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THE LIBRARY

THE UNIVERSITY
OF CALIFORNIA

LOS ANGELES

ATS

EAST ABUTMENT OF BLACKWELL'S ISLAND BRIDGE, NEW YORK

JOHfl S. PRELL '

Civil 6- Mechanical Engineer.

SAN FRANCISCO, CAL.

Cyclopedia

Civil Engineering

A General Reference Work

ON SURVEYING, RAILROAD ENGINEERING, STRUCTURAL ENGINEERING, ROOFS

AND BRIDGES, MASONRY AND REINFORCED CONCRETE, HIGHWAY

CONSTRUCTION, HYDRAULIC ENGINEERING, IRRIGATION.

RIVER AND HARBOR IMPROVEMENT, MUNICIPAL

ENGINEERING, COST ANALYSIS, ETC.

Editor-in- Chief
FREDERICK E. TURNEAURE, C. E., Dr. Eng.

DEAN, COLLEGE OF ENGINEERING, UNIVERSITY OF WISCONSIN

Assisted by a Corps of

CIVIL AND CONSULTING ENGINEERS AND TECHNICAL EXPERTS OF THE
HIGHEST PROFESSIONAL STANDING

Illustrated with over Three Thousand Engravings

EIGHT VOLUMES

CHICAGO
AMERICAN TECHNICAL SOCIETY

1908

OOPYBIGHT, 1908
BY

AMERICAN SCHOOL OF CORRESPONDENCE

AMERICAN TECHNICAL SOCIETY

Entered at Stationers' Hall, London
All Rights Reserved,

Engineering
limy

145

Editor-in- Chief
FREDERICK E. TURNEAURE, C. E., Dr. Eng.

Dean, CoJJege of Engineering, University of Wisconsin

Authors and Collaborators

WALTER LORING WEBB, C. E.

Consulting Civil Engineer

American Society of Civil Engineers

Author of "Railroad Construction, ' "Economics of Railroad Construction," etc.

FRANK 0. DUFOUR, C. E.

Assistant Professor of Structural Engineering, University of Illinois
American Society of Civil Engineers
American Society for Testing Materials

HALBERT P. GILLETTE, C. E.

Consulting Engineer
American Society of Civil Engineers
Managing Editor "Engineering-Contracting"

Author of "Handbook of Cost Data for Contractors and Engineers," "Earthwork
and its Cost," "Rock Excavation— Methods and Cost"

ADOLPH BLACK, C. E.

Adjunct Professor of Civil Engineering, Columbia University, N. Y.

EDWARD R. MAURER, B. C. E.

Professor of Mechanics, University of Wisconsin

Joint Author of "Principles of Reinforced Concrete Construction"

W. HERBERT GIBSON, B. S., C. E.

Civil Engineer

Designer of Reinforced Concrete

AUSTIN T. BYRNE

Civil Engineer

Author of "Highway Construction," "Materials and Workmanship"

713593

Authors and Collaborators— Continued

FREDERICK E. TURNEAURE, C. E., Dr. Eng.

Dean of the College of Engineering, and Professor of Engineering, University of

Wisconsin

American Society of Civil Engineers
Joint Author of "Principles of Reinforced Concrete Construction," "Public Water

Supplies," etc.

THOMAS E. DIAL, B. S.

Instructor in Civil Engineering, American School of Correspondence

Formerly with Engineering Department, Atchison, Topeka & Santa Fe Railroad

ALFRED E. PHILLIPS, C. E., Ph. D.

Head of Department of Civil Engineering, Armour Institute of Technology
•f

DARWIN S. HATCH, B. S.

Instructor in Mechanical Engineering, American School of Correspondence

CHARLES E. MORRISON, C. E., A. M.

Instructor in Civil Engineering. Columbia University, N. Y.
Author of "Highway Engineering."

ERVIN KENISON, S. B.

Instructor in Mechanical Drawing, Massachusetts Institute of Technology
^»

EDWARD B. WAITE

Head of Instruction Department, American School of Correspondence
American Society of Mechanical Engineers
Western Society of Engineers

EDWARD A. TUCKER, S. B.

A rchitectural Engineer

American Society of Civil Engineers

ERNEST L. WALLACE, S. B.

Instructor in Electrical Engineering, American School of Correspondence
American Institute of Electrical Engineers

A. MARSTON, C. E.

Dean of Division of Engineering and Professor of Civil Engineering, Iowa State

College

American Society of Civil Engineers
Western Society of Civil Engineers

Authors and Collaborators— Continued

CHARLES B. BALL

Civil and Sanitary Engineer

Chief Sanitary Inspector, City of Chicago

American Society of Civil Engineers

ALFRED E. ZAPF, S. B.

Secretary, American School of Correspondence

SIDNEY T. STRICKLAND, S. B.

Massachusetts Institute of Technology
Ecole des Beaux Arts, Paris

RICHARD T. DANA

Consulting Engineer

American Society of Civil Engineers

Chief Engineer, Construction Service Co.

ALFRED S. JOHNSON, A. M., Ph. D.

Textbook Department, American School of Correspondenc
Formerly Instructor, Cornell University
Royal Astronomical Society of Canada

WILLIAM BEALL GRAY

Sanitary Engineer

National Association of Master Plumbers

United Association of Journeyman Plumbers

^«

R. T. MILLER, Jr., A. M., LL. B.

President American School of Correspondence

GEORGE R. METCALFE, M. E.

Head of Technical Publication Department, Westinghouse Electric & Manufac-
turing Co.

Formerly Technical Editor, "Street- Rail way Review"
Formerly Editor "The Technical World Magazine"

MAURICE LE BOSQUET, S. B.

Massachusetts Institute of Technology

British Society of Chemical Industry, American Chemical Society, etc.

HARRIS C. TROW, S. B., Managing Editor

Editor of Textbook Department, American School of Correspondence
American Institute of Electrical Engineers

Authorities Consulted

THE editors have freely consulted the standard technical literature of
America and Europe in the preparation of these volumes. They desire
to express their indebtedness, particularly, to the following eminent
authorities, whose well-known treatises should be in the library of everyone
interested in Civil Engineering.

Grateful acknowledgment is here made also for the invaluable co-opera-
tion of the foremost Civil, Structural, Railroad, Hydraulic, and Sanitary
Engineers in making these volumes thoroughly representative of the very
best and latest practice in every branch of the broad field of Civil Engineer-
ing; also for the valuable drawings and data, illustrations, suggestions,
criticisms, and other courtesies.

WILLIAM G. RAYMOND, C. E.

Dean of the School of Applied Science and Professor of Civil Engineering in the State

University of Iowa; American Society of Civil Engineers.
Author of "A Textbook of Plane Surveying," "The Elements of Railroad Engineering."

**•

JOSEPH P. FRIZELL

Hydraulic Engineer and Water-Power Expert; American Society of Civil Engineers.
Author of " Water Power, the Development and Application of the Energy of Flowing
Water."

FREDERICK E. TURNEAURE, C. E., Dr. Eng.

Dean of the College of Engineering and Professor of Engineering, University of Wisconsin.
Joint Author of "Public Water Supplies," "Theory and Practice of Modern Framed
Structures," " Principles of Reinforced Concrete Construction."

^»

H. N. OGDEN, C. E.

Assistant Professor of Civil Engineering, Cornell University.
Author of "Sewer Design."

V»

DANIEL CARHART, C. E.

Professor of Civil Engineering in the Western University of Pennsylvania.
Author of "A Treatise on Plane Surveying."

^
HALBERT P. GILLETTE

Editor of Engineering-Contracting; American Society of Civil Engineers; Late Chief

Engineer, Washington State Railroad Commission.
Author of " Handbook of Cost Data for Contractors and Engineers."

^«

CHARLES E. GREENE, A. M., C. E.

Late Professor of Civil Engineering, University of Michigan.

Author of "Trusses and Arches, Graphic Method," "Structural Mechanics."

Authorities Consulted— Continued

A. PRESCOTT FOLWELL

Editor of Municipal Journal and Engineer; Formerly Professor of Municipal Engineer-
ing, Lafayette College.
Author of " Water Supply Engineering," "Sewerage."

LEVESON FRANCIS VERNON-HARCOURT, M. A.

Emeritus Professor of Civil Engineering and Surveying. University College, London;
Institution of Civil Engineers.

Author of "Rivers and Canals," "Harbors and Docks," "Achievements in Engineer-
ing," "Civil Engineering as Applied in Construction."

PAUL C. NUGENT, A. M., C. E.

Professor of Civil Engineering, Syracuse University.
Author of " Plane Surveying."

FRANK W. SKINNER

Consulting Engineer; Associate Editor of The Engineering Record; Non-Resident Lec-
turer on Field Engineering in Cornell University.
Author of " Types and Details of Bridge Construction."

HANBURY BROWN, K. C. M. G.

Member of the Institution of Civil Engineers.
Author of " Irrigation, Its Principles and Practice."

SANFORD E. THOMPSON, S. B., C. E.

American Society of Civil Engineers.

Joint Author of "A Treatise on Concrete, Plain and Reinforced."

*•

JOSEPH KENDALL FREITAG, B. S., C. E.

American Society of Civil Engineers.

Author of "Architectural Engineering," "Fireproofing of Steel Buildings."

AUSTIN T. BYRNE, C. E.

Civil Engineer.

Author of "Highway Construction," "Inspection of Materials and Workmanship Em-
ployed in Construction."

JOHN F. HAYFORD, C. E.

Inspector of Geodetic Work and Chief of Computing Division, Coast and Geodetic Survey;

American Society of Civil Engineers.
Author of "A Textbook of Geodetic Astronomy."

WALTER LORING WEBB, C. E.

Consulting Civil Engineer; American Society of Civil Engineers.

Author of "Railroad Construction in Theory and Practice," "Economics of Railroad
Construction, "etc.

Authorities Consulted— Continued

EDWARD R. MAURER, B. C. E.

Professor of Mechanics, University of Wisconsin.
Joint Author of " Principles of Reinforced Concrete Construction."
^»

HERBERT M. WILSON, C. E.

Geographer and Former Irrigation Engineer, United States Geological Survey; American

Society of Civil Engineers.
Author of " Topographic Surveying," " Irrigation Engineering," etc.

MANSFIELD MERRIMAN, C. E., Ph. D.

Professor of Civil Engineering, Lehigh University.

Author of " The Elements of Precise Surveying and Geodesy," "A Treatise on Hydraul-
ics," "Mechanics of Materials," " Retaining Walls and Masonry Dams, " "Introduction
to Geodetic Surveying," "A Textbook on Roofs and Bridges," "A Handbook for
Surveyors,'' etc.

DAVID M. STAUFFER

American Society of Civil Engineers; Institution of Civil Engineers; Vice-President,

Engineering News Publishing Co.
Author of ' Modern Tunnel Practice."

CHARLES L. CRANDALL

Professor of Railroad Engineering and Geodesy in Cornell University.
Author of "A Textbook on Geodesy and Least Squares."

N. CLIFFORD RICKER, M. Arch.

Professor of Architecture, University of Illinois: Fellow of the American Institute of

Architects and of the Western Association of Architects.
Author of " Elementary Graphic Statics and the Construction of Trussed Roofs."

JOHN C. TRAUTWINE

Civil Engineer.

Author of "The Civil Engineer's Pocketbook."
^»

HENRY T. BOVEY

Professor of Civil Engineering and Applied Mechanics, McGill University, Montreal.
Author of "A Treatise on Hydraulics."

WILLIAM H. BIRKMIRE, C. E.

Author of "Planning and Construction of High Office Buildings," "Architectural Iron
and Steel, and Its Application in the Construction of Buildings," " Compound Riv-
eted Girders," 'Skeleton Structures,' etc.

IRA O. BAKER, C. E.

Professor of Civil Engineering, University of Illinois.

Author of "A Treatise on Masonry Construction," " Engineers' Surveying Instruments,
Their Construction, Adjustment, and Use," ''Roads and Pavements."

Authorities Consulted— Continued

JOHN CLAYTON TRACY, C. E.

Assistant Professor of Structural Engineering, Sheffield Scientific School, Yale University.
Author of " Plane Surveying: A Textbook and Pocket Manual."

FREDERICK W. TAYLOR, M. E.

Joint Author of "A Treatise on Concrete, Plain and Reinforced.
^«

JAMES J. LAWLER

Author of " Modern Plumbing, Steam and Hot- Water Heating."

FRANK E. KIDDER, C. E., Ph. D.

Consulting Architect and Structural Engineer; Fellow of the American Institute of
Architects.

Author of "Architect's and Builder's Pocketbook," " Building Construction and Super-
intendence, Parti, Masons' Work; Part II, Carpenters' Work; Part III, Trussed Roof s
and Roof Trusses," " Strength of Beams, Floors, and Roofs."
^«

WILLIAM H. BURR, C. E.

Professor of Civil Engineering, Columbia University; Consulting Engineer; American

Society of Civil Engineers; Institution of Civil Engineers.
Author of " Elasticity and Resistance of the Materials of Engineering;" Joint Author of

"The Design and Construction of Metallic Bridges."

WILLIAM M. GILLESPIE, LL. D.

Formerly Professor of Civil Engineering in Union University.
Author of " Land Surveying and Direct Leveling," " Higher Surveying."
*»

GEORGE W. TILLSON, C. E.

President of the Brooklyn Engineers' Club; American Society of Civil Engineers; Ameri-
can Society of Municipal Improvements; Principal Assistant Engineer, Department
of Highways, Brooklyn.

Author of " Street Pavements and Street Paving Material."
^«

G. E. FOWLER

Civil Engineer; President, The Pacific Northwestern Society of Engineers; American

Society of Civil Engineers.
Author of " Ordinary Foundations."

WILLIAM M. CAMP

Editor of The Railway and Engineering Review; American Society of Civil Engineers.
Author of "Notes on Track Construction and Maintenance."
V

W. M. PATTON

Late Professor of Engineering at the Virginia Military Institute.
Author of "A Treatise on Civil Ensrineerin.-r."

Foreword

HE marvelous developments of the present day in the field
of Civil Engineering, as seen in the extension of railroad
lines, the improvement of highways and waterways, the
increasing application of steel and reinforced concrete
to construction work, the development of water power
and irrigation projects, etc., have created a distinct necessity
for an authoritative work of general reference embodying the
results and methods of the latest engineering achievement.
The Cyclopedia of Civil Engineering is designed to fill this
acknowledged need.

C. The aim of the publishers has been to create a work which,
while adequate to meet all demands of the technically trained
expert, will appeal equally to the self-taught practical man,
who, as a result of the unavoidable conditions of his environ-
ment, may be denied the advantages of training at a resident
technical school. The Cyclopedia covers not only the funda-
mentals that underlie all civil engineering, but their application
to all types of engineering problems; and, by placing the reader
in direct contact with the experience of teachers fresh from
practical work, furnishes him that adjustment to advanced
modern needs and conditions which is a necessity even to the
technical graduate.

€L The Cyclopedia of Civil Engineering is a compilation of
representative Instruction Books of the American School of Cor-
respondence, and is based upon the method which this school
has developed and effectively used for many years in teaching
the principles and practice of engineering in its different
branches. The success attained by this institution as a factor
in the machinery of modern technical education is in itself the
best possible guarantee for the present work.

€L Therefore, while these volumes are a marked innovation in
technical literature — representing, as they do, the best ideas and
methods of a large number of different authors, each an ac-
knowledged authority in his work — they are by no means an
experiment, but are in fact based on what long experience has
demonstrated to be the best method yet devised for the educa-
tion of the busy workingman. They have been prepared only
after the most careful study of modern needs as developed
under conditions of actual practice at engineering headquarters
and in the field.

C. Grateful acknowledgment is due the corps of authors and
collaborators — engineers of wide practical experience, and
teachers of well-recognized ability — without whose co-opera-
tion this work would have been impossible.

Table of Contents

VOLUME VI
BRIDGE ENGINEERING By Frank O. Dufour\ Page *11

Bridge Analysis — Early Bridges — Trusses — Girders — Truss and Girder Bridges

— Deck and Through Bridges — Truss Members — Lateral Bracing — Portals —
Sway Bracing — Knee-Braces — Kinds of Trusses — Chord and Web Character-
istics — Weights of Bridges — Loads (Dead, Live, Wind) — Principles of Analysis

— Resolution of Forces — Method of Moments — Stresses in Web and Chord
Members — Warren Truss under Live and Dead Loads — Position of Live Load
for Maximum Moments — Counters — Maximum and Minimum Stresses — Pratt,
Howe, Baltimore, Bowstring, Parabolic, and Other Trusses under Dead and Live
Loads — Engine Loads — Position of Wheel Loads for Maximum Shear and Mo-
ments — Pratt Truss under Engine Loads — Impact Stresses — Snow- Load
Stresses — Wind-Load Effects — Top and Bottom Lateral Bracing — Overturning

— Pratt Truss under Wind Loads — Girder Spans— Floor- Beam Reactions —
Plate-Girder Reactions — Bridge Design — Economic Considerations — Types of
Bridges for Various Spans — Economic Proportions of Members — Clearance
Diagram — Weights and Loadings — Specifications — Stress Sheet — Design of a
Plate-Girder Railway Span — Masonry Plan — Determination of Span — Ties and
Guard-Rails — Web and Flanges — Rivet Spacing — Lateral Systems and Cross-
Frames — Stiff eners — Web-Splice — Bearings — Design of a Through Pratt Rail-
way Span — Stringers — Floor- Beams — Tension Members — Intermediate Posts —
Lacing Bars — End-Posts — Pins — Transverse Bracing — Shoes and Roller Nests

HIGHWAY CONSTRUCTION . By A. T.Byrne and A.E.Phillips Page 267

Country Roads — Road Resistances to Traction — Axle Friction — Air Resistance

— Tractive Power and Gradients — Effects of Springs on Vehicles — Location of
Roads — Contour Lines — Levels — Cross- Levels — Bridge Sites — Mountain Roads

— Alignment — Zigzags — Construction Profile — Width and Transverse Contour

— Drainage Ditches and Culverts — Earthwork — Roads on Rocky Slopes — Earth
and Sand Roads — Grading Tools —• Rollers — Sprinkling Carts — Road Coverings

— Gravel Roads — Macadam Roads — City Streets — Catch-Basins — Pavement
Foundations — Stone- Block Pavements — Properties of Stones — Cobblestone
Pavement — Belgian Block Pavement — Brick Pavement — Paving Brick — Con-
crete Mixers — Gravel Heaters — Melting Furnaces — Wood Pavements — Asphalt
Pavements — Mixing Formulae — Footpaths — Curbstones — Artificial Stone —
Pavement Selection — Safety — Life of Pavements — Cost — Relative Economies

REVIEW QUESTIONS Page 399

INDEX Page 405

*For page numbers, see foot of pages.

tFor professional standing of authors, see list of Authors and Collaborators at
front of volume.

BRIDGE ENGINEERING

PART I

BRIDGE ANALYSIS

1. Introduction. The following treatment of the subject of
Bridge Analysis, while not exhaustive, is regarded as sufficiently
elaborate to develop and instill the principal theoretical considera-
tions, to illustrate the most convenient and practical methods of
analyzing the common forms of trusses and girders, and also to lay
a sufficient foundation for the analysis of such other trusses as are
not specifically mentioned or treated herein.

The necessary steps and operations required for a proper analysis
of the several types of bridges are fully demonstrated by sketches
and computations, the numerical values being mechanically obtained
by the use of a slide rule, which is a handy instrument for quickly
performing the operations of multiplication and division, and for
squaring and extracting the roots of numbers. The values given
may differ from the exact value by one unit in the second decimal
place (seldom more) and are sufficiently accurate for the purpose of
design. All bridge computers should be proficient in the use of the
slide rule.

The problems given in the back of this Instruction Paper,
exemplifying the practical application of the subject-matter treated
in the various articles, should be solved by the student as each article
is mastered.

HISTORY

2. Early Bridges. Early bridges were not bridges according
to the present conception of the term. They were simple pile
trestle bents placed at frequent intervals and connected by wooden
beams on which the floor was placed. The Pons sublicius, built over
the Tiber, at Rome, about 650 years before Christ was born, was of
this trestle type. Also the famous bridge b'uilt by Caesar across the
Rhine in 55 B. C. was of the same kind of construction. As civiliza-
tion progressed, the arch type was developed; and in 1390 the great

BRIDGE ENGINEERING

bridge at Trezzo over the River Adda was built of one span of 251
feet, which has never been eclipsed or equaled.

3. Truss Bridge Development. The first truss bridge is sup-
posed to have been originated by Palladio, an Italian, who used the
king-post truss (Fig. 1) about 1570. Its importance was not recog-
nized, and it became entirely for-
gotten until it was rediscovered in
1798 by Theodore Burr, an Ameri-
can, who used it in his construction.
About the same time, Burr invented
the truss that bears his name,
which was in reality a series of
king-post trusses (see Fig. 8). This
was found to be unstable under moving loads, and was therefore
stiffened by the use of an arch (Fig. 2), or was built somewhat as an
arch, there being considerable rise at the center of the span (Fig. 3).
By 1830 the principle of the double cross-bracing in the panel was
understood; and in quick succession came the patents of Long,
Howe, Pratt, and Whipple on forms of trusses which bear their
respective names.

It remained for Squire Whipple in 1847 to place the science of
bridge building on a rational and exact mathematical basis such

Fig. l. King-Post Truss.

Fig. 2. King-Post Truss Bridge Stiffened by Arch.

as is now used. Previous to this time, and indeed several years
afterwards — for Whipple's work did not become generally known
until a much later date — bridges were built, not from previously
computed strains, but by "judgment." All parts of a bridge were
made of the same size, and if one started to fail it was replaced by
a larger one; or small models were made and loaded proportionally,
broken members being replaced by larger ones. There is no doubt

BRIDGE ENGINEERING

that many of the bridges built at this period were very weak as well
as very strong. The failures are not remembered; but the sound
judgment of many of our earlier bridge engineers is recorded in the
wooden structures they left behind them, some of which have stood
the demands of traffic for over a century. After 1850, bridges were
built from computed stresses; wood was discarded; and the develop-
ment became rapid, until about 1870, when the introduction of sub-
diagonal systems brought the truss system to practically what it is
to-day.

DEFINITIONS AND DESCRIPTIONS

4. Trusses. A truss is a series of members taking stress in
the direction of their length, placed together so as to form a triangle

Fig. 3. Burr Trusts Bridge, Arched.

or system of triangles, which, when placed upon supports a certain
distance apart, will, in addition to their own weight, sustain certain
loads applied at the points where the members intersect. These
points are called panel points,

5. Bridge Trusses. A bridge truss is one in which the members
that carry the superimposed loads are in the same plane. Usually
this plane is vertical.

6. Truss Bridges. A truss bridge is a structure consisting of
two or more — usually two — bridge trusses connected by a system
of beams called the floor system, which transfer to panel points the
load for which the trusses are designed.

7. Girders. These are beams consisting of a wide, thin plate,
called a web plate, with shapes, usually angles and narrow, thin
plates called flanges, at the top and bottom edges. All are firmly
riveted together. (See Part IV, "Steel Construction.")

8. Girder Bridges. These consist of usually two, sometimes

BRIDGE ENGINEERING

three, girders connected as in the case of truss bridges by a system
of beams.

9. Deck Bridges. In cases where the floor system connects
the trusses at their tops, the bridge is called a deck bridge, since the
traffic moves on a deck, as it were (see Fig. 4).

Fig. 5. Through Bridge.

10. Through Bridges. In cases where the floor system con-
nects the bottoms of the trusses, the bridge is called a through bridge,
as the traffic moves through the space between the trusses (see Fig. 5).

11. Members of a Truss. Each truss consists of a top and
bottom chord, end-posts, and web members. The web members are
further divided into hip verticals, intermediate posts, and diagonals.
Fig. 6 shows these various classes, A- A being top chord, B-B

Fig. 6. Showing the Members of a Truss.

bottom chord, A-B end-posts, vertical members C-b intermediate
posts, A-a hip verticals, and A-b and C-b diagonals.

12. Pony=Truss Bridges. When the height of the trusses of
a through bridge is less than the height of the loads that go over
them, they are called pony trusses, and the bridge a pony-truss bridge.

BRIDGE ENGINEERING

13. Lateral Bracing. In all deck bridges, and in all through
bridges except pony-truss bridges, the chords which are not con-
nected by the floor system are connected by a horizontal truss system
called the lateral bracing. In all bridges the chords which are con-
nected by the floor system are connected by a horizontal truss system,
also called the lateral bracing. One of these systems is called the

String

Fig. 7. Through Bridge. Showing Top and Bottom Systems of Lateral Bracing, also
Portal Bracing and Floor System.

top lateral system, as it connects the top chords; and the other is called
the bottom lateral system, as it connects the bottom chords (see Fig. 7).

14. Portals. In through bridges, the end-posts of the pair
cf trusses are connected by a system of braces in order to preserve
the rectangular cross-section of the bridge. This is called the portal
bracing (see Fig. 7).

15. Sway Bracing and Knee=Braces. These serve the same
purpose as the portal braces, and are either small struts or systems

BRIDGE ENGINEERING

of cross-bracing placed at the intermediate posts. The former are
called knee-braces, and the latter sway bracing.

16. Floor Systems. In both highway and railway bridges,
there are beams running from the intermediate posts or hip ver-
ticals across to the like members opposite. These are called floor-
beams. In highway bridges, there are smaller beams running parallel
to the trusses and resting at their ends upon the floor-beams. These
are called floor-joists, and the plank or other floor rests directly upon
them. In railway bridges, two beams or girders per track run parallel
to the trusses and are connected at their ends to the floor-beams.
These are called track stringers (or simply stringers'). The ties rest
directly upon them. The various members of the floor system of
a railway bridge are shown in Fig. 7. The diagonals connecting the
top chords, and those connecting the bottom chords, are the top and
bottom laterals respectively.

CLASSES OF TRUSSES

17. Names. Trusses may be classified according to their
names, the character of their chords, and the system of wrebbing.
Table I gives the classification of the more important of these accord-
ing to name.

TABLE I
Chronological List of Trusses

NAME 1 YEAR OF
, ORIGIN

INVENTOR

COUNTRY

ILLUSTRATED IN

King-Post

1570

Palladio

Italy

Fig. 1

King-Post

1798

Theodore Bun-

America

Fig. 1

Burr

1798

Theodore Burr

America

Fig. 8

Warren

1838

England

Fig. 9

Howe

1840

William Howe

America

Fig. 10

Pratt

1844

Thos & Caleb Pratt

America

Fig. 11

Whipple
Bowstring
Baltimore

1847
1847
1877

Squire Whipple
Squire Whipple
Penn. R. R.

America
America
America

Fig. 12
Fig. 13
Fig. 14

Of the types of trusses listed in Table I, the Warren, Howe,
Pratt, Bowstring, and Baltimore are now built; and of these construc-
tions probably 90 per cent are Pratt trusses. The Baltimore truss
is used for long spans only.

18. Chord Characteristics. In most types of bridges the

BRIDGE ENGINEERING

Fig. 8. Burr Truss.

Fig. 9. Warren Truss.

Fig. 10. Howe Truss.

Fig. 11. Pratt Truss.

Fig. 12. Whipple Truss.

BRIDGE ENGINEERING

chords are parallel. When such is the case, the stresses increase
from the end toward the center, and there is a considerable difference
between any two adjacent panels of the same chord. This neces-
sitates different areas for each section. When the chords are not
parallel, as in the bowstring truss, the stresses in the chords are so
nearly equal that the same area is used throughout or nearly through-
out the entire chord. Also, the stresses in the diagonals are nearly
equal. These conditions would seem to indicate that this was a very
economical form of truss. Theoretically it is; but practical con-
siderations— such as the beveled joints and the posts which must be
constructed to withstand reversals of stress — customarily limit this
type to the longer spans.

19. Web Characteristics. The web systems of the Burr,
Warren, Howe, Pratt, and Bowstring trusses are called single sys-
tems; that of the Whipple truss is a multiple system; while those of
the Baltimore trusses are examples of webbing with sub-systems.
As the maximum economical panel length has been found to be about
25 feet, which makes the economical height of the truss about 30
feet, and as the length of the span should not be more than ten times
the depth, the span for trusses with simple systems of webbing is
limited to about 300 feet. In order to increase the limiting span,
multiple systems like that of the Whipple or similar ones were intro-
duced. Calculations of stresses in members of the Whipple truss are
somewhat unreliable on account of the fact that we are unable to
tell just how the effects of the loads are distributed. For this reason,
that type has gone out of use, and the sub-systems are used instead.
These allow spans of twice the above limit; and, indeed, trusses with
this type of webbing have been built up to and over 600 feet. This
style of webbing can be applied to the bowstring truss, almost all
long-span bridges being of this type with sub-systems of webbing.

WEIGHTS OF BRIDGES

20. Formulae. In order to obtain the stresses due to the
weight of the structure, the latter quantity must be known. As
this weight can be determined only after the structure has been
designed, it is evident that an assumption as to the weight must be
made. The best method is to use the actual weight of a similar
structure of like span which has been built. As the necessary data

BRIDGE ENGINEERING

for this is not always available, it is customary to use formulae to
derive an approximate weight of sufficient accuracy for purposes of
computation. Table II gives some of the most reliable formulee.

TABLE II
Formulae for Weights of Bridges

CLASS OF BRIDOB

WEIGHT or STEEL PER LINEAR FOOT OF SPAN

AuTHOB*

Heavy Interurban

Riveted

w=6QQ + l.&l + 27b + -Lbl(l + -rjm1)

E. S. Shaw

First-Class High-

/ \

way Riveted

M) = 300 + l + 22b-\ bl(l H 1 )

E. S. Shaw

First-Class High-

15 \ 1,000 /

way Pin

w = 34 + 226 + Q.lQbl + 0.71

J.A.L.Waddell

Light Country
Highway

w = 250 + 2.51

Author

Railroad Truss

E 50

w = (650 + 70

F. E. Turneaure

Railroad Truss

E 40

w = | (650 + 70

F.E. Turneaure

Railroad Truss •

E 30

w >> | (650 + 70

F E. Turneaure

Railroad Deck

Girder E 50

w = 124.0 + 12.01

Author

Railroad Deck

Girder E 40

w = 123.5 4- 10.01

Author

Railroad Deck

Girder E 30

w = 111.0 + 8.81 lAuthor

In the above formulae, w = Weight of steel per linear foot of span;
I = Length of span in feet; b = Breadth of roadway, including sidewalks.

In using the formulae of Table II, remember that a span has two trusses.
The weights for highway bridges do not include the weight of the wooden
floor, which may be assumed as 10 pounds per square foot of floor surface
All highway bridges have steel joists. The weights of railroad spans do not
include the weight of the ties and rails, which may be assumed at 400
pounds per track per linear foot of span. If solid steel floors are to be used,
700 pounds per linear foot of span are to be added to the weights computed
from the table.

All the weights given for railroad spans are for single track. Double-
track truss-spans are about 95 per cent'heavier; and double-track girder-spans
are 100 per cent heavier. Through girder spans are about 25 per cent heavier
than deck girder spans; and through spans are about 15 to 40 per cent heavier
than deck spans.

The spans on which Table II is based are of medium steel. Bridges
built of soft steel or wrought iron will weigh 10 to 15 per cent more.

* The author is indebted to the distinguished engineers whose names appear in
Table II, for permission in this connection to make use of the formulas given opposite
their names.

BRIDGE ENGINEERING

In order to give an idea of the relative weights of steel in different
classes of bridges, let it be required to compute the dead weight of a
100-foot span of each class. For heaviest highway bridges to carry
heavy interurban cars:
u> = 600-f!80+27 X16 + 16 *100 t + _HH = i 358 lbs. per linear ft.

Fig. 13. Bowstring Truss.

For heavy riveted highway bridges to carry heavy farm engines :

u = 300 + 100 + 22 X 16 + 16 100 1 + - = 870 Ibs. per linear foot.

Fig. 14. Two Forms of Baltimore Trusses.

For heavy pin-connected highway bridges to carry heavy farm or
traction engines:

w = 34 +22 X 16 + 0.16 X 16 X 100 + 0.7 X 100 = 710 Ibs. per linear ft.
For light country highway bridges to carry 100 pounds per square
foot of floor surface:

w = 250 + 2.5 X 100 = 500 Ibs. per linear foot.
If the total weight is required, the weight of the wooden floor
must be added. Take, for example, the last bridge:

BRIDGE ENGINEERING

Weight of steel = 500 X 100 = 50 000 pounds.

" floor = 100 X 16 X 10 = 16000 pounds.

Total dead load = 66 000 pounds.

The weight per linear foot for a railroad truss bridge of 100-foot

span is:

w = 650 + 7 X 100 = 1 350 Ibs. per linear foot.

This is about the same as that for a heavy interurban bridge
The reason for this is that in addition to the heavy rolling stock
of the electric road, the heavy highway traffic must be provided for.
A deck girder of 100-foot span weighs:

w = 124 + 12 X 100 = 1 324 Ibs. per linear foot.

21. Actual Weights of Railroad Spans. In case actual weights
can be obtained, a more exact analysis can be made. The weights
of bridges indicated in the accompanying tables and diagrams, are
based on actual constructions recently erected. These bridges rep-
resent the very best modern practice of engineers and manufacturers

The weights of through truss-spans made of medium steel
and designed for E 50 loading, are given in Fig. 15. The weights
include the weight of the ordinary open steel floor, and they also
include the weight of the ties and rails, which is taken at 400 pounds
per linear foot per track.

The weight of steel in medium steel deck plate-girder spans
designed for E 50 loading, is given in Table III.

TABLE HI

Weights of Deck Plate-Girders, Medium Steel

Loading E 50

SPAN
(in feet)

WEIGHT
(in pounds)

SPAN
(in feet)

WEIGHT
(in pounds)

5300

59500

7800

67300

11 800

76300

14500

94200

18800

105 500

23300

114200

27 400

100

123 600

32 400

105

146000

38800

110

161 700

45500

115

174 900

51 500

120

187000

The spans are the distance center to center of bearings; and the weights
do not include the weight of the ties and rails, which is to be taken at 400
pounds per linear foot per track. Intermediate spans may be interpolated.

BRIDGE ENGINEERING

22. Actual Weights of Highway Spans. The actual weights
of highway spans for heavy interurban trolley-cars and traffic, should
preferably be computed from the formulae of Shaw or Waddell (Table
II). ,The weights of country bridges, including floor, may be taken
from the diagram of Fig. 16.

LOADS

23. Classes of Loads. Those weights just given constitute
what is called the dead load of the bridge. The traffic which passes

(30000

lEQOOO

110000

IOQOOO

9QOOO

I
8QOOO

6QOOO

5QOOO

A - /

•tt

Fig. 15. Weights of Through Truss-Spans. Fig. 16.
Medium Steel, E 50 Loading.

Weights of Country Bridges, In
eluding Floor.

over the bridge is called the live or moving load. In addition to the
two classes mentioned, is the effect of the wind, which is designated
as the wind load. These loads vary with the class of bridge, be it
highway or railway, and with the purpose for which it is intended.

24. Live Loads for Highway Bridges. Highway bridges are
usually divided into several classes according to the traffic, which
may be that of heavy interurban cars, light trolley-cars, farm engines,
road rollers, teams, human beings, or some combination of these
loadings. The standard specifications of J. A. L. Waddell or of
Theodore Cooper are obtainable for a very small sum. Their pur-

BRIDGE ENGINEERING 13

chase is advised, and the reader is referred to them for further infor-
mation.

The trusses of country highway bridges are usually designed for
a live load of 100 pounds per square foot of roadway. This may be
considered good practice; and it is the law in some States. The
floor system of these same bridges should be of sufficient strength to
sustain 100 pounds per square foot of roadway, or a 12-ton farm
engine having 4 tons on the two rear wheels, which are 12 inches wide
and 6 feet apart, and 2 tons on each of the front wheels, which are 6
inches wide and 5 feet apart. The axles of this engine are spaced
8 feet center to center.

25. Live Loads for Railway Bridges. The loads for any par-
ticular railroad bridge are not always the same, on account of the
great variation in the weights and wrheel spacings of engines and
cars. It is customary to design the bridge for the heaviest in use
at the time of construction, or for the heaviest that could reasonably
be expected to be built thereafter.

As the computations with engines were formerly somewhat
laborious on account of the different weights and spacing of wheels,
it has been proposed by some engineers to use a uniform load, called
the equivalent load, which would give stresses the same, or very nearly
the same, as those obtained by the use of engine loads. However,
as these loads are different for each weight of engine, and also different
for the chord members, the web members, and the floor-beam reaction
of each different length of span, and as the labor of the computations,
using engine- wheel loads, has been greatly reduced by means of
diagrams, it does not seem as if this method would ever come into
very general favor except for long-span bridges, where the live load is
much smaller than the dead load.

The equivalent loads for Cooper's E 40 (see Fig. 85) are given
in Table IV.

Most railways specify that their bridges shall be computed by
using two engines and tenders followed by a train. The spacing
of the wheels, and the load which comes on each wheel of the engines
and tenders, are fixed by the railway company. The train is repre-
sented by a uniform load. Formerly there was a great diversity of
practice among the different roads in regard to the engine and. train
loads specified; but practice has of late years become quite uniform,

BRIDGE ENGINEERING

TABLE IV
Equivalent Uniform Loads

Loading E 40

SPAN
(in feet)

EQUIVALENT UNIFORM LOAD

SPAN
(in feet)

EQUIVALENT UNIFORM LOAD

Chords

Web

Floor-
Beam

Chords

Web

Floor-
Beam

9 000

12000

8200

6 330

7240

5240

9310

11 640

7960

6220

7 140

5200

9 340

11 330

7830

6 110

" 7060

5 140

9 340

11 080

7600

6040

6940

5 130

9210

10 860

7460

5 960 i 6 820

5120

9 030

10 670

7330

5880

6 720

5 110

8850

10 500

7120

5800

6620

5090

8650

10 350

6940

5730

6530

5080

8430

10 240

6 780

5690

6490

5080

8220

10 100

6630

5700

6450

5070

x20

8000

10 000

6 500

5620

6450

5070

8040

9 780

6 390

5560

6380

5060

8040

9580

6 290

5510

6340

5060

8010

9400

6200

5490

6320

5030

7960

9230

6 120

5460

6 300

5010

7890

9080

6040

5440

6290

4990

7780

8930

5970

5420

6270

4970

7660

8790

5900

5400

6250

4950

7540

8660

5830

5370

6230

4930

7420

8540

5770

5340

6200

4910

7300

8 430

5720

5310

6 180

4890

7220

8320

5680

5270

6 150

4870

7 140

8 190

5650

5250

6 130

860

7050

8080

5620

5250

6 110

830

6960

7980

5600

5210

6090

810

6870

7890

5 570

5 170

6060

780

6820

7820

5530

5 150

6040

760

6760

7750

5500

100

5 140 | 6 020

740

6700

7690

5460

125

5 100

5770

720

6630

7630

-5430

150

5010

5570

4 700

6560

7570

5400

175

4890

5350

4686

6530

7450

5340

200

4740

5240

4660

44 6 470 7 340

5300

250

4510

5030

4640

with an apparent tendency to standardize in accordance with the
classes of loading specified by Cooper. Cooper's Class E 50, which
represents the heaviest engines now in common use, was invented
by Theodore Cooper, a consulting engineer of New York City. It is
given in Fig. 17.

Lighter loadings for light traffic on the same general plan are
advocated by Mr. Cooper, and are given at length in his "General
Specifications for Iron and Steel Railway Bridges and Viaducts"
(1906 edition).

26. Wind Loads. Some designers require that the stresses due

BRIDGE ENGINEERING

005?

0~OOOQ (^

ooffos \^^ ~r

to wind shall be computed by using 30 pounds
per square foot of actual truss surface. This
requires that you know the size of the mem-
bers of the bridge before it is designed — which
is evidently an impossibility; or that an as-
sumption as to their size be made — which
allows a chance for a mistake in judgment,
especially in an inexperienced person. A more
logical method, and one used to a great extent,
is to assume a force of so many pounds per
linear foot to act on the top and bottom chords
and on the traffic as it moves across the bridge.

In highway through bridges, it is the usual
practice to take the wind load as 150 pounds
per linear foot of top and bottom chords, and
150 pounds per linear foot of the amount of
live load which is on the bridge.

For railroad bridges, it is customary to use
considerably higher values than those used in
highway practice — not that the wind blows
harder on railroad than on highway bridges,
but so that the bracing designed by the use of
these values may be sufficiently strong to stiffen-
the bridge not only against the wind, but also
against the vibrations caused by the rapidly
moving traffic. Good practice for through
bridges is to use 150 pounds per linear foot
of the top chord, 150 pounds per linear foot
of the bottom chord, and 450 pounds per linear
foot of live load on the bridge. This latter
force is supposed to act at a line 8.5 feet above
the base of the rail.

For deck bridges, for both highway and
railway use, the unit-loads on the moving or live
load remain the same, but the unit-loads on
the top and bottom chords are reversed.

In computations involving the live load, it
is always assumed that the live load moves over
the bridge from right to left.

BRIDGE ENGINEERING

THEORY

27. Principles of Analysis. The stresses in bridge trusses may
be determined by both algebraic and graphic methods. In some

P P P p p

a R

Fig. 18. Truss under Loads, Illustrating Principles of Analysis.

cases, one is more expeditious than the other. Algebraic methods
alone will be considered in this text.

The analysis of stresses is based upon the fact that the interior
stresses in a member or group of members hold in equilibrium the
exterior forces. That this is a fact, can easily be understood. Con-
sider a man pulling on a rope which is fastened at one end to an im-
movable object. There will be a
stress in the rope equal to, and
opposite in direction to, the pull
exerted by the man. In order to
prove this, cut the rope and ap-
ply a force equal and opposite to
the pull exerted by the man, where
the cut is made ; and the rope
and man will be in equilibrium.
Also, suppose that a truss under
loads, as indicated by the arrows,
Fig. 18, were cut along the section
a-a, and that forces Fy F3, F5
equal to the stresses S2, S3, and S5 were placed at the ends of the
members as indicated in Fig. 19, then that portion of the truss to
the left of the section would be in equilibrium. The interior stresses,
represented by F2, F3, and F5, would hold in equilibrium the exterior
forces p and R.

Forces Substituted for Stresses in
Truss of Fig. 18.

BRIDGE ENGINEERING

From inspection of Fig. 19, it will appear evident that, as the
position of the truss to the left of the section is in equilibrium, the
following statements are true :

1. The algebraic sum of the moments of the exterior forces and the
stresses in the members cut by the section, is equal to zero. This is true of the
moments taken about any or all points; for, if it were not, the portion of the
truss would begin to rotate about some point, and would continue until
equilibrium was established.

2. In a vertical plane, the algebraic sum of the components of the
exterior forces and the stresses in the members cut by the section is equal
to zero; for, if such were not the case, the portion of the truss shown would
move up or down with a constant acceleration.

3. The algebraic sum of the components of the exterior forces and the
stresses in the members cut by the section in a horizontal plane, is equal to
zero; for, if such were not the case, the portion of the truss would move either
to the right or to the left, with a constant acceleration.

4. From 2 and 3, above, it is evident that the algebraic sum of the
components of the exterior forces and the stresses in the members cut by
the section is equal to zero in any and all planes.

The section is not necessarily required to be a vertical line as in

Fig. 20.

Oblique Section Cutting
Truss.

Fig. 20a.

Circular Section
Cutting Truss.

Fig. 21.

Illustrating Resolution
of Forces.

Fig. 19. It may be oblique, as in Fig. 20; or it may be a circular
section, as shown in Fig. 20a. When the latter is the case, it is said
that the sum of the components of the forces around the point Ul
is in equilibrium in any plane that may be taken.

It is also evident that the forces in the members cut by the
section, and the exterior forces to the right, are in equilibrium.
This condition is very seldom utilized in the determination of stresses,
as that portion of the truss to the left of the section is almost always
considered.

28. Resolution of Forces. This method is one of the simplest

BRIDGE ENGINEERING

and at the same time least laborious. The forces are generally
resolved into their horizontal and vertical components, or parallel
and perpendicular to some member. In cases where two unknown
stresses occur, two equations can usually be formed, and these solved.
It should be assumed that the unknown stress acts away from
the section which cuts it. It will then solve out, with the proper
sign indicating the character of the stress — that is, whether it is
tensile or compressive. Tensile stresses are indicated by placing

6. P P

Fig. 24. Pig. 25.

Diagrams Illustrating Application of Method of Resolution of Forces in Analysis of

the plus ( + ) sign before them, while a minus ( — ) sign indicates
compression.

A few equations showing the application of the method of the
resolution of forces can be written after inspection of Figs. 21 to 25
inclusive. In all cases, Sx is the unknown stress, and is assumed
to be acting away from the section. The other stresses S1} S2, etc.,
are known, and their direction given them accordingly, it being
toward the section if the member is in compression, and away from
the section if the member is in tension. Forces or components
acting upward or to the right are considered plus ; those acting down-

BRIDGE ENGINEERING 19

ward or to the left are considered minus. For a fuller explanation,
see the instruction paper on Statics, Articles 17 to 23 inclusive.

In Fig 21, the sum of the vertical components is taken, and the

equation is :

+ R - p - p - S* cos (j) = 0;
whence,

<SX = — ( + R —p — p) sec <f>.

In Fig. 22, the section is oblique, and the sum of the vertical
components is taken :

+ R - p - p + Sx = 0;
whence,

Sx = - ( + R - p - p).

In both of the above cases, it will be noted that the chord
stresses do not enter into the equation, as their vertical components
are zero.

In Fig. 23, the sum of the horizontal forces is used in deter-
mining the stress <SX. Note that the exterior forces R and p do not
enter the equation, as they are not to the left of the section, and also
their horizontal components are zero.

+ Sl sin $ + S2 sin 0 + S3 sin <j> + Sx = 0;
whence,

SK = - (Sl + S2 + S3) sin <f>.

In Fig. 24, the sum of the vertical forces is again used. Here
the section cuts the member with the known tensile stress Sr

+ R — p — p - p — p - p + Sl cos (j) - Sx cos $ = 0;
whence,

Sx = + (R - 5p) sec $ + Si.

In Fig. 25, use is made of the fact that the sum of the components
of the forces about a point is zero when they are resolved in any plane.
Here they will be resolved perpendicular to the diagonal.

-S,sin-0 + SK = 0.

Sx = + S1 sin <£.

These are some of the most common cases which occur in the
determination of stresses in simple trusses. In all cases, follow this
method of procedure:

1. Pass a section cutting as few members as possible, one of which
must be the one whose stress is desired.

2. The stress in all the members cut, with but one exception, must be
known.

3. Write your equation, always placing it equal to zero.

4. Solve for your stress.

BRIDGE ENGINEERING

29. Method of Moments. The stresses in all members of a
truss can be determined by this method. By section 1 of Art. 28,
the point about which the moments are considered can be taken
anywhere. Fig. 26 represents the point as taken somewhere outside
of the truss at a distance a above the point V \. The equation will

then be:

-S, X a - S2 X b - S3 (a + A) + Rp ± P, X 0 + P2 X p = 0.

This involves three unknown quantities, and therefore two other

points should be taken, and two
more equations written. By the
use of the three equations, the
stresses can be determined.

It is customary to assume the
center of moments at such a place
that the moments of all the un-
known stresses, with one excep-
tion, are zero. This condition
requires that their lines of action
pass through the center of mo-
ments. Let it be required to
— - 53 determine the stress *S3. If the
center of moments is taken at
Uv then, as the lines of action
of <Sj and S2 pass through this

Fig. 26. Diagram illustrating Application point, their moments will be zero,

of Method of Moments in Analysis

and the following is true:
+ R X 2p - P, X p ± P2 X 0 - S3 X h = 0.

whence,

S, =--

X 2p - P, X p).

Likewise, if the top chord is curved, the center of moments can be
taken in such a position that only the unknown stress will enter into
the equation. If it is desired to determine the stress S2, Fig. 27,
the equation would be :

-S2 X I - R X a + P, (a . + p) + P2 (a + 2p) =0,
the center of moment being at 0, the intersection of the lines of stress
of Sl and Sy Solving the equation just stated,

S2 = -L j -Ra + P, (a + p) + P2 (a + 2p) 1

BRIDGE ENGINEERING

30. Stresses in Web Members. By reference to Articles 28 and
29, it is seen that several methods are presented for the solution of
stresses in web members. Each should be adapted to the case in
hand. The simplest method, and the one which is commonly used
in all trusses with parallel chords, is by the resolution of the vertical
forces. Fig. 21 is to be referred to. The equation given on page

19 is:

+ R - p — p - Sx cos $ = 0.

Fig. 27. Diagram Illustrating Application of Method of Moments In Analysis of Trusses.
Top chord curved. .

But R — p — p is equal to V, the vertical shear at the section, and
so the equation may now be written :

V - Sxcos<f> = 0 (1)

whence the following important rule is deduced:

'The algebraic sum of the vertical shear at the section and the ver-
tical components of the stress in all of the members cut by the section, is
equal to zero.

In trusses with horizontal chords and a simple system of webbing,
the equation may be put in the form:

S* = + V sec <f>;

and the statement that the stress in any web member is equal to the
shear times the secant of the angle that it makes with the vertical is

BRIDGE ENGINEERING

true. The practice of using this latter statement is not to be en-
couraged, as it leads to confusion in the signs of the stresses. Equa-
tion (1 ) should be written in all cases, and the stress will then solve with
its correct characteristic sign, indicating that the stress is either
tensile or compressive.

As an example, let it be required to determine the stresses in the
web members S2 and S3 of the Pratt truss shown in Fig. 28, the loads
being in thousands of pounds. First, a section should be passed,
cutting that member and as few others as possible. Next, the shear
at that section should be computed. Then the vertical components
of all the stresses cut by the section, and the vertical shear, should be

Fig. 28. Calculation of Stresses in Web Members of Pratt Truss.

equated to zero. Finally, solve the equation. Remember that the
unknown stress is to be assumed as acting away from the section,
and that forces or resultants acting downward are considered negative,
while those acting upward are considered positive.
To determine S2:

The vertical shear at the section a— a is :

+ 37.5 -2 X 10 - 5 = +12.5.
As the chord stresses do not exert a vertical component, the equation is:

+ 12.5 + S2 = 0

S2 = —12.5, which is a compressive stress of 12,500 pounds.
Note that in this case the angle which the member makes with the
vertical is zero, and the cosine and secant are unity.
To determine S3:

The vertical shear at the section b — b is

+ 37.5 -2X10-2X5= +7.5.
The equation is :

+ 7.5 - S3 cos 0 = 0

S3 = +7.5 -sec <£.

BRIDGE ENGINEERING

Sec </> is equal to V 3Q2 + 252 H- 30, which is equal to 1.302; and
therefore,

S3 = +7.5 X 1.302

= +9.765, which is a tensile stress of 9,765 pounds.

31. Stresses in Chord Members. The stresses in chord may be
obtained by either the method of moments or the method of resolu-
tion of forces, this latter being usually the resolution of horizontal
forces.

In accordance with the text of Article 29, the following rule may
be stated with regard to the solution of stresses in chord members
by the method of moments :

Pass a plane section cutting the member whose stress is to be computed,
and as few others as possible; then take the center of moments at such a point

/ I \ \ x ' , , \

i- V -A

Pig. 29. Calculation of Stresses in Chord Members by "Tangent" or "Chord-Increment"

Method.

that the lines of action of as many forces as possible, the unknown one excepted,
pass through that point; write an equation of the moments about this point of
the known loads and forces to the left of the section, assuming the unknown force
to act away from the section, and taking the known forces to act as given, the
tensile stresses to act away from the section, and the compressive stresses to act
towards the section; place the equation equal to zero, and solve.

The stress will solve out with its correct characteristic sign.

In the majority of cases a section can be made to cut three mem-
bers only, one of the three being the one whose unknown stress is
desired. In such cases, take the center of moments at the inter-
section of the other two, and proceed as before. As examples of this
latter case, note the centers of moments at U2, Fig. 26, and 0, Fig.
27, and also the equations resulting therefrom.

When the method of resolution of forces is used, it is usually
designated as the tangent method or the chord increment method. The
simplest application of this method is to trusses with horizontal
chords and vertical posts in the web members. Then the stress in
any chord member is equal to the product of the sum of the shears

BRIDGE ENGINEERING

in the panels up to that section, and the tangent of the angle which
the diagonals make with the vertical.

This can readily be proved by reference to Fig. 29. Let it be re-
quired to determine the stress in the chord member Sx. Pass the
section a— a. The stresses Sv S2, <S3, and S4 are now computed,
and are St= - Vl sec <f>; S2= +V2 sec <f>; S3= +V3 sec <f>; and S4 =
+ F4sec <£. Now noting the directions of the known stresses and

Lo L, LE L3 L* L5 L6

Fig. 30. Illustrating Method of Notation of Stresses and Members in a Through Bridge.

LI Lg l_3 I_4. I_5

Fig. 31. Illustrating Method of Notation of Stresses and Members in a Deck Bridge.

assuming Sx to act away from the section, the equation of the hori-
zontal component is:

+ <S, sin $ + S2 sin $ + S» sin <}> + St sin <£ + Sx =0.
Now, substituting the values of St, <S2, etc., and remembering that
sec 0 = - — -JT, the equation becomes:

. y Bin 0 sin ^ Bin .

. * cos ^ 2 cos ^> 3 cos

from which, .

sinj>

4 COS 0

r3 + T4) tan <£;

Sx = -2"F tan <j>.

From inspection of Fig. 29, it will be noticed that the stress in
any section of the chord is equal to that in the section to the left of

BRIDGE ENGINEERING

it, plus the increment (horizontal component) of the diagonal; hence
the name chord increment method.

32. Notation. The practice hitherto used in designating
stresses by Sv S2, etc., will now be discontinued, as it is inconvenient
in the extreme; moreover, it is not the method used in practical work.
The notation to be used is that given in Figs. 30 and 31, the former
being for a through and the latter for a deck truss.

The practical advantages of this system are very great. When
Ul U2 is noted, it is at once known to be the top chord of
the second panel; £72 L2 is known to be the second vertical ; while U2
L3 is at once recognized as the diagonal in the third panel. A stress
in a member, as well as the member itself, is designated by the
subscript letters at its ends. Thus U1 L2 may mean the member

Fig. 32. Calculation of Stresses in a Six-Panel Warren Truss Through Bridge.

itself or the stress in the member. The text will clear this up. In
analysis, the stress would be implied, while in design the member
itself would be intended.

33. Warren Truss under Dead Loads. The Warren truss has
its web members so built of angles and plates or of channels, that
they can take either tension or compression. The top chord is of
structural shapes, while the lower chord may be of built-up shapes
or simply of bars.

Let it be required to determine all of the stresses in the six-
panel truss of a through WTarren highway 120-foot span for country
traffic. The height is to be 20 feet. The outline is given in Fig. 32.
According to Fig. 16, the total weight of the span, including wooden
floor, is 76 000 pounds. Each truss carries one-half of this, or 76 000

26 BRIDGE ENGINEERING

-7- 2 = 38 000 pounds. As there are six panels, each panel load is
38 000 -7-6 = 6 333 pounds. This means that we must compute
the stresses in the above truss by considering that a load of 6 333
pounds is at points Lv L2, L3, L4, and L5. Of course there is some
weight at L0 and L6; but this does not stress the bridge, as it is
directed over the abutments or supports. The reactions at L0 and
L6 are each equal to (5 X 6 333) -=- 2 = 15 833 pounds.
The shears are next computed, and are:

Vl = +15833 - 0 = +15833

F2 = +15 833 - 6 333 = +9 500

F3 = + 15 833 - 2 X 6 333 = + 3 167

It is unnecessary to go past the center of the bridge, as it is symmetri-
cal. The Vl represents the shear on any section between L0 and Lx;
V2 represents the shear on any section between Ll and Z/2; and F3
represents the shear on any section between L2 and L3. The secant
of the angle <f> is :

H- 20 = 1.12.

The stresses in the web members are computed as follows:
For Lu £7r Pass section a— a. Assume stress acting away from
the section, as shown. Then,

F, + L0C7, cos 0 = 0;

L0Ul = — Fj sec 0;

L0U, = - 15 833 X 1.12 = - 17 700 pounds,

which shows that L0 Ul has a compressive stress of 17 700 pounds.
For Ul Lr Pass section 6 — 6. Assume stress acting away from
the section, as shown. Then,

F, — UlLl cos 0 = 0;

C/.L, = +T, sec 0;

UiLt = +15833 X 1.12 =+ 17 700 pounds,

which shows that UJL^ has a tensile stress of 17 700 pounds.
For LJJ2. Pass section c — c. Then,

F2 + LI C/2 cos 0 = 0;

Z^t/2 = — F2 sec 0;

L,C72 = - 9500 X 1.12 = - 10 640 pounds.

For U^LV Pass section d-d. Then,

+ 9500 - U2L2cos 0 = 0;

UjL, = +9500 X 1.12 = +10640.

BRIDGE ENGINEERING 27

For L2Uy Pass section e — e. Then,

+ 3 167 + L2U3cos<f> = 0;

L2U3 = -3 167 X 1.12 = -3 540.

For U3L3. Pass section / -/ . Then,

+ 3 167 - UaL3cos <£ = 0;

U3L3 = +3 167 X 1.12 = +3 540.

The computation of the stresses in the chords is made by the
method of moments, and is as follows:

For LJL^ Section b — b cuts UlLl and UJJ2, besides the mem-
ber whose stress is desired, and therefore the center of moments will
be taken at their intersection f7r The equation is :

+ 15 833 X 10 - LQL, X 20 = 0,
whence,

L0L, = ( + 15833 X 10) •*• 20;

= +7 917 = a tension of 7 917 pounds.

For LiLr Either section c — c or d — d may be used, and each
shows the center of moments to be at U2. The equation is :
+ 15 833 X 30 - 6 333 X 10 - L,L2 X 20 = 0;
L,L2 = ( + 15833 X 30 - 6333 X 10) -H 20;

= +20 583 = a tensile stress of 20 583 pounds.

For L^Ly Either section e — e or / — / may be used, and each
shows the center of moments to be at Ua. The equation is :

+ 15 833 X 50 - 6 333 X 30 - 6 333 X 10 - L2L3 X 20 = 0;

whence,

L2L3 = +26917.

The center of moments for UJJ2 is at Lt; for U2U3, it is at L2; and
for UaUt, it is at Ly The following equations can now be written :

+ 20 X U,U2 + 20 X 15 833 = 6; whence U1U2 = -15 833;
+ 20 X U2U3 + 40 X 15 833 - 20 X 6 333 = 0; whence U2U'3 = -25 333;
+ 20 X U,U4 + 60 X 15 833 - 40 X 6 333 - 20 X 6333 = 0; whence U,Ut
= -28500.

A diagram of half of the truss should now be made, and all the
stresses placed upon it. The dimensions should also be put upon this
diagram. The student should cultivate this habit, as it shows him
at a glance the general relation of stresses and the general rules of their
variations. Fig. 33 gives the half-truss, together with the stresses
and dimensions. The stresses in the members of the right half of
the truss are the same as those in the corresponding members of the
left half.

BRIDGE ENGINEERING

From inspection of the above diagram, it is seen that the chord
stresses increase from the end toward the center; that the web stresses
decrease from the end toward the center; and that all members slant-
ing the same way as the end-post L0U1 have stresses of that sign,

Fig. 33. Dimensions and Stress Diagram of Half a Six-Panel Warren Through Truss.

while all that slant a different way have an opposite sign. These
relations are true of all trusses with parallel chords and simple systems
of webbings.

34. Position of Live Load for Maximum Positive and Negative
Shears. The dead load, by reason of its nature, is an unchangeable
load. The stresses due to it are the same at any and at all times.

* e +y )

* y >

W Ibs. per lin. Foot.

Fig. 34. Calculating Maximum Positive and Negative Shears in Simple Beam under Live
Load. Conventional Method.

With the live load, the case is different. The live load represents
the movement of traffic upon the bridge. At certain times there may
be none on the bridge, while at other times it may fill the bridge
partially or entirely. In such cases the shears due to live load will vary.

BRIDGE ENGINEERING 29

Conventional Method. It has been found that the maximum
positive shear at any section of a simple beam occurs when the beam is
loaded from that section to the right support, and that the maximum
negative shear occurs at the same section when this beam is loaded from
the section to the left support. This can be proved as follows :

Let a beam be as in Fig. 34, and let a — a be the section under
consideration. The reaction 7?t is due to the load wy on the part y,
and to the load wx on the part x. That is,

Now the shear at the section a — a is Rt — wx; or,

w xl 2

wx I 2 ' \-wx\ + ^ = Fa.a
From inspection of this last equation, it is seen that wx

I
is the amount that is added to the reaction by loading the part x.

Also., that T— is less than unity, is evident. The amount in

brackets in the last equation represents the effect of the loading of
the segment x of the beam. As this is negative and will only reduce

the positive valued term — -~- , it is therefore proved that to get the

largest positive shear the beam should be loaded from the section to
the right support.

From further inspection of the equation, it will be seen that the
term in brackets, which represents the effect of the load on the seg-
ment x on the shear, is always negative; and that the term — — ,which

represents the effect of the load on the segment y on the shear, is
always positive. Hence, to get the largest negative shear at the
section, the load should be on the segment x. That is, the loading
should be from the section to the left support.

BRIDGE ENGINEERING

In a truss, the loads are placed at the panel points; and the
above rules in application, should be formulated as follows:

To get the maximum positive shear at a section or in a panel, load all
panel points to the right of it.

To get the maximum negative shear at a section or in a pan-el, load all
panel points to the left of it.

Example. Determine the maximum positive and the maximum
negative shears in the panels of the 7-panel Pratt truss shown in Fig. 35, the

\/
A

I e 3

Tat £0=

Fig. 35. Calculation of Shears in Panels of 7-Panel Pratt Truss.

live panel load being 40 000 pounds. (It will be noticed that the height of
the truss is not required.)

For maximum + V in 1st panel, load L,, L2, L3, L4, L5 and Le.

+ V " 2d ' < L2, L3, L4, L5) and L6.

+ V " 3d ' ' L3, L4, L6, andL6.

4-V " 4th ' ' L4, L5, and L6.

+ V " 5th ' ' LsandL6

+ V " 6th ' ' L6.

+ V " 7th ' ' no panel points at all.

The reaction produced by each of the loadings is equal to the
shear for that particular case, since the shear at any section or in any
panel is equal to the reaction minus the loads to the left of the section
or panel, and, according to the method of loading, there are no loads
to the left of the section ; therefore the reaction is equal to the shear.

For the first panel, the computation is made as follows, the
center of moments being, of course, at L7:

(fl, = + V'O X 7 X 20 = 40 X 20 + 40 X 2 X 20 + 40 X 3 X 20 + 40 X
4 X 20 + 40 X 5 X 20 + 40 X 6 X 20.

It will be seen that as 20 occurs in all terms of this equation, it
can be factored out by dividing both sides by 20, and the result will
be the same. The equation can now be written:

BRIDGE ENGINEERING 31

+ Ft X 7 = 40 + 40 X 2 + 40 X 3 + 40 X 4 + 40 X 5 + 40 X 6,
and can still be simplified by writing:

+ F, -y- (1+2+3 + 4 + 5+6)=+ 120.00,

which is the form customarily used, the panel length being taken as
a unit of measurement. The other shears are now easily computed
in a similar manner:

+ F2 = ~ (1 + 2 + 3 + 4+ 5) =+85.71
+ F3 = ^-(l+2 + 3 + 4) = +57.14
+ V4 = 1° (1 + 2 + 3) = +34.28
+ VS = 4° (1 + 2) = +17.14
+ F8 = 4° (1) = +5.71
+ F7 =^-(0) = +0

In computing the maximum negative shears, sometimes called
the minimum shears, the reaction is not the same as the shear, as
there are loads to the left of the section, and these must be sub-
tracted. The loadings are :

For maximum — V in 1st panel, load no points.

-V " 2d " " Lt.

-V " '3d " " LiandL2.

-V " 4th " " LIf L2, andL3.

- V " 5th " " Lu L2, L3, and L4.

- V " 6th " " Lu L.,, L3, L4, and L5.

- V " 7th " " L,, L,, L3, L4, L5, and L0.

It is evident that the maximum — V1 is equal to zero, there
being no loads on the span. The maximum negative shear in the
second panel is equal to the reaction produced by loading the panel
point Lv and the load at Lr Thus,
77?, = 40 X 6

- F2 = /?, - load at L,

-5.71

BRIDGE ENGINEERING

The other shears are next computed as follows:

40
_ V3 = ~ (6 + 5) - 2 X 40 = - 17.14 .

- V4 = — (6 + 5 + 4) - 3 X 40 = -34.28

5 + 4 + 3)-4X40 = -57.14
5 + 4 + 3 + 2) - 5 X 40= -85.71 •
- F7 = -- (6 + 5 + 4 + 3 + 2+ 1)-6X 40 = - 120.00

The maximum positive and the maximum negative live-load
shears should now be written side by side, and inspected, in order to
observe any existing relations which might help to lessen the labor
of future computations. The values are given in thousands of pounds
below:

LOCATION

MAX. + LIVE-LOAD SHEAR

MAX. — LIVE-LOAD SHEAR

+ 120.00

- 0.00

+ 85.71

- 5.71

+ 57.14

-17.14

+ 34.28

-34.28

+ 17.14

-57.14

+ 5.71

-85.71

+ 0.00

-120.00

It is at once seen that the negative shears are numerically equal in
value to the positive ones, but that they occur in reverse order. This
simplifies the labor required in the derivation of the negative shears;
for, after computing the maximum positive shears, these may be
written in reverse order, and the negative sign prefixed; the result
will be the maximum negative shears.

The above method for maximum live-load shears is called the
conventional method. It is the one that is almost universally used,
and its use will be continued throughout this text.

Exact Method. On account of the fact that the floor stringers
or joists transfer the loads to the panel points, it would be impossible
to have a full panel live load at one panel point and no load at the
panel point ahead or behind. In order to have a full panel load at
one point, the stringers in the panels, on both sides of the point must

BRIDGE ENGINEERING

be full-loaded, and this would give a load at the panel point ahead,
provided the bridge was fully loaded up to and not beyond the panel
point ahead, equal in value to one-half of a full panel load (see Fig.

Unifoi

Lyfre

7///////Y/

Fig. 36. Illustrating "Exact" Method of Calculating Live-Load Shears in Panels.

36). The uniform live load, in order to produce full panel loads at
L2, L3 and Lv will also produce one-half a panel load at Lv

By the methods of differential calculus, it can be proved that
the true maximum positive live-load shear occurs in a panel when the

m. panels -

Fig. 37. Calculating Maximum Positive Live-Load Shear in Panel.

live load extends from the panel point to the right into that panel
an amount (see Fig. 37) equal to

in which,

n •= Number of the panel point to the left of the panel under considera-
tion, counting from the right;
m = Total number of panels in the bridge;
p = Panel length.

Let the truss of Fig. 35 be considered, the live load being 2 000

34 BRIDGE ENGINEERING

pounds per linear foot of truss, and let it be required to determine the
true maximum positive live-load shear in the 5th panel from the right

end.

7-1

X 20 = 13.333 feet.

There will now be (4 X 20 + 13.333) X 2 000 = 186 666 pounds on
the truss; and the left reaction will be { 186 666 X (4 X 20+13.333)
-i- 2} -=-140 = 62200 pounds. From this must be subtracted the
amount of the load on the 13.333 feet, which is transferred to the
point Lr This is equal to the reaction of a beam of a span equal to
the panel length, loaded for a distance of 13.333 feet from the right
support with a uniform load of 2 000 pounds per linear foot. This

1 ^ ^^^
amounts to (13.333 X 2 OOOX i°|°°) - 20 - 8 890 pounds. The

true shear is now:

+ Va = +62 200 - 8 890 = +53 310 pounds.

The + F3, as computed by the conventional method, was + 57 140,
making a difference of 3 730 pounds between the two. If the true
shears were computed and compared with the conventional shears,
it would be found that the Vl would be the same, and that the
remainder of the conventional shears would be greater than the
corresponding true shears. The difference between any two corre-
sponding shears would increase from the left to the right end; that
is, the difference between the conventional and exact shears would
be greatest in the panel Z-5Z>6.

To get the maximum negative shear in any panel, load from the
left support and out into the panel under consideration an amount
p — x, and proceed in a manner similar to that above described.

As this method of exact or true shears is seldom employed,
problems illustrating its application will here be omitted.

35. Position of Live Load for Maximum Moments. In order
to obtain the maximum moment at any point, the live load must cover
the entire bridge. Let the beam of Fig. 34 be considered, and let
it be required to obtain the maximum moment at the section a — a.
The reaction, as before computed, is:

BRIDGE ENGINEERING 35

all terms of which are positive. The moment at the section is :

-jr

M = Rl x x - wx -jr-;

and substituting for R^ its value,

wx2 ( x \ wx2 wy"x

-r(^ + y)-^r h^r;
W3*(JL + JL_ l\ + ™y2 .

X \2l + I ~2j + 21

But y = I — or; therefore,

The first term of this equation represents the effect of the load
on the portion x, and the second term represents the effect of the
load on the portion y. The value of M will always be positive. The
quantity x varies between 0 and /. When x = 0, M is equal
to 0. When x = I, the moment is equal to + wy2x + 2. For
all values of x between 0 and /, the first term is positive; and the
second term being positive in all cases, it is therefore proved that for
maximum live-load moments at any point, the entire span should
be loaded, as loads on both segments add positive values to the
moment value.

36. Warren Truss under Live Load. In order to analyze a
truss intelligently, it is necessary to know its physical structure;
that is, it must be known what character of stress can be withstood
by the different members. The top chords of all trusses are built
to take only compression, and the bottom chords are built to take
only tension; while some web members of some trusses are built for
tension stresses, some for compression stresses, and some for both.
The characteristic of the Warren truss is that the web members are
built so as to be able to withstand either tension or compression.

Let it be required to determine the live-load stresses in the
Warren truss of Fig. 32. Let the live load per square foot of roadway,
which is assumed to be 15 feet wide, be 100 pounds. The live panel
load is then 100 X 15 X 20 -i- 2 = 15 000 pounds, and the live-load
reaction under full load is 2^ X 15 000 = 37 500 pounds.

36 BRIDGE ENGINEERING

As the live load must cover the entire bridge to give maximum
moments — and therefore maximum chord stresses, as the <:hord stress
is equal to the moment divided by the height of the truss — a simple
method for the determination of live-load chord stresses presents
itself. The live load and the dead load being applied at the same
points, and being different in intensity, the stresses produced will
be proportional to the panel loads. The maximum live-load chord
stresses (see Fig. 33) will then be equal to the dead-load chord stresses
multiplied by 15 000 -+ 6 333 = 2.371, and they are as follows:

Lot/! = -2.371 X 17700 = -42000
U1U2 = -2.371 X 15833 = -37530
U2U3 = -2.371 X 25333 = -60050
U3U4 = -2.371 X 28 500 = -67 600
L0L, = +2.371 X 7917= +18770
L,L2 = +2.371 X 20583 = +48800
L2L3 = +2.371 X 26917 = +63850

The next step in order is to determine the maximum positive
shears, and from these write the maximum negative shears. This
is done as follows:

+ Live-Load V — Live-Load V

(1 + 2 + 3 + 4+ 5) = +37 500 0

V2= (i + 2 + 3 + 4) =+25000 -2500

F 15_000 + 2 + 3) =+15000 -7500

6
= 15000 + 2) = + 7 500 - 15 000

= + 2500 -25000

y6= +0 -37500

The stresses produced by the positive shears are called the
maximum live-load stresses, and are :

+ L0Ul cos <£ + 37 500 = 0 .-. L0Ul = -37 500 X 1.12 = -42 000

- U,L, cos 0 + 37 500 = 0 .'. U,L, = +37 500 X 1 12 = +42 000
+ L1U, cos 0 + 25 000 = 0 .-. L,U2 = -25 000 X 1.12 = -28 000

- U2L2 cos <j> + 25 000 = 0 .-. U2L2 = +25 000 X 1.12 = +28 000
+ L2C73cos <£ + 15000 = 0 .-. L2C73 = -15000 X 1.12 = -16800

- U3L3 cos <j) + 15 000 = 0 .-. U3L3 = + 15 000 X 1.12 = +16 800

The stresses produced by the negative shears are called the
minimum live-load stresses, and are:

BRIDGE ENGINEERING

+ L0C7, cos <;

!> + 0 =0

.-. L0Ul = 0

— UlLl cos (j

!> + 0 =0

.-. t/'.L, = 0

+ L! f/2 COS <;

& - 2 500 = 0

.-. L^a = +2500 X 1.12 =

+ 2800

- [72L2 cos <;

& - 2 500 = 0

.-. U2L2 = -2500 X 1.12 =

-2800

+ L2t/3 cos c

b - 7 500 = 0

.-. L2U3 = +7500 X 1.12 =

+ 8400

- U3L3 cos ^

6 - 7 500 = 0

.-. U3L3 = - 7 500 X 1.12 =

-8400

These stresses, together with the dead-load stresses, should
now be placed together as a half-diagram, as is done in Fig. 38, the
stresses being rounded off to the nearest ten pounds and then ex-
pressed in thousands of pounds. No minimum live-load stress is
given for the chords, as this will evidently be zero in all cases, since
no position of the live load will cause a reversal of stress. It will be
seen that the stresses produced by the negative shears are of opposite

dl - 1583
U,MU - 37.53

— 28 50

Pig, 38. Dimension and Stress Diagram of Warren Half-Truss under Live Load.

sign from the stress produced by the dead load, and these tend to
decrease the dead-load stress by that amount; and in some cases
(see L2 U3 and U3 L3, Fig. 38) it will be so large as to overcome the
dead-load stress and therefore change the total stress from one kind
to another. Do not forget, in considering any combination of the
above stresses, that the dead load occurs with either the maximum
or the minimum live load, but not with both at the same time.

37. Counters. By reference to U3L3 (Fig. 38), it is seen that
when the live load is on the panel points Ll and L2 the total stress in
the member is + 3.54 + (— 8.40) = —4.86, a compressive stress of
4 860 pounds ; whereas, under dead load alone, the stress was + 3.54,
a tensile stress of 3 540 pounds. If the member U3 L3 had been built
of long, thin bars which could take only tension, and which con-
sequently would have doubled up under the resultant compression

BRIDGE ENGINEERING

brought upon them by the combined
action of the dead and minimum live-
load stresses, then this member could
not be used in this case, but some
other arrangement would be necessary
in order to insure the stability of the
truss.

In the Warren truss, no special ar-
rangement is necessary, as the web
members are built so as to take either
tension or compression; but with the
Pratt and Howe trusses some special
arrangement is necessary, as the diag-
onals are built to take one kind of
stress only. The case of the Pratt will
be considered first.

The Pratt truss has the diagonals
made of long bars which take tension
only, and the intermediate posts are
constructed so as to be able to take
compression only. It is not necessary
to consider the intermediate posts, for
the action of the web members is such
that the resulting stresses are always
compressive.

Let the 13-panel Pratt truss of
Fig. 39 be considered. The panel
length is 18 feet, the height 25 feet, the
dead panel load 22 000 pounds, and
the live panel load 58 500 pounds.
The secant is (182 + 252 )* -=- 25 =
1.231. The dead-load shears and the
maximum and minimum live-load
shears are placed directly below their
respective panels. Only those mem-
bers are shown full-lined in Fig. 39
which act under the dead load. Note
that the dead -load shears in the center

BRIDGE ENGINEERING 39

panel being zero, "the dead-load stress in the diagonals in the center
panel would be 0 X sec <f> = 0.

In the first four panels from either end, the live-load shear,
which is of a different sign from that of the dead-load shear, is
smaller than the dead -load shear, and therefore will not cause a
reversal of stress in the member in that panel. For example, take
U3L4; then, for dead-load stress,

-C/3L4 cos 0 + 66.0 = 0 .:U,La = +66.0 X 1.231 = +81.20

For live-load stress,

-£73L4 cos <f> - 27.0 = 0 .-.UaL, = -27.0 X 1.231 = -33.25

The total stress = + 81.20 - 33.25 = + 47.95, which is still
tension.

Considering Lg U10, the stress equations are :
For dead-load stress,

+ L9C710 cos (j) - 66.0 = 0 .'. L9C710 = +81.20

For live-load stress,

+ Lu[/10cos <j> + 27.0 = 0 •••L0UM = -33.25

The total stress, as before, is + 47.95, or a tension of 47 950 pounds.
An inspection of the center panel and the two panels on each
side of it, shows that the live-load shear is of a different sign from
the dead-load shear, and is also greater in value than the dead-load
shear. If the members shown in Fig. 39 were the only ones in the
panels, then the dead-load stresses would be:

- UtLs cos $ + 44.0 = 0 U4LS = +54.20

- C75L6 cos <j> + 22.0 = 0 f/5L6 = +27.10
+ L7[/S cos 0 - 22.0 = 0 L7US = +27.10
+ L8i7,eofl i - 44.0 = 0 L8U9 = +54.20

and the live-load stresses caused by the shear of opposite sign from
that of the dead-load shear, are:

-t/4L5cos 0 - 45.0 = 0 U4L, = -55.40

- U6LK cos 0 - 67.5 = 0 t/8L6 = -83.10
+ L7Uscos <f> + 67.5 = 0 L7US = -83.10
+ Ls[/9cos <f> + 45.0 = 0 LsUa = -55.40

As no diagonal acts under dead load in the center panel, we may
assume that U6 L7 acts under live load. The stresses which occur

in this are:

+ U,,L7 cos $ + 94.5 = 0 UKL7 = +116.30

- UbL7 cos 0 - 94.5 = 0 C78L7 = - 116.30

40 BRIDGE ENGINEERING

The above shows that compressive stresses will occur in the
diagonals which were built for tension only. These stresses are :

U4L5 = +54.20 - 55.40= - 1 200 pounds

U&L6 = +27.10 - 83.10= -56000 "

L7US = +27.10 - 83.10= -56000 "

LsUg = +54.20 - 55.40 = - 1 200 "

U6L7 = 0 - 116.30 = -116300 "

If some provision were not made for these stresses, they would cause
the members to crumple up and the truss to fail. In order to allow
for them, diagonals are placed in the panels, as shown by the dotted
and dashed lines. These members will take up the above stress;
and moreover, as they slope the opposite way from the main members,
they will be in tension.

In order to prove this, assume LbU6 to act when the live load is on
points L5, L4, Ly L2, and Lv Now, U5L6 will not be regarded, as its
stress will be zero. Then the stresses will be :
For dead load,

+ L5[76 cos <£ + 22.0 = 0 L&U6 = -27.10.

