Chapter IX: Part 9

These results are for the centre deflections of main girders, but Stone infers that the augmentation of stress for any member, due to causes included in impact allowance, will be the same percentage for the same ratios of live to dead load stresses. Valuable measurements of the deformations of girders and tension members due to moving trains have been made by S.W. Robinson (_Trans. Am. Soc. C.E._ xvi.) and by F.E. Turneaure (_Trans. Am. Soc. C.E._ xli.). The latter used a recording deflectometer and two recording extensometers. The observations are difficult, and the inertia of the instrument is liable to cause error, but much care was taken. The most striking conclusions from the results are that the locomotive balance weights have a large effect in causing vibration, and next, that in certain cases the vibrations are cumulative, reaching a value greater than that due to any single impact action. Generally: (1) At speeds less than 25 m. an hour there is not much vibration. (2) The increase of deflection due to impact at 40 or 50 m. an hour is likely to reach 40 to 50% for girder spans of less than 50 ft. (3) This percentage decreases rapidly for longer spans, becoming about 25% for 75-ft. spans. (4) The increase per cent of boom stresses due to impact is about the same as that of deflection; that in web bracing bars is rather greater. (5) Speed of train produces no effect on the mean deflection, but only on the magnitude of the vibrations.

A purely empirical allowance for impact stresses has been proposed, amounting to 20% of the live load stresses for floor stringers; 15% for floor cross girders; and for main girders, 10% for 40-ft. spans, and 5% for 100-ft. spans. These percentages are added to the live load stresses.

iii. _Dead Load._--The dead load consists of the weight of main girders, flooring and wind-bracing. It is generally reckoned to be uniformly distributed, but in large spans the distribution of weight in the main girders should be calculated and taken into account. The weight of the bridge flooring depends on the type adopted. Road bridges vary so much in the character of the flooring that no general rule can be given. In railway bridges the weight of sleepers, rails, &c., is 0.2 to 0.25 tons per ft. run for each line of way, while the rail girders, cross girders, &c., weigh 0.15 to 0.2 tons. If a footway is added about 0.4 ton per ft. run may be allowed for this. The weight of main girders increases with the span, and there is for any type of bridge a limiting span beyond which the dead load stresses exceed the assigned limit of working stress.

Let W_l be the total live load, W_f the total flooring load on a bridge of span l, both being considered for the present purpose to be uniform per ft. run. Let k(W_l+W_f) be the weight of main girders designed to carry W_l+W_f, but not their own weight in addition. Then

W_g = (W_l+W_f)(k+k^2+k^3 ...)

will be the weight of main girders to carry W_l+W_f and their own weight (Buck, _Proc. Inst. C.E._ lxvii. p. 331). Hence,

W_g = (W_l+W_f)k/(1-k).

Since in designing a bridge W_l+W_f is known, k(W_l+W_f) can be found from a provisional design in which the weight W_g is neglected. The actual bridge must have the section of all members greater than those in the provisional design in the ratio k/(1-k).

Waddell (_De Pontibus_) gives the following convenient empirical relations. Let w_1, w_2 be the weights of main girders per ft. run for a live load p per ft. run and spans l_1, l_2. Then

w_2/w_1 = ½ [l_2/l_1+(l_2/l_1)^2].

Now let w_1', w_2' be the girder weights per ft. run for spans l_1, l_2, and live loads p' per ft. run. Then

w_2'/w_2 = 1/5(1+4p'/p)

w_2'/w_1 = 1/10[l_2/l_1+(l_2/l_1)^2](1+4p'/p)

A partially rational approximate formula for the weight of main girders is the following (Unwin, _Wrought Iron Bridges and Roofs_, 1869, p. 40):--

Let w = total live load per ft. run of girder; w_2 the weight of platform per ft. run; w_3 the weight of main girders per ft. run, all in tons; l = span in ft.; s = average stress in tons per sq. in. on gross section of metal; d = depth of girder at centre in ft.; r = ratio of span to depth of girder so that r = l/d. Then

w_3 = (w_1+w_2)l^2/(Cds-l_2) = (w_1+w_2)lr/(Cs-lr),

where C is a constant for any type of girder. It is not easy to fix the average stress s per sq. in. of gross section. Hence the formula is more useful in the form

w = (w_1+w_2)l^2/(Kd-l^2) = (w_1+w_2)lr/(K-lr)

where K = (w_1+w_2+w_3)lr/w_3 is to be deduced from the data of some bridge previously designed with the same working stresses. From some known examples, C varies from 1500 to 1800 for iron braced parallel or bowstring girders, and from 1200 to 1500 for similar girders of steel. K = 6000 to 7200 for iron and = 7200 to 9000 for steel bridges.

