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.REESE LIBRARY
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UNIVERSITY OF CALIFORNIA.
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Wo. $Oy~£/~ . . Class No.
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ELEMENTAEY
THEORY AND CALCULATION
OF
IKON BKIDGES AND KOOFS.
BY
AUGUST EITTER, DR. PHIL.,
PROFESSOB AT THE POLYTECHNIC SCHOOL AT AIX-LA-CHAPELLE.
TE AN SLATED FROM THE GERMAN (THIRD EDITION)
BY
H. E. SANKBY,
LIEUTENANT ROYAL ENGINEERS.
E. & F. N. SPON, 46, CHAEING CEOSS
NEW YORK:
446, BBOOME STREET.
1879.
TEANSLATOR'S PEEFACE.
THE first edition of Professor Bitter's work, of which the
following is a literal translation, was published in 1862* to
advocate the use of the " Method of Moments " in calculating
the stresses in bridges and roofs. This " Method of Moments "
is in reality but an application of Kankine's " Method of Sec-
tions." The adaptation of the method to various cases is
explained and illustrated by means of numerical examples
comprising several of the forms of bridges and roofs in general
use as well as others not often met with.
Some interesting problems are discussed in the Eleventh
Chapter, and are possibly not generally known in this country.
It is required to determine the form a structure should have to
fulfil given conditions as regards the stresses. These problems
give a considerable insight into the manner in which the
stresses are distributed amongst the various bars of a structure,
and show also that comparatively small changes in the form
may produce great changes in the stresses. The effect of
changes of temperature on the deflection and on the stresses in
a " composite structure " is treated at some length in the Four-
teenth and Sixteenth Chapters. The theory of loaded beams is
only touched upon — in fact, only those cases are considered which
are required in the various examples.
The Sixteenth Chapter contains a very instructive example
of a composite structure consisting of a pair of braced girders
combined with suspension chains. It should be observed,
however, that Herr Hugo B. Buschmann, in a pamphlet * On
The substance of the first two chapters was published previously in the
' Zeitschrift des Architecten- und Ingenieur-Vereins fur das Konigreich Hannover/
vol. vii., No. 4.
v TBANSLATOE'S PBEFAOE.
the Theory of Combined Girder and Suspension Bridges/
takes exception to the manner in which the equations (§§ 56
to 61) giving the stresses in the girders produced by the
moving load are arrived at. In the preface to the third edition
Professor Bitter says : " Herr Buschmann maintains that these
equations depend on arbitrary assumptions, and thinks to prove
their unsoundness by remarking that under certain conditions
of loading, namely, when both ends of the girders are loaded,
they give a negative bending moment at the centre of the
girders, or, in other words, the girders would be bent upwards.
Thus the radii of curvature for the central part of the suspension
chains would be increased, and this requires a diminution of the
length of these chains, which is evidently absurd. This con-
clusion is, however, incorrect. Herr Buschmann overlooks the
fact that exceedingly small changes in the form of the suspen-
sion chains are under consideration, and that therefore not only
the vertical but also the horizontal displacement of each
element of the chain must be taken into account. Without
doubt if the chord of the arc whose radii of curvature are
increased did not alter, the length of this arc would be
diminished. But if at the same time the length of the
chord increases, not only may the length of the arc remain
unaltered, but it may even be lengthened ; and this is what in
reality occurs, owing to the horizontal displacement of each
point of the chains, not only in the case under consideration,
but also for all positions of the loads."
In some few instances Professor Eitter does not agree with
the more recent English practice, notably so in his estimate of
the wind-pressure on roofs. These instances have been pointed
out in notes added to the text and in an Appendix.
H. E. S.
Gibraltar, 1879.
CONTENTS.
CHAPTER I.
PAGE
1. Preliminary Remarks 1
2. Method of Moments 4
3. Calculation of the Stresses in a Eoof of 100 feet Span (Drill-shed of
the Welfenplat^Barracks, Hanover) 10
4. Roof Truss of 32 metres Span 14
CHAPTER II.
§ 5. Application of the Method of Moments to the Calculation of Bridges . . 17
§ 6. Parabolic Girder of 16 metres Span with a single System of Diagonals 20
§ 7. Derived Forms 26
§ 8. Theory of Parabolic Girders 31
CHAPTER III.
§ 9. Application of the Method of Moments to the Calculation of the
Stresses in Braced Girders having Parallel Booms 36
§ 10. Braced Girder of 16 metres Span, composed of single Right-angled
Triangles 39
§ 11. Derived Forms 44
§ 12. Remarks on the Degree of Accuracy of the Assumptions made with
regard to the Distribution of the Loads 48
§ 13. Braced Girder with Equilateral Triangles, or Warren Girder (Railway
Bridge over the Trent, near Newark) 52
§ 14. Multiple Lattice Girders , 59
CHAPTER IV.
§ 16, Sickle-shaped (Bowstring) Roof of 208 feet Span with a single system
of Diagonals (Roof over Railway Station, Birmingham) . . . . . . 76
§ 16. Derived Forms 87
§ 17. Apparent Failures of the Method of Moments . . .-. 93
§ 18. Theory of Sickle-shaped Trusses 95
Vi CONTENTS.
CHAPTER V.
PAGE
§ 19. Cantilever Roof \vith Stay, Span 6 metres ............ 105
§ 20. Cantilever Koof without Stay .................. HI
CHAPTER VI.
§ 21. Braced Arch of 24 metres Span ................ _H4
§ 22. Braced Arch of 40 metres Span (Bridge over the Theiss, at Szegedin) 126
§ 23. Stability of the Abutments of the Braced Arch .......... 147
§ 24. Theory of Hinged Bridges .................. •• 151
CHAPTER VII.
§ 25. Variation in the Stresses due to alterations in the Span ...... 157
a. Parabolic Girder .................... 158
6. Braced Girder with Parallel Booms ............ 160
c. Braced Arch and Suspension Bridge ............ 161
CHAPTER VIII.
§ 26. Suspension Bridge in Three Spans. Span of Central Opening, 120
metres ; Span of each Side Opening, 60 metres ........ 164
§ 27. Stability of the Central Piers .................. 180
§ 28. Stability of the Shore Abutments ................ 185
CHAPTER IX.
§ 29. On the Calculation of the Stresses in Domes ............ 188
§ 30. Dome of 100 metres Span .................... 188
§ 31. Generating Curve, for a Dome, requiring the least Quantity of Material 201
§ 32. Dome formed of Articulated Eibs and Rings ............ 203
CHAPTER X.
33. Continuous Girder Bridges .................. 213
34. Continuous Girder Bridge in Three Spans. Central Opening, 160
metres; Side Openings,- 130 metres .............. 217
35. To determine the Subdivision of the whole Span requiring the least
Quantity of Material .................... 227
CHAPTER XI.
36. Determination of the Turning Points and Lever Arms by Calculation 234
37. Application of the Method of Moments to find the Form a Structure
should have in order that it may fulfil given Conditions .. .. 241
CONTENTS. Vll
PAGE
§ 38. Girder in which the Minimum Stress in the Diagonals is Zero
(Schwedler's Girder) 243
§ 39. Girder having the Maximum Stress in its Diagonals equal 249
§ 40. Girder in which the Stresses in the Bow are equal throughout (Pauli's
Girder) 254
CHAPTER XII.
§ 41. Determination of the Cross-sectional Areas of the Bars in a Structure 264
§ 42. Bracing required to resist the Pressure of the Wind and Horizontal
Vibrations 266
§ 43. Intermediate Bearers 268
CHAPTEE XIII.
§ 44. On the Deflection of Loaded Structures . .. 282
§ 45. Deflection of Parabolic Arches and Girders 287
§ 46. Deflection of Braced Girders with Parallel Booms .... 290
CHAPTER XIV.
§ 47. Theory of Composite Structures 294
§ 48. Trussed Beams without Diagonals 301
§ 49. Influence of Changes of Temperature 305
CHAPTER XV.
§ 50. Resistance of Beams to Flexure 310
§ 51. Deflection of Beams 321
§ 52. Resistance of long Columns to Bending and Buckling 340
CHAPTER XVI,
§ 53. Compound Lattice and Suspension Bridge, Span 60 metres— Deter-
mination of the best Ratio between the Depth of the Girders and
the Height of the Arc of the Suspension Chains .. .. .. .. 347
§ 54. Calculation of the Stresses produced by Changes of Temperature . . 350
§ 55. Calculation of the Stresses produced by the Permanent Load . . . . 353
§ 56. Calculation of the Stresses produced by a Moving Load 355
§ 57. Determination of the Worst 'Condition of Loading for the Girders . . 359
§ 58. Calculation of the Stresses produced in the Booms of the Girders by
the Moving Load 362
§ 59. Calculation of the Stresses produced by the Permanent and Tem-
perature Loads in the Booms of the Girders . . . . ' 366
Vlll CONTENTS.
PAGE
§ 60. Calculation of the Shearing Stress produced by the Moving Load .. 367
§ 61. Calculation of the Shearing Stresses due to the Permanent and Tem-
perature Loads, and of the Maximum Stresses in the Braces . . 369
§]62. Calculation of the Stresses in the Wind-stays and Wind-braces .. 370
§ 63. Influence of the Extension of the Back-stays 373
§ 64. Kecapitulation of the Eesults of the Calculations 378
§ 65. Adjustment of the Vertical Bods connecting the Girders with the
Suspension Chains 381
§ 66. Eemarks on the degree of Accuracy of the Method employed . . . . 385
APPENDIX.
a. Loads on Koofs and the Reactions at the Abutments caused by the Wind-
pressure 389
b. Stability of Piers as regards Overturning 393
CALCULATION OF THE STRESSES
IN BRIDGES AND ROOFS.
ERRATA.
Page 9, line 14, for Ur, read Uv.
11 Y =- 32,300 Ibs., „ Y =- 11,700 Ibs.
l&l J»
12, „
24', ',', 12,000 x 37-5,
„ 21,000 X 37-5.
13, ,
7, „ Z2 X 5,
„ — Z2 x 5.
R
R
18, ,
R y
b> j? 2 '
»> 4/x-
21, ,
, — „ Fig. 23,
„ Fig. 24, and vice versa.
34,
• „ Fig. 46,
„ Fig. 47, and vice versa.
46,
17, „ Section 9,
„ § 9.
66,
- » Fig. 96,
„ Fig. 99, and vice versa.
82,
23, „ Section,
„ Chapter.
135,
15, add (Fig. 186) after Y8
159,
„ 24, for 7000,
„ 6000.
392,
> 8, „ —
»» ~
conditions.
This can be fulfilled in the case of bars which are under
direct tension or compression, for then the stress is uniformly
distributed over the whole sectional area ; but in the case of a
beam under bending stress, it cannot be complied with, because
the stresses are not uniformly distributed over the cross-section.
Therefore, in a good construction, the various parts should
be, if possible, either in direct tension or compression, and
bending stress should be avoided.
These views are more or less carried out in practice, and
the larger the structure, the nearer is the approximation. The
Vlll CONTENTS.
FACE
§ 60. Calculation of the Shearing Stress produced by- the Moving Load . . 367
§ 61. Calculation of the Shearing Stresses due to the Permanent and Tem-
perature Loads, and of the Maximum Stresses in the Braces . . 369
§] 62. Calculation of the Stresses in the Wind-stays and Wind-braces . . 370
§ 63. Influence of the Extension of the Back-stays 373
§ 64. Recapitulation of the Results of the Calculations 378
§ 65. Adjustment of the Vertical Rods connecting the Girders with the
Suspension Chains 381
§ 66. Remarks on the degree of Accuracy of the Method employed . . . . 385
APPENDIX.
I TI r in T?nnn+innfl n± thp. Abutments caused bv the Wind-
CALCULATION OF THE STEESSES
IN BEIDGES AND HOOFS.
FIRST CHAPTER
§ 1. — PEELIMINAKY KEMAKKS.
IN large bridges and roofs the design should be such that the
quantity of material employed is the smallest possible, not only
because in such cases the cost of materials is great in com-
parison to that of labour, but more especially because the dead
load is thereby unnecessarily increased, and the very success of
the undertaking may possibly depend on the smallness of this
load.
In a well designed structure, the maximum safe resistance
of the material should be called forth in every part, and no-
where should there be any unnecessary excess of material.
This can also be stated thus : When the structure is placed in
the worst conditions as to loading, the intensity of stress in
every part should be equal to what is considered the safe re-
sistance to the stress to which it is subject when under these
conditions.
This can be fulfilled in the case of bars which are under
direct tension or compression, for then the stress is uniformly
distributed over the whole sectional area ; but in the case of a
beam under bending stress, it cannot be complied with, because
the stresses are not uniformly distributed over the cross-section.
Therefore, in a good construction, the various parts should
be, if possible, either in direct tension or compression, and
bending stress should be avoided.
These views are more or less carried out in practice, and
the larger the structure, the nearer is the approximation. The
2 BRIDGES AND ROOFS.
endeavour to save material has led from the massive beams of
rectangular section to those of I and II section, and when
further the solid web was replaced by braces, those combinations
of bars were arrived at in which only direct tension or com-
pression exist. The iron roofs and the braced girders of
modern times are examples of such structures.
To comply rigidly with the above conditions, the joints
should be made with single bolts (pin joints). If a bar be con-
nected to another by a single bolt, it can turn freely about its
end, but if the joint be made with rivets, the end of the bar is
fixed, and will therefore be subject to a slight amount of bending
stress. Thus, with rivetted joints, the material is not employed
to the best account ; and, especially in the case of large, im-
portant structures, there is the disadvantage that the maximum
stresses are not accurately known, whilst if the structure were
theoretically correct, these stresses could be ascertained to the
greatest degree of accuracy. It is worth noticing that the
theoretical structures are also the easiest to calculate.
In all the following examples it will be assumed that the
joints are hinged, the connections being made by single bolts.
It will also be supposed that these joints are the only points of
loading. This distribution of the load can always be obtained
in practice by using bearers to bridge over the space between
the joints. Whether it is advisable to construct these bearers
as separate parts or to fuse them into the main structure, is a
question that will be considered further on. -
As regards the weight of the structure itself, it will be con-
sidered as evenly distributed over the span, and in accordance
with the above, concentrated at the joints; the degree of
accuracy of this assumption will be tested in the sequel.
[NOTE. — There is no doubt but that hinged connections made by means of
single pins would be theoretically more perfect than rivetted joints, if a per-
fectly uniform distribution of the stress were the only consideration. But it
is found that in structures subject to vibrations, the pins in many cases shake
loose, and the holes in the bars become elliptical, owing to the hammering
action that takes place between the pins and the faces of the holes, and this
action will always occur unless the pins are made a drawing fit in the holes.
This is the case, for instance, in the central joints of a railway bridge, where
(as will be seen) the stresses are constantly changing from tension to com-
§ 1. — PEELIMINAEY EEMARKS. 3
pression, and vice versa. It may be mentioned that this action occurred in the
Crumlin Viaduct, and that in consequence gusset plates had to be added. Pin
joints may however often be used with advantage, both with regard to economy,
simplicity of erection, and appearance in structures subject to a purely dead
load, or even to a live load, if unaccompanied by vibrations and rapid changes
in the nature of the stresses, as, for instance, in the case of roofs.
The objections raised to rivetted joints by Professor Bitter apply in
reality only to those as usually designed, for the arrangement of the rivets in
a joint can be such that little or no bending stress occurs in the bars connected.
This was pointed out by Professor Callcott Keilly in two papers read before the
Institution of Civil Engineers.* Premising the following definition — " The
mean fibre of a bar is the line passing through the centres of gravity of all
cross-sections, and is consequently one of the axes of gravity of the bar,"
the rules according to which rivetted joints should be designed are thus stated
by Professor Keilly : —
1st. The mean fibres of any two or more members of a truss connected by
a group of rivets must intersect at one point.
2nd. The group of rivets connecting the bars must be arranged symmetri-
cally round this point of intersection of the mean fibres of the bars ; or in
other words, the resultant resisting force of the group of rivets must occur at
the intersection of the mean fibres of the bars connected by the said group.
3rd. The first row of rivets in each bar, that is, the row on the side
towards which the stress is transmitted, must be symmetrical with the mean
fibre of that bar.
If the stress is uniformly distributed over the cross-section of any bar, the
resultant stress must lie in the mean fibre ; it is therefore evident that unless
the mean fibres of the bars connected intersect in a point, the stress, in some
of them at least, will not be uniformly distributed.
The resultant pull or thrust of a bar must evidently lie in the same straight
line as the resultant resistance of the rivets connecting the bar. If, therefore,
rule 2 be not complied with, the resultant pull or thrust will not pass through
the mean fibre, and evidently the stresses will not be uniformly distributed but
will be uniformly varying, and therefore more intense on one edge (the edge
nearest the resultant) than upon the other. In a similar manner the stress will
not be uniformly distributed if the first row of rivets be not symmetrical with
the mean fibre.
To obtain a theoretically perfect joint, every row of rivets should be sym-
metrical with the mean fibre. Such an arrangement can, however, only be
obtained when two bars cross at right angles. But the first row of rivets, for
instance, relieves the part of the bar beyond of a certain amount of stress ;
therefore- unless the second row of rivets be very much displaced, the greatest
intensity of stress at the section through this row will not reach the intensity
or stress in the bar before the leading rivets. This is evidently, a fortiori,
true of the 3rd, 4th, &c., rows of rivets. The rules given above are therefore
sufficient for practical purposes.
* 'Minutes of Proceedings,' vols. xxiv. and xxix.
B 2
4 BRIDGES AND ROOFS.
Rivetted joints have also this advantage over pin joints, that the ends of
the bars connected can be considered " fixed," and this materially increases
the resistance of those bars subject to compression. Pin joints are also as a
rule more expensive than ri vetted joints, but easier to put together by unskilled
labour.
It will be observed that the mean fibre need not necessarily be a straight
line ; but if it is curved no arrangement of the joints will make the stress
uniformly distributed at every cross-section. This is, for instance, the case if
the bow of a bowstring girder is curved between the joints. Fig. 1 represents
FIG. 1.
a portion of the top boom of such a girder, and the mean fibres of the various
bars are indicated by dotted lines. The thrust in the booms must evidently
act in the straight line joining A and B, and must therefore give rise to
bending stress, or, in other words, the stress will not be uniformly distributed.]
§ 2. — METHOD OF MOMENTS.
The method adopted to calculate the stresses in the various
structures given in the following examples is known as the
K
" method of moments," and it will be explained by means of
the roof represented by Fig. 1 (a).
§ 2. — METHOD OF MOMENTS. 5
The total load on this roof consists of the five single loads,
P, Q, E, S, T, which are to be considered as weights hung to the
top joints. These five loads produce the reactions D and K at
the points of support ; the sum of these reactions must be equal
to the whole load, and they can easily be determined by the
ordinary rules of statics. If, for instance, the span AB is
divided by the verticals through the weights into six equal
parts,
FIG. 2.
t-i
Imagine the combination of bars divided into two parts
by the line L L, then each part (Fig. 2, for instance) can
only be retained in equilibrium by applying to each bar
at the point of section a force
which represents the action of
the other part. This force must
lie in the direction of the "bar, for
otherwise the bar would rotate
round its end ; this force is, in
fact, what is called the stress in
the bar. Thus the stresses X, Y, Z
in the three bars, which have
been cut through, together with the remaining loads D, P, Q,
are in equilibrium. All these forces lie in the same vertical
plane, and they therefore must satisfy the. three following
conditions of equilibrium : —
1. The sum of the vertical forces acting upwards must be
equal to the sum of the vertical forces acting downwards.
2. The sum of the horizontal forces acting towards the right
must be equal to the sum of the horizontal forces acting towards
the left.
3. The sum of the statical moments of all the forces
tending to turn the part of the roof represented by Fig. 2
round any point from right to left, must be equal to the sum
of the statical moments of those forces tending to turn it from
left to right round the same point ; for the part of the roof
under consideration can be regarded as a lever, and any point
can be taken as the fulcrum.
6 BRIDGES AND EOOFS.
These three conditions can be expressed more concisely by
the equations
2(H) = 0, 2(V) = 0, 5(M) = 0,
where H and V are the resolved parts of any force horizontally
and vertically respectively, M the moment of a force round
any point, and 2 denotes that the forces or moments have been
added together algebraically, that is, the sign of each force or
moment is taken as plus or minus according to the direction in
which it acts. The three stresses X, Y, and Z will be contained
in each of these equations, and by solving them the values of
X, Y, and Z can be obtained. The stresses in all the members
of the roof can be similarly ascertained by taking other
sections.
This method can always be applied, but it has two serious
defects. The first is, that H and V contain the cosine and sine
of the angles the bars make with the horizontal, and these
angles must therefore be determined. The second is, and it is
more serious than the other, that in order to ascertain any one
stress all three equations have, as a rule, to be solved.
There is, however, a very simple method, which can be
applied to all cases, and which is free from the above defects.
Apart from this, the method has the advantage of requiring
only the application of the principle of the lever (in its more
general form the law of statical moments), and can there-
fore be easily understood by those who are acquainted but
with the very elements of mechanics. In fact only the last
equation, that of statical moments, need be used, for if to
obtain the stress in one bar moments are taken round the
point of intersection of the other two bars, an equation will be
arrived at containing only one unknown, the stress required,
for evidently the moments of the stresses in the other two bars
vanish.
The lever arms of the various forces will have to be de-
termined, and this can be done with sufficient accuracy from a
drawing to scale.
A general rule, framed from the above, can be thus stated : —
Consider the structure divided into two parts ly a section,
§ 2. — METHOD OF MOMENTS. 7
which if possible should only cut through three bars, and apply the
forces X, Y, Z to these tars to maintain equilibrium, these forces
being the stresses in the bars. Form the equation of moments for
either part of the structure, and if X is to be determined the point
of intersection of Yand Zis to be chosen as the point round which
to take moments, if Y the point of intersection of X and Z, and
if Z the point of intersection of X and Y.
FIG. 3.
For instance, in the above example to determine X, the
point round which to take moments would be E, the intersection
of Y and Z (Fig 3). The equation of moments is
x x - P . CE + D . AE = o,
or
P . CE - D . AE
To determine Y take moments round A, the point of inter-
section of X and Z, thus :
or
- Yy + P . AC + Q. AE = 0,
Y_P.AC + Q.AE
And to determine Z take moments round H, the point of inter-
section of X and Y, thus :
- Z z - Q . EL - P. CL + D . AL = 0,
or
z _ - Q . EL - P . CL + P . AL
z
This method can be directly applied to all structures in
8
BRIDGES AND EOOFS.
which it is possible to reach each bar by a section that does
not cut through more than three bars.
In some cases, however, in the truss shown in Fig. 4, for
instance, there may be bars which cannot be reached by sections
cutting only through three bars ; such are the bars F G, D G, D E.
FIG. 4.
But even in such a case the method may be directly applied
if all the bars cut through by the section (which may be curved
or straight) intersect in a point except the one the stress in
which is to be determined.
For instance, to find the stress V in the bar F G, take a
FIG. 5,
FIG. 6.
section a fiy and form the equation of moments about the
point H for the part cut out (Fig. 5).
or
FH
§ 2. — METHOD OF MOMENTS.
9
In the same manner the stress U in the bar D Gr can be ascer-
tained by taking a section a S 7 and forming the equation of
moments round the point H for the part cut out, thus :
U u — R r - o,
or,
Similarly the stresses in K J and L J can be found. The re-
maining bars can all be reached either by sections which only
cut through three bars, or else by sections which cut through
four bars but the stress in one of which is already known. In
both cases the method of moments can be applied.
Thus, when the stress U is known the stresses X, Y, Z in
the bars D F, D E, C E, can be found from the equations
X . DE + U r - Q . NO - P . MO + W . AO = 0,
Y . AD + U I + Q . AN + P . AM = 0,
A
obtained by taking moments round the points, E, A, and D
respectively.
This more complicated ex- FIG. 7.
ample shows the advantages of
the proposed method. They be-
come even more apparent when
it is considered that only the
beginner will require to make
separate figures for each calcula-
tion. The adept will easily form
the equations from the principal
drawing.
The general method having now been explained, its applica-
tion to various cases will be best seen by means of numerical
examples. It will be sufficient to give the complete calculations
for a few bars only, those which can be considered as the repre-
sentatives of others similarly situated. For the remaining bars
only the equation of moments and the results will be given.
It is of no consequence which direction of rotation is taken
as the positive one, but to avoid errors some direction should be
chosen ; it will be considered in the sequel that rotation from
10
BRIDGES AND HOOFS.
left to right is positive,* and that rotation from right to left is
negative.
Further, all stresses will be considered as pulling stresses
(this has already been done in the former examples), therefore
positive stresses will represent tension and negative stresses com-
pression.
This, it will be observed, is the reverse to the usual English practice.
[NOTE. — A great deal of clerical labour can be spared by rightly choosing
the scale by which the lever arms are to be measured. This remark refers
principally to structures divided into bays of equal length. It will probably
be best in this case to make the length of each bay unity, when it will be found
that the lever arms of the various loads are generally whole numbers. This
plan it will be seen has been adopted in several of the examples given. If the
loads on the structure are placed at equal intervals, the horizontal distance
between them should be taken as unity.]
§ 3. — CALCULATION OF THE STKESSES IN A ROOF OF
100 FEETf SPAN.
Drill-shed of the Welfenplatz Barracks, Hanover.
The weight of the roof covering and framing (Fig. 8) is
11*3 Ibs.f per square foot of horizontal surface covered, and
FIG. 8.
20 Ibs. more per square foot must be added for snow and wind
pressure.! The total load is therefore 31 -3 Ibs. per square foot
of horizontal surface.
The distance apart of the principals is 15 J feet, and since
the span is 100 feet, 15 J x 100 square feet of horizontal
surface is supported by each principal, and the load on each is
15 J X 100 x 31 -3 Ibs., or in round numbers 48,000 Ibs. The
* In the same direction as the hands of a watch,
t German feet and Ibs.
J This estimate of the snow and wind pressure does not agree with the latest
English practice. See Appendix.
§ 3. — KOOP TRUSS 100 FEET SPAN.
11
load on each of the eight divisions of the roof is therefore
6000 Ibs. It may be considered that one-half of the 6000 Ibs. on
each division is applied at each of the two adjacent joints, and
this can be effected by means of bearers or common rafters.*
The load on the seven central joints will therefore be 6000 Ibs.,
and on each of the end joints 3000 Ibs. Evidently the load on
the end joints will be taken up directly by the abutments. The
reaction at each abutment is altogether 24,000 Ibs., and sub-
tracting the 3000 Ibs. on the end joint, the pressure against
the combination of bars is 21,000 Ibs.
FIG. 9.
- a
The structure is therefore subject to the action of nine
exterior forces ; seven of 6000 Ibs. each acting downwards on
the central joints, and two of 21,000 Ibs. each acting upwards
on the end joints.
To find the stresses X, Y, Z, in the bars of the central bay
FIG. 10.
a
\6000 \JL
UU r J^,-
'^^
2l!000x
(Fig. 9) let the roof be divided into two parts by a section a @
* This distribution of the load requires the common rafters to be articulated
at each joint. They are, however, generally continuous, and this slightly alters
the distribution of the load, for then part of the load is transmitted directly to the
abutments by the common rafters. It is, however, usual in practice to adopt
the above distribution, the error being on the side of safety. See ' Lectures on
the Elements of Applied Mechanics/ by Morgan W. Crofton, F.K.S., and
' Instruction in Construction,' by Col. Wray, R.E.— TRANS.
12 BRIDGES AND ROOFS.
and the forces X, Y, Z applied to maintain equilibrium. To
obtain X, consider the part shown in Fig. 10 as a lever with its
fulcrum at D, the point of intersection of Y and Z ; then for
equilibrium the following equation of moments must hold : *
0 = X x 18-6 + 21,000 x 50 - 6000 x 12-5 - 6000 x 25 - 6000 x 37'5,
whence
X = - 32,300 Ibs.
Similarly to find Y take moments round A, the point of inter-
section of X and Z, thus :
0 = Y x 38-4 + 6000 x 12'5 + 6000 X 25 + 6000 x 37'5
Y = - 32,300 Ibs.
And to obtain % take moments round the point E :
0 = - Z x 15 + 21,000 x 37-5 - 6000 x 12'5 - 6000 x 25
Z = + 37,500 Ibs.
To find the stress in the vertical rod E F take an oblique
FIG. 11.
section 7 S (Fig. 11), and the equation of moments round A,
the point of intersection of the other two bars intersected by
7 8 will be
0 = - V x 37'5 + 6000 x 12-5 + 6000 X 25,
whence
V = + 6000 Ibs.
The equations for the similarly situated bars can be formed in like manner,
thus: — frfffd*
0 = X1 X 13-9 + 12,000 X 37' 5- 6000 X 12*5-6000 x 25.
Xj = — 40,400 Ibs. (fTurning point F).
0 = Y! X 23-5 + 6000 x 12'5 + 6000 X 25.
Y! = - 9570 Ibs. (Turning point A).
0 = - Z1 x 10 + 21,000 X 25 - 6000 x 12'5.
Zl = + 45,000 Ibs. (Turning point C).
* The method of finding the lever arms by calculation is given in the eleventh
section of this book.
f The "turning point" (Drehungspunkt) is the point with reference to
which the equation of moments is formed.
§ 3. — ROOF TRUSS 100 FEET SPAN. 13
0 = - yi X 25 + 6000 x 12-5 (Turning point A).
Vl = + 3000 Ibs.
0 = X2 x 9-3 + 21,000 x 25 - 6000 x 12-5.
X2 = - 48,400 Ibs. (Turning point H).
0 = Y2 X 9-3 + 6000 + 12-5 (Turning point A).
Y2 = — 8100 Ibs.
0 =TZ2 x 5 + 21,000 x 12-5 (Turning point J).
Z2 = 4- 52,500 Ibs.
The stress in X3 is to be found by means of the section \ /JL
(Fig. 12), which only intersects two bars. In such a case any
FIG. 12.
sitooox
D
point in the other bar can be chosen as turning point.
Thus taking moments round D,
0 = X3 x 18-6 + 21,000 x 50.
X3 = - 56,500 Ibs.
The central vertical bar is the only one the stress in which
cannot be found directly. To obtain this stress, that in one of
FIG. 13.
6000 j^'" \
50'
the adjacent bars must be known. Thus it has been found
that X = - 32,300 Ibs.; hence (Fig. 13) the equation of
moments about B to find the stress U is
whence
0 = - U X 50 - 6000 x 50 - ( - 32,300) x 37'2,
U = + 18,000 Ibs.
14 BRIDGES AND ROOFS.
The results of the above calculations are given in Fig. 14.
FIG. 14.
+ 52500
+45000
+ 31500
§ 4. — KOOF TRUSS OF 32 METRES SPAN.
The total load on the roof truss represented in Fig. 15 is
assumed to be 32,000 kilos., or 1000 kilos, per metre of span.
Proceeding as in the previous example, it is found that there is
on each central joint a load of 4000 kilos, and an upward
pressure of 14,000 kilos, at each abutment.
14000
It will be seen that to find the stresses in the bars V, U, X,
Y, and Z, the variation of the method of moments explained at
the end of § 2 must be employed.
For instance, to calculate the stress V the portion of the
roof shown in Fig. 16 must be considered, and taking moments
round C the equation
0 = - V x 4-308 - 4000 x 4
is obtained, whence
V = — 3714 kilos.
§ 4. — ROOF TRUSS 32 METRES SPAN.
15
Similarly the stress U can be found from the portion of the
roof shown in Fig 17, thus
0 = U x 3-2 - 4000 x 4, or U = 4- 5000 kilos.
Having determined U, the stress X can be found from
Fig. 18 by taking moments round E :
0 = X x 3-4465 + 14000 x 9'28 - 4000 (1'28 4- 5-28) + 5000 X 3-2,
or X = - 34,725 kilos.
Likewise Y can be found by taking moments round A :
0 = - Y x 6-4 + 4000 (4 4- 8) 4- 5000 x 3'2,
or Y = 10,000 kilos.
To determine the stress Z (Fig. 15), an oblique section pass-
ing to the left of the point E (Fig. 18) must be drawn, and by
taking moments round A,
0 = Z x 8-616 4- 4000 (4 + 8) 4- 5000 x 3*2,
or Z = - 7428 kilos.
FIG. 18.
4000
14000
The remaining nine bars of the left half of the roof can each
be reached by a section intersecting only three bars, and the
16
BRIDGES AND EOOFS.
stresses in them can therefore be calculated in the manner shown
in the previous numerical example. The results of these cal-
FIG. 19.
35000
30000
20000
culations are given in Fig. 19, and as the bars in compression
have been drawn with double lines, the signs have been
omitted.
SECOND CHAPTEE.
§ 5. — APPLICATION OF THE METHOD OF MOMENTS TO THE
CALCULATION OF BRIDGES.
One great advantage of the method described in the previous
pages is that the stress in any particular bar can be found at
once by means of a single equation. But there is yet another
advantage which adapts this method more particularly to the
calculation of Bridges. It is this : that from the inspection of
one equation of moments it is possible to ascertain what loads
on the bridge increase the stress in any particular bar and
what loads decrease it. Therefore to find the maximum stress
in a bar it is only necessary to leave out of the equation those
loads which diminish the stress. And to find the minimum
stress (which in some cases will be compression) those loads
which increase the stress must be omitted. It is unnecessary
to add that the above has reference to temporary loads only.
This does not apply to the previous examples, for — as can
be easily ascertained — the removal of any of the loads does not
increase the stress (either tension or compression) in any of the
bars. In the case, however, of the structures that are usually
adopted for bridges and also in some roof trusses (as will appear
further on), it is of great importance to ascertain ,the effect of
the variation of the loading,* for the greatest stress (either of
tension or compression) may not occur when the structure is
fully loaded.
[Throughout, the term greatest stress is used irrespective of
the sign of the stress, but the terms maximum and minimum
depend on the sign, thus the minimum stress may be the
greatest compression.]
* For example, the temporary load produced on a bridge by a train, or in
the case of roofs, by the snow or wind-pressure, applied to one side only.
C
18
BRIDGES AND EOOFS.
In the girder shown in Fig. 20, for instance, the stress S is
found by taking a section M N and forming the equation of
moments round 0
or substituting for D its value : | P + ^ + 5 ;
Q B
s =
The member containing P is negative, and the members
containing Q and K are positive. Evidently then the load P
FIG. 20.
^v. Jb
b
diminishes the stress and the loads Q and R increase it.
Hence the equation
Q , B
2*+4*
S (max.) = —
s
gives the greatest tension produced, and the equation
S (min.) =
gives the greatest compression.
For simplicity it has been considered that P, Q, E are
moving loads, and that the girder itself has no weight.
§ 5. — CALCULATION OF BRIDGES. 19
The equations for T and U are
or substituting for D its value and solving
CD
from which it appears that the greatest stresses in these bars
occur when the girder is fully loaded.
The equation to find the stress V is (Section a {L Turning-
point 0,)
- v
o OP
or substituting for W its value : ^R + ^ + 4
V =
whence as before
Y (max.) = +
for the greatest tension, and
V (min.) =
for the greatest compression.
c 2
20 BRIDGES AND ROOFS.
The above is expressed by the following rule : —
Consider that the structure is fully loaded and form the
equation of moments accordingly for the lar the greatest stresses in
which are to le found. Arrange this equation so that the effect of
each load can be easily ascertained. Then to find the greatest
tension leave out all the temporary loads that diminish the stress
and to find the least tension, or the greatest compression, leave out
all the temporary loads that increase the stress.
Or shorter thus : In the equation giving the greatest stress in
a lar (either tension or compression) the members containing the
moving loads must have the same sign.
The equation of moments for the fully loaded bridge gives
the greatest stress only in one case ; when the members con-
taining the moving loads have all the same sign.
The following numerical example will illustrate the above
rule.
§ 6. — PARABOLIC GIRDER* OF 16 METRES SPAN WITH A
SINGLE SYSTEM OF DIAGONALS.
The dimensions are given in Fig. 21. — The dead load on
the bridge, designed for a single line of railway, can be taken at
1000 kilos, per metre and the live load at 5000 kilos, per
FIG. 21.
metre. One half of this is carried by each girder, and the
length of each bay being 2 metres, 1000 kilos, dead load and
5000 kilos, live load act on each joint (Fig. 22).
To find the stress X, take a section a fi through the first
bay and form the equation of moments for the part shown in
Fig. 23 round the point C.
o = Xj x I + D x 2.
* Thus called because the bow is in the form of a polygon inscribed in a
parabola. — TRANS.
§ 6.— PARABOLIC GIRDER.
But when the bridge is fully loaded,
D = IpOO (i + f + f + | + f + f + £)
+ 5000 (i + | + f + f + f + | + |).
Therefore, substituting this value of D,
21
0 = X, x
1000 (i + £
5000 (i + f
FIG. 22.
1
lotoo 10 bo 10
Xi I X*
X5
JO
I I A .
looo ioi)o loToo lojoo w
. i X. i X.T i Xy i Xt I.
It will be observed that the seven members of this equation
due to the live load have all the same sign, and hence the
greatest stress in the bar Xx occurs when the bridge is fully
loaded. Solving : —
Xt (min.) = - 48000 kilos.
The stress Zj can also be obtained from Fig. 23 by taking
moments round B.
o = - Zx x o-8 + D x 2,
or substituting for D its value.
o = - z, x 0-8 + 1000 (i +f
+ 5000
FIG. 4s.
Here also the greatest stress occurs when the bridge is fully
loaded, therefore
Zj (max.) = + 52500 kilos.
To find the stress Vi take a section 78 and form the
equation of moments for the part shown in Fig. 24 round the
point B.
o = - Vt x 2-8-D x 0-8.
22 BRIDGES AND ROOFS.
and substituting for D its value
(* +
<* +
Here again it is evident that
0 = - Vj x 2-8 -
- 5000 (| +
25
is greatest when the bridge is fully
5000 loaded. Hence
Vj (min.) = — 6000 kilos.
The stresses X2, Y2, Z2 can be found
by cutting off the part of the girder
shown in Fig. 25. For X2 take
moments round E
0 = X2xl-5 + Dx4- 1000 X 2 - 5000 X 2,
or substituting for D.
0 = X2 x 1-5 ' + 1000 ft + f + . + j) x 4
+ 5000 (i + I + . + £) X 4
- 1000 x 2 - 5000 x 2.
The live load of 5000 kilos, acting at B is contained in two
members of this equation. One, 4- 5000 x f X 4, is the effect
produced by the part of the load transmitted to the abutment
A, and the other, - 5000 x 2 is the direct effect of the load.
According to the rule these two members must be united into
one, viz. 5000 (f x 4 — 2), the equation then takes the form : —
o = X2 x 1-5 + 1000 { (| + ... + £) 4 + a x 4 - 2) }
+ 5000 { (i + . . . + |) 4 + (I X 4 - 2) } .
It is easily seen that all the members multiplied by 5000
are positive, hence the greatest stress occurs where the bridge
is fully loaded, and
X2 (min.) = - 48000 kilos.
To find Y2 take moments round E and by substituting for D
its value
0 = Y2 x 1'68 - 1000 (i + .. . + |) 0-8 - 5000 (| + . . . + |) x 0'8
+ 1000 X 2-8 + 5000 X 2'8;
or arranging the equation according to the rule,
0 = Y2 x 1-68 - 1000 {(£+ ... + |)0-8- (2-8 - | 0-8) }
- 5000 (£ + . . . + I) 0-8 + 5000 (2-8 - | 0'8).
§ 6. — PARABOLIC GIEDER. 23
Of the seven members multiplied by 5000, and representing
the effect of the moving load, 6 are negative and 1 is positive.
Leaving out therefore the positive member (which diminishes
the stress)
0 = Y2 x 1-68 - 1000 { (i + . . • + f) 0-8 - (2'8 - $ 0'8) }
-5000(i +... + |)0-8,
whence
Y2 (max.) = + 6250 kilos.
Next leaving out the six negative members,
0 = Y2 x 1-68 - 1000 {(i + ... + |)0-8 - (2-8-f 0-8)}
+ 5000(2-8-|0-8),
whence
Y2 (min.) = - 6250 kilos.
(It appears that Y2 = 0 when the bridge is fully loaded.
This result will be explained further on when treating of the
theory of parabolic girders.)
FIG. 26.
5000
**» 1
10
£>
00 T
N!
r
~~"^J
>. xx
\
}E
The stress Z2 is found by taking moments round B and
arranging the equation as before.
0 = - Z2 x 0-835 + 1000 (i + . . . + 1) x 2 + 5000 (i + . . . + |) X 2,
from which it is evident that Z2 is a maximum when the bridge
is fully loaded. Hence
Z2 (max.) = + 50300 kilos.
To determine V2 take a section ?? 0 and form the equation of
moments for the part shown in Fig. 26 with S as turning point.
24 BRIDGES AND ROOFS.
Arranging this equation
0= - V8x8-1000 {(i + ... + f)4- (6-|.4)}
- 5000 (I + . . . + I) 4 + 5000 (6 - | x 4).
First leaving out the positive member multiplied by 5000
o = - v2 x 8 - 1000 { ft + . . . + f) 4 - (6 - 1 x 4) }
- 5000 ft + . . . + f ) 4,
or
V2 (min.) = - 7560 kilos.
Then leaving out the negative members multiplied by 5000
0 = - V2 X 8 - 1000 { (| + , . . + f) 4 - (6 - -i x 4) }
+ 5000 (6 - | x 4),
or
V3 (max.) = + 560 kilos,
These examples sufficiently illustrate the rule, and the
calculations for the remaining bars need not be given so fully.
The general equation of moments and the results for the remaining bars are
given below.
0 = X3 x 1-875 + 1000 { (i + . . . + .£) 6 + (f . 6 - 2) + £ . 6 - 4) }
+ 5000 { ft + . . . + |) 6 + (f . 6 - 2) + ft . 6 - 4) }
X3 (min.) = - 48000 kilos.
0 = Y3 x 5 • 47 - 1000 { ft + . . . + f ) 4 - (8 - f . 4) - (6 - J. . 4) }
- 5000 {(£ + ... + f)4 + 5000(8-f x 4) + 5000 (6 - 1 . 4) }
Y ( (max.) = + 6850 kilos.
3 I (H)in.) = - 6850 kilos.
0 = - Z3 x 1-474 + 1000 { ft + ... + f) 4 + ft . 4 - 2) }
Z3 (max.) = + 48900 kilos.
0 = - V3 X 30 - 1000 { (| + ... + f) 24 - (28 - f x 24) - (26 - f . 24) }
- 5000 ft + . . . + f ) 24 + 5000 (28 - f . 24) + 5000 (26 - f . 24)
v ( (max.) = + 1500 kilos.
3 I (min.) = - 8500 kilos.
0 = X4 x 2 + 1000 { (| + ... + |) 8 + (| . 8 - 2) + (f . 8 - 4) + ft. 8 - 6) }
+ 5000 { ft + . . . + 1) 8 + (| . 8 - 2) + (| . 8 - 4) + ft . 8 - 6) }
X4 (min.) = - 48000 kilos.
0 = Y4 x 21-2 - 1000 { ft + . . . + |) 24 - (30 - f . 24) - (28 - f . 24)
+ 5000 ft + . . . + |) 24 + 5000 { (30 - f . 24) + (28 - • . 24)
+ (26-1.24)}
/ (max.) .= + 7080 kilos.
4 I (min.) = - 7080 kilos.
§ 6. — PARABOLIC GIRDER. 25
0 = - Z4 x 1 ' 873 + 1000 { (i + . . . + f ) 6 + (£ . 6 - 2) + (£ . 6 - 4) }
+ 5000 { (| + ... + f)6 + (f . 6 - 2) + (J . 6 - 4) }
Z4 (max.) = + 48100 kilos.
(The following equations of moments are formed with reference to the part of
the girder lying to the right of the section line) :
0 = - V4 X 32 + 1000 { (i + . . . + |) 24 - (32 - f . 24) - (30 - £ . 24)
-(28-f.24) (-26-|. 24)}
+ 5000 (i + ... + f)24
- 5000 { (32 - f . 24) + (30 - 1 . 24)
+ (28 - |- . 24) + (26 - | . 24) }
v ( (max.) = + 1800 kilos.
4 1 (min.) = - 8800 kilos.
0 = - X5 x 1 ' 875 - 1000 { (I + . . . + |) 6 + (f . 6 - 2) + ( J . 6 - 4) }
- 5000 {(! + ... + f) 6 + (f. 6 -2) + (I. 6 -4)}
X5 (min.) = - 48000.
0 = Y5 X 21-88 + 1000 { (| + . . . + |) 24 - (30 - f . 24) - (28 - f . 24)
- (26 - | . 24) }
+ 5000 (i + ... + £) 24 - 5000 { (30 - f .24) + (28 - f.24)
-(26-*. 24)}
Y j (max.) = + 6850 kilos.
5 I (min.) = - 6850 kilos.
-5000 {(* + ... +*)8 + (*. 8-2) + (f. 8 -4) + (1.8 -6)}
Z5 (max.) = + 48100 kilos.
0 = - V9 X 10 + 1000 { (i + ... + |) 4 - (10 - f . 4)
-(8-f .4)-(6-1.4)}
+ 5000 (| + . . . + A) 4 - 5000 { (10 - | . 4)
= + 1500 kilos.
8500 kilos.
v ( (max.) = +
5 I (min.) = -
0 = - X. X 1 • 5 - 1000 { (i + . . . + | ) 4 + (1 . 4 - 2)
X6 (min.) = - 48000 kilos.
0 = Y. x 6 + 1000 {(| + ... + |) 4 - (8 - f . 4) - (6 - 1 . 4)}
+ 5000 (i + . . . + A) 4 - 5000 { (8 - f . 4) + (6 - | . 4) }
Y f (max.) = + 6250 kilos.
6 \ (min.) = - 6250 kilos.
0 = Z6 x 1-84 - 1000 { (i + . . . + f) 6 + (f . 6 - 2) + (| . 6 - 4) }
- 5000 { (i + ... + |) 6 + (f . 6 - 2) + (I • 6 - 4) }
Z6 (max.) = + 48900 kilos.
0 = - V6 x 4-8 + 1000 { (| + . . . + |) 0-8-(4-8-f . 0'8) - (2-8 ~J . 0'8)
+ 5000 (i + ... + |) 0-8 - 5000 { (4-8 - f . 0-8)
y ( (max.) = + 560 kilos.
6 I (min.) = - 7560 kilos.
26
BRIDGES AND EOOFS.
0009-,
09SL-
ooeu
ooss-
0081^
0088-
OOST +
OOS8-
09S+
0 = - X7 x 0-875 - 1000 (1 + . . . + 1) 2
- 5000 (| + . . . + 1) 2
X7 (min.) = - 48000 kilos.
0 = Y7 x 1-92 + 1000 {(I + ... + f)0-8
+ 5000(| + . .. + |)0-8
- 5000(2-8 - -1.0-8)
Y / (max.) = + 5470 kilos.
7 I (min.) = - 5470 kilos.
0 = Z7 x 1-43 + 1000.JQ. + ... + f)4 + (|.4 - 2)
— 5000 { (1 + ...+£) 4 + (1 . 4 — 2)
Z7 (max.) = + 50300 kilos.
0 = - V7 x 2 - 1000 x 2 - 5000 x 2
V7 (min.) = - 6000 kilos.
0 = - X8 x 0-875 - 1000 (1 + . . . + 1) 2
-5000(1 +... + 1)2
X8 (min.) = - 48000 kilos.
0 = Z8 x 0-8 - 1000 (i + ... + !) 2
- 5000 (| + . . . + 1) 2
Z8 (max.) = - 52500 kilos.
These results are shown in Fig 27.
§ 7. — DERIVED FORMS.
From the above calculations it is appa-
§ rent that the greatest stresses in the vertical
* and diagonal braces occur when the bridge
is partially loaded. It will be interesting to
ascertain according to what law the girder is
loaded when the greatest stresses obtain in the
braces. By noticing in each case what tem-
porary loads are left out of the general equa-
tion of moments it will be observed that any
diagonal brace, Y3 for instance, will be subject
to the greatest tension when all the joints
lying to the right of it are loaded, and will be
under the greatest compression when all the
joints lying to the left are loaded. This is
represented in Fig. 28 by the words " Ten-
sion " and " Compression."
Evidently if this diagonal were inclined
upward to the right instead of to the left the
§ 7. — PARABOLIC GIRDERS.
27
words in Fig. 28 would simply have to change places, and also
if the girder be looked at from behind (or else its image in a
looking-glass) the diagonal Y6 will appear in the same bay as
Y3 in Fig. 28 ; thus the arrangement of the moving load to pro-
duce the greatest stresses in Y6 will be as shown in Fig. 29.
If both diagonals are present in the same girder, as shown
in Fig. 30, and are so constructed as to be incapable of resisting
FIG. 28.
Compression. , Tension.
compression, they will come into play only when the loading is
such as to produce tension in them ; at other times they will be
subject to no stress just as if they were threads. In such a
FIG. 29.
Tension.
Compression.
girder therefore only the greatest tension given for the diagonal
bars in the above example need be considered. Thus in Fig. 30
the greatest tension in the diagonals of the third bay from the
FIG. 30.
left will be for the brace inclined upwards to the left the same
as that in Y3 (as found in the previous numerical example), and
for the brace inclined upward to the right the same as that
in Y6. Similarly the greatest tension in the other diagonal
braces of Fig. 30 can be written down from Fig. 27.
The vertical braces are always in compression in this case,
28
BRIDGES AND KOOFS.
FIG. 31.
as will appear at once from Fig. 31, for
since the diagonals are incapable of resisting
compression there would be nothing to op-
pose the vertical downward force produced
by the vertical brace if
it were in tension. The
greatest compression in
the verticals will be the
same as given in Fig. 27,
for only one of the dia-
gonals in each bay is in
tension at a time, and the other being there-
fore slack can be considered as absent.
Thus without any further calculations
the greatest stresses in a girder with crossed
diagonals can be written down from those
obtained in the previous example, and this
is done in Fig. 32.
If the diagonals are so constructed that
they can only take up compression (this is
sometimes the case in wooden girders),
it will appear by similar reasoning that
for the diagonals, only the greatest com-
pression, and for the verticals only the
greatest tension, found in the previous ex-
ample, will apply. As regards the minimum
stress or compression in the verticals, the
load each vertical supports at the top joint
can alone produce compression in it, for
the diagonals cannot do so, as they never
can be in tension. This load is either
1000 kil. or 1000 + 5000 kil., and there-
fore the greatest compression in the verti-
cals is
V (min.) = - 6000 kilos.
The greatest stresses in a girder of the above construction
§ 7. — PARABOLIC GIBBERS.
29
are given in Fig. 33, and the diagonals are shown in double
lines to express their incapability to resist tension.
In girders with a single system of diagonals, varying, how-
ever, from Fig. 27 in that the arrangement is symmetrical on
each side of the centre, the greatest stresses can be written
-45000
FIGS. 33, 34, AND 35.
•48000 —48000 -48000
-48000
-48000
-48000
-48000
+48100
-48000
+48100
-48000
+ 48100
• 48100
-48000
-4800Q
-48000
-48000
-48000
+ 48300
/
Centre.
+ 48100
down at once from the above, with the single exception of that
in the central vertical brace.
The stress in the central vertical of Fig. 34 evidently depends
on the tension in the adjacent parts of the lower boom at its
foot, and it must therefore always be in compression. This
compression will reach its greatest value when the tension in
the boom is a maximum, that is, when the bridge is fully
30
BRIDGES AND ROOFS.
-560
+ 7560
-560
§ 8. — THEORY OF PARABOLIC GIRDERS. 31
loaded ; and in this case each vertical brace has a compression
of 6000 kilos, to bear. Therefore, for the central vertical also,
V (min.) = - 6000 kilos.
In Fig. 35 it is obvious that the central vertical can only be
in a state of stress when there is a direct load on the top joint ;
this stress must therefore be compression, and its greatest
value is evidently
V (min.) = - 6000 kilos.
Lastly, if in the girder shown in Fig. 27 the signs of all
the stresses be changed, the greatest stresses for a parabolic
girder having the bow above, as shown in Fig. 36, will be
obtained. In fact, the whole of the reasoning and the equations
are precisely the same, except that all the signs must be
changed, and that maximum must be put for minimum, and
minimum for maximum. The derived forms shown in Figs. 37,
38, 39, 40, can be obtained from Fig. 36, as before.
§ 8.— THEORY OF PARABOLIC GIRDERS.
It appears from the above example that the stresses in a
parabolic girder can be found by the method of moments, even
when the theory of such girders is not known. Two properties
of these girders were discovered : the first is that the stress in
the horizontal boom is greatest when the bridge is fully loaded,
and is then equal throughout ; and the second, that when the
bridge is fully loaded the stress in
the diagonal braces is everywhere nil.
The last property is in reality con- i
tained in the first, for when X = Xj T j IP j -Xx
(Fig. 41), Y = O, or else the hori- |\ /
zontal forces at P would not be in
equilibrium.
It will be useful to investigate the
conditions upon which these properties depend. This knowledge
is not necessary to enable the calculations for any given parabolic
32 BKIDGES AND EOOFS.
girder to be made, but is required when the form of a girder
is to be found which will have these properties.
Consider a chain attached to the two fixed points A and B
(Fig. 42), and hanging in its curve of equilibrium. Let the
load be uniformly distributed over the span AB, and equal to
q per unit of length. Suppose that the chain is cut at its
FIG. 42.
lowest point S (where it is horizontal), and a horizontal
force H applied at the point of section to maintain equili-
brium. This force must be horizontal, since the part of the
chain at S is horizontal. Let the chain be also cut at any
other point P, applying a force T to maintain equilibrium. It
is evident that this force
FIG. 43. must lie in the direction of
the tangent at the point P.
The piece S P of the chain
(Fig. 43) is held in equili-
brium by three forces : viz.,
H, T and the resultant of
the load on the part S P.
This last force is equal to q%, where x denotes the hori-
zontal distance of P from S; and its point of application is
at a distance » from either P or S, since the load is uniformly
distributed over #.
Taking moments round P : —
Hy = qx.%. (1)
But since P is any point on the curve, this equation is true for
the point A, thus substituting I for «, and / for y.
H/-^.- (2)
§ 8. — THEORY OF PARABOLIC GIRDERS.
33
Dividing eq. (1) by eq. (2).
(3)
This is the equation to a parabola, and it is evidently also
applicable to the part S B of the chain. Thus the position of
all points of the chain can be determined by this equation, by
giving values to x and solving for y.
It is evident from Fig. 43 that the horizontal component
of T is everywhere equal to H, it is so therefore at the points
of attachment A and B ; it is also evident that the vertical
component V of T is q.x, and therefore equal to q.l at A and B ;
and, lastly, that
T = VH2 + V2.
If the manner of loading is altered, some points of the chain
FIG. 44.
Ft
FIG, 45.
may still remain on the parabola, and for these points equation
(3) will hold good. This is the case, for instance, when the loads
on each side of S are concen-
trated at points, so long as the
load at each point is equal to
the sum of half the distributed
load on the two adjacent bays
(Fig. 44); for the part SP of
the chain (Fig. 45) will still be
subject to the vertical load q.x
(the resultant of the four con-
centrated loads shown in the figure), and the point of applica-
ft*
tion of this resultant will still be at the distance horn S.
34
BRIDGES AND HOOFS.
The chain can be loaded as above by employing vertical
struts to transfer the uniformly distributed load (Fig. 44). The
unloaded portions of the chain are evidently straight, and the
form of equilibrium will therefore be that of a polygon inscribed
in a parabola. The above is even true if the vertex is not a
loaded point ; for consider the part P Pj of the chain cut out,
and equilibrium maintained by the forces T,Ti (Fig. 46). Taking
moments round P,
and this equation will not be altered if, instead of S and Q,
other points, such as S: and Q! (Fig. 47), are taken as loaded
points.
FIG. 4T.
u---
If the two points A and B cannot offer any horizontal re-
sistance but only vertical reactions, other means to resist the
horizontal pull H must be adopted, for instance, introducing
a horizontal boom. Thus a parabolic girder of the form shown
in Fig. 48 is obtained, which can carry a load, uniformly distri-
8. — THEORY OF PARABOLIC GIRDERS.
35
bated over the whole span, without requiring any diagonal
braces. The conditions, therefore, that must obtain, in prder
that the girder may have the properties mentioned above, can
be briefly stated thus : —
The feet of the verticals must lie in a parabola, the axis of
which is vertical and passes through the centre of the span.
FIG. 48.
111!
FIG.
I t
.]/
49.
The whole of the above reasoning remains true, if the girder
be turned upside down and all the forces reversed (Fig. 49) ; it
need not, therefore, be repeated.
This subject will be further considered under the head of
" Sickle-shaped Trusses " (Bowstring-roofs).
D 2
THIKD CHAPTER
§ 9. — APPLICATION OF THE METHOD OF MOMENTS TO THE
CALCULATION OF THE STRESSES IN BRACED GIRDERS
HAVING PARALLEL BOOMS.
The method of moments can also be employed in calculating
ordinary braced girders divided into rectangular bays. It is
hardly necessary to observe that the equation of moments
remains true although two of the three bars cut through are
parallel, their point of intersection being therefore at an infinite
distance, and consequently the lever arm of the stress in the third
bar being also infinite. All the lever arms in the equation of
moments are, however, infinite, and thus divide out of the
equation, enabling the required stress to be determined.* For
example, in the girder shown in Fig. 50, the stress Y in the
diagonal F G is to be found by taking a section a 13, applying
the forces X, Y, Z to maintain equilibrium, and forming the
FIG. 50.
equation of moments for one of the parts of the girder
(Fig. 51) with reference to the point of intersection of X and
* In this case these infinities are all equal ; they can therefore be considered
as a common factor, but generally it is a mathematical fallacy to treat infinite
factors in this manner. — TKANS.
§ 9. — GIRDERS WITH PARALLEL BOOMS.
37
Z. This point, which is at infinity, can be considered as lying
on the central horizontal line of the girder, and since X and Z
pass through it their lever arms are nil, but the lever arms of
all the vertical forces are evidently infinite. Now, if 0 were
the point of intersection of X and Z, at a distance x from the
point where a /3 cuts Y, the lever arm of Y would be x . sin <£,
and if O be considered to move off to infinity, x becomes
infinite, and the lever arm of Y is oo . sin <j>.
FIG. 51.
Therefore, the equation of moments is
or dividing out by oo .
0 = Y . sin f - D + + + O + q) + (p + q).
Here Y . sin <f> is the vertical component of Y, hence the
above equation is merely the expression of the law that for
equilibrium the sum of the vertical forces must be zero.
Thus the principal object of the method of moments (to
obtain an equation containing only one unknown) is, in this
special case, arrived at by resolving the forces vertically. This
shows the general applicability of the method of moments ;
for even in special cases like the present, in which a shorter
way of obtaining the required result exists, it can be used, and
even points to the shorter method.
38 BRIDGES AND ROOFS.
Substituting the value of D in the above equation, viz. :
,
and arranging it according to the rule, to ascertain the effect
of the moving load :
-«(* + * + f + * + !) + « {(i-f) + a -I)}-
The maximum value of Y can now be obtained by leaving out
the positive member produced by q, and the minimum value
by omitting the negative member.
There is no difficulty in determining the stresses X and Z
(Fig. 51). Let X be the length of a bay, and h the height of
the girder, take moments first round G and then round F,
thus:
0 = XA + (p + <?) { (| + f + f + A + f) . 3 A + (f . 3 A - A)
+ C1.3A-2A)}
0 =- Z A + O + 9) { (i + f + f + f + f + f) . 2 A + (f . 2 A - A) } .
From which it is evident that these bars are in the greatest
state of stress when the bridge is fully loaded.
FIG. 52.
\*
Lastly, to find the stress in the adjacent vertical to the right,
take a section 7 8 (Fig. 52). The point about which to take
moments is at infinity, and
0 = - V . cc - D . oo + (| + |) oo + (p + q) <x> + (p + 5) oo
§ 10. — GIRDEKS WITH PARALLEL BOOMS.
39
The only difference between this equation and the one pre-
viously obtained for Y . sin </> is that — V is written instead of
Y . sin (f>. Hence it is obvious that
-|) -(1-1)}-
The values Y . sin <£ and — Y are therefore identical ; hence
by first finding V, Y can be ascertained by dividing by
( — sin <£). This is expressed in the following rule: the re-
solved parts vertically of the stresses in a diagonal and vertical
meeting at an unloaded joint are of equal magnitude, but of
unlike sign.
§ 10. — BRACED GIRDER, OF 16 METRES SPAN, COMPOSED OF
SINGLE EIGHT-ANGLE TRIANGLES.
Apart from the difference of form, the dimensions of this
girder are the same as those of the parabolic girder, already
calculated (p. 20), that is, the span (16m.) and the depth (2m.)
are the same. The loads are also the same, viz., 1000 kilos.
dead load and 5000 kilos, live load, on each bay. It is also
assumed that the line of railway is on a level with the upper
boom ; these loads, therefore, act at the upper apices (Fig. 53).
2500
r
efoo
B
5000
I
1000
<D
50 00 50 00
I
1000
F
i
lOiOO
FIG. 53.
5000
1000
K
5000
1000
Jf
5000
I
1000
,0
50 00 25 00
1 I
*- 500
S
Calculation of Y0 and Zx.
Since V0 and D are the only vertical forces acting at A
(Fig. 54):
V0 + D = 0, or V0 = ~ D.
40
BRIDGES AND EOOFS.
V0 therefore reaches its greatest numerical value when D is
greatest ; that is evidently when the bridge is fully loaded. In
.,. '•_ 48000,.,
this case D = — ^— kilos., and hence :
a
V0 (min.) = — 24000 kilos.
Further, since Zx is the only horizontal force at A :
FIG. 54.
2500 5000 5000 5000 5000 5000 5000 5000 2500
)0
1000 1000 1000 10
)0 1000 1000 1000 500
D
Calculation of X8 and
Two vertical forces act at S (Fig. 53), viz., the load at this
point (the greatest value of which is 3000 kilos.) and V8.
Therefore,
V8 (min.) = - 3000 kilos.
And since X8 is the only horizontal force at S :
Calculation of Xx, Z2, Vx, Yr
(Section aj3. Fig. 55.)
Taking moments for the part of the girder shown in Fig. 55,
about the point C :
whence
0 = X, x 2 + (1000 + 5000) X (i + | + . . . + |) x 2,
X, (min.) = - 21000 kilos.
Therefore,
§ 10. — GIRDERS WITH PARALLEL BOOMS.
The equation of moments about the point D is
0 = - Z2 x 2 + (1000 + 5000) (| + .. . +1)2.
Z2 (max.) = + 21000 kilos.
Eesolving the forces vertically :
o = - v» - 1000 (* + . . . + j) - 5000 (i + . . . + 1),
Vi (min.) = - 21000 kilos.
41
or
The diagonal Yx makes an angle of 45° with the horizontal.
Therefore, the resolved part of Yx vertically is Yx . sin 45°,
or Y! . — -= • Hence, by the rule :
r! = +21000 X ^2,
\ (max.) = + 29700 kilos.
FIG. 55.
^500
jroQ K
R1
< \ D
XT, \
J^
4k
\
" c V
FIG. 56.
2500
Calculation of X2, Z3, V2, Y2.
(Section 7 S. jp7^. 56.)
or,
Taking moments about E :
0 = X2 x 2 + (1000 + 5000) { (i + . • • + I) 4 + (1 . 4 - 2) } ,
X2 (min.) = — 36000 kilos.
42 BRIDGES AND ROOFS,
Taking moments about F :
0=-Z3X 2 +(1000 + 5000) {(£ + ... + f)4 + (1.4 -2)}
Z3 (max.) = + 36000 kilos.
Eesolving the forces vertically :
0=-V2-1000{i + ... + f_(l- I)}
- 5000(| + . .. + I) + 5000(1 - I).
By leaving out of this equation, first the negative, then the
positive members, multiplied by 5000 :
V2 (max.) = - 1875 kilos.
V2 (min.) = - 15625 kilos.
Then multiplying these values by -y/2, and changing the
sign:
Y2 (max.) = + 22100 kilos.
Y2 (min.) = + 2650 kilos.
In a similar manner the stress in the remaining bars can be ascertained from
the following equations :
0 = X3 x 2 + (1000 + 5000) { (i + ... + £) 6 + (f . 6 - 2) + (1 . 6 - 4) }
X3 (min.) = - 45000 kilos.
0 = - Z4 x 2 + (1000 + 5000) { (* + ... + f) 6 + (| . 6 - 2) + (J . 6 - 4) }
Z4 (max.) = + 45000 kilos.
0 = - V. - 1000 { i + ... + I - (1 - f) - (1 - 1) }
- 5000 (i+... + £) + 5000 { (1 - |) + (1 - 1) }
v ( (max.) = + 375 kilos.
3 I (min.) = - 10875 kilos.
Y j (max.) = + 15400 kilos.
3 \ (min.) = - 530 kilos.
0 = X4 x 2 + (1000 + 5000) { (i + . . . + |) 8 + (f . 8 - 2)
X4 (min.) = - 48000 kilos.
0 = - Z5 x 2 + (1 000 + 5000) { (| + . . . + *) . 8 + (f . 8 - 2)
Z5 (max.) = + 48000 kilos.
0= - V4-1000{ i + ... + A _(l _!)_(!-. f)_ (i_
- 5000 (i + ... + !) + 5000 {(1 -
y ( (max.) = + 3250 kilos.
4 I (min.) = - 6750 kilos.
Y f (max.) = + 9550 kilos.
4 I (min.) = -4600 kilos.
§ 10. — GIRDERS WITH PARALLEL BOOMS. 43
The following equations are formed with reference to the part of the girder
situated to the right of the section line :
0 = - X5 x 2 - (1000 + 5000) { (1 + ... + |) 6 + (f .6 - 2) + (|.6 - 4)}
X5 (min.) = - 45000 kilos.
0 = Z6 x 2 - (1000 + 5000) { (I + . . . + f ) 6 + (f . 6 - 2) + (| . 6 - 4) }
Z6 (max.) = + 45000 kilos.
0= - V. + 1000 U + ...+*-(l-f)-(l-f)- (1-1)}
+ 5000 (£ + ... + 1) - 5000 { (1 - *) + (1 - *) + (1 - 1) }
v / (max.) = 4- 6750 kilos.
5\C ' :
min.) = — 3250 kilos,
max.) = + 4600 kilos,
min.) = - 9550 kilos.
0 = - X6 x 2 - (1000 + 5000) {(i + ... + |) 4 + (|.4 - 2)}
X6 (min.) = - 36000 kilos.
0 = Z7 x 2 - (1000 + 5000) { (i + ... + f) 4 + (|.4 - 2) }
Z7 (max.) = + 36000 kilos.
0 = - V6 + 1000 {* + ... +f - (1 - I) - (1 - I) }
+ 5000 (i + . . . + I) - 5000 { (1 - |) + (1 - 1) }
v ( (max.) = + 10875 kilos.
6 1 (min.) = - 375 kilos.
Y ( (max.) = + 530 kilos.
6 I (min.) = - 15400 kilos.
0 = - X7 x 2 - (1000 + 5000) (| + . . . + D.. 2
X7 (min.) = - 21000 kilos.
0 = Z8 x 2 - (1000 + 5000) (i + . . . + 1) . 2
Z8 (max.) = + 21000 kilos.
0 = - V7 + 1000 [i + ... + I - (1 - I) } + 5000 (| + ... + I)- 5000(1 - |)
v ( (max.) = + 15625 kilos.
7 I (min.) = + 1875 kilos.
Y ( (max.) = - 2650 kilos.