For live load,

+ Lst/6 cos 0 - 67.5 = 0 L5C76 = +83 10;

and the total stress in L5*76 will be - 27.10 +• 83.10 = + 56.00.

In a similar manner, the stresses in the other members are:LJJ5
= +1.2; L6U7=+ 116.30; U7LS =+ 56.00; and UgL9= + 1.2.
These diagonals are called counters or counter-bracing.

From a consideration of the foregoing, it is evident that:

(a) // the live-load shear in any panel is of opposite sign and greater
than the dead-load shear in the same panel, then a counter is required.

(5) The stress in a counter is equal to the algebraic sum of the dead-load
shear and the live-load shear of opposite sign times the secant of the angle it
makes with the vertical.

This is true for any truss with horizontal chords and a simple system
of webbing with diagonals and verticals.

38. Maximum and Minimum Stresses. Some specifications
require the member to be designed for the maximum stress, while
others take into account the range of stress. In this latter case it
is necessary to determine the minimum as well as the maximum stress.
Except where a reversal of stress occurs — and this does not happen
in trusses with horizontal chords — few specifications require any

BRIDGE ENGINEERING

but the maximum stresses to be com-
puted. For that reason, little space
will here be devoted to the minimum
stresses, their computation in succeed-
ing articles being thought to illustrate
them sufficiently.

(a) The maximum stress in a member
is equal to the sum of the dead-load stress
and the live-load stress of the same sign.

(b) The minimum stress is equal to
the sum of the dead-load stress and the live-
load stress of the opposite sign, or to the dead-
load stress alone, according to which gives
the smallest value algebraically. By this
latter statement it should be seen that if
the maximum stress is —58.60, then 0 or
+ 18.00 would be smaller than -3.00.

(c) It is evident that the minimum
in all counters and in all main members
in panels where counters are employed will
be zero, for when the counter is acting the
main member is not, and therefore its stress
is zero. The reverse is also true.

(rf) An exception to a is seen in the
case of the counters. Here it is evident
that the maximum stress is equal to the
algebraic sum of the dead-load shear and
the live-load shear of opposite sign times
the secant of the angle which the counter
makes with the vertical.

While it is true that in trusses with
horizontal chords the loading for maxi-
mum shears will give the maximum
live-load stress to be added to the
dead load for the maximum stress, it is
not always true that the loading for
minimum live-load shears will give the
stress to add to the dead-load stress to
get the minimum stress. However, the
loading for the minimum live-load
shears will give the live-load stress to
be added to the dead-load stress for
the minimum stress, except in the case

BRIDGE ENGINEERING

of verticals placed between panels each of which contains counters,
and in that case it may or may not do so. In such cases a loading
must be assumed — preferably the one for minimum shears — and
the shears in the panels on each side of the vertical must be com-
puted for the loading assumed.
If the resultant shear is the same
sign as the live load, then the
main diagonal acts; if it is of
different sign, then the counter
acts.

As an example, let it be re-
quired to find the minimum stress
in the vertical U5L5 of the truss
of Figs. 39 and 40. It is assumed
that the loading for minimum
shears will give the result. The
section a— a is then passed, and
the live load placed on L5 and all
points to the left. The shears will
then be as shown in Fig. 41. To
obtain the shear in the panel LtL5,

d.l. 22.0
11 58.5

dlv

4- 44.0

.+ 22.0

• 9.0

- 67.5

Totalv

—

Fig. 41. Stress Diagram for Vertical in
Truss of Fig. 40.

under this loading, it must be re-
membered that a load is at L5; and so the shear is the shear in the
panel L5L6 with the panel load at L5 added, or, —67.5 + 58.5 =
— 9.0. The diagonals now act as indicated by Fig. 41, and the total
stress in UbL5 is determined by passing a circular section around t/5,
and it is :

-Load at U, - U5L5 = 0.

As there is no load at U5, the stress in U5L5 is = 0. The same result
will occur if points Z/4 or L3 and to the left are loaded; but if points
L2 and to the left are loaded, the members U^L5 and UbL6 will act,
and the stress in U5L5 will then be equal to the shear on the section
a — a. The stresses are: Dead-load, — 22.0; and live-load, +
13.5, which gives a total of — 8.5; but as the maximum stress is
-22.0 - 126.0 = - 148.0, it is evident that 0 and not -8.5 is the
minimum.

The computation of the maximum stress is as follows:

Load points L6 and to the right. The shear on a — a is, for

BRIDGE ENGINEERING

dead load, +22.0; and for + live load, +126.0; and the equations of
the stresses are:

+ 22.0 + C/5L5 = 0 U5LS = - 22.0

+ 126.0 + C75L6 = 0 UbLs = - 126.0

Max. = -1480

TRUSSES UNDER DEAD AND LIVE LOADS

39. The Pratt Truss. The Pratt truss is used to perhaps a
greater extent than any other form; probably 90 per cent of all simple
truss spans are of this kind.

Let it be desired to determine the stresses in the 8-panel 200-foot
single-track span shown in Fig. 42, the height being 30 feet, the dead
panel load being 30 000 pounds, and the live panel load 62 400

pounds. The secant is ,("25* +~302) * -i- 30 = 1.302, and the cosine

is 0.7685. The dead-load reaction is 3£ X 30.0 = 105.0.
The dead-load shears are :

F, = +105.0
V, = + 75.0
V3 = + 45.0
V4 = + 15.0
Vt= - 15.0

The dead-load chord stresses may be tabulated as follows (see
Articles 27 and 29):

Dead- Load Chord Stresses

MEMBER

SEC-
TION

CEN-
TER OF
MO-
MENTS

STRESS EQUATION

STRESS

Z/0L, = L1L2

a — a

+ 105.0 X 25 -

Z/,L2 X 30 = 0

+ 87.5

L2L3

b-b

+ 105.0 X 50 -

30.0 X 25 - L2L3 X

30 = 0

+ 150.0

L3L4

c — c

+ 105.0 X 75 -

30.0 (25 + 50) - L3L4

X 30 = 0

+ 187.5

U,U2

a — a

+ 105.0 X 75 -

30.0 X 25 + C7,[/2 X

30 = 0

-150.0

UMS

b-b

+ 105.0 X 75 - 30.0 (25 + 50) +

U3Ut

c — c

U2U3 X 30 = 0
+ 105.6 X 100 - 30.0 (25 + 50 + 75)

-187.5

+ U,U4 X 30

= 0

-200.0

In determining dead-load stresses in web members, it is cus-
tomary to assume one-third of the dead panel loads as applied at the

BRIDGE ENGINEERING

upper chord points. This, as will be seen, makes no difference in the
stresses in the chords or in the diagonals, the stresses in the verticals
only being different from what is the case when all the dead load is
taken on the lower chord.

The stresses in the diagonals (see Articles 27,28, and 30) are:

Dead = Load Stresses in Diagonals

MEM-
BER

SEC-
TION

SHEAR ON
SECTION

STRESS

EQTT

ATION

STRESS

LUj

o — o

+ 105.0

+ LnU

i X

0.7685 =

-136.70

[7,1/2

a — a

+ 75.0

- U}L

n X

0.7685 =

+ 97.60

U L,

b-b

+ 45.0

- U2L

t X

0.7685 =

+ 58.60

U3L4

c — c

+ 15.0

- U3L

4 X

0.7685 =

+ 19.53

In determining the stresses in the verticals, it is to be remem-
bered that one-third the dead panel load (or 10.0) is at the panel

o a 'i b • 's'c 3'

Fig. 42. Outline Diagram of 8-Panel Single-Track Pratt Truss Span.

points of the upper chord, and two-thirds (or 20.0) is at the lower
chord. The stress in the hip vertical UlLl is determined by passing
a circular section around L,. It is solved thus :

-20.0 + U^L, = 0 U^ = +20.0
In a similar manner the stress in U4L4 is found to be :

-10 - t/4L4' = 0 C74L4 = -10.0

In order to find the stress in the remaining verticals, sections 1 — 1
and 2 — 2 are passed, cutting them, and the shears on these sections
computed. The shears are:

Ft- , - + 105.0 - 2 X 20 - 1 X 10 = + 55.0
F2_2 = + 105.0 -3X20-2X10= +25.0

The stress equations are written, remembering that as the verticals

BRIDGE ENGINEERING

make an angle of zero with the vertical, their cosine is equal to unity.

These equations are:

+ U2L2 + 55.0 = 0 U2L2 = -55.0

+ U3L3 + 25.0 = 0 U3L3 = -25.0

The live-load chord stresses will be proportional to the dead-
load chord stresses, as both loads cover the entire truss in exactly
the same manner. The ratio of the panel loads by which the dead-
load chord stresses are multiplied in order to get the live-load chord

stresses, is ^7^ = 2.08, and the chord stresses are :
oO 000

LaL, = L,L2 = + 87.5 X 2.08 = +182.0

L2L3 = +150.0 X 2.08 = +312.0

L3L4 = +187.5 X 2.08 = +390.0

VJJ* = -150.0 X 2.08 = 312.0

U2Ua = -1875 X 2.08 = -390.0

U3U4 = -200.0 X 2.08 = -416.0

As the entire bridge is to be loaded to get the maximum stress in
L0U19 it is therefore equal to the
dead-load stress times the above
ratio; or LJJl = -136.70 X 2.08
- 284.20.

The maximum live-load stress
in U^ is determined by passing
a circular section around Lv and
is solved (see Fig. 43) from the
equation :

+ U,L - 62.4 = 0 /. U.L, = +62.4
For UJjv the section a — a is
passed, and the points L2 and to
the right are loaded. The maxi- .

mum shear is:

+ V2 = + -^ (1+2 + 3 + 4 + 5 + 6)= + 163.8;

and the stress equation is :

+ 163.8 - C/iL2 X 0.7685 - 0 .'. E7,L2 = +213.2.

In a similar manner, pass section 6 — 6, and load points L3 and to
the right, and the shear and the stress equations for U2La are:

co 4

+ F3= +—— (1 + 2 + 3 + 4 + 5)= +117.0
8

+ 117.0 - U2L3 X 0.7685 = 0 .'. U2L3 = +152.4

BRIDGE ENGINEERING

For UJjv the section c — c is passed, and the panel points to the
right are loaded. The shear and stress equations are :

+ V4 = + ~ (1 + 2 + 3 + 4)=+ 78.0

+ 78.0 - C73L4 X 0.7685 = 0 /. U3Lt = +101.6

For the maximum stresses in the verticals, sections 1 — 1, 2 — 2,
and 3 — 3 are passed, and in each case the panel points to the left
of these loaded. The shears are
62.4

(1+2 + 3 + 4 + 5)= +117.0

62.4

V2-, = —^- (1 + 2 + 3 + 4) = +78.0
o

V,- 3= -^(1 + 2 + 3) = +46.8

The stress equations for U^L2 and U3L3 are simple, as only three
members are cut. They are :

+ 117.0 + U2L2 = 0 .-. £/2L2 = - 117.0

+ 78.0 + U3L3 = 0 /. U3L3 = - 78.0

It is seen that the section 3—3 cuts the member LJJ5, and
therefore the stress in this must

\ L,

be determined before the stress
equation can be written, as its ver-
tical component will enter into it.
However, by comparing the dead-
load shear in that panel, which is
— 15.0, and the live-load shear V3_3,
which is +46.8, it is seen that the
resultant shear is + ; and, as this
is of opposite sign from the dead-
load shear, a counter is required
and is acting. The stress in £75L4 is zero, and the diagonals act as
in Fig. 44, the section 3 — 3 then cutting three members. The
stress equation is + 46.8 + UtLt = 0, from which U,L4 - - 46.8.
Care should be taken not to add to this -46.8 the -10.0
derived as dead-load stress on page 44, in order to get the maximum
stress, as the —10.0 previously derived was the dead-load stress
in U4L4 when U3L4 and LJJ5 were acting. The dead-load stress which
goes with the live-load stress of —46.8 acts simultaneously with it,

Pig. 44. Calculation of Stress in Diag-
onal of Span of Fig. 42.

BRIDGE ENGINEERING

and is the dead-load stress in UtL4 when the members U,L4 and U4L5
are acting as in Fig. 44. The dead-load shear on the section 3 — 3
would then be the left reaction minus the loads at points Uv Uv U3,
LI} L2, L3and Z/4; or,

F3_3 = +105.0 -3X10-4X20= -5.0;
and

-5.0 + UtL4 = 0 .-.U4L4 = +5.0.

Remember that this + 5.0 can act only when the live load tends to
produce a stress of —46.8; and thus the total stress in U4Lt with live
load in that position is -46.8 + 5.0 = -43.8, while with dead load
only in the truss the stress is — 10.0.

The dead-load shears and the maximum + and — live-load
shears should now be written for inspection, in order to investigate
for counters and then for the minimum stresses. Those whose
derivation has hot been given should be easily computed by the stu-
dent at this time. The shears are

DEAD- LOAD

+ LIVE- LOAD

— LlVE-LOAD

F, + 105.0
V2 + 75.0
V3 + 45.0
Vt + 15.0

+ 218.4
+ 163.8
+ 117.0
+ 78.0

- 7.8
-23.4
-46.8

From a study of these it is seen that a counter is required in the
4th panel according to rule a, Article 37; and according to rule b of
the same article, the maximum stress is ( — 46.8 + 15.0) X 1.302 =
+ 41.4, the minimum stress for it and also U3L4 being zero according
to the same article. A counter is also required in panel 5, as the
truss is symmetrical.

The minimum live-load stress in UlLl is zero, and occurs when
no live load is at the point Lr

The minimum live-load stresses in the diagonals UJL2 and U2L3
occur when the truss is loaded successively to the left of the sections
a — a and b — b, in which case the shears are —7.8 and —23.4
respectively. The stress equations are

- U,L2 - 7.8 X 0.7685 = 0 .'. f/,L2 = - 10.16

-U2L3 - 23.4 X 0.7685 = 0 .'. U2L3 = -29.15

The minimum live-load stress in UJL2 is obtained by passing

BRIDGE ENGINEERING

section 1 — 1 and loading the panel points to the left. The live-load
shear is the same at this section as it is at the section 6 — 6 — namely,
— 23.4. The stress equation is

+ U2L2 - 23.4 = 0 .'. U2L2 = +23.4

To determine the minimum live-load stress in U3L3, proceed as
indicated on page 42. By loading points L3 and to the left, the live-
load shear in the 4th panel will be —46.8, and in the 3d panel under
this same loading it will be -46.8 + 62.4 = + 15.6. The sign of the
total shear in the two adjacent panels, and the members acting, are
shown in Fig. 45. The stress in U3L3 is then determined by using a

4- 10
/ 4U5

eo.o

6E4-

eo.o

©B4

EO.O

L_5

eo.o

d.lv

+ 4-5.O

+ I5.O

l.l.v.

+ 1 5.6

— 4-6.6

Totalv

—

d.l.v.

+ 15.0

- 15.0

l.l.v.

- 15.6

-78.0

Total v

—

Fig. 45. Fig. 46.

Stress Diagrams for Verticals in Span of Fig. 42.

circular section around U3, and is simply the dead load at U3, there
being no live-load stress in the member when the bridge is loaded as
has been done.

In finding the minimum live-load stress and also the minimum
stress in UtL4, the same method of procedure will be followed. Let
LI and to the left be loaded. Then the shear in the 5th panel is
— 78.0, and under this same loading the shear in the 4th panel is
-78.0 + 62.4 = -15.6. The sign of the total shear in each of the
adjacent panels is given in Fig. 46. It should be remembered that
a resultant shear with the same sign as the dead -load shear causes
the main diagonal to act, while a resultant shear of opposite sign to
that of the dead-load shear causes the counter to act. The members

SC «

Ill
ll

BRIDGE ENGINEERING

acting are shown, and a section 4 — 4 can be passed. The dead-load
shear at this section is 105 - 3 X 20 - 4 X 10 = +5.0; and
accordingly,

- U4L4 + 5.0 = 0.
Therefore,

UJjt = +5.0 = Dead-load stress in this case.
The live-load stress which acts at the same time is:

-U4L4 - 15.6 = 0 .-. U4L4 = -15.6,

the term — 15.6 representing the live-load shear on the section 4 — 4.
This is not the minimum stress, as will next be shown, but it illus-
trates the fact that the loading for minimum live-load shears does
not always give the minimum
live-load stress.

By loading Lv the live-load
shear in the second panel, and
likewise all others from this to
the right support, will be —7.8.
The total shears, together with
their sign, and also the members
they cause to act, are given in
Fig. 47. The minimum live-load
stress in U4L4 is found to be zero,
and the dead-load stress is — 10,
as is derived by passing a circular
section around U4, the equation

£0

LB
EO

d.lv.

+ 15.0

-15.0

l.l.v

- 7.8

Total v

Fig. 47. Stress Diagram for Vertical in
Span of Fig. 42.

being as follows:

-Live load at U4 - U4L4 = 0 .'. U4L4 = 0 for live load.
-Dead load at U4- U4L4 = 0 .'. U4L4 = -10.0 for dead load.

A diagram of half the truss should now be made, and all dead
and live load stresses placed upon it, and these should be combined so
as to form the maximum and the minimum stresses. Such a dia-
gram, together with all stresses, is given in Fig. 48.

The stresses are written in the following order: Dead load,
maximum live load, minimum live load, the maximum, and the
minimum. In the chord and end-post stresses, there is no minimum
live-load stress recorded, it being zero. Where pairs of stresses occur
simultaneously, a bent arrow connects them.

40. The Howe Truss. The physical make-up of the Howe truss

BRIDGE ENGINEERING

differs from that of
the Pratt in that
the diagonals are
made to stand com-
pression only, and
the verticals can
stand tension only.
In the Pratt truss
it was found that
none of the inter-
mediate posts could
be brought into
tension by any
loading. In the
Howe truss it will
be found that none
of the verticals can
be brought into
compression.

Let it be required
to determine the
stresses in a Howe
truss of the same
span, height, and
loading as the Pratt
truss of Article 39.
An outline diagram
is given in Fig. 49.

The dead-load
shears and the
maximum and min-
imum live-load
shears will be the
same as for the
Pratt truss, and
they are : .

BRIDGE ENGINEERING

DEAD-LOAD V

+ LIVE-LOAD V

-LIVE-LOAD V

V, + 105.0

+ 218.4

- 0

V, + 75.0

+ 163.8

- 7.8

V, + 45.0

+ 117.0

-23.4

V4 + 15.0

+ 78.0

-46.8

K5 - 15.0

+ 46.8

-78.0

Inspection of these shows that counters are required in the 4th
and 5th panels (see Article 37).

The dead-load lower chord stresses will be computed by the

Fig. 49. Outline Diagram of 8-Panel Single-Track Howe Truss Span.

tangent method (see Article 31), the section being y — ylf etc. The
tangent of <f> is 25 -+ 30 = 0.8333. The stresses may be conven-
iently tabulated as follows:

Dead-Load Chord Stresses (Lower Chord)

MEM-
BER

SECTION

STRESS EQUATION

STRESS

_ V|

- 105.0 X 0.8333 + L0L, = 0

+ 87.5

L,L?

- 2/2

-(105.0 + 75.0) 0.8333 + L,L2 = 0

+ 150.0

- 2/3

-(105.0 + 75.0 + 45) 0.8333 + L2L, = 0

+ 187.5

L3L4

- 2/4

-(105.0 + 75.0 +45.0 +15.0) 0.8333 + L3L4 = 0

+ 200.0

A simple method for the determination of the upper chord
stresses, is to pass a section and to equate the sum of the horizontal
forces. Pass section 1 — 1. The only horizontal forces are the
stresses in L0Ll and UJJ^; and as these are parallel, one must be equal
and opposite to the other. In a like manner the stresses in the other
sections of the top chord are found. The stresses are :
U,U2 = -L0L, = -(+ 87.5) = - 87.5
U2U3 = -L,L2 = -( + 150.6) = -150.0
U3U4 = -L2L3 = -( + 187.5) = -187.5

52 BRIDGE ENGINEERING

A consideration of the Pratt truss shows that this method can be
applied to it in determining the chord stresses.

As it is known that the diagonal web members are in compression
under the dead load which produces a positive shear in the left half
of the truss, it is evident that positive live-load shears will produce
compressive stresses, and negative live-load shears tensile stresses,
in the diagonals in the left half of the truss. Also, from Article 30,
the stress in a diagonal is V sec <£. The stresses can now be written
directly without the aid of the stress equation :

L0Ul = -105.0 X 1.302 = -136.70

L,C72 = - 75.0 X 1.302 = - 97.60

L2U3 = - 45.0 X 1.302 = - 58.60

LaUt = - 15.0 X 1.302 = - 19.53

Likewise the stresses in the verticals can be written directly, remem-
bering that here the secant is unity, and that the shear at the section
cutting the member is to be used, not forgetting that ^ of the dead
panel load is applied at the top panel points. The shears and
stresses are:

V. - , = +105.0 - 10 = +95.0 Z7,L, = +95.0

Fo_ 2 = +105.0 - 20 - 2 X 10 = +65.0 U.2L2 = +65.0

V3_ 3 = +105.0 -2X20-3X10= +35.0 U3LX = +35.0

The member U^L4 cannot be easily determined by passing a
section 4—4, for this cuts four members. It is determined by passing
a circular section about the point L4, the equation being + U4L4
- 20.0 = 0, from which C74L4 = + 20.0, which is equal to the dead
panel load at the point Z»4.

The live-load chord stresses are determined by multiplying the
dead-load chord stresses by the ratio of the live to the dead loads.
This has been found to be equal to 12.08. The live-load chord stresses
are found to be :

L0L, = +182.0 U,U2 = -182.0

L,L2 = +312.0 U2U3 = -312.0

L2L3 = +390.0 U3U4 = -390.0
L3L< = +416.0

As the character of the stresses which can be taken by the
diagonals and the verticals is known, the maximum and minimum
live-load stresses can be written without first writing the stress
equations. The maximum live-load stresses are:

BRIDGE ENGINEERING

L0Ul = -218.4 X 1.302 = -284.36 tf.L, = +218.4

L1U2 = -163.8 X 1.302 = -213.27 U2L2 = +163.8

L2U3 = -117.0 X 1.302 = -152.33 U3L3 = +117.0

L3U4 = - 78.0 X 1.302 = -101.56 f/4L4 = + 78.0

It should be noted that when Z/4 and all panel points to the right
are loaded, the shears and the members acting are as shown in Fig.
50. The dead-load shear on the section 4 — 4 is +15.0, less the load
at U4, or + 15.0 — 10.0 = +5.0; and the equation of stress is — U4L4
+ 5.0 = 0, from which J74L4 = +5.0. Thus it is seen that in this

EO.O

eo.p

d.l.v.

+ 15.0

-15.0

U.v.

-- 7.6

+ 15-6

Total v

d.l.v.

^-5.0

+ 15.0

U.v

— 7.8

Total v

Fig. 50. Fig. 51.

Stress Diagrams for Members of Howe Truss Span of Fig. 49.

case the dead-load stress is +5.0 when the live-load stress is +78.0.
The maximum stresses in the counters (see Article 37) are:

(-46.8 + 15.0) 1.302 = -41.4.
The minimum live-load stresses are now written as follows:

L,C72 = + 7.8 X 1.302 = +10.16
L2U3 = +23.4 X 1.302 = +29.15
L3U4 = 0
U3L< = 0

tf,L, - 0

U2L2 =-- -7.8

U3L3 | See discussion

f/4L4 \ following.

If live panel loads were placed at points Lv Lv and L3 the live-
load shear in c — c would be —46.8; and the dead-load shear being
+ 15, the counter would act, and the stress in UgLa would be tensile
and equal to the sum of the dead and live panel loads which are at
its lower end Lr If points L} and Z/2 had live panel loads on them,

54 BRIDGE ENGINEERING

the resultant shear in c - c would be -23.4 + 15.0 - -8.4; the
counter would act, and the stress .in UtL4 would be tensile and equal
to the dead panel load which is at Ly There being no live panel load
at I/g, the live-load stress in U^ would be zero under this loading.
If a live panel load be placed at L^ only, then the shears and the mem-
bers acting will be as shown in Fig. 51, and V3 3 for dead load =
+ 45.0 - the load at f/3, or - 45 - 10 - +35.0. The V3 3 for
live load = —7.8, and the stress equation — U3L3 — 7.8 = 0, from
which U3L3 = — 7.8. So this live-load compression stress of 7 800
pounds occurs at the same time as the dead-load tensile stress
of 45 000 pounds.

By loading various groups of panel points in succession and
determining the resulting live-load stresses in UtL4, it will be found
that under no loading can a negative live-load stress be produced.
The minimum live-load stress is therefore zero, and occurs when
there is no live load on the bridge.

The stresses should now be placed on an outline diagram similar
to that of Fig. 48, and the stresses in corresponding members com-
pared with those in that figure. This is left for the student.

41. Bowstring and Parabolic Trusses. A bowstring truss is
shown in Fig. 13, the full lines representing the main members, which
are the members under stress by the dead load. The dotted members
represent counters which may be stressed by the action of the live
load.

As before mentioned, the stresses in the chords and also in the
webbing are quite uniform. When the end supports and the panel
points lie on the arc of a certain curve, called a parabola, then, under
full load, the stresses in all panels of the lower chord are equal; the
stress in all verticals is tensile and is equal to the panel load at the
lower end ; and the stress in all diagonals is zero. Under partial load,
the stresses in the webbing are exceedingly small, and the chord
stresses remain almost equal.

If it is desired to have a parabolic truss, first decide upon the
length of span, the number of panels, and the height at
the center. The height of any vertical post is given by the
formula :

, rr

h = H

BRIDGE ENGINEERING

in which,

H = Approximate height at center;
d = Distance of vertical post from center;
I = Span;
h = Height of vertical post sought.

All distances are in feet. Suppose, as an example, that it was
desired to determine the heights of the vertical posts in an 8-panel
parabolic truss of a height approximately equal to 24 feet. One-half

Fig. 52. One-Half of 8-Panel Parabolic Truss.

the truss is shown in Fig. 52. At the center, d = 0, and the equation
reduces to h = H, which is 24 feet. For U3L3, d = 20; then,

4 X 24 X 202 -

h = 24 -

from which,
For t/2L2,

1602
h -= 22.5 feet.

d = 40

4 X 24 X 402
1602

h = 24 ^-~

For U,LV

d = 60

= 24 - 4 x 24 x^°a
1602

= 18.0 feet.

= 10.5 feet.

Inspection of the above results shows that the span or the center
height must become quite great before the clearance at UtLt will be
sufficient to allow the traffic to pass under a portal bracing at this
point. For this reason these trusses are usually built as through
trusses with bracing on the outside of the truss, which connects to
the floor-beams extended.

BRIDGE ENGINEERING

In the bowstring truss,
the panel points of the
top chord usually lie on
the arc of a parabola
which does not pass
through the supports.
For example, suppose
that it was decided to
have the span and pan-
els the same as shown in
Fig. 52, but the height
at Ll was to be '28 feet,
and at £4 36 feet. By
substituting these values
in the equation just
given, and solving for /,
the place will be deter-
mined where the para-
bolic curve cuts the
lower chord extended,
and the lengths of the
vertical posts may be
computed as before.
Substituting these re-
sults:

28 = 36 -

4 X 36 X 602

(-36 + 28) la = - 4 X 36
X 602

7 _ / 4 X 36 X 6"0"2

= 254.5,

which shows that the
arc cuts the lower chord
extended at a point
254.5 -T- 2 = 127.25 feet
from the center of the
span (see Fig. 53).

BRIDGE ENGINEERING

The other vertical posts are:
U,L, k = 30 _«*J6J^g_ 3;ul feet.

= 32.44 feet;

4 V *3fi V fiO

[/.L! /i = 36 - =5 — = 28.00 feet, which checks.

^54.5

The analysis of a bowstring truss will now be given. Both the
maximum and minimum stresses will be determined, as reversal of

Fig. 54. Outline Diagram of 5- Panel Bowstring Truss Span.

stresses is liable to occur in the intermediate posts. The loading for
minimum live-load stresses can be ascertained only by trial, care
being taken to compute the dead-load stresses for the arrangement
of web members caused by that particular live loading.

Let it be required to determine the maximum stresses in the
5-panel 100-foot bowstring truss
shown in Fig. 54, remembering
that the diagonals take only ten-
sion. The height of U^ is 20
feet, and of U^ 25 feet. The
dead panel load is 17 200 pounds,
and the live panel load is 50 000
pounds. The full lines show the
main members which act under
dead-load stress, and the dotted
lines show the counters which may
act under the action of the live load.

L,
11.47

Fig. 55. Resolution of Forces around

Panel Point in Bowstring Truss

of Fig. 54.

One-third of the dead
panel load, or 5 730 pounds, is taken as acting at the upper
panel points, while the remainder, 11 470 pounds, acts at the lower

BRIDGE ENGINEERING

Fig: 61. Fig. 62.

Analysis of Stresses in Various Members of the Bowstring Truss of Fig. 54.

BRIDGE ENGINEERING

Fig.
Analysis of Stresses in Various Members of the Bowstring Truss of Fig. 54.

60 BRIDGE ENGINEERING

ones. Articles 27, 28, and 29 should be carefully reviewed before
going further. The shear times the secant method cannot be con-
veniently employed for the live-load stresses in the members UtL2
and LJJ2, as the section will cut the member UJJ2, and the vertical
component of its stress must be reckoned with in the stress equation
The method of moments as illustrated in Fig. 27, Article 29, will be
used for these members.

The dead-load reaction is 2 X 17.2 - +34.4. The dead-load
chord stresses should first be computed.

By resolving the horizontal forces around Lv it is seen that L0Ll
= LjL2 (see Fig. 55). Passing the section a — a, taking the center
of moments, at Uv and stating the equation of the moments to the
left of the section, there results (see Fig. 56) :

+ 34.4 X 20 - L,L2 X 20 = 0 .-. L,L2 = +34.4

For LJjy the section b — b is passed; the center of moments is
at U2; and the equation of the moments to the left of this section
(see Fig. 57) is:
+ 34.4 X 2 X 20 - (11.47 + 5.73) 20 - L2L3 X 25 = 0 .'. L2L3 = +41.26.

By passing a vertical section cutting L^ and LJJV the stress in
L0Ul can be determined by taking the sum of the vertical forces to
the left and equating them to the vertical component of the stress
(see Fig. 58). The equation is:
+ 34.4 + L0U1 X 0.707 = 0, from which L0U1 = -34.4 X 1.414 = - 48.7.

A section a — a (Fig. 59) shows that the center of moments for
UJJ2 is at U2; and stating the moments of the stress, and the forces
to the left of the section, there results an equation in which an
unknown lever arm enters. This lever arm I is readily computed
to be 24.28 feet, and the equation can now be written:

+ 34.4 X 2 X 20 - (11.47 + 5.73) 20 + UJJ* X 24.28 = 0
.'. U,U2 = -42.51.

The stress in U2U3 is determined by passing a vertical section
in the 3d panel, and taking the sum of the horizontal forces. As there
is no dead-load stress in the members L2U3 and UJLZ, their compo-
nents will be zero. Therefore (see Fig. 60) it is evident that U2U3
must be equal and opposite to L2L3 and will be equal to —41.26.

By reference to Fig. 55, the stress in L1Ul is seen to be tensile and
equal to +11.47.

BRIDGE ENGINEERING 61

Pass a circular section around U2 and take the sum of the vertical
components, assuming that the stress in UJL2 acts away from the
section. The length of UJJ2 is 1/52 + 202 = 20.6, and therefore the
vertical component of UJJ2 will be (42.51 H- 20.6) X 5 = 10.32,
which acts upward. The stress equation of U^L^ (see Fig. 61) is:

+ 10.32 - 5.73 - U2L2 = 0 .'. U2L, = +4.59,

showing that a tensile stress occurs in U2L2 when all panel points are
loaded.

The simplest method of ascertaining the stress in U^ is to pass
a vertical section cutting members as shown in Fig. 62, and to equate
the horizontal forces and stresses. The horizontal component of
U&is:

~ X 20 = 41.30, which acts toward the left.
ZO.o

The equation of stress is, then :

-41.30 + 34.40 + C7,L2 sin 0 = 0; but sin <£ = 0.707;
t/,L2 = +6.90 X 1.414
= +9.76

All the dead-load stresses being computed, the next operation
will be to determine the live-load chord stresses. These are pro-
portional to the dead-load stresses in the same ratio as the live panel
load is to the dead panel load. This ratio is 50 -f- 17.2 - 2.907,
and the chord and end-post live-load stresses are:

Lot/, = -48.71 X 2.907 = -141.7

U1U2 = -42.51 X 2.907 = -123.6

U2U3 = -42.26 X 2.907 = -123.0

L0L2 = +34.40 X 2.907 = +100.2

L2L3 = +41.26 X 2.907 = +120.3

Also, the stress in U2L2 when the live load covers the entire bridge
is not 2.907 X 4.59, as it must be remembered that part of the dead
load is at the panel points of the upper chord. Taking a circular
section around U2 (see Fig. 61), and noting that there is no load at U2,
it is seen that the stress in U2L2 due to live load is simply equal to the
vertical component of the live-load stress of UJ72 and wjll be tensile.
It is:

U1U2 = (123.6" -J- 20.6) X 5 = +30.0.

The maximum live-load stress in UlLl is tensile, and equal to
the live panel load at Ll (see Fig. 55).

02 BRIDGE ENGINEERING

To obtain the maximum stress in U2L3, load L3 and L4. The

Cf\

shear F3 will then be — (1 + 2) - +30.0. The section will cut
o

the members as shown in Fig. 63, and the equation of stress will be :

25
+ 30.0- UtLa cos <£=0; but cos 0 = — -z==r =0.782;

.'. C/^ = +38.4.

If panel points Ll and L2 were loaded, it is evident that the stress
in L2U3 would be +38.4.

To obtain the maximum live-load stress in VJjv a section is
passed cutting UJ7V L^LV and U,L2 (Fig. 64). The center of
moments will be at the intersection of UJJ2 and L,I/2, and this point
lies some place to the left of the support L0. The lever arm of UtL2
will be the perpendicular distance from this point to the line UtL2
extended. The panel points L2, L3, and Lt are loaded. The left

reaction is then (1 + 2 + 3) — = + 60.0. The lever arms are

easily computed, and these, together with the members cut, are shown
in Fig. 64. The equation of stress is :

- 60.0 X 60.0+ t71L2X>70.8 = 0 .-. U,L2 = +50.80.

If a load were put on Ll only, then the reaction at L0 would be
4

— X 50 = 40; and the equation of stress would then be:
5

-40.0 X 60.0 + 50 X (60.0 + 20.0) + U,L2 X 70.8 = 0 .-. C/,L2 = -22.6.

As this is compression and greater than the dead-load stress, + 9.76,
a counter is required in that panel. In order to get the stress in the
counter, it must be inserted, U^ being left *out, and the dead and
live load stresses computed and their difference taken. Fig. 65 gives
the lever arms, center of moments, and the forces acting in this case.
The dead-load stress is:

-34.4 X 60.0 + (11.47 + 5.73) (60 + 20) - L.tf, X 62.5 = 0
.'.L^U, = -11.02;

and the live-load stress is:

-40 X 60 + 50 (60 + 20) - LJJ^ X 62.5 = 0 .-. L1f/2 = +25.60,

and the stress in the counter is the algebraic sum of these two, or
-11.02 + 25.60 - +14.58.

BRIDGE ENGINEERING

When a live panel load is at Lv LJJ2 is acting, as has just been
proved. As this load at Ll causes a negative shear in all panels to
the right, this negative shear in the center panel will cause L2U3 to
act. A section may now be passed as shown in Fig. 66, and the stress
equations for U2L2 written:

For dead load, +34.4 - 11.47 - 2 X 5.73 - U2L2 = 0 .'. U2L2 =- .-11.47
For live load, +40.0 - 50.0 - U2L2 = 0 .'. U2L2 = -10.00

Total = + 1.47

This is evidently not a maximum for U2L2, for when a full live load
was on the span, the stress was +30.0 due to live load and +4.59
due to dead load.

It might be well to consider what effect is produced by loading
L3 and L4. The loading of L2 and Ll need not be considered, since
it is evident that, as this causes the total shear in panel 2 to be positive
and the total shear in panel 3 to be negative, therefore UJj2 arid L2U3

Pig. 68. Stress Diagram of Half-Span of Parabolic Truss of Fig. 54.

will act, and this causes a tensile stress in U2L2 equal to the vertical
components of the dead and live load stresses in UJJ2 less the dead
panel load at Uy With a live panel load at L3 and Lv the left reaction

is — __(! + 2) = +30.0. The section, the live-load forces, the cen-
ter of moments, and the members acting are shown in Fig. 67. The
dead -load stress in U2L2 will be the same as when the truss has no
live load on it. The stress equation for the live load is:

-GO X 30 - (60 + 20 + 20) X L2Ut = 0 .'. L2U2 = -18.0.

BRIDGE ENGINEERING

The dead-load stress being +4.59, this
stress of — 18.0 causes a reversal of stress
in the vertical. For this reason the ver-
ticals of bowstring trusses are, like web
members of Warren trusses, built so as
to take either tension or compression.
The minimum stresses in the diagonals
will be zero, for when one diagonal in a
panel is acting, the other is not.

The diagram of half of the truss in Fig.
68 gives all the stresses.

It is to be noted by the student, that
in some cases one method for the deter-
mination of stresses is preferable to others
in that it saves labor of computation.
The analysis of the truss of Fig. 68 illus-
trates this fact.

42. The Baltimore Truss. Baltimore
trusses are of two classes — those in which
the half-diagonals, called sub-diagonals,
are in compression, and those in which
the sub-diagonals are in tension. The
latter class is the one most usually built,
as it is more economical on account of
many of its members being in tension, in
which case these members are cheaper
and easier to build than if they were com-
pression members. Fig. 14 shows botii
types of truss. The Baltimore truss does
not have a simple system of webbing, and
t'oi that reason the analysis is here pre-
sented. As the tension sub-diagonal truss
is the type in most common use, its analy-
sis will be given.

Let it be required to compute the
maximum stresses in the 14-panel 280-
foot span of Fig. 69. The height is 40
feet, the dead panel load 24 000 pounds,

BRIDGE ENGINEERING 65

and the live panel load 40 000 pounds. One-third of the dead
panel load is applied at the upper ends of the long verticals and also
of the half-verticals. These half-verticals are designated as sub-
verticals. Attention is called to the system of notation used for the
ends of the sub-verticals. The full lines in Fig. 69 represent the
main members, being stressed by dead load only. The heavy lines
indicate those members that take compression, the light lines those
that take tension, and the broken lines the counter-braces. In this,
as in nearly all Baltimore trusses, the diagonals make an angle of 45
degrees with the vertical.

The dead and the positive live-load shears in the various panels
should be computed. They are :

DEAD-LOAD V + LIVE-LOAD V

40
Vl +156.00 P! = (1 + 13) Z_ = +260.00

V2 +132.00 Y2 = (1 + ....12)~ = +223.00

40
Y3 + 108.00 V3 = (1 + .... 11) ^ = + 188.50

F4 + 84.00 F4 = (1 + 10)-^ = +157.20

F5 + 60.00 V5 = (1 + .... 9) ~ = +128.50

F6 + 36 00 F6 = (1 + 8) y = + 102.80

V7 + 12.00 Vr = (1 + .... 7) |5. = + 80.00

It is only necessary to determine the negative live-load shear in
panels 5 and 7, in order to ascertain if there is a counter required.
These shears are :

- Vs = (10 + 11 + 12+ 13) — - 4 X 40 = -28.60

40
-F7 = (8 + 9 + 10 + 11 + 12 + 13) y^- - 6 X 40 = -60.00

From a comparison of these with the dead-load shears, it is seen
(see Article 37) that a counter is required in panel 7 only.

The dead-load stresses are first to be computed. The stress in
any sub-vertical is found by passing a circular section around its
lower end, and equating the sum of the vertical forces, assuming
in this, as in all cases, that the unknown stress acts away from the

BRIDGE ENGINEERING

section. Take 3/jTO,, for example. Fig. 70 gives the section, the
forces acting, and the members cut. Then,

+ Mlm1 - 16.0 = 0 .'. Mimi = +16.0

As all sub-verticals have the same dead load at their lower end, it
follows that the dead-load stress in all sub-verticals is the same, a
tensile stress of 16 000 pounds.

The dead-load stresses in the sub-diagonals are determined by
resolving the forces around the joint at their lower end. The com-
ponents perpendicular to the diagonal are taken (see Fig. 71). Take

Fig. 70. Diagram for Calculating Stress in
Sub- Vertical of Baltimore Truss.

m2f72. The known forces or
stresses are the dead panel load
of 8.0 and the stress in m2M2,
which is 16.0 and which being
tensile acts away from the sec-
tion. The stress equation is:

Fig. 71. Diagram for Calculating Stress in
Sub-Diagonal of Baltimore Truss.

+ m2U2 - 8.0 sine <f> - 16.0 sine 0 = 0.
c£ = 45°, sine «£ = 0.707,and
m2U2 - 8.0 X 0.707 - 16.0 X 0.707 = 0
.\m,Ua = +16.96.

This equation may be put in another form by multiplying and dividing
the numerical values by 2:

(8.0 + 16.0)

X 1.414 = 0;

or,

which proves the well-known saying that the stress in the sub-diagonals
w equal to one-half the panel load, times the secant of the angle <f>. It
also shows that the vertical component of the sub-diagonal is equal to

BRIDGE ENGINEERING

one-half 'the panel load. This fact should be remembered, as it will
be frequently used further on.

In a similar manner, the stress in all the tension sub-diagonals
will be found to be the same, + 16.96, and
the stress in the compression sub-diagonal
mjji is -16.96.

The stress in the member L0ml and in
the upper half of any main diagonal (i. e.,
Ujrnv Ujn3, and ?73m4) is determined as in
the diagonals of the Pratt or Howe truss,
for the section passed cuts but one mem-
ber, which has a vertical component. Take
L0ml (see Fig. 72). Then +156.0 + LQml
cos 45° = 0, from which L0ml = —220.5. For U^ the section is
passed as in Fig. 73, and the equation of stress is + V3 — Ujm2 cos
45° = 0, or +108.0 - U^ X 0.707 = 0, from which Z7,ra2 =
+ 152.9.

In a similar manner,

U2m3 = +60.0 - 0.707 = +84.84;
U3m4 = +12.0 + 0.707 = +16.96.

The stresses in mlUl, w2L2, and m3L3 may be determined by

Fig. 72. Stress in Diagonal
of Baltimore Truss.

Fig. 73. Stress in Upper Half of Main
Diagonal of Baltimore Truss.

Pig. 74. Stress in Diagonal of Baltimore
Truss.

resolving the forces about mv mv and ra3; but a neater solution is to
pass a vertical section cutting the member whose stress is desired,
and to equate to zero the shear and the vertical components of all

BRIDGE ENGINEERING

the members cut (see Fig. 24, Article 28). The section for m1Ui is
passed as in Fig. 74. The equation of stress is then :
mlUl cos 45° + m,!/! cos 45° + F2 = 0;

24
but the vertical component of m^ is— = 12; and therefore,

mlUl X 0.707 + 12 + 132 = 0
.'. mlUl = -203.6.

For m^L2, the section is as shown in Fig. 75, and the stress

equation is :

-m2L2 X 0.707 + vert, component m2U2 + Vt = 0
-m^ X 0.707 + 12 + 84 = 0
.-. m2L2 = + 135.6.

[Fig. 75. Calculating Stress in Lower Half-Diagonal of Baltimore Truss.

In a similar manner, passing a section cutting U2U3, m3U3, mJL^
and M^Ly the stress equation may be written:
- m3L3 X 0.707 + 12 + 36 = 0
.-. m3L3 = +67.85.

The stresses in the verticals are best determined by resolving
the vertical forces at their lower end. Referring successively to
diagrams a, b, and c of Fig. 76, the stress equations are :

+ U1L1 - 16.0 - 12.0 = 0 .'. C7,L, = +28.0

+ U2L2 - 16.0 + 96.0 = 0 .'. U2L2 = -80.0

+ U3L3 - 16.0 + 48.0 = 0 .'. U3L3 = -32.0

96 and 48 being the vertical components of ra^ and m3L3 respec-
tively.

BRIDGE ENGINEERING

The chord stresses are easiest computed by considering the
resolution of horizontal forces at the panel points. As the diagonals
make an angle of 45° with the vertical, their horizontal and vertical

16.0 16.0

Fig. 76. Calculating Stresses in Verticals of Baltimore Truss of Fig. 69.

components are equal. For instance, the horizontal component of
the members Ljnv V \mv and U2m3 are equal to the shear in that
panel, which is their vertical component. At point I/0 (see Fig. 77),
there results:

+ L0Afj — horizontal component of AfjL0 = 0; or,

+ L0Ml — 156 = 0

/. L0M, = +156.0;

and from Fig. 70 it is evident that L0Mt =
Fig. 78), LtM2 is equal to MlLl,
less the horizontal component of
MJjv and the equation is :

-156 + 12 + L,M2 = 0
. L,M2 = + 144.0; and M2L2 = + 144.0.

At point L2 (see Fig. 79), L2M3
is equal to the sum of the horizon-
tal components of MJL2 and ra^;
that is,

+ LaAf, - 144.0 - 96.0 = 0

.-.L2M3= M3L3

At point Z/j (see

Fig. 77. Chord Stress in Baltimore Truss.

+ 240.0.
In a similar manner, at point L3, the stress equation is:

+ L,M4 - 240.0 - 48.0 = 0

.-. L3M4 = M4Lt = +288.0.

At the upper panel point Ul (see Fig. 80), there results the

equation : %

+ UlUt + hor. comp. f7jOT2 + hor. comp. m,?7, = 0;
17,17, + 108.0 + 144 = 0; or, ?7,t72 = -252.0.
For the member U2U3 (see Fig. 81), the equation is:

BRIDGE ENGINEERING

+ Ujj^ 4. U2U3 — hor. comp. m2U2 + hor. comp. U2m3 = 0
+ 252.0 + U2U3 - 12 + 60 = 0
.-. U2U3 = -300.0.

In a similar manner, by resolving the horizontal forces at U3, it
will be seen that the action of m3U3 will neutralize that of Uzmv as

L , m*^ L2

Fig. 78. Fig. 79.

Bottom Chord Stresses in Baltimore Truss.

they are equal and pull in opposite directions, and UaUt is equal to
U2U3 = -300.0.

The live-load stresses in the chords, the end-post, and the sub-
diagonals are all proportional to the dead-load stresses in the same

Fig. 80. Fig. 81.

Top Chord Stresses in Baltimore Truss.

ratio as the live panel load is to the dead panel load. This ratio is

— = 1.667. By reference to Fig. 70, it will be seen that the live-
load stress in the sub-verticals is +40.0 for each one. The following
stresses can now be determined :

Lomi = -220.5

m1Ul = -203.6

U,U2 = -252.0

U3Ua = -300.0

U3L\ = -300.0

L0L, = +156.0

L,L2 = +144.0

L2L3 = +240.0

L3L4 = +288.0

X 1.667 = -367.5

X 1.667 = -339.5

X 1.667 = -420.0

X 1.667 = -500.0

X 1.667 = +260.0

X 1.667 = +240.0

X 1.667 = +405.0

X 1.667 - +481.0

BRIDGE ENGINEERING

m.L, = - 16.96 X 1.667 = - 28.28
maU3 = m3U3 = +16.96 X 1.667 = +28.28

The vertical U^ will have its maximum live-load stress when
points M1 and Ll are loaded, for these are the only loads which cause
a stress in that member (see Fig. 76a). The equation is:

~ - 40 +

= 0,

from which,
C/,L, = +60.0.

The maximum
live-load stresses
in Ujnv U2m3, and
U3m4 are obtained
in a manner exact-
ly like that used
in obtaining dead-
load stress, only
the live-load posi-
tive shear is used.
The stresses are:

Fig.

Stress in Lower Half of Main Diagonal of Baltimore
Truss.

•U,ma X 0.707 -f- 188.5 = 0
U3m~3 X 0.707 + 128.5 = 0
U3mt X 0.707 + 80.0 = 0

.'. Ulm2 = +2665
. . U2m3 = +181 5
.'. U3mt = +113.1

In the determination of the maximum live-load stress in the
lower halves of the main diagonals, m2L2, ra^, and m4L4, one of the
peculiarities of this truss becomes apparent. A section being passed
as in Fig. 82, the panel point ahead of the section, and all between
the section and the right support, must be loaded. This of course
produces a stress in m2U2, and the vertical component of this enters
the stress equation." The shear in the section a — a under this load-
ing is:

V._. = + 188.5 - 40 = + 148.5;

and the stress equation is:

-»ijLa X 0.707 + -^ + 148.5 = 0
.-. m2L2 = +238.0.

If the truss had been loaded from the section to the right, there being
no load on M2, no stress would result in m2U2, and the stress in ra2L2

BRIDGE ENGINEERING

157 2
would have been mJL2 = — — '— = +222.2. In a similar manner,

by loading successively points M3 and to the right, and Mt and to
the right, the stress equations of m^ and ra4L4 are :

40
-m3La X 0.707 +^- + 128.5 - 40 = 0 .'. m3L3 = +153.3

-m4L4 X 0.707 + -~- + 80.0 - 40 = 0 .'. m4L4 = + 84.8

The maximum live-load stresses in the main verticals occur when
the panel points to the right of the section which cuts the member

under considera-

II. 0\ ^ 1 I -

tion are loaded.
There being no
load at the end of
the sub-vertical
just to the left of
the section, there
will be no stress

Fig. 83. Stress in Main Vertical of Baltimore Truss.

in the sub-diag-
onal which the sec-
tion cuts. The chords, of course, do not exert a vertical com-
ponent; and so the only unknown term of the stress equation is the
stress in the member itself. Fig. 83 showrs how the section should
be passed when U^L2 is considered. The stress equation is:
+ U2L2 + Fa_a = 0; + U2L2 + 128.5 = 0; .-. U2L2 = -128.5.

In a similar manner, by passing a section cutting U2U3, m3U3,
UgLg, L3mt, and loading Mt and to the right, it is seen that the stress
equation for UJj3 is:

+ U3L3 + 80.0 = 0 .'. U3L3 = -80.0

The components of m3U3 and L3m4 are zero, as can readily be proved
by solving for them under this loading.

Fig. 84 gives all the stresses, and they are written in order of
dead load, live load, and maximum.

43. Other Trusses. The analysis of the foregoing trusses will
enable one to solve any of the trusses of modern times. For the
solution of the Whipple (sometimes called the "double-intersection
Pratt") and others which are not mentioned in this text, the student

BRIDGE ENGINEERING

is referred to the text-
books of F. E. Tur-
neaure and Mans-
field Merriman.

ENGINE LOADS

44. Use of En=
gine Loads. It was

formerly the custom
for railroads to spec-
ify that the engine to
be used in computing
the stresses in their
bridges should be one
of their own which
was in actual use.
The engines of differ-
ent roads were usual-
ly different both in
regard to the weight
on the various wheels
and in regard to the
number and spacing
of the wheels. Of
late years, consider-
able progress has
been made towards
the adoption of a
typical engine load-
ing as standard.
These typical engines
(see Fig. 17, Article
25) vary only in re-
gard to the weights
on the wheels, the
number and spacing
of wheels being the
same in all engines.

BRIDGE ENGINEERING

- c r; ;= —

The distance between wheels is an
even number of feet, instead of an odd
number of feet and inches and frac-
tions thereof. For examples of load-
ings which are in almost universal use,
consult the specifications of Cooper or
Waddell.

The labor of computation of stresses
when engine loads are used is consid-
erably lessened by the use of the so-
called engine diagrams. Fig. 85 gives
a diagram which has been found very
convenient. The first line at the top
represents the bending moment of all
the loads about the point to the right
of it. All the loads are given in thou-
sands of pounds, and all the moments
are in thousands of pound-feet. The
practice of writing results in thousands
of pounds — or, in case of moments, in
thousands of pound-feet or pound-
inches — is to be recommended, as it
saves the unnecessary labor of writing
ciphers. Throughout this text this
practice has been extensively followed,
the stresses being written to the near-
est ten pounds or one-hundred pounds
as the case may be. For example,
6 433 may be written 6.43 or 6.4, the
few pounds which are neglected mak-
ing no appreciable difference in the
design. The distances are in feet.

As an example of the use of the first
line at the top, suppose that it is de-
sired to find the moment of all the
loads to the left of a certain point
when wheel 6 (the numbers of the
wheels are placed inside of the circles

BRIDGE ENGINEERING 75

representing the wheels) is just over the point. The moment will
be 1 640 000 pound-feet, which is obtained by reading off the 1 640
just to the right of the line through wheel 6.

When using the first line for values at sections in the uniform
load, the values given represent the moment of all wheel and uniform
loads about the points in the line or section to the left of the value
given. For example, if it is desired to find the moment about a
point in line 2, it will be 19 304 000 pound-feet, the value 19 304
appearing to the right of the line.

The line of figures below the wheels indicates the distances
between any two wheels.

The third line of figures indicates the distance from the first
wheel to the wheel to the right. For instance, 37 is the distance from
wheel 1 to wheel 7,

The values in the fourth line indicate the sum total of all the
loads to the left of the value given. For example, 245 signifies that
the loads 1 to 15 inclusive weigh 245 000 pounds.

The values in lines 5 and 6 are similar to those of lines 3 and 4,
except that the starting point is at the head of the uniform load.
For example, 40 in line 5, and 112 in line 6, indicate that it is 40 feet
from the head of the uniform load to the wheel 12, and that wheels
18 to 13 inclusive weigh 112 000 pounds.

The values in lines 7 to 16 indicate the value of the moment of
all the wheels from the zigzag line up to and including the one to the
left or the right, according as the value is to the left or the right of
the zigzag line. For example, 2 745, line 11, indicates that the
moments of wheels 8 to 14 inclusive about the zigzag line just under
wheel 15, is 2 745 000 pound-feet; or the value 1 704, line 14, shows
that the moments of wheels 13 to 18 about the zigzag line just under
wheel 12 is 1 704 000 pound-feet.

When line 4 of figures is under the uniform load, the values refer
to the vertical line to the right; 'thus 324 is the value of all loads to
the left of line 3 about that line.

For values of moments at points which fall in between wheels,
or at positions in the uniform load where the value of the moment
is not given, a very important principle of applied mechanics is used.
It is:

M. = M' + Wx +. '.

BRIDGE ENGINEERING

which,

M a = Moment at section desired;
M' = Value of moment at preceding vertical line;
W = Sum total of all loads to the left of and at the point where M'

is taken;
x = Distance from section under consideration to vertical line to

which M' is referred;
w = Uniform load on the distance x.

Let it be desired, for example, to determine the moment at a
a point c, 3 feet to the right of wheel

13. The position of the loads is
given in Fig. 86. The moment is:

Ma-. = 7 668 + 212 X 3
= 7 668 + 636

= 8 304 = 8 304 000 pound-
feet, there being no uni-
form load.

To illustrate the method when
applied to points in the uniform
load, assume the point to be 7 feet
to the right of line 2. The po-
sition is illustrated in Fig. 87. The
moment is :

6708

74'

ise

7668

•jft.H

79'

eie

•— —*.

14.

88'

Fig. 86. Calculation of Moment at a Point
under Engine Load.

= 19 304 + 304 X 7 +

72 X 2

= 21 481 = 21 481 000 pound-feet.

The use of the moment diagram is now apparent. Reactions
due to any position of the engines may be determined by dividing the
span into the value obtained for the moment at the right end of the
span. Likewise, if the moment of the reaction about any panel point
is determined and from it the moment of the wheel loads about that
same panel point are subtracted, then the result, divided by the
height of the truss, will give the chord stress. For example, if the
right end of an 8-panel 196-foot span truss, height 25 feet, came 7
feet to the right of the vertical line 2, then the moment at this point
(see Fig. 87) would be 21481000, and the reaction would be
21 481 000 -T- 196 = 109 600. This position of the loads would
cause the panel point L6 to come 3 feet to the right of wheel 13. The
moment of the reaction about LR is 109 600 X 6 X 24.5 = 16 111 200;

BRIDGE ENGINEERING

and the chord stress U6L6 for this position of the engine is:

16111200 - 8304000

— — — - 312 000 pounds.

In using the engine to determine the shear in any particular
panel, it must be remembered that the shear is not the left reaction
less all the loads to the left of the panel point on the right of the
section, as the loads in the panel under consideration are carried on
stringers, and these stringers transfer a portion of the loads to the

149-44

ie3 64

19304

7ft.

per L incorFootx/x

10'

119'

129'

304

384

Fig. 87. Calculation of Moment at Point under Uniform Load.

panel point on the left of the panel, and a portion to the panel point
on the right of the panel. Only that portion of the loads in the panel
which is transferred to the left panel point should be subtracted from
the reaction, as should all of the loads to the left of the panel under
consideration. If, in a 6-panel 120-foot span Pratt truss, the wheel
6 comes at Ly, the left reaction will be :

*-&(

16 364 + 3 X 284 +

32 X 2.0

143.6;

and the loads in the first two panels will be in position as indicated
by Fig. 88, the wheel 3 being 1 foot to the right of point Lr Let it
be required to determine the shear in the panel L^2 when the loads
are in this position. It will be the reaction 143.6 minus loads 1 and
2 and also that portion of the loads 3, 4, and 5 which will be trans-
ferred by the stringers to -the point Lr As the stringers are simple

BRIDGE ENGINEERING

Fig. 89. Shear Diagram for Panel under
Engine Load.

beams, the amount transferred to Lt will
be the reaction of the stringer L^Lr Re-
ferring to Fig. 89, the reaction is:
RLi = (20 X 9 + 20 X 14 + 20 X 19) + 20
= 42.0

The shear in the second panel is now
found to be :

V2 = 143.6 -(10 + 20 + 42.0) = +71.6
In the majority of cases where it is
necessary to determine the shear in a
panel, none of the loads will be in the
panel to fhe left of the one under consid-
eration. In this case the operation is
somewhat simplified, as the engine dia-
gram can be used directly. If the engine
be placed so that the third wheel is at
Lv wheel 16 will be just over the right
support, and the left reaction will be :
Ri = 12 041 * 120 = 100.3.

As there are no wheel loads in the first panel, the amount to be sub-
tracted from the reaction will be that proportion of the loads 1 and 2
which is transferred to I,,; and this (see Fig. 90) is 230 -=- 20 = 11.5.
The shear in the second panel is then 100.3 - 11.5 = +88.8.

From inspection of the resulting shear in the second panel when
wheel 6 is at L2 and when wheel 3 is at L2, it is seen that different
wheels at L2 will give different shears in the panel to the left. Evi-
dently there is some wheel which will give the greatest shear possible.
The same is true of the relation between wheels and moments- The

BRIDGE ENGINEERING

next two articles are devoted to subject-matter which will enable. one
to tell which of several wheels is the correct wheel at the point, without
the necessity of solving for the shear each time every wheel is at the
point.

45. Position of Wheel Loads for Maximum Shear. By methods
of differential calculus, it can be proved that, for any system, either
of wheel loads or wheel loads followed by a uniform load (see Fig. 91),
the correct wheel that should be at the panel point b in order to

6 at eo'=ieo'

Fig. 90. Determination of Shear in Panel under Engine Load.

give a very great or maximum shear in the panel a — b, is such a

W W

wheel that the quantity Q = — - G is positive when q = -

m m

(G + P) is negative. In these equations,

W = Total load on the truss;

m = Number of panels in the truss;

G = Load in panel under consideration; and

P = Load at panel points on right of panel.

If a load is directly over the panel point a, it is not to be included in
the weight G; neither is P included in the weight G, If a wheel load
should come directly over the right end of the truss, it should not be
considered in the quantity W.

The only way to determine which wheel is the correct one, is
to try wheel 1, then wheel 2, and so on, until the wheel or wheels are
reached that will give the Q and q signs of an opposite character.

BRIDGE ENGINEERING

The process should not be stopped there, but the next succeeding
wheels should be tried until Q and q again have the same sign.

As an example, let it be required to determine the position of the
wheel loads to produce the maximum positive shears in a 6-panel
120-foot Pratt truss. This work should be arranged in tabular form,
and Table V is found to be convenient.

TABLE V

Determination of Position of Wheel Loads for Maximi
(m-6)

Shear

PANEL
POINT

WHEEL

POINT

G + P

« REMARKS

if - 47.33

if = 48.67

gives a maximum

if _ 50.03

gives a maximum

if - 50.03

100

~j^- = 38.67

-^ = 40.83

gives a maximum

•^p- = 43.00

-IT - 25'67

-^p = 28.67

gives a maximum

192

T - 34'°

gives a maximum

212

•TT- = 37-°
o

—

116

-i^- = 21.50

gives a maximum

-l^2. = 23.67

70
-— - = 11.67
o

= 15.00

gives a maximum

-^L - 17.17

;;

v> •£
</) PS

5 5

S*
2 i

H .§

•<
?«

X >
Jfl *

BRIDGE ENGINEERING

A study of Table V shows the fact
that wheel 1 can never produce a maxi-
mum. It also shows that there are in
some cases two positions which will give
large values of the moment. In these
cases the shears for each position of the
engines must be determined in order to
tell which wheel at the panel point in
reality gives the greatest. In practical
work it is customary to use the first
position found, as the difference in the
shears resulting from the use of the two
positions is not large enough to affect the
final design.

Fig. 92 shows the engine diagram
on the truss in the correct position to
give the maximum shear in the second
panel. The weight of wheel 16 is not
included in the weight W, as it is directly
over the right support.

46. Position of Wheel Loads for
Maximum Moments. In this case the
methods of differential calculus are em-
ployed to determine which wheels will, if
placed at a point, give a maximum mo-
ment at that point. For any system,
either of wheel loads or of wheel loads
followed by a uniform load, that wheel

which will cause K — — L to be positive, and k = -^-^ —

m m

(L + P) to be negative, is the wheel. Here n is the number of the
panel under consideration, and is to be reckoned from the left
end; L = the load to the left of the point under consideration;
and the remainder of the letters signify the same as they do in
Article 45. In some cases there will be more than one position
of the loads which will satisfy the above condition. It is then
necessary to work out the actual moments created by the loads
in each position, in order to find out which is the largest. The

BRIDGE ENGINEERING

position of the loads
for the greatest mo-
ments should be de-
termined for all panel
points except the one
on the extreme right,
as the greatest moment
possible maybe caused
by wheels of the rear
engine being on the
point on the right-
hand side of the truss,
instead of the wheels
of the front engine be-
ing at the correspond-
ing point on the left-
hand side.

In general, it may
be said that there will
be a number of wheels
which, if placed at the
panel point in the cen-
ter of the span, will
satisfy the given con-
ditions. In this par-
ticular case, it is not
necessary to determine
all of the moments.
The greatest moment
possible will occur
when that one of the
heaviest wheels of the
second locomotive
which gives the heav-
iest load upon the
truss is at the point.
In case several of the
heavy wheels give the

BRIDGE ENGINEERING

same maximum load W, use the first wheel which gives this max-
imum W.

Let it be required to determine the position of the wheel loads
for maximum moments at the lower panel points of the 6-panel 120-
foot Pratt truss of Article 45. The necessary work can be con-
veniently arranged in the form of a table, as is done in Table VI.

TABLE VI
Determination of Position of Wheel Loads for Maximum Moments

PANEL
POINT

WHEEL

POINT

.+ !•

REMARKS

L\
L,

2
3
4
5

10
30
50
60

284 -5- 6 = 47.3
292 -5- 6 = 48.7
302 -f- 6 = 50.3
302 -5- 6 = 50.3

20
20
20

30
50
70
80

gives a maximum
gives a maximum
wheel 1 is off bridge

70
90
103

(271-5-6)X2 = 90.3
(290--6)X2 = 96.7
(300-r-6)X2 = 100.0

20
13
13

90
103
116

gives a maximum

8
9
10
11
12
13
14

116
129
142
142
142
142
142

(271n-6)X3 = 135.5

(298 4-6) X3= 149.0
(304 -5-6) X3= 151.3
(294n-6)X3 = 147.0
(284-6)X3=142.0
(274-h6)X3 = 137.0

13
13
10
20
20
20

129
142
152
162
162
162
162

^
0

gives a maximum
gives a maximum
gives a maximum
gives a maximum
gives a maximum

12
13
14

152

172

192
212

(271 -=-6) X4 = 180.6

(284n-6)X4=189.4
(294-=-6)X4 = 196.0
(304-r-6)X4 = 202.6

20
20
20

162

192
212
232

Note wheel 18 not
included,
gives a maximum
gives a maximum

One should carefully note that in certain positions, as when
wheels 11, 12, 13, and 14 are at L3, some wheels are to the left of the
left support; that is, they are not upon the bridge. In all such cases
they are counted neither in the quantity L nor in W.

In the case of L3, wheel 11, being the first large driver of the
second engine, will give the greatest moment, as it is the first driver
to come at the point when the maximum load of 304 000 pounds is on
the truss. Fig. 93 shows the engine diagram on the truss in correct
position to give the maximum moment at point Lr

47. Pratt Truss under Engine Loads. In order to exemplify
the use of the engine-load diagram, let it be required to determine the
stresses in the Pratt truss of Article 45 due to E 40 loading, the

BRIDGE ENGINEERING

height being 25 feet. The
secant is (^? + ~202)< -25 =
1.28.

The maximum positive
shears in the various panels
should first be computed.
These, written in reverse or-
der, will be the maximum
negative or minimum shears.
Table V should now be re-
ferred to, and an outline dia-
gram drawn to the same scale
as the engine used, on which
to place the engine diagram in
the correct position. The vari-
ous values can then be read
off the diagram at the right-
hand end of the truss. It will
be found convenient to lay off
to scale the first ten feet of
the lower chord of the truss
from the right support, mak-
ing the divisions one foot apart.
This will enable one to ascer-
tain the distance of the last
wheel load from the right sup-
port, or the amount of uni-
form load upon the bridge,
without scaling or further com-
putation. In case it is desired
to have the wheel loads appear
on the lower chord, as in Fig.
93, the outline of the truss
should be on tracing cloth or
transparent paper. This is not
to be advised, however, as er-
rors are likely to occur because
of failure to distinguish clearly

BRIDGE ENGINEERING 85

the various numerical values. It is far better to place the diagram
as in Fig. 92, in which case the outline of both the truss and the dia-
gram can be drawn on good stiff paper.

For wheel 3 at point Ll (see Articles 44 and 45), the left reaction
is as follows, there being four feet of uniform load upon the truss:

(42 v 2 0 \
16 364 + 284 X 4 + * j -4- 120 = 146.0;

and the proportion of loads in the panel which is transferred to
the point L0 by the stringers is 230 -=- 20 = 11.5. The shear is
therefore Ft = +146.0 - 11.5 - +134.5. The computation for
the shear when wheel 4 is at the point, will not be made; for, as has
been noted before, the result will not be much different from the
above.

For wheel 3 at L2, wheel 16 comes over the right support. The
left reaction is:

Ri = 12041 -H 120 = 100.3;

the proportional part of the loads which is transferred to Ll is 11.5;
and the shear is:

V2 = ( + 100.3 - 11.5) = +88.8.

For wheel 2 at L3, wheel 11 comes four feet from the right sup-
port. The left reaction is:

Ri = (5 848 + 172 X 4) + 120 = 54.5.

That part of wheel 1 which is transferred to L2 is 80 -=- 20 = 4.0,
and the shear is therefore:

V3 = ( + 54.5 - 4) = +50.5.

For wheel 2 at L4, wheel 9 comes over the right support. The
left reaction is:

R\ = 3 496 +• 120 = 29.1.

That part of wheel 1 which is transferred to Z/3 is 4.0, and the shear
is therefore:

V4 = ( + 29.1 - 4.0) == +25.1.

For wheel 2 at L5, wheel 5 is five feet from the right support,
and the left reaction is:

Ri = (830 + 90 X 5) -T- 120 = 10.7
and the shear is :

F5 = ( + 10.7 - 4.0) = +6.7.

86 BRIDGE ENGINEERING

If the dead panel load is 20 000 pounds, all the shears may now
be written as follows:

DBAS-LOAD V + LIVE-LOAD V - LIVE-LOAD V

+ 134.5 ± 0.0

+ 88.8 - 6.7

+ 50.5 -25.1

+ 25.1 -50.5

+ 6.7 -88.8

± 0.0 -134.5

A comparison of the shears in the third and fourth panels shows
that counters are required. The stress in these counters is:
U3L2 = U3L4 = +1.28 X (25.1 - 10.0) = +19.2

As it is known that positive shears cause a compressive stress
in L0Ul and tensile stresses in the diagonals, and that negative
shears produce compressive stresses in the intermediate posts, the
left half of the bridge being considered, the web stresses for dead and
live load can be determined without in all cases writing the stress
equations in order to determine the sign. It should be remembered
that one-third of the dead panel load, or 6 700 pounds, is applied at
the panel points of the top chord.

Dead-Load Stresses in the Diagonals —

L0Ut = -1.28 X 50 = -64.0
£7,1/2 = +1.28 X 30 = +38.4
U2L3 = +1.28 X 10 = +12.8

Dead-Load Stresses in the Verticals. For Z72Z/2, the section
passed will cut UJJ2, U^, and L2L3, and the shear on this section
will be 50 - 2 X 13.3 - 6.7 = + 16.7. The stress equation is
+ 16.7 + U2LZ = 0, from which UJ,, = - 16.7.

The dead-load stress in U^ is found by passing a circular sec-
tion around Uy Then - U3L3 - 6.7 = 0, from which U3L3 = -6.7.
In a similar manner, by passing a section around Lv the stress is
found to be + 13.3.

Live-Load Stresses in the Diagonals —

MAXIMUM MINIMUM

LoU, = -1.28 X 134.5 = -172.2 0

C7,L2 = +1.28 X 88.8 = +113.6 -1.28 X 6.7 = -8.6

U2L3 = +1.28 X 50.5 = + 64.7 0

Live-Load Stresses in the Verticals. The maximum stress in
UlLl occurs when one of the large drivers is at Lv and the loads ID

BRIDGE ENGINEERING

oooo

o (

the first panel are as near as possible one-half the sum of the loads
in panels 1 and 2 and the load at Lr This can be established as a
fact by use of the differential calculus. In the present case, this con-
dition is satisfied when wheel 4 is at Lr Then the weight of the
wheels in panel 1 is 50 000 pounds, and the sum total is 116 000 pounds.
If wheel 13 be placed at Lv the result will be the same, and then the
engine diagram can be used. Fig. 94 represents the engine diagram
in place, ready to use. According
to Article 44, the value 480 is the
moment of wheels 10 to 12 about
Lv Therefore 480 -r- 20 (20 is
the panel length) - 24.00, is that
amount of wheels 10 to 12 which
is transferred to L0. In like
manner, 529 -r- 20 - 26.45 is the
amount of wheels 14 to 16 trans-
ferred to L2. As the total weight
of the loads in the two panels is
116000 pounds, the amount
transferred to Lt must be 116.0
- (24.00 + 26.45) = 65.55, and
the stress in U^ is therefore
+ 65.55.

The maximum live-load stress
in LJJ2 occurs when the loading
is in a position to give the maxi-
mum shear in the third panel, as
the shear at a section cutting UJJV U2L2, and L2L3 is the same' as
that at a vertical section in the panel. The stress equation is + U2L2
+ 50.5 = 0, from which UJL^ = —50.5. In a similar manner, the
stress equation for the maximum live-load stress in U3L3 is + U3L3
+ 25.1 = 0, as U3Lt is working, and therefore U^L3 = — 25.1. As in
the case of the analysis of the Pratt truss under uniform load (see
Article 39), the dead-load stress of —6.7 cannot be added to this
stress of —25.1 to obtain the maximum; but the dead-load stress in
U3L3 must be obtained when diagonals U3L3 and UJL,i are in action.
In the manner explained in Article 39, this is found to be +3.30.

It will be found that as the'engines come on the bridge from the

Fig. 94.

Engine Diagram for Determina-
tion of Live-Load Stress in Vertical

of Pratt Truss.

88 BRIDGE ENGINEERING

left, the counters come into action in the case of UZL2; and in the
case of U3L3, both U2LZ and LJJ^ act, thus causing the live-load stress
in these verticals to be zero; and when this is the case, the dead-load
stress is — 6.7, which is the minimum.

Dead-Load Chord Stresses. The dead-load chord stresses can
be found by any of the methods previously given; but they will be
found by the tangent method as indicated below, the tangent being
20 -r- 25 = 0.8:

LoL, = +0.8 X 50 = +40.0 = L,L2
UJJ, = -(50 + 30) X 0.8 = -64.0
U2U3 = -(50 + 30 + 10) X 0.8 = -72.0
L2L3 = -UiU, = -(-64.0) = +64.0

Live-Load Chord Stresses. On account of the wheel loading, nc
ratio can be established between these stresses and the dead-load
chord stresses. The maximum moments at each point must be
determined, and these divided by the height of the truss will give the
chord stresses. For all points to the left of the center of the bridge,
the main diagonal will act. For points to the right of the center, an
uncertainty exists. The shear in the panels on either side of the
point under consideration should be determined when the loading is
in position to give the maximum moment at that point. This will
indicate which diagonals act, which fact will indicate for what chord
member that point is the center of moments.

When wheel 3 is at Llt four feet of uniform load are on the truss,
and the left reaction is:

Ri = (16 364 + 384 X 4 + ^-~) + 120 = 146.0.

The moment of this reaction about Lv less the moment of wheels 1
and 2 about Lv will be the moment at Ll due to this loading. The
moment of wheels 1 and 2 about Ll is taken from the diagram, where
it occurs in the first line of values just to the right of the vertical line
through wheel 3, and therefore:

Ml = 146.0 X 20 - 230 = 2 690 000 pound-feet.

When wheel 4 is at Lv there are nine feet of uniform load on
the truss, and the left reaction is:

Ri = (16 364 + 284 X 9 + ^~ ) -=- 120 = 158.0;
and in this case,

Ml = 158.0 X 20 - 480 = 2 680 000 pound-feet.

BRIDGE ENGINEERING 89

As this is less than when wheel 3 is at the point, wheel 3 gives the
greatest moment.

When considering the point L2 with wheel 6, the left reaction is:

02 y o

Ri = (16 364 + 284 X 3 + — £-=) H- 120 = 143.5
M2 = 143.5 X 2 X 20 - 1 640
= 4 100 000 pound-feet.

The conditions at L3 indicate that there are several wheels which
give large moments; but according to Article 46, wheel 11 gives the
maximum moment. When this wheel is at L3, wheel 1 is off the
truss, and 15 feet of uniform load are on the truss. The moment of

att the loads about the right support is 19 304 + 304 X 5 + ^ * 2

= 20 849, from which should be subtracted the moment of wheel 1
.about the right support. This moment of wheel 1 is 10 X 124 =
1 240, and the moment about L6 of all loads on the truss is 20 849
- 1 240 = 19 609. The left reaction is:

ft = 19 609 •*• 120 = 163.4
M3 = 163.4 X 3 X 20 - (5 848 - 10 X 64)
= 4 596 000 pound-feet.

In the case of Lt, the reactions and the moments for the two
positions are:

For wheel 12, ft = 16364 -s- 120 = 136.4

M4 = 136.4 X 4 X 20 - 6 708
= 4 204 000 pound-feet.

For wheel 13, ft = (16 364 + 5 X 284 + ^-j^) + 120 = 148.4
Mt = 148.4 X 4 X 20 - 7 668
= 4 204 000 pound-feet,

which shows that each wheel gives the same moment, and also that
the moment is greater than that at L2, the corresponding point on the
left-hand side of the center of the truss. As L2 is the center of
moments of UJJ2, then, if the center of moments for UtU5 falls at
L4 (that is, if LJJ5 acts), the stress in U4Ub will be greater than the
stress in ?7X?72 when wheel 6 is at L2. Of course, if the engine came
on the truss from the left, UjU2 would receive the same stress that
UJJ5 now receives. According to the shears, LJJ5 always acts, and
therefore the center of moments for UJJ5 does fall at Lt.

BRIDGE ENGINEERING

TABLE VII
Stresses In a Pratt Truss

LOWER CHORD

IM 0

The various moments are written
in order, as such action will facili-
tate the remainder of the computa-
tions.
M, = 2 690
M 2 = 4 100
M 3 = 4 596
M4 = 4 204

The chord stresses are now found
to be:

LL L L - l 2 69° l 107 5

q 10

iO 0

O iO

10 0
t^ O

. UPPER CHORD

f-t

1 1

cq q
1 1

ZiO

4 100

o 03

1 1

c^i -^

CO CO

1 1

U>U, 25 164.0
r rr 4 596

t/2C/3~ 25
II U 4 2°4 16S -

DIAGONALS

oq q o
1 +

(M O

OS 0

t/4(/5 2. [68..

L2L3 = - C7,L2 = - ( - 164.2) = + 164.2
L3L4 = -UtUs= -(- 168.2) = + 168.2

When the load comes on from the
left, the stresses in U1U2 and L^
will be - 168.2 and + 168.2 respec-
tively, which are the maximum live-
load stresses for these members.
Instead of placing the values of
the stresses on a truss outline, they
are sometimes put in tabular form,
as in Table VII.
48. Impact Stresses. WTien an
engine is at rest on a bridge, the
stresses in the members are the
same as those computed for that
loading. Wlien the loads move
across the bridge at any speed, the
vibrations and the shocks produced
by the counterweights in the drivers
and by other causes create stresses
in the various members in excess
of those computed by aid of the
engine diagram. The excess stresses

00 I- O
C\ -* O

I-H CD

10 q

•<* q cq
+ + 1

q oq

VERTICALS

5S§§

+ 1

00 t~
1 1

2 d §
1 1

<M r-

SS
I i

co g o

q cq
oo co

q <N q
1 I

<M O

1 I

KIND OF

STRESS

Maximum..
Minimum. .

ICC

BRIDGE ENGINEERING 91

are designated as impact stresses. This term, however, is mislead-
ing to a certain extent, as causes other than the impact or pounding
of the engine wheels help to produce the stresses referred to.

It is a well-known fact capable of mathematical demonstration,
that a load, if suddenly applied, will produce a stress equal to twice
that which it will produce as a static load; also, that as the ratio of
the weight of the load to the weight of the structure increases, the
vibrations produced by the impact will be less. These two facts are
the basis of most of the empirical formulae for impact stresses; and
empirical formulse are used to obtain these stresses, as the existing
conditions and producing causes are not such as to make them sus-
ceptible of mathematical treatment. The result of experiments
on actual bridges under the effect of passing engines and trains, have
been the basis of many formulse. One of these is :

where / = Impact stress in the member;

S = Live-load stress in the member caused by the engine

load when at rest;
L = Length of that part of the bridge which is loaded when

the stress S is produced; and
300 = A constant. value derived from experiments.