iv. _Wind Pressure._--Much attention has been given to wind action since the disaster to the Tay bridge in 1879. As to the maximum wind pressure on small plates normal to the wind, there is not much doubt. Anemometer observations show that pressures of 30 lb per sq. ft. occur in storms annually in many localities, and that occasionally higher pressures are recorded in exposed positions. Thus at Bidstone, Liverpool, where the gauge has an exceptional exposure, a pressure of 80 lb per sq. ft. has been observed. In tornadoes, such as that at St Louis in 1896, it has been calculated, from the stability of structures overturned, that pressures of 45 to 90 lb per sq. ft. must have been reached. As to anemometer pressures, it should be observed that the recorded pressure is made up of a positive front and negative (vacuum) back pressure, but in structures the latter must be absent or only partially developed. Great difference of opinion exists as to whether on large surfaces the average pressure per sq. ft. is as great as on small surfaces, such as anemometer plates. The experiments of Sir B. Baker at the Forth bridge showed that on a surface 30 ft. × 15 ft. the intensity of pressure was less than on a similarly exposed anemometer plate. In the case of bridges there is the further difficulty that some surfaces partially [v.04 p.0549] shield other surfaces; one girder, for instance, shields the girder behind it (see _Brit. Assoc. Report_, 1884). In 1881 a committee of the Board of Trade decided that the maximum wind pressure on a vertical surface in Great Britain should be assumed in designing structures to be 56 lb per sq. ft. For a plate girder bridge of less height than the train, the wind is to be taken to act on a surface equal to the projected area of one girder and the exposed part of a train covering the bridge. In the case of braced girder bridges, the wind pressure is taken as acting on a continuous surface extending from the rails to the top of the carriages, plus the vertical projected area of so much of one girder as is exposed above the train or below the rails. In addition, an allowance is made for pressure on the leeward girder according to a scale. The committee recommended that a factor of safety of 4 should be taken for wind stresses. For safety against overturning they considered a factor of 2 sufficient. In the case of bridges not subject to Board of Trade inspection, the allowance for wind pressure varies in different cases. C. Shaler Smith allows 300 lb per ft. run for the pressure on the side of a train, and in addition 30 lb per sq. ft. on twice the vertical projected area of one girder, treating the pressure on the train as a travelling load. In the case of bridges of less than 50 ft. span he also provides strength to resist a pressure of 50 lb per sq. ft. on twice the vertical projection of one truss, no train being supposed to be on the bridge.

19. _Stresses Permitted._--For a long time engineers held the convenient opinion that, if the total dead and live load stress on any section of a structure (of iron) did not exceed 5 tons per sq. in., ample safety was secured. It is no longer possible to design by so simple a rule. In an interesting address to the British Association in 1885, Sir B. Baker described the condition of opinion as to the safe limits of stress as chaotic. "The old foundations," he said, "are shaken, and engineers have not come to an agreement respecting the rebuilding of the structure. The variance in the strength of existing bridges is such as to be apparent to the educated eye without any calculation. In the present day engineers are in accord as to the principles of estimating the magnitude of the stresses on the members of a structure, but not so in proportioning the members to resist those stresses. The practical result is that a bridge which would be passed by the English Board of Trade would require to be strengthened 5% in some parts and 60% in others, before it would be accepted by the German government, or by any of the leading railway companies in America." Sir B. Baker then described the results of experiments on repetition of stress, and added that "hundreds of existing bridges which carry twenty trains a day with perfect safety would break down quickly under twenty trains an hour. This fact was forced on my attention nearly twenty-five years ago by the fracture of a number of girders of ordinary strength under a five-minutes' train service."

Practical experience taught engineers that though 5 tons per sq. in. for iron, or 6½ tons per sq. in. for steel, was safe or more than safe for long bridges with large ratio of dead to live load, it was not safe for short ones in which the stresses are mainly due to live load, the weight of the bridge being small. The experiments of A. Wöhler, repeated by Johann Bauschinger, Sir B. Baker and others, show that the breaking stress of a bar is not a fixed quantity, but depends on the range of variation of stress to which it is subjected, if that variation is repeated a very large number of times. Let K be the breaking strength of a bar per unit of section, when it is loaded once gradually to breaking. This may be termed the statical breaking strength. Let k_{max.} be the breaking strength of the same bar when subjected to stresses varying from k_{max.} to k_{min.} alternately and repeated an indefinitely great number of times; k_{min.} is to be reckoned + if of the same kind as k_{max.} and - if of the opposite kind (tension or thrust). The range of stress is therefore k_{max.}-k_{min.}, if the stresses are both of the same kind, and k_{max.}+k_{min.}, if they are of opposite kinds. Let [Delta] = k_{max.} ± k_{min.} = the range of stress, where [Delta] is always positive. Then Wöhler's results agree closely with the rule,