7 I (min.) = - 22100 kilos.
The diagonal Y8 does not meet any vertical at an unloaded
joint, for the joint K (Fig. 53) cannot be considered unloaded on
account of the reaction of the abutment. The rule for finding
Y is therefore not applicable in this case. The vertical forces
Y
acting at K are the resolved part of Y8 or —j= , the reaction
*v 2
W of the abutment and the stress in the last vertical, which
has already been found = 3000 kilos. Hence for equilibrium :
^L + W - 3000 = 0.
V2
44
BRIDGES AND ROOFS.
SS9SI+
SI801+
OS19 +
OSL9-
SI8
£1801-
ooois-
Y8 therefore obtains its greatest negative
value when W is a maximum, that is, when
| the bridge is fully loaded, in which case
$ 48000,.,
W = — JT — kilos. ; and, therefore,
A
§ Y8 (min.) = - 21000 X V 2 = - 29700 kilos.
§
+ The results obtained are shown in Fig. 57.
§ 11. — DERIVED FORMS.
If the above equations be examined, to
ascertain what positions of the live load
produce the greatest stress in the diagonal
braces, it will be found that the law already
found for parabolic girders (p. 26) holds
good, or the stress in any diagonal is a
maximum or a minimum when the joints on
one side only are loaded.
The stresses in a girder in which the dia-
gonals slope upwards from left to right
(instead of from right to left) can evidently
be obtained by looking at the girder of
Fig. 57 from behind.*
If the diagonals are to be tension-braces,
and unable to resist compression, the follow-
ing alterations will have to be made in the
arrangement of the original girder ; in the
bays where the diagonals are always in
compression, they must be changed for
diagonals sloping in the opposite direc-
tion, and in the bays where the diagonal
braces are subject alternately to tension and
to compression, two diagonals must be in-
troduced.
Figs. 58 and 59 represent two girders having opposite
diagonal systems, and the kind of stress in each brace is denoted
* Holding the page up to the light. — TRANS.
OOOTZ-
§ 11. — GIRDERS WITH PARALLEL BOOMS.
45
by the sign + , signifying tension, and — compression. Carrying
out the above alterations, Fig. 60 is obtained, in which the dia-
gonals are never in compression, and the greatest numerical
value of the tension in them can at once be written down by
means of Figs. 58 and 59, taking the values from Fig. 57.
FIG. 58.
A vertical brace can only be in tension when the diagonals
meeting it at an unloaded joint are in compression. This can
never occur in Fig. 60 ; and the verticals can, therefore, only
be in compression ; consequently, only the values of V (min.)
FIG. 59.
need be considered, and for the left-hand side of the girder
these values must be taken from Fig. 58, and for the right-
hand side from Fig. 59.
The stresses in the horizontal bars X and Z are greatest when
FIG. 60.
the girder is fully loaded, and when this is the case it is easily
seen that, in the left half of the girder, the diagonals sloping
upwards from right to left will be in tension, and in the right
half those sloping from left to right. Evidently, therefore, the
stresses in the booms can be obtained from Fig. 58 for the left
half, and from Fig. 59 forjihe right half of the new girder.
46
BRIDGES AND ROOFS.
ooots-
ouoiz-
Thus, without further calculation, the
stresses in a girder of the form shown in
Fig. 60 can be obtained. These stresses
have been given in Fig. 61.
If the diagonals can only resist com-
pression (as is often the case in wooden
structures), the stresses can be obtained
by an exactly similar process from Fig. 57.
These stresses are shown in Fig. 62.
If the line of railway is carried on the
lower apices, instead of on the upper, it can
be considered that both the live and the
dead loads are applied to the lower joints.
The stresses in the horizontal and diagonal
bars will not thereby be altered, and the
stress in the verticals can be found by
the rule of Section 9, namely, that the
diagonal and vertical braces meeting at an
unloaded joint have equal vertical stresses,
but of contrary sign. In this case the un-
loaded joints are the upper ones, and in
Fig. 63 the stress in any vertical can be
found by dividing the stress in the dia-
gonal meeting it at the top joint by <\/2,
and changing the sign. From Fig. 63 the
derived forms shown in Figs. 64 and 65 can
be deduced as before.
If the line is carried on the verticals
between the booms, the points of attach-
ment can also be considered as the points
of application of the live and dead loads.
All the upper as well as the lower joints
are therefore unloaded, consequently the
resolved part vertically of the stress in any
diagonal will be the numerical value of the
stress in the parts of both the verticals it
meets. For instance, in Fig. 66 the diagonal in the third bay
is subject to the maximum and minimum stresses
+ 15400 and - 530 kilos.
SZ.9SI-
00013-
uoo^s—
11. — GIRDERS WITH PARALLEL BOOMS.
FIG. 62.
47
-21000
V45000
+ 21000
-21000
36000
-36000
+ 45000
FIG. 64.
-45000
48000
-48000
+ 48000
-48000
+21000
•36000
+45000
+ 45000
-45000
+21000
-21000
+ 36000
+45000
+ 48000
+48000
-48000
21000
+3600"
+45000
+48000
+43000
48
BRIDGES AND ROOFS.
421000
+15625
+ 10815
+ 6150
-3250
+3250
-10815
These values divided by \/2 and with
changed signs give
- 10875 and + 375 kilos,
and these will be the stresses in the
upper part of the vertical to the left
and in the lower part of the vertical to
the right. The stresses in the verticals
in this figure as well as in the girders
shown in Figs. 67 and 68 can therefore
be obtained without difficulty, using the
rule given in § 9. As to the stresses in
the horizontal and diagonal bars it is
evidently immaterial whether the loads
be carried on the top or bottom joints or
between them. Lastly in girders with
symmetrically arranged diagonals all the
stresses can be written down from Figs.
57, 63, and 66, with the exception of the
stress in the central vertical. For this
reason only the central part of the girder
is shown in Figs. 69, 70, 71, 72, 73, and
74, and it is easy to see that the stress in
the central vertical will be either ± 6000
kilos, or 0 according as the end which
does not meet a diagonal is loaded or not.
§ 12. — EEMARKS ON THE DEGREE OF
ACCURACY OF THE ASSUMPTIONS
MADE WITH REGARD TO THE DIS-
TRIBUTION OF THE LOADS.
Some objections may be raised to the
above calculations, for the distribution of
the loads on which they are based is not
strictly true, and the results to be accurate
require a slight correction.
12. — REMARKS ON THE DISTRIBUTION OF THE LOADS. 49
FIG. 69.
21000
g +1563 5
-10875
-375
+3^50
3750
375
1-10875
1-375
+ 6750
^3250
i
-6750
3250
-10875
-21000
^21000 [^-24000
-45000
-48000
-48000
- 45000
+45000
45000
FlG. 70.
-45000
-45000
+ 45000 +48000 +48000
FlG. 71.
45000
-45000
-4.8000
-45000
•4-45000 -t-45000
FIG. 72.
-45000
- 45000
+ 45000
+ 48000 +48000 +45000
FIG. 73.
45000
45000
50
BRIDGES AND ROOFS.
In the first place, the weight of the girder itself acts on
the upper as well as on the lower joints, and not, as assumed,
at the points through which the line of rails passes only. The
correction in this case, however, will only apply to the vertical
braces ; for, as already seen, the stress in the remaining bars is
independent of the position of the rails. Taking any of the
verticals in Fig. 53 or Fig. 54 and distributing the load on it
in due proportion between the top and bottom, it is easily seen
that the method of moments could be applied to find the stress
in that vertical. But it is better to make the calculation as in
§ 9 and § 10 (i. e. taking the point of application of the dead
load the same as that of the live load), and then, if considered
necessary, apply a correction in the following manner : Imagine
FIG. 74.
* 48000
+ 48000
45000
a secondary vertical placed alongside of the main one, the object
of this vertical being to realize the assumption made by trans-
mitting the load on what has been considered the unloaded joint
to the loaded joint. This secondary vertical will be a strut when
the load has to be transmitted downwards, and a tie when it has
to be transmitted upwards. The stress in it will therefore be
negative when it is above the line of rails, and positive when it
is below. Now if the secondary vertical be considered amal-
gamated with the principal vertical, it is evident that the actual
stress in the latter can be found by adding to the stress already
determined the stress in the former.
To make this clearer by an example, let the true distribu-
tion of the dead load in Fig. 57 be |rds of the 1000 kilos, on the
top joints and Jrd on the bottom joints, whereas it was con-
sidered that the whole of the dead load was applied to the top
joints. The secondary vertical has therefore to transmit
333 kilos, from the lower to the upper joints, and is consequently
§ 12. — REMARKS ON THE DISTRIBUTION OF THE LOADS. 51
a tie with a stress of + 333 kilos. This must now be added
to the stress in all the verticals. For instance, in vertical V3
V3 (max.) = + 375 + 333 = 708 kilos.
V3 (min.) = - 10875 + 333 = - 10542 kilos.
In the girder shown in Fig. 63 the line of rails is attached
to the bottom joints. Supposing that the true distribution of
the dead load is ^rd on the upper joints and frds on the lower
joints, it is evident that the secondary verticals will be struts
transmitting 333 kilos, from the upper to the lower joints,
and therefore to obtain the correct stress in the verticals
— 333 kilos, must be added to the stresses already found. For
instance in vertical V3
V3 (max.) = + 3250 - 333 = + 2917 kilos.
V3 (min.) = - 6750 - 333 = - 7083 kilos.
In this case the correction is so small that it might be
neglected. But in larger bridges, where the dead load is large
in comparison to the live load, and is more equally distributed
on the joints, the correction becomes important.
There is a second correction to be made, in connection with
the distribution of the moving load. It will be remembered that
it was assumed that each bay was bridged over by secondary
girders,* so as to convey the dead and live loads on the line of
rails to the adjacent joints. It is evident that it is only when a
bay is fully loaded that the reaction at each end of the
secondary girder can be equal to half the load on one bay.
Now the stresses in the diagonal and vertical braces were
calculated on the supposition that all the joints on one side
of the brace were fully loaded, and all those on the other side
free of the moving load. With a uniformly distributed moving
load this obviously cannot occur.
Yet, when it is considered that in reality the moving load
is not uniform and continuous, but that, on the contrary, the
load is concentrated at the points of contact of the wheels with
* These secondary girders are supposed to be discontinuous. — TRANS.
E 2
52
BRIDGES AND ROOFS.
the rails, and that in the case when the distance between the
wheels is equal to the distance between the joints, the above
assumption is strictly true, it would appear that the error is
insignificant, unless indeed the number of bays is small. At
any rate the error affects only the diagonals and verticals, and
is on the safe side.
Both these sources of error disappear, the latter when the
number of bays is very great, and the first when there are no
verticals, in which case the calculations differ slightly from the
preceding ones. To illustrate this latter point, the following
example has been chosen.
§ 13. — BRACED GIRDER WITH EQUILATERAL TRIANGLES.
(WARREN GIRDER.)
(Railway Bridge over the Trent near Newark.)
Each girder (Fig. 75) is composed of 27 bays of equilateral
triangles having their apices alternately above and below.
The line of rails is on a level with the bottom boom, one half
of the load is carried directly on the lower joints, and the other
half is transmitted by means of vertical ties to the upper joints.
Thus one half of the dead as well as of the live load acts on the
lower joints, and the remaining half on the upper joints. The
FIG. 75.
whole girder is supported at A and B by bolts carried on cast-
iron frames which rest on the piers. The distance apart of these
points of support is 259 feet, and therefore the length of the
259
side of one of the triangles is -jj- = 18 5 feet. The depth of
the girder is —^- x tan. 60° = 9 • 25 X 1 • 73 feet. Taking 9 • 25
§ 13. — WABREN GIRDER.
53
feet as the unit of length, the side of the triangles will be
represented^ by 2, the height of the girder by 1'73, and its
length by 28.
The weight of the whole bridge is 589 tons, and since there
589
are four girders, the dead load on each girder is -j- = 147 ? tons.
Taking the moving load on each line at 1 ton per foot run,
2 x 259
it will amount to — = 129 • 5 tons on each girder. The
total load on one girder is therefore
147-25 + 129-5 = 276-75 tons.
276 ' 75
Thus the load on each joint is — ^— tons, or in round
numbers lO^tons. In the following calculations the live as well
as the dead load has been taken for simplicity at 5 tons on each
joint, although the proportion of the dead to the live load is as
147 * 25 : 129 • 5 ; this willj make no difference in the stresses in
the booms, and the greatest stresses in the diagonals will be
slightly increased. Besides, the live load cannot be considered
as accurately equal to 1 ton per foot run ; often it is taken
higher.
FIG. 76.
D
l
U
1 U
ll
is
ll
U
1,
ll
i.
N
la
1*
U
z
The only object of the vertical bars is to transmit part of
the load to the upper apices; they do not form an integral
part of the truss, and can therefore be omitted in the calcula-
tions, the upper joints being considered loaded instead. The
distribution of the load is therefore as shown in Fig. 76.
54
BRIDGES AND ROOFS.
Calculation of the Stresses X and Z in the Upper and
Lower Booms.
Cutting off the part shown in Fig. 77 by the section line
a/3, and taking moments first about the point M and then
about the point N, the following equations are obtained, denot-
ing by D the reaction at the abutment A :
0 = X4 x 1-73 + D x7-5 (1 + 2 + 3 + 4 + 5 + 6)
0= - Z4 x 1-73 + D x 8 -5(1 + 2 + 3 + 4 + 5 + 6 + 7)
-5(1 + 2 + 3 + 4 + 5 + 6 + 7).
Substituting for D its value
D = 5(A + A + -•• + «) + 5(A + A + ••• + tt),
FIG. 77.
ID
and arranging the equations so that the effect of each load may
be seen (according to the previous rule).
o = X4 x 1-73
+ 5 { (^ + . . . + fi) 7 + (|| . 7 - 1) + (1| . 7 - 2) + . . . + (|| . 7 - 6) }
+ 5 { (^ + . . . + |i) 7 + (H • 7 - 1) + (|| . 7 - 2) + . . . + (|| . 7 - 6) } .
0 = - Z4 x 1-73
+ 5 { (^ + . . . + |f ) 8 + (fi . 8 - 1) + (|| . 8 - 2) + . . . + (|| . 8 - 7)
In these equations all the members containing the live load
have positive signs, and the omission of any of them would con-
sequently diminish the numerical value of the stress.
Having thus shown that the booms obtain their greatest
stress when the bridge is fully loaded (and this, it may be
added, is true in the case of all lattice girders), the calculations
§ 13. — WARREN GIRDER. 55
can be simplified by putting for D its value when the moving
load covers the bridge, namely,
D = lOtfy + * + - .. + ft) = 135 tons,
and combining the members containing the live and dead loads.
The above equations then become
0 = X4 X 1-73 + 135 x 7 - 10(1 + 2 + ... +6)
0= -Z4x 1-73 + 135x8- 10 (1 + 2 + ... + 7),
whence
X4 (min.) = - 425 tons,
Z4 (max.) = + 462 tons.
The following equations, for the remaining parts of the booms, are obtained in
an exactly similar manner :
0 = X, x 1-73+135 x 1
Xj (min.) = — 78 tons.
0 = - Z! x 1-73 + 135 x 2 - 10 X 1
Ti^ (max.) = + 150 tons.
0 = X2 x 1-73 + 135 x 3 - 10(1 + 2)
X2 (min.) = - 216 tons.
0= -Z2 x 1-73 + 135 X 4 - 10(1 + 2 + 3)
Z2 (max.) = + 277 tons.
0 = X3 x 1-73 + 135 x 5 - 10(1 + 2 + 3 + 4)
X3 (min.) = - 338 tons.
0= -Z3 x 1-73 + 135 x 6- 10(1 + 2 + 3 + 4 + 5)
Z3 (max.) = + 381 tons.
0 = X5 x 1-73 + 135 x 9 - 10 (1 + 2 + . . . + 8)
X5 (min.) = - 494 tons. *
0 =. - Z3 x 1-73 + 135 x 10 - 10(1 + 2 + ... + 9)
Z5 (max.) = + 520 tons.
0 = X6 x 1.73 + 135 x 11 - 10 (1 + 2 + . . . + 10)
X6 (min.) = - 540 tons.
0 = - Z6 x 1-73 + 135 x 12 - 10 (1 + 2 + . . . + 11)
Z6 (max.) = + 555 tons.
0 = X7 x 1'73 + 135 x 13 - 10 (1 + 2 + . . . + 12)
X7 (min.) = - 564 tons.
0 = - Z7 X 1-73 + 135 x 14 - 10(1 + 2 + ... + 13)
Z7 (max.) = + 566 tons.
56 BRIDGES AND ROOFS.
Calculation of the Stresses Y and U in the Braces.
Since the braces make an angle of 60° with the horizontal,
the resolved part vertically of the stresses in them are
Y . sin 60° and U . sin 60°
or
Y x 0-866 and TJ x 0'866.
Kesolving the forces acting on the parts of the girder shown in
Figs. 78 and 79 vertically.
0 = Y4 x 0-866 _D + 5x6 + 5x6
0 = - U4 x 0-866 -D + 5x7 + 5x7.
FIG. 78.
P 1. I.
U U L .
I is L is /
A\~~ A A AcoV
X
j. U
I 1.
Substituting for D its value
and arranging the equations according to the previous rule
o = Y4 x 0-866
- 5 { A + A + . . . + ft - (i - ft) - (i - if) - . . . - (i - 1|) }
- 5 (A- + . . . + ||) + 5 { (1 - ||) + (1 - ||) + . . . + (1 _
0 = - U4 x 0-866
- 5 { *V + • • • + it - (i - !i) - (i - It) - • • • - (i - ID }
In these equations the members containing the moving load
do not all possess the same sign, therefore leaving out first the
positive and then the negative members,
o = Y4 x 0-866 -
-
X4 (max.) = + 91 tons.
§ 13.— WARREN GIRDER. 57
0 = Y, X 0-866 - 5 to + • • • + ft - A - A ~ • • • - A)
+ 5QV + A + --.+ 3V)
X4 (min.) = + 39 tons.
0 - - U4 x 0 • 866 - 5 ( ,V + . . . + |f - ^ - A - . . . - A)
+ 5 (A + * + .•• + A)
U4 (max.) = — 32 tons.
0 = - U4 x 0 • 866 - 5 to + . . . + fa - A _ _«_ _ . . . _ _i¥)
-5(A + A + --- + tt)
U4 (min.) = — 81 tons.
It appears that only Y4 (max.) and U4 (min.) need be
taken into consideration, and therefore the calculations for Y4
(min.) and U4 (max.) could have been spared. But it is ad-
FIG. 79.
I, i.
/
.
I L\ *
7
\
visable always to calculate both values, for sometimes the
stresses are of different signs, in which case both must be
retained.
The equations and the stresses in the remaining diagonals are found similarly
as shown below :
0 = Y, x 0-866 - 5 to+ ... + «) - 5 to + ... +-fj)
Y ( (max.) = + 156 tons
1 I (min.) = + 78 tons
0 = - U, x 0-866 - 5 to + ... -Ml - A) - 5 (A + ••• + ID
( (max.) = — 72 tons
I
(min.) = — 144 tons
0 = Y2 x 0-866 - 5 to + --• + « ~ & ~ A)
- 5 to + ... + if) + 5 (A +
Y / (max.) = -f 133 tons
2 \ (min.) = + 66 tons
58 BRIDGES AND ROOFS.
0 = - U2 x 0 • 866 - 5 ( A + . • - + ft " - A - A - A)
- 5 (A + . . . + It) + 5 (A + & + A)
IT ( (max.) = - 59 tons
2 I (min.) = - 122 tons
0 = Y3 x 0-866 - 5 (A + ... + «- A --'.- A)
Y f (max.) = + 112 tons
3 I (min.) = + 53 tons
0 = - U3 X 0-866 - 5 (A + ... + ft ~ & «... - A)
- 5 (A + ... + «) + 5 (A + ... + A)
jj f (max.) = — 46 tons
3 \ (min.) = — 101 tons
0 = Y5 x 0-866 -
f (max.) = +
5 I (min.) = +
= + 71 tons
24 tons
0 = - Us X 0-866 - 5 (A + ... + if - ^ - ... - A)
- 5 (^ + .. . + if) + 5 (A + ... + A)
YJ ( (max.) = — 17 tons
5 I (min.) = — 61 tons
0 = Y6 X 0-866 -5 (A +-.. +
Y f (max.) = + 52 tons
6 \ (min.) = + 9 tons
0 = - U6 X 0-866 -5 (A + ... + «-«-... -A)
- 5 (A + ... + «) + 5 Q| + ... + A)
»j T (max.) = — 0-8 tons
6 I (min.) = - 42 tons
0 = X7 x 0-866 - 5 (A + - - - + if - H - . . . - A)
- 5 (A + .-. 4- if) + 5 (If + ... + A)
2£ | (max.) = + 34 tons
7\(min.) = -7-4 tons
0 = - U7 x 0-866 - 5 (A + ... + B - ft - — - A>
- 5 (A + ... + W + 5 (B + ... + A)
^j | (max.) = + 16 tons
7 \ (min.) = - 25 tons.
Since the girder is symmetrical with respect to the central
line, the stresses in the corresponding braces in the other half
will be exactly the same, and need not therefore be calculated.
The verticals have to sustain a part of the dead load as
well as the 5 tons moving load. Half the weight of the line
of railway, which forms part of the dead load, is supported by
the lower joints, and the other half is transmitted by the
verticals to the upper joints. This weight is 24*75 tons,
14. — MULTIPLE LATTICE GIRDERS.
59
and since there are fourteen
bays, each vertical will have
24-75
FIG. 80.
FIG. 81.
= 0-88 ton to carry,
is there-
2 x 14
Each vertical
fore subject to a tension of
+ 5*88 tons.
The results of the above
calculations are collected
together in Fig. 80.
If all the signs of the
stresses be changed, those in
a similar girder turned upside
down (Fig, 81) are obtained.
[NOTE. — Since the loads are
equally distributed on the top
and bottom joints, it is immaterial,
so far as regards the booms and
diagonals, whether the line be
placed on a level with the bottom
or the top boom. In the first case
the verticals will be ties, and in
the second they will be struts.]
§14.
In order that the rela-
tion between braced girders
with a single triangulation,
and those having two or
more (or Trellis and Lattice
girders as they are some-
times called), may be clearly
shown, the girders in the
following examples will have
a span of 16 metres and a
total load of 48,000 kilos.,
so that they may be com-
pared with the girder of ^ 10.
60 BRIDGES AND EOOFS.
If the load on the girder shown in Fig. 57 were only half
what it was assumed to be, the stresses would be exactly a half
of those given in the figure. Taking two such girders with
halved stresses, one in which the diagonals are inclined upwards
from right to left as in Fig. 57, and the other from left to right
as in Fig. 63, and placing them exactly one behind the other,
so that all the corresponding bars, with the exception of the
diagonals, cover each other, a girder is obtained the crossed
diagonals of which are capable of taking up either tension or
compression. Wherever two bars coincide the stresses in them
are to be added, and the stresses thus obtained are those pro-
duced in the derived girder by the total original load, one-half
of which acts on the upper apices, and the other half on the
lower apices. In each of the verticals, except those over the
abutments, the stress vanishes, for the maximum stress in one
girder is added to the minimum stress in the other. To pro-
duce the above loading it is necessary, when the whole load is
applied at the level of the upper boom, to introduce verticals to
transmit half of it to the lower joints, in which case these
verticals will be in compression ; or if the line of railway is
attached to the lower boom, vertical ties must be used to convey
half the load to the top joints ; and lastly, if the line is placed
between the two booms a vertical will be required, the lower
half of which will be in compression, and the upper half in
tension. The verticals in compression will have a stress of
— 3000 kilos., and those in tension of + 3000 kilos. In this
manner the stresses given in Figs. 82, 83, 84 have been ob-
tained. (So as not to overload the diagrams with figures, the
stresses in the booms and diagonals have been omitted from
Figs. 83 and 84. They are the same as those in Fig. 82.)
Again, if two girders of the design shown in Fig. 57, with
halved stresses, be placed so that one overlaps the other by half
a bay, and if the stresses in the parts of the booms where they
overlap be added, the stresses in a braced girder of the form
shown in Fig. 85 will be obtained, but they will only be true if
the girder be supported as indicated in the figure. If the line
of railway is placed on the lower boom or between the two
booms, Figs. 63 and 66 can be employed in a similar manner
14. — MULTIPLE LATTICE GIRDERS.
61
to form the girders given in Figs. 86 and 87. (The stresses in
the booms and diagonals have been omitted, being the same as
those in Fig. 85.) If the diagonals in such a girder be so con-
structed that they can only resist tension, the form of the
girder and the stresses in it will be as shown in Figs. 88, 89, 90.
(The stresses in the horizontal and diagonal bars in Figs. 89 and
90 are the same as those in Fig. 88.)
FIGS. 82, 83, AND 84.
Centre
-4G&00
•10500
+28500
+ 40500
+46500
+46500
\
The girders in the last six figures have a length of 17 metres
(instead of 16 metres), and the stresses given are only true
if there are two points supports at each abutment.*
If, however, the original span and method of supporting the
girders is to be retained, the design shown in Fig. 91, made up
of the two simple systems of Figs. 92 and 93, can be employed.
The stresses given have been calculated on the supposition that
the live as well as the dead load is applied to the upper
* The girders shown in Figs. 85 to 90 are not of much practical use, for
although their length is 17 metres, the clear span is only 16 metres.— TRANS.
62
-1500
-1500
•H812
+5431 T
-181
+54.31^
-181
+3315'
-1625
+3315
-1625
-3315
+1625
-5437
+187
-5437
+187
-7812
-7812
-10500
it
at
-10500*1
a
e
o
r-
-12000^
c
o
-12000
FIG. 85.
BRIDGES AND ROOFS.
FIG. 86.
+10500
+10500
+ +7812
+ 5437
-187
+ +5437
I ~'8'
+ +3375
-1625
+ +3375
-162
5 -,,«
. +1625
+ 1625
+ -5437
+ 187
+ -5437
*181
-*• -7812
o
o
\j>
o
+ -1812
o
<J -1060
-J050Q
FIG. 87.
+10500
+10501
+ 7812
+ 7812-
+5437
-187
+ 337
-162
+ 337
-162
-3375
+ 162
-337
+1625
-5437
+ 187
-5437
+ 187
-7812
-7812
-10500
-10500
31500
3-1500
-7812
§ 11. — MULTIPLE LATTICE GIRDERS.
63
extremity of the verticals. The stresses in Fig. 92 are therefore
obtained by dividing those in Fig. 57 by two. The stresses in
Fig. 93 must, however, be calculated anew, taking the dead
load at 4000 kilos, and the moving load at 20,000 kilos.
These calculations are exactly similar to those given in
§ 10. It is to be observed that in this case, contrary to all
FIGS. 88, 89, AND 90.
<S O O g ^
7-10500^-71000 V-mOO 7-36000 7-40500 7-45000 *?-46500 ^-48000 ?-48000C?-48000 ?-4G500
0 +10500 -t-SJOOO ^28500 +36000 +40500 +45000 +45UOO -.-45000 +40500
the previous examples, the first and last verticals jare^ subject
to bending stress (for this reason they have been shown in
double lines in the figure). From Fig. 94 the three following
equations are obtained for the three bars in the first bay : —
0 = X x 2 + 12000 x 1 (turning point P)
X = - 6000 kilos.
0 = Y x A= - 12000, Y = 16971 kilos.
0 = — Z x 2 — 12000 (turning point 0)
Z = - 6000 kilos.
FIG. 91.
0 -13500
BRIDGES AND ROOFS.
FIG. 92.
xj -1500
—187
-aoe
'+3375
—1625
-2500
-3375
'& /+906
-4406
—5137
*
X~ / -7812
— 12000 /-2-1000
+ 7812
+5437
—187
-1825
-3375
-5437
—10500
—12000
FIG. 93.
0 -12000
+9156
+6625
+4406
—906
+2500
-4406
—12000
§ 14. — MULTIPLE LATTICE GIRDERS.
65
Both X and Z are negative, these bars are therefore in com-
pression. The first vertical is therefore held in equilibrium by
the four forces shown in Fig. 95 ; thus irrespective of the
12,000 kilos, direct compression in its lower half, it is in the
same condition as a beam supported at both ends and loaded in
FIG. 94.
FIG. 95.
o
\
:
.. V
\j
»
;\
V >7
12
000 |
-Qt— > £j
eooo
GOOO
10971
JOO
the centre with 12,000 kilos. The same figure evidently also
represents the stresses in the last vertical.
To avoid these bending stresses the first and last diagonal
can be placed as represented in Fig. 96, in which case the
equations for the three bars of the first bay will become (Figs.
97 and 98)
0 = X X 2 + 12000 X 1 (turning point P)
X = - 6000 kilos.
0 = Y x — °
— 12000
Y = + 13416 kilos.
0 - — Z x 2 (turning point J)
Z = 0.
The stresses in the bars of the last bay will be similarly
altered. This alteration will, however, not affect the stresses in
the other bars, and they remain the same as in Fig. 93. Com-
bining the design of Fig. 96 with that of Fig. 92, a girder of
the form shown in Fig. 99 is obtained. (Only a few of the
stresses are given, for the others coincide with those of Fig. 91.)
Starting with the girders of Figs. 91 and 99, a series of derived
forms can be obtained by altering the position of the loads and
the nature of the diagonals. Following, for instance, the
F
BRIDGES AND ROOFS.
-1500
/+7812
—24000
- §'
-f6625
/ -f4406
— 906
-2500
—4406
-9156
-12000
§ 14. — MULTIPLE LATTICE GIRDERS.
67
reasoning of § 11 and assuming that the diagonals can only
take up tension, Figs., 100 and 101 are obtained, in which only
a half of the girder is shown, for it is symmetrical about the
centre, and the stresses in the corresponding bars are equal.
Fig. 102 is obtained by replacing the verticals in Fig. 91
by diagonals inclined to the right at an angle of 45°. This is
a trellis girder with four triangulations, and can also be con-
sidered as made up of the four girders shown in Figs. 103,
104, 105, and 106.
FIG. 97.
FIG. 98.
GOOO
12000
13416
12000
The stresses given have been calculated on the supposition
that the span of the girders is 16 metres, their height 2 metres ;
the total dead load __ kilos, and the total live load
4
FIG. 100.
—16500 —25500 —33000 —
—43560 —46500 —48000 —48000 • -48000
+6000 +16600 4-25500 +33000 +-39000 +43500 +46500 j +46500
kilos., so that the girder of Fig. 102 may correspond with the
former cases.
The dead load, however, according to the more accurate
assumption, has been equally distributed betwe^Br-the— bottom
*afa?«!£\
UNIVERSITY
68
—16500 —25509
BRIDGES AND ROOFS.
FIG. 101.
—48000 I —48000
o 4-6000
FIG. 102.
+46500 | +46500
i I I 1 4 I i I 4 i 1 I 1 i I 1 i
I 1 1 I i I I I I i I ! i I I I
I i i i i I
+3312,6 -7187,5
-10187,5
2687,0
+8312,5
—2687,5
+10812,5 +12812,5 +7812,5 -2687,5
FlG. 104.
—10187,5 -11687,5 -71S7,5 +38123
-2687^ +7812^ +12812^ +10812,5 +8312,5
FIG. 105.
-4625 -10625 -10625 -4625
+9000
-9000
+12000
FIG. 106.
-12000
+9000
* £
+5875
+11875
+J187S
+5875
§ 14. — MULTIPLE LATTICE GIKDERS.
69
and top joints, but the live load, as in previous cases, is
exclusively applied to the upper joints.*
Therefore at each lower joint there is a dead load of
250 kilos, and at each upper joint a dead load of 250 kilos.,
together with a live load of 2500 kilos, (with the exception of
the end joints, which of course have only one-half the load to
carry). Fig. 107 is obtained by calculating the stresses in the
four single lattice girders with these assumptions and then
fuzing them together. The end verticals in Figs. 102, 103,
104, and 107 are represented by double lines to indicate that
they are under bending stress.
A comparison of this trellis girder of four triangulations
with the one of eight triangulations, shown in Fig. 108, will
FIG. 107.
-4000 —14500 —£3500 —31000 -31000 -41500 —44600 —46000 ! -46000
+6500 +17000 +26000 4-33500 +89500 +44000 +47000 +-48500
+48500
convince that when the stresses in a trellis girder of a still
greater number of triangulations have to be determined it is not
necessary to go through all the calculations of all the single
systems of which it is composed.
The stresses m the booms increase gradually from the abut-
ments to the centre, and the stresses in the diagonals decrease
gradually in the same direction, and these increments and
decrements vary according to a law which becomes all the
more evident the greater the number of triangulations and the
greater the number of stresses actually calculated.
* This assumption is not strictly accurate, for the weight of the line of way,
the longitudinal, and the cross-girders is applied to the same joints as the live load,
and it is only the weight of the truss itself that can be considered as equally dis-
tributed between the bottom and top joints. — TRANS.
70 BRIDGES AtfD ROOFS.
Thus, as soon as the calculations have been carried to a
certain point, the shorter method of interpolation may be
adopted.
The stresses in the diagonals of the girder of Fig. 108 are
on an average half those of the corresponding ones of Fig. 107,
which agrees with the number of diagonals being double, and the
stress in each of the new diagonals is very nearly the arith-
metical mean between the two adjacent diagonals. Further,
the number representing the stress in any part of the booms of
Fig. 107 is almost exactly the arithmetical mean of the two
numbers that take its place in Fig. 108. It therefore is not
FIG. 108.
-1187,5 -6812,5 -12062,5 -16937,5 -21437,8 -25562,5 -29812,5
ZA+3687,5 +9812,5 +11562,5" +19437,5 +28937,5 +28062,5 +31812^
actually necessary to calculate the stresses in each of the eight
single lattice girders, forming the girder of Fig. 108, but the
stresses already found for lattice girders with two or four
triangulations can be used, according to the degree of
accuracy required, to determine the required stresses by
interpolation.
The points of application and the values of the bending
forces acting on the end verticals are given in Figs. 109 and 110.
In the first figure the resolved parts vertically and horizontally
of the stresses are given instead of the stresses themselves, and
in the second figure only the bending forces on the vertical are
§ 14.— MULTIPLE LATTICE GIRDERS.
71
FIQ. 109.
j 1187,5
shown. The direct compression which also exists in these
verticals can be found from Fig. 108.
In Fig. 108 it was assumed that a dead load of 125 kilos,
was placed on each lower joint and a dead load of 125 kilos.,
together with a live load of 1250 kilos, on every upper joint.
If instead of this, the dead load on the lower and upper joints
had been taken as 1, and the live load as 0, the stresses given
in Fig. Ill would have been obtained.
These numbers are what may be called the stress-numbers
when the bridge has no moving
load upon it.
These numbers also apply to
any similar trellis girder with
eight triangulations if the span
is eight times the height. If the
dead load is p kilos, (or any
other unit of weight) on each
joint, the stress in kilos, can evi-
dently be obtained by multi-
plying the stress-numbers by p.
Fig. 112 gives the stress-
numbers obtained supposing the
dead load to be 0 and the live
load on the upper as well as on
the lower joints to be 1. To
obtain the stress in kilos, of
a geometrically similar girder
having a live load of m kilos, on
the top and bottom joints, the stress-numbers are to be multi-
plied by m.
The stress-numbers in Fig. Ill are simultaneous, whereas
those in Fig. 112 do not occur at the same time, but give the
greatest stresses due to partial loading. The functions of the
diagonals can therefore be best investigated from Fig. Ill,
and it will be observed that by multiplying the stress-numbers
of this figure by p + m, the stresses in the fully-loaded girder
are obtained.
Taking two vertical sections through this figure, one at the
FIG. no.
1125
1000
625
250
750
3687,5
72 BRIDGES AND ROOFS.
FIG. 111.
—28 —20 -12 -4
FIG. 112.
-12 -4
§ 14. — MULTIPLE LATTICE GIRDERS.
73
FIG. 113.
first top joint from the left and the other at the second top
joint and applying the requisite horizontal and vertical forces
to maintain equilibrium, Fig. 113 is obtained. It will be
seen that in each section the vertical forces distribute them-
selves equally between the points of crossing of the diagonals,
and that the total difference of the vertical forces is equal to
the total load on the bay. It will further be observed that the
horizontal forces increase from the
centre towards the booms, and are
proportional to their distance from
the centre. These laws would have
been even more clearly expressed
with a trellis girder of sixteen tri-
angulations.
The greater the number of tri-
angulations, the greater will be the
analogy between the functions of the
diagonals and those of the solid web
of a plate girder.
The stress-numbers in Figs. Ill
and 112 can be employed as follows
to determine the stress in any other
trellis girder with eight triangu-
lations, geometrically similar, and
having a dead load of p kilos, and
a live load of m kilos, on the
upper as well as the lower joints :
Multiply the stress-numbers in
Fig. Ill ~by p, and those in Fig. 112
by m. The sum of the numbers obtained will give the stress
required.
Or representing by Zp the stress produced by the dead
load, by Zm the stress produced by the live load, and by Z the
total stress :
Z = p Zp + m Zm.
As an example take a trellis girder of 64 metres span and 8
metres depth, the dead load being p = 1500 kilos, and the
74 BRIDGES AND ROOFS.
live load m = 2000 kilos, on each top and bottom joint; then
for the sixteenth brace inclined upwards from right to left,
the above equation becomes
Z = 1500 x 0-707 + 2000 x 1-768 = + 4596 kilos.
as the maximum stress or greatest tension, and
Z = 1500 x 0-707 - 2000 X 1'061 = - 1061 kilos.
as the minimum stress or greatest compression.
In the same manner the following stresses are obtained for
the remaining diagonals inclined upwards from right to left,
beginning at the left end of the girder :
+ 21036, + 19799, + 18562, + 17324,
+ 16175, -f 15027, + 13877, -f 12727,
+ 11667, + 10606, + 9546, + 8485,
(+ 3713, /+ 2828,
I- 1944, I- 2828,
and these stresses multiplied by — 1 will give the stresses in
the diagonals inclined upwards from left to right.
Similarly it is found that the different parts of the booms
are subject to the following stresses, commencing at the left
end of the girder :
11375, 37625, 62125, 84875, 105875,
135125, 142625, 158375, 172375,
184625, 195125, 203875, 210875,
216125, 219625, 221375.
These numbers taken with the positive sign give the stresses
in kilos, in the lower boom and with the negative sign those in
the upper boom.
Again if p = 125 kilos, and m = 625 kilos, in the girder of
16m span, the stresses in the diagonals inclined upwards from
right to left, beginning at the left end of the girder, are
+ 4508, + 4243, + 3977, + 3712,
+ 3475, + 3237, + 3000, + 2762,
+ 2552, + 2342, + 2132, + 1922'
— 66,
j + 1740, (+ 1558, (+ 1376, (+ 1193,
I- 193, I- 320, I- 447, \- 574,
+ 1039, (+ 884,
_ 729, I- 884,
§ 14. — MULTIPLE LATTICE GIRDEKS. 75
and the stresses in the parts of the booms commencing from the
left are
2437-5, 8062-5, 13312-5, 18187-5,
22687-5, 26812-5, 30562 "5, 33937 '5,
36937-5, 39587-5, 41812-5, 43687'5,
45187-5, 46312-5, 47062-2, 47437 '5.
It will appear by comparing these numbers with those
given in Fig. 108 that the stresses are much altered when the
live load is applied to the lower as well -as to the upper joints
instead of only to the upper joints.
( 76 ) .
FOUKTH CHAPTER
§ 15. — SICKLE-SHAPED (BOWSTRING) KOOF OF 208 FEET
SPAN WITH A SINGLE SYSTEM OF DIAGONALS.
(Roof over Railway Station, Birmingham)
The unit of length in Fig. 114 is 16 feet, and the dimensions
given must therefore be multiplied by 16 to obtain them in
feet. Accordingly the span is
2 x 6-5 X 16 = 208 feet,
the height of the upper bow is
(1 + 1-5) X 16 = 40 feet,
and that of the lower bow
1X16 = 16 feet,*
and lastly the horizontal length of each bay is
1 X 16 = 16 feet.
The'ordinates of the upper bow are everywhere 2*5 times
those of the lower bow.
The load on the roof has been taken at 40 Ibs. per foot
super of horizontal area covered, including snow and wind-
pressure.
The distance apart of the principals being 24 feet, the area
supported by each is
208 X 24 = 4992 sq. feet,
* In the actual roof at Birmingham, the height of the lower bow is 17 feet,
but 16 feet has been adopted here to simplify the calculations. Otherwise the
dimensions given are the same as those in the actual roof. The calculations have
been made for a roof with single diagonals sloping upwards from right to left ;
but under the head of " Derived Forms " will be found the stresses for a roof
similar to the one at Birmingham, with crossed diagonal ties.
15. — SICKLE-SHAPED ROOF.
77
and the corresponding load
is
4992 x 40 = 199680 Ibs.
The load on each of the
thirteen bays is therefore
199680
FIG. 114
13
= 15360 Ibs., or
7-5 tons nearly (2000 Ibs.
to the ton).
The weight of the prin-
cipal itself deduced from
the dimensions of its parts
is very nearly 1*5 ton for
each bay.
Half the load on each
end bay is taken up di-
rectly by the abutments,
and each of the twelve ^
central joints has 1 • 5 tons ~
permanent and 7*5 tons
variable * load to carry. «^
Calculation of the Stresses
X and Z in the Upper
and Lower Bows.
Cutting off the part of
the roof shown in Fig. 115
by the section line a ft and
taking moments first about
M and then about N, the
following equations are ob-
tained :
0 = X4x I'205 + Dx4
-1-5(1 + 2 + 3)
-7-5(1 + 2 + 3)
0 = -Z4 x 1-055 + D x3
-1-5(1 + 2)
-7-5(1 + 2).
* As \vill be seen in the sequel, Prof. Bitter under-
stands by the load being variable that any joint or
joints may be loaded and the rest unloaded. This, it
will be observed, is not the usual English practice in the case of roofs.— TRANS.
78 BKIDGES AND ROOFS.
Substituting for D its value :
D = 1-5 to + -ft + • - . + «) + 7-5 to + A + • • • + if),
and arranging the equations according to the previous rule so
that the effect of the variable load may be traced :
o = X4 x 1-205
+ 1 ' 5 { (A + - • • + A) 4 + (if . 4 - 1) + (H . 4 - 2) + (i| . 4 - 3) }
+ 7'5 {(if . 4 - 1) + (ft . 4 - 2) + (|| . 4 - 3)
0= - Z4 x 1-055
+ 1 ' 5 { (^ + • • - + W) 3 + (H • 3 - 1 ) + (H • 3 - 2)
The members' containing the variable load, 7 • 5 tons, are all
positive and therefore of the same sign as the members due to
the permanent load 1 • 5 ton. Hence the stresses X4 and Z4 are
FIG. 115.
greatest when the structure is fully loaded. Solving these
equations :
X4 (min.) = - 134-4 tons.
Z4 (max.) = + 128-0 tons.
It having thus been shown that the greatest stresses in the
bows occur when the roof is fully loaded, it is better to
substitute for D its corresponding value,
§ 15. — SICKLE-SHAPED ROOF. 79
in the original equations, which then become
0 = X4 x 1-205 + 54 x 4 - 9(1 + 2 + 3)
0 = - Z4 x 1-055 + 54 x 3 - 9 (1 + 2).
In a similar manner the equations for the remaining parts of the bows are ob-
tained as given below :
0 = X, x 0-347 + 54 X 1
K! (min.) = — 155-6 tons
0 = -Z, x 0-41 + 54 x 1
Z1(max.) = + 131 -7 tons
0 = X2 x 0*672 + 54x2-9x1
X2(min.) = - 147 -3 tons
0= -Z2 X 0-415 + 54 x 1
Z2(max.) = + 130 -2 tons
0 = Xs x 0-963 + 54 x 3 - 9 '(1 + 2)
Xs (min.) = — 140-2 tons
0 = - Z3 x 0-767 + 54x2-9x1
Zs(max.) = + 129-1 tons
0 = X5 x 1'382 + 54x5-9(1 + 2 + 3 + 4)
Xs (min.) = — 130-2 tons
0 = - Z5 X 1-272 + 54 X 4 - 9 (1 + 2 + 3)
Zs (max.) = + 127-3 tons
0 = X6 x 1-481 + 54 X6-9 (1 + 2 + 33-4 + 5)
Xe(min.)= ~ 127 "6 tons
0 = - Z6 x 1-419 + 54 x 5 - 9 (1 + 2 + 3 + 4)
Za(max.)= +126 -9 tons
0 = X7 x 1-491 + 54 X7-9 (1 + 2 + 3 + 4 + 5 + 6)
X7(min.) = -126 -7 tons
0 = - Z7 x 1-491 + 54 x6-9 (1 + 2 + 3 + 4 + 5)
Z7 (max.) = + 126-7
0 = X8 x 1'41 + 54 x 8 - 9 (1 + 2 + . . . + 7)
X8 (min.) = — 127 '6 tons
0 = - Z8 x 1-489 + 54x7-9(l + 2+... + 6)
Z8 (max.) = + 126 -9
0 = X9 x 1-244 + 54 x 9 - 9 (1 + 2 + . . . + 8)
X9(min.) = -130 -2 tons
0 = -Z9X 1-414 + 54x8 -9 (1 + 2 + ... + 7)
Z9 (max.) = + 127-3 tons
0 = X10 X 1-004 + 54 x 10 - 9 (1 + 2 + . . . + 9)
X10 (min.) = — 134-4 tons
0 = - Z10 x 1-265 + 54x9-9(l+2 + ... + 8)
Z10(max.) = + 128-0 tons
80 BRIDGES AND ROOFS.
0 = Xn x 0-706 + 54 x 11 -9 (1+2 + ... + 10)
Xn (min.) = - 140 -2 tons
0 = - Zu x 1-046 + 54 x 10 - 9 (1 + 2 + . . . + 9)
Zn (max.) = + 129*1 tons
0 = X12x 0-367 + 54 x 12 -9 (1 + 2 +... + 11)
X12 (min.) = - 147-3 tons
0 = - Z12 x 0-76 + 54 x 11 - 9 (1 + 2 + . . . + 10)
Z12 (max.) = + 130-2 tons
0 = X13 x 0-347 + 54 x 12 - 9 (1 + 2 + ... + 11)
Xi3 (min.) = — 155 '6 tons
0 = - Z13 X 0-41 + 54 x 12 - 9 (1 + 2 + . . . + 11)
Z13 (max.) = + 131-7 tons.
It appears from these results that the greatest stresses in
the symmetrically placed parts of the bow are equal. Now as
the only difference between the corresponding bays on each
side of the centre is in the direction of the diagonal, it follows
that the greatest stresses in the bows are independent of the
position of the diagonals. It therefore makes no difference in
the results if the point round which moments are taken is at
the right or left angle of the bay ; that is, whether the point lies
in the diagonal or not. But this cannot be the case unless the
stress in the diagonal is nothing. From this it follows that
the diagonals have no stress in them when the roof is fully
loaded.
This property of bowstring roofs will be further discussed
in the " Theory of bowstring trusses."
Calculation of the Stress Y in the Diagonals.
To determine the stress Y4 take moments about 0 for the
part of the roof given in Fig. 116. 0 is the point of inter-
section of X4 and Z4, and it is found by construction that its
distance to the left of A is 2, and that the lever arm of Y4 is
4 • 68. Hence the equation,
0 = Y4 x 4-68 - D x 2 + 1-5 { (3 + 2) + (2 + 2) + (1 + 2) }
+ 7-5 { (3 + 2) + (2 + 2) + (1 + 2) } .
§ 15. — SICKLE-SHAPED ROOF. 81
substituting for D its value
D = 1-5 (A + A + — -Hi) + 7-5 (A + A + — + tf):
and arranging the equation so as to be able to observe the
effect of the variable load,
0 = Y4 x 4-68 + 1-5 { [3 + 2 (1 - tf)] + [2 + 2 (1 - #)]
+ [1 + 2 (1 - t|)] - fa + A + . . . + A) 2 }
+ 7- 5 { [3 + 2 (1 - «)] + [2 + 2 (1 - H)] + P + 2 (1 - «)]J
which simplifies to the following :
0 = Y4 X 4-68 - 1-5 [(A + ... + A) 2 - (3 + 2 + 1) (1
On calculation it appears that the co-efficient of 1 • 5 is zero,
thus confirming the result obtained above, by means of the
greatest stresses in the booms, that a uniformly distributed
load, such as the weight of the principal itself, produces no
stress in the diagonals. The last equation can therefore be
written in the simpler form :
Leaving out first the positive then the negative members
containing 7 • 5 tons.
/ (max.) — + 11 ' 1 tons ;
4 \(min.) = -11-1 tons.
Q
82 BRIDGES AND ROOFS.
This result again shows that a uniformly distributed load
produces no stress in the diagonals, for the maximum and mini-
mum values of Y4 are equal.
The stresses in the remaining diagonals are obtained in a precisely similar
manner. The equations given below have been written in the simpl'est form, that
is, omitting the permanent load :
0 = Y2 x 0-92 - 7-5 (TV + .. . + H) 0-2 + 7'5 (l + ^f )
Y ((max.) = + 8 -3 tons
2 \(min.) = -8 -3 tons
0 = Y3 X 2-52 - 7-5 (-& + ... + ii) 0-75
y ( (max.) = 49-5 tons
3 I (min.) = — 9 • 5 tons
0 = Y5 x 8-3 - 7-5 GV + ... +
Y ( (max.) = + 12-6 tons
5 \(min.) = —12 -6 tons
0 = Y6 x 17-6 - 7-5 (J-s + .. . + ^) 15
Y / (max.) = + 13-8 tons
6 I (min.) = - 13 -8 tons.
The point about which to take moments for the diagonal in the central bay lies
at infinity ; it is therefore necessary to follow the rule given in the third section,
§ 9. The sine of the angle this diagonal makes with the horizontal is = 0-831.
The equation to find Y7 is therefore,
0 = Y7X 0-831 X oo -7-5 (A + ... + -£) oo
+ 7-5 (6 + 5 + 4 + 3 + 2 + 1) (1 + ft) ;
or since the finite vanishes in comparison to the infinite,
0 = Y7 x 0-831 X oo-
+ 7-5(6 + 5 + 4 + 3 + 2+1) £-
and dividing out by the common factor GO,
0 = Y7 x 0-831 - 7-5 (^ + .. . + A) + 7-5 (6 + . . . + 1) ^
Y r(max.) = + 14-6tona
7 \(min.) = - 14 -6 tons
In the following equations the points about which moments are taken fall to
the right of the section line :
0 = - Y8 x 16-1 + 7-5 (TV + ... + A) 28
T (max.) = + 15-0 tons
8 I (min.) = - 15'0 tons
§ 15. — SICKLE-SHAPED EOOF.
83
0 = - Y9 x 7-1 + 7'5 (^ + ... + A) 18
-7'5(8 + ...+ l)(H — 1)
Y ( (max.) = +14-6 tons
9 \(min.) = -14 -6 tons
0 = - Y10 x 3-68 + 7-5 (A + T23 + A) 15
Y ( (max.) = + 14 •! tons
10 \(min.) = - 14-1 tons
0 = - Yn x 1-82 + 7-5 (TV + -ft) 13-75
Y / (max-) = + 13-0 tons
11 \(min.) = -13-0 tons
0 = - Y12 x 0-65 + 7'5 x ^3 X 13-2
/ (max.) = + 11 '6 tons
12\(min.) = - 11 -6 tons.
Calculation of the Stresses V in the Verticals.
To find Vi take moments for the part of the roof shown in
Fig. 117, round the point of intersection of Xi and Z2. By
FIG. 117. FIG. 118.
construction it is found that this point is at a horizontal distance
of 0 * 1 to the right of A. Hence the equation of moments is :
o= - Y! x 0-9 + D x o-i,
or substituting for D and arranging the equation,
0= -V X0-9
The co-efficient of 7 • 5 is positive, and therefore Vx is greatest
when the variable load covers the roof. Solving the equation
V, (max.) = + 6 tons.
The stress V2 is to be found from Fig. 118. The point of
G 2
84 BEIDGES AND ROOFS.
intersection of X2 and Z3 is at a horizontal distance of 0*06 to
the right of A and the equation of moments is
0 = - V2 X 1-94 + D x 0-06 + 1-5 x 0-94 + 7'5 x 0-94,
substituting for D and arranging the equation,
0 = -V2X 1-94 + 1-5 [GV + .
Here again the co-efficient of 7*5 is positive, and the equation
can therefore be solved as it stands, whence,
V2(max.)= +6 tons.
For the next vertical, V3, the point about which to take
moments falls to the left of A, and the equation consequently
alters its form to a slight extent.
The equation of moments roun*d 0 in Fig. 119 is
0 = - V3 X 3-214 - D x 0-214 + 1'5 (1-214 + 2*214)
+ 7-5(1-214 + 2-214),
and substituting for D and arranging the equation
0 = - V3 x 3-214 - 1-5 [(^ + . . . + ii)0-214 - (2 + 1) (l + £^-4)]
- 7'5 to + » • + if) 0-214 + 7-5(2 + !)(! + ^).
FIG. 119.
I *
In this case one of the members containing 7 * 5 is positive
and the other negative. To determine Y3 (max.) the negative
member, and to find V3 (min.), the positive member must be
left out. The value of V3, keeping both these members in the
equation, will also be calculated ; this gives the value of V3
when the roof is fully loaded : the reason for doing this will
appear further on.
§ 15. — SICKLE-SHAPED ROOF. 85
The values obtained are
y J(max ) = + 8' 1 tons. y =
3\(min.) . = - 1-1 tons.
Similarly the following equations are obtained :
0 = - V4 x 4-91 - 1-5 (^ + .
-7-5 to +... + A)0-91 +7-5(8 + 2
/(max.) = + lO-Stons. y = 6t
4l(min.)= -3'8 tons.
0= - V. x 7-5 - 1-5 [to + ...+A)2-5-
. + ^2-5 + 7-5(4 + ..
v {(max.) = + 12-9 tons. ys = + 6 tons
5\(mm.) = -5-9 tons.
+ 7-5(5 + ...
v J(max) = + 14-5 tons. y6 = + 6 tons.
l(mm.) = — 7'5tons.
- 7'5(TV + ... + A) 24-5 + 7-5(6 + ... + l)(l + ^)J.
v jOaax.) = + 15-4tons. y = + 6t
M(mm.)= -8-4 tons.
In the remaining bays the point about which moments are taken is to
the right of the section line, and the signs of the equations are consequently
3d.
0 = V. x 60 + l-5[to + — + *)68 - (7 + ... + l)(ff - 1)]
-7-5(7 + ...
0 = V9 X 13- 5 + 1 • 5 [to + • • • + A) 22 '5 - (8 + . . . + 1) (22 - l)]
+ 7-5 to+ ... + A) 22'5 - 7>
{(max.) = + 15-6 tons. V9 = + 6 tons.
9l(min.)= -8 -6 tons.
-43- 7-5(9 + .. . + l)-l.
v {(max.) = + 14-8tons. y = + 6tons<
10\(min.)= -7- 8 tons.
0 = Vn X 3-3 + 1-5 [(^ + A) 14>3 - (10 + . . . + 1) G£ - 1)]
+ 7'5 to + A) 14<3 - 7-5 (10 + . . . + 1) (^ - 1).
86 BRIDGES AND ROOFS.
y ((max ) = +1S-5 tons. Vll = + 6 tons.
l(mm.) = - 6- 5 tons.
0 = V12 x 1-385 + 1-5 [^ X 13-385 - (11 + ... + 1) (Hj** - l)]
+ 7'5 x A x 13-385 - 7-5(11 + ... + 1) (^ - l).
It was recommended in § 12 to assume that both the per-
manent and variable loads were applied to the same joints, and
this assumption was made possible by the introduction of
secondary verticals, whose object was to convey to the supposed
loaded joints the part of the permanent load belonging to the
other joints. In the present case it was supposed that the whole
of the weight of the principal was applied to the top joints.
Now in reality, this load is distributed between the upper
and lower joints, but the upper joints have the greater propor-
tion to bear, and only about one-third or 0*5 ton of the
permanent load on each bay falls directly on each lower joint.
The secondary vertical introduced to transmit this load to the
upper joints is therefore a tie, and the tension in it is 0*5 ton,
and this stress must fce added to the stresses in the verticals
previously found.
The more accurate values of the stresses in the verticals are therefore
V,(max.) = + 6- 5 tons.
V2(max.;= +6 -5 tons.
y {(max.) = + 8-6 tons. y 6.5 tong
3\(niin.) = — 0-6 tons.
v ((max.) = + 11-8 tons. y 6.5 1
* \(min.) = -3-3 tons.
v ((max.) = +13-4 tons. y 6.5 ^
l(mm.) = — 5'4 tons.
v ((max.) = + 15-0 tons. y fi.5 1
bl(min.) = -7-0 tons.
v ((max ) = +15-9 tons. y = fi.5 ^
l(mm.) = — 7' 9 tons.
v |(max.) = + 16-3 tons. y 6.g ^
8l(min.) = -8-3 tons.
(max ) = +16-1 tons.
^ ((max.) = + 16'
9\(min.)= -8-1
r ((max.) = +15-
10\(mm.) = - 7-3
tons.
y
§ 16.— SICKLE-SHAPED BOOFS.
87
The whole of the results are given
in Fig. 120.
§ 16. — DERIVED FORMS.
The above calculations show that
the diagonals of a double bowstring
roof possessing only one single system
of diagonals are subject both to tension
and compression. On examining the
equation of moments for the stress in a
diagonal it will be seen that the maxi-
mum stress in it is reached when all the
joints to the right, and the minimum
when all the joints to the left of it, are 7
loaded.
If the diagonals were inclined up- £
wards from left to right, the reverse 7
would obviously be the case, and the
stresses that then obtain can easily be ^
found by looking at Fig. 120 as it *
were from behind ; or, what amounts to
the same thing, the stress in a diagonal
inclined upwards from left to right can
be found from that in the diagonal situ-
ated in the symmetrically placed bay
and inclined upwards from right to left.
If in any bay the diagonal inclined
to the left can only take up tension, a
second diagonal of like properties in-
clined to the right must be introduced,
and it 'will come into play only when
the first one is slack, and vice versa.
The stresses in these diagonals can be
obtained from Fig. 120 ; the maximum
-15.3
-7.3
-8,
+ 1G,3
+ 15,9
-7
-7,0
+ 13,4
-0,6
88 BRIDGES AND ROOFS.
stress in the diagonal inclined to the left can be found directly,
and the maximum stress in the diagonal inclined to the right
will be the same as that in the diagonal of the symmetrically
placed bay.
Before the stresses in the verticals of a roof with crossed
diagonals can be determined, it is necessary to ascertain which
of the diagonals is in tension under the partial loading, for the
section line must be parallel to the diagonal which is in tension
in order to cut through only three bars. When the roof is fully
loaded, the stress in all the diagonals is zero, and at the same
time the stress in the lower boom is greatest. The tension in
the verticals will then also be greatest.
For besides the permanent load p
,(Fig. 121) the stresses Z and Z; are the
only forces that can produce tension
in the vertical. The stresses Y and
Z-* 1 f j *z Y' in the diagonals, when they exist,
produce on the contrary compression,
for the resolved parts vertically of
the stresses in them act upwards. But with a full load Z
and 71 are greatest, and Y and Y' are nothing, and therefore
the tension in the verticals is greatest under these circum-
stances.
The above appears even more clearly by observing the
effect produced by unloading one of the joints when the full
load is applied. Unloading a joint can be considered as the
application of a vertical force acting upwards ; and since the
diagonals are under no stress when the structure is not loaded
at all, as well as when it is fully loaded, it follows that it is
only necessary to investigate the effect of a vertical force acting
upwards on the unloaded and weightless structure as represented
in Fig. 122. The vertical force'K produces the reactions D and
W at the abutments A and B, and for simplicity only those
diagonals have been shown which are brought into tension by
this force. To find which of the diagonals in any bay is in
tension, take a section through this bay, and form the equation
of moments for the part (Fig. 123) which does not contain K,
round 0, the .point of intersection of the directions of the
§ 16. — SICKLE-SHAPED ROOFS.
89
booms. It is then easy to see which of the diagonals must be
acting to maintain equilibrium. For instance, in the fourth
bay it is the diagonal inclined to the right, for the equation of
moments is (Fig. 123)
0 = Dd-Yy or Y= + — * •
The stress Y is therefore positive, or the diagonal is in tension
FIG. 122.
(the equation for the other diagonal would give a negative stress
or compression).
As soon as it has been determined by this means which of
the diagonals are in tension it can be decided by a similar
process whether any particular vertical is in tension or compres-
sion. Thus, for instance, for the third vertical (Fig. 124)
the equation of moments is
0 = D 5 + V v.
from which a negative value is obtained for V, showing that it
is in compression.
A different process must, however, be employed for the
90
BRIDGES AND KOOFS.
vertical, acted on directly by the force K, because the section
line would cut through four bars. Now it is easy to see that K
produces compression in every part of the lower bow, the
equation of moments to find Z (Fig. 125) being
Thus the parts of the bow acting on the foot of the vertical in
question (Fig. 122) being in compression will produce com-
pression in it.
FIG. 124.
FIG. 125.
It is thus seen that unloading any joint diminishes the stress
in all the verticals, from which it follows that the tension in the
verticals will be greatest when all the joints are loaded.
It now remains to be decided what joints should be un-
loaded in order that the stress in any vertical may be a
minimum. Take, for instance, the ninth vertical ; it is evident
that unloading the eighth joint will diminish the stress in it,
FIG. 126.
t
«
7
B
•• '
\
and the same effect will likewise be produced by unloading the
seventh, sixth, &c., to first joint ; and, what is an important
point, the unloading of each of these joints will bring the same
16. — SICKLE-SHAPED ROOFS.
91
system of diagonals into tension in the two bays adjacent to
the ninth vertical (Fig. 126). But if the unloading were
still further continued the compression in the ninth vertical
would be diminished. For since in the two adjacent bays
to this vertical the same set of diagonals is in tension (the
system inclined to the left), the equation already found for
V9 (min.) holds good, and this equation shows that the com-
pression is diminished by unloading the ninth, tenth, eleventh,
and twelfth joints. Hence the value of Y9 (min.) found above
is also true if the diagonals are crossed. There is, however, a
second minimum value of V9 ; for it can be shown in a similar
manner to the above that the stress in V9 is a minimum when
the joints 10, 11, and 12 are unloaded and the remainder loaded
(Fig. 127). Evidently in this case the ninth vertical is in the
same condition as the fourth vertical in Fig. 120, and therefore
the value of V4 (min.) obtains. Hence to find the greatest
compression in any vertical the values of the two minima must
be compared and the one whose absolute value is greatest
taken.
As regards the stresses in the bows, they are greatest when
the roof is fully loaded, and consequently when the stress
in the diagonals is nothing ; the arrangement of the diagonals
can therefore produce no alteration in the stresses in the
bows.
Thus without any new calculations the stresses already
found can be inscribed in Fig. 128, showing a bowstring roof
with crossed diagonal ties.
By similar reasoning it is easy to prove that iti the case of
crossed diagonal struts which are not capable of taking up
tension (this is the case in wooden structures) only the maxima
values of the stress in the verticals apply, and that in fact
92
BRIDGES AND ROOFS.
compression cannot occur in the verticals owing to the com-
pression in the diagonals. The stresses given in Fig. 129
require, therefore, no further comment.
It must however be observed, and this does not only apply
to this case but also to wherever crossed diagonals exist, that
the stresses found above are only true if no artificial stresses
exist in the bars. Such artificial stresses cannot occur in single
diagonal systems, for in this case every bar can be reached by a
FIG. 128.
FIG. 129,
section line cutting through only three bars. Thus if no
exterior forces are acting on the system, the equation of
moments for any bar, whose stress is Y, about the point of
intersection of the other two bars included in the section is
But if two diagonals cross each other in a quadrilateral the
section line must cut through four bars, and the stresses in the
two diagonals tend to turn the part cut off in opposite direc-
tions round the point of intersection of the other two bars.
§ 17. — APPARENT FAILURES OF METHOD OF MOMENTS. 93
From Fig. 130 the equation of moments is
which implies the condition
Y y'
r = 7
but the absolute values of Y and Y' are indeterminate.
Therefore, if by means of set screws or otherwise an
artificial stress Y is set up in one diagonal, the stress in the
other will immediately change, in the above proportion, to
Y'. This will alter the stresses in the verticals and parts of
FIG. 130.
the bow in the same bay, and they can easily be found by the
method of moments as soon as Y is known and Y' determined.
The stresses given above are therefore only true if, when
the structure is unloaded, all the bars are without stress,
Then only one of the diagonals (either a tie or a strut) will be
acting at any time, but if artificial stresses are introduced it
might happen that both diagonals would be acting at the same
time.
§ 17. — APPARENT FAILURES OP THE METHOD OF
MOMENTS.
There are cases in the employment of this method, and some
have occurred in the last example, in which it would appear
that although a result is obtained it can only be approximate.
94 BRIDGES AND ROOFS.
In every case the point about which moments are taken is
the intersection of two of the bars cut through by the section
line. When these bars are nearly parallel the accurate deter-
mination of this point and the measurement of the lever arms
is connected with difficulties. In all probability two distinct
computers would arrive at different results.
This would seem to be a great disadvantage of the method.
But on further consideration it will appear that it is possible on
the contrary to derive some use from the circumstance.
It is clear that limits to the error can be obtained by first
intentionally giving the lines too great and secondly too small a
convergence (Fig. 131), and calculating in each case the stress.
Thus two values are obtained, and evidently the true value
FIG. 131.
lies between them. By comparing these values with the
intentionally committed errors it is possible to ascertain to what
degree the stresses will be altered by small errors in the
carrying out of the work.
For the uncertainty apparent on the drawing is in
reality a representation of what actually occurs by errors in
the construction. As the workman deviates in one direction
or the other from the working drawings, so the stresses
will approach one or the other limit. Therefore it is possible
to ascertain the alterations produced in the stresses owing to
inaccuracies in the carrying out of the design.
A second objection, even less founded than the former, is
that the method does not depend entirely on calculation, but
must obtain some of its data by graphic means. But calcula-
§ 18. — THEORY OF SICKLE-SHAPED TRUSSES.
95
tions should only be made when they arrive quicker at the
result than other methods. If, therefore, the graphic method
is shorter than calculation it should be adopted, especially
as there is less liability to error in measuring than in calcu-
lating.
§ 18. — THEORY OF SICKLE-SHAPED TRUSSES.
It will be noticed that in the preceding numerical example
the stresses were obtained without knowing anything of the laws
respecting the distribution of the stresses in the structure. If,
however, it were required to determine the form of the structure,
it would be necessary to be acquainted with these laws. For
this reason it is proposed to extend the " Theory of Parabolic
Trusses," commenced in § 8.
In that paragraph the equilibrium of a loaded chain was
considered (Fig. 42). If this chain be imagined to rotate through
two right angles about the horizontal axis A B, the vertical
FIG. 132.
forces will be reversed in direction and Fig. 132 obtained. The
chain can be considered as negatively loaded, and evidently the
equation obtained for Fig. 42 remains true, namely :
Similarly for another parabolic chain (Fig. 133) loaded with a
positive load P per unit of length of the span, the equation
HF =
PZ2
holds good, and evidently the load P can be so chosen that the
horizontal thrust H will be the same as the horizontal pull H
96
BRIDGES AND KOOFS.
in the former case. The requisite condition can be obtained
by dividing one equation by the other, thus :
/_*
F~P'
or in words : the loads per unit of length of the span must be
as the heights of arc. ,
If both these chains are placed on the same abutments
(Fig. 134) the reaction will be entirely vertical, for the hori-
l?
FIG. 134.
zontal thrust of one chain neutralizes the horizontal pull of the
other. The vertical reaction at the abutment will therefore be
equal to
V - v = P l -p l = (P - p) l.
This reaction is equal to that produced by a straight girder of
span 2 1 and with a uniformly distributed load of P — p.