This formula was proposed by C. C. Schneider in 1887, and is
given in the "Transactions" of the American Society of Civil En-
gineers, Vol. 34, page 331. While it does not take into considera-
tion the relative weights of the bridge and the live-load loads, this
formula does make allowance for the time it takes to produce the
stress, by introducing L, the distance over which the engine passes
before causing the stress S. It is seen that the smaller the distance L,
the greater will be the impact stress for any given value of S. When
L becomes exceedingly small, the effect would be that of a suddenly
applied load, and the impact stress would equal the stress S. Table

VIII gives the values of j- -~ „„„ , which is called the impact co-

efficient, for different values of L. Values not given may be inter-
polated.

For example, by consulting Fig. 92, which gives the position of
the engines for the maximum live-load stress in V \L2, it is seen that
93 feet (the distance from* wheel 1 to the right support) is the loaded

101

BRIDGE ENGINEERING

[TABLE VIII
Values of the Impact Coefficient

300

L + 300

5
6

0.984
0.980
0.977

31
32
33

0.906
0.904
0.901

58
59

0.840
0.838
0.836

84
85

0.783
0.781
0.779

145
150

0.674
0.667

8
9

0.974
0 971

34
35

0.898
0 896

0.833

86
87

0.777
0 775

155

0.659

0.968

36
37

0.893
0.890

61
62

0.831
0 829

88
89

0.773
0.771

160
165

0.652
0.645

0.965

38
39

0.888
0.885

63
64

0.826
0 824

0.769

170
175

0.638
0.632

12
13

0.962
0.958

0.882

65
66

0.822
0 820

0.767

180
185

0.625
0.619

14
15
16
17
18

0.955
0.952
0.949
0.946
0.943

41
42
43

0.880
0.877
0.875
0.872

67
68
69
70

0.817
0.815
0.813
0.811

92
93
94
95
96

0.765
0.763
0.761
0.759
0.758

190
195
200

0]f)

0.612
0.606
0.600

0 588

0.940
0.937

45
46

0.870
0.867
0.865

0.809
0.806

97
98
99

0.756
0.754
0.752

220
230

0.577
0.566

0.935

48
49

0.862
0.860

73
74

0.804
0.802

100

0.750

250

0.546

OC-JC

22
23

0.932
0.929

0.857

75
76

0.800
0.798

105

0.741

270

0.526

Oci 7

24
25
26

0.926
0.923
0.920

51
52

0.855
0.852

77
78
79

0.796
0.794
0.792

110
115
120

0.732
0.725
0.714

290
300

0.508
0.500

0 915

0.850

0847

125
1 3O

0.706

OfiQS

40O

0 429

29
30

0.912
0.909

55
56

0.845
0.843

0.787
0.785

135
140

0.690
0.682

500
600

0.375
0.333

length. From Table VII, it is seen that the stress in U^ produced
by this loading is +118.6; and from Table VIII, the impact coefficient
for 93 feet is found to be 0.763. The impact stress is now computed :

/ = 0.763 X 118.6 = +90.6.
The maximum stress in UtL2 is now :

Dead-load = + 38.4
Live-load = +118.6
Impact = + 90.6

Maximum = +247.6

Table IX gives the necessary information for computing the
impact stresses, and also gives the impact stresses corresponding to
the maximum live-load stresses in the members of the truss of Article
47.

102

BRIDGE ENGINEERING

TABLE IX
Impact Stresses in a Pratt Truss

MEMBER

300

RMMARKS

L + 300

L0U,

-172.2

113

0.727

— 125 .2 4 ft. of uniform load on truss

U,Lr
U2L2
U3L3

+ 65.6
- 50.5
- 25.1

37
68

0.890
0.815
0.862

+ 58.4
- 41.3
- 21.6

See succeeding text.
Same as for U2L3
Same as for U3L4

U,L2
U2L3
U3L,

+ 118.6
+ 64.7
+ 32.0

68
48

0.763
0.815
0.862

+ 90.6

+ 52.7
+ 27.6

Wheel 16 at LR
Wheel 11 is 4 ft. from L6
Wheel 9 at L0.

17,17,

u,u,

-168.2
-183.8

114
114

0.724
0.724

-121.8
-133.2

j Wheel 13 at L4.
/ 5 ft. of uniform load on bridge.
/ 15 ft. of uniform load on bridge.
\ Wheel 1 off bridge.

L0L3
L2L3

+ 107.5
+ 168.2

113
114

0.727
0.724

+ 78.2
+ 121.8

Same loading as for L0Ul
Same loading as for U2U3

In the case of U1LV it should be noted that only the wheels 10 to
16 inclusive cause the stress (see Fig. 94), and that the loaded
length is the distance from wheel 10 to wheel 16.

Some specifications do not call for impact stresses. The unit-
stresses in these specifications are made low, and the sections designed
are large enough to withstand the additional stresses due to impact.
In cases where the impact stresses are required, they must be con-
sidered in computing the maximum and minimum stresses.

49. Snow=Load Stresses. In some localities the snowfall is
considerable, and its weight should be taken into account in com-
puting stresses. This should be done by considering it as an addi-
tional dead load of 15 pounds per square foot of floor surface for every
foot of snowfall. As it covers the. entire floor surface, the stresses
will be proportional to the dead-load stresses. Also it is evident that
the snow load should not be taken into account in railroad bridges
unless they have solid floors, as most of it falls through the open
spaces between ties and stringers.

As an example, let it be required to determine the snow-load
stresses in a member of a highway bridge, the dead-load stress in the
member being +84.0, the dead panel load being 12 000 pounds, and

103

94 BRIDGE ENGINEERING

the snow be'rig H feet deep on the roadway, which is 14 feet wide.
The snow panel load is:

|- (14 X 15 X li X 20) = 3 150 pounds.

In the above equation, 14 is the width of roadway; 15 is the weight
in pounds of one square foot of snow one foot deep; and 20 is the
length of one panel. One-half of the weight of snow must be taken,
as half is carried by each truss. The snow-load stress is then :

»X 84.0- +22.1.

In like manner, all snow-load stresses can be computed.

Most of the standard specifications which have been published
do not specify snow loads; and in fact it is not customary to include
the snow load in any designs except those for bridges in extreme
northern latitudes. It is hardly probable that the greatest load will
come upon a country bridge when it is covered with snow. Also,
in cities, the sidewalks are cleaned of snow; and so is the roadway
if the city is of large size.

WIND=LOAD EFFECTS

50. Top Lateral System Through=Bridges. The unit-loads for
this system are given in Article 26. Common practice is to take
150 pounds per linear foot of top chord, the end-post being con-
sidered part of the top chord in this computation.

In many of the longer-span modern bridges, the diagonals of
this system are designed to take either tension or compression ; but in
the majority of the shorter spans, 200 feet and under, while generally
consisting of angles or other stiff shapes, they are designed to take
tension only. The verticals or top lateral struts take compression.
This combination of tension diagonals and compression verticals
makes the so-called Pratt system of webbing; and indeed the lateral
systems, both top and bottom, are Pratt trusses in a horizontal posi-
tion. Fig. 95 shows the side elevation of the truss of Article 47, and
also the top and bottom laterals. The diagonals shown in full lines
act when the wind is right, and those shown by dotted lines act
when the wind is left. Wind right indicates that the wind is blow-
ing from the right hand when a person stands facing the righ< end of

104

BRIDGE ENGINEERING

0 the bridge. Wind left indicates that

5 the wind blows from a person's left

when standing as above described.

The wind load of 150 pounds is di-
vided between the two trusses, this
being exact enough for practical pur-
poses; for, by actual experiment, the
<§ difference between the readings of
g wind-pressure gauges placed at points
o opposite each other in the top chords
| of a through-bridge was only from 8 to
^ 10 per cent.

The problem, then, is one of a deck
5 Pratt truss with a dead panel load of
| 150 X 20 - 3.0 divided between the
| two chords. Fig. 96 showTs the distri-
^ bution of loads and the reaction, it
* being considered that the portal brac-
| ings and the end -posts (see Fig. 95)
£ are stiff enough to distribute the
g reaction equally between the bearing
-°f' / | >" I points L0, L0', L6, L9'. Each panel

5 load is indicated by an arrow, and
d 3 is equal to 3.0 -r- 2 = 1.5. The re-

•S action at each of the points L0, L0',
| L6, and L6' is 10x1.5 +4 = 3.75.
g The truss being symmetrical, the

6 stresses in like members on each side
S of the center will be the same. The
§ shears in the top system are:

g Fx = +2 X 3.75 - 2 X 1.5 = +4.5

Fa-a = +2 X 3.75 - 3 X 1.5 = +3.0
F2 = +2 X 3.75 - 4 X 1.5 = +1.5

and the secant ^ is (172 + 202 )* + 17
= 1.544. The stresses in the diag-
onals are:

UiU9 = +1.544
U2'Ua = +1.544

X 4.5 = +6.95
X 1.5 = +2.32.

106

BRIDGE ENGINEERING

The vertical U2fU2 = —3.0; and by passing a section 6 — 6 around
Utf, the stress in U3fU3 is found to be - 1.5.

In obtaining the chord stresses in this system, the case is the
same as if the reactions were applied at C7/ and U5', as the portal and
end-posts are not in the same plane as the lateral system. The tan-
gent method is the simplest to use in this ca'se. The tangent is
20 -r- 17 = 1.176, and the stresses (see Fig. 95) are:

USU,' = - 4.5 X 1.176 = -5.29

U2'U3' = -(4.5 + 1.5) X 1.176 = -7.06

U,U2 = 0

U2U3 = -USUS = +5.29

Fig. 97 is a diagram with the stresses caused by wind right and

W.R.* 0

W.L. - 5.29 W.L. - 7.O6

Fig. 97. Wind Stress Diagram of Pratt Truss of Fig. 95.

wind left indicated thereon. The stresses for wind left can easily
be written by inspection.

51. Bottom Lateral Bracing, Through=Bridges. Fig. 95 shows
the lower lateral system with the panel points loaded with the fixed
or dead wind load. In this case it is all taken as acting on one side,
it being assumed that the floor system protects the leeward truss.
The problem then becomes that of determining the stresses in a deck
Pratt truss of 6 panels of 20 feet each, the height being 1 7 feet. When
wind is right, the members shown by broken lines in Fig. 95 do not act.

The fixed wind load (Article 26) is 150 pounds per linear foot of
chord. The panel load will be the same as before, 3.0, but all will
be on one chord. The shears are :

V, = 2* X 3.0 = +7.5

Fa-a = +7.5

F2 == +7.5 - 3.0 = +4.5

Fb-b = +4.5

Va = +7.5 - 2 X 3.0 = +1.5

107

BRIDGE ENGINEERING

The secant being 1.544, as previously computed, the web
stresses are:

L0'Ll = +7.5 X 1.544 = +11.60 L/L, = -7.5
L/L2 = +4.5 X 1-544 = + 6.95 L2'L, = -4.5
L2'L3 = +1.5 X 1.544 = + 2.32 L3'L3 = -3.0

The stress in L3'L3 is determined by passing section c — c and
resolving the vertical forces at L3f (see Fig. 95).

By using the tangent method, the chord stresses are computed
as follows:

L/L/ =
L,'L3' =
L,L2 .
L2L3 -

-7.5 X 1.176 = -8.82
-(7.5 + 4.5) X 1.176 = -14.12'
-(7.5 + 4.5 + 1.5) X 1.176 = -15.88
-L/L/ = -(-8.82) = +8.82

-L/L.

-(.- 14.12) = +14.12

The wind load acting on the train is 450 pounds per linear foot.
It is evident that the train may cover the span either partially or
entirely, and therefore its action on the lower lateral system is the
same as if it were stressed by a live load of 450 pounds per linear foot
of truss.

The live panel load is 450 X 20 = 9.0. The maximum live-
load reaction is 5 X 9.0 +- 2 = 22.5, and the positive live-load shears
are:

V, = +22.5

72 = (1 + 2 + 3 + 4) ^ = + 15.0
V3 = (1 + 2 + 3 ) -jr- = +9.0

It is unnecessary to go further than the center, as only the maximum
stresses are required in the members: The web stresses are com-
puted as given below :

L/L, = -22.5
L2'L2= -15.0
L3'L3 = - 9.0

The maximum chord stresses due to this load of 450 pounds per
linear foot of train, occur when the train covers the entire span; and
they are directly proportional to the stresses produced by the fixed
load, in the same ratio as the live panel load is to the fixed panel load.

This ratio is }-— = 3.0. The chord stresses, therefore, are:

LO'L! = +22.5 X 1.544 = +34.75
L/L2 = +15.0 X 1.544 = +23.15
L2'L3 = + 9.0 X 1.544 = +13.91

108

BRIDGE ENGINEERING

L0'L/ =
26.46

- 8.82 X 3 =
-14.12 X 3 =
-15.88 X 3 =
+ 8.82 X 3 =
+ 14.12 X 3 =

4— ~~f

L/L/ -
-42.36
L2'L3' =
-47.64
L,L2 =
+ 26.46

+ 42.36

X t

Table X, Article 53,
gives the stresses in the
top and bottom lateral
systems for wind right
and wind left.

52. Overturning Ef=
feet of Wind on Truss.
When the wind blows on
the top chord, it tends to
overturn the truss. As
the truss is held down by
its own weight, the action
of the wind does not
overturn it, but causes
the dead-load reaction on
the windward side to be
less and that on the lee-
ward side to increase by
a like amount. The

2?o
amount is ± V = -y X

-,- . where 2w = the sum
b

of all .the wind panel
loads, h = the height of
the truss, and b = the
distance center to center
of. trusses. The effect
upon the leeward truss is
the same as if two ver-

100

BRIDGE ENGINEERING

tical loads, each equal to V and acting downward, were placed at the
hips U1 and U5 (see Fig. 98). The effect on the windward truss is
the same as if two vertical loads, each equal to V and acting upward,
were placed at the hips C7/ and C76'.

The stresses in the leeward truss will now be worked out. The
stresses in the windward truss are the same, but with opposite signs.

10 X 1 5 2*1
The truss is that of Article 47. Here V = — — X — - = 11000.

Fig. 99 shows the truss with the loads in the correct position, the

Fig. 99. Truss under Wind Loads.

reactions each being 11.00. Vl = +11.00, and V2 = +11.00 —
11.00 = 0. The shears in the 2d, 3d, 4th, and 5th panels are also
zero. As the shear in these panels is zero, the stress in the diagonals
and vertical posts is zero X secant $ = zero. The stress in the hip
verticals U1L1 and U&L5 is zero, as there are no loads at Ll and L5.
The stress in the end-post is -11.00 X 1.28 = -14.08. Taking
the center of moments at Uv the stress equation of L0Ll — LtL2 is :
-I^a X 25 + 11.00 X 20 - 0; whence LtL2 = +8.8. The stress
in all the lower chord members will be found to be + 8.8. By
summing the horizontal forces at the section a — a, noting that, as
f/jI/2 is zero, its component is also zero, there results: +LtL2 + UJJ2
= 0; whence UJJ2 = -LJL2 = - (+8.80) = -8.80. This is also
the stress in all members of the top chord.

It is now seen that the overturning effect of the wind on the
truss causes stresses only in the end-posts and chords. The wind on
the lower chord causes no overturning effect, as it is transferred
directly to the abutments.

53. Overturning Effect of Wind on Train. The wind blowing
upon the train tends to overturn it, and in so doing the pressure on

110

BRIDGE ENGINEERING

101

the leeward stringer is increased and that on the windward stringer
decreased by the same amount. This difference in pressures is
transferred to the floor-beam and then to the panel points (see Fig.
100), where its value is:

± L =
where W

W X (8.5 + a)

Panel load due to wind

on train;

8.5= A constant established
by the Specifications
(see Article 26, p. 15);
a = Distance from base of
rail to center line of
lower chord. It may
be taken as 3 feet in
most cases, as this is
approximately the
usual depth of floor).
b = Distance center to cen-
ter of trusses.

For the case in hand, W = 20
X 450 = 9 000. Therefore,

9 000 X (8.5

L = ±

Illustrating Overturning Effect of
Wind on Train.

= ± 6 090 pounds.

The action of the wind in tending to overturn the train is the
same as if the truss were under a live panel loading of L, the panel
load L acting upward on the windward and downward on the leeward
truss.

The chord stresses due to this will be proportional to the dead-
load stresses in the same ratio as this panel load L is to the dead panel

load.

6090

For the truss of Article 47, this ratio is -- -- = 0.303, and

20 000

the chord stresses caused by the overturning effect of the wind on the
train (see Table VII) are:

UlUt = -64.0 X 0.303 = -19.39
U2U3 = -72.0 X 0.303 = -21.82
L0L, = L,L2 = +40.0 X 0.303 = +12.12
L2L3 = +64.0 X 0.303 = +19.39

The stress in UlLl is +6.09, the panel load at Lv
The maximum positive shears are:

Vv = ^ (1+2 + 3 + 4 + 5)= + 15.22

111

102

BRIDGE ENGINEERING

V, = ^(

1 + 2 + 3 + 4) = +10.15

v,= ^(

1 + 2 +^3) = +6.09

y4= 12? (

1 + 2) = +3.05

and the maximum web stresses are found to be:

L0U1 = -15.22 X 1.28 = -19.50

UJjt = +10.15 X 1.28 = +13.00

U2L3 = + 6.09 X 1.28 = + 7.80

U3L4 = + 3.05 X 1.28 = + 3.91

U2L2 = -6.09
C73L3 = -3.05

It is unnecessary to compute the shears further than one panel past
the middle of the span, as only the maximum stresses are usually
required.

The wind stresses from various causes are grouped together and
given in Table X.

From Table X it is seen that large wind stresses occur in some
of the members. Most specifications require that the stresses due to
wind shall be neglected in the design unless they exceed 25 per cent
of the sum of the dead-load and live-load stresses.

The subject of wind stresses does not ordinarily receive the con-
sideration it should have; in fact, it appears to be common practice,
in the case of spans up to 200 feet, to neglect the action of the wind in
all members of the bridge except the top and bottom lateral diagonals,
the top struts, the portal, and the bending
effect in the end-post. For the last two
effects mentioned, see the next succeeding
article.

54. Portals and Sway Bracing. One-
half of the wind on the top chord is trans-
ferred to the hips Ul'Ul and U6'U6. From
there it is carried to the abutments by
means of the portal bracing and the end-
posts. Various styles of portal bracing are
in use, but few are so easily analyzed and
constructed as that of Fig. 101. This form

Fig. 101. Style of Portal
Bracing in Common Use on
Spans up to 250 Feet.

112

BRIDGE ENGINEERING

103

TABLE X
Wind Stresses in Pratt Truss

WEB MEMBERS

OVERTURNING

L,V,

UtL.,

V,L,

U,L,

U.L,

U,L,

Wind Right

on Truss

-14.08

0. 0

on Train

-19.50

-13.00

+ 7.80 +3.91

+ 6.09

-6.09

-3.05

Wind Left

on Truss

+ 14.08

on Train

+ 19.50

-13.00

-7.80! -3.91

-6.09

+ 6.09

+ 3.05

Maximum

+ Stress

+ 33.58

+ 13.00

+ 7.80

+ 3.91

+ 6.09

+ 3.05

Maximum

- Stress

-33.58

-13.00

-7.80

-3.91

-6.09

-3.05

MEMBER

L,,L,

L,Ln,

L*L3

u,u.

U,U3

Direct

1 .

Wind Right

+ 8.82

+ 14.12

+ 5.29

+ 26.46

+ 42 . 36

Wind Left

- 8.82

-14.12

-15.88

- 5.29

- 7.06

-26.46

-42.36

-47.64

Overturning Truss
Wind Right

+ 8.80

+ 8.80 - 8.80

- 8.80

Wind Left

- 8.80

+ 8.80

Overturning Train
Wind Right

+ 12.12

+ 19.39

-19.39

-21.82

Wind Left

-12.12

-19.39 ; +19.39

+ 21.82

Maximum

+ Stress

+ 20.92

+ 56.20

+ 84 . 67

+ 28.19

+ 35.91

Maximum

- Stress

-56.20

-77.40

-91.71

-33.48

-37.68

LATERAL SYSTEMS

MEMBER

Ut'U,

U,'U3

Ut'U,

U3'U3

L,/L,

L,'L2

L,'L,

Wind Right
on Truss

+ 6 . 95

+ 2.32

-3.0

-1.5

+ 1 1 . 60

+ 6.95

+ 2.32

on Train

+ 34.75

+ 23.15

+ 13.91

Wind Left

on Truss

-3.0

-1.5

on Train

Maximum

+ 6.95

+ 2.32

-3.0

-1.5

+ 46.35

+ 30 ..10

+ 16.23

The stresses -in Z/ L,, L,f L,, and L3' L3 are not given in the above table. These
members are the floor-beams, and the small stress due to wind Is neglected In their
design.

113

104 BRIDGE ENGINEERING

of portal is at present being used almost universally on all spans up
to 250 feet.

Let it be required to analyze a portal of this form, all the dis-
tances being as indicated in Fig. 101 ; and let :

w = Wind panel load of upper chord;
m' = Number of panels in upper chord;
Then,

P = (m' - 1) w;
and,

V = ± j (P + w} + w | A;

also,

Hl = H2 = { (P + w) + w \ + 2.

The stress in BC, the center of moments being at D, is:

. ^ = (P + w) a + HJ. = _ \ (P + W} + H J_\
a 1 2 a )

The stress in AB, the center of moments being at E, is:

g,._+j-±it_+.+tflJ,

a a

For the stress in BD, the center of moments is taken at C, and
the perpendicular distance c to BD is determined. The stress in
BD, then, is:

<J -LW *1

The stress in BE is:

*!.

It must be remembered that hv is not the height of the truss,
but is the length of the end-post from L0 to Uv

For the truss of Article 47, w — 1.5; mf = 4; and P = 4.5.
The value ht - (202 + 252)^ = 32.0 feet. The distance a must be
so chosen that BD will not interfere with engines or other traffic
which passes through the bridge. It will be assumed as 5 feet in

this case. Then V = (4.5 + 1.5 + 1.5) ~ = 14.08; and Hl = H2

75
= -^r «= 3.75; whence,

SBC = - (e.O + 3.75 X —} = - 26.25

114

BRIDGE ENGINEERING

105

SAB = 1.5 + 3.75 X - = +21.75

The dis-tance BD = ^BC + CD2 = ^8.5* -f- 5.02 = = 9.85.
Then, from similar triangles DCS and DEC, is obtained the
proportion :

CF^BC.

CD' BD'

= +3.75 X

4 . o

= +27.90

-3.75 X ~ = -27.90

-4 . o

When the wind blows from the other side, the stresses in the
diagonals are reversed, and those in the top are transposed. The
members shown by broken lines take no stress. When the wind
blows, the end-posts tend to bend as shown in Fig. 102. This is with-

Fig. 102. Illustrating Tendency
of End-Posts to Bend under
Wind Load.

Fig. 103. Bending Tendency

when Knd-Posts are Fixed

at Lower End.

stood by the cross-section of the post at the points E and D. The
bending moment caused at these points by the wind is Hl X I and
H2 X I. For the truss under consideration,

MD = ME = 3.75 X 27 X 12 = 1 215 000 Ib.-ins.

If the posts are fixed at the lower end, then they will tend to
bend as shown in Fig. 103, the post resisting the bending at two
points D and d. The section at each point withstands in this case
only half of the moment just computed, or 1 215 000 -r- 2 - 607 500
Ib.-ins. A further discussion of this will be given in Part II, on
"Bridge Design."

115

106

BRIDGE ENGINEERING

Various forms of sway bracing are used to connect the inter-
mediate posts and thus stiffen the cross-section of the bridge at those
points. The form of portal just given is often used, as is also the
form shown in Fig. 104. Here h is the height of the truss. The
braces BD are called knee-braces. Here w is the wind panel load of

the top chord, and

V =

2wh

*.-*.- -r

SBC = -(wo + HJ) -H a

SB'c> = + (w

»-£->

The stress in B 'B is the direct
compression due to wind right or
pig. KM. A Type of Portal and Sway left, and differs in accordance

Bracing in Frequent Use. . , , . . » i

with the position of the top strut.
There is also a bending moment at B' and B, which is:

^MBI = MB = — V'g + H2h.

The bending moment at D and D' is equal to H2l or H2l -=- 2,
according to whether or not the lower ends of the posts are fixed.

The determination of the stresses for the truss of Article 47 is
left to the student.

When the wind is from the other side of the truss, the signs of
the stresses in the knee-braces and the members C'B' and CB are
reversed.

55. Final Stresses. The class of stresses which go to make
up the maximum or minimum for which the member is designed, is
.determined by the specifications used. The dead-load and live-load
stresses are always included, and then those due to impact and wind
should be added if required. In computing the maximum stresses,
the algebraic sum should always be used. In a large majority of
cases, all stresses which go to make up the maximum have the

116

BRIDGE ENGINEERING 107

same sign, but some exceptions have been noted, as in the middle
vertical of a Pratt or Howe truss. The minimum stresses ure, with
rare exceptions, obtained by combining stresses with signs of opposite
character.

GIRDER SPANS

56. Moments and Shears in Floor=Beams. In any bridge the
floor-beam acts as a support for either the joists or stringers, and the
moments and shears occurring in it are due to the loads which come
on the joists or stringers. In a highway bridge the joists are spaced
so closely that the load which they transmit to the floor-beams may
be considered as uniformly distributed, providing the live load is a
uniform load, in which case,

, _ (2PL + Pn) X panel length in inches

V = (2PL + PD) •*• 2,

where M = Maximum moment in pound-inches;
V = Maximum shear;
PL = Live panel load;
PD = Weight of stringers and floor material in one panel.

It will be seen that these formulae are those for the maximum
moment and shear in a uniformly loaded beam, the total load being
2PL + PD •

As an example, let it be required to determine the maximum
moment and shear in the floor-beam of a highway bridge whose panels
are 20 feet long, and trusses 16 feet center to center, the live load
being 100 pounds per square foot of floor surface, the flooring weighing
10 pounds per square foot, and there being 5 lines of joists weighing
15 pounds per linear foot, and 2 lines of joists weighing 8 pounds per
linear foot.

Pi. = -- X 20 X 100 = 16 000 pounds.

PD = 5 X 20 X 15 + 2 X 20 X 8 + 16 X 20 X 10 = 5 020 pounds.

Therefore,

(2 X 16 000 + 5 020 ^ 20 X 12
M = — g—

= 1 110 600 pound-inches at center of floor-beam.
V = (2 X 16 000 + 5 020) - 2

= 18 510 pounds at ends of floor-beam.

117

108

BRIDGE ENGINEERING

In the case of a single-track railroad bridge, there are only two
stringers upon which the weight of the track, the engine, and the
train is supported. These join the floor-beam at points equally
distant from the center of same. The weight of the ties, rails, and
fastenings is usually taken at 400 pounds per linear foot of one
track. As regards the live load, the proposition reduces itself to
placing the wheel loads so that the sum of the reactions of
stringers in the adjacent panels will be a maximum on the floor-
beam under consideration. This is discussed in Article 47, page 87
(see Fig. 94).

In determining the values of the maximum moment and shear in
the floor-beam, the case is that of a beam symmetrically loaded with
two equal concentrated loads. Each load is equal to the dead weight
of one stringer, one-half the track weight in one panel, and the maxi-
mum sum total of the reactions due to the wheel loads on the stringers

in adjacent panels which, meet at
that point. This latter quantity is
called the floor-beam reaction. For
a general arrangement of the loads,
see Fig. 105. The distance a has
become standard for single-track
spans, and is 6 feet 6 inches.

Let it be required to determine
the maximum shears and moments
in the floor-beam of the truss of
Article 47.

The weight of the stringer may be obtained by the formula of
Table II, and is:

Stringer = 20 (123.5 + 10 X 20) H- 2 = 3 200 pounds.
The weight of one-half of the ties, rails, etc., in one panel is:

i Track = (400 X 20) -H 2 = 4 000 pounds.

The weight that comes from the engine wheels is given in Article
47, page 87 (see Fig. 94), and is 65.55. Each load is therefore the
sum of all the above weights, as follows:

3 200 + 4 000 + 65 550 = 73 750.

The maximum shear (see Fig. 105) is seen to be 73 750 pounds ;
and the maximum moment occurs at C and D, and is:

1 p

^ Stringers ^

1 -1
Floor Beam

. a.'6'.s

5 17 Ft

loC. of Truss?

Fig. 105. Arrangement of Loads for Cal-
culating Moments and Shears
in Floor-Beams.

118

BRIDGE ENGINEERING

109

M = 73 750 X (- " ) X 12 = 4 646 250 pound-inches.

For any particular engine the floor-beam reactions for different
length panels are easily tabulated for future reference. Table XI
gives the floor-beam reactions for panel lengths from 10 to 24 feet
inclusive.

TABLE XI

Floor=Beam Reactions
E 40 Loading

PANEL
LENOTH

MAXIMUM
FLOOR-BM.
REACTION

PANEL
LENGTH

MAXIMUM
FLOOR-BM.
REACTION

PANEL

LENGTH

MAXIMUM
FLOOR-BM.
REACTION

41 000

55000

65.55

43800

57000

67.10

46600

59000

69.20

44400

61 000

71.30

52200

63000

73 40

In many cases it is desirable to keep the dead-load shears and
moments separate from those of the live load; and this can easily be
done.

In neither of the above cases has the weight of the beam itself
been taken into account. This
should be done in the final design.
The method of procedure is to
compute the moment and shears
as above; then make a provisional
design of the beam. Next, com-
pute the weight of the beam thus
designed, and add the moments
and shears caused by this weight
to the other dead-load moments
and shears ; then re-design the beam
and compute its weight. If this

last weight varies 10 per cent from the previous weight, another re-
design should be made. The above proceeding belongs to Bridge
Design, Part II, and will there be treated.

57. Moments in Plate=Girders. Plate girders are of two classes
— namely (1) those which have the ties or floor laid directly upon the
upper flanges of the girders; these are called deck plate-girder bridges;

Fig. 106. Cross-Section of Deck Plate-
Girder Railway Bridge.

lie

110

BRIDGE ENGINEERING

and (2) those in which the webs of the girders are connected with
each other at intervals by floor-beams which in turn carry stringers
or joists in exactly the same manner as in the floor system of a
railroad or highway truss-bridge ; this latter type is called a through
plate-girder bridge. Figs. 106 and 107 show cross-sections of deck

and through plate-
girder bridges re-
spectively, for rail-
way service. Fig.

108 is a side view
of a deck plate-
girder bridge. Fig.

109 is a longitudinal
section of a through
plate-girder rail-
road bridge. The

section is taken down the middle of the track. The bridge shown has
5 panels. An odd number of panels should be chosen, as this does
not bring a floor-beam at the center of the span, and hence the great
moment which would then be caused is avoided.

The analysis of the shears and moments of a through plate-
girder is precisely the same as that for a truss bridge. The shear is

Fig. 107. Cross-Section of Through Plate-Girder Railway
Bridge.

*fl m \m &.

m m m.

m B8LJH

m w BB «

1 M

Fig. 108. Side View of Deck Plate-Girder Railway Bridge.

constant between any two panel points as 0-1 or 1-2, etc., and the
moments are computed for the points 1, 2, 3, and 4. ^

If wheel loads are used for moments, the relation that K = —

-Lmustbe. + , and that k = — - (L + P) must be-, holds

true when the loads are in correct position for maximum moments.
Here m = the number of panels, and n = the panel under considera-

120

BRIDGE ENGINEERING

111

tion and is to be reckoned from
the left end ; in fact, all terms
have the same value as men-
tioned in Article 46. A careful
review of Articles 44 and 46
should enable the student to
follow the example which will
now be given.

EXAMPLE. It is required
to determine the moments at
the points of floor-beam support
for a 5-panel through plate-
girder of 75-foot span. The
live loading is Cooper's E 40.

Dead-Load Moments.
Through plate-girders, on ac-
count of the heavy floor system
and the fact that the floor sys-
tem transfers its own weight
and that of the live load to the
girders as concentrated loads,
are about 40 per cent heavier
than deck plate-girder bridges
of the same span. The weight
of the entire span, therefore, is;

1.4 X 75(123.5 + 10 X 75) =
91 700 pounds.

Part of this 91 700 pounds (the
weight of the girders themselves)
acts as a uniform load ; the re-
mainder (the weight of the
floor-beams and stringers) acts
as concentrated loads at the
points where the floor-beams
join the web. Experience has
shown that the weight of the
floor for a single-track railroad
system is about 400 pounds

121

112

BRIDGE ENGINEERING

Wheel

TABLE XII
Position, Moments in a Through Plate-Girder

-- c

L + P

^ - 34.4

^j2 = 38.4

~ = 4QA

™* = 40.4

^ = 39.0

152 X 2 60 8

172 X 2 68 8

/ u

^-U

152 X3

1 Q

103

ino

172 X 3 103 2

1 ^

116

lUo

1 1 fi

192 X 3

lo
1 q

129

1 ID

129 X 4 1Q3 0

1 S

1 r»o

lUo

103

142X4

116

— 1 lo . O
o

116

152 X 4

142

129

152 X 4

142

— IZI . O

REMARKS

Maximum
Maximum
Maximum

Maximum

Maximum
Maximum

Maximum
Maximum
Maximum

per linear foot. The weight of the stringers and floor-beams for
this bridge is therefore 75 X 400 = 30 000 pounds, and 91 700 -
30 000 = 61 700 pounds acts as a uniform load. This 61 700 pounds
is distributed over two girders, and so gives 61 700 -r- (2X75) = say,
412 pounds per linear foot of one girder.

The dead load which is concentrated at each panel point is that
due to the weight of the steel floor and the weight of ties, rails, and
fastenings. It is, for one girder,

15 X (400 + 400) -f- 2 = 6 000 pounds.

122

!!§•-*

!5II

3* I*
is2?£

S S^^-E

^ l?fli

<• a»j

^ »5?

a in

3 «u«

BRIDGE ENGINEERING

113

The dead-load moments are now computed by the methods of
Strength of Materials, and are found to be :

M0 = Mt = 0;

Mt = M4 = +4 390 000 pound-inches;

M2 = Ms = +6 580 000 pound-inches.

Live-Load Moments. The positions of the wheels for maximum
moments are now determined (see Table XII).

The computations for the reactions are best arranged in tabular
form. Table XIII gives the values.

TABLE XIII
Reactions for a Through Plate-Girder

IH *
£ w^S
2 \a<o

gLp *

1 i 3

1 | 4
1 5

~~2 4

3 | 6
3 7

EQUATION FOR REACTION

REAC-
TION

R = (6 708
R = (7 668
72 = (8 728

+ 192 X 4) - 75
+ 212 X 4 - 10 X 78) * 75
+ 232 X 4 - 10 X 83 - 20 X 75) - 75

99.7
103.2
97.8

R = (4 632

+ 152 X 7) -H 75

76.0

R = (4 632
72 = (5 848

+ 152 X 6) -J- 75
+ 172 X 3) H- 75

73.9
84.9

4 6

4 7
4 8

R = (2 851
72 = (3 496
72 = (4 632

+ 129 X 4) -=- 75
+ 142 X 4) + 75
+ 152 X 2) H- 75

44.9
54.2
65.4

The live-load moments are computed as follows:

I Wheel 3, M = 99 .7 X 15 - 230 = 1 265 1)00 pound-feet.
Wheel 4, M =103.2 X 15 - 20' X 5 - 20 X 10 = 1247000
pound-feet.
Wheel 5, M = 97.8 X 15 - 20 X 5 - 20 X 10 = 1 167000
pound-feet.

Point 2 Wheel 4, M = 76.0 X 30 -
[Wheel 6, M = 73

Point 3

480 = 1 800 000 pound-feet.
X 45 - 1 640 = 1 785 000 pound-feet.

"[Wheel 7, M = 84.9 X 45 - 2 155 = 1 665 000 pound-feet,
f Wheel 6, M = 44.9 X 60 - 1 640 = 1 054 000 pound-feet.
54.2 X 60 - 2 155 = 1 097 000 pound-feet.

Point 4^ Wheel 7, M

I Wheel 8, M = 65.4 X 60 - 2 851 =1 073 000 pound-feet.

• The above values show that the greatest live-load moments are:

M at Point 1, by wheel 3 = 1 265 000 pound-feet
Mat " 2," " 4=1800000 " "
Mat " 3, " " 4 = 1800000 " "
Mat " 4, " " 3 - 1265000 " "

129

114 BRIDGE ENGINEERING

The last two values are obtained when the load comes on the bridge
from the left. Inspection of the results obtained at points 3 anti 4
when the load comes on from the right, shows that they are con-
siderably smaller than the results obtained at their symmetrical
points 1 and 2, and therefore it was not necessary to determine the
moments for any points to the right of the center. This is true of all
girder spans, deck or through.

The method of procedure when the girder is a deck plate-girder
is the same as that just illustrated, except that in the computation
of the dead-load moments there is no concentration of certain por-
tions of the dead load, the weight of the girders themselves being a
uniform load, as is also the weight of the ties and rails or, if it be a
highway bridge, the floor-joists which run transversely. Highway
spans are seldom built of deck plate-girders, it being preferable to
use the through girders, as then the girders themselves serve as a rail-
ing and keep the traffic confined to the roadway. The girder span is
usually divided into ten equal divisions, the points of division being
called the tenth-points. The shears and moments are computed for
the center point and those points which lie to the left of the center.
After the values are computed, they are laid off as ordinates, with the
corresponding tenth-points as abscissae. A curve is then drawn
through their upper ends, and the curve of maximum shears or
moments is the result. To get the maximum shear or moment at
any point other than a tenth-point, the ordinate is scaled at the
desired point.

EXAMPLE. Let it be required to determine the maximum
moments at the tenth-points of a 100-foot-span deck plate-girder.

Dead-Load Moments. The weight of steel in the span is 100
(123.5 + 10 X 100) = 112 350 pounds, and the weight of the track
is 400 X 100 = 40 000 pounds, making a total of 152 350 pounds,
or 152 350 •*• (2 X 100) = say, 762 pounds per linear foot per girder.
The dead -load moments are now determined according to the methods
of Strength of Materials, and are :

M, = 342 800 pound-feet

M2 = 609400 " "

A/3 = 799 840 " "

M4 = 914 100 " "

M5 = 952 200 ' ' "
Live-Load Moments. The determination of the wheel load

124

BRIDGE ENGINEERING

115

positions is made by the use of the formulae K = ( — — L) and

w m

k = (L -f P) ; only, in this case, n is the number of divisions

from the left support to the section, and m is the number of divisions
into which the girder is divided.

The determination of the wheel positions is given in Table XIV.

TABLE XIV
Wheel Positions, Moments in Deck Plate-Girder

WHEEL ]
POINT

Wn
m

L + P

REMARKS

1
1
1
1
1
1

2
3
4
5
6
7

10
20
20
20
20
13

258 X 0.1 = 25.8
261 X 0.1 = 26.1
254 X 0.1 = 25.4
242 X 0.1 = 24.2
240 X 0.1 = 24.0
230 X 0.1 = 23.0

20
20
20
20
13
13

30
40
40
40
33
26

+
+
+
+
+
+

—

Maximum

2
2
2
2

2
3
4
5

10
30
50
60

232 X 0.2 = 46.4
245 X 0.2 = 49.0
258 X 0.2 = 51.6
261 X 0.2 = 52.2

30
20

20
20

30
50
70
80

+
+
+

Maximum

3
3
3
3

3
4
5
6

50
70
80

232 X 0.3 = 69.6
232 X 0.3 = 69.6
245 X 0.3 = 73.5
261 X 0.3 = 78.3

20
20
20
13

50
70
90
93

+
+
+

Maximum

4
4
4
4
4

4
5
6

50
70-
90
103
106

212 X 0.4 = 84.8
232 X 0.4 = 92.8
245 X 0.4 = 98.0
258 X 0.4 = 103.2
261 X 0.4 = 104.4

20
20
13
13
13

70
90
103
116
119

+
+
+
+

Maximum

6
5
5
6
5
5
5
5
5

7
8
9
10
11
12
13
14

90
103
116
129
132
102
102
102
122

232 X 0.5 = 116.0
232 X 0.5 = 116.0
245 X 0.5 = 127.5
258 X 0.5 = 129.0
274 X 0.5 = 137.0
244 X 0.5 = 122.0
234 X 0.5 = 117.0
224 X 0.5 = 112.0
234 X 0.5 = 117.0

13
13
13
13
10
20
20
20
20

103
116
129
142
142
122
122
122
132

+
+
+
0

+
+
+
+

+
0

Maximum

While many wheels on point 1 satisfy the condition, the greatest
moment will occur when one of the large drivers is at the point, and
it is therefore unnecessary to examine the point for other wheels.
The same is true at the center point, 5, the maximum occurring under
one of the heavy driver wheels. The reactions and the computations
for the same are given in Table XV.

125

116

BRIDGE ENGINEERING

TABLE XV
Reactions for a Deck Plate-Girder

POINT

WHEEL AT
POINT

EQUATION FOR REACTION

REACTION

1
1
1
1

2
3
4
5

R = (12041 + 5 X 258) -.100
R = (12 599 + 4 X 261) H- 100
R = (11 984 + 4 X 254) -5- 100
R = (11 334 + 4 X 234 -T- 2 X 4a - 2) - 100

133.31
136.43
130.00
122.86

2
2

3
4

R = 12041 H-100
R = (12 041 + 5 X 258) -r- 100

120.41
133.31

3
3

4
5

fl = 10 816 - 100

R = 12 041 -5- 100

108.16
120.41

4
4
4

5
6

R = (8 728 + 4 X 232) - 100
fl = (10 816 + 4 X 245) -f- 100
fl = (12 041 + 4 X 258) -f- 100

96.56
117.96
130 . 73

6
5
5
5
5
5
6

7
8
9
10
11
12
13

# = (8 728 + 8 X 232) H- 100
R = 12041 -100
R =(12041 + 5X 258) •*• 100
R = (13 904 + 2 X 274) H- 100
R =(11 334+ 5X234 + 2X 52 - 2) - 100
R = (9 514 + 10 X 214 + 2 X 102 - 2) - 100
R = (7794 + 15 X 194 + 2 X 152 - 2) H- 100

105.84
120.41
133.31
144.52
125.29
117.54
109.29

TABLE XVI
Maximum Moments in a Deck Plate-Girder

POINT

WHEEL

POINT

EQUATION FOR MOMENT

MOMENT IN
PODND-FEET

1
1

2
3
4
5

M =

M =
M =
M =

133.31 X 10
136.43 X 10
130.00 X 10
122.86 X 10

- 80
- 5 X 20
- 5 X 20
- 5 X 20

1 253 100
1264000
1 200 000
1 128 000

2
2

3
4

J!f =

M =

120.41 X 20
133.31 X 20

- 230
- 480

2 178 200
2186200-

3
3

4
5

M =

108.16 X 30
120.41 X 30

- 480
- 830

2 764 800
2 782 300

4
4
4

5
6
7

M =

96.56 X 40
117.96 X 40
130.73 X40

- 830
- 11 640
- 2155

3 032 400
3 078 400
3 074 200

5
5
5
5
5
5
5

7
8
9
10
11
12
13

M =
M =
M =
M =
M =
A/ =
M =

105.84 X 50
120.41 X 50
133.31 X 50
144.52 X 50
125.29 X 50
117.54 X 50
109.29 X 50

- 2 155
- 2851
- 3496
- 4072
- 3068
- 2658
- 2248

3 137 000
3 169 500
3 169 500
3 154 000
3 196 500
3 219 000
3 216 500

120

BRIDGE ENGINEERING 117

Table XVI gives the computations of the live-load moments at
the tenth-points, the final results being in pound-feet.

Whenever any loads were off the left end of the bridge, the lines
7 to 16 of the engine diagram were used (Fig. 85). For example, with
wheel 10 at 5, wheel 1 would be off the left end. By looking in the
second space of line 8, there is found the quantity 13 904, which is
the moment of wheels 2 to 18 inclusive about a point directly
under wheel 18. Just to the right of the vertical line through
wheel 18, is the value 284, which is the weight of wheels 1 to 18 in-
clusive; but this must be decreased by 10, the weight of wheel 1, as
that wheel is off the span. As wheel 18 is 2 feet from the right end
of the girder, the moment about the point is 13 904 + 274 X 2. By
looking in the second space of line 16, the value 4 072 is found. This
is the value of the moment of loads 2 to 9 inclusive about a point
directly under wheel 10, and must be subtracted from the moment
of the reaction in order to get the moment at 5 for this loading.
See Articles 21 and 47 for further information regarding the use of
the values in lines 7 to 16 of the engine diagram.

By the help of differential calculus it can be proved that the
greatest possible moment does not occur at the middle of a beam loaded
either with concentrated loads or with concentrated loads followed by
a uniform load, but it occurs under the load nearest the middle of the
beam when the loads are so placed that the middle of the beam is half
way between the center of gravity of all the loads and the nearest load.

The wheel which produces this greatest moment is the same one
which produces the maximum moment at the middle of the beam.
The exact solution of this problem involves the use of quadratic
equations, but for all practical purposes the following rule will
suffice :

Place the loading so that the wheel which produces the maximum
moment at the middle of the beam is^at that point. Find the distance of the
center of gravity of all the loads from the right end. Move the loads so that
the middle of the beam is half way between the center of gravity as found
above and the load which produced the maximum moment at the middle
of the beam. Find the moment under that load, with the loads in the position
just mentioned.

For the case in hand, wheel 12 at 5 gives the maximum moment.
The moment at the right end of the span, wheel 12 being at 5, is:

9 514 + 10 X 214 + 2 X~102 -7-2 = 11 754 000 pound-feet.

197

118 . BRIDGE ENGINEERING

The center o; gravity is — - = 50.2 feet from the right sup-

Zo4

port, or 0.2 foot to the left of the center of the girder. Now place
whee' 12 one-tenth of a foot to the right of the center, and deter-
mine the moment under it. The reaction will be :

R = (9 514 + 9.9 X 214 + 2 X 9~92 + 2) H- 100 = 117.306;
and,

M = 117.306 X 50. 1 - 2 658 = 3 219 306 pound-feet.

In this particular case the difference between the greatest
moment possible and the greatest moment at the middle is not
sufficient to warrant the extra labor involved in computing it. In
general it may be said that if the greatest moment possible occurs within
six inches of the middle of the beam, it is not necessary to compute it,
the moment at point 5 being taken.

58. Shears in Plate=Girders. In the case of through plate-
girders, the maximum live-load shears are determined by placing the

wheels in such a position that Q = - - G is + , and q = —
(G + P)is-.

In these equations, m is the number of panels into which the
span is divided; and the other quantities are the same as given in
Article 45, which should now be reviewed.

For example, let it be required to determine the dead and live
load shears in the through plate-girder of Article 57, p. 111.

The weight of one girder = 61 700 -T- 2 = 30 850 Ibs.
" "4 the floor = 5 X 6 000 = 30 000 Ibs.

Total weight on one girder = 60 850 Ibs.

The dead-load shears are then computed by the methods given in
Strength of Materials, and are given as follows, it being remembered
that the concentrated load which comes at the end is one-half a panel
load, or 3000 pounds: '

T0 = 60 850 H- 2 = 30 475 Ibs. = end shear;

or» ocri

F, = 30475 - 3000 - —=21 305 Ibs.;

or* o c f\

V2 = 30 475 - 3 000 - 6 000 - 2 X — *— = 9 135 Ibs.

V3 = 30 475 - 3 000 - 2 X 6 000 - 2 X - = 3 135 Ibs.

128

BRIDGE ENGINEERING

119

where V0 = the shear at the end ; Vl = the shear just to the left of
point 1; V., = the shear just to the left of point 2; V3 = the shear
just to the right of point 2 ; and Vc = the shear at the middle of the
girder.

The determination of the wheel load position for maximum
live-load shears is given in Table XVII. By comparing the formulae
Q and K, it will be seen that for the first panel Q = K, and q = k,
as n = 1. The position of wheel loads for maximum moments at
point 1 is the same as for maximum shear in the first panel. Accord-
ing to Table XII, wheels 3; 4, and 5 at point 1 all give maximum
shears in the first panel. In this case, as in previous ones, only the
shear for the first position of the loading found- for any particular
point will be determined, as the difference between this and the other
cases is too small to warrant the additional labor necessary in com-
puting them. It is evidently unnecessary to go past panel 3, as only
the maximum shears are required.

TABLE XVII
Wheel Positions, Shears in a Through Plate-Girder

WHEEL

POINT

P+G

REMARKS

POINT

See Table XII, and

text above this table

142 -5- 5 = 28.4

Maximum

152 -s- 5 = 30.4

—

Maximum

152 -5- 5 = 30.4

—

116 Hh- 5 = 23.2

Maximum

116 H- 5 = 23.2 ! 20

—

For wheel 3 at point 1, the left reaction (see Table XIII) is 99.7.
That portion of wheels 1 and 2 which is transferred to point 0 is
230 -r- 15 = 15.33; and the shear in the first panel, therefore, is:

Vl = 99.70 - 15.33 = +84.37.
For wheel 2 at point 2, the left reaction is :

R = (3 496 + 142 X 5) -^ 75 = 56.10;
therefore,

80
F2= 56.10 --5- = + 50.67.

129

120 BRIDGE ENGINEERING

A computation with wheel 3 at point 2 will give a shear only 560
pounds greater, which difference would not influence the design to
any appreciable extent.

For wheel 2 at point 3, the left reaction is:

R = (2 155 + 116 X 1) -5- 75 = 30.30;
therefore,

>3 = 30.3 - ~= +24.97.
1 o

When the girder is a deck one, the computation of the dead
shears is very much simplified, as' all of the load is uniform.

Fig. 110. Beam under Wheel Loads Followed by Uniform Load.

Let it be required to determine the dead and live load shears
at the tenth-points of the deck plate-girder of Article 57.
The total weight of one girder and track is 76 175 Ibs.

V0 = 76 175 -*- 2 = +38 088 pounds.

V, = 38088 - 7~^ = +30 470 pounds.

V2 = 38 088 - — X 76 175 = +22 450 pounds.
V3 = 38 088 - |0 X 76 175 = + 15 230 pounds.

Vt = 38 088 - ~ X 76 175 =+ 7 618 pounds.

^s = 0.

The position of the wheel loads to produce the maximum shear
cannot be determined by the same relation as that used in structures
which have a system of floor-beams and stringers, for here not a
portion, but all of the load to the left of the section, must be sub-
tracted from the left reaction in order to give the shear.

180

BRIDGE ENGINEERING 121

The correct relation f6r the wheel load position will now be
deduced.

Let Fig. 1 10 represent a beam of span I loaded with a series of
wheel loads followed by a uniform load. Let P equal the weight of
the first wheel, W equal the weight of all the loads, and g the distance
from the center of gravity of all of the loads to the right abutment.
The distance between the first and second wheel centers is a, and the
first wheel is at the section b - b at a distance x{ from the left support.
Then,

;? ' W9--
R, =-r,

and

F'b_h = R,' - (loads to left of section) = ^X

Now, assume that the loads move forward the distance a. The
wheel 2 will be at section b-b, and Fig. Ill will represent the
position of the loads. Then,

„„ W (g + a).
«, • z

and

F"b_b = R," - P

_.W(g + a) .

- + - P.

It is now evident that in order to get the greatest shear at section
b-b, wheel 2 must be placed at the section whenever F"b_b.is
greater than F'b-b- Then,

F"b_h > F'b_b;

Wg ,Wa Wg

_ ---- r _P >__.

Now, canceling out the term — ^-, which appears on both sides of the

equation, there results:

For the engine under consideration, a = 8 feet, and P = 10 000
pounds, and the equation reduces to:

II'

131

122 BRIDGE ENGINEERING

which is to say that when the load on the girder is greater than \\
times the span, then wheel 2 should be placed at the section in order
to give the maximum shear.

For loading E 40, the following is true :

For all' sections up to and including the center of all spans, place
wheel 2 at the section to give the maximum shear.

In Fig. Ill it is immaterial whether or not any additional loads
come on the span at the right end when the loads move forward the

> a I

W
It S

Jr^

r /"K JL s~*\ ^

) I

JC A JO

mmm

1 u x ' »^

L___1__J

Fig. 111. Beam of Fig. 110 with Loads Moved Forward.

distance a, as they would only tend to increase the left reaction and
therefore the shear F"b_b. If the relation deduced is true for the
case when no extra loads come on at the right end, it will be true
when they do.

The live-load shears at the left end and at the tenth-points,
wheel 2 being at the section in all cases, are computed from the gen-
eral formula, which is:

V = R - I P,
in which,

R = Left reaction;

I P = All loads to left of section, and is equal to 10 000 pounds for
all sections except the end of the girder.

The computations and results can be conveniently placed in
tabular form, and are given in Table XVIII.

In order to illustrate the use of the relation W> 1 — I, let point

3 in the above span be taken. Place wheel 2 at point 3; then, as
wheels 1 to 13 are on the girder, the total weight W is 212. As I =
100, \\l = 125. Therefore, as 212 is greater than 125, wheel 2 is
the correct wheel.

132

BRIDGE ENGINEERING

123

TABLE XVIII
Maximum Shears in a Deck Plate-Girder

REACTION EQUATION

'IP

REMARKS

(13904 + 4X274)^100

150.00

Wheel 18, 4 ft. from rt. end

1
2

(12041+5X258)-r-100
10816--100

133.31
108.16

10
10

123.31
98.16

Wheel 16, 5 ft. from rt. end
Wheel 15 at right end

( 7 668 + 4X212) -100

85.16

75.16

Wheel 13, 4 ft. from rt. end

( 5848 + 4X172)^-100

65.36

55.36

Wheel 11, 4 ft. from rt. end

( 4632 + 2Xl52)-hlOO

49.36

39.36

Wheel 10, 2 ft. from rt. end

The curves of maximum live-load moments and shears are shown
in Fig. 112. They should always be drawn. From them the shear
or moment at any desired section
can be determined. For exam-
ple, let it be desired to determine
the maximum live-load shear and
moment at a point 24 feet from
the left end of the girder. By
drawing the ordinate, shown by
a broken line in Fig. 112, and
scaling, the following values are
found :

VM =• 88 000 pounds;

M,4 = 2 440 000 pound-feet.

A similar set of curves for
the dead-load shears and moments
should be made. The set for the
deck plate-girder in hand isshown
in Fig. 113. These are easily
constructed by laying off the max-
imum values of the shear at the
end, and the maximum value of
the moment at the center. The
shear curve is a straight line from the end to the center, while the
moment curve is a parabola from the center to the end.

The stresses in the lateral systems of plate-girders are computed
in a manner the same as that employed for the lateral systems of
trusses, the unit-load being taken according to the specifications used.

150

Ife.

•&

''>-.

•T

•i

-•

100

Dbg

2C(

/ -

\^-

500

«i

101

, |

/ u

r\j

•

f\.

-o

. g

•^

Fig. 112. Curves of Maximum Live-Load
Moments and Shears.

133

124

BRIDGE ENGINEERING

59. Stresses in Plate=Girders. The stresses in plate-girders
are treated in the Instruction Paper on Steel Construction, Part IV,
pages 251 to 263, and the student is referred to this treatise for infor-
mation regarding this subject.

The stress in the flange is seen to depend upon the distance from
center of gravity to center of gravity. This distance, in turn, depends

Pounds
Fig. 113. Curves of Dead-Load Shears and Moments in a 100-Foot Span Deck Plate-Girder.

upon the depth of the girder. Certain approximate rules have been
proposed in order to determine this, but the following formula will
give the width of the web plate in accordance with best modern
practice :

d =

0.005 I + 0.543'

in which

d = Width of the web plate, in the even inch;
I = Span, in feet.

For example, let it be required to determine the width of the
web plate of a plate-girder of 80-foot span center to center of end
bearings.

0.005 X 80 + 0.543 0.94

= 85.2 (say 86) inches.

If the resultant value had been 85 inches, the width would have been
taken as either 84 or 86. The reason for this is that the wide plates
kept in stock at the mills are usually the even inch in width and can
therefore "be procured more quickly than if odd -inch widths were
ordered, in which case the purchaser would be forced to wait until

134

BRIDGE ENGINEERING

125

they were rolled — often a period of several months. The distance
back to back of flange angles, the so-called depth of girder, is one-half
inch more than the width of the web. This is due to the fact that
each pair of flange angles extend one-fourth inch beyond the edge of
the web plate, so as to keep any small irregularities caused on the
edge of the web plate by the rolling, from extending beyond the backs
of the angles.

EXERCISES AND PROBLEMS

1. Determine the maximum positive shears in the first six panels of
a 9-panel 114-foot Pratt truss, the live panel load being 8.0. Use the exact
and also the conventional method.

ANSWER:

EXACT SHEARS

CONVENTIONAL SHEARS

+ 32 . 00

+ 24.54

+ 24.90

+ 18.05

+ 18.70

+ 12.48

+ 13.35

+ 7.85

+ 8.90

+ 4.50 '

+ 5.34

2. Find the maximum and minimum stresses in LJJ2 and U3L3 of an
8-panel 160-foot through Warren truss. Height 20 ft.; dead panel load 10.00,
all on lower chord; live panel load 12.00.

ANSWER: In L,U2: d. 1., -28.00; 1.1., - 35. 30 and +1.68;
max., -63.30;min.,"-26.32. In U3L5: d. 1., +16.80; 1.1., + 25. 20
and -5.04;max., +42.00;min., + 11.76.

3. In the truss of Problem 2, determine the maximum stress in L2L3
by the method of moments, and also by the tangent method.

ANSWER: d. 1. = +67.50; 1.1. = +81.00; max. - +148.50.

4. Determine the dead-load stresses in the members U2L2 and L4U6
of a 9-panel 180-foot through Warren truss. Height is 24 feet; dead panel
load is 10.0, one-third being at each panel point of the upper chord, and two-
thirds being at each panel point of the lower chord.

ANSWER: J72L2 = + 30 . 60 ; L,U, = - 1 . 80.

5. Determine the stress in the counter of a through Howe truss of 8
panels and 160-foot span. Height is 30 ft.; dead panel load, 9.6; live panel
load; 11.5.

ANSWER: —4.59.

135

126

BRIDGE ENGINEERING

6. In the truss of Problem 5, determine the maximum and minimum
stress in U2L2, L2U3, and L3C/4.

ANSWER:

UtLt

L2C/3

L3U4

d. 1.
1.1.
1.1.

+ 20 . 80
+ 30.30
-14.40

-17.30
-25.90

+ 5.18

- 5.76
-17.30
0.00

Max.
Min.

+ 15.10
+ 19.36

-43.20 -23.06
-21.20 ± 0.00

Fig. 114. Deck Parabolic Bowstring Truss.

7. In the deck parabolic bowstring truss of Fig. 114, determine the
maximum stress in LjL2, L,C/2, and U3L3. The dead panel load is 4.0, all
on upper chord; and the live panel load, 20.0.

ANSWER: Z^L, = +201.9; L,^ = +21.8; I73L3 = — 33.6.

Fig. 115. Through Bowstring Truss.

8. In the through bowstring truss of Fig. 115, determine the maximum
stress in f7,L2 and L^U^, the dead panel load being 5.0, and the live panel
load 15.0.

ANSWER: U,L2 = +33.50;

+38.0.

136

BRIDGE ENGINEERING

127

9. Determine the maximum and minimum stresses in the members
{/,£/„ f/,L2, UJLi2, and C73L3 of a 7-panel 175-foot through Pratt truss 30 feet
high. Dead panel load is 10.0, all on lower chord; live panel load is 15.0.

ANSWER :

U,L,

VtL+

C/2L2

U3L3

d. 1.
1.1.
1. 1.

+ 10.0
+ 15.0
0.0

+ 26.00
+ 41.70

- 2.78

-10.00
-21.40
+ 6.42

0.00

-12.85
0.00

Max.
Min.

+ 25.0
+ 10.0

+ 67.70
+ 23.22

-31.40

- 3.58

-12.85
0.00

10. Determine the maximum and minimum stresses in the members
C/jWij, m3L3, U2L2, and m2U2 of the deck Baltimore truss shown in Fig. 110.
Dead panel load, 30 000 Ibs.; live panel load, 50 000 Ibs. One-third of dead
panel load is applied at the lower ends of all the verticals.

ANSWER:

L\mt

m3L3

CTjZ,

W.IT,

d. 1.
1.1.
1.1.

+ 190 '8
+ 333 . 5
- 15.1

+ 84.8
+ 191.5
- 50.5

-110.0
-211.0
+ 10.7

+ 21.2
+ 56.6
0.0

Max.

Min.

+ 524 . 3

+ 175.7

+ 276.3
+ 34.3

-321.0
- 99.3

+ 56.6
+ 21.2

11. In the truss of Problem 10, determine the maximum stress in
M2t/2andL3L4.

ANSWER: M2U2 = -840.0; L3L4 = +960.0.

12. Determine the position of the wheel loads of Cooper's E40 loading
to produce the maximum positive live-load shears in the panels of a 7-panel
175-foot Pratt truss.

ANSWER : Lv wheel 4 ; L2, wheels 3 and 4 ; La, wheel 3 ; L4, wheel
3; L5, wheel 2; L6, wheel 2.

13. Determine the maximum positive live-load shears for the truss
of Problem 12.

ANSWER: V, = 192.8; F2 - 137.8; F3 = 90.8; F4 - 52. G;
F5 = 25.0; F0= 6.8.

137

128

BRIDGE ENGINEERING

14. Determine the position of the wheel
loads of Cooper's E 40 loading to produce max-
imum moments at the panel points of the truss
of Problem 12.

ANSWER; Lv wheel 4; L2, wheel 7;
L3, wheels 11 and 12; Lt, wheels 13 and
14.

15. Determine the maximum moments at
the panel points of the truss of Problem 12.
Loading, Cooper's E 40.

ANSWER: Ml = 4820000; M2 =
7745000; M3 = 9192000; M4 =
9 082 000, all in pound-feet.

16. Compute the maximum live-load web
| stresses in the truss of Problem 12, the height
H being 32 feet. Loading, E. 40.

17. Compute the maximum live-load chord
73 stresses in the truss of Problem 12, the height
| being 32 feet. Loading, E 40.

« 18. Compute the impact stresses for all

members of the truss of Problem 12.
2

19. Determine the maximum live-load

g1 shears at the tenth-points of a 65-foot span
deck plate-girder. Loading, E 40.

ANSWER: F0 = 103.0; Vl = 86; F2
= 69.7; F3 = 54.5; V4 = 40.8; F5 =
28.4.

20. Compute the shear due to impact in
the girder of Problem 19.

ANSWER: F0 = 84.7; Vl = 71.5; F2
= 58.8; F3 = 47.0; F4= 35.6; F5 =
25.4.

21. Compute the maximum live-load
moments at the tenth-points of the girder of
Problem 19. Loading, Cooper's E 40.

138

-US

BRIDGE ENGINEERING

129

ANSWER:

POINT

WHEEL

MOMENT

IMPACT MOMENT

6540

5520

11 320

9520

14860

12500

16850

14050

16860

14530

1 .45' from center

16920

14650

All moments are in thousands of pound-inches.

BRIDGE ENGINEERING

PART II

BRIDGE DESIGN

60. General Economic Considerations. The prime considera-
tion which influences the decision to build is cost. After the decision
to build has been made, the problem is one of a purely engineering
character, whereas in the first case it was one of either a political or
an engineering character, or both. The engineering problem is an
economic one, in which maximum benefits must be obtained at a
minimum cost.

A map of the proposed bridge site and the approaches, as well
as of the country for a considerable distance up and down stream,
should be made. This map should show the contours, the soundings,
the borings, the high and low water-mark elevations, and the excep-
tional flood line. On this map the bridge should be plotted in its
proposed location and also in various others. In the case of each of
these locations, various schemes taking into account different numbers
of piers and spans should be considered.

Several authors have attempted to present formulae having a
more or less theoretical derivation and purporting to indicate the
correct number of piers and spans for a minimum cost. The use of
these formulae should not be encouraged, since they do not in any case
give results close enough to serve for anything but a rough guide.

The cost of abutments will vary somewhat with the location and
the character of the approach. This variation is usually small, and
ordinarily an approximate location of the abutments can be quickly
made. As the number of abutments is in all cases constant, their
effect upon the problem of the location of the bridge is small, the main
proposition being that of the cost and the number of piers and spans.

The cost of the piers will usually not be constant, those closer to
the middle of the stream costing more on account of the depth of the
water and the more difficult character of the foundation. Piers

141

132 BRIDGE ENGINEER ING

should not be placed on a skew; neither should they be placed directly
in the maximum line of action of the current. If a skew is unavoid-
able, it should be as small as possible. The cost of piers should be
ascertained by the most careful estimates. In the case of small
bridges where there are only one or two piers, the matter is very simple,
but with a considerable number of piers the problem becomes very
complicated and requires weeks and sometimes months or years for
its solution.

The determination of the cost of the superstructure is a com-
paratively simple matter. In certain instances the class of bridge is
limited to some extent by the specifications. Cooper, in Article 2 of
his "Specifications for Steel Railroad Bridges and Viaducts" (edition
of 1906), gives the following:

Types of Bridges for Various Spans

SPANS

KIND OF BRIDOB

Up to 20 feet
20 to 75 "
75 to 120 "
120 to 150 "
Over 150 "

Rolled beams
Riveted plate-girders
Riveted plate- or lattice-girders
Lattice or pin-connected trusses
Pin-connected trusses

One railroad expresses a preference for plate-girders for all spans
from 20 to 115 feet; and for spans from there to 150 feet, riveted
trusses.

The question as to whether the bridge will be deck or through
is one which is decided by the controlling influences of water-way,
false work, time of erection, and extra cost of masonry. If the clear
height required for the water-way is sufficiently small, the deck
bridge should be chosen, as in this class the cost of false work is less,
the time of erection is less, and the cost of masonry is less by an amount
equal to the cross-section of the piers times the depth of the truss.
Deck bridges also cost less than through bridges of equal span.

The conditions permitting, girders should be used in preference
to trusses. While for equal spans girders are heavier and therefore
cost more, the steel work alone being considered, little or no false
work is required, and the time of erection is much less than in the

142

BRIDGE ENGINEERING 133

case of trusses. This makes the total cost of girder bridges less than
those in which trusses are used. Another item in favor of girders is
their great stiffness.

While pin-connected bridges cost less and are easier to erect,
their stiffness is not so great as that of riveted bridges, which cost
more. The time required for the erection of riveted bridges is also
greater than that for pin-connected bridges. This is on account of
the great amount of time required to make the riveted connections.
For long spans, say over 200 feet, it is necessary to use pin-connected
bridges, as the extreme size of the connection plates prohibits the use
of the riveted type. Also, it is unnecessary to use riveted long-span
trusses to obtain stiffness, as the weight of the pin-connected bridges
is so great when compared with the live load that sufficient stiffness
is obtained.

The cost of spans of different lengths and character may be
obtained directly from the bridge companies; or their weights may
be computed from the formulae given in Article 20, p. 9 (Part I,
"Bridge Analysis"), and multiplied by the unit price which your
experience indicates is correct, thus giving the total cost.

Evidently the solution of problems of this nature cannot be
made within the limits of this text, but the following example will
tend to indicate somewhat the manner of procedure in a problem of
this kind. For example, if the length between abutments is 1 400 ft.,
the cost of each abutment is $12 000, and the cost of each pier is
$15 000, then, if we have fourteen 100-foot plate-girder spans, each
costing $4 300, and thirteen piers, the total cost will be $279 200. On
the other hand, if nine piers and ten 140-foot truss spans, each cost-
ing $9 200, are used, the cost will be $251 000, showing a balance of
$28 200 in favor of the truss scheme. The live loading is E 50.

61. Economic Proportions. The depth of girders is given in
Article 59, Part I.

In the case of trusses, the effect of an increase in the height
is to increase the stresses in the web members and to decrease the
stresses in the chord members. This variation does not affect the
weights to any considerable extent; in fact, a variation of 20 per cent
in the height will not affect the weight more than 2 or 3 per cent.

The height of the bridge is usually fixed by some considerations
which are in turn determined by the specifications. The height must

143

134

BRIDGE ENGINEERING

Panel Lenqtl
istoes feet.

Doable Track

be sufficient to clear whatever traffic will pass through. It should
also be sufficient to prevent overturning on account of the wind
pressure on the truss or on the traffic. In addition, the height of the
bridge is influenced by the depth of the portal bracing. A deep
portal bracing is desirable, in that it stiffens the trusses under the
action of the wind and the vibration due to the passing traffic; but
a deep portal bracing increases the height of the truss and therefore

^ft the bending in the

end-posts due to
the wind. Judg-
ment on the part
of the engineer
should be used in
order to determine
the limiting height
for securin gamax-
imum amount of
benefit as regards
stiffness and a
minimum amount
of bad effect due
to the bending in
the end-posts. Fig.
117, which gives
the height for any
given length of
span, may be said
to represent the best modern practice (1908). Variations of a foot
or more from those given do not affect the weight to any appreciable
extent.

The distance from center to center of trusses for highway
bridges depends upon the width of the street or, if in the country,
the width of the roadway. Streets, of course, vary in width in differ-
ent localities, but country highway bridges usually have a roadway of
from 14 to 16 feet in the clear.

In the case of railroad bridges, the distance from center to center
of trusses depends upon whether the track is straight or on a
curve, and also upon whether the bridge is a deck or a through bridge.

Single Track

5 pan in f«ct.

Fig. 11

Curves Showing Relation between Height of Trusses
n Double
ay Bridg

and Length of Span in Double- and Single-Track
Railw

144

BRIDGE ENGINEERING 135

The actual amount varies in most cases, and is fixed by specification.
Some specifications require that when the track is straight, the dis-
tance from center to center of trusses shall be 17 feet; or that, in case
one-twentieth of the span exceeds the 17 feet, then one-twentieth of
the span shall be used.

For deck plate-girders the common practice appears to be to
space them as given below :

Width of Plate-Girder Bridges for Various Spans

. SPANS

DISTANCE CENTER TO CENTER OF
PLATE-GIRDERS

Up to 65 feet
65 to 80 "
80 to 115 "

6 feet 6 inches
7 feet 0 inches
7 feet 6 inches

For through plate-girders the spacing should be such that no
part of the clearance diagram will touch any part of the girder. In
case of double-track plate-girders with one center girder, great care
should be exercised in order that the center girder shall not be so deep
nor have so wide a flange as to interfere with the clearance diagram
(see Fig. 126).

On account of the wind on a train which runs on track placed
at the elevation of the top chord of deck bridges, the overturning
effect is exceedingly great, and special care should be taken that the
height and width are such as to prevent overturning.

In through bridges the clearance must be such as to allow the
clearance diagram to pass. Special attention should be paid to the
knee-braces and also to the portal braces. When the bridge is on a
tangent, the spacing of the trusses is a comparatively simple matter,
being just sufficient for the clearance diagram; but on curves, allow-
ance must be made for the tilt of the diagram due to the super-
elevation of the outer rail, and also allowance must be made for the
fact that the length of the cars between trucks forms a chord to the
curve, and as such the middle ordinance must be taken into account.
It is also necessary to allow for that part of the car which projects
over the trucks, as this will extend beyond the outer rail by an amount
greater than one-half the width of the clearance diagram. (See
Figs. 119 and 120.)

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BRIDGE ENGINEERING

137

62. The Clearance Diagram.
The clearance diagram is not
supposed to represent the outline
of the largest engine or car which
may run over the line, but repre-
sents the maximum amount of
space which may be taken up by
objects which are to be shipped
over the line. For instance, the
lower part of the clearance dia-
gram may allow for snow-plow
or ballast distributors, and the
upper part may take into account
the passage of such material as
carloads of lumber, piles, or tele-
graph poles. The standard clear-
ance diagram of the Lehigh
Valley Railroad is given in Fig.
118. This diagram is for the
clearance on straight track only.
On curves, the diagram tilts as
showrn in Fig. 119, and to allow for this tilting the Lehigh Valley
Railroad requires 2.V inches additional clearance on the inside of
curves for each inch of elevation of the outer rail. In addition to this
tilting effect, the clearance should also be increased on account of the

Fig. 119. Clearance Diagram on Curves,
Showing Tilting.

Fig. 120. Standard Car on Curve, Showing Necessity for Wider Spacing of Trusses.

length of the cars and their projection over the outer and inner rails.
Fig. 120 shows a standard car according to the specifications of the
American Railway Engineering & Maintenance of Way Association,
in such a position on a single-track span as to show the effect of the
curve upon the widening of the spacing, center to center of trusses.

147

138 BRIDGE ENGINEERING

This car is 80 feet long, 60 feet between centers of trucks, and is as
wide as the clearance diagram, 14 feet for single track. It is evident
that the trusses cannot be spaced so as to interfere with the clear-
ance line of the body of the car and its projecting ends. These
clearance lines are represented as broken lines in Fig. 120, and are
marked c-c. Note that the center of the track is seldom in the center
of the floor-beam. Also, it is evident that the sharper the curve,
the greater the required distance between trusses, and accordingly
the greater the floor-beams in length. This varies the moment in the
different floor-beams and therefore makes them more costly. The
stringers, also, are more costly, on account of the fact that their ends
are skewed. On account of the eccentricity of the track, one truss
takes more of the load than the other, and therefore the trusses are
not the same — a fact which further increases the cost.

From the above it is seen that almost all conditions incident to
the building of a bridge on a curve tend to increase the cost; and
hence a fundamental principle of bridge engineering: Avoid build-
ing bridges on curves.

63. Weights and Loadings. For the weight of steel in any
particular span, and for the loading required for any particular class
of bridge, see Articles 20 to 23, Part I. The weight of the ties and
the rails and their fastenings is usually set by the specifications at
400 pounds per linear foot of track. For highway bridges the weight
of the wooden floor is usually taken at 4^ pounds per square foot of
roadway for every inch in thickness of floor.

Highway bridges are divided into different classes according to
their loadings (see Cooper's Specifications). The decision as to the
class to be employed depends somewhat upon the distance to the
nearest bridge across the same stream. In case the nearest bridge
is only a few miles away and is of heavy construction, it is not actually
necessary to construct a heavy bridge at the proposed site, the heavier
traffic being required to pass over the other bridge. In case a heavy
bridge is not in the neighborhood, then one shpuld be constructed at
the proposed site. If the proposed site is on a road connecting adja-
cent towns of large size, then a heavy bridge should be constructed
and provision made for future interurban traffic, even if none is at
that time in view, since it will be more economical to do this than to
erect a new bridge in the future.

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140 BRIDGE ENGINEERING

In the case of railroad b'ridges, new ones are nearly always con-
structed to carry the heaviest main line engines. These are usually
of a class corresponding to Cooper's E 40 or E 50. In some localities
branch-line bridges are built for the same live loadings; but in the
majority of cases the branch-line bridges consist of the old bridges
from the main line.

64. Specifications. For any particular bridge the specifica-
tions are either written by the engineer in charge, or some of the very
excellent general specifications which are on the market in printed
form are used. Some railroads use these general specifications
with the addition of certain clauses which are desired by the chief
or bridge engineer. The principal differences in these general
specifications are in regard to the allowance for impact.

Whenever highway design is mentioned in this text, it is to be in
accordance with Cooper's Highway Specifications (edition of 1901).
Wherever plate-girder design is given, it is in accordance with Cooper's
Railway Specifications (edition of 1906); and wherever truss design
is given, it is in accordance with the general specifications of the
American Railway Engineering & Maintenance of Way Association
(second edition, 1906).

65. Stress Sheet. Before the sections are designed, the com-
puter makes a skeleton outline of the truss, and on this places the
dead -load and live-load stresses, and, in case the wind should be
considered, the wind-load stresses. This is sent to the designer. The
designer determines the various sections, and also the moments and
shears in the stringers and floor-beams. These are placed on a sheet
usually 17 by 23 inches. This is called a stress sheet. This sheet is
now given to the draftsman, who makes a shop drawing. The stress
sheets for railroad bridges are usually more elaborate than those for
highway bridges. Plate I is the stress sheet of a highway bridge;
and Plate II (Article 78) and Plate III (Article 93) are examples of
the best modern practice in the making of plate-girder and truss-
bridge stress sheets.

66. Floor System. Perhaps no part of bridge design is better
standardized than the construction of the open steel floors for
railroad bridges. The stringers are usually placed 6 feet 6 inches
apart, and consist of small plate-girders, or, if the panel length is
.short, of one or more I-beams. I-beams are economical in regard to

150

BRIDGE ENGINEERING

141

TABLE XIX
Safe Spans for I Beams

(Based on unit-stress of 10 000 Ibs. per square inch in extreme fibre)

WEIGHT
PER FOOT
(Lbs.)

1 £-

ENGINE CLASS E 40

ENGINE CLASS E 50

Safe span C to C of Bearings

1 Beam
per rail

2 Beams
per rail

3 Beams
per rail

1 Beam
per rail

2 Beams
per rail

3 Beams
per rail

9 in.
9 "
9 "
10 "
10 "
10 "
12 "
12 "
12 "
12 "
15 "
15 "
15 "
15 "
15 "
18 "
18 "
18 "
18 "
18 "
20 "
20 "
20 "
20 "
20 "
24 "
24 "
24 "
24 "
24 "

25
30
35
30
35
40

40
50
65
42
50
60
70
80
55
65
75
80
90
65
75
85
90
100
80
85
90
95
100

92
102
112
135
163
175
218
274
332
403
443
515
619
718
774
809
. 890
1023
1 063

1 180
1 277
1 453
1502
1 650
2 112
2182
2356
2427
2497

51
5
6
6

s
9
9
11
12
13
12
13
15
K;

K;
16

IS
21

t. 3i
9

0
3
6
6
9
9
9
3
6
3
0
9
0
9
6
9
6
0

9 •
lo •
11 •
18 '

12 '

r.', •

t. 9 in.
9 "
6 "
0 "
0 "
0 "
9 '
9 '

4f
6 '
5 '

7 '
7 '

9.'

b. 9 ii
0
3
9
0
6
9
6

I t
8

11
11
18

Ki
15
11
15

16
18

isi
17

20
20

21
20

88
23

25
26

27
27

t. Oin
• 0 '
' 9
6
0
0
3
3
6
0
0
3
9
6
3
9
9
0
6
9
9
3
0
3
6
9
3
3
9
' 0 '

3f
4
4 '
4
5
7
8
7
8

11
12

11
11

11
it
15

15
1C,
17
17

t. 6 in.
' 3 "
' 6 "
9 "
9 "
3 "
6 "
6 "
6 "
6 "
6 "
0 "
0 "
9 "
6 "
9 '
6 '
9 '
3 '
3 '

3 ft. 0 in.
" 6 "

" 9 "
" 0 "
' 9 "

15 '

18 '

15 '

16 '

Ml '
20 '

21 •
2(1 '

•jo •

22 '

o •

6 '

3 '
0 '
9 '
6 "
0 "
9 "
6 "

8 '

in •

HI '

9 '

lo •
11 •

0 "

0 "
3 "
3 "

9 "

3 "
9 "
6 "
6 "
6 "
9 "
3 "
6 "
0 "

•11 '
12 '

11 •

18 •
14 '

15 •
14 '

15 •

16 '

If, •

16 '

ir i

3
3
6
6
6
9
3
0
0
0
6
6
6
3
6
9
9
6
0
6 '
0 '
6 '

24 " 6 '

23 " 0 '
24 " 0 '

12 '
11 '
12

3 "

9 "
0 "
0 "
3 "
' 6 "

22
2::

•st
24

25
25

• 3 "

' 0 "
' 9 "
' 6 "
' 0 "
' 3 "

26 " 6 '
28 " 0 '
29 " 3 '
29 " 9 '
30 " 9 '
31 '• 0 '
32 " 0 '

13 ' 6 "
14 ' 0 "
14 ' 6 "
14 ' 9 "
15 ' 0 "
15 ' 3 "

MI •

20 '
21 '

21 '
22 '
22 '

DIAGRAM

ENGINES

§ Ills 1111

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first cost, but are disadvantageous on account of the eccentric con-
nections which necessitate heavy brackets to resist part of their re-
action. They are also somewhat undesirable on account of the fact
that, the ties deflecting, most of the load is carried by the inner
I-beam. However, I-beams for stringers and for short-span bridges
(see Fig. 121) are much used in present practice, and give good re-
sults. Figs. 121 to 127 show the standard open floor sections of the
I^ehigh Valley Railroad. Table XIX gives the required number of
I-beams, together with their weight, which are to be used for short-
span bridges or as stringers in panels of given length.