k_{max.} = ½[Delta]+[root](K²-n[Delta]K),

where n is a constant which varies from 1.3 to 2 in various qualities of iron and steel. For ductile iron or mild steel it may be taken as 1.5. For a statical load, range of stress nil, [Delta] = 0, k_{max.} = K, the statical breaking stress. For a bar so placed that it is alternately loaded and the load removed, [Delta] = k_{max.} and k_{max.} = 0.6 K. For a bar subjected to alternate tension and compression of equal amount, [Delta] = 2 f_{max.} and k_{max.} = 0.33 K. The safe working stress in these different cases is k_{max.} divided by the factor of safety. It is sometimes said that a bar is "fatigued" by repeated straining. The real nature of the action is not well understood, but the word fatigue may be used, if it is not considered to imply more than that the breaking stress under repetition of loading diminishes as the range of variation increases.

It was pointed out as early as 1869 (Unwin, _Wrought Iron Bridges and Roofs_) that a rational method of fixing the working stress, so far as knowledge went at that time, would be to make it depend on the ratio of live to dead load, and in such a way that the factor of safety for the live load stresses was double that for the dead load stresses. Let A be the dead load and B the live load, producing stress in a bar; [rho] = B/A the ratio of live to dead load; f_1 the safe working limit of stress for a bar subjected to a dead load only and f the safe working stress in any other case. Then

f_1 (A+B)/(A+2B) = f_1(1+[rho])/(1+2[rho]).

The following table gives values of f so computed on the assumption that f_1 = 7½ tons per sq. in. for iron and 9 tons per sq. in. for steel.

_Working Stress for combined Dead and Live Load. Factor of Safety twice as great for Live Load as for Dead Load._

----------------------+-------+----------+-----------------------------+
| Ratio | 1+[rho] |Values of f, tons per sq. in.|
| [rho] | ------- +-----------------------------+
| | 1+2[rho] | Iron. | Mild Steel.|
----------------------+-------+----------+----------------+------------+
All dead load | 0 | 1.00 | 7.5 | 9.0 |
| .25 | 0.83 | 6.2 | 7.5 |
| .33 | 0.78 | 5.8 | 7.0 |
| .50 | 0.75 | 5.6 | 6.8 |
| .66 | 0.71 | 5.3 | 6.4 |
Live load = Dead load | 1.00 | 0.66 | 4.9 | 5.9 |
| 2.00 | 0.60 | 4.5 | 5.4 |
| 4.00 | 0.56 | 4.2 | 5.0 |
All live load | [inf] | 0.50 | 3.7 | 4.5 |
----------------------+-------+----------+----------------+------------+

Bridge sections designed by this rule differ little from those designed by formulae based directly on Wöhler's experiments. This rule has been revived in America, and appears to be increasingly relied on in bridge-designing. (See _Trans. Am. Soc. C.E._ xli. p. 156.)

The method of J.J. Weyrauch and W. Launhardt, based on an empirical expression for Wöhler's law, has been much used in bridge designing (see _Proc. Inst. C.E._ lxiii. p. 275). Let t be the _statical breaking strength_ of a bar, loaded once gradually up to fracture (t = breaking load divided by original area of section); u the breaking strength of a bar loaded and unloaded an indefinitely great number of times, the stress varying from u to 0 alternately (this is termed the _primitive strength_); and, lastly, let s be the breaking strength of a bar subjected to an indefinitely great number of repetitions of stresses equal and opposite in sign (tension and thrust), so that the stress ranges alternately from s to -s. This is termed the _vibration strength_. Wöhler's and Bauschinger's experiments give values of t, u, and s, for some materials. If a bar is subjected to alternations of stress having the range [Delta] = f_{max.}-f_{min.}, then, by Wöhler's law, the bar will ultimately break, if

f_{max.} = F[Delta], . . . (1)

where F is some unknown function. Launhardt found that, for stresses always of the same kind, F = (t-u)/(t-f_{max.}) approximately agreed with experiment. For stresses of different kinds Weyrauch found F = (u-s)/(2u-s-f_{max.}) to be similarly approximate. Now let f_{max.}/f_{min.} = [phi], where [phi] is + or - according as the stresses are of the same or opposite signs. Putting the values of F in (1) and solving for f_{max.}, we get for the breaking stress of a bar subjected to repetition of varying stress,

f_{max.} = u(1+(t-u)[phi]/u) [Stresses of same sign.]
f_{max.} = u(1+(u-s)[phi]/u) [Stresses of opposite sign.]