The lower chain can be loaded negatively by means of ties
§ 18. — THEOEY OF SICKLE-SHAPED TEUSSES.
97
pulling upwards, the tension in them being equal to p per] unit
of length of span.
In the same manner part of the positive load on the upper
chain can be produced by means of ties pulling downwards. If
this part of the load be equal to the negative load on the lower
chain, namely p per unit of length of the span, there will still be
a load P — p on the upper chain, which will be designated by
Jc and which can be applied by external loads (Fig. 135) ; if the
FIG. 135.
-F-p
ties of the upper and lower chains be considered joined together,
the load p can be omitted, for its effect is exactly reproduced
by these vertical ties. A double bowstring truss without
diagonals has thus been built up, carrying a load on the top
equal to Jc per unit of length of the span.
The stresses in both bows as well as in the verticals can be
FIG. 136.
calculated from the magnitudes ?,/, F, Jc. For simplicity there
is a vertical to every unit of length of the span (Fig. 136), and
this is quite legitimate, for it was shown in § 8 that the load
could.be concentrated at points, and so long as the load at
H
98 BRIDGES AND EOOFS.
each point was equal to half the uniform load on the adjoining
intervals those points remained on the parabola.
P and p can be found from the equations ;
Putting ^ = - and reducing,
k
The load ~k produces therefore a tension - - in the verticals.
n - 1
In the preceding numerical example
!=/= J- = 2 5
n F 2-5 5" 2
Hence the tension in the verticals (or the negative load on the
lower bow) is in this case
and the load on the upper bow is
Thus, if the external load on each top joint is 7 '5 tons, the
tension in each vertical will be f x 7 • 5 = 5 tons, and the upper
bow is in the same condition as if loaded with § x 7*5 = 12 • 5
tons at each joint.
The load of 1 * 5 tons on each top joint due to the weight of
the truss itself produces a tension in the verticals = § x 1 . 5 =
1 ton, and the positive load on the upper bow is -J x 1 * 5 =
2 • 5 tons.
Lastly, if 7*5 + 1*5 =9 tons is the total load on each top
joint, the negative load per unit of length of span on the lower
bow is 6 tons, the positive load on the upper bow is 15 tons, and
the tension in each vertical is 6 tons.
If, however, part of the load is applied at the lower joints it
must be conveyed by secondary verticals to the top joints, and
the tension in these secondary verticals is to be added to that
in the main verticals. For instance, in the preceding example
§ 18. — THEORY OF SICKLE-SHAPED TRUSSES.
99
0 • 5 ton of the weight of the truss was considered as acting on
the lower joints ; 0 • 5 ton must therefore be added to the 6 tons
tension found above, and this coincides exactly with the value
obtained by the method of moments. The negative load
on the lower bow remains the same as before, namely 6 tons,
for the tension in the secondary verticals evidently does not
affect it.
The constant horizontal stress in the bows is :
which is tension in the lower bow and compression in the upper
bow. This is the same value that was obtained by the method
of moments (Z, = +126-7 tons, and X7 = - 126- 7 tons).
If - = 0, it follows that p = 0 and P = k ; that is, if the lower
n
bow becomes a horizontal straight line the loading of the upper
bow produces no tension in the verticals.
Further, if - becomes negative p also becomes negative ; that
n
is, the loading of the upper bow produces compression in the
verticals. For instance, if
In this case, therefore, one half of the load placed on the
top is transferred to the bottom bow (Fig. 137).
FIG. 137.
Generally, the above equations, &c., are true for a negative
as well as for a positive value of/
In all cases, therefore, when the load is uniformly distributed
H 2
100
BRIDGES AND EOOFS.
over the span the verticals alone are capable of maintaining
equilibrium. It is only when the load is unevenly distributed
that there is any tendency to deformation, and this is met by
the introduction of diagonals.
The law upon which the stresses in the diagonals of a double bowstring truss
depend can also be found, and is appended here for those readers who are
acquainted with the Calculus.
The diagonals together with the verticals make the truss perfectly rigid, and it
therefore behaves towards external forces in the same manner as a simple beam
FIG. 138.
supported at both ends. Thus, if a load Q be placed on it at a horizontal dis-
tance *, from the right abutment (Fig. 138) a reaction
will be produced at the other abutment A.
Take a vertical section M N through the truss
to the left of the weight Q, dividing the struc-
ture into two parts, one of which is shown in
Fig. 139. To maintain equilibrium forces must
be applied to this section. In order that the
algebraic sum of the vertical forces may be zero,
a vertical force V must be applied equal to D,
therefore
The force V alone would, however, with D produce a couple, and it is therefore
necessary for equilibrium to apply at the section a couple of equal moment. The
horizontal forces h h form such a couple. If the section line M N is indefinitely
near to one of the vertical braces, M and N are the only points at which a bar
can be intersected, and the horizontal forces must therefore be applied at these
points. The value of h can be found by taking moments about A, thus :
If, as before, the heights of the aro of the parabolas are F and / respectively,
the equations to these curves are,
-
F ~ P /'
From which the following values for M L and N L are obtained :
§ 18. — THEORY OF SICKLE-SHAPED TRUSSES. 101
Substituting these, as well as the value found above for V, in the equation of
moments ;
whence
2(F .-
And differentiating with respect to x,
dh Qlz
*= <*'•
dx 2(F-
This differential equation gives the rate of increase of h for an increase of the
abscissa x, that is when the point M moves towards the left.
The absolute value of h will evidently be greatest when the whole of the bow
J TT
from B to M is loaded with weights Q. Replacing Q by kdz and writing -3—
ax
for — - according to the previous notation used for a distributed load :
kl
/•« = ! + *
—^ • I z dz
*> J * = o
dx 2(F-
djl = kl_ (*+a?)«
dx ~ 2 (F -/)(/+ a)2- X 2
dH kl
"dx
If the loading were continued to the left of the point M, this negative
j TT
value of ~ — would approach 0, and to prove this it is only necessary to ascertain
as before the effect of a single load placed to the left of M, at a distance z from
the abutment A.
It is then found that all such loads make — - positive, and therefore - is
dx * dx
greatest when every point from A to M is loaded. The equations thus obtained
are:
o= - Q±
t _ QZa dh _ Qlz
~
2(F -/)(*-*
102
BRIDGES AND ROOFS.
dx ~
The general equation for •- — can therefore be put in the form :
fmaximum = +
I minimum = —
These results can -be employed in the following manner to determine the stresses
in the diagonals.
It will be remembered that the section line M N was taken indefinitely near to
a vertical. The point where the diagonal is cut will therefore be at the inter-
section with one of the booms, for instance, the lower one (Fig. 140). The three
forces H, V, H distribute themselves as follows ; at the point of intersection M
the force H is applied together with as much of the vertical force V as is necessary
to produce a resultant in the direction of the bow ; at the point of intersection
MTI
FIG. 140.
FIG. 141.
N with the diagonal a part of the horizontal force H, and as much of the vertical
force V as is necessary to give a resultant in the direction of the diagonal ; and
lastly at the point of intersection N with the lower bow the remainder of H, and
as much of the vertical force V as is necessary to produce a resultant in the
direction of the lower bow.
Had the section been taken at a distance d x further to the left, H and V would
have been replaced by
dx
§ 18. — THEORY OF SICKLE-SHAPED TRUSSES.
103
If both sections be taken simultaneously (Fig. 141) and the forces acting on
the part of the structure thus cut out be considered it will be observed that the
excess — . dx of the horizontal forces is the force that tends to move the upper
dx
bow to the right and the lower bow to the left.
If the breadth of the piece cut out be taken equal to the small quantity A
(instead of the indefinitely small quantity dx) — . A will still very nearly re-
present the distorting force, or substituting for — the value found above it will
be very nearly equal to
kl
.\
when a maximum or a minimum respectively.
This force distributes itself on the apices of the triangles formed by the
diagonals and verticals, and acts towards the left or towards the right according
to the position of the load.
FIG. 143.
FIG. 142.
Y=
If A is the length of one bay, the distorting force is applied to one apex only
and can be resolved into two components, one along the diagonal and the other
along the vertical (Fig. 143).
The component along the diagonal is
kl A
Y (max. or mm.) = db —= -r — —
4(F-/) cos a
or since = d, the length of the diagonal,
Y (max. or min.) = ±
_ ~
Therefore to find the greatest stress in any diagonal it is only necessary to
multiply its length by ± •
For instance in the case of the roof calculated in § 15,
kl 7'5 x 6-5
4(F-/) 4(2-5-1)
= 8-125.
104 BKIDGES AND ROOFS.
By measuring the lengths dv eZ3 . . . dlz of the diagonals and multiplying by
8' 125 the following table is obtained :
dz = 1-018, Y2 = 8-125 x 1-018 = 8'3
d3 = 1-163, Y3 = 8-125 x 1-163 = 9-5
c?4 = 1-861, Y4 = 8-125 X 1-361 = 11-1
d& = 1-55, Y5 = 8-125 x 1-55 = 12-6
d6 = 1-7, Y6 = 8-125 X 1-7 = 13'8
d,= I'8, Y7 = 8-125 x 1-8 = 14-6
d8= 1-835, Y8 =8-125 x 1-835= 14-9
d9 = 1-815, Y9 = 8-125 x 1'815 = 14-7
dw = 1-735,
dn = 1-605,
rf12 = 1-426, Y12 = 8-125 x 1-426=11-6.
Comparing these values with those given in Fig. 120 it will be seen that the
differences are very small.
The above law can be applied to the case of fish-bellied girders, by writing
-/for/;
dfH _ kl
~d^- ±4(F+/)*
It is also true in the special cases when the lower or the upper bow become
straight ; in the first case / = 0 and in the second F = 0, or — - = — , when the
lower bow is straight, and -^ - ** when the upper bow is straight.
For instance in the parabolic girder calculated in § 6,
<*H_**_ 2500x8
d^~47~ 4x2
and measuring the lengths c?2, d3 . . . d7 of the diagonals :
dz = 2-5 Y2 = 2500 X 2-5 = 6250
d3 = 2-741 Y3 = 2500 X 2'741 = 6850
d4 = 2-828 Y4 = 2500 X 2-828 = 7070
ds = 2-741 Y6 = 2500 X 2-741 = 6850
d6 = 2-5 Y6 = 2500 x 2-5 = 6250
dj = 2-183 Y7 = 2500 x 2- 183 = 5460
These stresses agree almost exactly with those given in Fig. 27.
It is possible to investigate a similar law for the stresses in the verticals, but
on account of their double function, first as braces and secondly as struts or ties
to convey the load from one joint to another, this law is very complicated and
consequently unsuited to practical purposes. Nor would the results agree with
those obtained by the method of moments as well as in the case of the diagonals.
For these general laws are based upon the supposition that the moving load
progresses gradually, whereas when using the method of moments it is considered
that the moving load advances by jumps from one joint to the next. It is there-
fore better in all cases to employ the method of moments to calculate the stresses
in the verticals.
( 105 )
FIFTH CHAPTEE.
§ 19. — CANTILEVER KOOF WITH STAY, SPAN 6 METRES.
The load, including snow and wind, is assumed to be 200
kilos, per square metre of horizontal area, covered. The dis-
tance apart of the principals is 4 metres. The load on each
principal is therefore
6 X 4 x 200 = 4800 kilos.
and the load on each of the 6 bays is 800 kilos.; of the 7
joints, the first and the last have 400 kilos, to support, and
FIG. 144.
the remaining five 800 kilos. (Fig. 144). The weight of the
truss itself being small, the whole of this load may be taken
as a variable load.*
Calculation of the Stress H in the Horizontal Bars.
The reactions W and P produced at the two points of
support A and C by a load Q are shown in Fig. 145. To
find the stress H in the bar M N, due to this load, the
equation of moments about the point O, for the part of
* As will be seen in the sequel, Prof. Bitter understands by the load being
variable that any joint or joints may be loaded and the rest unloaded. This, it
will be observed, is not the usual English practice in the case of roofs. — TRANS,
106
BRIDGES AND ROOFS.
the roof given in Fig. 146, would have to be formed. But the
position of Q has been so chosen that the resultant of Q and P
passes through O, and consequently H = o. It will also easily
be seen that all loads to the left of Q produce negative stresses,
and all loads to the right of Q positive stresses, in the bar
M N. Hence, when H is a minimum, the part of the roof over
FIG. 145.
Compression.
which "Compression" is written in Fig. 145 will be loaded,
and the remainder unloaded ; and when H is a maximum, the
loads will extend over the part marked " Tension."
FIG. 146.
FIG. 147.
The same result can be arrived at, however, by forming the
equation of moments, when all the joints are loaded, and arrang-
ing this equation so that the effect of every load can be seen.
When Q (distant 4 metres from the wall) is the only
load on the roof, a stress P is produced in the rod B C,
whose vertical component is S Q (Fig. 147) ; for the equation
' V OF TBE
UNIVERSITY
&fi
§ 19. — CANTILEVEK ROOF WITH STAY.
of moments about A shows that the vertical component of P
acts in the same manner as the reaction of the point of sup-
port B would if A B were a girder resting on two supports,
A and B. Since A B is to A C as 6 to 3, it follows that the
horizontal component of P is always twice as great as the
vertical component, and is in this case therefore equal to | Q.
Thus, the equation of moments to find H3 is (Fig. 148) —
or
H.x|=-Q|2-f
The increment to the stress H3 produced by Q is there-
fore composed of three parts. The first is the direct influence
of the load, and the other
two the indirect effect pro-
duced by the reactions.
If Q, however, were situ-
ated to the right of the sec-
FIG. 148.
tion line, the increment to
the stress would be com-
posed of two terms only,
both the indirect effect of the reactions. For instance, the
increment to the stress H3, produced by a load Q, 2 metres
from the wall, is to be found from the equation
or
Thus the equation to find H3, when all the joints are loaded, is
H3 x | = 800 (i • 4 - * . |) + 800 (* . 4 - f . |)
+ 800 (i . 4 - 1 . f - 1) - 800 (2 - f. . 4 + * . f )
+ 800 (3 - | . 4 + | . |) - 400 (4 - 1 . 4 + 2 . |).
Omitting the negative members from the right-hand side of
the equation,
H3 (max.) = + 2000 kilos. ;
and leaving out the positive members,
H3 (min.) = - 2000 kilos.
108 BRIDGES AND ROOFS.
The following equations for the remaining horizontal bars are obtained in a
similar manner :
H, x 1 = - 800 (1 - i . 6 + i . 1) - 800 (2 - $ . 6 + $ . 1)
- 800 (3 - i . 6 + 1 . 1) - 800 (4 - f . 6 + £ . 1)
- 800 (5 - | . 6 + | . 1) - 400 (6 - 1 . 6 4- 2 . 1)
Hj (max.) = 0, Ht (min.) = — 4800 kilos.
H2 x * = 800 (£ . 5 - * . f) + 800 (i . 5 - f . £ - 1)
- 800 (2 - i . 5 + 1 . J) - 800 (3 - | . 5 + * . |)
- 800 (4 - 1 . 5 + f . |) - 400 (5 - 1 . 5 + 2 . f)
H2 (max.) = + 640 kilos., H2 (min.) = - 3040 kilos.
H4 x i = 800 (i . 3 - * . i) + 800 (i . 3 - | . *)
+ 800 (| . 3 - 1 . i) + 800 (f . 3 - | . ± - 1)
_ 800 (2 - f . 3 + £ . i) - 400 (3 - 1 . 3 + 2 . i)
H4 (max.) = + 3733 kilos., H4 (min.) = - 1333 kilos.
H5 x i = 800 (i - 2 - £ . i) + SCO (i . 2 - § . £)
+ 800 (| . 2 - 1 . i) + 800 (| . 2 - f . £)
+ 800 (f . 2 - | . | - 1) - 400 (2 — 1 . 2 + 2 . £)
H5 (max.) = + 5600 kilos., H5 (min.) = - 800 kilos.
H6 x i = 800 (i . 1 - * • i) + 800 (| . 1 - | . i)
+ 800 (| . 1 - 1 . |) + 800 (| . 1 - A . i)
+ 800 (f. l--f.£) -400 (1-1.1 + 2.|)
H6 (max.) = + 8000 kilos., H6 (min.) = - 800 kilos.
For all the remaining bars the turning point lies in the line
A B ; and since the resultant W, of any load Q and the ten-
sion P produced by it in B C, always passes through A, it
follows that the greatest stress in all the remaining bars
occurs when every joint is loaded.
This total load of 4800 kilos, can be considered to act at the
centre of A B, and the vertical component of P will then be
i . 4800 = 2400 kilos. The horizontal component of P is twice
as great, or 4800 kilos. Consequently,
P = V24002 + 48002 = 5367 kilos.,
and this is the greatest tension in B C. From Figs. 149 and
150 the lever-arm of the stress X3, with respect to the point
M, is
/»
L M . cos a - \ . , - 0-4932 metre. '
'
§ 19. — CANTILEVER ROOF WITH STAY.
109
The equation of moments to determine X3 is therefore (Fig.
150)
0 = X3 x 0-4932 + 2400 x 3 - 800 (0 + 1 + 2 + f),
or
X, = - 7299 kilos.
FIG. 149.
'P
400
24:00
Similarly,
0 = X! x 0-822 + 2400 x5-800{
X! = - 2433 kilos.
0 = X2 X 0-6576 + 2400 x4- 800
X2 = - 4866 kilos.
0 = X4 x 0-3288 + 2400 x 2 - 800 (1 + f)
X4 = - 9732 kilos.
0 = X5 x 0-1644 + 2400 x 1 - 400 x 1
Xs = - 12166 kilos.
0 = X6 x 0-1644 + 2400 x 1 - 400 X 1
X, = - 12166 kilos.
2^3
FIG. 150.
800
...,-r"
,--'"'800
4800.
To find the stresses in the diagonals, moments will have to
be taken about the point B. The lever-arm of Y3 (Fig. 150)
with respect to this point is
B M . sin e = 3 . *" = 1'664 metre.
* The length of V3 (Fig. 149) is evidently % metre.— TRANS.
110
FIG. 152.
800
—1600
-2000
-2400
BRIDGES AND ROOFS.
and the equation of moments
0 = - Y3 x 1-664 + 800 (1 + 2 + 3),
or
Y3 = + 2884 kilos.
Similarly,
0 = - Y! x 3-536 + 800 (1 + 2 + 3 + 4 + 5)
Y! = + 3394 kilos.
0 = - Y2 x 2-561 + 800 (1 + 2 + 3 + 4)
Y2 = + 3124 kilos.
0 = - Y4 x 0-89 + 800 (1 + 2)
Y4 = + 2683 kilos.
0 = - Y5 x 0-316+ 800 x 1
Y5 = + 2530 kilos.
The stress in the verticals is also to
be found by taking moments about B.
Thus, for V3 the equation of moments is
(Fig. 151)
0 = V3 X 4 + 800 (4 + 3 + 2 + 1),
V3 = - 2000 kilos.
FIG. 151.
-2400
Similarly,
0 = ^x6 + 800 (f + 5 + 4 + 3 + 2 + 1)
V, = - 2400 kilos.
0 = V2 x 5 + 800 (5 + 4 + 3 + 2 + 1)
V2 = - 2400 kilos.
0 = V4 X 3 + 800 (3 + 2 + 1)
V4 = - 1600 kilos.
0 = V5 X 2 + 800 (2 + 1)
Va = - 1200 kilos.
0 = V6 X 1 + 800 x 1
V6 = - 800 kilos.
§ 20. — CANTILEVER ROOF WITHOUT STAY.
Ill
The reaction W of the fixed point of support A can be
found from its components H! and V1? and its greatest value
is,
W = V WTV? = V 48002 + 24002 = 5367 kilos.
or numerically the same as the tension in B C.
The stresses obtained are collected together in Fig. 152.
§ 20. — CANTILEVER KOOF WITHOUT STAY.
The dimensions and loading of the roof are shown in Fig.
153, and are the same as those of the similar roof given in
Fig. 144. Instead, however, of the toe being supported by a
FIG. 153.
stay, as in the former case, the roof is tied into the wall at the
point E.
The lever-arm with respect to A of the stress X, in the
bar tying the roof into the wall at E, is
A E . cos a = 1 .
= 0-9864.
And hence the equation of moments to determine X is
0 = X X 0-9864 - 800 (1 + 2 + 3 + 4 + 5 + |),
X = + 14599 kilos.
or
112
BRIDGES AND HOOFS.
FIG. 154.
-1200
—1600
-2000
—2400
To find Xx . . . X6, the corre-
sponding equations in the last para-
graph can be employed, by putting
P and its components = 0. Thus
the following equations are ob-
tained : —
0 = K! x 0-822 - 800 (1 + 2 + 3 + 4 + 1)
Xj = + 12166 kilos.
0 = X2 x 0-6576 - 800 (1 + 2 + 3 + |)
X2 = + 9732 kilos.
0 = X3 x 0-4932 - 800 (1 + 2 + |)
X3 = + 7299 kilos.
0 = X4x b-3288-800(l + f)
X4 = + 4866 kilos.
0 = X5 x 0-1644-400 x 1
X5 = + 2433 kilos.
0 = X6 x 0-1644 — 400 x 1
X6 = + 2433 kilos.
The same applies to the stresses
H! . . . H6, and the following equa-
tions are deduced from the former
ones by putting P and its compo-
nents = 0.
0 =
o =
o =
o =
o =
o =
H2xf
H2 = —
H3X|
-800 (1 + 2 + 3 + 4 + 5 +
- 14400 kilos.
- 800 (1 + 2 + 3 + 4 + |)
12000 kilos.
- 800 (1 + 2 + 3 + |)
- 9600 kilos.
- 800 (1 + 2 + |)
- 7200 kilos.
- 800 (1 + |)
- 4800 kilos.
- 400 x 1
- 2400 kilos.
The equations of moments to
find the stresses Yj ... V6 and
Y, . . . Y6 must be formed with re-
ference to the point B as in the
§ 20. — CANTILEVER ROOF WITHOUT STAY. 113
former case. In the preceding example the force P passed
through B; it had, therefore, no influence on the stresses
in the diagonals and verticals. Thus the stresses found for the
verticals and diagonals in the former example hold good in
this.
The reaction W can be found from its components Vi and
Hi, thus : —
W = Wi2 + Hx2 = */ 24002 + 144002 = 14599 kilos.
It is, therefore, numerically equal to the tension X.
The stresses in the various bars are collected together in
Fig. 154.
SIXTH CHAPTER.
§ 21. — BRACED ARCH OF 24 METRES SPAN.
The bridge is designed to carry a single line of railway, and
is supported by two braced arches. The moving load on the
bridge is taken at 4000 kilos, per metre run, of which, therefore,
one-half comes on each braced arch, and the length of a bay
being 3 metres, the moving load on each joint is 6000 kilos.,
or 6 tons (1000 kilos, to the ton). The dead load is estimated
at 1400 kilos, per metre run, or 700 kilos, for each arch ; that
is, 2100 kilos, on each joint, or approximately 2 tons.
The two halves of the arch are in contact at the point S
only (Fig. 155), and the connection is made by means of a
single bolt, thus forming a hinge.* Hinged joints are also
placed at the abutments A and Aj.
Preparatory to finding the greatest stresses, the effect of a
single load placed on the weightless structure will be investi-
gated.
A load Q placed anywhere on the right half of the arch
produces a reaction R at the hinge S (Fig. 156), between the
two halves of the arch. For the left half, the direction of this
force must pass through the point A, for otherwise rotation
round this point would take place. This force produces at A
a reaction R, acting in the direction A S ; this must be its
direction, or else rotation would ensue round S, besides which
action and reaction are equal and opposite. Let P be the
intersection of- the two forces R and Q, then it is easy to see,
* It would be a more rational form of construction if the hinge were situated
in the horizontal B Bt. But the above construction is more general, and in the
case of wooden structures the rational form would be difficult of execution. For
these reasons it has not been adopted here, but can easily be deduced by making
SO = 0 instead of 0-5.
§ 21.— BRACED ARCH. 115
by taking moments about this point, that the reaction D pro-
duced at the hinge Aj must pass through P, in order that
equilibrium may obtain. This reaction is also evidently equal
in magnitude and opposite in direction to the resultant of
K and Q.
Thus, to find the direction of the reactions at the abut-
ments due to a load Q placed on one half of the arch, the
line joining the hinge at the abutment of the other half with
FIG. 155.
FIG. 156.
the central hinge is produced to intersect the vertical through
the load, and from this point a line is drawn to the hinge
at the other abutment. The pressure at the central hinge
on the unloaded half is always directed to the hinge at its
abutment. (In the sequel the central hinge will be called
" the hinge," and the other two hinges the " abutments.")
The magnitude of the hinge- reaction E can be found by
resolving it into its horizontal and vertical components, and
then forming two equations of moments, one for each half
i 2
116 BRIDGES AND ROOFS.
of the arch. Thus, if H and V are these components, the
following equations are obtained from Fig. 156 :
0 = Vxl2-Hx4,
whence
V = ana H-
Thus having found the action of a single load Q on the
whole arch, it remains to determine the stresses this load pro-
duces on the various bars composing the structure. This is
best done by taking a section through any three bars, as before,
and writing the equation of moments for the part of the arch
comprised between this section and the hinge. As in former
cases, the moments are taken about the point of intersection of
two of the bars cut through. Whether any particular load
produces tension or compression in the bar under consideration,
can easily be determined by noticing in which direction the
load tends to make the part of the arch rotate. In this manner
the joints that must be loaded to produce tension in a bar, and
those which must be loaded to produce compression, can easily
be ascertained. The maximum stress is found by loading
all the former, and the minimum stress by loading the latter
only.
[NoTE. — It is necessary to know the direction in which the vertical com-
ponent V of the central hinge-reaction acts on each half of the arch. By ex-
amining the various figures given, it will be evident that this can always be
decided on by inspection, but it would, perhaps, be safer to assume some
direction as the positive one; for instance, let V be positive when it acts up-
wards against the left half of the arch (as in Fig. 160), then a negative value
of V would indicate the state of things in Fig. 157 or Fig. 167.]
Calculation of the Stresses X in the Horizontal Bars.
The equation of moments to find X will evidently, in every
case, be taken about the foot of the diagonal (Fig. 158). A
load on the left half of the arch produces a hinge-reaction in
§ 21. — BRACED ARCH.
117
the direction AI S, and the resultant of this reaction and the
load tends to turn the part of the arch between the section
line and the hinge from right to left — that is, in the same
direction as X tends to make it rotate. For equilibrium,
therefore, X must be negative.
A load on the right half of the arch produces a hinge-
reaction, which passes through the point round which moments
6
2
6
2
6
2
3
1
f^
^
^
/
are taken ; such a load will, therefore, have no effect on the
stress in X. Obviously, therefore, X is always in compression.
Hence, to find the greatest compression or minimum stress in
X, the whole of the left half of the arch must be considered
loaded, and the other half can be loaded or not, the result in
either case being the same. For simplicity, both halves will be
considered loaded (Fig. 157). The equations to obtain the
hinge-reaction are then
whence
Hx4-4xl2-8(9 +
= o,
H=r48.
Consequently, the equation of moments to determine Xx
(Fig. 158), with respect to the point E, is
or
0 = - X, x 3-5 - 48 X 3 + 8 (3 + 6) -f 4. X 9,
X1(min.) = -10 -29 tons.
118 BKIDGES AND ROOFS.
The following equations are obtained in the same manner :
X2 (min.) = — 19-2 tons
0 = -X3X 1-5-48 X 1 + 4 X3
X3 (min.) = - 24 tons
0 = -X4 x 0-5
X4 = 0.
FIG. 158.
Calculation of the Stresses Yin the Diagonals.
The stress in the diagonal Y2 will be calculated to illustrate
the method. The loads can be divided into three groups.
Those in the first group make Y2 positive, those in the second
FIG. 159.
negative, and lastly, those in the third exert no influence ; and
therefore, if acting alone, the stress in Y2 would be zero.
These groups are shown in Fig. 159, by the signs +, — ,
andO.
The loads on the third and fourth joints belong to the first
§ 21. — BRACED AECH.
119
group, for the resultants of these loads and the hinge-reactions
produced by them tend to turn the part of the arch between
the section line a ft and the hinge, from right to left. The
stress Y2 has the opposite tendency (Fig. 161), and is there-
fore made positive by these loads.
The load on the second joint is the only one belonging to
the second group. This load does not act directly on the part
of the arch under consideration, but by means of the hinge-
reaction produced by it, which acts in the direction Aa S, thus
tending to produce rotation from left to right, or, in other
words, making the stress Y2 negative.
The third group contains the loads on all the remaining
joints ; for either they produce no hinge-reaction (1st and 9th),
and have therefore no influence, or else they act indirectly
through a hinge-reaction in the direction A S, passing through
the point F, round which moments are taken, and consequently
producing no stress in Y2.
To determine Y2 (max.), therefore, the 3rd and 4th joints
are to be loaded, and the 2nd is to remain unloaded. (The
FIG. 160.
2i
other joints may be loaded or not ; they will, however, be con-
sidered as unloaded.) The hinge-reaction for this loading must
now be found from the equations. (Fig. 160.)
whence
V=3-75,
H = 23-25
120
BRIDGES AND EOOFS.
and the equation of moments for the part shown in Fig. 161
about F is therefore
0 = Y2 x 6-72 + 23-25 xO'5 + 3'75 X T5-1 x 1'5 - 8 (4-5 + 7 -5),
or
Y2 (max.) = + 11-94 tons.
1
< 1
.6
2
t'
It
F
To determine Y2 (min.) the 3rd and 4th joints must be
unloaded, 'and the second joint loaded (the remaining joints
will be considered un-
FIG. 161. loaded). The components
of the hinge-reaction can
be found from the equa-
tions :
0 = -Vxl2 + Hx4-lxl2
- 2 (9 + 6 + 3),
0 = -Vxl2-Hx4 + lxl2
+ 2 (9 + 6 + 3) + 6 x 3,
V = 0-75, H = 14-25;
whence, from Fig. 163, the equation of moments is
0 = Y2 x 6-72 + 0-75 x 1-5+ 14-25 x 0-5 - 1 X 1-5 - 2 (4-5 + 7'5),
therefore,
Y3(min.) = + 2-57 tons.
FIG. 162.
n ii
and
Similarly for the.remaining three diagonals.
Y, (max.).
Take every joint loaded, then
V = 0, H = 48,
0 = Y, x 10-25 + 48 x 0-5 - 4 x 1-5 ~ 8 (4-5 + 7-5 + 10-5),
Y, (max.) = + 15-8 tons.
§ 21. — BRACED ARCH.
121
Yj (min.) need not be considered, as no distribution of the load produces
compression in this diagonal.
Y3 (max.).
The 4th joint only is to be loaded, then
V = 2'25, H = 18-75,
and
0 = Y3 x 3-35 + 2-25 x 1-5 + 18'75 x 0'5 - 1 x 1-5 - 8 x 4-5.
Y3(max.)= +7-38 tons,
Ys (min.).
The 2nd and third joints only
are to be loaded, then
V = 2-25, H = 18-75,
and
0 = Y3 x 3-35 + 2-25 x 1'5
+ 18-75 x 0-5-1 x 1-5
-2 x 4-5.
Y3(min.) = -0'67tons,
Y4(max.).
Y4 (max.) need not be considered, as no distribution of the load produces
tension in this diagonal.
Y4 (min.).
To obtain Y4 (min.) joints 2, 3, and 4 are to be loaded, then
V = 4-5, H = 25-5,
and
0 = Y4 x 0-738 + 4-5 x 1-5 + 25-5 x 0'5 - 1 x 1-5.
Y4(min.) = -24 -4 tons.
Calculation of the Stresses Z in the Lower Bars.
The stress in the bar Z3 will be calculated, to illustrate the
method. In this case moments will be taken about the point
J (Fig. 164). A vertical through G, the point of intersection
of A J and A! S, gives the position of the load which produces
no stress in Z3, for the resultant E, of a load Q in this position
and its hinge-reaction D, passes through the point J. Any
load to the right of G will produce compression in Z3, for the
resultant E then passes to the right of J, and the tendency
is to turn the part of the arch under consideration (Fig. 166)
from left to right ; Z3 has the same tendency, and must there-
fore be negative to maintain equilibrium. Any load to the left
of G, on the contrary, produces tension in Z3, for the resultant
122
BRIDGES AND EOOFS.
E passes to the left of J. The vertical through G is therefore
what may be termed the loading boundary between the loads
producing tension and those producing compression.
Thus (Fig. 165) the components of the hinge-reaction when
Z3 is a maximum are to be found from the equations
Hx4-lxl2-2(9 + 6
0 = -Vxl2-Hx4+lXl2 + 2(9 + 6
V = 2-25, H = 18-75;
whence from Fig. 166 the equation of moments is
0 = Z3 X 2-37 - 2-25 x 6 + 18-75 X 0-5 + 1 x 6 + 2 x 3,
Z3(max.) = — 3-32 tons.
FIG. 164.
Tension.
FIG. 165.
To determine Z3 (min.) the hinge-reaction produced by the
loading shown in Fig. 167 must be found from the equations
= 2-25, H = 41-25;
§ 21. — BRACED ARCH.
123
and the equation of moments for the part of the arch shown in
Fig. 168, is therefore
0 = Z3 X 2-37 + 2-25 x 6 + 41-25 x 0'5 + 4 x 6 + 8 X 3.
Z3 (min.) = — 34-6 tons.
Similarly the stresses in the remaining bars Z can be calculated as follows :
FIG. 166.
I \2
<*;?
The loading boundary is at the 1st
joint, therefore only the minimum stress
need be considered. For this
V = 0, H = 48,
0 = Z1 x 4-27 + 48 x 0-5 + 4 x 12
+ 8 (9 + 6 + 3).
Z, (min.) = —50 -6 tons.
Z2.
(Loading boundary between the 2nd and 3rd joints.)
For the maximum stress,
V = 0-75, H = 14-25,
0 = Z2 x 3-32 - 0-75 X 9 + 14-25 x 0'5 + 1 x 9 + 2 (6 + 3).
Z2 (max.) = — 8-25 tons.
1
H=18,76
For the minimum stress,
V = 0-75, H = 45-75,
0 = Z2 x 3-32 + 0-75 x 9 + 45'75 X 0'5
+ 4 x 9 + 8 (6 + 3).
Z2 (min.) = — 41-45 tons.
Z4.
(Loading boundary between the 2nd
and 3rd joints.)
For the maximum stress,
V = 4-5,
FIG. 168.
H = 25-5,
0 = Z4 x 1-423 -4-5x3 + 25'5 X 0'5 + 1x3.
Z4 (max.) = — 1-58 tons.
124
BEIDGES AND ROOFS.
For the minimum stress,
V = 4-5, H = 34-5,
0 = Z4 X 1-423 + 4-5x3 + 34-5 x 0'5 + 4 X 3.
Z4(min.)= -30-0 tons.
Calculation of the Stresses U in the Vertical Bars.
The bar U3 will be chosen to exemplify the method. 7 B is
the section line, and F is the point about which to take
moments. As in the case of Y3 the loads divide themselves
into three groups relatively to the kind of stress produced in
FIG. 169.
U3, and in Fig. 169 these groups are indicated by the signs
+, — , and 0. The stress in U3 reaches its maximum value
when the second joint alone is loaded, and the hinge-reaction
can then be obtained from Fig. 162. The values already found
for its components are :
V = 0-75, H = 14-25,
and the equation of moments from Fig. 170 is :
0 = -U3X 7-5 + 0-75 X 1-5 +14-25x0-5-1 x 1'5 - 2 (4-5 + 7'5).
U3 (max.) = — 2-3 tons.
When U3 is a minimum the third and fourth joints are
alone loaded, and the values of the components of the hinge-
reaction already found from Fig. 160 are :
V = 3-75, H = 23-25;
and from Fig. 171 the equation of moments is :
0 = — U, X 7-5 + 3-75 X 1-5+ 23-25 x 0-5-1 x 1-5-8 (4'5 + 7'5).
U3 (min.) = — 10-7 tons.
§ 21. — BRACED ARCH. 125
In a similar manner the remaining stresses in the bars U can be found.
Uj (max.) need not be considered, for tension cannot be produced in this bar
by any distribution of the loads.
"Ul (min.) obtains when all the joints are loaded, then
V = 0, H = 48,
and
0 = - U, X 13-5 + 48 X 0-5 - 4 x 1-5 - 8 (4'5 + 7'5 + 10-5) - 4 x 13-5.
U, (min.) = — 16-0 tons.
U2.
Here again TJ2 (max.) need not be considered, and U, (min.) obtains when
every joint is loaded, then
V = 0, H = 48,
and
0 = - U2 X 10-5 + 48 X 0-5 - 4 x 1-5 - 8 (4-5 + 7'5 + 10-5).
U2(min.) = -15 -4 tons.
FIG. 170.
!M4,25
\a
¥=0,75
FIG. 171.
6 16
2 12 U
F
;;-.£>
*H=23/2B
V=3,75
U4.
U4 (max.) occurs when the 2nd and 3rd joints are loaded, and the 4th un-
loaded, then
V = 2-25, H = 18-75,
and
0 = -U4 X 4-5 + 2-25 X 1'5 + 18'75 x 0-5 - 1 x 1'5 - 2 X 4-5.
U4(max.) = + 0-5 tons.
Ut (min.) occurs when the 4th joint is loaded and the 2nd and 3rd unloaded,
then
V = 2-25
H = 18-75,
and
0= -U4 x 4-5 + 2-25 X 1-5 + 18 '75 X 0-5 - 1 x 1-5-8 x 4-5.
U4 (min.) = — 5-5 tons.
126
BRIDGES AND ROOFS.
The 5th vertical is divided in two at the hinge, and since the head of this
vertical is only connected with a horizontal bar, it follows that the only vertical
force that can come upon it is the load on the joint, which can never be more than
4 tons for each half. Hence
Us (min.) = — 4 tons.
The results obtained are collected together in Fig. 172.
-10.29
FIG. 172.
-24
§ 22. — BRACED ARCH OP 40 METRES SPAN.
(Bridge over the Theiss at Szegedin*)
This bridge, supported by two braced arches, is designed
for a single line of railway. The permanent load can be taken
at 2400 kilos, and the moving load at 4000 kilos, per metre
run, and one-half of this is supported by each arch.
The length of a bay being 2 metres, each joint has
2400 kilos, permanent load and 4000 kilos, moving load to bear,
or (taking 1000 kilos. = 1 ton) 2 -4 tons permanent and 4 tons
moving load. The two halves of the arch are connected
together at the centre by a hinge, and hinges are also provided
at the abutments. The form and dimensions of the structure
are given in Fig. 173.
* With the exception of some slight alterations in the dimensions and the
addition of a central hinge, the figure represents the bridge over the Theiss. The
object of the central hinge will appear in the " Theory of Braced Arches." That
the diagonals in the central bays of the existing bridge have been expanded into
a plate-web can hardly be considered a difference in the principle of the construc-
tion.
§ 22. — BRACED ARCH.
127
Calculation of the Stresses X in
the Horizontal Bars.
The bar X5 will be taken to
illustrate the calculations. The
first step is to determine which
loads create tension in X5 and
which compression, and to do
this the point must be found
where a load can be placed so as
to produce no stress in X5. The
vertical through the intersection
of A L and A! 8 produced (Fig.
174) gives this required loading
boundary, for a load Q placed
in this position produces a hinge-
reaction D acting in the direction
AJ S (this reaction must pass
through A! to prevent rotation
round that point), and these to-
gether give a resultant R which
must be directed to A, so that
the left half of the arch may not
rotate round this point, but by
construction the line C A passes
through L, and as this is the
point about which to take mo-
ments to determine X5, it follows
that the load Q can produce no
stress in X5. The reaction E,
due to a load to the right of Q
passes below L and tends there-
fore to turn the part L S of the
arch from left to right, and X5
will be positive since it acts in
the opposite direction.
On the contrary, every load
situated to the left of Q will
FIG. 173.
128
BRIDGES AND ROOFS.
produce a negative stress in X5, for the resultant E in this case
passes above L, and therefore tends to produce rotation in the
same direction as X5 does.
The vertical through C is therefore the required loading
boundary, and the maximum stress obtains when all the joints
to the right of it are loaded.
FIG. 174.
Compression.
Tension.
_,!
For X5 this loading boundary is situated at a distance of
16 metres from the left abutment, coinciding therefore with the
vertical bar U9. To find X5 (max.) therefore j^the joints 10,
11, 12 — 21 must be loaded, and the remainder unloaded
FIG. 175.
20"
(Fig. 175), and the components of the hinge-reaction for this
loading are obtained from the two following equations of
moments :
0 = V X 20 + H x 5 - 2-4 (^ + 18 + 16 + . . . + 4 + 2)
-4 (^+18 + 16 + ... + 4 + 2)
0 = V X 20 - H x 5 + 2'4 (^ + 18 + 16 + ... + 4 + 2)
+ 4 (22 + 18)
whence
and
§ 22. — BRACED AECH.
129
and the equation of moments for the part of the arch shown in
Fig. 176 about L is therefore :
0 = - Xs x 1'75 - 99-2 x 1-25 + 7'2 x 10
+ 2-4 Qf + 8 + 6 + 4 + 2) + 4 Qf + 8)
X5 (max.) =+ 34-29 tons.
FIG. 176.
41 2
2,J4 ?J4 2/|4 2J4 2J4 1
To find X5 (min.) the hinge-reaction must be determined
from Fig. 177 thus :
0 = - V X 20 + H X 5 - 2-4 (<^> + 18 + . . . + 2)
0= - V X 20 -H x 5 + 2-4(^> + 18+ ... + 2)
42,42,4 2,|4 1,|2 l/22,|42/[42J42/|42l42/l42/42
I t [H H.I I I I
and the equation .of moments is (Fig. 178) :
0= -X5 x 1-75-5-6 X 10-70-4 X l'25 + 2-4(iP + 8 + ...+ 2)
+ 4 (4 + 2)
X5(min.)= - 34 -29 tons.
130
BRIDGES AND ROOFS.
From this it appears that the numerical values of X5 (max.)
and X5 (min.) are identical, whence it follows that X5 = 0,
when the loads producing the maximum stress are on the
bridge together with those producing the minimum stress, that
is when the bridge is fully loaded (for the load on the 9th
joint has no effect). This property is easily explained by the
"Theory of parabolic girders," given in § 8, for in the pre-
sent example the arch has the form of a parabola, and it has
been shown that this is the curve of equilibrium (or linear
arch) for a load uniformly distributed over the span. Directly
therefore the bridge is fully loaded, neither the horizontal bars
nor the diagonals are necessary to maintain equilibrium, the
verticals, however, are required to transmit the loads to the
linear arch (Fig. 179).
Now the permanent load is uniformly distributed over the
span, and produces therefore no stress in the horizontal bars or
the diagonals. Thus in calculating the stresses in them, the
FIG. 179.
permanent load can be left out of consideration, and further it
is only necessary to obtain either the maximum or minimum
stress when the other can be found by changing the sign.
The calculation for X5 could therefore have been given in
the following form.
0=-Vx20 + Hx5
0 = - V X 20 - H X 5 + 4 (14 + 12 + ... + 2)
V = 5'6 H = 22-4
0 = - X5 X 1-75 - 5-6 X 10 - 22-4 x 1'25 + 4 (4 + 2)
X5 = ±34 -29 tons.
And the following calculations are made in a similar manner :
X,.
(Loading boundary in 7th bay.)
0=-Vx20 + Hx5
0 = -Vx20-Hx5 + 4(12 + 10 + ... + 2)
V = 4'2 H = 16-8
0 = - X, x 4-55 - 4-2 X 18 - 16'8 X 4'05 + 4.(10 + 8 + . . . + 2)
X = ±5- 20 tons.
§ 22. — BRACED ARCH. 131
X2.
(Loading boundary in 8th bay.)
0=-Vx20 + Hx5
0 = - V X 20 - H x 5 + 4 (14 + 12 + . .. + 2)
V = 5'6 H = 22-4
0= - X2 x 3-7 -5-6 x 16 - 22-4 x 3*2 + 4 (10 + ... + 2)
X2 = ±11 -16 tons.
X3.
(Loading boundary in 8th bay.)
V = 5-6 °H = 22-4
0 = - X3 x 2-95 - 5-6 x 14 - 22-4 x 2-45 + 4(8 + 6 + ... + 2)
X3 = ±18 -06 tons.
X4.
(Loading boundary in 8th bay.)
V = 5-6 H = 22-4
0 = - X4 x 2-3 - 5-6 X 12 - 22-4 x 1-8 + 4 (6 + 4 + 2)
X4 = ±25 -88 tons.
X..
(Loading boundary in 9th bay.)
0=-Vx20 + Hxo
0= -V x 20 -Hx5 + 4 (16+ 14 + ... + 2)
V = 7'2 H = 28'8
0 = - X6 x 1-3 - 7-2 x 8 - 28-8 x 0-8 + 4 (4 + 2)
X6 = ±43 -57 tons.
X7.
(Loading boundary in 9th bay.)
V = 7'2 H = 28-8
0 = - X7 x 0-95 - 7-2 x 6 - 28-8 x 0'45 + 4x2
X7 = ±50-70tons.
(Loading boundary in 10th bay.)
0=-Vx20 + Hx5
0=-Vx20-Hx5 + 4(18 + 16 + ... + 2)
V = 9 H = 36
0 = - X8 x 0-7 - 9 x 4 - 36 x 0-2 + 4 x 2
X8 = ± 50 -29 tons.
(Loading boundary in 10th bay.)
V = 9 H = 36
0 = - X9 x 0-55 - 9 x 2 - 36 x 0'05
X9 = ±36-0 tons.
K 2
132
BRIDGES AND ROOFS.
X,..
There never can be any stress in this bar, for no horizontal force can act
on its right extremity. Hence
X10 = 0.
Calculation of the Stresses Y in the Diagonals.
The diagonal marked Y5 will serve to illustrate the calcula-
tions.
The point about which moments will be taken is M
(Fig. 180), and the vertical through the intersection of A M
and A! S produced will give the loading boundary. For a
load Q placed in this position gives with the hinge-reaction D
a resultant E whose direction is E M A. If the load lies to
FIG. 180.
Tension.
Compression.
the right of E the resultant E, or if it is placed on the right
half of the arch the hinge-reaction passes below M, and con-
sequently tends to turn the part S a fi from left to right,
but Y5 has the same tendency, and must therefore be nega-
tive.
If on the contrary the load is to the left of E, the resultant
E, or if the load is to the left of a ft, the hinge-reaction D
passes above M, and in either case Y5 is evidently positive.
Thus if the part of the bridge lying to the right of the
vertical through E be loaded, Y5 will be a minimum ; and if
the part to the left be loaded, Y5 will be a maximum.
It has already been remarked when dealing with the
stresses in the horizontal bars, that the permanent load produces
no stress in the diagonals ; it seems therefore unnecessary to
carry the proof any further. Hence in the following calculations
22. — BRACED ARCH.
133
the permanent load will be left out of consideration (Fig. 181).
Also since the numerical values of the maxima and minima
stresses are equal, only the maximum stress will be calculated
in each case, the minimum stress being obtained by changing
the sign.
FIG. 181.
Thus to determine Y5 (max.) the equations to find the
hinge-reaction are (Fig. 182)
0 = - V X 20 - H X 5 + 4 (16 + 14 + . . . + 2)
V = 7'2 H = 28-8
and from the same figure the equation of moments is
and
0 = Y5 x 5-51 - 7-2 x 3-64 + 28'8 x 0*5
- 4 (0-36 -f 2-36 + 4-36 + 6'36)
Y5(max.)= + 11 '9 tons
Y5(min.)= - 11 '9 tons.
FIG. 182.
In a similar manner the stresses in the remaining diagonals can be found as
follows :
(Loading boundary in 7th bay.)
V = 4-2 H = 16-8 (see calc. for X,)
0 = Y! x 10-6 - 4-2 x 8-42 + 16'8 X 0'5 + 4 x 0'42
- 4 (1-58 + 3-58 + 5-58 + 7'58 + 9'58)
Y, = ±12 -92 tons.
134 BRIDGES AND EOOFS.
Y2.
(Loading boundary in 8th bay.)
V = 5-6 H = 22-4 (see calc. for X2)
0 = Y2 x 9-42 - 5-6 x 7-294 + 22'4 x 0'5 + 4 X 1'294
- 4 (0-706 + 2-706 + 4'706 + 6'706 + 8'706)
Y2 = ±12 -59 tons.
Y3.
(Loading boundary in 8th bay.)
V = 5'6 H = 22-4
0 = Y3 x 8-16 - 5-6 x 6-13 + 22-4 x 0-5 + 4 x 0-13
- 4 (1-87 + 3-87 + 5-87 + 7-87)
Y3 = ±12 -3 tons.
(Loading boundary in 9th bay.)
V = 7-2 H = 28-8 (see calc. for X6)
0 = Y4 x 6-834 - 7-2 x 4-923 + 28-8 x 0-5 + 4 x 0*923
- 4 (1-077 + 3-077 + 5-077 + 7-077)
Y4 = ± 12 -07 tons.
Y6-
(Loading boundary in 9th bay.)
V = 7-2 H = 28-8
0 = Y6 x 4-24 - 7-2 x 2-223 + 28-8 x 0-5
- 4 (1-777 + 3-777 + 5-777)
Y6 = ±11 -07 tons.
In the case of Y7 it is found that the point about which
to take moments is situated in the central bay, and this, as
will appear, makes the arrangement of the loading giving the
greatest stresses differ from that of the previous cases (Fig. 183).
There are, in fact, three groups of loads, two of these produce
compression and the third tension. For as will be seen from
the figure the line Ax S in this case passes below the point
N, and consequently the hinge-reaction produced by a load on
the left half of the arch also passes below N. Now a load
situated to the left of the section line \ p acts on the part
of the arch S X yu, by its hinge-reaction D, and tends therefore
to turn this part round N from left to right, thus making Y?
negative.
Thus the section line \ p is a second loading boundary ;
§ 22. — BRACED ARCH.
135
for, as in the former cases, a load placed between X //, and 7 p
produces tension in Y7.
Hence, to find the greatest stress in Y7, either the two com-
pression groups can be considered loaded or .else the tension
group.
FIG. 183.
Compression.
Tension.
Compressor).
y
N
^^' "
n
P "^^
In the latter case the equations to find the hinge-reaction
are (Fig. 184),
0=-Vx20 + Hx5
0 = - V X 20 - H x 5 + 4 (18 + 16 + 14)
V = 4-8 H = 19-2
FIG. 184.
g^i-Jte
T
and the equation of moments from Fig. 185 is,
0=Y7X 3-194 -4-8 x 0-571 + J9'2 x 0'5
-4 (1-429 + 3-429 + 5-429)
Y7 = ±10 -73 tons.
FIG. 185.
41 41 °>™
I J Nil
For Y8 the loads also form
three groups. The turning
point is in this case situated in Y
the right centre bay and the
hinge-reaction D passes below
F. The calculations are exactly
similar to those for Y7, and
0=-Vx20 + Hx5
0 = - V X 20 - H X 5 + 4 (18 + 16)
V = 3-4 H = 13-6
0 = Y8 X 2-51 + 3-4 x 1-6+13-6 X 0'5 - 4(5'6 + 3'6)
Y8 = ± 9-8 tons.
136
BRIDGES AND ROOFS.
For Y9, however, the position of the turning point is such
that only two groups are formed (Fig. 187). Here the section
line o- r is itself the loading boundary, for every load to the left
of <7 T acts on the part S cr T through its hinge-reaction D, which
Compression.
FIG. 186.
Tension.
Compression.
evidently makes Y7 negative. But every load to the right of
o- T on the part S a T, produces with its hinge-reaction a
resultant which tends to induce rotation from right to left,
and all loads on the right half of the arch acting by means of
FIG. 187.
Compression. Tension.
A A
their hinge-reaction W have the same effect. Consequently all
loads to the right of o- T make Y9 positive.
The loading boundary is therefore situated in the. 9th bay,
and (see calculation for X6)
V = 7'2 H = 28-8
Whence the equation of moments is
0 = Y9 X 2-47 + 7-2 x 5-33 + 28'8 x 0-5
Y9 = ±21'4tons.
Similarly it is found that the section line is the loading
boundary for Y10, therefore,
0=-Vx20 + Hx5
0=-Vx20-Hx5 + 4(18 + 16 + .
V = 9 H = 36
0 = Y10 x 5-324 + 9 x 20 + 36 x 0-5
Y10 = ±37-29tons.
§ 22.— BRACED ARCH. 137
Calculation of the Stresses U in the Verticals.
The effect of the permanent load on the vertical bars can
be deduced from Fig. 179. If half the permanent load is
applied to the top of the verticals and the other half to the
foot, the compression produced in each vertical will be 1*2
tons (with the exception, however, of the first and last verticals
which have only half the amount to sustain). To these stresses
must now be added those produced by the moving load.
The maximum and minimum stresses produced by the
moving load must therefore be found.
The vertical U5 will be taken to illustrate the method.
The loading boundary can be found by the construction
employed in Fig. 180, for in both cases M is the turning point.
Compression. FIG. 188. Tension.
A load Q placed on the vertical through E, gives with its
hinge-reaction D a resultant E which passes through M, and
hence Q can produce no stress in U5. The vertical through E
is therefore the loading boundary, and all loads to the right
produce tension, and all loads to the left compression. When
U5 (min.) obtains the bridge will be loaded with the com-
pression group, and then (see calculations for Y5)
V = 7'2 H = 28-8
Whence the equation of moments for the part of the arch
shown in Fig. 189 is (denoting by u5 the stress due to the
moving load alone),
0 = - «5 x 8-36 - 7'2 x 3-64 + 28*8 X 0'5
- 4 (0-36 + 2-36 + 4-36 + 6'36 + 8'36)
M5 (min.) = — 11-84 tons.
u5 (max.) can be deduced from this without further calcula-
tion in the following manner :
138 BRIDGES AND ROOFS.
If the moving load covers the whole bridge it is evident
that the stress in each vertical is — 4 tons, and hence u5
(max.) together with u5 (min.) must be equal to — 4 tons, or
w5 (max.) = - 4 - (- 11-84) = + 7'84 tons.
To obtain U5 (max.) and U5 (min.) - 1 • 2 tons must be added
to the values just found, thus : ,
U5 (min.) = - 11-84 - 1-2 = - 13-04 tons
U5 (max.) = + 7-84 - 1-2 = + 6-64 tons.*
FIG. 189.
The stresses in the remaining verticals can be found as follows
0 = - M, X 11-58 - 4-2 x 8-42 + 16'8 X 0-5 + 4 X 0-42 *
- 4 (1-58 + 3-58 + 5-58 + 7'58 + 9'58 + 11-58)
w, (min.) =- 15-82 M, (max.) = + 11-82
Uj (min.) = — 17 '02 tons
U, (max.) = + 10-62 tons.
Uf. .
0 = -w2X 10-706-5-6 x 7-294 + 22-4 x 0-5 + 4 x 1'294
- 4 (0-706 + 2-706 + 4-706 + 6-706 + 8'706 + 10-706)
M2 (min.) = — 15-08 uz (max.) = + 11-08
U2 (min.) = - 16-28 tons
U2(max.) = + 9 -88 tons,
* These stresses could have been obtained quicker thus : To find U5 (min.)
add to the vertical component of Y5 (min.) - (4 + 1-2) tons, and to find TJ5
(max.), add to the vertical component of Y5 (max.) — 1-2 tons. As, however,
this method cannot always be adopted, the longer one has been preferred.
t Strictly speaking, the load on the 1st vertical is only half the load on the
others, the remaining half being taken by the abutment. It has, however, been
considered fully loaded, as this would probably be the course pursued in practice.
§ 22. — BRACED ARCH. 139
0= - w3x 9-87 -5-6x6-13 + 22-4 x 0-5 + 4 x 0-13
- 4 (1-87 + 3-87 + 5-87 + 7'87 + 9'87)
w3 (min.) =- 14-2 u3 (max.) = + 10 • 2
U3 (min.) = — 15-4 tons
U3 (max.) = + 9-0 tons.
U4.
0 = -w4 x 9-077 -7'2 x 4-923 + 28-8 x 0-5 + 4 x 0-923
- 4 (1-077 + 3-077 + 5-077 + 7'077 + 9-077)
w4 (min.) = - 13 • 1 w4 (max.) = + 9 • 1
U4 (min.) = — 14-3 tons
U4 (max.) = + 7'9 tons.
U6.
0 = - w6 x 7-777 - 7-2 x 2-223 + 28'8 X 0-5
- 4 (1-777 + 3-777 + 5-777 + 7'777)
M6 (min.) = - 10-03 wa (max.) = + 6 '03
U6(min.) = - 11 -23 tons
U6(max.)=+ 4 -83 tons.
U7.
In determining the stress in Yt it was found that the loads
divided themselves into three groups. This is also true in
the case of U? with this difference, that the second loading
boundary is placed one bay more to the left on account of the
oblique direction of the section line <j> -fy (Fig. 190).
FIG. 190.
Tension. Compression. Tension.
<- ----------- -X -------- •>«- ---------------------- >
fl
When the loads producing compression are on the arch the
equations to obtain the components of the hinge-reaction are,
0 = — Vx20 + Hx5
0 = - V x 20 - H x 5 + 4 (18 + 16 + 14 + 12)
V = 6 H = 24
140 BRIDGES AND ROOFS.
and the equation of moments for the part S </> i|r is,
0 = - ur x 7-429 - 6 X 0-571 + 24 X 0'5
-4(1-429 + 3-4294- 5-429 + 7'429)
Uj min.) = — 8-38 w7 (max.) = + 4 -38
U7 (min.) = — 9-58 tons
U7 (max.) = + 3-18 tons.
In this case also there are two loading boundaries, and
0 = -Vx20 + Hx5
0 = - V x 20 - H x 5 + 4 (18 + 16 + 14)
- V = 4-8 H = 19-2
0 = — «8 X 7-6 + 4-8 x 1-6 + 19-2 x 0'5
-4(3-6+ 5-6 + 7-6)
«8 (min.) = — 6*57 «8 (max.) = + 2 '57
U8(min.) = -7 -77 tons
U8 (max.) = + 1-37 tons.
As in the case of Y9 the loads divide themselves again into
two groups ; the loading boundary is "however one bay more to
the left (Fig. 191).
Tension.
FIG. 191.
Compression.
To obtain the hinge-reaction,
and
0 = — Vx20 — Hx5 + 4(14+12
V = 5-6 H = 22-4
0 = - u9 x 9-33 + 5-6 X 5-33 + 22-4 X 0*5
«9 (min.) = — 8'4 u9 (max.) = + 4-4
U9 (min.) = — 9-6 tons
UD(max.) = + 3-2 tons.
§ 22. — BEACED ARCH.
141
Ui..
The loading boundary is in the 9th bay, and
V = 7-2 H = 28-8
0 = — MIO x 22 + 7-2 X 20 + 28-8 x 0'5
MJO (min.) = — 11-2 «1Q (max.) = + 7'2
U10 (min.) = — 12-4 tons
U10 (max.) = + 6*0 tons.
Uu.
The vertical in the centre is divided in two by the hinge,
and as at the top it is only connected to a horizontal bar, the
only stress that can exist in it is the compression produced by a
load placed on the top. The greatest load for each half is
0 • 6 ton permanent, and 2 tons moving toad. Hence
Un (min,) = —2- 6 tons.
Calculation of the Stresses Z in the Bow.
To find the stress in the bar Z5 a section line is drawn
through the 5th bay, and the equation of moments formed for
the part of the arch lying between the section line and the
hinge with reference to the point of intersection 0 of the
diagonal and horizontal bars (Fig. 192).
Tension.
FIG. 192.
Compression.
8
The vertical through the point of intersection F of A 0
and A! S is the loading boundary, for the resultant K of a load
Q in this position and its hinge-reaction D passes through O.
This loading boundary lies in the 6th bay.
In this case the permanent load will have to be taken into
142
BRIDGES AND ROOFS.
consideration ; every joint will therefore have a permanent load
of 2*4 tons, and those joints which have been called the loaded
joints will carry 4 tons of moving load besides. The dis-
tribution of the moving load producing the greatest tension is
shown in Fig. 192, and the equations to find the hinge-reaction
are,
0 = -V X 20 + H X 5-2-4(^> + 18 + 16 + ... + 2)
0 = - V X 20 - H x 5 + 2-4 (-2J> + 18 + 16 + . . . + 2)
+ 4(10 + 8 + . ..'+ 2)
V = 3 H = 60.
Consequently, from Fig. 193,
0 = Z5 x 2-218 -3 x 12 + 60 X 0'5 + 2-4 (if + 10 + ... + 2) + 4 x2
Z5 (max.) = — 39-86 tons.
FIG. 193.
0
4 t I4 t 1
s°
rTs
Again, to find Z5 (min.) (Fig. 192),
0 = V X 20 + H x 5 — 6-4(^0 + 18 + ... + 2)
0 = V X 20 — H x 5 + 2 -4 (^> + 18 + . . . + 2) + 4 (M + 18 + . . . + 2)
V = 3 H = 116
and from Fig. 194
0 = Z5 X 2-218 + 3 X 12 + 116 X 0-5 + 2-4 (^ + 10 + ... + 2)
+ 4 (11 + 10 + ... + 4)
Z5(min.) = -142 -70 tons.
For the sake of comparison the stress in Z5, when the moving
load covers the whole bridge, will also be calculated, thus,
0 = V X 20 + H x 5 - 6-4 CM + 18 + . . . + 2)
0 = V X 20 - H x 5 + 6-4 (^? + 18 + . . . + 2)
V = 0 2H = 128
0 = Z5 x 2-218 + 128x0-5 + 6-4 (i* + 10 + ... + 2)
Zs = -132 -7 tons.
§ 22. — BRACED ARCH.
143
From the above calculations it appears that the compression
in the bars forming the bow can be considerably greater with
an uneven than with a uniformly distributed load. In this a
parabolic arch-bridge differs from a parabolic girder-bridge, for
it was shown that in the latter the greatest compression in the
bow occurred when the bridge was fully loaded.
It also appears that it is not absolutely necessary to calculate
the maximum stress in the bow. For if the maximum and
minimum stresses produced by the moving load be added
together the result is the stress due to the moving load when it
covers the bridge. And it is evident that this stress is always
negative (for altering 6*4 to 4 in the last equation will not
change the sign of Z5). Consequently the absolute value of
the minimum stress produced by the moving load must be
greater than that of the maximum stress, and the compression
produced by the dead load still further increases the balance in
favour of the minimum stress. And since a greater section of
material is generally required to resist compression than the same
amount of tension, the maxima stresses might be neglected.
They will, however, be calculated, and for the following
reason. If the arch be imagined turned upside down it
becomes a suspension bridge, and the same calculations with
reversed signs would apply, the minima stresses becoming
the maxima stresses and vice versa. Now it is just possible, if
the dead load were small in comparison to the moving load,
that the minima stresses in the suspension bridge might
become negative, and would then probably determine the
section (the bars being long columns). For such cases there-
fore it is necessary to know what are the maxima stresses
inZ.
144 BRIDGES AND EOOFS.
The stresses in the remaining bars Z can be calculated as follows :
Zi-
(The loading boundary coincides with the 1st vertical.)
When the bridge is free of the moving load,
0 = Vx 20 + H x5-2-4(2^ + 18 + ... + 2)
0 = V X 20 -H X 5 + 2-4(^> + 18 + ... + 2)
V = 0 H = 48
0 = Zj x 4-968 + 48 X 0'5 + 2-4 (<y> + 18 + .. . + 2)
Z, (max.) = — 53-14 tons.
When the moving load covers the bridge,
V = 0 H = 128
0 = Zx X 4-968 + 128 X 0-5 + 6-4 (^> + 18 + . . . + 2)
Z, (min.) = -141 -71 tons.,
Z2.
(Loading boundary in the 2nd bay.)
V = 0'2 H = 48-8
0 = Z2X 4-186 -0-2 x 18 + 48-8 x 0'5
+ 2-40U + 16 + ...+2)
Z2 (max.) = — 51-41 tons.
V = 0-2 H = 127-2
0 = Z2 X 4-186 + 0-2 X 18 + 127'2 X 0'5
+ 6-4 (±J + 16 + ... + 2)
Z2 (min.) = — 139*89 tons.
(Loading boundary in the 4th bay.)
V = 1'2 H = 52-8
0 = Z3 x 3-464 — 1-2 X 16 + 52-8 x 0-5
+ 2-4(^ + 14 + ... + 2) + 4 x2
Z3 (max.) = — 48-73 tons.
V = 1'2 H = 123-2-
0 = Z3 X 3-464 + 1-2 X 16 + 123-2 x 0-5
+ 2-4(^+14+. .. + 2)
+ 4(if + 14 + . .. + 4)
Z3 (min.) = — 139-3 tons.
§ 22. — BRACED ARCH. 145
Z4.
(Loading boundary in the 5th bay.)
V = 2 H = 56
0 = Z4 x 2-805 - 2 x 14 + 56 x 0-5
+ 2-4 (^* + 12 + . . . + 2) + 4 x 2
Z4(max.) = -44 '77 tons.
V = 2 H = 120
0 = Z4 x 2-805 + 2 X 14 + 120 x 0'5
+ 2-4 (2_t+l2 + ...+ 2)
+ 4 (o + 12 + . . . + 2)
Z4 (min.) = - 140-3 tons.
Z6.
(Loading boundary in the 7th bay.)
V = 4-2 H = 64-8
0 = Z0 x 1-707 - 4-2 x 10 + 64'8 x 0-5
+ 2-4(^ + 8 + . .. + 2) + 4x2
Z6 (max.) = — 34-21 tons.
V = 4-2 H = 111-2
0 = Z6 x 1-707 + 4-2 x 10 + 111-2 x 0-5
+ 2-4(1-0 + . .. + 2) + 4(i5 + ... +4)
Z6 (min.) = -146 -2 tons.
Z7.
(Loading boundary in the 8th bay.)
V = 5-6 H = 70'4
0 = Z7 x 1-28 -5-6x8 + 70-4 x 0-5
+ 2-4(| + 6 + 4 + 2) + 4x2
Z7 (max.) = -28 -74 tons.
V = 5-6 H = 105-6
0 = Z7 x 1-28 + 5-6x8 + 105-6 x 0-5
+ 2-4 (f + 6 + 4 + 2) + 4 (| + 6 + 4)
Z7(min.) = -149 -9 tons.
Z8.
(Loading boundary in the 8th bay.)
V = 5-6 H = 70-4
0 = Z8 x 0-943 -5-6x6 + 70-4 x 0-5 + 2'4 (f + 4 + 2)
Z8 (max.) = — 24-6 tons.
V = 5-6 H = 105-6
0 = Z8 x 0-943 + 5-6x6 + 105-6 x 0'5
+ 2-4 (| + 4 + 2) + 4 (A + 4 + 2)
Z8(min.) = — 152 -8 tons.
L
146 BRIDGES AND EOOFS.
Z9.
(Loading boundary in the 9th bay.)
V = 7-2 H = 76-8
0 = Z9 x 0-698 -7-2x4 + 76'8 x 0-5 + 2-4 (| + 2)
Z9 (max.) = — 27-5 tons.
V = 7'2 H = 99-2
0 = Z9 x 0-698 + 7-2x4 + 99-2 x 0-5
+ 2-4 C| + 2) + 4 (| + 2)
Z9(min.) = - 149-0 tons.
A slight alteration occurs in the grouping of the loads in
this case, for in no position does the moving load produce
tension in Z10. This is shown in Fig. 195. The reason is that
the prolongation of the line Ax S happens to coincide with the
diagonal of the 10th bay, but every load to the left of the
section line e co acts indirectly on the part S e &> by means of its
hinge-reaction D, which passes through the turning point P,
and consequently produces no stress in Z10.