Solid floors consist of angles and plates, channels and plates, or
other shapes. They extend transversely across the bridge from truss

151

142

BRIDGE ENGINEERING

i. s'-n" il

to truss, the lower
chords, in case of
truss bridges, be-
ing made heavy
enough to act as
girders as well as
tension members.
Figs. 128 to 130
show sections of
solid floors. The
ballast is laid di-
rectly upon these
solid floors, which
are first covered
with a good damp-
proof paint. The
floors should also
be supplied with
good drainage fa-

Lb ^'V

lei ^ H?

t rs=^_U_ — 8x9x10'

T ^ i r" *

,-. c c T

Fig. 121. Sectional View Showing Open-Floor Construction of
Railroad Bridge of Short Span, Single Track. I-Beams
used for Stringers. Lehigh Valley Standard.

1 °

— 1

t [Hrr~~ii~- :rr5. 8 * 9 " lo'-°'

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=JL

nF-1 v

-g," — -

cilities.

Concrete is
sometimes laid di-
rectly upon the steel floor, and the ballast put upon this concrete,

Fig. 122. Section of Open-Floor Construction of Deck-Girder

and Truss Bridge, Single Track. Lehigh Valley Standard.

Plate-Girders used for Stringers.

Fig. 123. Floor Construction of a Through-Girder Bridge, Single Track.
Lehigh Valley Standard.

159

BRIDGE ENGINEERING

145

which has previously had a
"layer of some good waterproof-
ing applied on its upper sur-
face.

67. Practical Considera=
tions. The possibilities of the
rolling mill and the various
shops of a bridge company,
such as the drafting room,
forge, foundry, templet shop,
assembling shop, and riveting
and finishing shop, and also
the shipping and erect *r.g facil-
ities, should be well known in
order to make the most eco-
nomical use of them. This
requisite knowledge comes
only from experience. The
best way to obtain this experi-
ence without being actually
employed in the shops, is to go
into the shops every chance
that presents itself, keep your
eyes and ears open, and ask all
the questions you can. The
use to be made of handbooks
of the various steel manu-
facturers is given in Part I
of "Steel Construction," and
should be thoroughly studied
before going further. Some
one of these handbooks is in-
dispensable to persons design-
ing steel structures. That of
the Carnegie Steel Company
(edition 1903) is one of the
best, and will be frequently
referred to in the present text.

155

146 BRIDGE ENGINEERING

Copies may be procured from the Carnegie Steel Company, Frick
Building, Pittsburg, Pa. The usual price to students is 50 cents,
to others $2.00.

DESIGN OF A PLATE=GIRDER RAILWAY-SPAN

68. The Masonry Plan. In some cases the general dimensions
of the masonry are limited ; such a case, for example, would occur in
the crossing of a street or narrow waterway. Here the length of the
span and the distance above the street or the surface of the water,

Drain Piper I1.

Fig. 129. Solid-Floor Construction of Plates and Angles.

are the limited dimensions. The span and under-clearance may be
unlimited, as in the case of a country stream crossed by a roadway
which is a considerable distance above the surface of the water.
The term unlimited is not here used in its exact meaning, as the span
in this case is really limited by the cost, which rapidly increases with
the length of the span.

In some cases, as when the engineer is in a bridge office, the
masonry plans are sent in by the railroad. In such cases many of the
limited dimensions are fixed. The most usual dimensions to be fixed
are the elevation of base of rail, the elevation and size of the bridge
seat, and the length of the span under coping. These limit the
depth of the girder, or the depth of the floor if it be a through
girder, and also limit the length of the bearing plates at the end.
Fig. 131, the masonry plan of a road crossing, shows in general what
can be expected. All the dimensions usually fixed are given, and those
marked x and y may or may not be, but x should never be less than
3 feet.

69. Determination of the Class. As before mentioned, the
deck plate-girder should be used if possible, since its cost is less.
There are some cases, however — such as track elevation in cities —
where the additional cost required to elevate the track so as to use a

156

BRIDGE ENGINEERING 147

deck plate-girder will more than balance
the saving in its favor. In such cases the
through plate-girder is used.

The case whose design is under con-
sideration will be taken similar to that of
Fig. 131, and the span will therefore be a
deck one.

70. Determination of the Span, Cen=
ter to Center. Fig. 132 shows the various
spans — namely, under coping, center to cen-
ter of end bearings, and over all. The span
under coping is that span from under cop-
ing to under coping lines of the abutments,
and is so chosen as to give the required dis-
tance between the abutments at their base.
The span center to center is equal to the
span under coping plus the length of one
bearing plate. The span over all is the
extreme length of the girder. The length
of the bearing plate is influenced by the
width of the bridge seat, and also by the
maximum reaction of the girder. The
length should seldom be greater than 18
inches and never greater than 2 feet, as
the deflection of the girder will cause a
great amount of the weight to come on the
inner edge of the bearing plate and also on
the masonry, in which case the masonry is
liable to fail at that point and the bearing
plates are over-stressed.

Cast-steel bearings are now almost
universally used. They decrease the height
of the masonry, and distribute the pressure
more evenly and for a greater distance
over the bridge seat. When these castings
are used, the bearing area between them
and the girder may be made quite small,
thus doing away to a great extent with the

157

148

BRIDGE ENGINEERING

deteriorating effect due to the deflection of the girder as mentioned
above. Fig. 133 shows the end of a girder equipped with a cast-
steel pedestal. Table XX gives the length of the bearing on the

asg of Rail-

elevation j;

[/Center Line ot Road

ipan under Copinq

= ep'-o-

°o
Q

Elevation 756.

T
^—

Fig. 131. Masonry Plan of a Road Crossing.

masonry for various spans, Cooper's E '40 loading being used and
cast-steel pedestals being employed.

TABLE XX
Length of Masonry Bearings

SPAN

LENGTH OF BEARING

15 to 24 f(
25 to 44
45 to 69
70 to 79
80 to 89
90 to 115

ot-

12 inches
16
21
23
29
36

As an example, let it be required to determine the span center

^- Masonry Plate Masonry Bearino'r»F=i
r Span Under Copim) M

•K

\ Soon Center to Center Bearinq ""^

, ' \ Span Over AH /

Fig. 132. Diagram Showing Various Spans Considered in Bridge Construction.

loading being E 40. From Table XX.it is seen that the length of
the masonry l>caring will 1x3 21 inches, and therefore the span center

158

BR IDGE ENG IN E E R I XG

149

to center of bearings will be GO + 2 X ($ X 1 ft. 9 in.) = 61 feet
9 inches.

In Articles 71 to 77 the above girder will be designed ; and also
such information as is of importance regarding the subject-matter
of each article will be treated. The dead- and live-load shears are
computed by the methods of Part I, and are given in Fig. 134.

71. Ties and Guard=Rails. The length of ties for single-track
bridges is 10 feet.
For double-track
bridges the
length is in most
cases the same.
In some double-
track bridges,
however, either
each tie or every
third tie extends
entirely across
the bridge. In
other cases every
third tie on one
track extends to
the opposite
track, thus act-
ing as a support
for the foot-walk
which is laid

Upon them. It Fig 133. End of a Girder Equipped with a Cast-Steel Pedestal.

is the best prac-
tice to limit the length of the ties on double-track bridges to 10 feet,
since, if they extend into the opposite track in any way whatsoever,
unnecessary expense is incurred whenever repairs or renewals are
made, because both tracks must necessarily be disturbed to some
extent.

The size of the ties varies with the weight of the engines and the
spacing of the stringers or girders on which they rest. They are
usually sawed to size instead of hewn, and the following sizes may be
purchased on the open market— namely, 6 by 8, 7 by 9, 8 by 9, 9 by 10,

110

150

BRIDGE ENGINEERING

1600

Fig. 134. Shear and Moment Diagram.

160

BRIDGE ENGINEERING 151

and 10 by 12 inches. Larger sizes may be obtained on special order.

The elevation blocks (see Fig. 127) should be of length to suit the
width of the cover-plates and the spacing of the supports. They are
usually made of the best quality of white oak, since the cost of renewal
is great enough to demand that they be made of material as permanent
as possible.

The guard-rails should be placed in accordance with the specifica-
tions (see Articles 13 and 14). Some railroads specify that the guard-
rails shall be in 24-foot lengths unless the bridge is shorter than 24
feet, in which case one length of timber should be used. For method
of connection and other details, consult Figs. 121 to 127. The
guard-rails and the ties are usually made of Georgia long-leaf yellow
pine, prime inspection. Other wood, such as chestnut, cedar, and
oak, may be used.

In addition to the wooden guard-rail, a steel guard-rail usually
consisting of railroad rails is placed within about 8 inches of the
track rail.

In designing ties, the problem is that of a simple beam symmet-
rically loaded with two equal concentrated loads, the weight of the
rail and tie itself usually being
neglected. For the case in hand,
which is that of a deck plate-
girder, loading E 40, the con-
centrated load for which the tie
must be designed is, according
to Specifications (Article 23, 3d

r»ar^ Q 3*^ r»rmnrlc A^nrv-Hr,. Fig. 135. Distribution of Loading on Ties Of
partj, O 666 pounds. According Deck plate-Girder Bridge.

to Article 23, 100 000 pounds is

on four wheels. This gives 25 000 pounds on one wheel, and ac-
cording to Article 15, one-third of this, or 8 333 pounds, will come
on one tie. Fig. 135 shows the condition of the loading, the space
center to center of rail being taken as 4 feet 10 inches. Some
designers take this distance as 5 feet; but as the standard rail head
is about 2 inches, and the standard gauge 4 feet 8£ inches, the distance
here taken seems to be the more logical one.

The formula to be used in the design of this beam is that given

in "Strength of Materials," and is M - — • In this case M = W X

181

152

BRIDGE ENGINEERING

8 333 = 83 330 pound-inches. In the above formula, / = bd*+ 12,
and c = d ~- 2, and therefore - = — -. Substituting the value

of the moment in the above formula, and solving for S, there results'

499 980

For a 6 by 8-inch tie, the unit-stress wTould then be :
S = 4699^ = 1 310 pounds.

If a 7 by 9-inch tie is used, the unit-stress is found to be 880 pounds.

Since according to Article 15 of the Specifications, the unit-stress

cannot be greater than 1 000 per
square inch, it is necessary to use
a 7 by 9-inch tie. If the engine
loading had been E 50, the mo-
ment would have been 100000
pound-inches, and then the stress
in a 7 by 9-inch tie would be 1 060
pounds per square inch, and the
stress in an 8 by 9-inch tie would
be 930 pounds, which would neces-
sitate the use of the latter.

The guard-rails on this bridge
will be placed according to the
Lehigh Valley standard, and hence
their inner face will be 4 feet 1£
inches from the center of the
track.

Elevation blocks will not be
required, as the bridge is on a
tangent.
72. The Web. The economic depth of the web, according to

Article 59, Part I, will be:

61 ft. 9 in.

Fig. 136. End Rivets Transferring Shear
to Web.

Depth =

— =72.5.

0.005 X 61 ft. 9 in. + 0.543
The depth might be taken as 72 inches, but 74 inches will be decided
upon, as this will decrease the area of the flange and also will not
affect the total weight to any great extent- The unit-stress for shear

162

BRIDGE ENGINEERING

153

is 9 000 pounds per square inch (see
Specifications, Articles 40 and 41).

The maximum shear in the girder
occurs at the end, where it is 117 800
pounds. The area required for the web
is then 117800 -r- 9000 = 13.09 square
inches, and the required thickness is 13.09
-T- 74 = 0.177 inch. This latter value
cannot be used, since, on account of Ar-
ticle 82 of the Specifications, no material
less than f inch can be used. The web
plate will therefore be taken as 74 in. by
f in. in size.

Some engineers insist that the net
section of the web should be considered.
Consider Fig. 136, the shear being trans-
ferred to the web by the end rivets. The
web will not tend to shear along the
section B-B, in which case the rivet-holes
should be subtracted; but it will shear
along section A- A, a section which is
unaffected by the rivet-holes. The web
splice should come at one of the stiffeners,
and will therefore be considered in Arti-
cle 76.

73. The Flanges. This portion of
the girder is usually built either of two
angles or of two angles and one or more
plates. In heavy girders where the flange
areas are large, additional area is ob-
tained by using side plates or side plates
and four angles. Sometimes two chan-
nels are used in the place of side plates
and angles. Fig. 137 shows the different
methods of constructing the top flanges
of girders. The lower flanges are usually
of the same construction. Fig. 137 b has
the web extending beyond the upper sur-

163

154

BRIDGE ENGINEERING

et or less

jotorUss.

faces of the upper flange angles. This is done in order that the
ties may be dapped over it, and thus prevent the labor usually
required for cutting holes in the lower face of the tie in order to
allow for the projecting rivet-heads. Fig. 137 g is usually uneconom-
ical, since the thinness of the channel web requires a great many
rivets to sufficiently transmit the shear from the web to the flange,
and also since the cover-plates must be very narrow.

Specifications usually state that the flanges shall have at least
one-half of the total flange area in the angles, or that the angles shall
be the largest that are manufactured. The largest angles are not
usually employed, since their thickness is greater than three-quarters
of an inch and therefore the rivet-holes must be bored, not punched.
The reason for this is that the depth of the rivet-hole is too great in
proportion to its diameter, and on this account the dies used for
punching frequently break. Also, the punching of such thick material

injures the adjacent metal, which
makes it undesirable. In reality
the flange area of only the short-
span girders is small enough to
allow the flange area to be taken up
by the angles.

In choosing the thickness of the
cover-plates, care should be taken
so that the outer row of rivets will
not come closer to the outer edge
of the plate than eight times the
thickness of the thinnest plate. In

case eight times the thickness of the plate is greater than 5 inches,
then 5 inches should be the limit. Also, the distance between the
inner rows of rivets should not exceed thirty times the thickness of
the thinnest plate. These limitations are placed by Article 77 of the
Specifications, and Fig. 138 indicates their significance.

The determination of the required flange area depends upon the
distance between the centers of gravity of the flanges; and in order
to determine this exactly, the area and composition of the flanges
should be known. The above condition requires an approximate
design to be made, the supposition being that the flanges consist of
two angles and one or more plates as shown in Fig. 138. .

Fig. 138. Diagram Showing Relation be-
tween Thickness of Cover-Plates
and Position of Rivets.

164

BRIDGE ENGINEERING 155

The distance back to back of angles will be taken as 74 + 2 X
s = 74j inches. Article 74 of the Specifications requires j^ inch ;
but I ; inch is better practice, since the edges of the web plate are very
liable to overrun more than -^6 inch. Some specifications require
j inch.

In the computation of the approximate flange area, the center
of gravity of each flange will be assumed as one inch from the back
of the angles. The approximate effective depth is then 74j less 2
X 1 inch, which equals 72£ inches. The approximate stresses in
the flange areas are :

, 275 000 X 12
I* or dead load, - — = 4o 000 pounds.

, , 1 340 000 X 12
For live load,] - - = 222 000 pounds.

The approximate flange areas are now obtained by dividing
these amounts by the allowable unit-stresses for dead and live load,
which are (see Specifications, p. 8, Article 31): 20 000 and 10 000
pounds per square inch respectively; and the resulting areas are:

45 600

For dead load, — ^ = 2 . 28 square inches.
ZU UUU

222 000
For live load, -T-- = 22. 20 square inches.

These amounts give a total of 24.48 square inches as the approximate
net flange area required.

It will be assumed that one-half the total area, or 24.48 -r- 2 =
12.24 square inches, is to be taken up by angles. If 12.24 sq. in.
is distributed- over two angles, then 12.24 -f- 2 = 6.12 square inches
is the net area for one angle. Of course it is not to be assumed that
the area of the angle chosen must be exactly 6. 12, but that this 6. 12
square inches is the approximate area of the angle to be chosen, and
the net area of the angle (see Specifications, Article 149) must not be
2 £ per cent less than this, although it may be greater.

From Steel Construction, Part I, Table VII, or from the Car-
negie Handbook, p. 117, a 6 by 6 by f-inch angle gives a gross area of
8.44 square inches and a net area of 8.44 - 2 X (£ + i) X f =
6 . 94 square inches, J-inch rivets being used and so spaced that two
rivets are taken out of each angle (see Specifications, Article 63,
arid Fig. 139). A 6 by 4 by | flinch angle, giving a gross area of
7.47 and a net area of 6.66 square inches, one rivet-hole being out,

lli.'i

150

BRIDGE ENGINEERING

could have been used, but { jj inch is too thick to punch, and there-
fore the above angle is chosen.

The required net area of the cover-plate is now found to be
24.48 - 2 X 6.94 = 10. 60 square inches. Since the legs of the
angles arc G inches and the thickness of the web is | inch, the outer

edges of the angles are 12f inches
apart; and since the cover-plate
must extend somewhat over the
edges of the angle, and the width
of the cover-plate should be in the
even inch, the width of the cover-
plates must be at least 14 inches,
as shown in Fig. 139.

On account of the 1-inch rivet-
holes to be deducted, the real or
net width of the cover-plate " is : 2 X n + m = 14 — 2X1 = 12
inches. The thickness of all the cover-plates at the center is now:

Fig. 139. Calculation of Size of Angles
and Cover-Plate.

— ;,— - 0.885 inch— say,

kich.

A thickness of I of an inch is decided upon, for the reason that plates
are rolled only to the nearest sixteenth of an inch.

The approximate section at the center has now been determined,
and is:

2 Angles 0 by <> by f-inch = 13.88 sq. in. net.
Cover-plates | inch thick = lO.SOsq. in. net.

Total = 24.38sq. in. net.

This approximate section must now be examined, and, if it shows too
great an excess or a de- ^ 0875.. |£86

ficiency, must be revised.
In order to deter-
mine the effective depth
the distance between the
centers of gravity of the
flanges must first be com-
puted, the gross areas be-
ing used. Theoretically, perhaps, the net areas should be used;
but this is an unnecessary refinement, since the effect on the final
result is of no practical importance.

Fig. 140. Determination of Center of Gravity.

166

BRIDGE ENGINEERING 157

In computing the center of gravity (see Fig. 140), the axis is
taken at the center of the cover-plates, as this reduces the moment of
the cover-plates to zero. The distance of the center of gravity of the
angles from their back (Carnegie Handbook, p. 117, column 6) is
1 .78 inches'. The distance of this center of gravity from the center
of the cover-plate, is 1 .78 + 0.875 -^ 2 = 2.22 inches.

Gross area of the angles = 2 X 8.44 = 1G.88 sq. in.
" " " " cover-plates = Jx 14 = 12.25sq. in.

Total = 29.13 sq. in.

The center of gravity is now found to be 16.88 X 2.22 4- 29.13 =
1.286 inches from the center of the cover-plate, and 1.286-0.875
-^ 2 = 0:848 inch from the back of the angle. The effective depth
he is 74.25 - 2 X 0.848 = 72.554 inches, and the required flange
areas are:

275 000 X 12

72 . 554 X 20 000

1 340 OOP X 12

"727554 X 10 000

= 2.272 sq. in. for dead load.
= 22.200 sq. in. for live load.

Total = 24.472 sq. in.

The values of the moments, as taken from the curves, must be mul-
tiplied by 12 in order to reduce them to pound-inches.

A total of 24.86 square inches is given by the section approxi-
mately designed, and the difference between that and the section as
above determined is : (24 . 472 - 24 . 38) -=- 24 . 472 = 0 . 38 per cent,
and as this is less than 2-£ per cent (see Specifications, Article 149), it
may be used without any further change. If there should have been
a deficiency or an excess greater than 24 per cent, then it would have
been necessary to revise. In case a revision of section is necessary, the
size and thickness of the angles generally remain the same as those
taken in the approximate design, the thickness of the cover-plates
being decreased or increased as the case may be.

The total thickness of the cover-plates, | inch, is too thick to be
punched. In such cases as this, the section is made up of two or more
plates whose total thickness is equal to that required. If plates of
more than one thickness are decided upon, then their thickness
should decrease from the flange angles outward. For the case in
hand, one plate £ inch thick and one plate I inch thick will be decided
upon. The flange section at the center as finally designed is:

167

158

BRIDGE ENGINEERING

SHAPE

NET SECTION

GROSS SECTION

2 Angles 6 by

6 by f in.

13.88 sq. in.

16.88sq.

in.

1 Cover-plate

14 by f in.

4.50 "

5 .25 "

1 Cover-plate

14 by \ in.

6 . 00 "

7 . 00 ' '

Total 24. 38 ' '

29.13 "

The above is the section required at the center of the girder;
for any other point it will be less, decreasing toward the end, where
it will be zero. Evidently, then, the cover-plates will not be required
to extend the entire length. The following analysis will determine
where they should be stopped. If the load were uniform, the moment

Fig. 141. Diagram Showing Curve of Required Flange Areas.

curve would be a parabola. Although under wheel loading the curve
of moments is not a parabola, yet it is sufficient for practical purposes
to consider it as such. The curve of flange areas, like that of moments,
is to be considered a parabola (see Fig. 141).

Let a, = Net area, in square inches, of the outer cover- plate;

a2 = Net area, in square inches, of the next cover- plate;

«3, etc. = Net areas of the other cover-plates;

A = Net area of all the cover-plates and the flange angle.
Then, from the properties of the parabola,

where L = Length of cover-plate in question;
I = Length of span, center to center;
a = Net area of that cover-plate and all above it ; and
A = Total net area of the flange, I of the gross area of the web not being
considered in this quantity.

168

BRIDGE ENGINEERING 150

The lengths of the cover-plates for the section above designed (see Fig. 141)
are:

L, = 61.75 J ^§r = 26.45 feet.

L, = 61.75 = 40.00 feet.

One foot is usually added on each end of the cover-plate as theoretically
determined above. The results are also usually rounded off to the nearest
half-foot. This is done in order to allow a safe margin because of the fact
that the curve of flange areas is not a true parabola. The final measurements
of the cover-plates are:

14 in. by f in. by 28 ft. 6 in. long.

14 in. by ^ in. by 42 ft. 6 in. long.

In most cases the cover-plate next to the angle on the top flange only is
made to extend the entire length of the girder. Although this is not required
for flange area, it is done in order to provide additional stiffness to the flange
angles toward the ends of the span, and to prevent the action of the elements
from deteriorating the angles and the web by attacking the joint at the top
(see broken lines, Fig. 141, for length of first cover-plate extended).

EXAMPLES FOR PRACTICE

1. The dead-load moment equals 469 000 pound-inches; and the live-
load moment, 4 522 000 pound-inches. Design a flange section entirely
of angles, if the distance back to back of angles is 45} inches.

2. The dead-load moment is 3 340 000 pound-inches, and the* live-
load moment, 21 235 000 pound-inches. Design a flange section using G by
6-inch angles and three 14-inch cover-plates, the distance back to back of
flange angles being 78} inches.

3. In each of the above cases, design the flange section considering
that i of the web area is taken as effective flange area. (For demonstra-
tion of the methods to be employed in the solution of this problem, see the
succeeding text.)

While the section of a plate-girder is composite — that is, it con-
sists of certain shapes joined together, and is not one solid piece —
nevertheless these shapes are joined so securely that the section may
be considered as a solid one and its moment of resistance computed
accordingly. Let Fig. 142 be considered.

The moment of resistance of the section is:

in the derivation of which the moment of inertia of the flange about
its own neutral axis is considered as zero, and A equals the net area of
one flange. Now, as the values of he and h seldom differ by more

160

BRIDGE ENGINEERING

CG.Flanqe-

than one inch, for all practical purposes they may be considered* as
equal. The above expression then reduces to :

M = S x h ( A + -^ )

= S X h (net area of flange + one-sixth gross area of web)

Since the rivet-holes decrease the moment of resistance of the
web, one-sixth of the gross area cannot be considered, as is theoreti-
cally indicated in the above formu-
la. It is common practice to take
one-eighth, instead of one-sixth, of
the gross web area. Substituting
this value in the above equation,
and transposing, there results:

M
Area of flange -I- I gross web area = ^ •

The flange section will now be
designed for the moments previously
given, considering £ of the gross web
area as efficient in withstanding the
moment.

The gross area of the web is
74 x f = 27.75 square inches; and
^ of this is 3.47 square-inches. The
total approximate amount of flange
area required is, as in the first case,
24.48 square inches.

According to the above formula,
| of the web area, or 3.47 square
inches, may be considered as flange
area, and therefore 24.48 - 3.47 -
21.01 square inches, is the approxi-
mate area of the angles and cover-plates of the flange. The ap-
proximate area of one angle is then 21 .01 -=- (2 X 2) = 5.25 square
inches. A 6 by 6 by rVinch angle gives the gross area of 6.43 square
inches and, two rivet-holes being deducted, a net area of 5.305
square inches (see "Steel Construction," Part I, Table VIII, or
Carnegie Handbook, p. 117). As this is quite close to the approxi-
mate area determined above, this angle will be taken. The ap-

Fig. 142. Section of Plate-Girder.

170

BRIDGE ENGINEERING

161

proximate area of the cover-plates is 21.01 - 2 X 5.305 = 10.40
square inches. As before, the gross width of the cover-plate will

10 40
---- '— - —

= 0 . <S67 inch— say f inch.

be taken as 14 inches. The thickness

then

The gross area of the angles being 12.86 square inches, and that
of the cover-plates 12.25 square inches, the center of gravity of the
section is found, by a method similar to that previously employed,
to be 1 .10 inches from the center of the cover-plate, or 1.10 — 0.438
= 0 . 662 inch from the back of the flange angles. This makes the
effective depth 72.93 inches.

For this section, the true live-load flange stress is (1 340 000 X
12) -T- 72 . 93 = 221 000 pounds, and the actual dead-load flange stress
is (275000 X 12) -1-72.93 = 45400 pounds. The actual areas
required for the live and dead load are 22. 10 and 2.27 square inches,
which are obtained by dividing the above flange stresses by 10 000
and 20 000 pounds, respectively. The total required area is the sum
of the two areas above, and is equal to 24.37 square inches. The
total area required in the flange angles and cover-plates is therefore
24.37, less £ the gross area of the web, 3.47, which leaves 20.90
square inches. The same angles as decided upon before will be used.
This gives a required area for the cover-plates of 20.90 — 10.61 =
10.29 square inches. The required thickness is then 10.29 + (14
— 2) = 0.857 — say f inch. The following section of the flange will
therefore be decided upon:

SHAPE

NET SECTION

GROSS SECTION

2 Angles 6 in. by 6 in. by T9^ in.
1 Cover-plate 14 in. by H in.
1 Cover-plate 14 in. by \ in.

10. Gl sq. in.
4 . 50 "
6 . 00 "

12.86 sq. in.
5.25 "
7.00 "

Total =

21.11 "

25.11 "

As the total net area above is within 2^ per cent of the required net
area, that section will be taken (see Specifications, Article 149). Note
that in this case, the thickness of the cover-plates in the final design
is the same as that determined in the preliminary design.
Also note that the total net area is about 4 square inches, or 20 per

171

162 BRIDGE ENGINEERING

cent, less than in the flange as first designed, in which case none of the
area of the web was considered as withstanding the bending moment.
The ^-inch cover-plate on the top flange will extend the entire
length of the grider, and is therefore 62 feet 9 inches long. The
lengths of the other cover-plates are:

For i-inch plate at the bottom, L = 01 .75 -y' ^ ' '^ = 43 . 5 feet.

For each f-inch plate, L = 61.75 ^^lT = 28-5 feet-

One foot should be added to each of the above lengths at each end,
thus making the total lengths 45 feet (i inches and 30 feet 6 inches,
respectively.

EXAMPLES FOR PRACTICE

1. If the span is 63 feet center to center, compute the length of the
cover-plate. The section consists of two angles 6 by 6 by f in.; one cover-
plate 14 by \ in.; and one cover-plate 14 by % in.; two rivet-holes being taken
out of each angle and each cover-plate.

2. If the span is 87 ft. 9 in. Center to center, compute the length of the
cover-plate if the flange consists of two angles 6 by 6 by | in., and four cover-
plates 16 by •j'g- in., two rivet-holes being taken out of each angle and each
cover-plate.

In determining the area of plates, the tables in the Carnegie
Handbook, pp. 245 to 250, are convenient. In order to obtain plates

F F F F C F

Pig. 143. Diagram Illustrating Transference of Shear from Web to Flanges by Rivets.

whose widths are greater than 12f inches, see the note in the right-
hand column on page 250, Carnegie Handbook. For another pres-
entation of the above subject-matter, see "Steel Construction," Part
IV. pp. 252, 254, and 261.

The spacing of the rivets in the flanges is a matter of considerable
importance; the shear is transferred from the web to the flanges,
where it becomes flange stress. This is done by the rivets, each rivet
taking as much flange stress as is allowed by the Specifications. The
conditions are similar to those shown in Fig. 143, where V represents

178

BRIDGE ENGINEERING

163

an object exerting a pull on a long, thin plate A - A which has, at
various points along this length, small objects F attached to it by
means of pegs r - r. These small objects F hold the plate A- A in
equilibrium. Here V represents the shear which tends to cause the
movement; A- A, the web; r-r, the rivets; and 2F the amount of
flange stress taken by each rivet.

At section c - c the total amount in the web to be trans-
formed is 2F; at section b - b
it is 10F. From this it is seen
that enough rivets r-r must be
put in between the sections b -
b and c - c to take up 10F —
2F = 8F; hence it is proved
that the rivets between any two
sections of the flange take up
the difference in flange stress
between those two sections.

The discussion just given
will be the means of giving us the
number of rivets required between any two sections; but it does not
give us the rivet spacing between these two sections. In order to
determine the rivet spacing at any particular point, the following
analysis is presented (see Fig. 144).

Let A/\ = Moment at one section,

M2 = Moment at another section nearer center of girder than the section

where Af, occurs;

V = Shear at section where M , occurs;
v = Amount of flange stress one rivet can transfer; or it is the stress on

one rivet;

s = Distance between the two sections;
n = Number of rivets between the two sections.

Then,

Fig. 144. Determination of Rivet Spacing.

— - — Flange stress due to moment M,;

_ 2 = Flange stress due to moment M.,;

2 i = Difference of flange stress between the two sections;

( 2 _ — J \ -j- v = n, Number of rivets required in space «...(!)

173

164

BRIDGE ENGINEERING

If the above sections be taken close enough together so that
the number of rivets required will.be 1 (that is, n = 1), then V
can be considered as constant between the two sections, and then
the moment M2 = Mv + Vs (see Article 44, Part I). Substituting
in Equation 1, above, there results:

/M. + V* M, \

from which,

rhe

which is the formula for the rivet spacing in the vertical parts of the
flanges of any girder, providing the flange is not subjected to localized
loading. It is to be used for the rivet
spacing in both the top and bottom
flanges of through girders, but not in
the top flanges of deck plate-girders for
railroad service. It is to be used, how-
ever, in the bottom flanges of deck plate-
girders for railroad service. The dis-
tance Ae is not ordinarily used, the
distance between rivet lines being used
instead (see Fig. 145). The rivet spacing

Fig. 145. Determination of Rivet jn the cover.plates and horizontal legs of

the angles is made to stagger with that

in the vertical legs, and usually the staggering is with every other rivet
in the vertical flange. The term stagger signifies that the rivets in
the top flange are not placed opposite the rivets in the vertical legs
of the flange angles — or, that in case there are two lines of rivets
in the vertical legs of the angle, a rivet near the outer edge of the
cover-plate is placed in the same section where a rivet occurs near the
lower edge of the vertical legs of the angle, and vice versa.

EXAMPLES FOR PRACTICE

1. Determine the rivet spacing at a section where the shear is 147 200
pounds, the value of one rivet 4 920 pounds, and the effective depth of the
section 84} inches.

ANS. 2.82 inches.

2. Determine the stress on a rivet at a section where the shear is
299 400 pounds, the spacing 2^ inches, and the effective depth of the girder
84} inches.

ANS. 8 870 pounds.

174

BRIDGE ENGINEERING

165

The rivet spacing is usually determined at the tenth-points;
and a curve is plotted with the spacing as ordinates and the tenth-
points as abscissse. The rivet spacing at any intermediate point can
be determined from this curve. When one-eighth the gross section
of the web is considered as flange area, then only that proportion of
the shear which is transferred to flanges is to be considered in com-
puting the rivet spacing, on account of the fact that some of the
shear is transferred directly to the bending moment in the web.

In order to determine the distance between rivet lines, the
gauge, or distance out from the back of the angles to the place where
the rivets must be placed, must be known for different lengths of legs.
Table XXI gives the standard gauges, and also the diameter of the
largest rivet or bolt which is allowed to be used in any sized leg. No
gauges should be punched otherwise unless your large experience or
instructions from one higher in authority demand it, and this should
be so seldom that indeed it might be said never to be necessary.

TABLE XXI

Standard Gauges for Angles
(All dimensions given in inches)

MAXI-

MAXI- ||

MAXI-

MUM
RIVKT
oi{ BOLT

u •

MUM

RIVET
OR BOLT

MUM
RIVET OR
BOLT

-H

3
2i

.2*

*When thickness is J- inch or over.

175

166

BRIDGE ENGINEERING

Fig. 146. Determination of
Distance between

The distance between rivet lines for the girder being designed
(see Fig. 146), is, in the first case:

= 74.25 - (2 X 2i + 2J)
= 67.00 inches.

In the second case, where | of the web is
considered, the above distance is 74.25 —
(2 X 2J + 24) = 67.25 inches. The compu-
tations and the rivet spaces at the tenth-
point, and at the ends of the cover-plates in
the bottom chord of the plate-girder, are
shown for each case in Table XXII. The
value of v is the value of a |-inch rivet in
bearing in a f-inch web (see Specifications,
Article 40, and Carnegie Handbook, p. 195,
second table). This value is 4 920 pounds.

In the first column, 7.98— indicates that
the end of the cover-plate next to the flange

is 7.98 feet from the end of the girder, and that this section is taken
just to one side of that point, the side being that nearest the end of
the girder. In a similar manner, 7.98+ indicates that the section
'is taken to that side of the point which is nearest the center of the
girder. A like interpretation should be placed on 15.55— and
15.55 + , the point under consideration in this -case being the end of
the outer or top cover-plate.

In the fifth column, values are given which indicate that portion
of the shear which is transferred to the flanges. For example,
Q7 700

40 " 10 en = 74 700' and the difference between 97 70° and

74700 represents that portion of the shear which is taken up
directly by the web in the form of bending moment. An inspection
of the headings of the third and fourth columns will tend to make
this matter clearer.

Where there is local loading, as in the top flange, the rivets, in
addition to the stress caused by the transferring of web stresses, are
stressed by the vertical action of the angles being pressed downward
by the ties and the consequent upward pressure of the web. Accord-

176

BRIDGE ENGINEERING

167

TABLE XXII
Rivet Spacing in Bottom Flange

Flange Taking All the Moment

SECTION

TOTAL SHEAR
(Pounds)

(Inches)

(Pounds)

RIVET
SPACING
(Inches)

REMARKS

117800

4290

2.80

97700

4290

3.38

79300

4290

4.16

61 300 67

4290

5.38

44 200 67

4290

7.46

I See Specifica-

28 600 -

4290

11.52

\ tions, Art. 54

One-Eighth of Web Area Considered
hi ~ 67.25 inches; v = 4 920 pounds

NET FLANGE

REDUCED

RlV ET

SECTION

(Pounds)

WEB AREA
(Sq. Inches)

AREA
(Sq. Inches)

SHEAR
(Pounds)

SPACING
(Inches)

117800

14.08

10.61

88800

3.75

97 700

14.08

10.61

74 700

4.45

7.98-

90 000

14.08

10.61

67 900

4.88

7.98 +

90000

20.08

16.61

74800

4.42

79300

20.08

16.61

65800

5.02

15.55-

67500

20.08

16.61

56 000

5.81

15.55 +

67500

24.58

21.11

58 100

5.69

61 300

24.58

21.11 .

52 700

6.28

44 200

24.58

21.11

38000

8.71

28600

24.58

21.11

24 600

13.42

ing to Article 15 of the Specifications, the weight of one driver is
distributed over three ties (see Fig. 147).

W
Let -j- , = w, the load per linear inch caused by one wheel W, which

load is assumed to be uniformly distributed over the dis-
tance 7;

ws = t'i, the vertical load or stress that comes on one rivet in the
space s;

Vs
v = -T— , the stress on a rivet due to the distribution of flange

stresses when s is a space, and V the shear at that point.

When these two stresses act on the rivet, the maximum stress will
be vu, the ultimate amount that it is allowed to carry, and this will
act as shown in Fig. 147 : Then,

177

168

BRIDGE ENGINEERING

r- ) + (w*)2

from which,

which gives the spacing at any point in the girder flange under
localized loading. Note that if w equals zero — that is, if there is no
localized loading — there results:

which is the same as previously deduced for flanges with-
out localized load-
ings.

The rivet spacing
for the top flange
of the girder which
is being designed
is given in Table
XXIII. Here W
= 20000; / = (3X
7 + 3 X 6) = 39

. , 20 000

inches; w = — ~—

= 513; hr = 67
inches; and vw =
4 920 pounds, which
is the bearing of a
f-inch rivet in the
1-inch web. The
top cover-plate is
run theen tire length
of the span.

Fig 147. Rivet Spacing Determined by Stresses Distributed
under Localized Loading.

178

BRIDGE ENGINEERING

109

TABLE XXIII
Rivet Spacing in Top Flange

Flange Taking All the Moment

SECTION

TOTAL SHEAR
(Pounds)

RIVET SPACE
(Inches)

var-

262 GOO

117800

3 080 000

1 825

2.70

' '

97700

2 140 000

1 550

3.17

79 300

1 390 000

1 285

3 . 83

61 300

840 000

1050

4.68

" .

44200

435 000

835

5.88

28600

181 500

660

7.45

One-Eighth of Web Area Considered
= 513; ht = 67i inches; vu = 4 920 pounds.

SKCTION . w-

REDUCED
SHEAR
(Pounds)

(*)•

RIVET SPACE
(Inches)

var-

262 600

97600

2 100 000

1 538

3.20

262 600

81000

1 450 000

1315

3.74

262 600

65800

985 000

1 100

4.50

15.55-

262 600

56000

695 000

980

5.02

15.55 +

262 600

58 100

765 000

1 014

4.85

262 600

52 700

616 000

938

5.24

262 600

38 000

320 000

763

6.44

262 600

24600

134 000

629

7.82

The points other than the tenth-points referred to in the first column
are for sections taken just to the left and right of the top cover-plate.
The values of the reduced shears given in the third column are ob-
tained as has been previously explained. Although the rivet spacing
in the lower flange is considerably greater than that in the upper
flange, and accordingly a smaller number of rivets would be required,
yet the spacing in the lower flange is made the same as that in the
upper. Convenience in the preparing of plans and facility in manu-
facture make this action economical. Theoretical spacing greater
than 6 inches should be dealt with according to Article 54 of the
Specifications.

The values of the rivet spacing given in Tables XXII and XXIII
are plotted in Fig. 148. Note that the effect of the localized loading
is to decrease the rivet spacing, and also note that the effect increases
from the ends toward the center.

179

170

BRIDGE ENGINEERING

£9

"d 8

Note.-5ccond Cover Plate oi

Top Flange Extends Lntire
Length, of Girder.

o \ e 3 ^ 5

Fig. 148. Plotted Values of Rivet Spacing Given in Tables XXII and XXIII.

180

BRIDGE ENGINEERING

171

The size of the flange angles and the width of the cover-plates
for different spans, are a matter of choice. Once the size is deter-
mined, the thickness can be computed. The sizes very generally
adopted in practice are as follows:

SPANS

WIDTH OF COVEU-PLATE

15 to 20 feet

5x3^ inches

none

20'

6x4

' '

25'

6x6 '

' '

40'

6x6 '

inches

75'

100

6x6 '

100' 120

8x8

For another method of the presentation of this subject, see
"Steel Construction," Part IV, pp. 264 to 268.

EXAMPLE FOR PRACTICE

1. Determine the rivet spacing for the top chord of a plate-girder,
loading E 40, and 7 by 9-inch ties being used. The web is f inch thick;
distance from back to back of angles, 6 feet 6^ inches; flange angles, 6 by 6
by ^-inch; and cover- plate, 14 by jj-inch, two J-inch rivets'being out of each.
First, consider the flange as taking all the bending moment; and second,
consider one-eighth the gross area of the web. The total unreduced shear
is 80 000 pounds in both cases.

ANS. 3 . 21 inches; 3 . 76 inches.

74. Lateral Systems and Cross=Frames. There are two methods
in use in common practice in determining where the panels of the
lateral bracing shall fall — namely, (1) To choose the number of panels
so that the panel points come opposite the stiffness, and (2) to choose
the number of panels so that the placing of the panel points is inde-
pendent of the stiffener spacing. The lateral systems should be of
the Warren type; and in both of the above cases the angles that the
diagonals make with the girder should not be greater than 45 degrees.
Also, it is best to have all panels the same length and to have an equal
number of panels. This latter condition will simplify the drafting
very much, since one-half of one girder can be drawn and the other
half will be symmetrical, the opposite girder being similar to the one
drawn, but being left-handed.

The members of the lateral systems will take tension or com-
pression according to the direction the wind blows. Cross-frames
are placed at intermediate points to stiffen the girders. These are

181

BRIDGE ENGINEERING 173

diagonal bracings (see Plate II), and are placed at certain intervals
according to the judgment of the engineer. Good practice demands
that their number should be about as indicated below :

SPAN OF GIRDER

NUMBER OF CROSS-FRAMES

0 to 20 feet

2 .

20 to 35 "

35 to 70 "

70 to 85 "

85 to 110 "

The above is not intended to serve as a hard and fixed rule. Varia-
tions from the limits given are to be made as the case demands. In
all cases they are put at the panel points of the bracing, the top and
bottom parts acting as sub-verticals in the lateral system. Also, the
cross-frames should divide the span into equal parts if possible. In
cases where that is not possible, the shortest divisions should come
near the ends of the spans.

If the panel points are to be located at the stiffene'rs, the number
of panels is a function of the depth of the girder (see Specifications,
Articles 47 and 48) . In this case the number of panels is given by :

,. _ Span in inches.

Depth of girder in inches'

no fraction being considered. As an example, let it be required to
determine the number of panels in a girder 85 feet center to center
of bearings, the depth being 90| inches back to back of angles. Then,

85 X 12
N = qn~~9~~ = 11-3, or, say, 11 panels.

Each panel will then be 92.8 inches long. This, according to Article
47 of the Specifications, being greater than 5 feet, would not be allowed
as a space between two stiffeners; but one stiffener can be placed in
between, and then the panel points will come at every other stiffener.
The cross-frames should be five in number.

The arrangement of panels and cross-frames is shown in Fig.
149. Here the cross-frames are marked C. F., and the broken lines
represent the lowrer lateral system.

In case an even number of panels were desired, then ten would
be the number chosen and the general arrangement would be as
shown in Fig. 150. The length of a panel would be 85 X 12 4-

183

174

BRIDGE ENGINEERING

10 = 102 inches, or 8 feet 6 inches, which would allow of one stiffener
in between and still keep the stiffener spacing within the limit of
5 feet.

The cross-frames at the ends of the span are designated as end
cross-frames, and those in between are designated as intermediate
cross-frames.

In case the spacing of the stiffeners is not required to be such as
to coincide with the panel points of the lateral bracing, the panel
length will depend upon the spacing of the girders, being equal to or

Fig. 150.
Arrangements of Panels and Cross- Frames.

greater than the spacing in order to keep the angle which the diagonals
make with the girder less than 45 degrees. In this case,

Span in feet

N — — • — — •

Distance center to center of girders in feet

For the girder considered on page 1 74, the number of panels would be

11.3 — or, say, 11 panels — if odd numbers were to be used,

85
7.5
and 12 if even numbers were to be desired.

For the case in hand, the panel points of the bracing will be taken

at the stiffeners, and an even number of panels will be used. Then,

61.75 X 12

74.25

9. 98 (say 10).

The arrangement of the panels and cross-frames, and also the maxi-
mum stresses in the diagonals, are shown in Plate II, the stresses being
determined according to Article 50, Part I, "Bridge Analysis," and
Article 24 of the Specifications. All the wind is taken as acting on
one side of the bridge; and no overturning effect, either on the girder

184

BRIDGE ENGINEERING

- 175

or on the train, is considered. Also, note that the wind stresses in the
flanges a-re not considered. Should the student determine these, he
will find them too small to be considered according to Specifications,
Article 39.

Before designing the lateral diagonals which consist of one or
two angles, Articles 31, 33, 34, 35 (last portion), 38, 40, 63, and 83 of
the Specifications should be care-
fully studied. The upper lateral ^ e^ft. j

bracing is to be designed first.
Carnegie Handbook, pp. 109 to
119, is to be used.

The member U0U1 must be
designed for a compressive stress
equal to 23.20 + 0.8 X 20.6
= — 39.68, and a tensile stress of
20.6+0.8X20.6 = +37.08. The
length of the diagonal measured
from center to center of girders
is 1/6.5' +6.22 - 9 feet, or 108
inches. In reality the length is not

108 inches, as the cover-plate takes off a certain amount, as shown
in Fig. 151. The true length, which is to be taken as a column length
in designing, is 108 — 2y, and y is readily computed to be 9.70 inches,
thus making the true length 88.6 inches. The least allowable rectan-
gular radius of gyration is obtained from the relation that the greatest

Fig. 151. Determination of Length of
Diagonal in Lateral Bracing.

value of - = 120, and therefore the least value of r =

0.74.

r 120

It will be assumed that a 6 by 4 by r96-inch angle with an area of
5.31 square inches will be sufficient. Here the length equals 88 . 60
inches, and the least rectangular radius of gyration is 1 . 14; hence,

oo £*r\

P = 13 000 - 60 X ~~
1 .14

= 8 330 pounds per square inch,
on QAQ

The required area is — — = 4 . 73 square inches. As the angle
8 ooO

assumed has an area of 5.31 square inches, which is considerably
greater than the 4.73 square inches required, this angle cannot be
used, and other assumptions must be made until the area of the angle

185

170

BRIDGE ENGINEER ING

assumed and the required area as computed are equal or very nearly so.

A 6 by 4 by ^-inch angle with an area of 4.75 square inches will

now be assumed. The length is 88 .60 inches as before, and the least

rectangular radius of gyration is 1.15. The unit-1'oad P = 8340

39 360
pounds per square inch, and the required area is ' * = 4 . 72 square

o o4U

inches. As the area of the angle assumed and the required area as
computed are very close, this sized angle will be used.

The section must now be examined for tension, and in order that
both legs of the angle may be considered as effective section, both legs

must be connected at the end.

™, • i -11 u 36 700
1 he area required will be

18 000

2.04 square inches.

o o \o / \
[_ o_ _P\gx_ -X \

ui 1

Cover Plate -j (

Considering one rivet-hole is taken out of the
angle, the net area is 4 . 75 — 1 X
(1 + s ) X $ = 4 . 25 square inches,
which is amply sufficient.

If the 4-inch leg only were as-
sumed to be connected, the gross
area would be 4 X $ = 2. CO
square inches, and the net area
would then be 2.00 - 1 X (A +
s) X ^-inch = 1 .50 square inches,
which is not sufficient. If the
6-inch leg were connected, the
area would be sufficient. See Fig.
152 for method of connection and
rivets.

The number of rivets required (see Specifications, Article 38
and 40) is computed as follows: If the member were not subjected
to both tension and compression, the number of rivets required in
single shear would be:

39360_

(9 000 + 50 per cent of 9 000) X 0 . 6013
_ 39360
8100

= 4. 86 (say 5).

But according to Article 38 of the Specifications, this number must be
increased 50 per cent, and accordingly 4.86 X U =7. 29 (say 8)

Fig. 152. Method of Connecting Angle Legs
in Lateral Bracing and Cross-Frames.

186

JM .
31

S 5

K f
W j?

°J

M -0

w 2
SB 75
5 -2

^ §
Cd =
O *

§«

BRIDGE ENGINEERING 177

shop rivets are to be used. In the above formula, 0.6013 is the area
of the cross-section in square inches of a f -inch rivet. In order that
both legs should be connected, a clip angle as shown in Fig. 152 is
used ; and the same number of rivets must go through both legs of the
clip angle, since the stress in the vertical leg of the main angle is
transferred to the clip angle and from there into the connecting plate.

The above number of rivets makes the joint safe so that it will
not shear off in the plane between the connection plate and the hori-
zontal leg of the angle. The joint must also be designed so that
there will be sufficient rivets in bearing to prevent them from tearing
out of the connecting plate. The number required is:

= 39 360 _____

~ (15 000 + 50 per cent of 15 000) X | X f

- I39 3GO
= ~7ll80

= 5.34.

This 5.34 must be increased 50 per cent, making a total of 5.34
X 1.5 = 8.01 = say, 8 shop rivets as before.

The above rivets are shop rivets, since it is assumed that the span,
being a small one, will be riveted complete in the shop and shipped to
the bridge site ready to place in position without any further riveting.
In case the girders are shipped separately, then the lateral bracing
must be riveted up in the field ; and according to last part of Article
-40 of the Specifications, the rivets, being field rivets, must have the
allowed unit-stresses reduced one-third, which is equivalent to having
the number of shop rivets increased 50 per cent. This will make the
required number of field rivets 8 X 1.5 = 12.

As a rule, the connection plates are f inch thick, seldom more.
Also, the members of the upper lateral system are connected on the
lower side of the connection plate in order not to interfere with the
ties. Note that the use of the clip angles requires a smaller connec-
tion plate than would be necessary if these angles were not used, since
in the latter case all the rivets must then be placed in one row in the
horizontal leg of the angle.

The number of rivets required in the connection plate and the
flange of the girder must be sufficient to take up the component of
that member parallel to the girder. For the case in hand, the num-
ber (see Fig. 153) is: »

187

178

BRIDGE ENGINEERING

x _612
T ~{TO'
from which,

x = 5.5 (say 6) rivets.

Additional rivets should also be put in, in order to take up the compo-
nent of the other lateral diagonal which meets at this point.

The member LT/ Ul is to be designed for a maximum compressive
stress of 20 . 6 + 0 . 8 X 16 . 0 - - 33 . 4. A 6 by 4 by -, Vinch angle

with an area of 4.18 square
inches will be assumed. The
least rectangular radius of gyra-
tion is 1.1G. The unit allowable
load is:

P = 13 000 - 60 X f^rt = 8 420
1 . lo

pounds per square inch.

The required area is

8 420

3.97 square inches. As this is
very near the area assumed, and
as trials with other angles do
not give required areas which
come any closer, this angle will
be used.

The rivets required in single
shear are :

33400

6^ ft.

55 nv.

Fig. 153. Calculation of Rivets in Connection
Plate and Flange of Girder.

33 400

X 1.5 = 6.21 (say 7) shop rivets, and

8100
6.21 X 1.5 = 9.3 (say 10) field rivets.

The rivets required in bearing in the f-inch connection plate are:

—jj-ngh X 1.5 = 6 . 78 (say 7) shop rivets, and
« 6.78 X 1.5 = 10.17 (say 11) field rivets.

The above computations show the joint to be weakest in bearing,
and therefore 7 shop or 11 field rivets must be used. It is not neces-
sary to investigate this member for tension, as the computations for
the first diagonal indicate that the area will be sufficient, both legs
being connected.

The member UJJ2' must be designed for a maximum compres-
sive stress of 16.0 + 0.8 X 14.1 = -27.28. A 6 by 4 by f-inch

188

BRIDGE ENGINEERING 179

angle with an area of 3.61 square inches and a least rectangular
radius of gyration of 1.17 will be assumed. The unit-stress P, as
computed from the formula in the Specifications, is 8 400 pounds per

27 280

square inch; and the required area is --— ~ — = 3 . 23 square inches.

8 460

This angle will be used, as the given and required areas are close
together, and as the next smaller angle — a 6 by 3i- by §-inch angle
with an area of 3.42 square inches — gives a required area of 3.58
square inches, thus being too small.

The rivets required in single shear are :

27 280

— X 1.5 = 5 . 04 (say 5) shop rivets, and
o 1UU

5.04 X 1.5 = 7. G (say 8) field rivets.
The rivets required in bearing in a f-inch web are :

27 280

X 1.5 = 5 . 54 (say 6) shop rivets, and
7 ooO

5.54 X 1.5 = 8.3 (say 9) field rivets.

In order to make the joints safe, 6 shop or 9 field rivets should be used.
The member U2'U2 must be designed for a maximum compressive
stress of 9.6 + 0.8 X 8.0 - -16.00. A 3^ by 3 by f-inch angle
with an area of 2.30 square inches and a least rectangular radius of
gyration of 0.90 will be assumed. The unit-stress P is 7 090 pounds

. 16000
per square inch, and the required area is _ ft(~- = 2 . 26 square inches.

As the required and the 'actual areas are very close together, this
angle will be used.

The rivets required in single shear are :

X 1.5 = 2.96 (say 3) shop rivets, and
o lOU

2.96 X H = 4.44 (say 5) field rivets.

By computation similar to the above, it is found that 4 shop or 5
field rivets are required in bearing. Since the bearing requires the
most rivets to make the joint safe, 4 shop or 5 field rivets must be used.

If the Specifications would have allowed a 3! by 3^ by j5<pinch
angle with an area of 2.09 square inches, this angle would have
exactly fulfilled the requirements, the required area being 2.09
square inches.

The member U2Uar must be designed for a maximum compres-

189

180 BRIDGE ENGINEERING

sive stress of 8.0 + 0. 8X4. 1= -11.28. A 3 by 3 by 1-inch
angle with an area of 2.11 square inches and a least radius of
gyration of 0.91 will be assumed. In this case the unit-stress is
7 160, and the area required is 1.58 square inches. The required
area is considerably less than the area of the angle assumed; but it
must be used, since it is the smallest allowed by the Specifications,
which require that the material shall not be less than f-inch, and
from Table XXI it is seen that 3 inches is the smallest size leg in which
a | -inch rivet can be used.

The stresses in all the members of the lower lateral system are
less than the stresses in the member just designed, and therefore all
members of the lower lateral system will be made of one 3 by 3 by
f-inch angle.

For the last member designed in the upper lateral system, and
for all members in the lower lateral system, 3 shop or 5 field rivets
will be required at the ends. These are more than sufficient to take
up the stress, but it has been found that less- than three rivets do not
make a good joint.

The stress sheet, Plate II, shows the general arrangement of the
lateral system, the number of rivets in the connections and also in the
connection plates where they join the flanges.

The intermediate cross-frames do not lend themselves to a theo-
retical design, since the stresses which come upon them are not easily
ascertained. It is good practice to require that all members be of
the sizes as given below:

RIVETS

SPAN OF GIRDER
(in Feet)

ANGLES
(in Inches)

SHOP FIELD

30 to 65
65 to 110

3i-x 3* x f
4 x 4 x f

4 5

The angles in the intermediate cross-frames will therefore be 3\ by
3\ by §-inch.

The end cross-frames (see Fig. 154) act in a manner somewhat
similar to the portal bracing in a bridge, since they transfer all the
wind which comes on the top chord and on the train to the abutment.
This load, which acts at the level of the ties, is in this case (see Article
24 of the Specifications) :

190

BRIDGE ENGINEERING

181

P =

600 X 61 ft. 9 in.

18 525 pounds.

It is usually assumed that half of this is' transferred to the point a by
means of a-b, and from there down a-b' to the masonry. The other
half goes directly down b-ar to the masonry. This causes stresses as
shown in Fig. 154. Note that the stress in a-b will always be com-
pression; but the stresses in the diagonal will be either tension or
compression according to the direction the wind blows. The mem-
ber a-b will be a
3^ by 3o by f -inch
angle. To form
the connections at
its end, 3 shop or
5 field rivets will
be used.

The maxi-
mum compressive
stress for which
the diagonals are
to be designed is
12.70 + 0.8X12.70
= -22.86. Here
the length is 108

inches if the angle tends to bend one way; but if it bends as
shown by the broken lines in Fig. 154, the length will be one-half
of this. For this reason, angles with unequal legs should be used,
the longer leg extending outward. This allows the greatest rectan-
gular radius of gyration to be used.

A 4 by 3 by

and a radius of gyration of 1 .25 will be assumed. The unit-load P
is computed to be 8 750 pounds, and the required area is therefore
22 860 -r- 8 750 = 2.61 square inches. This does not coincide very
closely with the given area, but will be used since this angle comes
nearer to fulfilling the condition than any of the other sizes rolled.
The joint will require more rivets in bearing than in single shear.
It is not necessary to perform the complete computations in order to
determine this, since a comparison of the values of a rivet in single
shear and in bearing shows that the value in bearing is less than that

Fig. 154. Action of End Cross-Frames.

1 6 -inch angle with an area of 2.87 square inches

191

182

BRIDGE ENGINEERING

in single shear, and therefore the number of rivets required in bearing
will be greater than that number required in single shear. The
number of rivets required in bearing is:

22 860
4920-XTB =^4 ^°p rivets, and

4.00 X 1.5 = 6 field rivets.

75. The Stiffeners. According to Article 47 of the Specifica-
tions, these should be placed at certain intervals whenever the unit-
shear is greater than

5 = 10 000 - 75 X ~ = 10 000 - 14 800 = -4 800 pounds.

This negative
sign signifies that
whenever the
unit shearing
stress is greater
than zero, the
stiffeners must be
placed through-
out the entire
length of the
span at distances
not to exceed 5
feet.

The interme-
diate stiffeners
should have the
outstanding leg
long enough to
give good sup-
port to the flange angle (see Fig. 155). The filler bars or fillers
are put in so as to allow the stiffener angles to remain straight
throughout their entire length; otherwise they will have to be
bent as shown in Fig. 156. This bending is called crimping.
Stiffeners must also conform to Article 48 of the Specifications.
This would require a different sized stiffener at each point, and also
a different number of rivets in each stiffener. This is not usually
done in practice. In practice the stiffener for the first intermediate

Fig. 155. Use of Straight Stiffeners, with Filler Bars.

IQi

BRIDGE ENGINEERING

183

point is designed, and the remainder are made the same in size and
have the same number of rivets. An exception to this is where a
stiffener comes at a web splice. In this case the size is usually kept
the same, but the number of rivets is changed somewhat to conform
to the requirements of the splice design.

The second intermediate stiffener comes at the first tenth-point,
and is 6. 175 (say 6.2) feet from the end, since it is at the first panel
point, or opposite the first panel point, of the lateral system. The
first stiffener will be 3 . 1 feet from the end ; and scaling off the value
of the shear at this point (see Fig. 134), it is found to be 108000

Fig. 156. Crimping of Stiffener Angles
'where No Filler Bars are Used.

Fig. 157. Section of Intermediate Stiffener
Construction.

pounds. Here the length I to be used in the formula for the unit
allowable compressive stress is 74J — 2 X f = 72.75 inches, the f
being the thickness of the flange angle. The section of the
material which according to Article 48 of the Specifications is to be
considered as a column, is shown in Fig. 157. The assumed
column cannot bend about the axis B -B, but about the axis A- A,
and therefore the radius of gyration about the axis A - A must be
computed. The moment of inertia of the fillers and the web plates
about their own axes is considered as zero.

A 4 by 4 by ^-inch angle with an area of 3 . 75 square inches will
be assumed to be sufficient to withstand the stress. The moment
of inertia of this and the filler bars and the web portion is

/A_A = 2 (5.55 + 3.75 X 2.12- 4- 3.00 X 0.5632) = 46.70r

193

184

BRIDGE ENGINEERING

The radius of gyration, then, is, */ ' •-= 1.764, and the unit-stress

computed with this value and a length of 72 . 75 inches is 8 140 pounds.
The required area is now determined to be 108 000 -=- 8 140 - 13 . 27
square inches. The value 15.00 used in the above computation for
the radius of gyration is the value of the area of the angles, the filler
bars, and the web portion. A 5 by 3i-inch angle with the 5-inch leg
out would have given better support to the flange, but would not

make so good a job, as it would
have extended about £ inch be-
yond the curved part of the hori-
zontal leg of the flange angle.

The bearing determines the
number of rivets in this case.

&•-—

\fiic ater

Less thair

-8

Fig. 158. Rivets Placed in Two Rows to
Give Necessary Number and Spacing.

The number is 108 000 -=- 4 920
= 22 shop rivets in the web.

The angle must now be inves-
tigated in order to see if these 22
rivets can go in one row without
being closer together than 2|
inches, which is three diameters
of the f-inch rivet. The total
length in which these rivets must

be placed is 72.25 inches, and therefore we have 72.25 -r- 22 = 3.3
inches as a spacing. Since this is greater than 2| inches, 22
rivets can be placed in one row. If the spacing as determined
above had been less than 2f inches, it would have been neces-
sary to use two rows of rivets spaced as shown in Fig. 158; and then
the distance center to center would be more than 2f inches, although
the spacing in a vertical line would be less than that.

The four angles at the ends of the girders are called the end
stifleners. These are placed in pairs on opposite sides of the web
(see Plate II, Article 74).

The total end shear is 117 800 pounds, and this is assumed to be
carried by the two pairs of end-stiffener angles, each carrying one-
half. This amount would require lighter angles than the angles
used for intermediate stiffeners It is the customary practice tc
make them the same size and thickness as the intermediate stiffeners,

194

BRIDGE ENGINEERING

185

additional strength being allowed in order to withstand the effects
of the end cross-frame when in action.

The bearing determines the number of rivets required in each

117 800
pair of stiffeners. The number required is — -- = 12 shop

rivets.

Some engineers arbitrarily choose the stiffeners regardless of
the shear, enough rivets, however, being put in the end stiffeners to
take up all the shear; and the spacing in the intermediate stiffeners is
made the same. One noted engineering firm determines its stifFeners
according to the following:

FLANGE ANGLE

HORIZONTAL
LEO

THICKNESS

INTEKMF

A T i •:

4 in.

Any

3 x 3

(.'If

11.

3 X 3

in.

5 in.
Gin.

Any
Over | in.

4x4
4x4

1).

3* rX3i

;n x si

in.

6 in.

Less than | in.

5x3i

5 x3$

in.

8 in.

Any

6x6

6 x4

in.

No rational method has as yet been determined for ascertaining
the stresses in the stiffeners of plate-girders. Results obtained by
placing extensomcters on the stiffeners of actual plate-girders appear
to indicate that the stresses are very small, in fact in most cases not
being greater than 1 500 or 2 000 pounds per square inch.

PROBLEMS FOR PRACTICE

1. Design, according to Cooper's Specifications, the end stiffeners it
the shear is 150 000 pounds, the distance back to back of angles is 6 feet 6\
inches, the web f inch thick, and the flange angle 6 by 6 by Wnch. Use fillers.

2. Design the intermediate stiffeners for the girder of Problem 1,
above, where the shear is equal to 75 000 pounds. Use crimped stiffener angles.
Note that in this case the angles lie close against the web, no filler bars being
used in between.

76. The Web Splice. Web splices are required because of the
fact that wide plates cannot be rolled sufficiently long. Web splices
should be as few as possible, and good practice demands that they be
placed at the same points as the stiffener angle.

The tables on page 30 of the Carnegie Handbook give the extreme
length of plates which can be procured for any given width. The

195

186

BRIDGE ENGINEERING

length of plates for widths which are not given in these tables, should be
taken equal to the length of the next plate given whose width is less
than that of the desired plate. From the first table it is seen that a
74 by f-mch plate can be rolled up to 400 inches, or 33 feet 4 inches,
in length. Therefore, if the girder under consideration is spliced

at the center, the web plates will be required to be- ^ — = 30

feet 10?r inches, which value does not exceed the 33 feet 4 inches as

given above.

According to Articles 46 and 71 of the Specifications, a plate must

be placed on each
side of the web as
shown in Fig. 159,
and enough rivets
placed in each side
to take the total
shear. The total
thickness of both
plates, and also
their length, must
be sufficient to
stand the total
shear, but must
not be less than
f inch.

The total shear

Section A- A
Fig. 159. Splice Plates Placed on Each Side of Web.

at the center of
the girder under consideration (see Fig. 134, p. 150) is 28 600 pounds.

The area required

, . -, ,. . . . 28600
each 01 the two splice plates is — : — -

= 1 . 59 square inches ; and as their length is 62 . 25 inches, the
thickness must be 1 .59 H- 62.25 = 0.0255 inch, but they must be
made f inch thick according to the Specifications. The width should
be somewhat greater than twice the width of the stift'ener angle leg.
This would make the width in this case about 10 inches.

The bearing governs the numl>er of rivets required in this case,
and they are 28 600 -f- 4 920 - 5.81, say 6, shop rivets. More rivets
than this will be required by practical considerations, as indicated by

BRIDGE ENGINEERING

187

Article 54 of the Specifications or in order to make the spacing in the
stiffener angle the same as that in the other stiffeners. This detail
is to be left to the draftsman, the required number only being put
on the stress sheet

In case | of the gross area of the web is considered as efficient
flange area, then provision must be made in the splice for the bending
moment which the web takes. A very economical and efficient splice
is shown in Fig. 1GO. The horizontal plates take the stress due to the
moment, and the vertical plates take the stress due to the shear.

The web equivalent is 3.47 square inches and the total
moment is 1 615 000 pound-feet, which is composed of 275 000 pound-

o o

0 0 (

o o ' o

O O O I

0«°~9~

0 0 0 .0 \

o~o~ o"

o°o°o°o /

(

)

^ 6 Shop/

I, M Shop)

6o0o°o

^oV1 ]

000

0000 I

1 o o o

o o o /

o o

o o /

^Moment-Plates

Fig. 160. Splice Consisting of Vertical and Horizontal Plates.

feet due to dead load and 1 340 000 pound-feet due to live load.
Therefore that proportion of the 3.47 which is taken up by the dead
load is:

27.5 000

X 3.47 = 0.59 square inch;

1 615 000

and that proportion taken up by the live load is:
3.47 = 2.88 square inches.

The equivalent flange area is assumed to act at the center of
gravity of the flange ; and the bending moments equivalent to the
above areas are, for dead load;

0.55) X 20000 X 72.554 = 850 000 pound-inches;
and for live load :

2.88 X 10000 X 72.554 = 2 090 000 pound-inches.

107

188 BRIDGE ENGINEERING

These bending moments must be taken up by the horizontal splice
plates The stresses in these plates (see Fig. 160) are, for dead load:

"kjTojT = l^ "80 pounds;

and for live load,

2 090 000 _
54 . 25

While the allowable unit-stresses are a maximum at the center
of gravity of the flange and are those given by the Specifications, they
decrease rapidly towards the center of the girder, being zero at the
neutral axis of the entire section. The unit allowable stress at the
center of the horizontal plates will not be so great as the maximum
allowable, but will be proportional to the distance from center (see
Fig. 160). The horizontal plates 'will be taken 8 inches in width.
The unit-stresses are easily determined by means of the similar
triangles oab and oab'. The dead-load stress is determined from the

proportion :

54.25

20 OOP = 72.25'
2

and is 14 950 pounds. For live load, the unit-stress will be one-half
of this amount, or 7 475 pounds.

The area required for this plate is, for dead load, -^—^~ ; = 1 .05

OO £Q()

square inches, and for live load — * • ==5.16 square inches, making

/ 4 /O

a total of 6.21 square inches for both plates. Assuming two rivet-
holes out of the section, the net width is 8 — 2 (£ + £) = 6 inches;

and the required thickness for one plate is ~— = 0.52, say yV
inch.

The joint will be weakest in bearing in the |-inch web. The
number of rivets required is :

15670+ 38500

4920

= 11 shop rivets.

The design of the shear plate is as follows : The shear is 28 CGO

oo f*(\(\

pounds, and the required area is =3.18 square inches. As

010

the length of the plate is 46 j inches, the required thickness is — , -—

2 X 46. .20

198

BRIDGE ENGINEERING 189

= 0.034 inch, but on account of the Specifications it cannot be less
than f inch thick. It will, however, be made i9,. inch thick, since
it will then fill out even with the horizontal tension plates and no
filler will be required. Bearing in the web plate decides the number
of rivets, which is:

28600

^4920 = ° sh°P nVetS-

The width of this shear plate should be, as before, 10 inches. The
same conditions limiting the spacing of the rivets apply here as in the
case where the splice was designed for shear only. The length of the
horizontal plates should be sufficient to get in all the rivets, and this
is a detail which is left to the judgment of the draftsman.

PROBLEMS FOR
PRACTICE

1. A plate-
girder is 87 ft. 9 in.
center to center of
end bearings. The
dead-load moment

is 9125000 pound- .* f

inches, and the live- i .V '

load moment is }

38 205 000 pound-
inches, the total
shear at the sec-
tion being 202 700
pounds. The web is
90 by /..--inch, arid
the flange angles are F1* 1GK Proportions oMjtay^Aggto* Hearing Plate, and

6 by 6 by £-inch.

Design the web splice when no part of the web is considered as taking

bending moment.

2. For the girder of Problem 1, above, design the splice when i of the
gross area of the web is considered as effective flange area.

77. The Bearings. Articles 113 to 119 of the Specifications
should be carefully studied before proceeding; also Article 87. Article
70 should be referred to, and the remarks there made about bearings
should be read. In case the length of bearing is such as to allow a
simple |-inch plate, care must be taken that the bearing plate does
not extend past the flange angles more than 2 inches, or that the
masonry plate does not extend past the tearing plate over 2 inches.
Reference to "Steel Construction," Part II, p. 96, to Fig. 161,

109

190

BRIDGE ENGINEERING

and to the discussion which follows, will explain the reason of this.
M = — = 250 x I X -i;

12 12 :

1X9 .
16X 12'

and as s = 10 000,

250 JL _ 10 OOOJ<_9
2 16 X 12 '

from which,

I = 1.94, say 2 inches.

In case it is desirable to have a simple masonry plate instead of a
cast-steel pedestal, and to have the plate extend over the sides of the

angles a distance
greater than 2 in-
ches, then some
arrangement must
be made for sup-
porting the pro-
jecting portion.
Fig. 162 shows one
of the methods
most commonly
used. Notwith-
standing the brac-
ing of the gusset
plates, the mason-
ry plate is not ade-
quately supported,
the greater pro-
portion of the
stress coming
upon the ends.

The disadvan-
tage of having the
masonry plate too
long is plainly

shown by Fig. 163. Here the girder is shown deflected under a live
load, the rear end of the plate being tilted up and the greater part

Fig. 162. Arrangement where Masonry Plate is Used instead
of Cast-Steel Pedestal.

200

BRIDGE ENGINEERING

191

of the pressure coming upon the forward end. The use of this style
of plate is not to be recommended
for spans over 40 feet.

The design for the bearing
of the girder under consideration
will now be made. The total
reaction of one girder must now
be computed. This will be due
to the weight of the steel in the
girder, to the weight of the
track, and to the reaction pro-
duced by the E 40 loading when
wheel 2 is directly over the end
support. This reaction is :

Fig. 163. Effect of Having Masonry Plate
Too I,ong.

Weight of Steel,

(123.5 + 10 X 61.75) 61.75

400

Weight of Track, - - (61 . 75 + 1 . 75) *
Reaction Due to Engine Loading

Total

1 1 430 pounds

6350 "
= 99700 "
= 117 480 pounds.

. 117480
ihe square inches ot bearing surlace required is — 0^ —

470;

and, as the length is 1 foot 9 inches, or 21 inches, the total width of the

470
cast-steel pedestal will be — = 22.4, say 23 inches, or 1 foot 11

inches.

A bearing plate must be riveted to the lower flange where it
rests upon the pedestal. The pedestal must be so constructed as to
allow this bearing plate to set in it. Hand-holes should be provided
in the casting in order to allow the bolts which connect the casting to
the girder to be inserted. These bolts should be at least f inch in
diameter. Anchor bolts £ inch thick and at least 8 inches long should
be provided and fox-bolted to the masonry. The thickness of the
metal in all parts of the casting should be at least li inches. The
details of the pedestal are given in Fig. 164, the length of the bearing
being made 12 inches so as to allow one rivet to be driven in the
flange angle in the space between the end stiffeners.

Allowance should be made for a variation of 150 degrees in tem-

201

192

BRIDGE ENGINEERING

perature. The coefficient of expansion for steel per unit of length is
O.OOOOOG5, and the amount of expansion for 150 degrees of tempera-
ture will be:

0.0000065 X (61 ft. 9 in.) X 150 = 0.06 foot.

This is about f inch, and therefore the holes in the flanges at one end
of the girder should be made oblong and long enough to allow the

Fig. 1C4. Side and End Elevations Showing Construction of Pedestal and Connection
to Bearing Plate.

girder to move f inch, or f inch either backward or forward from a
central position. In determining the length of this slotted hole (see
Fig. 105), it must l)e noted that the -£-inch bolt takes up part of this
hole, and therefore its length should l>e I + f = say If inches. The
width of the hole should l>e sufficient to allow for any over-run in the
diameter of the bolt. It should be at least 1 ] inches wide.

PROBLEMS FOR PRACTICE

1. Determine the distance center
to center of bearings, and the size
of the masonry plate, for a plate-
girder of 40-foot span under coping,
the loading being E 40.

ANS. 41 ft. 4 in.; 350 square inches.
(NOTE — Interpolate values in Ta-
ble I, Cooper, p. 30.)

2. If the girder span is 78 feet
under coping, and the loading E 40,

determine the maximum end reaction and the width of the masonry
plate. Axs. 147 130 pounds; 24 J inches.

78. The Stress Sheet. Plate II (p. 1 72) shows the stress sheet for

Fig. 165. Slotted Bolt-Hole in Flange at
End of Girder to Allow for Contraction and
Expansion Due to Temperature Changes.

BRIDGE ENGINEERING

193

the girder which has just been designed. It represents the best modern
practice in that it gives, in addition to the sizes of all the sections,
the curves for the maximum shears and moments, the rivet-spacing
curve, and the number of rivets required in the different parts of the
structure. This general form has been adopted by one of the largest
bridge corporations in this country, and is to be recommended since
it gives the" draftsman all necessary data and thus prevents the loss
of time by an inexperienced man in recomputing certain results. The
results just referred to are the shears, the moments, the rivet spacing,
and the number of rivets required in the various parts. Formerly it
was not customary to give this information on the stress sheet, and
the draftsman was therefore required to do all this computation which
had previously been worked out by the designer but had not been
placed on the stress sheet in available form, and thus unnecessary loss
of time resulted.

DESIGN OF A THROUGH PRATT RAILWAY=SPAN

79. The Masonry Plan. The same remarks which are made
in Article 67 apply here. In this case the length of the masonry plate
is usually determined by considerations relative to the number and
length of the rollers in the bearing, and not by the bearing per square
inch upon the masonry, the size of the plate as determined by the
above considerations being usually much larger than if it had been
determined by the unit bearing stress. A preliminary design of the
masonry plate is usually made in a manner similar to that done in the
case of the plate-girder; or the length of the masonry plate may be
approximately determined from the following:

LENGTH OF MASONRY PLATE

FIXED END

ROLLER END

100 feet

23 inches

125 "

26 "

150 "

28 "

175 "

31 "

200 "

35 "

The above masonry plates are for single-track bridges, with or with-
out end floor-beams, the length being the same in either case.

B03

194 BRIDGE ENGINEERING

80. Determination of the Span. The determination of the span
is made in exactly the same manner as described in Article 68. Care
should be taken, in case end floor-beams are not used, to allow for
the pedestal stones, which are square stones resting directly upon the
bridge seat, and upon the top of which rest the masonry plates of the
stringers. Their height must, of course, be such as to keep the
stringers level. In case these stones are used, their size must be
determined ; and if it is greater than that of the bearing or masonry
plates, then their size determines the width of the bridge seat and
the span center to center of bearings.

81. The Ties. In the design of the ties, as well as in all the
design which follows, the Specifications of the American Railway

Fig. 166. Spacing of Stringers and Rails, and Position of Loads.

Engineering & Maintenance of Way Association will be followed.
Whenever reference is made to these specifications, the number of
the article will be enclosed in parentheses, as "(5)," which signifies
that Article 5 of the Specifications is to be referred to.

The stringers in the bridge in question will be taken 6 ft. 6 in.
center to center. The maximum loading (7) is such as to bring
8 333 pounds on one tie, and to this must be added 100 per cent for
impact, making a total of 16 667 pounds. In order to illustrate the
method of assuming the distance, center to center of rails, as 5 feet,
that distance will be used in this case. The maximum moment will
then be 9 X 16 667 = say, 150 000 pound-inches. The size of the
tie will be determined as in Article 71, the allowable unit-stress
being 2000 pounds per square inch (5). If a 7 by 9-inch tie is
used, the unit-stress will be 1 590 pounds. If a 6 by 8-inch tie is
vised, the unit-stress will be 2 340 pounds. It is evident that a 7
by 9-inch tie must be used. See Fig. 166 for spacing of stringers and

204

BRIDGE ENGINEERING 195

rails, and for position of the loads. Note that, although impact
is taken into account in this case, the size of the tie is the same as
that designed for the plate-girder, although the unit allowable stress
also differs.

82. The Stringers. The width, center to center of trusses, will
be assumed as 17 feet, since this is sufficient to clear the clearance
diagram in cases of single-track bridges of spans less than 250 feet.