The working stress in any case is f_{max.} divided by a factor of safety. Let that factor be 3. Then Wöhler's results for iron and Bauschinger's for steel give the following equations for tension or thrust:--

Iron, working stress, f = 4.4 (1+½[phi])
Steel, working stress, f = 5.87 (1+½[phi]).

In these equations [phi] is to have its + or - value according to the case considered. For shearing stresses the working stress may have 0.8 of its value for tension. The following table gives values of the working stress calculated by these equations:--

_Working Stress for Tension or Thrust by Launhardt and Weyrauch Formula._

------------------------+-------+-----------+--------------------+
| [phi] | [phi] | Working Stress f, |
| | 1 + ----- | tons per sq. in. |
| | 2 +--------------------+
| | | Iron. | Steel. |
------------------------+-------+-----------+--------------------+
All dead load | 1.0 | 1.5 | 6.60 | 8.80 |
| 0.75 | 1.375 | 6.05 | 8.07 |
| 0.50 | 1.25 | 5.50 | 7.34 |
| 0.25 | 1.125 | 4.95 | 6.60 |
All live load | 0.00 | 1.00 | 4.40 | 5.87 |
| -0.25 | 0.875 | 3.85 | 5.14 |
| -0.50 | 0.75 | 3.30 | 4.40 |
| -0.75 | 0.625 | 2.75 | 3.67 |
Equal stresses + and - | -1.00 | 0.500 | 2.20 | 2.93 |
------------------------+-------+-----------+--------------------+

[v.04 p.0550] To compare this with the previous table, [phi] = (A+B)/A = 1+[rho]. Except when the limiting stresses are of opposite sign, the two tables agree very well. In bridge work this occurs only in some of the bracing bars.

It is a matter of discussion whether, if fatigue is allowed for by the Weyrauch method, an additional allowance should be made for impact. There was no impact in Wöhler's experiments, and therefore it would seem rational to add the impact allowance to that for fatigue; but in that case the bridge sections become larger than experience shows to be necessary. Some engineers escape this difficulty by asserting that Wöhler's results are not applicable to bridge work. They reject the allowance for fatigue (that is, the effect of repetition) and design bridge members for the total dead and live load, plus a large allowance for impact varied according to some purely empirical rule. (See Waddell, _De Pontibus_, p.7.) Now in applying Wöhler's law, f_{max.} for any bridge member is found for the maximum possible live load, a live load which though it may sometimes come on the bridge and must therefore be provided for, is not the usual live load to which the bridge is subjected. Hence the range of stress, f_{max.}-f_{min.}, from which the working stress is deduced, is not the ordinary range of stress which is repeated a practically infinite number of times, but is a range of stress to which the bridge is subjected only at comparatively long intervals. Hence practically it appears probable that the allowance for fatigue made in either of the tables above is sufficient to cover the ordinary effects of impact also.

English bridge-builders are somewhat hampered in adopting rational limits of working stress by the rules of the Board of Trade. Nor do they all accept the guidance of Wöhler's law. The following are some examples of limits adopted. For the Dufferin bridge (steel) the working stress was taken at 6.5 tons per sq. in. in bottom booms and diagonals, 6.0 tons in top booms, 5.0 tons in verticals and long compression members. For the Stanley bridge at Brisbane the limits were 6.5 tons per sq. in. in compression boom, 7.0 tons in tension boom, 5.0 tons in vertical struts, 6.5 tons in diagonal ties, 8.0 tons in wind bracing, and 6.5 tons in cross and rail girders. In the new Tay bridge the limit of stress is generally 5 tons per sq. in., but in members in which the stress changes sign 4 tons per sq. in. In the Forth bridge for members in which the stress varied from 0 to a maximum frequently, the limit was 5.0 tons per sq. in., or if the stress varied rarely 5.6 tons per sq. in.; for members subjected to alternations of tension and thrust frequently 3.3 tons per sq. in. or 5 tons per sq. in. if the alternations were infrequent. The shearing area of rivets in tension members was made 1½ times the useful section of plate in tension. For compression members the shearing area of rivets in butt-joints was made half the useful section of plate in compression.