Thus to find Z10 (max.) the bridge can be considered
unloaded (or if one chooses, loaded up to the point P), and
when Z10 (min.) obtains the moving load will cover the bridge
(or else up to the section line e co only). Hence the following
equations :
V = 0 H = 48
0 = Z10 x 0-5498 + 48 x 0-5 + 2-4 x f
Z10 (max.) = — 48-02 tons.
V = 0 H = 128
0 = Z10 x 0-5498 + 128 X 0'5 + 2'4 x f + 4 x f
Z10 (min.) = — 128-05 tons.
The results obtained are collected together in Fig. 196.
§ 23. — STABILITY OF THE ABUTMENTS. 147
If the signs of all the stresses in Fig. 196 be changed, the
stresses in the suspension bridge formed by turning the arch
upside down will be obtained, if the abutment hinges become
the points of attachment. This suspension bridge is shown in
Fig. 197.
§ 23. — STABILITY OF THE ABUTMENTS OF THE BRACED AKCH.
The stability of the abutments can be tested by the method
of moments, and it also can be ascertained which distribution of
the moving load acts the most injuriously in this respect. The
force tending to overturn the abutments or piers is the hori-
zontal component of the thrust of the arch. The vertical com-
ponent of the same force, on the contrary, adds to the stability.
Both components are greatest when the bridge is fully loaded,
yet the excess of the moment of the horizontal component over
that of the vertical component may reach its maximum with
a partial load.
To decide this point the first step is to find the position
which a load must occupy on the bridge, so that it may have
no overturning effect on the pier. The axis about which the
pier tends to rotate is represented in Fig. 198 by the point F,*
and for the load Q to have no overturning effect the reac-
tion produced by it at the abutment A! must pass through F.
Evidently the vertical drawn through the intersection of F Al
and A S produced gives the required position of the load Q,
and it is also easily seen that the reaction for all loads to
the right of Q will pass inside F, and for all loads to the left of
Q outside F. The worst case for the pier is therefore when the
bridge is loaded from the left abutment up to the vertical
through I. The position of this vertical evidently depends on
the ratio =- of the height of the pier (up to the hinge Aj) to its
breadth.
As an example suppose that
* To allow for the compressibility of the material of which the pier is built,
moments should not be taken round F but round an axis nearer the centre of the
pier. See Appendix. — TRANS.
L 2
148
BKIDGES AND ROOFS.
FIG. 196.
FIG. 197.
-17,02;
-10,62
§ 23. — STABILITY OF THE ABUTMENTS.
149
then for the arch calculated in the preceding paragraph the
loading boundary IQ falls in the 18th bay. Fig. 199 repre-
sents the most unfavourable arrangement of the load as regards
the pier at AI. From this figure the equations to obtain
the hinge-reaction are :
0 = — V X 20 + H x 5 - 2-4 (2a + 18 + . . . + 2)
-4(*Q + 18+...+ 6)
0 = - V X 20 - H x 5 + 2-4 (22°- + 18 + . . . + 2)
+ 4(*4+_18 +...+ 6)
V = 0-6 H = 125-6.
FIG. 198.
FIG. 199.
numm mi
1 I '
Let M represent the overturning moment of the whole
fridge about the horizontal axis through F (Fig. 198), then :
= _o-6 (6 + 20) + 125-6 (£ + 5)
-2-4 \^~^) + (18 + 6) + (16 + 6) + . . . + (2 + 6) + |]
* • *+ (6 +
150
BRIDGES AND EOOFS.
To resist this there is the moment of stability of the pier, and
if, as in Fig. 200, there is a second arch abutting against the
other side of the pier, there will also be the moment of this arch
tending to turn the pier over in the contrary direction. This
arch, however, must be considered unloaded, for every load on
g-125,6
u n
i i 1 • i
I I
V V
II
FIG. 200.
i I I I 4 i ! I
,H=48
it produces a thrust passing above F, and consequently the most
unfavourable case for the pier will be when this arch has no
moving load upon it. It has been already found that in this
case
V = 0 and H = 48;
and hence, if Mj is the moment of the second span about F,
Let G be the weight of the pier ; then the condition of stability
is expressed by
By substituting in this equation the values of M and Mx the
requisite dimensions of the pier can be ascertained.*
* The pier could, however, fail by sliding if the friction between any of the
bed-joints were not sufficiently great. A similar case is discussed at p. 186. —
TRANS.
§ 24. — THEORY OF HINGED-BRIDGES. 151
§ 24 — THEORY OF HINGED-BRIDGES.
It is now proposed to consider from a more general point of
view the principles of construction of these bridges.
The stresses in any system of bars can be calculated by the
method of moments as soon as the direction and magnitude of
the reactions at the abutments are known. In girder-bridges
the abutments are so arranged that they can only produce
vertical reactions, and there can therefore be no uncertainty
as to their magnitude. But in the case of arched or sus-
pension bridges a horizontal reaction is added to the vertical
reaction, and it is only when this former can be determined
that the stresses can be calculated.
This horizontal reaction is indeterminate unless the con-
tinuity of the structure is interrupted at some point and a hinge
introduced, as will be proved by the following.
In the " Theory of parabolic girders " (§ 8) it was shown
that the parabola is the curve of equilibrium of an inverted
chain in the form of an arch, when the load is uniformly distri-
buted over the span ; and in this case both the horizontal and
vertical reactions are determinable. But the slightest alteration
either in the distribution of the load or in the form of the curve,
would make the chain collapse unless it is stiffened by some
means. This stiffening can be obtained in two different ways :
either by transforming the flexible chain into a stiff bow which
prevents deformation by its resistance to flexure, or else by
means of a system of braces composed of horizontal, vertical,
and diagonal bars, forming triangles with each other. In both
cases the flexible arch will be transformed into a stiff structure,
and the abutments will have to supply horizontal as well as
vertical reactions.
The magnitude of the vertical reactions can always be
determined ; this will appear by taking moments about the
abutment B (Fig. 201 or Fig. 202), thus :
o = V • 21 -
or
152
BRIDGES AND ROOF!?.
But the horizontal reactions are indeterminate, for the only
condition of equilibrium to which they are subject is that the hori-
zontal force acting at A shall be equal and opposite to that acting
at B, and this condition can evidently be satisfied by an infinite
number of values. This condition can also be expressed as
FIG. 201.
FIG. 202.
follows : the resultants D and W of the reactions at the abut-
ments must, for equilibrium, meet the vertical through Q at
the same point. The position of this point on the vertical is,
however, indeterminate, and depends on the magnitude of H.
The point P (Fig. 203) will lie above the horizontal through
the abutments when the horizontal reactions act inwards, and
below the horizontal (Fig. 204) when the horizontal reactions
act outwards. The nearer P is to the horizontal the greater
FIG. 203.
FIG. 204.
the horizontal reactions, and when this distance vanishes these
reactions become infinite.
[NOTE. — In reality H is not indeterminate, as will be evident by the
application of Canon Moseley's Principle of Least Resistance, which, as stated
by Professor Rankine, is as follows : —
" If the forces, which balance each other in or upon a given body or
structure be distinguished into two systems, called respectively active and
passive, which stand to each other in the relation of cause and effect, then will
§ 24. — THEORY OP HINGED-BRIDGES.
153
the passive forces be the least which are capable of balancing the active
forces, consistently with the physical condition of the body or structure."
Now in the present case the vertical passive forces are determinate, and
the principle therefore applies only to H, the constant horizontal thrust
in the arch. He must therefore have the least value consistent with
stability].
The actual magnitude of this horizontal reaction depends on
the attachments, on the resistance of the abutments, on the
changes of temperature, and in fact on several causes which can
hardly be allowed for by calculation.* Yet this is of the
very greatest importance, for the structure could fail either by
the horizontal reaction decreasing or increasing considerably.
Suppose, for instance, that the braced arch just calculated
were constructed without a hinge in the centre, and that by
the abutments giving way slightly, the horizontal reactions
vanished, the structure would then become an ordinary girder
FIG. 205.
(Fig. 205), and the stress in the bar X10 for instance, could be
found according to the previous method, by means of the equa-
tion of moments :
0 = X10 X 0-5 + 6-4 [(^ + A + . . . + 1$) 20
+ (ii . 20 - 2) + (f5 . 20 - 4) + ... + 0$ . 20 - 18)],
or
X, (min.) = - 1280 tons.
In the braced arch it was found that X10 = ±50*7 tons.
* Professor Rankine shows, both in his ' Applied Mechanics ' and in his
' Civil Engineering/ how braced arches without hinges are to be treated, allowing
for the yielding of the abutments, temperature, &c. — TEANS.
154
BRIDGES AND ROOFS.
or
Similarly for Z10 :
0 = ZIO X 0-5498 + 6-4 [(A + A + - - + M) 18
+ (ft . 18 - 2) + (±3 . 18 - 4) + . . . + (i* . 18 - 16)],
Z10 (max.) = + 1229 tons;
whereas for the arch :
Z10 (min.) = — 128-05 tons.
Thus, if the horizontal reactions vanish the stress in Z10
would be increased nearly ten times, and that in X10 more
than twenty times.
If, however, the abutments remain firm, and the arch expands
FIG. 206.
FIG. 207.
by an increase of temperature, the horizontal reaction will be
increased, and it is not difficult to see that in some parts the
stresses will become considerably greater.
As soon, however, as a hinge is introduced, as at S (Fig.
207), all the indeterminateness as to the magnitude of the
horizontal reaction disappears, and likewise the danger caused
thereby. It has already been shown that a load Q (Fig. 207)
placed on one half of the arch calls forth .a reaction at the
24. — THEORY OF HINGED-BRIDGES-.
155
abutment of the other half, which must of necessity pass through
the central hinge. Consequently by producing B S the position
of P can be fixed, and with it the horizontal reactions at the
abutments. If the abutments give way slightly, the hinge will
be slightly lowered, and it will rise a little when the arch
lengthens with an increase of temperature, but in no case will
the difference produced in the stresses be appreciable.
It has already been pointed out in § 8, and again at the
end of § 20, that there is no difference between the calculations
for an arch, that is when the convexity of the bow is turned
upwards, and those for a suspension bridge in which the
convexity is turned downwards. Thus Fig. 208 is obtained
from Fig. 207 by turning the arch upside down, and then
FIG. 208.
FIG. 209.
FIG. 210.
changing the direction of all the forces. It is also evident that
all the remarks made relatively to the arch also apply to the
suspension bridge.
It is hardly necessary to observe that hinge-bridges can be
constructed of a variety of forms. Two of these are represented
in Figs. 209 and 210. Fig. 209 can be regarded as the para-
156 BRIDGES AND ROOFS.
bolic girder of Fig. 39, the lower boom of which has been cut
through in the centre, and the resistance to tension of the boom
replaced by the horizontal reactions at the abutments. Fig. 210
is a similar form in which the resistance to tension of the
abutments is brought into requisition. Both structures can be
calculated in the manner explained in § 21 and § 22.
SEVENTH CHAPTER.
§ 25. — VARIATION IN THE STRESSES DUE TO ALTERATIONS
IN THE SPAN.
In the preceding chapters the equations of moments, &c., have
been given in extenso for each part of the structure, for it is
possible to employ these equations and the stresses obtained in
many ways for structures that are geometrically similar to
those that have been calculated, or as it may be expressed, for
structures that differ only in their unit of length.
Were it not that the loads alter according to the span, and
especially that the proportion between the permanent and
moving load changes, the equations and stresses found would
be directly applicable whatever the span. For it makes no
difference in the results whether the unit of length is a foot,
or a metre, or a yard, since the equations of moments depend
only on the proportion between the lever arms and not on their
absolute length.
If then, when the span increased, the permanent and the
moving load increased in the same ratio, it would only be
necessary to multiply the stresses already found by this ratio
to obtain the new stresses. But in general this cannot be
done, for the permanent load as a rule augments much more
rapidly as the span increases than the moving load, and con*
sequently an increase of span will affect the stresses in dif-
ferent parts of a structure differently. The problem is therefore
to find these new stresses from those already calculated, and to
do so by as short a way as possible.
The following notation will be used : p and m will represent
the permanent and moving loads on the structure that has
already been calculated, and pl and ml the permanent and
moving loads on the new structure.
158 BKIDGES AND HOOFS.
Now every stress can be divided into two parts, one pro-
duced by the permanent and the other by the moving load. If
the stress already found be thus divided, and the first part be
multiplied by the second by — 1 and the results added the re-
quired stress will evidently be obtained.
The various bars of a structure divide themselves into three
groups, with respect to the effect of the permanent and moving
loads upon them as follows : —
The first group contains all those bars the stress in which
depends entirely on the moving load. In this case the new
stress is obtained by multiplying the old stress by the
,. m>\
ratio - •
m
The second group comprises those bars in which the stress
is greatest when the structure is fully loaded. For them the
new stress is equal to the old stress multiplied by— - 1 •
p + m
And the third group consists of all the remaining bars, that
is those who obtain their greatest stress with a partial load. In
this case the stress produced by the permanent load must be
multiplied by ~ and that by the moving load by - 1 and the re-
sults added together to obtain the new stress.
The last group is the only one which ever requires new
calculations, and as a rule these calculations are very simple.
The stresses in the bars of the first and second groups can be
obtained without difficulty from the stresses already found. As
an illustration a few examples are appended —
a. Parabolic Girder.
Here the diagonals belong to the first, the horizontal bars to
the second, and the verticals to the third group.
Thus to find, from Fig. 27, the stresses in the diagonals of a
similar girder, 48 metres span, with a permanent load of 8000
kilos, and a moving load of 12,000 kilos, on each joint, the
stress in each of the six diagonals must be multiplied by
12000 = o.d
5000
§ 25.— VARIATION IN STRESSES DUE TO ALTERATIONS IN SPAN. 159
and the new stresses are
± 15000, ± 16400, d= 17000, ± 16400, ± 15000, ± 13130.
The stress in the horizontal bars must be multiplied by
8000 + 12000 _ 10
1000+ 5000 ~ ¥
to find the new stress, which is
- 160,000 kilos.
Similarly the stresses in the bow are to be multiplied by ^- thus :
+ 175000, + 167700, + 163000, + 160300.
The stresses in the verticals must be divided into two parts
as explained above. Now the permanent load produces a stress
of — 1000 kilos, in each vertical, and therefore the effect of the
moving load can be found by adding — 1000 to the stresses
given in Fig. 27 ; thus
For the Maxima Stresses,
( - 1000 \ / - 1000 \ / - 1000 \ / - 1000 \ Permanent load.
\ 0 / \ + 1560 / \ + 2500 / \ + 2800 / Moving load.
And for the Minima Stresses,
( — 1000 \ / — 1000 \ / — 1000 \ / — 1000 \ Permanent load.
\ - 5000 / \ - 6560 j \ — 7500 / \ - 7800 / Moving load.
and the new stresses are obtained from these by multiplying the
first figures in brackets by 8 and the second by 2*4, thus
For the Maxima Stresses,
i _ 8000 \ / - 8000 \ _
\ 0 / - \ +3740 / -
( _ 8000 \ _ / - 8000 \ _
\ + 7000 / ~ \ + 6720 / -
And for the Minima Stresses,
*n
12000 J
{ I ,25 5 \ = - 20000 I ~ ,!!X2 \ = - 23700
- — 26700
160 BRIDGES AND ROOFS.
Since the stresses in the remaining three verticals are
repetitions of the above, it is unnecessary to calculate
them.
b. Braced Girder with Parallel Booms.
Here the first group does not occur ; all the horizontal bars
belong to the second group, and the diagonals and verticals to
the third group.
As an example, let it be required to deduce from Fig. 57 the
stresses in a similar girder of 48 metres, and (as in the last
example) with a permanent load of 8000 kilos, and a moving
load of 12000 kilos, on each joint.
The stresses in the horizontal bars are to be multiplied by
8000 + 12000 _ 10
1000 + 5000 ~ ~3
to obtain the new stresses — thus,
70000, 120000, 150000, 160000,
which for the lower boom must be taken with a positive sign
and for the upper boom with a negative sign.
The stresses in the diagonals and verticals could be as
quickly calculated by introducing into the equations of moments
given at p. 39-43, the new values of the loads as by the present
method of dividing the stresses into two parts. This latter
course will, however, be adopted, for in so doing the stress pro-
duced by the permanent load alone and by the moving load
alone will be found, and thereby a better view of the functions
of these braces will be obtained. As an example, take the
diagonal and the vertical of the third bay. In § 10 it was found
that
Y3 = V3 . V2,
and that
0 = Y, x 0-707 - 1000 [Q- + | + . . . + f ) - (1 - -«) - (1 - |)J
- 5000 (i + | + . . . + |) + 5000 [(1 - f) + (1 - I)].
To find the stress produced in this diagonal by the per-
manent load alone, leave out the two members multiplied by
5000 and solve the equation, thus obtaining + 2,120 kilos.
§ 25. — VARIATION IN STRESSES DUE TO ALTERATION IN SPAN. 161
Then leave out the permanent load, and the stresses due to the
moving load alone are found to be
+ 13260 and - 2650.
The last two stresses are to be multiplied by
12000 -2-4
~
and the first by
8000
1000 " '
and the results added together, thus :
The stresses in the vertical V3 can be found by multiplying
the above stresses by -- — , thus :
( - 12000 \ _ 00500 / - 12000 \ _
\ - 22500 / - \ + 4500 / =
c. Braced Arch and Suspension Bridge.
The diagonal and horizontal bars in this case belong to the
first group, the verticals and the bars in the bow to the third
group, the second group has no representatives.
As an example let it be required to find the stresses in a
suspension bridge geometrically similar to that given in Fig.
197, and having a span of 120 metres.
The permanent load on each joint will be taken at 20 tons,
and the moving load at 12 tons. The stresses in the horizontal
bars (Fig. 197) are therefore to be multiplied by ^ = 3 to
obtain the new stresses, thus :
±15-6, ±33-48, ±54-18, ± 77'64, ±102-87, ± 130 '71,
± 152-1, ± 150-87, ± 108-0, 0.
Likewise the stresses in the diagonals are to be multiplied
by 3, thus
± 38-76, ± 37-77, ± 36-9, ± 36'21, ± 35'7 ± 33-21,
±32-19, ±29-4, ±64-2 ±111-87.
M
162 BRIDGES AND HOOFS.
For the verticals the values of ul9 u2 already found (p. 124),
and representing the effect of the moving load alone (taken with
contrary signs for a suspension bridge) can be used.
These values multiplied by the ratio ^ = 3 give for the
maxima stresses,
+ 47'46, + 45-24, + 42-6, + 39-3, + 35*52, + 30*09,
+ 25-14, +19-71, +25-2, + 33'6, +6;
and for the minima stresses
-35-46, -33-24, -30'6, - 27'3, - 23'52, -18-09, - 13'14,
-7-71, -13-2, -21-6, 0.
The stress in the verticals produced by the permanent load
is (with the exception of that in the eleventh vertical, which
has only one-half to bear) + 1*2 ton, and to obtain the
stress in the larger bridge due to the permanent load alone,
this must be multiplied by -— = 8 • 33, and the new stress is
1*2 x 8*33 = 10 tons, which must be added to the stresses
due to the moving load, thus :
+ 57-46 I +55-24 I + 52'6 I +49-3 I +45'52 I +40-09 I +35-14
-25-46 I -23-24 | -20'6 | - 17'3 | - 13'52 j - 8'09 j - 3'14
+ 29-71 I +35-2 | +43-6 I +11
- 2-29 I - 3-2 I -11-6 I + 5
The stresses in the chains can also be determined by split-
ting up the stresses as above, for the stress produced by the
permanent load alone, which is uniformly distributed over the
horizontal span, can be easily found from the formula given
in § 8, the chains being in the form of a parabola ; the stress
due to the moving load alone can then be found by subtraction
from the total stress. The result of thus splitting up the
stresses given in Fig. 197 is the following, where the upper
figures are due to the permanent, and the lower to the moving
load.
53-2 I 52-2 I 51-2 I 50'5 I 49'8 I 49'3 I 48'7 I 48'4 I 48'1 I 48
88-5 87-7J 88-1 89'8 92'9 96'9 101-2 104-4 100-9 80
§ 25. — VARIATION IN STRESSES DUE TO ALTERATION IN SPAN. 1 63
The first are to be multiplied by |^ = 8 '33, and the second
by ^ = 3, thus :
443 I 434-5 I 427 I 421 I 415 I 410 I 406 I 403 I 401 I 400
265 I 263 I 264 | 269 | 279 | 291 | 304 | 313 | 303 | 240
and the new stresses are obtained by adding these together,
thus:
708 | 697-5 | 691 | 690 | 694 | 701 | 710 | 716 | 704 | 640
M 2
( 164 )
EIGHTH CHAPTER
§ 26. — SUSPENSION BKIDGE IN THREE SPANS. SPAN OF
CENTRAL OPENING, 120 METRES. SPAN OF EACH SIDE
OPENING, 60 METRES.
Suspension bridges do not, unless special arrangements are
made, compare favourably with braced arches, as regards the
amount of materials employed ; for in the latter the points of
connection with the abutments are placed low down, and the
horizontal thrust acts against the abutments in the direction in
which they are strongest; whereas in the former, on the
contrary, the points of attachment are placed high up, and
the horizontal pull tends to turn the piers over in the direction
in which they are weakest; consequently, the quantity of
material in the piers will be much greater in one case than
in the other. Whereas, therefore, with a braced arch a com-
paratively small expenditure of material is required for the
abutments, especially if natural ones of rock can be obtained,
the quantity would be enormous with a suspension bridge, if
it were wished to attach the chains to the piers, as shown at E
and F (Fig. 211).
FIG. 211.
The comparison would, however, be less unfavourable to the
suspension bridge if there were several spans, as shown in
Fig. 211. The horizontal tensions neutralize each other at the
central piers A and B, at least when the spans are equally
loaded ; but there would be the same disadvantage at the land-
§ 26. — SUSPENSION BEIDGE IN THREE SPANS. 165
piers E and F.* In such a structure the horizontal and diagonal
bars would be under no stress when the bridge is uniformly
loaded, assuming the curves to be parabolas.
There would be no alteration, as regards this last point, if
the ends E C and D F were cut off and the chains attached at
the points C and ^D to abutments. This arrangement has the
advantage of lowering the points of attachment of the chains
atjathe^shore end, thereby increasing the stability of the abut-
ments. If, besides this, the points A and B are hung to Ax
and B! by means of vertical rods, the central piers will be
entirely relieved of all horizontal thrust, even when the load is
not uniformly distributed, for the reactions at Ax and Bx must
of necessity be entirely vertical (Fig. 212). The chains in the
FIG. 212.
B,
parts C A and B D act as land-ties to' the central opening, and
at the same time the material in them is employed to bridge
over the side spans.
(The connections at the points A, B, C, and D, shown in
Fig. 212, are only given by way of illustration, other and
better means of arriving at the same result will be discussed
farther on.)
Such a bridge can, on the whole, be represented by the com-
bination of four rods shown in Fig. 213. These rods are con-
nected together by smooth hinges ; they are supported directly
by the fixed points C and D, and by means of vertical rods at
A and B. It is also supposed that the rods are weightless.
Now it is evident that the direction of a force acting at each
end of an unloaded rod must be that of the rod itself; for
otherwise rotation would ensue. Therefore a load Q placed
at P can only produce a reaction K at C acting in the direc-
tion A C ; and, similarly, a load Qx at Px can only give rise to
* It will be observed that it is usual to place land-ties at E and F. This
would greatly diminish the quantity of material in the end piers. — TKANS.
166
BRIDGES AND EOOFS.
a reaction in the same direction. But if a load Q be placed
anywhere on the rod A C (Fig. 214), the reaction at C will be
vertical. For if in this case there were a horizontal thrust at
C, an equal horizontal thrust would be required at A and S.
But there can be no force acting at S ; for since both rods A 8
and B S are unloaded, this force would be required to have
FIG. 213.
simultaneously the directions A S and B S. Thus, a load
. placed on the rod C A has no effect on the remaining three
rods A S B D.
When the rod C A, therefore, is alone loaded, it behaves like
an ordinary beam supported at both ends, and when the rods
A S, S B are loaded they are in the same condition as if their
points of suspension A and B were fixed points.
The stresses in the bridge shown in Fig. 212 can now be
found.
a. Calculation of the Stresses in the Central Span A B.
The stresses in the bars of each half of the central span A B
can, in accordance with the above, since A and B may be
regarded as fixed points, be found by the method employed to
calculate those given in Fig. 197. The span is 120 metres, and,
assuming that the form is geometrically similar and the loads
the same, the stresses found for the suspension bridge in § 25 c,
will be those required.
§ 26. — SUSPENSION BKIDGE IN THKEE SPANS. 167
b. Calculation of the Stresses in the Side Span A C, Fig. 215.
It will be assumed that the parts A C and B D of the bridge
are, as regards their form and construction, geometrically
similar to each half of the suspension bridge of § 22, and
FIG. 215.
the same letters have been used to denote the corresponding
parts. The loads will also be taken the same as those given
in § 25 c, namely, 20 tons permanent,* and 12 tons moving
load on each joint. The method adopted for the calculations
of the braced arch of § 22 will be followed, and for each bar it
will be found which loads produce tension and compression
respectively. To do this, recourse must be had to the two
laws given above, which are :
1. A load on the central span requires a reaction E at the
points A and C, whose direction is A C.
2. A load on the side span A C produces vertical reactions
at the points A and B.
Calculation of the Stresses X in the Horizontal Bars.
The stress in X5 is to be found by taking a section aft, and
then forming the equation of moments, either for the part in
Fig. 216 or the part in Fig. 217, with reference to the point J.
Any load on Fig. 216 produces a vertical reaction W at A,
which tends to turn Fig. 217 from right to left round J.
* This assumes that the weight of the bridge is uniformly distributed, and
this is not far from the truth, as will appear by examining Figs. 238 and 239,
p. 181.— TRANS.
168
BRIDGES AND ROOFS.
X5 acts in the contrary direction, and is therefore positive.
The loads on the part C a @ belong, therefore, to the tension
group.
A load placed on Fig. 217 produces a vertical reaction D at
C, making X5 positive, as before ; consequently the loads on
A a ft also belong to the tension group.
FIG. 216.
FIG. 217.
A load on the principal span produces a reaction K at C,
which tends to turn the part in Fig. 216 from right to left,
thus making X5 negative ; therefore the loads on the principal
span belong to the compression group.
A load placed in any position on B D has no effect on A C,
and the part B D is marked accordingly in Fig. 218.
TENSION
FIG. 218.
COMPRESSION
NO EFFECT
-x >
Since when the load is uniformly distributed over the
whole bridge the stress in the horizontal bars is nothing, the
permanent load can be omitted from the calculations ; and,
further, the maxima and minima values of the stress produced
in the horizontal bars by the moving load are numerically
equal ; therefore it is only necessary to find one of them. The
central span must alone be loaded when X5 (min.) obtains, and
26. — SUSPENSION BRIDGE IN THREE SPANS.
169
the consequent horizontal tension at S (Fig. 219) is given by
the equation :
0 = - H x 15 + 12 (ip + 54 + 48 + . . . + 12 + 6)
H = 240.
FIG. 219.
IT
The horizontal component of K must evidently be equal to
H, and since the ratio of the vertical to the horizontal com-
ponent of E is as 15 : 60, or as 1 : 4 ;
Hence the equation of moments from Fig. 220 is
0 = - X5 x 5-25 + 240 x 3-75 - 60 x 30
Xs(min.) = -171 -4 tons.
FIG. 220.
As a check, X5 (max.) can be computed by taking the side
span A C alone loaded, and considering it as a girder, thus :
0 = - X5 X 5-25 + 12 K^y + .. . + •&) 30 + (A . 30 - 6)
+ (^ . 30 - 12) + fa . 30 - 18) + (.& . 80 - 24)]
X5(max.) = + 171 -4 tons,
which agrees exactly.
In a similar manner the following equations are obtained for the remaining
horizontal bars :
0 = - X! x 13-65 + 240 X 12-15 - 60 x 54
Xj = ±23-7 tons
170
BRIDGES AND ROOFS.
0 = - X2 x ll'l + 240 x 9-6 - 60 X 48
X2 = ±51 -9 tons
0 = - X3 x 8-85 + 240 x 7'35 - 60 X 42
X3 = ±85-4 tons
0 = - X4 X 6-9 + 240 x 5-4 - 60 X 36
X4 = ±125 -2 tons
0 = - X6 x 3-9 + 240 x 2-4 - 60 X 24
X6 = ±221 -5 tons
0 = - X7 x 2'85 + 240 X 1'35 - 60 x 18
X7 = ±265-3tons
0 = - X8 x 2-1 + 240 X 0-6 - 60 x 12
X8 = ±274 -3 tons
0 = - X9 x 1-65 + 240 x 0-15 - 60 X 6
X9 = ±196-4 tons
0 = — X10 X 1-5
Xin = 0.
Calculation of the Stresses Y in the Diagonals.
As in the preceding case, and for the same reason, it is
unnecessary to consider the permanent load, and it is only
requisite to calculate either the maximum or the minimum
FIG. 221.
FIG. 222.
value of the stress. The stress in Y5, for instance, can be found
by means of the equation of moments formed either for the part
shown in Fig. 221, or for that shown in Fig. 222, with respect to
the point L.
The reaction E at C, due to a load on the central span,
26. — SUSPENSION BKIDGE IN THREE SPANS.
171
evidently makes Y5 positive. A load placed anywhere on the
part A aft produces a vertical reaction D at the point C,
which makes Y5 negative. A load on C a ft requires a vertical
reaction W at A, which also makes Y5 negative. Hence
Fig. 223, showing the manner in which the stress in Y5 is
affected by the various loads.
COMPRESS/ON
FIG. 223.
TENSION
IW EFFECT
Thus to find Y5 (max.) the central span alone must be
loaded, and, as before (p. 169),
H = 240 and V = 60 ;
and from Fig. 224 the equation of moments is
0 = Y5 x 16-53 - 240 x 1-5 - 60 x 10-92
Y5 = ±61 -4 tons.
FIG. 224.
eio
Similarly, for the remaining diagonals :
0 = Yj x 31-8 - 240 x 1-5 - 60 X 25-26
Yj = ±59-0 tons
0 = Y2 x 28-26 — 240 X 1-5 — 60 x 21-88
, Y2 = ± 59-2 tons
0 = Y3 x 24-48 - 240 x 1-5 - 60 x 18'39
Y3 = ±59-8 tons
0 = Y4 x 20-5 - 240 x 1-5 - 60 x 14-77
Y4 = ±60-8 tons
THE
UNIVERSITY
172 BRIDGES AND ROOFS.
0 = Y6 x 12-72 - 240 X 1'5 - 60 x 6'67
Y6 = ±59 -7 tons
d = Y7 x 9-58 - 240 x 1-5 - 60 x 1'713
Y7 = ±48 -3 tons.
In the case of Y8 the turning-point is situated to the left of
C, consequently the sign of the moment of the vertical reaction
at D is reversed, and thus the loads producing tension extend
over the central span and up to the section line 7 S, as shown
in Fig. 225. In this case it is easier to calculate Y8 (min.),
and from Fig. 226, or else from Fig. 227, the following equation
of moments is obtained :
COMPRESSION
0 = - Y8 x 7-53 -12 (^
Y8 = ± 31 tons.
FIG. 225.
TENSION
64-8
NO EFFECT
The turning-point for Y9 is situated so far to the left that
the moment of K, the reaction due to the loads on the central
span, also changes its sign, and consequently the loads on this
span produce compression in Y9 (Fig. 228). Here again it is
easiest to calculate Y9 (min.). Thus from Fig. 229 :
0 = Y9 x 7-41 - 240 x 1-5 + 60 x 16 + 12 (22 -
Y9 = ±93-3 tons.
x 16)
§ 26. — SUSPENSION BKIDGE IN THREE SPANS. 173
Lastly, for Y10 all the loads on A C produce tension, and Y10
can therefore be obtained by considering the central span
loaded, and as before (p. 171) :
0 = Y10 x 15-97 - 240 x 1-5 + 60 x 60
Y10 = ±202 -9 tons.
FIG. 227.
16 | | r
U
Vfo
JEMSION
FIG. 228.
COMPRESSION
NO EFFECT
Calculation of the Stresses U in the Verticals.
The stress in each vertical can be divided into two parts ;
one part due to the permanent load, and the other to the
moving load. The first is the same for all the verticals, and is
equal to + 10 tons, if it is assumed that one-half of the total
permanent load (20 tons) is applied to the upper joints, and
the other half to the lower joints. Denoting the part of the
stress due to the moving load by u, it is evident that
u (max.) + u (min.) = -f 12 tons,
because when the moving load covers the bridge, it produces a
tension of 12 tons in each vertical (being applied to the lower
joints only). Thus, if u (min.) be calculated, u (max.) can be
found from the equation
M (max.) = + 12 — u (min.).
174
BRIDGES AND ROOFS.
And finally, the total stress in the verticals can be found by
means of the equations
U (max.) = u (max.) + 10
U (min.) = u (min.) + 10.
The turning-points for the verticals are the same as those
for the diagonals ; the loading boundaries will therefore in
general be the same, but the loads that produce tension in the
diagonals will produce compression in the verticals, and vice
versa. The bars U9 and U10, however, possess a second loading
boundary, which is determined by the section line itself, as
was also found to be the case with the corresponding diagonals,
but will be shifted one bay to the left, owing to the oblique
direction of the section line.
TENSION
FIG. 230.
COMPRESSION
NO EFFECT
\
FIG. 231.
240
\
The groups of loads for U5 are shown in Fig. 230 ; and from
Fig. 231 the equation of moments to find u5 (min.) is
0 = — us x 25-08 — 240 x 1-5 — 60 x 10-92
M5 (min.) = — 40 '5 ;
and substituting in the above equations
u5 (max.) = + 12 - (— 40'5) = + 52'5 tons
Us (max.) = + 52-5 + 10 = + 62-5 tons
U5 (min.) = - 40-5 + 10 = - 30'5 tons.
§ 26. — SUSPENSION BRIDGE IN THREE SPANS. 175
In the same manner the stresses in the remaining verticals can be found
thus:
x 34-74 - 240 X 1-5 - 60 x 25'26
M! (min.) = — 54 «! (max.) = + 66
U, (min.) = - 44 tons Ux (max.) = + 76 tons
0 = - u2 x 32-12 - 240 x 1-5 - 60 x 21-88
w2 (min.) = - 52-1 w2 (max.) = + 64'1
U2 (min.) = - 42 • 1 tons U2 (max.) = + 74 • 1 tons
0 = - u3 x 29-61 - 240 x 1-5 - 60 X 18-39
M3 (min.) = — 49-5 u3 (max.) = + 61 -5
U3 (min.) = — 39-5 tons U3 (max.) = + 71 '5 tons
0 = - M4 x 27-23 - 240 x 1-5 - 60 x 14-77
M4 (min.) = — 45-8 w4 (max.) = + 57 '8
U4 (min.) = - 35-8 tons U4 (max.) = + 67 '8 tons
0 = - u6 x 23-33 - 240 x 1-5 - 60 x 6'67
u6 (min.) = - 32-6 u6 (max.) = + 44 '6
U6 (min.) = - 22-6 tons U6 (max.) = + 54-6 tons
0 - - u7 x 22-29 - 240 x 1-5 - 60 x 1-713
u7 (min.) = - 20-8 «7 (max.) = + 32-8 .
U7 (rnin.) = - 10-8 tons U7 (max.) = + 42 '8 tons.
For the three following verticals it is best to find the value of u (max.)
thus:
0 = - u, x 22-8 - 12 (TV + & + ^) 64-8
u8 (max.) = + 20-5 M8 (min.) = — 8*5
U8 (max.) = + 30 • 5 tons U8 (min.) = -f 1 • 5 tons
0 = - u9 x 28 - 240 x 1-5 + 60 X 16
+ 12 [(22 - ^ . 16) + (28 - ft . 16)]
M9 (max.) = +31-2 u9 (min.) = — 19 • 2
U9 (max.) = +41-2 tons U9 (min.) = — 9*2 tons
0 = - MIO x 66 - 240 x 1-5 + 60 x 60 + 12 (66 - & . 60)
u10 (max.) = + 51-3 'ulo (min.) = — 39 '3
U10 (max.) = + 61-3 tons U10 (min.) = — 29 '3 tons.
Lastly, the only stress in the llth vertical is that due to a load hung under-
neath it. The greatest value this load can have is -^ tons moving load added to
Y tons permanent load ; hence
Un (max.) = + 11 tons.
176
BRIDGES AND ROOFS.
Calculation of the stresses Z in the chains.
The sign of Z5 (Figs. 232 and 233) depends on the sign of
the moments of the three forces, E, D, and W. From Fig.
' 232 it appears that the moments of K (the reaction due to a
load on the central span) and Z5 about 0 have different signs ;
therefore a load on the central span makes Z5 positive.
When the part Aa/3 (Fig. 233) is loaded, the reaction
D is produced at C, the sign of whose moment about 0 (Fig.
232) is the same as that of Z5, or Z5 is negative.
FIG. 232.
FIG. 233.
Again, if C a ft is loaded, the reaction W at A has (Fig.
233) a moment about 0, whose sign is the same as that of Z5 ;
therefore Z5 is again negative.
The greatest compression occurs, therefore, when the side
COMPRESSION
FIG. 234.
TENSION
NO EFFECT
span is fully loaded, and the greatest tension when the central
span is loaded, as shown in Fig. 234.
The stress in Z5 can be calculated in two different ways.
The first is as follows : — When Z5 (max.) obtains, the central
span is fully loaded, and the side span has only the perma-
§ 26. — SUSPENSION BRIDGE IN THREE SPANS. 177
nent load upon it. From Fig. 235 the equation of moments
is
= Z5 x 6-654 -H x 1-5- V X 36 + D x 36-20(6+ 12
30),
where H and V are the components of the reaction K, due to
the load on the central span. The corresponding values of H
and V, due to the moving load alone, have already been found
FIG. 235.
to be 240 and 60 respectively. To find the values now required,
these must be multiplied by the ratio
20+ 12 _ 8
12 • 3
Further, D is the vertical reaction at C, due to the permanent
load on the side span.
Substituting, the equation of moments becomes
0 = Z5 x 6-654 - 640 X 1-5 -.160 x 36 + 20 (^ +
- 20 (6 + 12 + . . . + 30)
Z5 (max.) = + 794 tons.
FIG. 236.
+ JL) 3G
D
H-400
=32(fo
*-•$
tio ^
10
^—^
10
X.
a
io^^_
< .---•" '4
^~^-~^
5^^^
^
\^
ft
•
,
feioo.
10 >
,12 |l2 ,
.10 ,
U j
40 >
42 J
10 N
** \P
,Z5
The value of Z5 (rain.) can be obtained from Fig. 236. H
and V are the components of the reaction E, due to the per-
178
BRIDGES AND EOOFS.
manent load alone on the central span, and D is the vertical
reaction produced by the total load on the side span. Hence
and
H =
V =
640 = 400
160 = 100
= Z5 x 6-654 - 400 x 1'5 - 100 X 36 + 32 (•&
- 32 (6 + 12 + . . . + 30)
Z5 (min.) = + 285 tons.
...+ •&) 36
The second method is to split up the stress in Z5 in two
parts, one p5 due to the permanent load alone, and the other
z5 due to the moving load alone. The value of p$ has already
been obtained in § 25 (for when the bridge is covered with a
uniform load, the side spans are precisely in the same con-
dition as either of the halves of the main span). It was found
that
ps = + 415 tons.
It is only necessary to calculate either z5 (max.) or 05 (min.),
for both together must be equal to the stress produced by the
moving load when it covers the whole bridge ; and this stress
can easily be found by comparison with p$ — in fact, by multi-
plying p5 by the ratio J§ = f ; therefore
or
z5 (max.) + z6 (min.) = £ x 415 = + 249,
z6 (min.) = + 249 - z5 (max.).
It is easiest to obtain z5 (max.), and it can be found from
the equation of moments for the part of the side span shown
FIG. 237.
,H=240
in Fig. 237 (the values of H and V will be those already found
when the moving load covers the central span). Hence
0 = z& x 6-654 - 240 x 1'5 - 60 x 36
z& (max.) = 379 tons,
§ 26. — SUSPENSION BRIDGE IN THREE SPANS. 179
or
zs (min.) = + 249 - 379 = - 130 tons.
Finally, adding p5 and z5 together,
Z5 (max.) = 415 + 379 = + 794 tons
Z5 (min.) = 415 — 130 = + 285 tons.
The same result is thus obtained by both methods; the
last, however, is the simpler, and will therefore be employed
for the calculation of the stresses in the remaining bars Z ;
thus : —
0 = *, x 14-904 - 240 x 1-5 - 60 x 60
zl (max.) =+265-5
zl (min.) = + 265-5 - 265-5 = 0
Zx (max.) = 443 + 265-5 = + 708-5 tons
Zj (min.) =443 + 0 = + 443 tons
0 = z2 x 12-56 — 240 X 1-5 — 60 x 54
22(max.) = 286-5
*2 (min.) = 261 - 286-5 = - 25'5
Z2 (max.) = 434-5 + 286-5 = + 721 tons
Z2 (min.) = 434-5 - 25-5 = + 409 tons.
0 = z3 x 10-39 - 240 x 1-5 - 60 x 48
z3 (max.) = 312
z3 (min.) = 256 - 312 = - 56
Z3 (max.) = 427 + 312 = + 739 tons
Z3 (min.) = 427 - 56 = + 371 tons
0 = Zt x 8-415 — 240 x 1-5 — 60 x 42
z^ (max.) = 342
z4 (min.) = 252 - 342 = - 90
Z4 (max.) = 421 + 342 = -f 763 tons
' Z4 (min.) = 421 - 90 = + 331 tons
0 = z6 x 5-121 — 240 x 1-5 — 60 x 30
z6 (max.) = 422
*6 (min.) = 246 - 422 = - 176
Z6 (max.) = 410 + 422 = + 832 tons
Z6 (min.) = 410 - 176 = + 234 tons
0 = zj x 3-84 — 240 X 1-5 — 60 x 24
Zj (max.) = 469
z, (min.) = 244 - 469 = - 225
Z7 (max.) = 406 + 469 = + 875 tons
Z7 (min.) = 406 - 225 = + 181 tons
N 2
180 BRIDGES AND ROOFS.
0 = z8 X 2-83 - 240 X 1-5 - 60 X 18
z8 (max.) = 509
*8 (min.) = 242 - 509 = - 267
Z8 (max.) = 403 + 509 = + 912 tons
Z8 (min.) = 403 — 267 = + 136 tons
0 = zsx 2-094 — 240 X 1-5 — 60 x 12
z9 (max.) = 516
z9 (min.) = 241 - 516 = - 275
Z9 (max.) = 401 + 516 = + 917 tons
Z9 (min.) = 401 - 275 = + 126 tons
0 = zu X 1-649 - 240 x 1-5 - 60 X 6
zlo (max.) = 437
z10 (min.) = 240 - 437 = - 203
Z10 (max.) = 400 + 437 = + 837 tons
Z10 (min.) = 400 - 203 = + 197 tons
The results of the above calculations are collected together
in Fig. 238. And the stresses in the central span are given in
Fig. 239, having been deduced from Fig. 197 and § 21.
§ 27. — STABILITY OF THE CENTRAL PIERS.
It was assumed, in the preceding calculations, that the con-
nections at the points of support were made as indicated in
Fig. 212. For these calculations to be true, it is necessary that
at the points A and B vertical forces only should act on the
bridge, and it therefore follows that these points should be
perfectly free to move in a horizontal direction. If such a
mode of attachment be adopted, the stability of the central
piers is a question that need not be considered (in so far as the
vertical forces on the bridge are concerned). The manner of
forming these connections shown in Fig. 212 is not, however,
the only one by which this advantage may be gained, and it
was only chosen as an illustration, and there are better ways of
arriving at the same result. For instance, the points Ai and 1^
can be placed below A and B, as shown in Fig. 240.
Nor is it necessary that the chains of the adjacent spans
§ 27. — STABILITY OF THE CENTRAL PIERS.
181
FIG. 239.
25,46
182
BRIDGES AND ROOFS.
should be attached to the same point A ; in fact, it is better
to place them apart the full width of the pier, and attach
them to the points au aZ) Fig. 241. The span is thus slightly
diminished. The freedom of these points to move horizontally
can be obtained in a variety of ways. Thus, in Fig. 241 an
unbraced parallelogram is formed by the three bars a\ ck, a^ blf
and a2 &2, the fourth side being the head of the pier bi b2.
(The stresses in these three bars are found immediately from
the former calculations, and are inscribed in the figure.)
Or the chains can be attached to the axis of two friction
rollers (Fig. 242), and the piers being carried up form the
FIG. 242. FIG. 243.
roller path. Or again, the chains may be fastened to a plate
placed upon rollers, the pier being carried up as in the former
case (Fig. 243).
But if the arrangement shown in Fig. 244 were adopted, the
stresses obtained in § 26 would no longer be true, and the
advantage of having no lateral thrust on the central piers would
§ 27. — STABILITY OF THE CENTRAL PIERS.
183
also have to be given up. It is true that when the bridge is
fully loaded the reaction at the central piers would be vertical,
but this would not be the case with a partial load.
FIG. 244.
FIG. 245.
This will become apparent by finding the reaction at the
fixed point 0, due to a single load Q (considering the structure
to have no weight). By
proceeding as in §§ 22 and
26, it will be found that the
forced (Figs. 244 and 245) <--
acts in the direction a^ P, *
and the force K2 in the
direction a2 C. These two
forces, together with the
reaction at 0, maintain the
bent lever a^ Oa2, in equili-
brium, and their resultant K
must therefore pass through
O. The horizontal compo-
nent h of K is the force that
tends to overturn the pier,
and will be greatest when
all the loads producing the
same effect as Q are on
the bridge. These loads extend from at to S, for a load situated
to the right of S has no overturning effect on the pier in
question, since it acts through its hinge-reaction av S, the hori-
zontal component of which is equal to H2 .
In the previous example the moving load from a± to S
was = 120 tons, and when this load is on the bridge it will be
found that
H! = 120 and V, = 90.
184 BRIDGES AND ROOFS.
Now taking moments about 0,
0 = H1a + V1^~H2a-V2
But it was found, p. 169, that
V2 = |H2.
Hence solving for H
but (Fig. 245)
Putting for H^ its value (120 tons), and assuming, for example,
that - = - ,
a 2
2A = 28-2 tons,
or 28 • 2 tons is the maximum horizontal thrust of the central
span of the bridge against either of the central piers.
The horizontal thrust in the contrary direction, produced
by loading a2 G would be very nearly as great.
The force h diminishes together with -, and becomes
nothing when - = 0 ; that is, when both arms of the bent
lever unite and form a single strut. This is the construction
of Fig. 241, and it or one of its modifications is to be preferred.
The vertical pressure on the central piers is, when the bridge
is fully loaded, equal to the load on the parts A S and A C
together, and is therefore
2 (32 X 10 + 32 X 10) = 1280 tons.
[NOTE. — It is easily seen that this must be the case when it is considered that
A C and A S, being equal, will balance about A when equally loaded. Or again,
thus :— In Fig. 235 it is shown that V, the vertical component of the pull pro-
§ 28. — STABILITY OP THE SHOKE ABUTMENTS. 185
duced at C by A S when fully loaded, is 160 tons. When A C is fully loaded,
the vertical reaction at C is also 160 tons. The two vertical forces acting at C
are therefore equal and opposite, and hence neutralize each other. Thus the
whole load on A C and A S is supported at A.]
§ 28. — STABILITY OF THE SHOBE ABUTMENTS.
The reaction E at C, due to a load on the central span
(Fig. 246), tends to overturn the abutment about its lower
edge E, and also to make it slip along its bed F E. Every
NO EFFECT
FIG. 246.
LOADS TE_NDING_ TOWcRTUPNP/EK
load on the side span C A produces but a vertical pressure D
at C, which is neutral as regards overturning, and helps the
abutment to resist sliding.
The moment of the overturning force will thus be greatest
when the central span is fully loaded, in which case the hori-
zontal component of E is
H (max.) = + 640 tons.
The vertical component of E passes through E, and therefore
(similarly to D) is neutral as regards overturning. Thus the
condition of stability is that the moment of G (the weight of
the abutment) about E is not less than the moment of the
horizontal pull, 2 H, of the whole bridge about the same point.
This is expressed by
G - > 2 x 640 x
[1]
from which the least dimensions of the pier to resist over-
turning can be found.
186
BRIDGES AND EOOFS.
But it must also be ascertained that the pier will not slide.
Both the components of E act injuriously in this respect,
H directly, and E indirectly in that it diminishes the pressure
on the base, and thereby also the resistance to sliding.
The reaction D, however, increases the resistance to sliding.
The danger of sliding will therefore be greatest when the
central span is fully loaded and the side span unloaded
(moving load), in which case D is equal to half the weight of
the part A C (Fig. 247). For these conditions of loading
H = 640,
V = 160,
D = 100 ;
and these values must be doubled to represent the effect of
the bridge.
Hence, if / is the coefficient of friction,
/(G + 2D - 2V)>2H,
or
/(G + 2 X 100 - 2 X 160) >2 X 640,
in order that the abutments may not slide.
FIG. 247.
[2]
To prevent failure, the value of G must be taken at least
as great as the greater of the values obtained from the two
conditions expressed in [1] and [2].
The chain 0 A must be securely attached to the abutment
pier, and this can be done by means of a chain built in the
masonry (Fig. 247) and anchored at F. The direction of this
' § 28.— STABILITY OF THE SHORE ABUTMENTS. 187
chain must not be horizontal at C ; for if such were the case, C
would rise when the central span was loaded and the side span
unloaded. In this case, V = 160 tons and D = 100 tons, and
the direction of the chain in the masonry must be such as to
supply the vertical force necessary for equilibrium. Thus if a
is the angle the chain makes with the horizontal at C, this
angle must at least be as great as the angle made with the
horizontal by the resultant of K and D (or of the three forces,
H, V, and D) when the bridge is loaded as above. The
tangent of this last angle is
V-P 160 - 100 3 n.nw--
nsr -eio- =82= I0937D;
therefore
tan a > 0-09375,
or
a > 5° 22'.
The greatest tension in the chain can be obtained by
making its horizontal component equal to the maximum
value of H, or 640 tons.
( 188 )
i '',.'•'
NINTH CHAPTER
§ 29. — ON THE CALCULATION OF THE STRESSES IN DOMES.
In all the preceding examples it could be assumed that
every joint was equally loaded as well as regards the permanent
as the moving load. If the weight of the structure itself were
not quite uniformly distributed over the span, the difference
in each case was small and did not affect to any appreciable
extent the values of the stresses found.
But in the case of domes, the error entailed by such an
assumption would be too great ; for the ribs or principals radiate
from the centre like the spokes of a wheel, and consequently the
loads on them increase considerably from the centre towards the
abutments.
The surface of a dome can be considered as generated by the
revolution of a properly shaped curve round the vertical axis,
and if the ribs are equally spaced the portion of this surface
contained between the vertical planes through two adjacent
principals will represent the load on each principal. Further,
if the bays formed by the bracing are equal, the loads on each
joint will vary as the length of the arc of the circle (seen on
plan) passing through the joint and contained between two
adjacent ribs. But the length of these arcs is proportional to
their distance from the vertical axis of the dome. Thus if
p is the load on the joint situated at the unit of distance from
the axis, p p will be the load on a joint placed at a distance p
from the axis. If, therefore, the load on any joint be known,
tlie load on any other joint can at once be found by simply
measuring its horizontal distance from the vertical axis.
Once the loads on the various joints are known, the stresses
can be found, and conveniently so, by the method of moments, as
will appear in the following example : —
§ 30.— DOME OF 100 METRES SPAN.
The exterior surface of the dome is a hemisphere of 51
metres radius, and contains 16,338 square metres. There are
30. — DOME OF 100 METKES SPAN.
189
eight ribs, each in the form of a quadrant of a circle, and each
rib supports 2042 square metres of the surface of the dome.
The load per square metre of the surface of the dome is
assumed to be 235 kilos, (consisting of the weight of the covering,
together with that of the snow and wind pressure). Each rib
has therefore 2042 x 235 = 480,000 kilos, nearly, or 480 tons
(1000 kilos, to the ton) to carry. The whole of this load will
be considered variable, not only on account of the snow and
wind pressure, but also because it is possible that part of the
covering might be removed for a time. The only permanent
load is the weight of the rib itself, which is estimated at 60 tons.
This load can be considered as equally distributed on the
exterior joints. Each rib consists of two concentric booms,
2 metres apart and connected together by triangular bracing,
dividing the rib into fifteen bays of equal length (Figs. 248,
249). The permanent load is therefore 4 tons on each exterior
joint.
To find the distribution the variable load on the joints, the
distance of these joints from the vertical axis must be measured.
These distances are as follows : —
DISTANCE — No. OF JOINT.
5-3
10-6
15-8
20-7
25-5
30
34-1
i
2
3
4
5
6
1
44-2
46-6
48-5
49-9
50-7
51
10
11
12
13
14
15
37-9
41-3
These numbers, as already seen, are proportional to the
variable load on each joint. Therefore if the whole of the
variable load on the rib, 480 tons, be divided by the sum
of all these numbers, 512, the quotient multiplied by each
number in succession will give the variable load acting on the
corresponding joint, thus :
LOAD — No. OP JOINT.
5 9-9 14-8 19-4 23'9 28'1 32 35-5 38'7 41-4
14-8
19-4
23-9
28'1
32
35-5
3
4
5
6
»
8
43-7
45-5
46-8
47-6
11
12
13
14
190
BKIDGES AND ROOFS.
A ring is attached to the feet of the ribs, so that the walls
supporting the dome may have no horizontal thrust to bear,
and the ribs are exactly in the same condition as if their lower
extremities were attached to fixed points.
FIG. 248.
11
n
Also, in order that the stresses may be independent of the
variations of temperature or of alterations in the tension of the
§ 30. — DOME OF 100 METKES SPAN.
191
ring, it will be assumed that the tops of the ribs are connected
together by a hinge.* Hinges are also placed at A and B.
The arrangement adopted is thus similar to that of the
braced arch of § 22, and the reasoning will consequently also be
similar.
The process to find the hinge-reaction at S will, for instance,
be the same. It will thus be found that when the dome is fully
loaded, or when it is quite unloaded, the vertical component of
this hinge-reaction is zero, and the reaction is therefore hori-
zontal. The magnitude of this horizontal force H can be found
by equating its moment about A to the moment of all the loads
about the same point. The lever arms of the loads can be
obtained by subtracting their distances from the centre given
above from 50, the half radius, thus :
LEVER ARMS — No. OF JOINT.
50
44-7
39-4
34-2
29-3
24-5
20
15-9
12-1
s
i
2
3
4
5
6
1
8
5-8 1 3-4
1-5
o-i
-0-7
10 11
12
13
14
8-7
Thus when the ribs are unloaded, H is found from the
equation
H x 50 = 4 (^> + 44-7 + 39-4 + ...+ 1-5 + 0-1 - 0-7) = 1056,
or
H = 21 -12 tons.
If, however, the full variable load is applied, the equation
becomes
H x 50 = 4(^ + 44-7 + 39-4 + ... + 1-5 + 0-1 -0-7)
+ 5 x 44-7 + 9-9 X 39-4 + . . . + 45-5 x 1-5
+ 46-8 X 0-1 -47-6 x 0-7.
The last products in the equation are the moments of the
* To meet the objection that might be raised, that the number of hinges
crossing each other at S would render the construction impossible, the ribs are
supposed to abut against a ball, which will act as a hinge for each rib.
192
BRIDGES AND ROOFS.
several variable loads about A, and as they will occur frequently
in the sequel, they are tabulated here :
223-5 | 390
j
1 j 2
506-2
3
568-4 ] 585-6
4 ] 5
562
6
508-8
1
336-7
240-1
148-6 [ 68-3
4-7
-33-3
9
10
1
11 12
13
14
429-6
or
The substitution of these values in the equation gives
H x 50 = 1056 + 223-5 + ... + 4'7 - 33-3 = 5595,
H = 111 -9 tons.
Calculation of the Stresses X in the Outer Boom.
The part of the outer boom situated between joints 5 and
6 will be taken to illustrate the calculations (Fig. 250); M
FIG. 250.
is the turning point, and the loading boundary is therefore
found by producing A M and B S to meet at E ; then the vertical
through E is the required boundary (compare § 22). This
§ 30. — DOME OF 100 METRES SPAN.
193
vertical is at a distance of 13 metres from the axis, and falls
between the second and third joints, the effect of the various
loads is therefore as shown in Fig. 250.
The stress in X will be a maximum when the joints 3, 4,
5, ... 14* are unloaded and the remainder loaded.
But the same result is obtained by considering every joint
loaded, and applying to joints 3, 4, 5, ... 14 vertical upward
forces equal to their respective variable loads.
To find the components of the hinge-reaction under these
conditions, the equations of moments of each rib about its point
of support for a full load are to be used, deducting, however,
from the equation for the left rib the moments of the twelve
unloading forces, thus :
For the right rib,
0 = Hx50 + Vx50- 5595
For the left rib,
0=-Hx50 + Vx50 + 5595 - 506 - 568 - 586
4-7 + 33-3
H = 72-7 V = 39-2.
Then from Fig. 251 the equation of moments to obtain
X (max.) is :
0 = - X x 2 - 72-7 X 8-9 + 39'2 x 26'7
+ 4 fc7 + 21-4 + 16-1 + 10-9 + 6 +
+ 5.X 21-4 + 9-9 x 16-1
X (max.) = + 4:70-9 tons.
FIG. 251.
\
* Since the vertical through the 14th joint passes to the left of the turning point
A the load upon it produces tension in X, and the same remark applies to the
joint near B, consequently there are in reality four groups of loads, but the error
entailed by taking only two groups as above is so small that it may be neglected.
O
194
BRIDGES AND ROOFS.
X (min.) obtains when the joints 3, 4, 5, ... 14 alone are
loaded, and in this case the components of the hinge-reaction
can be found from the following equations of moments :
0 = Hx50-Vx50- 1056
0=-Hx50-Vx50 + 1056 + 506 + 568
+ 586 + ... + 4-7 - 33-3
H = 60'3 V = 39'2;
and from Fig. 252 the equation of moments to find X (min.) is :
0 = - X x 2 - 60-3 x 8-9 - 39-2 x 26'7
+ 4 (2^l + 21-4 + 16-1 + 10-9 + 6 + 1
+ 14-8 X 10-9 + 19-4 x 6 + 23'9 x 1-2
X (min.) = — 500-6 tons.
FIG. 252.
Calculation of the Stresses Z in the Loiver Boom.
As an example, the stress in the part of the lower boom cut
through by the section line C a (Fig. 250) will be found. The
point 6 is the turning point, and the vertical through F, the
intersection of A 6 and B S (Fig. 253), is therefore the loading
boundary. This vertical is at a distance of 17*3 metres from
the axis, and is situated between the third and fourth joints.
When, therefore, the stress in Z is a maximum, the joints
4, 5 ... 14 are alone loaded, and the equations to find the hinge-
reaction are :
Hx50-Vx50- 1056
-Hx50-Vx50 + 1056
568 + 586
4-7 - 33'3
§ 30. — DOME OF 100 METRES SPAN.
and the equation to find Z (max.) is from Fig. 254 :
0 = Z x 2 - 55-3 x 8-74 - 34-2 x 30
+ 4 (M + 24-7 + 19-4 + 14-2 + 9-3 + 4-5)
+ 19-4 X 9-3 + 23-9 x 4-5
Z(max.) = + 436 -5 tons.
FIG. 253.
195
FIG. 254.
When Z (min.) obtains, the joints 4, 5, ... 14 must be un-
loaded, and to determine the hinge-reaction the moments due
to the loads on these joints must be deducted from the equation
of moments of the left rib, considered fully loaded, thus :
0 = Hx50 + Vx50- 5595
0=-Hx50 + Vx50 + 5559 - 568 - 586 - . .
H = 77'7 V = 34-2;
.-4-7 + 33-3;
o 2
196
BRIDGES AND ROOFS.
therefore the equation of moments to find Z (min.) obtained
from Fig. 255 is :
0 = Z x 2 - 77'7 X 8-74 + 34-2 x 30
+ 4 (^> + 24-7 + 19-4 + 14-2 + 9'3 + 4'5)
+ 5 x 24-7 + 9-9 x 19-4 + 14'8 x 14'2
Z(min.) = - 610 -5 tons.
FIG. 255.
Calculation of the Stresses Y in the Diagonals.
Of the two diagonals placed between the joints 9 and 10,
the one connected to joint 10 will be chosen to illustrate the
method of calculation.
It will be seen that the case that occurred in § 9 is repeated
here, namely, that the point about which moments are taken
is infinitely distant. The direction of the straight line con-
taining this point is that of the tangent to the circle (centre C)
at the point where the diagonal is cut (Fig. 256). The radius
at this point makes an angle of 58 J° with the vertical axis, and
the tangent therefore also makes an angle of 58 J° with the hori-
zontal. The only difference between this case and that of § 9
is, that in the latter case, the turning point was in the hori-
zontal, and in the present it is in the direction of the tangent.
The simplest way is to resolve every force acting on S /3
into two components, one parallel to the tangent, and the
other to the normal at the point where the diagonal is cut ;
evidently the moments of all the former components is zero.
Let the normal component of Y be denoted by N, then all
the loads that make N positive will also make Y positive.
Therefore the forces that are acting in the same direction as N
§30. — DOME OF 100 METRES SPAN.
197
make Y negative, and those that are acting in the opposite
direction make Y positive.
The loading boundary will thus be seen to be the vertical
through the intersection of B S, with a line through A drawn
parallel to the tangent. For the direction of the resultant of a
load placed in the vertical through J and its hinge-reaction, is
J A, and the resultant has therefore no component parallel to the
FIG. 256.
-
normal. All the loads to the right of the boundary make N
negative, and those to the left as far as the section line make N
positive. The loads on the other side of the section line, how-
ever, again make N negative, for they act indirectly on the
portion S /3 by means of their hinge-reactions. The loads there-
fore divide themselves into three groups, as shown in Fig. 256.
By construction, it is found that the distance of the vertical
through J from the axis is 12 metres, and that the loading
boundary falls between the second and third joint. Thus the
force N, and therefore also Y, is a maximum when the joints
3, 4, 5, 6, 7, 8, 9 are loaded. With this loading the equations
to find the components of the hinge-reaction are :
0 = Hx50-Vx50- 1056
0=-Hx50-Vx50 + 1056 + 506 + 568 + 586 + 562 + 509 + 430 + 337
H = 56-l V = 35.
198
BRIDGES AND EOOFS.
To find N, the resolved parts parallel to N of all the forces
acting on the portion of the rib shown in Fig. 257 must be
equated to zero, thus :
0 = N + 56'1 cos 31|° + 35 sin 3H° - 4 x 9*5 sin 31£°
-(14-8 + 19-4 + 23-9 + 28'1 + 32 + 35-5 + 38'7) sin 31J°.
Solving this equation :
N(max.) = 54 -3 tons;
and since Y makes an angle of 52° 35' with N,
FIG. 257.
To determine Y (rnin.) the joints 3, 4, 5, 6, 7, 8, 9, are to
be unloaded, and in this case the equations to find the hinge-
reaction become
0 = Hx50 + Vx50- 5595
0=_Hx50 + Vx50 + 55S5 - 506 - 568 - 586 - 562 - 509 - 430 - 337
H = 77 V = 35. .
§ 30. — DOME OP 100 METRES SPAN. 199
Hence, Fig. 258, the equation to find N is :
0 = N + 77 cos 31 £° - 35 sin 31|° - 4 x 9'5 sin 31£°
-(5 + 9-9)sin31£°;
or,
N(min.)= - 19'7tons;
and the corresponding value of Y is :
Y(min-> = si:^ = -32-5tons-
FIG. 258.
•V.35
These three examples show sufficiently the mode of calcu-
lating the stress in the various bars, and the calculations for
the remaining bars will not be given, as they would occupy too
much space.
Calculation of the Stress in the Ring.
In a case like the present, where the number of ribs is small,
the ring connecting their lower extremities will be a polygon,
and from Fig. 259 the following is the equation to find the
stress S in the sides of this polygon :
2 S sin 22£° = H
200
BRIDGES AND ROOFS.
If, however, the number of ribs is very large, the horizontal
thrust acting on the ring can be considered as uniformly dis-
tributed over its total length. Let p be this normal thrust
against the inside of the ring per unit of length, then, from
Fig. 260, if </> is a very small angle :
or,
S =pr.
FIG. 259.
FIG. 260.
In the preceding example :
P-r
H
111-9
r . _
4
51 X 0-7854
= 2-794 j1
and therefore,
S = 2-794 X 51 = 142-5 tons.
The difference in this case is so small that it does not matter
which method of calculation is adopted.
§ 31.— GENERATING CURVE FOR DOMES. 201
§ 31. — GENEKATING CURVE, FOR A DOME, REQUIRING THE
LEAST QUANTITY OF MATERIAL.
In the preceding example it was required to calculate the
stresses in a given dome. To simplify the calculations the form
was taken as that of a hemisphere, or the generating curve was
the quadrant of a circle. If, however, the form of the generating
curve necessitating the least quantity of material in the ribs
were required, the form of the linear arch (or curve of equi-
librium) to carry the unequally distributed loads would have to
be found. The principal boom would be made to this curve,
a hinge connecting each half would be placed at its vertex. To
meet the effect of partial loading a secondary boom, connected
to the principal boom by means of diagonals, should be provided.
Evidently the secondary boom and the diagonals would have
no stress in them when the whole load was on the structure,
and also the arithmetical values of the maxima and minima
stresses in them would be equal (precisely as in the horizontal
and diagonal bars of the braced arch of § 22).
There is no difficulty in finding the required curve if the
dome be sufficiently flat, and the number of ribs sufficiently
great, for the portions of the surface contained between two
adjacent ribs to be considered, without too great an error, as
plane triangles, and if the load can be assumed as uniformly
distributed over the area of this triangle. In this case the centre
of gravity of the triangle S P can be taken as the point of appli-
cation of the resultant load on the part covered by this triangle
(Figs. 261, 262, 263).
Let K therefore be the load per unit of horizontal surface
and n the number of ribs (consequently - - the very small
angle contained by the horizontal projection of two adjacent
ribs), then x . — . ~ is the area of the triangle S P, and, taking
n 2i
•moments about P (Fig. 263),
202
BRIDGES AND ROOFS.
This equation is also necessarily true for the triangle S A,
hence putting
x = I and y = /
H/ =
n 3
Dividing equation 1 by equation 2,
[2]
This is the equation to the cubical parabola, and if this
form be given to the ribs, the stress in the secondary boom
and the diagonals will be zero when the dome is fully loaded.
FIGS. 261, 262, AND 263.
i
S
In the above investigation it has been assumed that the
weight of the rib itself is very small, or at any rate distributed
in the same manner as the other loads. It would, however, be
more accurate to consider this load as uniformly distributed
along the horizontal projection of the rib, and if p is this
§ 32. — DOME FORMED OF ARTICULATED RIBS AND RINGS. 203
uniform load per unit of length of the span, the above equations
become
• ^ [!A]
P4 PA]
2/C7T
[3A]
If the height of the dome and the number of ribs were such
that the above assumption could not be made without sensible
error, the required curve would have to be found by following
the principles laid down for the determination of linear arches.
§ 32. — DOME FORMED OF ARTICULATED EIBS
AND EINGS.*
The skeleton of the dome given in Figs. 264 and 265 shows
half a regular eighteen-sided polygon in elevation, and a
regular octagon on plan. It is assumed that the various bars
FIG. 264.
are connected together by free joints. These joints lie in the
surface of a hemisphere of 10 metres radius, and the surface of
* See ' Berliner Zeitschrift fur Bauwesen,' 1866, " The Construction of
Domed Eoofs," by W. Schwedler.