The span which is to be designed in the following articles is a
through-Pratt with 7 panels of 21 feet each, making a total span, center
to center of bearings, of 147 feet 0 inches. See Plate III (p. 251) . Rivets
| inch in diameter will be used throughout, except in channel flanges.

The length of the stringers end to end will be 21 feet, and accord-
ing to Cooper's Specifications; p. 32, the maximum moment for the
live load will be 226000 pound-feet per rail. The coefficient of

impact (9) will be (— «AA) = 0.935, and therefore the moment

due to impact will be 0.935 X 226000 X 12 = 2535000 pound-
inches, making a total of 5 247 000 pound-inches due to live load.

The section modulus for any particular beam is equal to the
bending moment divided by the unit-stress, and this is equal to the
moment of inertia divided by one-half the depth of the beam. This
latter quantity is constant for any given beam, and for I-beams may
be found in column 11, Carnegie Handbook, p. 98.

On account of the cheapness of I-beams, they will be used for
stringers in this bridge; and sufficiently heavy shelf angles will be used
to take up any distorting influences due to the eccentric connections
which are unavoidable in this case. In case an I-beam had not been
decided upon, the stringers would have been small plate-girders with
a span of 21 feet and depth according to formula given. They would
have been computed in exactly the same manner as a plate-girder
span of 21 feet center to center of bearings.

Since the dead-load moment cannot be determined until the size
of the stringer is known, an approximate design must first be made
by using the live-load bending moment alone; and then, with the
size determined in this manner, the extra section modulus required
for the dead-load moment due to the weight of the beam and the
track must be computed. If this extra section modulus, added to the
one previously determined, is greater than that given by the beam in

205

196 BRIDGE ENGINEERING

question, a larger size beam must be used and a recomputation made.

5 247 000
The section modulus (17) required for live load only is —

16 000

= 327.9. As this is too large for one beam, two beams will be used,
thus giving a required section modulus of 164 for one beam. Two
24-inch 80-pound I-beams will be used, giving a total section modulus
of 2 X 174 - 348.

Assuming the rails and ties to weigh 400 pounds per linear foot,

400
the dead load per linear foot per stringer is 80 + - = 180 pounds.

The dead-load moment is therefore 18Q X 21 X 21 X 12 = ng ^

pound-inches. This requires an additional section modulus of
1 1 q f]f)f)

— = 7 .45. This, added to the 164 as determined above, makes
lo 000

a total required section modulus of 171.25, which, being less than
174 (which is that for one I-beam), indicates that the above chosen
beam is sufficient in strength, and it will therefore be used.

The number of rivets in the end connections will now be deter-
mined. The total end reaction for one I-beam is equal to the weight
of one-half the beam, one-eighth the track in the panel, and one-half
the maximum live-load reaction for one rail. These quantities are:

§ Live-Load Reaction = — - — = 25 700 pounds.

Impact = 25 700 X 0.935. =24 030 "

Weight of Track = 4~ X -2-J- = 530 "

Weight of Stringer = -- X 80 = 840 "

Total = 51 100 pounds.

The coefficient of impact is that for a loaded length of 21 feet.

From p. 177, Carnegie Handbook, sixth column, it is seen that
the longest connection angle which can be used with a 24-inch I-beam
is 20| , say 20 inches. In this case the thickness of the connection

angles must be ——-— —=0.23 inch; but according to (36), f

10 000 X 2.2

inch will be used. The angles chosen will be 6 by 3| by f-inch.
The 6-inch leg will be placed against the web of the floor-beam in
order to allow for sufficient room for rivets to be driven.

2oe

BRIDGE ENGINEERING 197

The rivets will tend to shear off at places between the webs of
the stringer and floor-beam and the connection angles. They will
also tend to tear out of the web of the stringer and out of the web of
the floor-beam. As the thickness of this latter is not known, the
determination of the rivets for this condition will be made under the
next article. The bearing value of a f-inch rivet in a $-inch plate

(19) is I X i X 24 000 = 10 500 pounds, and therefore51-11 - = 5

10 500

shop rivets are required in bearing in the web of the stringer. The
value of a |-inch shop rivet in single shear (18) is 0.6013 X 12 000 =
7 220 pounds, and the number of rivets required to prevent shearing

between the connection angles and the webs is ~y^7r = 7 shop rivets.
The value of a |-inch field rivet in single shear (18) is 0 . 6013 X 10 000

= 6 013 pounds, and therefore — — = 9 field rivets are required to

6 01 o

connect the connection angle to the web of the floor-beam. As men-
tioned above, the number of rivets
in bearing in the web of the floor-

beam will be determined in the
next article; and if the number
required for bearing is greater
than 9, then that number must
be used instead of 9. Fig. 167
shows the connection of the
stringer to the floor-beam web,
and also the number of rivets as
determined above, in their proper

9 Field

! I-bea

Floor 6 earn Web

/I-beam Web

F5hop
•Id

positions. The distance between Fig. 167. Connection of Stringer to Floor-
, „ , . Beam Web; also Number of Rivets.

the webs ol the stringers must

be such as to prevent their flanges from touching at the top.

The stringers should be connected at the bottom by a system of
lateral bracing of the Warren type. The size of these angles cannot
be determined by theoretical considerations, but is usually chosen to
be 3 i by 3 J by f-in. See Plate II (p. 172) for the general arrangement
of this bracing.

83. The Floor=Beams. All floor-beams should be of sufficient
depth to allow the use of small-legged connection angles at the ends

198 BRIDGE ENGINEERING •

where they join the end-posts. The thickness of the web should also
be greater than that which is theoretically computed, in. order that
sufficient bearing may be given so that the rivets for the stringer
connections will not require the stringers to be of too great a depth.
The depth of the floor-beam will, of course, vary somewhat with the
length of the panel and with the loading, but should not be less than
36 inches in any case. A considerable variation in the depth will
not affect the weight of the floor-beam or the bridge to an appreciable
extent. A good plan is not to exceed a depth of 5 feet, with panel
lengths of 25 feet and E 50 loading. In this bridge the depth of all
intermediate floor-beams will be taken as 48 inches. It is good
practice not to consider -| the web area when designing flanges of
floor-beams and stringers, and the design here given does not consider
the web as taking any bending moment.

The design of an intermediate floor-beam will now be made.
The loads for which it is designed are the floor-beam reaction due to
the live load (see Cooper,p. 32), the floor-beam reaction due to impact,
the dead weight of the stringers and track, and the weight of the beam
itself. The latter .weight is distributed uniformly over the entire
length of the beam, and the other loads act as concentrated loads
spaced 6 feet 6 inches apart at equal distances from the center.
The computation of the concentrated loads is as follows:

Live Load= 68 000 pounds

son

Impact = 68 000 X ( „- ) 60 500 "

o / ~r oUU

Dead Load of Stringer = 2 ( 21 x ^ X 8° ) . . 3 320 "

400
Dead Load of Track = -- X 21 4200- "

Total = 136 020 pounds.

The moment at points under the loads (see Fig. 168) is 136 020
X 5.25 X 12 - 8 575 000 pound-inches. This is due to the con-
centrated loads only. The weight of one floor-beam may be approxi-
mately determined by the same formula as used to determine the
weight of plate-girder spans; only, in place of the length of the span,
the length of the panel must be substituted. The total weight of the
above floor-beam, then, is:

TF= 0.45 X (123.5 +10 X 21) X 21 = 3 160 pounds.

208

BRIDGE ENGINEERING

199

The dead-load moment at the center duo to this weight will be:
Wl _ 3 160 x 17 x 12
8 8

80 700 pound-inches,

making a total moment at the center of the beam of 8 575 000 + 80 700

S
0

5'- 3" ( ©"-e" I 5-3"

,7'-0« *l

Fig. 168. Diagram Showing Loads on Floor-Beam.

= 8 655 700 pound-inches. Note that the dead-load moment at the

center of the beam is added to the concentrated-load moment at the

point where the concentrated load is applied. This will give the

total moment at the center of the beam as shown by Fig. 169, since

the concentrated-load moment, is

constant between the points of

application. The end shear • is

readily computed to be 136020

+ 1 580 - 137 600 pounds. The

curves of moments and shears are

shown in Fig. 169.

The total depth of the floor-
beam, back to back of angles, is

48i inches ; and the effective depth Fig m ghear and Moment Diagram
will, for approximate computa-
tion of the flange area, be taken as somewhat less, say 44^ inches,
since the flange angles will probably be 6 by 6-inch and the center
of gravity of most of these angles lies about If inches from

the back. The approximate flange stress is

pounds, and the required net area (17)

44.5
will

= 194 500
194 500
16000

209

200

BRIDGE ENGINEERING

= 12.2 square inches. In assuming the size of the angle, it is to be
remembered that when, as in this case, no cover-plates are used, no
rivet-holes will be taken out of the top flange, and only one rivet-
hole will be taken out of the vertical flange.

Two 6 by 6 by f-inch angles give a gross area of 7.11 square
inches each, and a net section of 7 . 1 1 — 0 . 625 = 6 . 485 square inches
each, or 12 . 97 square inches net for both. As this is near the required
area, these angles will be taken; and a recomputation will now be
made with the actual effective depth, in order to see if sufficient
variation in the areas occurs to require another angle to be taken.
The actual effective depth is now 48^ — 2 X 1.84 = 44. 57 inches;
and making computations with this, it will be found that a net area
of 12.10 square inches is required. As this is practically the same

as was determined at first, no
change will be made in the size
of the angle.

The web is to be designed
for a total shear of 137600
pounds. The required area (18) is

137600
1QOOO = 13.76 square inches,

and the required thickness is
l-^ = 0.286 inch; but on ac-

Fig.170. Calculation of Number of Rivets COUllt °f tlle Specifications (36),

throughaXFTotorBeanmlewe0bfStringer t inch must be used. The web

will accordingly be 48 by f-inch.

The determination of the number of rivets which go through the
connection angle of the stringer and the web of the floor-beam can
now be made. The value of a f-inch field rivet in bearing in a f-inch
plate (19) is f X f X 20 000 = 6 560 pounds, and the total number

required in one connection angle will be - — =11 field rivets

2t /\ o Oou

(see Fig. 170).

The pitch of the rivets in the flange can in this case be
determined by the use of the formula:

**,

.-*_

210

BRIDGE ENGINEERING 201

Since the flange is of the same cross-section throughout, the value
of the effective depth will not change, and it can therefore be used in
the above equation instead of considering the value of the distance
between rivet lines. The shear being practically constant from the
connection of the stringers to the end of the floor-beams, the rivet
spacing will be constant in this distance. It will be:

7880 X 44 . 57

137600 -2.51 inches,

the value of a f-inch shop rivet in bearing in the f-inch web being
| X f X 24 000 = 7 880 pounds. This is seen to be less than 2f
inches ; but, as the angles have 6-inch legs, this spacing can be used in
a horizontal direction ; and the distance from center to center of rivets,
which will be placed in rows on two gauge lines, will still be greater
than 2f inches.

The shear between the stringer connections is practically zero,
and therefore the spacing will be very large. Being over 6 inches, it
will be subject to (37).

The connection angles at the ends of the floor-beams will be
taken as 6 by 3^ by f-inch, the 6-inch legs being against the web of
the floor-beam. The other legs are chosen small in order that they
may fit into the channels which will very likely be required for the
posts; and according to the sixth column, p. 183, Carnegie Handbook,
only 8} inches is available for this purpose. This 8| inches is taken
from a 10-inch channel, since this is the smallest channel that can be
used which will give room for connection and yet be in accordance
with the Specifications. This is due to the fact that its web (36) is
greater than f inch. The rivets which connect the end angles to the
floor-beam web are shop rivets, and those which connect the end
angles to the posts are field rivets. Since the size of the post is not
known, the thickness of its metal, of course, cannot be used, and
therefore the number of rivets required in bearing in the post cannot
be determined at this time.

The number of shop rivets required through the end angles
and the floor-beam web is governed by the bearing of the rivets in the
f-inch web of the floor-beam. The value of a f-inch shop rivet in
bearing in a f-inch web (19), as has just been computed, is equal to

7 880 pounds, and the number of rivets required is — =?= 18.

211

202

BRIDGE ENGINEERING

bers may be

The number of field rivets required in single shear to connect
the end angles with the posts is ^ = 23. An even number of

rivets will, of course, have to be used, one-half going into each
of the 3i-inch legs. See Fig 171 for the position and the number
of rivets. It must be remembered that more than these num-
used by the draftsman on account of rivet spa-
cing which may be required by
conditions other than those of
design.

The design of the end floor-
beam is somewhat different from
that of the intermediate floor-
beams in that the load which
comes upon it is considerably
lighter, since this floor-beam
takes the dead load of only one-
half the panel and the live load
due to the maximum end reaction

Channel of
/Post

Web ot Floor

Fig. 171. Position and Number of Rivets
to Connect End Angles with Posts.

for a stringer length instead of
the floor-beam reaction for the
stringer length (see Cooper, p.
32).
The maximum end shear is computed as follows:

End Shear for 21-foot Span 51 400 pounds.

300 \

Impact = 51 400 X

\*-.i.

Dead Load of Stringers =

Dead Load of Track = -

- 300 / ' '

80 X 2 X 21

400 X 21
2X2

. 48 000

1680

2 100

Total 103 180 pounds.

The maximum moment due to the above load is 103 ISO X 5.25
X 12 = 6 500 000 pound-inches. The wreight of the beam may be
assumed as 3 160 pounds. This is the same as was computed for the
intermediate floor-beam, but will be used for this beam, since the size
of the web will be the same as in the others; and, although the flange
area will be less, the end connections will be somewhat heavier owing
to the connection of the beam to the end-post and the roller bearing,

212

BRIDGE ENGINEERING 203

and this additional weight will cause the total weight of the end
floor-beam to be about the same as that of the intermediate ones.
The total moment at the center will then be 6 500 000 + 80 700 -
6 580 700 pound-inches.

The depth of the end floor-beam will be somewhat greater than
the depth of the intermediate floor-beams. This is due to the fact
that it extends downward a greater distance, resting upon the bearing
plate, which comes directly upon the top of the rollers. The exact
depth cannot, of course, be determined until after the roller bearings
are designed; but it may be safely assumed as four or five inches
deeper than the intermediate floor-beams, and in case this is riot
enough, the draftsman can easily fill in the remaining distance with
filler plates, as this distance will not be very great. In case this depth
is too great, the flange angles may be bent upward at the end, or a
re-design may be made.

The depth will be assumed as 52 inches in this case. The
effective depth will be assumed as 48 inches, and this gives an approxi-
mate flange stress of

6580TOO =137000poundS)

and an approximate net flange area required of

137 000

1fi -_- = 8.57 square inches.

A 6 by 6 by -^g-inch angle gives a gross area of 5.06 square
inches, and a net area of -5. 06 — (f + |) r\ = 4.62 square inches.
A recomputation with the true effective depth requires 8.42 square
inches net. Two of these angles give 9.24 square inches; and as this
coincides very closely with the required area, it will be used; The
size of the web plate is 52 by f-inch.

The pitch or spacing of rivets in the flanges is :

7 880 X 48

~ 104 7CO = ' mches-

The maximum end shear as above computed is taken by two
stringers; and therefore the number of rivets required in bearing to
form the connection between the stringers, connection angles, and
the floor-beam web is, for each angle:

104 760

.TT = 8 field rivets.

2 X 6 560

213

204 BRIDGE ENGINEERING

The value 6 560 in the above equation is the value of a |-inch field
rivet in bearing in the f-inch web.

The number of rivets required in the end angles on the floor-
beam is :

104 760

-y^r= 14 shop rivets.

These rivets go through the web of the floor-beam. The connection
of the floor-beam to the end-post is made by means of field rivets and
a large gusset plate. This gusset plate is usually | inch in thickness.
The number of rivets through the end connection angles and this
gusset plate is governed by single shear, since the rivets will shear off
between the angles and the gusset plate before they will tear out of
the gusset plate, as the value of a rivet is greater in bearing than in
shear. The number required is:
104 760

6013

= 18 field rivets.

The general arrangement of the intermediate floor-beams is
shown in Fig. 172. The ends of the lower flange are bent up as
shown, in order to allow the I-bar heads or any other section of the
lower chord to have clearance. This makes it necessary for the floor-
beam web to be spliced at the ends, as shown. The distance which
this plate should extend above the floor-beam proper depends upon
the distance which the lower chord is bent up. In any case the length
of the connection on the post should be at least equal to the depth
of the floor-beam. Two splice plates, one on either side of the web,
are placed here in a manner similar to that of a splice as designed in
the plate-girder when shear only was considered. Here shear only is
considered, and the number of rivets which must be on each side of
the splice will be:

137 640
7880 = 18 shop rivets.

The 7 880 which occurs in the above equation is the value of a |-inch
rivet in bearing in a f-inch plate (19). Inspection of Plate II (p.
172) will make this design clearer. Plate II also shows the shape of
the end floor-beams.

The small shelf angle shown in Fig. 172 should have sufficient
rivets to prevent any twist of the stringers due to their being con-
nected on one side of their web only. This number is a matter of

214

BRIDGE ENGINEERING

205

judgment. Experience seems to indicate that enough rivets to take
up one-third of the total reaction of the stringers will be sufficient.
This will require shop rivets, and the number will be :

103 180
- = 5 shop rivets in single shear.

84. The Tension Members. Tension members usually consist
of long, thin, flat plates with circular heads forged upon their ends.

Fig. 17Z. General Arrangement of Rivets, Splices, Connections, etc., for Intermediate
Floor-Beams.

These circular heads have holes punched through their centers and
then very carefully bored. Through these holes are run cylindrical
bars of steel called pins. These pins connect them with other mem-
bers of the truss. See Carnegie Handbook, p. 212, for table of I-bars.
The I-bars given are standard I-bars; and while departures from these
widths and minimum thicknesses may be made, it may be done only
at great cost to the purchaser. Note that there are no standard
9-inch I-bars. The thicknesses given are the minimum thicknesses

215

206 BRIDGE ENGINEERING

for that width of bar, and do not indicate that thicker bars of that
width cannot be obtained; but on the contrary thicker bars of that
width can be obtained, and this should be done, the minimum thick-
ness as given in the table being avoided if possible.

It has been found that bars which have a ratio of thickness to
width of about one-sixth give good service and are easy to forge.
This relation gives us a rough guide which will enable us to determine
the approximate width and thickness of any bar of a given area.
Once the approximate dimensions are determined, the actual dimen-
sions can be chosen from the market sizes of the material (see Car-
negie Handbook, pp. 245 to 250).

An expression for the approximate de,pth of the bar will now be
derived by using the above relation.

Let A = Area of bar, in square inches;
d = Width of bar, in inches;
t = Thickness. of bar, in inches.
Then,

td = A;
also,

Substituting the value of t in the expression for A, there results:

d = VGA.

The stresses in all the members in the truss under consideration
are computed by the method described in Part I, and are placed on
the stress sheet, Plate III (p. 251). In the succeeding design, the
student should obtain his stresses from Plate III without his attention
being again called to the matter.

Table XXIV gives the tension members and their dead-load,
live-load, impact, total, and unit stresses (15), together with the
required area, the number of bars, the approximate depth of bars,
and the final sizes used.

The first seven columns in Table XXIV are self-explanatory.
The number of bars to be used in any particular case is a matter of
judgment. One fast rule is that an even number of bars should
always be used, except in the case of counters, where one is permis-
sible. This is due to the fact that the placing of one of the main
members in the center of the pin would create a large moment, and

216

BRIDGE ENGINEERING

207

i i

AREA*

SECTION
USED

HUHH

* NOTE — All areas in square inches, and all dimensions in inches.
Observe that the above table is not completely filled out with respect to the first two members given. This is on ac-
count of the requirements of the Specifications (80).

therefore, an ex-
cessively large pin
would be required,
and accordingly a
very large head on
the I-bar in pro-
portion to its width
— all of which are
very undesirable
and costly. In gen-
eral the number of
I-bars should be as
small as possible,
and they should be
so chosen that the
widths of the chord
members increase
from the ends to-
ward the center of
the truss, and the
widths of the diago-
nals decrease from
the ends toward the
center of the truss.
The area of one
bar is obtained by
dividing the total
stress by the num-
ber of bars and also
by the allowable
unit-stress. Thus,
for the member
LJJV for example,
the required area
of one bar is :
295 100

SECTION

USED*

. • <*oo*o *.

'. '. X X X X X X

: ;S ;-8§SS

AREA*;
ONK BAR

: :8SSS§S

; 1 • p> i-I >o *o bi oo

0 %
|ffl

; ; (N ^H (M Tt< Tfl 01

REQUIRED

AREA*

S£3Sg§§2

«OOOQOrH,-HfOCOO

III!!!!!

i""'".

ESSES (in poun
IMPACT

£ 5

;3 2 >-i 53

DEAD- LOAD

rH CO t~- .' Tf< 00 <N .'

<* rt T}< . C^ CO 00 .

M
H

35S&S-33&

. — 9 22
2 X 16 000

square inches.

217

208 BRIDGE ENGINEERING

The approximate depth of this bar is determined by taking the square
root of 6 times the area as above determined. It is:

d = V 6 X 9 . 22 = 7 . 44 inches.

As this is nearer 7 than 8 inches, a 7-inch bar will be chosen; and
looking in the first column, Carnegie Handbook, p. 248, for an area
which will be equal to or in excess of 9.22, it is found that a
If -inch bar satisfies this condition, and therefore the section of this
member consists of two bars 7 by If inches.

According to (80), the first two sections for the lower chord are
to be made of built-up members. This requires that instead of
I-bars they are to be made of angles and plates, or, in case the stress is
light, of channels. The depth of the section is limited by the size
of the greatest I-bar head. As the diameter of the I-bar head depends
upon the size of the pin, it cannot of course be determined accurately
before the pin is designed. It is customary to assume the largest
head, and to design the section so as to clear this. The size of the
largest head for bars of given width is given in the Carnegie Hand-
book, p. 212.

The design of the member L0L2 will depend upon the size of
the largest head of the 7-inch I-bar of the member L\Lr This is 17 \
inches; and in order that the head may have some clearance, it will
be necessary to add \ inch to the top and the bottom, making a total
of 18A inches. Since the flange angle, as in the case of plate-girders,
will extend over the plate about \ inch, the plate itself may be 18
inches wide and still give sufficient clearance.

The total stress is 234200 pounds, and the allowable unit-
stress (15) is 16 000 pounds per square inch. The required net
area, then, is:

234 200
160QO = 14. 64 square inches.

According to the Specifications, the thickness of the plate cannot be
less than f inch. The gross area of two 18 by f-inch plates is 13.5
square inches, and the gross area of four 3^ by 3^ by f-inch angles,
which are assumed to be sufficient, is 9 . 92 square inches, thus making
a total gross area of 23 . 42 square inches. If 5 rivet-holes are assumed
to be taken out of each web, and one rivet-hole taken out of each
angle, this will require a certain number of square inches to be

218

BRIDGE ENGINEERING

209

less

deducted from the section, and this is computed as follows:

Out of webs, 2 X 5 (| + £) X -jj- = 3.75 sq. in.
Out of angles, 4 (J + i) X f . . = 1 .50 " "

Total = 5 . 25 sq. in.

The net area of the section is now determined to be 23 .42 — 5 .25 =
18.17 square inches. This is somewhat greater than the required
net area, but must be used, for according to (39), these are the smallest
and thinnest angles that may be used.

Figs. 173 and 174 show the cross-section and the general detail
at L2. The width of the member cannot be determined until after
the section of the end-post is
computed, since it must fit inside
of the end-post, the horizontal
legs of the angles being cut off to
allow this. The end-post, Arti-
cle 87, is 14 1 inches inside. If it
is assumed that all the pin-plates
on the end-post are placed on
the outside, and all those at L0
on L0L2 are on the inside, then
the width of L0L2, back to back
of plates, must be 14 - (2 X 2
+ 2 X 1) = 121- inches or less,
J-inch clearance being allowed
between the sides of the angles and the web plates of the end-post
(see Fig. 173).

The total net section through the pin-hole at L2 (26) must be
1-|- X 18.17 = 22. 7 square inches, or 11.35 square inches for one
side. The plate which is to increase the section must be on the out-
side, since the intermediate post U2L2 and the two I-bars of member
UiL2 must go inside. The gross width of this plate is 1 1 } inches (see
Fig. 174), and the net width is 2w = \\\ - 5 = 6J inches. The
net area through the pin is:

Two 3^ by 3£ by f-in. angles = 4 . 96 square inches.

One 18 by \-\n. plate = 9 - 5 X \ = 6 . 50 "

Total = 11 .40 square inches

Since this is greater than the 11 .35 required, no plate will be necessary
to fulfil (26) in this respect.

Fig. 173. Cross-Section Showing Construc-
tion of Lower Chord Member.

219

210

BRIDGE ENGINEERING

«u\~~ iK\\

13 I j/*\

220

BRIDGE ENGINEERING 211

Sufficient bearing area must be provided at this point. The
total stress is 234 200 pounds, the total bearing area required is

234 200
' — = 9.76 square inches, and the total thickness for one side is

^4 UUU

:— - = 0 . 976 inches. Since the thickness of the web is $ inch, the
2X5

pin-plates must be 0.976 - 0.50 - 0.476 inch (say \ inch) thick.
A |-inch pin-plate must be used, and as the total thickness of the

bearing area is now 1 .00 inch, this pin-plate will take — '-

= 58 550 pounds. The joint is weakest in shear, and will therefore

require ---™~— = 8 + (say 9) shop rivets.

In case it is necessary to put the member U^ on the outer side
of L0L2, then the outer legs of the upper angles must be cut off to allow
t/^Lj to pass. This will decrease the section by an amount (3^ — f )
X f = 1 . 20 square inches. Considering the pin-plate, which is
(18$- — 2X3$) — i=llj inches, the | inch being allowed for
clearance between the edges of its flange angles, the total net
section through the pin-hole on one side will be:

One Angle 3£ by 3£ by f-in. = 2.48 square inches
One Cut Angle 3£ by 3£ by f-in. = 1.31 " "

One Web (18 - 5) J sq. in. =6.50 "

One Pin-Plate (11J - 5) f sq. in. = 2.34 "

Total = 12 . 63 square inches.

This is greater than 11 .35 as required, and is therefore safe.

The distance from the center of the pin to the end must now
be determined (26). The total net section of the body of the member
is 18.17 square inches, or 9.09 square inches for one side, and the
thickness of the web and the pin-plate is 1 inch. The distance from

9 09
the pin to the end of the member is then — - = 9^ inches, and the

distance to the center of the pin is 9|- + ~ -= llf, say 12 inches (see

Fig. 174). Rivets should be countersunk where necessary to prevent
interference with I-bars. For signs, see "Steel Construction," Part
III, p. 192, and Carnegie Handbook, p. 191.

221

212

BRIDGE ENGINEERING

At point L0 of this member, the pin is 6J inches in diameter, and,
as previously mentioned, the legs of the angles are cut (see Fig. 175).

The total bearing area required for one side is — '- — = 4 . 88, and the
required thickness is ^—^- = 0. 781 inch. Subtracting the thickness

O . -£«3

of the Wnch web from this gives 0.281 inch. A pin-plate f inch
thick must be used.

The net area through the pin (26) must be 1 1 . 35 square inches.

'/>'

*. /

v, //{'

I"H

/ ^

e Pm Plate

i j ^

•j^

"MSj*

i i

Fig. 175. Elevation and Section Showing Pin Connection at End of Truss.

This net area, remembering that the angle legs are cut and therefore
their area is that of a bar 3^ by f-inch, computed for one side, is as
follows :

Two Angles, legs cut, 3£ by 3J by f-in. = 2.62 square inches

One Web 18 X i - 6£ X \ sq. in. = 5.88 "

One Pin- Plate Hi X | — 6J X f sq. in. = 1.87 " "

Total 10 . 37 square inches.

This shows the section to be deficient, and the thickness of the pin-
plate must be made \ inch. This gives a net area through the pin
of 11 .62 square inches.

The distance between rivet lines (see Fig. 173) is \1\ inches,
and (44) the tie-plates must be 17 \ (say 18) inches long, and their

17 25

thickness — '- — = 0.346 inch (say f inch).
50

BRIDGE ENGINEERING

213

The lattice bars (45) must be 2^ inches wide, and (47) must be
double. From (45) and Table XXV, page 219, the thickness must
be r76 inch, the distance c being 17.25 X secant 45° - 2 ft. 0T\ in.

The design of the hip vertical U1L1 is also made in accordance
with (80) of the Specifications. It will be assumed that the section
consists of one 8 by f-inch plate, and four 3^ by 3| by f-inch angles,
since this is the lightest section that may be used according to the
Specifications, the 8-inch plate being chosen as it gives some clearance
between the inner edges of the legs of the angles.

The total stress in the member is 141 600 pounds, and the unit-
stress is 16000 pounds per square inch, thus requiring a net area of
8.85 square inches. The plate gives a
net area of 2 . 20 square inches, and the
four angles give a net area of 6.88 square
inches, making a total of 9.08 square
inches, one rivet-hole being taken out of
each angle, and two out of the web, at
any particular section. The net area is
somewhat greater than that required, but
must be used, as this is the minimum sec-
tion allowed by the Specifications. Fig.
176 shows a cross-section of this member
as above determined.

This member will be connected to the upper chord and end-post
by means of a pin which is 6| inches in diameter, the diameter of the
pin being determined later. The total stress is 141 600 pounds, and
this will be taken by two plates, one on either side of the member.
The net section of the member is 9.08 square inches, and the section
through the pin (26) must be 25 per cent in excess of this, making a
total of 11.35 square inches, or 5.68 square inches for each plate.
The total width of these plates will be taken as 12 inches, and this
(see Fig. 177) will make the required thickness:
5.1

Fig. 176. Cross-Section of Hip
Vertical.

5.68
2c

-=0.99 inch.

12.00 - 6.25

Fig. 177 shows the details of these pin-plates. Since the above
thickness is too great to be punched in one single piece, the above
thickness will be made up of two plates, each ^ inch thick. The
area at section A-A must be equal to that of the body of the bar. It

214

BRIDGE ENGINEERING

is 12.00 X $ = 6.00 square inches for one side, or 12.00 square
inches for both sides. As this is greater than the 8 .85 square inches
as above computed, the area at A- A is sufficient, as is also the width
of the plate, which was assumed as 12 inches.

One of the plates will be riveted directly to the member, and the
other will be riveted to it as a pin-plate. The section back of the pin
(26) must be equal to the net section in the body of the member. The
net section is 4.54 square inches for one side, and the total thickness

Fig. 177. Connection of Hip Vertical to Upper Chord and End-Post.

of the pin-plates is 1 . 125 inches, making the distance from the end

4 54

of the member to the pin — '—-- = 4% inches, and the distance to
1 .125

Ai
the center of the pin 4| + -^ = 1\ inches.

The joint between the plates and the main member will be weak
in shear, the rivets tending to shear off between the f -inch angles and
the plate, and also between the two plates themselves. As each side
takes one-half of the above stress, the number of rivets required to
connect the plates to the main member will be:

141 600 •*• 2

= 10 shop rivets,

and the number of rivets required to connect the inner -J-inch plate
to the outer one which is connected to the member itself will be:
141 600 H- 4

7220

5 shop rivets.

224

BRIDGE ENGINEERING

215

The distance from the center of the pin to the top of the main
part must be greater than one-half the diameter of the largest I-bar
head — that is, 17^ -j- 2 = say, 9 inches.

At the lower end, this member is connected to the bottom chord
by means of a couple of clip angles and four or five rivets. Only
sufficient rivets are required to prevent the sagging of the bottom
chord, since the floor-beam is connected to the hip vertical above the
lower chord, and hence no stress comes on the joint at the lower end
(see Fig. 178).

The width of the plate has been assumed as 8 inches. This
width is liable to be changed after the design of the intermediate

Fig. 178. Connection of Hip Vertical to Lower Chord.

posts has been made, since it will be economical to have all the inter-
mediate floor-beams of the same length; and therefore the width of
this plate will be changed so as to make the width of the hip vertical
the same as the width of the intermediate posts.

85. The Intermediate Posts. The post UyL2 must be designed
to stand a total stress of 163 600 pounds. Where possible, it is
economical to make the intermediate posts out of channels, as this
saves a large amount of riveting. As seen by the stress sheet, the
length of these posts is 30.1 feet center to center of end pins. It is

usually required that — must not be more than 100, and this con-
dition requires that the least radius of gyration cannot be less than
30.1 X 12

100

= 3.62.

225

216 BRIDGE ENGINEERING

From Carnegie Handbook, p. 101, it is seen that a 12-inch 30-
pound channel has a radius of gyration of 4.28, and will fulfil the
conditions. The area of two of these channels is 17 . 64 square inches.
The unit allowable stress (16) is:

P = 16 000 - 70 X X — - = 1° 090 pounds per square inch.

The required area is then determined to be 1OftQO - 16.2 square

inches; and as this coincides very closely with the area given, these

channels are efficient and will be used.

Fig. 179 shows the cross-section of this post. The radius of

gyration which was used above was the radius of gyration of the chan-
nels about an axis perpendicular
to their web. The radius of gyra-
tion of the entire section about
an axis perpendicular to the

Shop

,5 Shop 9 Shop

e Field

=fiii

I/- is Field

web will be the same as that of
one channel. In order to have
the sections safe, the radius of
gyration about the axis B-B must
be equal to or greater than the

about the axis B-B can be in-
creased or decreased by spacing the channels. The exact distance
which will make the two rectangular radii of gyration equal may
be determined by the methods of "Strength of Materials," or it
may be found in columns 14 and 15 of the Carnegie Handbook, p.
102. For any particular case it is equal to the value given in column
14, plus four times that given in column 15. For the channels under
consideration, it is equal to 7.07 + 4 X 0.704 = 9. 89 inches. Any
increase in this distance will only tend to increase the radius of
gyration about the axis B-B, and will make the post safer about
that axis.

Fig. 179 shows a diaphragm. The web of this diaphragm cannot
be less than f inch, and the size of the angles cannot be less than 3^
by 3 \ by f -inch, as this is the least allowed by the Specifications. The
function of this diaphragm is to transfer one-half of the floor-beam
reaction to the outer side of the post. The rivets which connect the

BRIDGE ENGINEERING 217

angles to the diaphragm web are shop rivets, and (see design of floor-

137 600
beam) must be- - ^-^—' — 9 in number. The rivets which connect

2X7 880

the diaphragm angle with the outer channel of the post are also shop
rivets, and are — -^ - = 10 in number, 5 on each side. The

— y\ / — — ' '

same rivets which connect the floor-beam to the post go through
the diaphragm angle on that side of the diaphragm next to the cen-
ter of the bridge, and must therefore be field rivets and take the

entire floor-beam reaction. These must be ----- = 23 in number,

6 013

12 on each side. The exact distance, back to back of the channels of
the post, cannot be determined until after the top chord has been de-
signed, since the post must slide up in the top chord and also leave
room on each side for the diagonal members of the truss. The
width is determined by the packing of the members at joint L2 (see
Fig. 174), and is found to be 9J inches. Since this is less than that
required above, the post must be examined for bending about an axis
parallel to the web of the channels.

According to the methods of "Mechanics" and "Strength of
Materials," with the help of the Carnegie Handbook, p. 102, the
moment of inertia about this axis is found to be 286.42, and the ra-
dius of gyration 3 .96. The unit allowable compressive stress is then
computed to be 9 580 pounds per square inch, and the required area

163 600

_ ---- = 17.10 square inches, which, being less than 17.64, shows

the section to be safe.

This member is connected to the top chord at its upper end by
a 5-inch pin. The total stress is 163 600 pounds, and the total bear-

163 600 s

ing area required is — -„„„ = 6.8 square inches, or 6 .4 square inches
^4 (JOU

for each side (19). The total thickness of the bearing area for each

side is — - = 0.68 inch. The thickness of the web of a 12-inch
5

30-pound channel is 0.513 inch, which leaves 0.68 - 0.513 = 0. 167
inch as the thickness of the pin-plate, but it must be made f inch
according to the Specifications. Fig. 180 shows the arrangement of
the plates and the rivets.

227

218

BRIDGE ENGINEERING

The sum total of the pin-plates and the channel web is 0.888
inch, and therefore on one side the stress transferred to the pin by

1 ^ 0.375 X 163 600
means of the pin-plate, which is 0 . 375 inch, is — X O~QQC —

= 34 600 pounds. This plate will tend to shear off the rivets between

34 600
it and the channel web, and therefore = 5 shop rivets are

required.

The stress that is shown on the stress sheet is the stress in the
post above the floor-beam. The stress in that part below the floor-

r~-
i
i
l

l '
j

EEZ-Ih^'

K1P

—

-- -

--jif---3

"71

tfi?

ih«i

,r
'i
1 1

i
1
,1

ife

M I •

iiilii

i>
•I

sW
p^p

A'l

.!fi

1 1 1 l

mtf

jfea

1 1

i°

!
0'
l

!°

°!

Fig. 180. Arrangement of Plates, Rivets. Pin, etc., at Connection of Intermediate Post
to Top Chord.

beam is equal to the vertical component of the diagonal in the panel
ahead of the post in question. In this case it is the vertical com-
ponent of the stress in U^, and is equal to 242 000 pounds, and this

. 242 000
requires a total bearing area of -^77^7^ = 10.1 square inches, and

a total thickness of

24000
v V" = 1.01 inches on each side, the pin being

5 inches in diameter. From this total thickness must be subtracted
the thickness of the web of the channel, and this leaves 1.01 — 0.513
= 0.497 inch as the total thickness of the pin-plates required. This
shows that we must use one ^-inch plate. The total thickness of the
bearing area is now 0.51 3 + 0.50 - 1.013 inches.

Each plate takes a total stress of

= 59 700

228

BRIDGE ENGINEERING

219

pounds; and the joint being weak in shear, the number of rivets

. , .„ ,
required will be

59 700

= 9 rivets in single shear. The detail will be

7220
similar to that in Fig. 180.

The distance, back to back of the channels in this post, will
probably not be greater than 12 inches, and this will make the dis-
tance between rivet lines about 9 inches. According to (44), the end
tie-plates must be at least 9 inches long and of course 12 inches wide.

The thickness cannot be less than — = 0.18 inch, but they will be

made f inch (36). Between the tie-plates the channels will be con-
nected by means of lattices. The Specifications (45) require that

they should not be less than 1\ inches in width and (1 .414 X 9) — - =

0.318 (say f) inch in thickness. Table XXV gives the thickness of
lacing bars for any distance between rivets.

TABLE XXV
Thickness of Lacing Bars

SINGLE LACING

(<=40:^=30°)

DOUBLE LACING

(i=^: 0 = 45o)

1 in.

Oft. 10 in.

i in.

1 ft. 3 in.

js in-

1 ft. 0^ in.

W in.

1 ft. 6} in.

f in. -

ft. 3 in.

1 in.

1 ft. lOJt in.

I7c in.

ft. 5i in.

T7« in-

2ft. 2±in.

i in.

ft. 8 in.

i in.

2 ft. 6 in.

T« in-

ft. 10^ in.

T* in-

2ft. 9|in.

« in-

2 ft. 1 in.

f in.

3ft. Hirf.

A width of 2} inches is chosen above, since according to Carnegie
Handbook, p. 183, a f-inch rivet is the largest which can be used in
the channel flange.

The post U^ must be designed for a total stress of 87000
pounds. It will be assumed that two 10-inch 20-pound channels

220 BRIDGE ENGINEERING

with a radius of gyration 3 .66 and an area of 5 .80 square inches each
will be sufficient. The length, as before, is 30.1 feet, and the unit-
stress is:

qrj i \/ 1 O

P = 16 000 - 70 X - — « 9 080 pounds,

o.ob

The required area is = 9.60 square inches. Since the

3 UoU

total area of the two channels is 11 .76 square inches, and the required
area is 9.6 square inches, it is seen that the}7 do not coincide very
closely. These channels, however, will be used, since the thickness
of the web is the thinnest allowed by the Specifications, and the
width of the channels is the smallest that can be used and still give
sufficient room to make the connections with the end connection
angles of the floor-beams.

The lower end of this post also has a diaphragm which must
transfer half of the stress to the outer channel of the post. The sides

of the diaphragm are the same
as in the posts previously de-
signed; and the number of rivets
required is computed in a simi-
lar manner and found to be as
indicated in -Fig. 181, which
shows the cross-section of this
post.

At the upper end the bear-
Fig. 181. Cross-Section of Intermediate . .

Post. mg area required on one chan-

87 1 00

nel is- = 1.814 square inches, and the thickness required

tL /\ <u4 UuU

is — ^ — == 0.363 inch, a 5-inch pin being used. As the web of the
o

channel is 0.382 inch thick, it will give sufficient bearing area
without pin-plates.

At the lower end, the vertical component of UJL3 is 157 500

pounds. The bearing area required on each side of the post is
-i r*j KAH o oo

= 3.28 square inches, and the thickness is — = 0.66

2i X 24 000 5

inch. The thickness of the channel web being 0 . 382 inch leaves 0 . 660

— 0.382 = 0.278 inch as the required thickness of the pin-plate;

BRIDGE ENGINEERING

221

but f inch must be used, making a total thickness of 0.382 + 0.375

= 0.757 inch. The plate will carry ~?j? X ~— = 39 000

39000
pounds, and this requires - = 6 shop rivets in single shear.

The distance, back to back of channels, will be the same as in

Fig. 182. End and Side Elevations Showing Detail of Construction at Lower End of
Intermediate Post.

C72L2, and therefore the tie-plates and lacing bars will be the same.
Fig. 182 gives a detail of the lower end of U3L3.

86. The Top Chord. The top chords of small railway bridges
may be made of two channels laced on their top and bottom sides.
This is not very good practice, since it leaves the tops of the channels
open and lets in the rain and snow, wjiich tends to deteriorate the
joints. It is better to add a small cover-plate, even if this does give

231

222

BRIDGE ENGINEERING

an excessive section. In case of stress such
as is demanded, the chords may consist of
two channels and a cover-plate. In this
case it is necessary to place small pieces
called flats upon the lower flanges of the
channel, in order to lower the center of
gravity of the section and to bring it near
the center of the web. This section makes
a very economical section in that it saves
much riveting. On account of channels
being made only up to 15 inches in depth,
' the use of this section is quite limited owing
to the fact that it is not deep enough to
allow the I-bar heads sufficient clearance,
for the I-bar heads in bridges of even ordi-
nary span will exceed this amount.

The most common section is that which
consists of two side plates, four angles, and
one cover-plate. Sometimes this section
has flats placed upon the lower angle in
order to lower the center of gravity, as ex-
plained above. According to (33), the sec-
tion should be as symmetrical as possible,
and the center of gravity should lie as near
the center of the web as is consistent with
economy.

In case the stress is great enough to
demand a heavier section than that above
described, additional plates are added upon
the sides of the original plates, and heavier
and larger cover-plates and angles are used.
Fig. 183 shows different types of chord sec-
tions.

Ir addition to the cover-plate being
designed to withstand the total stress, close
attention must be paid to (42). This clause
has been inserted on account of practical
considerations, since it has been found out

BRIDGE ENGINEERING 223

that if plates are made much thinner than the proportions here
required, they will crumple up and fail long before the allowable
unit of stress as computed from the formula has been reached. In
some cases — especially where the stress is light — the proportions laid
down in (42) and (36) will govern the design of the section, instead of
the required net area as determined by the formula for the allowable
unit compressive stress.

The design of the first section of the top chord will now be made.
Here, as in the case of the first sections of the lower chord, the
diameter of the head of the greatest I-bar determines the width of the
plates in the section The head of the 7-inch I-bar which constitutes
the member UtL2 is 17^ inches, and, allowing a clearance of | inch
on either side of the head, the total depth inside the chord should be
18| inches. As in the case of the lower chord, plates 18 inches wide
may be used.

The size of the angles to be chosen is a matter of judgment.
Usually any size should be chosen at first, and the preliminary design
will indicate at once what size should have been taken. For this case,
3^ by 3i by f-inch will be assumed at first.

For sections of this character, the radius of gyration is approxi-
mately equal to 0.4A, in which h is the height, or rather the width,
of the side plate. The approximate radius of gyration is r = 0.4 X
18 = 7.2 inches, and the length is equal to one panel length, or 21
feet. The allowable unit of stress (16) is:

P = 16 000 - 70 X 21?X212 = 13 550 pounds.

449 500
The required area is— ^-^n = 33.2 square inches. The correct

lo OOU

proportion for sections of this character is that 0 . 4 of the total area
should be taken up by the web. The area of the web would then be
0.4 X 33.2 = 13.28 square inches, and the thickness would be

L3J^L = o.37 inch. According to this, a f-inch plate should be used,
2 X 18 145

but (42) requires that it shall be.-^- = 0.483 inch or thicker.

Therefore an 18 by ^-inch plate must be used for the web.

The correct proportion for sections of this character is that the
width between plates should be about £ the width of the side plates.

233

224

BRIDGE ENGINEERING

This will give the required width between plates equal to | X 18 =

15.75 inches. The cover-plate (42) must not be thinner than —

the distance between the connecting rivet lines. The rivet lines are,
in this case, 15.75 + 2 X 2 = 19.75 inches apart, and therefore the

thickness of the cover-plate cannot be less than — — = 0.494 inch.

__ Neutral Axisr-v
Center Line of Pins.J

The cover-plate will therefore be taken as ^ inch thick. The width

of the cover-plate (see Fig. 184)
must be about 15.75 + 2 X 34
+ i = 23| inches (say 23 inches).
The cover-plate will be taken 23
by 4-inch.

The center line of pins will
be taken at the center line of the
web, and the center of gravity of
the section wrill be assumed as ^

inch above this. In order that

j the center of gravity may be near

. 1 j_» that assumed, the moment of the

— Ai ' -gh* I "™"! • cover-plate about the assumed

center of gravity axis should be
about equal to the moment of the
flats about the same axis. The
moment of the cover-plate about the assumed axis is:

OQ -y Q

(9.0 - 0.5 + 0.25 + 0.25) = -

approx.

I I

Fig. 184. Section of Top Chord.

and the moment of the flats about the same axis is:
A (9.0 + 0.5 + 0.25 + 0.5) = 10.25 A,

in which A is the area in square inches of both of the flats. Equating
these two expressions, and solving for A, there results:
23 X 9

A =

2 X 10.25

= 10.1 square inches.

Assuming the flats to be 4 inches wide, the thickness on each side
will be 1 .25 inches. As this is too thick to punch, the flats on each
side will be composed of two 4 by f-inch plates.
The total area is:

BRIDGE ENGINEERING 225

One cover-plate = 2.3 X i = 11.5 sq. in.

Two web plates = 2 X IS X i =18.0 " "
Two flats 4 X 1} =10.0 " "

Total 39.5 sq. in.

But the required area is 32.2 square inches, which is considerably
less than the area above given, and which does not include the angles
and hence we can use the smallest size angles, which are those pre-
viously assumed. The area of each of these angles is 2.48 square
inches, thus making the total area of the section 39 . 5 + 4 X 2 . 48 =
49.42 square inches. This is considerably in excess of the area as
required according to the formula for compression; but it is the least
allowed by the Specifications. Note that this is the case where
(42), instead of the formula for compressive stress, is the ruling factor
in the determination of the section.

The center of gravity of the approximate section must now be
determined, the moment of inertia and the radius of gyration about
the neutral axis must be computed, and the required area must be
determined by using this radius of gyration as computed. If the
required area as determined with the actual radius of gyration is less
than the approximate area, then the thickness of the angles or the
plates must be increased and the section then examined for its radius
of gyration and required area. If the area is sufficient, the section
is used ; if not, another recomputation is in order.

In the determination of the center of gravity of the section, the
moment is taken about the top of fhe cover-plate. The moments are
computed as follows:

Cover- plate (23 X $) X \ 2 . 88

Webs 2 (18 X i) X (9 + |) 175.60

Top angles 2 (2.48) X (1.01 + i) 7.50

Lower angles 2 (2.48) X (i + i + 18 + \ - - 1.01). . . . 89.30
Flats" 2 (4 X H) X 19$ 196 . 25

Total ...... 471.53

The center of gravity is now found to be i^-^r =9.55

~ 4 /\ .«w.4o

inches from the top of the cover-plate. The distance from the top of
the cover-plate to the middle line of the web is 9 + \ -f- % = 9.75
inches, and this leaves a distance of 9.75' — 9.55 = 0.2 inch from
the center line of the web to the neutral axis. This distance is gen-

235

226 BRIDGE ENGINEERING

erally represented by the letter e, and it is known as the eccentricity of
the section.

The moment of inertia about this axis must now be computed.
The relation used is that the moment of inertia about any axis is
eq-ual to the moment of inertia about some other axis, plus the product
of the square of the distance between the two axes by the area of the
section whose moment of inertia is desired. The moments of inertia
of the various parts of the section (see "Steel Construction," Part
IV, pp. 292 and 293) are computed and are as follows:

Cover-plates 955 . 26

Webs 486 . 72

Top angles 325 . 74

Lower angles I 359 . 74

Flats... ..-1017.37

Total. . .. 3 184.83

The radius of gyration is equal to the square root of the quotient
obtained by dividing the moment of inertia by the area. It is

184.83 =8Q4

\ 49.42

Using this value of the radius of gyration in the formula for the com-
pressive stress, there is obtained 13 800 pounds as the unit allowable

, 449 500

stress in compression, and this requires an area of =32.5

13 800

square inches. Since this is considerably less than the actual area of
the section, the section will not be changed but will be taken as first
assumed.

In order that the section should be safe about both axes, the
moment of inertia about the axis perpendicular to the cover-plate
should be equal to or greater than that as above computed. By com-
puting the moment of inertia about the axis perpendicular to the
cover-plate, it is found to be 3 256 . 3, which gives a radius of gyration
of 8 . 11 ; and since both of these are greater than those first computed,
it is seen that the section is safer about the axis perpendicular to the
cover-plate than it is about an axis perpendicular to the web plates.

There are small stresses in this member due to its own weight
and to the fact that the pins are not placed directly upon the neutral
axis (see "Strength of Materials," p. 82). These stresses are seldom
more than 1 000 pounds per square inch in the extreme fibre; and

BRIDGE ENGINEERING 227

since the section has such an excess of area, they will not be computed,
as it is evident that there is sufficient strength in the member to with-
stand them.

The section just designed is that for the top chord having the
greatest stress; and since this is the minimum section allowed by
the Specifications, it must be used in all the sections of the top chord.

The section as finally designed is:

One cover plate, 23 by J inch;
Two webs, 18 by £ inch;
Four angles, 3£ by 3£ by f-inch
Four flats, 4 by f-inch.

A pin 6|- inches in diameter will be used at the point Ur The
stress in the member UJJ2 is 378 200 pounds, and the bearing area

378 200
required is — - - = 15. 75 square inches, or 7 . 875 for each side.

7 875
This makes a total required thickness of ' = 1 . 265 inches for one

side. Since the thickness of the web plate is | inch, it will be necessary
to provide pin-plates whose total thickness must be 1.265 — 0.5 =
0 . 765 inch. Two f-inch plates will give a thickness of 0 . 75 inch ; and
since this is less than the required thickness by an amount not over
2 .V per cent, they may be used. The total thickness of the bearing
area is now 1 . 265 inches. The stress transferred to the two f-inch
plates is:

X 189 100 = 113 500 pounds.

The rivets required to keep the outer plate from shearing off the

113 500
other are - --- - =8 shop rivets, and the rivets required to keep both

— /\ / Z.£kj

of the §-inch plates from shearing off the web of the chord section are

113 500

-—--"— =16 shop rivets in single shear. The bearing of a |-inch

shop rivet on a i-inch plate is 10 500 pounds, and therefore the num-
ber of rivets required to keep these pin-plates from tearing the rivets

113 500

out of the vy-inch web plates is =11 shop rivets in bearing.

1U olJu

Fig. 185 shows the detail of this end of the top chord section. The pin-
plates should extend well back on the member, and at least one pin-

237

228

BRIDGE ENGINEERING

plate should go over the angle, and enough rivets, as computed above,
should go through the angles and this pin-plate. Experiments on full-
sized bridge members go to show that unless the pin-plates cover the
angles and extend well down on the member, the member will fail
before the unit-stress reaches that value computed by the formula
for compression.

Since the ends of the chord are milled at the splices, and therefore
butt up against each other and allow the stress to be transmitted

V®--®.

^O O O O

0 0 0 I

u, j

\ 0 0
®\ Q.

| -g Plate

® 1 o

'«> 0

o o

0..0

2 8 Plate'?

0 0

o o

\| gg 0

o o o •<

Fig. 185. Detail of Top Chord Section at Point Uv

directly, only sufficient rivets need be placed in the splice to keep the
top chord sections in line (55).

At the point U2, it is not necessary to put in a pin-plate to take
the stress in the upper chord; but it is only necessary to provide a
pin-plate to take up the difference in stress between the two chord
sections. This difference in stress is equal to the horizontal com-
ponent of the maximum stress in the member U2Ly This is 110 000
pounds, and the area required on each side for bearing is 2.3 square
inches; and as a 5-inch pin is used here, the thickness of the bearing

As this thickness is less than the thick-

2 3

area is -^- = 0.46 inch,
o

ness of the web plate, no pin-plates will be required.

At the point U3, a bearing area will be required to withstand the
horizontal component of the member U3L4. This is 56 300, and the

BRIDGE ENGINEERING

229

bearing area required on each side is

56300
24 000 X 2
1.18

= 1.18 inches. The

required thickness of the bearing area is ~ — = 0.24 inch, as a 5-

inch pin is used here also. As this thickness is less than the thickness
of the web plate, no pin-plate will be required.

The under parts of these members must be stiffened by tie or
batten plates, and these plates (44) must be equal in length to the
distance between rivet lines. This is 1 9 i inches. They will be made
20 inches long and 23 inches wide. The thickness of these plates (44)

must be

19.5
50

0.39 inch (say TV inch). The size of the tie-plates

will then be 20 in. by T77 in. by 1 ft. 11 in.

Since the distance between the rivet lines is greater than 15
inches, double latticing must be used (47); and according to Table
XXV the lacing must be \ inch
thick; also, according to (45), it
must be 2£ inches wide, as the
rivets used are f- inch in diam-
eter. The lattices will then be
2-fc by Hn.

87. The End=Post. Since the
minimum section as chosen for
the top chord is about 50 per
cent in excess of that required by
the compression formula, it will

Fig. 186. Calculation of End-Post.

be assumed to be sufficient for
the section of the end-post, and
it will now be investigated to see if it is safe.

In addition to the stress due to direct compression, the end-post
is stressed by its own weight, by eccentric loading due to the pin being
in the center of the web instead of at the center of gravity of the
section, and to a bending moment at the place where the portal brace
joins it. This is due to the bending action of the wind on the top
chord. These different stresses will now be computed; and since the
post is in all cases stressed by a combination of bending and compres-,
sive stresses, this fact should be considered in the design. In deter-
mining the stress in the end-post due to its own weight, the entire

239

230 BRIDGE ENGINEERING

weight must not be used in computing the bending action, but only
that component of it which is perpendicular to the end-post. The
length of the end-post is readily computed, and is as shown in Fig.
186. The general formula for accurately computing stresses due to
bending when the member is also subjected to compression, is:

My,

s = ~rr^T'

10E
in which,

S = Stress in pounds per square inch in the extreme upper fibre of the

beam;

M = Exterior moment causing the stress, and is considered positive if it
bends the beam downward, and negative if it bends the beam up-
ward;

j/,= Distance from the neutral axis to the extreme upper fibre;

7 = Moment of inertia of the section;

P = Direct compressive stress, in pounds;
I = Total length, in inches;

E = Modulus of elasticity of steel, which is usually taken as 28 000 000
pounds per square inch.

In this case the force causing the bending is that component of
the weight perpendicular to the end-post. This is Wl sin^, in which
W is the weight of the steel in the end-post; and this is computed and
is as follows :

Cover-plate 1 435 Ibs.

Web plates '. 2 245 "

Angles 1 250 "

Flats... ..1245"

6 175 Ibs.
Add 25 per cent for details 1 544 ' '

Total. ... 7 719 Ibs.
Substituting in the above formula the various values, there results :

„ = j X 7 719 X 36.7 X 0.572 X 12 X 9.55
,1Q_ 410500 X (36.7 X 12)2

10 X 28 000 000

= 800 pounds per square inch compression in the upper
fibre due to bending.

In the above equation, the stress in the member is 410 500 pounds;
the distance yl is the distance from the neutral axis to the top of the
cover-plate, and the coefficient of elasticity of steel is taken as
28 000 000.

240

BRIDGE ENGINEERING

231

In computing the stress due to the eccentric loading, the moment
is equal to the product of the total stress in the member by the dis-
tance" from the neutral axis to the center of gravity axis causing a
negative moment. Substituting in the above formula for combined
stresses, there results:

- 410500 X 0.2 X 9.55

S = -

3 185 -

410500 X (36.7 X 12)*

4 £70

10 X 28 000 000
= 270 pounds per square inch tension in the upper fibre.

In order to find the compression in the lower fibre, it is only necessary
to notice that the stresses are proportional to the distances from the
neutral axis. Accordingly (see Fig. 187),
the stress in the lower fibre due to the
weight is 895 pounds tension, and the
stress in the lower fibre due to the eccen-
tric loading is 302 pounds compression.

Before computing the stress due to the
bending moment caused by the wind on
the upper chord, it is necessary to in-
vestigate the post to see if it is fixed or
hinged at its lower end. This is very
important, since, if the post is found to
be hinged, the bending moment will be
one-half of that which will occur when
the post is not hinged.

An end-post is considered hinged when the product of one-half
of the total stress times the distance between the web plates is greater
than the product of the wind load acting at the hip, or joint Uv times

the length of the end-post.

Calculation of Stress
in Chord.

T u- i C .. 410500
In this case the first value is — X

15 = 3075000; and the product of the latter (see Article 29) is
12 600 X 36.7 X 12 = 5 550 000. Since the latter is greater than
the former, the post is hinged, and the bending moment at the foot
of the portal strut, which joins the end-post 28.2 feet from the end, is
6300 X 28.2 X 12 = 2 130000 pound-inches. The stress in the
extreme fibre due to this bending moment is:
2130000 x 11.5

S =

3256.3 -

410500 X (36.7 X 12)a
10 X 28 000 000

241

232 BRIDGE ENGINEERING

= 8 250 pounds per square inch tension or compression.
In computing this stress due to the wind moment, care must be taken
to take y1 equal to one-half the width of the cover-plate, and to take
'the moment of inertia as that about the axis perpendicular to the
cover-plate.

In computing the total stress on the extreme fibre, it must be
noted that the stresses due to weight and eccentric loading do not
stress the same extreme fibres as the stress due to wind, the former
stressing the extreme fibres on the top and bottom of the post, while
the latter stresses those on the inner and outer sides. The total

410 500
direct unit-stress is ———— = 8310 pounds per square inch; and

this, added to the 8 250 pounds per square inch due to the wind,
gives a total of 16 560 pounds per square inch on the extreme
fibre only.

oc 7 v 1 9

The allowable unit-stress is 16 000 - 70 X— - = 12 200

8.11

pounds per square inch when wind is not taken into account, and
(23) is H X 12 200 = 15 250 pounds per square inch when the wind
is taken into account. The difference between this and the actual
stress is 16560 — 15250 = 1 310 pounds per square inch, which
shows that the section is not strong enough. The section can be in-
creased by widening the cover-plate or by making the plates thicker ;
but as this excess is due to wind only, the section being amply suffi-
cient under the other stresses, and is fixed to some extent by the
floor-beam connection, no change will be made.

The pin at each end of the end-post will be the same — namely,
6^ inches in diameter — and therefore the pin-plates will be the same
at each end. The total stress in the post is 410 500 pounds, which

, 410 500 , „ 0

makes a required bearing area of = 17.2 square inches for

^4 UUU

both sides, or 8.6 square inches for one side, and the total required

8 6
thickness of — '— = 1.375 square inches for one side. Since the

thickness of the web plates is \ inch, this leaves a remainder
of 1.375 — 0.5 = 0.875 inch for the thickness of the pin-plates.
One plate f inch thick and one plate \ inch thick will be used.

The proportion of the total stress which is taken by the f-inch

242

BRIDGE ENGINEERING

233

plate is X — = 56 000 pounds; and that taken by the

1 . o75 2i

£-inch plate is ^7~X 205250 - '74 600 pounds. The number of

rivets required to transfer the stress from the f -inch plate to the ^-inch

56 000
plate is = 8 shop rivets in single shear; and the number of

rivets required to transfer the stress from both pin-plates to the web is

56 000 + 74 600 • , • • L •

— — — =18 shop rivets in single shear. As in the case of

the top chord, one pin-plate should extend over the angle, and the
number of rivets required in that pin-plate should go through the pin-
plate and the angles (see Fig.
188). The |-in. hinge plate is
used for erection purposes, and is
not considered as a pin-plate. It
is omitted at L0.

Since this section is the same
as that of the top chord, the tie-
plates and the lattice bars must
be the same size.

88. The Pins. The design of
the pins requires a simple but
quite lengthy computation. Sim-
ple Pratt railroad trusses for
single-track bridges usually have
the same arrangement of tension

and compression members; that is, the same tension members occupy
relatively the same positions with respect to the compression mem-
bers. Also,while theoretically a different sized pin will be required at
every joint, it is not customary to make them so. In practice the
pins at the joints Ul and Z/0 are made of the same diameter, and
those at the remainder of the joints are also made in diameter equal
to each other but different from those at U1 and L0, the pins at U\
and L0 usually being larger in diameter. On account of the above
conditions and facts, it is unnecessary to design the pins in spans
under 200 feet, since usually they are the same for any given span
and loading. Table XXVI gives the diameters of pins for spans of
100 up to 200 feet for loading E 50.

Fig.

Plates and Riveting at Upper
End of End-Post,

243

234

BRIDGE ENGINEERING

TABLE XXVI
Pins for SingIe=Track Bridges

Loading E 50

DIAMETER OF Pi:

U1 and L0

All Others

100 feet

4i inches

4 inches

125 "

5* "

150 "

6J "

5i "

175 "

6| "

52 «'

200 "

For E 40 loading, decrease the above values by J inch; for E 30 loading,
decrease them by f inch. The diameter of pins for spans not given in the
table can be interpolated from the given values. No pin should be less than
3£ inches in diameter.

The span of this bridge is 147 (say 150) feet, and the diameter
of the pins at Ul and Z/0 is 6^ — | = 6J inches; and the diameter
of the pins at the other panel points is 5| — J = 5 inches. It
should be noted that no pin is required at point Lv as the two mem-
bers which join here are built-up members and are riveted together.

The above table is for single-track bridges only. The diameters
of pins for double-track bridges are given in Table XXVII. These
values are for E 50 loading ; and for E 40 and E 30 loading, deduc-
tions must be made as required in the case of Table XXVI.

TABLE XXVII
Pins for Double-Track Bridges

Loading E 50

DIAMETER OF PIN

E7, and L0

All Others

100 feet

6 inches

5} inches

125 "

6j "

150 "

1\ "

175 "

9} "

• 81 "

200 "

9i "

No pin in a double-track bridge should be less than 4£ inches in diameter.

Pins for highway bridges are usually much less in diameter than
those for railway bridges, except in the case of first-class trusses for
heavy interurban traffic or for city bridges carrying paved streets.

244

BRIDGE ENGINEERING

235

where they should be taken equal to those given for E 30 loading.
Table XXVIII gives the diameters of pins for different length spans
of simple highway bridges designed for 16- ton road-rollers or farm
wagons and 100 pounds per square foot of roadway.

TABLE XXVIII
Pins for Country Highway Bridges

DIAMETER OF PIN

U , and Lower Chord

Upper Chord

50 feet

2£ inches

2 inches

100 "

2i "

150 "

3i "

2| "

200 "

3 "

89. The Portal. In order to have a clearance of 21 feet (2)
above the top of rail, it is necessary that the portal be as shown in
Fig. 189. The stresses are found
by methods of Article 54, Part I,
the wind load being computed
according to (10). It must be
remembered that the column is
hinged.

In case the members of the
portal braces bend about one
axis, their length will be equal
to the distance from one end to
the other. In case they bend
about the other axis as indicated
by the broken line in Fig. 189,
their length will be one-half of
what it was in the first case.

The portal struts or
diagonals will be designed first.
Their length is 8.5X1.414-
12 feet, or 144 inches. This is

Fig. 189. Portal Dimension and Stress
Diagram.

the total length. Although
the Specifications do not men-
tion it, the ratio of the length to the radius of gyration should
not exceed 120. This means that the radius of gyration in this

245

236 BRIDGE ENGINEERING

144
case should be greater than-— = 1.2. The section of the strut will

be composed of two angles placed back to back.

Two angles 3J by 3 by f-inch, with an area of 4.6 square inches
and r2 equal to 1 .72 — see Carnegie Handbook, p. 146, and (72)— will
be assumed to be sufficient to take the stress, and they must now be
examined to see if the assumption is correct.

The allowable unit-stress (23) is 25 per cent greater than in the
case of live or dead loads. This makes the unit-stress as computed
from the formula:

16 000 - 70 x

j \\ = 12 680 pounds per square inch.

*3S 500
The required area is ' = 3.05 square inches; and since this is

less than the given area, the angle will be amply sufficient. The re-
quired area is over one square inch less than the given area, but this
angle must be used, since it is the smallest angle allowed by the Specifi-
cations. Note that unequal legged angles should be used, as this will
make the radius of gyration about one axis larger than about the other;
and this will prove economical, since, when one axis is considered,
the length of the member is greater than when the other is considered.
The above angle should also be examined for tension, it being
considered that one rivet-hole is taken out of the section of each angle.
The net section of the two angles will now be 4.60 - 2 (| + £) X f

38 500
= 3 . 85 square inches ; and the area required for tension is ftnm

= 1 .93 square inches, which shows that the angle is amply sufficient.
It should be noted that these Specifications do not require that only
one leg of the angle shall be efficient unless both legs are connected. In
case this strut had been designed according to Cooper's Specifica-
tions, two angles 5 by 3 by ^-inch would have been required, and the
5-inch leg would have been placed vertically and the angle connected
by this leg alone. While it is not within the province of this work to
discuss the question of connecting angles by one or by both legs, yet
it might be said that tests made on angles connected with one leg
only, seem to indicate that the ultimate strength in tension is about
60 per cent of that obtained from the same angle when tested with
both legs connected.

246

BRIDGE ENGINEERING 237

While according to (20) the alternate strains in the wind bracing
do not have to be considered, since they do not occur very closely
together, yet in framing the connections it is required that the sum of
both positive and negative stresses shall be added. In this case the
stress for which the connections must be designed is 2 X 38 500
= 77 000. It must be remembered that in this case also, the
unit-stresses are increased 25 per cent over those allowed for live and
dead loads.

The number of rivets required in the end connections will be
governed by bearing in the connection plates, and these plates are
usually made f-inch thick. The number of rivets required is

77 000 77 000

TSSO^Oi = 8 Sh°p nV6tS' °r 6560XT* =

The portal bracing is riveted up in the shop and brought to the
bridge site, where it is connected to the trusses by field-riveted con-
nections at its end. Therefore the end of the portal struts which
connect with the top piece will have 8 shop rivets, and the other end
which connects with the end-post will have 10 field rivets. Since
the angles are small, all the above rivets must go in one line, and this
will cause the connection plate to be quite large. It will probably be
better to connect both legs of the angle by means of clip angles and
thus reduce the size of the connection plates.

The top part of the portal bracing will consist of two. angles.
Two angles 3^V by 3 by f-inch will be assumed and examined to
determine if the area is sufficient. The length of this strut is the
distance center to center of trusses, and is equal to 17 X 12 = 204

^04
inches. The least radius of gyration is therefore *"-— = 1 . 70. The

radius of gyration of the two angles assumed is 1 . 72 when referred to
an axis parallel to the shorter leg when the two angles are placed back
to back and One-half inch apart. The unit-stress is now computed :

P = Ae 000 - 70 X ^2) H = 9 625 pounds per square inch.

27 200

The required area is " = 2.825 square inches. This is con-

siderably less than the area given by the two angles; but as these are
the minimum angles allowable, they must be used. Since the stress
in this case is less than in the previous -case, and since the angles

247

238

BRIDGE ENGINEERING

used are the same, it is evident that these angles are safe in tension.
The number of rivets is determined by the bearing in the f-inch
connection plates, and is :

2 X 27 200

7880 X 1.25
2 X 27 200
6 560 X 1 . 25

6 shop rivets, and

= 10 field rivets.

As in the case of the lateral strut, this member should be connected
by both legs of the angle in order to reduce the size of the connection

plates. Fig. 190 gives the
details of the portal bra-
cing and its method of
connection to the end-
post. The full circles
represent shop rivets, and
the blackened circles rep-
resent field rivets. Some
engineers connect the
portal bracing to the top
cover-plate of the end-
post. This produces an
excessive eccentricity in

Fig. 190. Details of Portal Bracing and Connection the end-DOSt and is bad
to End-Post.

practice.

Those members of the portal bracing which do not take any
stress will be made of single angles, and the size of these angles will be
taken 3£ by 3 by f-inch.

90. The Transverse Bracing. This bracing will be the same
general style as the portal bracing, except that the top member will
consist of two angles placed at a distance apart equal to the depth
of the top chord, and these angles will be joined together by lacing.
As in the case of portal bracing, those members which do not take
stress will be made of one angle 3| by 3 by f-inch.

The general outline is shown in Fig. 191, and the stresses are com-
puted from (10) and by the methods of Article 54, Part I. In design-
ing this top member, the top angle only is supposed to take the stress.
The length in this case is 204 inches. Two 3% by 3 by f-inch angles
will be assumed as sufficient and will be examined. These angles

248

BRIDGE ENGINEERING

239

give a total area of 4.60 square inches. In examining these it will
be found that they are amply sufficient, in fact so much so that it will
be better to see if one single angle at the top will not be better.
According to the length, the smallest radius of gyration which can be
used is 1.7. In looking over the tables of angles, it is seen that the

T-£6"

Section ot BW

Fig. 191. General Outline of Transverse Bracing.

first angle to fulfil this condition is a 6 by 3^ by f-inch, and it has a
radius of gyration of 1.94. The allowable unit-stress is computed
as follows:

6 000 - 70 X

= 10 78° pounds per square jnch;

and the required area is -=~; = 0.85 square inch. This is con-

249

240

BRIDGE ENGINEERING

siderably smaller than the area of the angle, which is 3.97 square
inches; but since this is the smallest possible angle which will fulfil
the conditions of the Specifications, and since it is much smaller than
the two angles as first assumed, it will be used. Fig. 192 gives a cross-
section of this member. Since this angle is
joined to the cover-plate by one leg, the joints
will be weak in single shear, and the number of
rivets required will be:

2 X 9 100

2 X 9 100
6013 X H

3 field rivets.

According to (45), the width of the latticing
must be 1\ inches; and according to Table XXV,
the thickness must be -,7B inch, the distance c be-
ing 1 foot 11 inches.

The length of the knee-bracing is 144 inches;
but on account of the small stress, one angle will
be used. One 4 by 3 by 4-inch angle, with an

*, ' *

area of 2 . 48 square inches and a radius of gyra-
tion 1.26, will be assumed as sufficient. The radius of gyration
is greater than the minimum allowable, which is 1.2. The allowable
unit-stress is:

Pig. 192. Cross-Sec-

tion of Top Member or

Transverse Bracing.

P = ( 1C 000 - 70 X y-Q?;) H = 10000 pounds per square

inch.

The required area is

12300

= 1 . 23 square inches. The required

10000

area is much less than the given area; but this angle must be used,
since it is the only one allowed on account of its radius of gyration.
Two of the minimum sized angles might have been used; but their
total area, 4.60 square inches, is much in excess of that of the angle
used.

This angle must be examined for tension. The net area is 2.48
- (I + i) X I = 2 . 1 square inches. The required net area in

12 300

tension is ~L = 0.615 square inch, which shows this angle

ID uuu /\ i j

to be amply sufficient.

250

BRIDGE ENGINEERING

241

The number of rivets required will be governed by the shear,
since the angle is connected by one leg only; and it is:

2 X 12 300

7 220 X H
2 X 12 300
6013 X H

= 3 shop rivets, and

= 4 field rivets.

91. The Lateral Systems. The stresses in these systems must
be computed according to (10) and Article 54, Part I. They are given
on the stress sheet, Plate III (p. 251). Since according to (68) these
members must be constructed of rigid shapes, it is customary, in com-
puting the stresses, to assume that one-half the shear is taken by each
of the diagonals in any given panel; that is, one diagonal is in tension,
and the other diagonal is in compression. The
stresses given on the stress sheet are computed
by making this assumption. Also, since both
diagonals in each panel are considered as acting
at the same time, the stresses in all the verticals
are zero.

The section of the upper lateral members will
be made up of two angles placed apart a distance
equal to the depth of the top chord. Fig. 193
shows the section. The radius of gyration about
the axis parallel to the long leg will be consider-
ably larger than that about an axis parallel to the
shorter leg. In fact, it is so much greater that the
strut will not need to be examined with respect
to this axis. The diagram of the first panel is
given in Fig. 194. The radius of gyration is to be taken about
the horizontal axis if the entire length is to be taken; and the radius
of gyration is to be taken about the vertical axis if one-half the
length is taken, in which case it will bend as shown by the broken
line in Fig. 194. The members are designed for the latter
conditions only, since they are amply safe in regard to the first
condition if they satisfy the latter. The length in this latter con-
dition is 13.5 feet, wrhich requires a radius of gyration not less than
13.5 X 12

Fifi.

Section of
Latei
mber.

Upper Lateral

120

1.35.

Two angles 5 by 3 by f-inch, with a total area of 5 . 72 square

251

242

BRIDGE ENGINEERING

inches and a radius of gyration equal to 1.61, will be assumed and
investigated to determine if they are sufficient.

1Q (% \x 19

The unit-stress is computed to be P = (16 000- 70 X — p^") X

A 7OO

1 \ = 11 000 pounds per square inch, and the required area is =

1 1 UUU

0.61 square inch. The required area is very much less than the
given area; but the angle chosen must be used, since this is the
smallest one which conforms to the requirements of the Specifications.

Fig. 194. Outline Diagram of First Panel
in Upper Lateral System.

F«

Fig. 195. Outline Diagram of First Panel
in Lower Lateral System.

The width of the lattices (46) must be 2| inches; and according
to Table XXV, the thickness must be -^ inch, the distance c being
23 inches.

Single shear governs the number of rivets required. In accord-

ance with (20) and (23), their number is

= 2 rivets. Field

6 013 X 1 j

rivets 3 in number are used in all places, since the lateral system is
riveted up after the trusses are swung into place.

Since this is the minimum sized angle which will give a radius of
gyration greater than 1 . 35, it must be used in the remainder of the
panels of the top chord. Four angles of the minimum size might
have been used, and would have been satisfactory, except that the
area would have been excessive.

The stresses in the lower lateral system are computed according
to (10), a similar assumption to that for the upper lateral system being
made — namely, that both diagonals in each panel are stressed at the
same time, one taking tension and the other taking compression.
Fig. 195 shows the first panel of the lower lateral system. These

252

BRIDGE ENGINEERING 243

diagonals are connected to the stringers wherever they cross them,
and also to each other where they cross in the center. This reduces
the length which must be used in computing the cross-section of the
member. In this case it is the distance C-A, and is equal to 90 inches.
Since the angle is free to move about either axis, angles with even
legs should preferably be employed, since this will give greater

economy. The radius of gyration must be greater than - = 0.75.

One angle 3 1 by 3^ by f-inch, with an area of 2 .48 square inches
and a radius of gyration of 1.07, will be assumed and investigated.

The allowable unit-stress is P = (16 000 - 70 X — ) 1| = 12650

30 500
pounds per square inch, and the required area is = 2 . 38 square

inches. This is nearly equal to the given area, and therefore the
angle chosen will be taken for the section.

This angle must now be investigated for tension, one rivet-hole
being taken out of the section. The net area is 2.48 — (J + i) f =

on £Af)

2.10 square inches. The required net area is ^ -- - =1.53

square inches, which shows the angle to be sufficiently strong.

Single shear determines the number of rivets to be required.
These are :

All rivets in the lower lateral system are field rivets, since this system
also must be riveted up in the field after the trusses are swung into
place.

The total stress in the second panel is 21 500 pounds, and a
3£ by 3 by f-inch angle, with an area of 2.30 square inches and a
least radius of gyration of 0 . 90, will be assumed and examined. The

90
allowable unit-stress in compression is 1^ (16 000 — 70 X •) =

21 500
1 1 250 pounds per square inch, and the required area is = 1 .91

square inches. Since this is less than the given area, and since the
size of the angle (72) is the smallest allowable, this angle must
be used.

253

244

BRIDGE ENGINEERING

It is required that this member shall have a net area of
91 c^oo
- = 1.08 square inches in tension. The net area of the

angle, one rivet-hole being taken out, is 1.92 square inches, which
shows the angle to be safe in tension.

The number of rivets required is determined by single shear,

Fig. 196. Two Types of Bearings.

since they tend to shear off between the member itself and the con-

2 X 21 500

necting plates. The number required is 77:^ — = 6 field rivets.

6 Olo X lj

Since the above angle is the smallest that can be used, and since
the remaining angles of the panel of the lateral bracing have smaller
stresses than the one just designed, it is evident that this size angle
must be used in all panels of the lower lateral system other than the
first.

92. The Shoes and Roller Nests. For bridges of short spans
and for plate-girders whose spans require rocker bearings to be pro-
vided (80), several different classes of bearings are in use. Two such
bearings are shown in Fig. 196 (a and b). The type illustrated by a
is seldom used on any spans except plate-girders. That shown in b

BRIDGE ENGINEERING

245

may be used on either plate-girders or small truss spans; it is the
invention of Mr. F. E. Schall, Bridge Engineer of the Lehigh Valley
Railroad, who uses it on plate-girders. It has given very great satis-
faction ; and for simplicity of design and also for economy it is to be
recommended. Some railroads have used a bearing which consisted
of a lens-shaped disc of phosphor-bronze, the faces of which fitted
into corresponding indentations in both the masonry and the bearing
plates. One advantage of this bearing is that it allows movements
due to the deflection of the girder, and also lateral deflection of the
floor-beam. It is claimed to have given satisfaction.

A bearing which is used on both short-span and long-span bridges

Fig. 197. Bearing Adapted to Bridges of Both Short and Long Span.

is shown in Fig. 197. This class of bearing will be used. The end
reaction of the bridge proper is equal to the vertical component of the

30 1
stress in the end-post, and is -^— - X 410 500 = 336 500 pounds,

336 500
which requires a bearing area (19) on the masonry, ot — ^- — =561

()U1/

square inches. According to the table on page 193, the masonry
plate will be 28 inches long.

