalent eccentricity of 2'79 in it gives a total

COLUMNS AND STRUTS 181

eccentricity of 2*91 in. and the resulting deflection due to end thrust is
0-3 in. instead of 0'29 in. as above calculated. The effect of this
deflection of 0*12 in. (due to dead weight) is so slight in cases of light
struts, it may be neglected. Total eccentricity = equivalent eccentricity
-f deflection due to end thrust = 2'79 + 0'30 = 3'09 in.

B.M. at centre of strut due to 15 tons end thrust = 15 x 3' 09 = 46 '35
„ „ due to dead weight = 21*10

and total B.M. = 67'45 inch-tons.

fl f X 194
Moment of resistance of section = — - = — p — = 67'45; whence

/ - ±2*1 tons per sq. inch.

Direct compression = ,77'., = + 2 '2 tons per square inch.

/. Maximum stress at upper edge of section = + 2'1 + 2'2 = 4 3
and minimum „ lower „ = — 2-1 -f- 2*2 = O'l

tons per square inch, compression in both cases.

Note. — If the tensile stress caused by bending is found to more
than annul the direct compression, the calculation should be revised,
deducting the rivet holes in calculating the I of the section. The
maximum stress of 4-3 tons per square inch is rather low, but making
allowance for the fluctuating action of wind load, a lighter section is
not commendable. It should be noted also that corrosion is more rapid
in light lattice work of this description, and the
working stress should be lower than in heavier parts
of the same structure, in order to secure approxi-
mately equal length of life.

EXAMPLE 28. — Three girders are connected with
a column 20 feet high, as shown in plan in Fig. 134.
The girder connected with the flange imposes a load of
36 tons, and those connected with the web, loads of 20
and 12 tons. The column carries, in addition, a
central load of 40 tons. Required a suitable section.

This illustrates a very general case of eccentric loading. The
eccentricity in the plane Y-Y is relatively large.

Taking moments about o, and assuming a 12" x 12" beam section,
36 x 12" + (12 + 20 + 40) x 6" = 108 X go, whence go = 8", or the
centre of gravity of the total load is situated in a plane passing through
g, 2 in. from the axis X-X. The eccentricity in the plane X-X is
very slight, as will be found by taking moments about Y-Y, and for
practical purposes, the e.g. of the loads may be assumed at g on YY.
The intentional eccentricity e' = 2 in. Radius of gyration of section
about X-X = 5' 07 in. Hence—

£ + 0-6 = «^ + 0-6 = 1-07

Treating the column as fixed -ended, the length of the equivalent
round-ended column = 10 ft. or 120 in.

Ratio -=r— -=24. Referring to Fig. 113, the safe load per
T D'07

182

STRUCTURAL ENGINEERING

square inch for ratio 24, from curve No. 1, is 11,200 Ibs. As the
value of 'A2 + 0-6 is here a little higher than 1,-the safe load will be
a little under 11,200 Ibs., say 11,000 Ibs.

23'6 (sectl. area) x 11000
/. Safe load on column = - 2240

116 tons.

The total load is actually 108 tons.

Foundation Blocks for Columns. — It is very essential that founda-
tion blocks for columns may not undergo any appreciable subsidence,
as the stresses in the members of the structure carried by the columns
would thereby be seriously affected. The subsoil must therefore
provide a firm bearing surface, and the safe pressure on the soil not
be exceeded. On gravel and hard clay foundations the safe pressure
may be 4 to 5 or G tons per square foot. Grillage foundations are
preferable to concrete blocks where the safe pressure is below 1J tons
per square foot. In the case of columns carrying
central loads only, the foundation block simply requires
to be of sufficient area to reduce the bearing on the
soil to the required safe limit, and of sufficient depth
to resist shearing of the block along a vertical plane
beneath the edge of the base plate. In the case of
columns subject to side loads, the centre of pressure
may fall very near one edge of the foundation block,
especially where the vertical load on the column is
small. In Fig. 135 a column 20 feet high is bolted
to a concrete base 4 feet square and 3 feet deep. The
column carries a vertical load of 12 tons, including its
own weight, and is subject to a horizontal wind
pressure of 1 ton at the top, and £ ton at the
centre. The resultant horizontal pressure = Ij tons
acting at a point 21 feet above the foundation. The
weight of the foundation block at 140 Ibs. per
cubic ft. = 3 tons, and total vertical load on foundation = 15 tons.

Hence centre of pressure^? is situated T~ of 21' = 1' 9" from c, or 3 in.

from edge of base. The resulting maximum intensity of pressure on

2 X 15
foundation at b = ^ -—-£- (2 — ~) = 3'4 tons per sq. ft. (see page 332).

Provided the foundation will resist this pressure without appreciably
yielding, the column will not tilt. If, however, this pressure exceeds
what may be safely put upon the soil, the bearing area of the base
block must be increased.

Suppose a column of the same height to carry an inclusive vertical
load of 50 tons, and to be subject to the same horizontal pressures.
Using the same size of base block, the centre of pressure is

~~ of 21' = 0-49', say 6 in. from c, or 18 in. from I, and intensity of

2 X 53
pressure on foundation at 1) = *.* (2 — If) = 5'8 tons per sq. ft.

tr s\ *t

FIG. 135.

COLUMNS AND STRUTS

183

On a hard foundation this would not be an excessive pressure, and it will
be seen from this example that a lightly loaded column may require a
foundation block quite as large as a much more heavily loaded column,
when subject to lateral pressure. In practice there is usually 1 to 3
feet of earth over the top of the block, and a little additional stability
is derived from its weight.

Foundation Bolts for Columns.— In cases where the centre of
pressure due to the resultant of all the vertical and horizontal loads
above the base plate falls within the base of the bolt holes, there
will be no uplift on the foundation bolts. Frequently, however,
with columns subject to lateral wind loading, the centre of pressure
falls beyond the bolts on the leeward side of the column, and the
windward bolts must have a sufficient section to safely resist the
resulting uplift. Considering, Fig. 136, height of column above base

FIG. 137.

plate = 24 feet, and above ground-level 22 feet. Longitudinal spacing
of columns 20 feet. Roof span, 40 feet, with rise 10 feet. Suppose
the columns to carry the side of the building and to be exposed to a
maximum horizontal wind pressure of 30 Ibs. per square foot. Then,
total wind pressure on side per 20 ft. length = 20 X 22 x 30 Ibs.
= 5'9, say 6 tons, acting at 13 feet above base plate. Assume the
effective horizontal wind pressure against the roof as 1*5 tons acting
at 29 feet above base plate, weight of roof per 20 ft. length, 7 tons,
and weight of each column 2 tons. Lastly, take the weight of roof
girder, framing, and corrugated sheeting for one side of building per
20 ft. run as 2^ tons.

If the connection of the roof with the columns be very stiff and
rigid, the lateral wind pressure tends to tilt the structure as a whole,
with the result that the pressure on the windward columns is reduced,

184 STRUCTURAL ENGINEERING

and that on the leeward columns increased. Let R = upward reaction
beneath column A. Take moments about B —

R x 40' + 6 x 13' -f 1*5 X 29' = 16 (total wt. tons) x 20' + 3
(vertl. compt. of wind) x 30'

.*. R = 7 '02, say 7 tons.

With a firm attachment at C, the column will bend as shown in
Fi£. 136 A, with reversal of curvature at P, situated at one half the
height of the column above the base plate, or at a somewhat less height
varying with the depth of the roof truss connection. As the exact
position of P is doubtful of location, assume it in the highest possible
position, i.e. 12 feet above base. The total horizontal pressure applied
above this level = (§| of 1*5) horizontal component of wind reaction
at 0 + (Jf of 6) proportion of wind pressure on face of building,
applied above P = 4-31 tons. This pressure is transferred to P, where
it acts as a concentrated load applied at 12 feet above the base. The
wind load on the face below level of P = J§ of 6 = 2-73 tons applied
at 7 feet above the base. The dimensions of the base plate and
spacing of bolts are shown in Fig. 137. As the strip of plate between
the leeward bolts L, L, and outer edge 00, is relatively weak, it
appears reasonable to take moments about LL in considering the uplift
on the windward bolts W, W. Let T = tension in each windward
bolt, then, taking moments about LL,

2T X 2J' + 7 tons X lj' = 4'81 X 12' + 2-73 X 7',

whence T = 13 -4 tons. Employing a working stress of 8 tons per

13*4
sq. inch, sectional area = -g— = 1-67 sq. in., and diameter of bolt

having this net section at bottom of thread = If in.

With the usual type of connection of V-roof s to columns, the degree
of rigidity assumed above will seldom be realized, and the roof will
tend to rack over as in Fig. 102, A. In such a case the lateral pressure
on the roof will be applied at the head of the column instead of at P,
and the pressure on the face of the structure will all be applied half-
way up the covering. In cases where a fair amount of rigidity may be
attained by the use of knee braces, etc., the point P will be situated
somewhere between the head and centre of the column, but its exact
location is practically impossible.

Compound Columns. — Compound columns consisting of two or
three beam sections connected by tie-plates or bracing are largely used
for supporting crane tracks and roof in works buildings. With tie-
plates as habitually employed, the distribution of load amongst the
individual beam sections will be very imperfect, and each beam will
practically carry the load resting immediately upon it. Thus, in the
column in Fig. 95, E, the two outer beams support much heavier loads
than the central one. Tie-plates, as here shown, may only be con-
veniently riveted to the outside halves of the flanges of the outer
beams as well as to both halves of the central beam flanges, and will
therefore be of little value in transferring a portion of the outer loads
W3 and W3 to the central beam. If stiff diagonal angle or channel

COLUMNS AND STRUTS

185

bracing be employed instead of tie-plates, the load distribution will be
more equal, although still incapable of exact determination. The tie-
plate compound column is uneconomical and much inferior to a com-
pound column with continuous plates.

Live Loads on Columns. — No hard-and-fast rule can be followed
in the treatment of live loads. These are so variable in character and
effect that each case must be considered on its own merits. The most
suitable procedure will be to increase the moving load by a suitable
percentage, dependent on the character of the load", and treat the
resulting increased load as so much equivalent dead load. In extreme
cases, where columns are liable to sudden shocks, it will be necessary to
increase the live load causing such shocks by nearly, or quite, 100 per
cent. Where the live load is more gradual in rate of application a
lower percentage will serve, but its estimate must be the result of
careful judgment. In designing columns for supporting crane girders,
the crane load should invariably be doubled.

Maximum Rivet Pitch in built-up Steel Columns. — If the pitch of
rivets connecting the outer plates with the main members of built-up
or compound columns be too great, there will be
risk of local failure of individual plates by buckling,
as shown in Fig. 138. In a column composed of
separate elements, such as two or more plates con-
nected by angles or channels with other rolled
section bars, the strength of the column will be
represented by that of its weakest component.
The outer plates of built-up columns possessing a
smaller radius of gyration than the other com-
ponent sections, are the most liable to fail by local
buckling, whilst from their position in the cross-
section (farthest from the neutral axis) they are
further subject to the maximum intensity of stress
due to both bending and direct compression. The

tendency to buckle of the local unsupported 3T \.

lengths of plate, such as ab, will be most marked FIG. 138.

in the case of short columns very heavily loaded, or
longer columns subject to relatively heavy lateral loading. In both
these cases the maximum allowable compression of, say, 9 to 10 tons
per square inch may be approached, and the vertical pitch I of the

rivets must be such that the ratio - does not exceed the value corre-

sponding with this working stress ; r being the radius of gyration for
the thinnest plates employed. Again, a length of plate, such as ab, can
be very little stronger than a round-ended column, since the only
fixation tending to constrain it to act as a fixed-ended column is that
derived from the grip of the outer edges of the rivet heads, whilst the
plate is considerably weakened across the section of the rivet-holes,
especially where the rivets may be four or six abreast.

The ratio - for a safe buckling stress of 24,000 Ibs. per square inch
for mild steel columns with rounded ends is 65 (from Euler's formula),

186 STRUCTURAL ENGINEERING

and r for f in. plates = 0-108 in., for J in. plates = 0144 in., and for
f in. plates = 0'18 in.

Hence, maximum safe pitch of rivets in outer plates—

f" thick = 65 x 0-108" = 7'02" say 7"
£" „ = 65 X 0-144" = 9-36" „ 9f
f" „ = 65 X 0-180" = 11-70" „ llf

Plates f in. thick will seldom be employed in compound columns. In
order to keep well within any possible risk of local failure, a maximum
pitch of 6 in. may be adopted for f in. plates, 5 in. for J in. plates, and
4 in. for f in. plates, since other influences, such as the transverse pitch
of the rivets, also enter into consideration. That these values will be
amply safe is also borne out by the results of tests of columns in which
failure has taken place by local buckling of the plates. In practice,
4 inches is very commonly adopted as the rivet pitch for any ordinary
thickness of plate.

CHAPTER VI.

PLATE GIRDERS.

PLATE girders consist of a combination of plates and angles riveted
together in the manner shown in Fig. 139. The web plate is
necessarily continuous from end to end of the girder, but may vary in
thickness. The flanges are usually composed of a pair of flange angles
continuous for the full length of the girder, and one or more plates,

FIG. 140.

not necessarily extending the full length of the girder. Different
forms of flanges are occasionally used, especially in America, and
examples of such are given in Fig. 140.

Economic Limiting Span. — The principal difference between plate
and lattice girders lies in the construction of the web. The continuous
webs of plate girders require no complicated and expensive connections
with the flanges, but for very large girders the depth of web necessitates
special stiffeners, or very thick web plates, which add considerably to
the weight and cost of the girders. These considerations limit the
span for which plate girders may be economically used to about
100 feet, above which some form of lattice girder would be more
economical.

Depth. — The stresses in the flanges, and consequently the flange
area, vary inversely as the depth of the girder, but the sectional area of
the web plate increases rapidly with an increasing depth of beam. The
present practice, which has been found to give economical proportions
of flanges and web, is to make the depth of plate girders equal to ^ to
•J3 of the span, according to the purpose for which they are to be used.

Breadth of Flange. — If girders be unsupported laterally, a sufficient
width of flange must be adopted to resist any side pressure that may be
imposed on the girder. No general rule can be given for the flange

187

188

STRUCTURAL ENGINEERING

width, on account of the very varying lateral forces to which the
girders may be subjected ; each case must be considered for its special
conditions of loading. To prevent local buckling, flange plates should
not project more than 2 to 3 inches beyond the flange angles, or they
K A should be stiffened by knee web stiffeners at frequent

intervals.

Estimated Weight. — The weight of plate girders
depends upon the system of loading, the unit
stresses adopted, and other factors, so that no
accurate general formula can be deduced, but the
weight as obtained from the formula given in
Chapter II. may be used as a first approximation.

Sectional Area of Flanges. — In Fig. 141 the
modulus figure for a plate girder is shown, and
demonstrates the small part played by the web and
the vertical legs of the angles in resisting the bend-
ing stresses. Except for very light girders, the re-
sistance to bending of the web and vertical legs
of the angles is neglected, and the flange plates and horizontal legs of
the angles made sufficiently strong to resist the whole of the bending
stresses.

Let D = distance between centres of gravity of flanges.
/ = maximum intensity of stress in the flanges.
A = total area of each flange.

Since the thickness of the flange is very small compared with its
distance from the neutral axis, the stress in the flanges may be con-
sidered as equally distributed over the whole of the flange section.
The flange area may then be considered as concentrated at its centre of
gravity, and the effective, depth of the girder equal to the distance
between the centres of gravity of the flanges.

The moment of inertia of the flanges

FIG. 141.

Ix is so small as to be negligible, and the moment of inertia may be

AD2
considered = —^— -

AD2

2
=/AD

The required net area of each flange may therefore be obtained
from

B.M. =/AD

A _ B.M.
A -

PLATE GIRDERS

189

D must be measured in the same units of length as the bending
moment.

EXAMPLE 29. — To find a suitable section of flange at the middle of the
span for a plate girder 60 feet span, 5 feet deep, carrying a distributed
load of 2 tons per foot run, the stress not to exceed 8 tons per square
inch.

The bending moment at the middle of the span

= 60 X 2 X 60

8
= 900 ft.-tons

The net area of flange required
900

5X8

= 22-5 sq. in.

Suppose a flange width of 18 inches and 4" x 4" x }" angles be
adopted. The area of the horizontal legs of the angles, allowing for
| in. rivets,

= (8 - 2 x ft) X j
= (say) 3 square inches.

The area of plate will therefore

= 22*5 — 3 = 19'5 square inches.
19-5

Thickness of plate

18 - 2 x

= 1-2 in.

This thickness may be made up by using two f in. plates, or two
f in. plates, and one J in. plate, as in Fig. 142.

Thickness of Web ^Plate. — It is usual when designing plate girders to
assume that the web resists the whole of the vertical shearing force, if
the flanges be parallel. The action of the stresses in the web is a
matter of great controversy, and various methods have been suggested

H°Z Fl flales 13

FIG. 142.

FIG. 143.

for calculating the required thickness of the web. The shear in the
web is accompanied by tensile and compressive stresses acting at right
angles to each other, and in parallel girders, at 45° to the ^ vertical,
Fig. 143. The web plate along ab is therefore in compression, and
may fail in a similar manner to a long strut.

Let t = thickness of plate.
/ = length along ab.

190

STRUCTURAL ENGINEERING

The least radius of gyration

>-=x/^= —

V 12 X t Vl'2

The ratio

The ends may be assumed fixed, and the safe intensity of stress
obtained from the curve in Fig. 109. The actual intensity of stress
along the line ab is equal to the intensity on a vertical section of the
web due to the vertical shear at the section, i.e. the vertical shear
divided by the sectional area of the web. If the stiffeners of the
girder be placed closer together than the depth of the girder, the length
I will be reduced to the distance between the stiffeners measured along
a line at 45° with the vertical.

Another, and perhaps the most widely adopted, method of design-
ing the web is to limit the intensity of shear stress on the vertical
plane to a fixed safe value. The thickness of the web will by this

method = -^y

/«L'

where S = vertical shear at the section ;
D = depth of web ;
/, = safe intensity of shear stress.

/, may vary for mild steel between 2-0 and 2'5 tons per square inch.

To allow for corrosion the minimum thickness of web plate should
be f in. Theoretically, the thickness of web plate should increase
towards the supports, but the maximum thickness is usually employed
throughout the length to overcome the difficulty of flange connection.
In cases where the web changes thickness, packings must be placed
under the flange angles.

Stiffeners. — To strengthen the web against buckling stiffeners are

riveted to the girder at intervals
along its length. The usual forms
of stiffeners are shown in Fig. 144.
a is a single angle stiffener used to a
large extent in America, but seldom
employed in this country. A tee
stiffener as at c is the more usual
English type, and has the advantage,
when used at a web joint, of being
riveted directly to both portions of
the web. At d is shown a gusset
stiffener, used where the flange has
a large overhang beyond the flange
angles. It serves to support the plates
of the tension flange and reduces
FIG. 144. the buckling tendency of the plates

in the compression flange. It is also

very effective in stiffening the girder laterally. The stiffeners a, ft, and c
are carried over the vertical legs of the flange angles either by placing

PLATE GIRDERS 191

packing plates equal in thickness to the flange angles between the web
and stiffeners or by joggling the ends of the stiffeners an amount equal
to the thickness of the flange angles. The former method adds
material to the girder, but, except in cases where the flange angles are
very thick, is less costly than the second method.

Although formerly the spacing was much greater, it is now cus-
tomary to space the stiffeners at distances not exceeding the depth of
the girder. The modification of the stresses in the web due to the
introduction of stiffeners is at present very uncertain, but there is no
doubt that stiffeners spaced at distances greater than the depth of the
girder, do, to a certain extent, stiffen the girder ; the closer spacing,
however, conforms better with the theory of the principal stresses in
the web. In deep girders the joints in the web plates will necessitate
covers and stiffeners at intervals of less than the depth, and in no case-
should the spacing exceed 5 feet.

Pitch of Rivets in Flange Angles. — It has already been proved that
the intensity of horizontal shear along any horizontal plane of a beam

AYS
is equal to — r-, and the total stress for one foot length of the beam,

12 AYS
supposing the intensity to remain constant, = — j- — . The rivets

through the vertical legs of the angles must resist the horizontal shear
stress between the flanges and the web. The moment of inertia of the
flanges may be assumed = 2AR2

where A = the area of each flange ;

R = distance of the centre of gravity of the flanges from the

neutral axis
= half of the depth of the girder

= — = Y, in the above formula.
Substituting these values for I and Y—

Ax -X S

12AYS_ .,/ 2

2Ax(-

the depth D of the girder is measured in inches. If D be measured in
feet, the horizontal shear per foot length = jy, that is = the average

vertical shear per foot depth.

Let R = resistance of one rivet.

S
Then the number of rivets required per foot length = =jyp.

R will be the resistance of one rivet in double shear, or the bearing
resistance of one rivet in the web plate or angles, whichever is the
smaller. It is obvious that the pitch of the rivets through the flange

192

STRUCTURAL ENGINEERING

FIG. 145.

plates would, by calculation, be greater than the pitch of the rivets

through the web, but for practical reasons the same pitch is adopted.

Lateral Bracing. — "When used
in bridge construction plate
girders require stiffening laterally
to resist the wind pressure. Two
examples of the methods of stiff-
ening are shown in Fig. 145.
The floor in a sufficiently stiffens
the lower flanges of the girders,
and brackets c, riveted to the webs
and cross girders at intervals, re-
sist the overturning action of the
wind pressure. 1) is an example
of a bridge having the main
girders under the rails. The
stiffening in this case consists of
the floor on the top of the girders,
and a system of diagonal angle
bracing on the lower flanges,

and diagonal angle bracing between the girders.

EXAMPLE 30. — Design of plate girders for 25-ton crane, 60 feet

span.

Let it be supposed that the specification requires the following

working stresses and particulars to be adopted.
Working stresses not to exceed —

Tension .... 7

Compression ... 7

Vertical shear in web 2J „ „

Rivet shear ... 5 „ „

Bearing pressure .8 „ „

Double shear to be taken equal to 1} times single shear.

Flanges. — The tension flange to be calculated on the net section,
i.e. the gross section of the plates and horizontal tables of the angles
minus the area of the greatest number of rivet holes in any trans-
verse section, and the area of any other rivets that may occur within
2J inches of such section. The diameter of the rivet holes to be taken
as — inch larger than the nominal diameter of the rivets. Where the
flanges are parallel, the compression flange to have the same gross area
as the tension flange.

jfgj. — The minimum thickness of web to be f inch. The shearing
resistance of the web to be calculated on the gross area of the web at
any vertical section.

Depth. — The depth of the girders at the centre of the span to be not
less than -^ the span.

Let Fig. 146 represent the effective depths of the main girders.
The loads supported by the girders consist of —

(1) Load to be lifted

(2) Dead load of crab

(3) „ „ girders.

tons per square inch

PLATE GIRDERS 193

The load raised (25 tons), when lifted suddenly, will exert a pressure
on the girders far exceeding its actual dead weight, and in an extreme

case equal to double its dead h eo* ^

weight. It is therefore neces- .3- -j ^ 1

sary to increase the dead load

lifted to the probable force it

would, at any time, exert on FIG. 146.

the girders. In the present

case, 100 per cent, will be added for this dynamic action, making the

equivalent dead load on the crab equal to 50 tons. The weights of

crabs vary, and the weight would, in an actual design, be obtained after

the crab had been designed. Seven tons being a probable weight for

such a crab, will be adopted. The axles of the crab will be assumed

to be at 5 feet centres. The total load on each wheel will then be

—T — = 14 J tons. An estimate of the dead weight of the girders

may be obtained from the formula given in Chapter II. Dead weight

, . , WL

of girder = — ,

where L = span of girder in feet,

W = an equivalent distributed load producing a maximum
bending moment equal to the maximum bending
moment produced by the live loads.

In this case the wheel loads may be assumed concentrated at the
middle of the span, when the maximum bending moment would —

= WL = 2 X 14j X 60

T 4

= 427'5 ft.-tons.

The equivalent distributed load to produce a similar bending
moment —

= WL _ W x 60 _ 427.5

8 8

/. W = 57 tons.

The approximate weight of each girder will therefore

- WL _ 57 X 60
530 530

= 6-4 (say) 6J tons.

The bending moment diagrams may now be constructed by the
methods illustrated in Chapter III.

In Fig. 147 a is the curve of bending moments for the assumed
distributed weight of the girder, b is the curve of maximum bending
moments produced by the axle loads, c is the curve of total bending
moments obtained by adding together the ordinates of a and &.

The horizontal flange stress at any section is obtained by dividing
the bending moment at the section, as scaled off from the curve c, by

194

STRUCTURAL ENGINEERING

the effective depth of the girder at the section. The horizontal stress
thus obtained will be the direct stress for the horizontal portions of the

25 30 J5

FIG. 147.

60 fJ

The direct stress in the inclined portions of the lower flange
will be to the horizontal stress as the inclined length is to the horizontal
length. If ab, Fig. 148, represent the horizontal stress at any section
of the flange, and 0 be the angle of inclination of the flange, then the
inclined stress will be represented by the length ac,

i.e. the inclined stress = horizontal stress x ^

= horizontal stress x sec. 0

The vertical component be of the inclined stress is part of the
vertical shear at the section resisted by the flange. The web in such
a case will only be called upon to resist the maximum vertical shear at
the section minus the shear be resisted by the flange. The horizontal

FIG. 149.

and inclined stresses in the flanges at sections 5 feet apart are tabulated
in the following table, columns 4 and 5. Column 7 of the same table
shows the vertical shearing force resisted by the inclined portion of the
lower flange at such sections.

The maximum vertical shearing force diagram, Fig. 149, has been

PLATE GIRDERS

195

constructed by the method in Chapter III., and the vertical shearing
forces at the different sections are tabulated in column G of the following
table.

Distance
from end.

Effective
depth.

Maxiomm
B.M.

Horizontal
flange
stress.

Inclined
flange
stress.

Maximum
veitical
shear.

Vertical
fchear taken
by flange.

Vertical
shear taken
by web.

ft.

ft. in.

ft.-tous.

tons.

2 0

—

30-56

—

30-56

2 9

125

45-5

45-9

27-54

6-82

20-72

3 6

241

68-8

69-5

24-65

10-32

14-33

4 3

330

77-6

78-4

21-57

11-64

9-93

5 0

393

78-6

(79-4

18-85
18-85

11-79

7-06
18-85

5 0

430

86-0

—

15-96

—

15-96

5 0

441

88-2

—

13-06

—

13-06

The two sets of figures given at the 20-feet section are for the
stresses immediately to the left and right of the section.

A diagram of the stresses in the flanges may now be drawn from
the data given in columns 4 and 5.

Fig. 150 is a diagram for both flanges ; the left-hand half represents

toofats

OS M IS 20 £S 30 ZS 20 IS 10 S O

FIG. 150.

the stresses in the upper flange, and the right-hand half the stresses in
the lower flange.

00. O

At the centre the required flange area = —=- = 12'6 sq. inches.

Suppose a width of 10 inches be selected for the flange, and
4" x 4" x i" angles, and |" rivets be adopted. The net area of the
horizontal tables of the angles = 2(4 - {f)i = 3'125 sq. inches.

196 STRUCTURAL ENGINEERING

The net area of the flange plates must therefore = 12-6 — 3*125
= 9-475 sq. inches.

The net width of plates = 10 - 2 x jg = 8£ in.
The total thickness of plates therefore —

= »J«? = 1-16 in.
8-125

This thickness may be made up by using either one ~ in. and one
| in. plate, or two f in. and one yg in. plate. For the upper flange the
former sections will be adopted, and for the lower the latter. The area
of the flanges is slightly in excess of the theoretical requirements, as the
exact area cannot be obtained with practical sections. The plates
should be arranged in order of thickness, the thickest being placed next
the angles, and the thinnest on the outside.

The stresses in the different plates and angles are as follows : —

2 angles = 3'125 x 7 = 21-875 tons.

f in. plate = 8*125 X f X 7 = 35-54 „
A „ = 8-125 x T7e X 7 = 24-88
•fs „ 8-125 x T9e X 7 = 32-00 „

f 8-125 x | X 7 = 21-32 „

As the flange stress decreases towards the ends of the girder, a
proportionate reduction of flange area could be made. It would,
however, be impracticable to have a gradual change of section, or many
small changes, and so the same sectional area is maintained until the
stress has diminished to such an extent that the outermost plate may
be discontinued. The positions where the changes of section may take
place are readily obtained from the flange stress diagram, Fig. 150.

Set out ab = the stress in the angles

be = „ ,, | in. plate

Through #, c and 6? draw horizontal lines cutting the curve at/? and /.

At the section I the total stress in the flange = ac* which is equal to
the resistance of the angles and the f in. plate. Between I and the end
6f the girder the ~ in. plate is unnecessary, and may be discontinued.
In practice the plate is continued for about one foot beyond the section
at /. At the section through p the f in. plate could be stopped, but it
would be inadvisable as the lateral resistance of the girder would be
seriously reduced. The lengths of the f in. plates on the lower flange are
found in a similar manner on the right-hand portion of Fig. 1 50. For
parallel girders, where the effective depth is constant, the bending moment
diagram is also, to a different vertical scale, the flange stress diagram,
and may be used for such to obtain the lengths of the flange plates.
The lengths of the flange plates will be-

in. in. ft. in.

top flange 10 x f x 60 0

10 x & X 47 8
bottom flange 10 x r§ X 60 0 (horizontal projection)

10 x | X 51 3

10 x g X 42 4

Plates exceeding 40 feet in length are charged extra per foot of

PLATE GIRDERS

197

COM.
fi/ate

FIG. 151.

length over 40 feet, and it is more economical to employ cover plates
when the length is much greater than the ordinary rolling limit. In
the present case the plates 60 feet and 51 feet 3 inches would be
jointed, the extra on the other plates exceeding 40 feet in length not
being equal to" the cost of covers Joints in the two flanges, or in the
flange angles, should not be in the same vertical section. A convenient
position for the joint in the f- inch top flange plate
would be about 5 feet on the left of the centre.

The cover plates should have a strength equal to
that of the jointed plate, and must be connected to
the flange by rivets having an equal resistance to that
of the cover. The section of the top flange is shown
in Fig. 151, and it will be seen that the only available
position for the covers is under the top tables of the
angles. The width of covers will be 3J inches, and,
deducting the rivet area, the section = 2(3J — ~) x thickness.

The maximum stress in the f inch flange plate = 35*54 tons. There-

35*54

fore the thickness of covers = - --- -T-QI -- f->\ =

7 X 2(6-2 — fj)^

The shearing resistance of a | inch rivet in single shear = 3 tons.
The number of rivets required at each side of the joint will

therefore = — ^ — = 12.

The bearing resistance of the rivets in the f" plate =12 x f" X f" X 8
= 52 '5 tons, or far above the requirements.

The pitch of the rivets = 4 inches (calculated later).

The length of the covers will therefore be 4 feet.

Suppose the ~ inch plate in the bottom flange to be jointed 5 feet
to the right of the centre. A ~ inch plate riveted to the under side of
the flange will be equal to the strength of the plate jointed.

The maximum stress in the plate = 24*88 tons.

k) 4 .00

The number of rivets in single shear required = ^ = 9.

As the rivets are in pairs, ten rivets must be used at each side of
the joint. The bearing resistance is again in excess of requirements.
The length of the cover plate to each side of the joint will be 20 inches.
The same plate may be utilised as a cover for the joint in the f inch

mcn> or> sav> 1 incn-

PIG. 152.

PIG. 153.

plate, by making such joint 20 inches from the joint in the T7g inch
plate (Figs. 152 and 153), and continuing the cover plate the necessary
additional length.

Stress in the | inch plate = 21*32 tons. 21-32

Number of rivets required at each side of joint =
rivets in single shear.

= 8 for

—

198 STRUCTURAL ENGINEERING

Bearing resistance of rivets = 8(|xf)x8 = 21 tons.

This is slightly below the stress in the plate, but, as the bearing

21'32 X 8
pressure would only be — -^ = 8-12 tons per square inch on the

fJomf m bocfc angle 1'ivet, the Small CXC6SS

intensity may be al-
lowed. The additional
length of cover plate
to the right of the joint
will be 16 inches.

The joints in the flange angles are covered by bent plates or
wrappers as in Fig. 154.

The stress in each angle = 10*937 tons.

The thickness of the covers = ^p— 2 — US = °'35 in<
f inch bent plates would be used.

TVT ^ t ' • • i v- • j 10'937

Number of rivets in single shear required = — - — = 4

10-937

bearing „ = 7 3 — = 5

s X s X «

The joints arid arrangement of rivets are shown in Fig. 154, and
the positions of the joints on the girders are shown on Fig. 157, in the
elevation.

Web Plate. — The maximum vertical shear on the web occurs at the
ends of the girder, where it is equal to 30*56 tons. The section of the
web will be = 24 in. x thickness of web.

Since the allowable shearing stress = 2J tons per square inch the

net area of web required = '— j— =13-5 square inches. The thickness
will therefore = -~- = 0*56 inch. The nearest practical size being

YQ in. such thickness would be adopted. The required thickness of web
decreases very rapidly from the end to a section 20 feet from the end,
owing to the depth increasing whilst the shear decreases. At this
section the required thickness would be less than the minimum
thickness specified. The shape of the web plate renders it convenient
to have the web joints at these sections, and the central 20 feet portion
of the web may be reduced to | in. thickness. To allow of the flange
angles being kept straight ~ in. packings must be placed between them
and the f in. web.

If the covers and rivets at the web joints were designed to resist
the actual vertical shear at the joints, the thickness of covers and the
number of rivets required would be reduced to unpractical sizes. The
covers, therefore, will be of the minimum thickness, J in. with a double
row of rivets at either side of the joint. Packings ^ in. thick are
required under the inner halves of the covers owing to the change of
thickness of the web plate.

Pitch of Rivets. — The horizontal shear per foot length of girder is
•

given by the expression jy The minimum pitch will occur where the
shear is the greatest, i.e. at the ends of the girders. The number of

PLATE GIRDERS

199

rivets required through the vertical tables of the angles per foot length,
at the ends of the girder

S
DR

30-56
2 x 5-25

= (say) 3 for shear

and

, -_„— o = (say) 4 for bearing in web.

8 A 10 A O

A pitch of 3 in. is therefore necessary at the ends. As the pitch is
inversely proportional to jy it would increase towards the middle of the

span, being at 5 feet from the end equal to 4'7 in. and at the centre
equal to 12 in. For practical reasons the pitch is kept constant through-
out the length of the girder or has a practical minimum number of
changes. In the present case it would be advisable to have a pitch of
3 in. for a distance of 5 feet from either end, and the remaining portion
pitched at 4 in. From Fig. 150 it will be seen that the rate of increase
of stress in the inclined portions of the lower flange is, for all practical
purposes, equal to the rate of increase of the horizontal stress, and
therefore the pitch of the rivets in the inclined flange will be made the
same as for the upper flange.

Stiff eners. — 6" x 3" x f" tees will be used as stiff eners and spaced at

intervals of 5 feet. Packings ^ in. thick will be placed between them

e joggling th
Rail. — A 70 Ibs. bridge rail riveted to the flange at 1 ft. 3 in. and

and the web to save joggling the tees over the flange angles.

1 ft. 4 in. pitches will be used for the crab to travel on.
To check the Assumed Weight of Girder.

No.

Description.

Section.

Length.

Total length.

Weight
per ft.

Weight.

in. in.

ft. in.

ft.

Ibs.

Flange plate

10 X f

60 0

21-25

1,275

j J

10 X T9S

47 8

47-67

19-13

912

10 xj

60 0

14-38

892

10 X |

51 3

51-25

12-75

653

j j

10 XI

42 4

42-3

12-75

540

covers

8* X 1

4 0

11-9

10 X^

4 8

4-67

14-88

Flange angles

4 X 4 X \

60 0

120-0

12-75

1,530

4 X 4 X &

60 3

120-5

12-75

1,536

,, covers

3i X 3£ X i

3 0

8-45

102

packings

4XA

20 0

1-13

Web plate

60 x|

20 0

76-5

1,530

(60to24)X196

20 0

80-33

3,213

covers

12 X I

4 4

17-3

15-3

265

packings

6X&

4 4

17-3

1-92

stiffeners

6 X 3 X i

4 11

49-17

11-0

585

9 J

4 2

16-67

}

183

3 5

13-67

150

2 8

10-67

117

packings

6X$'

4 4

43-33

10!2

442

3 7

14-33

146

2 10

11-33

116

i y

2 1

8-33

14,559

Rivets, say 5 per cent. . . . 728

Total weight . . . .
= 6-83 tons.

15,287

200

STRUCTURAL ENGINEERING

The assumed weight was therefore 6'83 —6*5 = 0'33 ton too small ;
but such a small difference in weight would not produce stresses
warranting any change of the designed sections.

End Girders or Cradles. — The ends of the main girders are supported
on lateral girders, to which are fixed the wheels for the longitudinal
motion of the crane. The general arrangement of connections is shown
in Fig. 157. The main girders are carried through to the web CD of
the end girder, the web AB being in three parts, each connected to the
main girders by vertical angles. The maximum loading of the end
girders will occur when the crab, fully loaded, is at the end of its travel.
It may be assumed that the whole of the weight of the crab and load is
then carried by the adjacent end girder. The load from each main
girder will be half the weight of one main girder, plus half the weight
of the crab plus half the pressure due .to the load lifted

= i(6'83 + 7 + 50) = 31-91 tons (say) 32 tons.

Fig. 155 is a diagram of loading, bending moments, and shear forces.
The weakest section of the end girder will be where the flange plates
stop, 18 in. from the centre line of the main girders.

32 forts 32 fans

I —

•• — ^-+-- —

\^/

|s hv

B.M. Diagram \

5.F. 'Diagram.

FIG. 155.

f'X

•t II
FIG. 156.

Bending moment at that section = 32(2' 9" - 1' 6") = 40 ft. -tons.
Suppose the section, Fig.^156, be assumed for the end girder. The
modulus of section = 182'3 inches3.

The maximum stress =

= 2'6 tons Per square inch.

This is considerably less than the allowable stress, but the sections
being the minimum no reduction can be made.

The maximum shear intensity will occur at the section through the
wheel axles. The shear on each web = 1C tons. The average shear

1 n

intensity = f 26-25 -^4-7513 = (sa^ ^ ^ons Per S(luare
being the diameter of the axle hole.

A f in. stiffening plate is riveted to the web, to which the wheel
bearings are attached.

Connections. — The load from the main girders will be equally
divided between the two webs of the end girders. The rivets at y will
be subject to a vertical shear of 16 tons.

PLATE GIRDERS

201

202

STRUCTURAL ENGINEERING

number of rivets required

1 £>

in double shear = v -e = 4
5'25

in bearing = 7 9

I X ft X 8

The rivets Y' (Figs. 157 and 158) will be subject to a horizontal
shearing force, due to the bending moment to be resisted by the con-

FIG. 158.

nection, in addition to the vertical shear. The rivets at Z will be put
into shear by the bending action, and the couple formed by the shear-
ing resistance of these rivets at the upper and lower flanges of the end
girder will act in opposition to the bending moment.

Resistance of rivets at Z to shear = 4 x 3 = 12 tons.
Moment of resistance = 12 x 26 '25 = 315 in.-tons.

The bending moment at the connection = 16 x 33" = 528 in.-tons.
The bending moment to be resisted by the rivets at Y'

= 528 -315 = 213 in.-tons.

Let/, = the horizontal shear stress per sq. in. on the outside rivet,

a = sectional area of the rivets.
Then the moment of resistance of the whole system of rivets

+ 2/i2 + yf + etc.). (See Chap. IV.)

---*' X 2{(|)2 + 32 + (4i)2 + 62 + (74)a 4- 102}

= 5S-7/.

213
/. /, = g-v-- = 8 '9 6 tons per square inch.

Total horizontal shearing force on the outside rivets

= 3-96 X 1'2 = 4-75 tons.
Each rivet will be subject to a vertical shearing stress of — = 1*23

PLATE GIRDERS 203

tons, due to the vertical shearing force on the section. The actual
shearing force will be the resultant of the horizontal and vertical forces

= V^-To2 +T232 = 4-9 tons.

The resistance of one |in. rivet in double shear = 5'95 tons. There-
fore the shearing resistance of the system is sufficient to transmit the
bending moment.

The force of 4*9 tons on the outside rivet would produce a bearing

4*9

stress of 7,, &, = 14-9 tons per square inch, which far exceeds the

t * I
safe bearing pressure. The bearing area may be increased by riveting

an extra f" plate to the web. The bearing pressure would then be
reduced to 7'45 tons per square inch.

The horizontal intensity of shear stress on rivet (6)

= — x 3-96 = 2-97 tons per square inch

Horizontal shear on (6)

= 2-97 x 1-2 = 3-56 tons.
Resultant shear on (6)

= 3'5tfa + 1-232 = 3-77 tons.
Bearing stress on the web without the bearing plate

Q -rT 7

= / M- = 11-49 tons per square inch

S X 8

Bearing stress on the web and the bearing plate = 5'75 tons per
square inch.

The bearing stresses for the remaining rivets on the web and bear-
ing plate will be found to be —

Rivet (5) = 4'72 tons per square inch.
„ (4) = 3-29 „
„ (3) = 2-6 „
„ (2) = 2'1 „
3, (1) = 1*87 „ „ „

If pb be the bearing stress of any rivet on the web and bearing
plate, the stress transmitted to the bearing plate = f " X |" X pb. The
bearing plate must be riveted to the web by a system of rivets whose
resistance is equal to the pressure transmitted to the bearing plate.

Total pressure transmitted to the bearing plate

= | X | X (2(7-45 + 5-75 + 4'72 + 3'29 + 2'6 + 2'1) + 1-87}
= 17 '3 tons.

Bearing resistance of one rivet

= X X 8 = 2-62 tons.

204 STRUCTURAL ENGINEERING

The number of rivets required

_ 17-3 _ „
2-62

The lower half of the group of rivets at Y will be in tension due
to the bending moment of 213 in. -tons.
The moment of resistance of the system

°i0* 2 : -{(I)2 + & + (W + <W

= 172-8/,

213
.*. ft = -.Ca.o = 1*23 tons per square inch.

The maximum tension in the rivets is therefore well below the
working stress.

EXAMPLE 31. — Design of Plate Girder Railway Bridge.— Span
80 feet. The bridge to carry a double track of rails, ballast, and plate
flooring carried by longitudinal rail bearers and cross-girders supported
by two main girders, Figs. 160 and 160A. The maximum live load to
consist of locomotives covering the span, the heaviest axle load being
19 tons, and the driving axles 8 feet apart. Working stresses to be
fixed by the Range Formula, Table 24, Chapter II.

Longitudinal Rail Bearers or Stringers. — Assume the cross -girders
to be spaced at 8-feet intervals. Then the maximum stress in the
flanges of the rail bearers will occur when the 19-ton axle load is at
the centre of the span of 8 feet.

Load on each wheel = 9-5 tons.

B.M. at centre = 9'5 X 8 = 19 foot-tons.

The distributed load on each stringer due to track, ballast, flooring,
and own weight, amounts to nearly 2-5 tons (see below), and the
additional B.M. due to this load

= 2'5 x § = 2-5 foot-tons.

Total B.M. = 19 + 2-5 = 21'5 foot-tons.

It should be noted this is an outside estimate of the bending
moment, since the concentrated wheel load is to some extent distributed
over the stringer through the agency of the rail, sleepers, and ballast.
The extent of this distribution cannot however be accurately calculated,
as it depends on the relative stiffness of the rail and stringer, positions
of adjacent axle loads, and other factors.

Assume the depth of stringer as 1 ft. 6 in.

Total flange stress = ^ = 14-33 tons.

1 '0

2'5
Flange stress due to dead load = j^ = l'G7 tons.

1 • A 7

Percentage of dead load stress = X 100 = 1T6

PLATE GIRDERS 205

The working stress from Table 24 for 11-6 per cent, of dead load stress
= 5 tons per square inch. The flange area required = ,° = 2*87 sq. in.

Using 4" x 3J" x J" angles and | in. rivets,

Area of horizontal tables of angles = (8-2 Xyf) X | = 3-06 sq. in.

From Table 24, the coefficient for estimating the dead load
equivalent to 89 per cent, of moving load is 1*9.

Hence, equivalent maximum shear when the axle load is entering
or leaving the span = (9*5 x 1'9) -f 1-25 = 19*3 tons.

Adopting a web plate J in. thick, the average shear stress on the

19 '3

web = — — - = 2-15 tons per square inch,
lo X 0*0

A somewhat thinner web would be strong enough, but it is desirable
to make the webs of stringers and cross-girders of about equal thick-
ness, in order to ensure the same length of life against corrosion.

Weight of one stringer.

2 Flange angles 4" x 3J" x f X 8' 0" = 190 Ibs.

2 „ „ 4" x 3i" X i" X 13' 0" = 310 „

1 Web plate 18" x J" X 8' 0" = 245 „

2 Tee stiffeners 5" x 3" x f X 1' 6" = 29 „
2 „ packings 5" x f X 10" = 15 ,

789 „
Rivets, say 3 per cent. = 24 „

Total =813 „

Cross-Girders. — The span of the cross-girders equals 27 feet ; the
distance between the centres of main girders.
Dead load on each cross-girder —

Weight of ballast @ \ ton per foot run . . . . 8,960 Ibs.

rails = 4 x f @ 86 Ibs. per yard . . 917

„ sleepers = 6 @ 125 Ibs. each . . . 750

„ chairs, etc., 12 @ 50 Ibs. each . . . 600

Weight of permanent way . . 11,227

asphalte, 24' x 8' x H" 3,600

floor plating = 24' x 8' X T7/ . . . 3,360

„ fender plates, say 300

Assume weight of cross-girder = 2 J tons . . . 5,040

Total distributed load on cross-girder .... 23,527 ,,

= (say) 10 '5 tons.

A portion of the above weight is actually applied by the stringers
as concentrated load at the junctions with the cross-girders, ^but the
error due to considering it as distributed load is very small, since the
live load is relatively large.

The weight of the stringers imposes four concentrated dead loads
= 813 Ibs. each.

206 8TBUGTUBAL ENGINEERING

Live load on cross-girders —

Four concentrated loads of 9*5 tons each.
The loading on the cross-girder is indicated in Fig. 159.

=§1 -ai *! =2!

sjS jsjS §S jsj>

^~mmmfm^^^lohldi^^loa^^j^^wn^^^^'^^^^^^^

j— - 27-0' — - H

FIG. 159.

Bending moment at centre due to —

10*5 X 27
Distributed dead load = - -1— - - = 35-44 ft.-tons.

813x2x10-5-813x5
Concentrated dead loads = rnrr - = 5 "8 ft.-tons.

Concentrated live loads = 9'5 x 2 x 10-5 — 9'5 x 5 = 152 ft. -tons.
Stress in flanges, assuming an effective depth of 2 ft. 6 in., due to —

•ri 11 j ?>5'44 + 5-8
Dead load = - - = 1G'5 tons.

2'5

1 ?»9

Live load = 4nr = GO'S fcons.

2*0

Percentage of dead load stress = gn.g i 1^ X 100 = 21-3 per cent.
Working stress from Table 24 = 5- 7 tons per square inch.
The area of flange required = -- — =r= — - = 13-56 sq. in.

Assume a flange width of 14", 4" x 4" x i" angles and f" rivets.
Area of horizontal tables of angles = (8 - 2 x yf)£- = 3-06 sq. in.

-| o • f /* O «/\ />

Thickness of plates = -r-r- ~^r*\ = 0'86 in.

i-t — ^ x i^)

Say two plates, y and f" thick for the lower flange. For the upper
flange, one plate 15" x J" and one 9" x J" will be used, leaving a 3"
margin on each side to which to rivet the floor plates.

The coefficient to obtain the equivalent of the live load hi terms of
the dead load is, from Table 24, = 1 G7.

The maximum shear is therefore —

from live load = 19 x 1*67 = 31-73 tons.
from distributed load = 5*25 „

813 x 2
from concentrated dead loads = = 0'72 ,,

Total shear at ends = 37*70

PLATE GIRDERS 207

Assuming the thickness of web to be J", the average intensity of

orf ,r*/\

shear stress in the web = ~ - ^ = 2'51 tons per square inch.

Designing the web as a fixed-ended strut of length equal to the
distance between the lines of rivets through the flange angles, measured
along a line inclined at 45°-

length of strut = (30 - 4J)V2 = 36'06 in.
least radius of gyration = /.—.-

I = 86'06 = 249-5
r 1

2\/l2

Safe intensity for this ratio, taken from the curve of safe loads on fixed-
ended struts (Fig. 109), is 5000 Ibs., or 2-23 tons per square inch.
Hence J inch thickness for the web is insufficient.

The average intensity of shear stress on a ^ inch plate = 2 -24 tons
per square inch. The safe intensity = 2*67 tons per square inch.
A T9g inch plate will therefore be adopted.

Main Girders. — The dead and live loads may be assumed as uniformly
distributed on the main girders.

Dead weight per 8-foot length —

Permanent way ............ 11,227 Ibs.

Asphalte .............. 3,600 „

Floor plating .... ........ 3,360 „

Fender plates ............. 300 „

4 stringers .............. 3,252 ,.

1 cross-girder (from calculated weight as designed) . 5,000 „

26,739 „
Weight per foot run .... 3,342 „

Dead load per foot run per girder ...... 1,671 „

0-75 ton.

The equivalent distributed live load for a span of 80 feet may be
taken at 2 tons per foot run (see Table 22).
The estimated dead weight of each girder

= WL _ (0-75 + 2)80 x 80

510 ~ 510

= 34-5 tons.

Bending moment at centre due to

dead load = (0'76 X 80)80 + 84-5 x 80

= 945 ft.-tons.
live load = *XJ

8
= 1600 ft.-tons.