20. _Determination of Stresses in the Members of Bridges._--It is convenient to consider beam girder or truss bridges, and it is the stresses in the main girders which primarily require to be determined. A main girder consists of an upper and lower flange, boom or chord and a vertical web. The loading forces to be considered are vertical, the horizontal forces due to wind pressure are treated separately and provided for by a horizontal system of bracing. For practical purposes it is accurate enough to consider the booms or chords as carrying exclusively the horizontal tension and compression and the web as resisting the whole of the vertical and, in a plate web, the equal horizontal shearing forces. Let fig. 37 represent a beam with any system of loads W_1, W_2, ... W_n.

The reaction at the right abutment is

R_2 = W_1x_1/l+W_2x_2/l+...

That at the left abutment is

R_1 = W_1+W_2+...-R_2.

Consider any section a b. The total shear at a b is

S = R-[Sigma](W_1+W_2 ...)

where the summation extends to all the loads to the left of the section. Let p_1, p_2 ... be the distances of the loads from a b, and p the distance of R_1 from a b; then the bending moment at a b is

M = R_1p-[Sigma](W_1p_1+W_2p_2 ...)

where the summation extends to all the loads to the left of a b. If the loads on the right of the section are considered the expressions are similar and give the same results.

If A_t A_c are the cross sections of the tension and compression flanges or chords, and h the distance between their mass centres, then on the assumption that they resist all the direct horizontal forces the total stress on each flange is

H_t = H_c = M/h

and the intensity of stress of tension or compression is

f_t = M/A_th,
f_c = M/A_ch.

If A is the area of the plate web in a vertical section, the intensity of shearing stress is

f_x = S/A

and the intensity on horizontal sections is the same. If the web is a braced web, then the vertical component of the stress in the web bars cut by the section must be equal to S.

21. _Method of Sections. A. Ritter's Method._--In the case of braced structures the following method is convenient: When a section of a girder can be taken cutting only three bars, the stresses in the bars can be found by taking moments. In fig. 38 m n cuts three bars, and the forces in the three bars cut by the section are C, S and T. There are to the left of the section the external forces, R, W_1, W_2. Let s be the perpendicular from O, the join of C and T on the direction of S; t the perpendicular from A, the join of C and S on the direction of T; and c the perpendicular from B, the join of S and T on the direction of C. Taking moments about O,

R_x-W_1(x+a)-W_2(x+2a) = Ss;

taking moments about A,

R3a-W_12a-W_2a = Tt;

and taking moments about B,

R2a-W_1a = Cc

Or generally, if M_1 M_2 M_3 are the moments of the external forces to the left of O, A, and B respectively, and s, t and c the perpendiculars from O, A and B on the directions of the forces cut by the section, then

Ss = M_1; Tt = M_2 and Cc = M_3.

Still more generally if H is the stress on any bar, h the perpendicular distance from the join of the other two bars cut by the section, and M is the moment of the forces on one side of that join,

Hh = M.

22. _Distribution of Bending Moment and Shearing Force._--Let a girder of span l, fig. 39, supported at the ends, carry a fixed load W at m from the right abutment. The reactions at the abutments are R_1 = Wm/l and R_2 = W(l-m)/l. The shears on vertical sections to the left and right of the load are R_1 and -R_2, and the distribution of shearing force is given by two rectangles. Bending moment increases uniformly from either abutment to the load, at which the bending moment is M = R_2m = R_1(l-m). The distribution of bending moment is given by the ordinates of a triangle. Next let the girder carry a uniform load w per ft. run (fig. 40). The total load [v.04 p.0551] is wl; the reactions at abutments, R_1 = R_2 = ½wl. The distribution of shear on vertical sections is given by the ordinates of a sloping line. The greatest bending moment is at the centre and = M_c = 1/8wl^2. At any point x from the abutment, the bending moment is M = ½wx(l-x), an equation to a parabola.

23. _Shear due to Travelling Loads._--Let a uniform train weighing w per ft. run advance over a girder of span 2c, from the left abutment. When it covers the girder to a distance x from the centre (fig. 41) the total load is w(c+x); the reaction at B is

R_2 = w(c+x)×(c+x)/4c = w/4c(c+x)²,

which is also the shearing force at C for that position of the load. As the load travels, the shear at the head of the train will be given by the ordinates of a parabola having its vertex at A, and a maximum F_{max.} = -½wl at B. If the load travels the reverse way, the shearing force at the head of the train is given by the ordinates of the dotted parabola. The greatest shear at C for any position of the load occurs when the head of the train is at C. For any load p between C and B will increase the reaction at B and therefore the shear at C by part of p, but at the same time will diminish the shear at C by the whole of p. The web of a girder must resist the maximum shear, and, with a travelling load like a railway train, this is greater for partial than for complete loading. Generally a girder supports both a dead and a live load. The distribution of total shear, due to a dead load w_l per ft. run and a travelling load w_l per ft. run, is shown in fig. 42, arranged so that the dead load shear is added to the maximum travelling load shear of the same sign.