The total bearing area for one of the vertical plates is:

255

246 BRIDGE ENGINEERING

and the total required thickness is:

1^=1.12 inches,

a 6^-inch pin being used at L0. Since the vertical plates will be made
f inch thick, this leaves a remainder of | inch to be made up of pin-
plates.

The amount of stress which is carried by the f-inch pin-plate is

= 56 100 pounds. These plates will tend to shear

56 100
off the rivets at a plane between the plates, and therefore = 8

shop rivets will be required to fasten them to the vertical plate.

Since the length of the masonry plate is 28 inches, and the total

561
area required is 561 square inches, the required width is — — = 20

inches. The actual width will be greater than this, since it must be
sufficient to allow for the connecting angles and also for the bearings
of the end floor-beam. The connecting angles should be f inch thick,
and should not be less than 6 by 6 inches; and the plates to which they
are connected should not be less than f inch in thickness, and likewise
they should not be greater, on account of the punching. The bottom
plate should extend outward about 3 inches, in order to allow suf-
ficient room for the anchor bolts, which should be f inch in diameter
and should extend into the masonry at least 8 inches.

In addition to the reaction of the bridge proper, the masonry
plate must be of sufficient area to give bearing for the end reaction of
the end floor-beam. The maximum end reaction (see Article 83, p.
197) is 104 740 pounds. The bearing area required on the masonry

104 740

is — — — — = 175 square inches; and assuming that the base of the
uUU

bearing will be 12 inches long (see Fig. 197), the required length will
be 14.6 inches. Usually, however, the bearing is extended the entire
length of the masonry plate, which is 28 inches in this case.

The distance from the center of the pin to the top of the masonry
will be the same for both the fixed and the roller end. This distance
should be such that the angles of the shoe will clear the bottom chord
member and allow the floor-beam to rest upon the plate as shown.
Since the first section of the bottom chord is 18A inches deep, the top
of the angles of the two must be at least 9j inches from the center line

256

BRIDGE ENGINEERING

247

of pins. This requires that the distance from the center line of the
pin to the base of the angle shall be at least (9j + 6) = 15j inches,
or more.

The tops of all floor-beams are at the same height, and the
bottoms of the intermediate floor-beams must be on a level with the
bottom of the first section of the lower chord (see Fig. 174). This
requires that the bottom of the intermediate floor-beams shall be
9j inches below the center line of pins, and this brings the top of the
floor-beams (48-f — 9|) = 39 inches above the center line of the pins.
Since the end floor-beam is 52 \ inches deep, back to back of angles, the

Fig. 198. Type of Bearing Construction where End Floor-Beam Does Not Rest Directly on
Bearing or Masonry Plate. Grillage of Iron Bars Used instead of Cast-Steel Pedestal.

lower flange will be (52 J — 39) = 13| inches below the center line of
pins. In case the end floor-beam does not rest directly upon the
bearing plate or the masonry plate, the intervening space is filled out
with a grillage of iron bars or a cast-steel pedestal, as shown in Figs.
197 and 198.

The small plates upon the side of the shoe, going entirely around
the pin, are called the shoe hinge-plates. These do not take any stress,
and require only sufficient rivets to hold them in position. They are
used during erection to keep the end-post in line; and after erection
their function is to keep the end-post on the shoe, and to prevent it
from having any upward motion due to the vibration of the structure.

257

248 BRIDGE ENGINEERING

The rivets through the vertical legs of the shoe angles are in
double bearing in the f-inch angles, in single bearing in the vertical
plate, and in double shear. A rivet in double shear has a less value
than in bearing in the plates. This value is 14440 pounds, and
therefore the number of shop rivets required through the vertical legs
of the angles is :

336 500
2^04440 =12nVetS-

The rivets which go through the horizontal leg of the angle and
through the cap plate and cap angles, do not take stress. The num-
ber of rivets put in is that demanded by the detailing, the rivets in the
horizontal legs of the angles usually staggering with those in the verti-
cal legs. The cap plate tends to keep the vertical plates in line, and
to keep out the dust and dirt and other deteriorating influences of
the elements.

Wherever the rivet-heads tend to interfere with other members
or project beyond surfaces which are required to be flat — as, for
example, the bottom of the masonry or bearing plates — they must be
countersunk (see Carnegie Handbook, p. 191, and "Steel Construc-
tion," Part III, p. 192).

The space for the anchor bolts, that for the connection angles,
and that for the bearing of the end floor-beam, require that the total
width of the masonry plate for the fixed end shall be 2 X f + 14£ + 2
X 6 + £ + 3 + 12 - 3 feet 7$ inches.

The design of the roller end requires that the length of the
masonry bearing, the size of the vertical plates and angles, and also
the number of rivets shall be the same as that for the fixed end. The
width of the masonry plate is determined by the length of the rollers
and their connections at the end.

The rollers (60) are required to be 6 inches in diameter, and the
unit-stress (19) per linear inch is 6 X 600 - 3 600 pounds, which
requires :

336 500

—n-fnfr = 9»- 0 linear inches,
o oUU

This is for the reaction of the bridge alone; and in addition to this,
there are required for the floor-beam reaction :
104 740

3600

= 29.0 linear inches.

258

BRIDGE ENGINEERING

249

The total number of linear inches is 93.5 + 29.0 = 122. 5; and if 5

122 5

rollers are used, they must be at least — ^- = 24.5 inches long. The

masonry plate is only 28 inches long, and therefore cylindrical rollers
cannot be used, since they would occupy a space 30 inches or over.
Segmental rollers (see Fig. 199)
must be used.

The determination of the
sizes of the angles which go at the
end of the rollers, and also of the
guide-plates, is a matter of judg-
ment and experience. Those
sizes indicated in Fig. 198, repre-
sent good engineering practice,
and will be used.

The distance from the center
line of pins to the top of the ma-
sonry can now be determined,
and is 1G| • + f -f 6 + f = 23f
inches.

On account of putting in Fig. 199. Segmental Rollers Used for Bear-
ings in Space under 30 Inches.

sufficient connections and angles

as shown in Fig. 198, the masonry plate must be considerably wider
than that theoretically determined. According to Fig. 198, the total
width must be as follows, and the width should be computed in two
parts, as the plate is not symmetrical about the center line of the truss:
From center line to outer edge:

^ + I + 6 + £ + (3 - $ = 2f ) + (3$ - | = 3i) + 3

= 1 ft. Hi in., (say, 1 ft. 11 in.).
From center line to inner edge:

— + | + 6 + 1 + 12 + £ + 2$ + 3i + 3 = 3 ft. Oi in., say, 3 ft. 0 in.

Total width 4 ft. 11 in.

Allowing guide-plates and guide-bars of dimensions as shown in
Fig. 198, and assuming | inch as clearance at the ends, the total length
of the rollers is:

(4ft. 11 in. ) - 2(3 + 3i + | + | + J) = 44. 5 inches.
This shows them to be amply long enough, as only 22 inches is

259

250 BRIDGE ENGINEERING

theoretically required. Here, as in most cases for single-track spans
up to 200 feet in length, the width of the masonry plate is determined
by the detail and not by the unit bearing stress.

The guide-plates are small bars riveted to the top of the bottom
plates, and serve to keep the rollers in line. The guide-bars are con-
nected to rollers at their ends, and serve to keep the rollers equi-
distant, therefore causing them to roll easier and keeping them from
becoming worn by contact with each other.

The expansion (57) must be allowed for at the rate of | inch for
every 10 feet in length of span. This makes a total allowed for tem-

147
perature of expansion of --•_- X & = If (say 2) inches. No slotted

holes are to be provided for the anchor bolts, since they do not go

i i i ! '

•. M (-7
•<•.: ""•;!-£"*-,.' I

-14

Fig. 200. Binding of Insufficiently Spaced Fig. 201. Computation of Spacing for Seg-
Segmental Rollers. mental Rollers.

through that part of the bridge which slides. The shoe slides over
the rollers, and is kept in place by the angles at the end, which are
riveted to the masonry plate (see Fig. 198).

Unless sufficient room is allowed between the segmental rollers,
they will tend to bind when the bridge has reached the extreme posi-
tion for expansion or contraction (see Fig. 200). This distance can be
computed from proportions as indicated in Fig. 201, and from the
following formulae*:

in which e is the amount allowed for the change of temperature, and
D is the diameter of the rollers, both being taken in inches. The
angle <f> is in degrees. In the present case, e'is2 inches; D is 6 inches;

* Derived by Prof. Frank B. McKibben of Lehigh University, and published in
Engineering Xews, December, 1896.

252

BRIDGE ENGINEERING

and <£, computed from the above formula, is 9° 30'. Substituting in
the equation giving the value for y, there is obtained 1.02 inches (say
\\ inches) for the distance between rollers. Rollers must not be less
in thickness than the total expansion allowed for temperature.

Since there are 5 rollers, there are 4 spaces between them. Also,
since the rollers must occupy a space of 28 inches, the length of the
masonry plate, each roller must be:

OQ _ 4 v 1 4-

— -1=4.6 inches (say 4£ inches) wide,
o

The width of the guide-bars must be such as to allow freedom

Fig. 202. Details of End Floor-Beam Connections.

of motion for the rollers. The maximum width allowable is given by
the formula :*

W = -^-cos <£,

in which <£ and D are indicated above. This requires the bar to be .

W =^~X 0.985 = 2.96 (say 2^) inches wide.

93. The Stress Sheet. Plate III shows the stress sheet of the
bridge which has been designed in the preceding articles. This
sheet represents the best current practice among the larger bridge

* Derived by Prof . Frank B. McKibben of Lehigh University, and published in
Engineering News, December, 1896.

BRIDGE ENGINEERING

253

corporations. It will
be noted that very
few details are given
upon the sheet; also
that few rivets are
noted, and that
sketches showing the
manner in which the
parts go together are
entirely wanting. The
shears and moments
for the stringers and
floor-beams, as well as
the reactions and the
number of rollers re-
quired, are given. This
is to save the drafts-
man the trouble of
recomputing values
which have necessarily
been determined by
the designer.

The details of the
various members, and
also • the manner in
which the different
members are con-
nected, are left to the
draftsman, who is un-
der the direct super-
vision of the engineer
in charge of the draft-
ing room, upon whom
rests the responsibility
for good details. The
figures given in the
text indicate the best
current practice. Figs.

n _j i 10 £

-1 " -1 a

ffl

IP ••

. Details of

-H

254 BRIDGE ENGINEERING

202 to 204 show details of the end floor-beam connections, and also
the packing of the members of the upper and the lower chord. The
arrangement here given may be said to be standard for single-track
Pratt truss spans up to 200 feet in length.

BIBLIOGRAPHY

The following books are recommended to the student in case it
is desired to pursue further the study of the subjects of Bridge
Analysis and Bridge Design :

The Theory and Practice of Modern Frame Structures. JOHNSON;
BRYAN, and TURNEAURE. John Wiley & Sons, New York, N. Y.

Roofs and Bridges. MERRIMAN and JACOB Y. John Wiley & Sons, New
York, N. Y.

Design and Construction of Metallic Bridges. BURR and FALK. John
Wiley & Sons, New York, N. Y.

Influence Lines for Bridges and Roofs. BURR and FALK. John Wiley
& Sons, New York, N. Y.

Details of Bridge Construction. FRANK W. SKINNER. McGraw Pub-
lishing Company, New York, N. Y.

Steel Mill Buildings. Milo S. Ketchum. Engineering News Publish-
ing Company, New York, N. Y.

Statically Indeterminate Stresses. HIROI. Engineering News Pub-
lishing Company, New York, N. Y.

Stresses in Frame Structures. A. JAY DuBois. John Wiley & Sons,
New York, N. Y.

Die Zusatzkrdfte und Nebenspannungen eiscrner Fachwerkbrucken. FR.
ENGESSER. Julius Springer, Berlin, Germany.

Bridge Drafting. WRIGHT and WING. Engineering News Publishing
Company, New York, N. Y.

It must not be presumed that the above is a complete list of the
books which have been published relating to the theory and practice
of Bridge Engineering; neither must it be presumed that the obtaining
of information relative to bridges is limited to textbooks on the
subject. One of the best sources of information is found in the cur-
rent engineering periodicals and the "Proceedings" of the various
technical societies. The great advantage of these sources is that they
give the most up-to-date information, and usually they are very pro-
fusely illustrated.

264

ROAD CONSTRUCTION IN THE PHILIPPINES

Section of 35-mile road built by American engineers, connecting the seaport of Dagupan with
the mountain village of Baguio. province of Beuguet. island of Luz6n, and affording a cool and
healthful retreat from the heat and malaria of the lowland repions. Dagupan lies 120 miles north or
Manila, with which it is connected by rail. This view reveals some of the engineering dlfflcoluea to
be overcome, masonry and concrete work of the type shown being necessary at many points.

o £
l

§ 1

2 2

8 =

w 2

H g

fc. «

O <
"Z

O x

£ 6

HIGHWAY CONSTRUCTION

PART I.

COUNTRY ROADS.

GENERAL CONSIDERATIONS.

Object of Roads. The object of a road is to provide a way
for the transportation of persons and goods from one place to another
with the least expenditure of power and expense. The facility with
which this traffic or transportation may be conducted over any given
road depends upon the resistance offered to the movement of vehicles.
This resistance is composed of: (1) The resistance offered by the
roadway, which consists of (a) "friction" between the surface of
the road and the wheel tires; (6) resistance offered to the rolling of
the wheels, occasioned by the want of uniformity in the road surface,
or lack of strength to resist the penetrating efforts of loaded wheels,
thus requiring the load to be lifted over projecting points and out of
hollows and ruts, thereby diminishing the effective load the horse
may draw to such as it can lift. This resistance is called "resistance
to rolling" or "penetration;" (c) resistance due to gravity called
"grade resistance;" (2) The resistance offered by vehicles, termed
"axle friction;" (3) Resistance of the air.

The road which offers the least resistance to traffic should com-
bine a surface on which the friction of the wheels is reduced to the
least possible amount, while offering a good foothold for horses, to
enable them to exert their utmost tractive power, and should be so
located as to give the most direct route with the least gradients.

Friction. The resistance of friction arises from the rubbing of
the wheel tires against the surface of the road. This resistance to
traction is variable, and can be determined only by experiment.
From many experiments the following deductions are drawn :

(1) The resistance to traction is directly proportional to the
pressure,

267

HIGHWAY CONSTRUCTION

(2) On solid, unyielding surfaces it is independent of the width
of the tire, but on compressible surfaces the resistance decreases as
the width of the tire increases (but there is no material advantage
gained in making a tire more than 4 inches wide).

(3) It is independent of the speed.

(4) On rough, irregular surfaces, which give rise to constant
concussion, it increases with the speed.

The following table shows the relative resistance to traction of
various surfaces:

TABLE 1.
Resistance to Traction on Different Road Surfaces.

Pounds per ton.

In terms of load.

Earth road — ordinary condition
Gravel

50 to 200
50 to 100

A to TV
• :nr to ?V

Sand

100 to 200

1 tO T1*

Macadam

30 to 100

5t to £

Plank Road

30 to 50

JL to A

Steel Wheelway

15 to 40

i to A

Traction Resistance.

These coefficients refer to the power required to keep the load
in motion. It requires from two to six or eight times as much force
to start a load as it does to keep it in motion, at two or three miles
per hour. The extra force required to start a load is due in part
to the fact that during the stop the wheel may settle into the road
surface, in part to the fact that the axle friction at starting is greater
than after motion has begun, and further in part to the fact that
energy is consumed in accelerating the load.

Resistance to Rolling. This resistance is caused (1) by the
wheel penetrating or sinking below the surface of the road, leaving
a track or rut behind it. It is equal to the product of the load mul-
tiplied by one-third of the semi-chord of the submerged arc of the
wheel; and (2) by the wheel striking or colliding with loose or pro-
jecting stones, which give a sudden check to the horses, depending
upon the height of the obstacle, the momentum destroyed being
oftentimes considerable.

The rolling resistance varies inversely as some function of the

868

HIGHWAY CONSTRUCTION

diameter of the wheel, as the larger the wheel the less force required
to lift it over the obstruction or to roll it up the inclination due to the
indentation of the surface.

The power required to draw a wheel over a stone or any ob-
stacle, such as S in Fig. 1, may be thus calculated. Let P represent
the power sought, or that which would just balance the weight on

the point of the stone, and the
slightest increase of which
would draw it over. This
power acts in the direction
C P with the leverage of B C
or D E. Gravity, represented
by W, resists in the direction
C B with the leverage B D.
The equation of equilibrium

s
Fig. 1.

will be P X C B = W X B D, whence

BD VCD'-BC2

CB~ CD-AD

I^et the radius of the wheel = C D = 26 inches, and the height
of the obstacle = A B = 4 inches. Let the weight W = 500 pounds,
of which 200 pounds may be the weight of the wheel and 300 pounds
the load on the axle. The formula then becomes

P-500

26- 4

500

13.85
"22

314.7 pounds.

The pressure at the point D is compounded of the weight and
the power, and equals

CD
CB

500 X

591 pounds,

and therefore acts with this great effect to destroy the road in its
collision with the stone, in addition there is to be considered the
effect of the blow given by the wheel in descending from it. For
minute accuracy the non-horizontal direction of the draught and
the thickness of the axle should be taken into account. The power
required is lessened by proper springs to vehicles, by enlarged wheels,
and by making the line of draught ascending.

HIGHWAY CONSTRUCTION

The mechanical advantage of the wheel in surmounting an
obstacle may be computed from the principle of the lever.

Let the wheel, Fig. 2, touch the horizontal line of traction in
the point A and meet a protuberance B D. Suppose the line of
draught C P to be parallel to A B. Join C D and draw the perpen-
diculars DE and D F. We
may suppose the power to be
applied at E and the weight at
F, and the action is then the
same as the bent lever E D F
turning round the fulcrum at
D. HenceP:W::FD :DE.
ButFD :DE ::tanFCD:l,
and tan F C D = tan 2
(DAB); therefore P - W
tan 2 (DAB). Now it is obvious that the angle DAB increases
as the radius of the circle diminishes; and therefore, the weight W
being constant, the power required to overcome an obstacle of given
height is diminished when the diameter is increased. Large wheels
are therefore the best adapted for surmounting inequalities of the
road.

There are, however, circumstances which provide limits to the
height of the wheels of vehicles. If the radius AC exceeds the
height of that part of the horse to which the traces are attached,
the line of traction C P will be inclined to the horse, and part of the
power will be exerted in pressing the wheel against the ground. The
best average size of wheels is considered to be about 6 feet in diameter.
Wheels of large diameter do less damage to a road than small
ones, and cause less draught for the horses.

With the same load, a two- wheeled cart does far more damage
than one with four wheels, and this because of their sudden and
irregular twisting motion in the trackway.

Grade Resistance is due to the action of gravity, and is the
same on good and bad roads. On level roads its effect is immaterial,
as it acts in a direction perpendicular to the plane of the horizon, and
neither accelerates nor retards motion. On inclined roads it offers
considerable resistance, proportional to the steepness of the incline.

270

HIGHWAY CONSTRUCTION

The resistance due to gravity on any incline in pounds per ton

2000

is equal to — ^ r-

rate ot grade

The following table shows the resistance due to gravity on dif-
ferent grades.

TABLE 2.
Resistance Due to Gravity on Different Inclinations.

Grade 1 in 20 30 40 50 60 70 80 90 100 200 300 400

Rise in feet per mile . . .264 176 132 105 88 75 66 58 52 26 17 13
Resistance in Ib. per ton . 1 12 74£ 56 45 38 32 28 25 22 11 J 1\ 5£
The additional resistance caused by inclines may be investigated
in the following manner: Suppose the whole weight to be borne on
one pair of wheels, and that the tractive force is applied in a direction
parallel to the surface of the road.

Let A B in Fig. 3 represent a portion of the inclined road, C
being a vehicle just sustained in its position by a force acting in the
direction C D. It is evident that the vehicle is kept in its position
by three forces; namely, by its own weight W acting in the vertical
direction C F, by the force F applied in the direction C D parallel
to the surface of the road, and by the pressure P which the vehicle
exerts against the surface of the road acting in the direction C E

perpendicular to same. To
determine the relative magni-
tude of these three forces,
draw a horizontal line A G
and the vertical one B G;
then, since the two lines C F
and B G are parallel and
are both cut by the line A B,

Fig. 3. they must make the two

angles C F E and A B G

equal; also the two angles C E F and A G B are equal; therefore, the
remaining angles F C E and B A G are equal, and the two triangles
C F E and A B G are similar. And as the three sides of the former
are proportional to the three forces by which the vehicle is sustained,
so also are the three sides of the latter; namely, AB or the length
of the road is proportional to W, or the weight of the vehicle; B G,

271

6 HIGHWAY CONSTRUCTION

or the vertical rise in the same, to F, or the force required to sustain
the vehicle on the incline; and A G, or the horizontal distance in
which the rise occurs, to P, or the force with which the vehicle presses
upon the surface of the road. Therefore,

W:AB ::F:GB,
and

W :AB : :P : A G.

If to A G such a value be assigned that the vertical rise of the
road is exactly one foot, then

and

F = - = - = W -sin A

AB i/AG'+l

p = W-AG = W-AG

AB l/AG2+l

in which A is the angle BAG.

To find the force requisite to sustain a vehicle upon an inclined
road (the effects of friction being neglected), divide the weight of the
vehicle and its load by the inclined length of the road, the vertical
rise of which is one foot, and the quotient is the force required.

To find the pressure of a vehicle against the surface of an inclined
road, multiply the weight of the loaded vehicle by the horizontal
length of the road) and divide the product by the inclined length of
the same; the quotient is the pressure required.

The force with which a vehicle presses upon an inclined road
is always less than its actual weight; the difference is so small that,
unless the inclination is very steep, it may be taken equal to the
weight of the loaded vehicle.

To find the resistance to traction in passing up or down an
incline, ascertain the resistance on a -level road having the same surface
as the incline, to which add, if the vehicle ascends, or subtract, if
it descends, the force requisite to sustain it on the incline; the sum
or difference, as the case may be, will express the resistance.

Tractive Power and Gradients. The necessity for easy
grades is dependent upon the power of the horse to overcome the
resistance to motion composed of the four forces, friction, collision,
gravity, and the resistance of the air.

All estimates on the tractive powder of horses must to a certain

272

HIGHWAY CONSTRUCTION

extent be vague, owing to the different strengths and speeds of animals
of the same kind, as well as to the extent of their training to any
particular kind of work.

The draught or pull which a good average horse, weighing 1,200
pounds, can exert on a level, smooth road at a speed of 2£ miles per
hour is 100 pounds, equivalent to 22,000 foot-pounds per minute,
or 13,200,000 foot-pounds per day of 10 hours.

The tractive power diminishes as the speed increases and, per-
haps, within certain limits, say from f to 4 miles per hour, nearly
in inverse proportion to it. Thus the average tractive force of a
horse, on a level, and actually pulling for 10 hours, may be assumed

approximately as follows:

TABLE 3.
Tractive Power of Horses at Different Velocities.

Miles per hour.

Tractive
Force. Ll>.

Miles per hour.

Tractive
Force. Lb.

333 33

Ill 11

250

100

200

90 91

166 66

83 33

142 86

71 43

125

62 50

The work done by a horse is greatest when the velocity with
which he moves is f of the greatest velocity with which he can move
when unloaded; and the force thus exerted is 0.45 of the utmost
force that he can exert at a dead pull.

The- traction power of a horse may be increased in about the
same proportion as the time is diminished, so that when working
from 5 to 10 hours, on a level, it will be about as shown in the following

table:

TABLE 4.

Hours per day Traction (pounds) Hours per day Traction (pounds)

10 100 7 146|

9 llli 6 166^

8 125 5 200

The tractive power of teams is about as follows

1 horse =1

2 horses 0.95 X 2 - 1.90

3 " 0.85 X 3 = 2.55

4 " ' ... 0.80 X 4 - 3.20

273

HIGHWAY CONSTRUCTION

Loss of Tractive Power on Inclines. In ascending in-
clines a horse's power diminishes rapidly; a large portion of his
strength is expended in overcoming the resistance of gravity due to
his own weight and that of the load. Table 5 shows that as the
steepness of the grade increases the efficiency of both the horse and
the road surface diminishes; that the more of the horse's energy is
expended in overcoming gravity the less remains to overcome the
surface resistance.

TABLE 5.

Effects of Grades Upon the Load a Horse can Draw on Different
Pavements.

Grade.

Earth.

Broken Stone.

Stone Blocks.

Asphalt.

Level

1.00

1 :100

.80

.66

.72

.41

2 :100

.66

.50

.55

.25

3 100

.55

.40

.44

.18

4 100

.47

.33

.36

.13

5 100

.41

.29

.30

.10

10 100

.26

.16

.14

.04

15 100

.10

.05

.07

20 100

.04

.03

Table 6 shows the gross load which an average horse, weighing
1,200 pounds, can draw on different kinds of road surfaces, on a
level and on grades rising five and ten feet per one hundred feet.

TABLE 6.

Description of Surface.

Level.

5 per cent
grade.

10 per cent
grade.

\sphalt

13 216

Broken stone (best condition)
(slightly muddy)
' " (ruts and mud)

6,700
4,700
3,000

1,840
1,500
1,390

1,060
1,000
890

" (very bad condition) . .
Earth (best condition)
(average condition)
(moist but not muddy).
Stone-block pavement (dry and clean)
" (muddy)
Sand (wet)

1,840
3,600
1,400
1,100
8,300
6,250
1 500

1,040
1,500
900
780
1,920
1,800
675

740
930
660
600
1,090
1,040
390

" (dry)

1 087

445

217

The decrease in the load which a horse can draw upon an incline
is not due alone to gravity; it varies with the amount of foothold

274

HIGHWAY CONSTRUCTION 9

afforded by the road surface. The tangent of the angle of inclination
should not be greater than the coefficient of tractional resistance;
therefore it is evident that the smoother the road surface, the easier
should be the grade. The smoother the surface the less the foothold,
and consequently the load.

The loss of tractive power on inclines is greater than any inves-
tigation will show; for, besides the increase of draught caused by
gravity, the power of the horse is much diminished by fatigue upon
a long ascent, and even in greater ratio than man, owing to its anatom-
ical formation and great weight. Though a horse on a level is as
strong as five men, on a grade of 15 per cent, it is less strong than
three; for three men carrying each 100 pounds will ascend such a
grade faster and with less fatigue than a horse with 300 pounds.

A horse can exert for a short time twice the average tractive
pull which he can exert continuously throughout the day's work;
hence, so long as the resistance on the incline is not more than double
the resistance on the level, the horse will be able to take up the full
load which he is capable of drawing.

Steep grades are thus seen to be objectionable, and particularly
so when a single one occurs on an otherwise comparatively level road,
in which case the load carried over the less inclined portions must
be reduced to what can be hauled up the steeper portion.

The bad effects of steep grades are especially felt in winter,
when ice covers the roads, for the slippery condition of the surface
causes danger in descending, as well as increased labor in ascending;
the water of rains also runs down the road and gulleys it out, destroy-
ing its surface, thus causing a constant expense for repairs. The
inclined portions are subject to greater wear from horses ascending,
thus requiring thicker covering than the more level portions, and
hence increasing the cost of construction.

It will rarely be possible, except in a flat or comparatively level
country, to combine* easy grades with the best and most direct route.
These two requirements will often conflict. In such a case, increase
the length. The proportion of this increase will depend upon the
friction of the covering adopted. But no general rule can be given
to meet all cases as respects the length which may thus be added,
for the comparative time occupied in making the journey forms an

275

HIGHWAY CONSTRUCTION

important element in any case which arises for settlement. Disre-
garding time, the horizontal length of a road may be increased to
avoid a 5 per cent grade, seventy times the height.

Table 7 shows, for most practical purposes, the force required
to draw loaded vehicles over inclined roads. The first column ex-
presses the rate of inclination; the second, the pressure on the plane
in pounds per ton; the third, the tendency down the plane (or force
required to overcome the effect of gravity) in pounds per ton; the
fourth, the force required to haul one ton up the incline; the fifth, the
length of level road which would be equivalent to a mile in length of
the inclined road — that is, the length which would require the same
motive power to be expended in drawing the load over it, as would
be necessary to draw over a mile of the inclined roadj the sixth, the
maximum load which an average horse weighing 1,200 pounds can
draw over such inclines, the friction of the surface being taken at
-fa of the load drawn.

TABLE 7.

Rate of grade.
Feet per 100
feet.

Pressure on
the plane in
Ib. per ton.

Tendency
down the
plane in Ib.
per ton.

Power in Ib.
required to
haul one ton
up the plane.

Equivalent
! length of level
road. Miles.

Maximum
load in Ib.
which a horse
can haul.

0.0

2240

.00

45.00 1.000

6270

0.25

2240

- 5.60

50.60

1.121

5376

0.50

2240

11.20

56.20

1.242

4973

0.75

2240

16.80

61.80

1.373

4490

2240

22.40

67.40

1.500

4145

1.25

*2240

28.00

73.00

1.622

3830

1.50

2240

33.60

78.60

1.746

3584

1.75

2240

39.20

84.20

1.871

3290

2240

45.00

90.00

2.000

3114

2.25

2240

50.40

95.4'0

2.120

2935

2.50

2240

56.00

101.00

2.244

2725

2.75

2240

61.33

106.33

2.363

2620

2239

67.20

112.20

2.484

2486

2238

89.20

134.20

2.982

2083

2237

112.00

157.00

3.444

1800

2233

134.40

179.40

3.986

1568

2232

156.80

201.80

4.844

1367

2232

179.20

224 . 20

4.982

1235

2231

201.60

246.60

5.840

1125

2229

224.00

269.00

5.977

1030

* Near enough for practice; actually 3239.888.

Pressure on the plane = weight X nat cos of angle of plane.

Axle Friction. The resistance of the hub to turning on the
axle is the same as that of a journal revolving in its bearing, and has

276

HIGHWAY CONSTRUCTION

nothing to do with the condition of the road surface. The coefficient
of journal friction varies with the material of the journal and its
bearing, and with the lubricant. It is nearly independent of the
velocity, and seems to vary about inversely as the square root of the
pressure. For light carriages when loaded, the coefficient of friction
is about 0.020 of the weight on the axle; for the ordinary thimble-
skein wagon when loaded, it is about 0.012. These coefficients are
for good lubrication; if the lubrication is deficient, the axle friction
is two to six times as much as above.

The traction power required to overcome the above axle friction
for carriages of the usual proportions is about 3 to 3£ Ib. per ton of
the weight on the axle; and for truck wagons, which have medium
sized wheels and axles, is about 3i. to 4£ Ib. per ton.

Resistance of the Air. The resistance arising from the
force of the wind will vary with the velocity of the wind, with the
velocity -of the vehicle, with the area of the surface acted upon, and
also with the angle of incidence of direction of the wind with the
plane of the surface.

The following table gives the force per square foot for various
velocities :

TABLE 8.

Velocity of wind in miles
per hour.

Force in Ibs. per sq. ft.

Description.

1 .107

Pleasant Breeze

20
25

1.968)
3.075/

Brisk Gale

30
35

4.428\
6.027J

High Wind

40
45

7.872?

9.963\

Very High Wind

50 12.300 Storm

Effect of Springs on Vehicles. Experiments have shown
that vehicles mounted on springs materially decrease the resistance to
traction, and diminish the wear of the road, especially at speeds
beyond a walking pace. Going at a trot, they were found not to
cause more wear than vehicles without springs at a, walk, all other
conditions being similar. Vehicles with springs improperly fixed
cause considerable concussion, which in turn destroys the road
covering.

277

12 HIGHWAY COXSTRUCTIOX

LOCATION OF COUNTRY ROADS,

The considerations governing the location of country roads are
dependent upon the commercial condition of the country to be
traversed. In old and long-inhabited sections the controlling ele-
ments will be the character of the traffic to be accommodated. In
such a section, the route is generally predetermined, and therefore
there is less liberty of a choice and selection than in a new and sparsely
settled district, where the object is to establish the easiest, shortest,
and most economical line of intercommunication according to the
physical character of the ground.

Whichever of these two cases may have to be dealt with, the same
principle governs the engineer, namely, to so lay out the road as to
effect the conveyance of the traffic with the least expenditure of
motive power consistent with economy of construction and main-
tenance.

Economy of motive power is promoted by easy grades, by the
avoidance of all unnecessary ascents and descents, and by a direct
line; but directness must be sacrificed to secure easy grades and to
avoid expensive construction.

Reconnoissance. The selection of the best route demands
much care and consideration on the part of the engineer. To obtam
the requisite data upon which to form his judgment, he must make
a personal reconnoissance of the district. This requires that the
proposed route be either ridden or walked over and a careful examina-
tion made of the principal physical contours and natural features of
the district. The amount of care demanded and the difficulties
attending the operations will altogether depend upon the character
of the country.

The immediate object of the reconnoissance is to select one or
more trial lines, from which the final route may be ultimately deter-
mined.

When there are no maps of the section traversed, or when those
which can be procured are indefinite or inaccurate, the work of
reconnoitering will be much increased.

In making a reconnoissance there are several points which, if
carefully attended to, will very considerably lessen the labor and
time otherwise required. Lines which would run along the imme-

278

HIGHWAY CONSTRUCTION 13

diate bank of a large stream must of necessity intersect all the tribu-
taries confluent on that bank, thereby demanding a corresponding
number of bridges. Those, again, which are situated along the
slopes of hills are more liable in rainy weather to suffer from washing
away of the earthwork and sliding of the embankments; the others
which are placed in valleys or elevated plateaux, when the line crosses
the ridges dividing the principal water courses will have steep ascents
and descents.

In making an examination of a tract of country, the first point
to attract notice is the unevenness or undulations of its surface, which
appears to be entirely without system, order, or arrangement; but
upon closer examination it will be perceived that one general prin-
ciple of configuration obtains even in the most irregular countries.
The country is intersected in various directions by main water courses
or rivers, which increase in size as they approach the point of their
discharge. Towards these main rivers lesser rivers approach on
both sides, running right and left through the country, and into these,
again, enter still smaller streams and brooks. The streams thus
divide the hills into branches or spurs having approximately the same
direction as themselves, and the ground falls in every direction from
the main chain of hills towards the water courses, forming ridges
more or less elevated.

The main ridge is cut down at the heads of the streams into
depressions called gaps or passes; the more elevated points are called
peaks. The water which has fallen upon these peaks is the origin
of the streams which have hollowed out the valleys. Furthermore,
the ground falls in every direction towards the natural water courses,
forming ridges more or less elevated running between them and
separating from each other the districts drained by the streams.

The natural water courses mark not only the lowest lines, but
the lines of the greatest longitudinal slope in the valleys through which
they flow.

The direction and position of the principal streams give also
the direction and approximate position of the high ground or ridges
which lie between them.

The positions of the tributaries to the larger stream generally
indicate the points of greatest depression in the summits of the ridges,

279

U HIGHWAY CONSTRUCTION

Fig. 4. Contour Lines.

880

HIGHWAY CONSTRUCTION 15

and therefore the points at which lateral communication across the
high ground separating contiguous valleys can be most readily made.

The instruments employed in reconnoitering, are : The compass,
for ascertaining the direction; the aneroid barometer, to fix the ap-
proximate elevation of summits, etc. ; and the hand level, to ascertain
the elevation of neighboring points. If a vehicle can be used, an
odometer may be added, but distances can usually be guessed or
ascertained by time estimates or otherwise, closely enough for pre-
liminary purposes. The best maps obtainable and traveling com-
panions who possess a local knowledge of the country, together with
the above outfit is all that will be necessary for the first inspection.

The reconnoissance being completed, instrumental surveys of
the routes deemed most advantageous should be made. When the
several lines are plotted to the same scale, a good map can be pre-
pared from which the exact location of the road can be determined.

In making the preliminary surveys the topographical features
should be noted for a convenient distance to the right and the left of
the line, and all prominent points located by compass bearings. The
following data should also be obtained: the importance, magnitude,
and direction of all streams and roads crossed; the character of the
material to be excavated or available for embankments, the position
of quarries and gravel pits, and the modes of access thereto; and all
other information that may effect a selection.

Topography. There are various methods of delineating upon
paper the irregularities of the surface of the ground. The method
of most utility to the engineer is that by means of "contour lines."
These are fine lines traced through the points of equal level over the
surface surveyed, and denote that the level of the ground throughout
the whole of their course is identical; that is to say, that every part
of the ground over which the line passes is at a certain height above
a known fixed point termed the datum, this height being indicated
by the figures written against the line.

The intervals between the lines vertically are equal and. may
be.l, 3, 5, 10 or more feet apart; where the surface is very steep they
lie close together. These lines by their greater or less distance apart
have the effect of shading, and make apparent to the eye, the
undulations and irregularities in the surface of the country.

281

ir,

HIGHWAY CONSTRUCTION

— 60.OO

- 54.4-

- 54.62

Fig. 4 shows an imaginary tract of country, the physical features
of which are shown by contour lines.

Map. The map should show the
lengths and direction of the different por-
tions of the line, the topography, rivers,
water courses, roads, railroads, and other
matters of interest, such as town and
county lines, dividing lines between property,
timbered and cultivated lands, etc.

Any convenient scale may be adopted;
400 feet to an inch will be found the most
useful.

Memoir. The descriptive memoir
should give with minuteness all information,
such as the nature of the soil, character of
the several excavations whether earth or
rock, and such particular features as can-
not be clearly shown upon the map or
profile.

Special information should be given re-
garding the rivers crossed, as to their width,
depth at highest known flood, velocity of
current, character of banks and bottom,
and the angle of skew which the course
makes with the line of the road.

Levels. Levels should be taken along
the course of each line, usually at every 100
feet, or at closer intervals, depending upon
the nature of the country.

In taking the levels, the heights of
all existing roads, railroads, rivers, or
canals should be noted. "Bench marks"
should be established at least every half
mile, that is, marks made on any fixed
object, such as a gate post, side of a house,
or, in the absence of these, a cut made
on a large tree. The height and exact

— 54.80

— 55.10-

—55.00-

Fig. 5. Preliminary Profile

282

HIGHWAY CONSTRUCTION 17

description of each bench mark should be recorded in the level book.

Cross Levels. Wherever considered necessary levels at right
angle to the center line should be taken. These will be found useful
in showing what effect a deviation to the right or left of the surveyed
line would have. Cross levels should be taken at the intersection of
all roads and railroads to show to what extent, if any, these levels
will have to be altered to suit the levels of the proposed road.

Profile. A profile is a longitudinal section of the route, made
from the levels. Its horizontal scale should be the same as that of
the map; the vertical scale should be such as will show with distinct-
ness the inequalities of .the ground. '

Fig. 5 shows the manner in which a profile is drawn and the
nature of the information to be given upon it.

Bridge Sites. The question of choosing the site of bridges is
an important one. If the selection is not restricted to a particular
point, the river should be examined for a considerable distance above
and below what would be the most convenient point for crossing; and
if a better site is found, the line of the road must be made subordinate
to it. If several practicable crossings exist, they must.be carefully
compared in order to select the one most advantageous. The follow-
ing are controlling conditions: (1) Good character of the river bed,
affording a firm foundation. If rock is present near the surface of
the river bed, the foundation will be easy of execution and stability
and economy will be insured. (2) Stability of river banks, thus
securing a permanent concentration of the waters in the same bed.
(3) The axis of the bridge should be at right angles to the direction
of the current. (4) Bends in rivers are not suitable localities and
should be avoided if possible. A straight reach above the bridge
should be 'secured if possible.

Final Selection. In making the final selection the following
principles should be observed as far as practicable.

(a) To follow that route which affords the easiest grades. The
easiest grade for a given road will depend on the kind of covering
adopted for its surface.

(6) To connect the places by the shortest and most direct route
commensurate with easy grades.

283

18 HIGHWAY CONSTRUCTION

road is encumbered with useless ascents, the wasteful expenditure of
power is considerable.

(d) To give the center line such a position, with reference to
the natural surface of the ground, that the cost of construction shall
be reduced to the smallest possible amount.

(e) To cross all obstacles (wThere- structures are necessary) as
nearly as possible at right angles. The cost of skew structures
increases nearly as the square of the secant of the obliquity.

(/) To cross ridges through the lowest pass which occurs.

(gr) To cross either under or over railroads; for grade crossings
mean danger to every user of the highway.

Examples of Cases to be Treated. In laying out the line
of a road, there are three cases which may have to be treated, and
each of these is exemplified in the contour map, Fig. 4. First, the
two places to be connected, as the towns A and B on the plan, may
be both situated in the same valley, and upon the same side of it ; that
is, they are not separated from each other by the main stream which
drains the valley. This is the simplest case. Secondly, although
both in the same valley, the two places may be on opposite sides of
the valley, as at A and C, being separated by the main river. Thirdly,
they may be situated in different valleys, separated by an intervening
ridge of ground more or less elevated, as at A and D. In laying out
an extensive line of road, it frequently happens -that all these cases
have to be dealt with.

The most perfect road is that of which the course is perfectly
straight and the surface practically level; and, all other things being
the same, the best road is that which answers nearest to this de-
scription.

Now, in the first case, that of the two towns situated on the
same side of the main valley, there are two methods which may be
pursued in forming a communication between them. A road follow-
ing the direct line between them, shown by the thick dotted line A B,
may be made, or a line may be adopted which will gradually and
equally incline from one town to another, supposing them to be at
different levels; or, if they are on the same level, the line should keep
at that level throughout its entire course, following all the sinuosities
and curves which the irregular formation of the country may render

284

HIGHWAY CONSTRUCTION 19

necessary for the fulfillment of these conditions. According to the
first method, a level or uniformly inclined road might be made from
one to the other; this line would cross all the valleys and streams
which run down to the main river, thus necessitating deep cuttings,
heavy embankments, and numerous bridges; or these expensive
works might be avoided by following the sinuosities of the valley.
When the sides of the main valley are pierced by numerous ravines
with projecting spurs and ridges intervening, instead of following the
sinuosities, it will be found better to make a nearly straight line
cutting through the projecting points in such a way that the material
excavated should be just sufficient to fill the hollows.

Of all these, the best is the straight or uniformly inclined, or
level road, although at the same time it is the most expensive. If
the importance of the traffic passing between the places is not suffi-
cient to warrant so great an outlay, it will become a matter of consider-
ation whether the course of the road should be kept straight, its surface
being made to undulate with the natural face of the country; or
whether,' a level or equally inclined line being adopted, the course
of the road should be made to deviate from the direct line, and follow
the winding course which such a condition is supposed to necessitate.

In the second case, that of two places situated on opposite sides
of the same valley, there is, in like manner, the choice of a perfectly
straight Hne to connect them, which would probably require a big
embankment if the road was kept level, or steep inclines if it followed
the surface of the country; or by winding the road, it may be carried
across the valley at a higher point, where, if the level road be taken,
the embankment would not be so high, or, if kept on the surface,
the inclination would be reduced.

In the third case, there is, in like manner, the alternative of
carrying the road across the intervening ridge in a perfectly straight
line, or of deviating it to the right and left, and crossing the ridge
at a point where the elevation is less.

The proper determination of the question which of these courses
is the best under certain circumstances involves a consideration of
the comparative advantages and disadvantages of inclines and
curves. What additional increase in the length of a road would be
equivalent to a given inclined plane upon it; or conversely, what

20 HIGHWAY CONSTRUCTION

inclination might be given to a road as an equivalent to a given de-
crease in its length ? To satisfy this question, the comparative force
required to draw different vehicles with given loads must be known,
both upon level and variously inclined roads.

The route which will give the most general satisfaction consists
in following the valleys as much as possible and rising afterward by
gentle grades. This course traverses the cultivated lands, regions
studded with farmhouses and factories. The value of such a line
is much more considerable than that of a route by the ridges. The
water courses which flow down to the main valley are, it is true,
crossed where they are the largest and require works of large dimen-
sions, but also they are fewer in number.

Intermediate Towns. Suppose that it is desired to form a
road between two distant towns, A and B, Fig. 6, and let us for the
present neglect altogether the consideration of the physical features
of the intervening country, assuming that it is equally favorable
whichever line we select. Now at first sight, it would appear that
under such circumstances a perfectly straight line drawn from one

town to the other would be

"X the best that could be chos-

NN en. On more careful exam-

NN ination however, of the lo-

SNN cality, we may find that

xsv there is a third town, C,

situated somewhat on one
side of the straight line

which we have drawn from A to B ; and although our primary object
is to connect only the two latter, that it would nevertheless be of
considerable service if the whole of the three towns were put into
mutual connection with each other.

This may be effected in three different ways, any one of which
might, under the circumstances, be the best. In the first place, we
might, as originally suggested, form a straight road from A to B,
and in a similar manner two other straight roads from A to C, and
from B to C, and this would be the most perfect way of effecting the
object in view the distance between any of the two towns being
reduced to the least possible. It would, however, be attended with

286

HIGHWAY CONSTRUCTION 21

considerable expense, and it would he requisite to construct a much
greater length of road than according to the second plan, which would
be to form, as before, a straight road from A to B, and from C to con-
struct a road which should join the former at a point D, so as to be per-
pendicular to it. The traffic between A or B and C would proceed to
the point D and then turn off to C. With this arrangement, white
the length of the roads would be very materially decreased, only a
slight increase would be occasioned in the distance between C and
the other two towns. The third method would be to form only the
two roads A C and C B, in which case the distance between A and B
would be somewhat increased, while that between A C or B and C
would be diminished, and the total length of road to be constructed
would also be lessened.

As a general rule it may be taken that the last of these methods
is the best and most convenient for the public; that is to say, that
if the physical character of the country does not determine the course
of the road, it will generally be found best not to adopt a perfectly
straight line, but to vary the line so as to pass through all the prin-
cipal towns near its general course.

flountain Roads. The location of roads in mountainous
countries presents greater difficulties than in an ordinary undulating
country; the same latitude in adopting undulating grades and choice
of position is not permissible, for the maximum must be kept before
the eye perpetually. A mountain road has to be constructed on the
maximum grade or at grades closely approximating it, and but one
fixed point can be obtained before commencing the survey, and that
is the lowest pass in the mountain range; from this point the survey
must be commenced. The reason for this is that the lower slopes
of the mountain are flatter than those at their summit; they cover a
larger area, and merge into the valley in diverse undulations. So
that a road at a foot of a mountain may be carried at will in the
desired direction by more than one route, while at the top of a moun-
tain range any deviation from the lowest pass involves increased
length of line. The engineer having less command of the ground,
owing to the reduced area he has to deal with and the greater abrupt-
ness of the slopes, is liable to be frustrated in his attempt to get his
line carried in the desired direction. '

287

22 HIGHWAY CONSTRUCTION

It is a common practice to run a mountain survey up hill, but
this should be avoided. Whenever an acute-angled zigzag is met
with on a mountain road near the summit, the inference to be drawn
is that the line being carried up hill on reaching the summit was
too low and the zigzag was necessary to reach the desired pass. The
only remedy in such a case is by a resurvey beginning at the summit
and running down hill. This method requires a reversal of that
usually adopted. The grade line is first staked out and its horizontal
location surveyed afterwards. The most appropriate instrument for
this work is a transit with a vertical circle on which the telescope may
be set to the angle of the maximum grade.

Loss of Height. Loss of height 'is to be carefully avoided in a
mountain road. By loss of height is meant an intermediate rise in a
descending grade. If a descending grade is interrupted by the intro-
duction of an unnecessary ascent, the length of the road will be in-
creased over that due to the continuous grade by the length of the
portion of the road intervening between the summit of the rise and
the point in the road on a level with that rise — a length which is double
that due on the gradient to the height of the rise. For example,
if a road descending a mountain rises at some intermediate point to
cross over a ridge or spur, and the height ascended amounts to 110
feet before the descent is continued, such a road would be just one
mile longer than if the descent had been uninterrupted; for 110 feet
is the rise due to a half-mile length at 1 : 24.

Water on Mountain Roads. Water is needed by the work-
men and during the construction of the road ; it is also very necessary
for the traffic, especially during hot weather; and if the road exceeds
5 miles in length, provision should be made to have it either close
to or within easy reach of the road. With a little ingenuity the
water from springs above the road, if such exist, can be led down to
drinking fountains for men, and to troughs for animals.

In a tropical country it would be a matter for serious consider-
ation if the best line for a mountain road 10 miles in length or up-
wards, but without water, should not be abandoned in favor of a
worse line with a water supply available.

Halting Places. On long lines of mountain roads halting
places should be provided at frequent intervals.

HIGHWAY CONSTRUCTION 23

Alignment. No rule can be laid down for the alignment of a
road; it will depend both upon the character of the traffic on it and
upon the "lay of the land." To promote economy of transportation
it should be straight; but if s-traightness is obtained at the expense
of easy grades that might have been obtained by deflections and
increase in length, it will prove very expensive to the community
that uses it.

Where curves are necessary, employ the greatest radius possible
and never less than fifty feet. They may be circular or parabolic.
The parabolic will be found exceedingly useful for joining tangents
of unequal length, and for following contour lines; its curvature
being least at its beginning and* ending, makes the deviations from
a straight line less strongly marked than by a circular arc.

When a curve occurs on an ascent, the grade at that place must
be diminished in order to compensate for the additional resistance of
the curve.

The width of the wheel way on curves must be increased. This
increase should be one-quarter of the width for central angles between
90 and 120 degrees, and one-half for angles between 60 and 90 degrees.
Excessive crookedness of alignment is to be avoided, for any unneces-
sary length causes a constant threefold waste; first, of the interest
of the capital expended in making that unnecessary portion; secondly,
of the ever recurring expense of repairing it; and thirdly, of the time
and labor employed on travelling over it.

.The curving road around a hill may be often no longer than the
straight one over it, for the latter is straight only with reference to
the horizontal plane, while it is curved as to the vertical plane; -the
former is curved as to the horizontal plane, but straight as to the
vertical plane. Both lines curve, and we call the one passing over
the hill straight only because its vertical curvature is less apparent
to our eyes.

The differen.ce in length between a straight road and one which
is slightly curved is very small. If a road between two places ten
miles apart were made to curve so that the eye could nowhere see
farther than one-quarter of a mile of it at once, its length would
exceed that of a straight road between the same points by only about
four hundred and fifty feet.

289

24 HIGHWAY CONSTRUCTION

Zigzags. The method of surmounting a height by a series of
zigzags or by a series of reaches with practicable curves at the turns,
is objectionable.

(1) An acute-angled zigzag obliges the traffic to reverse its
direction without affording it convenient room for the purpose. The
consequence is that with slow traffic a single train of vehicles is
brought to a stand, while if two trains of vehicles travelling in opposite
directions meet at the zigzag a block ensues.

(2) With zigzags little progress is made towards the ultimate
destination of the road ; height is surmounted, but horizontal distance
is increased for which there is no necessity or compensation.

(3) Zigzags are dangerous. In case of a runaway down hill
the zigzag must prove fatal.

(4) If the drainage cannot be carried clear of the road at the
end of each reach, it must be carried under the road in one reach only
to appear again at the next, when a second bridge, culvert, or drain
will be required, and so on at the other reaches. If the drainage can
be carried clear at the termination of each reach, the lengths between
the curves will be very short, entailing numerous zigzag curves, which
are expensive to construct and maintain.

Final Location. The route being finally determined upon, it
requires to be located. This consists in tracing the line, placing a
stake at every 100 feet on the straight portions and at every 50 or

25 feet on the curves. At the tangent point of curves, and at points
of compound and reverse curves, a larger and more permanent stake
should be placed. Lest those stakes should be disturbed in the
process of construction, their exact distance from several points
outside of the ground to be occupied by the road should be carefully
measured and recorded in the notebook, so that they may be replaced.
The stakes above referred to show the position of the center line of
the road, and form the base line from which all operations of con-
struction are carried on. Levels are taken at each stake," and cross
levels are taken at every change of longitudinal slope.

Construction Profile. The construction or working profile
is made from the levels obtained on location. It should be drawn to a
horizontal scale of 400 feet to the inch and a vertical scale of 20 feet
to the inch. Fig. 7 represents a portion of such a profile. The

290

HIGHWAY CONSTRUCTION

figures in column A represent the elevation of the ground at every
100 feet, or where a stake has been driven, above datum. The
figures in column B are the elevations of the grade above datum.
The figures in column C indicate the depth of cutting or height of
fill; they are obtained by taking the difference between the level of
the road and the level of the surface of the ground. The straight line

-CLEARING-ACRES

-EARTH £ GRAVEL CLHTING-CYDS

£ 2.50.' 100

P •. o o ' in x 7 7 I ^00:|00 * * i \/
OoQSoJjo'o'oo °l

M ? ? i : i i ? i ? ? 1 5 1 i

SSSSSSeiMJlRSiiiifl

A g i ) a

7 [DATUM MEAN TIDE | ? ^

SSSlr''v•'>in]'nS^<I>^?i
<oa>coooi--ogNfflKioin!o

STATION NUMBERS

Fig.

at the top represents the grade of the road; the upper surface of the
road when finished would be somewhat higher than this, while the
given line represents what is termed the sub-grade or formation level.
All the dimensions refer to the formation level, to which the surface
of the ground is to be formed to receive the road covering.

At all changes in the rate of inclination of the grade line a heavier
vertical line should be drawn.

Gradient. The grade of a line is its longitudinal slope, and
is designated by the proportion between its length and the difference
of height of its two extremes. The ratio of these two qualities gives
it its name; if the road ascends or falls one foot in every twenty feet
of its length, it is said to have a grade of 1 : 20 or a 5 per cent grade.
Grades are of two kinds, maximum and minimum. The maximum
is the steepest which is to be permitted and which on no account is to be
exceeded. The minimum is the least allowable for good drainage.
(For method of designating grades see Table 9).

Determination of Gradients. The maximum grade is fixed
by two considerations, one relating to the power expended in ascend-
ing, the other to the acceleration in descending the incline.

There is a certain inclination, depending upon the degree of
perfection given to the surface of the road, which cannot be exceeded

HIGHWAY CONSTRUCTION

without a direct loss of tractive power. This inclination is that in
descending which, at a uniform speed, the traces slacken, or which
causes the vehicles to press on the horses; the limiting inclination
within which this effect does not take place is the angle of repose.

TABLE 9.

American method.
Feet per 100 feet.

English method.

Feet per mile.

Angle with the horizon.

\ 1 400

13.2

0°

36"

200

26.4

150

39.6

'_'_'

100

52.8

ll»

66§

79.2

57*

92.4

.50

105.6

2i ]

441

118.8

132

36£ 145.2

:;i

33| 158.4

30f 171.6

.-,]

28$ 184.8

26| 198

25 211.2

1 7

23* 221.4

221

237.6

:il

250.8

264

13?

316.8

14?

369.6

12*

422.4

475.2

1 10

528

•12

The angle of repose for any given road surface can be easily
ascertained from the tractive force required upon a level with the
same character of surface. Thus if the force necessary on a level
to overcome the resistance of the load is ^ °f l^ weight, then the
same fraction expresses the angle of repose for that surface.

On all inclines less steep than the angle of repose a certain
amount of tractive force is necessary in the descent as well as in
the ascent, and the mean of the two drawing forces, ascending and
descending, is equal to the force along the level of the road. Thus
on such inclines, as much mechanical force is gained in the descent
as is lost in the ascent. From this it might.be inferred that when a
vehicle passes alternately each way along the road, no real loss is

292

HIGHWAY CONSTRUCTION 27

occasioned by the inclination of the road; such is not, however,
practically the fact with animal power, for while it is necessary in
the ascending journey to have either a less or a greater number of
horses than would be requisite if the road were entirely level, no
corresponding reduction can be made in the descending journey.
On inclines which are more steep than the angle of repose, the load
presses on the horses during their descent, so as to impede their
action, and their power is expended in checking the descent of the
load; or if this effect be prevented by the use of any form of drag or
brake, then the power expended on such a drag or brake corresponds
to an equal quantity of mechanical power expended in the ascent,
for which no equivalent is obtained in the descent.

The maximum grade for a given road will depend (1) upon the
class of traffic that will use it, whether fast and light, slow and heavy,
or mixed, consisting of both light and heavy; (2) upon the character
of the pavement adopted; and (3) upon the question of cost of con-'
struction. Economy of motive power and low cost of construction are
antagonistic to each other, and the engineer will have to weigh the
two in the balance.

For fast and light traffic the grades should not exceed 2 per
cent; for mixed traffic 3 per cent may be adopted; while for slow
traffic combined with economy 5 per cent should not be exceeded.
This grade is practicable but not convenient.

Minimum Grade. From the previous considerations it would
appear that an absolutely level road was the one to be sought for, but
this is not so; there is a minimum or least allowable grade which the
road must not fall short of, as well as a maximum one which it must
not exceed. If the road was perfectly level in its longitudinal direc-
tion, its surface could not be kept free from water without giving it
so great a rise in its middle as would expose vehicles to the danger of
overturning. The minimum grade commonly used is 1 per cent.

Undulating Grades. From the fact that the power required
to move a load at a given velocity on a level road is decreased on a
descending grade to the same extent it is increased in ascending the
same grade, it must not be inferred that the animal force expended
in passing alternately each way over a rising and falling road will
gain as much in descending the several inclines as it will lose in ascend-

293

28 HIGHWAY CONSTKUCTION

ing them. Such is not the case. The»animal force must be sufficient,
either in power or number, to draw the load over the level portions
and up the steepest inclines of the road, and in practice no reduction
in the number of horses can be made to correspond with the decreased
power required in descending the inclines.

The popular theory that a gentle undulating road is less fatiguing
to horses than one which is perfectly level is erroneous. The asser-
tion that the alternations of ascent, descent, arid levels call into play
different muscles, allowing some to rest while others are exerted,
and thus relieving each in turn, is demonstrably false, and con-
tradicted by the anatomical structure of the horse. Since this doc-
trine is a mere popular error, it should be utterly rejected, not only
because false in itself, but still more because it encourages the building
of undulating roads, and this increases the labor and cost of trans-
portation upon them.

Level Stretches. On long ascents' it is generally recom-
mended to introduce level or nearly level stretches at frequent inter-
vals in order to rest the animals. These are objectionable when
they cause loss of height, and animals will be more rested by halting
and unharnessing for half an hour than by travelling over a level
portion. The only case which justifies the introduction of levels
into an ascending road is where such levels will advance the road
towards its objective point; where this is the case there will be no
loss of either length or height, and it will simply be exchanging a
level road below for a level road above.

Establishing the Grade. . When the profile of a proposed
route has been made, a grade line is drawn upon it (usually in red) in
such a manner as to follow its general slope, but to average its irregular
elevation and depressions.

If the ratio between the whole distance and the height of the line
is less than the maximum grade intended to be used, this line will be
satisfactory; but if it be found steeper, the cuttings or the length
of the line will have to be increased; the latter is generally preferable.

The apex or meeting point of all curves should be rounded off
by a vertical curve, as shown in Fig. 8, thus slightly changing the
grade at and near the point of intersection. A vertical curve rarely
need extend more than 200 feet each way from that point.

294

HIGHWAY CONSTRUCTION

Let A B, B C, be two grades in profile, intersecting at station B,
and let A and C be the adjacent stations. It is required to join the

Fig. 8.

grades by a vertical curve extending from A to C. Imagine a chord
drawn from A to C. The elevation of the middle point of the chord
will be a mean of the elevations of the grade at A and C, and one-
half of the difference" between this and the elevation of the grade at
B will be the middle ordinate of the curve. Hence we have

M = -

in which M equals the correction in grade for the point B. The
correction for any other point is proportional to the square of its
distance from A or C. Thus the correction A+ 25. isTVM; at
A + 50 it is \ M; at A + 75 it is T96 M; and the same for corre-
sponding points on the other side of B. The corrections in this case
shown are subtractive, since M is negative. They are additive
when M is positive, and the curve concave upward.

WIDTH AND TRANSVERSE CONTOUR.

A road should be wide enough to accommodate the traffic for
which it is intended, and should comprise a wheelway for vehicles
and a space on each side for pedestrians.

The wheelway of country highways need be no wider than is
absolutely necessary to accommodate the traffic using it; in many
places a track wide enough for a single team is all that is necessary.
But the breadth of the land appropriated for highway purposes
should be sufficient to provide for all future increase of traffic. The
wheelways of roads in rural sections should be double; that is, one
portion paved (preferably the center), and the other left with the

205

30 HIGHWAY CONSTRUCTION

natural soil. The latter if kept in repair will for at least one-half
the year be preferred by teamsters.

The minimum width of the paved portion, if intended to carry
two lines of travel, is fixed by the width required to allow two vehicles
to pass each other safely. This width is 16 feet. If intended for
a single line of travel, 8 feet is sufficient, but suitable turnouts must be
provided at frequent intervals. The most economical width for any
roadway is some multiple of eight.

Wide roads are the best; they expose a larger surface to the
drying action of the sun and wind, and require less supervision than
narrow ones. Their first cost is greater than narrow ones, and that
nearly in the ratio of the increased width.

The cost of maintaining a mile of road depends more upon the
extent of the traffic. than upon the extent of its surface, and unless
extremes be taken, the same quantity of material will be necessary
for the repair of the road whether wide or narrow, which is subjected
to the same amount of traffic. The cost of spreading the materials
over the wide road will be somewhat greater, but the cost of the
materials will be the same. On narrow roads the traffic, being
confined to one track, will wear more severely than if spread over a
wider surface.

The width of land appropriated for road purposes varies in the
United States from 49^ feet to 66 feet; in England and France from
26 to 66 feet. And the width or space macadamized is also subject
to variation; in the United States the average width is 16 feet; in
France it varies between 16 and 22 feet; in Belgium 8J feet seems
to be the regular width, while in Austria from 14^ to 26 J feet.

Transverse Contour. The center of all roadways should
be higher than the sides. The object of this is to facilitate the flow
of the rain water to the gutters. Where a good surface is. maintained
a very moderate amount of rise is sufficient for this purpose. Earth
roads require the most and asphalt the least. The rise should bear
a certain proportion to the width of the carriageway. The most
suitable proportions for the different paving materials is shown in
table 10.

Form of Transverse Contour. All authorities agree that
the form should be convex, but they differ in the amount and form

HIGHWAY CONSTRUCTION 31

of the convexity. Circular arcs, two straight lines joined by a circular
arc, and ellipses, all have their advocates.

TABLE 10.

Kind of Surface. Proportions of the

Carriageway. Width.

Earth Rise at center fo

Gravel -fa

Broken Stone fa

For country roads a curve of suitable convexity may be obtained
as follows: Give £ of the total rise at \ the width from the center
to the side, and f of the total rise at ^ the width (Fig. 9).

Excessive height and convexity of cross-section contract the
width of the wheelway, by concentrating the traffic at the center,
that being the only part where a vehicle can run upright. The force'
required to haul vehicles over such cross-sections is increased, be-

cause an undue proportion of the load is thrown upon two wheels
instead of being distributed equally over the four. The continual
tread of horses' feet in one track soon forms a depression which holds
water, and the surface is not so dry as with a flat section, which allows
the traffic to distribute itself over the whole width.

Sides formed of straight lines are also objectionable. They
wear hollow, retain water, and defeat the object sought by raising
the center.

The required convexity should be obtained by rounding the
formation surface, and not by diminishing the thickness of the
covering at the sides.

Although on hillside and mountain roads it is generally recom-
mended that the surface should consist of a single slope inclining
inwards, there is no reason for "or advantage gained by this method.
The form best adapted to these roads is the same as for a road under
ordinary conditions.

With a roadway raised in the center and the rain water draining
off to gutters on each side, the drainage will be more effectual and

297

32 HIGHWAY CONSTRUCTION

speedy than if the drainage of the outer half of the road has to pass
over the inner half. The inner half of such a road is usually sub-
jected to more traffic than the outer half. If formed of a straight
incline, this side will be worn hollow and retain water. The inclined
flat section never can be properly repaired to withstand the traffic.
Consequently it never can be kept in good order, no matter how
constantly it may be mended. It is always below par and when
heavy rain falls it is seriously damaged.
DRAINAGE.

In the construction of roads, drainage is of the first importance.
The ability of earth to sustain a load depends in a large measure upon
the amount of moisture retained by it. Most earths form a good
firm foundation so long as they are kept dry, but when wet they lose
their sustaining power, becoming soft and incoherent.

The drainage of roadways is of two kinds, viz., surface and sub-
surface. The first provides for the speedy removal of all water
falling on the surface of the road ; the second provides for the removal
of the underground water found in the body of the road, a thorough
removal of which is of the utmost importance and essential to the
life of the road. A road covering placed on a wet undrained bottom
will be destroyed by both water and frost, and will always be trouble-
some and expensive to maintain; perfect subsoil drainage is a neces-
sity and will be found economical in the end even if in securing it
considerable expense is required.

The methods employed for securing the subsoil drainage must
be varied according to the character of the natural soil, each kind of
soil requiring different treatment.

The natural soil may be divided into the following classes:
silicious, argillaceous, and calcareous; rock, swamps, and morasses.

The silicious and calcareous soils, the sandy loams and rock,
present no great difficulty in securing a dry and solid foundation.
Ordinarily they are not retentive of water and therefore require no
underdrains; ditches on each side of the road will generally be found
sufficient.

The argillaceous soils and softer marls require more care; they
retain water and are difficult to compact, except at the surface;
and they are very unstable under the action of water and frost.

298

HIGHWAY CONSTRUCTION

The drainage of these soils may be effected by transverse drains
and deep side ditches of ample width. The transverse drains are
placed across the road, not at right angles but in the form of an
inverted V with the point directed up hill; the depth at the angle
point should not be less than 18 inches below the subgrade surface,
and each branch should descend from the apex to the side ditches
with a fall of not less than 1 inch in 5 feet. The distance apart of
these drains will depend upon the wetness of the soil; in the case of
very wret soil they should be at intervals of 15 feet, which may be
increased to 25 feet as the ground becomes drier and firmer.

The transverse drains are best formed of unglazed circular tile
of a diameter not less than 3 inches, jointed with loose collars. The
tiles are made from terra cotta or burnt clay, are porous, and are
superior to all other kinds of drains. They carry off the water with
greater ease, rarely if ever get choked up, and only require a slight
inclination to keep the water moving through them.

Fig. 10. Tile Drain.

Fig. 11. Silt Basin.

The tiles are made in a variety of forms, as horseshoe, sole,
double sole, and round, the name being derived from the shape of
the cross-sections. Round tile is superior to all other forms. The
inside diameter of these tiles varies from 1£ to 6 inches, but they are
manufactured as large as 24 inches. Pieces of the larger pipe serve
as collars for the smaller ones. They are made in lengths of 12,
14 and 24 inches, and in thickness of shell from j'of an inch to 1 inch.

The collar which encircles the joint of the small tile allows a
large opening, and at the same time prevents sand and silt from

HIGHWAY CONSTRUCTION

entering the drain. Perishable material should not be used for
jointing. When laid in the ditch they should be held in place by
small stones. Connections should be made by proper Y-branches.
The outlets may be formed by building a dwarf wall of brick or
stone, whichever is the cheapest or most convenient in the locality.
The outlet should be covered with an iron grating to prevent vermin
entering the drain pipes, building nests and thus choking up the
waterway. (See Fig. 12.)

Fig. 12. Outlet.

Silt-basins should be constructed at all junctions and wherever
else they may be considered necessary; they may be made from a
single 6-inch pipe (Fig. 11) or constructed of brick masonry.

The trenches for the tile should be excavated at least 3 feet
wide on top and 12 inches on the bottom. After the tiles are laid
the trenches must be filled to subgrade level with round field or
cobble stones; stones with angular edges are unsuitable for this
purpose. Fine gravel, sand, or soil should not be placed over the
drains. Bricks and flat stones may be substituted for the tiles,
and the trenches filled as above stated.

As tile drains are more liable to injury from frost than those
of either brick or stone, their ends at the side ditches should not
in very cold climates be exposed directly to the weather, but may
terminate in blind drains, or a few lengths of vitrified clay pipe
reaching under the road a distance of about 3 to 4 feet from the
inner slope of the ditch.

Another method of draining the roadbed offering security from
frost is by one or more rows of longitudinal drains. These drains
are placed at equal distances from the side ditches and from each
other, and discharge into cross drains placed from 2oO to 300 feet

HIGHWAY CONSTRUCTION

apart, more or less, depending on the contour of the ground. The
cross drains into which they discharge should be of ample dimensions.
On these longitudinal lines of tiles the introduction of catch basins
at intervals of 50 feet will facilitate the removal of the water. These
catch basins may be excavated three or more feet square and as deep
as the tiles are laid. After the tiles are laid the pit is filled with gravel
and small stones.

Fig. 13.

Fall of Drains. It is a mistake to give too much fall to small
drains, the only effect of which is to produce such a current through
them as will wash away or undermine the ground around them, and
ultimately cause their own destruction. When a drain is once closed
by any obstruction no amount of fall which could be given it will

Fig. 14.

again clear the passage. A drain with a considerable current through
it is much more likely to be stopped from foreign matter carried into
it, which a less rapid stream could not have transported.

A fall of 1 inch in 5 feet will generally be sufficient, and 1 inch
in 30 inches should never be exceeded.

Fig. 15.

Side Ditches are provided to carry away the subsoil water
from the base of the road, and the rain water which falls upon its
surface; to do this speedily they must have capacity and inclination

301

HIGHWAY CONSTRUCTION

proportionate to the amount of water reaching them. The width
of the bed should not be less than 18 inches; the depth will vary with
circumstances, but should be such that the water surface shall not
reach the subgrade, but remain at least 12 inches below the crown
of the road. The sides should slope at least 1£ to 1.

The longitudinal inclination of the ditch follows the configura-
tion of the general topography, that is, the lines of natural drainage.
When the latter has to be aided artificially, grades from 1 in 500 to
1 in 800 will usually answer.

Fig. 10.

In absorbent soil less fall is sufficient, and in certain cases level
ditches are permissible. The slopes of the ditches must be protected
where the grade is considerable. This can be accomplished by sod
revetments, riprapping, or paving.

These ditches may be placed either on the road or land side of the
fence. In localities where open ditches are undesirable they may be
constructed as shown in Figs. 13 to 17, and may be formed of stone

Fig. 17.

or tile pipe, according to the availability of either material. If for
any reason two can not be built, build one.

Springs found in the roadbed should be tapped and led into the
side ditches.

Drainage of the Surface. The drainage of the roadway
surface depends upon the preservation of the cross-section, with
regular and uninterrupted fall to the sides, without hollows or ruts
in which the water can lie, and also upon the longitudinal fall of the

302

HIGHWAY CONSTRUCTION 37

road. If this is not sufficient the road becomes flooded during heavy
rainstorms and melting snow, and is considerably damaged.

The removal of surface water from country roads may be effected
by the side ditches, into which, when there are no sidewalks, the
water flows directly. When there are sidewalks, gutters are formed
between the roadway and footpath, as shown in Figs. 13 to 17, and
the water is conducted from these gutters into the side ditches by
tile pipes laid under the walks at intervals of about 50 feet. The
entrance to these pipes should be protected against washing by a
rough stone paving. In the case of 'covered ditches under the footpath
the water must be led into them by first passing through a catch
basin. These are small masonry vaults covered with iron gratings to
prevent the ingress of stones, leaves, etc. Connection from the
catch basin is made by a tile pipe about 6 inches in diameter. The
mouth of this pipe is placed a few feet above the bottom of the catch
basin, and the space below it acts as a depository for the silt carried
by the water, and is cleaned out periodically. The catch basins may
be placed from 200 to 300 feet apart. They should be made of
dimensions sufficient to convey the amount of water which is liable
to flow into them during heavy and continuous rains.

If on inclines the velocity of the water is greater than the nature
of the soil will withstand, the gutters will be roughly paved. In all
cases, the slope adjoining the footpath should be covered with sod.

A velocity of 30 feet a minute will not disturb clay with sand and
stone. 40 feet per minute will move coarse sand. 60 feet a minute
will move gravel. 120 feet a minute should move round pebbles 1 inch
in diameter, and 180 feet a minute will move angular stones If inches
in diameter.

The scour in the gutters on inclines may be prevented by small
weirs of stones or fascines constructed by the roadmen at a nominal
cost. At junctions and crossroads the gutters and side ditches re-
quire careful arrangement so that the water from one road may not
l>e thrown upon another; cross drains and culverts will be retjuired
at such places.

Water Breaks to turn the surface drainage into the side ditches
should not be constructed on improved roads. They increase the
grade and are an impediment to convenient and easy travel. Where

303

HIGHWAY CONSTRUCTION

it is necessary that water should cross the road a culvert should be
built.

On the side hill or mountain roads catch-water ditches should
be cut on the mountain side above the road, to cut off and convey the
drainage of the ground above them to the neighboring ravines. The
size of these ditches will be determined by the amount of rainfall,
extent of drainage from the mountain wrhich they intercept, and by the
distances of the ravine water courses on each side.

The inner road gutter should be of ample dimensions to carry
off the water reaching it; when in soil, it should be roughly paved with
stone. When paving is not absolutely necessary, but it is desirable
to arrest the scouring action of running water during heavy rains,
stone weirs may be erected across the gutter at convenient intervals.
The outer gutter need not be more than 12 inches wide and 9 inches
deep. The gutter is formed by a depression in the surface of the
road close to the parapet or revetted earthen protection mound. The
drainage which falls into this gutter is led off through the parapet,
or other roadside protection at frequent intervals. The guard stones
on the outside of the road are placed in and across this gutter, just
below the drainage holes, so as to turn the current of the drainage
into these holes or channels. On straight reaches, with parapet
protection, drainage holes with guard stones should be placed every
20 feet apart. Where earthen mounds are used and it may not be
convenient to have the drainage holes or channels every 20 feet, the
guardstones are to be placed in advance of the gutter to allow the
drainage to pass behind them. This drainage is either to be run off
at the cross drainage of the road, or to be turned off as before by a
guard stone set across the gutter.

At re-entering turns, where the outer side of the road requires
particular protection, guard stones should be placed every 4 feet.
As all re-entering turns should be protected by parapets, the drainage
holes through them may be placed as close together as desired.

Culverts are necessary for carrying under a road the streams
it crosses, and also for conveying the surface water collected in the
side ditches from the upper side to that side on which the natural
water courses lie.

Especial care is required to provide an ample way for the water

304

HIGHWAY CONSTRUCTION 39

to be passed. If the culvert is too small, it is liable to cause a washout,
entailing interruption of traffic and cost of repairs, and possibly may
cause accidents that will require payment of large sums for damages.
On the other hand, if the culvert is made unnecessarily large, the
cost of construction is needlessly increased.

The area of waterway required depends (1) upon the rate of
rainfall; (2) the kind and condition of the soil; (3) the character
and inclination of the surface; (4) the condition and inclination of
the bed of the stream; (5) the shape of the area to be drained, and
the position of the branches of the stream ; (6) the form of the mouth
and the inclination of the bed of the culvert; and (7) whether it is
permissible to back the water up above the culvert, thereby causing
it to discharge under a head.

(1) It is the maximum rate of rainfall during the severest storms
which is required in this connection. This varies greatly in different
sections of the country.

The maximum rainfall as shown by statistics is about one inch
per hour (except during heavy storms), equal to 3,630 cubic feet per
acre. Owing to various causes, not more than 50 to 75 per cent of
this amount will reach the culvert within the same hour.

Inches of rainfall X 3,630 = cubic feet per acre.

Inches of rainfall X 2,323,200 = cubic feet per square mile.

(2) The amount of water to be drained off will depend upon the
permeability of the surface of the ground, which will vary greatly
with the kind of soil, the degree of saturation, the condition of the
cultivation, the amount of vegetation, etc.

(3) The rapidity with which the water will reach the water
course depends upon whether the surface is rough or smooth, steep
or flat, barren or covered with vegetation, etc.

(4) The rapidity with which the water will reach the culvert
depends upon whether there is a well-defined and unobstructed
channel, or whether the water finds its way in a broad thin sheet.
It the water course is unobstructed and has a considerable inclination,
the water may arrive at the culvert nearly as rapidly as it falls; but
if the channel is obstructed, the water may be much longer in passing
the culvert than in falling.

(5) The area of waterway depends upon the amount of the area

305

40 HIGHWAY CONSTRUCTION

to be drained ; but in many cases the shape of this area and the posi-
tion of the branches of the stream are of more importance than the
amount of the territory. For example, if the area is long and narrow,
the water from the lower portion may pass through the culvert before
that from the upper end arrives; or, on the other hand, if the upper
end of the area is steeper than the lower, the water from the former
may arrive simultaneously with that from the latter. Again, if the
lower part of the area is better supplied with branches than the upper
portion, the water from the former will be carried past the culvert
before the arrival of that from the latter; or, on the other hand, if
the upper part is better supplied with branch water courses than
the lower, the water from the whole area may arrive at the culvert
at nearly the same time. In large areas the shape of the area and
the position of the water courses are very important considerations.

(6) The efficiency of a culvert may be very materially increased
by so arranging the upper end that the water may enter into it without
being retarded. The discharging capacity of a culvert can be greatly
increased by increasing the inclination of its bed, provided the channel
below will allow the water to flow away freely after having passed
the culvert.

(7) The discharging capacity of a culvert can be greatly increased
by allowing the water to dam up above it. A culvert will discharge
twice as much under a head of four feet as under a head of one foot.
This can be done safely only with a well constructed culvert.

The determination of the values of the different factors entering
into the problem is almost wholly a matter of judgment. An estimate
for any one of the above factors is liable to be in error from 100 to
200 per cent, or even more, and of course any result deduced from
such data must be very uncertain. Fortunately, mathematical exact-
ness is not required by the problem nor warranted by the data. The
question is not one of 10 or 20 per cent of increase; for if a 2-foot pipe
is sufficient, a 3-foot pipe will probably be the next size, an increase
of 225 per cent; and if a G-foot arch culvert is too small, an 8-foot will
be used, an increase of ISO per cent. The real question is whether
a 2-foot pipe or an S-foot arch culvert is needed.

Valuable data on the proper size of any particular culvert may
be obtained (1) by observing the existing openings on the same

HIGHWAY CONSTRUCTION 41

stream; (2) by measuring, preferably at time of high water, a cross-
section of the stream at some narrow place; and (3) determining the
height of high water as indicated by drift and the evidence of the
inhabitants of the neighborhood.