208 STRUCTURAL ENGINEERING

Assuming the effective depth of the girder = 8i- ft.
The flange stresses due to

dead load = — = 111-2 tons.
8*5

live load = — = 188-2 tons.
•8*5

1 11*2
Percentage dead load stress = Hi -2 + 188"? x 10° = 37<1 Per cent<

Working stress from Table 24 = 7 tons per square inch.

999-4
Flange area required = •"'„ = 42*77 sq. inch.

Assume a breadth of flange of 24", 6" x G" xj" angles, and |" rivets.
Area of horizontal tables of angles = (12 — 4xj|)J = 4'125 sq. in.
Thickness of plates required in tension flange
= 42-77- 4-125

24 - 4 X if

Say three J in. plates and one f in. plate, or three f in. plates. For
the compression flange the rivet holes need not be deducted.

Area of horizontal tables of angles = 12 x J = 6 sq. in.

A O •rfr7 (*

Thickness of plates = — - = 1*53 in.

2-t

One f in. and two J in. plates may be used.

Thickness of Wei.— Assume a web joint at each 8 feet section along
the girder, with double angle and plate stiffeners at the joints and
intermediate tee stiffeners. The horizontal length of unsupported web
will be 2 feet.

The length of the web column = 24" x *J*2 = 34 in.

The shear at the ends of the girders

due to dead load = 0-75 X 40 4- 17-25 = 47'25 tons.
„ live load = 2 x 1'44 x 40 = 115'2

Total = 162-45 „
Assume a ^r in. web plate.
Average intensity of shear in web

= T7v ^ o = 2-83 tons per square inch.
102 X jg

Least radius of gyration of plate

= -4l== 0-162
Vl2

5- A—

Safe stress for this ratio from Fig. 109 = 6400 Ibs. = 2-85 tons per
square inch.

PLATE GIRDERS

209

FIG. 160.

210

STRUCTURAL ENGINEERING

In a girder of this size the web thickness might be reduced towards
the middle of the span with an appreciable saving in weight.

Pitch of Rivets in Main Angles. — Maximum shear per foot of depth

of web = — 3-r— = 19'1 tons. Number of rivets required per foot run

8*0 jg.j

of angles, at 5 tons per rivet in double shear = — — = 4. A3 in.

single pitch or 6 in. double pitch is suitable, the latter being adopted.
The bearing stress on the web plate, with this pitch, is 9'7 tons per
square inch. A 5 in. double pitch would reduce the bearing stress to
8 tons per square inch.

..-°. Effective Span BO ft

Plan of Flange Riveting.

Abutment
Masonry.

! O O D i

!5o o o <b !; o o i

jo o o;oj

mo! o o o !OMO; o <j

Ins. 12 e o /t P 3 4 s

6 7 a Ft

FIG. 160A.

Web Joint. — The first vertical joint in the web occurs at J-J,
Fig. 160A, 4 feet from the bearing pin.

Shear at J-J = fg X 162-45 = 146'2 tons.

PLATE GIRDERS

211

FIG. 161.

212 STRUCTURAL ENGINEERING

Number of rivets in double shear required on each side of joint

Bearing resistance of one f inch rivet in ^ inch plate, at 8 tons per
square inch

= i X & X 8 = 3-94 tons.

Number of rivets required to limit bearing on web to 8 tons per
square inch

146-2

= ^94 =

Pitching the rivets in the web covers at 4J in. provides 39 on each
side of the joint. Web cover plates 14" x J" may be used.

The lengths of flange plates will be readily deduced by the method
of the previous example.

The detailed arrangement is shown in Figs. 160 and 160A. Fig.
160 is a half cross section of the bridge and part longitudinal section
showing the connection of stringers to cross-girders. Fig. 160A shows
the detail at end of girder. The over-all length is 82 feet 6 in. and
pin bearings are placed 80 feet apart for carrying the structure. An
expansion roller bearing similar to that in Fig. 196 would be employed
under one end. The face of the abutment between the main girders is
built up close to the end cross-girder at A-A, and the ballast carried
over the gap by a short length of floor plate P. The fender plates are
omitted in Fig. 160A.

Where the end cross-girder cannot be placed close to the end of
the main girders, additional rail bearers carry the floor and track
between the end cross-girder and the abutment, their outer ends resting
on bed stones built into the abutment.

Plate Girder Deck Bridge. — Fig. 161 is a half cross-section and part
longitudinal section at the middle of the span of a plate girder bridge
of 60 feet span. The girders, three in number, are placed under the
track and so spaced that an almost equal load is carried by each girder.
Angle bracing placed at 5 feet intervals stiffens the girders laterally
against wind pressure. The floor is formed of steel troughing, the
overhanging ends of which are supported on the bottom flanges of
lattice parapet girders.

CHAPTER VII.

LATTICE GIRDERS.

Types of Lattice Girders. — Girders having open-work webs consist-
ing of ties and struts are classed generally as lattice girders, in contra-
distinction to plate girders with continuous webs. They are
constructed in various forms and often referred to by distinctive
names, according to the arrangement of the lattice bars and uniformity
or variation of depth. Fig. 162 shows the more usual types employed.

XKNNIXIXJTI/ITK

xxxxxxxxxxxx

/
/K

FIG. 162.

No. 1 is the Linville or N-girder, called in America, the Pratt truss.
The vertical members are struts and the diagonal ones ties. No. 2, the
Howe truss, has the diagonals reversed, so that they become struts,
whilst the verticals are in tension. The Howe truss is seldom con-
structed entirely in steel but is more suitable for composite girders

213

214 STRUCTURAL ENGINEERING

having the sloping struts and upper boom of timber and the vertical
ties and lower boom of steel.1 The N-truss is preferable, since the
shorter members are in compression. No. 3 is the double N, also known
as the Whipple truss, formed by inserting additional verticals and
diagonals midway between those of No. 1. The advantages resulting
from this arrangement are the reduced length of the segments of the
upper boom or chord, thereby reducing the tendency to buckling and
enabling it to ba made of lighter section ; lighter compressive stresses
in the vertical struts and shorter spacing of floor beams or cross-girders,
thereby requiring shorter, and consequently lighter, longitudinal
girders or bearers beneath the road or railway. Types 1 and 3 are
frequently built with sloping ends, as indicated by the dotted lines.
Nos. 4 and 5 are respectively singly and doubly braced hog-backed
lattice girders, often referred to as lattice- bow girders. Generally
speaking, they are used for larger spans than the parallel girders with
resulting economy in weight of material, principally due to the shorter
length of the vertical struts near the ends of the girder where the
compressive stresses are greatest. The curved upper boom also relieves
the verticals of a proportion of the compressive stress, by resisting part
of the shear, whilst in the parallel types of girder, the horizontal upper
boom resists the direct stress only and takes practically no part in
resisting the vertical shear. Nos. 6 and 7 are two forms of the
Baltimore truss, an almost exclusively American type. The normal
outline is that of an N-girder, with additional members V, V, known as
sub-verticals, inserted midway along the main panels. These serve as
suspenders for intermediate floor or cross-beams and so achieve the
same result as the double system of bracing in No. 3. The stresses in
the sub-verticals may be transferred to the main panel points either by
ties T, T, as in No 6, or by struts S, S, as in No. 7. No. 8, known as
the Pennsylvania truss, is also of American origin and aims at combin-
ing the advantages of the Baltimore and double-N trusses by employing
sub-verticals V, V, together with that of the lattice-bow type by
possessing a greater depth at the centre than near the ends. In large
span trusses, additional members shown by dotted lines are frequently
added for the purpose of stiffening the main struts S, S, near the centre of
their length. These dotted members, however, have no part in resist-
ing the primary stresses in the truss.

Nos. 9 and 10 are respectively the single and double Warren girders,
both ties and struts being inclined, usually at angles varying between
60° and 45° with the horizontal. Vertical members as shown by the
dotted lines are occasionally inserted in No. 9 for the support of
intermediate loads or for stiffening the segments of the upper boom.
No. 11, known as the multiple lattice type, although formerly largely
employed for main girders of long span, is now almost confined to
parapet girders and main girders of foot-bridges. It possesses several
systems of bracing, the bars of which are riveted to each other at the
intersection points, the intention being to increase the rigidity. The
effect, however, is to render it impossible to estimate at all accurately
the stresses in the various systems. When constructed with flat bars
only at close spacing, vertical stiffeners S, S, of T or channel section
1 Mins. Proceedings Inst. C. E., vol. cxxviii. p. 222, Plate 5.

LATTICE GIRDERS 215

are necessary, and the girder approximates closely in character to a
plate girder, but involves more workmanship (see Fig. 198). Nos. 12
and 13 are respectively upright and inverted bowstring girders. In
girders of this type, the stress in the booms is nearly uniform through-
out, whilst the stresses in the web bracing is also much more uniform
than is the case in parallel girders. The web members may therefore be
made of equal section with very little sacrifice of economy. Bowstring
girders are usually built with crossed diagonals in every panel, in
which case the diagonals are designed for resisting tension only. If
built with a single diagonal in each panel, the diagonals must be capable
of resisting both tension and compression if the girder be required to
carry a travelling load. No. 14 is a modification of the bowstring
girder known as the Pauli or lenticular truss. Relatively few important
spans have been bridged on this principle. It possesses the same
general advantages as the bowstring type.

The essential points of difference between lattice and plate girders
are as follow. In the lattice girder, the various members are arranged
so that each is subject to stress only in the direction of its length, and
the arrangement of the joints should be such that the members are
further subject to direct tension or compression only. The direct
bending stress, which in a plate girder is resisted mainly by the flanges,
acts as direct compression or tension in the booms or chords of a lattice
girder. The compression flange of a plate girder being attached to the
web at short intervals of a few inches only, as determined by the rivet
pitch, is less liable to buckle than the segments of the compression
boom of a lattice girder, which are unsupported for the whole panel
width. The stresses set up by the shearing force in a plate web are
transferred to the flanges along innumerable lines in the web, whereas
in a lattice girder these stresses are localized and constrained to act as
direct tensile or compressive forces along a few well-defined lines,
represented by the axes of the ties and struts. The load may be
applied to a plate girder at any number of points along its length, but
in a lattice girder it should only be applied as a number of concentrated
loads at the panel points or intersections of the bracing bars with the
booms. Any application of load between the panel points causes local
bending of the boom segments, which must be of correspondingly
stronger section to resist such bending in addition to the direct
compression coming upon them. Curved members are economically
inadmissible in lattice structures, since the direct stress acting through
their ends sets up more or less bending moment on the central section,
the amount depending on the sharpness of the curve given to the
member. The upper booms of hog-backed lattice girders are often
curved in outline to avoid the more complex joints and plate profiles
which result from a polygonal outline. The curvature in such cases is,
however, small, and the increment of stress due to bending correspond-
ingly small also, whilst the girder has a neater appearance than if of
polygonal outline. The common practice of introducing curved members
in lattice structures, especially in roof trusses, presumably for aesthetic
reasons, is, however, bad construction, and cannot be too strongly
condemned.

Spans of Lattice Girders. — Considerable divergence is met with as

216

STRUCTURAL ENGINEERING

regards the length of span for which any particular type of truss B
most suitable. Plate girders are rarely used beyond 100 ft. span, ard
about 80 ft. may be said to be their practical economic limit. The
various types of parallel lattice girders with single system of bracing
are generally employed for spans of from 80 to 150 ft., although in
roof construction, light Warren and N-girders of from 20 to 50 ft.
span are in very general use. Parallel and lattice bow girders with
double systems of bracing are necessary for larger spans from about
150 to 350 ft. and upwards. In a few instances these last have been
erected over much larger spans, notably the Kuilenberg bridge over me
river Lek in Holland of 492 ft. span, the Ohio river bridge of 519 ft.
span, and the Oovington and Cincinnati bridge over the Ohio riier
having one span of 550 ft., and two of 490 ft. Bowstring girders
have been widely employed for spans of from 80 to 300 ft., whilst
the lenticular trusses of the Saltash bridge have a span of 455 ft.

Stresses in Braced Girders. — For the purpose of computing the
stresses in braced girders, they may conveniently be divided into two
classes — parallel girders, and girders of varying depth. The stresses
in any parallel braced girder may be readily written down from a
consideration of the shearing force for the loading and span in question.
Thus, in Fig. 163, let AK represent an N-girder of eight panels carry-

E +16 f +IS G +12 H +7 K

> INN

/ {.

/* '

zl*

^ 2 2 2 2 -IS Z -12 2 -7 2 0

FIG. 163.

ing a load of 2 tons at each lower joint. The reaction at each support
is 7 tons, and the upper figure shows the shearing force diagram for the
span. The shearing force is constant, and equal to 7 tons from A to B.
It then falls to 5 tons across the panel BC, 3 tons across CD, and 1 ton
across panel DE. Since the function of the bracing bars is to resist
the shearing force, these shears of 7, 5, 3, and 1 tons will be the vertical
components of the stress in the four diagonal ties from A to the centre
of the span. These figures may be at once written down in a vertical
position across the ties in question, the minus sign indicating tension,
since the effect of the shearing force is to cause tensile stress in the
diagonals which slope downwards from the ends towards the centre of
the span. If the ties be assumed to slope at 45°, the horizontal com-
ponent of the stress in each tie will be equal to the vertical component,
and on the right-hand half of the girder the corresponding horizontal
stress in each tie may be written down, taking care to write the figures
horizontally. From these the stresses in the segments of the upper

LATTICE GIRDERS

217

and lower booms are readily obtained by summation. Thus, the com-
pression in KH = 7 tons, caused by the horizontal tension or pull in
the diagonal K7a. HG receives a thrust of 7 tons directly from KH,
and an additional compression of 5 tons from the tie H#, or a total
compression of 12 tons. Similarly, GF receives the thrust of 12 tons
from HG, plus 3 tons from the tie G/, giving a total compression of
15 tons. Finally, the compression in FE = 15 + 1 = 16 tons. The
loading in this case being symmetrical, the stresses on the left-hand
side of the centre will be similar to those on the right.

In the lower boom the stress in hk is nothing, since the vertical K&
has no horizontal component, gh resists the horizontal pull of 7 tons
exerted by Kh. fg resists the combined horizontal tensions in H# and
gh = 5 + 7 = 12 tons, and e/1 resists the horizontal tensions in/G and^.
The compression in vertical Dd is 1 ton caused by the downward pull
of 1 ton in the tie De. In Cc the compression is 3 tons due to the
tension in tie Cd. Similarly, the compressions in B& and Act are 5 and
7 tons respectively. Finally, the direct or actual stresses acting along
the sloping direction of the ties is equal to the horizontal tension
X ratio of slope to horizontal length of the tie, which for an inclination
of 45° = V2. Thus, the real or direct tensions in the ties are —

eF =

= 5^/2, and hK = 7^2 tons.

Generally, it is unnecessary to draw the shear force diagram, since
the shearing force in each panel is readily estimated by successively
subtracting the panel loads from the reaction at either end of the
girder. In the subsequent examples the shear diagram is therefore
omitted.

If the above girder be inverted, it becomes a Howe truss. The
diagonals are then in compression, and the verticals in tension.
Assuming the same loads, the stresses are as follow (Fig. 164). The

A +15 +12 +7

2 2 -16 2 -15 2 -12 2 -7

FIG. 164.

loading being the same, the shearing force in each panel is as before,
but the sign of the stresses in the diagonals is changed to plus to
denote compression. Inserting the corresponding horizontal stresses
in the diagonals on the right-hand side, and summing up the boom
stresses as before, it will be seen that the lower boom stresses are equal
in amount to those of the upper boom in the previous example, whilst
those in the upper boom equal those in the lower boom in the previous
case. The stresses in corresponding vertical members are a little higher
than before, but the end verticals, shown dotted, may be omitted, since
the final shear of 7 tons is transmitted directly to the abutment through
the inclined end strut AB, Fig. 164. The middle vertical here acts as
a suspender for the central load of 2 tons, and is in tension to that

218

STRUCTURAL ENGINEERING

amount. The stresses in the diagonals are the same as before, but
being compressive instead of tensile, and the diagonals being the longer
members, their sectional area, and consequently their weight, will be
appreciably greater than that of the vertical struts in the N-truss under
similar conditions of span and load. It is principally this feature
which places the Howe truss at a disadvantage as regards economy of
material.

Stresses in an N-truss of Seven Panels carrying a Load of 6 tons
at each Lower Joint. Ties inclined 45°. — In Fig. 105 the total load is

G +36 H +36 L +3O

+/&

"•o o'

\ f

S? «vj

*Vl \Q

«o o

/*.

0 -6

-12

-18

B 'X

D '\

?/ \K

666 -36 6 -SO 6 -18 6 O

FIG. 165.

36 tons, which, being symmetrically disposed, gives a reaction or shearing
force of 18 tons at each end of the span. The compression in the end
vertical is therefore 18 tons. This member pressing upwards causes a
vertical tension of 18 tons in the first diagonal. At B the diagonal
AB exerts an uplift of 18 tons on the lower end of strut BO, but the
load of 6 tons at B is pulling in the opposite direction, and so reduces
the compression in BO to 18 —6 = 12 tons. This upward pressure of
12 tons in BC stretches the second diagonal OD with a vertical tension
of 12 tons. CD then exerts an uplift of 12 tons on the lower end of
strut DE, which is again reduced by the 6 tons load at D, giving a net
compression of 12 — 6 = 6 tons in DE. This compression in DE
causes a vertical tension of 6 tons in tie EF, which just balances the
load of 6 tons at F. There is therefore no stress in FG-, as might be
expected, since there is no diagonal member attached at G to carry the
stress forward. If the shear-force diagram be drawn for this case of
loading, it will exhibit no shear in the panel GH, thereby indicating no
stress in any diagonals which may be inserted in that panel. In an
actual girder as built, two diagonals, FH and GK, would be employed,
but these will suffer no stress under a symmetrical system of loads such
as under consideration. If, however, one half, say the left-hand half
of the girder, be more heavily loaded than the other, the tendency of

the load will be to distort the
girder, as shown in Fig. 1C 6, when
the diagonal FH would be put
into tension and GK into com-
pression. Similarly, if the right-
hand half be the more heavily
loaded, GK will be in tension and FH in compression. The duty of
these central diagonals is therefore to make the girder capable of
carrying an unsymmetrical system of loads, such as most usually occurs
in practice. In Fig. 165 the horizontal stresses are written against the
corresponding diagonals on the right-hand half of the girder, from
which the boom stresses are readily obtained. The stress of 36 tons
in HL is transmitted directly through GH, there being no increment

LATTICE GIRDERS

219

applied by the diagonal FH. As before, the direct stresses in the sloping
bars are obtained by multiplying the horizontal stresses by *J 2 or 1'414.

Stresses in an N-truss loaded with 16 tons at each Lower Joint
and 2 tons at each Upper Joint. Diagonals inclined 50° with
Horizontal. — This example represents more closely the case of a
practical girder having the flooring of a bridge attached to the lower
joints, whilst a proportion of the dead weight of the girder with
overhead bracing is applied directly at the upper joints.

It should be noticed (Fig. 167) the loads on the two end lower
joints are neglected, since they are applied immediately over the

2 2 2 2 2 +120-9 2 +113-4 2 +9O-7 2 +52-9 2

\ !

? a

i \H-

N. ,
«Vj 55

1 V»-

?v?

-7-5
/

-22-7

-37-8
/

-52-9
A

B \

D \

N \

16 <6 & « -113-4 K -9Q-7 & -S2-9 16 O

FIG. 167.

supports and do not give rise to any stress in the members of the
girder. The total load is 130 tons, giving a vertical reaction or shear
of 65 tons at each support. The conapressive stress in the end strut is
therefore 65 tons. Two tons of this is caused by the 2 tons load at A,
leaving a net uplift of 63 tons as the vertical tension in tie AB. At
B the tie AB exerts an upward vertical pull of 63 tons, which alone
would cause a similar compression in strut BO. The downward acting
load of 16 tons at B, however, reduces this to 63 —16 = 47 tons for
the compressive stress in BC. Two tons of this is again due to the
load at C, leaving 45 tons vertical tension in CD. By successive sub-
traction of the loads, the vertical stresses in the other members are
obtained as indicated on the left-hand side of Fig. 167. The vertical
EF here suffers a compression of 2 tons due to the load at E, which is
transmitted from E to F, and thence as tension to the two ties HF and
KF. At joint F, the downward acting forces are 2 -f 16 = 18 tons,
whilst the tie HF exerts an uplift of 9 tons, giving
18 —9 = 9 tons as the vertical tension in tie FK
necessary to maintain equilibrium. The diagonals in
this case being inclined at 50° with the horizontal, the
horizontal stress in any diagonal will be less than the
vertical stress. In Fig. 168 the diagonal CD is shown
separately. The horizontal stress in CD will bear the
same ratio to the vertical stress as the length BD

does to the length BO, or = ]g, whioh in a trianSle of

50° = 0-84. /. horizontal stress = vertical stress X 0'84, and multiply-
ing the vertical stresses already written down on the left-hand side of
the girder by 0*84, the corresponding horizontal stresses are obtained

as indicated on the right-hand half. The fraction g^ = 0'84 may be

obtained directly from trigonometrical tables, it being = cotangent 50°,
or by scaling off BD and BC from the elevation of the girder and

FIG. 168.

220

STRUCTURAL ENGINEERING

dividing the one by the other. From the horizontal stresses in the
diagonals, the stresses in top and bottom booms are readily obtained by
summation as before. The direct stresses in the ties are obtained from
the vertical stresses as follows. In Fig. 168 the ratio of direct stress

01)
(i.e. acting along the slope CD) to vertical stress in diagonal CD = ^3,

which for 50° = 1-8.

direct stress

_. ,. ,

= 1-8' Or direct streSS = Verfcl0al StreSS X

Multiplying the vertical stresses in the diagonals by 1'3, the direct
stresses are 63 x 1'3 = 81-9, 45 x 1'3 = 58-5, 27 X 1'3 = 35'1, and
9 x 1*3 = 11*7 tons tension in AB, CD, LN, and HF respectively.

Stresses in N-girder Unsymmetrically Loaded. — Fig. 169 repre-
sents an N-girder of 8 panels with ties at 45°. The upper and lower

40 12

FIG. 169.

joints carry loads as indicated. Such a condition of loading would
correspond with the case of a main bridge girder when the rolling load
extends from the left-hand abutment up to and loading panel point F.
The joints B, D, and F then carry both dead and live load, and the
remaining lower joints dead load only, composed of part weight of
main girders, cross-girders, flooring, and permanent way. The loads
on the upper joints represent part weight of main girders and weight
of over-head wind bracing.

The reaction at left-hand end of span

= f (half upper loads) + (J + S + f ) X 40 + (| + f + f + 1) X 12

= 117-5 tons. Total load = 25 (top) -f 168 (bottom) = 193 tons.

.'. Reaction at right-hand end = 193 — 117'5 = 75'5 tons.

Compression in left-hand vertical = 117'5 tons. Two tons of this
is due to the load at A, .*. vertical tension in AB = 115'5 tons. Sub-
tracting 40 tons at B, compression in BC = 75'5 tons. Tension in
CD = 75-5 - 3 = 72-5. Compression in DE = 72-5 - 40 = 32-5.
Tension in EF = 32-5 -3 = 29'5. At F a load of 40 tons is carried,
of which 29-5 tons is supported by the tie EF, leaving 40 -29'5 = 10'5
tons tension in vertical FG, which, for this position of the rolling load,
acts as a tie instead of a strut. At G the vertical FG exerts a down-
ward pull of 10-5 tons on the end of tie GH, which is augmented by
the load of 3 tons at G, giving a vertical compress ion of 10'5 + 3 = 13'5
tons in GH, which now acts as a strut instead of a tie, provided it be
built of a suitable section to resist compression. The 3 tons load at K
is transmitted to H, causing a compression of 3 tons in KH. At H
the downward forces are the 3 tons thrust in KH, 13'5 tons in GH,

LATTICE GIRDERS

221

and the 12 tons load at H, all tending to produce tension in HL and
totalling 28'5 tons. At L the 3 tons load and vertical tension of 28-5
tons in HL both produce compression in LM = 31*5 tons. Similarly,
tension in MN = 31-5 -f 12 = 43'5, compression in NO = 43'5 + 3
= 46-5, tension in OP = 4G'5 + 12 = 58-5, compression in PQ = 58*5
+ 8 = 61-5, tension in QR = 61'5 -f- 12 = 73-5, and lastly, compression
in RS = 73-5 -f 2 = 75 '5 tons, which agrees with the previously
determined reaction at S.

With diagonals at 45°, the horizontal stresses in the diagonals are
equal to the vertical stresses. In Fig. 170 these are written against

+115-5 +188 +217-5 +2O4 + 2O4 +175-5 +132 -¥73-5

\
-115-5

-72-5

-29-5

-28-5

-43-5

-58-5

-1/5-5

-188

-217-5 -I75-S -f32

FIG. 170.

-735

-73 -5

the corresponding bars, from which the top and bottom boom stresses,
as indicated, are readily obtained. The direct stresses in the diagonals
are equal to the horizontal stresses multiplied by v% the ratio of
inclined length of tie to horizontal breadth of panel. It will be
noticed that the stress in diagonal GH is compressive. This is due to
the rolling load being in the position which develops the maximum
compression in GH, which compression is more than sufficient to
neutralize the permanent tensile stress set up in GH by the dead load
alone. The tie GH therefore undergoes a reversal of stress when the
rolling load passes this position. A similar reversal would take place
in the tie HL under the action of a rolling load advancing from the
right. Usually, the diagonal ties consist of a pair of flat bars, Fig.
193, which are incapable of resisting compression, and in such cases
the panel GFHK would be counter-braced by inserting a second tie,
FK, sloping in the opposite direction to GH. The stresses in the
panel GFHK are then as shown in _ j 5 5 3

Fig. 171. The main tie GH, being-
incapable of resisting any compression,
falls into a state of no stress and
becomes relatively slack. The upper -
3 tons load at G is transmitted down
GF as compression, whilst at F, the
downward forces being 43 tons, and upward pull of tie EF 29-5 tons,
the difference, 18*5 tons, determines the vertical tension in the counter-
tie FK. This, increased by the 3 tons at K, gives a compression of
16-5 tons in KH,and adding the 12 tons at H,^the vertical tension in
HL = 16-5 + 12 = 28-5 tons, as previously obtained in Fig. 169.
From L to the right-hand support the stresses are then the same as
already determined in Fig. 170. For this position of the rolling load
the stress in the other counter-tie KM is zero, it only coming into
action when the rolling load extends from the right-hand abutment up
to and loading joint M. The girder may be built without counter-ties,
provided the members GH and HL are constructed of a suitable

A
/

SM1>

40 12

FIG. 171.

222

STRUCTURAL ENGINEERING

section for resisting both tension and compression, and this is occa-
sionally done, although greater economy is realized by inserting counter-
ties. It should be noticed the tension in the counter-tie is the same in
amount as the compression which would be produced in the main tie
of the same panel if not counter-braced.

Stresses in N-truss with Double System of Bracing.— Suppose the
girder in Fig. 172 to be loaded with 2 tons and 8 tons respectively at

2+90 2+90 2 + 85 2+75 2+6O 2+4O 2

fit
\

-IS

6 8-33 &-7S 8-60 8-40 8 -IS S O

FIG. 172.

each upper and lower joint. The N-truss, with double system of
bracing, may be regarded as consisting of two separate girders, super-
posed as indicated by the light and heavy lines in the figure. Con-
sidering the heavily-lined system first, the total load is 40 tons on tbe
lower joints -f- 14 tons on the upper joints, counting the two end loads
of 2 tons as carried by this system. The load being symmetrically
disposed the reaction at each end for this system alone is 27 tons. The
vertical stresses are written down on the left-hand side in the same
manner as those in Fig. 167, giving, for the bars taken in order from
left to right, + 27, - 25, -f 17, - 15, -f 7, and - 5 tons, and for the
central vertical -f 2 tons. At the central lower joint the 8 tons load
4- 2 tons compression from the central vertical make up 10 tons, which
is resisted by the vertical tension of 5 tons in each of the two
central ties.

Considering the lightly-lined system, the total load is 48 tons on
lower joints + 12 tons on upper joints = 60 tons, giving a reaction at
each end of 30 tons. An additional compression is set up in the end
vertical, since this member is common to both systems, the total
compression in the end vertical being, therefore, 27 + 30 = 57 tons,
which of course equals half the total load on the complete girder.
The vertical stresses are written down as before, taking care not to
deduct the end upper load of 2 tons, since this has already been
allotted to the other system. These stresses are, from left to right,
4- 30, - 30, + 22, - 20, + 12, - 10, and + 2 tons. Assuming the
diagonals to slope at 45°, the horizontal components of stress are equal
to the vertical in all the ties excepting the two end ones. These have
an inclination of one horizontal to two vertical, and therefore the
horizontal stress for these two members equals Jialf the vertical stress.
The summation of the boom stresses is effected as in the previous
examples, the end section of the upper boom receiving a compres-
sion of 25 -h 15 = 40 tons, applied by the end ties of both systems,
since both are attached at the upper end of the boom. The
direct stresses in the 45° diagonals equal the horizontal stresses
X V2. For the direct stress in the two end diagonals, proceed as
follows—

•VJ
V

\
\

LATTICE GIRDERS 223

inclined length of diagonal _ ^5
horizontal length of diagonal 1

Hence the inclined or direct stress = horizontal stress x >J 5

= 15 X \f5 = 33-54 tons.

Double system N-girders occasionally have the end members
disposed as in Fig. 173. Assuming this to be a modified arrangement
of the previous girder, the number of panels z+ss *+ eo 2+75 2
and loads remaining the same, the stresses in the
end panel will be altered as follows : AC now
acts simply as a suspender for the 8 tons load
at C, and AB being reversed, acts as a strut
and transmits the vertical pull of 20 tons in -is a-is a-4o B
diagonal A8, + 8 tons in AC, together with the pIG> 173.

2 tons load at A, making 30 tons in all, to the
abutment at B. The remaining stresses in the web are unaltered. The
segment EA of the upper boom resists the horizontal pull of 25 tons
in EF. At A the segment AG receives a pressure of 25 tons from EA,
a horizontal pressure of 15 tons from BA, and resists the horizontal
pull of 20 tons in the tie A8, making 60 tons compression in all.
From B to F the lower boom resists the outward horizontal pressure
of 15 tons in strut AB. F8 resists the combined pull of 15 tons and
25 tons in FC and FE respectively. The remaining boom stresses are
unaltered. The advantage of this arrangement is a simpler junction
at E, where three members only, instead of four, have to be joined.

Note. — In the N-girder as usually constructed, it will be noticed in
Figs. 163, 165, 167, 170, and 172, that the end segment of the lower
boom is not stressed under the action of vertical loads. This segment,
however, is always stiffened in such a manner as to be capable of
resisting compression as well as tension. It comes into action in cases
where application of the brakes on a train is made during its passage
over the bridge. The longitudinal racking tendency thus created
compresses the end segment of the lower boom in advance of the train,
and puts into tension the end segment in rear of the train, the opposite
effect resulting when a train passes in the reverse direction. Heavy
end pressure, due to wind or the momentum only of the rolling load,
also stresses these members. The amount of stress produced may be
roughly calculated from the weight of rolling load and coefficient of
friction, but it is not capable of exact determination. The end
segments of the lower boom are, in practice, usually carried through of
the same cross-section as obtains in the second panel, and in girders
having lower booms of vertical plates only, these are braced together to
form a lattice box strut. As this appears to be a satisfactory
provision, the longitudinal stress is evidently not excessive.

Stresses in Warren Girder with Symmetrical Load. — Fig. 174
represents a Warren girder with single system of bracing, carrying
10 tons at each lower, and 2 tons at each upper joint. The total load
is 62 tons, giving a reaction of 31 tons at each end. The vertical
stresses are written against the web members on the left-hand half of
the girder, commencing with 31 tons in the end strut, and subtracting
each load in order. The corresponding horizontal stresses on the

224 STRUCTURAL ENGINEERING

right-hand side are obtained as follows. In any inclined bar AB,
Fig. 174A,

+ 108 + SS. +*0 !V

Z 2 2 -J3 2 JS 2 J5 2 A

A A A N HAL K HAG F E/|\D

J o, o> V N \. _S_ + 2. .12 +19 _29 + 3L

/? u A i\ £ u J* •& •& ^ vJ Js

/ V V V Q v p v o \

10 10 IO _I03 10 _79_ 1O _3/

FIG. 174. Fig. 174A.

horizontal stress __ BC
vertical stress AC

If the bars be assumed inclined at 60° with the horizontal,

BQ 1

AC = V3

, , , vertical stress

.-. horizontal stress = — -.— -

V«>

Dividing the vertical stresses by V3> the horizontal stresses are obtained
as indicated. The end segment F of the upper boom receives a

horizontal thrust of —7- tons from the strut D and an additional

Vo
29
thrust of -j^ tons caused by the tie E pulling from the opposite side

of the vertical Y, making -^ -f -^ or -^ tons in all. Segment
K receives the direct thrust of - / -y tons from F H — ~ from G H — ^

V o Y o Y o

96
from H, or a total of —r$ tons. Similarly the compression in

N = -^r- from K + -^from L + -Vrj from M = —r- tons.

V'J V" V «' V«^

In the lower boom, the outward horizontal thrust of —^ tons in

V«5
31
D produces a tension of — TTT in 0. Segment P resists the direct pull

V o

31 29

of ~~ tons in 0 + horizontal pull of — in E + outward thrust of

19 79

-jr> in G, or a total of -j~ tons. Similarly the tension in segment

79 17 7 103

Q = 7^ + 73 + 7a = 7^

Finally, the direct stresses in the inclined bars are obtained by
doubling the horizontal stresses, since for an inclination of GO0, the
inclined length AB = twice the horizontal length BO. The resulting

direct or inclined stress in D therefore = -rt x 2 = 35'8 tons ; in E,

LATTICE GIRDERS 225

33-5 tons ; in G, 21'9 tons ; in H, 19'6 tons ; in L, 8'1 tons, and in
M, 2-9 tons.

Stresses in Warren Girder with TJnsymmetrical Load. — Fig. 175,
A, represents a singly braced Warren girder with a greater intensity of

222222

A/\ ,/X, ~^\ ^7\~^7\ "7\
„/* V* \/ \/ \/ \/T \,

18 18 18 6 6 T

+ C& +J3* +144 +J£ +64

•tt VJ v3 J3 v3

B A A A A A /\

+ 4A _42 +25. _23 +£_ _3. _J5, + _/7_ _23_ + 25_ _ _$L +32.

«. «^s «^« *^« «^« ^ *^

FIG. 175.

load covering the left-hand portion of the span, up to and including the
central lower joint. The reaction at X = J x 12 tons (on upper joints)
-f 18(§ + | + f) -\- G(§ -f £), being the proportions of the loads on
the lower joints borne by X, and = 45 tons. Total load = 78 tons,
/. reaction at Y = 78 — 45 = 33 tons. Commencing at X, the vertical
stresses are written down against all the inclined bars from X to Y,
since in this case the stresses will not be symmetrical. The vertical
stress in the last inclined strut at Y must, of course, equal the reaction
of 33 tons at Y, which affords a check on the calculation. The
corresponding horizontal stresses are inserted in Fig. 175, B, from
which the boom stresses are readily obtained by summation. As before,
the direct stresses in the inclined bars are obtained by doubling the
horizontal stresses for an inclination of 60°.

Stresses in Warren Girder with Double System of Bracing and
Unsymmetrical Load. — Fig. 176 represents a double Warren girder

\ /

+17 -13

'X?

&" "5^

Nxs/

*0 *0

*X'

^. K/F\

X !

-17 2 -41 2 -49 2 -45 2 -33 2 ~/3

FIG. 176.

with unequal loading on the upper joints, and uniform load on the
lower joints, the lattice bars sloping at 45°. The girder possesses two
distinct systems of bracing, AHKLM, etc., and ABCDE, etc. Con-
sidering first the system AHKLMNOP, the reaction at A due to the
loads carried by this system = 6 (at H) -f- J(3 x 2) at K, M, and 0
+ § X 12 (at L) + J x 6 (at N) = 19 tons. Hence reaction at G due
to these loads = 33 - 19 = 14 tons. The vertical stresses in the
diagonal bars of this system are therefore as indicated. Considering

226

STRUCTURAL ENGINEERING

system ABCDEFG, reaction at A = i(2 + 2) at 0 and E + f X 12
(at B) -f J x 8 (at D) 4- £ x 6 (at F) = 17 tons, and reaction at G due
v to these loads = 30 - 17 = 18

ooz •* — * tons. The vertical stresses

for this system are 4- 17, — 5,
4- 3 tons, etc. The inclina-
tion being 45°, the horizontal
stresses
stresses.

9SI-SI-

9£/-

PO/+

91- PO/-

91-

9U-

941-91-
' 8P-

9PZ +

P9Z

equal the vertical
Writing these down
horizontally on the respective
bars, the stresses in the hori-
zontal booms readily follow.
The direct stresses in the lattice
bars are obtained by multiply-
ing the horizontal stresses by
V2. It will be noticed that
the member LM, which under
a symmetrical load would be
in tension, is put into com-
pression to the extent of
1 x V2 = 1'41 tons, under
the system of loads considered,
and would therefore require
to be designed as a strut.

Stresses in Baltimore
Truss with Unsymmetrical
Loads. — Fig. 177 indicates a
Baltimore truss of eight main
panels, subdivided into sixteen
panels by the sub-verticals
supporting the intermediate
loads marked by the circles.
The loads assumed represent
fairly closely the case of a
large span girder carrying a
rolling load extending from X
up ito and loading the main
panel point P, the remainder
of the girder carrying dead
load only.

Reaction at X = (J X 7
X 8) tons, due to loads on
upper joints 4- 32(jj| + 1

4" TiT

. _u • 10 i 10 i i<>/ wvyAj°> due
to loads on lower joints = 264
tons. Total load = 464 tons,
.-. reaction at Y = 464 - 264
= 200 tons. The sub-verticals V act as suspenders for transferring the
intermediate loads (ringed) to the main joints of the girder. Each is

LATTICE GIRDERS 227

consequently in tension to the extent of the load supported at its lower
end. The tension of 32 tons in YI is shared equally by the struts A and B,
giving a vertical compression of 16 tons in each. The tension of 32 tons
in Y2 is divided between the upper half C of the main tie of the second
panel and the sub-tie D of the same panel, giving 16 tons vertical
tension in each. Similarly the other tensions are divided between the
pairs of diagonal members supporting the sub-verticals. The stresses
in the main members of the N-system may now be taken out. These
members are shown by heavy lines. Commencing at X, the upward
reaction is 264 tons, which produces a vertical compression of 264 tons
in strut A. As 16 tons of this has already been written against this
member, the remainder, 248 tons, is added on. This will also be the
vertical compression in E. The vertical F acts as a suspender for the
32 tons load at its lower end, and also resists the downward vertical
thrust of 16 tons in B, and therefore suffers a tension of 32 -f 16 = 48
tons. At the first upper joint the strut E exerts an uplift of 248 tons,
whilst ties F and C exert a downward pull of 48 -f 16 = 64 tons, which
together with the 8 tons of load, make a total downward force of
72 tons. The excess upward thrust of 248 - 72 = 176 tons must be
resisted by the vertical tension in tie C, over and above the 16 tons
tension already caused in it by the intermediate load on V2. This
vertical tension of 176 tons is written against both upper and lower
halves of the main tie C. At the second lower joint the vertical
tension of 176 tons in C is balanced by the downward thrust in strut
G + the 32 tons load, giving 176 — 32 = 144 tons compression in G.
The other panels are similarly treated. At the central upper joint the
vertical tensions of 16 and 24 tons in H and K, together with the
8 tons load cause a compression of 48 tons in L. At the central lower
joint the downward force is 88 tons, consisting of the 48 tons thrust
in L + the 40 tons load. The upward pull in M = 32 tons, leaving
88 - 32 = 56 tons to be resisted by the vertical tension in N. The
last main vertical R resists the vertical thrust of 4 tons in the last sub-
strut, and also acts as a suspender for the load of 8 tons at its lower end,
its tension being therefore 4 4- 8 = 12 tons. At the upper end of the
inclined end strut T, the downward forces are 176 tons in tie S -f 12
tons in R -f 8 tons of load on the joint, giving a vertical compression
in T of 196 tons, which added to the 4 tons due to the sub- vertical V3
gives 200 tons compression in the lower half of T, which checks with
the reaction of 200 tons at Y.

The horizontal stresses in all the inclined bars, assuming their
inclination as 45° are written on the lower diagram of Fig. 177, from
which the indicated boom stresses are easily summed up. As before,
the direct stresses in all the inclined bars equal the horizontal stresses
X V2.

Stresses in Parallel Lattice Cantilever with Symmetrical Load.
—The stresses in parallel cantilevers are readily obtained by the method
of the previous examples. Assuming the loads indicated in Fig. 178,
the vertical stresses are written down on the left-hand half of the
girder, commencing with the 60 tons load at the outer end, and
terminating with the supporting leg L, in which the compression is
100 tons, or half the total load, which is here symmetrically disposed

228

STRUCTURAL ENGINEERING

about the centre. The horizontal stresses in the diagonals (for bars at
45°) are shown on the right-hand half, from which the boom stresses
readily follow. The upper boom is, of course, in tension, and the lower
in compression. There will be no stress in the dotted diagonals of the
central panel, these coming into action under an unsymmetrical load.

2 -300 2 -2/0 2 -ISO 2 -6O 2

t /

so o'

g e

& §

tt §

& y

-90

-so

-70

-60

+ ,' Vvv

60 i.

? i

3 t

5 a

\+30ff , '

a +joo f

3 + ZtO I

} -f/3<7 «

} +6O 6O

5 ;<\

-^/

IXN

\N\

' m m Vv

/ /

FIG. 178.

The lower dotted members, if the girder be supported on a braced pier
as shown, will be required to resist longitudinal displacing forces, such
as end wind pressure and longitudinal racking force, caused by the
application of brakes to a rolling load traversing the girder. These
forces would, in the absence of members m, m, tend to rack the
structure as shown in the smaller figure.

Stresses in Parallel Cantilever with Unsymmetrical Load. — In
Fig. 179 the same cantilever is taken with greater loads on the lower
joints of the left-hand half. Proceeding as before, the vertical stresses
may be written down for both halves until arriving at the upper points
A and B. Inserting then the corresponding horizontal stresses in all the

c A B o

Z -100 2 -ZK> 2 -3*2 2 -3OO 2 -&O 2-130

-100

-114
/

-128
S

§ /

•f / 1 <0

-«. Ij

o'-~^

? -x^

-80
\

-60

W+IOO 12 +214 12+342 12+434 12

+484
H F

K L

n r'

8+30O B +ZIO 3 +130 B +6O 6O

FIG. 179.

diagonals outside the points A and B, the boom stresses are summed
up working from the outer ends towards the pier. The stress in AC is
thus found to be 342 tons of tension, and that in BD, 210 tons. Next,
the combined horizontal pulls of 210 and 90 tons in BD and BE
respectively will create a tension in AB of 300 tons. As this is
insufficient to balance the combined horizontal pull of 342 4- 142
= 484 tons exerted by AC and AG, the difference *= 484 - 300, or
184 tons, will give the horizontal tension in AF, which thus comes into
action as a tie, whilst HB remains lax. The 184 tons of horizontal
stress in AF will be accompanied by a vertical stress of 184 tons (if
inclined at 45°), inserting which, the remaining vertical stresses may be
computed. At A the downward pull on the upper end of strut
AH = 142 tons in AG + 184 tons in AF, which, with the 2 tons load
at A, make up 328 tons of compression in AH. Adding the 12 tons

LATTICE GIRDERS

229

load at H, the compression in the pier leg HK = 328 + 12 = 340 tons.
At B the downward pull of 90 tons in BE, together with the 2 tons
load at B, causes a compression of 92 tons in BF. Lastly, at F the
vertical uplift in tie AF is 184 tons, whilst the downward forces are the
thrust of 92 tons in BF + the 8 tons load at F, or 100 tons. The
resultant uplift is, therefore, 184 — 100, or 84 tons, which must be
resisted by 84 tons of tension in the pier leg FL. This leg, for the
loading in question, would therefore require to be anchored down by
foundation bolts capable of resisting 84 tons of tension.

The loads on a large cantilever would greatly exceed those assumed
above, and would result in an excessive uplift or tension in the pier
legs, according as one or the other half of the girder were the more
heavily loaded. In order to avoid such excessive uplift on the
foundation, the legs forming the supports may be placed further apart,
as in Fig. 180. The modification in the stresses will then be as follows.

-2/4 C -Z8SJ • A -24Z$ B -2IO D -ISO -6O

§ ^

^ ^

& &

/ '

i >«>\

? K

* ' \

+ '\

! '\

+ 100 / +214 +342 6
00 fe 12 ^

+342 H+Z8SJ F+24Z§
IZ IZ 88

E+2/0 +13O +6O

a a &

Moment 342 [ g>

K \SI4- Moment ^

3 r

FIG. 180.

Assuming the same loads, but placing the supports at G and E instead
of at H and F, the reactions in the two legs GK and EL are first found
by taking moments about either.

Taking moments about G, and denoting the panel width by unity,
the moment of the loads to the left of G = 100 X 3 + 14(2 + 1) = 342.

Moment of loads to right of G = CO x 6 + 10(5 + 4 + 3 + 2)
+ 14 X 1 = 514.

The excess of the right-hand moments = 514 — 342 = 172, which,
tending to pull down the portion of the girder to the right of G, will
have to be resisted by an upward thrust in leg EL, applied at a
leverage GE = 3 panel lengths.

Hence, compression in EL = —- = 57^ tons.

The total load on the girder = 256 tons, /. compression in leg
GK = 256 - 57J = 198f tons.

Commencing again with the vertical stresses, on reaching point C,
the upward thrust in GC = 198J tons in GK - the 12 tons load at
G = 186f tons. This is partly resisted by the downward pull of 128 tons
to the left of C + the 2 tons load at C = 130 tons. The difference of
W>§ — 130 = 56| tons, will be resisted by the vertical tension in tie
CH. It will be noticed the inclined member in this panel has been
reversed as compared with the previous example. If the member AG
be retained it would take a vertical compression of 56§ tons, whilst CG
would take 130 tons compression instead of 186§ tons. The tie from
C to H is the preferable arrangement. The remaining vertical stresses
call for no special comment. The horizontal stresses in the booms,

230

STRUCTURAL ENGINEERING

assuming the diagonals inclined at 45°, are summed up as usual. The
figures for the horizontal stresses in the diagonals, having the same
value as the vertical stresses, have been omitted in Fig. 180 for the
sake of clearness.

Stresses in Lattice Girders of Variable Depth. — In girders of the
types shown in Fig. 162, Nos. 4, 5, 8, 12, 13, and 14, the curved or
inclined boom resists a portion of the vertical shearing force, so that
the inclined ties and struts are not called upon to resist the whole
shear, and the variation in depth of any particular girder will determine
what proportion of the total shear is borne by the booms or flanges,
whilst the remainder only will be resisted by the web members. It is
impossible, therefore, to write down the vertical stresses in the lattice
bars merely from an inspection of the loads, as in the case of parallel
girders, and a slightly modified method of analysis becomes necessary.

Stresses in Hog-backed Lattice Girder with Symmetrical Load.—
Fig. 181 represents the elevation of one half of a girder of 80 feet span
having 8 panels of 10 feet breadth, a central depth of 12 feet, and end
depth of 6 feet, carrying the loads indicated on upper and lower joints.
Under symmetrical loading only one half of the girder need be con-
sidered.

G -AW H -151-0 K -169-6

Fig. 181.

The reaction at each end =4x2 + 3x24 + ^x24 = 92 tons.
Bending moment atG = (92-l)x 10 = 910 foot-tons.

„ „ H = 91 X 20 - 26 X 10 = 1560 foot-tons.

„ „ K = 91 X 30 - 26(20 + 10) = 1950 foot-tons.

„ „ L = 91 x 40 - 26(30 + 20 + 10)= 2080 foot-tons.

The curve of the upper boom being drawn in from A to E, the
depths intercepted at G, H, and K are 8J, 10^, and 11 J feet respectively.
Dividing the B.M. at each section by the vertical depth of the girder,
the horizontal boom stress is obtained. Thus —

Horizontal stress in AB =

BC =

T)F =

B.M. at G
depth BG
B.M. at H

910
~8J~
1560

depth OE~ 10 J

depth DK
B<M- at L

2080

depth EL " 12

= 107-1 tons.

= 151-0

= 1G9-G

= 173-3

These horizontal stresses are now written against the respective
members, with a plus sign to denote compression, whence the horizontal

LATTICE GIRDERS 231

stresses in the diagonal ties at once follow. Thus, the horizontal
compression of 107*1 tons in AB can only be caused by the hori-
zontal pull applied by the tie AG, since the member AF is vertical and
has no horizontal component of stress. The horizontal tension in AG is
therefore 107*1 tons. Similarly the increase of horizontal compression
from 107*1 tons in AB to 151 tons in BC is caused by the horizontal
pull of tie BH = 151 - 107*1 = 43*9 tons. Also horizontal tension in
CK = 169*6 - 151 = 18*6 tons, and in DL = 173*3 - 169*6 = 3*7
tons. The lower boom stresses now follow, being summed up as in
a parallel girder. The vertical stresses in the ties are next required, and
are obtained by multiplying the horizontal stress by the ratio

vertical length

•r — = — .11 J..1 for each tie. Inus —
horizontal breadth

Vertical stress in AG = 107*1 x ~ = 107*1 X £• = 64*2 tons.