204 BRIDGES AND ROOFS.
which is equal to 2irr2 == 2 x 3-1416 x 102== 628 -32 square
metres. If therefore the load on the dome be p = 200 kilos,
per square metre, the sum of the loads on all the joints
= G = p . 2 TT rz = 200 X 628'32 = 125664 kilos.
It can be assumed, without any great error, that the sum of
the loads on all the joints in any one of the horizontal rings
(Fig. 265) is proportional to the radius of the circle circum-
FIG. 265.
scribing the corresponding octagon (with the exception, however,
of the lowest ring, which has only half the load to bear). The
radii of these five circles are :
rl = r sin 10° = 10 X 0' 17365 = 1'7365 metre
r2 = rsin30° = 10 X 0-5 = 5'0
r3 - r sin 50° = 10 X 0'76604 = 7'6604 „
rt = r sin 70° = 10 X 0'93969 = 9-3969 „
rs = r sin 90° = 10 X 1 = 10-
Now, x the load on a ring whose radius is unity, can be
found from the following equation :
1-7365 x + 5 x + 7-6604 x + 9'3969 x + -/• = 125664 ;
§ 32. — DOME FORMED OF ARTICULATED RIBS AND RINGS. 205
whence :
# = 4364 -3 kilos.
Thus the loads on the various rings are :
xr^ - 4364-3 x 1'7365 = 7578 '6 kilos.*
xr, = 4364-3 x 5'0 = 21821-5 „
ar3 = 4364-3 X 7-6604 = 33432-3 „
x r4 = 4364-3 x 9 • 3969 = 41010 '9 „
=21821.5
and as each ring contains eight joints, these loads must be
divided by 8 to obtain the load on one joint ; f thus :
Ql=
where Q1? Q2 . . . Q5 denote the loads on the five joints of a rib.
Calculation of the Stresses produced ly the Full Load.
Imagine that two sections are taken through the dome by
means of vertical planes, as shown in Figs. 266 and 267, and that
equilibrium is maintained by forces applied to the end of, and in
the direction of, each bar that has been cut through.
The upper ring exerts a horizontal pull KI on the rib,
* It is evident that if a lantern or any other load were placed on the top, the
load on the top ring would have to be increased accordingly.
t If the number of ribs had been 16 instead of 8, these loads would have to
be divided by 16, and so on ; otherwise the calculations are the same. A small
number of ribs has been chosen to obtain distinct figures.
206
BRIDGES AND ROOFS.
which is the resultant of the stresses Xx in the bars of this
ring. This is expressed by the equation
2X1sine = K1 or
in which
— —
2 sin e '
= 22-5°.
FIG. 267.
The same occurs at each joint, and the stresses in the ring
and their resultant are connected by a similar equation.
§ 32. — DOME FORMED OF ARTICULATED RIBS AND RINGS. 207
Further, let the part A B of the rib be cut through, and
equilibrium maintained by an applied force ~Dl (Fig. 268), and
let this force be replaced by its horizontal and vertical com-
ponents ; then, by equating the algebraic sum of the horizontal
forces, and also that of the vertical forces, to zero,
Y! = 947 kilos, and K! = - Ht ;
now, from the figure,
V» V,
tan a, = =± , sm ttl = -i- ;
H, Dt
and since a± = 20°, the following values are obtained :
Q47
H! = £2- = + 2602 kilos,
tan «
947
K, = - 2602 kilos.
X, = - 2sin22.50 = - 3400 kilos.
FIG. 269.
FIG. 268.
947 2728
,-- I\
The same process applied to the part of the structure
shown in Fig. 269, gives:
V2 = 947 + 2728 = + 3675 kilos.
= + 5717 kilos.
sin a2 sin 40°
K2 = 2602 - H2 = - 1778 kilos.
1778
208 BRIDGES AND ROOFS.
And similarly for Fig. 270 :
FIG. 270.
4179
V3 = 947 + 2728 + 4179
= + 7854 kilos.
_ Y, 7854
3 tan a3 tan 60°
= + 4535 kilos.
D
sin o3 sin o3
= + 9069 kilos.
B3 = 2602 + 1778 - H3
= - 155 kilos.
X -155
3 2 sin 22-5°
= - 203 kilos.
And, lastly, from Fig,
271:
V4 = 947 + 2728 + 4179 + 5126 = + 12980 kilos.
V4 12980
4 ~ tan a4 ~ tan 80°
= + 2289 kilos.
^ = -= = . QAO = + 13180 kilos.
sin o4 sin 80°
K4 = 2602 + 1778 + 155 - H4 = + 2246 kilos.
~ = + 2935 kilos.
2 sin 22-5°
FIG. 271.
2728
94T
5126
§ 32. — DOME FORMED OF ARTICULATED RIBS AND RINGS. 209
V4, the resolved part vertically of the stress in the bar E F,
is produced by the vertical reaction of the point of support;
and H4, the resolved part horizontally, is due to the tension in
the bottom ring. Hence
R5 = H4 = + 2289 kilos.
FIG. 272.
Minima Stresses in the Rings.
The whole load on the dome will be considered variable ;
at the same time, however, the loads must always be sym-
metrical with the vertical axis, or the structure would collapse.
If the load on the top ring be removed, it is easy to see
that the stress in that ring becomes zero; and likewise that
when the load on the second
ring is removed, the stress in
that ring also becomes zero;
and so on for the remaining
rings. From this it is evident
that the load on any one ring
has no influence whatever on
the stresses in the rings above
it, or, in other words, produces
no stress in them ; but the load
on any ring produces tension
in all the rings below it. Con-
sequently, the minimum stress
or greatest compression will
occur in any ring when it alone
is loaded. In Fig. 272 the
third ring is represented as loaded, and E3 is the resultant of
the stresses in that ring. For equilibrium, the resultant of E3
and Q3 must lie in the direction of the part of rib just below
the joint C. Hence
R3 (min.) = —
Q3
tan a3
4179
tan 60°
= - 2413 kilos.
210 BRIDGES AND ROOFS.
and similarly for the second and fourth rings :
K2 (min.) = -- ?L_ = - 2728Q = - 3251 kilos.
tan a2 tan 40°
E4 (min.) = - -i- = - - = - 904 kilos.
tan a4 tan 80°
The minima stresses in the rings are therefore
Q04
The first and fifth rings are not considered, for in the fifth
compression can never occur, and the top ring is always in
compression ; the value already obtained (Xj. = — 3400 kilos.)
is therefore the minimum stress required.
Maxima Stresses in the Eings.
From what has been already said, it is evident that the
maximum stress or greatest tension occurs in any ring when
all the rings above it are loaded, itself unloaded, and the
rings below it either loaded or unloaded. Thus in Fig. 273 the
resultant K3 of the tensions in the ring at C reaches its maxi-
mum value when the tw.o upper rings are loaded, and can be
found from the equation (Fig. 274)
v v
K3 = H2 - H3 = -la. - -11-;
tan a2 tan o3
but
V2 = V3 = 947 + 2728 = 3675,
hence
3675 3675
tan 40° tan 60C
and therefore
= + 2258 kilos. ;
2258
X, (max.) = + -r-o = + 2050 kilos.
§ 32. — DOME FORMED OF ARTICULATED RIBS AND RINGS. 211
To find the maximum stress in the second ring, the top
ring alone must be loaded, and consequently,
V2 = V, = 947,
947
947
2 " tan 20° ~ tan 40°
= + 1473,
1473
X2(maX.)=-2sin22.5o
= + 1925 kilos.
FIG. 273.
2728
FIG. 274.
.
Lastly, to determine X4 (max.) the three upper rings should
alone be loaded, and the equations are —
V4 = V3 = 947 + 2728 + 4179 = 7854,
7854 7854
X4 (max.) = +
tan 60° tan 80°
3150
= + 3150,
2 sin 22 -5
— - = + 4116 kilos.
In the above calculations it has always been considered
that the load on any ring was equally distributed amongst the
joints, or, in other words, that the loading was symmetrical
P 2
212 BRIDGES AND ROOFS.
with the axis of the dome. The joints being free, any depar-
ture from this symmetrical loading would immediately bring
the structure down. To enable the dome to resist unequal
loading, either the covering must possess sufficient stiffness to
prevent any deformation, or else the free joints must be replaced
by fixed joints, and both the ribs and rings strengthened, so that
they may prevent deformation by their resistance to bending.
The determination of the bending stresses thus called into play
cannot, however, be accomplished by elementary means.
( 213 )
TENTH CHAPTER.
§ 33. — CONTINUOUS GIBBER BRIDGES.
It was seen, in treating of braced arches, that by intro-
ducing hinges the stresses in the various bars could be kept
between easily controllable limits, and also that the dangers
arising from a slight giving way of the abutments, or by
changes of temperature, could be totally avoided. But hinges
can also be employed with advantage in girder bridges (those
that require only vertical reactions at the abutments or piers)
when the span is great, and there are two or more openings in
succession to be bridged over.
It is found that in such cases a great saving of material is
effected by using a continuous girder, instead of several span-
ning each opening separately. But in these structures, as in
braced arches without hinges, there is the danger of a very
slight alteration in the position of the supports producing very
great differences in the stresses ; in braced arches the danger
lies in the horizontal displacement of the abutments, but in the
present case a vertical displacement becomes critical. There-
fore the same reasons that were given with reference to braced
arches in § 24 would point to the advisability of breaking
the continuity of the girder by means of hinges, and thus
making the stresses in the structure independent of small
vertical displacements of the points of support. In the case of
braced arches, the crown and the abutments were found to be
the best places for the hinges ; but with girder bridges the
best positions are on each side of the central piers, so that the
portions of the girder over the piers may act as supports to
the other parts (Fig. 275).
The part of the girder resting on either of the piers is to be
regarded as supported at two points; and in order that there
may be no chance of overturning with a partial load, the dis-
214
BRIDGES AND ROOFS.
tance of these two points, or the breadth of the pier, must not
be less than a certain dimension which will now be found. The
worst distribution of the load, as regards the left pier portion,
is that shown in Fig. 276, in which only the parts B C and
C E are loaded, and the remainder of the bridge unloaded
(moving load).
FIG. 275.
5=
M
FIG. 276.
l ir 1 „ -I ,,. -- -
!f=^
"trp
TW — i
1 l
C
t
i
The equation of
moments about the point B is then
o = G>
+ q) x z +
(p 4- g) ^ . g - p x (z + &) - p
c.+»>(i£),
where p is
of length.
the permanent, and q the moving, load per unit
Solving this equation, and putting n for the
q
ratio - 5
P
Now, since the ratio n generally increases as the span
diminishes, it follows that very small spans would require pro-
portionately very wide piers. To obviate this, the part of the
girder over the pier can be anchored down to the masonry
by tension rods. With a partial load, a tension, K, is pro-
duced in these rods, the moment of which about B ( = K b)
helps to maintain equilibrium. The equation of moments
then becomes
0 = dp + (?) x z + (p + 9) ^ - p x (z + 6) - p <L+^! _ K 6.
& z
If this arrangement be adopted, however, the weight Q of the
§ 33. — CONTINUOUS GIRDER BRIDGES.
215
pier must be such as to prevent overturning. This condition is
expressed by (Fig. 277)
j,
If, however, Q becomes greater than is thought advisable,
the distance b of the two points A B can be increased by
using double piers.
The central and abutment portions being simply supported
at the ends can be constructed either as para-
bolic girders (described in the second chapter),
or as braced girders with parallel booms (de-
scribed in the third chapter). The portions
over the piers could also be given this latter
form ; but a variety of the parabolic form may
also be adopted. This variety can be deduced as
follows : —
When two or more equal and symmetrical chains (either
hanging or arch-shaped), having the same load per unit of
length of the span, are so placed next each other that the
second abutment of the first chain is the first abutment of the
second chain, and the second abutment of the second chain is
FIG. 278.
'JUT
the first abutment of the third chain, and so on, the horizontal
tensions or thrusts balance each other at the common abutments,
and the reactions are entirely vertical, so that these abutments
might be replaced by tension rods. Instead of a single rod,
two separated by a horizontal bar, A B, might be used, as
already seen in § 27 (Figs. 278 and 279).
216
BRIDGES AND HOOFS.
If the part L A B M be separated from the two parts K L
and M N, the horizontal forces required for equilibrium can
be obtained by joining K L, M N, and L M by means of horizontal
tie-rods. The vertical forces required for the part L A B M are
equal, and opposite to those required for the parts K L and
M N, and equilibrium will therefore be maintained if the latter
parts be placed on the former in their original position.
The stresses in the chain have not been altered by the
introduction of the horizontal tie-rods, and consequently a bridge
constructed as shown in Fig. 280 will, when the load is uni-
FIG. 280.
formly distributed, require no diagonals. Further, since the
reasoning for an arch chain applies also to a hanging chain,
what has been said above will apply to the structure shown in
Fig. 281, the tie-rod becoming a compression bar.
1
FIG. 281.
The stresses in the chains can be found from the formula
given in § 8. The laws given at page 33 are also applicable
namely, that the resolved part vertically of the stress at any
point is equal to the load on the bridge between that point
and the centre, and that the resolved part horizontally of the
stress is constant. It is only an alteration in the height of the
arc that changes this horizontal stress.
Evidently the height of arc of the three ordinary parabolic
girders in Figs. 280 and 281 can be altered without affecting
the portions of the bridge over the piers, for only the vertical
reactions are transmitted to these latter ; the horizontal stress
in the parabolic girders will, however, be altered. The hori-
§ 34. — CONTINUOUS GIRDER BRIDGE IN THREE SPANS. 217
zontal stress in the pier portions depends solely on the height
of arc of the parabolas of which they are formed.
The parabolic girders of Figs. 280 and 281 can be trans-
posed without making any difference in the other parts. Thus
FIG. 282.
a structure of the form shown in Fig. 282 is obtained, which
can be adopted with advantage if head room under the bridge
is of importance. It is hardly necessary to remark that the
FIG. 283.
parabolic girders can also be replaced by braced girders with
parallel booms, and that this will not affect the pier portions
(Fig. 283).
§ 34. — CONTINUOUS GIRDER BRIDGE IN THREE SPANS. CEN-
TRAL OPENING, 160 METRES; SIDE OPENINGS, 130 METRES.
The weight of the bridge (Fig. 284) is estimated at 8000
kilos, per metre run, and each girder has half of this to
carry. The length of each bay being 10 metres, each joint has
40,000 kilos., or, reckoning 1000 kilos, to the ton, 40 tons dead
load to bear. The moving load is taken at 4000 kilos, per metre
run, which is equivalent to 20 tons moving load per joint.
The three parabolic girders placed between the piers
(Fig. 284) have each a span of 100 metres and a height of
12 • 5 metres, and the stresses in them can be found as explained
in the second chapter (Fig. 39). • It will therefore only be
218
BRIDGES AND ROOFS.
necessary to show how to find the stresses in the parts resting
on the piers.
The pressure D produced at C by the girder CE will be
greatest when the girder is fully loaded, and least when it is
unloaded. In the first case :
and, in the second case :
FIG. 284.
0
> c
Either one or the other of these values will have to be
substituted for D, according as it tends to increase or decrease
the stress in any bar, and according as the maximum or the
minimum stress in that bar is to be determined.
Thus, to find Xi draw a section line a/3 (Figs. 285 and 286)
and form the equation of moments with reference to the point
B, thus :
0=-X! x 11-266 + D x 30 + 40(10 + 20 + ^) + 20 (10 + 20 + *f).
From this equation it appears that D, as well as the loads on
the points F, G, C, make Xx positive. To find Xx (max.), there-
FIG. 286.
D
Y.
10
§ 31— CONTINUOUS GIRDER BRIDGE IN THREE SPANS. 219
fore, the girder C E and the joints F, G, C must be loaded, and
by substituting the greatest value of D the equation becomes :
0 = .- Xt x 11'266 + 300 x 30 + 60(10 + 20 + ^)
X! (max.) = + 1038-5 tons.
To determine Yl moments must be taken about P (Fig. 286).
It is not necessary to consider the permanent load, for, as
already remarked in § 33, a
uniformly distributed load
produces no stress in the
diagonals. Further, it is
only necessary to calcu-
late either Y! (max.) or Yx
(min.), as both values are
numerically equal but of
contrary sign, since the
moving load when covering
the bridge is uniformly distributed. But to show that the
method is quite independent of such previous knowledge, the
calculations both for Yx (max.) and Yx (min.) will be made, taking
the permanent load in each case into consideration.
The equation of moments is :
0 = - Y! x 15-852 + D x 4 + 40 (f - 6 - 16) + 10 x 4 - 20(6 + 16).
To obtain Yj (max.) the greatest value of D must be sub-
stituted, and the negative terms due to the moving load
omitted, thus :
0= - Y, x 15-852 + 300 x 4 + 40 (f - 6 - 16) + 10 X 4;
or,
Y^max.)^ +27 -76 tons.
And to find Y! (min.), D must be given its least value and the
positive terms due to the moving load must be left out, thus :
0 = - Y! x 15-852 + 200 x 4 + 40 (| - 6 - 16) - 20(6 + 16)
Y,(mm.) = -27 -76 tons.
The stress in the bar Zx can be found .from the following
equation of moments, formed with reference to the point F :
0 = Z, x 7-692 + D x 20 + 40 (2f + 10) + 20 (^ + 10);
220
BRIDGES AND ROOFS.
from this equation it appears that all the loads produce com-
pression, therefore putting D = 300
or,
0 = Z, x 7-692 + 300 x 20 + 40(^ + 10) + 20 (^ + 10);
Zt (min.) = - 936 tons.
Vi is found by taking moments about 0 (Fig. 287). Half the
dead load will be considered as applied to the lower joints, and
the other half to the upper joints ; the equation of moments is :
0 = Vx x 18-4615 + D X 1-5385 + 40
- 8-4615 -
Y! is therefore greatest when
D is least, and the joint C un-
load ed. Consequently :
0 = Vx x 18-4615 + 200 x 1-5385
D
- 20(8-4615 + 18-4615)
Y! (max.) = + 49-2 tons;
and YI is least when D is
greatest and the joint C alone
is loaded, hence :
0 = Y! X 18-4615 + 300 x 1-5385 + 40 (^ - 8-4615 -
+ 20 x^.
Vj(min.) = +10-8 tons.
(The stresses produced by the moving load alone are :
+ 29-2 and —9-2,
which, added to the stress due to the dead load alone, or + 20,
gives the values found above).
For the remaining bars the following equations and results are obtained :
0 = - X2 X 7 ' 1 + 300 x 20 + (40 + 20) (f + 10)
X2 (max.) = + 1014 tons
0 = - Y2 X 6-138 + (200 +100) 1-5385 + 40 (l^ - 8-4615)
+ 20 x1-^-5- 20 x 8-4615
=± 27-57 tons
§ 34. — CONTINUOUS GIRDER BRIDGE IN THREE SPANS. 221
0 = Za X 3-526 + 300 x 10 + 30 X 10
Z2 (min.) = - 936 tons
0 = V2 X 10 - 20 x 10 - 20 x 10
V2 (max.) = + 40 tons V2 (min.) = + 20 tons
0 = - X3 x 3-325 + 300 x 10 + 30 x 10
X3 (max.) = + 992-5 tons
0 = Z3 x 3-526 + 300 X 10 + 30 x 10
Z3 (min.) = — 936 tons.
The calculations to find the stresses in the diagonal, and in
the two verticals immediately above the pier, vary slightly
from the above, on account of the reaction of the pier. This
reaction W (Fig. 288) is the pressure the fixed point B exerts
against the structure.
To determine W, C D can be considered as a lever having
its fulcrum at the other point of support A. It is evident that
200
when W is to be as great as possible, all the joints to the right
of A must be loaded, and those to the left unloaded, but since
the load on the joint B is taken up directly by the pier, it has
no influence on the magnitude of W. Hence from Fig. 288 the
equation of moments is :
0 = - W x 12-5 + 20 x 12-5 + 60 (22'5 + 32'5 + ^)
+ 300 x 42-5 - 40 (10 + 20 + ^) - 200 x 30
W (max.) = 782 tons.
A glance at Fig. 289 will show that this value of W makes
V0 a minimum, for the increment added to W by a load on the
part of the structure shown in the figure is greater than the
load that produces it, and it has also a greater lever-arm with
reference to the turning-point.
222 BRIDGES AND ROOFS.
The equation of moments for V0 (min.) is therefore :
0 = V0 X 26 + 782 x 26 + 300 x 4 + 30 x 4 - 60(6 + 16)
V0 (min.) = — 782 tons.
The same loading makes Y0 a maximum. For since X0
and Z0 are parallel, the equation of moments reduces to the
FIG. 289.
300
condition that the sum of the vertical forces should vanish.
Hence v, the resolved part vertically of Y0 (max.), is equal to the
difference between W and the loads on the part of the structure
FIG. 290.
shown in Fig. 290, and this difference is greatest when this
part is fully loaded; for the reaction produced at B by any
such load is greater than the load itself. Consequently
0 = v - 782 + 20 + 60 + 60 + 30 + 300
v = 312 tons ;
and since the diagonal makes an angle of 45° with the
vertical :
Y0(max.) = _0 = 312 */2 = + 441 12 tons.
§ 34. — CONTINUOUS GIRDER BRIDGE IN THREE SPANS. 223
As already observed, Y0 = 0 when the bridge is fully loaded,
hence the numerical value of Y0 (min.) must be equal to that
of Y0 (max.). Therefore :
'8
As to the stress in the vertical U0, it is evident that the
only vertical force acting upon it at the top, besides the per-
manent load of 20 tons, is the vertical component of the stress
in the top boom, and this component is evidently greatest with
a full load. Consequently :
- U0 = 20 + (60 + 60 + 30 + 300)
U0 (min.) = — 470 tons.
(The maximum stresses in both the verticals over the pier
are negative, and it is therefore not necessary to consider them.
By reversing, however, the loading in Fig. 288, it is easily
found that U0 (max.) = - 320 tons.)
The stresses in X0 and Z0 are, as in the case of the other
horizontal bars, equal to the horizontal stress in the fundamental
parabolic chain. Consequently :
X0 (max.) = + 936 tons
Z0 (min.) = — 936 tons.
The value of this horizontal stress depends, as already
remarked, on the height of the arc of the parabola E K L F,
Fig. 281, to which the part C D belongs. In the present
case the vertex of the parabola is 8-0128 metres below the
horizontal K L (Fig. 281), and the height of arc is there-
fore :
/ = 12-5 + 8*0128 = 20-5128 metres.
The corresponding span is :
2 I = 160 metres.
The stress in the horizontal bars can therefore be obtained
from the formula (see p. 32) :
*
~ 936 tons-
2x20-5128
224
BRIDGES AND ROOFS.
(The stress in the horizontal bars of the central parabolic
girder is only :
(4 + 2) x 502
2 x 12-5
= 600 tons.)
The results obtained by the above calculations are collected
together in Fig. 291. By changing the signs of the stresses
given in this figure, those for a similar girder turned upside
down are obtained. The points A and B will, however, still be
FIG. 291.
A > 4936 B,
-782
-936 -936 -936 |~^936 | -936 -936 -936
the points of support. But in this case Ax, Bj will generally
be chosen as points of support, and the stresses in the two
verticals over the pier will consequently be altered. These new
stresses can either be determined directly, or else by em-
ploying secondary verticals, a method which has been used
FIG. 292.
+936 _ +936 A +936 B +936 +936 493fl
before (see § 12). The stresses thus obtained are given in
Fig. 292.
From Figs. 291 and 292 several derived forms can be
obtained, as was done in previous cases (see §§7, 11, and 16),
but only the alterations that can be made in the construction
of the central bay over the pier will be considered.
If, for instance, there are two diagonals in the central bay,
both of which are capable of resisting either tension or com-
pression (Fig. 293), the stresses in each diagonal will be exactly
§ 34. — CONTINUOUS GIRDER BRIDGE IN THREE SPANS. 225
one-half of the stress found above for the diagonal, and the
stresses in the two verticals will each be the arithmetical mean
of those already obtained (it is easy to satisfy oneself of this by
imagining two such girders, with halved stresses, placed one
behind the other, one with the central diagonal inclined to
the right, and the other with this diagonal inclined to the left).
FIG. 293. FIG. 296.
•r936 4936 4936 +93 6
-936
-936
If, however, both diagonals can only take up tension or can
only resist compression, the stresses given in Figs. 294 and 295
respectively will be those required. The inability of the dia-
gonals to resist tension is expressed in Fig. 295 by double
lines.
In a similar manner the structures shown in Figs. 296, 297,
and 298 can be derived from Fig. 292.
In all the preceding calculations the load on the joints
over the piers has, for simplicity, been taken the same as on
the other joints, though accurately speaking the load on these
joints is slightly greater on account of the pier being a little
wider than a bay.
Q
226
BRIDGES AND ROOFS.
This width is necessary to prevent the pier girders over-
turning when subject to the action of a partial load. By sub-
stituting in the formula of § 33, viz.
the values
it is found that,
[• -f •*).+• V (• + z)2
a? = 50, * = 30, n = |g,
b > 11 -38 metres.
The width of the pier assumed above, namely 12*5 metres, is
therefore a little in excess (all the more so as the permanent
load of the central girders is a little less than that of the
pier girders, although in the calculations it has been taken as
equal).
Continuous Girder with Parallel Booms.
The whole of the continuous girder may be constructed
with parallel booms as shown in Fig. 299, or else only
parts resting on the piers may be so designed. The stresses in
FIG. 299.
these latter can be obtained by an exactly similar process
to that followed above, and these stresses are given in Figs.
300, 301, 302, 303, and 304. To form some idea of the rela-
Dt O
FIG. 300.
+576 A, +936 B ,±576
•576
-936 A -936 A _
A Jo
-576
-264:
tive quantities of material required by the two designs, the
span and height of the girders are the same as in the former
§ 35. — SUBDIVISION OF THE WHOLE SPAN, ETC. 227
example. It must also be observed that all the figures refer to
girders carrying the line of railway on their lower booms, and
also that the hinges connect the lower booms. Further, in
D +264 +57(>
FIG. 301.
A + <J3G B +.93G
.5 76'
I 0 -264 -576 -936 -576 -264,
Fio.*302.
J>76 +936 + 576'
-4
-936 & -.930 A -93 ft
W
FIG. 303.
+V36 +576'
FIG. 304.
+936 +576'
-936
-.936
-936
Fig. 302 both the diagonals of the central bay are capable of
resisting either tension or compression ; in Fig. 303 they can
only resist tension, and in Fig. 304, compression.
§ 35. — TO DETEKMINE THE SUBDIVISION OF THE WHOLE
SPAN REQUIRING THE LEAST QUANTITY OF MATERIAL.
By comparing the stresses given in Fig. 291 it appears, first,
that the stresses in the diagonals and verticals are small compared
with those in the booms ; and secondly, that the stresses in the
curved part of the bow do not differ materially from each other or
from the stress in the horizontal boom. Now, as the quantity of
Q 2
228 BRIDGES AND ROOFS.
material can be taken as nearly proportional to the stress, it
follows that by far the largest quantity is contained in the
booms, and that it is nearly equally distributed between them.
These remarks also apply to the parabolic central girder, as will
at once appear by reference to § 6.
Therefore it cannot be far from the truth to assert that the
quantity of material in the bridge is proportional to that con-
tained in the horizontal boom.
The above problem resolves itself therefore into the follow-
ing : To find what subdivision of the span gives the least
quantity of material in the horizontal boom.
To solve this problem it is first necessary to find the most
advantageous position of the hinges in the central span, and
also the most advantageous position of the hinges in both the
side spans.
a. Subdivision of the Central Span.
Let the parts C E (Fig. 306) and C A (Fig. 307) be cut out
of Fig. 305 and equilibrium maintained by applying the forces
H and Hx respectively, which are the stresses in the booms.
Taking moments about S for C E and about Aj for C A.
H. h=px.x-px . [1]
[2]
Now the sectional area of the booms can be found by divid-
ing the stress in them by S, the safe stress of the material per
unit of area. Therefore if F is the sectional area of the boom
C E, and F! that of C A,
F-5
~ '
and by substituting the values of H and Hj from equations 1
and 2,
F -
"
px*
2AS
2TS"
§ 35 — SUBDIVISION OF THE WHOLE SPAN, ETC. 229
By multiplying these sectional areas by the length of the cor-
responding boom the quantity of material in C E and C A
respectively is obtained,, thus,
[5]
[6]
[7]
and the amount of material in A E is
and (M -f MI) is to be a minimum.
FIG. 305.
FIG. 306.
If in this equation x really represents the value that makes
(M + Mx) a minimum, it is evident that the addition of a very
small quantity + A to x (or, in other words, replacing x by
[x ± A] ) must have the effect of increasing (M + Mx). But
this can only be the case if the first of the three terms added
to the expression in bracket by changing x to x ± A, namely,
± A ( - P - 2 lx + 6 a:2) + A2 (6 x - 0 ± 2 A3,
is equal to zero. For otherwise, by judiciously choosing A
the first term (which would then be large in comparison to the
other two) could be made negative ; and consequently (M + Mx)
230 BRIDGES AND ROOFS.
would be diminished. Therefore the condition for a minimum
is that
s» = 0;* [8]
/v» 7 t_ |B fp
whence the following values for j and — j — are obtained :
[9]
[10]
or the lengths A C and C E must be approximately in the
ratio 0-4 : 0-6, or 2 : 3.
By substituting the value of x from equation 8 in equation
7, the quantity of material in the horizontal boom from A to
E is obtained :
M+M1 = 0-47184^1. [11]
b. Subdivision of the Side Spans.
The quantity of material J in the horizontal boom D F
(Fig. 305) can be found by substituting z for x in equation 5,
thus:
And Ji, the amount of material in B D, can be obtained from
equation 6 by writing Zx for I and z for x.
or replacing l± by a — z
The quantity of material in the horizontal boom from B to
G is therefore
2J + J
* Or, in other words, the first differential -coefficient of the expression in
brackets of equation 7 must be equated to zero.
§ 35. — SUBDIVISION OF THE WHOLE SPAN, ETC. 231
If z ± A is written for 0, the terms in brackets are increased
by the expression
± A ( - 4a2 + 8 a z + 6*2) + A2 (4 a + 6 z) ± 2 A3 ;
and as before the value of z that makes 2 J -f- J\ a minimum
can be found from the equation
- 4 a2 + 8 a z + 6 z* = 0 ; [15]
whence the following values for z and a — 2 0 are obtained :
. * = ! a(-l + V~2-5) = 0-3874 a [16]
a -2z = 0-2252 a; [17]
from which
_l- = f = 0-6324 • [18]
is obtained.
By substituting in equation 14 the value found for z in
equation 16, the least quantity of material in the boom from
B to Gr is found to be
2 J+ J1 = 0-16706-^. [19]
c. Proportion of the Central Span to the Side Spans.
The preceding numerical values for •=• and =- are quite in-
l /x
dependent of the span of their respective openings, and there
fore also of the ratio
according to which the whole span is divided into three spans,
a, 2 /, a.
The converse, however, is not true, and in fact the most
advantageous division of the whole span depends on the sub-
division of the single spans.
It will be considered that the single spans have been sub-
divided in the most economical manner in accordance with
equations 9 and 18.
232 BEIDGES AND KOOFS.
In this case the quantity of material required for half the
whole span L is found by adding equations 11 and 19 together,
thus :
M + M1 + 2J + J1 = Jj-g (0-47184 J3 + 0-16706 a3);
or writing L — a for I,
(0'47184 (L -a3) + 0-16706a3). [20]
As before by changing a into a ± A it will be found that if
the quantity of material is to be a minimum the condition
- 3 x 0-47184 (L - a)2 + 3 x 0'1670Ga2 = 0 [21]
must obtain. Whence
L-a
or writing I instead of (L — a) and n instead of ^y
n = 0-8403. [22]
Applying the results obtained in equations 9, 18, and 22
to the previous numerical example, it will be found that the
whole span, 420 metres, is subdivided as shown in Fig. 308.
FIG. 308.
It will be observed that the span of the central parabolic
girder is rather smaller than that of the two side ones. This
would slightly increase the expense of execution, and owing
also to the unsymmetrical arrangement and to the unequal
loading of the hinges, the diagonals in the bays immediately
above the central piers would be in a state of stress even with a
distributed load, thereby slightly adding to the quantity of
material.
UNIVERSITY
^££CALIFOV^>
§ 35. — SUBDIVISION OF THE WHOLE SPAN,'
When it is considered also that the calculations in the above
investigation are only approximate, it seems better to choose the
simpler ratios
130 x 50 z 50
= 160' 7 = 80' ^^80'
making the girders symmetrical with respect to the central
piers.
Equations 1 and 2 also give the greatest stress in the
booms of a girder with parallel booms. It is true that in such
girders the stress in the booms decreases from the centre to the
points of support, but this is more or less compensated by the
dimensions of the diagonals and verticals increasing towards
the abutments.
It can therefore be assumed that the quantity of metal in
such girders is nearly the same as that in parabolic girders —
an assumption which will be found justified by comparing
Fig. 27 with Fig. 57, and Fig. 291 with Fig. 300. The premises
are therefore approximately the same as in the case of parabolic
girders, and consequently equations 9, 18, and 22 may be con-
sidered as approximately true for girders with parallel booms.
( 234 )
ELEVENTH CHAPTER.
§ 36. — DETEKMINATION OF THE TURNING POINTS AND LEVEE
ARMS BY CALCULATION.
Although a drawing to scale of the structure under considera-
tion can generally be made, upon which the turning points can
be found by construction, and the lever arms, with ample ac-
curacy, by measurement, yet cases may occur where it is necessary
to obtain these data by calculation and without the help of a
drawing. In the following it will be shown that this is by no
means difficult, and that when the structure is composed of
straight bars the required results can be obtained simply by the
comparison of two similar triangles. The examples have been
chosen from the various structures already considered, and will
therefore render the former calculations more complete.
Boof of § 3.
To find the lever arm x of the stress X in Fig, 10, the simi-
larity of the two triangles D M C and ADC can be employed
thus (Fig. 309) :
OS
D~C
AD
To
D
and putting
C D = 20, A D = 50,
A C = J 502 + 202,
as given in Fig. 8.
20X50 =18-6.
The lever arm y of the stress Y can be ascertained by
comparing the two similar right-angled triangles ALD and
E F D, obtaining the equation
y - EF-
AI> ~ E~D'
§ 36. — TURNING POINTS AND LEVER ARMS.
or substituting
235
AD = 50, EF = 15, ED = Vl2-52 + 152,
• •• . 50 x 15
y =
- = 38-4.
Parabolic Girder (§ 6).
To ascertain the stresses V2 and] Y3 the position of the
turning point S (Fig. 26) had to be determined. This can be
done from the equation
2A
= tan a,
obtained from Fig. 310 by comparing the similar right-angled
triangles SDE and SFG. :
FIG. 310.
Now, according to Fig. 21,
\ = 2 metres, u = 1-5 metre, v = 1*875 metre.
Therefore
x — ( u ~ — - \ A = 4 metres.
\ v - u )
Again, from the similarity of the two right-angled triangles
SDH and G D F, the following equation is obtained :
y... ° •
SD~GD'
236 BRIDGES AND ROOFS.
from which, by substituting
SD=#+2A=8 metres,
and
G D = V v2 + A2 = Vl-8752 + 22.
8x 1-875
y = . — = 5-47 metres.
Lastly, to find the lever arm D J = z (for the moment of
the stress Z3 about D), the following equation is deduced from
Fig. 310,
SD 8
= 1 ' 474 metre.
Sickle-shaped Truss (§15). .
The equation for as found above can also be put in the form
* + 2A = —
This equation can be adapted to another parabola whose
ordinates are n times those of the former, by writing nv for v
v
and n u for u. The ratio - in the denominator of the above
u
equation becomes -— , or remains unchanged. Consequently
the intersection of the chord E G, with the horizontal through the
points of support is independent of the height of the arc of the
parabola.
From this it follows that in the sickle-shaped truss of
Fig. 311 the intersection of the chords H N and M J lies in the
horizontal through the points of support. The position of 0
can therefore be found, as in the preceding case, from the
equation
x + 3A x + 4A
Or substituting from Fig. 114 the values
A = 1, u = 0-710, v = 0-852,
0-710
0-852-0-710
-3 = 2.
§ 36. — TURNING POINTS AND LEVEE ARMS.
237
The position of the point L can be determined by means
of the similar triangles N R L and M Q L, thus :
w w + X w A
— = , or - = ;
V U + Z V U -}- Z — V
but from Fig. 114, z = 1'065, therefore
0-852
= 0-9231.
To obtain the lever arm y (for the stress Y4 in Fig. 116) the
similarity of the two right-angled triangles O P L and M Q L
gives the equation
y y
OL ML'
in which
O L = 2 + 4 + 0-9231 = 6*9231,
and
M L = VO-8522 + 0-92312 = 1-256.
= 4-7.
The value of y thus found is
_ 6-9231 x 0-852
y ~ l-256~
The position of O being known, the lever arm t (for the
stress Z4) can be found by comparing the similar right-angled
triangles J G N and J K 0, thus :
t OR
238 BEIDGES AND ROOFS.
and since OB = 5, and 0 J = \/52 + 0'712,
t =
If, however, the position of 0 were not known it would be
better to find t from the equation
t_ \
z~ JM'
The lever arm s (for the stress X4) can be obtained from
the equation
s_OQ
in which
r = 1-278, O Q = 6, and O H = */62 + 2-132 ,
whence
6x1-278
=1-205.
But if the position of 0 were not known this lever arm
could be more easily obtained from the equation
5 _ A.
r = NH*
To calculate the stress V3, it is necessary to know the
point of intersection of the two parabolic chords N F and M J.
From Fig. 312,
N T = p tan a = z + p tan e,
or
p =
tan a — tan e
Here
* = 1-065 and tan e =
<4-K
or, according to Fig. 114,
0-852 - 0-710
tan e = - - - = 0-142,
36. — TURNING POINTS AND LEVER ARMS.
therefore
1-065
p -
= 3-22.
0-4:725 -0-142
The horizontal distance of U from A is therefore
0- = 3-22 -3 = 0-22.
239
\
Q
Braced Arch (§ 22).
The position of the loading boundary at C (Fig. 174) can be
determined from Fig. 313 by the equations
C N = (I + #) tan e = (7 - *) tan a,
or
__ /tan a — tan e\
Vtan a + tan e)
FIG. 313.
A K N
Putting (from Fig. 174),
and
240
BRIDGES AND ROOFS.
The loading boundary through F (Fig. 192) can be found
from the same equation by substituting for tan a = y —
0-6875, thus:
0-6875-0-25
The intersection of the chords of the parabola with the hori-
zontal upper ~boom can be determined by the method adopted for
parabolic trusses (page 236). Thus the position of the point M
in Figs. 180 and 182 can be found from the equation (Fig. 314),
w + A'
FIG. 314.
but from Fig. 173
therefore,
and
inj"3
A = 2, « = l-75, 0 = 2-3,
2 x 1-75
2-3- 1-75
- = 6-36,
= 16-36.
The position of the loading boundary through E can now
ME 5.5
be obtained by putting tan a — -r-^ = j^ = 0*336 in the
previous equation, thus :
37. — STRUCTURES FULFILLING GIVEN CONDITIONS. 241
§ 37. — APPLICATION OF THE METHOD OF MOMENTS TO FIND
THE FORM A STRUCTURE SHOULD HAVE IN ORDER THAT IT
MAY FULFIL GIVEN CONDITIONS.
In all the preceding examples the form of the structure was
given, and the method of moments was employed to find the
stresses produced in the various bars by the application of
known loads. It will now be shown how the same method may
be employed to determine the form a structure should have in
order that it may fulfil certain given conditions.
The form and dimensions of the parabolic girder cal-
culated in § 6 were given, and in determining the stresses it
was found that when the bridge was fully loaded the stress
in all the diagonals vanished, a property which was explained
and found to belong to all parabolic girders in the subse-
quent "theory of parabolic girders," § 8. The operation could,
however, be reversed, and it might be required to ascertain
what form must be given to a girder in order that it may possess
the above property. If for instance the span, the number of
bays, and the depth of the girder are given, from which (Fig.
315) the points A, E, B, and likewise the positions of the loads
7)
FIG. 315.
Q
X I
on the upper boom are determined, the only unknowns, if the
girder be symmetrical, are the heights hi, h2, h3.
^ can be found by forming the equation of moments for the
part of the girder shown in Fig. 316, the turning-point being 0^
the intersection of the directions of the stresses X and Z. Now
by the conditions this point must have such a position that
Y = o. Hence the equation
0 = - D^ + Q { (#, + A) + Ot + 2 A) + (ar, + 3 A) } .
R
242
But
BRIDGES AND KOOFS.
substituting and solving for osi
xl = 12 X.
The height hT can now be calculated from the equation
hi _ #! + 3 A _ 15
h ~ xl + 4 A ~ 16 *
In a similar manner the following equation is obtained from
Fig. 317 :
0 = - D*2 + Q {(a?2 + A) + (*2 + 2 A) } .
FIG. 316.
9 Q 9
O,
FIG. 317.
9 9
7Q
And again putting D = -^ and solving for $2>
A
X* = 2 A ;
whence h2 can be found from the equation,
^2 _ #2 + 2 A _ 4^
^ ~ a;2 + 3A ~ 5'
or replacing ^ by its value in terms of h,
X
Y-o
§ 38. — STKUCTURES FULFILLING GIVEN CONDITIONS. 243
Lastly, to determine h3 (Fig. 318),
or
As an example put h = 2 metres, as in
Fig. 21, then h, = 1-875 metre, fe2 = l'5
metre and h3 — 0'875 metre, or the dimen-
sions given in Fig. 21. If further, A. = 2 jo
metres ; xl = 24 metres, x2 = 4 metres, and
#3 = 0*8 metre; thus assigning to Olt O2, and O3 the same positions that were
obtained graphically in § 6, and by calculation in § 37.
§ 38. — GIRDER IN WHICH THE MINIMUM STRESS IN THE
DIAGONALS is ZERO. (Schwedler's Girder.)
If the symmetrical parabolic girder of Fig. 35 be compared
with the symmetrical girder with parallel booms of Fig. 69, it
will be observed that in the first the maximum and the minimum
stress in each diagonal are numerically equal but of opposite
signs, whereas in the second, if the diagonals are inclined up-
wards from the centre towards the ends, the maximum stress has
the largest numerical value, and the minimum stress is positive in
all the bays except in those near the centre (and in fact the mini-
mum stress in these latter diagonals would also become positive
if the permanent load were sufficiently large in comparison to
the moving load). It might therefore be expected that there
exists an intermediate form of girder in which the minimum
stress in all the diagonals is nothing.
This form of girder will now be found ; it will be assumed
that the number of bays is eight (as in § 37), the depth of
the girder at the centre = h, the length of a bay = A,, and the
span = 8 X. If the girder be symmetrical hl9 h2, h3 are the
only dimensions required to determine its form (Fig. 319).
To find ^! the girder must be so loaded that the stress Y is
a minimum. This condition of loading is given in Fig. 319.
Taking moments for the part of the girder shown in Fig. 320
R 2
244
BRIDGES AND ROOFS.
about the point 01? the intersection of the direction of the stresses
X and Z, and putting Y = o,
but
0 = - D a?j + (p + m) { (a?, + A) + (o^ + 2 A) + (37!"+ 3 A) } ;
substituting and solving for
Now hi can be found from the equation
^ _ xl + 3 A
7i ~ a;, + 4 A '
or replacing #! by its value
15m
FIG. 319.
[1]
[U]
FIG. 320.
f i 1 !
A IP i 1 i
In the second bay (counted from the centre) the stress in the
diagonal will be a minimum when the girder is loaded, as shown
in Fig. 321, and
D «
§ 38. — STRUCTURES FULFILLING GIVEN CONDITIONS. 245
Taking moments about 02 for the part of the girder shown
in Fig. 322, and remembering that the stress in the diagonal
is zero,
0 = - D#2 + (p + m) { <>2 + A) + <>2 + 2 A) } ;
whence
A
8(p
lD \m
r i i
~~4j>-«
FIG. 321.
[2]
Ml I 1 1 I
^JH
0,
but h2 can be found from the equation
Ag _ #2 + 2 A
^~^2 + 3A'
or substituting for %2,
h% 1 6 p + 6 m
[2AJ
Similarly %3 and h3 can be obtained as follows (Figs. 323 and
324) :
20p-m
_
A? ~ z3 + 2 A ~ 48 jo + 6m
[3]
[3A]
By means of the foregoing equations the effect of altering the proportion
between the permanent and moving loads can be studied.
246 BEIDGES AND ROOFS.
Thus if m — 0 the girder is parabolic, as is shown by the following results
or, = 12 A, a;2 = 2 A, #3 = 0-4 A
*i _ 15 ^2 _ * ha _ _?
h ~ 16 ' hv ~ 5 ' h2 ~ 12 '
which are the values obtained in the last example, § 37.
FIG. 323.
FIG. 324.
Suppose
then
™ -r- 94. r 24
^1-72 ^ = f! 1« = f!
A 'A 7' A 39
and
Thus, if, for example, A = 1,
A! _ 75 A2 _ 38
h ~76' ^ ~45'
21
These heights can be plotted above the horizontal A B, as in Fig. 325, or one-
half can be plotted above and the other half beneath, as shown in Fig. 326. In
FIG. 325.
both cases tension alone will occur in the diagonals. (If, however, the dia-
gonals were inclined the other way, as in Fig. 327, they would always be in com-
pression, and the maximum stress would be nothing.)
Again, if — = - it will be found that xl = oo, and -1 = 1, and when ->« '
§ 38. — STRUCTURES FULFILLING GIVEN CONDITIONS. 247
xl becomes negative, and ^ > h. In this last case the depth of the girder in
the centre would be less than it is in the adjoining bays. For other obvious
reasons it would not be advisable to construct the girder thus ; the conditions
cannot therefore be complied with in this case.
FIG. 327.
*,
Further, let - = - , then from equation IA, -i = — but, instead of this,
p o n 47
-^ = 1 would be taken. The equations for hz and h3 are still applicable. For
from equation 2A, -2 = — , or h2 — 0'9 h ; and from equation 3A, ^ = - ; whence
«1 10 "2 **
Thus if the girder be 8 metres deep and the span 64 metres, the permanent
load p = 12,000 kilos., and the moving load m = 16,000 kilos., the dimensions
FIG. 328.
8'"
FIG. 329.
-211050
-22-1000
-224000
+210000
+ 210000
obtained would be those given in Fig. 328. The stresses can be found in the
manner described in the Second Chapter, and are given in Fig. 329. If the
248
BRIDGES AND ROOFS.
level of the rails is above instead of below, the girder would be of the form shown
in Fig. 330.
The stresses in the verticals given for both girders are computed on the assump-
tion that the whole of the permanent load is concentrated at the joints carrying
the track. If, however, it is supposed that half the permanent load is applied to
the upper joints, and the other half to the lower ones (in accordance with § 12),
— 6000 kilos, must be added to the stress in all the verticals of Fig. 329, and the
stresses in these verticals, from the centre outwards, will become :
- 1700° , / ~ 1000° ,
+ 9000' 1+18000'
In Fig. 330, however, + 6000 kilos, must be added to the stresses in the
verticals, thus :
-22000, -42000, -48670, -43000.
When — = 4, it appears that #2 = oo (equation 2), and when - > 4, x£ is
negative, and A2 > li^ . In this case, therefore, it will be necessary to make h2
as well as At equal to ht and the four central bays will be rectangular. If, for
instance, p = 1000 kilos., m = 5000 kilos., and \ = h = 2 metres (as in the girder
calculated in § 6), it will be found that x3 = 6-4 metres, and A3 = 1-615 metre.
Fig. 331 therefore represents the form of the girder and Fig. 332 gives the stresses
in it, those in the four central bays coinciding with those given in Fig. 61.
FIG. 331.
FIG. 332.
— 36000 -45000
-48000
— A 8000
+ 36000
+45000
+ 45000
Lastly, if m > 20 p, x3 is negative (equation 3), and 7?3 > hv As it cannot
be considered advisable to make the depth cf the girder diminish towards the
centre, in this case the above conditions cannot be fulfilled in any bay.
§ 39. — STRUCTURES FULFILLING GIVEN CONDITIONS. 249
§ 39. — GIRDER HAVING THE MAXIMUM STRESS IN ITS
DIAGONALS EQUAL.
The problem discussed in § 38, namely, that the minimum
stress in all the diagonals should be zero, is only a special case
of the following : To determine the form of a girder in which
the maximum stress in the diagonals is equal to some given
quantity.
Since by turning the girder upside down all the stresses
change their sign — and consequently^ the maximum stress
becomes the minimum, and vice versa — it is of no consequence
whether the maximum or the minimum stresses be assumed
equal to a given quantity. But as the minima stresses are
usually negative, it is perhaps as well to make the assumption
with regard to the maxima stresses. This will now be done,
and it will be supposed that the girder is symmetrical, that
there are eight bays, and that the maximum stress in the
diagonals is equal to Y.
FIG. 333.
The stress in the diagonal
of the second bay will be a
maximum when the girder is
loaded, as shown in Fig. 333,
and the equation of moments
(Fig. 334) is
Or by substituting for y its value :
y = (x + 2 A) sin a,
0 = Y (x + 2 A) siii a - D x +p (x + A).
250
BRIDGES AND ROOFS.
(This equation could also have been arrived at by resolving
Y into its horizontal and vertical components, when the moment
of the horizontal component would have vanished.)
Now,
[1]
therefore substituting and solving for #,
x p + 2 Y sin a
A ~ 2'5p + 2-625 m — Ysin a
a?! can be determined in the same manner from Fig. 335 (the
five joints to the right of the section line being loaded) thus :
0 = Y fa + 3 A) sin a, - Dt a?1 + p { fa + A) + fa + 2 A) }
»+ 1-875 m.
And lastly, to find x.2 (Fig. 336) (the four joints to the right of
the section line being loaded) :
0 = Y fa + 4 A) sin a2 - D2*2 +p { fa + A) + fa + 2 A) + fa + 3 A) {
D2 = 3'5^ + l-25m.
xi 6p + 4 Y sin a2
A 0-5 £> + 1-25 m — Ysin a2
[3]
§ 39. — STKUCTURES FULFILLING GIVEN CONDITIONS. 251
The form of the girder can be obtained from these three
equations in the following manner. Some value being assumed
for z, sin a can be found from the equation
sin o =
V *2 + A2
and then x is known from equation 1, whence zl (Fig. 334) can
be determined from the equation
zl = x + 2\
Z X + A
a?! can now be found from equation 2 by substituting
and therefore z2 can be obtained from (Fig. 335)
22 3?i + O A p--,
7, = ^+^'
Likewise a?2 can be obtained from equation 3 by substi-
tuting
rin«.= , *2 .
V Z22 + A2
and A can then be found from the following equation (Fig.
336):
A = *• + **. [6]
*2 a?a + 3 A
As an example, suppose that in the girder of § 6 it was wished to diminish
the greatest compression in the diagonals, thereby increasing the greatest tension.
For instance, let it be assumed that the maximum stress in every diagonal should
be 8000 kilos., then in the above equations, m = 5000, p = 1000, and Y = + 8000.
Assuming that z = 5 (taking the length of a bay as unity), the following values
are obtained by following the steps indicated above ; sin a = 0*447, x = 0-677,
*! = 0-798, sin ox = 0-625, xl = 3'06, zz = 0'958, sin a2 = 0'693, xz = 23-28,
h — 0 • 996. The form of the girder obtained is shown in Fig. 337, and the stresses
in the various bars are given in Fig. 338.
As another example, let Y = 0, the above equations then give the form of a
girder the diagonals of which are always in compression, and if at the same time
- = - , it will be found that
P *
x _ 16 xl _ 16 xz _ 16
A oi. A. JLo A. o
and
_
~*
252
BRIDGES AND EOOFS.
By multiplying these three last equations together,
h _ 28 55 138 _ 92
~z ~ 25 ' 42 "77~ ~ 35 '
From this equation z can be found as soon as some value is given to h. If, for
instance, h = 1 :
35
138 35_15
"77" *92~ 22'
FIG. 337.
25
28
+42000
4-42000 +45110
FIG. 339.
+46970
-48030
+4(5970
In the girders shown in Fig. 327 and Fig. 339 the diagonals are always in
compression. If these girders were reversed the diagonals would always be in
tension. It appears, therefore, that there are two solutions to the problem of
§ 38, and that the form of the girder obtained depends on the direction in
which the diagonals are inclined. If the direction be that shown in Fig. 327,
the conditions can only be complied with, it was seen, to a certain extent when
the ratio - lies between - and 20, and not at all when -> 20. But with the direc-
p 3 p
tion of the diagonals chosen in this § a girder can always be designed meeting
the imposed conditions. Even in the extreme case, when — = co or p = 0, the
above equations give 'results that can be practically applied. Then -^ , xa , a;3,
§ 39. — STRUCTURES FULFILLING GIVEN CONDITIONS. 253
all become zero, and the truss takes the triangular form shown in Fig. 340.
If, on the contrary, — = oo or m — 0, the limiting form is the parabolic girder.
It is obvious that it makes no difference whether the ordinates be plotted above
or below the horizontal through the abutments, or whether a part be placed
FIG. 340.
above and the remainder below. The diagonals in the girder thus obtained will
always be in compression, and if it be reversed the diagonals will always be in
tension.
254 BRIDGES AND ROOFS.
Again, let Y = 0, p = 1000 kilos., m = 5QOO kilos., and £ = A = 1, then by
solving equations 1 to 6 the girder given in Fig. 341 is obtained, and of which
Fig. 342 is a variation. By comparing this girder with that of Fig. 338, in which
the maxima stresses in the diagonals are equal, with the Schwedler's girder of
Fig. 332, and with the parabolic girder of Fig. 39, the influence the form of the
girder has on the stresses will become apparent.
The stresses in the girder of Fig. 341 are given in Fig. 343, and by multiplying
these stresses by — 1, those for the girder of Fig. 344 are obtained, in which the
diagonals are always in tension. In either case it is assumed that both the
moving and permanent loads are applied to the joints situated in the horizontal
boom. If half the permanent load is concentrated on the upper joints and the
remainder on the lower joints, — 500 must be added to the stresses in the verticals
of Fig. 343 and + 500 to those of Fig. 344.
If in Fig. 333 the loads were applied to the upper extremity of the verticals
instead of to the lower, the values of Y sin o and Y sin at , which appear in equa-
tions 1 and 2, taken negatively, would each be the stress in the vertical to the right
of the corresponding diagonal. The above equations can therefore be employed
to determine the form of a girder in which the greatest compression in the ver-
ticals is equal to some given quantity, by making the vertically resolved part of
Y equal to this quantity.
§ 40. — GIRDER IN WHICH THE STRESSES IN THE Bow
ARE THE SAME THROUGHOUT. (Pauli's Girder.)
In the symmetrical parabolic girder of Fig. 34 the stresses
in the bow increase from the centre towards the ends ; but in
the girder of Fig. 70, with parallel booms, the stresses in the
booms, on the contrary, diminished from the centre towards
the ends. It is evident, therefore, that some intermediate form
must exist, in which the stresses in the bow are equal through-
out. It has been seen that the stresses in the booms are
greatest when the girder is fully loaded ; the moving load
need not, therefore, be separated from the permanent load.
Let Q be the total load on each loaded joint (Fig. 345), then
the girder having eight bays, the reaction at each abutment
will beD = ^ = 3-5 Q.
£l
The stress in the bow in any one of the bays situated to the
left of the centre can be found from the equation of moments,
formed with reference to the part of the girder to the left of a
§ 40. — STRUCTURES FULFILLING GIVEN CONDITIONS. 255
vertical section cutting through the bay in question, the inter-
section of the diagonal and the top boom being the turning-
point.
Thus, from Fig. 345, are obtained the following equations
to find the stresses Z1? Z2, Z3, and Z4 :
z, ft = D A
Z3 p3 = D 3 \ - Q (2 A + A)
Z4 p4 = D 4 A - Q (3 A + 2 A + A).
Let M! . . . M4 be the sum of the moments on the right-hand side
of these equations, then, after substituting for D,
M1 = 3-5QA, M2 = 6QA, M3 = 7'5QA, M4 = 8QA.
Now, according to the conditions, the stresses Zx . . . Z4 must be
equal, say, to Z : hence
Suppose, for instance, that A = 1 and Q = 6000 kilos., then M, = 21000,
M2 = 36000, M3 = 45000, M4 = 48000. If, therefore, the stress Z is to be equal
to 36000 kilos, throughout the bow,
36000 _ 1 _ 45000 _ 5 _ 48000 _ 4
_-,
21000
36000
1_
12
36000
36000
36000
To find the form of the girder by construction, describe
circles with radii px ... p4 from the corresponding turning-points
as centres, and draw the lower boom in each bay a tangent to its
circle. Starting from the abutment A, Fig. 346 is thus obtained,
and the depths hi^.h^ can be then found by measurement.
256
BEIDGES AND EOOFS.
The form of the girder can also be obtained by calculation,
instead of by construction, in the following manner: — From
Fig. 345 two expressions can be obtained for cos alt cos a2,
cos a3, and cos a4, and ^ ... h^ can be found by equating these
expressions. Thus, in the first bay,
[1]
or,
p.]
1
A VA2-Pl2
and hi can be obtained from this equation, for pl = ^ is
known.
For the second bay,
P» A
h* V A2 + (
solving this quadratic equation :
= cos a2 ;
[2]
FIG. 346.
In a similar manner the following equations are obtained
for the third and fourth bays :
Ps _
= COS 03.
[3]
[III.]
W
§ 40. — STRUCTURES FULFILLING GIVEN CONDITIONS. 257
If the value chosen for Z were such that the lever-arm
pi = -—^ = — became equal to X, then, from equation I., it
Zi Lt
appears that — = oo. Therefore the conditions can no longer
X
be complied with when ^ > X, or Z ^ D.
M
In the previous numerical example — - = D = 21000, if then Q remains equal
A.
to 6000 kilos., it would be impossible so to construct the girder that stress in the
bow should everywhere be equal or less than 21000 kilos.
If Z has such a value that the lever-arm p2 = — becomes
Z
equal to X, equation II. takes the indeterminate form _ ? = oo x 0.
X
In such a case, h2 must be found from equation 2, which,
when X is substituted for p2, takes the form,
and solving for h2
h _ A2 + V
or,
^? - i (— 4. ^\
A 2VAj \y'
This case occurred in the preceding numerical example (Fig. 346), for Z was
TVT TVT
takenat 36000 kilos., and it was found that — 2 = 36000, therefore pz = -£ — \ = 1.
A. Z
Hence, finding the value of ^ from equation I., viz. ht = 0'7182, and substi-
tuting in equation HA.,
then, from equation III., h3 = 1-2811; and finally h4 = 1-3364 from equation IV.
The same course would have to be pursued if, in any other
bay, the lever-arm of Z became equal to X.
For instance, if Z = 48000 kilos., and the other data remain the same as in the
previous example, namely, A = 1 and Q = 6000 kilos, (so that the moments are the
S
258
BRIDGES AND ROOFS.
same, viz. Mj = 21000, M2 = 36000, M3 = 45000, M4 = 48000), the values ob-
tained for the lever-arms are : (see Fig. 347)
21
36
45 15
48
~48~16' P2~48~4'
Then, from equations I., II., III.,
^ = 0-4865, Jiz = 0-7823, h3 = 0'951.
But equation IV. takes the indeterminate form, — = oo x 0 ; ht must therefore be
A.
calculated from the equation :
FIG. 347.
If the lever-arms pi . . . p4 are halved, the stress Z is
doubled; but if at the same time Q is halved, Z does not
alter. For instance, the dimensions of the girder given in Fig.
348 would apply when Q. = 6000 kilos, and Z = 96000 kilos., or
when Q = 3000 kilos, and Z = 48000 kilos., and generally
in all cases when Z = 16 Q.
FIG. 348.
If the girder be reversed, the new stresses can be found by
multiplying the old ones by — 1. By reversing, therefore,
Fig. 348, the girder shown in Fig. 349 is obtained, in which the
tension in the bow is everywhere the same, and is = -j- 48000
kilos, if Q = 3000 kilos.
The stresses given in Figs. 349 and 350 have been cal-
culated on the supposition that the total load on each
§ 40. — STEUCTUEES FULFILLING GIVEN CONDITIONS. 259
loaded joint is 3000 kilos, (composed of a permanent load
p 1000 T ., -,P . T , m 5000 T ., x
±- = kilos., and of a moving load, - = - kilos.). By
imagining these two girders united to form Fig. 351, a girder
is obtained in which the stress in the lower as well as in the
upper bow is everywhere equal to 48,000 kilos. The ratio
of the depth to span (1 : 8) and the loads (permanent load
p = 1000 kilos., and moving load m = 5000 kilos.) are the same
as those of the parabolic girder of § 6. The stresses in the
horizontal booms in Fig. 351 destroy each other mutually;
this boom is therefore omitted as being unnecessary. The
FIG. 349.
__ 48000
-48000
—48000
+48000
FIG. 351.
_4gOOO
48000
+48000
stresses in the verticals correspond to the assumption that the
points of application of the permanent as well as of the moving
load lie in the horizontal through the abutments.
If the girder is to have crossed diagonals, and not two
half-diagonals meeting each other at the centre of the verticals,
the depths hit hz...h± will have to be slightly altered ; but as
s 2
260
BRIDGES AND ROOFS.
may be expected these alterations will be small, and, in fact,
Fig. 351 is an approximate form.
To find the accurate form, the first step is to obtain the
dimensions of the girder shown in Fig. 352 (which will become
the lower half of the required girder), when the stress in the
bow is throughout equal to the given stress.
The equations of moments are :
Z P! = D A
Z Ps = D (2 A + «,) - Q { (A + 1*3) + u3 }
Z Pi = D (3 A + «4) - Q { (2 X + O + (A + M4) + w4 } .
[5]
[6]
[7]
[8]
FIG. 352.
•D
r x
* «* _ *-
4 to,
As there is no difference in Figs. 345 and 352 in the first
bay, hi can be found from equation L, and by substituting for
pi its value from equation 5.
[V.]
To determine w2 and p2, the following equations are obtained
from Fig. 353 :
a,
W2 A
Solving the first equation for u2,
= tan 02,
i =
§ 40. — STRUCTURES FULFILLING GIVEN CONDITIONS. 261
and finding z2 from the second by substituting for %,
2M2
Consequently,
= * cos e =
2 Aj hz cos 62
" Aj + A,
[10]
Substituting these values of u2 and /o2 in equation 6, and
solving for — ,
h,= (2D-Q)A,
A 2 Z A! COS e2 - D A
[VI.]
From this equation ^2 can be found when hi has been calcu-
lated from equation V., and the angle e2 is known. But
COS €o =
V A2
[11]
so that cos €2 depends on h2. Now cos 62 is less than unity
and greater than cos e1? therefore an approximation to h2 can
be found by assuming for cos e2 some value between these
limits (or one of the limits may be first assumed, say the limit 1),
The value thus found for &2, substituted in equation 11, will
give a nearer approximation to cos e2, from which a more
accurate value of 7i2 can be obtained, and so on until the
required degree of accuracy has been arrived at.
262
BEIDGES AND EOOFS.
For the third bay, equations similar to 9, 10, 11 are obtained
— namely,
2 A2 A3 cos e3
COS 6, =
[13]
[14]
By substituting these values of u3 and p3 in equation 7, and
solving for — ,
A 2 Z A2 cos e3 - (2 D - Q) A
Similarly, for the fourth bay,
2 7z3 A4 cos €4
P* = — * — T^
COS 64 =
[15]
[16]
[17]
and combining equations 15 and 16 with equation 8 :
\ 2 Z A3 cos e4 - 3 (D - Q) A
[VIIL]
Again, assuming that A = 1 and Z = 16 Q, as in the former numerical ex-
ample (Fig. 348), the value of hl9 obtained from equation V., is 0'2242, as
before. Putting this value of At in equation VI., and assuming cos e2 — 1, hz is
found to be 0'3661 as a first approximation. Equation 11 then gives 0' 99 as a
FIG. 354.
nearer value for cos e2, which, substituted in equation VI., gives hz = 0'3734.
A repetition of the same operation gives cos e2 = 0*9889, and hz — 0*3742 as
a third approximation. In a similar manner, from equations VII. and 14,
hs = 0-4745, and from equations VIII. and 17, h4 = 0-4942. (See Fig. 354.)
§ 40. — STRUCTURES FULFILLING GIVEN CONDITIONS. 263
The stresses given in Figs. 355 and 356 have been cal-
culated on the supposition that the total load on a loaded
joint is 3000 kilos, consisting of a permanent load, -^ = — - —
. A £
TI - i A m 5000i n /A • •
kilos., and of a moving load, — = — ^— kilos. (A comparison is
_j 2t
thus obtained with the girders of Figs. 349 and 350.)
FIG. 355.
-48000
— 48000
-1-48000
The combination of these two girders gives the girder shown
in Fig. 357, in which the loading (permanent load p = 1000 kilos.,
and moving load m = 5000 kilos.) and the ratio of depth to span
(0 * 9884 : 8) are the same as those of the parabolic girder of § 6.
( 264 )
TWELFTH CHAPTER
§ 41. — DETERMINATION OF THE CEOSS-SECTIONAL ABE AS
OF THE BAKS IN A STRUCTURE.
Each bar of a structure can be regarded as a bundle of
rods firmly bound together, each rod having a cross-section equal
to the unit of area. To obtain, therefore, the number of units
of area which the cross-section of any lar of a structure must
contain, it is only necessary to divide the stress in the bar ty what
is considered to be the safe stress on a unit of area.
So long as the stress is within the limits of elasticity, it can
be considered safe. For instance, a wrought-iron rod • whose
cross-section has an area of 1 square millimetre can, on an
average, have a stress of 15 kilos, applied to it without the
limit of elasticity being exceeded ; but any increase in the stress
would produce a set. Fifteen kilos, per square millimetre can
be, therefore, considered as the limit of safety ; but in practice
it is usual to allow only 6 or 8 kilos.* per square millimetre for
wrought iron, and it ?s only in special cases, where the risk
may be run, that the limit should be approached nearer.
Thus, to obtain the area of the cross-sections in square
millimetres of the various parts of the structures considered
in the preceding chapters (supposing them to be made of
wrought iron), the calculated stresses in kilos, should be divided
by 6.
For instance, the cross-sectional areas in square millimetres
required for the braced girder of Fig. 61 are as follows : —
1. For the top boom :
3500, 6000, 7500, 8000, 8000, 7500, 6000, 3500.
2. For the bottom boom :
0, 3500, 6000, 7500, 7500, 6000, 3500, 0.
* This is equivalent to 3' 8 and 5'1 tons per sq. inch. — TRANS.
§ 41. — DIMENSIONS OF THE BAKS IN A STRUCTUEE. 265
3. For the verticals :
4000, 3500, 2604, 1813, 1125, 1813, 2604, 3500, 4000.
4. For the diagonals :
4950, 3683, 2567, 1592, 1592, 2567, 3683, 4950.
5. For the counter-braces : *
88, 767, 767, 88.
It must, however, be carefully remembered that the resist-
ance to compression cannot always be taken as equal to the
resistance to tension; on the contrary, in many cases the
resistance to compression is far less, when, in fact, the bar is a
long column and is liable to fail by buckling. This point will
be considered more fully in the sequel. (See " Kesistance of
Long Columns to Buckling.")
Further, it must be observed that these sectional areas are
those due to the stresses in the main structure only ; secondary
structures may, however, be fused into the main structure,
altering the sections accordingly.
The stresses obtained in the preceding chapters were calcu-
lated under the following assumptions : — 1. That, except the
reactions at the abutments, all the exterior forces acting, on
the structure are vertical forces. 2. That the joints are the
only points of application of those forces.