24. _Counterbracing._--In the case of girders with braced webs, the tension bars of which are not adapted to resist a thrust, another circumstance due to the position of the live load must be considered. For a train advancing from the left, the travelling load shear in the left half of the span is of a different sign from that due to the dead load. Fig. 43 shows the maximum shear at vertical sections due to a dead and travelling load, the latter advancing (fig. 43, a) from the left and (fig. 43, b) from the right abutment. Comparing the figures it will be seen that over a distance x near the middle of the girder the shear changes sign, according as the load advances from the left or the right. The bracing bars, therefore, for this part of the girder must be adapted to resist either tension or thrust. Further, the range of stress to which they are subjected is the sum of the stresses due to the load advancing from the left or the right.

25. _Greatest Shear when concentrated Loads travel over the Bridge._--To find the greatest shear with a set of concentrated loads at fixed distances, let the loads advance from the left abutment, and let C be the section at which the shear is required (fig. 44). The greatest shear at C may occur with W_1 at C. If W_1 passes beyond C, the shear at C will probably be greatest when W_2 is at C. Let R be the resultant of the loads on the bridge when W_1 is at C. Then the reaction at B and shear at C is Rn/l. Next let the loads advance a distance a so that W_2 comes to C. Then the shear at C is R(n+a)/l-W_1, plus any reaction d at B, due to any additional load which has come on the girder during the movement. The shear will therefore be increased by bringing W_2 to C, if Ra/l+d > W_1 and d is generally small and negligible. This result is modified if the action of the load near the section is distributed to the bracing intersections by rail and cross girders. In fig. 45 the action of W is distributed to A and B by the flooring. Then the loads at A and B are W(p-x)/p and Wx/p. Now let C (fig. 46) be the section at which the greatest shear is required, and let the loads advance from the left till W_1 is at C. If R is the resultant of the loads then on the girder, the reaction at B and shear at C is Rn/l. But the shear may be greater when W_2 is at C. In that case the shear at C becomes R(n+a)/l+d-W_1, if a > p, and R(n+a)/l+d-W_1a/p, if a < p. If we neglect d, then the shear increases by moving W_2 to C, if Ra/l > W_1 in the first case, and if Ra/l > W_1a/p in the second case.

26. _Greatest Bending Moment due to travelling concentrated Loads._--For the greatest bending moment due to a travelling live load, let a load of w per ft. run advance from the left abutment (fig. 47), and let its centre be at x from the left abutment. The reaction at B is 2wx²/l and the bending moment at any section C, at m from the left abutment, is 2wx²/(l-m)/l, which increases as x increases till the span is covered. Hence, for uniform travelling loads, the bending moments are greatest when the loading is complete. In that case the loads on either side of C are proportional to m and l-m. In the case of a series of travelling loads at fixed distances apart passing over the girder from the left, let W_1, W_2 (fig. 48), at distances x and x+a from the left abutment, be their resultants on either side of C. Then the reaction at B is W_1x/l+W_2(x+a)/l. The bending moment at C is

M = W_1x(l-m)/l+W_2m{1-(x+a)/l}.

If the loads are moved a distance [Delta]x to the right, the bending moment becomes

M+[Delta]M = W_1(x+[Delta]x)(l-m)/l+W_2m{1-(x+[Delta]x+a)/l}
[Delta]m = W_1[Delta]x(l-m)/l-W_2[Delta]xm/l,

and this is positive or the bending moment increases, if W_1(l-m) > W_2m, or if W_1/m > W_2/(l-m). But these are the average loads per ft. run to the left and right of C. Hence, if the average load to the left of a section is greater than that to the right, the bending moment at the section will be increased by moving the loads to the right, and vice versa. Hence the maximum bending moment at C for a series of travelling loads will occur when the average load is the same on either side of C. If one of the loads is at C, spread over a very small distance in the neighbourhood of C, then a very small displacement of the loads will permit the fulfilment of the condition. Hence the criterion for the position of the loads which makes the moment at C greatest is this: one load must be at C, and the other loads must be distributed, so that the average loads per ft. on either side of C (the load at C being neglected) are nearly equal. If the loads are very unequal in magnitude or distance this condition may be satisfied for more than one position of the loads, but it is not difficult to ascertain which position gives the maximum moment. Generally one of the largest of the loads must be at C with as many others to right and left as is consistent with that condition.