±(jT 10

BH = 43*9 X SSr = 43*9 X ?-| = 37*3 „

" CK = 18'6 x HE = 18'6 x if = 19'2 "

DTT 1 1-L

- - DL= 3'7XEC= 8'7xir ^ "

These stresses are written vertically against the respective members.
The stresses in the verticals easily follow. In AF the compression is
of course equal to the reaction of 92 tons. At G the vertical uplift of
04-2 tons in AG is partly balanced by the downward pull of the 24 tons
load at G, leaving a compression of 64*2 — 24 = 40'2 tons in BG.
Similarly compression in CH = 37*3 — 24 = 13*3 tons. At K the
vertical uplift in tie CK is only 19*2 tons, whilst the load to be
supported at K is 24 tons. The difference = 24 — 19*2 = 4*8 tons is
therefore taken by DK in tension. At the centre L the uplift in both
ties DL and ML = 4*3 + 4* 3 = 8*6 tons, and the load supported at L
is again 24 tons. The difference 24 — 8*6 = 15*4 tons is therefore
taken by EL in tension.

It should be noted that whether the vertical DK is in compression
or tension depends on the outline given to the upper boom. A very
slight increase of depth at CH, say to lOf instead of 10^ feet, would
result in a horizontal stress in BC of 146*2 tons and in CK of 23'8
tons, giving a vertical stress in CK of 25*4 tons. This being greater
than the 24 tons load at K would result in DK being subject to 1*4
tons of compression instead of 4*8 tons tension.

It remains to convert the horizontal stresses in the ties and inclined
segments of the upper boom into the corresponding inclined or direct
stresses. This is done in each case by multiplying the horizontal stress
. ,, inclined length m, . .. , ,

in the member by the ratio SS^aTbreadth- The lnclmed lengths
may be calculated or scaled off, provided the elevation of the girder is
carefully drawn to scale. The inclined lengths are as follows : — AB,
10*3 ft. ; BC, 10*2 ft. ; CD, 10-1 ft. ; DE, 10*02 ft. ; AG, 11*66 ft. ;
BH, 13*12 ft. ; CK, 14*38 ft. ; DL, 15*24 ft. Hence direct stresses are,

232 STRUCTURAL ENGINEERING

AB = 107-1 X ^ = 110-3 tons. AG = 107'1 X -1-— = 124'8 tons.

BC = 151 X ^ = 154-0 „ BH = 43-9 X —^ = 57-6 „

-IA.-I 1/1'^ft

CD = 169-6 X — = 171-3 „ CK = 18'6 X -— = 26-7 „
DE = 173-3 X ^=173-7 „ DL = 8'7 X -^- = 5'G „

Any condition of unsymmetrical loading may be treated in the same
manner by calculating the bending moment at each panel point through-
out the girder, as exemplified in the following case.

Stresses in Cantilever of varying Depth with Unsymmetrical
Load. — Fig. 182 represents the outline of a cantilever girder supported
on piers 100 feet apart and overhanging 250 feet on either side. Each
arm contains five panels of 50 feet horizontal breadth. The loads
indicated are assumed as acting at the various panel points and re-
present closely the state of loading for a steel girder of these dimen-
sions, forming one of a pair for carrying a double line of railway,
when the left-hand half from P, up to and including joint N is loaded
with the rolling load plus dead load, whilst the right-hand half is
assumed to carry dead load only.

B.M. at A = 47 x 50 = 2350 foot-tons.

„ B = 47 x 100 + 103 x 50 = 9850 foot-tons.

„ 0 = 47 x 150 + 103 X 100 + 109 x 50 = 22,800 foot-tons.

„ D = 47 x 200 + 103 X 150 + 109 X 100 + 113 x 50

= 41,400 foot-tons.
„ E = 47X250 + 103X 200 + 109 X 150 + 113x100 + 117x50

= 65,850 foot-tons.
„ F = 22 X 250 + 53 x 200 + 59 X 150 + G3 x 100 + G7 X 50

= 34,600 foot-tons.

„ G = 22 x 200 + 53 X 150 + 59 X 100 + 63 X 50 = 21,400 foot-tons.
„ H = 22 x 150 + 53 x 100 + 59 X 50 = 11,550 foot-tons.
„ K = 22 x 100 + 53 x 50 = 4850 foot-tons.
„ L = 22 x 50 = 1100 foot-tons.

In the following table are entered the bending moments, depth of
girder at each panel point, and horizontal stresses in booms, obtained by
dividing the B.M. by depth of girder. The horizontal stresses in
diagonals are obtained by successive differences of the horizontal
stresses in the boom segments, and the vertical stresses in diagonals by
multiplying the horizontal stress by the ratio,

vertical height of diagonal
horizontal breadth of diagonal.

The resulting stresses are indicated on Fig. 182, the compression in
any vertical being obtained by subtracting the load at its lower end
from the vertical tension in the diagonal tie attached at that point.

LATTICE GIRDERS

233

B.M. in foot-
tons.

Depth.
Feet.

Horizontal stress in boom
segments in tons.

Horizontal stress in
diagonals.

Vertical stress
in diagonals.

PA, 58-8

•

2,350

AB, 58-8

OB, 120-3

132-3

9,850

BC, 179-1

KG, 146-6

205-2

22,800

CD, 325-7

SD, 134-3

241-7

41,400

DE, 460-0

TE, 88-8

213-1

65,850

120

EF + EM, 548-8

EM, 260-5

416-8 •

34,600

120

EP, 288-3

FV, 50-5

121-2

21,400

FG, 237-8

GW, 72-8

131-1

11,550

GH, 165-0

HX, 76-8

107-5

4,850

HK, 88-2

KY, 60-7

66-8

1,100

KL, 27-5

LZ, 27-5

The compression of 288*3 tons in MV is obtained by summing up
from Z inwards or is taken directly from the calculation, being

B.M. at F _,, . . ,TXT B.M. at E

= — To^s — • The coinpressive stress in MN = — r^/c-fi — = 548*8

1^0 It. 1ZU 1C.

tons. The difference 548*8 -288*3 == 260*5 tons gives the horizontal
tension in EM which comes into action when the left-hand cantilever
arm is the more heavily loaded, NF being then inoperative. The
pressures on the supports are, for the

left-hand pier = vertical compression in EN + load at N
= 836*7 + 108*0 = 944*7 tons, and for the
right-hand pier = compression in FM — vertical uplift in EM

-f load at M
= 298*9 -312*6 + 58*0 = 44*3 tons.

These may be verified if desired, by taking moments about either
pier. The following table gives the calculated inclined lengths of the
sloping ties and segments of the upper boom, from which the tabulated
direct stresses have been calculated as follows. Direct stress in any tie,
say SD,

ST) 10'"*

= horizontal stress x «^ = 134*3 X -^ = 276'6 tons.

Direct stress in any segment of the upper boom, say DE,
= horizontal stress x

T-vTjl i*1

rr = 460*0 x -T-- = 536*3 tons.

JJT-J

The horizontal and vertical stresses in members which are themselves
horizontal or vertical, constitute of course the direct stresses for those
members. The direct stress in any sloping member of any girder may,
if preferred, be obtained as follows.

Let H = horizontal stress. V = vertical stress, then direct stress
= \/H2 -f- V2. In large girders, the lengths of the longer struts as OS,

234

STRUCTURAL ENGINEERING

DT, NM, and EN being unavoidably great, these members would be
stiffened by auxiliary struts ss, shown by dotted lines, in order to

6:862+

9-ZI2-

2502-

I .

*J bD

reduce their tendency to buckle. Such members, however, do not enter
into the calculations for the primary stresses.

LATTICE GIRDERS

235

DIRECT STRESSES IN INCLINED MEMBERS OF CANTILEVER GIRDER

IN FIG. 182.

Member.

Horizontal stress.

Inclined length.

Horizontal length.

Direct stress.

tons.

ft.

tons.

58-8

64-0

75-3

58-8

52-2

61-7

179-1

52-2

188-0

325-7

53-9

351-1

460-0

58-3

536-3

237-8

58-3

277-2

165-0

53-9

177-8

88-2

52-2

92-1

27-5

52-2

28-7

27-5

64-0

35-2

120-3

74-3

178-7

146-6

86-0

252-1

134-3

103-0

276-6

88-8

130-0

230-8

260-5

156-2

100

406-9

50-5

130-0

131-3

72-8

103-0

149-9

76-8

86-0

132-1

60-7

74-3

90-2

Bowstring Girders.— The bowstring girder consists of one curved
and one straight boom with verticals and diagonals. In the upright
type, Fig. 162, No. 12, the curved boom is in compression and the
horizontal boom in tension. These stresses are reversed if the girder
be inverted as in Fig. 162, No. 13. The upright form is employed for
through spans and the inverted form for deck spans. If the outline of
the girder be made parabolic, then the horizontal stress in the curved
boom is uniform throughout under the action of a uniformly distributed
load. In Fig. 183, the parabola ACB
represents the B.M. diagram for a
uniformly distributed load. If the
depth cd of the girder be everywhere
proportionate to the ordinates of the
parabola ACB, then the B.M. CD, at
any point, divided by the correspond-
ing girder depth cd, will give a con-
stant quotient for the horizontal flange

stress whatever section be considered. This being the case, there is
evidently no stress in the diagonal bracing, since the horizontal boom
stress does not increase from panel to panel and the diagonals do not
therefore apply any increment of horizontal stress to the booms. The
diagonals being in a state of no stress, it is obvious that the whole of the
shearing force is borne by the curved boom. The verticals under a
uniform load simply act as suspenders for the panel loads applied at
their lower ends and are in tension. The horizontal boom adb resists
the outward thrust of the ends of the curved boom and is subject to a
tensile stress equal in amount to the constant horizontal compression in
the curved boom. These conditions of stress will obtain very closely in

FIG. 183.

236 STRUCTURAL ENGINEERING

girders of circular outline provided the circular curve does not deviate
far from the parabola. In girders the outlines of Avhich are not
parabolic there will, however, be small stresses in the bracing, the
magnitude of which will vary with the extent of deviation of the
outline from the parabola. Under an unsymnietrical system of loads,
for which most practical girders must be designed, the diagonals will
suffer considerable tensile or compressive stresses according to their
direction of slope, as will be seen from the following examples.

Stresses in Parabolic Bowstring: Girder under Uniform Load. —
Fig. 184, represents a bowstring girder of 80 feet span having 8 panels
of 10 feet width, a central depth of 10 feet, and carrying 2 tons and
8 tons respectively at each upper and lower joint. The depths of the
girder at B, 0, and D are respectively 4'375, 7*5, and 9 -375 feet, being
the correct values of the parabolic ordinates. The reaction at A is 35
tons. Then

B.M. at B = 35 X 10 = 350 ft.-tons.

„ C = 35 X 20 - 10 X 10 = 600 ft.-tons.

„ D = 35 X 30 - 10(20 + 10) = 750 ft.-tons.

„ E = 35 X 40 - 10(30 + 20 + 10) = 800 ft.-tons.

350

.*. Horizontal boom stress in AB = T^T = 80 tons.

4-375

600

» » n^ ~~ ~77F == 55

/ «.)

rn- 75°

"9-875"

Since the compression in each segment of the upper boom is 80
tons, no increment of horizontal stress is applied by the diagonals,
whence their stress is zero. The outward thrust of 80 tons in AB
creates a tension of 80 tons in AF and this stress is passed on unaltered
throughout the lower boom, since the diagonals do not affect it. The
stress in each vertical is 8 tons of tension. Finally, the inclined lengths
of AB, BC, CD and DE are respectively lO'O, 10'4, 10*2 and 10*02 feet,
whence the direct stresses are

10*9
in AB = 80 x -j^~ = 87*2 tons.

10*4
„ BC = 80 X -J0- = 83*2 „

10*2
„ CD = 80 X -j^j- = 81*6 „

and „ T)E = 80 x ^2 = 80vl 0 „

It will be noticed the greatest direct stress occurs in AB near the
support, whereas in parallel girders the greatest boom stress is in the
central panels, and further, that the direct stress in the curved boom

LATTICE GIRDERS

237

being only slightly greater near the ends than at the centre, the cross-
section adopted for AB may be used throughout the boom with little

-80

FIG. 184

sacrifice of economy, whilst the practical advantage is considerable.
For the same reason, the parabolic girder is a more economical type
than the parallel girder, span for span. In a large span, the dead
weight of the girder creates a considerable proportion of the total B.M.
In the parallel type the heaviest portions of the booms are near the
centre, that is, in the most disadvantageous position for creating B.M.,
and therefore stress, in the girder. In the bowstring girder, the weight
of the booms being practically uniformly distributed, or actually a little
greater towards the supports, the B.M. and stress due to dead weight is
relatively less. The bracing of bowstring girders is also usually lighter
than that of parallel types.

Stresses in Parabolic Bowstring Girder with Unsymmetrical Load.
— In Fig. 185, the girder of the previous example is supposed loaded

with an additional 12 tons at panel points F, Gr, H and K.

The reaction at A = £ X 14 (upper loads) -f 20(J + f + f -f f)
+ 8(f + f + J) = 68 tons, and at S = 118 - 68 = 50 tons.

B.M. at F = 68 X 10 = 680 ft.-tons.

G = 68 x 20 - 22 x 10 = 1140 ft.-tons.
„ H = 68 X ?>0 - 22(20 + 10) = 1380 ft.-tons.
„ K = 68 x 40 - 22(30 + 20 -f 10) = 1400 ft.-tons.

P = 50 X 30 - 10(20 + 10) = 1200 ft.-tons.
„ Q = 50 x 20 - 10 x 10 = 900 ft.-tons.
„ R = 50 x 10 = 500 ft.-tons.

Dividing these bending moments by the depths at the corresponding
sections, the horizontal boom stresses marked in the figure are obtained.
At B the horizontal stress of 155*4 tons in AB is not balanced by that
of 152 tons in BC, therefore the diagonal BGr must exert a horizontal
thrust of 155-4 —152 = 3-4 tons. Similarly CH and DK are found to
be in compression to the extent of 4 -8 and 7 '2 tons respectively. The

238

STRUCTURAL ENGINEERING

horizontal thrust of 128 tons in LM increases to 140 tons in EL, so
that the increment of 140 —128 = 12 tons must have been applied by
a horizontal tension of 12 tons in diagonal LK, Similarly 8-0 and 5*7
tons of tension respectively are found to exist in MP and NQ. These
horizontal stresses are converted into the corresponding vertical stresses

, , u . , . ,T , ,, , . vertical height c ,
as usual, by multiplying them by the ratio, , — = — . . T -rrr of each

' horizontal breadth

diagonal.

The stresses in the vertical members then follow. BF obviously
acts as a suspender for the 20 tons load at F. At G, the load of 20
tons + the vertical downward thrust of 1-5 tons in BG, together
produce a tension of 21-5 tons in CG. At H, 20 + 3-6 = 23-6 tons
tension in I)H. At K the downward forces are 6 -8 tons vertical
thrust in DK + 20 tons load, making 26*8 tons, which is partly
resisted by the vertical tension of 11-3 tons in LK, leaving 26-8 —11-3
= 15'5 tons tension, in EK. Similarly LP takes 2 tons and MQ, 5-5
tons of tension, whilst NR simply suspends the 8 tons load at R. The
direct stresses in the diagonals and upper boom segments are obtained
as in previous examples and are as follow : —

Upper boom.

AB = 169*4 tons, compn.
BC = 158-8
CD = 150-1
DE = 140-3
EL = 140-3
LM = 130-5
MN = 124-8
NS = 124-6

Diagonals.

BG = 3'7 tons, compn.

CH = 6-0

DK = 9-9

LK = 16-5 tons, tension.

MP = 10-0

NQ = 6-2

With a single system -of diagonals inclined as in Fig. 185, each
diagonal will be subject to both tension and compression under varying
positions of the unsymmetrical portion of the load. If the diagonals be
reversed in direction, then for the same arrangement of loads as in Fig.
185, the horizontal and vertical stresses will be as indicated in Fig. 186.

^^ ? Y r Nj ? xj \*

-ISZ-O 20 -147-2 20 -1400 eo-/40-0 g -I2Q-O g

-120-0 g -114-3

FIG. 186.

Here, again, the diagonals may be subject to either tension or com-
pression according to the position of the unsymmetrical load.

The B.M. at each section is as before and the horizontal stresses in
the boom segments and diagonals readily follow.

Stresses in Bowstring Girder with Crossed Diagonals in every
Panel. — A third arrangement, and one frequently adopted, is to cross-
brace every panel with flat tie-bars capable of resisting tension only.
Assuming the same loads and dimensions as in the two previous

LATTICE GIRDERS

239

cases, the horizontal and vertical stresses would then be as indicated
in Fig. 187.

In this case the diagonals being incapable of resisting compression,
become lax under any tendency of the load to create compressive stress
in them, whilst those diagonals inclined in the opposite direction come
into action for resisting the tension. Thus, in Fig. 187, the dotted
diagonals which ivould suffer compression if of suitable section to resist
it, are in a state of no stress, whilst the full line diagonals are put in
tension under the disposition of loads considered. The dotted diagonals,

of course, come into action when the right-hand portion of the girder
is the more heavily loaded. It is a mistake to cross-brace every panel
with two diagonals capable of resisting compression since the bracing is
then redundant, and an indeterminate amount of compression exists in
one diagonal and an indeterminate amount of tension in the other,
making it impossible to accurately compute the stresses.

Effect of Uniform Rolling Load on Parabolic Bowstring Girder.—
In each of the three preceding cases, the first four panel points from the
left-hand support were supposed loaded with 12 tons over and above
the symmetrical dead load in Fig. 184. This additional load may be
regarded as a uniform rolling load which has advanced from the left
abutment up to the central panel. The maximum horizontal stress in
the diagonals occurs in KL, Figs. 185 and 187, or in EP, Fig. 186, that
is, in the diagonal of the panel immediately in advance of the rolling
load, and is equal to 12 tons of tension or compression according to the
inclination of the diagonal. If the load be supposed to advance another
panel so that P now becomes loaded with an additional 12 tons, and a
similar analysis of the stresses be made, the diagonal MP will be found
to be the most heavily stressed, and further, the amount of the horizontal
stress in MP will again be 12 tons. This result is peculiar to the bow-
string girder of parabolic outline and may be stated as follows. Under
the action of a uniform rolling load advancing from one abutment, the
maximum horizontal stress in any diagonal occurs when the head of the
load reaches the panel in which the diagonal is situated, and the amount
of the maximum horizontal stress is the same for every diagonal in the
girder. When the rolling load covers the whole span, the loading is
again symmetrical and the B.M. at the centre of the span due to 12 tons
of rolling load on each lower joint, is then

= 42 x 40 - 12(30 + 20 -f 10) = 960 ft. -tons.

Dividing this B.M. by the central depth of 10 ft., the horizontal
boom stress at the centre, due to rolling load covering the span
= = 96 tons. It was shown above that the maximum horizontal

240 STRUCTURAL ENGINEERING

stress in each diagonal, as the rolling load reached ifc, was 12 tons.
The number of panels is 8 and ^ = 12 ; or stated generally, the maximum
horizontal stress in each diagonal

= Maximum horizontal boom stress when rolling load covers the
whole span -f- number of panels.

This relation furnishes a ready method of calculating the maximum
stresses in the diagonals due to the passage of a uniform rolling load.
A strictly mathematical investigation gives the maximum horizontal
stress in each diagonal

_ Maximum horizontal boom stress
Number of panels + 1

The discrepancy is due to the fact that it is impossible to fully load
any one panel point with a uniform load without, at the same instant,
partially loading the next panel point in advance. In the above
analysis the assumption is made that the panel point at the head of
the advancing load is fully loaded, whilst the next panel point in
advance is unloaded. This could only be effected by a uniform load

extending from, say, A to B in Fig.
188, together with a concentrated
load at B equal to half the panel
length of uniform load, in which
case B is fully loaded, whilst C is

unloaded. The diagonal BD then suffers its maximum horizontal stress.
As this represents more closely the practical condition of loading,
especially in the case of train loads on bridges, where the uniform train
load is headed by the heavy concentrated axle loads of the engine, the
results are more nearly correct than if a perfectly uniform load be
assumed throughout.

It is common practice to construct the upper boom with an actually
curved outline. In designing the cross-section of the various segments,
B it should be noted there will be both bending and
direct stress to provide for. Thus in Fig. 189 the
calculated direct stress P acts along the straight
dotted line AB. If the boom be curved, the B.M.
at its centre = P x d, the case being similar to that
of a deflected column in compression. The intensity
FIG. 189. of compressive stress is thereby augmented on the
under side of the boom, and slightly relieved on the
upper side. The weight of the boom segment itself sets up a small
amount of B.M. acting contrarily to that of Pd, and so tends to equalize
the stresses at upper and lower faces. Although the employment of
curved booms is theoretically disadvantageous, the practical advantage
resulting from greater facility of construction, especially in riveted
girders, more than compensates.

Design for Lattice Crane Girder. — The preceding methods will now
be applied for obtaining the maxima stresses in the members of the
lattice crane girder shown in outline in Fig. 190, and in detail in
Fig. 191. Lattice girders for travelling cranes are usually of the
Warren type with intermediate verticals for reducing the panel spans

LATTICE GIKDERS

241

of the upper boom over which the crab travels, and which are therefore
subject to bending moment in addition to direct compression. In this
example, the span is 46 ft. 6 in., depth of girder 5 ft., and breadth of
panels 5 ft. The useful crane load is 70 tons, weight of crab assumed
as 18 tons, and weight of each main girder 10 tons, of which 1 ton is
apportioned to each upper joint, the remaining 1 ton being divided
between the end supports, and consequently not appearing in the

-137-7

FIG. 190.

calculations for the stresses in the lattice members, since it does not
contribute to the bending moment. The distance between centres of
crab axles is 10 ft. The load lifted will be doubled and treated as
equivalent dead load. Then equivalent crab load = 70 x 2 -f 18
= 158 tons, i.e.1^- = 39 '5, say 40 tons per wheel. It should be noted
that by doubling the crane load an outside allowance is made for shock
due to possible slipping of the tackle, and in designing the sectional
areas of members a relatively high working stress may be taken,

The following table shows the stresses caused by placing the crab
in five different positions ; first, with the left-hand axle over point A,
and right-hand axle over D, afterwards moving it 5 ft. at a time, so
that the left-hand axle comes successively over 0, D, F, and G. The
transit over the right-hand half of the span will give rise to similar
stresses in corresponding members, and need not be treated. As the
method of calculation for each position is a repetition of the one before,
it is here stated for one position only of the load, namely, when the
left-hand axle is at D, and right-hand axle at G.

STRESSES IN MEMBERS OP LATTICE CRANE GIRDER.

Direct stress when left-hand axle of crab is at,

jMemoer.

+ 82-0

-f 72-3

+ 62-3

+ 52-2

+ 26-7

+ 74-4

+ 1007

+ 86-G

+ 72-4

+ 58-2

+ 90-0

+ 139-1

+ 147-8

+ 156-4

+ 125-0

- 67*3

- 58-5

- 50-0

- 41-9

- 33-9

-102-7

-120-4

~~ I37'7

-114-9

- 92-1

- 76-3

-116-8

— 156-9

-156-9

- 15-5

- 63-2

- 54-3

- 45-3

Sfi'2

-f 39-9

+ 27-8

+ 72'i

+ 59-9

+ 47-8

+ 17-9

- 26-4

- 14-3

- 58-5

- 46-4

- 19-3

- 31-5

+ 12-8

+ 0-7

+ 45'0

-1- 1-0

+ 4I>0

+ 1-0

+ i-o

+ 41-0

+ 1-0

+ 41 '°

+ 1-0

242 STRUCTURAL ENGINEERING

Reaction at X = jjjjof 40 + of 40 + 4J = 53-1 tons.

B.M. at A = 53-1 x 3J = 172*6 ft.-tons.
Vertical depth of girder at A = 3- 5 9 ft.

1 79*fi

/. Horizontal stress in XA or XB = -g^- = 48-1 tons.

Vertical stress in XB = 48'1 x ff" = 13-6 tons tension, 28 in. and
99 in. being respectively the vertical height and horizontal length of
XB. Since the downward vertical pull in XB and thrust in XA
together make up the reaction of 53'1 tons, vertical compression in
XA = 53-1 - 13'6 = 39-5 tons.

Writing these vertical stresses against XB and XA, Fig. 190, the
remaining vertical stresses for this position of the load are easily
written down. Thus, at A, 1 ton of the vertical compression of
39-5 tons in XA is due to the dead load at A, leaving a vertical tension
of 38-5 tons to be applied by AB. The 1 ton of dead load at C is
transmitted down CB as compression, so that at B there is an uplift
of 38-5 4- 13-6, exerted by ties AB and XB, = 52-1 tons - 1 ton due
to the thrust in BC, giving 51-1 tons as the vertical down-thrust or
compression in BD. At D the 41 tons of load creates 41 of the 51*1
tons compression in BD, whence the remaining 10*1 tons is due to the
vertical tension in DE. The other vertical stresses will be readily
followed, care being taken not to omit the loads transmitted down the
intermediate verticals from top to bottom joints. The stresses beyond
G are not required for insertion in the table, but are followed through
in order to check with the reaction at Y. On arriving at L, the
vertical thrust of 33 '9 tons in KL -f 1 ton in LM = 34-9 tons. This
represents the combined vertical tensions in LN and LY. The separate
tensions in these members are not required, as they will not be the
maxima stresses. They may be found, if desired, from the B.M. at N
in a similar manner as adopted for the stresses in XB and XA. Adding
the last 1 ton load at N, the pressure on Y = 34'9 -f 1*0 = 35-9 tons.
The total load on girder = 89 tons, and reaction at Y = 89 — reaction
at X = 89 — 53-1 = 35*9 tons, which of course agrees with the 35'9
tons pressure applied by members NY and LY. The inclination of
the lattice bars, excepting XA and XB, is 45°, and the horizontal
stresses equal the vertical stresses for bars AB, BD, DE, and EG. The
horizontal stress in XB and XA has already been calculated, and equals
48*1 tons. From the horizontal stresses the flange stresses are readily
computed, and finally, the direct stresses for the inclined members are
obtained in the usual way.

Direct stress in XB = 48-1 x W " = 50 tons»

XA = 48-1 X *^r = 62-3 tons,

O «J

103 in. and 50*5 in. being the inclined lengths of XB and XA
respectively. The direct stresses in AB, BD, DE, and EG = horizontal
stresses x ^2. The direct stresses are entered in column D of the

LATTICE GIRDERS 243

table, and the stresses due to the other positions of the crab being
similarly calculated and inserted, the maxima stresses are indicated
by the figures in heavy type.

Fig. 191 shows the sections adopted, and general arrangement of
the girder. The upper boom from 0 to G consists of one horizontal
plate 16" x f" two vertical channels 9" x 3J" X 25*39 Ibs., and two
vertical plates 7j" x f". From C to the end of the girder the two
vertical plates are suppressed. The C.G. of the section is 3*7 in. from
the upper surface, moment of inertia = 498*4, and modulus of section

for compressive stress at upper surface therefore = -^ — = 135. The

o*7

segments of the upper boom act as beams when the rolling load travels
over them, and are subject to bending stress in addition to the direct
compression indicated in the table. With 40 tons midway between two

panel points, the B.M. = — ^— = 50 foot - tons = 600 inch - tons.

This, it will be noted, is an outside estimate, since the continuity of
the upper boom will tend to diminish the B.M., which is here calculated
as for a simply supported span of 5 ft.

Therefore compressive stress due to bending only = f§§ = 4*45 tons
per square inch.

Total sectional area = 34*3 sq. in. Maximum direct compression in
segment DG (from table) = 156 '4 tons. Hence direct compression per

sq. in. = -g^- = 4'56 tons, and total compressive stress per square inch

at upper edge of section due to bending and direct compression = 4*45
-f 4*56 = 9*01 tons. Remembering that the rolling crane load was
originally doubled, this stress represents the outside maximum which
may occur under a sudden shock, due to a possible slip of tackle whilst
slinging the load. Under normal working conditions the maximum
compression in the top boom will not exceed 5 to 5J tons per square inch.
The compression members here are all relatively " short columns," and
no reduction in working stress is necessary on the score of slenderness.
The lower boom consists of two vertical plates 8" x f ", two
angles 7" x 3J" x 17*81 Ibs. with 7 in. leg vertical, and two horizontal
plates 3^" x i". Sectional area = 24 gq. in. Maximum stress

= 156*9 tons in EH, and maximum stress per sq. in. = * = 8*7

tons, after deducting 6 sq. in. for six f in. rivets — four in the vertical
and two in the horizontal portion of the boom. Each vertical member
suffers 41 tons of compression as an axle passes over it. These consist
of two T x 3" x 17'56 Ibs. channels, back to back. Sectional area

41*0
= 10*32. Maximum working stress = TTJTOS = say 4 tons per square

inch. Diagonal EG has the same section as the verticals. Its maximum
stresses are 45 tons compression and 31 §5 tons tension, and maximum

working stress = -— - = 4*4 tons per square inch compression. Diagonals

-LU'oSi

AB, BD, and DE all have the same section, consisting of two 7" X 3"
channels, with one 7" X i" plate between. Maximum stress in

244

STRUCTURAL ENGINEERING

FIG. 191.

LATTICE GIRDERS

245

AB = 63' 2 tons tension. Gross sectional area =13*82 sq. in., and
deducting, say, 5 rivet holes, four through gusset plates and one
through webs of channels and sandwich plate, nett sectional

/?o.o

area = 11 sq. in., giving a maximum working stress = -ry- = 5'8 tons

per square inch. In BD maximum compression = 72' 1 tons,and maximum

72'1
working stress = -. " ~ = 5' 3 tons per square inch. DE, with a maximum

tension of 58'5 tons, is obviously of ample section. In XB the vertical
angles and horizontal plates only are carried through, the two 8" x f"
vertical plates being suppressed on the left of joint B. The maximum
tension is 67'3 tons. The gross section is 14 sq. in., and deducting 6
rivet holes, nett section = 10*5 sq. in. Maximum working stress

= y-rv = 6*4 tons per square inch. The short strut XAis enclosed between

two deep f in. gusset plates, to which it is riveted throughout its length.
A considerable proportion of its stress will therefore be transmitted by
the gusset plates. Its maximum stress is 82 tons. The strut, apart
from the gussets, consists of two 7" X 3" channels, having a sectional
area of 10'32 sq. in. As sufficient rivets are provided to pass quite half
the stress to the gussets this section will be ample. The riveted
connections have been designed on a basis of 3 tons single shearing
resistance per f in. rivet, and a maximum bearing stress of 10 tons per
square inch. All gussets are f in. thick. On the cross-section a light gantry
is shown on the right-hand side, carried by a lattice girder 2 ft. 6 in.
deep, bracketed out from the main girder.

Practical Arrangement of Details of Lattice Girders. — Fig. 192

fc9-
A f

T T

JL I L

FIG. 192.

illustrates sections most commonly employed for the booms of lattice
girders.

Sections A and B are used for the flanges of light girders in roof
construction, or for parapet girders having webs of closely spaced
intersecting flat bars. These girders are now practically obsolete for
large spans. C is the usual flange section for girders of medium length.
The bracing consists of flats, channels, angles, or tees riveted on either
side of the vertical web plate P. The compression booms of large span

246 STRUCTURAL ENGINEERING

girders are generally of section E, the number of vertical and horizontal
plates being increased according to the sectional area required. Section
D is occasionally employed for the compression boom in light girders,
the channels being laced together top and bottom. For the tension
boom any of the above forms inverted may be used, whilst section F,
consisting of vertical plates with or without the dotted angles, is very
generally employed, and possesses the advantage of not accumulating
dirt as does the trough section E. The tension booms of pin- jointed
trusses consist of several flat eye-bars placed side by side as in Fig.
197. The bracing bars of lattice girders are of flat section in the
case of ties, and angle, tee, or channel section for the struts of small
girders. In larger girders the struts may be of rolled beam section, or
any of the column sections indicated in types 1, 7, 10, and 11, Chapter
V. Type 10 is most commonly employed in English practice, whilst in
pin-jointed girders of American design the struts are almost invariably
formed of two channels laced together as in Fig. 197.

Fig. 193 shows the usual arrangement of details adopted in English
practice for riveted main girders of bridges. The figure shows the
elevation and cross-section of one panel of a main girder for a double-
line railway bridge of a type suitable for spans of from 120 ft. to 150
ft. The upper boom A is of trough section similar to Fig. 192, E, a
joint being indicated at J. This joint is a full butt, with internal and
external covers to both horizontal and vertical plates. An alternative
arrangement is to make the horizontal and vertical plates break joint,
which however makes the work more difficult to handle in the shop.
The under side of the boom is laced as shown at L, and the boom
is stiffened transversely by plate diaphragms d, d. The lower boom B
consists of two parcels of vertical plates, the arrangement of a joint in
this boom being indicated at P. One or two plate diaphragms similar
to d in Fig. 192, F, are usually inserted in each panel, whilst angles as
shown dotted in Fig. 192, F, are often provided. If the girder be
finished with a vertical end post, these angles are necessary in the end
panels to allow of the vertical plates being laced together in order to
resist longitudinal compression due to application of brakes. The
vertical struts V are of four angles, back to back, with flat lacing bars.
The ties T are of two or four flats, according to the section required,
and are here shown butt- jointed to the gussets #, #, with double covers.
This joint is often lapped, but the butt joint is preferable. The
riveting should be arranged with one rivet only on the leading section
of the tie-bar, to avoid weakening the tie by more than one rivet
hole. The cross-girders G are attached to the lower ends of the
verticals V, which pass through the lower boom and carry a pair of
suspension plates S, between which the web of the cross-girder is
riveted. The vertical angles forming the strut are also carried through
and riveted to the cross-girder web. This is probably the most
desirable method of attachment of cross-girders, and its adoption is a
strong argument in favour of the open type of lower boom. Other
advantages are that the centre of gravity of cross-section is at the centre
of depth, and the boom does not accumulate dirt and water. In Fig.
192 the centres of gravity of the various boom sections are indicated at
eg. It is important in setting out the skeleton lines intended as the

LATTICE GIRDERS

247

axes of the members of a lattice girder that these lines intersect at the
centres of gravity of cross-section of the members, otherwise secondary
bending stresses will be set up in the region of the joints. In Fig. 193
the chain dotted line x-x, drawn on the upper boom in elevation, is

projected from the e.g. of the boom section, indicated by the small
circle on the cross-section. This line is sometimes referred to as the
gravity line or gravity axis of the boom, and it will be seen that the
centre lines or axes of the ties and struts all intersect on this line. The
details of jointing in larger girders with double systems of bracing are
similar to those just mentioned.

Fig. 194 shows a vertical section through the end post or strut of

248

STRUCTURAL ENGINEERING

a girder similar in type to that of Fig. 193. The end post P consists
of two side plates, one transverse plate, and eight angles arranged as
indicated on the horizontal cross-section. The upper boom may be
either lap- or butt-jointed to the end post. It is here shown lap-jointed,
and the end tie T is connected to gusset plates G, packing being required
as shown by the dotted shading. The two vertical plates of the lower
boom B are riveted to the lower end of the post P, and laced together
top and bottom by lacing bars L, L, for the length of the first panel of

FIG. 194.

the girder. The lower end of the end post carries a base or bolster
plate, which is bolted to the upper casting C of the deflection bearing.
The inner and outer faces of the end post are covered by end plates E,
having hand-holes H for convenience of painting the interior.

End Bearings for Girders.— Girders exceeding 70 feet span should
be provided with pin and roller bearings to allow of free deflection and
expansion and contraction under changes of temperature. The usual
arrangement is a combined roller and pin bearing B, Fig. 195, beneath

LATTICE GIRDERS

249

one end, and some form of pin bearing P, under the other end of the
girder. Considerable variety exists in the detailed design of bearings.
Figs. 194 and 196 show two
common types. In Fig. 194 the
girder is carried on two steel cast-
ings C and D, having machined
concave and convex surfaces re-
.spectively. In Fig. 196, which
is the more usual type employed
in English practice, the upper

casting C bears on a steel pin FIG. 195.

carried by the intermediate casting

D, which transmits the load to the nest of rollers R, these again resting
on a lower casting L. The rollers should not be less than 4 in.
diameter, and vary from 4 in. to 12 in. The safe load on rollers may

Ins. is

7 Ft:

FIG. 196.

be 1000 Vd Ibs. per lineal inch of bearing between rollers and castings.
Hence, a bearing having, say, eight rollers of 9 in. diameter and 2 ft.
long, may be safely loaded with —

24 x 8 X 100QV9
2240

= 257 tons.

The lower casting should be at least the same depth as the rollers,

250 STRUCTURAL ENGINEERING

and its base area increased if necessary to distribute the pressure over
a suitable area of masonry. The pressure between lower bed-plate and
masonry should not exceed 400 Ibs. per square inch. Built-up pedestals
are sometimes employed instead of castings, but the riveting is usually
cramped and difficult. The rollers are carried in a pair of light bar
frames, and either the rollers or castings are flanged to prevent lateral
movement.

Girders of less than 70 ft. span should have a steel bolster plate not .
less than f in. thick, or a casting 2 in. to 3 in. thick attached to the
lower flange under the end of the girder, such bolster or casting sliding
on a second cast bad-plate bolted to the masonry. One end only of
the girder is free to slide, the other end being suitably prevented from
moving longitudinally by a flange or stop on the bed-plate or by sub-
stantial bolts.

Fig. 196 also shows the detail of plating at the end of a bowstring
girder of 200 ft. span. The construction is clearly indicated, and calls
for no special remark.

Fig. 197 shows the detailed arrangement of one panel of a pin-
connected truss of 160 ft. span. The upper boom A consists of two
vertical plates and four angles laced together at upper and lower faces.
The lower boom consists of eye-bars E, the number and sectional area
being proportioned to the stress in each panel. The diagonal ties T
are also eye-bars, two or four being employed as required. The members
meeting at each panel point are assembled on a steel pin, the diameter
varying from 4 in. or 5 in. in small spans, to 9 in. or 10 in. in large
spans. Joints J in the upper boom are usually placed near the pin
instead of at centre of panel. The vertical struts V, Y, in girders of
moderate span consist usually of two channels laced together on inner
faces. A horizontal section through strut V is shown at S, from which
the assemblage of the members meeting on the lower pin P will be
readily traced. The right-hand view is an end elevation and part
vertical section through strut V, and shows the connection of a cross-
girder G to the inner face of the strut, as well as the overhead transverse
or sway bracing B. A diaphragm plate D is inserted in the plane of
the web of the cross-girder between the two channels forming the strut
V. The panel width, here 20 ft., being considerable, the cross-girders
are much more heavily loaded than when spaced at 7 ft. or 8 ft., and
require a correspondingly greater depth. They are cut away at X to
clear the pin end and internal eye- bars of the lower boom. This
necessitates the employment of a hitch plate H, which is riveted to the
U-plate fitted between the lower ends of the channels of strut V. The
hitch plate is required to resist the tendency of the horizontal diagonal
wind braces to bend the lower end of the strut inwards. These wind
braces are omitted in the figure, but are attached to the hitch plate,
which is so shaped as to form a junction plate or gusset for the
horizontal wind ties. A similar arrangement obtains at Y, where the
sway bracing B is cut away to clear the upper boom. The upper system
of horizontal wind braces are attached to the upper boom. The use of
pin connections obviously modifies the section of boom which may be
adopted. The upper boom plates bear edgewise on the pins, and in
order to obtain sufficient bearing area, a considerable portion of the

LATTICE GIRDERS

251

FIG. 197.

252

STRUCTURAL ENGINEERING

upper boom section is made up of vertically disposed plates, especially
in the case of very large girders. Fig. 102, G, shows the section of
the upper boom of the 439 ft. 9£ in. spans of the Bellefontaine bridge,1
which in the central panel is made up of one horizontal plate 43" x J",
two vertical inside plates 25^" x £•", two vertical outside plates
25i" X £", four side plates 12" x {", eight angles G" x 4" x f", and
four horizontal flats 4" x 1". The pins are 9 in. diameter, so that
this disposition of plating gives 56£ sq. in. of bearing area on the
pin. Further consideration of pin-jointed trusses is beyond the scope
of this work, but a comparison of Figs. 193 and 197 will illustrate the
noticeable differences in detailed design of pin- jointed and riveted
girders.

Fig. 198 shows the construction of a lattice girder such as commonly
used for foot-bridges over railway tracks. The boom sections are

ooooo

ooo

6 o o o o o o o o o~

o o o o o

O O !

*Ff

FIG. 198.

similar to Fig. 192, C, whilst the bracing consists of flat bars F,
riveted at the intersections with washers for packing. Angle, tee, or
channel stiffeners S are riveted to the web bracing and fitted well up
to the flanges at suitable intervals. Parapet girders of deck bridges
in open country are usually of this type, an example being shown in
Fig. 161. For bridges over town streets open-work parapet girders
are generally prohibited.

1 Engineering, Sept. 20, 1895.

CHAPTEE VIII.

DEFLECTION.

IN all structures changes of length of the separate members are produced
by the compressive or tensile stresses in the members, and although the
change in length of any individual member is comparatively very small,
yet the compounded elongations and contractions of the successive
members produces an appreciable displacement of certain points in
the structure. The vertical component of such displacement is termed the
deflection of that point. The importance of the magnitude of the
deflection will vary according to the class of structure under considera-
tion. For bridges and girders in general, it is usual to specify a limiting
ratio of deflection to span to ensure that no serious changes of length
take place in any of the members, and the deflection under test loads is
often measured for that purpose ; but such factors as the relative stiffness
of the connections, workmanship, etc., may so alter the theoretical
distribution of stress in the members that the deflection can in no-
wise be assumed as a measure of the strength of the structure, or
even an evidence that no member of the structure is being unduly
stressed.

However, in many structures such as the arms of swing-bridges,
cantilevers, arches, etc., that are being erected by building out from
the ends, it is of great importance to estimate to a close degree of
accuracy the deflection that will occur, to ensure the seatings being
placed at the correct levels, the overhanging ends meeting accurately,
etc.

Live loads produce greater deflections than static loads of the same
magnitude owing to the dynamic action of such loads. The deflection
produced by a suddenly applied load to, say, a crane girder, would be
double the deflection produced by a similar static load, but would be of
only momentary duration, and as the vertical vibration ceased such
deflection would be reduced to the magnitude of the static deflection.
The increase in deflection produced by the live loads on bridges will
vary according to the span, velocity of the moving loads, etc., and it is
impossible to establish an exact law for these conditions, but by making
due allowance for impact when calculating the stresses in the members,
a close approximation to the deflection will be obtained.

Deflection due to Static Loading.— Providing the elastic limit is
not exceeded, the change in length in any member will be —

253

254 STRUCTURAL ENGINEERING

"i . •

where L = the original length of the member

I = the change in length of the member
s = the intensity of stress
E = the modulus of elasticity of the material.
Or if Si = the total stress in the member

and A = the sectional area of the member

then ,=jg

/, L and A, and Sx and E, being measured in the same units.

Consider the action of the force P applied gradually to the girder
in Fig. 199. To produce the changes in lengths of the members form-

ing the girder a certain amount of work
is performed. If A be the deflection
of the point of application of P, the

r" p R ^ work performed during that deflection

FIG. 199. = IAP. As this is the only external

work done, it must be equal to the

total internal work done in the girder. The work done in straining
any member is equal to half the product of the stress in the member,

Q 7

and the change in length = ^-.

'-SI

Therefore the internal work performed on any member

2EA

and the total internal work is the sum of the work performed on each
member.

Therefore

A = PE

Q 2T

Calculating -V- for each member and dividing the sum of such

terms by the product of P and E, the deflection of the loaded joint is
obtained.

Let an additional load R be placed anywhere on the girder, and tbc
stress in any bar which was stressed Si under the action of P only, be
now equal to S2. The total work performed on such bar

where I = the change in length due to the load P

DEFLECTION 255

The work performed on the bar by P therefore

The total internal work performed by P is the sum of the work
done in the members

2EA

Therefore

2EA

and

A — _

where A is the deflection of the point of application of P.

Since the load R has been assumed anywhere on the girder, the
above expression is applicable to any system of loading, but attention
must be paid to the signs of Sj and S2, and the algebraic sum of the

Q Q T

terms ~r~ taken. If instead of considering the work performed by

the whole force P, the work performed by a unit portion of such load
be considered, the deflection will be obtained from the above expression
by substituting for Si the stress in the members due to the unit load
and replacing P by unity. Let S be the stress in any member due to a
unit portion of the load P.

Then A = ]

~E^ A

The deflection of any other point in the girder may be obtained by
substituting for S the stress in the members due to unit load at such
point.

EXAMPLE 32. — To find the deflection at the middle of the span of a
braced girder.

Let Fig. 200 represent the girder and the static loads at the panel
points.

M I

30 forts.

FIG. 200.

The stresses in columns 1 and 2 of the following table have been
calculated by the methods described in Chapter VII., and E assumed as
14,000 tons. In practice the sectional areas of the members are taken
from the completed design.

256

STRUCTURAL ENGINEERING

Member.

Stress, S*.

Stress, S.

Length, L.

Sectional

. .., A

S2SL

area, A.

tons.

ft.

sq. in.

- 0

+ o

-105

_ 1

58-33

-180

— 1

120-0

-225

— 11

177-6

+ 105

+ 5

40-4

+ 180

+ 1

80-0

+ 225

•ft!

120-5

+240

160-0

+ 105

+ 5

42-0

-105N/2

- K'2

20N/2

87-5 N/2

+ 75

+ i

37-5

- 75x/2

- K/2

20v/2

88-2 x/2

+ 45

+ £

37-5

- 45N/2

- K/2

20x/2

75x/2

+ 15

+ £

- 15./2

- K/2

20N/2

50s/2

Total for the half truss . . . 1,325

„ whole ,, ... 2,650

Multiplying by 12 to reduce L to inches 12

2?2SL = 31>800

Horizontal Displacement. — The horizontal displacement of any
point in a structure may be found by a similar method. If a unit
horizontal load be applied at such point, the work performed by this
load will be half the product of the displacement and unity. The total
internal work will

_ •<

2EA

where S3 = the stress in the members due to the horizontal unit load.
Let H = the horizontal displacement

Then lu^S^

= E *

EXAMPLE 33. — To find the total displacement o/ the left-hand end
P of the cantilever in Fig. 182. The outline of the girder is reproduced
in Fig. 201.

In the following table suitable sectional areas for the members have
been inserted. In a bridge of this type, the superstructure would be
securely fixed to one of the piers at M or N and allowed to slide
horizontally on the other. It will be assumed in this example that N
is the fixed point and horizontal movement may take place at M. This
assumption in no way affects the vertical deflection of the point P, but
has a direct bearing on the horizontal displacement. If M be fixed and

DEFLECTION

257

N capable of horizontal movement, the horizontal displacement of P
includes the contraction of the member MN, but by fixing N the
contraction of MN does not affect the horizontal displacement of P.
It will be noticed that the stress in the members DT and ET due to

FIG. 201.

the vertical unit load at P is opposite in sign to the actual stress in

S SI
those members, and therefore the values of -~- will be negative and

must be subtracted from the sum of the positive values in the other
members. The unit horizontal load creates stresses in the members of

q Q< T

the lower boom only, and the values of -^-r— for all other members will

be zero.

Member.

Sectional
area, A.

Length,

'
: Total stress,
82-

Unit load

stress, S.

S2SL
A

Unit load
stress, S3.

S3S2L
A

sq. in.

ft.

tons.

ton.

180

+548-8

+2-08

+ 318-2

+ 152-4

153

+460 +2-22 + 333-7

+150-3

110

+325-7 +2-14 + 316-8

+148-1

+ 179-1

+ 1-81 + 257-3

+ 142-1

+ 58-8 +1-25

+ 193-4

+ 154-7

120

58-3

-536-3 -2-59

+ 674-8

—

53-9

-351-1 -2-31

+ 539-7

—

CB 43

52-2

-188

-1-89

+ 431-3

—

BA 15

52-2

- 61-7

-1-305 + 280-2

—

AP 17

- 75-3 -1-88

+ 532-9

—

280

120

+836-7

+ 3-5

+1255-1

—

+ 122-1 i -0-33 - 0-8

—

+ 152-7 +0-14 + 33-2

—

+ 118-2 +0-46 + 99-7

—

+ 48-3 +0-61 + 98-2

—

130

-230-8 +0-36 - 282-2

—

103

-276-6 -0-16 + 95

—

-252-1 ! -0-57 + 294-2

—

74-3

-178-7

-0-83 + 3763

—

100

156-2

-406-9 -3-25 +2065-6

—

180

100

+ 548-8 +2-2 + 5707

—

+ 8474-3

+747-6

258 STRUCTURAL ENGINEERING

Vertical displacement of P

= A _ 8474-3 x 12

14,000
= 7-26 in.
Horizontal displacement of P

= H = 747>6 X 12
14,000

= 0-64 in.

Therefore when the bridge is loaded so as to produce the stresses S.2
in the members, the point P would be displaced to Pl5 as shown in Fig.
x 201, the scale for the displacements

- 1 - 1 3~i — I being greatly exaggerated.

|l I Na£lAA Beams with Solid Webs.— The

I — I deflection of beams having solid webs

x may be found by the application of

FIG. 202. £ne foregoing principles. Consider

the work done on any section x-x, Fig. 202, by a unit load placed at
the point at which the deflection is required.

Let s = the skin stress due to any system of loading,
sl = the skin stress due to the unit load.

Then the stress on an elementary strip situated at a distance yl from
the neutral axis and of area a

= as^ due to the total loads
= as^ „ unit load

Let Z = the length of the strip.

Then the extension due to the unit load

The work performed by the unit load on the strip

The work performed on the whole cross-section

= 2E:S~^r~

But 20?/i2 = the moment of inertia of the section = I.
Let M = the bending moment at the section due to the total loads.
m = unit load.