To comply with these assumptions, it is generally necessary
to add to the main structure intermediate bearers, which span
the distance between the joints, and concentrate the load at the
joints ; and also a system of bracing, to resist the pressure of
the wind or any other horizontal force.
Some of the bars of these secondary structures will run
parallel, and close to some of the bars of the main structure.
These parallel bars can either be left separate, or else, as it
were, fused together. In the latter case, the stress, and there-
from the cross-section of the resulting bar, can be found by
forming the algebraical sum of the stresses in each of the com-
ponent bars.
* The second diagonal in the central bays is called a counter-brace.
266 BEIDGES AND EOOFS.
§ 42. — BRACING REQUIRED TO RESIST THE PRESSUEE OF THE
WIND AND HORIZONTAL VIBRATIONS.
When a train passes over a bridge, horizontal forces are
brought into play by the oscillation of the locomotive and of
the carriages, and the pressure of the wind is also increased
proportionately to the surface of the train exposed. To resist
these forces, a system of horizontal bracing is introduced, pro-
ducing in reality a horizontal girder, the booms of which are
formed by those of the main girders.
The stresses in this horizontal girder can be obtained in the
manner explained in the Third Chapter ; for the main girders
being always parallel to each other, this horizontal girder will
always have parallel booms. The distinction made between the
permanent load (uniformly distributed) and the moving load (at
times unequally distributed) will also have to be made in this
case, for the train, in progressing along the bridge, adds con-
tinually to the wind- pressure and to the horizontal oscillations.
It is not far from the truth to assert that the proportion
between the permanent and moving load for the horizontal
girder is the same as that for the main girder. Thus, if it
happens that the width of the bridge is equal to the depth of
the vertical girders, the stresses in the horizontal girder can at
once be found, as explained in the Seventh Chapter.
There is only one point of dissimilarity between the two
girders : in the vertical girder the loads always act downwards,
but in the case of the horizontal girder they act sometimes on
one side and sometimes on the other. The latter must, there-
fore, be constructed symmetrically with reference to the central
line, and the stresses in the booms are to be marked with the
sign ±, for each part of the boom will have alternately to
resist tension and compression of equal magnitude.
Further, if (as in Fig. 61) the diagonals are designed to
resist tension only, counter-braces will have to be introduced
in every bay, and not only in the four central bays, as in Fig.
61. There is, in fact, no difference in this case between a
§ 42. — BRACING TO RESIST WIND, ETC. 267
brace and a counter-brace, for the bar that acts as a brace
when the wind is blowing on one side of the bridge, will be a
counter-brace when the wind is blowing in the opposite direc-
tion. Therefore, also, the greatest stresses in the diagonals
of any bay are equal.
In Fig. 61 the line of railway is on a level with the upper
boom; it is, therefore, possible (to prevent lateral distortion)
to brace the two vertical girders together. The lower booms
can also be braced together to form a second horizontal girder ;
and it can be assumed that by means of the transverse bracing
the load on the two horizontal girders is equally distributed
between them.
In continuation of § 41, let it be required to find the sec-
tional areas of the various bars in these two horizontal girders.
The total vertical load on the bridge was assumed to be
(see pp. 20 and 39)
p + m — 6000 kilos, per metre run.
Supposing that the load on the horizontal girders is
pl -f ml = 857 kilos, per metre run,
the requisite sections for these girders can be found (assuming
that the breadth and height of the bridge are equal) by multi-
plying those already obtained in § 41 by the ratio
Pi + mi _ 857 _ 1 e
p + m ~ 6000 ~~ 7'
remembering that the greatest sectional area obtained for any
two symmetrical bars must alone be retained. Thus, for the
left half of the girders the sectional areas, in square milli-
metres, are:
1. For the booms (either one side or the other):
500, 857, 1071, 1143.
2. For the cross-braces (corresponding to the verticals),
571, 500, 372, 259, 161.
3. For the diagonals (braces and counter-braces J :
707, 526, 367, 227.
268
BKIDGES AND KOOFS.
The stresses in the bars forming the transverse bracing can
be obtained by simply resolving the horizontal force, 857
kilos.,* acting on each joint along them (supposing them to be
placed at each joint). Thus in Fig. 358, if these braces can
only resist tension, the stress in them is (alternately)
857 x V 2 = + 1210 kilos.
and the area of their cross-section will be
= 202 square millimetres.
Each vertical in the main girders receives an increase of
857 kilos, compre scion, which should be added to the stresses
already found, although this probably would not be done in
practice. The sectional area already found for each vertical
857
should, therefore, in this case be increased by -^- = 143 square
millimetres (Fig. 359).
FIG. 358.
FIG. 359.
857
The numbers given in Figs. 360 and 361 express in square
millimetres the cross-sectional area of each bar of the girder,
and they are arrived at by combining the results just obtained.
These can be considered as the final sections, if the inter-
mediate bearers between the joints are constructed separately.
§ 43. — INTERMEDIATE BEARERS.
If the joints of the main structure are so far apart that
they do not offer a sufficient number of points of support, it
* The actual horizontal force on each top joint is 2 x 857 kilos. , according to
the assumptions, but one-half of this is resisted by the top horizontal girder, leav-
ing only 857 kilos, to be communicated to the lower horizontal girder. — TEANS.
§ 43. — INTERMEDIATE BEARERS.
269
becomes necessary (to fulfil the condition that the load on
the main structure is concentrated at the joints) to introduce
intermediate bearers that will span the distance between the
joints, and transmit the load to them, and that will also furnish
at the same time a sufficient number of points of support.
These intermediate trusses bear the same relation to the
loads upon them that the main girder does to its loads ; they
can therefore be similarly constructed as a combination of bars.
If the number of points of support offered by one set of
4000
6857
FIG. 360.
8571
9H3
9143
500
4357
7071
8643
8643
FIG. 361.
intermediate bearers be not sufficient, another set of a secondary
order must be introduced. If even then the points of support
are not near enough, a third order must be added, and so on
until the required number of points of support is obtained.
The triangles formed by the bracing of this last set may be so
small that the material saved in the void spaces would not
cover the extra expense for workmanship, and it then would be
better to use a plate-web instead of the bracing.
When all these intermediate bearers are placed in their
proper positions with reference to the main structure, it will
be found that several of the bars run side by side. These
bars may, as it were, be fused together, and the stress in the
270
BKIDGES AND HOOFS.
resulting bar will be the sum of the stresses in the bars of
which it is composed.
In several cases it is possible and advisable to design the
intermediate girders or bearers geometrically similar to the
main structure. If this be done, the stresses in the resulting
structure, however complicated, can be found by splitting it
up into its primary forms, and often the stresses are easier
determined in this manner than by employing the method of
moments, as will appear from the following example : —
The truss of Fig. 362 can be considered as made up of
the primary forms shown in Figs. 363, 364, 365, and 366.
Now Fig. 363 can be regarded as a parabolic girder, having
only one loaded point. As already explained, the laws relating
§ 43. — INTERMEDIATE BEARERS. 271
to parabolic girders are independent of the number of loaded
points. The equations of § 8, namely,
can therefore be used to find stresses in the bars Xo and Z0 by
L
substituting for 0. Thus (Fig. 363)
Now, since in this case the primary forms are geometrically
similar, the ratio j- is constant for all, and the stresses corre-
L
spending to Z0 and X0 in the intermediate bearers of the first,
second, and third order can be obtained by dividing the values
found above by 2, 4, and 8 respectively.
Let the three systems of intermediate bearers be now
framed into each other, and also into Fig. 363, so as to produce
Fig. 362. The stresses in the different parts of the bars A C
and B C will then evidently be as follows :
i + i + i);
and the stress in the horizontal bar A B is :
To find the stresses in the remaining bars, a similar process
can be applied to Fig. 365, after Fig. 366 has been framed
into it ; and again, to Fig. 364, after "both the previous figures
are combined with it.
Let 2 1 = 32 metres,/ = 6 • 4 metres, and 2p I = 32,000 kilos. ;
272
BRIDGES AND ROOFS.
then Fig. 367 gives the various stresses, omitting, however, the
system of Fig. 366. These stresses have been calculated on
the supposition that the points of loading lie in the horizontal
A B, but this only affects the stresses in the verticals.
It is easily seen that Fig. 368 is but a variation of the
above.
FIG. 367,
4ft
+ 5000
+10000
•4-20000
+ 35000
+ 50UO
4-10000
+20000
+35000
FIG. 368.
In both these examples the position, as well as the form, of
the intermediate bearers corresponded to those of the main
structure. Figs. 369, 370, 371, and 372, however, represent
a case in which the form of the intermediate bearers is similar
to that of the main structure, but in which the position is
different. Apart from this difference in position, the stresses
can be calculated as in the previous example.
§ 43. — INTERMEDIATE BEARERS.
273
Fig. 371 is the roof truss of § 4, and the stresses then
obtained could have been found as follows : —
The data were 2 1 = 32 metres, the height of roof / = 6 • 4
metres, and the total load 2jp I = 32,000 kilos. The stresses
FIG. 369.
in the main triangle, Fig. 369, can be obtained as in the
previous case (Fig. 363), and using the same symbols,
= ~
274 BRIDGES AND ROOFS.
Fig. 373 represents one of the intermediate bearers of the first
order, and the stresses given have been calculated by the method
of moments. These stresses once determined, those in the
intermediate bearers of the second order can be found by dividing
by two, those in the third order by dividing by four, and so on.
Fig. 371 is obtained by combining the intermediate bearers of
FIG. 373.
•fioooo
first and second order with the main triangle, and Fig. 372 is
likewise formed by the addition of the bearers of third order.
The stresses in each case can be found by adding together the
stresses in the separate systems where the bars coincide. Thus
in Fig. 71 the stresses in the two bars meeting at the abutments
are
20000 + 10000 + 5000 = + 35000 kilos.
- (21540 + 10770 + 5385) = - 37695 kilos.
And the stresses in the same bars in Fig. 372 are
20000 + 10000 + 5000 + 2500 = 37500 kilos.
- (21540 + 10770 + 5385 + 2692-5) = - 40387 '5 kilos.
The stresses already given in Fig. 19 can therefore be con-
sidered as made up in the manner shown in Fig. 374.
Another way of subdividing the distance between the joints
of the main structure is shown in Fig. 378, and Figs. 376 and
377 show the manner in which Fig. 378 is derived from 375.
The load on this roof truss can be considered as due to a
loaded horizontal beam, severed over each loaded joint, and
supported at these points by vertical columns. It is easy to see
that each joint receives half the load on the adjacent bays of the
beam, and consequently the stresses produced in each inter-
mediate bearer by its load can be calculated by the method of
moments in the manner indicated with reference to Fig. 373.
§ 43. — INTERMEDIATE BEARERS.
275
4-20000
FIG. 375.
FIG. 376.
FIG. 377.
FIG. 378.
T 2
276
BRIDGES AND EOOFS.
• If, as in the former roofs, 2 1 = 32 metres, /= 6'4 metres,
and 2 p I = 32,000 kilo's., the stresses obtained are those given
in Fig. 379.
The roof truss of § 3 could also be calculated in a similar
manner, and the stresses given in Fig. 14 can be considered as
made up as shown in Fig. 380.
FIG. 37&.
4- 5000
'•+ 5000
+ 5000
420000.
+35000
FIG. 380.
In all the above structures the application of the intermediate
trusses increased the stress in the coinciding bars of the main
truss because the stresses to be added together were of the same
sign, and for the same reason it was also unnecessary to dis-
tinguish between the permanent and moving loads. It is, how-
ever, advantageous, if possible, so to introduce the intermediate
§ 43. — INTERMEDIATE BEAKERS.
277
bearers that the stress in the coinciding bars may be of opposite
sign, so that when fuzed together the stresses may partially
neutralize each other, thus effecting a saving of material.
For instance, in a parabolic girder of 100 metres span (see
§ 34) the secondary trusses could be best arranged as shown in
Fig. 381, when the compression booms of the secondary trusses
will coincide with the tension boom of the main girder.
The tension in this latter is, with the numerical values,
given in § 34.
H =
2)50*
2 x 12-5
FIG. 381.
= + 600 tons.
\(^3^t^&±^^^^
50"1
The secondary trusses are small girders of 10 metres span ;
they have to carry the whole of the moving load and about half
of the permanent load. For example, let them be parabolic
. , . , . , depth ft 1 . , .
girders in which — - — = ^— = •= , then the compression in
span 2X 5
their horizontal booms is
-•-»*-
Therefore when the bridge is fully loaded the tension in
the main lower boom would be
H + h = 600 - 25 = 575 tons,
instead of 600 tons.
The cross-section of the main booms could not, however, be
diminished, for if only one bay were unloaded the stress in the
part of the lower boom belonging to that bay would scarcely
be reduced. But the whole of the material for the upper booms
278
BRIDGES AND ROOFS.
of the secondary trusses is saved. Since the quantity of material
in their curved booms is very little in excess of that required
for the upper booms, the material in the whole of the secondary
trusses will be to that in the main girder as the horizontal stress
in the secondary trusses is to twice the horizontal stress in the
main girder, or in the ratio,
25 _ j_
2 X 600 ~ 48 '
If the main girder is reversed as in Fig. 382, the secondary
trusses will also have to be reversed to obtain the same ad-
vantage.
FIG. 382.
^
It is hardly necessary to remark that the parabolic form
has been chosen for the secondary girders only as an
illustration, and that they could be constructed as lattice
girders, or, if the void spaces between the bars become too
small, as plate girders. In this latter case the secondary girders
take the form of strengthening ribs to prevent the horizontal
boom from bending, and it follows that the best position for
this rib is above the boom when the line of railway is on a
level with the top boom, and below the boom when the line is
level with the lower boom.
In lattice girders with crossed diagonals an intermediate
point of support can be obtained by making the intersection
of the diagonals a point of loading, the load being conveyed to
it by means of a vertical. (See Fig. 384.)
By introducing triangular intermediate girders, the number
of points of support can be increased exactly in a similar
manner to that adopted for the roof trusses. Thus from the
fundamental form of Fig. 383, the girders shown in Fig. 384,
385, and 386 are derived,
As to the manner of calculating the stresses in these girders.
The stresses in Fig. 384 would first be found by the method of
moments, and for the stresses in the booms the intersection
§ 43. — INTERMEDIATE BEARERS.
279
of the diagonals would be chosen as the turning-points. The
stress in any diagonal can be found by remembering that
the vertical component of the required stress is in this case half
the vertical force at any section (in other words, half the shear-
ing force). The stresses in Fig. 385 can then be obtained by
following the method already explained for roof trusses.
FIG. 383.
FIG. 385.
FIG. 386.
It must be observed that it makes a difference in the cal-
culated stresses whether the girder of Fig. 385 be considered
to have seventeen loaded joints (which it really has) or only
the nine belonging to the main structure (Fig. 384). (In the last
case the weight of the intermediate trusses is supposed to be
transmitted to the joints of the main girder.)
Evidently the assumption that, as the moving load proceeds,
one joint is fully loaded before the next receives any load at all,
is, as pointed out at the end of § 12, not strictly true, but it is
also evident that the greater the number of loaded joints — that
is, the nearer they are to each other — the less will be the result-
280
BRIDGES AND ROOFS.
ing error. It follows that the stresses given in Fig. 387
obtained under the supposition that the girder has only nine
loaded joints, will differ slightly from those that would be ob-
tained if it were considered to have seventeen loaded joints, the
latter being the more accurate values. It will also be observed,
when it is assumed that there are nine loaded joints and the
moving load has arrived over the centre of one of the inter-
mediate girders, that the next following joint of the main girder
has already received a part of its load, for it acts as a point of
support to the intermediate girder ; and when the moving load
has arrived at the end of one of the intermediate girders, the
joint of the main girder at this point has not yet received its fall
load, for it acts as a point of support to the next intermediate
girder which is as yet unloaded.
For the sake of comparison, the dimensions, &c., in Figs. 387
and 388 are the same as those of the girder calculated in § 1 0,
namely, depth = 2 metres, span = 16 metres, total load on the
girder = 48,000 kilos, (consisting of, permanent load = 8000
kilos., moving load = 40,000 kilos.).
The stresses calculated from these data for the girders of
Figs. 384 and 385 are given in Figs. 387 and 388.
FIG. 387.
-21000
45000
+ 21000
+ 45000
S4000
'1000
FIG. 388.
-48000
-48000
+21000
-45000
By comparing these two girders with those of § 14, it will
be seen that the number of points of support in - a simple
§ 43. — INTERMEDIATE BEAEER8. 281
girder such as that shown in Fig. 383 can be increased in two
different ways, viz. either by increasing the number of triangula-
tions or else by introducing intermediate trusses as explained
above.
The first method would evidently be the best if the sectional
area required to resist compression were always proportional to
the stress ; for then it would be possible to increase the
number of points of support indefinitely without adding to the
quantity of material, whereas in the second method every new
diagonal and vertical requires extra material. But this advan-
tage is only apparent, because with multiple latticing the
compression braces become very thin and are therefore liable to
bend or buckle, and hence require much larger cross-sections in
proportion.
The advantages and disadvantages of each system must
however be considered specially in each case, for this is a point
which cannot be decided generally.
NOTE. — In practice it is usual to rivet the braces together where they inter-
sect. This, in one respect, is not right, for the braces are thereby impeded from
acting independently, but the great advantage is obtained that the tendency to
buckle of the compression braces is greatly reduced, and the objection to multiple
lattice girders, mentioned above, is avoided.
( 282 )
THIRTEENTH CHAPTEE.
§ 44. — ON THE DEFLECTION OP LOADED STRUCTURES.
It has been found by experiment that the amount of alteration
in the length of a bar is proportional to the stress, so long as
the stress is within the limits of elasticity, and this whether
the stress be compression or tension.
Thus if S is the alteration of length in one unit of length
due to a stress S per unit of area,
Sac S;
or
'=!' . w.
where E is a constant quantity.
E is called the modulus of elasticity, and from equation I.
it is evident that ^ is the elongation or shortening in one unit
of length produced by the unit stress per unit of area. Another
definition of E can be deduced from equation I., namely, that it
is the stress per unit of area that will lengthen a bar to double
its original length.
Its value for wrought iron is about 20,000 when expressed
in millimetres and kilogrammes (equivalent to 28,450,000 when
the English inch and the Ib. avoirdupois are the units). Thus
every millimetre of the length of a w,rought-iron rod, whatever
be the section, increases its length by ^ITOTFO- millimetre when a
tension of 1 kilo, per square millimetre is applied to it. If,
however, the rod were subject to a tension of 6 kilos, per square
millimetre, each millimetre would increase its length by swoir or
^5\re millimetre.
The actual increase of length is obtained by multiplying
the original length I by S, thus :
A = 75. [IL]
§ 44. — DEFLECTION OF LOADED STRUCTURES. 283
A wrought-iron rod, therefore, of 10 metres length, subject to
a tension of 6 kilos, per square millimetre, increases its length by
A = 10000 X aoTJoo = 3 millimetres-
Negative stress or compression produces negative elongation
or shortening in the same proportion as above ; therefore equa-
tions I. and II. can be used for compression as well as for tension.
If the stresses in a loaded structure are known, it is a purely
geometrical problem to find the change of form and the con-
sequent deflection ; for the structure can be imagined taken to
pieces and then remade, using the altered lengths of the
several bars.
A few examples will now be given to show how the applica-
tion of the above laws can be transformed into a geometrical
problem. It is also of use to know the deflection of various
simple structures under a known load, for this will enable the
proportion to be found in which the load on a complex structure
subdivides itself between its component simple structures
whence the stresses in these latter can be deduced.
It will be assumed that in all cases the cross- section of any
bar is proportional to the stress in it, so that the elongation or
shortening of each part of the structure will be S for each unit
of length.
Thus in Fig. 389, owing to the shortening a B of the two bars
A C and C B due to a load at C, the point C is lowered to Cx .
FIG. 389.
The position of Ct can be found by describing arcs from A and B
with radii equal to a- a B. If E C = a $, E C^ can be drawn at
284
BEIDGES AND ROOFS.
right angles to A C without appreciable error, because a S is
small, and the same is true for the other side.
The triangles C E 01 and C D A being similar,
CE
or
[l]
When the abutments are relieved from thrust by means of
a tie rod A B, the point C is further lowered by an amount S2
due to the increase of length Z S of A D. Suppose that at first
A C and B C do not alter their length, then the position of C2
can be found by describing arcs from Aj and Bx with radii a
(Fig. 390).
FIG. 390.
44
D
BB,
From the similarity of the triangles C F C2 and C D A,
CC2 s2 I
or
OF /5 h'
= 4
' The total depression of the point C is therefore
,=,+..=
[2J
[3]
If the load, instead of being applied at C, is hung at D (Fig.
391) and transmitted to C by means of the tension rod C D, the
point D will be lowered by an amount s', which is the sum
§ 44. — DEFLECTION OF LOADED STRUCTURES, 285
of the depression of the point C and of the increase in length
a- of the C D, thus
The above equations are also true for the reversed position
of the structure.
FIG. 391.
C
r
In a similar manner the depressions of the points C and D
in Fig. 392 can be determined by first finding the part due to
the shortening of A C and D B :
FIG. 392.
ol
, !•
A&
~K
t'C>
Zfi^
h
— »- i
1
and then that due to the shortening of C D (Fig. 393)
be
The total depression is therefore
286
BRIDGES AND ROOFS.
When the points A and B are connected together by a
tension rod (Fig. 394) a further depression of
is produced, and in this case C and D will be depressed by an
amount
/a?
s = «i + s + <r = S
FIG. 393.
1 _ U
" "~c * "T~ * - 6 ^ * "-B
FIG. 395.
Lastly, when the loads are hung at E and F, Fig. 395,
these points are lowered by an amount equal to <r2, due to the
§ 45. — DEFLECTION OF LOADED STEUCTUEES. 287
lengthening of the tie rods, added to the depression of the
point C or D found above. But cr2 = h 8. Consequently
<r2 = [a2 + 6c + (6 + c) c
[7]
This method of investigation can be extended to the case
when there are three or more points of loading.
§ 45. — DEFLECTION OF PAEABOLIC AECHES AND
GIEDEES.
The effect of the load on the inverted parabolic chain in
Fig. 396 is to shorten the length of arc, and thus reduce its
height from /to/p The deflection s: at the crown can therefore
be found from the equation (Fig. 396)
*,=/-/,. [8]
FIG. 396.
f
When the ratio ^ of the height of the arc to the span is
£ I
small, it can be shown that the length of the arc of the original
parabola is given by the equation
.* The equation to the parabola is
l-xl
f~l*'
Differentiating with respect to a?,
*y -9f*
J~X-**l*' [But
288 BRIDGES AND EOOFS.
Consequently the length of arc of the compressed parabola
is
or substituting for S its value
and solving this equation for/i ,
32 x J2
The error entailed by leaving out the expression -^ ( 1 + f ^
and those that follow it is small, and hence, although ^ is as
sumed large,
approximately.
Substituting this value of/! in equation 8,
[9]
But the differential of the arc S is given by
H®
! + *£*
or approximately, if ^ is small,
Integrating between the limits — I and + I,
§ 45. — DEFLECTION OF LOADED STRUCTURES. 289
For example, let I = 20,000 millimetres, / = 5000 millimetres, 5 = -^g^, then
Sj = 18 (1 + Jj.) = 18 '75 millimetres, and this is the amount the crown of the
parabolic arch of § 22 sinks when the load produces a compression of 6 kilos, per
square millimetre in the bow.
Evidently equation 9 is applicable to a suspension bridge,
and in this case B will be the increase of length per unit of
length.
If the points A and B are connected together by a tie, the
crown will sink by a further amount s2, due to the extension
of this tie. The structure then becomes a girder, since the re-
actions at the abutments are vertical.
FIG. 397.
s2 can evidently be found by assuming that the tie alone
extends by an amount 2 IS, and that the length of the bow
remains unchanged. Therefore equating the length of arc of
the original parabola to that of a parabola having a height of
arc /2 and a span 2 / (1 + S),
and solving this equation approximately for/2,
The depression of the crown due to the extension of the tie
is therefore
[10]
The total depression of the crown when the bow is com-
pressed will evidently be
f»j [11]
u
290 BKIDGES AND ROOFS.
If the load is communicated from the lower boom to the
upper boom by means of vertical rods, the extension of these
rods will lower the centre of the tie by a further amount Sf.
But since the above results were obtained on the supposition
•p
that the ratio j was small, S/ can be neglected in comparison
to s, more especially as in actual girders the greatest tension in
the verticals occurs with a partial load.
Equation 11 is also true for a parabolic girder having the
bow underneath.
Example. — In the parabolic girder calculated in § 6, the span was 16 metres,
and the height of arc 2 metres ; let 5 = -^Q^ , then the deflection at the crown is
3 6 80002
S = 2 '20000' 2000 =14
§ 46. — DEFLECTION OF BEACED GIRDERS WITH
PARALLEL BOOMS.
In girders with parallel booms the deflection s (in the
centre) is composed of two parts, one entirely due to the booms
and the other to the braces.
To find the first part, s1? the centre line of the girder can
be considered as bent into the arc of a circle of radius p (Fig.
398). Since «! is small, the length of arc differs but by a very
small quantity from the chord; the length of the chord is
therefore 2 I, and the equation to the circle is
but Si2 is small in comparison to 2 p slt Hence,
8l = ~ [12]
Further, the lengths of arc of the outer circle (formed by
the lower boom) and the centre circle are as their radii, or
o 4- -
T 2 _ 2 I (1 + S) .
~7~~~ 2~7 ;
§ 46. — DEFLECTION OP LOADED STRUCTURES. 291
whence
Substituting this value of p in equation 12,
[13]
[14]
The deflection s2, due to the braces alone, can be obtained by
considering that the booms remain unaltered, and then finding
FIG. 398.
/
Iff
\
\
21(1+6)
the change of form produced in each of the right-angled tri-
angles formed by a vertical, a diagonal, and a horizontal bar,
by the lengthening of the diagonal and the shortening of the
vertical (Fig. 399).
FIG. 399.
The point C of the triangle A B C is lowered by an amount a
in consequence of this change of form ; and cr can be considered
as composed of two parts. The first, e, is due to the lengthen-
ing of the diagonal, and the second, X, to the shortening of the
vertical. The extension of the diagonal is c S, and e can there-
fore be obtained from the equation (Fig. 400)
u 2
292 BEIDGES AND ROOFS.
The shortening of the vertical is X, and evidently
A=/5.
Hence,
or, since c2 = f2 + &2,
FIG. 400.
If w is the number of bays between
the abutments and the centre, the de-
flection $2 will be n times er. Hence
or, since na = 1,
"-- N '7
.ijA.
By adding together the values of sl and' s2 (from equations
14 and 15), the "total deflection at the centre of the girder
shown in Fig. 399 is found to be
[16]
If the bays are square,
- = 1 and s =
a
and if at the same time there are eight bays,
- = 4 and s = 7 1 5 .
Example. — The deflection at the centre of the girder calculated in § 10 can
be found by means of this formula. The span is 16 metres, and assuming that
s = aorroo millimetres,
s = 7 x 8000 x 20&JO = 16-8 millimetres.
If the braces were constructed much stronger than neces-
sary, the alteration in their length, and consequently the
deflection due to them, might be neglected. The deflection s
§ 46. — DEFLECTION OF LOADED STBUCTUEES. 293
(equation 14), due to the booms alone, would then be the deflec-
tion at the centre of the girder, assuming that the deflection
curve is the arc of a circle. These conditions are approximately
fulfilled in a plate-web girder ; the deflection of such a girder
can therefore be found approximately from equation 14. This
equation also contains the general law of the deflection of a
beam of equal depth and symmetrical cross-section throughout ;
for the deflection at the centre can never be greater than when
the bending or curvature at every point is a maximum — that is
when the deflection curve is a circle. This is actually the case
in a beam of uniform strength ; that is, when the form of the
beam is so proportioned to the load that the greatest stress in
every cross-section is constant.
( 294 )
FOUETEENTH CHAPTER
§ 47. — THEORY OP COMPOSITE STRUCTURES.
The two following questions can be answered by means of
the equations obtained in the preceding chapter, or by means of
others similar to them : —
1. In a structure composed of two distinct systems con-
nected together, what strength and stiffness should each system
have in order that they may work evenly together ?
2. What is the proportion of the total load carried by each
system ?
As the common points of loading of the two simple systems
gradually sink owing to the increase of the load, the stresses
in each will augment, and alterations in the lengths of the
various bars, &c., will take place. So soon, however, as the
limit of elasticity in one of the systems is reached, the load
could not safely be increased, however remote the stresses in
the other system may be from their limit of elasticity. This
second system might therefore be made of weaker and less
elastic material without diminishing the safe resistance of the
whole structure, or, better still, the quantity of material in
it might be reduced, so that in both systems the limit of elas-
ticity would be reached at the same time. All the material
thus saved is not only unnecessary to the structure, but is
positively harmful, in that it adds to the dead load. Such
combined structures should therefore, if possible, be so designed
that the simple systems composing them are equally stiff.
The proportion of the load carried by each system can,
however, be found, whether this condition be complied with or
not, by equating the deflection of each system at the points
where they are connected together, and these are also the
points where the loads are transmitted from one system to the
other.
§ 47. — THEORY OF COMPOSITE STRUCTURES.
Si
295
An equation is thus obtained, giving -i, the ratio of the
£2
alterations of length per unit of length ; and from this it is
easy to obtain the ratio ^i, giving the proportion of the
H2
load carried by each system.
It appears from the previous investigations, that the general
equation giving the deflection of a loaded point can be written
s = A5,
where A is a constant, depending on the form and dimensions
of the structure (the value of this constant can be obtained for
the various cases considered in the last chapter, from equations
1 to 16). The deflection of each simple system at some point
where they are connected together will therefore be A! ^ and
A2 S2 respectively. Hence,
or
=
S2 A,
[III.]
Now the loads producing the alteration in length Si and S2
per unit of length in each of the simple systems respectively
can be found by the methods already explained; and since
the total load on the structure is known, the part carried by
each system can easily be found.
For example, let two simple systems, one like Fig. 389 and
the other like Fig. 398, be combined together as shown in
Fig. 401. If both systems are made of the same material,
296 BRIDGES AND ROOFS.
the condition that they may reach the elastic limit together
can be found by putting ^ = S2. Hence, from equation III.,
A, = A2,
or from equations 1 and 14,
a2_ I*
, *"/''
whence
/ i2
[The value thus found for $- may be termed "the econo-
mical ratio of the depths " (with regard only to the quantity
of material, not necessarily as regards the cost).]
If the elastic limit is not reached simultaneously, or if each
system is made of a different material, it appears from
equation III. that
The economical ratio of the depths can be found in this case by
putting for 8X and S2 their values at the limit of elasticity, for
the material of which the corresponding system is made.
From equation 18 the ratio — can be found, as follows :
Q2
Let the load Qx acting on the simple system formed by the
rods A E and B E, produce a stress Si per unit of area in each
of the rods, and let Fj be the sectional area of each rod ; then
therefore, the shortening per unit of length in these two rods
is
St _ Qi« „,
5l-E~-2F1AE1
Again, if F2 is the sectional area of either of the booms of
the girder C D at the point E, the stress S2 per unit of area
produced by the load Q2 is
§ 47. — THEORY OF COMPOSITE STRUCTURES.
297
and the consequent alteration of length per unit of length in
this girder is
s _ S2 __ Q2 1 r
2 — T?~ — O T? -ft? ' L^"J
E2 2F2/E2
By substituting these values of Sx and S2 in equation 18, the
following equation is obtained : —
Q2 ~~ fl3 ' f" ' F2 ' E2 '
from which, since Qi + Q2 = Q is known, the distribution of
the load can be ascertained. If the structure is made of the
same material throughout and complies with the condition
expressed by equation 17, equation 21 becomes
Q2
[22]
In the structure shown in Fig. 402, composed of Figs. 391
and 398, the condition that the limit of elasticity may be
FIG. 402.
reached in each system at the same time when the material
is the same in both, is (equations 4 and 14),
l~£#* [23]
The general equation for the distribution of the load can be
obtained in the manner indicated in the previous example,
and is
Qi _ i3 A2 F_i E^
"E2'
[24]
Q2 2 a'/2
which, combined with equation 23, gives for the special case
expressed by that equation
5i = 2«.Fi=i-/Fj.
298
BRIDGES AND ROOFS.
The last three equations are also true for the reverse
arrangement shown in Fig. 403.
The three corresponding equations for the structure given in
Fig. 404 can be obtained in a similar manner, by means of
equations 5 and 14 ; but in this case the deflection of the girder
at the points C and D must be considered, and it is equal to the
b2
deflection at the centre of the girder multiplied by 1 - — , for
i
the deflection varies as the square of the distance from the
centre. Hence the condition that each system may reach the
elastic limit at the same time (supposing both to be made of
the same material) is
/_ /«-&«
h ~ a? + be'
[26]
The general equation for the distribution of the load is
7\~ ~ ft, „ /V»2 _1_ A /A ' TG'" * T?~ ' L ' J
§ 47.— THEORY OF COMPOSITE STRUCTURES.
299
which, when combined with equation 26, gives for the special
case expressed by that equation
I=S-F;- • ™
Exactly in a similar manner, the following equations are
obtained for Figs. 405 and 406, by means of equations 7 and
14.
f I* - 62
Q2
F,
, E2
[29]
[30]
[31]
FIG. 405.
Mr h
Q
FIG. 406.
1«
Q
C b D
In all the preceding cases it was assumed that the altera-
tion in length of each unit of length was the same throughout
the same system, or, in other words, that the cross-section of
each bar was proportional to the stress in it.
If a structure is composed of several simple systems, each having a different
degree of stiffness, the total load on the structure will be distributed unequally
between them, whatever the load may be, and those which owing to their want of
stiffness reach their elastic limit last, evidently contain too much material.
300
BRIDGES AND ROOFS.
FIG. 407.
A solid beam can be considered as composed of an infinite number of lattice or
plate girders of unequal depths (Fig. 407), for imagine the beam divided into
horizontal layers, then any two layers equidistant from the central line will
represent the booms, and the material between them the
braces or the plate web.
These imaginary girders are not equally stiff, and
therefore there is a waste of material in a solid beam.
Now,' all these imaginary girders have the same
deflection, and from equation 14 it appears that if the
deflection remains constant the alteration of length of
the booms, and consequently the stress in them, is pro-
portional to the height of the girder. The stress in each
j r-1— ~~^"i, layer is therefore proportional to its distance from the
•*•' centre, so that if the outer layer, at a distance w from
the centre, is subject to a stress S per unit of area, the
stress in a layer whose distance is u from the centre will be
If the stress S is known at any cross-section of the beam, the stress in any layer
in the same cross-section can be obtained from this equation, and the stress in the
layer symmetrically placed on the opposite side of the centre has the same
numerical value, but with a contrary sign.
The total stress in such a layer is, if A be its thickness and * its breadth,
If such a beam be bent by the forces Kt, K, . . . . acting at right angles to
its axis, the maximum stress S at any section through P can be found by the
FIG. 408.
A
FIG. 409.
M
method of moments. Take a section a £ (Figs. 408 and 409), and, to maintain
equilibrium, apply to each layer the total stress in it. In order that there
may be no rotation about P, the algebraical sum of the moments, M, of all the
§ 48. — TRUSSED BEAMS WITHOUT DIAGONALS. 301
exterior forces acting on the part B a j8 about P must be equal to the sum of the
moments of all the stresses about the same point. Therefore
where 2 is a symbol indicating that the sum has been taken of all the separate
Q
moments due to the stress in the various layers. The factor - is common to all
these moments, and can therefore be placed outside the symbol, or
Now 2 (zw2 A) is the " moment of inertia "* of the cross-section about the
axis round which moments were taken, that is, about a horizontal axis in the
plane of the section and passing through its centre. The moment of inertia is
usually denoted by I. Hence
or
The value of I for any form of cross-section can be obtained by dividing up the
area into very small parts (or elements), and multiplying the area of each
element by the square of its distance from the horizontal axis through the centre
of gravity, and then adding together the products thus obtained.!
§ 48. — TRUSSED BEAMS WITHOUT DIAGONALS.
It was shown, § 8, that the diagonals of parabolic trusses only
come into play with a partial load, and also that they then
become an indispensable part of the structure, that is, if the
various bars are connected by single bolts, and are thus
only capable of taking up stress in the direction of their
length. If, however, the diagonals are omitted, one of the
* The moment of inertia is a term belonging to the dynamics of a rigid
body, and the following is the definition : —
If the mass of every particle of a body be multiplied by the square of its
distance from a straight line, the sum of the products so formed is called the
" moment of inertia " of the system about that line, which is also called the axis.
In the present case, it will be observed, the mass of every particle or element
lias been taken as unity ; also from the symmetry of Fig. 407 it is apparent that
the axis is horizontal and passes through the centre of gravity of the section. As
will be seen in the sequel, the resistance to flexure at any cross-section of a
body always depends on the moment of inertia of that cross- section about a
horizontal axis in the plane of and passing through the centre of gravity of the
cross-section. — TRANS.
t The continuation of this subject will be found in the Fifteenth Chapter.
302
BRIDGES AND EOOFS.
booms must be stiffened to enable the structure to resist the
deformation that would otherwise take place with a partial
load.
The stresses will, however, no longer be the same as those in
a simple parabolic girder, and the structure must, in fact, be
regarded as a combination of two systems, and the distribution
of the load must be found according to the principles laid down
in § 47.
Although it is not recommended to leave out the diagonals
(except when the load is uniformly distributed, as in the
trussed purlins of a roof), yet these theoretically imperfect
structures occur so frequently that the influence of the omission
will be illustrated by a few examples.
A theoretical parabolic girder of three bays should have
one diagonal at least in the central bay (Fig. 410). Without
this diagonal the deformation shown in Fig. 411 would occur
with a partial load. This can, however, be avoided by replacing
the three bars AC, CD, and D B by a continuous stiff beam
(Fig. 412). The structure is then no longer a simple truss, for
even when fully loaded a part of the load will be carried
directly by the stiff beam, and the remainder will be communi-
cated by the two verticals to the rods A E, E F, and F B, and
this part will be smaller the stiffer the beam is.
The structure consists of the two simple systems shown in
Figs. 413 and 414 ; and to. find the stresses produced by. the
two loads Q (Fig. 412), the ratio
Q2 (l-n)Q
48.— TKUSSED BEAMS WITHOUT DIAGONALS.
303
must be determined by following the method indicated in
§47.
As soon as n is found, the distribution of a partial load can
also be ascertained. Thus, for instance, if the point C alone is
IA _\ f\ FIG. 414.
r-"11) '^J .
J
loaded with a weight Q (Fig. 415),' the following distribution
takes place. One part,
Q and
forms the load on Fig. 416, and the remaining part,
is the load on Fig. 417.
That this really is the case is easy to see, in the following
way : — Evidently a load at D produces the same effect at C as
a load at C produces at D (by symmetry). Now, since the hori-
zontal stress in the bars A E and B F is always equal, the stress
in both verticals must also always be equal. A single load at
C therefore produces half the stress in the verticals that the
two loads at C and D together do ; or, in other words, the loads
on C and D (Fig. 416) due to Q at C or at D are | Q and — Q.
The loads on Fig. 417 are evidently found by subtraction.
304
BRIDGES AND ROOFS.
The exterior forces on the structure being known, the
stresses can be found by the method of moments ; the stresses
in the bars A E, E F, F B, due to the load at-C, are given in
D
Fig. 415. These stresses will remain the same when the point
D is loaded instead of the point C, and they will be doubled
when both C and D are loaded.
The stresses in these bars, when the truss is fully loaded,
are therefore to those that the same load would produce in a
parabolic truss of the same form in the ratio of nil. Now
the number n depends on the relative stiffness of each simple
system, and can only be 1 when there are free joints at C and
D. Thus it is evident that the error committed by treating
FIG. 418.
-UnA
the structure of Fig. 412 as a simple parabolic truss is greater
the smaller n is, or, in other words, the stiffer the beam is.
Obviously, if the truss be reversed (Fig. 418) the stresses
§ 49.— INFLUENCE OF CHANGES OF TEMPEKATUEE. 305
can be found in exactly the same manner. The stresses in the
structure of Fig. 419 can be calculated in a similar manner,
obtaining the number m from § 47, in the same way as n.
Fift. 419.
§ 49. — INFLUENCE OF CHANGES OF TEMPERATURE.
The results just obtained for composite structures are not
practically useful, because the influence of the changes of tem-
perature was not taken into account. As will be seen, the
distribution of the load between the two simple systems, and
also the economical ratio of their depths, depend on these
variations of temperature.
It is found by experiment that all bodies, with one or two
exceptions, expand as the temperature increases and contract
when the temperature diminishes; and it is also found that
the expansion or contraction is proportional to the increase or
decrease of temperature. The ratio of the alteration of length
due to one degree of temperature to the original length is
called the coefficient of linear expansion. Thus, for wrought
iron, if the temperature is reckoned in degrees Centigrade, this
coefficient is
a = 0-0000122; [32]
that is to say, a wrought-iron bar increases or decreases its
length by 10)01020%00ths for an increase or decrease of 1° C.
Since the alteration of length is proportional to the change
of temperature, the ratio of the alteration of length due to f
of temperature to the original length will be
A = a*. [33]
(In other words, A is the amount which a unit of length is
expanded or contracted by a change of temperature of t°.)
306 BRIDGES AND ROOFS.
For example, if t = + 41° C., the elongation of a wrought-
iron bar is = 41 x T^TOTOO o = 2oVoth °f its original length.
If the increase of temperature were only 20 • 5° C., the elon-
gation would only be ^^th ; and if the temperature were to
decrease by the same amount, the consequent shortening would
be the same as the previous elongation.
The alterations of form produced by variations of tempera-
ture are quite distinct from those due to elasticity. If, there-
fore, an increase of temperature coincides with tension in any
bar, the increments of length due to both causes are to be
added together to obtain the total elongation ; and if a decrease
of temperature occurs with compression, the separate decre-
ments of length are to be added together. But the total
elongation or shortening will be equal to the difference between
the increment and the decrement, if compression occurs with
increase of temperature, or tension with decrease of tempera-
ture.
In the combination of bars shown in Fig. 389 an increase
of temperature of t° would, according to equations 1 and 33,
raise the point C by an amount
<r=A^; [34]
and this point would sink by the same amount with a decrease
of*0.
The braced arch of § 22 can, as far as the influence of temperature on the
height of the hinge S is concerned, be regarded as a combination of two rods
similar to the above. From the dimensions of the arch h = 5000 millimetres,
a2 = 200002 + 50002,* and assuming that A = -^^ (for an increase of 20 -5° C.)
1 /200002 + 50002\
" = 4000 ( 8000— ) = 21'25 'mlllmetreS-
Consequently the hinge S is raised 21-25 millimetres when an increase of
temperature of 20 '5° C. takes place, and therefore also sinks 21-25 millimetres
with a decrease of 20 -5° C. If the variations of temperature were doubled,
the rising and the sinking of the hinge would be 42 • 5 millimetres. Supposing
that the decrease of temperature of 41° C. occurred at the same time that the
total load was on the bridge, which latter, according to § 45, equation 9, produces
a deflection sr = 18*75 millimetres, the total deflection would be
• 18-75 + 42-5 = 61-25 millimetres.
* See Fig. 175, page 127.
§ 49.— INFLUENCE OF CHANGES OF TEMPERATURE. 307
If Sl denotes the decrease in a unit of length due to com-
pression in the two rods A E and B E (Fig. 401), produced by
a load at C, and A the shortening per unit of length in the
same bars owing to a coincident decrease of temperature, the
total deflection of the point E is, according to equation 1 :
a2
* = (A + A) j > [35]
g p
and this deflection is to be equated to -—-, the deflection in
the centre of the horizontal beam (equation 18J, thus :
Solving this equation for , and putting a2 = Z2 -j- h2 :
The influence of the variations of temperature on the dis-
tribution of the load can be ascertained by finding the effect
produced by the change of temperature on the structure, sup-
posed weightless, and upon which no exterior loads are acting.
If A is the shortening per unit of length due to a decrease
of temperature, the point E (Fig. 401), if free, would be lowered
by an amount — =— , but the actual amount is less, owing to
the horizontal beam CD; or, in other words, the stiffness of
this beam produces a vertical force P acting upwards at E.
Both struts are elongated by this force, and this elongation
must be deducted from the shortening produced by the de-
crease of temperature. Therefore, if Si is the elongation per
unit of length produced by the force P alone, the actual
depression of the point E isl
* = (A — &i)^. [38]
x 2
308 BRIDGES AND ROOFS.
But, according to equation 19, ^ = 00 -^ , • Hence:
A£JI f i hi
The deflection at the centre of the beam due to the force P
from equation
from equation 20,
8 Z2
is, from equation 14, s = -^-, and substituting S2 =
p/3
Hence by equating the values of s in equations 39 and 40 :
_ P/3
~2E2F2/2'
And solving for P :
2 A Ex Ft -
P -- - - T421
Evidently P is the correction to be applied to the distribution
of the load between the two systems found from equation 21 :
in fact, a decrease of temperature diminishes the part of the
load carried by the struts by the amount P, and consequently
increases the load on the beam by the same amount. An
increase of temperature has the reverse effect.
By substituting in the above equation - = 0-6, - = 0*8, - = 3, also
E! = E2 = 20000 (supposing that both systems are made of wrought iron),
F! = F2 = 10000 square millimetres, and A = ± -jcW (corresponding to a
change of temperature of 20 -5° C.), the value of P is found to be P = ± 10700
kilos.
By introducing the same data into equation 21, it is found that
5l -4-608;
§ 49. — INFLUENCE OF CHANGES OF TEMPERATURE. 309
or if the total load is 80000 kilos.
Qi
Q, = -9*=r Q = tf?| x 80000 = 65740 kilos.
Q2 = - pr Q = K-^'X 80000 = 14260 kilos.
'
Thus, when the temperature decreases 20-5° C., the load carried by the struts
becomes
65740 - 10700 = 55040 kilos.
and by the beam
14260 + 10700 = 24960 kilos.
And when the temperature increases 20° '5 C. the struts carry
65740 + 10700 = 76440 kilos.
and the beam
14260 - 10700 = 3560 kilos.
(This subject will be further discussed in the sixteenth chapter.)
( 310 )
FIFTEENTH CHAPTER.
§ 50. — RESISTANCE OF BEAMS TO FLEXURE.
If one end of a horizontal beam be fixed, and a weight hung
to the free end, the beam, originally straight, will be bent
into the form of a curve, whose convexity is upwards (Fig. 420).
If the beam be regarded as a bundle of fibres, whose direction
is parallel to the length of the beam, and which are cemented
together, so that they cannot slip over one another, it is easy
to see that as soon as the beam is bent the upper fibres will be
FIG. 420.
FIG. 421.
lengthened, and the lower ones shortened. Between the upper
and the lower layer of fibres there must therefore be a layer
in which the fibres are neither lengthened nor shortened ; this
layer A B (Fig. 421) can be called the neutral surface.
The greater the distance of a fibre from this neutral surface
the greater will be its elongation if it is above the neutral sur-
face, and the greater its shortening if below. It can be assumed
that the sections made by planes at right angles to the neutral
surface before bending remain plane after bending,* and are
still at right angles to the neutral surface, which is now curved.
In Fig. 422, M and N are two of these planes very near to each
other ; originally they were parallel, but when bending took
* This has been ascertained by experiment. See 'Civil Engineering,' by
Prof. Rankine. — TRANS.
§ 50. — RESISTANCE TO FLEXURE.
311
FIG. 422.
place they converged in the directions C D and E F. Now, since
the portions of the fibres lying between the two planes were
originally all equal, the alteration in length of each fibre can
be found by drawing a plane G H parallel to E F, and at a
distance N M from it equal to the original length of the fibres,
the distances between the
planes C D and Gr H are
evidently the alterations in
length of the fibres. Hence
from Fig. 422 for any fibre
LQ:
GC
or in words : The alteration in
length of any fibre is propor-
tional to its distance from the
neutral surface. But, accord-
ing to the laws of elasticity,
the intensity of stress is pro-
portional to the alteration of
length, so long as the stress
is below the limit of elasticity, hence the stress in any fibre is
proportional to its distance from the neutral surface so long as
the stress in it is below the limit of elasticity.
Therefore, if s is the stress (per unit of area) in a fibre L Q
FIG. 424.
K
at a distance u from the neutral surface (Figs. 423 and 424), and
if S is the stress (per unit of area) in the fibre EC, at a distance w :
or s = !S-
[43]
312 BRIDGES AND ROOFS.
The total stress in any fibre can be found by multiplying its
area by s, the stress per unit of area. If the beam be imagined
divided into an infinite number of layers of fibres parallel to the
neutral surface N N, and if/ be the cross-sectional area of such
a layer whose distance from N N is u, the total stress in this
layer is evidently :
*f = $U-f. • [44]
This stress will be tension when the layer is above the
neutral surface, that is, when u is positive ; and will be com-
pression when the layer is below the neutral surface, that is,
when u is negative.
Let a section of the beam be made at N by a plane per-
pendicular to the neutral surface. For equilibrium a force will
FJG 425 kave to be applied to each
fibre acting in the direction
of its length, and equal to
the stress in it (Fig. 425).
These forces can be con-
sidered horizontal, for it
B must be assumed that the
amount of bending is very
small. They alone, how-
ever, could not maintain
equilibrium ; for, to resist
^ the vertical force K acting
downwards, a vertical force V, acting upwards, must be ap-
plied; and since the section plane N is very nearly vertical,
V can be considered as acting along it. It is this force which
prevents the part B N of the beam from slipping down along
the section plane, and it is called the resistance to shearing.
Eesolving the forces acting on B N vertically :
V = K. [45]
The only horizontal forces acting on BN are the direct
stresses in the fibres. Those above the neutral surface act from
§50. — RESISTANCE TO FLEXURE. 313
right to left, and those below from left to right. The algebraical
sum of these forces must be zero, therefore from equation 44 :
2(IM/) = 0; [46]
8
and since — is a common factor it can be omitted ; hence :
w
*(/«) = <>. [47]
This equation shows that the sum of the products of the
area of each elemental layer into its distance from the neutral
surface is zero.
But, as is well known, if x be the distance of the centre of
gravity from the neutral surface,
x 5 (/) = 5 (/ w) = 0, or a = 0 ;
that is, the neutral surf ace passes through the centre of gravity of
the section.
The third condition of equilibrium is that the sum of the
moments about any axis should vanish. Let the intersection of
the neutral surface with the section plane be this axis, which will
be perpendicular to the paper at N (Fig. 425), and is called
the neutral axis. The moment of K about this axis is K a;
(called the moment of flexure or else lending moment), and the
moment of the stress in the layer of fibres, whose distance is
C! •/*
u from the neutral surface, is — — u. Hence :
w
K*. [48]
or in words : The moment of resistance of the fibres is equal to
the moment of flexure of the lending force.
R
The common factor — can be placed outside the sign of
summation, thus
L x. [49]
But the expression 2(/w2) is the moment of inertia* of
* See footnote, page 301.
314
BRIDGES AND ROOFS.
the cross-section about the neutral axis, and it is generally
denoted by I. Therefore,
-I = Ktf. [50]
w
Or, writing M, instead of KOJ, to represent more generally
the moment of flexure,
- 1 = M.* [51]
w L J
S is the stress per unit of area in the fibre whose distance
is w from the neutral surface. If, therefore, w is the greatest
distance that occurs, S will be the greatest stress, and can be
found from
when
[52]
[53]
I = *(/««)
is known.
To find the value of I when the section is rectangular, con-
sider a layer of fibres at a distance u from the neutral axis
(Fig. 426), the area of this layer is/ = b A, and consequently,
I = 6 2 (A w2) ;
or, according to the notation of the Integral Calculus :
a 6
,
12
[54]
FIG. 426.
b
ih
The section shown in Fig. 427 can be considered as the dif-
* This equation has been obtained on the supposition that the stress
nowhere exceeds the limit of elasticity. This formula is therefore not applic-
able at the point of rupture, and this is borne out by experiment. — TRANS.
§ 50. — RESISTANCE TO FLEXURE.
315
ference of the two rectangular sections B H and b h, and hence
its moment of inertia about the neutral axis is
B H3 6 A3
12
12 '
[55]
The same value of I obtains for the section shown in Fig. 428.
The moment of inertia of all sections that can be resolved into
rectangles may be obtained in a similar manner by means of
equation 54 so long as they are symmetrical with respect to the
neutral axis. For instance, for the section shown in Fig. 429,
which is the sum of two rectangles :
12 + 12 '
and for the section of Fig. 430 :
B H3 _ 6A3 __ 6t V
~ ~W ~ 12 ~i¥" '
[56]
[57]
FIG. 429.
[NoTE. — It is sometimes necessary to find the moment of inertia of irre-
gular figures whose contour is not determined by any simple curve, the
section of a rail for instance.
In such cases the moment of inertia can only be obtained approximately
by finding that of a figure composed of rectangles, as shown in Fig. 430A, and
the greater the number of rectangles the nearer will be the approximation.
It becomes therefore necessary to know the moment of inertia of a rect-
angle about any axis parallel to one of its sides. Let the moment of inertia of
the rectangle A B (Fig. 430s) be required about the axis Y 0. Take an
elemental strip P P' of the rectangle, of width 8 x. The moment of inertia of
316
BRIDGES AND ROOFS.
this element about Y 0 is x2 . b 8 x. The moment of inertia of the rectangle
A B will therefore evidently be
Xc
a?bdx
b .
FIG. 430A.
FIG. 430B.
pV-^r-; -:::±.i i ?
iO
Applying this to the case of Fig. 430A, the amount of inertia of that
section about the axis N N is found to be
-/3) + o(/3 - d») +p(d* -63)
+ «3) + « (c3 - a3) + s (e* - c3) + t (h3 -e3)}.
The axis N N passes through the centre of gravity of the section, and the
position of this point must be found by the usual methods. A beautiful
graphic method of finding the moment of inertia and the centre of gravity of
any section is given in * Graphic Statics,' translated from the German by
Lieut. G. S. Clarke, RE.]
If the section is circular (Fig. 431) :
for evidently the moment of inertia of a circle is the same
about any diameter ; and since u2 + v2 = x2,
) -
= 2 (f**);
or,
§ 50. — KESISTANCE TO FLEXUKE.
317
2 (/#2) can evidently be found by replacing / by the area
of the elemental annulus (Fig. 432) whose radius is a?, and
breadth A. The area of this annulus is 2 x TT A ; therefore,
Or in the notation of the Integral Calculus :
TR
2;(/*2) = 2,r *»rfa = -B«;
\) o
FIG. 431.
FIG. 432.
and if the diameter of the circle is D,
[58]
The annular section shown in Fig. 433 can be regarded as
the difference between two circles
whose diameters are D and d respec-
tively. Hence, in this case :
cl D
[59]
If the section remains constant
throughout the length of the beam,
- will also be constant, and it appears (from equation 50) that
in this case the greatest intensity of stress will reach its
greatest value in the beam at the point where the bending
318
BRIDGES AND ROOFS.
moment is amaximum. This is evidently the case when
x = I ; and if SL is the greatest stress in the beam :
[60]
If the beam has a rectangular section (Fig. 434), w = - , and from equation
54, I = — - ; equation 60 therefore takes the form
12
6
FIG. 434.
I
[61]
B
+K
As an example, let K = 125 kilos., I = 800 millimetres, 6 = 20 millimetres,
h = 100 millimetres, then the greatest stress in the beam is,
6KJ 6 X 125 x 800
TV = 20 X 1002
= 3 °8<
mlllimetre'
The intensity of the stress is independent of the nature of the material. By
comparing it however with the stress per square millimetre considered safe
for the material in question, it can be ascertained whether the resistance of
the beam is sufficient or not, or to what extent the load might be increased so
that the safe stress should just be arrived at. Thus if the above beam were made
of wrought iron, the load K could be doubled (250 kilos.), in which case the
greatest stress per square millimetre would be 6 kilos, (see page 264).
FIG. 435.
I, A I
B,
The greatest stress in a beam supported at both^ends
(Fig. 435), and loaded at any point A, can be found from equa-
§ 50. — EESISTANCE TO FLEXURE. 319
tion 60 by writing for K the reaction at one of the abutments,
and for / the distance of the load from that abutment. For if the
part A B1 be considered fixed (encased in a wall, for instance),
it is evident that A B is in the Fame condition as the beam of
Fig. 434, with this difference, however, that the bending force
in this case acts upwards instead of downwards, ,and conse-
quently the greatest tension occurs in the lowest fibre. The
reaction K at the abutment A is found by taking moments
about the other abutment, thus :
. ,
and substituting in equation 60
§!l-«£i . **&- r621
» <+'- z^:
(Since Kx ^ = K I the same equation would be obtained if the
greatest stress in the part A Bx were found).
[NOTE. — If M is the bending moment at any section of the beam distant x
from BH , it is easily seen that
so long as the section is situated between B, and A, and
when the section lies between A and B.]
When a beam, supported at both ends, is loaded uniformly
and p is the load per unit of length, the reaction at either abut-
f) L
ment is K = •—- , and p x is the load on the part B M = x
(Fig. 436). If the part A M of the beam be imagined fixed
the part B M becomes a beam fixed at one end and acted upon
by two loads, namely, the reaction K acting upwards at B,
and the load px acting downwards at the centre of BM.
Taking moments about M :
M-K.-p. = «.-!. [63]
320
BRIDGES AND ROOFS.
Substituting in equation 51, it appears that the greatest
stress in a section whose distance from one of the abutments
is % can be found from the equation
• " 2 v
FIG. 436.
[63A]
>L-x
M\ 30
: z
I px
The product os (L — %) is greatest when % = — . Hence the
greatest stress in the beam is given by the equation :
-if *!=¥• [<*] '
By putting L = 2 1 and os = I - z in equation 63, the general
equation for the moment of flexure at any section is obtained,
distances being measured from the centre of the beam, thus :
and when z = 0 :
[65]
[66]
Equation 65 can be exhibited graphically by taking the
different values of z as abscissa, and the corresponding values
FIG. 437.
of M as ordinates. The curve shown in Fig. 437 is thus
obtained.
§ 51. — DEFLECTION OF BEAMS. 321
The shearing force (Fig. 436) at the section through M is :
for this is the vertical force that would have to be ^applied at
this section to maintain equilibrium.
Putting as before L = 2 1 and x = I — z :
V=pz. [66A]
§ 51. — DEFLECTION OF BEAMS.
The curved line A B (Fig. 422), into the form of which the
neutral surface is bent, is called the curve of deflection.
The element of arc M N of this curve can be regarded as
the arc of a circle whose centre is at 0, the intersection of C D
and E F. This circle is, in fact, the circle of curvature of the
curve at MN, and p its radius is the radius of curvature.
Since the two triangles C Gr M and M N 0 are similar :
CG _MG
MN~ON"
r\ pi
Now ^p- is evidently the elongation per unit of length of
the fibre E C, the stress in which is S per unit of area, therefore,
according to equation I :
CG _ S
MN~E'
Further M G = w and 0 N = p. Hence :
S to S E
_ = - or - = - . [67]
sit p w p
H
Substituting this value of - in equation 51 :
~ = M- [68]
The radius of curvature of the element of arc M N of the
deflection curve A B (Fig. 438) is therefore :
El El
322
BRIDGES AND ROOFS.
As already pointed out (p. 312) the curvature is assumed to
be very small, and consequently instead of the arc M N in the
equation
x — v M N
M N = p <j>. or (f> = -
P
its horizontal projection M P = A can be substituted. There-
fore :
TT fl ,*\ A
[70]
K
Now, consider the horizontal projection A Q = x of the arc
AM resolved into its elements A, and let x be replaced in the
above equation by its successive values differing each by A ; by
adding together all the equations thus obtained the sum of all
the small angles <j> will be found, thus :
[71]
S ($) is evidently &>, the angle which the radius of curvature
M 0 makes with the vertical (or the angle the tangent to the
§ 51. — DEFLECTION OF BEAMS. 323
curve at M makes with the horizontal). Let the beam be
prism, that is, let its cross-section be the same throughout;
rr
then I is constant and =^f is a common factor of all the terms
Ji* 1
contained under the sign of summation in the right-hand
expression. Hence :
° = ITI 2[(/ ~* A)] = Wi C'2 (A) " 2(* A)]' [72]
but
2(A) = # and 2 (x A) = ^ •
Therefore :
If # = Z, « becomes a, the angle the tangent to the curve at
the end B of the beam makes with the horizontal, and
K/2
a =
[74]
2EI
In the triangle MNP
e = A tan w ;
but since the curvature is small, &> is a small angle, and con
sequently
tan a> = w,
or
e = co A. [75]
Substituting in equation 73
again replacing x by its successive values differing from each
other by A, and summing up the values of e thus found :
2 («) = p 2 (a A) - * 2 (^ A)]. [77]
Now X (e) is evidently equal to y, the vertical projection of
the arc A M, and further,
Y 2
324 BBIDGES AND ROOFS.
therefore equation 77 becomes :
K
[78]
According to the notation of the Differential Calculus,
A = dx, f = dy, « = — ^, and ^ = dw = d (-^ } •
Equation 70 therefore takes the form
Assuming as before that I is constant, integrating twice and remembering
\,t when x — o, -
d
integration is zero.
that when x — o, — — o, and y = o, so that the constant occurring in each
From a reference to Fig. 438 it will be seen that when x = Z,
y = s, and therefore the deflection of the end B is (equation
78):
'
Dividing this last equation by equation 60,
and substituting — = 5, and 2 w = h, (assuming that the centre of gravity of the
.hi
cross-section is equidistant from the top and bottom fibres),
It is evident that this equation is also true in the case of the beam represented
by Fig. 435, when the weight Q is hanging at the centre. This equation can
therefore be employed to find what alteration will be produced in the equation
found for the compound system of Fig. 401, when the beam, instead of being of
uniform strength, has an equal section throughout. Thus, putting f 52 , instead
of 82,
and equation 21 becomes
§ 51. — DEFLECTION OF BEAMS. 325
[NOTE. — The equation
CG _MG
MN~"ON
is quite independent of the manner in which the beam is loaded or supported,
and is in fact a geometrical property depending only on the curvature. Equation
68, viz.
is therefore perfectly general and is applicable to the case of any beam under
bending stress.
Now, it is shown in works on geometry that for any curve referred to rect-
angular axes,
dx2
Let the straight line through the abutments be taken as the axis of a?, then the
tangent of the angle the tangent to the curve at the point-(», ?/) makes with the
axis of x is - — But by a previous assumption the curvature is very small, and
therefore this angle will also be very small for every point of the deflection curve ;
consequently ( -j- ) may be neglected in comparison to 1. Hence in the present
case,
dx*
Therefore,
Vy M^
dx* ±El'
The positive or negative sign being taken according as — increases or
diminishes with .v.
This is the differential equation to the deflection curve, and when M and I
are known for any point (#, y\ the equation to the curve can be fouud if the
integrations can be effected.]
The beam instead of being horizontal before being bent may
make a very small angle &> with the horizontal, Fig. 439. In
this case the angle a the tangent at B makes with the hori-
zontal, consists of two parts ; one part is the angle o> and the
other is the deflection angle that would obtain if the beam
326
BKIDGES AND KOOFS.
had been originally horizontal, and which can be found from
equation 74. Therefore :
a - « + 5^- • [80]
+ 2EI
In the same manner the deflection s can be considered as
FIG. 439.
made up of two parts ; one part is I tan co = I co (since w is a
small angle) and the other part can be obtained from equation
79. Hence :
K?3
2El
[81]
The above equations (80 and 81) can also be adapted to the
case when the loaded point B is not at the end of the beam.
If, in Fig. 440, C is the end of the beam, the part B C will
remain straight if there be no load on it. The angle made with
FIG. 440.
the horizontal will be the same at C as at B, and is consequently
equal to a (equation 80). The deflection at C is equal to the
deflection at B added to a tan a = a a (since a is a small angle).
Therefore :
K'3 - F821
• a a. [oAj
§ 51. — DEFLECTION OF BEAMS. 327
And substituting for a its value from equation 80 :
f = (I + a) a> H 6ET~~ — ' ^
Or adopting the notation of Fig. 441, that is, writing x
instead of /, I — x instead of a, and s instead of cr :
[84]
[85]
6EI
If there is also a load at C (Fig. 442), the deflection-angle
and the deflection at C will be increased by an amount which
can be found from equations 74 and 79, thus :
0 = « + iTST
Q/2
2EI ' 2EI
6EI
3EI
[86]
[87]
FIG. 441.
FIG. 442.
These equations are evidently also true if K and Q act up-
wards instead of downwards, when, however, their signs must
be changed.
In the last equations put o> = 0, Q = 0 and p d x instead
of K, further d a instead of a and d s instead of s, then :
2EI '
[88]
&HAJL
d s = ^-= . [89]
6El
328
BRIDGES AND ROOFS.
These equations give the effect, on a beam originally hori-
zontal, of the element p d x of a distributed load.
Integrating between the limits x = ^ and x = x2 for the
case shown in Fig. 443 :*
[90]
6EI
[91]
If the whole span is loaded with a load p per unit of length
the limits of the integration are ^ = 0, x2 = I and then :
--
6EI'
s- -
8ET
[92]
[93]
• Lastly, if the original inclination to the horizontal at the
point of fixing A is w and there is a load Q at C besides the
uniformly distributed load :
™/3 0/2
[94]
2EI'
pi* Q/3
[95]
FIG. 444.
f
l-x
Fig. 444 represents a beam supported at both ends. If the
left half be considered fixed in a wall, it is evident that the right
* The load is supposed to be uniformly distributed so that p is a constant. —
TRANS.
§ 51. — DEFLECTION OF BEAMS.
329
half is in the same condition as the fixed beam of Fig. 442,
when a) = 0 and Q = — K, and the deflection can therefore be
obtained by substituting these values of to and Q in equation
87 and writing — s instead of s, thus :
[NOTE.— If x = 0 the beam is loaded with a central load 2 K, and
or replacing 2 K by Q,
"GET
6EI
The equation to the deflection curve of a beam supported at both ends
and loaded in the centre, can be found from the equation given at p. 325,
namely :
cPy _ :M
do? ~ El*
If the beam is of the same section throughout, E I is constant, and
(Fig. 444A) taking 0 X and 0 Y as axes of reference, the moment of flexure
at any point P is
so long as x is less than I.
Y
FIG. 444A.
ft li t I t*
t=i=4 JSL
Hence:
- x
~ 2 El
The negative sign being taken because -^ diminishes as x increases.
Integrating
T^ = — r^ & + constant.
ax 4 Hi L
330 BRIDGES AND ROOFS.
ISTow. when x = 7, — • = 0, because the tangent to the curve is horizontal
Ct SO
at the centre of the beam. Therefore :
0 _ — -4rr I2 + constant ;
or
Integrating again,
but when y = 0, x = 0, hence constant = 0 and
— the equation required. It will be observed that this equation is only true up to
the centre of the beam, for at this point M changes from -— to — -- Q (x - Z);
2 2
but evidently the two halves of the deflection curve are exactly similar.
The greatest deflection occurs at the centre of the beam where x = I, and
writing s for the greatest value of y,
6EI'
or the value found above.]
The deflection due to a uniformly distributed load can be
found from equation 96 by first finding the deflection due to
an element of load pdw, thus writing d s instead of s, and p d x
instead of K :
. [97]
6EI
FIG. 445.
Then the deflection at the centre in Fig. 445 is obtained by
integrating this equation between the limits x1 and X2, thus :
j _ p [2 I3 Q2 - *.) - F (*,» - V) + j Qr24 - V)] ^ [98]
6EI
§ 51. — DEFLECTION OF BEAMS.