On mountain roads or roads subjected to heavy rainfall culverts
of ample dimensions should be provided wherever required, and it
will be more economical to construct them of masonry. In localities
where boulders and other debris are likely to be washed down during
wet weather, it will be a good precaution to construct catch pools at
the entrance of all culverts and cross drains for the reception of
such matter. In hard soil or rock these catch pools will be simple
\vell-like excavations, with their bottom two or three feet below the
entrance sill or floor of the culvert or drain. Where the soil is soft
they should be lined with stone laid dry; if very soft, with masonry.
The size of the catch pools will depend upon the width of the drainage
works. They should be wide enough to prevent the drains from
being injured by falling rocks and stones of a not inordinate size.

The use of catch pools obviates the necessity of building culverts
and drains at an angle to the axis of the road. Oblique structures
are objectionable, as being longer than if set at. right angles and by
reason of the acute- and obtuse-angled terminations to their piers,
abutments, and coverings.

Materials for Culverts. Culverts may be of stone, brick, vitri-
fied earthenware, or iron pipe. Wood should be absolutely avoided.

For small streams and a limited surface of rainfall either class
of pipes, in sizes varying from 12 to 24 inches in diameter, will serve
excellently. They are easily laid, and if properly bedded, with the
earth tamped about them, are very permanent. Their upper surface
should be at least 18 inches below the road surface, and the upper
end should be protected with stone paving so arranged that the water
can in no case work in around the pipe.

When the flow of water is estimated to be too great for two lines
of 24-inch pipes, a culvert is required. If stone abounds, it may be
built of large roughly squared stones laid either dry or in mortar.
When the span required is more than 5 feet, arch culverts either of
stone or brick masonry may be employed. For spans above 15 feet
the structure required becomes a bridge

307

HIGHWAY CONSTRUCTION

Earthenware Pipe Culverts. Construction. In laying the
pipe the bottom of the trench should be rounded out to fit the lower
half of the body of the pipe with proper depressions for the sockets.
If the ground is soft or sandy, the earth should be rammed carefully,
but solidly in and around the lower part of the pipe. The top surface
of the pipe should, as a rule, never be less than 18 inches below the
surface of the roadway, but there are many cases where pipes have
stood for several years under heavy loads with only 8 to 12 inches of
earth over them. No danger from frost need be apprehended, pro-
vided the culverts are so constructed that the water is carried away
from the level end. Ordinary soft drain tiles are not in the least
affected by the expansion of frost in the earth around them.

The freezing of water in the pipe, particularly if more than half
full, is liable to burst it; consequently the pipe should have a suffi-
cient fall to drain itself, and the outside should be so low that there
is no danger of back waters reaching the pipe. If properly drained,
there is no danger from frost.

Jointing. In many cases, perhaps in most, the joints are
not calked. If this is not done, there is liability of the water being
forced out of the joints and washing away the soil from around the
pipe. Even if the danger is not very imminent, the joints of the
larger pipes, at least, should be calked with hydraulic cement, since
the cost is very small compared with the insurance against damage

Fig. 18.

thereby secured. Sometimes the joints are calked with clay. Every
culvert should be built so that it can discharge water under a head
without damage to itself.

308

HIGHWAY CONSTRUCTION

Although often omitted, the end sections should be protected
with a masonry or timber bulkhead. The foundation of the bulk-
head should be deep enough not to be disturbed by frost. In con-
structing the end wall, U is well to increase the fall near the outlet
to allow for a possible settlement of the interior sections. When
stone and brick abutments are too expensive, a fair substitute can
be made by setting posts in the ground and spiking plank to them.
When planks are used, it is best to set them with considerable inclina-
tion towards the roadbed to prevent their being crowded outward
by the pressure of the embankment. The upper end of the culvert
should be so protected that the water will not readily find its way

along the outside of the pipes, in case the mouth of the culvert should
become submerged.

When the capacity of one pipe is not sufficient, two or more
may be laid side by side as shown in Fig. 19. Although the two
small pipes do not have as much discharging capacity as a single
large one of equal cross-section, yet there is an advantage in laying
two small ones side by side, since the water need not rise so high
to utilize the full capacity of the two pipes as would be necessary
to discharge itself through a single one of large size.

Iron Pipe Culverts. During recent years iron pipe has been
used for culverts on many prominent railroads, and may be used on
roads in sections where other materials are unavailable.

In constructing a culvert with cast-iron pipe the points requiring

HIGHWAY CONSTRUCTION

particular attention are (1) tamping the soil tightly around the pipe
to prevent the water from forming a channel along the outside, and
(2) protecting the ends by suitable head walls and, when necessary,
laying riprap at the lower end. The amount of masonry required
for the end walls depends upon the relative width of the embankment
and the number of sections of pipe used. For example, if the em-
bankment is, say, 40 feet wide at the base, the culvert may consist of
three 12-foot lengths of pipe and a light end wall near the toe of
the bank; but if the embankment is, say, 32 feet wide, the culvert
may consist of two 12-foot lengths of pipe and a comparatively heavy
end wall well back from the toe of the bank. The smaller sizes of
pipe usually come in 12- foot lengths, but sometimes a few 6-foot

Broken Stones
or Bricks

Fig. 20. Section of Pipe Culvert

lengths are included for use in adjusting the length of the culvert,
to the width of the bank. The larger sizes are generally 6 feet long.

EARTHWORK.

The term "earthwork" is applied to all the operations per-
formed in the making of excavation and embankments. In its
widest sense it comprehends work in rock as well as in the looser
materials of the earth's crust.

Balancing Cuts and Fills. In the construction of new roads,
the formation of the roadbed consists in bringing the surface of the
ground to the adopted grade This grade should be established so as

310

HIGHWAY CONSTRUCTION 45

to reduce the earthwork to the least possible amount, both to render
the cost of construction low, and to avoid unnecessary marring the
appearance of the country in the vicinity of the road. The most
desirable position of the grade line is usually that which makes the
amount of cutting and filling equal to each other, for any surplus
embankment over cutting must be made up by borrowing, and surplus
cutting must be wasted, both of these operations involving additional
cost for labor and land.

Inclination of Side Slopes. The proper inclination for the
side slopes of cutting and embankments depends upon the nature of
the soil, the action of the atmosphere and of internal moisture upon
it. For economy the inclination should be as steep as the nature
of the soil will permit.

The usual slopes in cuttings are:

Solid rock . 1 to 1

Earth and Gravel 3^ to 1

Clay .3 or G to 1

Fine sand 2 or 3 to 1

The slopes of embankment are usually made 1 -\- to 1 .
Form of Side Slopes. The natural, strongest, and ultimate
form of earth slopes is a concave curve, iii which the flattest portion
is at the bottom. This form is very rarely given to the slopes in con-
structing them; in fact, the reverse is often the case, the slopes being
made convex, thus saving excavation by the contractor and inviting
slips.

In cuttings exceeding 10 feet in depth the forming of concave
••lopes will materially aid in preventing slips, and in any case they will

Fig. 21 . Cross-Section for Embankment.

reduce the amount of material which will eventually have to be re-
moved when cleaning up. Straight or convex slopes will continue
to slip until the natural form is attained.

A revetment or retaining wall at the base of a slope will save
excavation.

311

46 HIGHWAY CONSTRUCTION

In excavations of considerable depth, and particularly in soils
liable to slips, the slope may be formed in terraces, the horizontal
offsets or benches being made a few feet in width with a ditch on
the inner side to receive the surface water from the portion of the
side slope above them. These benches catch and retain earth
that may fall from the slopes above them. The correct forms for ths
slopes of embankment and excavation are shown in Figs. 21 and 22.

Covering of Slopes. It is not usual to employ any artificial
means to protect the surface of the side slopes from the action of the
weather; but it is a precaution which in the end will save much labor

Fig. 22. Cross-Section for Excavation.

and expense in keeping the roadways in good order. The simplest
means which can be used for this purpose consists in covering the
slopes with good sods, or else with a layer of vegetable mould about
four inches thick, carefully laid and sown with grass seed. These
means are amply sufficient to protect the side slopes from injury
when they are not exposed to any other cause of deterioration than
the wash of the rain and the action of frost on the ordinary moisture
retained by the soil.

A covering of brushwood or a thatch of straw may also be used
with good effect; but from their perishable nature they will require
frequent renewal and repairs.

Where stone is abundant a small wall of stone laid dry may be
constructed at the foot of the slopes to prevent any wash from them
being carried into the ditches.

Shrinkage of Earthwork. All materials when excavated
increase in bulk, but after being deposited in banks subside or shrink
(rock excepted) until they occupy less space than in the pit from
which excavated.

Rock, on the other hand, increases in volume by being broken
up, and does not settle again into less than its original bulk. The
.increase may be taken at 50 per cent,

312

HIGHWAY CONSTRUCTION 47

The shrinkage in the different materials is about as follows:

.Gravel 8 per cent

Gravel and sand 9 " "

Clay and clay earths 10 " '

Loam and light sandy earths 12 " "

Loose vegetable soil. . . 15 " "

Puddled clay 25 "

Thus an excavation of loam measuring 1,000 cubic yards will
form only about 880 cubic yards of embankment, or an embankment
of 1,000 cubic yards will require about 1,120 cubic yards measured
in excavation to make it. A rock excavation measuring 1,000 yards
will make from 1,500 to 1,700 cubic yards of embankment, depending
upon the size of the fragments.

The lineal settlement of earth embankments will be about in
the ratio given above; therefore either the contractor should be
instructed in setting his poles to guide him as to the height of grade
on an earth embankment to add the required percentage to the fill
marked on the stakes, or the percentage may be included in the
fill marked on the stakes. In rock embankments this is not necessary.

Classification of Earthwork. Excavation is usually classi-
fied under the heads earth, hardpan, loose rock, and solid rock. For
each of these classes a specific price is usually agreed upon, and an
extra allowance is sometimes made when the haul or distance to
which the excavated material is moved exceeds a given amount.

The characteristics which determine the classes to which a given
material belongs are usually described with clearness in the speci-
fications, as:

Earth will include loam, clay, sand, and loose gravel.

Hardpan will include cemented gravel, slate, cobbles, and boul-
ders containing less than one cubic foot, and all other matters of an
earthy nature, however compact they may be.

Loose rocfc will include shale, decomposed rock, boulders, and
detached masses of rock containing not less than three cubic feet,
and all other matters of a rock nature which may be loosened with a
pic-k, although blasting may be resorted to in order to expedite the
work. ~

Solid rock will include all rock found in place in ledges and

813

48 HIGHWAY CONSTRUCTION

masses, or boulders measuring more than three cubic feet, and which
can only be removed by blasting.

Prosecution of Earthwork. No general rule can be laid
down for the exact method of carrying on an excavation and dis-
posing of the excavated material. The operation in each case can
only be determined by the requirements of the contract, character
of the material, magnitude of the work, length of haul, etc.

Formation of Embankments. Where embankments are to be
formed of less than two feet in height, all stumps, weeds, etc. should
be removed from the space to be occupied by the embankment.
For embankments exceeding two feet in height stumps need only
be close cut. Weeds and brush, however, ought to be removed and
if the surface is covered with grass sod, it is advisable to plow a fur-
row at the toe of the slope. Where a cutting passes into a fill all
the vegetable matter should be removed from the surface before
placing the fill. The site of the bank should be carefully examined
and all deposits of soft, compressible matter removed. WTien a bank
is to be made over a swamp or marsh, the site should be thoroughly
drained, and if possible the fill should be started on hard bottom.

Perfect stability is the object aimed at, and all precautions neces-
sary to this end should be taken. Embankments should be built in
successive layers, banks two feet and under in layers from six
inches to one foot, heavier banks in layers 2 and 3 feet thick. The
horses and vehicles conveying the materials should be required to
pass over the bank for the purpose of consolidating it, and care
should be taken to have the layers dip towards the center. Embank-
ments first built up in the center, and afterwards widened by dump-
ing the earth over the sides, should not be allowed.

Embankments on Hillsides. When the axis of the road
is laid out on the side slope of a hill, and the road is formed partly
by excavating and partly by embanking, the usual and most simple
method is to extend out the embankment gradually along the whole
line of the excavation. This method is insecure; the excavated
material if simply deposited on the natural slope is liable to slip,
and no pains should be spared to give it a secure hold, particularly
at the toe of the slope. The natural surface of the slope should be
cut into steps as shown in Figs. 23 and 24. The dotted line AB

314

A PECULIAR EXAMPLE OF HIGHWAY CONSTRUCTION

By means of the truck elevators here shown, the roadwav is carried over the steep bluffs of the Pal-
isades of the Hudson at Hoboken, N. J.

HIGHWAY CONSTRUCTION

represents the natural surface of the ground, C E B the excavation,
and ADC the embankment, resting on steps which have been cut
between A and C. The best position for these steps is perpendicular
to the axis of greatest pressure. If A D is inclined at the angle of
repose of the material, the steps near A should be inclined in the

Fig. 23. Method of Construction on Hillsides.

opposite direction to A D, and at an angle of nearly 90 degrees
thereto, while the steps near C may be level. If stone is abundant,
the toe of the slope may be further secured by a dry wall of stone.
On hillsides of great inclination the above method of construc-
tion will not be sufficiently secure; retaining walls of stone must
be substituted for the side slopes of both the excavations and em-
bankments. These walls may be made of stone laid dry, when stone

Fig. 24. Hillside Road with Retaining and Revetment Walls.

can be procured in blocks of sufficient size to render this kind of con-
struction of sufficient stability to resist the pressure of the earth.
When the stones laid dry do not offer this security, they must be laid
in mortar. The wall which forms the slope of the excavation should
be carried up as high as the natural surface of the ground. Unless
the material is such that the slope may be safely formed into steps
or benches as shown in Fig. 23, the wall that sustains the embank-
ment should l)e built up to the surface of the roadway, and a parapet

315

50 HIGHWAY CONSTRUCTION

wall or fence raised upon it, to protect pedestrians against accident.
(See Fig. 24.)

For the formula for calculating the dimensions of retaining walls
see instruction paper on Masonry Construction.

Roadways on Rock Slopes. On rock slopes when the in-
clination of the natural surface is not greater than one perpendicular
to two base, the road may be constructed partly in excavation and
partly in embankment in the usual manner, or by cutting the face
of the slope into horizontal steps with vertical faces, and building
up the embankment in the form of a solid stone wall in horizontal
courses, laid either dry or in mortar. Care is required in proportion-
ing the steps, as all attempts to lessen the quantity of excavation by
increasing the number and diminishing the width of the steps require
additional precautions against settlement in the built-up portion
of the roadway.

When the rock slope has a greater inclination than 1 : 2 the
whole of the roadway should be in excavation.

In some localities roads have been constructed along the face
of nearly perpendicular cliffs on timber frameworks consisting of
horizontal beams, firmly fixed at one end by being let into holes
drilled in the rock, the other end being supported by an inclined
strut resting against the rock in a shoulder cut to receive it. There
are also examples of similar platforms suspended instead of being
supported.

Earth Roads. The term "earth road" is applied to roads
where the surface consists of the native soil; this class of road is the
most common and cheapest in first cost. At certain seasons of the
year earth roads when properly cared for are second to none, but
during the spring and wet seasons they are very deficient in- the im-
portant requisite of hardness, and are almost impassable.

For the construction of new earth roads, all the principles pre-
viously discussed relating to alignment, grades, drainage, width, etc.,
should be carefully, followed. The crown or transverse contour
should be greater than in stone roads. Twelve inches at the center
in 25 feet will be sufficient.

Drainage is especially important, because the material of the
road is more susceptible to the action of water, and more easily

316

HIGHWAY CONSTRUCTION

destroyed by it than are the materials used in the construction of the
better class of roads. When water is allowed to stand upon the
road, the earth is softened, the wagon wheels penetrate it and the
horses' feet mix and kneed it until it becomes impassable mud. The
action of frost is also apt to be more disastrous upon the more per-
meable surface of the earth road, having the effect of swelling and
heaving the roadway and throwing its surface out of shape. It may

Fig. 25. Bush Hooks.

in fact be said that the whole problem of the improvement and
maintenance of ordinary country roads is one of drainage.

In the preparation of the wheelway all stumps, brush, vegetable
matter, rocks and boulders should be removed from the surface and
the resulting holes filled in with clean earth. The roadbed having

Fig. 26. Axe Mattock.

Fig. 27. Bush Mattock.

been brought to the required grade and crown should be thoroughly
rolled, all inequalities appearing during the rolling should be filled
up and re-rolled.

Care of Earth Roads. If the surface of the roadway is prop-
erly formed and kept smooth, the water will be shed into the side
ditches and do comparatively little harm; but if it remains upon the
surface, it will be absorbed and convert the road into mud. All
ruts and depressions should be filled up as soon as they appear.
Repairs should be attended to particularly in the spring. At this
season a judicious use of a road machine and rollers will make a

317

52 HIGHWAY CONSTRUCTION

smooth road. In summer when the surface gets roughed up it can
be improved by running a harrow over it; if the surface is a little
muddy this treatment will hasten the drying.

During the fall the surface should be repaired, with special
reference to putting it in shape to withstand the ravages of winter.
Saucer-like depressions and ruts should be filled up with clean earth
similar to that of the roadbed and tamped into place.

The side ditches should be examined in the fall to see that they
are free from dead weeds and grass, and late in winter they should
be examined again to see that they are not clogged. The mouths of
culverts should be cleaned of rubbish and the outlet of tile drains
opened. Attention to the side ditches will prevent overflow, and
washing of the roadway, and will also prevent the formation of ponds
at the roadside and the consequent saturation of the roadbed.

Holes and ruts should not be filled with stone, bricks, gravel
or other material harder than the earth of the roadway as the hard
material will not wear uniform with the rest of the road, but produce
bumps and ridges, and usually result in making two holes, each
larger than the original one. It is bad practice to cut a gutter from
a hole to drain it to the side of the road. Filling is the proper course,
whether the hole is dry or contains mud.

In the maintenance of clay roads neither sods nor turf should
be used to fill holes or ruts; for, though at first deceptively tough,
they soon decay and form the softest mud. Neither should the ruts
be filled with field stones; they will not wear uniformly with the rest
of the road, but will produce hard ridges.

Trees and close hedges should not be allowed within 200 feet
of a clay road. It requires all the sun and wind possible to keep its
surface in a dry and hard condition.

Sand Roads. The aim in the improvement of sand roads is to
have the wheelway as narrow and well defined as possible, so as to
have all the vehicles run in the same track. An abundant growth
of vegetation should be encouraged on each side of the wheelway,
for by this means the shearing of the sand is, in a great measure,
avoided. Ditching beyond a slight depth to carry away the rain
water is not desirable, for it tends to hasten the drying of the sands,,
which is to be avoided. Where possible the roads should be over-

318

HIGHWAY CONSTRUCTION

hung with trees, the leaves and twigs of which catching on the
wheelway will serve still further to diminish the effect of the wheels
in moving the sands about. If clay can be obtained, a coating 6
inches thick will be found a most effective and economical improve-
ment. A coating of 4 inches of loose straw will, after a few days'
travel, grind into the sand and become as hard and as firm as a
dry clay road.

The maintaining of smooth surfaces on all classes of earth roads
will be greatly assisted and cheapened by the frequent use of a roller
(either steam or horse) and any one of the various forms of road
grading and scraping machines. In repairing an earth road the
plough should not be used. It breaks up the surface which has
been compacted by time and travel.

TOOLS FOR GRADING.

Picks are made of various styles, according to the class of
material in which they are to be used. Fig. 28 shows the form

Fig. 29. Clay Pick.

usually employed in street work. Fig. 29 shows the form generally
used for clay or gravel excavation.

The eye of the pick is generally formed of wrought iron, pointed
with steel. The weight of picks ranges from 4 to 9 Ib.

Pig. 30. Shovels.

Shovels are made in two forms, square and round pointed,
usually of pressed steel.

Ploughs are extensively employed in grading, special forms
being manufactured for the purpose. They are known as " grading
ploughs," " road ploughs" " township ploughs," etc. They vary

319

HIGHWAY CONSTRUCTION

in form according to the kind of work they are intended for, viz. :
loosening earth, gravel, hardpan, and some of the softer rocks.

These ploughs are made of great strength, selected white oak,
rock elm, wrought steel and iron being generally used in their con-
struction. The cost of operating ploughs ranges from 2 to 5 cents
per cubic yard, depending upon the compactness of the soil. The
quantity of material loosened will vary from 2 to 5 cubic yards per
hour.

Fig. 31 shows the form usually adopted for loosening earth.
This plough does not turn the soil, but cuts a furrow about 10

Fig. 31. Grading Plow.

inches wide and of a depth adjustable up to 11 inches.

In light soil the ploughs are operated by two or four horses; in
heavy soils as many as eight are employed. Grading ploughs vary
in weight from 100 to 325 Ib.

Pig. 32. Hardpan Plow.

Fig. 32 illustrates a plough specially designed for tearing up
macadam, gravel, or similar material. The point is a straight bar
of cast steel drawn down to a point, and can be easily repaired.

320

HIGHWAY CONSTRUCTION

r,r,

Scrapers are generally used to move the material loosened by
ploughing; they are made of either iron or steel, and in a variety
of form, and are known by various names, as " drag," " buck,"
" pole," and " wheeled". The drag scrapers are usually employed
on short hauls, the wheeled on long hauls. Fig. 33 illustrates the
usual form of drag scrapers.

Drag scrapers are made in three sizes. The smallest, for one
horse, has a capacity of 3 cubic feet; the others, for two horses,

Fig. 33. Drag Scraper.

have a capacity of 5 to 7-£ cubic feet. The smallest weighs about
90 lb., and the larger ones from 94 to 102 Ib.

Buck scrapers are made in two sizes — two-horses, carrying 1\
cubic feet; four-horses, 12 cubic feet.

Pole scraper, Fig. 34, is designed for use in making and leveling
earth roads and for cutting and cleaning ditches; it is also well

Fig. 34. Pole Scraper.

adapted for moving earth short distances at a minimum cost.

Wheeled scrapers consist of a metal box, usually steel, mounted
~on wheels, and furnished with levers for raising, lowering, and

321

HIGHWAY CONSTRUCTION

dumping. They are operated in the same manner as drag scrapers,
except that all the movements are made by means of the levers, and
without stopping the team. By their use the excessive resistance to

Fig. 35. Wheeled Scraper.

traction of the drag scraper is avoided. Various sizes are made,
ranging in capacity from 10 to 17 cubic feet. In weight they range
from 350 to 700 Ib.

Wheelbarrows. The wheelbarrow shown in Fig. 36 is con-
structed of wood and is the most commonly employed for earth-
work. Its capacity ranges from 2 to 2J cubic feet. Weight about
50 Ib.

The barrow, Fig. 37, has a pressed-steel tray, oak frame, and
steel wheel, and will be found more durable in the maintenance

Fig. 36. Wooden Barrow.

department than the all wood barrow. Capacity from 3^ to 5 cubic
feet, depending on size of tray.

The barrow, Fig. 38, is constructed with tubular iron frames
and steel tray, and is adaptable to the heaviest work, such as

322

HIGHWAY CONSTRUCTION

moving heavy broken' stone, etc., or it may be employed with ad-
vantage in the cleaning department. Capacity from 3 to 4 cubic
feet. Weight from 70 to 82 Ib.

Fig. 37. Steel Tray Barrow.

The maximum distance to which earth can be moved economic-
ally in barrows is about 200 feet. The wheeling should be per-
formed upon planks, whose steepest inclination should not exceed 1
in 12. The force required to move a barrow on a plank is about oV
part of the weight; on hard dry earth, about y¥ part of the weight.

Fig. 38. Metal Barrow.

The time occupied in loading a barrow will vary with the
character of the material and the proportion of wheelers to shovel-
lers. Approximately, a shoveller takes about as long to fill a barrow
with earth as a wheeler takes to wheel a full barrow a distance of
about 100 or 120 feet on a horizontal plank and return with the
empty barrow.

Carts. The cart usually employed for hauling earth, etc., is
shown in Fig. 39. The average capacity is 22 cubic feet, and the
average weight is 800 Ib. These carts are usually furnished with
broad tires, and the body is so balanced that the load is evenly
divided about the axle.

323

58 HIGHWAY CONSTRUCTION

The time required to load a cart varies with the material. One
shoveller will require about as follows: Clay, seven minutes; loam,
six minutes; sand, five minutes.

Fig. 39. Earth Wagon.

Dump Cars. These cars are made to dump in several different
ways, viz., single or double side, single or double end, and rotary
or universal dumpers.

Dump cars may be operated singly or in trains, as the magni-
tude of the work mty demand. They may be moved by horses or

Pig. 40. Dump Cart.

small locomotives. They are made in various sizes, depending upon
the gauge of the track on which they are run. A common gauge is

HIGHWAY CONSTRUCTION

20 inches, but it varies from that up to the standard railroad gauge
of 662 inches.

Dump Wagons. (Fig. 40.) The use of these wagons for mov-
ing excavated earth, etc., and for transporting materials such as sand,
gravel, etc., materially shortens the time required for unloading the
ordinary form of contractor's wagon; having no reach or pole con-
necting the rear axle with the center bearing of the front axle, they
may be cramped short and the load deposited just where required.
They are operated by the driver, and the capacity ranges from 35
to 45 cubic feet.

Mechanical Graders are used extensively in the making and
maintaining of earth roads. They excavate and move earth more
expeditiously and economically than can be done by hand; they are
called by various names, such as "road machines," "graders,"
"road hones," etc. Their general form is shown in Fig. 41.

Briefly described, they consist of a large blade made entirely
of steel or of iron, or wood shod with steel, which is so arranged by
mechanism attached to the frame from which it is suspended that it
can be adjusted and fixed in any direction by the operator. In their
action they combine the work of excavating and transporting the

Fig. 41. Mechanical Grader.

earth. They have been chiefly employed in the forming and main-
tenance of earth roads, but may be also advantageously used in pre-
paring the subgrade surface of roads for the reception of broken
stone or other improved covering.

A large variety of such machines are on the market. The
"New Era" grader excavates the material from side ditches, and

325

60 HIGHWAY CONSTRUCTION

automatically loads the material into carts or wagons. Briefly de-
scribed, the machine consists of a plough which loosens and raises
the earth, depositing it upon a transverse carrying-belt, which con-
veys it from excavation to embankment. This carrier is built in four
sections, bolted together, so it can be used to deliver earth at 14,
17, 19, or 22 feet from the plough. The carrier belt is of heavy
3-ply rubber 3 feet wide.

The plough and carrier are supported by a strong trussed frame-
work resting on heavy steel axles and broad wheels. The large
rear wheels are ratcheted upon the axle, and connected with strong
gearing which propels the carrying-belt at- right angles to the direc-
tion in which the machine is moving.

The wheels and trusses are low and broad, occupying a space
8 feet wide and 14 feet long, exclusive of the side carrier. This
enables it to work on hillsides where any wheeled implements can
be used. Notwithstanding its large size it is so flexible that it may
be turned around on a 16-foot embankment. Pilot wheels and
levers enable the operator to raise or lower the plough or carrier at
pleasure.

. As a motive power 12 horses — 8 driven in front, 4 abreast, and
4 in the rear on a push cart — are usually employed.

When the teams are started, the operator lowers the plough and
throws the belting into gear, and as the plough raises and turns the
earth to the side the belt receives and delivers it at the distance for
which the carrier is adjusted, forming either excavation or embank-
ment.

When it becomes necessary to deliver the excavated earth beyond
the capacity of the machine (22. feet or 1\ feet above the plough),
the earth is loaded upon wagons, then conveyed to any distance.
Arranging the carrier at 19 feet, wagons are. driven under the car-
rier and loaded with \\ to H yards of earth in from 20 to 30
seconds. When one wagon turns out with its load, another drives
under the carrier, and the machine thus loads 600 to 800 wagons
per day. It is claimed that with six teams and three men it is capa-
ble of excavating and placing in embankment from 1000 to 1500
cubic yards of earth in ten hours, or of loading from 600 to 800

326

HIGHWAY CONSTRUCTION 61

wagons in the same time, and that the cost of this handling is from
1^ to 2^ cents per cubic yard.

Points to be Considered in Selecting a Road Machine. In

the selection of a road machine the following points should be care-
fully considered:

(1) Thoroughness and simplicity of its mechanical construction.

(2) Material and workmanship used in its construction.

(3) Ease of operation.

(4) Lightness of draft

(5) Adaptability for doing general road- work, ditching, etc.

(6) Safety to the operator.

Care of Road Machines. The road machine when not in use
should be stored in a dry house and thoroughly cleaned, its blade
brushed clean from all accumulations of mud, wiped thoroughly dry,
and well covered with grease or crude oil. The axles, journals, and
wearing parts should be kept well oiled when in use, and an extra
blade should be kept on hand to avoid stopping tHe machine while
the dulled one is being sharpened.

Surface Graders. The surface grader, Fig. 42, is used for re-
moving earth previously loosened by a plough. It is operated by
one horse. The load may be retained and carried a considerable

Fig. 42. Surface Grader.

distance, or it may be spread gradually as the operator desires. It
is also employed to level off and trim the surface after scrapers.

The blade is of steel, J-inch thick, 15 inches wide, and 30
inches long. The beam and other parts are of oak and iron.
Weight about 60 Ib.

327

HIGHWAY CONSTRUCTION

The road leveller, Fig. 43, is used for trimming and smoothing
the surface of earth roads. It is largely employed in the Spring
when the frost leaves the ground.

Fig. 43. Road Leveller.

NO. 3, MO. 4. NO. 5.

O.S.

No. 6.

No. /. No. 7.

Fig. 44. Draining Tools.

The blade is of steel, ^-inch thick by 4 inches by 72 inches, and
is provided with a seat for the driver. It is operated by a team of
horses. Weight about 150 Ib.

HIGHWAY CONSTRUCTION

Draining=tools. The tools employed for digging the ditches
and shaping the bottom to fit the drain tiles are shown in Fig. 44.
They are convenient to use, and expedite the work by avoiding
unnecessary excavation.

The tools are used as follows: Nos. 3, 4 and 5 are used for
digging the ditches; Nos. 6 and 7 for cleaning and rounding the

Fig. 45. Reversible Roller.

bottom of the ditch for round tile. No. 2 is used for shoveling out
loose earth and levelling the bottom of the ditch; No. 1 is used for
the same purpose when the ditch is intended for "sole " tile.

Fig. 46. Watering Cart.

Horse Rollers. There is a variety of horse rollers on the
market. Fig. 45 shows the general form. Each consists essentially
of a hollow cast-iron cylinder 4 to 5 feet long, 5 to G feet in

64 HIGHWAY CONSTRUCTION

diameter, and weighing from 3 to 6 tons. Some forms are provided
with boxes in which stone or iron may be placed to increase the
weight, and some have closed ends and may be filled with water or
sand.

Sprinkling-carts. Fig. 46 shows a convenient form of sprink-
ling cart for suburban streets and country roads. Capacity about
150 gallons.

ROAD COVERINGS.

Road coverings consist of some foreign material as gravel,
broken stone, clay, etc., placed on the surface of the earth road.
The object of this covering, whatever its nature, is (1) to protect the
natural soil from the effect of weather and travel, and (2) to furnish
a smooth surface on which the resistance to traction will be reduced
to the least possible amount, and over which vehicles may pass with
safety and expedition at all seasons of the year. . Where an artificial
covering is employed, the wheel loads coming upon its surface are
distributed over a greater area of the roadbed than if the loads
come directly upon the earth itself. The loads are not sustained by
the covering as a rigid structure, but are transferred through it to
the roadbed, which must support both the weight of the covering
and the load coming upon it.

Gravel Roads. Gravel is an accumulation of small more or
less rounded stones which usually vary from the size of 'a small pea
to a walnut. It is often intermixed with other substances, such as
sand, clay, loam, etc., from each of which it derives a distinctive
name. In selecting gravel for road purposes the chief quality to be
sought for is the property of binding.

Gravel in general is unserviceable for roadmaking. This is
due mainly to the fact that the surface of the pebbles is smooth, so
that they will not bind together in the manner of broken stone.
There is also an absence of dust or other material to serve as a
binder, and even if such binding material is furnished it is difficult
to effectively hold the rounded and polished surface of the pebbles
together.

In certain deposits of gravel, particularly where the pebbly
matter is to a greater or less extent composed of limestone, a con-
siderable amount of iron oxide has been gathered in the mass.

330

HIGHWAY CONSTRUCTION 65

This effect is due to the tendency of water which contains iron to
lay down that substance and to take lime in its place when the
opportunity for so doing occurs. Such gravels are termed ferru-
ginous. They are commonly found in a somewhat cemented state,
and when broken up and placed upon roads they again cement, even
more firmly than in the original state, often forming a roadway of
very good quality.

When no gravel but that found in rivers or on the seashore can
be obtained, one-half of the stone should be broken and mixed with
the other half; to the stone so mixed a small quantity of clay or
loam, about one-eighth of the bulk of the gravel, must be added:
an excess is injurious. Sand is unsuitable. It prevents packing in
proportion to the amount added.

Preparing the Gravel. Pit gravel usually contains too much
earth, and should be screened before being used. Two sieves should
be provided, one with meshes of one and one-half inches, so that all
pebbles above that size may be rejected, the other with meshes of
three quarters of an inch, and the material which passes through it
should be thrown away. The expense of screening will be more
than repaid by the superior condition of the road formed by the
cleaned material, and by the diminution of labor in keeping it in
order. The pebbles larger than one and a half inches may be
broken to that size and mixed with clean material.

Laying the Gravel. On the roadbed properly prepared a layer
of the prepared gravel four inches thick is uniformly spread over the
whole width, then compacted with a roller weighing not less than
two tons, and having a length of not less than thirty inches. The
rolling must be continued until the pebbles cease to rise or creep in
front of the roller. The surface must be moistened by sprinkling
in advance of the roller, but too much water must not be used.
Successive layers follow, each being treated in the above described
manner until the requisite depth and form has been attained.

The gravel in the bottom layer must be no larger than that in
the top layer; it must be uniformly mixed, large and small together,
for if not, the vibration of the traffic and the action of frost will
cause the larger pebbles to rise to the surface and the smaller ones
to descend, and the road will never be smooth or firm.

331

66 HIGHWAY CONSTRUCTION

The pebbles in a gravel road are simply imbedded in a paste
and can be easily displaced. It is for this reason, among others,
that such roads are subject to internal destruction.

The binding power of clay depends in a large measure upon
the state of the weather. During rainy periods a gravel road be-
comes soft and muddy, while in very dry weather the clay will con-
tract and crack, thus releasing the pebbles, and giving a loose
surface. The most favorable conditions are obtained in moderately
damp or dry weather, during which a gravel road offers several
advantages for light traffic, the character of the drainage, etc.,
largely determining durability, cost, maintenance, etc.

Repair. Gravel roads constructed as above described will
need but little repairs for some years, but daily attention is required
to make these. A garden rake should be kept at hand to draw
any loose gravel into the wheel tracks, and for filling any depres-
sions that may occur.

In making repairs, it is best to apply a small quantity of gravel
at a time, unless it is a spot which has actually cut through. Two
inches of gravel at once is more profitable than a larger amount.
Where a thick coating is applied at once it does not all pack, and if,
after the surface is solid, a cut be made, loose gravel will be found;
this holds water and makes the road heave and become spouty
under the action of frost. It will cost no more to apply six inches
of gravel at three different times than to do it at once.

At every one-eighth of a mile a few cubic yards of gravel
should be stored to be used in filling depressions and ruts as fast as
they appear, and there should be at least one laborer to every five
miles of road.

Broken Stone Roads. Broken stone roads are formed by pla-
cing small angular fragments of stone on the surface of the earth
roadbed and compacting into a solid mass by rolling. This class
of road covering is generally called a Macadam or Telford road
from the name of the two men who first introduced this type into
England.

The name of Telford is associated with a rough stone founda-
tion, which he did not always use, but which closely resembled that
which had been previously used in France. Macadam disregarded

HIGHWAY CONSTRUCTION 67

this foundation, contending that the subsoii, however bad, would
carry any weight if made dry by drainage and kept dry by an im-
pervious covering. The names of both have ever since been
associated with the class of road which each favored, as well as with
roads on which all their precepts have been disregarded.

Quality of Stones. The materials used for broken-stone pave-
ments must of necessity vary very much according to the locality.
Owing to the cost of haulage, local stone must generally be used,
especially if the traffic be only moderate. If, however, the traffic is
heavy, it will sometimes be found better and more economical to
obtain a superior material, even at a higher cost, than the local
stone; and in cases where the traffic is very great, the best material
that can be obtained is the most economical.

The qualities required in a good road stone are hardness and
and toughness and ability to resist the disintegrating action of the
weather. These qualities are seldom found together in the same
stone. Igneous and siliceous rocks, although frequently hard and
tough, do not consolidate so well nor so quick as limestone, owing
to the sandy detritus formed by the two first having no cohesion,
whilst the limestone has a detritus which acts like mortar in binding
the stones together.

A stone of good binding nature will frequently wear much
better than one without, although it is not so hard. A limestone
road well made and of good cross-section will be more impervious
than any other, owing to this cause, and will not disintegrate so
soon in dry weather, owing partly to this and partly to the well-
known quality which all limestone has of absorbing moisture from
the atmosphere. Mere hardness without toughness is not of much
use, as a stone may be very hard but so brittle as to be crushed to
powder under a heavy load, while a stone not so hard but having a
greater degree of toughness will be uninjured.'

By a stone of good binding quality is meant one that, when
moistened by water and subjected to the pressure of loaded wheels
or rollers, will bind or cement together. This quality is possessed to
a greater or less extent by nearly all rocks when in a state of dis-
integration. The binding is caused by the action of water upon the
chemical constituents of the stone contained in the detritus produced

68 HIGHWAY CONSTRUCTION

by crushing the stone, and by the friction of the fragments on each
other while being compacted; its strength varies with the different
species of rock, but it exists in some measure with them all, being
greatest with limestone and least with gneiss.

The essential condition of the stone to produce this binding
effect is that it bj sound. No decayed stone retains the property of
binding, though in some few cases, where the material contains iron
oxides, it may, by the cementing property of the oxide, undergo a
certain binding.

A stone for a road surface should be as little absorptive of
moisture as possible in order that it may not suffer injury from the
action of frost. Many limestones are objectionable on this account.

The stone used should be uniform in quality, otherwise it will
wear unevenly, and depressions will appear where the softer material
has been used.

As the under parts of the road covering are not subject to the
wear of traffic, and have only the weight of loads to sustain, it is
not necessary that the stone of the lower layer be so hard or so
tough as the stone for the surface, hence it is frequently possible by
using an inferior stone for that portion of the work, to greatly reduce
the cost of construction.

Size of Stones. The stone should be broken into fragments
as nearly cubical as possible. The size of the cubes will depend
upon the character of the rock. If it be granite or trap, they should
not exceed 1| inches in their greatest dimensions; if limestone, they
should not exceed 2 inches.

The smaller the stones the less the percentage of voids. Small
stones compact sooner, require less binding, and make a smoother
surface than large ones, but the size of the stone for any particular
section of a road must be determined to a certain extent by the
amount of traffic which it will have to bear and the character of the
rock used.

It is not necessary nor is it advisable that the stone should be
all of the same size; they may be of all sizes under the maximum.
In this condition the smaller stones fill the voids between the larger
and less binding is required.

Thickness of the Broken Stone. The offices of the broken

834

HIGHWAY CONSTRUCTION

stone are to endure friction and to shed water; its thickness must
be regulated by the quality of the stone, the amount of traffic, and
nature of the natural soil bed. Under heavy traffic it is advisable
to make the thickness greater than for light traffic, in order to pro-
vide for wear and lessen the cost of renewals.

When the roadbed is firm, well drained, and not likely to be
affected by ground water, it will always afford a firm foundation
for the broken stone, the thickness of which may be made the mini-
mum for good construction. This thickness is four inches. When
this thickness is employed the stone must be of exceptionally fine-
quality and the road must be maintained by the " continuous "
method. With heavy traffic the thickness should be increased over
the minimum a certain amount, say 2 inches, to provide for wear.
Where the foundation is unstable and there is a tendency on the
part of the loads to break through the covering, the thickness of
the stone must be made the maximum, which is 12 inches. In such
a case it may be advisable to employ a Telford foundation. Where
the covering exceeds six inches in thickness, the excess may be
composed of gravel, sand or ledge stone, the choice depending
entirely on the cost, for all are equally effective.

Foundation. The preparation of the natural soil over which
the road is to be constructed, to enable it to sustain the superstruc-
ture and the weights brought upon it, requires the observance of
certain precautions the neglect of which will sooner or later result
in the deterioration or possible destruction of the road covering.
These precautions vary with the character of the soil.

Soils of a siliceous and calcareous nature do not present any
great difficulty, as their porous nature generally affords good natural
drainage which secures a dry foundation. Their surface, however,
requires to be compacted; this' is effected by rolling.

The rolling should be carried out in dry weather, and any de-
pressions caused by the passage of the roller should be filled with
the same class of material as the surrounding soil. The rolling
must be repeated until a uniform and solid bed is obtained.

The argillaceous and allied soils, owing to their retentive
nature, are very unstable under the action of water and frost, and
in their natural condition afford a poor foundation. The prepara-

335

70 HIGHWAY CONSTRUCTION

tion of such soil is effected by drainage and by the application of a
layer of suitable material to entirely separate the surface from the
road material. This material may be sand, furnace ashes, or other
material of a similar nature, spread in a layer from 3 to 6 inches
thick over the surface of the natural soil.

When the road is formed in rock cuttings it is advisable to
spread a layer of sand or other material of light nature, so as to
fill up the irregularities of the surface as well as to form a cushion
for the road material to rest on.

Spreading the Stone. The stone should be hauled upon the
roadbed in broad-tire two-wheeled carts and dumped in heaps and
be spread evenly with a rake in a layer which should be of a depth
of 4J- inches.

Watering. Wetting the stone expedites the consolidation,
decreases crushing under the roller, and assists the filling of the
voids with the binder. It should be applied by a sprinkler and
should not be thrown on in quantity or from the plain nozzle of a
hose.

Excessive watering, especially in the earlier stages, tends to
soften the foundation, and care should be exercised in its appli-
cation.

Binding. As the voids in loosely spread broken stone range
from 35 to 50 per cent of the volume, and as no amount of rolling
will reduce the voids more than one-half, it is necessary, in order to
form an impervious and compact mass, to add some fine material
which is called the binder. It may consist of the fragments and
detritus obtained in crushing the stone. When this is insufficient,
as will be. the case with the harder rocks, the deficiency may be
made up of clean sand or gravel. The proportion of binder
should slightly exceed the voids in the aggregate; it must not be
mixed with the stones, but should be spread uniformly in small
quantities over the surface and rolled into the interstices with the
aid of water and brooms.

As the quality of the binding used is of vital importance, the
employment of inferior material, such as road scrapings or material
of a clayey nature, should be avoided, even if the initial cost of the
work should be greater when a good binding material is used.

336

HIGHWAY CONSTRUCTION 71

Stone consolidated with improper binding material may present
a good appearance immediately after being rolled and be otherwise
an apparently good piece of work, still in damp weather a consider-
able amount of "lick-Ing up" by the wheels of the vehicles will take
place, which reduces the strength of the coating and causes the sur-
face to wear unequally.

By the application of an immoderate quantity of binding of any
description the stone coating will become unsound or rotten in con-
dition, and if the binding be of an argillaceous nature, it will expand
during frost, owing to its absorbent properties, and cause the dis-
placement of the stones. The surface will become sticky, which
seriously affects the tractive power of horses, while the road itself
will suffer by the irregular deterioration of the surface.

The use of such material as mentioned for binding enables
rolling to be accomplished in much less time than when proper bind-
ing is used, and the cost of consolidating the stone may be reduced
by 25 per cent; but, on the other hand, the stone coating which will
probably contain under these circumstances from 30 to 40 per cent
of soft and soluble matter, and possibly present a smooth surface
immediately after being rolled, will quickly become "cupped" by
the wheel traffic, a bumpy surface being the result. This is caused
by the irregular wear, while the lasting qualities or "life" of the
coating will be shortened, giving unsatisfactory results to those
traveling over the road, and the work of renewing the surface of
the road in this manner may prove a failure on economical grounds.
There can be no doubt, and it is now being more generally recog-
nized, that sand as a material for binding in connection with rolling
operations, when applied in a limited but sufficient quantity, pro-
motes the durability- of the stone coating, while the general results
are equally satisfactory; a firm, compact, and smooth surface is
obtained, and the subsequent maintenance of the road is minimized.

A great amount of rolling is necessary when sand is employed
as a binding material, but economy is promoted, and the results are
more satisfactory when sand is used than by the use of the material
which gives to the stone an appearance only of having been properly
consolidated. If clean sand be used in combination with the screen-
ings from the crusher a very satisfactory surface will be obtained.

72 . HIGHWAY CONSTRUCTION

If the use of motor vehicles equipped with pueumatic tires be-
comes general, it is possible that some other description of binding
material will be necessary. The pumping action of suction created
by pneumatic tires, especially when propelled at a high speed, causes
a considerable movement of the fine particles of the binding material,
which on being displaced will convert the covering into a mass of
stones. This objection can probably be overcome by watering.

Compacting the Broken Stone. Three methods of compacting
the broken stone are practiced : (1 ) by the traffic passing over the road ;
(2) by rollers drawn by horses ; (3) by rollers propelled by steam.

The first method is both defective and objectionable. (1) It is
destructive to the horses and vehicles using the road. (2) It is waste-
ful of material ; about one-third of the stone is worn away in the oper-
ation. (3) Dung and dust are ground up with the stone, and the
road is more readily affected by wet and frost.

Steamrollers were first successfully introduced in France in 1860,
since which time they have been almost universally adopted on account
of the superiority and economy of the work done. Their use shortens
the time required for construction or repair, and effects an indirect
saving by the reduced wear and tear of horses and vehicles. They are
made in different weights ranging from 3 to 30 tons. For the compact-
ing of broken stone roads the weights in favor are from ten to fifteen
tons; the heavier weights are considered unwieldy and their use is
liable to cause damage to the underground structures that may be in
the roadway.

The advantage of steam rolling may be summed up as follows:

(1) They shorten the time of construction.

(2) A saving of road material, (a) because there are no loose
stones to be kicked about and worn ; (b) because there is no abrasion
of the stone, only one surface of the stone being exposed to wear; (c)
because a thinner coating of stone can be employed; (d) because no
ruts can be formed in which water can lie to rot the stone.

(3) Steam-rolled roads are easier to travel on account of their
even surface and superior hardness and they have a better appearance.

(4) The roads can be repaired at any season of the year.

(5) Saving both in materials and manual labor.

338

HIGHWAY CONSTRUCTION

PART II

STREETS AND HIGHWAYS

CITY STREETS

The first work requiring the skill of the engineer is to lay out town
sites properly, especially with reference to the future requirements of a
large city where any such possibility exists. Few if any of our large
cities were so planned. The same principles, to a limited extent, are
applicable to all towns or cities. The topography of the site should be
carefully studied, and the street lines adapted to it. These lines
should be laid out systematically, with a view to convenience and
comfort, and also with reference to economy of construction, future
sanitary improvements, grades, and drainage.

Arrangement of City Streets. Generally, the best method of
laying out streets is in straight lines, with frequent and regular inter-
secting streets, especially for the business parts of a city. When there
is some centrally located structure, such as a courthouse, city hall,
market, or other prominent building, it is very desirable to have several
diagonal streets leading thereto. In the residence portions of cities,
especially if on hilly ground, curves may with advantage replace
straight lines, by affording better grades at less cost of grading, and by
improving property through avoiding heavy embankments or cuttings.

Width of Streets. The width of streets should be proportioned
to the character of the traffic that will use them. No rule can be laid
down by which to determine the best width of streets; but it may safely
be said that a street which is likely to become a commercial thorough-
fare should have a width of not less than 120 feet between the building
lines — the carriage-way 80 feet wide, and the sidewalks each 20 feet
wide.

In streets occupied entirely by residences a carriage-way 32 feet
wide will be ample, but the width between the building lines may be as
great as desired. The sidewalks may be any amount over 10 feet

341

HIGHWAY CONSTRUCTION

which fancy dictates. Whatever width is adopted for them, not more
of it than 8 feet need be paved, the remainder being occupied with
grass and trees.

Street Grades. The grades of city streets depend upon the
topography of the site. . The necessity of avoiding deep cuttings or
high embankments which would seriously affect the value of adjoining
property for building purposes, often demands steeper grades than
are permissible on country roads. Many cities have paved streets
on 20 per cent grades. In establishing grades through unimproved
property, they may usually be laid with reference to securing the most
desirable percentage within a proper limit of cost. But when improve-
ments have already been made and have been located with reference
to the natural surface of the ground, giving a desirable grade is fre-
quently a matter of extreme difficulty without injury to adjoining
property. In such cases it becomes a question of how far individual

interests shall be sacrificed to the
general good. There are, how-
ever, certain conditions which it is
important to bear in mind :

(1) That the longitudinal
crown level should be uniformly
sustained from street' to street
intersection, whenever practicable.

(2) That the grade should
be sufficient to drain the surface.

(3) That the crown levels at
all intersections should be ex-
Fig. 47. tended transversely, to avoid form-
ing a depression at the junction.

Arrangements of Grades at Street Intersections, The best ar-
rangement for intersections of streets when either or both have much
inclination, is a matter requiring much consideration, and is one upon
which much diversity of opinion exists. No hard or fast rule can be
laid down; each will require special adjustment. The best and sim-
plest method is to make the rectangular space aaaaaaaa, Fig. 47,
level, with a rise of one-half inch in 10 feet from AAAA to B, placing
gulleys at AAAA and the catch basins at ccc. When this method is
not practicable, adopt such a grade (but one not exceeding 2| per cent)

342

HIGHWAY CONSTRUCTION

that the rectangle AAAA shall appear to he nearly level ; hut to secure
this it must actually have a considerable dip in the direction of the
slope of the street. If steep grades are continued 'across intersections,
they introduce side slopes in the streets thus crossed, which are trouble-
some, if not dangerous, to vehicles turning the corners, especially the
upper ones. Such intersections are especially objectionable in rainy

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Fig. 48.

weather. The storm water will fall to the lowest point, concentrating
a large quantity of water at two receiving basins, which, with a broken
grade, could be divided between four or more basins.

Fig. 48 shows the arrangement of intersections in steep grades
adapted for the streets of Duluth, Minn. From this it will be seen
that at these intersections the grades are flattened to three per cent for
the width of the roadway of the intersecting streets, and that the grade
of the curbs is flattened to eight per cent for the width of the intersecting
sidewalks. Grades of less amount on. roadway or sidewalk are con-
tinuous. The elevation of block-corners is found by adding together
the curb elevations at the faces of the block-corners, and 2£ per cent of

343

7(5

HIGHWAY CONSTRUCTION

the sum of the widths of the two sidewalks at the corner, and dividing
the whole by two. This gives an elevation equal to the average eleva-
tion of the curbs at the corners, plus an average rise of two and one-
half per cent across the width of the sidewalk.

Accommodation summits have to be introduced between street
intersections — first, in hilly localities, to avoid excessive excavation;
and second, when the intersecting streets are level or nearly so, for the
purpose of obtaining the fall necessary for surface drainage.

The elevation and location of these summits may be calculated
as follows : Let A be the elevation of the highest corner; B, the eleva-
tion of the lowest corner; D, the distance from corner to corne?;
and II, the rate of the accommodation grade. The elevation of the
summit is equal to

D-R + A + B.

The distance from A or B is found by subtracting the elevation of
either A or B from this quotient, and dividing the result by the rate
of grade. Or the summits may be located mechanically by specially
prepared scales. Prepare two scales divided to correspond to the rate
of grade; that is, if the rate of grade be 1 foot per 100 feet, then one
division of the scale should equal 100 feet on the map scale. These
divisions may be subdivided into tenths. One scale should read from
right to left, and one from left to right.

To use the scales, place them on the map so that their figures
correspond with the corner elevations; then, as the scales read in op-

Curb Le</el_

— — ••* ~"~ £oftom of Gutter **'*••«*

Fig. 49.

posite directions, there is of course some point at which the opposite
readings will be the same: this point is the location of the summits;
and the figures read off the scale its elevation. If the difference in
elevation of the corners is such as not to require an intermediate sum-
mit for drainage, it will be apparent as soon as the scales are placed
in position.

When an accommodation summit is employed, it should be form-
ed by joining the two straight grade lines by a vertical curve, as

344

HIGHWAY CONSTRUCTION

described in Part I. The curve should be used both in the crown of
the street and in the curb and footpath.

Where the grade is level between intersections, sufficient fall for
surface drainage may be secured without the aid of accommodation
summits, by arranging the grades as shown in Fig. 49. The curb is
set level between the corners; a summit is formed in the gutter; and
receiving basins are placed at each corner.

Transverse Grade. In transverse grade the street should be
level; that is, the curbs on opposite sides should be at the same level,
and the street crown rise equally from each side to the center. But in
hillside streets this condition cannot always be fulfilled, and opposite

Fig. 52.

sides of the street may differ as much as five feet; in such cases the
engineer will have to use his discretion as to whether he shall adopt a
straight slope inclining to the lo\ver side, thus draining the whole street
by the lower gutter, or adopt the three-curb method and sod the slope
of the higher side.

In the improvement of old streets with the sides at different levels,
much difficulty will be met, especially where shade trees have to be
spared. In such cases, recognized methods have to be abandoned, and
the engineer will have to adopt methods of overcoming the difficulties
in accordance with the conditions and necessities of each particular
case. Figs. 50, 51, and 52 illustrate several typical arrangements in

345

78 HIGHWAY CONSTRUCTION

the case of streets in which the opposite sides are at different levels.
Transverse Contour or Crown. The reason for crowning a pave-
ment— i. e., making the center higher than the sides — is to provide
for the rapid drainage of the surface. The most suitable form for the
crown is the parabolic curve, which may be started at the curb line,
or at the edge of the gutter adjoining the carriage-way about one foot

Fig. 53.

from the curb. Fig. 53 shows this form, which is obtained by dividing
the ordinate or width from the gutter to the center of the street into ten
equal parts, and raising perpendiculars the length of which will be
determined by multiplying the rise at the center by the respective
number of each perpendicular in the diagram. The amounts thus
obtained can be added to the rod readings; and the stakes, set at the
proper distance across the street, with their tops at this level, will give
the required curve.

The amount of transverse rise, or the height of the center alx>ve
the gutters, varies with the different paving materials, smooth pave-
ments requiring the least, and rough ones and earth the greatest. The
rise is generally stated in a proportion of the width of the carriage-way.
The most suitable proportions are:

Stone blocks, rise at center, Js width of carriage-way.

Wood » " " " TJn " "

Brick " " " " ,,V "

Asphalt" " " " tlff

Sub-Foundation Drainage of Streets. The sub-foundation
drainage of streets cannot be effected by transverse drains, because of
their liability to disturbance by the introduction of gas, water, and
other pipes.

Longitudinal drains must be depended upon entirely; they may
be constructed of the same materials and in the same manner as road
drains. The number of these longitudinal drains must tlepend upon

346

HIGHWAY CONSTRUCTION 79

the character of the soil. If the soil is moderately retentive, a single
row of tiles or a .hollow invert placed under the sewer in the center of
the street will generally be sufficient; or two rows of tiles may l)e em-
ployed, one placed outside each curb line; if, on the other hand, the
soil is exceedingly wet and the street very wide, four or more lines
may be employed. These drains may be permitted to discharge into
the sewers of the transverse streets.

Surface Drainage. The removal of water falling on the street
surface is provided for by collecting it in the gutters, from which it is
discharged into the sewers or other channels by means of catch-basins
placed at all street intersections and dips in the street grades.

Gutters. The gutters must be of sufficient depth to retain all the
water which reaches them and prevent its overflowing on the footpath.
The depth should never be less than 6 inches, and very rarely need be
more than 10 inches.

Catch=basins are of various forms, usually circular or rectangular,
built of brick masonry coated with a plaster of Portland cement.
Whichever form is adopted, they should fulfil the following conditions:

(1 ) The inlet and outlet should have sufficient capacity to receive
and discharge all water reaching the basin.

(2) The basins should have sufficient capacity below the outlet
to retain all sand and road detritus, and prevent it being carried into
the sewer.

(3) They should be trapped so as to prevent the escape of sewer
gas. (This requirement is frequently omitted, to the detriment of
the health of the people.)

(4) They should be constructed so that the pit can easily be
cleaned out.

(5) The inlet should be so constructed as not easily to be choked
by leaves or debris.

(6) They must offer the least possible obstruction to traffic.

(7) The pipe connecting the basin to the sewer should be easily
freed of any obstruction.

* The bottom of the basins should be 6 or 8 feet below the street
level ; and the water level in them should be from 3 to 4 feet lower than
the street surface, as a protection against freezing.

The capacity and number of basins will depend upon the area of
surface which they drain

347

80 HIGHWAY CONSTRUCTION

In streets having level or light longitudinal grades, gullies may be
formed along the line of the gutter at such intervals as may be found
necessary.

Catch-basins are usually placed at the curb line. In several cities,
the basin is placed in the center of the street, and connects to
inlets placed at the curb line. This reduces the cost of construction
and cleaning, and removes from the sidewalk the dirty operations of
cleaning the basins.

Catch-basins and gully-pits require to be cleaned out at frequent
intervals; otherwise the odor arising from the decomposing matter
contained in them will be very offensive. No rule can be laid down
for the intervals at which the cleaning should be done, but they must
be cleaned often enough to prevent the matter in them from putrefying.
There is no uniformity of practice observed by cities in this matter; in
some, the cleaning is done but once a year; in others, after every rain-
storm; in still others, at intervals of three or four months; wyhile in a
fewr cities the basins are cleaned out once a month.

FOUNDATIONS

• The stability, permanence, and maintenance of any pavement
depend upon its foundation. If the foundation is weak, the surface
will soon settle unequally, forming depressions and ruts. With a good
foundation, the condition of the surface will depend upon the material
employed for the pavement and upon the manner of laying it.

The essentials necessary to the forming of a good foundation are :

(1) The entire removal of all vegetable, perishable, and yielding
matter. It is of no use to lay good material on a bad substratum.

(2) The drainage of the subsoil wherever necessary. A per-
manent foundation can be secured only by keeping the subsoil dry;
for, where \vater is allowed to pass into and through it, its weak spots
will be quickly discovered and settlement will take place.

(3) The thorough compacting of the natural soil by .rolling with
a roller of proper weight and shape until it forms a uniform and un-
yielding surface.

(4) The placing on the natural soil so compacted, a sufficient
thickness of an impervious and incompressible material to cut off all
communication between the soil and the bottom of the pavement.

The character of the natural soil over \vhich the roadwTay is to be
built has an important bearing upon the kind of foundation and the
manner of forming it; each class of soil will require its own special

348

HIGHWAY CONSTRUCTION 81

treatment. Whatever its character, it must be brought to a dry and
tolerably hard -condition by draining and rolling. Sand and gravels
which do not hold water, present no difficulty in securing a solid and
secure foundation; clays and soils retentive of water are the most
difficult. Clay should be excavated to a depth of at least 18 inches
below the surface of the finished covering; and the space so excavated
should be filled in with sand, furnace slag, ashes, coal dust, oyster
shells, broken brick, or other materials which are not excessively absorb-
ent of water. A clay soil or one retaining water may be cheaply and
effectually improved by laying cross-drains with open joints at inter-
vals of 50 or 100 feet. These drains should be not less than 18 inches
below the surface, and the trenches filled with gravel. They should
be 4 inches in internal diameter, and should empty into longitudinal
drains.

Sand and planks, gravel, and broken stone have been successively
used to form the foundation for pavements; but, although eminently
useful materials, their application to this purpose has always been a
failure. Being inherently weak and possessing no cohesion, the main
reliance for both strength and wear must be placed upon the surface-
covering. This covering — usually (except in case of sheet asphalt)
composed of small units, with joints between them varying from one-
half an inch to one and a-half inches — possesses no elements of cohe-
sion; and under the blows and vibrations of traffic the independent
units or blocks will settle and be jarred loose. On account of their
porous nature, the subsoil quickly becomes saturated with urine and
surface waters, which percolate through the joints; winter frosts up-
heave them; and the surface of the street becomes blistered and broken
up in dozens of places.

Concrete. As a foundation for all classes of pavement (broken
stone excepted), hydraulic-cement concrete is superior to any other.
When properly constituted and laid, it becomes a solid, coherent mass
capable of bearing great weight without crushing. If it fail at all, it
must fail altogether. The concrete foundation is the most costly, but this
is balanced by its permanence and by the saving in the cost of repairs to
the pavement which it supports. It admits of access to subterranean
pipes with less injury to the neighboring pavement than any other, for
the concrete may be broken through at any point without unsettling
the foundation for a considerable distance around it, as is the case with

349

82 HIGHWAY CONSTRUCTION

sand or other incoherent material; and when the concrete is replaced
and set, the covering may be reset at its proper level, without the un-
certain allowance for settlement which is necessary in other cases.

Thickness of Concrete. The thickness of the concrete bed must
be proportioned by the engineer; it should be sufficient to provide
against breaking under transverse strain caused by the settlement of
the subsoil. On a well-drained soil, six inches will be found sufficient;
but in moist and clayey soils, twelve inches will not be excessive. On
such soils a layer of sand or gravel, spread and compacted before pla-
cing the concrete, will be found very beneficial.

The proportions of the ingredients for concrete used for pavement
foundations are usually:

1 part Portland cement
3 parts Sand
7 parts Broken Stone.
Or,

1 part Natural Hydraulic Cement

2 parts Sand

5 parts Broken Stone.

The question is sometimes raised as to whether Natural or Port-
land cement should be used. Natural cement is more extensively
employed on account of its being cheaper in price than Portland.
There is no advantage gained in using Portland cement. Concrete
should not be laid when the temperature falls below 32° F.

The concrete foundation, after completion, should be- allowed to
remain several days before the pavement is placed upon it, in order
that the mortar may become entirely set. During setting, the con-
crete should be protected from the drying action of the sun and
wind, and should be kept damp to prevent the formation of drying
cracks.

STONE BLOCK PAVEMENTS

Stone blocks are commonly employed for pavements where traffic
is heavy. The material of which the blocks are made should possess
sufficient hardness to resist the abrasive action of traffic, and sufficient
toughness to prevent them from being broken by the impact of loaded
wheels. The hardest stones will not necessarily give the best re-
sults in the pavement, since a very hard stone usually wears smooth
and becomes slippery. The edges of the block chip off, and the

350

HIGHWAY CONSTRUCTION

upper face becomes rounded, thus making the pavement very
rough.

The stone is sometimes tested to determine its strength, resistance
to abrasion, etc. ; but, as the conditions of use are quite different from
those under which it may be tested, such tests are seldom satisfactory.
However, examination of a stone as to its structure, the closeness of its
grain, its homogeneity, porosity, etc., may assist in forming an idea of
its value for use in a pavement. A low degree of permeability usually
indicates that the material will not be greatly affected by frost.

Materials. — Granite. Granite is more extensively employed for
stone block paving than any other variety of stone; and because of this
fact, the term "granite paving" is generally used as being synonymous
with stone block paving. The granite employed should be of a tough,
homogeneous nature. The hard, quartz granites are usually brittle,
and do not wear well under the blows of horses' feet or the impact of
vehicles; granite containing a high percentage of feldspar will be inju-
riously affected by atmospheric changes; and granite in which mica
predominates will wear rapidly on account of its laminated structure.
Granite possesses the very important property of splitting in three
planes at right angles to one another, so that paving blocks may readily
be formed with nearly plane faces and square corners. This property
is called the rift or cleavage.

Sandstones of a close-grained, compact nature often give very
satisfactory results under heavy traffic. They are less hard than
granite, and wear more rapidly, but do not become smooth and slip-
pery. Sandstones are generally known in the market by the name
of the quarry or place where produced as "Medina," "Berea,"
etc.

Trap rock, while answering well the requirements as to durability
and resistance to wear, is objectionable on account of its tendency to
wear smooth and become slippery; it is also difficult to break into
regular shapes.

Limestone has not usually been successful in use for the construc-
tion of block pavements, on account of its lack of durability against
atmospheric influences. The action of frost commonly splits the
blocks; and traffic shivers them, owing to the lamination being
vertical.

851

HIGHWAY CONSTRUCTION

TABLE 12.

Specific Gravity, Weight, Resistance to Crushing, and
Absorption Power of Stones.

MATERIAL

SPECIFIC
GRAVITY

WEIGHT
Pounds
per cu. ft.

RESISTANCE
TO CRUSHING
Pounds
per sq. in.

PERCENTAGE

OF WATER
ABSOHBEU

Granite —
Maximum

2.80

176

35.000

0.155

Minimum
Trap —
Maximum
Minimum
Sandstone —
Maximum

2.60

3.03
2.86

2.75
2.23

163

178
189

170
137

12,000

24.000
19.000

18.000
5,000

0.086

0.019
0.000

5.4PO
0.410

Limestone —
Maximum
Minimum
Brick Paving —

2.75
1.90

1.95

175

118

20.000
7,000

20.000

5.000
0.200

Minimum

2.55

10.000

Cobblestone Pavement. Cobblestones bedded in sand possess
the merit of cheapness, and afford an excellent foothold for horses;
but the roughness of such pavements requires the expenditure of a
large amount of tractive energy to move a load over them. Aside from
this, cobblestones are entirely wanting in the essential requisites of a
good pavement. The stones being of irregular size, it is almost impos-
sible to form a bond or to hold them in place. Under the action of the
h affic and frost, the roadway soon becomes a mass of loose stones.
Moreover, cobblestone pavements are difficult to keep clean, and very
unpleasant to travel over.

Belgian Block Pavement. Cobblestones were displaced by pave-
ments formed of small cubical blocks of stone. This type of
pavement was first laid in Brussels, thence imported to Paris, and from
there taken to the United States, where it has been widely known as
the "Belgian block" pavement. It has been largely used in New York
City, Brooklyn, and neighboring towns, the material being trap-rock
obtained from the Palisades on the Hudson River.

The stones, being of regular shape, remain in place better than
cobblestones; but the cubical form (usually five inches in each dimen-
sion) is a mistake. The foothold is bad; the stones wear round; and
the number of joints is so great that ruts and hollows are quickly
formed. This pavement offers less resistance to traction than cobble-
stones, but it is almost equally rough and noisy.

Granite Block Pavement. The Belgian block has been gradually

353

HIGHWAY CONSTRUCTION 85

displaced by the introduction of rectangular blocks of granite. Blocks
of comparatively large dimensions were at first employed. They were
from 6 to 3 inches in width on the surface, from 10 to 20 inches in
length, with a depth of 9 inches. They were merely placed in rows
on the subsoil, perfunctorily rammed, the joints filled with sand, and
the street thrown open to traffic. The unequal settlement of the
blocks, the insufficiency of the foothold, and the difficulty of cleansing
the street, led to the gradual development of the latest type of stone-
block pavement, which consists of narrow, rectangular .blocks of
granite, properly proportioned, laid on an unyielding and impervious
foundation, with the joints between the blocks filled with an imper-
meable cement.

Experience has proved beyond doubt that this latter type of
pavement is the most enduring ai d economical for roadways subjected
to heavy and constant traffic. Its advantages are many, while its
defects are few.

Advantages.

(1) Adaptability to all grades.

(2) Suits all classes of traffic.

(3) Exceedingly durable.

(4) Foothold, fair.

(5) Requires but little repair.

(6) Yields but little dust or mud.

(7) Facility for cleansing, fair.
Defects.

(1) Under certain conditions of the atmosphere, the surface of
the pavement becomes greasy and slippery.

(2) The incessant din and clatter occasioned by the movement
of traffic is an intolerable nuisance; it is claimed by many physicians
that the noise injuriously affects the nerves and health of persons who
are obliged to live or do business in the vicinity of streets so paved.

(3) Horses constantly employed upon it soon suffer from the
continual jarring produced in their legs and hoofs, and quickly wear
out.

(4) The discomfort of persons riding over the pavement is very
great, because of the continual jolting to which they are subjected.

(5) If stones of an unsuitable quality are used — for example,

353

86 HIGHWAY CONSTRUCTION

those that polish — the surface quickly becomes slippery and exceed-
ingly unsafe for travel.

Size and Shape of Blocks. The proper size of blocks for paving
purposes has been a subject of much discussion, and a great variety of
forms and dimensions are to be found in all cities.

For stability, a certain proportion must exist between the depth,
the length, and the breadth. The depth must be such that when the
wheel of a loaded vehicle passes over one edge of the upper surface
of a block, the block will not tend to tip up. The resultant direction
of the pressure of the load and adjoining blocks should always tend
to depress the whole block vertically ; where this does not happen, the
maintenance of a uniform surface is impossible. To fulfil this require-
ment, it is not necessary to make the block more than six inches deep.

Width of Blocks. The maximum width of blocks is controlled
by the size of horses' hoofs. To afford good foothold to horses draw-
ing heavy loads, it is necessary that the width of each block, measured
along the street, shall be the least possible consistent with stability.

Gutter formed of 3 rows of*
ti/ocks, Set long/tuciirialy

Fig. 54.

If the width be great, a horse drawing a heavy load, attempting to find
a joint, slips back, and requires an exceptionally wide joint to pull him
up. It is therefore desirable that the width of a block shall not exceed
3 inches ; or that four blocks, taken at random and placed side by side,
shall not measure more than 14 inches.

Length of Blocks, The length, measured across the street,
must be sufficient to break joints properly, for two or more joints in
line lead to the formation of grooves. For this purpose the length
of the block should be not less than 9 inches nor more than 12 inches.

Form of Blocks. The blocks should be well squared, and must
not taper in any direction ; sides and ends should be free from irregular
projections. Blocks that taper from the surface downwards (wedge-
shaped) should not be permitted in the work; but if any are allowed,
they should be set with the widest side down.

854

HIGHWAY CONSTRUCTION

Alanner of Laying Blocks. The blocks should be laid in parallel
courses, with their longest side at right angles to the axis of the street,
and the longitudinal joints broken by a lap of at least two inches (see
Figs. 54 and 55). The reason for this is to prevent the formation of
longitudinal ruts, which would happen if the blocks were laid length-
wise. Laying blocks obliquely and "herring-bone" fashion has been

tried in several cities, with the idea that the wear and formation of ruts
would be reduced by having the vehicle cross the blocks diagonally.
The method has failed to give satisfactory results; the wear was ir-
regular and the foothold defective; the difficulty of construction was
increased by reason of labor required to form the triangular joints; and
the method was wasteful of material.

Fig. 56.

The gutters should be formed by three or more courses of block,
laid with their length parallel to the curb-

At junctions or intersections of streets, the blocks should be laid
diagonally from the center, as shown in Fig. 56. The reasons for

855

88 HIGHWAY CONSTRUCTION

this are: (1) To prevent the traffic crossing the intersection from
following the longitudinal joints and thus forming depressions and
ruts; (2) Laid in this manner, the blocks afford a more secure foot-
hold for horses turning the corners. The ends of the diagonal blocks
where they abut against the straight blocks, must be cut to the re-
quired bevel.

The blocks forming each course must be of the same depth, and
no deviation greater than one-quarter of an inch should be permitted.
The blocks should be assorted as they are delivered, and only those
corresponding in depth and width should be used in the same course.
The better method would be to gauge the blocks at the quarry.
This would lessen the cost considerably ; it would also avoid the in-
convenience to the public due to the stopping of travel because of the
rejection of defective material on the ground. This method would
undoubtedly be preferable to the contractor, who would be saved the
expense of handling unsatisfactory material ; and it would also leave
the inspectors free to pay more attention to the manner in which the
work of paving is performed.

The accurate gauging of the blocks is a matter of much impor-
tance. If good work is to be executed, the blocks, when laid, must be
in parallel and even courses; and if the blocks be not accurately gauged
to one uniform size, the result will be a badly paved street, with the
courses running unevenly. The cost of assorting blocks into lots of
uniform wjdth, after delivery on the street, is far in excess of any ad-
ditional price which would have to be paid for accurate gauging at the
quarry.

Foundation. The foundation of the blocks must be solid and
unyielding. A bed of hydraulic-cement concrete is the most suitable,
the thickness of which must be regulated according to the traffic; the
thickness, however, should not be less than 4 inches, and need not be
more than 9 inches. A thickness of 6 inches will sustain traffic of 600
tons per foot of width.

Cushion Coat. Between the surface of the concrete and the base
of the blocks, there must be placed a cushion-coat formed of an incom-
pressible but mobile material, the particles of which will readily adjust
themselves to the irregularities of the bases of the blocks and transfer
the pressure of the traffic uniformly to the concrete below. A layer
pf dry, clean sand 1 to 2 inches thick forms an excellent cushion-coat-

356

HIGHWAY CONSTRUCTION 89

Its particles must be of such fineness as to pass through a No. 8 screen;
if coarse and containing pebbles, they will not adapt themselves to the
irregularities of the bases of the blocks; hence the blocks will be sup-
ported at only a few points, and unequal settlement will take place when
the pavement is subjected to the action of traffic. The sand must also
be perfectly free from moisture, and artificial heat must be used to dry
it if necessary. This requirement is an absolute necessity. There
should be no moisture below the blocks when laid ; nor should water
be allowed to penetrate below the blocks; if such happens, the effect of
frost will be to upheave the pavement and crack the concrete.

Where the best is desired without regard to cost, a layer half an
inch thick of asphaltic cement may be substituted for the sand, with
superior and very satisfactory results.

Laying Blocks. The blocks should be laid stone to stone, so that
the joint may be of the least possible width; wide joints cause increased
wear and noise, and do not increase the foothold. The courses should
be commenced on each side and worked toward the middle; and the
last stone should fit tightly.

Ramming. After the blocks have been set, they should be well
rammed down ; and the stones which sink below the general level
should be taken up and replaced with a deeper stone or brought to
level by increasing the sand bedding.

The practice of workmen is invariably to use the rammer so as to
secure a fair surface. This is not the result intended to be secured,
but to bring each block to an unyielding bearing. The result of such
a surfacing process is to produce an unsightly and uneven roadway
when the pressure of traffic is brought upon it. The rammer used
should weigh not less than 50 pounds and have a diameter of not less
than 3 inches.

Joint Filling. All stone block pavements depend for their water-
proof qualities upon the character of the joint filling. Joints filled
with sand and gravel are of course pervious. A grout of lime or cement
mortar does not make a permanently waterproof joint; it becomes
disintegrated under the vibration of traffic. An impervious joint can
be made only by employing a filling made from bituminous or asphaltic
material; this renders the pavement more impervious to moisture,
makes it less noisy, and adds considerably to its strength.

Bituminous Cement for Joint Filling. The bituminous materials

357

90 HIGHWAY CONSTRUCTION

employed are: (1) The tar produced in the manufacture of gas,
which, when redistilled, is called distillate, and is numbered 1, 2, 3, 4,
etc., according to its density; this, material under the name of paving-
pitch is extensively used, both alone and in combination with other
bituminous substances; (2) Combinations of gas tar or coal tar with
refined asphaltum; (3) Mixtures of refined asphaltum, creosote, and
coal tar.

The formula for the bituminous joint rilling used in New York
City, is:

Refined Trinidad asphaltum 20 parts.

No. 4 coal-tar distillate 100 parts.

Residuum of petroleum 3 parts.

In Washington, D. C., coal tar distillate No. 6 is used alone.

In Europe a bituminous cement much used is composed of coal-
tar, asphaltum, gas tar, and creosote oil, in the proportion of 100
pounds of asphaltum to 4 gallons of tar and 1 gallon of creosote.
These proportions are varied somewhat, according to the quality of the
asphaltum employed. The mixture is melted, and is boiled from
one to two hours in a suitable boiler, being then poured into the joints
in a boiling state. This mixture is impervious to moisture, and pos-
sesses a degree of elasticity sufficient to prevent it from cracking.

The mode of applying the bituminous cement is as follows : After
the blocks are rammed, the joints are filled to a depth of about two
inches with clean gravel heated to a temperature of about 250° F. ;
then the hot cement is poured in until it forms a layer of about one inch
on top of the gravel; then more gravel is filled in to a depth of about
two inches; then cement is poured in until it appears on top of the
gravel, more gravel being next added until it reaches to within half an
inch of the top of the blocks; this remaining half-inch is filled with
cement, and then fine gravel or sand is sprinkled over the joints.

In some cases the joints are first filled with heated gravel; the
cement is poured in until the sand beneath and the gravel between
the blocks will absorb no more, and the joints are filled flush with the
top of the pavement. This method is open to objection; for, if the
gravel is not sufficiently hot, the cement will be chilled and will not flow
to the bottom of the joint, but, instead, will form a thin layer near the
surface, which under the action of frost and the vibration of traffic,
will be quickly cracked and broken up; the gravel will settle, and the

358

A GOOD EXAMPLE OF ROAD-BUILDING ON A STEEP, ROCKY HILLSIDE
View near Eide, on Hardanger Fjord, Norway.

HIGHWAY CONSTRUCTION

blocks will be jarred loose, the surface of the pavement becoming a
series of ridges and hollows.

The quantity of cement required per square yard of pavement will
vary according to the shape of the blocks, the width of the joints, and
the depth of the sand bed. With well-shaped blocks, close joints, and
a half-inch sand bed, the quantity will vary from 3V to 5 gallons; with
ill-shaped blocks, wide joints, and a heavy sand bed, 10 to 12 gallons

Fig. 57.

would not be an excessive amount to use to secure the result obtained
by employing well-shaped blocks and close joints.

Stone Pavement on Steep Grades. Stone blocks may be em-
ployed on all practicable grades; but on grades exceeding 10 per cent,
cobblestones afford a better foothold than blocks. The cobblestones
should be of uniform length, the length being at least twice the breadth
— say stones 6 inches long and 2^ to 3 inches in diameter. These
should be set on a concrete foundation, laid stone to stone, and the

Fig. 58.

interstices filled with cement grout or bituminous cement; or a bitu-
minous concrete foundation may be employed and the interstices be-
tween the stones filled with asphaltic paving cement. Should stone
blocks be preferred, they must be laid, when the grade exceeds 5 per
cent, with a serrated surface, by either of the methods shown in Figs.
57 and 58. The method shown in Fig. 57 consists in slightly tilting
the blocks on their bed so as to form a series of ledges or steps, against
which the horses' feet being planted, a secure foothold is obtained.
The method shown in Fig. 58 consists in placing between the rows of

359

92 HIGHWAY CONSTRUCTION

stones a course of slate, or strips of creosoted wood, rather less than
one inch in thickness and about an inch less in depth than the blocks;
or the blocks may be spaced about one inch apart, and the joints filled
with a grout composed of gravel and cement. The pebbles of the
gravel should vary in size between one-quarter and three-quarters of
an inch.

BRICK PAVEMENTS

Characteristics of Brick Suitable for Paving. These are:

(1) Not to be acted upon by acids.

(2) Not to absorb more than 1-600 of its weight of water in
48 hours.

(3) Not susceptible to polish.

(4) Rough to the touch, resembling fine sandpaper.