Then

DEFLECTION
M//

259

s =

Substituting these values in the above expression, the work per-
formed on the section

= _!_ M.ml
~ 2E^~T~

The work performed by the unit load on the total length of the
beam will be obtained by substituting dx for I and integrating. This
internal work must equal the external work, or

For beams having a constant cross-section throughout their length
I will be constant, and

- 1 (
"ElJ

M.mdx

EXAMPLE 34.— To find the deflection at the end of a cantilever of
constant cross-section when supporting a load P at its outward end.

Let / = the length of the beam in inches. Then at any section
distant x from P M = — P

m = - x

A =

. x . dx

= JL *L
El' a

The following general formulae for beams of constant cross-section
may be deduced in a similar manner.

TABLE 27. — DEFLECTION OF BEAMS OF CONSTANT CROSS-SECTION.

Maximum deflection.

Cantilever.

Single concentrated load at)
the end /

Uniformly distributed load)
over the entire length . . /

Beam simply supported at the
ends.

Single concentrated load at]
a distance 6 from one end]

Uniformly distributed load)
over the entire length. . /

A =

3EI

8EI

5WZ3
384EI

260 STRUCTURAL ENGINEERING

where A = deflection in inches.
W = total load in tons.
/ = span in inches.

E = modulus of elasticity in tons per square inch.
I = moment of inertia.

Beams of varying Cross-section. — If the moment of inertia of the
cross-section of a beam is not constant throughout the entire length of
the beam the deflection will be

A-i«?.*

For the plate girder of Fig. 203, in which the moment of inertia

_____ i of the central length is I, and for the

remaining portions Ix

FIG. 203.

To find the deflection at the

centre if the girder carries a load of w tons per inch run. For any
section distant x from one support

M = wlx — \wxi
m = %x

and

(l (W ar3\ (h dx*

,.E=2Jiiiir^L+2jo!i^

CHAPTER IX.
ROOFS.

ROOFS may be divided generally into two classes : (1) roofs on which
the covering lies in a horizontal plane, (2) roofs on which the covering
is inclined to the horizontal. The supporting structure for the former
class is composed of systems of beams, the design of which was treated
in Chapter IY. The form of the structure in the second class will
vary according to the requirements of the building, the span and the
nature of the covering. Roofs of this latter class consist of three
distinct parts : the covering c, purlins /?, and some form of framed
principal or truss t (Fig. 204). The purlins serve as supports to the
covering between the principals, and their design is similar to that of
the beams treated in Chapter IV. The principals, which are spaced at
intervals along the roof, transmit the loads from the purlins to the
walls or other supports. Fig. 205 illustrates forms of principals

FIG. 204.

suitable for the spans indicated. Principals A to H are types of Y
trusses, a class which includes the great majority of roofs. The number
of members composing the truss will vary according to the span. To
obtain the maximum efficiency, the number of connections should be
kept a minimum consistent with a reasonable length, say 10 feet, of
the struts. By introducing more members into the forms E and F,
those principals may be used for larger spans than those specified. G
is an unsymmetrical Y truss, the rafters of which form a right angle
at the apex. The steeper sides of such roofs are glazed and made to
face north to prevent overheating of the interior of the building.
Types L and K are employed on railway platforms. 0, P, and S are
arched principals used for very large spans. R is a form of principal
formerly very generally adopted for large spans at railway stations, but

261

262

STRUCTURAL ENGINEERING

the present practice is to employ small span trusses resting on a system
of girders, as at T. The principals are here shown resting on longi-
tudinal girders /, which are themselves supported by the main cross
girders I. An alternative method is to have the ridges of the roof
running transversely, and the shoes of the principals resting directly

Span up fo 50 ft:

5pan up h 50 ft

^K/ VV.

5 pan up to 70ft

FIG. 205.

on the main girders. Such roofs are known as " ridge and furrow "
roofs.

Proportion of Rise to Span. — The slope of a roof is fixed according
to the material to be used for the covering, the climate, or any special
conditions. High-pitched roofs throw off rain and snow better than
low-pitched roofs, and the wind is less liable to strip off the covering
or blow the rain into the joints, but the extra length of slope increases
the cost of the covering and exposes a greater area to the wind pressure.
The most usual proportions of rise to span adopted for V trusses are
1 to 3, or 1 to 4.

ROOFS
TABLE 28.— ROOF SLOPES.

263

Rise

Angle of

Gradient of

Rise

Angle of

Gradient of

Span

slope.

Span

slope.

18° 25'

3 tol

53° 0'

f to 1

26° 35'

2 to 1

56° 20'

\ to 1

33° 42'

l|tol

63° 30'

i tol

•

45° 0'

1 to 1

The rise of the centre tie of the principal is made ^ to ^ of
the span.

Spacing of Principals. — The weight of the principals, and the weight
and depth of the purlins, are directly proportional to the distance
between the principals. If the spacing of the principals be too small,
uneconomical sections have to be used, whilst too large a spacing would
necessitate sections and connections of unwieldy proportions and ex-
cessively heavy purlins. An economical design for both purlins and
principals for ordinary span Y roofs is provided by spacing the principals
from 10 feet to 14 feet apart. Where circumstances compel increased
spacing, the purlins require to be made of very heavy sections, or in
the form of small lattice girders.

Load on Roof Principals. — The dead load consists of —

(1) The weight of the roof covering and purlins ;

(2) „ „ principal;

(3) „ „ snow.

The only live load is that of the wind pressure. The weight of the
covering may be estimated from the weights of materials given in
Chapter II., and the weight of each purlin computed after designing
such member.

Weight of Principals. — The dead weight of a principal is dependent
upon a number of conditions, the most important of which are the
following : (1) span, (2) spacing, (3) nature of covering, (4) the angle
of slope of the rafters, (5) the working stresses adopted in the members.
The formulas usually quoted as giving the approximate weight take
into consideration only the first two of the above conditions, but the
differences in weight of the roof coverings play such an important
part in determining the weight of the principals that a constant has
been added to the following formula to provide for these differences of
weights of coverings. Three values for the constant are given: for
heavy, medium, and light coverings, but other values between these
limits may be adopted to suit the class of covering employed.

The approximate weight of mild steel principals in pounds

= cDL(l + iL)

where D = the distance between the principals in feet ;
L = the span of the principals in feet ;
c — a constant for the nature of the covering ;
= J for heavy slated roofs ;
= J for glazed roofs ;
= {£ for light corrugated roofs.

264

STRUCTURAL ENGINEERING

The above formula is deficient, as no account is taken of the
conditions (4) and (5) stated above, and it is only recommended as
giving a first approximate estimate of the weight, which should be
checked on the completion of the preliminary design, and if found
seriously in error, corrected.

Snow Load. — The maximum weight of snow in England is usually
assumed to be 5 Ibs. per square foot of the horizontal area covered by
the roof. Many designers prefer to omit this load as being unlikely
to occur when the roof is subjected to the maximum wind pressure.

Wind Pressure. — The force exerted by the wind on roofs is very
indeterminable, and only an arbitrary pressure can be assumed. In
extreme situations the pressure may be modified, but an average allow-
ance of 40 Ibs. per square foot of projected vertical surface, acting
horizontally, may be assumed. The component of the wind pressure
normal to the roof slope only, will produce stress in the purlins and
principals ; the component acting along the slope does not, except in
a very slight degree, due to friction, affect the stresses in the roof
members. The normal wind pressure on the slope of a roof may be
calculated from the following empirical formula.

Button's Formula. —

Let PN = the normal pressure on the slope,
P = „ horizontal „ „

the inclination of the slope.

i =

'84 cos * ~

Then PN = P(sin
P = 40 Ibs. per square foot, the curve of normal

By assuming
pressures (Fig. 206) has been plotted.

iKhnaHon (
raftws.

o to'

Ratio if fi/se to Span

I I

FIG. 206.

Application of the Loads. — The area of the slope supported by any
purlin is equal to the distance between the principals multiplied by
half the distance between the purlins to either side of the purlin under
consideration. The purlin may be designed as a beam simply supported

ROOFS

265

at the ends and carrying a distributed load equal to the sum of the
dead weights of the covering and snow and the normal wind pressure
on the above area.

To prevent bending stresses in the rafters the purlins should
always be connected to the principals at the junctions of the struts and
rafters. The principals will then be subjected to loads equal to the
reactions of the purlins, at the bearings of the purlins on the rafters,
plus their own dead weight, which is assumed to act at the same points
in the rafters as the purlin loads.

The dead load will produce constant stresses in the members of the
principals, but the total stresses in the members will vary according to
the direction of the wind. The maximum stress in any member will
be ascertained by considering the wind acting on each slope of the
roof separately, and combining the maximum stress produced by the
wind with that due to the dead loads.

Reactions. — The reactions of the loads at the shoes of the principals
are found in a similar manner to the reactions of simple beams
supporting concentrated loads. The dead load reactions of a symmetrical
truss equally loaded on either slope, will be equal to half the total dead

FIG. 207.

load. If the truss be unsymmetrical, or the dead loads be of varying
intensity, the reactions may be found by the method of moments or
graphically. Consider the truss in Fig. 207. Taking moments about
the left-hand support —

— x b+P Xa

rfT 2 d' d

is equal to the total load minus R2, therefore Rx = (2P + 28) - R2.

266

STRUCTURAL ENGINEERING

The reactions are found graphically by the following method. Set

P P -I- S S

out the loads -, P, — ^ — , S, and ^ on a vertical line as in Fig. 207, e.

Select any pole 0, and connect it by straight lines with the ends of
each load in the load line Im. Between the load lines of ^ an^ P,

Fig. 207, /, draw a parallel to the line in Fig. e, joining the pole
0 to the junction of these loads in the load line. From the inter-
section of this parallel and the load line of P, draw a parallel to the
line in Fig. e, connecting 0 with the junction of the loads P and

T> I Q Q

— 2 — . Continuing this method, a point h in the load line ~ is

obtained. Join g and h, and from 0 draw a parallel cutting the load
line in k. The load line will then be divided in the same ratio as the
reactions. R! will be equal to Ik and R2 equal to km. Fig. e is
known as the polar diagram, and Fig. 207, /, the funicular polygon.

The position of the resultant of the loads may be obtained by
drawing from g and h lines parallel to 01 and Om respectively. The
point in which these lines intersect will be a point in the line of action
of the resultant. The direction of the resultant will be parallel to the
line joining I to m, in this case vertical.

Wind Load Reactions. — Let R, Fig. 208, be the resultant normal
wind pressure acting on the left-hand slope of a roof. If the principal

be securely fixed to the supports at
each shoe, the reactions must be
parallel to the resultant normal
pressure R, and taking moments
around h

R, x b = R x a

R^ = R - Ra

The reactions may be found
FIG. 208. graphically as follows. In the line

of the resultant set out ef to scale

to represent the resultant R. On a line through c, parallel to the
resultant, make cd equal to ef. Join hd cutting ef in g. Then eg will
be equal to the reaction at c , and gf equal to the reaction at h.

To allow for expansion in large span trusses, one end of the truss is
supported on roller bearings. Since such bearing cannot resist any
horizontal force the reaction at the bearing must be vertical. Suppose
the left-hand shoe of the principal (Fig. 209) be capable of moving
horizontally, then the reaction R! must be vertical. By mechanics it
is proved that for a system of forces to be in equilibrium the forces
must either be all parallel or they must all pass through one point.
Knowing, then, the direction of R and R1? the direction of R2 may be
found by producing R and R! until they cut, and joining the point of
intersection to the right-hand shoe. The magnitudes of the reactions
are obtained from the triangle of forces abc —

ROOFS

267

R2 = ac

The dotted lines show the construction if the left-hand shoe be fixed
and the right-hand free.

Wind Reactions for Curved Roof. — At the change of slope of the
roof in Fig. 210 two pressures p and pl will be produced by the wind

FIG. 209.

FIG. 210.

acting on the different slopes. The resultant pressure P at the joint is
obtained by a parallelogram of forces. Constructing polar and funicular
diagrams a point q is obtained, through which the resultant wind
pressure must act. But the resultant is equal in magnitude and
direction to the line joining g and #, the ends of the load line in the
polar diagram, and therefore is completely known. If the ends of the
principal be securely fixed the reactions will be parallel to the resultant
gb, and may be obtained by drawing oa parallel to Jcl.

Then

R! = ab
R2 = ag

If either end be free the reactions are found, after obtaining the
resultant 11, by the method of Fig. 207.

Reactions for Cantilever Roof. — Let D
represent the resultant dead load on the
cantilever roof of Fig. 21]. The hori-
zontal member de cannot transmit any
vertical load to the connection at d, there-
fore the whole vertical dead load must be
borne by the connection at c. The resultant
I) produces a turning moment about d,
which must be resisted by a force at c
having an equal but opposite moment about
d. The horizontal component of the reaction at c will therefore

be equal to D x T. The total reaction RD at c may be found by

— a — •»

FIG. 211.

268

STRUCTURAL ENGINEERING

combining the vertical and horizontal components by means of a
parallelogram of forces. The reaction at d may be found by taking
moments about c.

Rx x b = D x a

Since the directions of D and Rj are known the reactions may be
found graphically. The three forces D, Rj, and RD not being parallel
must pass through one point. The direction of RD will therefore be
ce, and the magnitudes of RD and Rj may be found by means of a
triangle of forces.

The directions of the reactions for the wind pressure W are Rw
and R!.

Reactions for Double Cantilever Roof. — The dead load on the double
cantilever roof (Fig. 212) produces a reaction in the central column
equal to the total dead load on the principal. Let R represent the
resultant normal wind pressure acting on the left-hand slope of the roof.
No single force applied at a can produce equilibrium, as the resultant
R has a turning moment about a equal to R X b. To produce
equilibrium a force having an equal and opposite moment must act
about a. This force is the resistance to bending offered by the vertical
member ca. To find the stresses in all other members of the truss

FIG. 212.

FIG. 213.

a horizontal force R^ is assumed to act at c, its magnitude being equal to
(see Fig. 218).

Reactions for Cantilever Roof on Ttvo Supports. — If the dead load on
the truss of Fig. 213 be symmetrically disposed, the dead load reactions
in the columns will each be equal to half the total dead load. The wind
acting on the left-hand slope produces reactions at a and c parallel to
the resultant of the wind pressure. Taking moments around a,

RN X b = R2 x d

and

/. R2 = RN x

T? T? i

-til = ±1N -f~

After obtaining the reactions the stress in each member of the frame
may be found either graphically or by calculation. The graphic method
is probably the quicker, and the closing line serves as a check upon the
accuracy of the diagram.

ROOFS

269

Stress Diagrams. — At each connection of the frame there are a
number of forces acting which may be represented by the sides
of a polygon. At the left-hand support (Fig. 214) there are four forces

produced by the dead load, viz. R1? — , the stress in the rafter and the

stress in the tie. These forces may be denoted by the letters or figures
to either side of them. In Fig. 214, w, a polygon has been constructed

p
for the above forces ; ab = the reaction, be = the load -~ , and the lines

cl and a\ being drawn respectively parallel to the members Cl and Al
of the truss, represent the stresses in those members. Proceeding now
to the joint CD21, there are two known forces, Cl and CD, and two
unknown forces, D2 and 2-1. From the point c, Fig. 214, m, set off cd
= the load P to the same scale as ab and fo, and from the points d and
1 draw parallels to D2 and 1-2 respectively, intersecting in point 2.

FIG. 214.

Wind Load Diagram.

The length 1-2 represents the stress in the member 1-2, and 2-d
represents the stress in the upper portion of the rafter. By following
this method of construction for the joints DE32, EF43, and FGA4, a
complete dead load diagram for the whole frame will be obtained. A
check on the accuracy with which the diagram has been drawn is
afforded when the joint at the right-hand support is reached. The
point 4 on the diagram m has already been fixed when considering the
previous joint, but its position is checked by the lines drawn horn a and/
parallel to A4 and F4 respectively, which must pass through the point 4.
The above construction requires modification when more than two
unknown forces act at a point. When dealing with the stress diagram
for Fig. 231 such a case will be considered. In practice it is usual to
complete the load line bg before commencing the other part of the
diagram. It will be observed that the dead loads at the shoes do not
affect the stresses in the members of the principal, and it is usual to
omit them from the diagram.

270 STRUCTURAL ENGINEERING

Wind Load Diagram. — Let the maximum wind pressure act on the

W W

left-hand slope of the roof, producing forces ^-, W, and -~- at the joints

in the rafters. The reactions may be found by the method previously
described for principals with fixed ends. Set out the load line abcdga,
Fig. 214, n, parallel to the normal wind pressure. Commencing with
the joint at the left-hand shoe, draw from a a line parallel to Al and
from c a line parallel to Cl. Where these lines intersect gives the
position of the point 1. The positions of the points 2 and 3 are found
in a similar manner. Since no load acts on the right-hand slope one
letter only is required on that slope to denote the forces. The letter
G has been used, so that the force at the apex will be DG and the
reaction will be denoted by GA. After establishing the position of the
point 3 the next joint to be considered is 3G4. As G4 is parallel to G3
the point 4 must fall on the line #3, and as it must also be on a line
parallel to 3-4 and passing through the point 3, the only position
possible for the point 4 is coincident with 3. Therefore no stress due to
the wind load occurs in the bar 3-4. Through a draw #4 parallel to
A4, which must pass through the point 4 already determined.

Maximum Stresses in the Members. — The total stress in any member
will be the algebraic sum of the dead load and wind load stresses. The
maximum stress in a number of the members will occur when the wind
pressure acts from the left, and in the other members when the wind
pressure is exerted on the right-hand slope. In the present case the
maxima stresses in members symmetrically placed will be equal, and
there is no necessity to draw another diagram for the wind pressure
acting on the right-hand slope.

Character of Stresses. — It is also necessary for purposes of design to
know the character of the stresses in the members. The direction of
action of at least one force at each connection is known, and therefore
from the polygon of forces for that connection the direction of the
remaining forces may be determined. Consider the polygon abcla of
the forces acting at the left-hand shoe. The direction of the reaction
a-b is upwards, and that of the force b-c downwards. The remaining
forces must continue in the same direction round the polygon, i.e. c to
1, 1 to a. Transferring these directions on to the line diagram of the
principal, it will be seen that the stress in the rafter acts towards the
joint at B, whilst the stress in the member Al acts away from
the joint, thus indicating that the rafter is in compression, and the
member Al in tension. At the joint CD21 the direction of the force
CD being downwards, the direction of the remaining forces in the
polygon will be d to 2, 2 to 1, and 1 to c. Transferring these directions
o-n to the principal it is observed that all the members meeting at this
joint are in compression. Continuing this process at the other joints
the nature of the stresses in all the members is determined.

Calculation of Stresses by the Method of Sections. — This method
of calculating the stresses in any frame is based upon the following
principle, proved in mechanics : if a structure be in equilibrium, the
algebraic sum of the moments of all the forces acting in one plane, and
to either side of any section, about any point in that plane, must
be zero.

ROOFS 271

Suppose the roof principal in Fig. 215 be subject to the vertical
loads indicated in the figure. To find the stress in any member, say
AB, draw a section X-X cutting such member. The forces acting to
the left of the section X-X are : the reaction at A and the stresses in
the members AB and AD, cut by the section. For these forces to be
in equilibrium the sum of their moments about any point in the plane
of the principal must be zero. By taking moments about any point #
in AD, Fig. 215, c, the moment of the stress in AD will be zero, and
there remains then only the unknown moment of the force in AB.
The portions of the figure to the left of the sections XX, YY, and ZZ
are reproduced to a larger scale in Figs, c, d, and e.

Let SAB, SBC, SAD, etc. = the stresses in the respective members.

Denoting all clockwise moments as positive and anticlockwise
moments as negative, then for the stress in AB, taking moments about
the point #, Fig. 215, c.

21 x 6' + SAB x 2' 8" = 0

21 X 6

2-67
= — 47 '2 cwts.

The above value of SAB being negative indicates that its moment
about b is anticlockwise, and therefore the force in AB acts towards the
joint at A. Hence the stress in the member AB is compressive.

To calculate the stress in the member AD take moments about the
point a, Fig. 215, c.

21 X 4-5 + SAD X 2'33 + SAB X 0 = 0

/. SAD = - 40-6 cwts.

The negative sign again indicates an anticlockwise moment. The
force in AD will therefore act from the joint at A and be of a tensile
character.

To find the stress in the member BC take a section YY, cutting
that member and the least number of other bars. The forces to the
left of the section YY are : the reaction at A, the load at B, and the
stresses in the members BC, BD, and AD, cut by the section. The
forces in the members BD and AD acting through the point D, their
moments about D are zero. The only unknown moment is therefore
that due to the stress in BC. Taking moments about D, Fig. 215, rf,

21 X 15 - 14 X 7J + SBC X 6-75 = 0

SBC = ~ 31'1 cwfcs-

The value of SBC being negative, the force in BC must act towards
the joint at B, and therefore BC is a compressive member.

To find the stress in BD. Take moments about A, Fig. 215, d.

21 x 0 + 14 x 7'5 + SBD X 7'125 + SBC X 0 + SAD X 0 = 0
.-. SBD = - 14-7 cwts.

The moment of SBD about A being anticlockwise, the member BD
will be in compression.

272

STRUCTURAL ENGINEERING

To find the stress in CD. Take the section ZZ cutting the members
BC, CD, DE, and DF. The forces on the left of the section are :
the reaction at A, the load at B, and the stresses in the members BC,
CD, DE, and DF. As the principal is loaded symmetrically, the
stresses in DE and DF are equal to the stresses in BD and AD

<L ~vv* ^~ i

5| «,„.

^.-r, — i

FIG. 215.

respectively. If the loading were not symmetrical the forces in DE
and DF would be calculated in a similar manner to the forces in BD
and AD.

Taking moments about A, Fig. 215, e,

21 X 0 + 14 x 7-5 + SBC X 0 4- SDE x 5-41 + SDF X 1'67 4- SCD X 15 = 0

J4 x 7-5 4- 14'7 X 5-41 -f 40'G X l'G7 + SCD X 15 = 0

.-. SCD = - 1C -8 cwts.

The moment of SCD about A being anticlockwise, the member CD is
in tension.

ROOFS

273

Wind load stresses for all the members may be obtained in a
similar manner.

It was shown in the wind diagram of Fig. 214 that no wind stress
occurred in the member 3-4 when the wind acted on the left-hand slope
of the roof. This is clearly demonstrated by the above method of
calculation. Take a section XX, Fig. 216, and moments about the

FIG. 216.

right-hand shoe, of all the forces acting to the right of the section —

(Ro + SA4 + SG3) X 0 + S3_4 x e = 0
.*. S3_4 x e = 0

that is, there is no stress in the member 3-4.

Fig. 217 is another example of the graphical method of determining

the dead load and wind load stresses in a roof principal. The com-
pressive stresses are indicated by the heavier lines.

In Fig. 218 are drawn the stress diagrams for a double cantilever
roof. The dead load diagram presents no difficulties, but the bending
action in the member 3-4 makes it impossible to obtain a closed wind
load diagram of the direct forces in all the members. The magnitude
of the bending moment in 3-4 has already been found in Fig. 212 to be
R X e. By calculating the horizontal force at the apex necessary to

274

STRUCTURAL ENGINEERING

balance this bending moment and plotting it at d£ on the wind diagram,
a closing point 4' is obtained. The stresses produced by the wind in
the member 3-4 are, the direct compressive stress 3-4' and the stresses
due to the bending moment d4' x /. All other stresses in the diagram
are the direct stresses in the respective members.

?- 5.6

Wind Diagram Dead Load Die

FIG. 218.

Three-hinged Arch Roofs. — Roofs of this description are frequently
used for exhibition and other buildings requiring very large span roofs.

The pins or hinged joints allow the apex
of the roof to rise or fall when, owing to
changes in temperature the expansion or
contraction affects the length of the arch.
The increase in stress in the members of
rigid and two-hinged arches due to
changes in temperature is considerable
and very tedious to estimate, but the use
of the third hinge at the crown allows of
changes in the length of the arch without
affecting the stresses in the frame.
Suppose a vertical load W be supported by a three-hinged arch and the
reactions be E1 and R2, Fig. 219.

Let Vj and V2 be the vertical components of the reactions.
Then the vertical forces being in equilibrium

w = v, + v2

FIG. 219.

ROOFS

275

Taking moments around the base hinges

Wx = V2Z or V2 = —-

and

or «

Taking moments about the crown, of the forces to the right

(1)

where H = the horizontal component of either reaction.
Substituting for Y2

Wx
2D

H =

To produce equilibrium the horizontal components, H, of the
reactions Ra and R2 must be equal. Let a = the angle of inclination

of the reaction R2. Then tan a = ^r

±1

V 2D

From the equation (1) above Tf = ~r-

.*. tan a =

The reaction R2 must therefore pass through the crown hinge.
Knowing the lines of action of two of the forces acting on the arch, the
direction of the third, Rlf may be obtained, since all three forces must
pass through the same point. Producing R2 through the crown hinge it
cuts the load line at a. Joining the left-hand hinge to a gives the direction
of Rj. The magnitude of the reactions may be obtained graphically
by a triangle of forces.

The same construction holds good if the load be inclined to the
vertical. Taking moments of the forces to the right, about the crown,
Fig. 220,

FIG. 220.

FIG. 221.

276 STRUCTURAL ENGINEERING

R.J must therefore pass through the crown.

If each half of the arch be loaded, the reactions may be found
graphically by treating each load separately and combining the
reactions so found, Fig. 221.

The reaction at the left support

due to Wl = R/'
„ » W> R/

The total reaction = R,

The reaction Rg is obtained in a similar manner.

If the loads be equal and similarly disposed
then Yj = T2 = Wx = W
and the vertical shear at the crown

= v, - W, = 0

The force on the hinge at the crown inusfc therefore act horizontally,
and the reactions and thrust at the crown may be found graphically as
follows, Fig. 222. Through the crown draw a horizontal line to cut
the load line in a. Join a to the base hinge. Then ab is the line of
action of the reaction. The magnitudes of R! and H are again obtained
by means of a triangle of forces.

FIG. 222. FIG. 223.

If the two halves of the arch be unequally loaded the thrust at the
crown will be inclined, Fig. 223. Its direction and magnitude may be
found by the following construction. Obtain the reactions R! and R2
by the above methods and draw the force diagram Wi, W2, RI, and R2.
The thrust on the crown is then equal to «£, the components of which
are, the horizontal component of either reaction and the difference of
the vertical components of the reactions.

In Fig. 224 are drawn the dead load stress diagram for the left-
hand half of a three-hinged arch and the wind stress diagram for the
whole arch, the wind assumed acting on the right-hand slope. The
positions and magnitudes of the resultants of the dead and wind loads
are found by means of polar and funicular diagrams. The dead load
on the two portions of the arch being equal, the thrust on the centre
pin due to that load will be horizontal. The magnitudes of the reaction
ab and the central thrust am are obtained on the dead load diagram.
The dead load stresses in the members of the right-hand portion will be
similar to the stresses in the corresponding members of the left-hand
portion. Sinoe the wind exerts no pressure on the left-hand half of
the truss, the reaction R15 due to the wind, passes through the centre
pin. R-2 passes through the right-hand pin and the intersection of RI
and the resultant wind pressure. The reactions on the wind stress diagram
are obtained by drawing parallels to Ra and R2 from I and W, the ends

ROOFS

277

of the resultant, respectively. These lines intersecting very obliquely
at «, the magnitudes of the reactions should be checked by calculation,

Oeaaf.Load Stress

D/aqram.
Left- Hand Haff

MM! Load Stress

Diagram.
Mnet oft Right

FIG. 224.

as a small error in the position of a will render a closed stress diagram
impossible.

The character of the stress in some of the members will vary

278

STRUCTURAL ENGINEERING

according to the direction of the wind. For example, the stresses in
the member 10- A are —

Due to dead load . compressive ;

„ wind on the right compressive ;

left . . . tensile.

The maximum compression in this member will therefore occur
when the direction of the wind is right to left, and will be the sum
of the dead and wind load stresses. The minimum compression or the
maximum tension will be the difference of the stresses due to the dead
load and the wind acting on the left-hand slope. The nature of the
stresses in the other members of the same panel are given in the
following table.

Member.

Stress due to

Maximum
compression
= stresses in
columns.

Maximum
tension
= stresses iii
columns.

Dead load.

Wind on right.

Wind on left.

11-H

1+3

2-1

compression

tension

compression

10-11

tension

compression

3-1

1+2

11-12

compression

tension

1+2

3-1

Ends of Roofs. — Where the ends of roofs are not supported on
walls, special provision must be made for covering in the ends by

either employing vertical
framing, an example of
which is given in detail in
Fig. 227, or providing an
inclined covering supported
on special end principals.
The usual arrangement of
these principals is shown in
Fig. 225, but for larger
spans additional principals
may be inserted at a and b.
A similar method of appor-
tioning the loads and deter-
mining the stresses in the
members of these hip and

half principals is followed as for ordinary V principals. The first
common principal pp must be specially designed for the additional
loading at its apex imposed by the end principals, the vertical com-
ponents of such loads only being considered, as the horizontal com-
ponents of the wind loads are transmitted through the wind bracing of
the roof. The members c of the end principals are not, theoretically,
stressed, and are only employed to add lateral stiffness to the lower
portions of the hip and half trusses.

Wind Bracing. — Wind bracing in roofs is employed to counteract

FIG. 225.

ROOFS

279

the overturning moment of the wind acting on the ends. The bracing,
together with the purlins and rafters, form girders in the planes of the
roof slopes which transmit the horizontal wind forces to the eaves. It
is usual to brace only the first three or four bays, as this forms a
sufficiently stiffened frame to resist the wind moment.

Consider the roof in Fig. 226 whose ends are covered by vertical
sheeting or glazing carried by the horizontal members at the purlin
levels.

The wind pressure, assumed acting horizontally and with an intensity

FIG. 226.

of 40 Ibs. per square foot, transmitted to the purlins by the members
in the end screen will be, to

«, - ^ X 5 x 40 = 5400 Ibs.

b, - Sfi X 5 x 40 = 3600 „

c, - i/ x 5 x 40 = 1800 „

d, - | X § X 40 = 450 „ say 448 Ibs.

As the frame formed by the bracing, purlins, and rafters is re-
dundant, the assumption will be made that the bracing systems in each
bay take an equal proportion of the loads. The horizontal compres-
sion in the purlins and ridge will diminish by one-quarter of the total
stress at each of the first four common principals.

The horizontal component of the stress in the tie e and the
corresponding member in the other slope

= (448 - 336)i = 56 Ibs.

This stress is transmitted to the purlin c, making the total compres-
sive stress in that member

= 1800 + 56 = 1856 Ibs.

The tie / will be stressed by the force transmitted by the system
from the ridge and the difference of the stresses in the purlin c in the
first two bays. The horizontal component of the stress in/

= 56 + (1800 - 1350) = 506 Ibs.

280 STRUCTURAL ENGINEERING

The horizontal stresses in the other web members of the first bay
will be —

in b = 3600 + 506 = 4106 Ibs.

„ g = 506 + (3600 - 2700) = 1406 Ibs.

„ a = 5400 + 1406 = 6806 Ibs.

„ h = 1406 + (5400 - 4050) = 2756 Ibs.

The direct stress in the ties will be

in e = M = 56X1-32

= 74 Ibs.

„ /= 506 X 1'32 = 668 Ibs.
„ g = 1406 x 1-32 = 1856 Ibs.
„ h = 2756 X 1-32 = 3628 Ibs.

The stresses in the bracing bars of the other bays will be similar to
the above.

It is evident that such small stresses in the braces would require
theoretical sections below practical limits. If round bars be employed,
the diameter should not be less than f inch, and flats of less than
2" x |" are undesirable. Such sections would admit of a less number
of bays being braced, but the increase in stress in the lower purlins by
decreasing the number of braced bays would be a greater disadvantage.

The stresses in the three intermediate rafters are —

in Jc = the component down the slope of the stress in e

}> ' == ?? 59 75 /

.» m = » » )* ff

>» ^ == »> » »> ""

The stresses in the last braced rafter are—

in o = the component down the slope, of the stress in e

„ p — sum of components down the slope, of the stresses in e and/

The maximum stress in the rafters will therefore occur in s and

= (56 + 506 + 1406 + 2756) X -

1 A"3

= 4724 x i^ = 4055 Ibs.

There will be a horizontal shearing stress at the shoe connections
of the first four principals equal to the horizontal component in the
lower braces,

i.e. = 2756 Ibs.

The components down the roof slope, of the stresses in e, f, and g,

ROOFS 281

create tension in the end rafter, the maximum amount of which occurs
in t, and

-i A.O

= (56 + 506 + 1406) X ^ = 1689 Ibs.

The effect of this tension is merely to neutralize a portion of the
dead load compression in t.

Counterbracing, shown by dotted lines, is inserted to withstand the
frictional wind force on the slopes when the wind blows in the opposite
direction.

If the roof has sloping ends, the wind pressure transmitted through
the wind bracing will be the horizontal component of the normal
pressure on such slopes.

Wind Screen. — Fig. 227 shows the general arrangement of framing
for the wind screen of a railway station covered by a V roof. The
entire wind pressure on the screen, and the weight of the screen itself,
is transmitted through the main girder, which is situated at about the
middle of the depth, to the end columns. The glazing of the screen
above and below the main girder is secured to horizontal tees H, or
double-angle bars, which are themselves supported by vertical canti-
levers, V, Y.

The method of design is as follows. The horizontal tee bars are
considered as beams simply supported at the ends, and of span equal
to the distance apart of the cantilevers. The forces acting on each
tee are, the weight of the glazing on the area of the screen supported
by the tee, acting vertically, and the horizontal wind pressure on the
same area. The maximum intensity of stress in the tee will be the
sum of the intensities produced by the vertical and horizontal loads.

The loads on the cantilevers will consist of the vertical and
horizontal forces transmitted from the tees at their connections with
the cantilevers and the horizontal wind pressure on the apex beams, A.
The reactions and stresses in the members of the cantilever may be
found by the graphical or mathematical methods previously described.

The loads on the main girder G- will consist of: (1) the forces,
both vertical and horizontal, equal to, but opposite in direction, to
the reactions of the cantilevers ; (2) the dead weights of the canti-
levers, main girder, and glazing directly connected to the girder ; and
(3) the horizontal wind pressure on such glazing. The girder is made
rectangular in shape and braced on all four sides, so forming two
vertical and two horizontal girders. The horizontal loads on the main
girder are transmitted through the web members of the top and bottom
girders, and all vertical forces through the web members of the vertical
side girders. The horizontal corner members m will form the flanges
of both the horizontal and vertical girders, and the stress in them will
be the sum of the stresses produced by the horizontal and vertical loads.
The unequal loading of the top and bottom girders and the^side girders
tends to produce a racking action on the whole girder, which must be
braced diagonally to counteract such action. The reactions of the main
girder will have both horizontal and vertical components, and the con-
nections to the columns and the columns themselves must be designed
to withstand such forces. Braced columns would preferably be

STRUCTURAL ENGINEERING

ROOFS

283

employed in such situations. The apex beams A will be designed as
part of the roof, subject only to the usual wind and dead loads.

Design of Members. — The design of the individual members of
roof principals follows the same general rules as those given for braced
girders. The axes of the members at each connection must pass
through the same point, otherwise bending stresses are produced in the
members and connections.

Struts. — The assumption has been made when determining the
stresses, that all the members of the frame are pin-jointed, but in
practice riveted or bolted connections predominate. However, when
designing compression members the assumption is still adhered to.
The design of pin-ended struts has been fully discussed in Chapter V.,
and will not be entered into again here. The usual forms of struts
employed for roofs are tees, single and double angles, beam sections,
single and double channels, double flats with distance pieces, and
occasionally tubing.

Ties may be made of any convenient section, those most generally
employed being flat, round, or angle section. The strength of ties is
calculated on the section of minimum area. When round bars are
used eyes or forks are forged at the ends for connection to the other

FIG. 228.

members. Various proportions have been suggested for the eyes, the
most commonly adopted being those given in Fig. 228, in terms of the
diameter D. The diameter of pin and the thickness t of eye or fork
may be obtained as follows —

Let P = the pull in the tie in tons.

/, = the safe shearing stress on the pin in tons per square inch.
/6 = „ bearing „ „ „

di = the diameter of the pin.

Assuming that the shearing resistance of the pin is 1J times that
of single shear,

284 STRUCTURAL ENGINEERING

then P=

The bearing resistance of the pin
= P = djft

Assuming /. = 5 tons per square inch

fb — $ 55 55

and ft = 6 „ „

where/* is the intensity of stress in the member,

then d^ in the above proportions = 0'83 D

and * „ „ =0-7D.

The breadth b of the connecting plates k is found from

P»(ft-4)*x/i

••••-!+*

Riveted connections are designed by the methods explained in
Chapter IV. The centre of gravity of the system of rivets or bolts
connecting one member to another should lie on the axis passing
through the centre of gravity of the member, otherwise bending
stresses will be produced.

FIG. 229.

In order to avoid obstacles ties have occasionally to be cranked, in
which case the maximum stress in the member is the sum of the
direct and bending stresses. In the tie in Fig. 229 suppose the pull
= P tons. The intensity of direct tension at any vertical section
= P -f- sectional area, A.

At any section between a and I the force P acts at a leverage of x
inches from the axis of the bar, thus producing a bending moment
= P# inch-tons. The moment of resistance of the section =/Z.

Px
/. P.E = /Z, and maximum stress due to bending =/=-»

This stress will be compressive along the edge ab, and tensile along
edge cd. Hence maximum tensile stress along edge cd, due to both
bending and direct pull

_Px P
= Z +A
which must not exceed the safe tensile resistance of the material.

ROOFS

285

Purlins are designed as simply supported beams of span equal to
the distance between the principals. The load consists of the normal
wind pressure and dead load on the area apportioned to the purlin. It
is usual to arrange the principal axes XX and YY of the purlin (Fig.
230) parallel and normal to the slope of the roof. The wind pressure
acting normally on the purlin causes bend-
ing about the axis XX. The vertical dead
load has a normal component ob, also
causing bending about XX, and a lateral
component Z>Y causing bending about YY.

Suppose the loads on the purlin, Fig.
230, be—

total wind pressure = 1-5 tons

total dead load = 1 ton __

and the span =12 feet. FlG 230

Resolving the dead load normally and parallel to the roof slope,
then the total load normal to the slope = 1*5 -f 0'88 = 2*38 tons, and
the load parallel to the slope = 0'44 ton.

Bending moment about the axis X-X

= ?j*§_Xj^<_12 _ 42.8 in _tong

Bending moment about Y-Y

0-44 X 12 x 12

= 7'92 in.-tons.

Suppose the purlin to consist of a 7" x 4" rolled beam section whose
modulus of section

about X-X = 11-2 ins.3
„ Y-Y= 1-7 „

The maximum intensity of stress due to bending

about

X-X = - § = 8'8 tons per square inch
1 1 "Z

Y Y - 7-92 _ ,.fi

„ I- I — -— ; -~ -41) „ „

1*1

and the total maximum intensity

= 3-8 _j_ 4-0 = 8-4 tons per square inch.

This stress being rather high a larger section would be preferable.

The error made by assuming the dead load to act normally to the
axis X-X of the purlin is demonstrated by the following calculation.

Then the total bending moment about X-X

= (±fi + 1) X 18 X 12 = 45 in..toM.
8

286 STRUCTUEAL ENGINEERING

The maximum intensity of stress

= — '— = 4 tons per square inch,

which appears to be a low working stress, but it has already been
shown above that such section would actually be stressed to 8 -3 tons
per square inch.

It is evident that steep roof slopes require purlins having a
comparatively large moment of resistance about the axis Y-Y.

EXAMPLE 35.— Design of a roof truss of 70 feet span.

The following conditions to be complied with : —

One shoe to be fixed, the other to rest on a roller bearing.

The spacing of the principals to be 12 feet between the centre
lines.

The covering to consist of slates resting on boarding and common
rafters.

The wind pressure to be assumed as acting horizontally with an
intensity of 40 Ibs. per square foot.

All members to be designed upon the maxima stresses, using the
following working stresses :—

Purlins ..... 6 tons per square inch.

Tension members . . 6 „ „

Compression members . The working stress to be two-thirds the

safe dead load stress for round-ended
struts of mild steel as given in the
curve of Fig. 209.

Shear stress .... 5 tons per square inch.

Bearing ...... 8 „

A suitable type of truss to adopt for such a span is the French truss
(Fig. 231). Making the rise | of the span the distance between the
purlins is 9*8 feet, and the area of slope supported by each inter-
mediate purlin = 9-8 x 12 = 117*6 sq. ft.

Dead Load. — From the weights of the materials given in Chapter
II. the dead loads are found to be —

Covering =16 Ibs. per square foot.
Approximate weight of principal = 46 cwts.

Snow = 37*5 cwts. per principal.

Wind Load.—'From Fig. 206 the normal wind load for this slope
= 23-5 Ibs. per square foot.

Purlins. — The loads on the intermediate purlins are : —
Dead load of covering

= !<L2L™ = ifi-8

snow = | X 37-5 = 47 „

Total = 21-5 „

ROOFS 287

Normal wind load

The dead load acting normally to the slope

= 21-5 x 0-88 = 18-9 cwts.
The dead load component parallel to the slope

= 21-5 x 0-44 = 9-5 cwts.

Maximum bending moment on the purlin due to the loads acting
normally to the slope

= (24'7 + * 12 X 12 = 39-24 in-torn.

and due to the load acting parallel to the slope

For convenience in fixing the common rafters, the purlin will be
composed of a rolled steel zed bar and a 9" x 3" timber runner.

Try a 7" X SJ" X 8J" zed bar.

Modulus about X-X = 12'745 ins.3

„ Y-Y= 3-52 „
and maximum intensity of stress due to bending

+ |^ = 5-2 tons per square inch.

12*745 «V52

This section is therefore adopted for the purlin.

A similar calculation will show that a pair of 6" x 4" x f" angles
will be strong enough for supporting the ridge. The load on the
purlins at the shoes being only half that on the intermediate purlins,
a 7" X 4" x i" angle will be suitable.

Load on Principal — Dead load on each intermediate joint —

Covering and snow load =21-5 cwts.

Weight of purlin = 2*0 „
Proportion of weight of principal = ^ = 5*75 „

Total . . = 29*25 „
say 30 cwts.

Wind load on each intermediate joint = 24-7, say 25 cwts.

Stress Diagrams. — The stress diagrams for the dead load and for
the wind acting on either slope, the left-hand shoe being supported on
roller bearings, are drawn in Fig. 231. When drawing the diagrams
a difficulty is met with when the joint D-E-5-4-3-2 is reached. Here
there are three unknown forces, and the usual procedure provides for only
two. The points 4, 5 and 6 on the dead load diagram may be obtained
in the following manner. Select on the line e-5 any point 5', and draw
lines 5'-6', o'-4', and G'-4' parallel to 5-6, 5-4, and 6-7 on the principal,

288

STRUCTURAL ENGINEERING

so obtaining the point 4'. The correct position of the point 4 is on
the line 3-4. By moving the triangle f/-4'-G' in a direction parallel
to e-5' until the apex 4' rests on the line 3-4, the correct positions of
4, 5, and 6 are obtained. It will be noticed that in all three diagrams
the stress lines 1-2 and 5-6, and 8-9, 12-13 are in the same straight

Wind oft /eft

FIG. 231.

line. This is due to the joints C-D-2-1 and E-F-G-r> being equally
loaded, and in all such cases the points f> and 0 may be obtained by
producing the stress line 1-2 until it cuts the lines *-5 and /-(I.

The stresses in the members as scaled from the diagrams are given
in the following table.

EOOFS
TABLE OF STRESSES.

289

Member.

Stresses.

Length of
struts.

Dead load.

Wind on left.

Wind on right.

Total maximum
stress.

0-1

cwts.
263

cwls.
123-2

cwts.
70-4

cwts.
386-2

ft.
9-8

D-2

250

123-2

70-4

373-2

&-5

237-5

123-2

70-4

360-7

F-6

223

123-2

70-4

346-2

G-8

223

62-8

132-8

355-8

H-9

237-5

62-8

132-8

370-3

K-12

250

62-8

132-8

382-8

L-13

263

62-8

132-8

395-8

1-2

26-5

52-5

4-8

3-4

49-6

103-6

8-6

5-6

26-5

52-5

4-3

8-9

25-6

24-4

4-3

10-11

104

8-6

12-13

25-6

24-4

4-3

A-l

236

104-8

340-8

A-3

202-5

73-6

276-1

A-7

129

11-2

189

A-ll

202-5

10-8

128

330-5

A-l 3

236

10-8

156-8

392-8

2-3

33-5

30-4

63-9

4-5

33-5

30-4

63-9

9-10

33-5

63-5

11-12

33-5

63-5

4-7

77*5

63-2

5-2

140-7

—

6-7

110-5

5-2

204-5

8-7

110-5

0-8

96-8

207-3

10-7

77-5

0-8

69-2

146-7

-—

The total stress in any member varies according to the direction of
the wind, the maximum stress being the sum of the dead load stress,
and the larger stress produced by the wind acting on either slope.
Although not subject to quite the same stresses, corresponding members
in the principal have been made of the same section for symmetry.

Design of Ties. — Each of the tension members will be designed as
flat and round sections, so furnishing alternative designs.

Members J.-13 and A-l.

Maximum stress = 392*8 cwts. = 19'64 tons.
Net sectional area required = —-* — = 3'27 sq. in.

Sections — Round, say 2j in. diameter.
Area = 3-094 sq. in.

19-64
Stress = Q.Q(U = 6-3 tons per square inch.

This stress, although slightly exceeding the working stress, may be
allowed.

290 STRUCTURAL ENGINEERING

Flat — The rivet connections of the flat sections reduce the sectional
area of the member, and the stress must be calculated on the minimum
sectional area.

Say 6" x 3" section, with a maximum of two f" rivets at any section.

Area = (6 - 2 x jf)f = 3*094 sq. in.
Stress = <r, = 6'3 tons per square inch.

The resistance of one f in. rivet in single shear

= 0-6x5 = 3 tons.
The resistance in double shear

= 3x1-5 = 4-5 tons.
The number of rivets required at the connections

for shear = 1^=5
4*5

for beariDg = rff^ = 4

Five rivets must therefore be used.

For the rivets to be in double shear there must be a connecting
plate at either side of the member, and the rivets must have a bearing
value in such plates equal to the stress in the member. The thickness
of the connecting plates must therefore be not less than

19*64

= °'28 ln" Say f i

5xfX8x2
Members A-ll and A-3.

Maximum stress = 330-5 cwts. = 16 '52 tons.
Area required = — g — =2*75 sq. in.

Sections — Round, If in. diameter.

Area = 2- 7 6 sq. in.
Stress = 5*9 tons per square inch.

Flat, allowing one f in. rivet in the section. Say, 5" x f ".

Area = 2*6 sq in.
Stress = 6'3 tons per square inch.
4 — | in. rivets required at the connections.

Member A-7.

Maximum stress =189 cwts. = 9 '45 tons.

Area required = -- = 1-57 sq. in.

ROOFS 291

Sections — Round, 1 J in. diameter.

Area = 1'767 sq. in.
Stress =5*3 tons per square inch.

Flat, allowing one f in. rivet in the section, 4" x i".

Area = 1*53 sq. in.
Stress = 6*1 tons per square inch.
3-f- in. rivets required at the connections.

Members 2-3, 4-5, 9-10, and 11-12.

Maximum stress = 63*9 cwts. = 3*195 tons.

Area required = — g— = 0*532 sq. in.

Sections — Round, f in. diameter.

Area = 0*6 sq. in.
Stress = 5*3 tons per square inch.

Flat, allowing one f in. rivet in the section. Say 2J" x f".

Area = 0*54 sq. in.
Stress =5*9 tons per square inch.
2— | in. rivets required at the connections.

Members 10-7 and 4-7.

Maximum stress = 146*7 cwts. = 7*335 tons.
Area required = 1*222 sq. in.

Sections — Round, Ij in. diameter.

Area = 1*227 sq. in.
Stress = 6 tons per square inch.

Flat, allowing one f in. rivet in the section. Say 3J" x £".

Area = 1*28 sq. in.
Stress = 5*73 tons per square inch.

2-f in. rivets required at the connections.
Members 8-7 and 6-7.

Maximum stress = 207*3 cwts. = 10*36 tons.
Area required = 1*73 sq. in.

Sections — Round, 1J in. diameter.

Area = 1*767 sq. in.
Stress = 5*86 tons per square inch.

Flat, allowing one J in. rivet in the section. Say 4" x ^j".

Area = 1*72 sq. in.
Stress = 6 tons per square in.
3-f in. rivets required at the connections.

Design of the Struts. — Rafters. — Maximum stress = 395*8 cwts.
19 '79 tons. Suppose two angles placed together be selected for

292 STRUCTURAL ENGINEERING

the rafters. Try two 5" x 3j" x J" angles with the 5 in. tables
placed together.

The least moment of inertia of the section = 14-46.

The sectional area = 8 sq. in.

= 1*34 ins.

1-34

The safe dead load for round-ended struts for this ratio, taken from

Fig. 109, = 8000 Ibs. per sq. in.

The safe working stress = f X 8000 = 5333 Ibs. = 2'41 tons per. sq. in.
The safe load on the rafters = 2'41 x 8 = 19-28 tons.

This is slightly below the maximum stress in the rafters, but as the

19*79
actual stress would only be — ^ — = 2*47 tons per square inch, this

section may be adopted.

Struts 10-11 and 3-4. — Maximum stress = 104 cwts. = 5'2 tons.
Try a 4" x 4" x J" T.

r = 0-814 in.

I = 8'6 X 12 = g

r 0-814

Safe dead load from Fig. 109 = 2'23 tons per square inch.

Safe load on the strut = f X 2-23 = 1-48 tons per square inch.

Safe total load on the strut = 1-48 x 3-758 = 5'56 tons.

This section may therefore be adopted.