331
If the whole span be loaded, the limits of the integration are
= 0 and x2 = I, in this case, therefore :
s- *
'""El'
[99]
Since the load on the left half of the beam produces the same
deflection as that on the right half, it is obvious that if the load
on one side be removed the deflection in the centre will be
FIG. 446.
halved. The deflection at the centre of a beam loaded as indi-
cated in Fig. 446 will therefore be :
p [2 I3 Q2 - Q - 1 Q23 - V) +
s —
12EI
[100]
If the left half of a beam supported at three equidistant
points as shown in Fig. 447 be imagined fixed in a wall, the
right half is evidently in the same condition as the beam of
Fig. 442 when co = 0, s = 0, and Q acts upwards or is negative.
By substituting these values in equation 87,
_
3EI'
and solving for Q,
Further the sum of the reactions at the three points of
support must equal 2K. Hence:
[102]
332 BRIDGES AND EOOFS.
Thus the reactions at the three points of support are found.
[NOTE. — Obviously the three points of support are in a straight line, for it
was assumed that s = 0. It is important to remember this, for the reactions
depend on the relative positions of the three points of support.]
Evidently the load on the left span produces the same re-
action at the central support as the load on the right span.
Therefore in the case represented in Fig. 448, the reaction W
at the central support is :
W = |P = K-Q; [103]
or
The reactions of the end supports can now be obtained by
taking moments, thus :
V-K(' + X\ W
V-Kl"2TJ"y
and
FIG. 448.
V
W
V
I
• >
I
«,
x B
fcS t
A
A
A
I
A
6'
K
or substituting for W its value :
The bending moment at a point N (Fig. 449), between A
and B, and situated at a distance z from A, is :
M = K (a? - *) - V (Z - *). [107]
Putting M = 0, substituting for V, and solving for z :
*+«-]>
which, when
becomes
§ 51. — DEFLECTION OF BEAMS.
= u, and
333
[109]
This equation determines the position of the point where
the bending moment is zero.
The value of the bending moment for every point in the
beam is represented graphically in Fig. 449, for the bending
moment at every point from A to B is given by equation 107,
and this equation represents a straight line, M and z being the
variables.
Let it be supposed that the beam is subject to a uniformly
distributed moving load, then equation 109, when put in the
[110]
FIG. 450.
AT
A
,
(max)
FIG. 451.
determines what parts of the beam should be loaded to produce
a maximum or a minimum bending moment at N (see Figs. 450
and 451).
33 i BRIDGES AND ROOFS.
[NOTE. — The truth of this can be shown as follows : — It will be found
that by substituting for Y its value from equation 105, and putting as before
I — x I — z
— - — = u and = v, that the equation
M = KO-z)-V(Z-*)
becomes :
TIT M/K/K 4 A
M = —— (5 --- M- ) .
4t3 \ V /
Therefore M is positive so long as
or
Thus, evidently any load situated between C and B produces a positive
bending moment at N. A little consideration will show that all loads on the
other span also produce positive bending moments at K Hence Fig. 450.
Again M is negative when
that is, when the load is placed between B and A. Therefore the greatest
negative or minimum bending moment occurs at N when the part B A is
loaded as in Fig. 451.]
When the reactions produced by a single load K have been
found by means of equations 104, 105, and 106, the reactions due
FIG. 452.
:• :.li'.".'i"' !'•:•• ••••••.'!
to a uniformly distributed load can be obtained by writing pdos
instead of K and integrating between the limits ^ and x2 (Fig.
452). Thus :
or
§ 51. — DEFLECTION OF BEAMS.
335
If several portions of the girder are thus loaded, the total
reaction is found by adding together the reactions produced by
each part separately.
Again, when the load is uniformly distributed over both
spans, the reaction at either outer support can be found if p d x
be written instead of K in equation 101 and the integration
performed between the limits 0 and Z, thus :
and
Pt = Zpl — 2 . %pl= Ipl [HlB]
If, therefore, p is the uniformly distributed load per unit of
length, and m is a uniformly distributed moving load on the
part o?2 — #! of the beam represented in Fig. 452, the reaction
Vis,
Fig. 453 represents a beam uniformly loaded and supported
at three equidistant points, the central support being at a dis-
tance s from the horizontal straight line joining the outer points
of support.
Supposing the central support to be removed, the deflection
s at the centre of the beam would be found from equation 99.
*>/*
[113]
~ " E I '
FIG. 453.
P
A
7
I
BS>t>^ ^flTB3E3IE
^ ^
Again, the upward deflection s2 produced by an upward
force P acting on the unloaded beam, can be found from
equation 96 by putting x = 0 and 2 K = P, thus ;
6EI"
[1141
336 BRIDGES AND ROOFS.
Now s is evidently equal to the difference of these deflec-
tions, therefore
[NOTE. — This result can also be obtained as follows :
If, in equation, 95 — Q be written for Q, — s for s, and o> = 0,
= _
~8EI 3EI'
This equation evidently applies to either half of the beam in Fig. 453,
also
.
Therefore,
j?Z4 j)/4 PP
S~SEI 3EI+6EI
2*EI 6EI'
Equation 115 can be used to find the distribution of the
load in Fig. 401, when the beam has the same cross-section
throughout and is uniformly loaded. For the force P, taken in
the opposite direction, deflects the apex of the struts by an
amount
and by equating these two values of s
_ .
2E1F1V~ 24E2 I2 6E2V
If the cross-section of the beam is rectangular, F2 its area,
and h2 its height,
_F2A22
~12~'
according to equation 54, and the above equation solved for
P becomes
p_ _ W _
~ a* V E_2 F/ [H8]
T I3 ' h^ ' Et • F,
This equation gives the part of the load supported by the
struts, supposing that the temperature does not alter and that
§ 51.— DEFLECTION OP BEAMS. 337
originally, before loading, the three points C, E, D were in a
straight line.
If, for instance, ^ = | , ^? = i , |^ = 1, and ^ = 2, the equation gives
I ' A, &i Jbj
P =
Tfl
If -^ = 0, P = £p J, or the same value that was obtained (equation 111 B)
for the reaction at the central support of a uniformly loaded beam resting on
three equidistant points placed in a horizontal straight line.*
When the temperature decreases the portion of the load
carried by the beam is increased, the struts being relieved of
the same amount, and the reverse occurs when the temperature
increases. This " temperature load " P can be found by equating
the deflection of the centre of the beam (equation 39) to the
deflection of the apex of the struts (equation 114), thus :
TTl T 2
Substituting for I = * 2 , and solving for P,
\2i
2 A Ej Ft ^
E2 ' F2 * a3 ' A22
If E!= E2 = 20,000 kilos., F^ 10,000 square millimetres, F2 = 20,000 square
millimetres,- = 0'8, — = 0*6, -* = 3, and A = 77^:7. (corresponding to a de-
ci a n2 4000
crease of temperature of 20° • 5 C.), it will be found that P = 5873 kilos. And if at
the same time the beam is subject to a uniformly distributed load 2p I (see
equation 118),
P = 1 • 128 pf — 5873 kilos.
for a decrease of 20° '5 C. ; and similarly
P = l-128pf + 5873 kilos.
for an increase of 20° • 5 C.
It was shown, page 321, that the general equation for the
radius of curvature of the elastic curve is
P=~, [121]
F
* Since =r = 0, the resistance of the struts is infinite, and they therefore act
as a fixed point of support. — TRANS.
338
BRIDGES AND ROOFS.
where I is the moment of inertia, and M the bending moment
for the cross-section at which p is taken.
Thus when the ratio ^ is constant for all sections p is con-
stant, and the deflection, curve is a circle.
Now in a beam of equal section throughout, I is constant.
A prismatic beam can therefore only bend in the shape of a
circle when M is constant. This would be the case, for instance,
in Fig. 422, if instead of the single force K, a couple acted at
the free end of the beam.
If, however, both M and I vary, and Mx, Ix are the moments
of bending and inertia respectively at any given section, for
instance at one of the ends of the beam, then the equation
1 - IL
M ~M^
[122]
or
expresses the relation that must exist in order that the
deflection curve may be a circle.
For example, in the case represented in Fig. 454, the
general condition takes the form,
[123]
FIG. 454.
If 0= fc, as in Figs. 455 and 456, the equation to the curve
to which the beam must be formed in elevation is
[124]
§ 51. — DEFLECTION OF BEAMS.
Again, if u = h,
339
K X
& = 7' [125J
showing that the beam should be triangular on plan when of
constant height.
FIGS. 455 AND 456.
FIGS. 457 AND 458.
A
The deflection of the point of loading can be found from the
equation to the circle, viz. (Fig. 459),
P = 2ps-s*.
But since the amount of bending is very FIG. 459.
small, s2 can be neglected and I can be
regarded as the length of the beam.
Hence,
• =~ [126]
2p
Substituting for p from equation 67
<2S [127]
If the section is symmetrical with
reference to the neutral axis, as in the case
represented in Fig. 454, so that 2 w = h,
S
and further if S be written for ^ (see equation I., page 282),
[128]
z 2
340 BRIDGES AND EOOFS.
In a beam supported at both ends and loaded in the centre, the deflection
curve will evidently be a circle, if each half of the beam be of the form shown
either in Figs. 455 and 456, or in Figs. 457 and 458, the thin end being placed
at the abutment. In the calculations connected with Fig. 401, it was assumed
that the deflection curve was a circle. This assumption therefore requires that
the beam should have either of the forms indicated above when its cross- section
is rectangular.
Since -=-= is constant it can be replaced by its value at the
point of fixing, viz. =?- , and equation 121 then becomes
whence, from equation 126,
KP
~2EI/
Hence, in a beam of the form shown in Figs. 457 and 458,
in which the curvature is constant, the deflection is 1 • 5 times
greater than that of a prismatic beam of the same depth
(see equation 79).
§ 52. — KESISTANCE OF LONG COLUMNS TO BENDING
AND BUCKLING.
If the straight prismatic beam shown in Fig. 460 be subject
to forces K K producing compression, the points of application
being at the centre of gravity of the end sections, and the
forces acting in the direction of the length of the beam, the
stress will be uniformly distributed over the area of every
cross-section of the beam, and if F is the area of the cross-section
the stress per unit of area will be
S, = |. [129]
Let it be supposed that by any means whatever the column
is bent until the height of arc of the curve formed is /
(Fig. 461), and further that the bent column is acted upon by
§ 52. — KESISTANCE OP LONG COLUMNS.
341
the two forces K, and that these forces are of themselves able
to maintain the column in this bent condition.
The stress at any cross-section can in this case be considered
as made up of two parts : the first is the uniformly distributed
compression Sx (from equation 129), and the second is the
A
K
FIGS. 460 AND 461.
C
K
bending stress. The fibres on the concave side will evidently
be compressed by the bending, and since the maximum bending
moment
M = K/ [130]
occurs at C, the greatest compression will also be at this point.
Substituting for M its value from equation 52, the greatest
intensity of compression S2 due to the bending, is
[131]
And the greatest compression per unit of area in the column
is evidently
8 = 8, + S2. [132]
From the above it follows that there are two different ways
in which a column can resist the action of a compressive force
K. The first is illustrated by Fig. 460, and in this case the
compression is uniformly distributed over the section and equal
to Si per unit of area. The second way is shown in Fig. 461,
and in this case the maximum compression attains the greater
value Sx + S2 per unit of area. In determining the
342 BRIDGES AND ROOFS.
section F of the column, it becomes a question whether the
column resists the force K in the first or in the second manner.
If it be in the first way F can be found directly from equation
129, by substituting for Si the safe resistance to crushing ; but
if it be in the second way the section must be such that the
sum Si + S2 is equal to the safe resistance to crushing.
In the case of long thin columns the smallest accidental
curvature is sufficient to enable the compressive force to
produce bending. Therefore in such a case the second mode
of resistance obtains (unless the column be so supported along
its length as to preclude the possibility of its bending), and
therefore to the direct compression B! per unit of area must be
added the compression S2 due to bending, giving to the lever
arm /its greatest possible value.
Now a value can be assigned to f of which it may be said
with certainty that if the column is sufficiently strong to resist
safely the force K, the height of the arc to which the column
may be bent will not reach /. For let it be supposed that the
curvature at every point of the column is the same ; that is, the
column will be bent into the arc of a circle. Now if the bend-
ing be so great that the greatest compression due to it alone is
equal to the elastic limit, it is evident that by adding the direct
compression the elastic limit will be overstepped. If the force
K could produce such a state of things the column must be
considered too weak. If therefore the corresponding value of
/ be substituted in equation 131 the value of S2 found will
obviously be greater than the compression due to the bending
produced by the force K would be in a column of sufficient
strength, and consequently the section F found by using this
value of S2 will be rather greater than required.
If in equation I. (§ 44) s means the elastic limit of com-
pression, 8 will be the shortening per unit of length at the
elastic limit, and the sought value of/ can therefore be found
Q1
from equation 127 by substituting & for ^ and / for s, thus :
§ 52.— KESISTANCE OF LONG COLUMNS. 343
Then if in equation 131 / be replaced by this value, and K
by the value obtained from equation 129,
5Z2F
8,= — .^. [134]
Whence the greatest compression in the column is (equation
132)
[135]
or putting the whole length of the column 2 I = L and repre-
senting the ratio — by n,
1
[136]
In this equation Sx is the uniform compression per unit of
area that would occur were it not for the bending, and n is the
number of times the greatest intensity of compression S may
be made to exceed Si by the bending — or in other words, S is
the safe stress per unit of area of the cross-section that can be
applied- to the long column under consideration.
The number n therefore gives the fraction of the safe resistance
to crushing that can be applied to a long column, so that it may
le safe as regards bending.
Substituting the value of 8 at the elastic limit for various
materials, the following formulae are obtained :
Cast iron 8 =
Wrought iron S =
Wood 5 =
In these equations I is the moment of inertia of the section
of the column about that axis through the centre of gravity of
the section, which is perpendicular to the plane of bending,
that is, at right angles to the direction in which bending takes
place easiest. It is in fact the minimum bending moment of the
section.
15
loood
15
n
n
n
1 FL*
[137]
[138]
[139]
1 5333 I '
1 FL*
20000
1-8
1 10666 I
1 FL*
1000
1 ' 4444 I •
344 BRIDGES AND ROOFS.
If, for instance, the section is rectangular, H the greater and B the smaller
side, I = - • ; if, however, H is the smaller of the two dimensions, I = •
12 -l —
In the last case, therefore,
F BH 12
and the above formulae become
Cast iron, n = 1 + 0-00225 ; [140]
Wrought iron, n = 1 + 0' 001 125 g; [141]
Wood, n = 1 + 0-0027 (g)'; [142]
from which the following table has been constructed :
^ = 10 20 30 40 50
H
Oast iron, n = 1-225 1-9 3-025 4'6 6-625
Wrought iron, n = 1-1125 1-45 2-0125 2-8 3-8125
Wood, n = 1-27 2'08 3'43 5'32 7'75
Thus if 6 kilos, per square millimetre be taken as the safe resistance to crushing
of wrought iron, then the safe compression per square millimetre on a long column
of the same material of rectangular section whose length is twenty times its least
dimension H, is
S,=5 = JL = 4.
and if for instance H = 10 millimetres, and B = 40 millimetres, the greatest
safe load that could be placed on it is,
K = FS! = 400 x 4-14 = 1656 kilos.
If the section is a hollow rectangle as in Fig. 427,
F BH-6ft
and the formula for wrought iron becomes,
[H3]
Thus for a rectangular wrought iron tube the thickness of which is
of the exterior dimensions,
»-.! +0-00062 (-); [144]
§ 52.— RESISTANCE OF LONG COLUMNS.
345
and if at the same time H is -g^th of the length,
n = 1-248.
Such a tube could therefore only be loaded with St =
= 4'8 kilos- Per
square millimetre.
For a circular tube of exterior diameter D and interior diameter d,
- (D2 - d2)
16
D2 + e?2'
and equation 138 becomes in the case of wrought iron,
n = 1 + 0-0015
Let - = 0-9, ^ = 20, then n = 1-3315 and St = 4-5 kilos.
If, however, d = 0, and D - 20, it will be found that n = 1'6, and S, =
3 '75 kilos.
If it be supposed that one half of the column of Fig. 461 is
firmly fixed (encastre), the state of stress of the other half
— and consequently the greatest compression at Fm 462
C — will not be altered thereby. Therefore the
general equation 136 is also true in the case
represented in Fig. 462.* And writing 2 1
instead of L,
If a round wrought-iron column be loaded in this manner,
and the length B C = Us twenty times the diameter, it will
/»
be found that n = 3 • 4, and Sj = — — = 1 • 76 kilos. For instance,
3*4
if the area of the section is 100 sq. millim., the safe load is
K = 176 kilos. If the same column were placed in the con-
ditions of Fig. 461, it could carry safely a load K = 375 kilos.
It appears from the above that the sectional
areas obtained in §§ 41 and 42 for the com-
pression braces, can only be adopted if by their construction
the value of n for them differs very little from 1.
* This is the case of a long column having one end fixed and the other free
but not "guided." If the free end were guided the strength of the column
would be materially increased. This subject will be found very fully treated
In * Der Constructeur,' by Prof. Rouleaux. — TEAKS.
346 BRIDGES AND ROOFS.
It also appears that, as a rule, a greater section is required
to resist compression than to resist the same amount of tension,
and that the greater the ratio of the length to the least
dimension of the cross-section, the greater must be the section
to resist compression, but not proportionately.
And lastly, that if the length remains the same and also
the form of section, the less the load to be borne, the greater,
proportionately, will the section be.
From this last remark it follows that, if possible, bars in
compression should not be split up into several isolated parts.
In this respect, therefore, the simplest forms of construction
are the best ; for instance, braced girders with a single triangu-
lation are to be preferred to trellis girders. Trellis girders
have, however, this advantage (already pointed out in § 43),
that owing to the greater number of points of support ob-
tained, there is a great saving of material in the longitudinal
girders. When deciding on the depth of a girder, it should be
remembered that the resistance of the compression braces
diminishes rapidly as the depth increases.
Further, it appears that the design of the structure should
be such that the compression braces are as short as possible.
For this reason, girders in which the verticals are in compres-
sion and the diagonals in tension, are generally to be preferred
to other forms.
No general rule can, however, be framed by means of which
it can be decided what form, what number of triangulations,
and what depth a girder should have in order that the bridge
may contain the least quantity of material.
( 347 )
SIXTEENTH CHAPTER
§ 53. — COMPOUND LATTICE AND SUSPENSION BRIDGE —
SPAN 60 METRES.
Determination of tlie best ratio between the depth of the girders
and the height of the arc of the suspension chains.
THE general design of the bridge is shown in Fig. 463.
In this figure, however, the points A and B are represented as
fixed points, so as to remove at first from the calculations
the consideration of the back-stays. The arrangement of the
bridge* in this respect is shown in Fig. 491.* The material is
wrought iron.
FIG. 463.
A I IB
The two lattice girders are continuous between the abut-
ments, and the section of the booms is the same throughout.
Each girder is connected to a suspension chain, by means of
vertical rods attached to each top joint of the girdet and to a
point in the chain. The suspension chains are in the form of
a parabola, or, more strictly speaking, of a polygon inscribed
in a parabola.
The total load on the bridge is 0'575 kilo, per millimetre
run, consisting of a permanent load p = 0*375 kilo., and a
moving load m = 0*2 kilo, per millimetre run.
The first step is to decide what the proportion between the
depth of the girder and the height of arc of the chain should
* This bridge was constructed by a German firm to be sent out to the Brazils,
and the author, at the request of the manufacturer, furnished the following
calculations.
348 BRIDGES AND EOOFS.
be, in order that the quantity of material in the structure may
be a minimum. Although the greatest stress will be reached
in certain parts of the girder when the moving load is unevenly
distributed, yet for the present inquiry this can be ignored, and
the moving load considered as covering the bridge.
When the temperature varies, the length of the suspension
chains will alter, and consequently the distribution of the load
between the two systems will also alter. The changes of tem-
perature must therefore be taken into account.
If S is the greatest elongation per unit of length produced
in the chains by the load alone, the deflection of their lowest
points will be, according to § 45,
+ f^2)- C14?]
In the present case, as will be seen, the ratio — is small ;
I
therefore
iz
ff = f 5 — (approximately). [148]
h
And if, further, A is the elongation per unit of length due to
the greatest increase of temperature (Fig. 464), the deflection
due to both causes is
A)!-.
FIG. 464.
The deflection in the centre of a prismatic beam, subject
to a uniformly distributed load, is from equation 99, § 51,
[150]
E^
in which equation q is the load per unit of length, Ix the
moment of inertia of the section (supposed symmetrical about
the neutral axis), and Ex is the modulus of elasticity of the
material. (Fig. 465.)
§ 53. — COMPOUND LATTICE AND SUSPENSION BEIDGE. 349
The greatest bending stress, Sl} in the beam can be found
from equation 64, § 50,
Substituting the value of 1^ obtained from this equation in
equation 150,
'•=*= C152]
FIG. 465.
I ........ .
sr
and writing Sj, the greatest elongation per unit of length occur-
ring in the beam instead of -^ ,
According to § 49, the economical ratio -i is found by
h
equating the two deflections, thus : —
| (5 + A) - = £ 8l - • [154]
Whence
. T = |(fTlj- C155]
If it be supposed that the bridge is erected at the mean
temperature, A will depend on the difference between the
greatest temperature that occurs and the mean temperature.
Since the chains are of wrought iron, if this difference of tem-
perature is 41° C., A = ^Vrr J but if the difference is 20° -5 0.,
A = ^Vir- Further^ the modulus of elasticity of wrought
iron is 20,000 kilos, per square millimetre, and the safe stress
can be taken at from 5 to 10 kilos, per square millimetre ;
therefore S and Si vary between 73 £ ^ and WcrW- Assuming
350
BEIDGES AND ROOFS.
that the height of the arc of the chains h = 4 metres, the
following table can be formed from equation 155 : —
5
(Chain).
2OOOO
¥oooo
2"ootTo
(Girder).
2OOOO
20000
"aOlTOO'
2OOOT)
i
(Temperature-elongation).
metres.
2-424
1-667
2-051
1-48
71
234
1-48
1-111
In using this table it must be remembered that, although
both the chains and the girders are made of wrought iron, yet,
for the following reasons, the value of S1 should be taken
smaller than 8. 1. ^ in the girders depends on the resistance
to compression, whereas in the chains 8 depends only on the
resistance to tension. 2. Because even when the bridge is fully
loaded and the temperature is at its highest, the maximum
stresses in the girders are not reached ; for they are also subject
to the effect of the horizontal pressure of the wind, and of the
unequal distribution of the moving load. 3. Because the points
of attachment of the chains, although considered fixed, really
approach each other slightly owing to the extension of the land
ties, and this has the same effect as if the elasticity of the
chains were increased.
It would appear, therefore, that if h = 4 metres, then hi =
1 • 5 metre is a good value for the depth of the girders, this
value being the arithmetic mean of those in last column, when
the two first are omitted.
§ 54. — CALCULATION OF THE STRESSES PRODUCED BY
CHANGES OF TEMPERATURE.
When the temperature diminishes, the chains shorten and
their lowest points consequently rise. This induces stresses in
the chains and girders, which are to be added to those produced
§ 54. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 351
by the loads ; but they will be calculated separately by con-
sidering that the structure is totally unloaded.
The ends, AL and B1? of the girders can be considered
as capable of resisting an upward as well as a downward
reaction (see § 66). The shortening of the chains due to a
diminution of temperature will be accompanied by an upward
bending of the girders. (Fig. 466.) The resistance thus pro-
FIG. 466.
duced will have the same effect on the chains as a uniformly
distributed load — say, k per unit of length ; and the chains will
be prevented from rising to the same height that they other-
wise would do. The actual amount s the lowest points of the
chains rise, is therefore equal to the difference between the up-
ward deflection si9 produced by the diminution of temperature,
and the deflection s2, due to the load k per unit of length.
If the shortening per unit of length due to a decrease in
temperature be represented by A,
s, = f A ^ • [156]
The horizontal tension in the chains, due to the load k
per unit of length, is, according to § 8,
TT
H=
[157]
If, therefore, E is the modulus of elasticity of the material
of the chains, and F the sectional area of both chains at
their lowest point, then 8 the elongation due to h is
H klz
352 BRIDGES AND ROOFS.
whence the deflection of the lowest points of chains is
72 I 74
and the actual upward deflection is therefore
72 I 74
[160]
The upward deflection of the girders must be equal to s, and
since it is produced by an upward uniform load Jc per unit of
length,
k i*
Equating the two values found for s,
or
2
If F! is the sum of the effective sectional areas of the four
booms of the two lattice girders (Fig. 467),
FIG. 467.
And substituting in the above equation
/f(1 + T-|-ir-5
[164]
[165]
§ 55. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 358
For example, let A = ^Vir, E = ET = 20,000 kilos., h =
4000 mm., F = 7500 sq. mm., F! = 15,000 sq. mm., I =
30,000 mm.; then k = 0' 074,896 kilo, per millimetre, or
nearly 75 kilos, per metre run.
If A were negative, Jc would also be negative ; and thus it
is seen that the action of an increase of temperature is to unload
the chains and load the girders by the amount Jc per unit of
length. Thus, if the bridge be constructed at the mean tem-
perature, the girders will -be unloaded by the amount of 75 kilos,
per metre run, and the chains loaded by the same amount when
the temperature is 41° C. below the mean ; and when the tem-
perature is 41° C. above the mean, exactly the reverse will take
place. Therefore the load Jc, which can be called a tempera-
ture load, is applied to the chains when the temperature
decreases, and to the girders when the temperature increases
and produces stresses which must be added to those due to the
ordinary loading. In the chains this temperature stress is
by equation 157
I 72
S = — —- = 1 • 1234 kilos, per sq. millimetre ; [166]
2i s h
and for the booms of the girders, from equations 151 and 164,
k i*
Si = - — = 2-996 kilos, per sq. millimetre. [167]
§ 55. — CALCULATION OF THE STRESSES PRODUCED BY THE
PERMANENT LOAD.
Let the uniformly distributed load on the bridge be p per
unit of length, and let n p be the portion carried by the chains :
(1 — n) p will therefore be the load on the girders (Figs. 468,
469, and 470). Now, similarly to equation 159, the deflection
of the lowest points of the chains is
and the deflection of the girders at the centre is (equations
161 and 164)
[169]
E, F, h\
2 A
354
BRIDGES AND EOOFS.
Therefore, equating these two values of s,
, npl*
or
EFA2
6
r 20 ' E ' F ' A2
FIGS. 468, 469, AND 470.
[170]
[171]
I ! 1 I 4 I 1 I 1 4 4 1 1 I 1 1
As before, let § = 1, ~ = 2, ^ = ^; then
rj r /z, 4 '
oVo
= 0-887656.
[172]
If, therefore, the dead load per metre run is 375 kilos, (or
p = 0*375 per millimetre run), it is distributed as follows:
On the chains,
np = 0-33287 kilo, per millimetre run ;
= 332 • 87 kilos, per metre run. [173]
On the girders,
(1 — ri)p = 0-04213 kilo, per millimetre run ;
= 42-13 kilos, per metre run. [174]
§ 56. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 355
And the stresses produced in the chains and in the booms of
the girders are respectively,
S =-
2 Jb h
= 4-993 kilos, per sq. millimetre ;
[175]
g, = C1 n]pl* = i -685 kilo, per sq. millimetre. [176]
§ 56. — CALCULATION OF THE STRESSES PRODUCED BY A
MOVING LOAD.
It was shown in § 8 that if the curve of equilibrium of a
chain is a parabola, the load must be equally distributed over
the span or horizontal projection. Now, in the present case
the deflection of the girders is but small ; it may therefore be
assumed that the chains retain always their parabolic form. It
follows that the chains must in all cases be uniformly loaded,
even if the load be concentrated at one or more points on the
girders.
In Fig. 471 let the elements dz of the span, at equal dis-
tances z from the centre, be loaded with q per unit of length ;
then q dz will be the load on each element. Further, let q dn
FIG. 471.
qd
represent the uniformly distributed load that the chains have
in consequence to bear (Fig. 472). dn can be found, as before,
by equating the deflections of the chains and of the girders.
According to equation 159, the deflection of the chains is
The deflection of the girders is equal to the deflection that the
two loads q dz, would produce of themselves (Fig. 473) minus
2 A 2
356
BRIDGES AND EOOFS.
the deflection due to the upward uniformly distributed load
q dn per unit of length (Fig. 474).
From equation 97 the first part is
[178]
and from equation 99 the second part is
K qdn. I4
Hence the actual deflection is
[179]
qdn .1
! [180]
and equating the two values found for s,
*3) - & ^^ ; [181]
or,
[182]
BFA»/
§ 56. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 357
Substituting from equation 164
and integrating the right-hand side between the limits zl and
z2 , and the left-hand side between the limits n-i and n2 ,
[183]
or,
2 I3 (^ - «!> - I (*23 - *!3) +
[184]
Thus, when the bridge is loaded as shown in Fig. 475, the
uniformly distributed load the chains have to carry is (n2 - %) q
per unit of length of the span.
q (>2 - z
(Putting 0! = 0, z2 = Z, and % = 0, w2 = n, so that the uni
formly distributed load covers the bridge,
E ' F
or the same value that was obtained from equation 172.)
Evidently each of the loads q (^2 — ^1) has the same ef-
fect on the chains; each of them, therefore, produces a load
of (
Per um^ length of the span. This is shown in
Figs'. 476 and 477. In this case the deflection of the
girders is compounded of the downward deflection due to the
358 BRIDGES AND ROOFS.
load q (z2 — Zj), minus the upward deflection due to the uni-
formly distributed upward load (~-^~ - ) 2 produced by the
chains. (Figs. 478 and 479.)
FIG. 476.
<•
I)
niiiiiiiiiiiiiiiiiniiiiiiiiiiiiiiuii
T^"
M
V
q (z.2 -
FIG. 477.
FIG. 478.
FIG. 479.
Therefore the bending moment at a point distant x to
§ 57. — COMPOUND LATTICE AND SUSPENSION BKIDGE. 359
the left of the centre, is the difference between the bending
moment due to the load q (z2 - zj Fig. 478,
M1=D(/-ar); [186]
and the bending moment due to the loading shown in Fig. 479,
The resulting bending moment is therefore
2. [188]
Similarly, it will be seen that the shearing force at the
same point is
V = V, - V2 = D - ^ qx. [189]
In both these equations, D is to be replaced by its value
derived from Fig. 478, viz. :
[190]
It will be observed that the direction of M and V, when the
bridge is fully loaded, has been taken as the positive direction.
§ 57. — DETERMINATION OF THE WORST CONDITION OF
LOADING FOR THE GIRDERS.
To find what conditions of loading produce the greatest
bending moment M, and the greatest shearing force V,
respectively at any given section of the girders, the points must
first be found where a load must be placed so that M = 0 and
V = 0 respectively ; for it is evident that these points separate
the loads that produce positive from those that produce negative
values of M and Y respectively.
The values of M and V found in equations 188 and 189
can be regarded as the sum of the increments due to each element
360
BRIDGES AND ROOFS.
of the load q (z2 — zj. Let d M and d V represent the increments
produced hy a load q d z situated at a distance z from the centre
(Fig. 480). Then proceeding as in the former case (equations
188 and 189), the following equations are obtained :
[192]
FIG. 480.
which become, when d n is replaced by its value obtained from
equation 182
21
[193]
Evidently the position of the load which produces no
bending moment at the section under consideration, can be
found by putting d M = 0. Thus (writing u instead of z as a
distinction) :
BFAV
2 (/ + ar)
= 2 /2 + 2 / u — v?.
[195]
substituting
F h 2
= - - and solving,
[196]
2(1 + ar)~
§57. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 361
Only one root of this quadratic equation applies to the
question under consideration ; for the other root would make
-y > 1, which is evidently inadmissible in the present case.
L
The same remark refers to equation 197 below.
Proceeding in the same manner for d V, and representing
by v the value of z, obtained by putting d V = 0
or,
ETTl 7 "1 CT
i -C i /ii 1 ' 0
As before, put ^ = l,^r = 2, TT~ = ~T~ ; then the equa-
tions for u and v become
/TCQ ^ I /IT 7
[199]
Both these equations refer to the case when the load is
placed to the right of the section under consideration. If %
and Vi are the distances to the left of the centre of the points
where a load must be placed to produce no bending moment
or shearing stress respectively at a section situated between
them and the centre, it will be found that
- 768 x
'
362
BRIDGES AND KOOFS.
The following table has been computed from the above four
equations : —
X
1
u
T
T'
V
T
Vl
- 1
- 0-2616
- 0-2616
- 0-75
(
- 0-18
- 0-0594
- 0-7041
0
- 0-5
- 0:0594
+ 0-5715
- 0-4694
. .
+ 1
- 0-4082
,
0
- 0-25
+ 0-1358
- 0-24
,
+ 0-1464
- 0-2
+ 0-1919
- 0-125
+ 0-29535
- 0-0612
+ 1
+ 0-4118
0
+ 0-5715
+ 0-5715
+ 0-0612
+ 0-4118
+ 1
+ 0-125
+ 0-29535
-1- 0-2
-f 0-1919
+ 0-24
+ 0-1464
+ 0-25
+ 0-1358
+ 0-4082
0
+ 0-4694
+ 1
+ 0-5
- 0-0594
.
+ 0-5715
+ 0-7041
,
0
+ 0-75
- 0-18
.
— 0-0594
+ 1
- 0-2616
•
- 0-2616
§ 58. — CALCULATION OF THE STRESSES PRODUCED IN THE BOOMS
OF THE GIRDERS BY THE MOVING LOAD.
The above table shows that for all values of x between
4- 0* 0612 Z and — 0'0612Z there are two zero-points for the
bending moment (calling, for shortness, the point where a load
must be placed to produce no bending moment, the zero-point),
and that for all other values of x there is only one zero-point.
When x = 0, that is, for the centre of the girders, the two
zero-points lie at equal distances, u = % = 0*5715 /, to the
right and to the left of the centre. These two zero-points
separate those loads that produce positive from those that
produce negative bending moments at the centre of the
girders. For instance, if the moving load were distributed
as shown in Fig. 481, the positive bending moment at the
centre would be at its maximum, and its value can be found
OKTHK
UNIVERSITY
§ 58. — COMPOUND LATTICE AND SUSPENSION BRI
as follows: — The first step is to find the part of the load carried
by the chains. This can be done by means of equation 183,
and in the present case the limits of the integration evidently
are % = 0 and n2 = n ; zl = 0 and z2 =0 • 5715 I It will then
be found that
n = 0-6983. [203]
FIG. 481.
0-5715 1 0-5715 f
Now, since the moving load is m = 0 • 2 kilo, per millimetre
run, in the present case the uniformly distributed load on the
chains will be
nm = 0-6983 x 0-2 = 0-1396 kilo, per millimetre run of the span.
This is also the upward uniformly distributed load on the
girders.
If Mj and M2 denote the bending moments at the centre of
the girders due to the loading shown in Figs. 482 and 483
respectively, the resultant bending moment at the centre is
M = M! - M2 [204]
and
Mx = mzl —
nm I2
[205]
[206]
Further Sw, the stress per unit of area due to the bending moment
M, can be found from the equation,
or,
2M
[207]
364 BRIDGES AND EOOFS.
and by substituting the above numerical values :
Sw = °'0237*2 = 0-948 kilo, per sq. millimetre. [208]
FjA,
From equation 176 it will be found that the same stress
would be produced by a uniform load on the girders of 23 * 7 kilos,
per metre run ; whereas when the moving load covers the bridge
the part supported by the girders is only
(1 — -887656) x 200 = 22*5 kilos, per metre run.
mz FIG. 482. mz
FIG. 483.
M2
As a further example, let it be required to find the maxi-
mum bending moment at the section whose distance is
aj=0'0612Z to the left of the centre. As will appear from
FIG. 484.
I 0-4118J
the table, p. 362, the moving load will in this case cover the
shaded portion in Fig. 484.
§ 58. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 365
Now, as explained at p. 357, the value of n for the part of
the load from the centre to 0*4118 I is half the value of n
when the load extends to 0*4118 I on each side of the centre.
The limits in equation 183 can therefore be taken as % = 0,
n2 = 2 ri, 0j = 0, and z2 = 0'4118 Z; and substituting these
values in equation 184, it will be found that
n' - -270172.
The same applies to the part of the load from the centre to the
left abutment, and from equation 185,
n" = 0-443828 ;
and hence the value of n for the distribution of the moving
load shown in Fig. 484 is
n-n' + n" = 0'7U"
Then, proceeding as in the previous case, it will be found
that the stress in the booms, at the section under consideration,
is:
Sm = ° ""7" =0-98 kilo, per sq. millimetre. [209]
1-344 1-344
The value of Sm for all values of x can be similarly ob-
tained, and the result is expressed graphically in Fig. 485,
which shows that the greatest bending moment occurs when
366 BRIDGES AND ROOFS.
§ 59. — CALCULATION OF THE STRESSES PRODUCED BY THE
PERMANENT AND TEMPERATURE LOADS IN THE BOOMS
OF THE GIRDERS.
It now remains to find the stresses in the booms of the
girders produced by the permanent load and the temperature
load.
The part of the permanent load carried by the girders
will evidently be uniformly distributed over the span. There-
fore, if M0 is the bending moment at the centre, the bending
moment Mx at a distance x from the centre can be found from
the equation (see equation 65) :
M, = M0 (l - £) . [210]
This is the equation to a parabola, as shown in Fig. 486.
FIG. 486.
x
— -x-
The stress Sp evidently also follows the same law as the
corresponding bending moment Mx; and since the value of
Sp at the centre is 1'685 kilo, per square millimetre (see
equation 176),
S» = 1-685 l-. [211]
Further, the stress St due to the temperature load, can be
similarly obtained, since this load is also uniformly distributed
over the span. And by equation 167 the value of this stress
at the centre is 2 • 996 kilos., therefore,
S, = 2-996 (l-£V [212]
§ 60. — COMPOUND LATTICE AND SUSPENSION BKIDGE. 367
The following table has been constructed from equations
211 and 212, and from the result obtained in § 58 : —
st
resses in the Booms
of Girders produced
by
Values of
X
T
Moving Load.
Permanent Load.
Temperature Load.
Total Stress.
Sm
SP
St
Sm + Sp + St
0
0-95
1-685
2-996
5'63
0-0612
0-98
1-68
2-98
5-64
0-125
1-07
•66
2-95
5-68
0-2
1-15
•62
2-88
5-64
0-24
1-20
•59
2-82
5-61
0-25
1-22
•58
2-81
5-61
0-408
1-34
•40
2-48
5-24
0-5
1-344
•26
2-25
4-85
0-6
1-28
1-08
1-92
4-28
0-75
1-0
0-74
1-31
3-05
1
0
0
0
0
This table shows that the maximum stress occurs when
x = '1257, and that it is then equal to 5 '68 kilos, per square
millimetre. It must, however, be observed that these stresses
will be further increased by the pressure of the wind, and also
by the extension of the land ties. The effect of both these
causes will be treated of in § 62 and § 63.
§ 60. — CALCULATION OF THE SHEARING STRESS PRODUCED BY
THE MOVING LOAD.
The table at the end of § 57 shows that for every value
of x there is only one zero-point. This zero-point forms a
loading boundary ; there is, however, a second loading boun-
dary, which is situated at the section under consideration itself,
for a load can produce no shearing force at the section imme-
diately below it.
Thus, for instance, Fig. 487 shows thje arrangement of the
load that gives the greatest shearing force VM at the section
whose distance is x = 0 * 75 I to the left of the centre. Now,
from equation 189,
Vra = D — n m x ;
368
BRIDGES AND ROOFS.
and it will be found by means of equation 184 that,
2n = 0-7376.
«
Therefore (see equation 190) :
(0-75?- 0-0594 O/ 0-75 / + ()•• 0594^
Vm~ ~2~ v~ ~zi ;
- 0-3688 x 0-2 x 0-75 = 1250 8 kilos.
0-75?
If the value of VM, for a series of values of x, be calculated,
and the results plotted, the curve shown in Fig. 448 will be
obtained. This curve shows that the vertical shearing force
VM due to the moving load, has one maximum and one minimum
2031-6
2031-6
value on each side of the centre and a maximum value at the
centre. The greater maximum (strictly speaking, it is not a
maximum) occurs at the abutments, and is 2031 • 6 kilos. ; the
smaller maximum is at the centre, and its value is 1500 kilos.
The minimum is at a distance J I from the centre, and is equal
to 1103-9 kilos.
§ 61. — COMPOUND LATTICE AND SUSPENSION BKIDGE. 369
§ 61. — CALCULATION OF THE SHEARING STRESSES DUE TO THE
PERMANENT AND TEMPERATURE LOADS, AND OF THE
MAXIMUM STRESSES IN THE BRACES.
The permanent load on the bridge produces a uniformly
distributed load of 42 • 13 kilos, per metre run on the girders
(see equation 174). The shearing force at any section of the
bridge can therefore be found by means of equation 66A, and
by substituting the various values it will be found that
VP = 1263-9^.
[213]
The temperature load being also uniformly distributed, the
shearing stress Y«atany section can be found from the same
equation. This load amounts to 75 kilos, per metre run.
Hence :
V, = 2250^. [214]
The results obtained are embodied in the following table : —
Vertical Shearing Force produced by
X
I
Moving Load.
Permanent Load.
Temperature Load.
Total.
Vm
VP
Vt
Vm + Vp + Vt
0
1500
0
0
1500
0-25
1419-5
316-0
562-5
2298-0
0-5
1103-9
631-9
1125-0
2860-8
0-75
1250-8
947'9
1687-5
3886-2
1-0
2031-6
1263-9
2250
5545-5
If the girder is divided into square bays by vertical braces,
and if in each bay two diagonals are placed which can only
resist tension, the verticals will be in compression. On refer-
ring to the above table, it will be seen that the compression in
the vertical over the abutments in each girder is 55*5'5 =
2772 • 75 kilos., and that the stress in the succeeding verticals
gradually decreases as far as to the centre, where its value is
2 B
370
BRIDGES AND ROOFS.
1500
= 750 kilos. The tension in the diagonals is
Zi
10 60 '7 kilos, in those of the central bay, and it increases
5545-5 x
towards the abutments, where it is
kilos.
2
= 3921-3
§ 62. — CALCULATION OF THE STRESSES IN THE WIND-STAYS
AND WIND-BRACES.
A parabolic form can be given with advantage to the wind-
stays, connecting them by horizontal rods to the lower joints
of the girder, as shown in Fig. 489. The lower booms are
braced together, and thus a combination of a girder with a sus-
pension chain, similar to the main structure, is obtained.
FIG. 489.
As before, the first step is to find the economical height of
arc of the wind-stay, and this can be done by writing / instead
of h, and /x = 2 '25 metres (the breadth of the bridge) instead
of lii in equation 155, thus :
f(5
[215]
where B is the safe extension of the wind-stays, A the exten-
sion due to temperature, and ^ the safe extension in the girder
per unit of length ; and these values are to be taken indepen-
dently of the extensions produced by the vertical loads. The
following table has been computed from the above equation by
giving different values to Si, &, and A.
§ 62. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 371
81
5
A
/
20660
aoooo
4000
3-7125
20OOO
2OHOO
2OOO
5-4
aoooTS"
"20oTTo
Tooo
4-3875
2"0000
TOOOO
4000
4-725
"20000
20060
2oVo
6-4125
2OOOO
"JToFo"^
4000
6-075
2000TT
120000
20OO
8-1
In choosing a value for / from this table, it should be con-
sidered whether it is probable, or even possible, that all the
unfavourable circumstances can occur simultaneously. If they
do occur simultaneously, then one of the larger values of/
must be taken.
Now, it is most unlikely that a high wind will be blowing
when the temperature is at its maximum, and further, a rise
of temperature of 20° • 5 C. (A = ^oW) above the mean is really
an ample allowance. It would therefore appear that / can
be made 4 metres, although it must be admitted that a greater
height of arc would be preferable, if the breadth of the abut-
ments will allow of it. It will be observed, however, that a
greater height of arc can be obtained, as indicated in Fig. 490.
FIG. 490.
:K
•^
[NOTE. — It is thought that this arrangement of the wind-stays is open to
the following objections. The bars connecting the centre joints of the lower
booms to the chains, are struts, and would therefore require a larger scantling
than the corresponding ties in the arrangement shown in Fig. 489. Further,
these struts are in unstable equilibrium — that is, with the slightest displace-
ment of the end attached to the chain a tendency to turn about the other end
would arise. These struts would therefore have to be braced up to the side of
the main girders.]
2 B 2
372 BRIDGES AND EOOFS.
The distribution of the wind-pressure between the chain
and the horizontal girder can be found from equation 171, by
writing / instead of Ti, /j instead of ht , <f> the sectional area of
the wind-stay instead of F, and \ F! instead of Fx (the reason
of this last alteration is that only the two bottom booms are
connected to form the horizontal girder), thus :
" E ' <j> ' f*
[216]
If E = E!, F! = 15000 sq. millimetres, <j> = 1250 sq. milli-
metres,/ = 4000 millimetres,/! = 2250 millimetres, it will be
found that
n = 0-5393. [217]
Let it be assumed that the wind-pressure is w = 0 • 2 kilo, per
millimetre run ; then the part carried by the wind-stay will be
0 • 5393 X 0 • 2 = 0 • 10786 kilo, per millimetre run ; [218]
and by the horizontal girder,
(1 - 0-5393) X 0-2 = 0-09214 kilo, per millimetre ran. [219]
The stress at the centre of the wind-stay can be found by
equation 175, thus :
g _ 0- 10786 lz _ 9<7 k.log^ per ^ jnfliijnetre . r_220]
and, from equation 176, the stress produced in the lower booms
of the lattice girders is
= 4-9 kilos, per sq. millimetre. [221]
The stresses produced by temperature in the wind- stays and
in the horizontal girder can be found in the manner explained
in § 54. Thus, if the range of temperature on each side of the
mean is 41° C., it will be found from equation 165 that
k = 0 • 0512 kilo, per millimetre ran ; [222]
§ 63. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 373
that is to say, when the temperature is 41° C. above the mean,
the girder will have a temperature load of 51 • 2 kilos, per metre
run, and when the temperature is 41° C. below the mean the
wind-stay will have a temperature load of the same amount
(supposing that there is no temperature load at the mean tem-
perature).
These loads produce the following stresses :
In the wind-stay,
S' = 4 • 608 kilos, per sq. millimetre ; [223]
in the lower booms of the girders,
S\ = 2 • 73 kilos, per sq. millimetre. [224]
The maximum horizontal load on the girder, due to the
wind and change of temperature conjointly, is
(1 _ n) w + k = 0 • 14334 kilo, per millimetre. [225]
The maximum shearing stress at the abutments is therefore
[(1 - n) w + K] I = 4300 kilos. [226]
It will be seen from the table at page 369 that the maximum
shearing stress at the abutments on each of the vertical girders
5545*5
is — - — = 2773 kilos. If, therefore, the braces in the hori-
zontal girder are arranged in the same manner as those in the
vertical girders, the stresses in them will be to those in the
vertical girders as 4300 : 2773, and their sections will have to
be made larger accordingly.
§ 63. — INFLUENCE OF THE EXTENSION OF THE BACK-STAYS.
In the previous calculations, the points of attachment of
the chains A and B (Fig. 491) were considered as absolutely
fixed. If, however, A and B are placed at the top of vertical
columns A A! and B Bx capable of free rotation about their
lower points Ax and B15 back-stays A C and B D must be intro-
374
BRIDGES AND ROOFS.
duced to keep these columns in position. The extension of
these stays will allow the points A and B to move slightly
towards the centre, and the lowest point of the chain A B will
FIG. 491.
be slightly lowered in consequence, and this will necessitate a
correction in the distribution of the loads already found.
FIG. 492.
If the original angle of inclination of the back-stay A C
(Fig. 492) is 45°, and its extension per unit of length 8, the
horizontal displacement of the point A is evidently
and the deflection at the centre of the main chains due to
this displacement is (similarly to equation 148),
s = I (2 5 a) - (approximately). [227]
Let D represent the extension per unit of length in the
§ 63. — COMPOUND LATTICE AND SUSPENSION BEIDQE. 375
main chains that would produce the same deflection (supposing
A and B to be fixed points), then, according to equation 148,
~ [228]
Equating these two values of s,
[229]
or substituting a = 4 metres and I = 30 metres,
[230]
Applying this result to equation 165, it will be seen that to
find the temperature load transmitted from the main chains
to the girders, when the extension of the back-stays per unit
of length is the same as that of the main chains, it is only
necessary to multiply the previous result by 1 -f rV. (It will
be observed that equation 230 is true for negative, as well as
for positive values of 8.)
It was found, by means of equation 165, that an increase
of temperature of 41° C. produced a temperature load of 75
kilos, per metre run on the girders. Therefore, owing to the
simultaneous extension by temperature of the back-stays, this
load will be increased by ^ x 75 = 20 kilos, per metre run,
and the total load will be 95 kilos, per metre run. The stress
in the booms of the girders at the centre corresponding to the
former temperature load was 3 kilos, per sq. millimetre ; it
will now be increased to
S't = (1 + T45> 3 = 3 ' 8 kilos. per sq. millimetre. [231]
The former vertical shearing force at the abutments, due
to temperature, was 2250 kilos. It now becomes
V, = (1 + T4T) 2250 = 2850 kilos. [232]
It will be observed that there is an important difference
between the extension in the back-stays, due to elasticity, and
that due to temperature ; for the maximum increase of load due
376 BRIDGES AND ROOFS.
to the latter can be taken either positively or negatively, and
can occur under any conditions of loading of the bridge ;
whereas the former can only be taken in the positive sense,
and its maximum effect can only occur when the chain is fully
loaded, and this happens precisely when the temperature is a
minimum.
The effect of the extension of the back-stays on' the distri-
bution of the permanent and moving loads can be found as
follows : —
The stress in the back-stays, when the angle of inclination
is 45°, is \/% times greater than the horizontal stress in the
main chains. If, however, their section be made \/2 times
greater than that of the main chains at the lowest point, the
stress per unit of area will be the same, and therefore also the
extension per unit of length.
In this case, as will be seen from equation 230, the exten-
sion of the back-stays has the same effect as if the extension
of the main chains had been increased in the ratio of 1 to
1 _j_ .^ or what amounts to the same, as if the modulus of
elasticity of the main chains had been diminished in the ratio
of 1 4- -fa to 1. The distribution of the loads, when the ex-
tension of the back-stays is taken into account, can therefore
evidently be found by substituting If E for E in the various
equations already obtained. Thus, from equation 171, the co-
efficient of load' distribution for the permanent load becomes
Wl = 0 • 86184 (instead of n = 0 • 887656). [233]
[NOTE. — This value of ni can also be obtained as follows : —
The deflection of the centre of the chains consists of two parts, one due to
the extension of the back-stays and the other to the extension of [the chains
themselves, and their amount is given in equations 227 and 168 respectively.
Now, since it is assumed that the sectional area of the back-stays is */2
times that of the main chains, and consequently the value of 8 is the same in
both, d in equation 227 can be replaced by its value OTT obtained from
equation 158, by writing n± p instead of k. Hence the total deflection at the
centre of the main chains is
nlPl*a
* EFA2 T
§ 63. — COMPOUND LATTICE AND SUSPENSION BEIDGE. 377
This deflection must be equated to that at the centre of the girders given by
equation 169, thus :
zn\Pl " , a "l V " _ A V
* E F W "* 8 E F A2 EJ
from which it will be found that
1
Now,
therefore,
w, =
= 0-86184.]
The part of the load taken by the girders will be therefore
increased in the ratio,
1 -n,_ 0-13816 _
1-n "0-11234"
or 23 per cent. ; and the stresses in the booms and the shearing
forces will be increased in the same proportion. The stress 8P
in the centre of the booms was 1*685 kilos, per sq. millimetre ;
it now becomes
S p = 1 • 685 X 1 ' 23 = 2 • 07 kilos, per sq. millimetre. [235]
The vertical shearing force at the abutments was 1263-9 kilos. ;
it is now
VV = 1263 • 9 X 1 ' 23 = 1554 • 3 kilos. [236]
In applying this correction to the stresses produced by
the moving load, it is to be observed that the most unfavour-
able arrangements of the load will be slightly altered ; or, in
other words, that the zero-points will be shifted. It will be
found, for instance, that instead of u = 0V5715Z in Fig. 481,
u = 0 • 685 I gives the position of the load when the stress STO in
the booms at the centre of the girders is a maximum, as will
378 BRIDGES AND EOOFS.
appear from equations 196 and 198, in which E must be
replaced by |f E. As before, it is found that in this case
n = 0-7609,
and
S'm = 1 • 12 kilos, per sq. millimetre (instead of 0 • 95 kilos). [237]
Similarly, when x = %l, the corrected value of n is 0'4082, and
S'M= 1-445.
The maximum shearing force Vm was 2031 '6 kilos., to
which corresponded v = 0*2616? (see tables, pages 362 and
369). The corrected value is v = - 0 • 245 Z, and
V « = 2373 kilos. [238]
§ 64. — RECAPITULATION OF THE RESULTS OF THE
CALCULATIONS.
In the preceding calculations the following dimensions, &c.,
were assumed or found : —
Permanent load p = 375 kilos, per metre run.
Moving „ m = 200 „ „ „
Wind-pressure to = 200 „ „ „
Sum of the sectional areas of the j F = 750Q min^res.
main chains )
Sum of the sectional areas of the back- j F ^ 3 = 10600 sq. millimetres.
stays )
Sectional area of each of the wind-stays <f> = 1250 sq. millimetres.
Sum of the sectional areas of the four\ -^
. ,. . , f * i =
booms of the girders ;
Height of arc of the main chains . . h = 4 metres.
Depth of the girders ^ = 1 * 5 metres.
Height of arc of the wind-stays . . / = 4 metres.
Width of the bridge fl = 2 • 25 metres.
Span of the bridge 2^ = 60 metres.
The tension in the lower boom of either of the girders will
be greatest when the moving load is in its most unfavourable
position, the temperature highest, and the wind-pressure greatest
(the wind blowing against the other girder). From equations
§ 64. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 379
235, 237, 231, 219, and 224, it will be found that the stresses
due to these various causes are :
S p = 2 • 07 kilos, per sq. millimetre (permanent load).
S'm = 1 • 12 kilos. „ „ (moving load).
S't = 3 '8 kilos. „ „ (temperature load).
S'w = 4: -9 + 2* 73 „ „ (wind-pressure occurring with highest '
temperature).
Adding these together, the greatest tension in the lower
booms is found to be
S („,«*.) =14-62 kilos, per sq. millimetre. [239]
The compression in the upper booms will also reach its
greatest value under these circumstances, and can be found by
adding together the first three stresses, amounting to 7 kilos,
per sq. millimetre (the wind-pressure has no effect on the upper
booms).
The maximum vertical shearing force at the abutments is
found by adding together the values given in equations 236,
238, 232, thus :
V'p = 1554 kilos (permanent load).
V'm = 2373 kilos (moving load).
V'» =2850 kilos (temperature load).
or
V(ma,.) = 6777 kilos. [240]
And this is evidently also the maximum stress in the bars
A A! or B B! (Fig. 463) to which the ends of the girders are
attached.
These bars, it was seen, have also to act as struts; it is
therefore necessary to find the minimum stress in them. Now
the vertical shearing force produced by the moving load alone
when covering the whole bridge, is evidently „=-= x 1554 =
829 kilos. Thus the minimum added to the maximum shear-
ing force produced by the moving load will be equal to
829 kilos.
380 BEIDGES AND EOOFS.
Therefore,
V '» = 829 - 2373 = - 1544 kilos.
The minimum stress in the bars A A! or BBx is therefore
V(»to.) = + 1554 - 1544 - 2850 = - 2840 kilos. ; [241]
and they must consequently be strong enough to resist a thrust
of 2840 kilos.
The stress in the main chains reaches its maximum value
when the bridge is fully loaded, and the temperature is lowest.
The total load on the bridge is 575 kilos, per metre run, and
of this the chains carry, according to equation 233, 0*862 x
575 = 495*56 kilos, per metre. The temperature load is 95
kilos, per metre (see page 375). In the most unfavourable case,
therefore, the chains have to carry 590 • 56 kilos, per metre of
the horizontal projection, and the corresponding stress at the
lowest points of the chains is, from equation 166,
If the section of the chains is constant, the stress at the
abutments will be (see § 8)
8 • 86 \f 1 + (-38o)2 = 9-13 kilos, per sq. millimetre. [243]
The stress in the back-stays will, however, evidently be
8 * 86 kilos, per sq. millimetre, since their section is V 2 times
that of the main chains.
If it be assumed that the weight of the chains is 4500 kilos.,
then the sum of the stresses on all the verticals connecting the
chains and the girders is
(max. load on the chains)
590-56 X 60 - 4500 = 30934 kilos. ;
30934
and the stress in each will therefore be — — , where N is their
number.
The wind-stays are under the worst conditions when the
§ 65. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 381
wind-pressure is greatest and the temperature is lowest. Hence
from equations 218 and 223 the maximum stress in the centre
of the wind-stays is,
9'7 + 4<6 = 14'3 kilos, per sq. millimetre.
From equation 226 it will be seen that the greatest shearing
force taken up by the horizontal wind-braces, is 4300 kilos.
And lastly, the maximum tension in each of the horizontal
rods connecting the wind-stays with the girders, is
(107-86 + 51-2) x 60 _ 9540
~~NI = NI *
according to equations 218 and 222, NI being their number.
§ 65. — ADJUSTMENT OF THE VERTICAL KODS CONNECTING
THE GIRDERS WITH THE SUSPENSION CHAINS.
When investigating the effect of the permanent load, it was
assumed that the structure was weightless and the girders
perfectly straight. The permanent load was then applied, and
the centre of the bridge consequently deflected by a certain
amount. Now, on account of the compound nature of the
structure, it is only by chance that this will be the actual
deflection of the structure when subject to its own weight, and
it is only in this case that the distribution of the load found by
means of equation 171, will be the true one. In fact, it is
evident that by shortening the vertical rods connecting the
girders with the suspension chains, the part of the load carried
by the former will be diminished, and the reverse effect will be
obtained by lengthening these rods.
It therefore becomes a question whether by altering the
length of these rods the stresses cannot be more uniformly dis-
tributed between the various parts of the structure.
The deflection at the centre of the girders produced by
the permanent load alone, as found from equation 169, is (sub-
382 BRIDGES AND ROOFS.
stituting however n± for n (from equation 233) to allow for the
extension of the back-stays),
. 0-13816 x 0-375 X 300004
S = * 20000X15000X1500* = 51'81 md"°'^= P**]
and from equation 235 the stress in the booms corresponding to
this deflection is S'p = 2 * 07 kilos, per sq. millimetre. But as
already explained, this will only be the actual stress in the girders
if, when put up, the deflection is 51*81 millimetres, supposing
that the girders are perfectly straight when in the condition of
no stress. But if, after erection, the girders are made straight
again by shortening the vertical rods by means of set-screws,
the stress S'p = 2 '07 kilos, per sq. millimetre will disappear,
and the maximum stress (equation 239) will be reduced from
14*62 to 12*55 kilos, per sq. millimetre. At the same time
the part of the permanent load, viz. 51*81 kilos, per metre,
formerly carried by the girders, will be supported by the sus-
pension chains.
The tightening of the screws may however be continued
until, for instance, an upward deflection of 51 *81 millimetres has
been obtained ; the maximum stress in the booms would thereby
be further reduced by 2 * 07 kilos, and would become 10 *48 kilos,
per sq. millimetre, and the total increase of load on the chains
would then be 2 x 51 '81 = 103*62 kilos, per metre of the
horizontal projection.
Tightening up the set-screws, although it diminishes the
stress due to a positive bending moment, evidently increases by
the same amount the stress due to a negative bending moment ;
that is, when the girder is bent upwards.
The limit to which the set-screws may be tightened up
with advantage is therefore reached, when the greatest positive
bending moment is equal to the greatest negative bending
moment.
• The 'greatest negative bending moment under the original
circumstances occurs at the centre of the bridge, for although,
according to the table at page 367, the moving load produces its
maximum effect at a distance x = \ I from the centre, yet it
§ 65. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 383
will be found that the other causes are sufficient to make the
negative bending moment greatest at the centre.
If 8 is the stress in the lower booms at the centre produced
by the moving load alone when covering the whole bridge,
it is evident that -Jr = |^ . But S, = 2-07 kilos, per sq.
millimetre. Hence
S = |£ox 2-07 = 1-105 kilos. [245]
Now the sum of the maximum and minimum stresses
produced by the moving load must be equal to 8. Hence
fiW) = S'm - S = 1 • 12 - 1 • 105 = - 0-015 kilos, per sq. millimetre [246]
To this negative stress must now be added the negative
stress due to the lowest temperature, — 3 '8 kilos, (equation
231), the negative stress produced by the wind-pressure at the
lowest temperature, - (4'9 - 2'73) = - 2'17 kilos, (equa-
tions 221 and 224), and lastly, the positive stress, +2*07 kilos.
produced by the dead load, thus :
S(«fa.) = -0-015 -3-8 -2-17 + 2-07
= — 3-915 kilos, per sq. millimetre. [247]
Therefore, if the set-screws be tightened up until the girder
is straight S(min.) will be increased to
- 3-915 - 2-07 = 5' 985 kilos, per sq. millimetre ;
and if the tightening be further continued until the upward
deflection of the girder is 51 *81 millimetres S(m^».) will become
- 5-985 - 2-07 = — 8-055 kilos, per sq. millimetre.
It thus appears that the set-screws may be tightened up
with advantage until the centre of the girder deflects upwards
51 -81 millimetres. For under the original conditions the stress
at the centre in the lower booms varied from + 14 '62 kilos.
to — 3*915 kilos.; whereas now these limits will be + 10*48
kilos, and - 8 • 055 kilos. ; and further, the stresses in the suspen-
sion chains will only be increased to 10-41 kilos, per square
384 BRIDGES AND ROOFS.
millimetre at the centre, and to 10*73 kilos, at the points of
attachment, representing a load of 590 '56 + 103 '62 = 694*18
kilos, per metre of horizontal projection.
fNoTE. — The ratio of the greatest stresses in the booms of the girders is :
10-48 5-24
8-055 4 '
and since the ratio of the resistance of wrought iron to tension to its re-
sistance to compression is approximately the same, it appears that the set-
screws should not be tightened any further.]
The total maximum load on the rods connecting the chains
and the girders will now be increased to
30934 + 60 X 103-62 = 37151 kilos. ;
and the stress in each will therefore be — ^r— .
The vertical shearing force at the end of the girders will be
diminished by 30 x 103*62 = 3109 kilos., and therefore (equa-
tions 240 and 241) :
V (max.) = + 6777 - 3109 = 3668 kilos. [248]
V (min.) = - 2840 - 3109 = - 5949 kilos. [249]
Therefore under the new conditions, the bars A Al or
BBi in Fig. 463 must be capable of bearing a thrust of
5949 kilos.
Since the deflection of 51 * 81 millimetres at the centre of
the girders corresponds to a load on the girders of 51 *81 kilos,
per metre, or of 375 kilos, per metre on the whole bridge, it
follows that the upward deflection of 51*81 millimetres will
disappear when a load of 375 kilos, per metre has been applied
to the bridge. Therefore, to ensure the above distribution of
the stresses the following can be specified : " The deflection at
the centre of the girders is to be zero when a load of 375 kilos,
per metre run is placed on the bridge at the mean temperature."
A temperature load on the girders of 51*81 kilos, will
evidently produce the same effect. Now an increase of tem-
perature of 41° C. loads the girders with 95 kilos, per metre ;
§ 66. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 885
therefore, the temperature at which the temperature load will
be 51 *81 kilos, per metre, is :
= 210-240.
[250]
The desired result can therefore be obtained if, when the
temperature is 21° *24 C., the girders are made straight by
means of the set-screws.
§ 66.— EEMAEKS ON THE DEGREE OF ACCURACY OF THE
METHOD EMPLOYED.
The method adopted to calculate the coefficient of load-
distribution n is only approximate. But, as will be seen, the
errors involved are very small, and to a certain extent they
balance each other; they are therefore of no practical im-
portance. To prove this, it will be sufficient to consider the
simpler case given in Fig. 463, in which the points of attach-
ment of the chains are considered fixed, and the difference
between the value of n found from equation 172, and its cor-
rected value, can be considered as a measure of the error.
It will be observed that equation 170, obtained by equating
the deflection at the centre both of the girders and the chains,
is not strictly accurate. For if the bottom ends of the rods
connecting the girders with the chains were free, they would,
FIG. 493.
when the chains deflected, be in a parabola (Fig. 493), whereas
their points of attachment to the girders would be in an elastic
curve (Fig. 494). These two curves cannot cover each other,
2 c
386
BRIDGES AND ROOFS.
and the actual deflection curve of the girders will lie between
them. It would therefore appear more accurate to equate the
mean deflections instead of the ordinates a- and s at the centre.
FIG. 494.
The mean deflection will be the mean ordinate of the curve,
and this mean ordinate can be defined to be the height of the
rectangle on the same base and of the same area as the area
enclosed between the curve and the axis of x. In the present
case, since the base of both curves is the same, = 2 Z, the result
can be attained by equating the areas themselves.
The area of the part of the parabola a I e in Fig. 493 is :
[251]
and by substituting the value of s from equation 168 :
J =
npl*
2 E F A2 '
[252]
Again, the height of arc of the elastic curve a 7 ft (Fig. 494)
is, according to equation 169 :
[253]
And the equation to this curve is : *
But the area comprised between the curve a ft y and the axis
of x is :
-,/V
[255]
* This equation can be obtained by a process similar to that employed at
p. 329.— TRANS.
§ 66. — COMPOUND LATTICE AND SUSPENSION BRIDGE. 387
Hence, substituting for a and y, and integrating between the
limits :
Equating both values of J :
[256]
Whence n = 0*88352 (instead of 0' 887656) and 1 - rc, =
0-11648 (instead of 0' 112344). This correction therefore
diminishes n, but even for 1 — n the error is only 3 * 7 per cent.
Secondly, the extension of the vertical rods connecting the
chains with the girders was not taken into account. If the
elongation of these rods is S per unit of length, their lower
extremities will lie in a parabola whose height of arc is Sh
(supposing that the suspension chains do not alter). Since
this parabola has its convex side upwards, B h must be sub-
tracted from the deflection s produced by the lengthening of
the chains, and if, further, S is also the extension per unit of
length of the chains, the actual deflection at the centre of the
span is
- 5A = f 5 Sh;
[257]
or more accurately, replacing the member § -~ previously
omitted (see equation 147) :
[258]
FIG. 495.
-x--
inn
2 c 2
388 BRIDGES AND HOOFS.
Substituting
FE 2EFA
and reducing,
This value of s — 8 h must be written instead of s in equa-
tion 168 ; it will then be found, by means of equation 170,
that
[260]
From this equation n = 0 • 88884 (instead of 0 • 887656). Th us
the second error partially neutralizes the first.
If both corrections be made, the more accurate formula for
finding n is :
whence n = 0-88474 (instead of 0-887656), showing that the
former value of n was only ^ per cent, in error, and that the
value of 1 - n is 2 • 6 per cent, in error. Obviously, therefore,
the simpler method possesses ample accuracy for practical
purposes.
LOADS ON ROOFS. 389
APPENDIX.
a. LOADS ON EOOFS AND THE KEACTIONS AT THE ABUTMENTS
CAUSED BY THE WiND-PKESSUEE.
THE manner of estimating the loads on roofs followed by Professor Eitter does
not accord with the best and more recent English practice. Professor Bitter
assumes that all the loads on a roof are vertical and evenly distributed over the
surface. Now this is obviously erroneous as regards the wind-pressure, for it
cannot act vertically on a roof nor on both sides at the same time. This manner
of estimating the loads on roofs was, however, adopted by Tredgold, but in his
time little was known about the pressure of wind.