This criterion may be stated in another way. The greatest bending moment will occur with one of the greatest loads at the section, and when this further condition is satisfied. Let fig. 49 represent a beam with the series of loads travelling from the right. Let a b be [v.04 p.0552] the section considered, and let W_x be the load at a b when the bending moment there is greatest, and W_n the last load to the right then on the bridge. Then the position of the loads must be that which satisfies the condition

x W_1+W_2+... W_{x-1}
--- greater than ------------------------
l W_1+W_2+... W_n

x W_1+W_2+... W_x
--- less than ------------------------
l W_1+W_2+... W_n

Fig. 50 shows the curve of bending moment under one of a series of travelling loads at fixed distances. Let W_1, W_2, W_3 traverse the girder from the left at fixed distances a, b. For the position shown the distribution of bending moment due to W_1 is given by ordinates of the triangle A'CB'; that due to W_2 by ordinates of A'DB'; and that due to W_3 by ordinates A'EB'. The total moment at W_1, due to three loads, is the sum mC+mn+mo of the intercepts which the triangle sides cut off from the vertical under W_1. As the loads move over the girder, the points C, D, E describe the parabolas M_1, M_2, M_3, the middle ordinates of which are ¼W_1l, ¼W_2l, and ¼W_3l. If these are first drawn it is easy, for any position of the loads, to draw the lines B'C, B'D, B'E, and to find the sum of the intercepts which is the total bending moment under a load. The lower portion of the figure is the curve of bending moments under the leading load. Till W_1 has advanced a distance a only one load is on the girder, and the curve A"F gives bending moments due to W_1 only; as W_1 advances to a distance a+b, two loads are on the girder, and the curve FG gives moments due to W_1 and W_2. GB" is the curve of moments for all three loads W_1+W_2+W_3.

Fig. 51 shows maximum bending moment curves for an extreme case of a short bridge with very unequal loads. The three lightly dotted parabolas are the curves of maximum moment for each of the loads taken separately. The three heavily dotted curves are curves of maximum moment under each of the loads, for the three loads passing over the bridge, at the given distances, from left to right. As might be expected, the moments are greatest in this case at the sections under the 15-ton load. The heavy continuous line gives the last-mentioned curve for the reverse direction of passage of the loads.

With short bridges it is best to draw the curve of maximum bending moments for some assumed typical set of loads in the way just described, and to design the girder accordingly. For longer bridges the funicular polygon affords a method of determining maximum bending moments which is perhaps more convenient. But very great accuracy in drawing this curve is unnecessary, because the rolling stock of railways varies so much that the precise magnitude and distribution of the loads which will pass over a bridge cannot be known. All that can be done is to assume a set of loads likely to produce somewhat severer straining than any probable actual rolling loads. Now, except for very short bridges and very unequal loads, a parabola can be found which includes the curve of maximum moments. This parabola is the curve of maximum moments for a travelling load uniform per ft. run. Let w_e be the load per ft. run which would produce the maximum moments represented by this parabola. Then w_e may be termed the uniform load per ft. equivalent to any assumed set of concentrated loads. Waddell has calculated tables of such equivalent uniform loads. But it is not difficult to find w_e, approximately enough for practical purposes, very simply. Experience shows that (a) a parabola having the same ordinate at the centre of the span, or (b) a parabola having the same ordinate at one-quarter span as the curve of maximum moments, agrees with it closely enough for practical designing. A criterion already given shows the position of any set of loads which will produce the greatest bending moment at the centre of the bridge, or at one-quarter span. Let M_c and M_a be those moments. At a section distant x from the centre of a girder of span 2c, the bending moment due to a uniform load w_e per ft run is

M = ½w_e(c-x)(c+x).

Putting x = 0, for the centre section

M_c = ½w_ec^2;

and putting x = ½c, for section at quarter span

M_a = 3/8w_ec^2.

From these equations a value of w_e can be obtained. Then the bridge is designed, so far as the direct stresses are concerned, for bending moments due to a uniform dead load and the uniform equivalent load w_e.