Struts 1-2, 5-6, 8-9, and 12-13.— Maximum stress = 52-5 cwts.

= 2-625 tons.
Try a 2J" x 2£" x f" T.

r = 0-457 in

j = 4-3X12

r 0-457

Safe dead load from Fig. 109, = 3 tons per square inch.
Safe load on the strut = f x 3 = 2 tons per square inch.
Safe total load on the strut = 1-556 x 2 = 3-112 tons.

The strength of this section is rather higher than is required, but
for practical reasons it would be inadvisable to use a smaller section.
One rivet at each connection of this strut would be sufficient to resist
the shear, but not less than two should be used at any connection.

The complete design for the principal with the flat sections as ties
is drawn in Fig. 232. The details of the connections for the round
ties are also illustrated on the same figure. As an example of the
design of these details, consider the tie A-13.

The maximum stress = 19*64 tons.

The resistance to shear of the connecting bolt = 1 j(j ^\f» J

ROOFS

293

294 STRUCTURAL ENGINEERING

where dv = the diameter of the bolt
and /, = the safe intensity of shear ;

.-. 19-64 = |(^2 x 0*7854 x 5)
d^ — 1-82 in., say Ij in.

The bearing resistance of the bolt = \\ x t x fb

where t = the thickness of the eye of the tie
and/6 = the safe bearing intensity ;

.-. 19-64 = If X / X 8

.-. t = 1-309, say If in.

Making the thickness of each connection plate half the thickness of
the eye, the bearing resistance in the plates will be the same as in the
eye.

The tensile resistance of the two plates = (b — d2)tft

where & = the width of the plates

d.2 = the diameter of the bolt hole (say 2")

/ = the safe intensity of tensile stress in the plates ;

/. 19-64 = (& - 2)lf x 6

# = 4-37 in., say, 4J in.

Weight of Principal. — The weight of the principal as calculated
from the design is 4-05 tons, or 0*25 ton less than the estimated
weight. The dead load at the purlin connections was therefore 0'03
ton in excess of the actual load, but the stresses in the members pro-
duced by such load would not warrant any changes in the sections of
the members.

A vertical suspender 1J" x f" is inserted at the centre to prevent
sag in the middle tie bar and minimise the bending stress due to its
own weight. It takes no part in resisting the primary stresses.

Roof Details.— In Fig. 233 are shown various roof details. Detail
A is a method of fixing patent glazing at the apex of a roof with
equally inclined slopes. Detail G shows an arrangement for fixing
glazing on the steeper slope and slates on the other slope at the apex
of a northern light roof. Details B, C, and D are connections of
patent glazing to timber and steel purlins. Details E and F show the
junction and attachment of glazing and corrugated sheeting to steel
purlins. Detail H indicates a method of securing corrugated sheeting
to angle purlins, and also the ridge of a corrugated sheeted roof. A
built-up ventilator is shown in detail K. The left-hand portion is a
section at the principal, and the right a section between the principals.
The louvres are not of sufficient strength to support themselves for a
span equal to the distance between the principals, so are supported at
intermediate positions by vertical tee bars fixed to the purlins. Detail
L shows the attachment of the weather boards on a cantilever roof
suitable for a loading or station platform. Weather boarding attached
to an end principal is shown in detail M.

ROOFS

295

FIG. 233.

CHAPTEE X.

MISCELLANEOUS APPLICATIONS AND TALL BUILDINGS.

Design for Lattice Eoof Girder .—Span SGfeet. Depth ±ft. G in. To
carry the feet of principals of two adjacent roof spans of 50 feet. Rooj
principals 12 feet apart. The general arrangement is shown in Fig.
235, A The roof pitch is 1 to 2, and the loading is assumed as
follows : —

Covering and snow at 20 Ibs. per square foot of area covered
= 12 x 50 x 20 = 12,000 Ibs. for one principal.

Weight of principal = IDL^l + ^) = f x 12 x 50(1 + *§)

= 2712 Ibs.

Normal wind pressure on one roof slope per 12 feet length, at 25 Ibs.
per square foot = 28 feet (length of slope) x 12 X 25 = 8400 Ibs.

Reactions due to wind pressure

at a = 8400 Ibs. x f J = 5787 Ibs.
and at I = 8400 - 5797 = 2613 Ibs.

The central girder I provides the smaller reaction for the right-hand
roof span and the larger reaction for the left-hand span when the wind
blows from the right. The total inclined wind load on girder b at the
points of support of the principals is therefore the total normal wind
pressure of 8400 Ibs. On the outer girders at a and e the inclined wind
load will be equal to the larger reaction of 5787, say 5800 Ibs.

The vertical component of the inclined pressure of 8400 Ibs.

= 8400 x YC = 840° x cos 26i° = 8400 X 0-894

= 7509, say 7500 Ibs.

The total vertical loading concentrated at points h and k Fig. 235, B,
is therefore for the central girder,

due to covering and snow .... 12,000 Ibs.
„ two half principals .... 2,712 „
„ vertical wind load .... 7,500 „

22,212 Ibs., say 10 tons.

The principals at m and n being supported directly by the columns,
do not affect the stresses in the girder.

The weight of the girder will be assumed as 3 tons, distributed

296

MISCELLANEOUS APPLICATIONS

297

equally amongst the upper joints. The vertical loading and stresses
are then as indicated in Fig. 234.

The horizontal component of the normal wind pressure causes lateral
bending of the girder, and although some portion of the lateral bending
moment will be resisted by the longitudinal roof members and transmitted

0 +10-06

-10-06

-10 06 -19-83 - 29 31

FIG. 234.

by diagonal wind bracing (where provided) to the columns, the amount
of the horizontal wind load so resisted is very uncertain, and it is only
reasonable to make the girders strong enough to resist the whole.

The horizontal component of wind pressure at h and /<;, Fig. 235, B,
= 8400 x sin 26^° = 8400 X 0'446 = 3746, say 3750 Ibs. = 1- 67 tons.
Hence lateral bending moment between h and Ic,

3750 X 12' x 12
2240

= 241 inch- tons.

Flanges. — The flanges over the five central panels consist of two
4i" x o" x y angles and two 12" x f" plates. The outer plate is
suppressed over the two end panels.

For the moment of inertia of the flanges about YY, Fig. 235, C,

Iy for one angle = 2*55, from section book ;
.'. IY „ „ = 2-55 + (3'5 x 3'752) = 51'5

IY „ four angles = 51'5 x 4: = 206 inch units.

- J_

123 X 4 = 210

IY „ four 12" x f plates =
Total IY for both flanges = 206 -f 216 = 422

241 X 6"
Stress due to lateral bending = .^ — = 3*43 tons per square inch.

Sectional area of top flange =16 square inches. Maximum com-
pression in central bay = 29*61 tons, and compression per square inch

29*61
due to vertical loading = — TTT- = 1*85 tons. Hence maximum intensity

of compression in top flange due to vertical and lateral loading
= 3 '43 -f- 1-85 = 5*28 tons per square inch.

The net section of the lower flange after deducting four f in. rivet
holes = 14J square inches, and maximum tension due to vertical loading

= — 14. 1 or — = 2*1 tons per square inch. Adding the stress due to

lateral bending, maximum intensity of tension in lower flange =2*1
-f- 3*43 = 5*53 tons per square inch. The actual value is a little
higher than this, since no rivet holes were deducted in calculating IY

STRUCTURAL ENGINEERING

MISCELLANEOUS APPLICATIONS 299

for the lateral bending. These working stresses are sufficiently low to
allow a fair margin for the impact effect of the wind pressure. It is,
however, very improbable that the maximum wind pressure would ever
be applied instantaneously. It will be noticed further that the maximum
wind load has been assumed on the left-hand roof span, whereas this
would be sheltered to some extent by the right-hand roof.

The moment of inertia of the upper flange section about XX, Fig.
235, D, works out at 29 units, and least radius of gyration = Vf|

7 J-ft

= 1-35 in. The panel length is 48 in., whence - = ^— = 3G, and average

T JL'oO

safe load on top flange supposing the ends rounded = 13,700 Ibs. or 6*1
tons per square inch. The average working stress is much below this,
so that the flange is amply safe against buckling.

Struts. — Struts 1 and 2, Fig. 235, H, consist of two angles
3J" x 2^" x y tied together by bolts and tube separators s, s, Fig. 235,
E. The maximum compression = 11*32 tons. Assuming one-half
resisted by each angle, compression in one angle = 5'G6 tons. This is
eccentrically applied since the angle is riveted to the gusset plate by one
leg only. Fig. 235, F, shows the dimensions concerned. The eccentricity
is 1*45 in., y-y being the axis through the e.g. of the angle section. The

sectional area = 2-75 square inches, and direct compression = -^=

= 2*06 tons per square inch. B.M. due to eccentricity of load
= 5*66 x 1'45 = 8*2 inch-tons, 1^ for the section = 1'43, hence com-

pressive stress due to

uare inch. B.M. due to eccentricity of
inch-tons, \y for the section = 1*43, hence
bending

x 0*7

=4*0 tons per square inch

and maximum intensity of compression = 2'06 -f 4*0 = 6*06 tons per
square inch. As the struts are not long relatively to the radius of
gyration, this intensity is not excessive, and they are further stiffened
by the separators. Strut 3, beneath the shoes of the principals, is
combined with gusset stiffeners to give lateral rigidity, and to assist in
transferring the lateral wind load to both flanges of the girder. It
consists of four 2J" x 2J" x f" angles, with f in. gusset plates. Struts
4 and 5 have very little compression to resist, but cannot well be made
of smaller sections than 3" x 2J" x f" for practical convenience in
riveting. They may be subject to somewhat higher stresses under the
action of a rolling gust of wind.

Ties. — These have been proportioned for a working stress of 6 tons
per square inch on the net section. Those in the first three panels
from the column, having practically the same stress, will be designed
for 15*09 tons. Using two flats, stress in one bar = 7*55 tons. Net

7*r>5
sectional area = ~TT- = 1*26 sq. in. Adding for one rivet-hole

f" x i" = 0'375, gross section = 1-26 + 0*375 = l'G35 sq. in. A
3J" x i" flat gives 1*75 sq. in. The ties in the three central panels
have very little stress. Two flats 3" x |" have been employed. The
central panel is counterbraced. Fig. 235, H, shows the general
elevation of half the girder, and G a part sectional plan with detail of

300 STRUCTURAL ENGINEERING

connection to a box section column. Sole plates 2' 8" x 12" x J" are
placed 12 ft. from each column, to which the feet of the adjacent roof
principals are bolted. The number of rivets are arranged on a basis of
4 tons shearing stress and 8 tons bearing stress per square inch
respectively. The connection to the column is made through two
vertical angle cleats 3" x 3" x J" bolted to the column by ten f in.
through bolts. All gusset plates are J in. thick.

Design for Crane Jib. — Length, 45 ft. Load, 3 tons. To lift at
2i ft* Per second and slew at 6 ft. per second, with 1 //. per second
acceleration for both lifting and slewing (Fig. 236). Lowest position of
jib inclined 30° with the horizontal. Inclination of backstays 15°, and of
rope 20°.

The nominal load lifted = 3 tons. Allowing 0*25 ton for the
weight of chain, bob and hook, total weight lifted = 3^ tons. This
will be doubled to provide against shock due to possible slipping of
tackle, giving an equivalent load lifted = GJ tons.

W y f 91 y 1

Accelerating force in lifting = - ~J = ^-~^- = 0-1 ton.

g 32

Maximum tension in chain = 6-5 + 0*1 = 6*6 tons.

In Fig. 236, A, AB inclined 30° represents the jib, and A^ and AT
the chain. Setting off AT, ATj, each = 6*6 tons, the resultant AR is
obtained from the parallelogram ATRTj. Draw RB parallel to the
backstay AS, and AB scaling 25 J tons gives the direct compression in
the jib due to the tension in the two portions of the chain.

Approximate Weight of Jib. — Adopting 2J" x 2J" x f" angles and
2^" x |" flats for the bracing on upper and lower faces, and 2J" x 2J"
X 0*3" angles and 2J" X f" flats for the side bracing, an approximate
estimate of the weight of the jib runs out at T75 tons.

Stress due to Weight of Jib. — With the jib in an inclined position,
its own weight will give rise to both direct compression and bending
stress in the main angles forming the section of the jib. In Fig. 236,
A, set off AE = 45 ft., and draw the vertical CD through M the middle
point of AE. Join EC. AC and EC give the directions of the
reactions at the pulley and hinge ends of the jib respectively, due to its
own weight. Note that the e.g. of the jib has been assumed at M.
Although the lower half is wider and consequently somewhat heavier
than the upper half, the additional weight concentrated around the
pulley head will practically neutralize the effect of the greater weight
of the lower half on the position of the e.g.

Make CD = 1'75 tons and draw DF parallel to the backstay AS.
CF = the reaction at the hinge E, and DF scaling 2- 9 6 tons = pull in
backstay due to weight of jib. DF is transferred to AG, and resolved
along and perpendicularly to the jib, giving AH = 2*85 tons and
HK = 077 ton respectively. The component AH acting along the jib
represents the direct compression applied by the backstay due to weight
of jib, whilst the component AK acting at right angles to the jib
causes bending moment at any section.

Considering the central section at M. The weight of the jib being
distributed and not actually concentrated at M, the compression due to

MISCELLANEOUS APPLICATIONS

301

302 STRUCTURAL ENGINEERING

its own weight will be cumulative from A towards E, consequently the
weight of the upper half AM must be taken into account in calculating
the compressive and bending stresses on section M. LN = J weight of
jib = 0-875 ton is resolved at LP and LQ. LP = 0'44 ton = the
additional direct compression accumulated from A to M over and
above that applied by the backstay. LQ = 0"76 ton = the component
of the weight of AM causing bending moment at section M in the
opposite sense to that caused by AK. Then at section M, direct com-
pression due to weight of jib = AH + LP = 2'85 + 0'44 = 8-29 tons,
and bending moment due to weight of jib = 0'77 X 22' 5' — 0'76
X 11-25' = 8-77 foot-tons = 105*24 inch-tons.

The section adopted at M is shown at Fig. 236, B, consisting of
four steel angles 3J" x 3J" x 0'425". From the section book the
sectional area of one angle = 2*8 sq. in. and I,. = 3'22.

Hence for the four angles Ix = (3'22 + 2'8 x 10-952) x 4 = 1356.

Stress due to bending = 105'^ * jjL = + 0'93 ton per square inch.

loob

Direct compression = 26*25 tons due to chain tensions -f 3 '2 9 tons

29*54
due to weight of jib = 29*54 tons, or ~7j^r = 2-64 tons per square inch.

The direct compression is slightly greater than this on account of the
slight batter of the main angles. As the difference between slope
length and axial length of jib is very small, this has been neglected.

Hence maximum compression due to load lifted, weight of jib,
accelerating force in lifting and bending of jib in vertical plane
= 0*93 -j- 2'64 = 3'57 tons per square inch.

Slewing. — The force necessary for accelerating the load during
slewing acts at the pulley end of the jib horizontally, and gives rise
to lateral bending moment.

Accelerating force for load = - — ^ x =0*1 ton.

Considering the central section of the jib, the force necessary for
accelerating the upper half of the jib also acts horizontally and creates
additional lateral bending moment on the central section. The centre
of inertia of the upper half of the jib is situated at 0*77 X 45 ft.
= 34*6 ft. from the slewing axis, here taken as the foot of the jib.
The acceleration at outer end of jib = 1 ft. per second, therefore at
0-77 of the jib length, the acceleration = 0*77 ft. per second, and
accelerating force for upper half of jib = 0-875 x 0*77 X ^ = 0*021
ton, acting at 34*6 — 22'5 = 12'1 ft. from central section. Hence,
bending moment at central section due to accelerating the load and
weight of upper half of jib = O'l x 22'5' + 0'021 x 12-1' = 2-5 ft.-tons
= 30 in.-tons. To this must be added the bending moment caused by
wind pressure on the side of the jib and on the load lifted.

Bending Moment due to Wind Pressure. — A safe outside allowance for
wind pressure will be made by assuming 40 Ibs. per square foot acting
on the total projected area of one side of the jib, taken as a continuous
surface. Only the ends and central portion are actually plated, and the

MISCELLANEOUS APPLICATIONS 303

above allowance will cover the wind pressure acting on the latticing of
the leeward side.

Mean breadth of upper half of jib = 19 in., and e.g. is 10*5 ft. from
centre of jib.

Therefore wind pressure on upper half of jib = 22J x jf X 40
= 1425 Ibs., acting at a leverage of 10'5 ft., and bending moment at
central section = 1425 X 10'5 = 14,9 G2 ft.-lbs.

The wind pressure on the load lifted, assuming 30 sq. ft. of effective
surface in the case of bulky loads, = 30 X 40 = 1200 Ibs., acting at a
leverage of 22*5 ft. from central section, and bending moment due to
this pressure = 1200 X 22*5 = 27,000 ft.-lbs.

Hence total bending moment due to lateral wind pressure = 14,962
-f 27,000 = 41,962 ft.-lbs. = 225 in.-tons.

The lateral deflection caused by the wind pressure and accelerating
forces still further increases the lateral bending moment, since the
direct compression along the jib then acts eccentrically. This increase
is, however, small — about 20 to 25 in.-tons, and is neglected, since
ample allowance has been made for wind.

Total lateral bending moment = 225 in.-tons due to wind -f 30 in.-
tons due to accelerating forces = 255 in.-tons.

In Fig. 236, B, the moment of inertia of one angle about yy = 3-22,
hence IY for four angles = (3-22 + 2'8 X 16-32) X 4 = 2988, and stress
due to bending

= 255t* 18'75 = ± 1-60 tons per square inch.

Hence, maximum intensity of compression at central section of jib
from all causes = 1'60 4- 3-57 = 5*17 tons per square inch. This
intensity occurs in the upper leeward angle at k.

The stresses on the lower end of the jib are next considered.
These are caused by direct compression, due to load and weight of
jib, and B.M. due to wind pressure and accelerating forces.

Direct Compression. — This has been already determined, and = due
to load 26-25 tons, and due to weight of jib = 2-85 (axial component
AH applied by backstay) -f 2PL (axial component of total jib weight)
= 2-85 -f 0-88 = 3-73 tons.

Total direct compression = 26-25 -f 3*73 = 29*98 tons.

Bending Moment dm to Wind Pressure.— Projected area of jib
= 45 x || in., and total presssure = 45 X jf X 40 = 2850 Ibs.

B.M. at foot of jib = 2850 X 22'5' = 64,125 ft.-lbs.

B.M. due to wind on surface of load = 1200 X 45' = 54,000 ft.-lbs.

Total B.M. = 64,125 + 54,000 = 118,125 ft.-lbs. = 632 in.-tons.

Sending Moment due to Accelerating Forces. — For the load = O'l ton
X 45' = 4-5 ft.-tons.

For whole weight of jib, centre of inertia is --^ ft. from foot, and

accelerating force = 1-75 x -r^ X -^ = 0-032 ton.
Hence B.M. = 0'032 X ^ = 0-83 ft.-tons.

V 6

304 STRUCTURAL ENGINEERING

Total B.M. = 4-5 -f 0'83 = 5*33 ft.-tons = 04 in.-tons.
Total B.M. due to wind pressure and accelerating forces = 632
+ 64 = 696 in.-tons.

The cross-section at foot of jib is shown at Fig. 236, c.

Iy = (3-22 + 2-8 X (27*55)2) X 4 = 8514
Stress due to bending = — arTJ — = - ^'46 tons per square inch.

29*98
Direct compression = =2*54 tons per square inch, and

maximum intensity of compression at foot of jib = 2*46 -f 2'54 = 5-00
tons per square inch. This intensity occurs at the outer edges m, m, of
the leeward angles.

The general design of the jib is shown in Fig. 236, D. The middle
portion of the sides is plated for a length of about 10 ft. to give
greater stiffness against buckling. The details at hinge and pulley
ends is shown at E and F. At E cast-steel bracketed eyes K are
bolted through the angles, and side plates with packing plates inserted
as indicated by the shading. These hinges bear on a 4-in. pin passing
through a pair of bearings Y, V, Fig. 236, c, bolted to the crane
carriage or framework.

Shearing and Bearing Stresses on Pin. — Direct compression down

29*98
each side of jib = —~— = 14-99, say 15 tons. Distance centre to

centre of bearings Y, Y, = 4 ft. Total B.M. on side of jib = 696 iii.-
tons. Hence pressure on leeward bearing, or uplift on windward
bearing, due to B.M. = ~, =14-5 tons. Adding the direct pressure
of 15 tons, total pressure on leeward bearing = 29 '5 tons. Bearing
area = 4" x 4" = 16 sq. in., and bearing pressure per square inch

29*5
= -J77- — 1*85 tons. Distance between outer faces of bearing = 4 ft.

4 in. Shear force on pin due to lateral bending = ~> = 13*4 tons.
Adding the direct pressure, total shear force at leeward section of pin
= 13*4 4- 15*0 = 28*4 tons. Sectional area of pin = 12*57 sq. in., and

28*4
mean shear on pin = TvTF? = 2*26 tons per square inch. The bearing

and shear stresses should be low, since the pin undergoes considerable
wear and tear.

The pulley end carries a 30 in. diameter pulley and chain guard,
the pin diameter being 2J in., and the pulley eye bushed with bronze.
The sizes of lacing bars and angles indicated will be found ample for
resisting the stresses caused by the lateral loading of the jib, treated
as a cantilever. It is unnecessary to calculate these in the case of a
light jib of this type.

Design for Riveted Steel Tank.— Capacity, 20,000 gallons. The
tank to be carried on girders and columns, so that when filled the water-
level shall be 25/£. above ground-level.

Such conditions are representative of those required for locomotive
feed tanks, etc. Fig. 237 shows the general arrangement and details. Side
and end elevations are shown at A, and an enlarged plan at B, on which

MISCELLANEOUS APPLICATIONS

305

306 STRUCTURAL ENGINEERING

the arrangement of covers and joints is indicated. The sides consist of
three plates, 8', 12', and 8' long x 6' 3" high ; the ends of one plate 12' 6"
X 6' 3", and the bottom of seven plates 4 ft. wide, running transversely
and turned up to form a butt joint at h with the side plates. The joints
in the bottom are covered by tee bars inside, turned up to form vertical
stiff eners for the sides. Vertical joints in the sides occur at J, J, in
the plan. All outer joints and horizontal inner joints are covered by
flat straps. The end and bottom corners are curved, and forged bosses
are placed at the four bottom corners, the detail of which is shown at
C. The tank is supported on four longitudinal rolled steel beams L,
4 ft. 2 in. centre to centre, resting on three transverse beams T, carried
by six columns 18 ft. high. Stays are inserted as indicated subsequently.

The principal features of the design are as follows.

Thickness of Plates. — Span 4 ft. Head of water 7 ft. 6 in. Load
per square foot on bottom due to water pressure = 62'5 x 7*5 = 469 Ibs.
or ~ = 39 Ibs. per inch width of plates.

Max B.M., assuming the plates simply supported at the joints,

+ **m 39_X_4_XAX_12 m in>_ton>

8 8 X 2240

Assuming a working stress of 9 tons per square inch.

Moment of resistance = Jxlx£2x9 = 0-418
whence thickness t = 0-528 in., say f in.

It may be noticed that a somewhat thinner plate would result if the
plate be supposed to act as a fixed instead of a simply supported beam.
The actual strength of flat plates employed under such conditions lies

2£»72

between the two above-mentioned limits, so that - - is an outside

estimate of the maximum bending moment on the plate. The side
plates being subject to a maximum head of 6 ft. 3 in. would require a
theoretical thickness of J in., but for practical reasons would be made
the same thickness as the bottom plates. A small additional stress
will be caused by the weight of plate, which has been neglected since
amply covered by the calculation.

Bottom Transverse Tee-covers. — These transfer the load on the
plates to the longitudinal bearing girders. Their span is 4 ft. 2 in.,
and their actual resistance lies between that of a simply supported and
a fixed beam. Each rib carries the weight of a volume of water 7 ft.
6 in. deep x 4' x 4' 2"

2240

B.M. at centre = 3'49 x ^L>Llg = 21-81 in.-tons.
8

The effective beam section resisting this moment is shown at Fig.
237, D, and is assumed to consist of the inner tee stiffener, outer cover
J in. thick, and portion of f in. plate between. The modulus of this
section after deducting two |f in. rivet holes = 4'3 ins.3

21'81
Hence maximum bending stress = ,.a = 5-07 tons per square inch.

MISCELLANEOUS APPLICATIONS 307

Stays. — Transverse and longitudinal stays are taken across the tank
at 4 ft. and 4 ft. 2 in. intervals respectively, at the centre of height of
the sides. Fig. 237, L,is a diagram of intensity of water pressure against
the side per 4 ft. width of tank.

db = 4 x 62J x 1\ = 1876 Ibs. per 4 ft. width,

and the horizontal breadth of triangle dbc gives the intensity of pressure
at any depth. The total pressure against the side per 4 ft. width

= db X 1 = 1876 X 3J' = 7035 Ibs.

acting through the centre of pressure at level g.
Taking moments round a—

tension in stay x 3' 9" = 7035 x 2' 6"

whence tension in stay due to horizontal water pressure = 4690 Ibs.
= 2*1 tons.

This tension will be augmented due to the overhang at the sides.
The increase of tension from this cause will be calculated by taking
moments about e.

Weight of volume of water cfea per 4 ft. width

= 4xl*2*gx02* = 1-05 tons.

Adding for weight of side of tank per 4 ft. width, 950 Ibs. or 0'43
ton, total overhanging weight per 4 ft. width = 1'05 + 0'43 = 1-48
tons. The common centre of gravity is 10^ in. to the left of e.

Taking moments about e —

tension in stay due to overhang x 3f = 1'48 X |'

whence additional tension in stay = 0*35 ton.

Total tension in stay = 2'1 4- 0*35 = 2*45 tons.

In Fig. 237, E, which shows an enlarged cross-section of one half of
the tank, flat stays L and T are employed. Round stays better resist
corrosion, but sag more severely under their own weight. The longi-
tudinal stays L, being 30 ft. 6 in. long, are arranged to rest on the
transverse stays T in order to relieve them of part of the bending stress
due to their own weight. Adopting flat bars 3" x £", the bending
stress due to the weight of the transverse stay T, and the 4 feet lengths
of the longitudinal stays L, L, resting on it, works out to 0-93 ton per
square inch.

Direct tension = =1-63 tons per square inch

0X3
and maximum tension =1'63 + 0'93 = 2'56 tons per square inch.

Smaller stays might be used, but this size will provide a margin
for corrosion ; f in. bolts at K will be suitable.

The upper half of the sides above the stays, Fig. 237, L, resists
the horizontal water pressure P by cantilever action, when not stayed
across the top of the tank.

Pressure P per 4 ft. width of side = ]- the pressure for the 7 ft.

308 STEUCTUEAL ENGINEERING

G in. depth = I X 7035 = 1759 Ibs. applied at 1 ft. 3 in. above the
level of attachment of stays.

Hence B.M. on side = -- — =11-8 in.-tons.

This is resisted by the section Fig. 237, D, the modulus of which
is 4-3,

11*8
/. bending stress = -775- = 2*8 tons per square inch.

The stress due to this bending action will be somewhat higher on
the ribs intermediate between the vertical joints, since the effective
beam section is there reduced by the absence of the outside J in. plate.

There will be direct tension across the tank bottom, due to the
reaction required to balance the horizontal forces acting on the sides,
which = total horizontal pressure on side — horizontal pull in stay due
to this pressure = 7035 - 4690 = 2345 Ibs. per 4 ft. width of tank.
This creates a negligibly small tension in the plates, which would be
entirely annulled if the stays were attached at g.

A close pitch of riveting, say 2J in., will be required to ensure water-
tightness, and all joints will be ca*ulked. Rivets J in. diameter will be
suitable.

The weight of the tank as designed runs out at 16-8 tons.

Longitudinal Girders. — The load applied at each bearing-point of
the tank on the two central longitudinal girders = 3| tons. These
girders are continuous, and one-half the B.M. diagram is drawn at
Fig. 237, F, and the characteristic point m found by the method of
Example 9, Fig. 47, Chapter III. The maximum moment occurs at
the centre, and = 22'3 ft.-tons = 267*6 inch-tons.

Using 9^" X 9J" x 51 Ibs. B.F. beams, the modulus = 52-2, and

0^*7*^*

working stress = -r^r = 5*12 tons per square inch. The B.M. on the

two outer girders is less than on the central ones, but all four are
necessarily of the same depth.

Transverse Girders. — The central reaction of the longitudinal girder,
calculated from the bending moment diagram F, by the method of
Example 9, is 16 tons. Each of the two central longitudinal girders
therefore applies a load of 16 tons to the central transverse girder.
This loading is shown at Fig. 237, G. The two outer loads do not
affect the bending moment, being directly over the column.

Maximum B.M. on transverse girder = 16 x 4£ X 12

= 800 inch-tons.

Using a 12J" X 12" X 85 Ibs. B.F. beam, the modulus = 115, and
working stress = r-ry = 6*95 tons per square inch.

Columns. — The maximum load comes on the two central columns,
and equals the sum of the central reactions from one central and one
outside longitudinal girder, plus a portion of the weight of the beams.
If the B.M. diagram for the outside longitudinal girder be drawn in a

MISCELLANEOUS APPLICATIONS 309

similar manner to that at F for the inner girder, the central reaction
will be found to be 12-3 tons.

Hence load on one central column

= from inner longitudinal girder . . . 16-00 tons

„ outer „ „ ... 12*30 „

™ of weight of longitudinal girder . . 0*85 „

J weight of transverse girder .... 0*27 „

Vertical component of stress in wind tie

(determined below) 2'50 „

Total . . 31*92 „

say 32 tons.

The column, if free to bend as at Fig. 237, H, will act as a round-
ended column 36 ft. long. The stiffness of the connection with the
central transverse girder and the action of the diagonal wind braces,
will tend to cause bending as at K, in which case the equivalent round-
ended column would have a length of 18 feet. The real strength will
lie between these two extremes, but it will be advisable to design the
column as at H, since the connections are not likely to be very rigid,
and the wind pressure acts on a relatively large surface.
Using an llj" X llf X 75 Ibs. B.F. beam-
Sectional area = 21 -9 sq. in., least radius = 2-66 in., and -

°fi x 12
= — =163. The safe load per square inch for this ratio, from

Fig. 109 = 3200 Ibs.

Total safe load = 820p*21'9 = 31 '3 tons.

.224:0

The maximum load to be carried is 31-92 tons.
Wind Pressure. — The wind pressure on side of tank at 35 Ibs. per
square foot

J^= 3-5 tons.

3*5
Frictional coefficient with tank empty = ^g = 0-21. Tanks

presenting a large area to the wind pressure may require bolting to the
pillars or girders by suitable brackets and bolts.

Length of wind ties = 24 feet.

Horizontal component of wind stress in central ties = f x 3-5 tons
= 2-2 tons.

Inclined stress = 2'2 x 7^ = 4'2 tons, and vertical stress = £J X 4-2

LZ O

= 2-5 tons, which gives the pressure to be added as above to the load
on the leeward central pillar.

For the diagonal braces a small angle section, say 2-f X 2f X |",
flat, or round bars may be used.

Three holes are provided in the tank bottom for supply, draw-off,
and overflow pipes.

310

STRUCTURAL ENGINEERING

TALL BUILDINGS.

These comprise special types of construction, principally developed
in New York and Chicago, in which the framework, consisting of steel
columns united by horizontal girders and joists, is carried to much
greater heights than is usual in ordinary types. Some of the most
notable are the Park Row buildings, New York, 382 feet high ;
Philadelphia City Hall tower, 537 feet ; the Fisher building, Chicago,
eighteen storeys high ; Great Northern Hotel, Chicago, fourteen storeys ;
Masonic Temple, twenty storeys, the Ivings building, New York, of
twenty-nine storeys, and the Singer building, New York, with a tower
612 feet in height. The distinguishing feature of these buildings is
the lightness of the covering employed for the walls, which, together
with that of the steel frame, enables such heights to be attained, with-
out unduly loading the foundations, which in New York and Chicago
are of generally poor bearing power. The foundations for the columns
are consequently usually constructed of the grillage type already
referred to.

Two principal methods of construction are in vogue. 1. Those
buildings in which the exterior walls are self-supporting, and the girders
and floors are carried by columns independently of the walls. 2. The
more usual construction, in which both walls and floors are carried by
girders, and ultimately by the columns. The latter method enables
any portion of the walls to be commenced independently as convenience
allows, and renders it possible to provide more surely for practically
uniform settlement of foundations. In the case of an existing party
wall, W, Fig. 238, the footings of which may not be further loaded or

FIG. 238.

interfered with, exterior columns such as A are frequently carried on
cantilever girders, G, balanced by the load on one or more neighbour-
ing columns as B.

Live Loads on Floors. — The following live loads are prescribed in
Chicago for the various storeys of ordinary high buildings, where the
lower storeys are used for stores or shops, and the upper ones for offices
and flats.

Cellars 102' 5 Ibs. per square foot.

Basement to fourth storey 119 „ „

Fourth to sixteenth storeys 68 „ „

Above sixteenth storey 37 „ „

MISCELLANEOUS APPLICATIONS

311

In New York the following loads are prescribed : —

Each storey of an hotel or flat ... 60 Ibs. per square foot.

Offices below the first storey .... 150

Offices above „ „ .... 75

Shops and stores for heavy merchandise 150

light „ 120

Schools, floors of ........ 75

Public assembly halls, floors of ... 90 „ „

Koofs of less than 20° inclination are required to be capable of
supporting 50 Ibs. per square foot, and roofs of more than 20° inclina-
tion, 30 Ibs. per square foot, in addition to the dead weight. In
proportioning the foundations of the columns the following regulations
obtain in New York. The maximum column load is assumed equal to
the total dead load plus 75 per cent, of the live load in the case of
stores, schools, churches, and public halls, and equal to the dead load
plus 60 per cent, of the live load for office buildings, hotels, and flats.
This reduction in the total nominal load is made having regard to the
improbability of all the floors carrying the maximum live load at any
one time. Similarly, in designing the floors, each individual floor is
designed for the maximum live load, bat the main girders which
transfer the floor loads to the columns are again unlikely to be fully
loaded at any instant, and may be designed for loads somewhat below
the nominal maximum. Further, such reductions from the nominal
maximum to the effective live load
may be greater in proportion to
the number of storeys, since there
will be less likelihood of all being
loaded together as the number in-
creases.

Details of Construction. — The
steelwork of columns is covered
by hollow fireproof terra-cotta tiles
2 in. to 3 in. thick, bonded to-
gether with an air space left between the tiles and column faces.
Examples of such coverings are shown in Fig. 239.

Floors. — The floor girders are similarly protected by the systems of
terra-cotta floors in general use. Fig. 240 gives a typical example of

FIG. 239.

FIG. 240.

these floors by the Pioneer Fireproof Construction Company. The
floor consists of flat arches of tile voussoirs, abutting on the parallel

STRUCTURAL ENGINEERING

joists, forming the primary framing of the floor. Table 21) gives
particulars and weights of such types of floors. The weight of concrete
and timber covering forming the floor surface requires adding to the
weights stated in the table, and a further 5 to 8 Ibs. per square foot if
the under surface be plastered. In the best systems the vertical webs
of the hollow tiles run transversely to the carrying joists, these being
known as " end-systems," whilst those having the webs parallel to the
joists are called "side-system" arches, the former being lighter for
equal strength.

TABLE 29. — WEIGHT OF HOLLOW TILE FLOOR ARCHES.

. Span of arch.

Deptb.

Weight per sq. ft.

ft.

5 to G
6 „ 7
7 „ 8
8 „ 9

In.
8
9
10
12

Ibs.
27
29
33
38

Tie-rods T, Fig. 240, are employed for resisting the end thrust of
such floor arches. The spacing of tie-rods should not exceed twenty
times the width of flange of floor beams. The approximate thrust
in flat tile arches is given by —

T = Ibs. per linear foot of arch.

Where w = load per square foot on arch, L = span in feet, and
d = depth in inches from top of arch to bottom of supporting beams.
The spacing of the tie-rods being decided, the diameter may be
calculated from the above thrust, 7 tons per square inch being allowed
on the net section of bars.

Partitions are constructed of wire lathing or expanded metal tied to
verticals of light channel section filled in with concrete and plastered
on the face ; or of slag wool and plaster slabs, or terra-cotta blocks
2 in. to 4 in. thick. The last named are the soundest and most

^ .. . ie- _^| fireproof. For thicknesses of 3, 4,

kfffi^r*^^^ 5, and G inches these weigh 16, 19,

F^IL- — _ . ••• •••JL^^3' -.-.-.-..j^ 22, and 23 Ibs. per square foot re-

j spectively.

I • j Roofs are covered usually with

3 in. book-tiles laid on T-bars as

FIG. 241. in Fig. 241, the T-bars being sup-

ported by the beams and girders

forming the roof framing. An outside coating of cement, tar and
gravel is applied over the tiles. The weights of the floors and roof
in the Fisher building, Chicago, are as follows : —

MISCELLANEOUS APPLICATIONS

313

Ibs. per sq. ft.

Floor — I in. Maple floor 4

Deadening, cinder concrete on top of floor arch . 15

15 in. hollow tile floor arch 41

Stsel joists and girders 10

Plaster on ceiling 5

Total .... 75

Ibs. per eq. ft.

Roof— 3 in. Book tiles 22

6 -ply tar and gravel roof 6

T-bars 4

Steel roof framing 8

Total .... 40

Chimney Stacks are usually constructed of a steel tube lined with
firebrick to a height of 60 or 70 ft., and with hollow tiles thence to
the top, a 2 in. air space being left
between the steel and lining.
Fig. 242 shows the sliding joint
J provided between the top of
the stack and the roof in the
Fisher building, which allows of
expansion and contraction with-
out dislocating the roof tiles.

Exterior walls are of brick or
terra-cotta, or brick with terra-
cotta facing, and are most usually FIG. 242.
carried by the horizontal girders

of the outer framing. The weight of the steel framing for buildings
of 16 to 20 storeys varies from If to 2 Ibs. per cubic foot of the
building, and the cost from 2^ to 3 pence per cubic foot, or £14 8s. to
£17 5s. per ton of steel, representing from } to I the total cost of
the building.

Wind Bracing. — In the case of tall buildings of light construction,
the walls and partitions being thin and not bonded together, the steel
framing must provide all resistance to the wind. In buildings of usual
proportions of height to width of base, the dead weight is sufficient to
resist bodily overturning, and the effect of the lateral wind pressure is
to create bending moment on the columns and horizontal shearing stress
on the connections, or to increase the compression in the leeward
columns and relieve that in the windward columns if the building is
braced diagonally from top to bottom.

In the case of exceptionally tall buildings which are practically
narrow towers, the wind pressure may create tension in the windward
columns exceeding the compression due to the dead weight of the
building, and so produce an actual uplift on the foundations of the
windward columns, necessitating their being securely anchored down.

The usual methods of bracing against the action of wind pressure
are illustrated below. In Fig. 243 diagonal ties are introduced in a

314

STRUCTURAL ENGINEERING

suitable number of vertical panels between the columns. This is
probably the most efficient method. The framing is thereby converted
into a series of cantilevers fixed at foundation level and subject to
horizontal loads p, p. The principal objection to this system is its
interference with window and door openings, for which reason diagonal
bracing is more frequently placed in some of the internal partition
walls than in the outer walls. This difficulty is avoided in the G12 ft.
tower of the Singer building by arranging the bracing as in Fig. 244.

FIG. 243.

FIG. 244.

The corner bays A, A, are braced for the whole height in alternate
long and short panels H and ^, the longer panels extending through
two storeys. The bracing bars thus intersect near the floor levels F, F,
leaving uninterrupted openings for windows W in all storeys.

Fig. 245 shows the more usual methods adopted by employing deep
and stiff girders G between the columns at the floor levels, or by intro-
ducing stiff brackets B in the
case of shallower girders.
The effect of this type of
bracing is to create local
bending stresses in the
columns at their junctions
with the horizontal girders,
and the general effect on the
framing is shown in Fig. 246,
the distortion being greatly
exaggerated. The rigidity of
the connections is here relied
on to distribute the bending action of the wind pressure equally amongst
those columns which are efficiently braced by deep girders, and this assump-
tion is the only" one on which any approximate calculation of the wind
stresses can be based. Points of contra-flexure p, p, occur at or near

FIG

MISCELLANEOUS APPLICATIONS

815

the centre of each storey length I of columns, where the bending moment
due to wind pressure will be zero. Assuming wind pressures P, P, to
act horizontally at the centre of each storey,
and the building to be three columns deep as
in Fig. 246, the bending moment on each column

P I P/

at the horizontal section A will be ^ x z = -TT.

O 2i O

At section B, one-third of the total wind
pressure above section B may be supposed
acting at each of the points^, and the bend-
ing moment on each column at section B

2P / PI
= IT X 2 = "a"" Similarly at C, the bending

q~P 7 T)7

moment on each column section = — x 5 = m

o '2i 2i

and at D, — x ^ = ±j--

This method of

FIG. 246.

transferring the wind load to the foundations
is often referred to as " table-leg " principle.

The column section is augmented from
roof to basement in order to meet the in-
creasing direct compression and bending moment, and the ultimate
effect of this type of bracing is to increase the intensity of compression
alternately on windward and leeward faces of every column so braced.

fans.

+ 6 /
-4

+ J /

-Z

-s /

-4

+ 3 _/

FIG. 247.

FIG. 248.

To be efficient, the rigidity of the horizontal girders should be large
relatively to that of the columns, and this result is obtained in modern
practice by giving such girders a liberal depth.

When diagonal bracing is employed, the skeleton of the building

316

STRUCTURAL ENGINEERING

acts wholly or partially as a braced frame or girder according as the
bracing extends throughout the whole system of columns, or is only
employed in certain bays. Fig. 247 indicates the stresses in one frame
of the four upper storeys of a building four columns deep from back to
front with diagonal bracing throughout, assuming the wind stress to
be equally shared by the diagonal braces in each storey ; and Fig. 248
the stresses in the same building when the central bay is unbraced.
The diagonals are assumed inclined at 45°. In Fig. 247 it will be
seen that the vertical tension in the diagonals of the windward bay AB
is augmenting at the rate of 1 ton per storey, so that the uplift at the
foundation of column A, if carried down through twelve storeys, would
amount to (1 + 2 + 3 + 4 . . . +12) = 78 tons. The foundation of
column D would be superloaded to the same extent under the maximum
wind pressure. The compression in columns B and 0 is augmented at
the rate of 1 ton per storey, but the ultimate load on their foundations
is not affected, the compression in each storey length of columns B and
C being balanced by the vertical tension in the wind tie attached at its
lower end. In Fig. 248 the bays AB and CD constitute separate canti-
levers, CD receiving its wind load through the horizontal girders

traversing the bay BC. The
uplift .on columns A and C
at a depth of 12 storeys =
(li + 3 + 4£ + G . . . +18)
= 117 tons, whilst an increased
pressure of 117 tons comes on
the foundations of columns B
and D at that depth.

In buildings of exceptional
height, the uplift on founda-
tions of windward columns may
exceed the pressure due to the
dead and live loads on the
structure. In such cases the
columns require efficient an-
chorage. As an example, ten
of the main columns of the
Singer building are anchored
down in the manner shown in
Fig. 249. The maximum uplift
in this instance is 413 tons.
Four steel bolts B, 4 in. in
diameter, pass through steel
cross-heads H bearing on heavy
gussets G riveted to the base of
the column C. The lower edges
of the gussets and column plates
are planed to bear on the cast-
steel shoe S. The bolts pass
between the beams L, L, of the grillage, below which they are
attached to a second steel casting embedded in the concrete 4 ft.
below the grillage. From this casting a series of steel eye-bars are

FIG. 249.

MISCELLANEOUS APPLICATIONS 317

embedded for a further depth of about 40 ft. in the concrete filling of
the foundation shaft. These anchor bars are diminished in number
towards the lower end. an adhesive force of 50 Ibs. per square inch
between their faces and the concrete being allowed in designing their
sectional area.

CHAPTEE XI.

MASONRY AND MASONRY STRUCTURES.

Masonry. — The following classes of masonry are principally employed
in masonry structures.

1. Unsquared or Random Rubble for unimportant and temporary
walling.

2. Squared Rubble, built either in courses or uncoursed. Built in
courses, this class of masonry is perhaps most widely used for general
work. For small walls, the individual stones are relatively small and
the courses of moderate height, whilst for retaining walls, heavy piers and
abutments of bridges, the scale of the construction may be magnified to
any desired degree. Squared uncoursed rubble is probably superior to
coursed rubble for work demanding great strength, since vertical bond-
ing is obtained in addition to horizontal bonding. It is, however, more
difficult to ensure really first-class work in this than in coursed rubble
masonry, and largely for this reason uncoursed rubble is not so exten-
sively used. Probably no class of masonry is so liable to be scamped as
rubble masonry, since in the absence of very thorough and constant
inspection, it is an easy matter to give the face work an excellent
appearance whilst the backing of heavy walls may be practically devoid
of bond and simply consist of a mass of small and relatively useless
material.

3. Rubble Concrete. — This is principally employed for the bulk of
the heaviest engineering structures, such as masonry dams. It may be
compared to rough uncoursed rubble on the largest scale, in which the
individual stones consist of masses of rock ranging from a few hundred-
weights to 7 or 8 tons in weight, whilst they are set in concrete instead
of in ordinary mortar. Such work requires the external faces covered
with large squared rubble facing set in cement mortar.

4. Ashlar Masonry is built of blocks of large size very carefully worked,
in courses usually of equal height and laid with joints seldom exceed-
ing one-eighth of an inch in thickness. It is the most expensive class
of masonry, and its place in engineering work is limited to the facing of
works to which it may be desired to give a highly finished appearance,
and to those portions of a structure which must of necessity be given
very accurate shapes and faces. Such, for example, are internal facing
of docks, locks, gate sills, quoins, copings, overflows for dams, arches
and the water face of quay walls. Piers are frequently given ashlar
quoins, whilst occasionally ashlar lacing courses are laid at intervals of
20 ft. to 30 ft., in rubble-stone piers, to bring up the work to a dead

318

MASONRY AND MASONRY STRUCTURES 319

level in order to more uniformly distribute the pressure before com-
mencing a further rise. Fig. 2G7 illustrates such an example. The
blocks in ashlar facing are usually worked smooth, but may be draughted
with rock face or other treatment as desired, depending largely on the
degree of ornamental appearance or otherwise required.

Although the above names are ordinarily accepted as implying a
certain class of masonry in general building construction, it is unwise
for an engineer to undertake to designate a particular class of masonry
by a special name without at the same time stating in the specification
a careful description of the kind of masonry intended to be built.
Class names for masonry vary considerably in different parts of the
country, and in different countries. The names applied to various
classes of masonry in America, for instance, would be almost unrecog-
nizable by an Englishman. Different qualities of coursed rubble are
variously referred to as " ranged rockwork," "random range work," and
ashlar is commonly known as " dimension stone masonry." " Rip-rap "
is a purely American term applied to rough irregularly shaped stones
used for pitching the slopes of earthen dams, as distinct from " paving,"
which is usually understood to imply roughly rectangular hammer-
dressed stone laid dry by hand in regular courses.

Piers, abutments and retaining walls are frequently built either
entirely in blue brick or with blue brick facing and stock brick hearting,
or with masonry facing and brick hearting. In laying masonry and in
writing specifications for masonry three essential aims should be borne
in mind. 1. Uniformity of bearing on each course or horizontal section.
2. Absolute solidity with absence of voids. 3. The most perfect bonding
attainable so that the mass may be, as nearly as possible, monolithic in
character. Whilst it is impossible in a work of this scope to give
complete specifications at length, some of the essential points which
should be included in specifications for masonry will now be noticed.
In the following notes on specifications, illustrations are inserted for
the purpose of rendering more clearly the intended arrangement and
bonding of the work, but it will be understood that such drawings do
not usually accompany specifications.

Outline Specifications for Masonry.— Stone.— The stone employed
for general rubble masonry should be of an approved kind, sound and
durable and free from all flaws, seams, cracks and discolorations. The
size of stones employed will depend on the magnitude of the work and
the sizes conveniently obtainable from the quarries. In general such
stones will be from 6 in. to 14 in. thick, 2 feet to 5 feet long, 10 in. to
36 in. wide, the larger blocks being used for the heavier classes of work.
Strongly acute angles on stones should be entirely prohibited, and any
angles less than 80° are undesirable. Blocks of the softer kinds of
sandstone and limestone should be thicker in proportion to length and
breadth than those of harder varieties, to mimimize danger of cracking
under heavy pressure. All stones in squared rubble masonry should
have the top and bottom beds parallel to the natural quarry bed and be
approximately rectangular in shape.

Face Dressing. — Fig. 250 shows various kinds of face dressing on
stones. A and B are examples of ashlar faced masonry, with rebated
joints, the rebate in A being formed entirely on one block, and in B

320

STRUCTURAL ENGINEERING

half on each adjacent block. The faces are finished smooth by rubbing
after the final tooling or sawing. Deep rebated joints are employed
where heavy characteristic lines are desired, the deep shadows imparting

FIG. 250.

boldness of appearance to the work. Fig. 250, C, shows a block finished
with fine and coarse axed face, on left- and right-hand sides respectively.
At D, the face is " droved " or " tooled," being worked with a broad flat
chisel. E is a rock-faced block with draughted margin. This type of

MASONRY AND MASONRY STRUCTURES 321

dressing without the draught is always employed for squared rubble
masonry, the draughts being occasionally cut on the quoins. The face
F is " broached " and draughted, whilst Gr is a rebated and rusticated
block. The more ornamental face dressings are seldom employed on
engineering work.