(5) To give a clear, ringing sound when struck together.

(6) When broken, to show a compact, uniform, close-grained,
structure, free from air-holes and pebbles.

(7) Not to scale, spall, or chip when quickly struck on the edges.

(8) Hard but not brittle.

Tests of Paving Brick. To ascertain the quality of paving brick,
they are generally subjected to four tests, namely: (1) Abrasion by
impact (commonly called the "Rattler" test); (2) absorption; (3)
transverse or cross-breaking; (4) crushing. With a view to securing
uniformity in the methods of making the above tests, the National
Brick Manufacturers' Association has adopted and recommends the
following:

Rattler Test

1. Dimensions of the Machine. The standard machine shall
be 28 inches in diameter and 20 inches in length, measured inside the
chamber.

Other machines may be used, varying in diameter between 26 and
30 inches, and in length between 18 and 24 inches; but if this is done,
a record of it must be attached to the official report. I^ong rattlers
may be cut up into sections of suitable length by the insertion of an
iron diaphragm at the proper point.

2 Construction of the Machine. The barrel shall be supported
on trunnions at either end; in no case shall a shaft pass through the
rattling chamber. The cross-section of the barrel shall be a regular
polygon having 14 sides. The heads shall be composed of gray cast-

HIGHWAY CONSTRUCTION 93

iron, not chilled or case-hardened. The staves shall preferably be
composed of steel plates, since cast-iron peans and ultimately breaks
under the wearing action on the inside. There shall be a space of one-
fourth of an inch between the staves, for the escape of dust and small
pieces of waste. Other machines may be used, having from twelve to
sixteen staves; but if this is done, a record of it must be attached to
the official report of the test.

3. Composition of the Charge. All tests must be executed on
charges containing but one make of brick or block at a time. The
charge shall consist of 9 paving blocks or 12 paving bricks, together
with 300 pounds of shot made of ordinary machinery cast-iron. This
shot shall be of two sizes, as described below; and the shot charge shall
be composed one-fourth (75 pounds) of the larger size, and three-
fourths (225 pounds) of the smaller size.

4. Size of the Shot. The larger size shall weigh about 1\ pounds
and be about 2i inches square and 4J- inches long, with slightly round-
ed edges. The smaller size shall be cubes of \\ inches on a side, with
rounded edges. The individual shot shall be replaced by new .ones
when they have lost one-tenth of their original weight.

5. Revolutions of the Charge. The number of revolutions of a
standard test shall be 1,800; and the speed of rotation shall not fall
below 28 nor exceed 30 revolutions per minute. The belt-power shall
be sufficient to rotate the rattler at the same speed, whether charged
or empty.

(>. Condition of the Charge. The bricks composing a charge
shall be thoroughly dried before making the test.

7. Calculation of the Results. The loss shall be calculated in
per cents of the weight of the dry brick composing the charge; and no
result shall be considered as official unless it is the average of two
distinct and complete tests made on separate charges of brick.
Absorption Test

1. The number of bricks for a standard test shall be five.

2. The test must be conducted on rattled brick. If none such
are available, the whole brick must be broken in halves before treatment.

3. Dry the bricks for 48 hours at a temperature ranging from
230° to 250° F. before weighing for the official dry weight.

4. Soak for 48 hours completely immersed in pure water.

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94 HIGHWAY CONSTRUCTION

5. After soaking, and before weighing, the bricks must be wiped
dry from surplus water.

6. The difference in the weight must be determined on scales
sensitive to one gram.

7. The increase in weight due to water absorbed shall be ca.-
culated in per cents of the initial dry7 weight.

Cross=Breaking Test

1. Support the brick on edge, or as laid in the pavement, on
hardened steel knife-edges, rounded longitudinally to a radius of
twelve inches and transversely to a radius of one-eighth inch, and
bolted in position so as to secure a span of six inches.

2. Apply the load to the middle of the top face through a hard-
ened steel knife-edge, straight longitudinally and rounded transversely
to a radius of one-sixteenth inch.

3. Apply the load at a Uniform rate of increase till fracture
ensues.

4. Compute the modulus of rupture by the formula

' 2bd?'

in which / = modulus of rupture, in pounds per square inch ;
w = total breaking load, in pounds;
/ = length of span, in inches = 6 ;
b = breadth of brick, in inches-
d = depth of brick, in inches.

5. Samples for test must be free from all visible irregularities of
surface or deformities of shape, and their upper and lower faces must
be practically parallel.

6. Not less than ten brick shall be broken, and the average of all
shall be taken for a standard test.

Crushing Test

1. The crushing test should be made on half-bricks, loaded
edgewise, or as they are laid in the street. If the machine used is
unable to crush a full half-brick, the area may be reduced by chipping
off, keeping the form of the piece to be tested as nearly prismatic as
possible. A machine of at least 100,000 pounds' capacity should be
used; and the specimen should not be reduced below four square
inches of area in cross-section at right angles to direction of load.

2. The upper and lower surfaces should preferably be ground to

HIGHWAY CONSTRUCTION 95

true and parallel planes. If this is not done, they should be bedded,
while in the testing machine, in plaster of Paris, which should be
allowed to harden ten minutes under weight of the crushing planes
only, before the load is applied.

3. The load should be applied at a uniform rate of increase to
the point of rupture.

4. Not less than an average obtained from five tests on five
different bricks shall constitute a standard test.

Properties of Paving Bricks. Paving bricks range in weight
from 5?, to 7\ pounds; in specific gravity, from 1.91 to 2.70; in resist-
ance to crushing, from 7,000 to 18,000 pounds per square inch; in
resistance to cross-breaking, R = 1,400 to 2,000 pounds; in absorption,
from 0.15 to 3 per cent in 24 hours. The dimensions vary according to
locality and the requirements of the specifications. The "standard"
bricks are 21 X 4 X 8 inches, requiring 58 bricks to the square yard,
and weighing 7 pounds each; "repressed", 2£ X 4 X8£ inches, requir-
ing 61 to the square yard, and weighing 6V pounds each; "Metropoli-
tan", 3X4X9 inches, requiring 45 to the square yard, and weighing
9 2 pounds each.

Advantages of Brick Pavements. These may be stated as follows:

(1) Ease of traction.

(2) Good foothold for horses.

(3) Not disagreeably noisy.

(4) Yields but little dust and mud.

(5) Adapted to all grades.

(6) Easily repaired.

(7) Easily cleaned.

(8) But slightly absorbent.

(9) Pleasing, to the eye.

(10) Expeditiously laid.

0 1) Durable under moderate traffic.

Defects of Brick Pavements. The principal defects of brick
pavements arise from lack of uniformity in the quality of the bricks,
and from the liability of incorporating in the pavement bricks of too
soft or porous a structure, which crumbles under the action of traffic
or frost.

Foundation. A brick pavement should have a firm foundation.
As the surface is made up of small, independent blocks, each one must

363

HIGHWAY CONSTRUCTION

be adequately supported, or the load coming upon it may force it
downwards and cause unevenness, a condition which conduces to the
rapid destruction of the pavement. Several forms of foundation have
been used — such as gravel, plank, sand, broken stone, and concrete.
The last mentioned is doubtless the best.

Sand Cushion. The sand cushion is a layer of sand placed on
top of the concrete to form a bed for the brick. Practice regarding
the depth of this layer of sand varies considerably. In some cases it
is only half an inch deep, varying from this up to three inches. The
sand cushion is very desirable, as it not only forms a perfectly true and
even surface upon which to place brick, but also makes the pavement
less hard and rigid than would be the case were the brick laid directly
on the concrete.

The sand is spread evenly, sprinkled with water, smoothed, and
brought to the proper contour by screeds or wooden templets, properly
trussed, mounted on wheels or shoes which bear upon the upper sur-
face of the curb. Moving the templet forward levels and forms the
sand to a uniform surface and proper shape.

The sand used for the cushion-coat should be clean and free from
loam, moderately coarse, and free from pebbles exceeding one-quarter
inch in size.

Manner of Laying. The bricks should be laid oh edge, as closely
and compactly as possible, in straight courses across the street, with
the length of the bricks at right angles to the axis of the street. Joints
should be broken by at least 3 inches. None but whole bricks should
be used, except in starting a course or making a closure. To provide
for the expansion of the pavement, both longitudinal and transverse
expansion-joints are used, the former being made by placing a board
templet seven-eighths of an inch thick against the curb and abutting
the brick thereto. The transverse joints are formed at intervals
varying between 25 and 50 feet, by placing a templet or building-lath
three-eighths of an inch thick between two or three rows of brick.
After the bricks are rammed and ready for grouting, these templets are
removed, and the spaces so left are filled with coal-tar pitch or asphal-
tic paving cement. The amount of pitch or cement required will vary
between one and one and a-half pounds per square yard of pavement,
depending upon the width of the joints. After 25 or 30 feet of the
pavement is laid, every part of it should be rammed with a rammer

HIGHWAY CONSTRUCTION

weighing not less than 50 pounds; and the bricks which sink below the
general level should be removed, sufficient sand being added to raise
the brick to the required level. After all objectionable brick have
been removed, the surface should be swept clean, then rolled with a
steam roller weighing from 3 to 6 tons. The object of rolling is to
bring the bricks to an unyielding bearing with a plane surface; if this
is not done, the pavement will be rough and noisy and will lack dura-

Fig. 59.

bility. The rolling should be first executed longitudinally, beginning
at the crown and working toward the gutter, taking care that each
return trip of the roller covers exactly the same area as the preceding
trip, so that the second passage may neutralize any careening of the
brick due to the first passage.

The manner of laying brick at street intersections is shown in
Figi 59.

365

HIGHWAY CONSTRUCTION

Joint Filling. The character of the material used in filling the
joints between the brick lias considerable influence on the success and
durability of the pavement. Various materials have been used — such
as sand, coal-tar pitch, asphalt, mixtures of coal-tar and asphalt, and
Portland cement, besides various patented fillers; as "Murphy's
grout", which is made from ground slag and cement. Each material
has its advocates, and there is much difference of opinion as to which
gives the best results.

The best results seem to be obtained by using a high grade of
Portland cement containing the smallest amount of lime in its composi-
tion, the presence of the lime increasing the tendency of the filler to
swell through absorption of moisture, causing the pavement to rise
or to be lifted away from its foundation, and thus producing the roaring
or rumbling noise so frequently complained of.

The Portland cement grout, when uniformly mixed and carefully
placed, resists the impact of traffic and wears well with brick. When a
failure occurs/repairs can be made quickly; and, if made early, the
pavement 'will be restored to a good condition. If, however, repairs
are neglected, the brick soon loosens and the pavement fails.

The office of a filler is to prevent water from reaching the founda-
tion, and to protect the edges of the brick from spalling under traffic.
In order to meet both of these requirements, every joint must be filled
to the top, and must remain so, wearing down with the brick. Sand
does not meet these requirements. Although at first making a good
filler, being inexpensive and reducing the liability of the pavement to
be noisy, it soon washes out, leaving the edges of the brick unprotected
and consequently liable to be chipped. Coal-tar and the mixtures of
coal-tar and asphalt have an advantage in rendering a pavement less
noisy and in cementing together any breaks that may occur through
upheavals from frost or other causes ; but, unless made very hard, they
have the disadvantage of becoming soft in hot weather and flowing
to the gutters and low places in the pavement, there forming a black
and unsightly scale and leaving the high parts unprotected. The
joints, thus deprived of their filling, become receptacles for water, mud,
and ice in turn; and the edges of the brick are quickly broken down.
Some of these mixtures become so brittle in winter that they crack
and fly out of the joints under the action of traffic.

The Portland cement filler is prepared by mixing two parts of

366

HIGHWAY CONSTRUCTION

cement and one part of fine sand with sufficient water to make a thin
grout. The most convenient arrangement for preparing and dis-
tributing the grout is a water-tight wooden box carried on four wooden
wheels about 12 inches in diameter. The box may be about 4 feet
wide, 7 feet long, and 12 inches deep, furnished with a gate about 8
inches wide, in the rear end. The box should be mounted on the
wheels with an inclination, so that the rear end is about. 4 inches lower
than the front end.

The operation of placing the filler is as follows : The cement and
sand are placed in the box, and sufficient water is added to make a
thin grout. The box is located about 12 feet from the gutter, the end
gate opened, and about 2. cubic feet of the grout allowed to flow out
and run over the top of the brick (care being taken to stir the grout
while it is being discharged). If the brick are very "dry, the entire
surface of the pavement should be thoroughly wet with a hose before
applying the grout; if not, absorption of the water from the grout by
the bricks will prevent adhesion between the bricks and the cement
grout. The grout is swept into the joints by ordinary bass brooms.
After about 100 feet in length of the pavement has been covered the
box is returned to the starting-point, and the operation is repeated
with a grout somewhat thicker than the first. If this second applica-
tion is not sufficient to fill the joints, the operation is repeated as often
as may be necessary to fill them. If the grout has been made too thin,
or the grade of the street is so great that the grout will not remain long
enough in place to set, dry cement may be sprinkled over the joints and
swept in. After the joints are completely filled and inspected, allowing
three or four hours to intervene, the completed pavement should be
covered with sand to a depth of about half an inch, and the roadway
barricaded, and no traffic allowed on it for at least ten days.

The object of covering the pavement with sand is to prevent the
grout from drying or settling too rapidly; hence, in dry and windy
weather, it should be sprinkled from time to time. If coarse sand is
employed in the grout, it will separate from the cement during the
operation of filling the joints, with the result that many joints will be
filial with sand and very little cement, while others will be filled with
cement and little or no sand; thus there will be many spots in the pave-
ment in which no bond is foimed between the bricks, and under the
action of traffic these portions will quickly become defective.

367

100 HIGHWAY CONSTRUCTION

The coal-tar filler is best applied by pouring the material from
buckets, and brooming it into the joints with wire brooms. In order to
fill the joints effectually, it must be used only when very hot. To
secure this condition, a heating tank on wheels is necessary. It should
have a capacity of at least five barrels, and be kept at a uniform tem-
perature all day. One man is necessary to feed the fire and draw the
material into the buckets; another, to carry the buckets from the heat-
ing-tank to a third, who pours the material over the street. The latter
starts to pour in the center of the street, working backward toward the
curb, and pouring a strip about two feet in width. A fourth man,
with a wire broom, follows immediately after him, sweeping the sur-
plus material toward the pourer and in the direction of the curb. This
method leaves the entire surface of the pavement covered with a thin
coating of pitch, which should immediately be covered with a light
coating of sand; the sand becomes imbedded in the pitch. Under the
action of traffic, this thin coating is quickly worn away, leaving the
surface of the bricks clean and smooth.

Tools Employed in Construction of Block Pavements. The
principal tools required in constructing block pavements comprise
hammers and rammers of varying sizes and shapes, depending on the
material and size of the blocks to be laid; also crowbars, sand screens,
and rattan and wire brooms. Cobblestones, square blocks, and brick
require different types of both hammers and rammers for adjusting
them to place and forcing them to their seat. A cobblestone rammer,
for example, is usually made of wood (generally locust) in the shape of
a long truncated cone, banded with iron at top and bottom, weighing
about 40 pounds, and having two handles, o.ne at the top and another
'on one side. A Belgian block rammer is slightly heavier, consisting
of an upper part of wood set in a steel base; while a rammer for granite
blocks is still heavier, comprising an iron base with cast-steel face, into
which is set a locust plug with hickory handles. For laying brick, a
wooden rammer shod with cast iron or steel and weighing about 27
pounds is used. A light rammer of about 20 pounds' weight, consist'
ing of a metallic base attached to a long, slim wooden handle, is used
for miscellaneous work, such as tamping in trenches, next to curbs, etc.

Concrete=Mixing Machine. Where large quantities of concrete
are required, as in the foundations of improved pavements, concrete
can be prepared more expeditiously and economically by the use of

888

HIGHWAY CONSTRUCTION

101

mechanical mixers, and the ingredients will he more thoroughly mixed,
than by hand. Thorough incorporation of the ingredients is an essen-
tial element in the quality of a concrete. When mixed by hand, how-
ever, the incorporation is rarely complete, because it depends upon the
proper manipulation of the hoe and shovel. The manipulation,
although extremely simple, is rarely performed by the ordinary laborer
unless he is constantly watched by the overseer.

Several varieties of concrete-mixing machines are in the market.
A convenient portable type is illustrated in Fig. 60. The capacity of

Fig. 60. Concrete Mixing Machine.

the mixers ranges from five to twenty cubic yards per hour, depending
upon size, regularity with which the materials are supplied, speed, etc.

Gravel Heaters. Fig. 61 illustrates a device commonly employed
for heating the gravel used for joint filling in stone-block pavements.
These heaters are made in various sizes, a common size being 9 feet
long, 5 feet wide, and 3 feet 9 inches high.

Melting Furnaces, for heating the pitch or tar for joint filling, are
illustrated by Fig. 62. Various sizes are on the market.
WOOD PAVEMENTS

Wood pavements are formed of either rectangular or cylindrical
blocks of wood. The rectangular blocks are generally 3 inches wide,

102

HIGHWAY CONSTRUCTION

9 inches long, and G inches deep; the round blocks are commonly 6
inches in diameter and 6 inches long.

The kinds of wood most commonly used are cedar, cypress, juni-
per, yellow pine, and mesquite; and recently jarrah from Australia,
and pyinyado from India, have been used.

The wood is used in its natural condition, or impregnated with
creosote or other chemical preservative.

The blocks of wood are laid either .on the natural soil, on a bed
of sand and gravel, on a layer of broken stone, on a layer of concrete,

Melting Furnace.

or, sometimes, on a double layer of plank. The joints are filled either
with sand, paving-pitch, or Portland-cement grout.

Advantages. The advantages of wood pavement may be stated
as follows:

(1) It affords good foothold for horses.

(2) It offers less resistance to traction than stone, and slightly
more than asphalt.

(3) It suits all classes of traffic.

(4) It may be used on grades up to five per cent.

(5) It is moderately durable.

(6) It yields no mud when laid upon an impervious foundation.

(7) It yields but little dust.

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HIGHWAY CONSTRUCTION 103

(8) It is moderate in first cost.

(9) It is not disagreeably noisy.

Defects. The principal objections to wood pavement are:

(1) It is difficult to cleanse.

(2) tnder certain conditions of the atmosphere it becomes
greasy and very unsafe for horses.

(3) It is not easy to open for the purpose of gaining access to
underground pipes, it being necessary to remove rather a large surface
for this purpose, which has to be left a little time after being repaired
before traffic is again allowed upon it.

(4) It is absorbent of moisture.

(5) It is claimed by many that wood pavements are unhealthy.
Quality of Wood. The question as to which of the various kinds

of wood available is the most durable and economical, has not
been satisfactorily determined. Many varieties have been tried. In
England, preference is given to Baltic fir, yellow pine, and Swedish
yellow deal. In the United States the variety most used (on account
of its abundance and cheapness) is cedar; but yellow pine, tamarack,
and mesquite have also been used to a limited extent, and cypress and
juniper are being largely used in some of the Southern States.

Hardwoods, such. as oak, etc., do not make the best pavements, as
such woods become slippery. The softer, close-grained woods, such
as cedar and pine, wear better and give good foothold.

The wood employed should be sound and seasoned, free from sap,
shakes, and knots. Defective blocks laid in the pavement will quickly
cause holes in the surface, and the adjoining blocks will suffer tinder
wear, the whole surface becoming bumpy.

Chemical Treatment of Wood. The great enemy of all wood
pavements is decay, induced by the action of the air and water. Wood
is porous, absorbs moisture, and thus hastens its own destruction.
Many processes have been invented to overcome this defect. The
most popular processes at present are creosoiing and modifications of
the same, known as the "creo-resinate" and "kreodine" processes.
These consist of creosote mixed with various chemicals which are
supposed to add to the preserving qualities of the creosote.

Creosoting. This process consists in impregnating the wood with
the oil of tar, called creosote, from which the ammonia has been ex-
pelled, the effect being to coagulate the albumen and thereby prevent

371

104 HIGHWAY CONSTKUCTION

its decomposition, also to fill the pores of the wood with a bituminous
substance which excludes both air and moisture, and which is noxious
to the lower forms of animal and vegetable life. In adopting this pro-
cess, all moisture should be dried out of the pores of the timber. The
softer woods, while warm from the drying-house, may be immersed at
once in an open tank containing hot creosote oil, when they will absorb
about 8 or 9 pounds per cubic foot. For hardwoods, and woods which
are required to absorb more than 8 or 9 pounds of creosote per cubic
foot, the timber should be placed in an iron cylinder with closed ends,
and the creosote, which should be heated to a temperature of about
120° F., forced in with a pressure of 170 pounds to the square inch.
The heat must be kept up until the process is complete, to prevent the
creosote from crystallizing in the pores of the wood. By this means
the softer woods will easily absorb from 10 to 12 pounds of oil per cubic
foot.

The most effective method, however, is to exhaust the air from tht
cylinder after the timber is inserted ; then to allow the oil to flow in ; and
when the cylinder is full, to use a force pump with a pressure of 150 to
200 pounds per square inch, until the wood has absorbed the requisite
quantity of oil, as indicated by a gauge, which should be fitted to the
reservoir tank.

The oil is usually heated by coils of pipe placed in the reservoir,
through which a current of steam is passed.

The quantity of creosote oil recommended to be forced into the
wood is from 8 to 12 pounds per cubic foot. Into oak and other hard
woods it is difficult to force, even with the greatest pressure, more than
2 or 3 pounds of oil.

The advantages of this process are : The chemical constituents of
the oil preserve the fibers of the wood by coagulating the albumen of
the sap; the fatty matters act mechanically by filling the pores and thus
exclude water; while the carbolic acid contained in the oil is a powerful
disinfectant. .

The life of the wood is extended by any of the above processes, by
preserving it from decay; but such processes have little or no effect on
the wear of the blocks under traffic.

The process of dipping the blocks in coal tar or creosote oil is
injurious. Besides affording a cover for the use of defective or sappy
wood, it hastens decay, especially of green wood; it closes up the ex-

372

HIGHWAY CONSTRUCTION

105

terior of the cells of the wood so that moisture cannot escape, thus
causing fermentation to take place in the interior of the block, which
quickly destroys the strength of the fibers and reduces them to punk.

Expansion of Blocks. Wood blocks expand on exposure to
moisture; and, when they are laid end to end across the street, the
curbstones are liable to be displaced, or the courses of the blocks will
be bent into reserve curves. To avoid this, the joints of the courses
near the curb may be left open until expansion has ceased, the space
being temporarily filled with sand. The rate of expansion is about
1 inch in 8 feet, but Caries for different woods. The time required for
the wood to become fully expanded varies from 12 to 18 months. By
employing blocks impregnated with the oil of creosote, this trouble will
be avoided. 'Blocks so treated do not contract or expand to any appre-
ciable extent.

The comparative expansion of creosoted and plain wood blocks
after immersion in water for forty-eight hours, in percentage on orig-
inal dimensions, was:

Expansion of Wood Paving Blocks

On length of block..
Oil width " " .
On depth " " .

Manner of Laying. The blocks are set with the fiber vertical,
and the long dimension crosswise of the street, the longitudinal joints
being broken by a lap of at least one-third the length of the block; the

blocks should be laid so as to have the least possible width of joint.
Wide joints hasten the destruction of the wood by permitting- the fibers
to wear under traffic, which also causes the surface of the pavement

373

106 HIGHWAY CONSTRUCTION

to wear in small ridges. The most recent practice for laying blocks
on 3 per cent grades, has been to remove from the top of one side of
each block a strip J inch thick and H inches deep, extending the length
of the block. When the blocks are laid and driven closely together,
there is a quarter-inch opening or joint extending clear across the street
in each course. These joints are filled with Portland cement grout.
Fig. 63 shows a section of pavement having this form of joint.

Filling for Joints. The best materials for filling the joints are
bitumen for the lower two or three inches, and hydraulic cement grout
for the remainder of the depth. The cement grout protects the pitch
from the action of the sun, and does not wear down very much below
the surface of the wood.

ASPHALT PAVEMENTS.

Asphaltic Paving Materials. All asphaltic or bituminous pave-
ments are composed of two essential parts — namely, the cementing
material (matrix) and the resisting material (aggregate). Each has a
distinct function to perform; the first furnishes and preserves the co-
herency of the mass; the second resists the wear of traffic.

Two classes of asphaltic paving compounds are in use, — namely,
natural and artificial. The "natural" variety is composed of either
limestone or sandstone naturally cemented with bitumen. To this
class belong the bituminous limestones of Europe, Texas, Utah, etc.,
and the bituminous sandstones of California, Kentucky, Texas, Indian
Territory, etc. The "artificial" consists of mixtures of asphaltic
cement with sand and stone dust. To this class belong the pavements
made from Trinidad, Bermudez, Cuban, and similar asphaltums.
For the artificial variety, most hard bitumens are, when properly
prepared, equally suitable. For the aggregate, the most suitable mate-
rials are stone-dust from the harder rocks, such as granite, trap, etc.,
and sharp angular sand. These materials should be entirely free from
loam and vegetable impurities. The strength and enduring qualities
of the mixture will depend upon the quality, strength, and proportion
of each ingredient, as well as upon the cohesion of the matrix and its
adhesion to the aggregate.

Bituminous limestone consists of carbonate of lime naturally
cemented with bitumen in proportions varying from 80 to 93 per cent
of carbonate of lime and from 7 to 20 per cent of bitumen. Its color,
when freshly broken, is a dark (almost black) chocolate brown, the

374

HIGHWAY CONSTRUCTION 107

darker color being due to a large percentage of bitumen. At a tem-
perature of from 55° to 70° F., the material is hard and sonorous, and
breaks easily with an irregular fracture; at temperatures between 70°
and 140° F. it softens, passing with the rise in temperature through
various degrees of plasticity, until, at between 140° and 160° F., it
begins to crumble; at 212° it commences to melt; and at 280° F. it is
completely disintegrated. Its specific gravity is about 2.235.

Bituminous limestone is the material employed for paving pur-
poses throughout Europe. It is obtained principally from deposits
at Val-de-Travers, canton of Neufchatel, Switzerland ; at Seysell, in
the Department of Ain, France; at Ragusa, Sicily; at Limmer, near
Hanover; and at Vorwohle, Germany.

Bituminous limestone is found in several parts of the United
States. Two of these deposits are at present being worked — one in
Texas, the material from which is called "lithocarbon"; and one on the
Wasatch Indian Reservation. These deposits contain from 10 to 30
per cent of bitumen.

The bituminous limestones which contain about 10 per cent of
bitumen are used for paving in their natural condition, being simply
reduced to powder, heated until thoroughly softened, then spread while
hot upon the foundation, and tamped and rammed until compacted.

Bituminous sandstones are composed of sandstone rock impreg-
nated with bitumen in amounts varying from a trace to 70 per cent.
They are found in both Europe and America. In Europe, they are
chiefly used for the production of pure bitumen, which is extracted by
boiling or macerating them with water. In the United States, exten-
sive deposits are found in the Western States; and since 1880 they have
been gradually coming into use as a paving material, so that now up-
wards of 150 miles of streets in Western cities are paved with them.
They are prepared for use as paving material by crushing to powder,
which is heated to about 250° F. or until it becomes plastic, then spread
upon the street arid compressed by rolling; sometimes sand or gravel
Js added, and it is stated that a mixture of about 80 per cent of gravel
makes a durable pavement.

Trinidad Asphaltum. The deposits of asphaltum in the island
of Trinidad, W. I., have been the main source of supply for the asphal-
tum used in street paving in the United States. Three kinds are found
there, which have been named, according to the source, lake-pitch,

375

108 HIGHWAY CONSTRUCTION

land or overflow pitch, and iron pitch. The first and most valuable
kind is obtained from the so-called Pitch Lake.

The term land or overflow pitch is applied to the deposits of
asphaltum found outside the lake. These deposits form extensive
beds of variable thickness, and are covered with from a few to several
feet of earth ; they are considered by some authorities to be formed from
pitch which has overflowed from the lake; by others to be of entirely
different origin. The name cheese pitch is given to such portions of the
land pitch as more nearly resemble that obtained from the lake.

The term iron pitch is used to designate large and isolated masses
of extremely hard asphaltum found both within and without the bor-
ders of the lake. It is supposed to have been formed by the action of
heat caused by forest fires, which, sweeping over the softer pitch, re-
moved its more volatile constituents.

The name epuree is given to asphaltum refined on the island of
Trinidad. The process is conducted in a very crude manner, in large,
open, cast-iron sugar boilers.

The characteristics of crude Trinidad asphaltum, both lake and
land, are as follows: It is composed of bitumen mixed with fine sand,
clay, arid vegetable matter. Its specific gravity varies according to the
impurities present, but is usually about 1.28. Its color, when fleshly
excavated, is a brown, which changes to black on exposure to the at-
mosphere. When freshly broken, it emits the usual bituminous odor-
It is porous, containing gas cavities, and in consistency resembles
cheese. If left long enough in the sun, the surface will soften and melt,
and will finally flow into a more or less compact mass.

Refined Trinidad Asphaltum. The crude asphaltum is refined
or purified by melting it in iron kettles or stills by the application of
indirect heat.

The operation of refining proceeds as follows : During the heat-
ing, the water and lighter oils are evaporated; the asphaltum is lique-
fied; the vegetable matter rises to the surface, and is skimmed off; the
earthy and siliceous matters settle to the bottom; and the liquid asphal-
tum is drawn off into old cement or flour barrels.

When the asphaltum is refined without agitation, the residue
remaining in the still forms a considerable percentage of the crude
material, frequently amounting to 12 per cent; and it was at one time
considered that the greater the amount of this residue the better the

876

HIGHWAY CONSTRUCTION

109

quality of the refined asphaltum. Since agitation has been adopted}
however, the greater part of the earthy and siliceous matters is retained
in suspension; and it has come to be considered just as desirable for
a part of the surface mixture as the sand which is subsequently added.
The refined asphaltum, if for local use, is generally converted into
cement in the same still in which it was refined.

The average composition of both the land and lake varieties is
shown by the following analyses :

Average Composition of Trinidad Asphaltum

CONSTITUENTS

LAKE

LAND

Hard

Soft

Water

Per Cent
27.85
26. H8
7.63
38.14

Per Cent
34.10
25.05
6.35
34.50

Per Cent
26.62
27.57
8.05
37.76

Inorganic matter
Organic non-bituminous matter
Bitumen ...

When the analyses are calculated to a basis of dry
substances, the composition is : Inorganic matter
Organic matter not bitumen

100.00

36.56

10.57
52.87

38.00
9.64
53.36

37.74
10.68
51.58

Bitumen

The substances volatilized in 10 hours at 400° F . ..
The substances soften at —
" " flow at

100.00

3.66
190° F.
200° F.

12.24
170° F.
185° F.

0.86 to 1.37
200° to 251)3 F.
2 10° to 328° F.

The characteristics of refined Trinidad asphaltum are as follows:
The color is black, with a homogeneous appearance. At a tempera-
ture of about 70° F., it is very brittle, and breaks with a conchoidal
fracture. It burns with a yellowish-white flame, and in burning emits
an empyreumatic odor, and possesses little cemsntitious quality. To
give it the required plasticity and tenacity, it is mixed while liquid with
from 16 to 21 pounds of residuum oil to 100 pounds of asphaltum.

The product resulting from the combination is called asphalt
paving-cement. Its consistency should be such that, at a temperature
of from 70° to 80° F., it can be easily indented with the fingers, and on
slight warming be drawn out in strings or threads.

Artificial Asphalt Pavements. The pavements made from Trini-
dad, Bermudez, California, and similar asphaltums, are composed of
mechanical mixtures of asphaltic cement, sand, and stone-dust.

The sand should be equal in quality to that used for hydraulic
cement mortar; it must be entirely free from clay, loam, and vegetable

377

110 HIGHWAY CONSTRUCTION

impurities; its grains should be angular and range from coarse to fine.

The stone-dust is used to aid in filling the voids in the sand and
thus reduce the amount of cement. The amount used varies with the
coarseness of the sand and the quality of the cement, and ranges from
5 to 15 per cent. (The voids in sand vary from .3 to .5 per cent.)

As to the quality of the stone-dust, that from any durable stone is
equally suitable. Limestone-dust was originally used, and has never
been entirely discarded.

The paving composition is prepared by heating the mixed sand
and stone-dust and the asphalt cement separately to a temperature of
about 300° F. The heated ingredients are measured into a pu^-mill
and thoroughly incorporated. When this is accomplished, the mix-
ture is ready for use. It is hauled to the street and spread with iron
rakes to such depth as will give the required thickness when compacted
(the finished thickness varies between 1^ and 2j inches). The re-
duction of thickness by compression is generally about 40 per cent.

The mixture is sometimes laid in two layers. The first is called
the binder or cushion=coat ; it contains from 2 to 5 per cent more cement
than the surface-coat; its thickness is usually -i inch. The object of the
binder course is to unite the surface mixture with the foundation, which
it does through the larger percentage of cement that it contains, which,
if put in the surface mixture, would render it too soft.

The paving composition is compressed by means of rollers and
tamping irons, the latter being heated in a fire contained in an iron
basket mounted on wheels. These irons are used for tamping such
portions as are inaccessible to the roller — namely, gutters, around man-
hole heads, etc.

. Two rollers are sometimes employed; one, weighing 5 to 6 tons
and of narrow tread, is used to give the first compression; and the
other, weighing about 10 tons and of broad tread, is used for finishing.
The amount of rolling varies; the average is about 1 hour per 1,000
square yards of surface. After the primary compression, natural
hydraulic or any impalpable mineral matter is sprinkled over the sur-
face, to prevent the adhesion of the material to the roller and to give
the surface a more pleasing appearance. When the asphalt is laid
up to the curb, the surface of the portion forming the gutter is painted
with a coat of hot cement.

Although asphaltum is a bad conductor of heat, and the cement

378

§1

Ell

551

}M Qg

Si!

I
H

HIGHWAY CONSTRUCTION 111

retains its plasticity for several hours, occasions may and do arise
through which the composition before it is spread has cooled; its con-
dition when this happens is analogous to hydraulic cement which has
taken a "set," and the same rules which apply to hydraulic cement in
this condition should be respected in regard to asphaltic cement.

The proportions of the ingredients in the paving mixture are not
constant, but vary with the climate of the place where the pavement
is to l>e used, the character of the sand, and the amount and character
of the traffic that will use the pavement. The range in the proportion
is as follows :

Formula for Asphaltic Paving Mixture

Asphalt cement 12 to 1.5 per cent.

Sand 70 to 83 " "

Stone-dust 5 to 15 " "

A cubic yard of the prepared material weighs about 4,500 pounds, and
will lay the following amount of wearing-surface:

2i inches thick 12 square yards.

2 " " 18 "

H " " 27 "

One ton of refined asphaltum makes about 2,300 pounds of asphalt
cement, equal to about 3.4 cubic yards of surface material.

Foundation. A solid, unyielding foundation is indispensable
with all asphaltic pavements, because asphalt of itself has no power of
offering resistance to the action of traffic, consequently it is nearh
always placed upon a bed of hydraulic cement concrete. The concrete
must be thoroughly set and its surface dry before the asphalt is laid
upon it; if not, the water will be sucked up and converted into steam,
with the result that coherence of the asphaltic mixture is prevented,
and, although its surface may be smooth, the mass is really honey-
combed, so that as soon as the pavement is subjected to the action of
traffic, the voids or fissures formed by the steam appear on the surface,
and the whole pavement is quickly broken up.

Advantages of Asphalt Pavement. These may be summed up
as follows:

(1) Ease of traction.

(2) It is comparatively noiseless under traffic.

(3) It is impervious.

(4) It is easily cleansed.

(5) It produces neither mud nor dust.

(6) It is pleasing to the eye.

379

112 HIGHWAY CONSTRUCTION

(7) It suits all classes of traffic.

(8) There is neither vibration nor concussion in traveling over it.

(9) It is expeditiously laid, thereby causing little inconvenience
to traffic.

(10) Openings to gain access to underground pipes are easily
made.

(11) It is durable.

(12) It is easily repaired.

Defects of Asphalt Pavement. These are as follows:

(1) It is slippery under certain conditions of the atmosphere.
The American asphalts are much less so than the European, on account
of their granular texture derived from the sand. The difference is
very noticeable ; the European are as smooth as glass, while the Ameri-
can resemble fine sandpaper.

(2) It will not stand constant moisture, and will disintegrate if
excessively sprinkled.

(3) Under extreme heat it is liable- to become so soft that it will
roll of creep under traffic and present a wavy surface; and under ex-
treme cold there is danger that the surface will crack and become
friable.

(4) It is not adapted to grades steeper than 2-V per cent, although
it is in use on grades up to 7.30 per cent.

(5) Repairs must be quickly made, for the material has little
coherence, and if, from irregular settlement of foundation or local vio-
lence, a break occurs, the passing wheels rapidly shear off the sides of
the hole, and it soon assumes formidable dimensions.

The strewing of sand upon asphalt renders it less slippery; but in
addition to the interference of the traffic while this is being done, there
are further objections — namely, the possible injury by the sand cutting
into the asphalt, the expense of labor and materials, and the mud
formed, which has afterwards to be removed.

Although pure asphaltum is absolutely impervious and insoluble
in either fresh or salt wafer, yet asphalt pavements in the continued
presence of water are quickly disintegrated. Ordinary rain or daily
sprinkling does not injure them when they are allowed to become per-
fectly dry again. The damage is most apparent in gutters and adja-
cent to overflowing drinking fountains. This defect has long been
recognized; and various measures have been taken to overcome it, or

HIGHWAY CONSTRUCTION

113

at least to reduce it to a minimum. In some cities, ordinances have
been passed, seeking to regulate the sprinkling of the streets; and in
many places the gutters are laid with stone or vitrified brick (see Figs.

Fig. 64.

64 and 65), while in others the asphalt is laid to the curb, a space of
12 to 15 inches along the curb being covered with a thin coating of
asphalt cement.

Asphalt laid adjoining center-bearing street-car rails is quickly
broken down and destroyed. This defect is not peculiar to asphalt.
All other materials when placed in similar positions are quickly worn.
Granite blocks laid along such tracks have been cut into at a rate of
more than half an inch a year. The frequent entering and turning off
of vehicles from car tracks is one of the severest tests that can be

j Asphalt

Curb

Brick \ Gutter

Fig. 65.

applied to any paving material; moreover, the gauge of trucks and
vehicles is frequently greater than that of the rails, so one wheel runs
on the rail and the other outside. The number of wheels thus travel-
ing in one line must quickly wear a rut in any material adjoining the
center-bearing rail.

To obviate the destruction of asphalt in such situations, it is usual
to lay a strip of granite block or brick paving along the rail. This
pavement should be of sufficient width to support the wheels of the
widest gauge using the street.

381

114 HIGHWAY CONSTRUCTION

The burning of leaves or making of fires on asphalt pavements
should not be permitted, as it injures the asphalt, and the paving com-
panies cannot be compelled to repair the damaged places without
compensation.

Asphalt Blocks. Asphalt paving blocks are formed from a mix-
ture of asphaltic cement and crushed stone in the proportion of 8 to 12
per cent of cement to 88 and 92 per cent of stone. The materials are
heated to a temperature of about 300° F., and mixed while hot in a
suitable vessel. When the mixing is complete, the material is placed
in moulds and subjected to heavy pressure, after which the blocks are
cooled suddenly by plunging into cold water.

The usual dimensions of the blocks are 4 inches wide, 3 inches
deep, and 12 inches long.

Foundation. The blocks are usually laid upon a concrete founda-
tion with a cushion-coat of sand about \ inch thick. They are laid
with their length at right angles to the axis of the street, and the longitu-
dinal joints should be broken by a lap of at least 4 inches. The blocks
are then either 'rammed with hand rammers 01 rolled with a light steam
roller, the surface being covered with clean, fine sand ; no joint filling is
used, as, under the action of the sun and traffic, the blocks soon become
cemented.

The advantages claimed for a pavement of asphalt blocks over a
continuous sheet of asphalt are: (1) That they can be made at a
factory located near the materials, whence they can be transported to
the place where they are to be used and can be laid by ordinary paviors,
whereas sheet pavements require special machinery and skilled labor;
^2) that they are less slippery, owing to the joints and the rougher
surface due to the use of crushed stone.

Asphalt Macadam— Bituminous Macadam. Recently it has been
proposed to use asphalt as a binding material for broken stone.
There are two patented processes — the Whinery and the Warren —
which differ slightly in details.

The advantages claimed for these methods are: (1) The first
coat will be materially less; (2) it will offer a better foothold for horses;
(3) it will be at least as durable as the ordinary sheet asphalt ; (4) it will
not shift under traffic and roll into waves; (5) it will not crack; (6) it

g 5

HIGHWAY CONSTRUCTION

115

can be repaired more cheaply and with less skilled labor than can the
ordinary sheet asphalt.

Tools Employed in Construction of Asphalt Pavements. The

*:*-~s -..-

V>^^r-^

Fig. 66. Steam Holler.

FiK- 67. Asphalt Tools

tools used in laying sheet asphalt pavements comprise iron rakes;
hand rammers; smoothing irons (Fig. 67); pouring pots (Fig. 69):

383

116

HIGHWAY CONSTRUCTION

hand rollers, either with or without a fire-pot (Fig. 68); and steam
rollers, with or without provision for heating the front roll (Fig. 66).
These rollers are different in construction, appearance, and weight

Fig. 68. Hand Rollers.

Fig. 69. Pouring Pots.

from those employed for compacting broken stone. The difference
is due to the different character of the work recjuired.

The principal dimensions of a five-ton roller are as follows :

•Front roll or steering-wheel 30 to 32 inches diameter.

Rear roll or driving-wheel 48

Width of front roll 40

" rear " 40 "

Extreme length 14 feet.

height 7 to 8 feet.

Water capacity 80 to 100 gallons.

Coal " 200 pounds.

HIGHWAY CONSTRUCTION 117

FOOTPATHS— CURBS— GUTTERS.

A footpath or walk is simply a road under another name— a road
for pedestrians instead of one for horses and vehicles. The only
difference that exists is in the degree of service required; but the con-
ditions of consruction that render a road well adapted to its object are
very much the same as those required for a walk.

The effects of heavy loads such as use carriageways are not felt
upon footpaths; but the destructive action of water and frost is the
same in either case, and the treatment to counteract or resist these
elements as far as practicable, and to produce permanency, must be
the controlling idea in each case, and should be carried out upon a
common principle. It is not less essential that a walk should be well
adapted to its object than that a road should be; and it is annoying to
find it impassable or insecure and in want of repair when it is needed
for convenience or pleasure. In point of economy, there is the same
advantage in constructing a footway skilfully and durably as there is
in the case of a road.

Width. The width of footwalks (exclusive of the space occupied
by projections and shade trees) should be ample to accomodate com-
fortably the number of people using them. In streets devoted entirely
to commercial purposes, the clear width should be at least one-third
the width of the carriageway; in residential and suburban streets, a
very pleasing result can be obtained by making the walk one-half the
width of the roadway, and devoting the greater part to grass and shade
trees.

Cross Slope. The surface of footpaths must be sloped so that
the surface water will readily flow to the gutters. This slope need not
be very great; £ inch per foot will be sufficient. A greater slope with a
thin coating of ice upon it, becomes dangerous to pedestrians.

Foundation. As in the case of roadways, so with footpaths, the
foundation is of primary importance. Whatever material may be used
for the surface, if the foundation is weak and yielding, the surface will
settle irregularly and become extremely objectionable, if not danger-
ous, to pedestrians.

Surface. The requirements of a good covering for sidewalks are :

(1) It must be smooth but not slippery.

(2) It must absorb the minimum amount of water, so that it
may dry rapidly after rain.

385

118 HIGHWAY CONSTRUCTION

(3) It must not be easily abraded.

(4) It must be of uniform quality throughout, so that it may
wear evenly.

(5) It must neither scale nor flake.

(6) Its texture must be such that dust will not adhere to it.

(7) It must be durable.

Materials. The materials used for footpaths are as follows:
Stone, natural and artificial; wood; asphalt; brick; tar concrete; and
gravel.

Of the natural stones, sandstone (Milestone) and granite are ex-
tensively employed.

The bluestone, when well laid, forms an excellent paving material.
It is of compact texture, absorbs water to a very limited extent, and
hence soon dries after rain ; it has sufficient hardness to resist abrasion,
and wears well without becoming excessively slippery.

Granite, although exceedingly durable, wrears very slippery, and
its surface has to be frequently roughened.

Slabs, of whatever stone, must be of equal thickness throughout

their entire area; the
edges must be dressed
true to the square for the
whole thickness (edges
must not be left feather-
ed as shown in Fig. 70) ; and the slabs must be solidly bedded on the
foundation and the joints filled with cement-mortar.

Badly set or faultily dressed flagstones are very unpleasant to
walk over, especially in rainy weather; the unevenness causes pedes-
trians to stumble, and rocking stones squirt dirty wrater over their
clothes.

Wood has been largely used in the form of planks ; it is cheap in
first cost, but proves very expensive from the fact that it lasts but a
comparatively short time and requires constant repair to keep it
from becoming dangerous.

Asphalt forms an excellent footway pavement; it is durable and
does not wear slippery.

Brick. Brick of suitable quality, well and carefully laid on a
concrete foundation, makes an excellent footway pavement for resi-

HIGHWAY CONSTRUCTION 119

dential and suburban streets of large cities, and also for the main
streets of smaller towns. The bricks should be a good quality of
paving brick (ordinary building brick are unsuitable, as they soon
wear out and are easily broken) . The bricks should be laid in parallel
rows on their edges, with their length at right angles to the axis of the
path.

Curbstones. Curbstones are employed for the outer side of foot-
ways, to sustain the coverings and form the gutter. Their upper edges
are set flush with the foot walk pavement, so that the water' can flov,
over them into the gutters.

The disturbing forces which the curb has to resist, are : (1) The
pressure of the earth behind it, which is frequently augmented by
piles of merchandise, building materials, etc. This pressure tends to
overturn it, break it transversely, or move it bodily on its base. (2)
The pressure due to the expansion of freezing earth behind and be-
neath it. This force is most frequent where the sidewalk is partly
sodded and the ground is accordingly moist. Successive freezing and
thawing of the earth behind the curb will occasion a succession of
thrusts forward, which, if the curb be of faulty design, will cause it. to
incline several degrees from the vertical. (3) The concussions and
abrasions caused by traffic To withstand the destructive effect of
wheels, curbs are faced with iron; and a concrete curb with a rounded
edge of steel has been patented and used to some extent. Fires built
in the gutters deface and seriously injure the curb. Posts and trees
set too near the curb, tend to break, displace, and destroy it.

The use of drain tiles under the curb is a subject of much differ-
ence of opinion among engineers. Where the subsoil contains water
naturally, or is likely to receive it from outside the curb-lines, the use
of drains is of decided benefit; but great care must be exercised in
jointing the drain-tiles, lest the soil shall be loosened and removed,
causing the curb to drop out of alignment.

The materials employed for curbing are the natural stones, as
granite, sandstone (bluestone), etc., artificial stone, fire-clay, and cast
iron.

The dimensions of curbstones vary considerably in different
localities and according to the width of the footpaths; the wider the
path, the wider should be the curb. It should, however, never be less
than 8 inches deep, nor narrower than 4 inches. Depth is necessary

387

120

HIGHWAY CONSTRUCTION

to prevent the curb turning over toward the gutter. It should never
be in smaller lengths than 3 feet. The top surface should be beveled
off to conform to the slope of the footpath. The front face should be
hammer-dressed for a depth of about 6 inches, in order that there may
be a smooth surface visible against the gutter. The back for 3 inches

Fig. 71.

from the top should also be dressed, so that the flagging or other paving
may butt fair against it. The end joints should be cut truly square,
the full thickness of the stone at the top, and so much below the top as
will be exposed , the remaining portion of the depth and bottom should
be roughly squared, and the bottom should be fairly parallel to the
top. (See Figs. 71 and 72).

Artificial Stone. Artificial stone is being extensively used as a

footway paving material. Its manufacture is the subject of several
patents, and numerous kinds are to be had in the market. When
manufactured of first-class materials and laid in a substantial manner,
with proper provision against the action of frost, artificial stone forms
a durable, agreeable, and inexpensive pavement.

HIGHWAY CONSTRUCTION

121

Fig. 73. Tamper.

The varieties most extensively used in the United States* are
known by the names of granolithic, monolithic, ferrolithic, kosmocrete,
metalithic, etc.

The process of manufacture is practically the same for all kinds,
the difference being in the materials em-
ployed. The usual ingredients are Port-
land cement, sand, gravel, and crushed
stone.

Artificial stone for footway pave-
ments is formed in two ways — namely,
in blocks manufactured at a factory,
brought on the ground, and laid in the
same manner as natural stone; or the raw
materials are brought upon the work, pre-
pared, and laid in place, blocks being formed by the use of board
moulds.

The manner of laying is practically the same for all kinds. The
area to be paved is excavated to a mini-
mum depth of 8 inches, and to such great-
er depths as the nature of the ground may
require to secure a solid foundation. The
surface of the ground so exposed is well
compacted by ramming; and a layer of
gravel, ashes, clinker, or other suitable material is spread and consoli-
dated; on this is placed the concrete wearing surface, usually 4 inches
thick. As a protection against the lifting
effects of frost, the concrete is laid in
squares, rectangles, or other forms hav-
ing areas ranging from 6 to 30 square
feet, strips of wood being employed to
form moulds in which the concrete is

placed. After the concrete is set, these strips are removed, leaving
joints about half an inch wide between the blocks. Under some
patents these joints are filled with cement; under others, with tarred
paper; and in some cases they are left open.

Tools Employed in Construction of Artificial Stone Pavements.
Tampers (Fig. 73). Cast iron, with hickory handle; range from 6
by 8 inches to 8 by 10 inches.

Fig. 74. Quarter-Round.

389

122

HIGHWAY CONSTRUCTION

Quarter-Round, (Fig. 74).
for forming corners and edges.

Fig. 7ti. Cutter.

Made of any desired radius. Used

Jointer (Fig. 75). Used for
trimming and finishing the joints.

Cutter (Fig. 76). Used for cut-
ting the concrete into blocks.

Gutter Tool (Fig. 77). Used
for forming and finishing gutters.

Imprint Rollers (Figs. 78 and
79). Here are shown two designs
of rollers for imprinting the surface
of artificial stone pavements with
grooves, etc.

SELECTING THE PAVEMENT

The problem of selecting the best pavement for any particular
case is a local one, not only for each city, but also for each of the various
parts into which the city is imperceptibly divided; and it involves so
many elements that the nicest balancing of the relative values for each
kind of pavement is required, to arrive at a correct conclusion.

In some localities, the proximity of one or more paving materials
determines the character of the pavement ;
while in other cases a careful investigation
may be required in order to select the
most suitable material. Local conditions
should always be considered; hence it is
not possible to lay down any fixed rule as to what material makes the
best pavement.

The qualities essential to a good pavement may be stated as
follows :

(1) It should be impervious.

(2) It should afford good foothold for horses.

(3) It should be hard and durable, so as to resist wear and dis-
integration.

(4) It should be adapted to every grade.

(5) It should suit every class of traffic.

(6) It should offer the minimum resistance to traction.

(7) It should be noiseless.

(8) It should yield neither dust nor mud.

390

HIGHWAY CONSTRUCTION

123

(9) It should be easily cleaned.

(10) It should be cheap.

Interests Affected in Selection. Of the above requirements,
numbers 2, 4, 5, and 0 affect the traffic and determine the cost of haul-
age by the limitations of loads, speed,
and wear and tear of horses and ve-
hicles. If the surface is rough or the
foothold bad, the weight of the load
a horse can draw is decreased, thus ne-
cessitating the making of more trips or
the employment of more horses and
vehicles to move a given weight. A
defective surface necessitates a reduc-
tion in the speed of movement and
consequent loss of time; it increases the
wrear of horses, thus decreasing their
Fig. ra. imprint Roller. jjfe service and lessening the value of

their current services ; it also increases
the cost of maintaining vehicles and
harness.

Numbers 7, 8, and 9 affect the
occupiers of adjacent premises, who
suffer from the effect of dust and
noise; they also affect the owners of
said premises, whose income from
rents is diminished where these disad-
vantages exist. Numbers 3 and 10 af-
fect the taxpayers alone — first, as to
the length of time during which the
covering remains serviceable; and sec-
ond, as to the amount of the annual

repairs. Number 1 affects the adjacent occupiers principally on
hygienic grounds. Numbers 7 and 8 affect both traffic and occupiers.
Problem Involved in Selection. The problem involved in the
selection of the most suitable pavement consists of the following
factors: (1) adaptability; (2) desirability; (3) serviceability; (4) dura-
bility; (5) cost.

Adaptability. The best pavement for any given roadway will

Attt

124 HIGHWAY CONSTRUCTION

depend altogether on local circumstances. Pavements must be adapt-
ed to the class of traffic that will use them. The pavement suitable
for a road through an agricultural district will not be suitable for the
streets of a manufacturing center; nor will the covering suitable for
heavy traffic be suitable for a pleasure drive or for a residential district.

General experience indicates the relative fitness of the several
materials as follows:

For country roads, suburban streets, and pleasure drives — broken
stone. For streets having heavy and constant traffic — rectangular
blocks of stone, laid on a concrete foundation, with the joints filled
with bituminous or Portland cement grout. For streets devoted to
retail trade, and where comparative noiselessness is essential — asphalt,
wood, or brick.

Desirability, The desirability of a pavement is its possession of
qualities which make it satisfactory to the people using and seeing it.
Between two pavements alike in cost and durability, people will have
preferences arising from the condition of their health, presonal pre-
judices, and various other intangible influences, causing them to select
one rather than the other in their respective streets. Such selections
are often made against the demonstrated economies of the case, and
usually in ignorance of them. Whenever one kind of pavement is
more economical and satisfactory to use than is any other, there should
not be any difference of opinion about securing it, either as a new
pavement or in the replacement of an old one.

The economic desirability of pavements is governed by the ease
of movement over them, and is measured by the number of horses or
pounds of tractive force required to move a given weight — usually one
ton — over them. The resistance offered to traction by different pave-
ments is shown in the following table:

Resistance to Traction on Different Pavements

TRACTIVE RESISTANCE

KIND OF PAVEMENT

Pounds per ton

In terms oftheUxid

Asphalt (sheet)

to 70

cV to

Brick

" 40

Tsi "

Cobblestones

" 100

Stone-block . " .

" 80

Wood-block rectangular

" 50

eV "

Wood-block round

" 80

Vo"

inr

392

HIGHWAY CONSTRUCTION 125

Serviceability. The serviceability of a pavement is its quality of
fitness for use. This quality is measured by the expense caused to the
traffic using it — namely, the wear and tear of horses and vehicles, loss
of time, etc. No statistics are available from which to deduce the
actual cost of wear and tear.

The serviceability of any pavement depends in great measure
upon the amount of foothold afforded by it to the horses — provided,
however, that its surface be not so rough as to absorb too large a per-
centage of the tractive energy required to move a given load over it.
Cobblestones afford excellent foothold, and for this reason are largely
employed by horse-car companies for paving between the rails; but
the resistance of their surface to motion requires the expenditure ot
about 40 pounds of tractive energy to move a load of 1 ton. Asphalt
affords the least foothold; but the tractive force required to overcome
the resistance it offers to motion is only about 30 pounds per ton.

Comparative Safety. The comparison of pavements in respect of
safety, is the average distance traveled before a horse falls. The
materials affording the best foothold for horses are as follows, stated
in the order of their merit :

(1) Earth, dry and compact.

(2) Gravel.

(3) Broken stone (macadam).

(4) Wood.

(5) Sandstone and brick.
(G) Asphalt.

(7) Granite blocks.

Durability. The durability of pavement is that quality which
determines the length of time during which it is serviceable, and does
not relate to the length of time it has been down. The only measure
of durability of a pavement if the amount of traffic tonnage it will bear
before it becomes so worn that the cost of replacing it is less than the
expense incurred by its use.

As a pavement is a construction, it necessarily follows that there
is a vast difference between the durability of the pavement and the
durability of the materials of which it is made. Iron is eminently
durable; but, as a paving material, it is a failure.

Durability and Dirt. The durability of a paving material will
vary considerably with the condition of cleanliness observed. One

393

126 HIGHWAY CONSTRUCTION

inch of overlying dirt will most effectually protect the pavement from
abrasion, and indefinitely prolong its life. But the dirt is expensive,
it injures apparel and merchandise, and is the cause of sickness and
discomfort. In the comparison of different pavements, no traffic
should he credited to the dirty one.

Life of Pavements. The life or durability of different pavements
under like conditions of traffic and maintenance, may be taken as
follows :

Life Terms of Various Pavements

MATERIAL,

IKK TKKM

Granite block
.Sandstone

to no years

Asphalt.

' 14

Wood
Limestone

• 3 '•

Brick

Macadam

. ? ,

Cost. The question of cost is the one which usually interests
taxpayers, and is probably the greatest stumbling-block in the attain-
ment of good roadways. The first cost is usually charged against the
property abutting on the highway to be improved. The result is that
the average property owner is always anxious for a pavement that costs
little, because he must pay for it, not caring for the fact that cheap
pavements soon wear out and become a source of endless annoyance
and expense. Thus false ideas of economy always have stood, and
undoubtedly to some extent always will stand, in the way of realizing
that the best is the cheapest.

The pavement which has cost the most is not always the best; nor
is that which cost the least the cheapest ; the one which is truly the cheap-
est is the one which makes the most profitable returns in proportion to the
amount expended upon it. No doubt there is a limit of cost to go be-
yond which would produce no practical benefit ; but it will always be
found more economical to spend enough to secure the best results, and
this will always cost less in the long run. One dollar well spent is
many times more effective than one-half the amount injudiciously
expended in the hopeless effort to reach sufficiently good results. The
cheaper work may look as well as the more expensive for the
time, but may very soon have to be done over again.

Economical Benefit. The economic benefit of a good roadway is

HIGHWAY CONSTRUCTION 127

comprised in its cheaper maintenance; the greater facility it offers for
traveling, thus reducing the cost of transportation; the lower cost of
repairs to vehicles, and less wear of horses, thus increasing their term
of serviceability and enhancing the value of their present service; the
saving of time; and the ease and comfort afforded to those using the
roadway.

First Cost. The cost of construction is largely controlled by the
locality of the place, its proximity to the particular material used, and
the character of the foundation.

The Relative Economies of Pavements — whether of the same
kind in different condition, or of different kinds in like good condition
— are sufficiently determined by summing their cost under the following
headings of account:

(1) Annual interest upon first cost.

(2) Annual expense for maintenance.

(3) Annual cost for cleaning and sprinkling.

(4) Annual cost for service and use.

(5) Annual cost for consequential damages.

Interest on First Cost. The first cost of a pavement, like any
other permanent investment, is measurable for purposes of comparison
by the amount of annual interest on the sum expended. Thus, assum-
ing the worth of money to be 4%, a pavement costing $4 per square
yard entails an annual interest loss or tax of $0.16 per square yard.

Cost of Maintenance. Under this head must be included all out-
lays for repairs and renewals which are made from the time when the
pavement is new and at its best to a time subsequent, when, by any
treatment, it is again put in equally good condition. The gross sum
so derived, divided by the number of years which elapse between the
two dates, gives an average annual cost for maintenance.

Maintenance means the keeping of the pavement in a condition
practically as good as when first laid. The cost will vary considerably
depending not only upon the material and the manner in which it is
constructed, but upon the condition of cleanliness observed, and the '
quantity and quality of the traffic using the pavement.

The prevailing opinion that no pavement is a good one unless,
when once laid, it will take care of itself, is erroneous; there is no such
•pavement. All pavements are being constantly worn by traffic and
by the action of the atmosphere; and if any defects which appear are

305

128 HIGHWAY CONSTRUCTION

not quickly repaired, the pavements soon become unsatisfactory and
are destroyed. To keep them in good repair, incessant attention is
necessary, and is consistent with economy. Yet claims are made that
particular pavements cost little or nothing for repairs, simply because
repairs in these cases are not made, while any one can see the need of
them.

Cost of Cleaning and Sprinkling. Any pavement, to be con-
sidered as properly cared for, must be kept dustless and clean. While
circumstances legitimately determine in many cases that streets must
be cleaned at daily, weekly, or semi-weekly intervals, the only admis-
sible condition for the purpose of analysis of street expenses must be
that of like requirements in both or all cases subjected to comparison.

The cleaning of pavements, as regards both efficiency and cost,
depends (1) upon the character of the surface; (2) upon the nature of
the materials of which the pavements are composed. Block pave-
ments present the greatest difficulty; the joints can never be perfectly
cleaned. The order of merit as regards facility of cleansing, is: (1)
asphalt, (2) brick, (3) stone, (4) wood, (5) macadam.

Cost of Service and Use. The annual cost for service is made up
by combining several items of cost incidental to the use of the pave-
ment for traffic — for instance, the limitation of the speed of movement,
as in cases where a bad pavement causes slow driving and consequent
loss of time; or cases where the condition of a pavement limits the
weight of the load which a horse can haul, and so compels the making
of more trips or the employment of more horses and vehicles ; or cases
where conditions are such as to cause greater wear and tear of vehicles,
of equipage, and of horses. If a vehicle is run 1,500 miles in a year,
and its maintenance costs $30 a year, then the cost of its maintenance
per mile traveled is two cents. If the value of a team's time is, say,
$1 for the legitimate time taken in going one mile with a load, and in
consequence of bad roads it takes double that time, then the cost to
traffic from having to use that one mile of bad roadway is $1 for each
load. The same reasoning applies to circumstances where the weight
of the load has to be reduced so as to necessitate the making of more
than one trip. Again, bad pavements lessen not only the life-service
of horses, but also the value of their current service.

Cost for Consequential Damages. The determination of conse-
quential damages arising from the use of defective or unsuitable pave-

HIGHWAY CONSTRUCTION 129

meats, involves the consideration of a wide array of diverse circum-
stances. Hough-surfaced pavements, when in their best condition,
afford a lodgment for organic matter composed largely of the urine
and excrement of the animals employed upon the roadway. In
warm and damp weather, these matters undergo putrefactive fer-
mentation, and become the most efficient agency for generating and
disseminating noxious vapors and disease germs, now recognized as
the cause of a large part of the ills afflicting mankind. Pavements
formed of porous materials are objectionable on the same, if not even
stronger, grounds.

Pavements productive of dust and mud are objectionable, and
especially so on streets devoted to retail trade. If this particular
disadvantage be appraised at so small a sum per lineal foot of frontage
as SI. 50 per month, or six cents per day, it exceeds the cost of the best
quality of pavement free from these disadvantages.

Rough -surfaced pavements are noisy under traffic and insufferable
to nervous invalids, and much nervous sickness is attributable to them.
To all persons interested in nervous invalids, this damage from noisy
pavements is rated as being far greater than would be the cost of sub-
stituting the best quality of noiseless pavement; but there are, under
many circumstances, specific financial losses, measurable in dollars
and cents, dependent upon the use of rough, noisy pavements. They
reduce the rental value of buildings and offices situated upon streets
so paved — offices devoted to pursuits wherein exhausting brain work
is required. In such locations, quietness is almost indispensable,
and no question about the cost of a noiseless pavement weighs against
its possession. When an investigator has done the best he can to
determine such a summary of costs of a pavement, he may divide the
amount of annual tonnage of the street traffic by the amount of annual
costs, and know what number of tons of traffic are borne for each cent
of the average annual cost, which is the crucial test for any comparison,
us follows:

(1) Annual interest upon first cost $

(2) Average annual expense for maintenance and renewal . . .

(3) Annual cost for custody (sprinkling and cleaning)

(4) Annual cost for service and use

(5) Annual cost for consequential damages

Amount of average annual cost

Annual tonnage of traffic

Tons of traffic for each cent of cost

Gross Cost of Pavements. Since the cost of a pavement depends

397

130 HIGHWAY CONSTRUCTION

upon the material of which it is formed, the width of the roadway, the
extent and nature of the traffic, and the condition of repair and clean-
liness in which it is maintained, it follows that in no two streets is the
endurance or the cost the same, and the difference between the highest
and lowest periods of endurance and- amount of cost is very con-
siderable.

The comparative cost of the various street pavements, including
interest on first cost, sinking fund, maintenance, and cleaning, when
reduced to a uniform standard traffic of 100,000 tons per annum for
each yard in width of the carriageway, is about as follows:
Comparative Cost of Various Pavements

MATERIAL i ANNUAL COST

PER SQ. YD.

Granite blocks $0.25

0.40
0.35

Asphalt street ... 0.40

IJrit-k

Wood ...,

REVIEW QUESTIONS.

PRACTICAL TEST QUESTIONS.

In the foregoing sections of this Cyclopedia nu-
merous illustrative examples are worked out in
detail in order to show the application of the
various methods and principles. Accompanying
these are examples for practice which will aid the
reader in fixing the principles in mind.

In the following pages are given a larg». num-
ber of test questions and problems which afford a
valuable means of testing the reader's knowledge
of the subjects treated. They will be found excel-
lent practice for those preparing for Civil Service
Examinations. In some cases numerical answers
are given as a further aid in this work.

399

REVIEW QUESTIONS

ON THE SUBJECT OF1

BRIDGE ENGINEERING

I' ART I

1. Write a short history of early bridges.

2. Define: Truss, bridge truss, truss bridge, girders, and
girder bridges.

3. Draw an outline of a through bridge, and also an outline of
a deck bridge.

4. Make an outline diagram of a truss, and write the names
of the various parts on the respective members.

5. Make an outline diagram of a Warren, Howe, Pratt,
bowstring, and Baltimore truss.

6. Compute the weight of steel in a 130-foot highway bridge
whose trusses are 16 feet center to center, given W = 34 + 22& +
0.1GW + 0.7/.

7. Compute the weight of steel in a deck plate-girder span of
100 feet. Loading, E 50. Given W - 124.0 + 12. 01.

8. What are equivalent uniform loads?

9. What is Cooper's E 40 loading?

10. Prove that the stress in a diagonal of a horizontal chord
truss with a simple web system is V sec <£.

1 1 . Prove that the chord stress is M -=- h, where M is the moment
at the point, and h is the height of the truss.

12. Prove that the load must be on the segment of the span to
the right of the section to produce the maximum positive shear.

13. Compute the maximum positive and negative live-load
shears in a 13-panel Howe truss, the live panel load being 40 000
pounds.

401

REVIEW QUESTIONS

T H K H U B J K tJ T OK

BRIDGE ENGINEERING

A K T II

1 . Write an essay of 200 words on the economic considerations
governing the decision to build and the decision as to what kind of
bridge to employ.

2. What determines the height and width of railroad truss
bridges?

3. Draw a clearance diagram for a bridge on a straight track,
and state what allowance should be made if the bridge is on a curve.

4. Describe a stress sheet, and tell what should be on it.

5. Make a sketch of a cross-section of a deck plate-girder,
showing the cross-ties, guard-rails, and rails in place.

6. Make a sketch showing how tracks on curves are con-
structed.

7. WTiat is the span under coping, the span center to center of
bearings, and the span over all?

8. Design a tie for Cooper's E 50 loading.

9. If the end shear of a plate-girder is 394 500 pounds, design
the web section, it being 108 inches deep.

10- If the dead-load moment is 8 489 000 pound-inches and
the live-load moment is 30 610 000 pound-inches, design the flange,
if the distance back to back of flange angle, is 7 feet 6^ inches, it being
assumed that the web does not take any bending moment.

11. If, in the girder of Question 10, above, the web were 90 by
-j^-inch, design the flange, considering ^ of the gross area of the web
as effective flange area.

402

REVIEW QUESTIONS

O N T H K SU H,l K C T O F

11 1 U 11 WAY CONSTRTJC T ION

1 . Upon what does the ease with which a vehicle can he moved
on a road depend?

2. What kind of a road surface offers the greatest resistance to
traction?

3. How may the power required to draw a vehicle over a pro-
jecting stone be calculated?

4. What effect has gravity on the load a horse can pull?

5. Under what condition is the tractive power of a horse de-
creased?

6. What are the best methods for improving sand roads?

7 State briefly how earth is loosened and transported and the
conditions under which each method is most advantageous?

8. What are the essential requisites for securing a good gravel
road?

9. How should gravel roads be repaired?

10. State the considerations that control the maximum grade.

11. How should different grades be joined?

12. What considerations control the width of a road?

13. What is the essential quality of a stone used for road
covering?

14. What should be the shape and size of broken stone?

15. For a light traffic road what thickness should the layer of
broken stone have?

16. How should the foundation for the broken stone be pre-
pared?

403

REVIEW QUESTIONS

ON THE SUBJECT OF1

HIOHWAY CONSTRUCTION

1. How should the natural soil be prepared to receive a
pavement ?

2. In ramming blocks in the pavement, what point requires
to be watched ?

3. How is a sand cushion prepared for use ?

4. What influences the durability of a granite?

5. How are rectangular stones laid on steep grades ?

6. How is the surface and sub-surface drainage of streets
provided for?

7. W hat are the principal objections to wood pavements?

8. What determines the best width for a street?

9. In filling the joints with gravel and bituminous cement,
what should be the condition of the material?

10. What controls the maximum grade for a given street?

11. What varieties of wood give the most satisfactory results?

12. To what tests are stones intended for paving subjected?

13. Do cobblestones form a satisfactory pavement?

14. What properties should a stone possess to produce a sat-
isfactory paving block?

15. How are expansion joints formed in a pavement?

16. What is the most suitable material for the foundation of
a pavement?

17. Under what class of traffic may wood be used?

18. Upon what does the durability of a pavement depend?

19. What materials are employed for filling the joints be-
tween the paving blocks?

404

INDEX

The page numbers of this volume ivill be found at the bottom of the
pages; the numbers at the top refer only to the section.

Page

Bridge Engineering

Asphalt pavement

374

weights

advantages of

379

bridge design

141

defects of

380

clearance diagram

147

foundation

379

economic considerations

141

materials for

374

economic proportions

143

tools used in construction of

383

floor system

150

Asphaltic paving mixture, formula for

379

practical considerations

155

Axle friction

276

specifications

150

stress sheet

150

weights and loadings

148

Baltimore truss

problems

135-139

Belgian block pavement
Bituminous limestone

352
374

Bridge trusses, definition of
Bridges, loads for

Bowstring truss

live loads

Brick pavements

360

wind loads

absorption test

361

Bridges, weights of

advantages of

363

formula; for

cross-breaking test

362

highway spans

crushing test

362

railroad spans

defects of

363

Broken-stone roads

332

foundation for

363

rattler test

360

Bridge analysis

Catch-basins

347

early bridges

City streets

341

truss bridge development

arrangement of

341

Bridge engineering 11-264

asphalt pavement

374

bibliography

264

Belgian block pavement

352

bridge analysis

brick pavements

360

definitions

catch-basins

347

descriptions

cobblestone pavement

352

girder spans

117

curbstones

387

history

drainage

346

loads

footpath

385

stresses

foundations

348

theory

grades

342

trusses 16

i, 53

granite block pavement

352

Note. — For page numbers see foot of pages.

405

INDEX

Page

City streets

gutters

347

Deck bridges, definition of

stone-block pavements

350

Drainage of ro;ids

298

1 ransverse contour

346

Drains, fall of

301

width of

341

wood pavements

369

Cobblestone pavements

352

Earth

313

Concrete

349

Earth roads

316

Concrete-mixing machine

368

Earthenware pipe culverts

30g

Country roads

2«7

Earthwork

310

drainage

298

balancing cuts and fills

310

culverts

304

classilicat ion of

3 1 3

fall of drains

301

embankments on hillsides

side ditches

301

prosecution of

314

of surface

302

shrinkage of

3 1 2

water breaks

303

slopes

311

earthwork

310

Embankments on hillsides

314

general considerations

Engine loads in computing stresses

axle friction.

276

maximum moments, position of wheel

efTect of springs on vehicles

277

loads for

friction

267

maximum shear, position of wheel

loss of tractive i>ower on inclines 274

loads for

object of roads

267

resistance of air

277

Final stresses

1 1 0

resistance to rolling

268

Floor-l>eams, moments and shears in

117

tractive i>ower and gradients

272

Floor systems

location of

278

Footpaths

38B

bridge sites

283

Forces, resolution of

cross levels

283

examples

284

final selection

283

Girder bridges, definition of

intermediate towns

286

Girder spans

117

levels

282

moments in plate-girders

1 19

map

282

moments and shears in floor-l>eam

117

memoir

282

shears in plate-girders

128

profile

283

stresses in plate-girders

134

reconnoissance

278

Girders, definition of

topography

281

( ; rades

road coverings

330

establishing

294

transverse contour

296

maximum

29!

minimum

293

width

295

undulating

293

Creosoting

371

Gradient

291

Culverts

304

Grading tools

319

jointing

308

carts

323

materials for

307

draining-tools

329

Curbstones

387

dump cars

324

Note. — For page numbers see foot of pages

406

INDEX

III

Page

Pago

Grading tools

Mountain roads

287

dump wagons

325

alignment

289

horse rollers

329

construction profile

290

mechanical graders

325

establishing grade

294

picks

319

final location

290

ploughs

319

gradient

291

road machine

327

halting places

288

scrapers

321

level stretches

294

shovels

319

loss of height

288

sprinkling carts

330

maximum grade

29 1

surface graders

327

minimum grade

293

wheelbarrows

322

undulating grades

293

Granite block pavement,

352

water on

288

Gravel heaters

369

y.ig/ags

290

Gravel roads

330

Gutters

347

Parabolic truss

Pavements

348

Hard pan

313

asphalt

374

Highway bridges, live loads for

Belgian block

352

Highway construction . 267-398

brick

360

city streets

341

cobblestone

352

country roads

267

concrete foundation for

349

drainage

298

cost

394

earthwork

310

granite block

352

grading tools

319

selecting

390

mountain roads

287

stone block

350

pavements

348

wood

369

road coverings

330

Plate-girder .railway-span design

1 56

Highway spans, actual weights of

bearings

199

Howe truss

cross-frames

181

determination of class

1 56

Inclines, loss of tractive power on

274

determination of span

157

Iron pipe culverts

309

flanges

163

lateral systems

181

masonry plan

156

Knee-braces, definition of

stifTeners

1 92

stress sheet

202

Lateral bracing, definition of
Live load, position of, for maximum

ties- and guard-rails
web, economic depth of
web splice

1 59
162
195

positive and negative shears
Ix)os3 rock

38
313

Plate-girders
moments in

1 19

shears in

128

Maximum and minimum stresses

stresses in

134

Mechanical graders

325

Pony-truss bridge, definition of

Melting furnaces

369

Portals, definition of

Moments in plate-girders

119

Pratt truss

48, 53, 93

Note. — For page numbers see foot of pages

407

INDEX

Page

Surf;

Swaj

Railroad spans, actual weights of

Railway bridges, live loads for

Road coverings

broken-stone

332 Tabl

gravel

330

Road machines

327

Roads

country

267

drainage of

298

earth

316

mountain

287

sand

318

Roadways on rock slopes

'316

Sand roads

318

Shears, maximum live-load

Shears in plate-girders

128

Shoes and roller nests

254

Side slopes

311

Slopes

311

covering of

312

form of

311

inclination of

311

Snow-load stresses

103

Solid rock

313

Steam-rollers

338

Stone block pavements

350

Street grades

342

Stresses

snow-load

103

wind-load

104

Stresses in bridge trusses

in chord members

counters

engine loads

live-load moments

live-load shears

maximum and minimum

moments method

notation used

35 1

resolution of forces method

27 1

Warren truss under dead load

35 1

Warren truss under live load

in web members

31 i

Stresses in plate-girders

134

Note. — For page numbers see foot of

pages.

Surface graders

Sway bracing, definition of

Page

327

bridges, formulae for weights of 19

bridges, types of for various spans 142
dead-load chord stresses 53, 61

dead-load stresses in diagonals 54

deck plate-girders, weights of 21

floor-beam reactions 119
force required to draw loaded vehicles

over inclined roads 276
force of wind per sq. ft. for various

velocities 277

grade data 292
grades, effects of, upon load horse can

draw on different pavements 274
gross load horse can draw on different

kinds of road surfaces 274
impact coefficient, values of 102
impact stresses in a Pratt truss 103
lacing bars, thickness of 229
loads, equivalent uniform 24
masonry bearings, length of 158
maximum moments in a deck plate-
girder 126
maximum moments, determination of

position of wheel loads for 93
maximum shears in a deck plate-
girder 133
maximum shear, determination of

position of wheel loads for 90
pavements, comparative cost of

various 398

pavements, li.'e term of various 394

pins for country highway bridges 245

pins for double-track bridges 244

pins for single-track bridges 244
plate-girder bridges,, width of, for

various spans 145

reactions for a deck plate-girder 126

reactions for a through plate-girder 123
resistance due to gravity on different

inclinations 271
resistance to traction on different

pavements 392

408

INDEX

Tables

resistance to traction on different

road surfaces 268

rise, suitable proportions for different

paving materials 297

rivet spacing in bottom flange 177

rivet spacing in top flange 179

safe spans for I-beams 151

specific gravity, weight, resistance to
crushing, and absorption
power of stones 352

standard gauges for angles 175

stresses in a Pratt truss 1 00

tension members 217

tractive power of horses at different

velocities 273

Trinidad asphaltum, average com-
position of 377
trusses, chronological list of 16
wheel position, moments in deck

plate-girder 1 25

wheel position, moments in a through

plate-girder 122

wheel position, shears in a through

plate-girder 129

wind stresses in Pratt truss 113

Through bridges, definition of 14

Through Pratt railway-span design 203

determination of span 204

end-post 239

floor-beams 207

intermediate posts 225

lateral systems 251

masonry plan 203

pins 243

portal 245

shoes and roller nests 254

Note. — For page numbers see foot of pages.

Through Pratt railway-span design

stringers

tension members

top chord

transverse bracing
Transverse grade
Trinidad asphaltum
Truss bridge development
Truss, members of
Truss bridge, definition of
Trusses, classes of

chord characteristics

web characteristics
Trusses, definition of
Trusses under dead and live loads

Baltimore

bowstring

Howe

parabolic

Pratt
Trusses under engine loads

Pratt

W
Warren truss

under dead loads

under live load
Water breaks
Web splice
Wind-load stresses

bottom lateral bracing

overturning effect of wind on train

overturning effect of wind on truss

portals and sway bracing

top lateral system
Wind loads
Wrood pavemenis

Page

205

215

231

248

345

375

35
45
303
195
104
107
110
109
112
104
24
369

409

UNIVERSITY OF CALIFORNIA LIBRARY

Los Angeles
This book is DUE on the last date stamped below.

JUL7E 1

Form L9-100m-9,'52(A3105)444

EBgintrrine
Library

14?

000688173 4

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