Although the scantlings obtained for wooden roofs by means of Tredgold's
assumption, coupled with a large factor of safety, have been proved by experience
to be sufficiently strong, it cannot be inferred that this would be the case for
iron roofs, at any rate, not for those of large span. And further, in iron roofs one
end should be left free to move, to allow for the expansion and contraction
produced by changes of temperature, a circumstance which affects the stresses
due to the wind-pressure. A proper distribution of material is also of greater
importance in an iron than in a wooden roof.
It is therefore necessary to arrive at some more accurate estimate of the loads
to be borne by roofs.
These loads consist of : —
Permanent load, such as the weight of roof-covering, framework, and in some
cases of the weight of a ceiling, lantern, &c. ;
Variable load, the wind-pressure, and in some countries the weight of snow.
The weight of roof-covering, purlins, ceiling, &c., can always be readily
obtained.* The only permanent load which is difficult to ascertain is that due
to the weight of the truss itself. It can be found approximately if the weight of
some similar structure is known. Or else approximate calculations can be made
considering the roof truss to have no weight, and the scantlings thus obtained
will give the required weight near enough for practical purposes, a correction
then being made to the scantlings to allow for the weight of the roof truss.
The allowance to be made for the weight of snow will depend entirely on the
locality in which the roof is to be erected. In this country it is not likely that
snow will attain a greater depth than 6 in. on a roof when a strong wind is
blowing, and this depth will also diminish as the pitch increases. An allowance
of 5 Ibs. per sq. ft. of horizontal surface covered would therefore seem ample,
* See Hurst's * Architectural Surveyor's Handbook,' or Molesworth's 'Pocket
Book of Engineering Formulae.'
390 BRIDGES AND ROOFS.
and it may also be assumed that the snow is uniformly distributed over the
roof.
The following theory of the pressure of wind on roofs is due to Professor
Unwin, and readers desirous of further information on the subject are referred to
his works.
According to the mathematical definition, a perfect fluid can exert but a
normal pressure on any body immersed within it, whether the body be at rest or
in motion relatively to the fluid. Air, as is proved by experiment, is almost a
perfect fluid, for, but a very slight tangential action is exerted on any body in
motion within it. This tangential action is so insignificant that it need not be
taken into account in the present case. It can therefore be assumed that the
wind-pressure acts normally to the slope of the roof.
Let A B (Fig. 496) represent a plane surface perpendicular to the plane of the
paper, and upon which air is impinging in
FIG. 496. a direction making an angle » with A B.
Let Pn be the normal pressure per unit of
area on the surface ; this force can be re-
solved into its components, Pv and P/,,
and evidently
Pn sin*' = Pfc.
Now, Hutton by experiments made with
his whirling machine measured the force
Ph for diiferent values of »', and found that
PA =
where P is the normal pressure per unit of area on a plane placed at right angles
to the direction of motion of the air. Hence
Pn = coseo » P sin » * >84 cos * ,
= Psinf 1>8*C08'-1. [a]
To apply this to the case of a roof, it remains to be determined what values
should be given to i and P. The following data will be of use.
On one occasion during five years, the greatest pressure recorded at the
Koyal Observatory, Greenwich, reached 41 Ibs. per sq. ft. At Bidston, near
Birkenhead, a very exposed situation, the wind blew for one hour at the rate of
92 miles per hour, equivalent to 42^ Ibs. per sq. ft., and momentarily the
pressure rose to 80 Ibs. per sq. ft.
Although these pressures were recorded on anemometers placed in very
exposed situations, and though it is probable that such wind-pressures are
never reached in ordinary situations, yet, until this is actually proved, it would
be unwise to make any reduction. Further, it is certain that the wind does not
always blow horizontally, but since neither the limits of deviation are known,
nor whether the intensity of pressure is changed or not when the direction of the
wind is thus altered, it is probably best to assume that the wind blows horizontally,
at the same time making an allowance by slightly increasing the value of P.
It thus appears reasonable to assume P = 50 Ibs.* per sq. ft., for it must
be remembered that the pressure of 80 Ibs. per sq. ft., registered at Bidston,
* Colonel Wray, R.E., in his ' Instruction in Construction,' assumes P = 50 Ibs.
per sq. ft., but Professor Unwin takes P == 40 Ibs. per sq. ft.
LOADS ON ROOFS.
391
occurred but momentarily and in a very exposed situation. To meet this greater
wind-pressure, it may, however, be advisable to increase the scantling of those
bars which are most affected by it. Since the wind has been assumed to blow
horizontally, i will be the angle made by the slope of the roof with the horizontal,
or, in other words, the pitch of the roof.
The wind-pressure can only act on one side of the roof at one time, and, owing
to want of information on the subject, is assumed to be uniformly distributed
(except, of course, in curved roofs). This assumption is, however, not altogether
unfounded, for although, no doubt, eddies are produced by the walls of the
building, &c., yet it is well known that a cushion of air is formed against the
side of the roof which tends to equalize the pressure over the surface.
The following table * will be found of use in calculating the wind-pressure on
roofs and the stresses caused thereby :—
i (pitch of roof).
Pn
P»f
P*t
0
Ibs. per sq. ft.
Ibs. per sq. ft.
Ibs. per sq. ft.
5
6-3
6-1
0-5
10
12-1
12-0
2-1
20
22-6
21-3
7-8
30
33-0
28-5
16-5
40
41-6
31-9
26-8
50
47-6
30-6
36-5
60
50-0
25-0
42-5
70
51'3
17-5
48-1
80
50-5
8-8
49-8
90
50-0
0
50-0
The inaccuracy in the values of PM for 60°, 70 p, and 80°, is due to the
empirical equation a being only approximate.
To resist the wind-pressure the supports of the roof must supply horizontal as
well as vertical reactions, and these have to be determined before the stresses
produced by the wind can be found.
FIG. 497.
Y
V
"Fx
H'
&
A !
i B
cu-
Let Fig. 497 represent a body supported on two points in a horizontal straight
line, and acted upon by a force F inclined to the vertical. This force can be
replaced by its vertical and horizontal components X and Y, and the reactions at
* This table has been taken from l Instruction in Construction,' by Colonel
Wray, R.E.
t It is shown in a pamphlet by Professor Unwin, « On the Effect of Wind-
Pressure on Roofs,' that these values of Pv and P& agree very well with some
experiments made by Froude and Wenham.
392
BKIDGES AND KOOFS.
the two points can also be replaced by their components V, H and V, H'. The
three conditions of equilibrium are :
X — H - H' = 0, resolving horizontally.
Y - V — V = 0, resolving vertically.
V (a -f- 6) — Y a = 0, moments round B.
Whence
H + H' = X.
y- a Y
a+b
V'= ~-6Y.
The vertical components V and V are therefore determinate, but the
horizontal components H and H' are indeterminate, the only condition necessary
fore quilibrium being that their sum is equal to the horizontal component X of
F. This will be easily understood when it is considered that if the support A
were smooth (mathematically), the whole of X would have to be supplied by B,
and vice versa. The values of H and H' depend therefore on the nature of the
supports.
Now, in iron roofs it is necessary to allow for the expansion and contraction
arising from changes of temperature, and to do this, it is usual to fix one end of
the truss, leaving the other free to move. In small roofs it is sufficient if the
shoe at the free end simply rests on the template, but in large roofs a special
arrangement of friction-rollers is generally provided.
Evidently the horizontal component of the reaction at the free end of the roof
can never exceed the resistance to motion of that end. This resistance is p. V
where /* is the coefficient of friction, and V the vertical reaction at the free end.
In small roofs where no rollers are provided, /* will usually be the coefficient of
friction of iron (cast) against stone, and may be taken at from '4 to *6. It
may happen that, owing to a change of temperature, the free end of the roof is
just on the point of motion. The full frictional resistance would thereby be
called into play, and possibly the wind-pressure might produce a horizontal com-
ponent equal and opposite to this resistance. For instance, let Fig. 498 represent
a roof of which the end A is fixed and the end B is free to move. When the
FIG. 498.
Tf
temperature increases, horizontal forces h are generated at A and B, and by the
above assumption, when the end B is on the point of motion h = p. V = H'. Evi-
dently, therefore, the horizontal reaction at A is
H + h = H + H' = X,
STABILITY OF PIEKS AS REGAEDS OVERTURNING. 393
and at B it is zero. A similar case may occur when the wind is blowing from the
left and the temperature diminishing. It is possible that this distribution of the
horizontal reactions may produce greater, stresses in some of the bars than the
more normal distribution, and should therefore be considered. Thus in small iron
roofs there are four cases to consider, namely, —
2 W^d °- I1 ft ' I Horizontal reaction at free end equal to friction.
3. Wind on right, i ,
4. Wind on left. } Honzo*tal reactlon at free end zero'
In large roofs with a roller end, p. is so small (being the coefficient of rolling
friction) that it may be neglected in the present inquiry. It may therefore be
assumed that the horizontal component at the free end is always zero, and only
two cases need be considered, namely, —
1. Wind on right, ) _
2 Wind on left f Horizontal reaction at free end zero.
This will also meet the case of a roof truss having one end (the free end) sup-
ported by a column, and the other by a wall. *
In calculating the stresses in a roof it is best first to find the stresses
produced by the weight of the roof-covering and framing when those occasioned
by snow can generally be found by simple proportion. The stresses due to the
wind are then to be ascertained as indicated above, and may be found by the
" Method of Moments." If the results thus obtained are collected in a tabular
form, the greatest tension or compression in each bar is easily found by inspection.
b. STABILITY OF PIERS AS REGARDS OVERTURNING.
IN the calculations made both at p. 149 and p. 185 to ascertain the stability of
the piers as regards overturning, Professor Eitter takes moments about the lower
edge F of the pier. Since the resultant compression on the bed-joint at F does
not appear in these equations of moments, it must act at F, or in other words, the
total compression on the bed-joint is concentrated on the outer edge F. The in-
tensity of pressure on this edge would therefore be so great that the material of
which the pier is composed would be crushed. (Mathematically speaking, the
edge is a line, and the pressure would therefore be infinite.) It is evident
therefore that moments should not be taken round the outer edge of the pier, but
about some axis inside the pier represented by the point E, Fig. 502, the position
of this axis being such that the greatest intensity of pressure shall not exceed the
safe resistance to crushing of the material (or of the mortar). It is proposed to
find the position of E when the pier is rectangular on plan.f
In large structures of this kind the tenacity of the mortar should not be taken
into account, for unequal settlements are liable to occur, which dislocate the
joints. The pier will therefore be regarded as built with " uncemented blocks."
Now consider a body rectangular on plan (Fig. 499) pressed against a plane
* See last paragraph p. 35 of ' Lectures on the Elements of Applied Mechanics,'
by Professor M. W. CroftoD, F.E.S., in which this is pointed out.
f For further information on this subject see ' Applied Mechanics ' and * Civil
Engineering,' by Professor Eankine ; * Engineering and Architecture,' by Eev.
Canon Moseley ; ' Instruction in Construction,' by Colonel Wray, E.E.
2 D
394
BRIDGES AND EOOFS.
surface by a force P, the side of the body in contact being also plane. For
simplicity, let the direction of P be as indicated in the figure, so that E the
centre of pressure is situated on G F which is parallel to and equidistant from
A B and D C. The pressure evidently cannot be uniformly distributed, unless E
FIG. 499.
bisects G F, but a fair assumption to make is that it varies uniformly. Or in
other words, the pressure will reach its greatest intensity along the edge A D,
and it will diminish uniformly towards the edge B 0. And further, since E is
equidistant from A B and D C, the intensity of pressure at all points on any line
parallel to A D will be the same, and hence the pressure along G F represents
that over the whole area.
Thus, if the position of E be such that the pressure at F is nothing, the triangle
GF K (Fig. 500) will represent the intensity of normal pressure at every point in
the line G F. The resultant normal pressure will evidently pass through the
centre of gravity of this triangle, hence
GE=iGF.
FIG. 500.
FIG. 501.
If however, G E < i G F (Fig. 501), the pressure will vanish at a point F',
such that
and from F' to F there will be no pressure.
Evidently the maximum intensity of pressure is twice what it would be were
the pressure uniformly distributed over G F', for if H bisects G K the area of the
rectangle H F' is equal to that of the triangle G F' K.
The position of E to fulfil the condition that the maximum intensity of
STABILITY OP PIERS AS REGARDS OVERTURNING.
395
pressure K G should not exceed the safe resistance to crushing of the material,
can probably be easiest ascertained by the graphic method.
For instance, let L M G F (Fig. 502) represent a pier rectangular on plan
subject to a thrust P. The weight of the pier is W, and acts in the vertical
through the centre of gravity C of the pier ; 0 T and O S represent P and W in
magnitude and direction, then by completing the parallelogram of forces the
resultant B is obtained, represented by 0 U. The resolved part O X of R at
right angles to G F is the total normal pressure on the bed-joint and the inter-
section E of O U and G F is the centre of pressure. Thus all the elements for
finding the maximum intensity of pressure are known.
FIG. 502.
L M _ ^
w
As a numerical example let the dimensions of the pier be: the breadth
b = 20 feet, and the thickness G F = 6 feet. Let also W = 52 tons, P =
18 tons, the angle T OC = 24°, and the height of O above G F = 7 feet. It is
then found by measurement that O X = 60 tons and G E = 1 foot, or G F' =
3 feet as previously explained. Hence the greatest intensity of stress along the
edge of the pier represented by G is
= 2.
&.GF'
60
20 . 3
= 2 tons per square foot.
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With 112 illustrations. Third edition, 4to, cloth, 12s. 6d.
io CATALOGUE OF SCIENTIFIC BOOKS
Graphic Statics. — A New Method of Graphic
Statics, applied in the construction of Wrought-Iron Girders, practically
illustrated by a series of Working Drawings of modern type. By
EDMUND OLANDER, of the Great Western Railway, Assoc. Mem. Inst.
C.E. Small folio, cloth, los. 6d.
Heat Engine. — Theory and Constriction of a
Rational Heat Motor. Translated from the German of RUDOLF DIESEL by
BRYAN DONKIN, Mem. Inst. C.E. Numerous cuts and plates, 8vo, cloth, 6s.
Hot Water. — Hot Water Supply: a Practical
Treatise upon the Fitting of Circulating Apparatus in connection with
Kitchen Range and other Boilers, to supply Hot Water for Domestic and
General Purposes. With a Chapter upon Estimating. By F. DYE.
With illustrations ) crown 8vo, cloth, 3^.
Hot Water.— Hot Water Apparatus: an Ele-
mentary Guide for the Fitting and Fixing of Boilers and Apparatus for
the Circulation of Hot Water for Heating and for Domestic Supply, and
containing a Chapter upon Boilers and Fittings for Steam Cooking. By
F. DYE. 32 illustrations, fcap. 8vo, cloth, is. 6d.
Household Manual. — Spons Household Manual :
a Treasury of Domestic Receipts and Guide for Home Management.
Demy 8vo, cloth, containing 975 pages and 250 illustrations, price 7*. 6d.
PRINCIPAL CONTENTS :
Hints for selecting a good House — Sanitation — Water Supply — Ventilation and Warming
— Lighting — Furniture and Decoration — Thieves and Fire — The Larder — Curing Foods for
lengthened Preservation— The Dairy— The Cellar— The Pantry— The Kitchen— Receipts for
Dishes — The Housewife's Room — Housekeeping, Marketing — The Dining-Room — The
Drawing-Room— The Bedroom— The Nursery— The Sick-Room— The Bath-Room— The
Laundry— The School-Room— The Playground— The Work- Room— The Library— The
Garden — The Farmyard — Small Motors — Household Law.
House Hunting. — Practical Hints on Taking a
House. By H. PERCY BOULNOIS, Mem. Inst. C.E., City Engineer,
Liverpool, Author of * The Municipal and Sanitary Engineer's Hand-
book,' ' Dirty Dustbins and Sloppy Streets,' &c. i8mo, cloth, u. 6d.
Hydraulics. — Simple Hydraulic Formula. By
T. W. STONE, C.E., late Resident District Engineer, Victoria Water
Supply. Crown 8vo, cloth, 4^.
Hydraulic Machinery.— Hydraulic Steam and
Hand-Power Lifting and Pressing Machinery. By FREDERICK COLYER,
M. Inst. C.E., M. Inst. M.E. Second edition, revised and enlarged. With
88 plates, 8vo, cloth, 28*.
Hydraulic Machinery. — Hydraulic Machinery.
With an Introduction to Hydraulics. By ROBERT GORDON ELAINE,
Assoc. M. Inst. C.E., &c. With 272 illustrations, 383 pp. 8vo, cloth, 14^.
PUBLISHED BY E. & F. N. SPON, LIMITED. n
Hydraulic Motors. — Water or Hydraulic Motors.
By PHILIP R. BJORLING. With 206 illustrations, crown 8vo, cloth, gs.
CONTENTS :
i. Introduction — 2. Hydraulics relating to Water Motors — 3. Water-wheels — 4. Breast
Water-wheels — 5. Overshot and High-breast Water-wheels — 6. Pelton Water-wheels — 7.
General Remarks on Water-wheels — 8. Turbines — 9. Outward-flow Turbines — 10. Inward-
flow Turbines — n. Mixed-flow Turbines — 12. Parallel-flow Turbines — 13. Circumferential-
flow Turbines — 14. Regulation of Turbines — 15. Details of Turbines — 16. Water-pressure or
Hydraulic Engines — 17. Reciprocating Water-pressure Engines— 18. Rotative Water-
pressure Engines — 19. Oscillating Water-pressure Engines — 20. Rotary Water-pressure
Engines — 21. General Remarks and Rules for Water-pressure Engines — 22. Hydraulic Rams
— 23. Hydraulic Rams without Air Vessel in Direct Communication with the Drive Pipe —
24. Hydraulic Rams with Air Vessel in Direct Communication with the Drive Pipe — 25.
Hydraulic Pumping Rams — 26. Hydraulic Ram Engines — 27. Details of Hydraulic Rams —
28. Rules, Formulas, and Tables for Hydraulic Rams — 29. Measuring Water in a Stream
and over a Weir — Index.
Hydropathic Establishments. — The Hydro-
pathic Establishment and its Baths. By R. O. ALLSOP, Architect.
Author of ' The Turkish Bath.' Illustrated with plates and sections, 8vo,
cloth, 5j. CONTENTS :
General Considerations — Requirements of the Hydropathic Establishment — Some existing
Institutions — Baths and Treatments and the arrangement of the Bath-House — Vapour Baths
and the Russian Bath — The Douche Room and its appliances — Massage and Electrical
Treatment — Pulverisation and the Mont Dore Cure — Inhalation and the Pine Cure — The
Sun Bath.
Ice Making. — Theoretical and Practical Ammonia
Refrigeration, a work of Reference for Engineers and others employed in
the management of Ice and Refrigeration Machinery. By ILTYD L.
REDWOOD, Assoc. Mem. Am. Soc. of M.E., Mem. Soc. Chemical Indus-
try. With 25 pages of tables. Square i6mo, cloth, 4-r. 6d.
Indicator. — Twenty Years with the Indicator. By
THOMAS PRAY, Jun., C.E., M.E., Member of the American Society of
Civil Engineers. With ilhtstrations, royal 8vo, cloth, los. 6d.
Indicator. — A Treatise on the Richards Steam-
Engine Indicator and the Development and Application of Force in the
Steam-Engine. By CHARLES T. PORTER. With illustrations. Fourth
edition, revised and enlarged, 8vo, cloth, gs.
Induction Coils. — Induction Coils and Coil
Making : a Treatise on the Construction and Working of Shock, Medical
and Spark Coils. By F. C. ALLSOP. Second edition, with 125 illustra-
tions, crown 8vo, cloth, 3-r. 6d.
Iron. — The Mechanical and other Properties of Iron
and Steel in connection with their Chemical Composition, By A. VOSMAER,
Engineer. Crown 8vo, cloth, 6s.
CONTENTS :
The metallurgical behaviour of Carbon with Iron and Steel, also Manganese — Silicon —
Phosphorus — Sulphur — Copper — Chromium — Titanium — Tungsten — Aluminium — Nickel —
Cobalt — Arsenic — Analyses of Iron and Steel, &c.
12 CATALOGUE OF SCIENTIFIC BOOKS
Iron Manufacture . — Roll- Turning for Sections in
Steel and Iron> working drawings for Rails, Sleepers, Girders, Bulbs,
Ties, Angles, &c., also Blooming and Cogging for Plates and Billets.
By ADAM SPENCER. Second edition, with 78 large plates. Illustrations
of nearly every class of work in this Industry. 4to, cloth, I/, icxr.
Locomotive. — The Construction of the Modern
Locomotive. By GEORGE HUGHES, Assistant in the Chief Mechanical
Engineer's Department, Lancashire and Yorkshire Railway. Numeroiis
engravings, 8vo, cloth, qs.
CONTENTS :
The Boiler— The Foundry— the Use of Steel Castings— Brass Foundry— The Forge-
Smithy, including Springs — Coppersmiths' Work — The Machine Shop — Erecting.
Lime and Cement. — A Manual of Lime and
Cement, their treatment and use in construction. By A. H. HEATH.
Crown 8vo, cloth, 6s.
Liquid Fuel. — Liquid Fuel for Mechanical and
Industrial Purposes. Compiled by E. A. BRAYLEY HODGETTS. With
wood engravings. 8vo, cloth, 5^.
Machinery Repairs. — The Repair and Mainten-
ance of Machinery ; a Handbook of Practical Notes and Memoranda for
Engineers and Machinery Users. By T. W. BARBER, C.E., M.E.,
Author of ' The Engineers' Sketch Book.' With about 400 illustrations,
8vo, cloth, IO.T. 6d.
Mechanical Engineering. — Handbook for Me-
chanical Engineers. By HENRY ADAMS, Professor of Engineering at
the City of London College, Mem. Inst. C.E., Mem. Inst. M.E., &c.
Fourth edition, revised and enlarged. Crown 8vo, cloth, *js. 6d.
CONTENTS :
Fundamental Principles or Mechanics — Varieties and Properties of Materials — Strength
of Materials and Structures — Pattern Making — Moulding and Founding— Forging, Welding
and Riveting — Workshop Tools and General Machinery — Transmission of Power, Friction
and Lubrication — Thermodynamics and Steam — Steam Boilers — The Steam Engine — Hy-
draulic Machinery — Electrical Engineering — Sundry Notes and Tables.
Mechanical Engineering. — The Mechanician:
a Treatise on the Construction and Manipulation of Tools, for the use and
instruction of Young Engineers and Scientific Amateurs, comprising the
Arts of Blacksmithing and Forging ; the Construction and Manufacture
of Hand Tools, and the various Methods of Using and Grinding them ;
description of Hand and Machine Processes ; Turning and Screw Cutting.
By CAMERON KNIGHT, Engineer. Containing 1147 illustrations > and
397 pages of letter-press. Fourth edition, 4to, cloth, i8j.
PUBLISHED BY E. & F. N. SPON, LIMITED. 13
Mechanical Movements. — The Engineers' '6 "ketch-
Book of 'Mechanical Movements, Devices ', Appliances, Contrivances, Details
employed in the Design and Construction of Machinery for every purpose.
Collected from numerous Sources and from Actual Work. Classified and
Arranged for Reference. With 2600 Illustrations. By T. W. BARBER,
Engineer. Third edition, 8vo, cloth, los. 6d.
Metal Plate Work.— Metal Plate Work: its
Patterns and their Geometry. Also Notes on Metals and Rules in Men-
suration for the use of Tin, Iron, and Zinc Plate-workers, Coppersmiths,
Boiler-makers and Plumbers. By C. T. MILLIS, M.I.M.E. Second
edition, considerably enlarged. With numerous illustrations. Crown
Svd, cloth, 9-r.
Metrical Tables. — Metrical Tables. By Sir G. L.
MOLESWORTH, M.I.C.E. 32mo, cloth, u. 6d.
Mill-Gearing. — A Practical Treatise on Mill- Gear-
ing> Wheels, Shafts, Riggers, etc. ; for the use of Engineers. By THOMAS
Box. Third edition, with 1 1 plates. Crown 8vo, cloth, 7*. 6d.
Mill - Gearing. — The Practical Millwright and
Engineer's Ready Reckoner; or Tables for finding the diameter and power
of cog-wheels, diameter, weight, and power of shafts, diameter and
strength of bolts, etc. By THOMAS DIXON. Fourth edition, i2mo,
cloth, 3*.
Mineral Oils. — A Practical Treatise on Mineral
Oils and their By-Products, including a Short History of the Scotch Shale
Oil Industry, the Geological and Geographical Distribution of Scotch
Shales, Recovery of Acid and Soda used in Oil Refining, and a list of
Patents relating to Mineral Oils. By ILTYD I. REDWOOD, Mem. Soc.
Chemical Industry. 8vo, cloth, 15^.
Miners' Pocket-Book. — Miners Pocket-Book; a
Reference Book for Miners, Mine Surveyors, Geologists, Mineralogists,
Millmen, Assayers, Metallurgists, and Metal Merchants all over the
world. By C. G. WARNFORD LOCK, author of ' Practical Gold Mining,'
' Mining and Ore-Dressing Machinery,' &c. Second edition, fcap. 8vo,
roan, gilt edges, I2J. 6d.
CONTENTS :
Motive Power — Dams and Reservoirs — Transmitting Power — Weights and Measures _
,
Copper Smelting— Treatment of Ores— Coal Cleaning— Mine Surveying— British Rocks-
Geological Maps — Mineral Veins — Mining Methods — Coal Seams — Minerals — Precious
Stones— Metals and Metallic Ores— Metalliferous Minerals— Assaying— Glossary— List of
Useful Books— Index, &c., &c., &c.
14 CATALOGUE OF SCIENTIFIC BOOKS
Mining and Ore-Dressing Machinery. — By
C. G. WARNFORD LOCK, Author of ' Practical Gold Mining.' Numerous
illustrations, super-royal 4to, cloth, 2$s.
Mining. — Economic Mining; a Practical Hand-
book for the Miner, the Metallurgist, and the Merchant. By C. G.
WARNFORD LOCK, Mem. Inst. of Mining and Metallurgy, Author of
'Practical Gold Mining.' With illustrations, 8vo, cloth, 2is.
Municipal Engineering. — The Municipal and
Sanitary Engineer's Handbook. By H. PERCY BOULNOIS, Mem. Inst.
C.E., Borough Engineer, Portsmouth. With numerous illustrations.
Second edition, demy 8vo, cloth, 15^.
CONTENTS :
The Appointment and Duties of the Town Surveyor — Traffic — Macadamised Roadways-
Steam Rolling — Road Metal and Breaking — Pitched Pavements — Asphalte — Wood Pavements
— Footpaths — Kerbs and Gutters — Street Naming and Numbering— Street Lighting — Sewer-
age— Ventilation of Sewers — Disposal of Sewage — House Drainage — Disinfection — Gas and
Water Companies, etc., Breaking up Streets — Improvement of Private Streets — Borrowing
Powers — Artirans' and Labourers' Dwellings — Public Conveniences — Scavenging, including
Street Cleansing — Watering and the Removing of Snow — Planting Street Trees — Deposit ot
Plans — Dangerous Buildings — Hoardings — Obstructions— Improving Street Lines — Cellar
Openings — Public Pleasure Grounds — Cemeteries — Mortuaries — Cattle and Ordinary Markets
— Public Slaughter-houses, etc.— Giving numerous Forms of Notices, Specifications, and
General Information upon these and other subjects of great importance to Municipal Engi-
neers and others engaged in Sanitary Work.
Paints. — Pigments, Paint and Painting. A
Practical Book for Practical Men. By GEORGE TERRY. With illus-
trations, crown 8vo, cloth, "js. 6d.
Paper Manufacture. — A Text-Book of Paper-
Making. By C. F. CROSS and E. J. BEVAN. With engravings, crown
Svo, cloth, I2s. 6d.
Perfumery. — Perfumes and their Preparation,
containing complete directions for making Handkerchief Perfumes,
Smelling Salts, Sachets, Fumigating Pastils, Preparations for the care of
the Skin, the Mouth, the Hair, and other Toilet articles, with a detailed
description of aromatic substances, their nature, tests of purity, and
wholesale manufacture. By G. W. ASKINSON, Dr. Chem. With 32
engravings, Svo, cloth, I2s. 6d.
Perspective. — Perspective, Explained and Illus-
trated. By G. S. CLARKE, Capt. R.E. With illustrations, Svo, cloth,
3*. 6d.
Phonograph. — The Phonograph, and How to Con-
struct it. With a Chapter on Sound. By W. GiLLETT. With engravings
and full working drawings, crown Svo, cloth, 5^.
PUBLISHED BY E. & F. N. SPON, LIMITED. 15
Popular Engineering. — Popular Engineering,
being interesting and instriictive examples in Civil, Mechanical, Electrical,
Chemical, Mining, Military and Naval Engineering, graphically and
plainly described, and specially written for those about to enter the
Engineering Profession and the Scientific Amateur, with chapters on
Perpetual Motion and Engineering Schools and Colleges. By F. DYE.
With 700 illustrations, crown 4to, cloth, Js. 6d.
Plumbing. — Plumbing, Drainage, Water Supply
and Hot Water Fitting. By.JOHN SMEATON, C.E., M.S.A., R.P.,
Examiner to the Worshipful Plumbers' Company. Numerous engravings,
Svo, cloth, 7-r. 6d.
Pumping Engines. — Practical Handbook on
Direct-acting Pumping Engine and Steam Pump Constmction. By
PHILIP R. BJORLING. With 20 plates, crown 8vo, cloth, 5-r.
Pumps. — A Practical Handbook on Pump Con-
struction. By PHILIP R. BJORLING. Plates, crown Svo, cloth, 5^.
CONTENTS :
Principle of the action of a Pump — Classification of Pumps — Description of various
classes of Pumps — Remarks on designing Pumps — Materials Pumps should be made of for
different kinds of Liquids — Description of various classes of Pump-valves — Materials Pump-
valves should be made of for different kinds of Liquids — Various Classes of Pump-buckets —
On designing Pump-buckets— Various Classes of Pump-pistons— Cup-leathers— Air-vessels-
Rules and Formulas, &c., &c.
Pumps. — Pump Details. With 278 illustrations.
By PHILIP R. BJORLING, author of a Practical Handbook on Pump
Construction. Crown Svo, cloth, Js. 6a'.
CONTENTS :
Windbores — Foot-valves and Strainers — Clack-pieces, Bucket-door-pieces, and H-pieces
Working-barrels and Plunger-cases — Plungers or Rams — Piston and Plunger, Bucket and
Plunger, Buckets and Valves — Pump-rods and Spears, Spear-rod Guides, &c. — Valve-swords,
Spindles, and Draw-hooks— Set-offs or Off-sets — Pipes, Pipe-joints, and Pipe-stays — Pump-
slings — Guide-rods and Guides, Kites, Yokes, and Connecting-rods — 1_ Bobs, T Bobs,
Angle or V Bobs, and Balance-beams, Rock-arms, and Fend-off Beams, Cisterns^ and Tanks
— Minor Details.
Pumps. — Pumps and Pumping- Machinery. By
F. COLYER, Mem. Inst. C.E., Mem. Inst. M.E. Part L, second edition,
revised and enlarged, with 50 plates, Svo, cloth, I/. &s.
CONTENTS :
Three-throw Lift and Well Pumps— Tonkin's Patent " Cornish " Steam Pump — Thorne-
will and Warham's Steam Pump — Water Valves — Water Meters — Centrifugal Pumping
Machinery — Airy and Anderson's Spiral Pumps — Blowing Engines — Air Compressors-
Horizontal High-pressure Engines — Horizontal Compound Engines — Reidler Engine — Ver-
tical Compound Pumping Engines — Compound Beam Pumping Engines — Shonheyder's
Patent Regulator — Cornish Beam Engines — Worthington High-duty Pumping Engine —
Davy's Patent Differential Pumping Engine — Tonkin's Patent Pumping Engine — Lancashire
Boiler— Babcock and Wilcox Water-tube Boilers.
1 6 CATALOGUE OF SCIENTIFIC BOOKS
Pumps. — P^l,mps, Historically, Theoretically, and
Practically Considered. By P. R. BJORLING. With 156 illustrations.
Crown 8vo, cloth, Js. 6d.
Quantities. — A Complete Set of Contract Documents
for a Country Lodge, comprising Drawings, Specifications, Dimensions
(for quantities), Abstracts, Bill of Quantities, Form of Tender and Con-
tract, with Notes by J. LEANING, printed in facsimile of the original
documents, on single sheets fcap., in linen case, $s.
Quantity Surveying. — Qiiantity Surveying. By
J. LEANING. With 68 illustrations. Third edition, revised, demy 8vo,
cloth, 15-r.
CONTENTS :
A complete Explanation of the London Schedule of Prices.
Practice. Form of Schedule of Prices.
General Instructions. Analysis of Schedule of Prices.
Order of Taking Off. Adjustment of Accounts.
Modes of Measurement of the various Trades. Form of a Bill of Variations.
Use and Waste. Remarks on Specifications.
Ventilation and Warming, i Prices and Valuation of Work, with
Credits, with various Examples of Treatment. Examples and Remarks upon each Trade.
Abbreviations. The Law as it affects Quantity Surveyors,
Squaring the Dimensions. with Law Reports.
Abstracting, with Examples in illustration of Taking Off after the Old Method.
each Trade. Northern Practice.
Billing. ! The General Statement of the Methods
Examples of Preambles to each Trade. recommended by the Manchester Society
Form for a Bill of Quantities. of Architects for taking Quantities.
Do. Bill of Credits. Examples of Collections.
Do. Bill for Alternative Estimate. . Examples of " Taking Off" in each Trade.
Restorations and Repairs, and Form of Bill. Remarks on the Past and Present Methods
Variations before Acceptance of Tender. of Estimating.
Errors in a Builder's Estimate.
Railway Curves. — Tables for Setting out Curves
for Railways, Canals, Roads, etc., varying from a radius of five chains
to three miles. By A. KENNEDY and R. W. HACKWOOD. Illustrated,
32mo, cloth, zs. 6d.
Roads. — The Maintenance of Macadamised Roads.
By T. CODRINGTON, M.I.C.E., F.G.S., General Superintendent of
County Roads for South Wales. Second edition, 8vo, cloth, •js. 6d.
Scamping Tricks. — Scamping Tricks and Odd
Knowledge occasionally practised upon Public Works, chronicled from the
confessions of some old Practitioners. By JOHN NEWMAN, Assoc. M.
Inst. C.E., author of * Earthwork Slips and Subsidences upon Public
Works,' ' Notes on Concrete,' &c. Crown 8vo, cloth, 2s. 6d.
Screw Cutting. — Turners Handbook on Screw
Cutting, Coning, etc., etc., with Tables, Examples, Gauges, and
Formulae. By WALTER PRICE. Fcap 8vo, cloth, is.
PUBLISHED BY E. & F. N. SPON, LIMITED. 17
Screw Cutting. — Screw Cutting Tables for En-
gineers and Machinists, giving the values of the different trains of Wheels
required to produce Screws of any pitch, calculated by Lord LINDSAY.
Oblong, cloth, 2s.
Screw Cutting. — Screw Cutting Tables, fgr the
use of Mechanical Engineers, showing the proper arrangement of Wheels
for cutting the Threads of Screws of any required pitch, with a Table for
making the Universal Gas-pipe Threads and Taps. By W. A. MARTIN,
Engineer. Second edition, oblong, cloth, u.
Sewerage. — Sewerage and Sewage Disposal. By
HENRY ROBINSON, Mem. Inst. C.E., F.G.S., Professor of Civil
Engineering, King's College, London, &c., with large folding plate.
Demy 8vo, cloth, 12s. 6d.
Slide Valve. — A Treatise on a Practical Method
of Designing Slide- Valve Gears by Simple Geometrical Construction, based
upon the principles enunciated in Euclid's Elements, and comprising the
various forms of Plain Slide- Valve and Expansion Gearing ; together with
Stephenson's, Gooch's, and Allan's Link-Motions, as applied either to
reversing or to variable expansion combinations. By EDWARD J. COW-
LING WELCH, Mem. Inst. M.E. Crown 8vo, cloth, 6s.
Soap. — A Treatise on the Manufacture of Soap and
Candles, Lubricants and Glycerine. By W. LANT CARPENTER, B.A.,
B.Sc. With illustrations, new edition, revised, crown 8vo, 12s. 6d.
Stair Building. — Practical Stair Building and
Handrailing by the Sqtiare Section and Falling Line System. By W. H.
WOOD. Folding plates, post 410, cloth, los. bd.
Steam Boilers. — Steam Boilers, their Manage-
ment and Working on land and sea. By JAMES PEATTIE. With
illustrations, crown 8vo, cloth, $s.
Steam Engine. — A Practical Treatise [on the
Steam Engine, containing Plans and Arrangements of Details for Fixed
Steam Engines, with Essays on the Principles involved in Design and
Construction. By ARTHUR RIGG, Engineer, Member of the Society of
Engineers and of the Royal Institution of Great Britain. Demy 4to,
copiously illustrated with woodcuts and 103 plates, in one Volume.
Second edition, cloth, 2$s.
This work is not, in any sense, an elementary treatise, or history of the steam engine, but
is intended to describe examples of Fixed Steam Engines without entering into the wide
domain of locomotive or marine practice. To this end illustrations will be given of the most
recent arrangements of Horizontal, Vertical, Beam, Pumping, Winding, Portable, Semi-
portable, Corliss, Allen, Compound, and other similar Engines, by the most eminent Firms in
Great Britain and America. The laws relating to the action and precautions to be observed
in the construction of the various details, such as Cylinders, Pistons, Piston-rods, Connecting-
B
1 8 CATALOGUE OF SCIENTIFIC BOOKS
rods, Cross-heads, Motion-blocks, Eccentrics, Simple, Expansion, Balanced, and Equilibrium
Slide-valves, and Valve-gearing will be minutely dealt with. In this connection will be found
articles upon the Velocity of Reciprocating Parts and the Mode of Applying the Indicator,
Heat and Expansion of Steam Governors, and the like. It is the writer's desire to draw
illustrations from every possible source, and give only those rules that present practice deems
correct.
Steam Engine. — The Steam Engine considered as
a Thermodynamic Machine, a treatise on the Thermodynamic efficiency
of Steam Engines, illustrated by Tables, Diagrams, and Examples from
Practice. By JAS. H. COTTERILL, M.A., F.R.S., Professor of Applied
Mechanics in the Royal Naval College. Third edition, revised and
enlarged, 8vo, cloth, i$s.
Steam Engine. — Steam Engine Management ; a
Treatise on the Working and Management of Steam Boilers. By F.
COLYER, M. Inst. C.E., Mem. Inst. M.E. New edition, i8mo, cloth,
3j. 6d.
Steam Engine. — A Treatise on Modern Steam
Engines and Boilers, including Land, Locomotive and Marine Engines
and Boilers, for the use of Students. By FREDERICK COLYER, M. Inst.
C.E., Mem. Inst. M.E. With 36 plates. 4to, cloth, I2s. 6d.
Sugar. — Tables for the Quantitative Estimation of
the Sugars, with Explanatory Notes. By Dr. ERNEST WEIN ; translated,
with additions, by WILLIAM FREW, Ph.D. Crown 8vo, cloth, 6s.
Sugar. — A Handbook for Planters and Refiners ;
being a comprehensive Treatise on the Culture of Sugar-yielding Plants,
and on the Manufacture, Refining, and Analysis of Cane, Palm, Maple,
Melon, Beet, Sorghum, Milk, and Starch Sugars ; with copious
Statistics of their Production and Commerce, and a chapter on the
Distillation of Rum. By C. G. WARNFORD LOCK, F.L.S., £c. ;
L. E. R. NEWLANDS, F.C.S., F.I.C., Mem. Council Soc. Chemical
Industry; and J. A. R. NEWLANDS, F.C.S., F.I.C. Upwards 0/200
illustrations and many plates, 8vo, cloth, I/. IQJ.
Surveying. — A Practical Treatise on the Science of
Land and Engineering Surveying, Levelling, Estimating Quantities, etc.,
with a general description of the several Instruments required for Sur-
veying, Levelling, Plotting, etc. By H. S. MERRETT. Fourth edition,
revised by G. W. USILL, Assoc. Mem. Inst. C.E. 41 plates, with illus-
trations and tables, royal 8vo, cloth, 12s. 6d.
Surveying and Levelling. - - Surveying and
Levelling Instruments theoretically and practically described, for construc-
tion, qualities, selection, preservation, adjustments, and uses, with other
apparatus and appliances used by Civil Engineers and Surveyors. By
W. F. STANLEY. Second edition. 350 cuts, crown 8vo, cloth, 7-r. 6^.
PUBLISHED BY E. & F. N. SPON, LIMITED. 19
Tables of Logarithms. — A B C Five-Figure
Logarithms for general iise. By C. J. WOODWARD, B.Sc. Containing
Mantissse of numbers to 10,000. Log. Sines, Tangents, Cotangents, and
Cosines to 10" of Arc. Together with full explanations and simple
exercises showing use of the tables. qs.
Tables of Squares. — Barlows Tables of Squares,
Cubes, Square Roots, Cube Roots , Reciprocals of all Integer Numbers up to
10,000. Post 8vo, cloth, 6s.
Telephones. — Telephones, their Construction and
Fitting. By F. C. ALLSOP. Fourth edition, revised. With 210 illustra-
tions. Crown 8vo, cloth, 5-r.
Tobacco Cultivation. — Tobacco Growing, Curing,
and Manufacturing ; a Handbook for Planters in all parts of the world.
Edited by C. G. WARNFORD LOCK, F.L.S. With illustrations. Crown
8vo, cloth, "js. 6d.
Tropical Agriculture. — Tropical Agriculture: a
Treatise on the Culture, Preparation, Commerce and Consumption of the
principal Products of the Vegetable Kingdom. By P. L. SIMMONDS,
F.L.S. , F.R.C.I. New edition, revised and enlarged, 8vo, cloth, 2.1s.
Turning. — The Practice of Hand Turning in Wood,
Ivory, Shell, etc., with Instructions for Turning such Work in Metal as
may be required in the Practice of Turning in Wood, Ivory, etc. ; also
an Appendix on Ornamental Turning. (A book for beginners.) By
FRANCIS CAMPIN. Third edition, -with -wood engravings, crown 8vo,
cloth, 3*. 6d.
Valve Gears. — Treatise on Valve- Gears, with
special consideration of the Link-Motions of Locomotive Engines. By
Dr. GUSTAV ZEUNER, Professor of Applied Mechanics at the Confede-
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Edition, by Professor J. F. KLEIN, Lehigh University, Bethlehem, Pa.
Illustrated, 8vo, cloth, I2s. 6d.
Varnish. — The practical Polish and Varnish- Maker ;
a Treatise containing 750 practical Receipts and Formulae for the Manu-
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workers in Wood and Metal, and directions for using same. By H. C.
STANDAGE (Practical Chemist), author of 'The Artist's Manual of
Pigments.' Crown 8vo, cloth, 6s.
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20 CATALOGUE OF SCIENTIFIC BOOKS
Ventilation. — Health and Comfort in House Build-
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and with some Actual Experiments by J. DRYSDALE, M.D., and J. W.
HAYWARD, M.D. With, plates and woodcuts. Third edition, with some
New Sections, and the whole carefully revised, 8vo, cloth, js. 6d.
Warming and Ventilating. — A Practical
Treatise upon Warming Buildings by Hot Water, and upon Heat and
Heating Appliances in general ; with an inquiry respecting Ventilation,
the cause and action of Draughts in Chimneys and Flues, and the laws
relating to Combustion. By CHARLES HOOD, F.R.S. Re-written by
FREDERICK DYE. Third edition. 8vo, cloth, i$s.
Watch work. — Treatise on Watchwork, Past and
Present. By the Rev. H. L. NELTHROPP, M.A., F.S.A. With 32
illustrations , crown 8vo, cloth, 6s. 6d.
CONTENTS :
Definitions of Words and Terms used in Watchwork — Tools — Time — Historical Sum-
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Trains, etc. — Of Dial Wheels, or Motion Work — Length of Time of Going without Winding
up — The Verge — The Horizontal — The Duplex — The Lever — The Chronometer — Repeating
Watches— Keyless Watches — The Pendulum, or Spiral Spring — Compensation — Jewelling of
Pivot Holes — Clerkenwell — Fallacies of the Trade — Incapacity of Workmen — How to Choose
and Use a Watch, etc.
Water Softening. — Water Softening and Purifi-
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HAROLD COLLET. Crown 8vo, cloth, 5-r.
Waterworks. -- The Principles of Waterworks
Engineering. By J. H. TUDSBERY TURNER, B.Sc., Hunter Medallist
of Glasgow University, M. Inst. C.E., and A. W. BRIGHTMORE, M.Sc.,
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examples of Wells. By ERNEST SPON, Assoc. Mem. Inst. C.E.
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Wood-working Factories. — On the Arrange-
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SPONS' DICTIONARY OF ENGINEERING,
CIVIL, MECHANICAL, MILITARY, & NAVAL,
WITH
Technical Terms in French, German, Italian, and Spanish.
In 97 numbers, Super-royal 8vo, containing 3132 printed pages and 7414
engravings. Any number can be had separate : Nos. I to 95 is. each,
post free ; Nos. 96, 97, 2s., post free.
COMPLETE LIST OF ALL THE SUBJECTS :
Nos.
Abacus .. .. .. ..I
Adhesion .. .. .. I
Agricultural Engines .. I and 2
Air-Chamber .. .. ..2
Air- Pump .. .. .. ..2
Algebraic Signs .. .. ..2
Alloy 2
Aluminium .. .. ..2
Amalgamating Machine . . 2
Ambulance .. .. ..2
Anchors .. .. .. ..2
Anemometer . . . . 2 and 3
Angular Motion . . . . 3 and 4
Angle-iron.. .. .. 3
Angle of Friction . . . . 3
Animal Charcoal Machine . . 4
Antimony, 4; Anvil .. -.4
Aqueduct, 4 ; Arch . . 4
Archimedean Screw .. ..4
Arming Press .. .. 4 and 5
Armour, 5 ; Arsenic .. 5
Artesian Well 5
Artillery, 5 and 6 ; Assaying .. 6
Atomic Weights .. .. 6 and 7
Auger, 7; Axles .. .. 7
Balance, 7 ; Ballast .. .. 7
Bank Note Machinery .. .-7
Barn Machinery .. .. 7 and 8
Barker's Mill 8
Barometer, 8 ; Barracks . . 8
Barrage
Battery
Bell and Bell-hanging
Belts and Belting . .
Bismuth
Blast Furnace
Blowing Machine
Body Plan..
Boilers
Bond
Bone Mill
Boot-making Machinery
Boring and Blasting
Brake
Bread Machine
Brewing Apparatus
Brick-making Machines
Bridges
Buffer
Cables
Cam, 29 ; Canal . .
Candles
Cement, 30; Chimney
Coal Cutting and Wash ng Ma-
chinery
Coast Defence
Compasses
Construction
Cooler, 34 ; Copper
Cork-cutting Machine
Nos.
8 and 9
9 and 10
.. 10
10 and ii
.. II
11 and 12
.. 12
12 and 13
r3» J4> 15
15 and 16
.. 16
.. 16
16 to 19
19 and 20
.. 20
20 and 21
.. 21
21 tO 28
.. 28
28 and 29
..29
29 and 30
30
31* 32
.. 32
32 and 33
•- 34
• • 34
22
CATALOGUE OF SCIENTIFIC BOOKS
Nos.
Corrosion 34 and 35
Cotton Machinery . •• 35
Damming . . . . ; 35 to 37
Details of Engines . 37, 38
Displacement .. . ..38
Distilling Apparatus . 38 and 39
Diving and Diving Bells .. 39
Docks 39 and 40
Drainage .. .. ..40 and 4 1
Drawbridge .. . ..41
Dredging Machine . ..41
Dynamometer .. . 411043
Electro-Metallurgy . 43, 44
Engines, Varieties . 44, 45
Engines, Agricultural . I and 2
Engines, Marine .. . 74» 75
Engines, Screw .. . 89, 90
Engines, Stationary . 91, 92
Escapement .. . 45 , 46
Fan .. .. . ..46
File-cutting Machine . ..46
Fire-arms .. .. . 46, 47
Flax Machinery .. . 47, 48
Float Water-wheels . . . 48
Forging .. .. . ..48
Founding and Casting . 48 to 50
Friction, 50 ; Friction, Angle of 3
Fuel, 50; Furnace .. 50, 51
Fuze, 51 ; Gas 51
Gearing 51, 52
Gearing Belt .. .. 10, n
Geodesy .. .. .. 52 and 53
Glass Machinery .. .. -.53
•Gold, 53, 54; Governor.. .. 54
Gravity, 54 ; Grindstone . . 54
Gun-carriage, 54 ; Gun Metal . . 54
Gunnery .. .. .. 541056
Gunpowder .. .. .. 56
Gun Machinery .. .. 56, 57
Hand Tools ' ,.. .. 57,58
Hanger, 58; Harbour .. ..58
Haulage, 58, 59 ; Hinging . . 59
Hydraulics and Hydraulic Ma
chinery
Ice-making Machine
India-rubber
Indicator
Injector
Iron
Iron Ship Building
Irrigation ..
59 to 63
.. 63
.. 63
.. 63 and 64
.. ' .. 64
64 to 67
.. 67
.. 67 and 68
Nos.
Isomorphism, 68 ; Joints .. 68
Keels and Coal Shipping 68 and 69
Kiln, 69 ; Knitting Machine .. 69
Kyanising .. .. .. ..69
Lamp, Safety .. .. 69, 70
Lead .. .. .. ..70
Lifts, Hoists .. .. 70, 71
Lights, Buoys, Beacons .. 71 and 72
Limes, Mortars, and Cements .. 72
72,
73,
Locks and Lock Gates
Locomotive
Machine Tools
Manganese
Marine Engine
Materials of Construction
Measuring and Folding .
Mechanical Movements .
Mercury, 77 ; Metallurgy
Meter
Metric System
Mills
Molecule, 79 ; Oblique Arch
Ores, 79, 80 ; Ovens
Over-shot Water-wheel ..
Paper Machinery ..
Permanent Way ..
Piles and Pile-driving
Pipes
Planimeter
Pumps
Quarrying ..
Railway Engineering
Retaining Walls ..
Rivers,
Roads
Roofs
Rope-making Machinery
Scaffolding . . .
Screw Engines
Signals, 90; Silver
Stationary Engine
Stave-making & Cask Machinery 92
Steel, 92 ; Sugar Mill ,. 92,93
Surveying and Surveying Instru-
73
73
74
..74
74 and 75
75 and 76
.. 76
76, 77
•• 77
77,78
.. 78
78,79
.. 79
.. 80
80, 81
.. 81
81, 82
82 and 83
83,84
.. 84
84 and 85
.. 85
85 and 86
.. 86
>, 87 ; Riveted Joint . . 87
.. ' .. .. 87,88
89
90
91
92
ments
Telegraphy
Testing, 95 ; Turbine
Ventilation
Waterworks
Wood-working Machinery
Zinc
95:
93. 94
94, 95
•• 95
96, 97
96, 97
96,97
96,97
PUBLISHED BY E. & F. N. SPON, LIMITED. 23
In super-royal 8vo, 1168 pp., with 2400 illustrations > in 3 Divisions, cloth, price 13*. 6<a?.
each ; or i vol., cloth, a/. ; or half-morocco, a/. 8*.
A SUPPLEMENT
TO
SPONS' DICTIONARY OF ENGINEERING.
EDITED BY ERNEST SPON, MEMB. Soc. ENGINEERS.
Abacus, Counters, Speed ;
Indicators, and Slide i
Rule.
Agricultural Implements
and Machinery.
Air Compressors.
Animal Charcoal Ma-
chinery.
Antimony.
Axles and Axle-boxes.
Barn Machinery.
Belts and Belting.
Blasting. Boilers.
Brakes.
Brick Machinery.
Bridges.
Cages for Mines.
Calculus, Differential and
Integral.
Canals.
Carpentry.
Cast Iron.
Cement, Concrete,
Limes, and Mortar.
Chimney Shafts.
Coal Cleansing and
Washing.
Coal Mining.
Coal Cutting Machines.
Coke Ovens. Copper.
Docks. Drainage.
Dredging Machinery.
Dynamo - Electric and
Magneto-Electric Ma-
chines.
Dynamometers.
Electrical Engineering,
Telegraphy, Electric
Lighting and its prac-
tical details/Telephones
Engines, Varieties of.
Explosives. Fans.
Founding, Moulding and
the practical work of
the Foundry.
Gas, Manufacture of.
Hammers, Steam and
other Power.
Heat. Horse Power.
Hydraulics.
Hydro-geology.
Indicators. Iron.
Lifts, Hoists, and Eleva-
tors.
Lighthouses, Buoys, and
Beacons.
Machine Tools.
Materials of Const
tion.
Meters.
Ores, Machinery and
Processes employed to
Dress.
Piers.
Pile Driving.
Pneumatic Transmis-
sion.
Pumps.
Pyrometers.
Road Locomotives.
Rock Drills.
Rolling Stock.
Sanitary Engineering.
Shafting.
Steel.
Steam Navvy.
Stone Machinery.
Tramways.
Well Sinking.
24 CATALOGUE OF SCIENTIFIC BOOKS.
In demy 4to, handsomely bound in cloth, illustrated with 220 full page plates^
Price ls.
ARCHITECTURAL EXAMPLES
IN BRICK, STONE, WOOD, AND IRON.
A COMPLETE WORK ON THE DETAILS AND ARRANGEMENT
OF BUILDING- CONSTRUCTION AND DESIGN.
BY WILLIAM FULLERTON, ARCHITECT.
Containing 220 Plates, with numerous Drawings selected from the Architecture
of Former and Present Times.
The Details and Designs are Drawn to Scale, -|", J", £", and Full size
being chiefly used.
The Plates are arranged in Two Parts. The First Part contains
Details of Work in the four principal Building materials, the following
being a few of the subjects in this Part: — Various forms of Doors and
Windows, Wood and Iron Roofs, Half Timber Work, Porches,
Towers, Spires, Belfries, Flying Buttresses, Groining, Carving, Church
Fittings, Constructive and Ornamental Iron Work, Classic and Gothic
Molds and Ornament, Foliation Natural and Conventional, Stained
Glass, Coloured Decoration, a Section to Scale of the Great Pyramid,
Grecian and Roman Work, Continental and English Gothic, Pile
Foundations, Chimney Shafts according to the regulations of the
London County Council, Board Schools. The Second Part consists
of Drawings of Plans and Elevations of Buildings, arranged under the
following heads : — Workmen's Cottages and Dwellings, Cottage Resi-
dences and Dwelling Houses, Shops, Factories, Warehouses, Schools,
Churches and Chapels, Public Buildings, Hotels and Taverns, and
Buildings of a general character.
All the Plates are accompanied with particulars of the Work, with
Explanatory Notes and Dimensions of the various parts.
Specimen Pages, reduced Jrom the originals.
A^irtcforol Examjdu- Window. 7Z
26
CATALOGUE OF SCIENTIFIC BOOKS
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SPONS' ENCYCLOPEDIA
INDUSTRIAL ARTS, MANUFACTURES, AND COMMERCIAL
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following :—
Acids, 207 pp. 220 figs.
Alcohol, 23 pp. 1 6 figs.
Alcoholic Liquors, 13 pp.
Alkalies, 89 pp. 78 figs.
Alloys. Alum.
Asphalt. Assaying.
Beverages, 89 pp. 29 figs.
Blacks.
Bleaching Powder, 15 pp.
Bleaching, 5 1 pp. 48 figs.
Candles, 18 pp. 9 figs.
Carbon Bisulphide. ,
Celluloid, 9 pp.
Cements. Clay.
Coal-tar Products, 44 pp.
14 figs.
Cocoa, 8 pp.
Coffee, 32 pp. 13 figs.
Cork, 8 pp. 17 figs.
Cotton Manufactures, 62
pp. 57 figs.
Drugs, 38 pp.
Dyeing and Calico
Printing, 28 pp. 9 figs.
Dyestuffs, 1 6 pp.
Electro-Metallurgy, 13
pp.
Explosives, 22 pp. 33 figs.
Feathers.
Fibrous Substances, 92
pp. 79 figs.
Floor-cloth, 1 6 pp. 21
figs.
Food Preservation, 8 pp.
Fruit, 8 pp.
Fur, 5 pp.
Gas, Coal, 8 pp.
Gems.
Glass, 45 pp. 77 figs.
Graphite, 7 pp.
Hair, 7 pp.
Hair Manufactures.
Hats, 26 pp. 26 figs.
Honey. Hops.
Horn.
Ice, 10 pp. 14 figs.
Indiarubber Manufac-
tures, 23 pp. 17 figs.
Ink, 17 pp.
Ivory.
Jute Manufactures, 1 1
pp., II figs.
Knitted Fabrics —
Hosiery, 15 pp. 13 figs.
Lace, 13 pp. 9 figs.
Leather, 28 pp. 31 figs.
Linen Manufactures, 16
pp. 6 figs.
Manures, 21 pp. 30 figs.
Matches, 17 pp. 38 figs.
Mordants, 13 pp.
Narcotics, 47 pp.
Nuts, 10 pp.
Oils and Fatty Sub-
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Paint.
Paper, 26 pp. 23 figs.
Paraffin, 8 pp. 6 figs.
Pearl and Coral, 8 pp.
Perfumes, 10 pp.
Photography, 13 pp. 20
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Pigments, 9 pp. 6 figs.
Pottery, 46 pp. 57 figs.
Printing and Engraving,
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Rags.
Resinous and Gummy
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Rope, 1 6 pp. 17 figs.
Salt, 31 pp. 23 figs.
Silk, 8 pp.
Silk Manufactures, 9 pp.
ii figs.
Skins, 5 pp.
Small Wares, 4 pp.
Soap and Glycerine, 39
pp. 45 figs.
Spices, 1 6 pp.
Sponge, 5 pp.
Starch, 9 pp. 10 figs.
Sugar, 155 pp. 134
figs.
Sulphur.
Tannin, 18 pp.
Tea, 12 pp.
Timber, 13 pp.
Varnish, 15 pp.
Vinegar, 5 pp.
Wax, 5 pp.
Wool, 2 pp.
Woollen Manufactures,
58 pp. 39 figs.
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SECOND EDITION,
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WORKSHOP RECEIPTS
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SYNOPSIS OF CONTENTS.
Alloys
Bleaching
Bookbinding
Bronzing
Candle-making
Cements and Lutes
Cleansing
Crayons
Drawings
Dyeing
Electro-plating
Engraving
Etching
Explosives
Fireworks
Fluxes
Fulminates
Glass
Graining
Gunpowder
Iron & Steel Tem-
pering
Lathing and Plas-
tering
Marble Working
Painting
Paper
Paper-hanging
Papier-M&ehe
Pavements
Photography
Plating
Polishing
Pottery
Recovering Waste
Metal
28
CATALOGUE OF SCIENTIFIC BOOKS
Crown 8vo, cloth, 485 pages, with illustrations, 5*.
WORKSHOP RECEIPTS,
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SYNOPSIS OF CONTENTS.
Acidimetry and Alkali-
Disinfectants.
lodoform.
metry.
Dyeing, Staining, and
Isinglass.
Albumen.
Colouring.
Ivory substitutes.
Alcohol.
Essences.
Leather.
Alkaloids.
Extracts.
Luminous bodies.
Baking-powders.
Fireproofing.
Magnesia.
Bitters.
Gelatine, Glue, and Size.
Matches.
Bleaching.
Glycerine.
Paper.
Boiler Incrustations.
Gut.
Parchment.
Cements and Lutes.
Cleansing.
Hydrogen peroxide.
Ink.
Perchloric acid.
Potassium oxalate.
Confectionery.
Iodine.
Preserving.
Copying.
Pigments, Paint, and Painting : embracing the preparation of
Pigments, including alumina lakes, blacks (animal, bone, Frankfort, ivory,
lamp, sight, soot), blues (antimony, Antwerp, cobalt, cseruleum, Egyptian,
manganate, Paris, Peligot, Prussian, smalt, ultramarine), browns (bistre,
hinau, sepia, sienna, umber, Vandyke), greens (baryta, Brighton, Brunswick,
chrome, cobalt, Douglas, emerald, manganese, mitis, mountain, Prussian,
sap, Scheele's, Schweinfurth, titanium, verdigris, zinc), reds (Brazilwood lake,
carminated lake, carmine, Cassius purple, cobalt pink, cochineal lake, colco-
thar, Indian red, madder lake, red chalk, red lead, vermilion), whites (alum,
baryta, Chinese, lead sulphate, white lead — by American, Dutch, French,
German, Kremnitz, and Pattinson processes, precautions in making, and
composition of commercial samples — whiting, Wilkinson's white, zinc white),
yellows (chrome, gamboge, Naples, orpiment, realgar, yellow lakes) ; Paint
(vehicles, testing oils, driers, grinding, storing, applying, priming, drying,
rilling, coats, brushes, surface, water-colours, removing smell, discoloration ;
miscellaneous paints— cement paint for carton-pierre, copper paint, gold paint,
iron paint, lime paints, silicated paints, steatite paint, transparent paints,
tungsten paints, window paint, zinc paints) ; Painting (general instructions,
proportions of ingredients, measuring paint work ; carriage painting — priming
paint, best putty, finishing colour, cause of cracking, mixing the paints, oils,
driers, and colours, varnishing, importance of washing vehicles, re-varnishing,
how to dry paint ; woodwork painting).
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Crown 8vo, cloth, 480 pages, with 183 illustrations,
WORKSHOP RECEIPTS,
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SYNOPSIS OF CONTENTS.
Alloys.
Iridium.
Rubidium.
Aluminium.
Iron and Steel.
Ruthenium.
Antimony.
Lacquers and Lacquering,
Selenium.
Barium.
Lanthanum.
Silver.
Beryllium.
Lead.
Slag.
Bismuth.
Lithium.
Sodium.
Cadmium.
Lubricants.
Strontium.
Caesium.
Magnesium.
Tantalum.
Calcium.
Manganese.
Terbium.
Cerium.
Mercury.
Thallium.
Chromium.
Mica.
Thorium.
Cobalt.
Molybdenum.
Tin.
Copper.
Nickel.
Titanium.
Didymium.
Niobium.
Tungsten.
Enamels and Glazes.
Osmium.
Uranium.
Erbium.
Palladium.
Vanadium.
Gallium.
Platinum.
Yttrium.
Glass,
Potassium.
Zinc.
Gold.
Rhodium.
Zirconium.
Indium.
Electrics.— Alarms, Bells, Batteries. Carbons, Coils, Dynamos, Micro-
phones, Measuring, Phonographs, Telephones, &c., 130 pp., 112 illustrations.
30 CATALOGUE OF SCIENTIFIC BOOKS
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Waterproofing — rubber goods, cuprammonium processes, miscellaneous
preparations.
Packing and Storing articles of delicate odour or colour, of a deliquescent
character, liable to ignition, apt to suffer from insects or damp, or easily
broken.
Embalming and Preserving anatomical specimens.
Leather Polishes.
Cooling Air and Water, producing low temperatures, making ice, cooling
syrups and solutions, and separating salts from liquors by refrigeration.
Pumps and Siphons, embracing every useful contrivance for raising and
supplying water on a moderate scale, and moving corrosive, tenacious,
and other liquids.
Desiccating — air- and water-ovens, and other appliances for drying natural
and artificial products.
Distilling— water, tinctures, extracts, pharmaceutical preparations, essences,
perfumes, and alcoholic liquids.
Emulsifying as required by pharmacists and photographers.
Evaporating — saline and other solutions, and liquids demanding special
precautions.
Filtering — water, and solutions of various kinds.
Percolating and Macerating.
Electrotyping.
Stereotyping by both plaster and paper processes.
Bookbinding in all its details.
Straw Plaiting and the fabrication of baskets, matting, etc.
Musical Instruments — the preservation, tuning, and repair of pianos,
harmoniums, musical boxes, etc.
Clock and Watch Mending — adapted for intelligent amateurs.
Photography — recent development in rapid processes, handy apparatus,
numerous recipes for sensitizing and developing solutions, and applica-
tions to modern illustrative purposes.
PUBLISHED BY E. & F. N. SPON, LIMITED. 31
Crown 8vo, cloth, with 373 illustrations, price $s.
WORKSHOP RECEIPTS,
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the previous Series, as follows, viz. : —
Anemometers.
Barometers, How to make.
Boat Building.
Camera Lucida, How to use.
Cements and Lutes.
Cooling.
Copying.
Corrosion and Protection of Metal
Surfaces.
Dendrometer, How to use.
Desiccating.
Diamond Cutting and Polishing. Elec-
trics. New Chemical Batteries, Bells,
Commutators, Galvanometers, Cost
of Electric Lighting, Microphones,
Simple Motors, Phonogram and
Graphophone, Registering Appa-
ratus, Regulators, Electric Welding
and Apparatus, Transformers.
Evaporating.
Explosives.
Filtering.
Fireproofing, Buildings, Textile Fa-
brics.
Fire-extinguishing Compounds and
Apparatus.
Glass Manipulating. Drilling, Cut-
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Powdering, &c.
Glass Manipulations for Laboratory
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Labels. Lacquers.
Illuminating Agents.
Inks. Writing, Copying, Invisible,
Marking, Stamping.
Magic Lanterns, their management
and preparation of slides.
Metal WTork. Casting Ornamental
Metal Work, Copper Welding
Enamels for Iron and other Metals,
Gold Beating, Smiths' Work.
Modelling and Plaster Casting.
Netting.
Packing and Storing. Acids, &c.
Percolation.
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Preserving Food, Plants, &c.
Pumps and Syphons for various
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Repairing Books.
Rope Tackle.
Stereotyping.
Taps, Various.
Tobacco Pipe Manufacture.
Tying and Splicing Ropes.
Velocipedes, Repairing.
Walking Sticks.
Waterproofing.
32 CATALOGUE OF SCIENTIFIC BOOKS.
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SPONS'
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CONTENTS.
Mechanical Drawing — Casting and Founding in Iron, Brass, Bronze,
and other Alloys — Forging and Finishing Iron — Sheetmetal Working
— Soldering, Brazing, and Burning — Carpentry and Joinery, embracing
descriptions of some 400 Woods, over 200 Illustrations of Tools and
their uses, Explanations (with Diagrams) of 116 joints and hinges, and
Details of Construction of Workshop appliances, rough furniture,
Garden and Yard Erections, and House Building — Cabinet-Making
and Veneering — Carving and Fretcutting — Upholstery — Painting,
Graining, and Marbling — Staining Furniture, Woods, Floors, and
Fittings — Gilding, dead and bright, on various grounds — Polishing
Marble, Metals, and Wood — Varnishing — Mechanical movements,
illustrating contrivances for transmitting motion — Turning in Wood
and Metals — Masonry, embracing Stonework, Brickwork, Terracotta
and Concrete — Roofing with Thatch, Tiles, Slates, Felt, Zinc, &c.—
Glazing with and without putty, and lead glazing — Plastering and
Whitewashing — Paper-hanging — Gas-fitting — Bell-hanging, ordinary
and electric Systems — Lighting — Warming — Ventilating — Roads,
Pavements, and Bridges — Hedges, Ditches, and Drains — Water
Supply and Sanitation— Hints on House Construction suited to new
countries.
E. & F. N. SPON, Limited, 125 Strand, London.
New York : SPON & CHAMBEBLATN.
LONDON : PRINTED BY WILLIAM CLOWES AND SONS, LIMITED, STAMFORD STREET
AND CHARING CROSS.
THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW
AN INITIAL FINE OF 25 CENTS
WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO SO CENTS ON THE FOURTH
DAY AND TO $1.OO ON THE SEVENTH DAY
OVERDUE.
NOV 6
BEG. OR. NOV
LD 21-100m-7,'33
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