General clauses regarding measurement for payment by the cube
yard (or cube foot in case of ashlar), conditions of inspection and execu-
tion according to drawings, usually precede a detailed specification for
masonry.

Heavy Squared Coursed Rubble (occasionally referred to as Bridge
Masonry, Coursed Blockstone, and Block in Course). — Foundation
courses which are laid immediately on a concrete foundation slab, to be
of selected stones not less than 18 in. thick and to have a bed area of
not less than 15 square feet. No course to be less than 14 in. nor more
than 24 in. in height, and each course to be continuous both through
and around the wall. If courses of unequal height are permitted, the
deeper ones to be at the bottom diminishing regularly towards the top.
Face stones to have rock-faced surfaces, the edges being dressed to
straight lines conforming to size of courses, and no part of rock face to
project more than 3 in. in heavy courses or 2 in. in medium courses.

In Figs. 251 the elevation and plans of two consecutive courses of
part of a rubble masonry pier are shown, together with a vertical section
on XY. The beds and vertical joints of face stones to be dressed back
at least 12 in. from face of wall and the under beds to extend to the
extreme back of stone as at B, Fig. 252. Overhanging stones as at C
leave spaces A which are liable to be indifferently packed. Stretchers
to have a length not less than 2J times their height, and no stone to
have a less width than 1^ times its height. The face bond to show not
less than 12 in. lap. The size and disposal of headers and bond stones
will depend on the thickness of wall. Headers should preferably be not
less than 3 feet to 4 feet long. For walls less than 4 feet thick the
headers should extend from back to front, giving the arrangement shown
in plan in Fig. 253. For walls exceeding about 4 feet in thickness,
there should be an equal number of headers built into both back and
face, so arranged that a face header shall be roughly midway between
two headers in the rear as shown in plan in Figs. 251, 254, and 257.
In the case of retaining and wing walls which are backed by earth, the
back face may of course be left much rougher than the exposed front
face, but this should not be allowed to excuse defective bonding at the
back of the wall. There should be one header to every two stretchers,
and they should sensibly retain their full face size for the whole length
of stone beyond the dressed surfaces, as A in Fig. 255. At B is shown
an inferior header, the dressed joints (shaded) not extending far enough
into the wall, whilst the tail end tapers off, requiring packing up with
rubbish or leaving voids beneath the stone. This is the commonest
defect in rubble masonry, and cannot be detected from the face appearance
once the course is covered up. In walls of 6 feet to 7 feet thickness the
inner ends of headers should overlap laterally.

The hearting or packing stones to be of large well-shaped stones of
the same general thickness as the face stones. No voids exceeding 6 in.
width to be left between these stones, and all voids to be thoroughly filled

322

STRUCTURAL ENGINEERING

Elevation.

FIG. 251. Sechon X-Y.

S |H| S ; S |H

FIG. 257,

MASONRY AND MASONRY STRUCTURES 323

with spalls bedded in cement mortar. The superposition of the larger
hearting stones in each two consecutive courses should have regard to
the preservation of the best possible bond and minimum number of
coincident vertical joints. (See plans of courses in Figs. 251.) Where
provision is to be made for leading off drainage water through masonry,
the positions of weep holes or inlaid pipes will be either shown on draw-
ings or personally indicated by the engineer in charge.

Other points to notice are that heavy hammering or dressing of
stone on the wall itself should be prohibited ; stones accidentally broken
or moved after setting, to be removed and properly replaced ; each
stone to be laid in a full bed of mortar, and joints not to exceed J in.
to f in. when laid. Stones and masonry to be kept free from dirt,
and all stones and adjoining masonry to be wetted with clean water
just before laying, especially in hot weather. If the masonry is to be
extended by building future work, the old masonry should be stepped
back uniformly to ensure a break in vertical bond between old and
new work of at least 12 in. Quoins in massive work are usually
draughted 2 in. to 3 in. wide, the width of draught being proportioned
to the scale of the work, and they may be further finished with rebated
joints, heavy rock face, or by rusticating, according to the degree of
finished appearance required.

The above notes relate more particularly to the massive types of
masonry structures. For work intermediate between these and ordinary
walls occurring in general building construction, the following class of
squared rubble is employed.

Ordinary Squared Coursed Rubble. — Fig. 256 shows the general
face appearance of such work, and Fig. 257 a plan of a typical course.
The main points of difference are the generally smaller size of stone,
and less regularity of arrangement. The usual height of each course is
12 in. to 14 in. The stones in each course are not all the full height of
the course, but all headers and bond stones should be so. The number
of headers may be specified as 1 to 3, 1 to 4 stretchers, etc., or as so
many per square yard of surface, or per 6, 10, or 12 ft. run of the course.
The smaller face stones should not have a less thickness than one-third
the course height, and preferably not more than two stones should be
allowed to the course. In highly finished work of this class, greater
regularity of appearance on the face is obtained, as in Fig. 258.
Specially skilled masons are required for executing really good work of
this character. Exceptionally good examples may be seen in the masonry
of the overflows and training walls at the Langsett reservoir near
Sheffield, executed by Mr. William Watts, M.Inst.C.E. The general
remarks on backing for heavy masonry apply equally with regard to
the backing in this case. Figs. 259 and 260 are other examples of
regularly faced rubble, although the bond in Fig. 259 is necessarily
inferior.

Rubble Concrete. — This class of masonry is practically confined to
the hearting of reservoir dams, breakwaters, etc. Fig. 261 indicates
the general disposition of such work when laid. Large irregular
blocks called plums or displacers, with fairly flat but rough beds, and
weighing from 1 ton to 8 tons, constitute the bulk of the work. They
are bedded on a thick layer of cement concrete, or very coarse mortar,

324

STRUCTURAL ENGINEERING

consisting of 2 parts sand, 4 crushed rock spoil, small enough to

pass a

1 part cement.

FIG. 261.

FIG. 262.

The blocks are usually slung on
to the work from an overhead
cableway, and are arranged to
break joint both horizontally
and vertically, no two stones
being nearer than about 6 in.,
to allow of perfect ramming of
the joints with concrete. The
stones are settled on their beds
by slowly working them back-
wards and forwards by crow-
bars, and by heavy hammering
with wooden malls. It is im-
portant in such work to use as heavy stone as is procurable, since
the stability depends on the weight of the mass, whilst the closer
together the stones can be laid, the greater the ultimate mean density
of the mass, and the less the amount of cement required. The stones
should not be of too irregular a shape, or very large voids require to be
filled with concrete. Smaller stones are used for partially filling the
joints. A fair proportion of stones, as indicated
in section m Fig. 261, should bond transversely,
as well as longitudinally. All such overhang-
ing parts as A in Fig. 262 should be removed,
otherwise the e.g. of the block is thrown ap-
preciably away from the centre of the bed,
causing unequal settlement and tilting on the soft bed of concrete.
Work of this character over large areas is necessarily interrupted, and
where new masonry is joined to old, the surface of the concrete is
picked over, carefully brushed and washed and covered with cement
mortar before commencing the new work. In this connexion, the late
Mr. G. F. Deacon considered that a hydraulic lime was probably superior

to cement, since the much slower setting
enabled new work to be bonded to work
several days old whilst still in a plastic
state.1 The face of such work is finished
in ashlar, or heavy, square, rock-faced
rubble. Fig. 263 shows the typical ap-
pearance of the outer face work of the
Yyrnwy dam.

Ashlar. — The stones are cut to exact
dimensions, with surfaces of beds and
joints dressed back to the full depth of
block, and joints should not exceed ^ in.
Where a considerable area is faced with

ashlar, the kind of bond is usually specified, and the stones laid and bonded
with the regularity of brickwork. Courses may be of equal or unequal
height, preferably the former. In the highest class work the beds and
joints are required to be plane throughout. As this entails heavy waste
in dressing, other specifications allow depressions below the level of the
1 Mins. Proceedings Inst. C.E., vol. cxlvi. p. 27.

FIG. 263.

MASONRY AND MASONRY STRUCTURES 325

beds, provided they do nob exceed, say, 6 in. in width, are not Dearer
than 4 in. to an edge, and do not aggregate more than one-quarter
the whole area of bed. To produce ashlar blocks quite free from
such depressions or plug-holes considerably increases the expense of
labour in cutting, and reduces the general bulk of the finished
blocks.

Arched Masonry Blocks require cutting to accurate radial lines,
and if large, are usually worked to drawings supplied. The plane of the
natural quarry bed should be perpendicular
to the direction of pressure acting through
the arch. In arches with ashlar ring-stones,
and brick or rubble sheeting, the bond
with the sheeting should not be less than
from 6 in. to 12 in., depending on size of
work. The ring-stones, or face voussoirs,
V, Fig. 264, should preferably be cut with FlG

square heads, in order to obtain a satis-
factory bond with the stones of the head wall, H.

Bonding of Brick and Masonry Arches. — Fig. 265 illustrates various
methods of arranging and bonding the bricks and masonry in arches.
A shows a 13^ in. brick arch with bricks laid alternately header and
stretcher. B shows the same bond applied to an 18 in. arch. Although
the bond is satisfactory throughout the depth of the arch, the dis-
advantage of laying the bricks in this way is that the joints become
excessively wide at the extrados unless tapered or radius bricks are
used. For this reason, arches exceeding 18 in. are generally built in
duplicate rings, each 9 in. thick, as at C, or in separate 4^ in. rings, as
at D. With either of these methods, the bond is deficient, and unless
very carefully built, unequal settlement on striking the centering, is
liable to cause one or more rings to fall away from the others, as shown
at F, when the load comes on a much reduced thickness of arch. In
order to secure some degree of through bonding, the arrangements at
G and E are often adopted. That at G shows the bond inserted in the
22 J in. portions of the lining of the Totley tunnel. The inner and
outer 9 in. are built in English bond as at B and C, whilst the central
4J in. constitutes an independent ring, which is bonded alternately with
the inner and outer rings at sections where the joints JJ are flush
throughout the whole depth. At E bond stones S or specially con-
structed brick voussoirs V are inserted at flush joints. In stone arches
any desired arrangement of bond may be adopted, since the stones are
cut to taper shapes. The expense is, however, much greater than in
the case of brick arches, and excepting where architectural effect is to be
secured, arches built of stone throughout are seldom employed in
ordinary structural work. The most usual construction for arches of
from four to eight half bricks in thickness, which comprise the majority
of engineering arches and tunnel linings, is to build the exposed soffit
ring in blue brick, and the backing rings in good stock brick. In the
best work, the outside faces are of masonry bonded alternately header
and stretcher with the brickwork. At D a label course, L, of stone
is shown. This is sometimes added in large brick arches to secure a
more finished appearance, and is usually from 4 in. to 6 in. thick. At

326

STRUCTURAL ENGINEERING

H is a masonry arch regularly bonded header and stretcher, the face
stones being rebated and rock-faced. This is a suitable treatment for
large span arches, where appearance is of importance. K illustrates a

FIG. 265.

heavy masonry arch ring having every second voussoir square-shouldered
to bond with the head wall. The rough square rubble arch at M is
occasionally adopted for rough or temporary work.

Piers. — Masonry piers are usually of rock-faced square rubble
masonry of the kind already described. Where, however, the work is
imposing by reason of exceptional magnitude or where considerable
architectural effect is desirable, all external faces are built in ashlar
with varying degrees of enrichment in the matter of rebates, mould-
ings, etc. Fig. 266 shows one of the piers for the aqueduct over the
river Loire at Briare in France, and is an excellent example of highly
finished and well-treated engineering architecture. This structure
carries the Loire canal over the river Loire in fifteen spans of 131*25 ft.,

MASONRY AND MASONRY STRUCTURES 327

FIG. 266.

FIG. 267.

328

STRUCTURAL ENGINEERING

and owing to its prominent situation and imposing appearance, has
been given a high degree of architectural finish. 1

Fig. 267 illustrates the construction of one pier of the Mussy
viaduct, also in France. This, one of the largest masonry viaducts
erected, comprises 18 arched spans of 82 feet each, the maximum height
from foundation to rail level being 180-25 ft. The piers are of rubble
masonry, with ashlar lacing courses at intervals of about o5 ft. A
plan of one of these courses L is shown in the figure, all the main
blocks being tied together by iron cramps, the detail of which is also
shown. The lower figure indicates the face treatment, and the
structure constitutes a fine example of massive bridge masonry on the
largest scale.2

Masonry- Faced Concrete Blockwork. — Fig. 268 shows the type of
construction followed in the re-constructed north Tyne breakwater.

The blocks C are of concrete
alone, and vary from 5^ to 19
cube yards content. The facing
blocks are covered with an outer
face F of Aberdeen granite, this
masonry facing being built in-
side, and forming one end of the
timber box in which the blocks
were moulded. The blocks are
regularly bonded alternately
header and stretcher, and are
further dowelled by filling the
vertical semi-cylindrical grooves
D with 4 to 1 cement concrete,
the blocks consisting of 6 to 1 concrete. An extremely durable outer
face having the appearance of a masonry structure is thus secured.3

Footing Courses.— The lowest portions of masonry structures
usually rest on footing courses, which project some distance beyond the
faces of the superstructure in order to distribute the pressure due to the
weight of the structure over a larger area. These footing courses
again rest on a mass of concrete laid in the bottom of the excavation for
the foundation. The concrete serves the following purposes. By being
made of still greater area than the footing courses it reduces the in-
tensity of pressure to the safe bearing power of the soil ; and, secondly,
it fills up all irregularities of the foundation and provides a smooth and
level surface on which to commence building the masonry. The thick-
ness given to the concrete bed depends largely on a variety of practical
considerations, although, as will be seen later, it must have not less than
a certain minimum thickness fixed by the weight and disposition of load
on the superstructure. It is, for example, often cheaper or more con-
venient to put in a considerable depth of concrete instead of building
additional masonry on a thinner layer, whilst in other cases the lower
portion of the structure may require to be formed by depositing con-
crete below water-level until a suitable height is reached on which to

1 Annaks des Fonts et Chaussdes, 1898, 2nd Trimestre.

2 1 did., 1901, 1st Trimestre.

3 Mins. Proceedings Inst. C.E., vol. clxxx., p. 133.

FIG. 268.

MASONEY AND MASONRY STRUCTURES

329

FIG. 269.

commence the masonry. Footing courses of masonry should consist of

specially selected large and well-shaped stones. Hardness is essential,

since they are subject to the heaviest pressure.

The successive offsets or projections should be

made very gradually. If the footing courses

project too far beyond the superstructure or

beyond each other, they are liable to be cracked

when bearing on concrete or to tilt up as in

Fig. 269, if bearing directly on the foundation

earth. In the latter case the advantage of the

footing courses is quite lost, since the effective

bearing area is reduced by the breadth AB.

No stone in a footing course should project more than 5 to J of its
length, and such stones should further have a liberal depth. (See pre-
ceding specification.) Piers, abutments, wing-walls and retaining walls
are frequently built of brickwork, and for these structures each footing
course should not be less than four bricks in thickness, and more for
heavy works. The projection of each course should not exceed 2j in.,
and the upper course of each ledge or offset should be all headers as at
H, H, Fig. 270. The bonding is shown in the figure for a wall executed
in English bond, three-quarter
bricks being inserted at A, A,
and whole bricks at B, B, in
order to properly break joint.

Pressure on Foundations. —
The bearing power of soils
naturally varies widely, and any
stated values can only be re-
garded as general standards.
Before commencing any impor-
tant structure, and especially
where the bearing capacity of
the foundation is an unknown or
doubtful quantity, careful exami-
nation and testing are desirable.
If the foundation be near the
surface, the bearing power may
be conveniently tested by erect-
ing a strong timber platform carried on four legs of 12 in. square cross-
section, and gradually and uniformly loading it until any ^ desired
maximum settlement takes place. The settlement of each leg is noted
at intervals of one or two days by taking level readings. For founda-
tions at considerable depths below the surface, a tank resting on a
plate of 2 or 3 square feet area may be placed at the bottom of
trial excavations and gradually filled with water. Whether a con-
siderable amount of settlement may be allowed depends on the kind
of structure to be erected. An appreciable amount of settlement is
not objectionable, provided it takes place uniformly over the whole
area of the foundation. Where a definite amount of settlement is
anticipated, the original level of the foundation is adjusted so that the
desired permanent level may be established when the total weight of

FIG. 270.

330

STRUCTURAL ENGINEERING

the completed structure comes upon the foundation. It is, however,
desirable to minimize settlement in such structures as piers and abut-
ments for bridges, chimneys, retaining walls, and dams. In the former,
settlement entails subsequent trouble with girder bearings, cracking of
arches, and frequent packing up of the track ; whilst in retaining walls
and dams, settlement, if of appreciable amount, is necessarily unequal,
and results in vertical displacement, creation of unknown shearing
stresses, or heeling over of the structure. In the case of tali framed
buildings, some amount of settlement is usually unavoidable. The
accuracy with which the bearing area of the foundations of the support-
ing columns may be proportioned to their loads, however, realizes a
high degree of uniformity of settlement. In the tall buildings of
Chicago a settlement of 6 in. to 12 in. is general, and cannot be avoided,
owing to the compressibility of the clay in which they are founded.

TABLE 30.— PERMISSIBLE SAFE PRESSURES ON FOUNDATIONS AND

MASONRY.

Material.

Tons per sq. ft.

Very hard rock . ...

25 to 30

Ordinary rock

8 , 16

Dry clay of considerable depth

4 , 6

Moderately dry clay ...

2 , 3

Soft clay . .

1 2

Compact gravel and coarse sand .

6 , 8

Clean dry sand laterally confined

Alluvial soils .... . .

i to 1

Good red brickwork in cement mortar .

Blue
Squared rubble freestone masonry in cement mortar .
Ashlar freestone masonry in cement mortar
Granite ashlar

15
8
15
25

Freestone ashlar bearing blocks

Granolithic „ „

Distribution of Pressure on Foundations. — Any masonry structure
symmetrical about its vertical axis, such as the rectangular and
pyramidal piers in Fig. 271, will have its centre of gravity G vertically
over the centre of gravity M of the area of foundation abed. The
point P, where the vertical through G cuts the foundation level, is the
centre of pressure on the foundation, and in these cases obviously coincides
with M. Further, in such cases the pressure per square foot on the
foundation area will be uniform, provided the upward resistance of the
soil is also uniform. If W = total weight of structure in tons, and A
= area of foundation in square feet, then pressure per square foot on

foundation = -v- tons. If, as usually happens, the area abed be too

small to suitably distribute the weight W, it must be increased by
either footings, a projecting concrete base, or grillage beams until the

MASONRY AND MASONRY STRUCTURES
W

331

unit pressure -v- does not exceed what may be safely imposed on the

soil. Thus, if the pier A be 24 ft. high, 30 ft. wide, and 5 ft. thick,
and built of masonry of 140 Ibs. to the cube foot, its weight will be
225 tons. Suppose it to carry a further central load of 300 tons
applied by girders resting on it, the total load W = 525 tons. The
foundation area A, without footings = 30 x 5 = 150 sq. ft., and the
pressure per square foot = fff = 3'5 tons. If it is undesirable to load
the foundation soil beyond 2-5 tons per square foot, the necessary area

5^5
with footings will be -^p = 210 sq. ft., which might be obtained by

making the bearing area a rectangle 31' 6" x 6' 8". It is often con-
venient to represent the intensity of pressure on foundations diagram-
matically. If eh be made = 2J tons to any convenient scale, then the
pressure per square foot being uniform, the ordinates of the rectangle
ffffh indicate the pressure per square foot at any point along the
foundation from e to/, and efgli is a pressure intensity diagram.

In the case of an unsymmetrical structure, such as the reservoir
dam in Fig. 272, the centre of gravity G is not vertically over the

FIG. 271.

FIG. 272.

centre of area M of the foundation abed, and the centre of pressure P
vertically below G- no longer coincides with M, but falls to one side of
it. The weight of the structure is here concentrated more towards ad
than fo, and the intensity of pressure on the foundation will be greater
at the inner face ad than at the outer face lc. The pressure intensity
diagram efgli will now be a trapezium, having eh considerably greater
than fff. If the values of the intensities of pressure eh and fff be
calculated, the straight line hg will cut off ordinates representing the
pressure per square foot at other points along the base. It should be
noticed that by drawing a straight line from h to ff, it is assumed that
the ground beneath the dam is of uniform character. If, for instance,
in the neighbourhood of the points H, H, H, the foundation were
appreciably harder than in other parts, the weight would bear more

332 STRUCTURAL ENGINEERING

intensely on these points, and the intensities of pressure on the base
would then be represented by some undulating line as 7c/, cutting off
deeper ordinates beneath H, H, H, and shallower ordinates beneath the
intervals of softer ground S, S. The mean pressure would remain the
same, that is, the area efgh would equal the area eflk. Local defective

places such as S, S, are made
good in all carefully prepared
foundations, and the intensity of
pressure may reasonably be ex-
pected to vary practically uni-
formly in well-executed works.

General Expression for In-
tensities of Pressure on Rec-
FIG. 273. tangular Foundations. — To

ascertain the values of the pres-
sure intensities eh and fg in Fig. 273, let b = breadth of foundation
in feet, W = total weight or vertical pressure on foundation in tons
per foot run acting at the centre of pressure P, distant mb feet from e.
Let x and y tons per square foot be the intensities of pressure at e and
/ respectively.

Considering a one-foot length of the structure, the mean pressure on

foundation = — j-^ tons per square foot, and the total pressure per foot

/T* L I At

run = — ~- x #, which must = W tons .... (1)

Also, taking moments about e, W x mb = moment of area efgh
about e. Dividing the area efgh into a rectangle and triangle by the
dotted line gk, its moment about e

3* I 'ii 9W

But from (1), ^- x & = W, whence y = -~ -
Substituting in (2)

2W
from which x = - -, (2 - 3m) . . (3)

and by substituting in (1)

(4)

These results are important, and are capable of general application
in all cases of rectangular surfaces in contact under a known pressure,

MASONRY AND MASONRY STRUCTURES

333

acting through a known centre of pressure. The factor m = ^r, so that

when the position of the centre of pressure P is known the value of m
is also fixed. Applying this result to the dam in Fig. 272, suppose the
base to be 60 ft. wide, the weight of the dam per foot run to be
200 tons, and the centre of pressure P to be 21 feet from the inner
face. Then m = §£, I = 60, and W = 200 tons. Therefore pressure
per square foot

at inner face = x —

2 X 200
60

(2 - 3 X M) = 6J tons

2 x 200
and at outer face = y = — — (3 X fj - 1) = §

ton.

FIG. 274.

These, of course, are the pressures at opposite edges of the foundation
clue simply to the dead weight of the dam, no account having been
taken of any water pressure acting against it.

It will be seen that here the centre of pressure falls to one side
of the centre of the foundation area, entirely on account of the shape
of the section of the dam,
the material being bulked
up more to the left- than to
the right-hand side. In
many structures the centre
of pressure is caused to fall
nearer to one edge of the
foundation by the action of
external forces. In Fig. 274
the pier A, under the action
of its own weight only, would
have the centre of pressure
on its foundation at P, the
centre of base. When subject
to a lateral wind pressure F, the resultant pressure on the base takes
the direction OP15 the centre of pressure being thereby displaced from
P to PX. In the retaining wall B the earth pressure E causes the
centre of pressure to be displaced from P to Px, and in the case of
the dam C the centre of pressure falls at Pa under the action of the
horizontal water pressure W, instead of at P when the reservoir is
empty.

The intensities x and y have particular values, according as the
fraction m is greater than, equal to, or less than ^. Four distinctive
cases occur, which are illustrated in Fig. 275. In each case a total
vertical load of 16 tons per foot run acts on a foundation 8 ft. wide.

Case 1. — Centre of pressure at middle of foundation, m = J ;
W = 16 tons ; I = 8 ft. Inserting these values in equations (3) and
(4), x — y = 2 tons per square foot, or the intensity of pressure is
uniform over the whole foundation area.

Case 2. — Centre of pressure 3 ft. from one edge of foundation.
m = g, or is greater than ^. Hence x — + $i tons, and y = + ^ ton
per square foot.

334

STRUCTURAL ENGINEERING

Case 3. — Centre of pressure 2f ft. from one edge, or | of 8 ft.
m = |, whence x = + 4 tons per square foot, and y = 0.

Case 4. — Centre of pressure 2 ft. from one edge, or less than J of
8 ft. m = ^, whence # = -f 5, and y = — 1 ton per square foot.

In the last case the value of «/, being negative, signifies the intensity
y to be a tensile instead of a compressive stress, and its value is
accordingly plotted on the opposite side of the horizontal line EF to
that of x. As the centre of pressure moves from the centre towards
the left-hand edge of the foundation area, the pressure per square foot
on the left steadily increases, whilst that on the right decreases. In
Case 3 the right-hand intensity diminished to zero, and in Case 4 it has
passed beyond zero and become an uplift instead of a downward
pressure. Assuming the foundation to be appreciably compressible —
clay, for instance — in Case 1 uniform settlement will take place, and a
wall standing on this 8 ft. base will settle from the level E'F to some
level e'f, the vertical depth EV representing the settlement produced by
a pressure of 2 tons per square foot on the subsoil in question. In

Case 2 an intensity of 2 tons per square foot exists at QR. Q# = EV
therefore represents the settlement beneath this point of the foundation.
If AB be produced to 0, then at this point the intensity of pressure on
the foundation would be zero, and the settlement also zero. Joining
Cq and producing to e, the depths Ee to F/* represent the unequal
settlement taking place in this case, which is everywhere proportional
to the intensity of pressure. Similarly in Case 3, QR = 2 tons per
square foot. Q# = EV, and Ee represents the settlement at E, whilst
that at C is nothing. In Case 4, QR = 2 tons per square foot.
Qg = EV, and eC/ determines the settlement Ee at E, whilst an uplift
F/ occurs at F, the point C on the base of the wall neither depressing
nor rising. The effect of the unequal settlement in Cases 2, 3, and 4
is to cause the structure to heel over towards the side where the greater
intensity of pressure exists, and in each case the structure rotates
slightly about the point C as centre. This action is especially apparent
in the case of heavy masonry structures on yielding foundations.

Masonry, concrete, and mortar possessing but small tensile strength, it
is generally admitted as undesirable to allow tensile stress to exist in such

MASONRY AND MASONRY STRUCTURES

335

16 fans.

materials. In Case 4 tension exists from C to F, and whether
EF represent a horizontal joint in a masonry structure, a horizontal
section of a monolithic concrete structure, or the actual foundation
level, whilst it is undesirable to have tension existing from F to C, it
does not necessarily follow that the structure is unsafe. Provided the
intensity of compression EA at the opposite face does not exceed
the safe compressive resistance of the mortar, concrete, or subsoil,
the structure may be perfectly safe. If, however, the maximum
intensity of tension FB exceed the tensile resistance of the mortar
or concrete, then in a masonry
structure the joint will open for
some distance in from F ; in a con-
crete structure cracks will be de-
veloped ; and at a foundation level
a certain amount of uplift will take
place. The immediate effect of
such actions is to reduce the

FIG. 276.

effective breadth of joint, as in 0-5-23
Fig. 276, from EF to EF'. ^ Em-
ploying the same values as in the
previous case, suppose the joint to
open for a length of 1 ft. The
reduced breadth EF' = 7 ft., m = f , and x = 5'23 tons per square
foot compression, and y = 0'65 ton per square foot tension. If this
tension is incapable of causing further opening of the joint or crack at
F', and the compression of 5'23 tons does not exceed the safe resistance
of the material, the structure will still be safe. If further opening
occurs at F', EF' is still further reduced, the intensity x increased, and
safety ultimately depends on the extent of this increase in the value of
x. In the case of piers, retaining- walls, and arches the existence of
tension at one face of the joints does not necessarily render the
structure unsafe ; but in masonry or con-
crete dams any opening of joints or cracks,
however slight, on the water face is accom-
panied by entry of the water, and consequent
application of an upward hydrostatic pres-
sure H in Fig. 277, which creates a new
resultant pressure PA with centre of pres-
sure at Pj, in place of the previous resultant
PR with centre of pressure at P. This
resultant, acting on the reduced bearing

surface EF^ gives rise to further tension at Fj which extends the
width of opening FF15 introduces increased upward water pressure
with further displacement of the resultant pressure PA and still
further reduction in the width of bearing surface. The intensity
of pressure at E thus rapidly increases until failure takes place by
crushing of the material near E, or bodily overturning of the portion
CDEF. In the design of these structures, therefore, it is important to
ensure the centre of pressure at any horizontal section falling within
the middle one-third of the width of the section, and so to avoid the
formation of cracks or open joints on the water face.

FIG. 277.

STRUCTURAL ENGINEERING

RETAINING WALLS.

A face of earth exposed to the action of the weather eventually
assumes a more or less uniform slope AB, Fig. 278, the inclination of
which with the horizontal is called the angle of repose or natural slope
of the particular soil in question. The natural slope of different kinds of
earth varies very widely, and cannot be closely
specified for any particular variety. In the
case of the more homogeneous and uniform
varieties, as fine dry sand and gravel free from
large stones, the angle of repose is fairly
constant. The presence of varying amounts
of water in the same soil, however, greatly
modifies the angle of slope, and the nature of
soils in general is so varied that although many
may be classed as similar, yet it is seldom any
two of the same class exhibit the same properties. Any table of angles
of repose and unit weights of stated varieties of earth can only be
accepted as representing approximate average values, and if more
accurate ones be required they should be obtained by actual measure-
ment of a sample of the earth involved. The following are fairly
representative values of these quantities : —

TABLE 31. — NATURAL SLOPE AND UNIT WEIGHT OF EARTH, ETC.

FIG. 278.

Material.

Natural slope.

Weight.

Sand, dry

deg.
30 to 35

Ibs. per cub. ft.
90

„ wet
Vegetable earth, dry
,, ,, moist ....
wet

26 „ 30
29
45 to 49
17

118
90 to 95
95 „ 110
110 120

Loamy soil, dry

80 ,, 100

Clay drv

120

damp

120 to 130

„ wet
Gravel, stone predominating
Gravel and sand .

16
45
26 to 30

135
90
100 to 110

These values being so variable, it is useless, in the theoretical design
of retaining walls, to expect the same degree of accuracy which may be
relied on in dealing with a constructive material like steel, whose
properties are much more closely established. Moreover, the earth
behind a high retaining wall is seldom uniform from top to bottom.
Two or three distinct beds of different classes of earth may be present,
which further increases the difficulty of applying mathematical procedure
in estimating the probable earth pressure against the wall

Earth Pressure. — In Fig. 278, in order to hold up the triangular
mass of earth ACB, a wall is required of sufficient stability to safely
resist the pressure P exerted on it by the earth. Many mathematical
formulae have been devised with a view to calculating the magnitude of

MASONRY AND MASONRY STRUCTURES

337

the earth pressure behind retaining walls, but they are necessarily based
on the fundamental assumption that the earth is perfectly uniform
throughout, and consequently do not provide for the variations
constantly encountered in practice. Some rule, however, is necessary as
a guide in the choice of a suitable section, and probably all formula? in
connexion with earth pressure at least err on the side of safety, since
the varieties of earth met with in practice are generally less uniform
and more cohesive than the ideal granular earth assumed in theory.

Rankine's Theory. — This theory has for many years been adopted
principally in England. It is based on a mathematical analysis of the
conditions of equilibrium of a particle in the interior of a mass of earth,
assumed to be perfectly dry, uniform and granular. It makes no allow-
ance for adhesion or for friction between the earth and back of the wall.
Stated briefly, if A = angle of repose of earth, II = height of wall in
feet, and w = weight of a cubic foot of earth in Ibs., then for a wall
retaining earth with a level upper surface, the horizontal earth pressure
per foot run of the wall

- sn

pounds.

Modifications of the formula are employed for giving the pressure in
cases of surcharged walls. It may be stated, however, that earth
pressures calculated by this formula are 25 to 30 per cent, in excess of
the actual pressures behind existing walls.

Rebhann's Method. — The following simple graphical method, due
to Professors Rebhann and Haseler, for determining the magnitude of
the earth pressure behind a
wall has been very generally
adopted on the continent of
Europe for many years past.
It has the advantage of being-
applicable alike to cases of hori-
zontal or surcharged upper sur-
face, and certainly gives results
both economical and satisfac-
tory in practice. The general
construction is shown in Fig.
279. BC represents the back
of the wall and CD the upper FIG. 279.

surface of the earth. Draw BD

making an angle A with the horizontal, equal to the angle of repose of
the earth. On BD describe a semicircle BGD. From C draw CE
making an angle = 2 A with the back of the wall. At E draw EG
perpendicular to BD to cut the semicircle in Gr. With centre B and
radius BGr, cut BD in K. Draw KL parallel to CE, cutting the upper
surface in L. Make KM = KL and join LM. The triangle KLM
is called the earth pressure triangle. Suppose it to represent a
triangular prism of earth one foot thick, as shown at T. Then the
resultant pressure P per foot run behind the wall, including the effect
of friction between the earth and wall, is given by the weight of this

838

STRUCTURAL ENGINEERING

prism of earth, or P = area KLM in square feet x 1 foot x weight
of earth per cubic foot.

It is unnecessary here to state the proof of this construction, for
which the reader may be referred to a paper by Professor G. F. Charnock.1
It may be stated briefly, however, that the method is based on Coulomb's
theory, and that the object of the construction is to determine the
position of point L, and consequently the location of the line BL. This
line BL marks what is termed the plane of rupture for maximum earth
pressure behind the -wall. In other words, if the wall be supposed to
yield, the triangular mass of earth BCL will tend to break away along
the plane BL, and the wall must be sufficiently stable to resist the
pressure caused by the tendency of the mass BCL to slide down the
incline LB. The actual earth pressure acts perpendicularly to the back
of the wall at two-thirds its depth H from the surface ; but the stability
of the mass of earth BCL is obviously partially maintained by the
frictional force F existing between the earth and back of the wall. The
resulting pressure P on the wall, as given by the earth-pressure triangle,
is therefore compounded of a certain horizontal pressure Q and a vertical
force F, depending on the coefficient of friction between the earth and
wall. It may reasonably be assumed that the coefficient of friction
between the earth and back of wall will be at least equal to that between

the particles of the earth itself, since the
backs of most walls are usually very rough.
The inclination of the resultant pressure
P may therefore be safely taken as making
an angle with the perpendicular to the
back of the wall equal to the angle of
repose of the earth. Possibly in the case
of a concrete wall constructed between
moulding boards back and front, the above
amount of friction might not be realized,
and a suitable reduction in the inclination
of P might be made accordingly.

The above construction is inapplicable
if the angle of repose exceed 45°. The
following modification may be adopted in
such cases, although the angle of repose
will seldom approach 45° in practice. In
Fig. 280, CF is the horizontal surface of

FIGS. 280, 281. the earth. Draw BE making the angle

of repose A, with BH and CH at right

angles to BE. On GEE describe a semicircle, and with centre H and
radius HE cut CH in P. BP is the plane of rupture, and producing it
to L, draw LK parallel to CE, which makes an angle of 2A with the
back of the wall. Cut off KM = KL, and KLM is the earth -pressure
triangle. (The coincident intersection at E of BE and the dotted line
CE in Fig. 280 is accidental.) For a surcharged wall as in Fig. 281,
first draw BFj parallel to the upper surface CF, and repeat the same

1 Stability of Retaining Walls, by Prof. G. F. Charnock. Read before the
Association of Yorkshire Students of the Inst. C. E., March 10, 1904.

MASONRY AND MASONRY STRUCTURES

339

construction excepting that OH15 perpendicular to BE, is used as the
diameter for the semicircle instead of carrying it through to intersect
BH.

An application of the method to the design of an actual wall will
now be given.

EXAMPLE 36. — Design a suitable section for a retaining wall, 20ft.
high, of rubble masonry weighing 140 Ibs. per cubic foot, the anqle of
repose being 27° and weight of earth 110 Ibs. per cubic foot. The
maximum pressure on foundation not to exceed 2J tons per square foot.

In Fig. 282, for a clear height of 20 ft. the total height of wall from
coping to foundation is taken as 24 ft., allowing for a concrete base

FIG. 282.

3 ft. thick, with the level of foundation 4 ft. below the lower ground-
level, and the section indicated is assumed. In all practical walls there
may be a slight counter-pressure p against the front of the toe. It is,
however, very small compared with the back pressure, and is frequently
reduced to zero by shrinkage of the earth. Also in walls with stepped
or battered backs, tjie shaded block of earth E, which apparently rests
on the wall and adds to its stability, has often been found to be quite
out of horizontal contact with the wall, owing to compacting under
pressure and subsequent shrinkage, or settlement of the wall away from
the earth, and for these reasons the effect of the pressure jt? and weight
of block E will be neglected. Assuming the concrete base as of
practically the same weight as the masonry, the total sectional area
including the base = 143| square feet, and weight of wall per foot run
= 143| x 1 X 140 Ibs. = 9 tons. Applying the construction for the
earth pressure, the area of the resulting earth-pressure triangle ABD

340 STRUCTURAL ENGINEERING

= \ x 14 J x 13 square feet, and the resultant pressure against the
wall = \ x 14J x 13 X 110 = 10,367 Ibs. = 4'63 tons per foot run of
wall. The centre of gravity of the wall section (including the base) is
situated on the vertical line OW, 3 ft. 9 in. from the back of the wall.
This may be found by dividing the section into a convenient number
of parts and taking moments about the back of the wall, or by
suspending a cardboard template of the section. Mark the point F at
| of 24 = 16 ft. from the surface, and draw FS" perpendicular to back
of wall. FO making 27° with FN gives the direction of the resultant
pressure behind the wall. From 0, where FO cuts the vertical through
the centre of gravity of the wall section, set off OP = the pressure of
4-63 tons, and OW = weight of wall, 9 tons, to any convenient scale.
OR to the same scale gives the resultant pressure on the foundation.
The centre of pressure C, in which OR cuts the foundation level, is
situated 3 ft. 6 in. from the outer edge of foundation, and the fraction
m to be used in calculating the intensity of pressure on foundation

Q' £\"

= -p-p = ^. The resultant vertical pressure on foundation acting

through C = 0V, the vertical component of OR, which, it should be
noted, consists of the wall weight OW + the frictional component WV
of the inclined pressure WR. OY scales off 11-2 tons. The intensities
of pressure on the foundation are then —

oT\T o v 1 1 -9

at H = -y(2 - 3m) = *.5 (2 - 3 x •&) = 2'11 tons per sq. ft.

9 y 1 1 '2

and at L = -y (8w» - 1) = 9<5 (3 X -ft - 1) = 0'25 ton per sq. ft.

Setting off ELK = 2-11 and LM = 0'25, and joining KM, the area
HKLM shows by its ordinates the varying intensity of pressure on the
foundation. As the maximum intensity of 2' 11 tons per square foot is
within the specified limit of 2 '5 tons, a slightly more economical
section might be adopted. For further illustrating the application
of principles, however, this section will be adhered to. The horizontal
component RV of the resultant pressure OR scales off 4'2 tons, and
tends to cause horizontal sliding of the wall on the foundation. This
horizontal component should not exceed the frictional resistance of the
wall to sliding. The following coefficients of friction may be applied
in checking the liability to sliding : —

Masonry and brickwork joints, dry .... 0'6 to 0*7

„ „ „ wet mortar . 0-47

„ „ on dry clay .... 0'50

„ „ on wet „ . . . . 0*33

In the case of retaining walls on fairly dry foundations there is
little risk of sliding, since the foundation level is invariably sunk
some feet below the surface, and the resistance to displacement of the
earth in front of the wall assists the frictional resistance of the wall
itself. Walls on wet clay foundations require careful examination in this
respect.

The total vertical pressure = 11 '2 tons. Assuming an average

MASONRY AND MASONRY STRUCTURES

341

foundation with coefficient of friction of 0*4, the f Fictional resistance
to sliding = 1T2 X 0*4 = 4*48 tons, which exceeds the horizontal
pressure of 4'2 tons, the resistance of the earth in front being
unconsidered. It remains to ascertain if the thickness of the concrete
base is sufficient to allow of a projection of 18 in. in front of the wall.
This portion of the base is shown enlarged in Fig. 283. It is subject
to an upward pressure from the foundation, represented by the portion
HKM of the area of the pressure intensity diagram. The intensity
hk scales off T71 tons. HK = 2*11 tons. The mean intensity

2-11 4- 1'71
over H/i = - — ^ — ~ = 1*91 tons per square foot. The area of

foundation beneath ~H.h = 1/5 sq. ft., and total upward pressure
= 1-91 x 15 = 2'865 tons, acting through the e.g. G of area HKM.
G is 9-4 in. to the left of hk. The B.M. on the section hf of the
concrete cantilever HF//& = 2-865 X 9'4 = 27 inch-tons. The cross-
section of this cantilever is shown at S. Its moment of resistance
= J X 12 x 36 X 36/6 = the B.M. of 27 X 2240 inch-lbs.,

C H

FIG. 283.

PIG. 284.

whence fb = 22*3 Ibs. per square inch tension at h and compression at
/. The safe transverse bending stress fb for concrete should not exceed
about 50 Ibs. per square inch. In a deep foundation the B.M. on//i
would be appreciably reduced by the action of the weight of the
concrete block ~Fh and the earth above it, which here has not been
considered. The section fh is also subject to a vertical shear of 2-865

m, 2-865 X 2240 . ,

tons. The mean shear = — r-^- ^-^, — = 14*9 Ibs. per square inch,

12 X ob

and maximum shear intensity = 14-9%X 1*5 = 22'4 Ibs. per square inch.
The safe shear on concrete should not exceed 60 Ibs. per square inch.

Surcharged Walls. — The following construction may be applied in
the case of walls retaining a bank of earth BCD, Fig. 284. Join
CE and draw BF parallel to CE. Join FE and proceed as before to
construct the earth-pressure triangle GHK, setting off the angle 2A
at F instead of at B.

Many walls, as in Fig. 285, are subject to additional pressure due to
weight of traffic, buildings, etc., in streets carried along the upper
level. It is impossible to estimate correctly the effect of such loads on

STRUCTURAL ENGINEERING

walls. Such additional loading is usually reduced to an approximately
equivalent layer of earth and the wall then treated as being surcharged

to that extent. Thus, if the
traffic along the street in Fig.
285 be taken as equivalent to
a load of 150 Ibs. per square
foot, this may be increased to,
say, 250 Ibs. to allow for
•*. vibration, and if the earth
weigh 100 Ibs. per cubic foot
an additional layer LL, 2 ft.
6 in.deep,being superimposed,
the effective depth behind the
wall will then be LH.

Needless to say, buildings
should not be erected imme-
diately in rear of existing
retaining walls unless pro-

FIG. 285. vision has been made at the

time of building the wall.

A more usual occurrence is when railway cuttings have to be made
through towns subsequently to the erection of the buildings, in which
cases suitable provision for the super-loading may be made as above.
The passage of heavy railway traffic along a cutting as at K makes
it desirable to ensure a good margin of stability in the supporting walls.
Types of Walls. — Retaining walls generally are of one or other
of the types shown in Fig. 286. A is a wall with battered face and

FIG. 286.

stepped back, which is a satisfactory section on an unyielding founda-
tion such as hard gravel or rock. 0 and D are sections of leaning
walls with buttresses or counterforts at intervals behind the wall.
Theoretically such counterforts should make for economy in material,
but their practical value is doubtful, especially in masonry walls, where
they are very liable to break away from the wall and so become useless.
Their employment is more satisfactory in concrete walls, where they are
moulded en ~bloc with the body of the wall. B is a leaning wall of
similar section to A. Curved walls as at C are not now very frequently
employed. The approximate position of the e.g. is marked in each
case, and it will be noticed that in type A the centre of pressure P

MASONRY AND MASONRY STRUCTURES

343

falls much nearer the outer edge of the base than the back, with a
consequently large variation in pressure intensity from front to back of
foundation. This is of little moment on hard foundations where
the settlement is very slight. In types C, D, and E the centre of
gravity of the wall is thrown well back, and the resultant pressure may
be designed to intersect the base at or near its centre, giving nearly
uniform pressure, and therefore uniform settlement on a yielding
foundation. These types are more suitable for clay or other soft
foundations where heavy pressure at the outer toe would result in
heeling over of the wall towards the front. Panelled walls as at E are
very generally used for the sides of railway cuttings through towns.
The buttresses F, roof and back arching and inverts are built of rubble
masonry or blue brick, and the backing either of rubble or concrete.
This type of wall is most suitable under conditions of heavy backing
and soft foundation. Whether the back of a wall is built in steps or
uniformly battered matters little as regards the stability, steps however
being usual in brickwork, and forming convenient ledges during
building for the support of staging. Where sliding forward of the
wall is apprehended, the base is often sloped upward towards the
front.

Example of Panelled Re-
taining Wall. — In designing a
panelled section such as E, Fig.
2 8 6, a length equal to the spacing
of the panels must be considered
instead of a one-foot run of
the wall, since the section is
not uniform throughout. The
position of the e.g. of one bay
may be found as follows.

Suppose Fig. 287 to represent
a proposed section for a panelled
wall. First equalize the curved
surfaces of the recess A by the
dotted rectangular boundary
lines. Next locate the e.g. of

the section abcfe, supposing the p

wall to be built solid through- FIG. 287.

out. Divide abcfe into two

portions abed and cdef. The e.g. of abed is at m, and of cdef at n.

= 123J, and of cdef = 11 X 20 = 220 sq. ft.,

Area abed = 13 x

8 + 11

220

mp _ Z'ZV
and total area abcfe = 343£ sq. ft. Divide mn in p so that - j^^-

p is the e.g. of the whole section abcfe. The e.g. of the panel recess
ghlk is at q. Next —

cub. ft.

Cubic content of one bay of 1 1 ft. of the solid section = 343£ X 1 1 = 3778-5
panel recess = 25 X 8 x 6 -

=1200-0

bay of the hollow wall = difference

= 2578-5

344 STRUCTURAL ENGINEERING

These volumes being proportional to the weights of the corresponding
masses may be considered as forces acting respectively through centres
of gravity /?, q, and r, r being the c.g. of the hollow wall. Taking
moments abaut the back of the wall and denoting by x the unknown
distance of r from </,

Moment of solid section about cf = 3778'5 X 5*8 = 21915
„ panel recess „ cf = 1200 X 8 = 9600
„ hollow wall „ cf = 2578-5 X x = 2578'5z

But the moment of the solid wall = sum of moments of panel re-
cess + hollow wall ;

.-. 2578-5Z + 9600 = 21,915, whence x = 4-77 ft.

Join qp and mark the position of r on qp produced distant 4'77 ft.
from cf. r is the required c.g. of the 11 ft. bay of the wall. This
calculation assumes the wall of uniform material throughout. If the
facing and backing be of materials of appreciably different unit weights,
a suitable mean value may be assumed without sensibly affecting the
exact result, which would otherwise be very tedious to work out. With
the materials in ordinary use, the above determination is sufficiently
close.

Continuing, the weight of one 11 ft. bay of the wall at 140 Ibs.

per cubic foot = — 8'5 x 14Q = 1G1 tons.

Assuming a face batter of 1 in 1C and earth backing at 100 Ibs. per
cubic foot having a natural slope of 25°, the area of the earth pressure
triangle = 124 sq. ft. The resultant pressure against wall per II foot

run = - — 2240 — ~ = 61 tans. Setting off OP = 61 tons and

OW = 161 tons, the line of resultant pressure OR cuts the base 4 ft.
6 in. from e. Draw ON perpendicular to ef and RN parallel to ef.
ON the normal pressure on foundation =187 tons per 11 foot run of
wall, or 17 tons per foot run. Hence for pressure intensities on

foundation, m = ^~ = ^, I = 11 ft., and W = 17 tons.

2 x 17
Intensity at e = — — (2 - 3 x ^) = 2'4 tons per square foot.

These intensities may be more nearly equalized if desired by extending
the base of the wall at the front. In these walls the back pressure is
transmitted to the buttresses by the horizontal arching forming the
back of the recesses, and from the buttresses to the foundation by
the inverted arches at the base. It is assumed the pressure on
the foundation per foot run of the wall is rendered uniform through the
action of the inverts. This distributing effect of the inverts will be
very closely realized in a well-executed wall. If, as is sometimes the
case, the panels be not provided with inverts as in Fig. 288, the

MASONRY AND MASONRY STRUCTURES

345

pressure must be considerably more intense immediately beneath
the buttresses, and the basal concrete slab S will yield as shown in
an exaggerated form by the curved lines. In such a case the greater
part of the pressure on the foundation per length L should be taken as
coming on a width W very little greater than the width of footing
of the buttress. A specially thick slab, or steel reinforcement should
be used in such cases, but inverts are desirable.

Methods of relieving Pressure behind Walls. — Various means are
employed for relieving the pressure on retaining walls, especially such
as may exhibit signs of weakness after erection. In Fig. 289 tiers of

FIG. 288.

FIG. 289.

arches are built behind the wall, which reduces the depth of earth in
contact with the wall, and throws a portion of the weight of the earth
on to the footings F of the piers carrying the arches. Such arches
may be part of the original design, or may be inserted subsequently to
the building of the wall. This method is occasionally adopted where
buildings have unavoidably to be erected immediately behind an existing

•mo 11

In Fig. 290 tie-rods, chains, or cables are passed through the wall
and taken back to anchor-plates A or raking piles P driven well in

FIG. 290.

rear of the natural slope. Both are expedients involving no great
difficulty of execution in applying to existing walls. An extension of
the concrete base B in Fig. 292, or the subsequent insertion of a
concrete block 0, is occasionally effected in order to counteract forward
sliding of a wall.

In Fig. 291 the heavy backing H is benched back, and the space

346

STRUCTURAL ENGINEERING

immediately in rear of the wall filled in with lighter material L. In
all cases where the backing is filled in after the building of the wall,

FIG. 291.

FIG. 292.

FIG. 293.

the material just behind the wall should be carefully deposited so as to
exert the minimum pressure. Carefully stacked rubble R, as in Fig.
293, is largely self-supporting, and will considerably relieve the pressure

on the wall, at the same time acting as
a drain. It is advisable to pack the space
immediately behind a wall for a width of
2 or 3 ft. with dry stone filling, as in
Fig. 292, and to provide water outlets or
weep-holes H at frequent intervals. In
very wet soils a properly constructed drain
should be laid behind the wall with
efficient discharge pipes passing through
or beneath the wall. The walls on either
side of deep cuttings are frequently
strutted by steel or arched masonry ribs
R, as in Fig. 294. An invert V is re-
quired in cases where the subsoil is liable to flow under pressure.

Walls of Variable Section. — Triangular wing-walls of bridge abut-
ments retain a gradually diminishing height of the embankment behind

Cross Sections
C

FIG. 294.

Pfa.

FIG. 295.

them, and are usually built as in Fig. 295, with battered face and
inclined steps at the back, the steps ending vertically at regular

MASONRY AND MASONRY STRUCTURES

347

intervals. Successive vertical sections «, b, c, d, of the wall are shown
at A, B, C, and D.

A similar case arises as in Fig. 296, where a long retaining wall W,
is built in order to avoid a heavy cutting C in the side of a hill H. The

FIG. 296.

wall W being of variable height, the back is conveniently stepped as
indicated by the dotted lines on the elevation.

MASONRY DAMS.

Reservoir dams of masonry or concrete are classed under the name
of masonry dams. The modern practice in regard to these works is
to construct them of rubble concrete with squared rubble masonry
facing. Many existing high dams, however, are composed entirely of
rubble masonry. Masonry dams are further classed as " gravity dams "
and "arched dams." The former resist the water pressure by the
action of dead weight only. The latter may only be employed on
sites where a reliable rock abutment is provided by the sides of the
valley across which the dam is built. They are curved in plan, and
resist the pressure behind them by acting as horizontal arches, and
are therefore of much lighter section than gravity dams. Arched dams
are not often employed, and are subject to complicated stress conditions
which cannot be accurately estimated, and the remarks here will be
confined to the consideration of gravity dams. A dam is virtually a
retaining wall for water, but since the
magnitude of the water pressure may be
accurately calculated, the design of a
dam is capable of more rigid treatment
than that of a retaining wall for earth.
Further, most dams are works of much
greater magnitude than retaining walls,
and it is the more necessary to employ
the minimum quantity of material con-
sistent with safety, since a relatively
small increase in sectional area involves
heavy additional cost. Dams possessing
the minimum section consistent with
safety are said to be designed of econo-
mical section.

Pressure of Water on the Inner Face of a Dam. — In Fig. 297 let
H feet be the depth of water above any horizontal section ss of a dam.
The pressure per square foot at B = w X H Ibs., where w = weight of

FIG. 297.

348

STRUCTURAL ENGINEERING

1 cubic foot of water. If BC be made = wTL pounds to scale, since the
pressure at the surface A is nothing, by joining AO the triangle ABC
forms a diagram of intensity of pressure at any desired horizontal level.
Thus, at the depth h feet, the pressure per square foot = &c Ibs., to the
same scale as BC represents ^#H. The resultant pressure per foot run
of the dam is represented by the area ABC, and

= JBC x AB = ±wR x H =

Ibs.

This resultant pressure acts through the e.g. G of the triangle ABC
in a direction perpendicular to the surface AB. Gp at right angles to
AB therefore determines p, the centre of pressure on the inner face,
which is obviously situated at two-thirds of the depth AB below the
surface. The inner surface of a high dam cannot, for reasons to be
presently explained, be made vertical for the whole depth, and is there-
fore given a slight straight or curved batter. As this batter is invariably
small, the following construction is sufficiently accurate for all practical
purposes. To obtain the resultant water pressure above any horizontal
section ss in Fig. 298, join AB, and set off BC = wTL at right angles
to AB. Join AC, then resultant pressure = area ABC = JwH X AB Ibs.

FIG. 298.

FIG. 299.

The centre of pressure p is obtained by projecting the c.g. G of triangle
ABC perpendicularly on to AB, and the direction of action of the
resultant pressure is sensibly Qp. For the slight batters employed in
practice, the line AB differs very slightly from the actual inner face of
the dam, and any closer estimate is both tedious and unnecessary.

Section of Dam Wall. — The stability of any proposed section
requires to be examined under two conditions: first, when the
reservoir is empty, and the material of the dam subject to the action
of its own weight only ; secondly, when filled with water, and subject
to the combined effect of the water pressure and weight of masonry.

1. Reservoir Empty. — The ideal theoretical section for a darn
would be a right-angled triangle ABC, Fig. 299, which forms the basis
from which to deduce a practical section. In any section, the c.g. G,
of the material A.hs above any horizontal section hs, being projected
vertically on to hs, determines the centre of pressure p on hs. In a
triangular section dam, p will obviously always fall at one-third the
horizontal breadth from h, or hp = ^hs. For this position of the
centre of pressure, it has already been shown that the intensity of

MASONRY AND MASONRY STRUCTURES

349

pressure at s is zero. Hence in a triangular section dam when the
reservoir is empty, no compression exists at the outer or down-stream
face, and a slight wind pressure acting on AC would be sufficient to set
up a small tensile stress in the material near the outer face. In
a practical section the following modifications of the theoretical
triangular section become necessary. In Fig.
300, the dam must be carried a few feet above
the water level L. The upper edge must be
given a reasonable practical width by adding
a mass of masonry A to the original triangular
section. Most dams carry a roadway along the
top, in which cases a width of several feet is
needed. The weight of the additional masonry
at A so affects the position of the centres of
pressure on all horizontal sections below a
certain horizontal level M, as to entail the
addition of a second mass of masonry, B, to
the left of the original triangular section, in
order to avoid creating tension on the outer face when the reservoir
is empty. Lastly, in very high dams a third addition to the section is
required at C to suitably increase the breadth of section, so that the
intensity of pressure at the inner or outer face in the lower part of
the dam may be kept within a safe limit, according as the reservoir is
empty or full. The outline given to the outer face may be a series of
straight batters, a continuous curve, or a combination of both. Whether
one or another of these be adopted makes little difference in the prac-
tical economy of the section. A continuously curved outer face has a
somewhat neater appearance, but is slightly more costly to build. In
recent practice, the overflow is usually allowed to run down the face of the

FIG. 300.

FIG. 301.

dam, after passing through a suitable number of arched spans beneath
the roadway as in Fig. 301, such a dam being called an " overflow

350

STRUCTURAL ENGINEERING

FIG. 302.

dam." This arrangement obtains in the Vyrnwy, Burrator, Elan and
Derwent Valley, and most of the modern dams, whilst in the older
ones, the overflow is carried down a stepped fall beyond one end of the
dam. The cross-section here shown is then necessary, the crest termi-
nating in a blunt angle, whilst the toe is more gradually drawn out to
conduct the water with reduced velocity into the overflow pool.

The effect of adding an additional mass of masonry A, Fig. 300,
to a triangular section will now be considered. In Fig. 302, ABC
represents the original right-angled triangular
section and ADE the addition necessary for
carrying a roadway, or for providing a satis-
factory top width AD for resisting the thrust
of ice and effects of weathering. AD will
generally vary between 5 ft. and 20 ft. G is
the e.g. of the mass A.hs, and g that of ADE.
If any horizontal section hs be taken at a
moderate depth below the top, and the points
G and g be projected vertically on to it at a
and b, Ga and gb are the lines of action of
the weight of the portions of masonry A.hs
and ADE respectively, and a and b their
respective centres of pressure on hs. As
before, a falls at ^hs from h, whilst b falls to
the right of a. The line of action of the

resultant weight of ADs^ will therefore fall between G« and gb as at PQ,
and the centre of pressure P due to the whole weight of masonry
above hs will consequently also fall to the rigid of a, and therefore

within the middle third of hs. Hence no
tension will be developed at the horizontal
section hs.

If now, in Fig. 303, the same construc-
tion be repeated for a horizontal section
hs taken considerably lower down, the point
b is found to fall to the left of a. The
centre of pressure P due to the whole weight
above hs, now falls to the left of a and
therefore outside the middle third of hs. In
this case tensile stress will be developed in
the material at s and for some distance in
from the outer face. The width of the section
hs must therefore be increased on the side
towards h, and this is done by giving a slight
batter to the inner face as shown by the
dotted line. There will evidently be one particular horizontal section for
which P falls neither within nor without the middle third of Jis, the con-
dition being that g and G must be situated on the same vertical line.

In Fig. 304 mark M the middle point of BC, and join AM. The
centres of gravity of all triangles such as ABC lie on AM. Project g
vertically to G on AM and make GH = JAG. Through H draw the
horizontal hs. The resultant weight of masonry above hs now acts
along #G, which being produced gives the centre of pressure at p, such

FIG. 303.

MASONRY AND MASONRY STRUCTURES

351

FIG. 304.

that hp = ^/is. For all horizontal sections above hs the centre of
pressure will fall within the middle third, whilst for those below hs, p
will fall outside the middle third towards the
inner face of the dam. Hence lis, which may
be termed the " critical plane," marks the level
at which the inside batter AF must be com-
menced in order to avoid creating tension at
the outer face sC. Practically this batter will
be commenced a little above the level As, as it is
undesirable for the compression to be reduced
quite to zero at s, as would be the case in Fig.

304. The batter required to effect this adjust-
ment is relatively slight and is readily ascer-
tained after one or two trials.

Stability of a Proposed Section.— Line of
Resistance.— The following method of examina-
tion of the stability of a proposed section has been adopted in the
design of mcst existing dams, and is often referred to as the middle-
third or trapezium method. The procedure will be first outlined,, after
which the extent to which the results obtained agree with the probable
actual stresses in the dam, will be noticed.

The section 0066 in Fig. 305 is assumed as that for a dam retain-
ing 100 feet depth of water, and constructed of material weighing 150
Ibs. per cubic foot. The section is divided into blocks or layers by the
horizontal planes 1-1, 2-2, etc., 20 feet or other convenient distance
apart, and the thickness of the section perpendicular to the plane of the
paper is taken as one foot.

1. Reservoir Empty.— Considering first the reservoir empty, Fig.

305. Mark the positions a, 0, c, d, #,/, of the centres of gravity of the
masses 0011, 1122, etc. . . . 5566. The centre of pressure on the
plane 1-1, due to the weight of masonry above 1-1, is obtained by
projecting a vertically to A on 1-1. The centre of pressure on any
other plane follows similarly by projecting the e.g. of all the masonry
above the plane vertically on to the plane. Thus, for the centre of pressure
on plane 2-2, the e.g. of the mass 0022 is required. Its position may
be most conveniently and accurately found from the known positions of
a and 0, and the weights of the first two masses 0011 and 1122 by
taking moments about the vertical line 0V.

Weight of 0011 per ft. run =

1122

10 X 10 X 150

2240
10 -4- 15-5 20 X 150

2 2240

Moment of 0011 about 0V = 6'7 X 5' = 33*50

= 6-7 tons.

17-08 tons.

1122

0V = 17-08 X6f = 113-87

Sum of moments = 147'37
Sum of weights 0011, 1122 = 6-7 + 17 08 = 23*78 tons.

Hence distance of e.g. of 0022 from 0V = OQ.4> =62 ft.

352

STRUCTURAL ENGINEERING

MASONRY AND MASONRY STRUCTURES 353

It is unnecessary to determine the height of the e.g. of 0022, since
being situated on a vertical 6-2 ft. from 0V, it must, when projected
on to 2-2, fall at B, also 6 -2 ft. from 0V. For the centre of pressure
on plane 3-3,

Weight of 2233 = 15'5 + *9'5 x *> *™ = 30-13 tons.
2 224:0

Moment of 0022 about 0V (from above) = 147'37
„ 2233 „ 0V = 30-13 Xll'= 331-43

Sum of moments above 3-3 = 478*80
Total weight above 3-3 = 23*78 + 30*13 = 53*91 tons.

A 7Q«Q

Hence distance of e.g. of 0033 from 0V = — — = 8-88 ft.

53*91

This e.g. projected on to 3-3 falls at C, 8*88 feet from 0V. In a
similar manner, by adding the moment of 3344 to the moment of 0033
and dividing by the total weight of 0044 above plane 4-4, the position
of D, the centre of pressure on plane 4-4 is obtained. The resulting
distances of D, E, and F from 0V are marked in Fig. 305. The curve
connecting these centres of pressure A, B, C . . . F, is known as the
Line of Resistance or Line of Resultant Pressure for the reservoir
empty. It intersects any horizontal section of the dam at the centre of
pressure due to the total weight of masonry above that section. From
the positions of A, B, 0, etc., the intensities of pressure perpendicular
to the horizontal planes 1-1, 2-2, 3-3, etc., may be calculated.

For example, the centre of pressure F on the basal section is
18'8 ft. from 0V, and therefore 23-8 ft. from the inside face at X.
The base is 67 ft. wide, and the total weight of masonry above the

23*8
base 231*76 tons, m = -»=- = 0*355, and intensities of vertical

pressure are

2 v 2^1 *7fi
at X = - ~--(2 - 3 x 0*355) = 6*47 tons per square foot.

and at Y = 2 x ^1>76(3 X 0*355 - 1) = 0*45

These are plotted at Xz and Y«/, the ordinates between XY and xy
giving the intensities of vertical pressure on the base. The intensities
on the other horizontal planes are plotted to the same scale in Fig. 305.
The dotted lines H/i and K& mark the limits of the middle third of the
width of the dam at each horizontal section, and the centres of pressure
A, B, G, etc., will be seen to fall just within these limits. With the
reservoir empty, there is therefore no tension acting perpend icular to
any horizontal section plane, whilst the maximum vertical compression
is that of 6*47 tons per square foot at X.

2. Reservoir Full. — Fig. 306 indicates the method of finding the
centres of pressure on the horizontal planes 1-1, 2-2, etc., when the
reservoir is full and the water pressure acting against the inner face
of the dam. Points A, B, C, etc., have been transferred from Fig. 305.
Considering any horizontal plane as 3-3, the weight of the mass of

2 A

354 STRUCTURAL ENGINEERING

masonry 0033 above 3-3 acts along the vertical through C. P3 repre-
sents the direction of the resultant water pressure acting on the inner
face for the depth L3. Producing P3 to meet the vertical through
C at c', the resultant pressure on the plane 3-3 now acts in the inclined
direction c'G^ The centre of pressure is thus displaced from C to C^
being nearer to the outer face than the inner, with the result that the
maximum intensity of vertical pressure on any horizontal plane occurs
at the outer face instead of at the inner, as was the case with the
reservoir empty.

The directions a'A.^ VB19 etc., of the inclined resultant pressures
are obtained from the right-hand diagram. The weights of masonry
above the several horizontal planes 1-1, 2-2, etc., are set out to scale
on the vertical line OM, and the corresponding water pressures along
their respective directions drawn from 0 towards W. In this example
the first two water pressures act horizontally, whilst the other four act
at a slight inclination with the horizontal. These inclinations have
been deduced in the manner referred to in Fig. 298, which is sufficiently
accurate for the small batters prevailing on the inside faces of masonry
dams. The water pressure

62*o x 52
Px = - - oTA = 0*35 ton acting horizontally at |Ll below L

P3 = - - = 28*25 tons acting perpendicular to L3 at |L3

below L.

Similarly P4 = 58*94, P6 = 100'8, and P6 = 139*51 tons. For plane
3-3, the masonry weight 0033 = 53*91 tons = OM3 and water pressure
P3 = 28*25 tons = OW3. Join W3M3 which gives the direction and
magnitude to scale of the resultant inclined pressure on plane 3-3.
Hence, through c' draw c'd parallel to W3M3, to intersect horizontal
3-3 in d. Repeating this construction for the five other horizontal
planes, the centres of pressure A1? Bj, 0^ ... Yl are obtained. The
curve connecting these points is the Line of Resistance for the reservoir
full. From the positions of A1? B15 Ox, etc., the altered intensities of
pressure perpendicular to the horizontal planes may now be calculated.
On the basal section 6-6, the centre of pressure F! is 24 ft. from
the outer toe Y, the breadth of section is 67 ft., and the resultant
vertical pressure at Fx = 240 tons, being the vertical component of
W6M6. This vertical component is a little greater— about 8 tons —
than the total masonry weight of 231*76 tons, since it includes the
downward water pressure on the inside battered face from 2 to 6. The
vertical components are obtained by projecting the inclined resultants
on to the vertical load line. Hence m being ff, the intensities of
vertical pressure on the base are —

at X = JL* 24(Ys x ~ - 1 ) = 0*534 ton per square foot

and at Y = 2 2 - 3 x = C-C3

MASONRY AND MASONRY STRUCTURES

355

The centres of pressure A» Bt, d, . . . F, again fall just within the
line KJc marking the outer limit of the middle third. With the reservoir
full, there is therefore no tension acting perpendicular to any horizontal
section plane, whilst the maximum vertical compression is that of 6-63
tons per square foot at Y. Farther, since the lines of resistance ABC
. . . F and A: ^ Cx . . . Fx fall only just within the limits of the middle
third, the section is " economical," since with a more liberal width the

FIG. 307.

lines of resistance would fall well within the middle third limits with
correspondingly reduced vertical stresses.

It is generally conceded that the safe maximum vertical pressure
on any horizontal section should not exceed about 7 tons per square
foot for the materials usually employed in masonry dams. If the
section in Fig. 305 be continued with the same inside and outside
batters to a greater depth than 100 ft. below top water-level, then this

356 STRUCTURAL ENGINEERING

maximum safe pressure of 7 tons per square foot will be realized at a
depth of about 110 ft., and for a dam of usual material to be continued
for a still greater depth, this depth of about 110 ft. marks the level at
which the width of the dain must be further increased in order to keep
the maximum vertical pressure on horizontal planes in the lower portion
of the dam within the above limit of 7 tons.

Dams not exceeding about 110 ft. in height are often alluded to as
"low dams," and those above 110 ft. as " high dams." If, for example,
the section in Fig. 305 be continued with the same batters to a depth
of 150 ft. below water-level, the new width of base is 102 ft., and the
centres of pressure, both for reservoir empty and full, still fall within
the middle third of the base. The intensities of pressure are, however,
9 '3 6 tons per square foot at inner face when empty, and 9*89 tons per
square foot at outer face when filled. These pressures greatly exceed
7 tons, and demonstrate the need for increased width of section at
lower levels of the dam.

Section of Dam for 150 ft. Depth of Water.— The method will now
be applied to determining a suitable section for the dam down to a
depth of 150 ft. In Fig. 307 the previous section is retained down to
100 ft. depth, but below this level the inside and outside batters are
increased as shown to give a basal width of 120 ffc. It should be
noticed the increased outer batters from YZ to 7*Yl cause the middle
third of the base X^ to move considerably to the right, whilst the
additional weight of masonry YSYiZ has relatively little effect on the
position of the e.g. of the total section, which moves very slightly
towards the right, as compared with its position if the outer batter
were continued from Y to S. The centre of pressure for the reservoir
empty will therefore fall outside the middle third of the base unless
the inside batter XXj be also increased. This causes the middle third
to move back somewhat towards the left, and counteracts the effect of
the increased outside batters YZ and ZYj. The additional weight of
masonry 6677 = 319-4 tons, and its e.g. is 39*3 ft. from the vertical 0V.
Sum of moments of masonry about 0V down to 6-6 = 4,347

Additional moment of 6677 about 0V = 319-4 x 39-3 = 12,552

Sum of moments down to 7-7 = 16,899
Total masonry weight above 7-7 = 231-76 + 319-4 = 551-16 tons.

16 899
Distance of c.g. of 0077 from 0V = ' = 30-66 ft., and distance

t)0-L" J.O

from inner face at Xl = 30-66 -f 14-00 = 44-66 ft. Width of middle
third = -if£ = 40 ft. The centre of pressure G for reservoir empty
therefore falls 4-66 ft. within the middle third of the base. For the

44-
intensities of vertical pressure on the base, m = ^~ and

intensity at ^ = 2 * 551-16/2 _ 3 x 44} \ = g.39 tong per gq ffc
120 \ 120/

and intensity at Y, = ^f^S X fg - l) = 1-07 „ „
Water pressure against inner face = -^ X ° * 77,— = 314 tons,

MASONRY AND MASONRY STRUCTURES

357

assumed acting at right angles to LX! at 50 ft. above the base. Com-
bining OM = 551*16 tons, and OW = 314 tons, the inclined resultant
is WM, having a vertical component wM. = 577 tons. Drawing g'Q^
parallel to WM, the centre of pressure for reservoir full falls at Gj,
51 ft. from YU or well within the middle third, m = ^, and intensities
of vertical pressure are

at Xl =

3 x — r - 1 ) = 2-64 tons per square foot

l^iO \ l^iO

and at Yx =

-8 X

S)—

The following table gives the vertical pressures on horizontal planes
at inner and outer faces as deduced by the above method.

RESERVOIR EMPTY.

Section plane.

pv at inner face.

Pv at outer face.

1-1

tons.
6-70

0-5

ft.
10

0-67

2-2

23-78

6-2

15-5

2-45

0-61

15-5

3-3

53-91

10-2
2975

29-5

3-52

0-14

4-4

102-46

15
43

4-54

0-22

5-5

169-46

20
6T

5-63

0-31

6-6

231-76

23-8

6-47

0-45

7-7

551-16

44-66
120"

120

8 39 1-07

RESERVOIR FULL.

Section plane.

pv at inner face.

pv at outer face.

tons.

4-ft

ft.

1-1

6-70

*r O

TSF

0-59

0 75

6.K

2-2

23-78

15-5

0-79

2-28

3-3

56-00

11-5
29-5

29'5

0-65

3-15

4-4

106-00

0-57

4-36

5-5

176-00

20-5

0-49

5-69

240-00

24
67

0-54

6-63

7-7

577-00

51
120

120

2-64

6-97

358

STRUCTURAL ENGINEERING

FIG. 308.

Actual Maximum Compression on the Material. — The intensities
of vertical compression on horizontal planes as calculated above by the

' ' trapezium rule " do not represent
the actual maximum compression
on the material of the dam. In
Fig. 308 the curved lines represent
generally the directions along which
the maximum intensity of compres-
sion acts at any point in the dam.
These lines, called " isostatic lines,"
or lines of principal stress, com-
mence at right angles to the inner
face, or in the same direction as
the lines of action of the water
pressure, and at the down-stream
face are nearly parallel to the outer
profile of the dam. The maximum
pressure at any point P on the
down-stream face will therefore act
on some plane PQ perpendicular
to the direction of the line of stress
passing through P. If the direc-
tions of the isostatic lines were closely determinable, the positions of
the planes on which the maximum pressure is exerted would be at
once known. The directions of these lines of stress cannot, however,
be exactly ascertained for any proposed section. That they follow
generally the curves indicated in Fig. 308 is borne out both by theory
and experiment.1 The extent to which the dam may be supposed
continuous with the rock foundation considerably modifies these
directions, afc and below the base of the dam.

Theoretically, if A be the inclination of the down-stream face to the
vertical, and pv the intensity of vertical compression on the horizontal
plane at P, as obtained by the preceding method, it may be proved that
the maximum intensity of compression at P, acting normally on a plane

PQ at right angles to the outer face = ^ 2.. The maximum stresses

in the dam are sometimes calculated by this formula, which, however,
implies that the outermost lines of stress are parallel to the outer
face of the dam, which, whilst sensibly true for the upper portion of
the dam, is probably considerably at variance with the conditions
existing near the toe. Here, wide variations exist in the outline given
to the down-stream toe of existing dams, and if the rapidly changing
inclination to the vertical be accepted as the governing factor in
calculating the maximum pressure, such calculated pressures near the
base must be in excess of the existing ones. Applying this in the case
of the dam in Fig. 307, the resulting maximum pressures at the
outer edges of the horizontal planes 2-2, 3-3, etc., for reservoir full,
are as given in the third column of the following table.

1 Mins. Proceedings Inst. C. E., vol. olxxii. p. 107, and pi. 5.

MASONRY AND MASONRY STRUCTURES

359

Horizontal section.

pv tons per sq. ft. by
trapezium law.

p» -7- cos2 A.

Maximum p, tons
per sq. ft. calculated
from dentilated

sections.

2-2

2 28

2-44

2-72

3-3

3-15

4-50

4-00

4-4

4-36

6-03

5-61

5-5

5-69

7-87

7-37

6-6

6-63

9-17

8-61

7-7

6-97

9-64

9-16

The maximum intensities of compression at the outer face, especially
towards the base, as given by this method, are probably appreciably
greater than those actually existing in the dam.

From the results of experiments made with model dams under
conditions approximating to those in actual dams,1 the maximum
intensity of pressure occurs at some distance in from the down-stream
face and the actual variation of pressure intensity on horizontal
sections, instead of following strictly the trapezium law, is indicated
more exactly by the ordinates to a curve such as EFG, Fig. 309, instead

FIG. 309.

of the straight line CD. This divergence from the variation of pressure
intensity as indicated by CD is apparently due to the relative weakness
of the acute-angled down-stream toe ABH, which is incapable of resisting
a maximum intensity of vertical pressure at B, and yields accordingly,
thus throwing the point of application / of maximum pressure intensity
some distance in from the down-stream face. This further shows the
fallacy of the assumption that the material of a dam behaves according
to the laws of bending for simple beams, since a plane section such as
KH will not remain plane after stressing, but will take up some curved
outline, as shown exaggeratedly by the dotted line. It would appear

1 Mins. Proceedings Inst. C. E., vol. clxxii. pp. 89, 107.

360 STRUCTURAL ENGINEERING

farther that the inherent weakness of a very acute-angled toe might
be corrected, and the material disposed with greater economy, by
bending in the lower portion of the down-stream face, as in Fig. 310,
which is a section of the Ban dam in the south of France. By so
doing the toe is, as it were, better held up to its work, with the result
that the point of application of the maximum pressure will approach
more nearly the outer edge of the base. The original section proposed
by Colonel Pennycuick for the Periyar dam in India also exhibits this
reversed batter or hump on the down-stream profile, which, in the
opinion of Colonel Pennycuick, is a preferable outline to the one
actually adopted for t'he dam.1

Another method of estimating the maximum pressures is as follows.
In Fig. 307 the inclined resultant pressures acting at points Bj, Cx, D1?
etc., have been transferred from Fig. 306. Any resultant pressure as
e'Ei is taken as acting on a dentilated section 5-5, the steps of which
are parallel and normal to e'Ej. The total effective breadth of the
normal faces of the steps is ab = 49 ft. The resultant pressure e'^
= 201 tons, and b^ = 18 feet, whence m = Jf, and the resulting
intensities of pressure for reservoir full are 7*37 and 0'84 tons per
square foot respectively at outer and inner faces at level 5-5. These
are plotted at bd and ac, the intensities for the other sections being
shown by the trapeziums plotted on the inclined base lines passing
through Bj, Cj, Dx, etc. The maximum values at the down-stream face
given by this method are inserted in the fourth column of the
preceding table. Prof. Gaudard of Lausanne considers this method
should give an excess of security.

The maximum stresses in the profile in Fig. 307 are well within the
safe compression which may be put upon the materials of which
modern dams are built, and keeping in view other causes of stress, such
as variable temperature, partial penetration of water into the interior,
lateral bending, etc., the extent of which cannot be approximately
estimated, any closer estimate of the direct compression due to the
water pressure and weight of masonry alone is of little importance.

Figs. 311 to 320 show the profiles of several of the more important
and recent masonry dams.

Vyrnwy Dam.— The Vyrnwy dam2 (Fig. 311) built in 1881-00,
for the water supply of Liverpool, has a length of 1172 feet and a
maximum height of 144 feet from the deepest point of foundations to
the crest. The maximum depth of water against the inner face of the
dam is 84 feet. It is an overflow dam (see Fig. 301), straight in plan,
and built of rubble concrete. Special care was taken to ensure the
concrete having a high specific gravity as well as to render the
masonry as watertight as possible. A system of drains formed
along the beds of rock at the foundation level, collect the water
from springs beneath the dam and conduct it into a small drainage
tunnel formed within the dam, from which it is discharged on the
down-stream face. This precaution was taken to avoid any possible
upward pressure on the base of the dam. The maximum pressure on
the material is approximately 7j tons per square foot.

1 Mins. Proceedings Inst. C. E., vol. cxv. p. 87. Ibid., vol. clxxii. p. 146.
8 Ibid., vol. cxxvi.

MASONRY AND MASONRY STRUCTURES 361

FIG. 312.

362 STRUCTURAL ENGINEERING

Assuan Dam.1 — The Assuan dam (Fig. 312) built across the river
Nile is 6400 feet in length and contains 180 si nice-ways at different
levels for discharging the impounded water. The sluices are opened in
April, May, and June, the issuing water supplementing the flow of the
Nile during the dry season, when in years of small discharge the
natural flow is inadequate for the irrigation of the country below
the site of the dam.

The dam is founded on granite rock and is constructed of rubble
erranite faced with coursed rubble masonry laid in cement mortar.
The maximum height above foundation level is 127 feet, and the head
of water above the bottom of the lowest sluice-ways is GOf feet. A
maximum pressure of 5*8 tons per square foot on the masonry at the
upstream face, with the reservoir empty, has been provided for, and in
the event of the water rising to the level of the roadway, the maximum
pressure on the down-stream faoe, with reservoir full, would be 4 tons
per square foot.

Thirlmere Dam2 (Fig. 313). — This dam has been constructed
across the outlet of Thirlmere for raising the level of the lake a
maximum height of 50 feet, the impounded water being used for the
supply of Manchester. The dam consists of two portions 310 and 520
feet long, divided by a ridge of rock. The maximum depth is 110 feet.
Fig. 313 shows the section at this maximum depth, and together with
those of the Vyrnwy and New Croton dams indicates to what a great
extent the foundations may add to the height and cost of dams when a
considerable depth of pervious material has to be excavated in order to
reach a sound rock foundation. The dam is of masonry-faced rubble
concrete, with a roadway 16 feet wide along the top.

Chartrain Dam3 (Fig. 314).— Built in 1888-92, the Chartrain dam
impounds water for the supply of the town of Roanne in the Loire
valley, in France, and also assists the controlling of floods in the district
below the dam. It is curved to a radius of 1312 feet in plan and has a
length of 720 feet, with a capacity of 990 million gallons. It is
founded on porphyry rock, has a maximum height of 177 feet, and
retains a maximum head of water of 151 ft. 9 ins. The dam is
constructed of rubble granite masonry in lime mortar, the inner face
being rendered with 1 to 1 cement mortar for a thickness of 1*2 inches.
The maximum pressure on the masonry at the inner toe, with the
reservoir empty, is 9 tons, and at the outer toe, with reservoir full, 9 '4
tons, per square foot.

Villar Dam, Spain4 (Fig. 315).— Built in 1870-78 for the increased
water supply of Madrid. The length is 546 feet, and in plan the dam
is curved to a radius of 440 feet. The maximum height is 170 feet.
For a length of 197 feet from one end the dam is utilized as a waste
weir over which the roadway, 14 ft. 9 ins. wide, is carried by an iron
bridge of twelve spans. The dam is constructed of rubble masonry,
the maximum pressure on which is estimated at 6 '5 tons per square foot.

1 Mins. Proceedings Inst. C. E., vol. clii. p. 78.

2 Ibid., vol. cxxvi. p. 3.

3 Les Reservoirs dans le Midi de la France, Marius Bouvier, pp. 18 to 21.
Congres International de Navigation interieure, Paris, 1892.

4 Mins. Proceedings Inst. C. E., vol. Ixxi. p. 379.

MASONRY AND MASONRY STRUCTURES 363

FIG. 318.

ROC*'

364 STRUCTURAL ENGINEERING

Ban Dam, France1 (Fig. 316).— Erected across the River Ban in
the lower Rhone valley in 1866-70. The length is 541 feet and the
dam is builfc with a convex face up-stream to a radius of 1325*6 feet, in
plan. It is built of rubble masonry founded on granite. The maximum
head of water retained is 148 feet, with a capacity of 407 million
gallons. The maximum pressure on the masonry is estimated to be 10
tons per square foot.

Periyar Dam, India2 (Fig. 317). — This dam was erected in
1887-95, for impounding water in the valley of the Periyar, which
water is diverted through a tunnel into the neighbouring valley of the
Vaigai and there utilized for irrigation purposes. The dam is
constructed of concrete faced with masonry and has a maximum height
of 173 feet. The width at the base is 143 feet, with a slope of 1 to 1
from the top of the outer toe to a point about halfway up the outer
face. The outer profile then becomes steeper, terminating with a top
thickness of 12 feet carrying a roadway. The overflow level is 162 feet
above the base, and the estimated flood level 11 feet higher.

La Grange Dam, California (Fig. 318). — This dam, built in
1891-04, has a length of 310 feet and is curved to a radius of 300 feel.
The maximum height on the up-stream face is 125 feet, and width of
base 90 feet. It is an overflow dam, and its outer profile was designed
to coincide with the slope of the overflow water when flowing 5 feet
deep over the crest of the dam. The cross section corresponds very
closely with that of the waste-weir of the New Croton dam (Fig. 320)
for a similar height, but the outer toe is made much stronger to with-
stand the shock of the falling flood water, which has already risen to a
height of 12 feet above the crest of the dam. The La Grange Dam is
built of uncoursed rubble masonry, and the water is employed for
irrigation purposes.

New Croton Dam, New York3 (Fig. 319).— The New Croton
masonry dam constitutes a portion only of the dam across the Croton
River for augmenting the water supply of New York. It is remarkable
for its great height from foundation to crest. The maximum height
above the deepest point of the foundation is 289 feet, although the
maximum head of water impounded is only about ] 40 feet. The dam
is of rubble masonry faced with ashlar above the level of the ground.
The waste weir, Fig. 320, 1000 feet long, is built as a continuation of
the main dam and curves round until at right angles with the axis of
the dam, thus facing the side of the valley and diminishing in height
from 150 feet to 13 feet. The great depth to which the foundations of
the darn have been carried was necessitated by the existence of 80 feet
of soft stratum overlying the rock, whilst a further maximum depth of
about 57 feet of unsound rock had to be removed, in order to reach a
sufficiently solid rock bed on which to found the dam.

Intensity and Distribution of Shear Stress. — During recent years
considerable attention has been drawn to the question of shear stress in
masonry dams. An accurate estimate of the shear stress would be
possible only if the actual variation of the compressive stresses were

1 Annales des Ponts et Chaussees, 1875, 1st Trimestre.

2 Mins. Proceedings Inst. C. E., vol. cxxviii. p. 140.

3 Transactions. Ant: Soc. C. E., 1900, vol. xliii.

MASONRY AND MASONRY STRUCTURES

365

known. A brief examination, however, under an assumed distribution
of vertical compressive stress probably worse than obtains in an actual
dam will show that the shear stress does
not exceed what may be safely resisted by
the material. Adopting the section in
Fig. 306, the intensities of vertical compres-
sion at inner and outer faces on plane 6-6
for reservoir full, are respectively 0*54 and
6' 63 tons per square foot. Assuming
uniform variation from X to Y, Fig. 321,
the intensity at intermediate points will be
given by the ordinates of the trapezium
66XY. Dividing the profile of the dam
into vertical sections by planes #-«, b-b,
etc., 10 ft. apart, the total vertical shear
on ah = the upward pressure against «6,
less the weight of the masonry ahG

= 5'72 t 6'68 X 10' - 5-39 = 56-36 tons.

Total shear on l-b
_ 4-81 + 6-63

2
= 92'84 tons.

X 20' - 4 X 5-39

Similarly total shear onc-c = 109-44 tons
<W= 106-16 „

e-e= 82-75 „
/-/= 32-00 „
0V = 8-65

FIG. 321.

These values are plotted vertically to scale at ACB. Dividing
each total shear by the vertical depth of the corresponding section,
the mean shear on the vertical planes is obtained. Thus —

Mean shear on a-a =

c-c =

56-36
16-1

l^T
109-44

48-3

e-e =

64-4

82-75
89-0

= 3-5 tons per square foot

= 2-88 „
= 2-27

= 0-93

/-/ = 82>0° =0-30
105-0

75-0

366

STRUCTURAL ENGINEERING

These values are plotted vertically above AB, giving the line AD,
and the intensity of shearing stress at any point being equal on
horizontal and vertical planes, the curve AD indicates the variation of
shear intensity along the horizontal plane G-6. The intensity of shear
on plane 6-6, i.e. on the 67 ft. base of the 100 ft. dam, is therefore zero
at the inner face, and increases to a maximum of 4*2 tons per square
foot, or practically twice the mean intensity, at the outer face.

Note that the mean intensity of shear on the base = horizontal
water pressure above plane 6-6-4-67 sq. ft. = 139 tons -4- 67 = 2*08
tons per square foot. Since the vertical pressure intensity is not quite
accurately represented by the trapezium 66XY (see Fig. 321), the
maximum shear will be less than 4-2 tons per square foot. In an
actual dam there is little doubt that the distribution of shear stress on
horizontal planes in the upper three-fourths of the height closely

agrees with that indicated by
a curve such as AD. Near the
base, however, the intensity and
distribution of shear may be
greatly modified, according to
the extent to which the base
of the dam is intimately bonded
with the neighbouring rock
foundation. In Fig. 322, if
the dam be supposed firmly
bonded to the mass of rock R
on its up-stream face, the resist-
ance of this mass to the hori-
zontal displacing water pressure
P, will set up considerable
tension across the vertical
plane AV. So long as this
tension does not cause rupture
in the neighbourhood of AV,

FIG. 322. proportionately large shearing

stresses will be created on the

basal plane at A, and for some distance to the right of A. For the
dam acting independently of the rock mass R, the shear distribution
on the base would be represented by a curve AB, similar to AD in
Fig. 321. The effect of the shear set up at A by continuity between
the dam and rock foundation R will be to tend to equalize the shear
intensity on the base, so that it will be represented by some such curve
as DE. If the rock on the up-stream face be much fissured or
jointed, it will offer little or no resistance to tension at A, and the
maximum shear BC on the basal section may approach twice the
mean shear.

If the dam be securely bonded with massive and un jointed rock,
the shear on the base will approximate to the mean shear. It is
doubtful if, on the majority of actual sites, much tensional resistance
may be expected from the rock foundation for any considerable distance
on the up-stream face, since the continuity must be interrupted sooner or
later by some system of bedding and joint planes as LLL, even in the

MASONRY AND MASONRY STRUCTURES 367

soundest stratified rock. Although it has often been stated that no
apparent cracks have developed at the inner toe in existing dams, it
should be remembered that an exceedingly fine crack is sufficient to
destroy the stress connection across the plane AV, and that such a
crack would not be readily discernible by a diver, whilst the emptying
of the reservoir for purposes of inspection would tend to close up
such possibly existing cracks, by reason of the maximum intensity of
compression reverting to the inner toe when the reservoir is empty.

Relatively few dams are founded at base level, the majority
penetrating more or less deeply into the rock as at AFC. (See also
Figs. 311 to 320.) The joint along AFC cannot but be regarded as a
possible plane of weakness towards the up-stream face, and in the event
of its opening for some distance from A towards F, the entry of water
under the pressure due to the head in the reservoir would introduce a
new and dangerous force beneath the dam. For this reason it appears
desirable to continue the inner face vertically into the rock for some
distance, since the penetration of water into a vertical crack is not of
great moment since its horizontal pressure is resisted by the reaction of
the rock mass R! on the down-stream face.

The total horizontal water pressure against the 150-feet dam in Fig.
307 is 314 tons, and the mean shear on the 120-feet base = fit = 2'61
tons per square foot. The maximum shear would probably amount to
4-5 or 4' 6 tons per square foot under unfavourable conditions of bond
with the surrounding rock. This, however, represents an outside figure,
since recent research appears to indicate that the shear intensity is not
uniform over the vertical planes a-a, b—b, etc., of Fig. 321, but is greater
in the upper portion of the dam than near the base. Until more definite
information is obtained on this point, however, the existence of shear
intensities of the above magnitude must be regarded as possible.

The following appears to be the concensus of expert opinion on the
subject of masonry dams at the present time (1911).

1. If a dam be designed according to the middle-third theory, so
that the maximum vertical pressure on any horizontal plane does not
exceed about 7 tons per square foot, corresponding with a possible
maximum pressure on certain inclined planes of from 10'5 to 11 tons
per square foot, and the lines of resistance for the reservoir empty and
full be contained within the middle-third limits, the actual stresses in
the dam will not exceed the above values.

2. The maximum shearing stresses accompanying these compressive
stresses will be safely within the shearing resistance of the materials
employed.

3. That tensile stress may occur in the region of the inner toe, the
maximum intensity of which may exceed the average intensity of shear-
ing stress on the base, but will be governed by the strength of the bond
between the dam and foundation on the up-stream face. Such tension,
when existent, will act generally in directions horizontal or slightly
inclined with the horizontal. On other than a very sound foundation,
it appears, however, impossible for such tension to exist.

The safety of the middle-third method of treatment is further
substantiated by reference to the records of failures of actual dams.
"With the exception of those failures directly due to subsidence or sliding

3G8

STRUCTURAL ENGINEERING

on faulty foundations, practically all others have been directly traceable
to the violation of one or other of the above-stated conditions, and no
failure of a high dam has been recorded where those conditions have
been satisfactorily complied with. It may be remarked that many
gravity dams have been given a slight curvature in plan with the convex
face up-stream, principally with a view to strengthening the dam against
longitudinal bending. Such dams do not, however, come within the
category of " arched dams," since they all possess " gravity " profiles.

The weight per cubic foot of the material employed in modern dams
varies from 142 to 1GO Ibs. The Vyrnwy dam, built of heavy clay slate
from the Lower Silurian formation, and in which special care was" taken
to obtain very dense concrete, has a specific gravity of 2*595. The
Burrator dam, built of granite rubble concrete weighs 150 Ibs. per cubic
foot, the granite alone weighing 165 Ibs. per cubic foot.

MASONRY AND CONCRETE ARCHES.

Arches of stone, brickwork or concrete are generally classed as
masonry arches. They differ from steel arched structures in two
important respects. 1. The dead load bears a greater ratio to the
useful or live load and is usually less uniformly distributed. 2. Little
or no tension may be permitted in the material. Arches of small span
are seldom designed from first principles, since they are simply repetitions
of types of well-established proportions, and the arch thickness may
safely follow some empirical rule. Moreover, small span arches usually
have a very large margin of safety. In the case of large span arches,
more scope exists for the application of economical principles of design,
and their dimensions are more carefully proportioned to the existing
stresses.

Disposition of Load on Arches. — The spandril spaces of small span
or very flat arches are usually filled up with earth, concrete or masonry,
to the level of the road or railway to be carried. Fig. 323, shows the

FIG. 323.

general construction of arched bridges of spans up to about 30 or 35
ft. The masonry or concrete of the abutment A is finished to a slope
tangent to the back of the arch ring R, and the remainder of the
spandril F filled with earth. S is a section of the wing-wall W. This

MASONRY AND MASONRY STRUCTURES

369

construction if adopted for large span arches would impose an excessive
dead load upon the arch. The spandril space F is therefore left hollow,
and the upper platform carried on jack arches J, turned between longi-
tudinal bearing walls as in Fig. 324, or by a series of transverse arches

resting on piers standing on the back of the main rib, Fig. 325, which
illustrates the 277-ft. arch at Luxembourg.1

The Pont Adolphe at Luxembourg, completed in July, 1903, has a
span of 277 ft. 9 in., and a rise of 101 ft. 6 in. It consists of two
separate arches in masonry, 18 ft. 6 in. wide, built side by side, having
their axes 36 ft. 11 in. apart. The intervening space is bridged over
by a reinforced concrete platform, the width between parapets being
52 ft. 6 in. The thickness of the arch at the crown is 4 ft. 8J in., and
at the joint where the greatest tendency to rupture occurs, the thickness
is 7 ft. 10^. ins. The masonry of the arch ring has a crushing strength

FIG. 325.

of 1280 tons per square foot. At the date of completion, it constituted
the largest existing masonry arch. Its span has, however, since been
exceeded by that of the masonry arch at Plauen, in Saxony, of 295 ft.
2J in. span and 59 ft. rise. The thickness at crown is 4 ft. 11 in.
The width between parapets is 52 ft. 6 in., and the spandrils, which are
hollow, contain a system of transverse and longitudinal vaulted spaces.
The masonry has a crushing resistance of 1670 tons per square foot.
This is at the present time (1911), the largest masonry arch in the

1 La Eevue Technique, May 25, 1904.

370

STRUCTURAL ENGINEERING

world. The Luxembourg arch was designed by M. Sejourne, and the
Plauen arch by Herr Liebold. The largest masonry arch in England is
the Grosve^ior bridge over the river Dee at Chester, the span of which
is 200 ft.

This type of construction is more generally followed on the continent,
whilst in English practice the hollow spandril spaces are masked by
continuous head or spandril walls. The magnitude and distribution
of the load may be closely estimated for any proposed outline of arch
when the arrangement of the superstructure has been decided. It
remains, then, to ascertain whether such load may be safely carried by
the outline of arched rib adopted, or whether the proposed design
requires modification.

Reduction of Actual Load to Equivalent Load of Uniform
Density. — It is convenient to convert the actual load consisting of
varied materials into an equivalent load of uniform density equal to
that of the material of the arch ring.

In Fig. 326 suppose the arch ring to be of masonry weighing 150
Ibs. per cubic foot, the concrete backing B, 140 Ibs., earfch filling F,

FIG. 326.

100 Ibs., concrete C, 140 Ibs., and wood pavement P, 40 Ibs. per cubic
foot. Further, suppose that a live load of 200 Ibs. per square foot of
roadway is to be provided for. Draw several verticals as ae, cutting
the layers of materials in points #, c, d. To avoid confusion of lines
the equivalent load area is drawn on the right of the centre line, a'b'
= yf§ of ab gives the depth of material of 150 Ibs. density required at A
to equalize the depth ab of 140 Ibs. density. Similarly for the other
materials, b'c' = \^ of be, c'd' = ^ of cd, and d'e' = -ffo of de. Repeating
the construction for each vertical, the equivalent load area is obtained,
each square foot of which represents 150 Ibs. of load per foot width of
the arch. The live load will be represented by an additional layer of
vertical depth e'f = f— 5 or 1^ ft. to scale. This equivalent load" area
GHKL may be conveniently subdivided to give the loading on short
segments of the arch ring.

Line of Resultant Pressures. —Let abed, Fig. 327, be the equivalent
load area for one half of a proposed arch under symmetrical loading,
deduced as above. Dividing abed into a number of panels &-1, 1-2, 2-3,
etc., by vertical lines 1, 2, 3, the centres of gravity of these panels,
marked by the small circles, may be readily obtained, each panel

MASONRY AND MASONRY STRUCTURES

371

approximating closely to a trapezium in outline. If the right-hand half
of the arch be supposed removed, the left-hand portion cd may be kept
in equilibrium by the application of a horizontal thrust T applied at
the crown. Taking moments round the springing point d, and calling
r the rise of the arch —

T X r =

from which T may be obtained. Set out the loads W1? ~W2, up to W6 to
any scale on a vertical line OwG and the horizontal thrust T to the
same scale at OP. Join P to wlt w.2 , . . w6. The figure OPw6 con-
stitutes a polar diagram, having P as the pole. Produce T to cut the
vertical through Wx and continue thence a line parallel to Pu\ to cut
the vertical through W2, a second line parallel to Pw2 to cut vertical
W3 and so on, until the last line drawn parallel to Pw6 emerges from the
arch ring at d. The polygonal line from T to d, which would approxi-
mate more nearly to a curve if the number of panels were increased, is
called the Line of Resultant Pressures, Line of Resistance or Linear
Arch for the loads, span and rise here assumed.

! i
! i

I r

I I i
-\ ' I 1— -•

!»* |W4 W, JW. jW,

^*yj !

^_J

FIG. 327.

Its direction indicates the direction of the resultant pressure on
any section of the arch, and the magnitude of the pressure on any
section Q is obtained from the length of the corresponding parallel ray
Pw3 of the polar diagram, measured to the same scale as the loads and
thrust T. The emergent line at d gives the direction of the inclined
thrust against the abutment, the magnitude being Pw6. The radial
lines P?#i, P^2» etc. of the polar diagram, which determine the thrust
at various points along the arch, all have the same horizontal component
OP = T, so that the horizontal thrust at any point in the arch is con-
stant, and the reaction R at the abutment is compounded of a horizontal
thrust T and a vertical reaction W equal to the sum of the loads

w,, w. . . w6.

It is necessary for the design of an arch to determine first the line
of resultant pressure due to the proposed load, span, and rise. It is
obvious that for a symmetrical load the right- and left-hand halves of
the complete line of resistance will be similar, and that only one-half
need be drawn, as in Fig. 327. If the loading be unsymmetrical, as,
for instance, when one half of the span carries the dead load only, and

372

STRUCTURAL ENGINEERING

the other half both dead and live load, the line of pressures will be also
unsymmetrical, and the thrust at the crown will no longer act horizontally.
As this condition of loading is the one which, practically speaking, most
severely stresses the arch, it is customary to examine any proposed
design, (1) under this condition of loading, and (2) when the arch
carries both dead and live load over the whole span. As the live load
on masonry arches bears a relatively small ratio to the dead load, the
former condition of loading is usually found to create the maximum
stresses in the material.

A line of resultant pressure passing through three fixed points A,
B, and C (Fig. 328), for any system of loads, may be drawn as follows.

FIG. 328.

Let abed be the equivalent load area for a proposed arch carrying
live load on the right-hand half span, together with dead load over the
whole span. Divide up the load area as before, and draw the lines of
action 1, 2, 3 ... 12 of the panel loads. Set out the loads to scale
on a vertical line OW, and select any point P as a pole. Join Pi, P2,
. . . Pi 2, and draw the corresponding link or funicular polygon AEF,
having its sides parallel to the rays Pi, P2, ... Pi 2 of the polar
diagram. AEF would be the line of resultant pressure for an arch
springing from A and F, having a central rise EG, and carrying the
proposed loads. The linear arch passing through A, B, and C has a
rise BD greater than EG, and will consequently have a less horizontal

thrust than the arch AEF, in the ratio -™, since the horizontal thrust

varies inversely as the rise. Join AF, and draw PH parallel to AF.
Draw HPi parallel to AC and equal to h (the original polar distance

EC
of P) x |vfy PI is then the correct pole position from which to draw

a new system of rays whose directions will give the required linear
arch passing through ABC. P^ to the scale of the loads is the
constant horizontal thrust acting in this arch, and OPT and PjW the

MASONRY AND MASONRY STRUCTURES

373

reactions at A and C respectively. It will be noticed that ABC is also
a bending moment diagram for the system of loads 1, 2, 3, . . . 12 on
the span AC.

In order to draw the line of pressures for a proposed arch, the
positions of points A, B, and C must be known beforehand. These
positions, however, may only be accurately fixed by providing the arch
with hinged or pin joints at the crown and springing points. Many

arches have been so erected, and it is only in such arches that an exact
estimate of the stresses is possible. Fig. 320 shows some of the methods
of applying hinges or articulations to masonry and concrete arches. At
A a number of cast-steel shoes are embedded in the ends of each semi-
arch and abut on steel pins F, a narrow joint J-J being left at crown
and springing to allow of slight movement under extremes of tempera-
ture. This arrangement permits of giving the arch rib an appearance

374 STRUCTURAL ENGINEERING

of continuity if desired, the joints being scarcely noticeable, whilst they
are sometimes masked by decoration. The concrete filling immediately
behind the casting should be of smaller aggregate to ensure thorough
ramming into the pockets of the castings. At B is a similar arrange-
ment, the bearings being secured by long tie-rods taken back into the
concrete. The hinges are more exposed, and become a noticeable
feature in this design. C shows the detail of the built-up articulations
employed for a concrete arched bridge of 164 ft. span erected over the
Danube at Munderkingen in 1893. For masonry ribs the arrangement
at B may be adopted, the castings being attached to the terminal
voussoirs by rag-bolts, or the ends of the ribs adjacent to the hinges
may be of concrete, or if of stone, the terminal stones are carefully cut
to fit suitable pockets in the castings. Frequently the arrangement at
D has been employed. Lead plates P, from I to | the depth of the
arch rib are inserted, the abutting voussoirs V, V, being preferably of
granite, basalt, or hard sandstone. The thickness of the lead is from
| in. to 1 in. for spans up to 150 ft., and the maximum pressure on it
may be 1500 Ibs. per square inch. Lead begins to yield slightly under
a pressure of about 1000 Ibs. per square inch, so that in the event of the
line of pressure approaching the edge of the plate, the increased intensity
of compression causes the lead to yield slightly, with consequent increase
of bearing surface and automatic reduction of pressure per square inch.

In addition to ensuring more accurate location of the line of pressure,
the provision of hinges has the important effect of annulling the stresses
due to change of temperature. Expansion and contraction, which in
straight girders is provided for by roller bearings, creates in rigid arches
a considerable amount of bending stress which is incapable of exact
determination. In a three-hinged arch the two semi-ribs rise and fall
slightly after the manner of a toggle joint, and consequently suffer no
stress under change of temperature. The stresses induced by expansion
and contraction are, however, relatively slight in a rigid masonry arch
as compared with those in structural steel arched ribs, since heat does
not penetrate a mass of masonry to the same extent as built-up steel
members composed of thin plates and section bars.

Line of Pressure in a Rigid Masonry Arch. — In an arch provided
with three hinged joints, the centres of the pins fix the position of
the line of pressure. In a rigid arch the abutting surfaces at crown
and springing have considerable depth, and it is possible to draw
several lines of pressure in the same arch, the outlines of which will
vary with the positions assumed for the points A, B, and C, in Fig.
328. Of any two possible lines of pressure abc and ABC, Fig. 330, due
to the same loads over a horizontal span L, the rise bd is greater than
the rise BD. Since the horizontal thrust is inversely proportional to
the rise, the line of pressure abc will possess a less horizontal thrust
than the line ABC. Hence the horizontal component Hrt of the inclined
thrust Ra at the abutment will be less than the horizontal component
HA of the inclined thrust RA. Many other pressure curves might be
drawn by varying the positions of A, B, and C. Of these, one will
possess a less horizontal thrust than all the others, and by the principle
of least resistance, that particular pressure curve will come into operation
which entails the least resistance on the part of the abutments. In

MASONRY AND MASONRY STRUCTURES

375

FIG. 330.

other words, the reactions at the abutments must be the least possible
consistent with satisfying certain other conditions relating to stress
intensity. The vertical reaction V, at either abutment (Fig. 330), is
fixed by the load distribntion ; consequently, by combining the least
possible value of H with V, the least value of the inclined thrust R
follows. Hence the particular pressure curve sought in a rigid arch is
that one which emerges from the arch ring at the greatest inclination
with the horizontal, and which satisfies the further following condition.

In order to avoid creating compressive or tensile stresses exceeding
the safe resistance of the material of the arch ring, the curve of pressure
must not cut any section
of the arch outside certain
pre-determined limits. If
tension is to be entirely
avoided, the pressure curve
must fall within the middle
third of the arch thickness.
It does not, however, follow
that the arch will be unsafe
if tension exist at certain
points, provided the maxi-
mum intensity of compres-
sion is still within the safe

limit. In masonry arches with good cement joints, a tension of 3 tons
per square foot should not endanger the joints, and even if some of
these should slightly open, the stability of the arch is not impaired,
provided the compression at the opposite edge of the joint be not
excessive. In continuous concrete ribs also a similar amount of tension
may be safely allowed.

If 32 tons per square foot be taken as the maximum safe compression
at the edge of a joint nearest to the centre of pressure, and the pressure
curve be allowed to approach within three-tenths of the breadth of the
joint from that edge, the tension at the opposite edge of the joint
would amount to 2'91 tons per square foot, or 45 Ibs. per square inch,
which in good work is quite permissible. It will therefore be assumed
that for a maximum pressure of 32 tons per square foot, the pressure
curve of least resistance must be contained within the central two-fifths
of the arch thickness. In the case of ashlar masonry arches, where the
maximum compression per square foot may be limited to 20 to 25 tons, the
corresponding tensile stresses for the centre of pressure at three-tenths
the breadth, would be 1*82 and 2 '28 tons per square foot respectively.

Method of Drawing the Curve of Least Resistance for a Rigid
Masonry Arch. — The curve of least resistance may be obtained by a series
of trials by varying the positions of points A, B, and C, Fig. 330, within
the prescribed limits. This is, however, very tedious, and the following
procedure, originally due to Prof. Fuller, gives a direct determination.

In Fig. 331 let DEFG be the equivalent load area, and verticals
1, 2, 3, etc., the lines of action of the weights of the various segments
(ten) into which DEFG- is divided. Taking AC, the line joining the
centres of the springing beds, as the effective span, assume any point B
and draw the line of resistance ABC for the loads 1, 2, 3, etc., and span

376

STRUCTURAL ENGINEERING

MASONRY AND MASONRY STRUCTURES 377

AC. The method of drawing ABC has already been given in Fig. 328.
The polar diagram from which ABC has been drawn is shown at PKZ,
Fig. 331. On the trial outline of the arch, mark the limits between
which it is desired to confine the line of least resistance. These are
indicated by the dotted curves, and have here been taken to include
the middle half of the arch thickness. On AC produced, select any two
points X and Y, and join BX and BY. Project points A, 1, 2, 3, etc.,
horizontally on to the lines BX and BY, so obtaining points X, I, II,
III, etc. Through these latter draw a new series of vertical lines, which,
for the sake of clearness, are carried through to the lower figure.
Where any original vertical as 2-2 cuts the middle half limits of the
arch in d and e, project these points horizontally to d' and e' on vertical
II-II. Similarly / and g on 9-9 will project to /' and cf on IX-IX.
Transfer the heights above XY of all the points so obtained to the
lower figure, and connect them by the irregular boundaries hkl and
mno. These boundaries form a distorted outline of the middle half
limits of the arch, having the same degree of horizontal distortion as
was given to the line of resistance A1234 . . . BC, by projecting it on
to the straight lines BX and BY. Since the line of resistance sought
will have ordinates in the same ratio as those of ABC, it will be
represented on the lower distorted diagram by two straight lines
included between the irregular boundaries hkl and mno. Further, the
line of least resistance will be represented by the two lines most steeply
inclined to the horizontal, which may be drawn between hkl and mno.
These are shown by s^m and s^. If two lines such as sjn and s^p cannot
be drawn within hid and mno, the arch thickness must be increased.

It remains, then, to re-proportion these lines horizontally by project-
ing their points of intersection with verticals I, II, III, etc., back on to
verticals 1, 2, 3, etc. The polygonal system of lines vs^w connecting these
points is the required curve of least resistance. This line of resistance
vs-iW is transferred to the upper figure, taking care to place it in the
same relative position to the horizontal line XY as in the lower figure,
when it will be found to lie within the middle half limits of the arch.

To obtain the reactions due to this curve of pressure, join VW
intersecting BR in r. The rise of the original line of resistance
ABC = BR. The corresponding rise of VsW = sr. The horizontal
thrust being inversely proportional to the rise, the horizontal thrust

T> D

for the linear arch YsW = PL x -— , PL being the horizontal thrust

scaled from the polar diagram PKZ, originally employed for drawing
the linear arch ABC. From L draw LPl parallel to VW, and mark

BR
P! at a horizontal distance from the load line KZ = PL x — . Join

PjK and PjZ. P: is the correct pole position for the line of resistance
VsW, and the reactions at V and W due to this line of resistance are
respectively PjK and PjZ, measured to the scale of the vertical loads.
These reactions are equal and opposite to the inclined thrusts at Y and
W, which are required in designing the abutments for the arch. If
PI be joined to the points 1, 2, 3, etc., on the load line KZ, the rays of
the new polar diagram so formed will be parallel to the segments of the
line of resistance VsW.

378 STRUCTURAL ENGINEERING

The intersection of the line of least resistance WW with any
section as yy, determines the centre of pressure p on yy, whilst the
length of the corresponding ray PXT of the polar diagram gives the
magnitude of the thrust on the section yy. From these data
the intensities of pressure or tension (if any) at opposite faces of the
section yy may be calculated in the usual manner. Generally the most
heavily stressed sections will be those at which the line of least
resistance approaches most nearly to the outer or inner faces of the
arch ring. In the figure these sections occur at q, V, and W. These
sections are often referred to as the joints or p lanes of rupture, since at
these sections the tendency to failure of the arch ring is greatest.

Design for Three-hinged Concrete Arched Bridge. — Span 150 ft.
Rise 15ft. To carry an equivalent distributed live load of 150 Ibs. per
square foot. Clear width between parapets 30ft.

The general arrangement is shown in the half longitudinal and
transverse sections in Fig. 332. The spandrils are hollow, the platform
being carried on jack arches of 4 ft. 6 in. span turned between
longitudinal spandril walls L, L, stiffened by two transverse walls T, T.
A 3 -in. asphalte roadway is laid over the upper surface of the
concrete backing of the jack arches. Between vertical section No. 2
and the crown the filling is solid.

In the upper figure the equivalent load area for one half of the
span, inclusive of the live load, is shown by abbcd . . . 2fg. The
weight of the longitudinal walls, platform and load is considered as
evenly distributed across the width of the arch sheet, the two vertical
projections at b and d representing the weight of the intermediate
portions of the transverse walls T, T, equalized for the full breadth of
the arch. The arch thickness is taken as 3 ft. 6 in. at crown and
springing, and 4 ft. 6 in. at the centre of each semi-rib. Considering
first the live load over the whole span, the load area and line of
resistance will be similar for each semi-span, and only one half of the
line of resistance need be drawn. The load area is here divided into
five equal panels of 15 ft. width by verticals 1, 2, 3, 4, and 5.
Preferably about ten panels should be taken, so that the resulting line
of resistance may approximate more closely to a curve, and the figure
be drawn to a much larger scale than is here permissible. The weights
of the panels 0-1, 1-2, 2-3, 3-4, and 4-5 per foot width of the arch,
estimated at 135 Ibs. per cubic foot, are respectively 4*77, 6'03, 7*11,
7*86, and 9*45 tons, and their centres of gravity are indicated by the
small circles.

The moments of these loads about the springing point A

= 4-77 X 67' = 319-6

6-03 x 52-3' = 315-4

7-11 x 37'5' = 266-7

7-86 x 22' = 173-0

9-45 X 6-5' = 61-5

Sum of moments = 1136-2 ft.-tons.

1136*2
Dividing by the rise, 15 ft., horizontal thrust at crown = — ^

= 75-8 tons.

MASONRY AND MASONRY STRUCTURES

379

380 STRUCTURAL ENGINEERING

On the polar diagram HPi = 75'8 tons to scale is drawn horizontally,
and the panel loads set off in order to the same scale, between H and
K. Lines drawn parallel to the polar rays of this diagram PiHK,
determine the line of resistance (indicated by similar dotted lines) on
the semi-arch in the lower sectional elevation. The direction and
magnitude of the thrust at the hinge A is given by PiK, which scales
off 84 tons. It should be noted that if a greater number of panels
were taken the line of resistance would sensibly approximate to a curve
tangent to the polygonal lines in the figure.

Over the abutment is provided an arched passage 10 ft. wide for
accommodation of traffic, towage, etc., along the river-bank. The
weight of masonry in this arch, together with that of the trapezoidal
abutment and live load from B to 0 equals 19 tons per foot width, and
the common centre of gravity is situated on the vertical Vr. Drawing
xy parallel to PiK and equal to 84 tons, and yz vertical =19 tons, xz
= 94 tons gives the direction and magnitude of the resultant pressure
on EF. (A smaller scale has been employed for the triangle xijz than
for the polar diagrams.) Drawing rp parallel to xz, the centre of
pressure p on EF falls practically at the centre of EF, whence the
intensity of pressure on EF = fJ=9'4 tons per square foot. In
designing an abutment of this type, the position and inclination of EF
may, after a few trials, readily be adjusted so that the resultant thrust
acts normally to EF and sensibly through its middle point.

Considering next the left-hand half span as carrying dead and live
load whilst the right-hand half span carries dead load only, the loads
per foot width of the arch are as indicated in the upper figure. The
right-hand loads are obtained by deducting the live load per panel

15 X 150
= — 224Q ~ sav> 1 ^on' fr°m ^he left-hand loads. These loads are

set off to scale from H to "W", commencing with 8'45 tons and following
in order from right to left. Selecting any pole P the polar diagram
PHW is drawn, the rays being indicated by chain-dotted lines.
Commencing at X, the corresponding linear arch XYZ is drawn having
its segments parallel to the polar rays of PHW. Join ZX and draw
PQ parallel to XZ. The correct pole position P2 for the linear arch
passing through X, m and A will be situated horizontally opposite to
Q. To calculate the horizontal thrust QP2, the rise Y'R = 35 ft., mn
= 15 ft., and horizontal thrust in the polar diagram PHW = 30 tons.
Hence horizontal thrust or polar distance —

QP2 = 30 X — = 30 x ff = 70 tons.

Join P2 to the load intervals on HW and draw in the line of
resistance XwA having its segments parallel to the rays of the polar
diagram P2HW. These are indicated by full lines. The actual curved
line of resistance will be sensibly tangent to the polygonal line XwA.
This line of resistance due to the unsymmetrical loading will be found
to be slightly raised on the left-hand or more heavily loaded semi-span,
and slightly depressed on the right-hand semi-span, as compared with
the line of resistance in the lower figure for a state of symmetrical
loading, The line of resistance approaches most nearly to the outer face

MASONRY AND MASONRY STRUCTURES 381

of the arch at the section ss^ At this section ssl = 50 in., st = 18 in.,
and the pressure on the section per foot width = P2S = 74 tons.
Hence intensities of pressure at ssj are

at extrados = np2 - jp = 32'? fcons P^ square foot
and at intrados = 2-*s x — - 1 = 2-9

— - 1 ) =

This maximum pressure of 32*7 tons per square foot, or 509 Ibs. per
square inch, represents the usual limiting compressive stress allowed on
concrete arches. In Prussia and the United States the official allow-
ance is 500 Ibs. per square inch, whilst this has been occasionally
exceeded in large span arches.

The reactions at A and X are respectively equal to P2W and P2H.
P2W scales off 78 tons, and is combined with the vertical abutment
weight of 19 tons at x'y'z', x'z' = 88 tons, giving the resultant pressure
on the base EF of the abutment, r'q parallel to x'z! gives the centre of
pressure <?, which again falls very nearly at the centre of EF, giving
for this disposition of load an intensity of pressure on EF slightly
greater than f§ = 8*8 tons per square foot.

Note. — The directions of the final thrusts on EF for the two cases
of loading considered are very slightly different, since the live load
bears only a small ratio to the dead load. A much larger scale drawing
should be made to enable these directions to be accurately ascertained.

In order to annul stresses due to changes of temperature, the ends
of the longitudinal spandril walls L, L, must be discontinuous and just
out of contact with the face of the abutment to allow of freedom
of movement under the slight rise and fall of the semi-ribs. In an
arch of these dimensions an interval of 1 in. to 1J in. is ample, the
magnitude of the movement at J being very small, as will be seen from
the following calculation. The length AmX = 155 ft. The increase
in length for a variation in temperature of 100° F. would be 155 x 12
X 0*0000055 x 100 = 1*02 inches, assuming the temperature of the
ribs to rise 100° F. throughout, which is most unlikely. The increase
in rise corresponding with this increase in length is therefore only a
small fraction of an inch, whilst the horizontal closure at J will be still
less (about £) by reason of the shorter leverage about the hinge A.
The relatively thin continuous road bed of concrete above J is
sufficiently flexible to accommodate itself to this slight movement. In
laying the concrete at J and over the crown, thin sheets should be
placed over the clearance spaces to prevent mortar or stones from
falling into and blocking them up.

The hinged joints may be of one of the types shown in Fig. 329.
The total breadth of arch at springing = 32 ft. 6 in. Allowing 4 ft. 6 in.
for intervals between pins, 14 steel pins 2 ft. long, and 3 in. diameter,
would give a projected bearing area of 28 x 12 x 3 = 1008 square
inches.

The maximum total thrust at springing = 84 tons X 32*5 = 2730
tons, and intensity of compression on hinges = fjf§ = 2*7 tons per
square inch.

382

STRUCTURAL ENGINEERING

Skew or Oblique Arches. — When one road or railway crosses
another obliquely by means of an arched bridge, it becomes necessary
to build a skew arch if the angle of obliquity of crossing is more than
a few degrees. In a square arch the bed joints of the masonry are
parallel to the springing line and the pressure in each foot width of
the arch acts normally to the joints. In a skew arch, Fig. 333, with
joints j,j\ parallel to the springing s, s, the pressure P on any foot width
of the arch may be resolved into a normal pressure N, and a tangential
or horizontal component T, which tends to cause lateral sliding of the
arch courses. This component T obviously increases with the angle of
skew or obliquity of crossing, and in order to prevent dislocation of the
masonry the bed joints are arranged in directions sensibly perpendicular
to that of the oblique pressure P. They therefore wind across the
surface of the arch in helical, or, as they are commonly but erroneously
called "spiral" curves. The ultimate thrust at the springing is
delivered on to checked masonry skewbacks s, s, Fig. 334, and the

PIG. 333.

FIG. 334.

lateral displacing effect of the horizontal component T has eventually
to be resisted by the abutments X and Y, Fig. 333.

Skew arches constructed of masonry throughout are very expensive,
since each stone, being of considerable size, has an appreciable amount
of twist on both bed and end joints, and requires its faces dressed to
twisted or helicoidal surfaces. For these reasons bricks are almost
universally employed, excepting where architectural effect is desired.
The amount of twist on a surface the size of the face of a brick is small
enough to be readily compensated for in the thickness of the mortar
joint, and the twist on a course of brickwork consequently augments
by frequent small steps instead of constantly as in the case of accurately
worked large stones. In the best structural skew arches the sheeting
is of brick, with ring stones or voussoirs at the faces bonded with the
brickwork, each ring stone being 9 in., 12 in., 15 in., etc., in breadth on
the soffit in order to bond with 3, 4, 5, etc., bricks, the size adopted
being in pleasing proportion to the dimensions of the arch.

The angle of stow asc. Fig. 333, is fixed by the exigencies of the
crossing. The angle of inclination of the checks or steps on the skew-
backs or springers then requires to be accurately ascertained, so that

MASONRY AND MASONRY STRUCTURES 383

the bed joints or " heading spirals " may traverse the arch as nearly as
possible at right angles to the pressure. Upon this angle, known as the
angle of skeiv bade, depends the shape of the springers and ring stones.
The checks on the springers being cut to the correct inclination,
the courses of brickwork, commencing from the skewback checks,
automatically follow the correct curves as they are laid on the
centering.

Fig. 335 shows the method of drawing the development of the soffit
or intrados of a skew arch, from which the angle of skewback and
shapes of springers may be determined. 0123 ... 12 represents the
elevation of the arch looking along the axis PQ. The square span is
40 ft., and angle of skew 23° 30'. The outline of the arch is a circular
segment having a rise of 15 ft. DEFG is the plan of the soffit. The seg-
ment 0123, etc., is divided into any number (12) of equal parts bypoints
1, 2, 3, etc. These being projected on to the plan, represent horizontal
lines running along the inner surface of the arch. If the curved surface
be supposed opened out or developed into a flat sheet by rotating it
about the line EF, the joints and outlines of ring stones and springers
drawn upon such development, will show their true shapes and incli-
nations. The square length 0-12' of the development will be the real
length of the arc 0123 ... 12. This may be stepped off in short
portions, but is preferably calculated. The angle 12 C 0 = 147° 28',
and radius 120 = 20 ft. 10 in. Hence, length of arc 0123 ... 12

14.7° 28'

= 36QO X 2 x 3-1416 X 20f = 53' 7|" bare.

This is set out at 0-12' and divided into twelve equal parts by points
1', 2', 3', etc. Vertical lines through these points determine the
developed positions of the lines 1,, 2X, 3a, etc., on the plan DEFG.
Projecting the points lx, 215 3^ where the dotted verticals cut DE and
FG, horizontally on to verticals 1', 2', 3', etc., points on the development
of the opposite edges of the soffit are obtained, which being joined give
the curved boundaries Ed and F# of the development.

Accurate Method of Drawing the Joints. — The correct directions
for the bed joints so that the lines of pressure may every where cut them
at right angles, are shown on the right-hand half pqgd of the develop-
ment. The method of drawing them (sometimes called the French
method) is as follows. Several curves aa, bb, cc, shown dotted, similar
to pd and qg are marked on the development. These represent developed
lines of oblique pressure, and the joints are then carefully drawn in so
that they intersect these curves, and the face curves pd and qg, perpen-
dicularly. As this results in a system of tapered courses, and necessitates
several breaks of bond in order to preserve the same width of voussoirs
on opposite faces of the arch, the construction is tedious and expensive
in stonework, and inapplicable for brickwork arches.

Approximate Method.— A system of parallel courses of uniform width
may be substituted for the exact courses without seriously affecting
the condition of perpendicularity of pressure. This method, known as
the English method, although in very general use, is illustrated on the
left-hand half EF gp of the development. F# is joined and divided

384

STRUCTURAL ENGINEERING

MASONRY AND MASONEY STRUCTURES 385

into a suitable number of voussoir widths. In this example, J?g measures
56 ft. 3 in., and represents practically the real length measured round
the oUiqw edge FG of the soffit. Adopting a width of voussoir of 15 in.,
so that each ring stone will bond with five courses of brickwork,
56' 3"
1> 3,, = 45, gives the number of voussoirs. This should be an odd

number if the appearance of a key-stone be desired on the face of the
arch, although it will be noticed there is no real key-stone in a skew
arch in the same sense as in a square arch, the central course at q on
one face running over to ql some distance from the crown on the
opposite face. Dividing F# into 45 equal parts, E is joined to that
division e which locates Ee as nearly as possible perpendicular to F#.
All the other joints are then ruled in parallel to Ee. A comparison of
the two halves of the development shows that the directions of the
joints so obtained are most in error at the springing, whilst at the crown
they almost coincide with the ideal joints. The angle FEe = 19J° is
the angle of skewback to which the checks on the springers are to
be cut.

If it be desired to insert the joints on the plan of the soffit DEFG,
although these curves are not of much practical value, they may be
drawn most accurately as follows. Produce any joint as Id both ways
to intersect the springing line at m and centre line pq of the develop-
ment. This point falls off the figure. Where 1dm intersects any
vertical as 2' on the development, in /, project Z horizontally to L
on the corresponding vertical 2 of the plan. The curve mLK continued
through to the centre line PQ, determines the outline in plan of a com-
plete bed joint from springing to crown. Every joint on the plan
contained between DE and FG, will be a portion of this curve, and by
making a template in stiff paper having the outline wKT, and sliding
it vertically along mT by successive equal intervals mm^ m^m^ mzm^
the joints are rapidly and neatly drawn in. For the other halves, here
omitted, the template is reversed, and slided along DG. The points
mlt m2, w3, etc., are the intersections of the bed joints produced, with
the springing line mT. In the example, seven of these intervals occur
between E and F. EF = 26 ft. 2 in., and mm^ mm^ etc., therefore equal
26' 2" -r 7. Five of the intermediate brickwork courses are indicated
at B on the development. The springers may be cut with separate
checks for each brick course, or be dressed to a butt heading joint with
the whole five bricks, the latter requiring much larger blocks of stone
in the rough. One of the intermediate springers is shown at S, the
arch being assumed as 22J in. thick. There being seven springers in
the length 26 ft. 2 in., the length of each stone will be 26' 2" -4- 7
= 3' 8 J" bare, or a little less to allow for joints. The length of the base
XY of each check = 3' 8f"-f- 5 = 9" bare, and drawing XZ at 19j°
slope, and YZ perpendicular to XZ, the correct template for one check
is obtained.

The end springers at E and F are respectively acute and obtuse at
the outer corners. They are shown in plan and elevation in Fig. 336,
to the same scale as S in Fig. 335. In practice, only the exposed faces,
beds and joints are worked smooth, the non-fitting portions of the
stones being left rough. The end springers are usually not cut with

2 c

386

STRUCTURAL ENGINEERING

separate checks, and the springer E, Fig. 336, is dressed square as at
abc, or obtuse, if the angle dbc, which depends on the angle of skew, is
very acute.

**c

FIG. 336.

TALL CHIMNEYS.

The stability of tall chimney stacks depends upon the weight of the
brickwork of the shaft and the pressure of the wind upon the outer
surface. In calm weather, the centre of pressure at any horizontal
section coincides with the centre of gravity of the section, and uniform
intensity of compression exists over the whole section. The application
of lateral wind pressure causes displacement of the centre of pressure at
every horizontal section, and this displacement should be restricted
within such limits as to prevent the occurrence of tensile stress in the
mortar joints. The effective wind pressure on the face of a tall chimney
varies considerably with the section of the stack. Colonel Moore l
the following table of coefficients to be applied in
effective wind pressure on the faces of various solids.

gives
calculating the

TABLE 32.— COEFFICIENTS OF EFFECTIVE WIND PRESSURE ON SOLIDS.

Solid.

Wind acting.

Sphere

0-31

Cube ...

0-81

Normal to face.

0-66

Parallel to diagonal of face.

Cylinder (height = diameter) . . .
Cone (height = diameter of base)

0-47
0-38

Normal to axis.
Parallel to base.

t K = Ratio of the effective pressure on a body to the pressure on a
thin plate of area equal to the projected area of the body on a plane
normal to the direction of the wind pressure.

It appears reasonable, therefore, to assume the effective pressure on
a circular chimney stack as about one-half that on a thin plate of area

1 Sanitary Engineering, Col. E. C. S. Moore, R.E., p. 752.

MASONRY AND MASONRY STRUCTURES 387

equal to the projected area of the chimney, that is, the area seen in
elevation, and the following effective horizontal wind pressures may be
employed for chimneys of various sections : —

On square chimneys, 50 Ibs. per square foot,
„ octagonal „ 32 „
„ circular „ 25 „ „

Let W = Weight of brickwork in tons above any horizontal section.
A = Area of horizontal section in square feet.
I = Moment of inertia of horizontal section in feet units.
R = Outside radius and r = inside radius of horizontal section

in feet.
P = Total wind pressure in tons against surface of chimney

above the horizontal section considered.
h = Height in feet of centre of pressure of wind above the

horizontal section considered.

Then, direct compression on section due to dead weight of brick-

work = -v- tons per square foot.

Bending moment due to wind pressure = P x h foot-tons. Stress
due to bending = ± - - tons per square foot, the plus sign

denoting compression at the leeward face and the minus sign tension at
the windward face of the section. The tensile stress due to the bend-
ing action of the wind pressure should not exceed the direct compression
due to the dead weight above any horizontal section. It is desirable
that the stress be not reduced quite to zero on the windward face, and a
residual compression of 100 to 200 Ibs. per square foot should remain
after deducting the tensile bending stress from the permanent direct
compression.

The moment of inertia of a hollow circular section = 0*7854
(R4 — r4), and of a hollow square section = ^(D4 — d4), D and d being
the breadths of outer and inner faces respectively.

EXAMPLE 37. — Fig. 337 shows the results of the application of the
above rules to the design of a circular chimney stack 1GO feet high
above ground-level. The dimensions and other data for the calculations
are given in the various columns opposite the respective horizontal
sections to which they refer. The weight of brickwork is taken as f 112
Ibs. per cubic foot. This is a little on the safe side as regards the
stability, since brickwork will usually weigh from 115 to 118 Ibs. per
cubic foot. As the calculations for the intensities of compression at any
horizontal section are similar, only those for one section need be stated.

Considering the horizontal section at 60 feet below the top, weight
of brickwork above the section = 41*9 -f 64'5 = 106*4 tons. Outer
diameter =11 ft. 4J in., inner diameter = 8 ft. 4£ in., and sectional
area = 46'54 sq. ft.

Hence direct compression = = 2'29 tons per square foot.

388 STRUCTURAL ENGINEERING

Projected area of the chimney above the section

= 282 -f- 318 = 600 sq. ft.

Height of centre of pressure above the section = 28*5 feet. The

ritions of the various centres of pressure are indicated on the figure
small circles.

Area
in

Weigh/-
in
Tons.

Diamekr,

Sectional
Elevation.

Foot
Units.

Projected
Wind
Area.

Compression in
Tons /> Sq. Ft

Outside.

Inside .

Windward.

LeeivanJ.

a' 3'

6' o'

41-9

i
30'

14-5'

28*

30-62

99$'

76}'.

*94

060

Z-IS

6' 9s-'

28 -5 '

64 -S

30'

3/0

46 54

ll' 4?"

if
il

see

O-4Z

4- -16

42'

90-6

36S

i j

6S-/S

n' 11$'

p'^4'

1.1

sa'

— hr

O-4I

563

a' sf

166-0

1
40'

! i

SS9

! i

89 91

is' o'

/o'e"

lass

0-/0

7-96

9'9'

Z37-0

30'

473

I53-O

IS' 9*

9' 9 '

\
*

4375

O-94

6-9O

576 O

ia 7-0

m f

490-0

p^Y^i):! ]0,

784 0

r. .*..*.! T •.-."•. ^

siezi

1-21

Z-39

F^p=q

FIG. 337.

Bending moment due to wind pressure =

600 X '25 X 28-5

2240
= 190-84 ft.-tons.
I, for the section = 582 foot units, outer radius = 5'7 ft., and stress

due to bending = ^—^ - = ±1-87 tons per square foot.

Hence compression on leeward face = 2'29 -f 1-87 = 4' 16, and

MASONRY AND MASONRY STRUCTURES 389

compression on windward face = 2*29 — 1*87 = 0*42 ton per square
foot.

At the ground-level the section of the pedestal, which is assumed
as square for 30 feet above the ground, is a hollow square of 15 ft. 9 in.
and 9 ft. 9 in. outer and inner faces. Its moment of inertia = 4375,
and total wind pressure against the shaft = 61,750 Ibs, This total
pressure consists of the pressure on the circular portion of the shaft
above the pedestal, taken at 25 Ibs. per square foot, and the pressure on
the face of the pedestal taken at 50 Ibs. per square foot.

The point of application of the resultant of these two pressures is at
60 feet above ground-level.

Hence bending moment = — 224Q — = 1654 ft. -tons
and stress due to bending = — 8 = + 2'98 tons per square foot.

The total weight of chimney above ground-level (excluding the
inner fire-brick lining of the pedestal which is not bonded with the
pedestal and consequently rests on the footings) equals 600 tons, and
sectional area is 153 sq. ft.

Therefore direct compression at ground-level = f§| = 3'92 tons per
square foot, giving 3 '9 2 + 2'98 = 6' 9 tons per square foot at leeward
face and 3'92 — 2'98 = 0*94 ton per square foot at windward face.

At the base of the footings the total vertical load is the weight of
stack, pedestal, fire-brick lining and footings = 787 tons, to which must
be added the weight of earth resting on footings, which, at 100 Ibs. per
cubic foot, amounts to 61 tons. The area of base of footings = 24'
X 24' = 576 sq. ft.

Direct pressure = — ^F~~ = l'*7 ^ons Per square foot,
o/o

. Bending moment due to wind at this level = 22^0 — foot-tons,

and I = ^ X 244 = 27,648.

61750 y 68 y 12
Stress due to bending = 2240 X 27648 = ± °'81 t0nS per Sq* ft'

Compression at leeward face = 1*47 + 0'81 = 2'28 „ ,,
„ at windward face = 1'47 - 0-81 = 0'66

A similar calculation at the foundation level gives the intensities of
pressure on the soil as 2*39 and 1*21 tons per square foot at leeward and
windward edges respectively. The total vertical load at foundation
level equals 787 tons due to brickwork in stack, lining and footings,
+ 490 tons in concrete base (at 140 Ibs. per cubic foot) +136 tons due
to weight of earth on footings and upper ledges of concrete block
= 1413 tons. The resultant wind pressure acts at a leverage of 78 feet
and I for the 28 ft. square base = ^ x 284 = 51,221.

Direct pressure = -- = 1'8 tons per square foot,

2 c 2

390 STRUCTURAL ENGINEERING

as X 7
51221

27'6 tons X 78' x 14
and stress due to bending = - - = ± 0'59 ton per

square foot.

Compression at leeward edge of foundation = 1-8 + 0'59 = 2-39
tons per square foot, and at windward edge = 1-8 — 0'59 = 1*21 tons
per square foot. The side of the pedestal at which the flue enters
should be suitably thickened to compensate for the weakening of the
horizontal section at this level.

INDEX

(JVos. refer to pages.")

ABSORPTION of water by bricks, 3

, percentage, of stones, 3

Aerating shed for Portland cement, 9
Aggregates, percentage voids in, 13
Analyses of iron and steel, 19
Arcbed masonry, 325

bridge, design of 3-binged, 378

Arcbes, bonding of, 325

, equivalent load on, 370

, line of least resistance in rigid

masonry, 374, 375

, load on, 368

, masonry and concrete, 368, 378

, pin-jointed, 274, 373, 378

, skew, 382

Ashlar masonry, 318, 324

Asphalte, 14, 27, 34

Assuan dam, 362

Axis, neutral, position of, 90

, physical, of columns, 134

BALLAST, weight of, 27, 28

Baltimore truss, stresses in, 226

Ban dam, 364

Beams, deflection of, 258

, fixed, 84, 87

, shear stress in, 105

, in, intensity and dis-
tribution of, 107

Bearings for girders, 248

Beech, 16

Bending moment, 50

and shearing force, relation

between, 51, 106

diagrams for rolling loads, 64,

66, 70, 74

— , maximum, due to rolling
loads, 71, 73

on cantilever, 50, 51

on beams supported at both

ends, 53, 54, 55, 56, 58, 64, 66, 70, 74

on balanced cantilevers, 59,

on continuous girders, 80,

, positive and negative, 53

Boarding, weight of, 32

Bolts, weight of, 32

Bond in masonry and brick arches, 325

Boucherie's process of preserving

timber, 18
Bowstring lattice girders, stresses in,

235

Bracket, connection to pillar, 121
Bricks, manufacture and properties of,

Brickwork, specification for, 4
Bridge, plate girder, design of, 204
, 3-hinged concrete arched, design

of, 378

Bridges, dead load on, 27
, equivalent distributed load on

main girders, 35

, highway, live load on, 36, 37

, live load on, 34

— , weight of cross-girders in, 30
Buckled plates, 24, 28

CANTILEVERS, balanced, bending mo-
ment and shear force on, 59, 62

, bending moment on, 50, 51

, shear force on, 50, 51

Cantilever girders, stresses in, 227, 232

, deflection of, 256

Cast iron, 19

Cement, 5

, aeration of, 9

, aerating shed for, 9

, natural, 5

— , Portland, 5

.composition and manufacture

of, 5

, effects of frost on, 8

, influence of fine grinding on

strength of mortar, 6

, influence of fine grinding on

weight of, 6

, Le Chatelier test for, 8

, modulus of elasticity of, 8

, tensile strength of, 7

, tests for, 7

, weight and specific gravity

of, 6

392

INDEX

Characteristic points, 78, 79

, f USe of, 80, 83, 86, 87, 88

Chartrain dam, 362

Chequered plates, mild steel, dimen-
sions of, 23

Chimney stacks, tall, 386
Column formulae, Euler's, 146

, Moncrieff's, 151

— , Rankine's, 148

, straight-line, 149

Column sections, properties of, 143

, selection of type of, 158

Columns, caps and bases of, 165

, compound, 184

, connections to, 159

, curves of safe loads on, 154, 155,

156, 158

, defects in practical, 132

, eccentrically loaded, 157

• , fixed-ended, 150

, foundation bolts, 183

, foundations, 170, 182, 183, 317

Columns, ideal and practical, 132

, live load on, 185

, method of application of load on,

134

— , methods of casting, 20

, of supporting ends of, 136

, numerical examples, 174-184

, physical axis of, 134

• , rivet pitches in, 185

Compression, intensity of, in masonry

dams, 357, 358, 360, 362, 364
Concrete, cement, 10

, bulk of ingredients for, 12

, gauging and mixing, 10

Gilbreth gauging and mixing

plant for, 11

percentage voids in aggregates, 13
structures, methods of building,

13 328

weight of, 26
Connections, riveted and bolted, 116

— subject to bending stresses, 121
Continuous beams, 76, 80, 83
, pressures on supports of, 81,

82
Corrugated sheets, dimensions of, 22

— , weight of, 33
Crane jib, design of, 300

lattice girders, design of, 240

plate girders, design of, 192

Creosoting, 18
Cross-girders, weight of, 30
Crushing strength of timber, 16
weight of granite, 3

— of limestone, 3

— of sandstone, 3

DAMS, masonry, 347

, , pressure of water on, 347

Dams, masonry, stability of, 348, 351
Dead load, general considerations of, 25

• on bridges, 27

• on roofs, 32

Deflection, 253
Depth of girders, 110
Design of crane jib, 300
of crane plate girders, 192

— of crane lattice girders, 240
of floor for warehouse, 124

— of lattice roof girder, 296
of masonry dam, 351

— of plate girder railway bridge, 204
of retaining walls, 339, 343

— of roof truss, 286

— of steel tank, 304

of tall chimney stack, 387

— • of 3-hinged concrete arch, 378
Details of lattice girders, 245—252
Distribution of shear stress in beams,

107

Dry rot in timber, 17
Dynamic formula, Fidler's, 44

EARTH pressure behind retaining walls,
337

Elm, 16, 27

Ends of columns, methods of support-
ing, 136

Equivalent distributed load, 35, 73

Euler's formula, 146

FACTOR of safety, 41
Fidler's dynamic formula, 44
Fittings for corrugated sheeting, weight

of, 33

Fixed beams, 84, 87
Flashing, lead and zinc, weight of, 34
Flats, rolled steel, 23
Floor, warehouse, design of, 124
Floors, live load on, 34, 310, 313
Footing courses, 321, 328
Foundations for columns, 170, 182, 183,

310, 316

, griUage, 170, 310, 316

, pressure on, 329, 330

GILBRETH, gauging and mixing plant

for concrete, 11
Girders, cross, weight of, 30

— , lattice, 213

, main, weight of, 30

, plate, 187

Glass, weight of, 27, 33
Glazing bars, weight of, 33

INDEX

393

Granite, composition and properties of,

2,3

, crushing weight of, 3

, percentage absorption of, 3

, weight of, 3, 27

Graphic method for modulus of section,

for moment of inertia, 92

Gravity dams, 347

Greenheart, 16, 27

Grillage foundations, 170, 310, 316

Gutters, weight of, 33

Gyration, radius of, 140

HOG-BACKED lattice girder, stresses in,

230

Hutton's formula, 264
Hydraulic limes, 5

INERTIA, moment of, 92, 93, 94, 102
Iron and steel, analyses of, 19

, physical properties of, 19

, cast, 19

, test for transverse strength

of, 20

• , ultimate strength of, 20

, weight of, 27

, wrought, 20

, bending tests for, 21

, market forms of, 21

, sizes of plates, 22

, tensile strength of, 21

JACK arches, 30, 369, 378

Jarrah, 16, 27

Joints in tension members, 119, 120, 283

KYAN'S process for preserving timber, 18

LA GRANGE dam, 364

Larch, weight of, 27

Lattice girders, crane, design of, 240

, details of, 245-252

, deflection of, 255

, spans of, 215

— , stresses in, 216-240
— , types of, 213
, weight of, 31

Launhardt-Weyrauch formula, 42

Least radius of gyration, 141

Lime, 4

, calcining and slaking of, 4

mortar, 5

Limes, hydraulic, classification of, 5

Limestone, composition and properties
of, 2, 3

, crushing weight of, 3

, percentage absorption of, 3

• , weight of, 3

Live load, general considerations of, 25

on bridges, 34

on columns, 185

on cross-girders, 36

on floors, 34, 310, 313

on highway bridges, 36, 37

• on rail bearers, 36

on troughing, 86

Load, dead and live, general consider-
ations of, 25

, , on bridges, 27

, , on roofs, 32

, due to wind, 36

, equivalent distributed, on bridges,

on arches, 368

on columns, application of, 134

on floors of tall buildings, 310, 313

on roller bearings, 249

Louvre blades, weight of, 33

Luxembourg arch, 369

MASONRY and concrete arches, 368

arches, bond in, 325

dam, design of, 351

dams, 347

, general deductions on, 367

, maximum compression in,

358, 360, 362, 364

, shear stress in, 364

faced concrete blocks, 328

footings, 321, 328

piers, 326

, specifications for, 319-325

types of, 318

, weight of, 26

Mild steel, 22

Modulus of elasticity of Portland

cement, 8

of section, 96, 98, 99, 102, 115

Moment, bending. See Bending

of inertia, 92, 93, 94, 102

of resistance, 89, 91

Moncrieff's column formula, 151
Mortar, cement, influence of fine

grinding on, 6

, , modulus of elasticity of, 8

, lime, proportions of constituents

of, 5

394

INDEX

Neutral axis, position of, 90

plane, 90

New Croton dam, 364
N-Girders, stresses in, 216-223
Nuts, weight of, 32

OAK, American, 16, 27

, English, 15, 27

Oblique arches, 382
Overflow dams, 349, 361, 363

PANELLED retaining walls, 343
Parabola, construction of, 52
Parapet girder, 252
Periyar dam, 364
Piers, masonry, 326
Pine, American, 15, 27

, Baltic, 15

•, Dantzic, 16

— , pitch, 15, 27
Pin-jointed bridge girder, 250
Pitch of rivets in girders, 109
Plate girders, depth of, 187

• , deflection of, 260

, economic span of, 187

, flange area, 188

for crane, design of, 192

, railway bridge, design of, 204

, rivet pitch in, 191

, web thickness, 189

, weight of, 30

Plates, buckled, 24, 28

, mild steel, sizes of, 23, 24

, wrought iron, sizes of, 22

Plauen arch, 369

Points, characteristic, 78, 79, 80, 83, 86,
87, 88

— of contra-flexure, 76
Portland cement, 5

Pressure of earth behind walls, 337

— on foundations, 829, 330
Principals roof, weight of, 32
Properties of bricks, 3

of rolled sections, example of, 114

of stones, 3

Purlins, design of, 284

QUICKLIME, 4

RADIUS bricks, 4

— of gyration, 140
Rail bearers, weight of, 30
Rails, weight of, 28
Railway track, weight of, 28
Range formula, Stone's, 45, 49
Rankine's column formula, 148

formula for earth pressure, 337

Rebhann's construction for earth pres-
sure, 337

Resistance, line of. in arches, 370, 374,
375

, , in dams, 353

, moment of, 89, 91

Retaining walls, 336

, design of, 339, 343

, methods of relieving pres-
sure on, 345

, surcharged, 341

, types of, 342

Ridging, galvanized, weight of, 33

Rivet heads, weight of, 32

pitch in columns, 185

in girders, 109, 191

Rolled sections, mild steel, 23, 31

Roller bearings for girders, 249

Rolling loads, 35, 37

, bending moment and shear

force due to, 64, 66, 70, 74

Roof coverings, 32

details, 294

girder, design of, 296

truss, design of, 286

Roofs, 261

— design of members, 283

dead load on, 32

reactions on, 265-268

stress diagrams for, 269

stresses in wind bracing, 278

types of, 262

Round steel bars, sizes of, 24
Rubble concrete, 318, 323

masonry, 318, 321, 323

Rueping process of creosoting, 18

SANDSTONE, composition and properties

of, 2, 3

— , crushing weight of, 3

, percentage absorption, 3

, weight of, 3

Sand, weight of, 27
Sectional area of members, 40
Section modulus, 96, 98, 99, 102, 115
Shear force, 50

— diagrams, 50-83
for rolling loads, 64, 66,

70, 74

— on cantilevers, 50, 51, 59, 62
on continuous girders, 82, 84

INDEX

Shear force, positive and negative, 53

stress in beams, 105

, intensity and dis-
tribution of, 107

in dams, 364

Sheets, corrugated, dimensions of, 22
Sizes of plates, 22, 23, 24
of round bars, 22, 24

— of slates, 33

of steel troughing, 28

• of timber, 15, 16, 17

Skew arches, 382

Slag, weight of, 27

Slates, sizes and weight of, 38

Snow load, 34

Specification for brickwork, 4

Specifications for masonry, 319-325

Steel, 22

— , mild, bending tests for, 23
, , sizes of rolled sections, 23,

24, 31, 143, 144, 145

, , troughing, 24, 28

, tensile strength of, 23

, weight of, 27

Stone, durability of, 1

, face dressing of, 319

, porosity of, 1

, resistance to compression, 1

, selection of, 1

, specification for, 319

, weight of, 1, 2, 3, 27

Stone's " Range " formula, 45
Straight line column formula, 149
Strength, crushing, of timber, 16

, shearing, of timber, 17

, tensile, of Portland cement, 7

, , of steel, 23

, , of wrought iron, 21

, transverse, of cast iron, 20

, , of timber, 16

, ultimate, of cast iron, 20

Stresses by method of sections, 270

— in lattice girders, 216-240

in roof trusses, 269-283

Structures, intensity of wind pressure

on, 89

Struts. See. Columns
Surcharged retaining walls, 341

TALL buildings, 310

, column anchorage, 316

— , details of, 311
— , load on floors of, 310, 313
• •, wind bracing in, 313

— chimneys, 386
Tank, steel, design of, 304
Teak, 16, 27
Test, Le Chatelier, for Portland

cement, 8

Tests, bending, for mild steel, 23
, , for wrought iron, 21

Tests, tensile, for Portland cement, 7
Thirlmere dam, 362
Three-hinged arches, 274, 378

, pin joints for, 373

Ties, riveted joints in, 119, 120
Timber, 14

, decay of, 17

, destruction of, 17

, preservation of, 18

, seasoning of, 17

, scantlings of, 15

, selection of, 15

, shearing strength of, 17

— , weight and strength of, 16, 27
Tramway rails, weight of, 28
Troughing, steel, 24, 28

VILLAR dam, 362

Voids, percentage, in aggregates, 13

Vyrnwy dam, 360

WARREN girders, stresses in, 223-226
Weight of asphalte, 27, 34

of ballast, 27

of boarding, 32

of bolts, nuts, and rivet-heads, 32

of concrete, 26

of corrugated sheeting, 33

of cross-girders, 30

of fittings for corrugated sheeting,

of galvanized ridging, 33

of glass, 27, 33

of glazing bars, 33

of granite, 2, 3, 27

of gutters, 33

of hollow tile floors, 312

of iron and steel, 27

of lead and zinc flashing, 34

of limestone, 2, 3

of Louvre blades, 33

of masonry, 26

of plate and lattice girders, 30

of Portland cement, 6

of rail bearers, 30

of rails, 28

of railway track, 28

— of rolled steel section bars, 31,

143, 144, 145

— of roof principals, 32, 263

of sand, 27

of sandstone, 3

of slag, 27

of slating, 33

of steel floor plates, 28

of timber, 16, 27

306

Weight of troughing, 28 .

of water, 27

of York paving flags, 27

Wet rot, 17
Weyrauch formula, 42
Wind bracing in roofs, 278

in tall buildings, 313

load, 36

INDEX

Wind pressure, intensity of, on struc-
tures, 39

on roofs, 264

screen, 281

Wohler's experiments, 42
Working stresses, 40

by " Range " formula, 46

Wrought iron, 20

THE E!NT)

PRINTED BY WILLIAM CLOWES AND SONS, LIMITED, LONDON AND BKCCLES.

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