27. _Influence Lines._--In dealing with the action of travelling loads much assistance may be obtained by using a line termed an _influence line_. Such a line has for abscissa the distance of a load from one end of a girder, and for ordinate the bending moment or shear at any given section, or on any member, due to that load. Generally the influence line is drawn for unit load. In fig. 52 let A'B' be a girder supported at the ends and let it be required to investigate the bending moment at C' due to unit load in any position on the girder. When the load is at F', the reaction at B' is m/l and the moment at C' is m(l-x)/l, which will be reckoned positive, when it resists a tendency of the right-hand part of the girder to turn counter-clockwise. Projecting A'F'C'B' on to the horizontal AB, take Ff = m(l-x)/l, the moment at C of unit load at F. If this process is repeated for all positions of the load, we get the influence line AGB for the bending moment at C. The area AGB is termed the influence area. The greatest moment CG at C is x(l-x)/l. To use this line to investigate the maximum moment at C due to a series of travelling loads at fixed distances, let P_1, P_2, P_3, ... be the loads which at the moment considered are at distances m_1, m_2, ... from the left abutment. Set off these distances along AB and let y_1, y_2, ... be the corresponding ordinates of the influence curve (y = Ff) on the verticals under the loads. Then the moment at C due to all the loads is

M = P_1y_1+P_2y_2+...

[v.04 p.0553]

The position of the loads which gives the greatest moment at C may be settled by the criterion given above. For a uniform travelling load w per ft. of span, consider a small interval Fk = [Delta]m on which the load is w[Delta]m. The moment due to this, at C, is wm(l-x)[Delta]m/l. But m(l-x)[Delta]m/l is the area of the strip Ffhk, that is y[Delta]m. Hence the moment of the load on [Delta]m at C is wy[Delta]m, and the moment of a uniform load over any portion of the girder is w × the area of the influence curve under that portion. If the scales are so chosen that a inch represents 1 in. ton of moment, and b inch represents 1 ft. of span, and w is in tons per ft. run, then ab is the unit of area in measuring the influence curve.

If the load is carried by a rail girder (stringer) with cross girders at the intersections of bracing and boom, its effect is distributed to the bracing intersections D'E' (fig. 53), and the part of the influence line for that bay (panel) is altered. With unit load in the position shown, the load at D' is (p-n)/p, and that at E' is n/p. The moment of the load at C is m(l-x)/l-n(p-n)/p. This is the equation to the dotted line RS (fig. 52).

If the unit load is at F', the reaction at B' and the shear at C' is m/l, positive if the shearing stress resists a tendency of the part of the girder on the right to move upwards; set up Ff = m/l (fig. 54) on the vertical under the load. Repeating the process for other positions, we get the influence line AGHB, for the shear at C due to unit load anywhere on the girder. GC = x/l and CH = -(l-x)/l. The lines AG, HB are parallel. If the load is in the bay D'E' and is carried by a rail girder which distributes it to cross girders at D'E', the part of the influence line under this bay is altered. Let n (Fig. 55) be the distance of the load from D', x_1 the distance of D' from the left abutment, and p the length of a bay. The loads at D', E, due to unit weight on the rail girder are (p-n)/p and n/p. The reaction at B' is {(p-n)x_1+n(x_1+p)}/pl. The shear at C' is the reaction at B' less the load at E', that is, {p(x_1+n)-nl}/pl, which is the equation to the line DH (fig. 54). Clearly, the distribution of the load by the rail girder considerably alters the distribution of shear due to a load in the bay in which the section considered lies. The total shear due to a series of loads P_1, P_2, ... at distances m_1, m_2, ... from the left abutment, y_1, y_2, ... being the ordinates of the influence curve under the loads, is S = P_1y_1+P_2y_2+.... Generally, the greatest shear S at C will occur when the longer of the segments into which C divides the girder is fully loaded and the other is unloaded, the leading load being at C. If the loads are very unequal or unequally spaced, a trial or two will determine which position gives the greatest value of S. The greatest shear at C' of the opposite sign to that due to the loading of the longer segment occurs with the shorter segment loaded. For a uniformly distributed load w per ft. run the shear at C is w × the area of the influence curve under the segment covered by the load, attention being paid to the sign of the area of the curve. If the load rests directly on the main girder, the greatest + and - shears at C will be w × AGC and -w × CHB. But if the load is distributed to the bracing intersections by rail and cross girders, then the shear at C' will be greatest when the load extends to N, and will have the values w × ADN and -w × NEB. An interesting paper by F.C. Lea, dealing with the determination of stress due to concentrated loads, by the method of influence lines will be found in _Proc. Inst. C.E._ clxi. p.261.

Influence lines were described by Fränkel, _Der Civilingenieur_, 1876. See also _Handbuch der Ingenieur-wissenschaften_, vol. ii. ch. x. (1882), and Levy, _La Statique graphique_ (1886). There is a useful paper by Prof. G.F. Swain (_Trans. Am. Soc. C.E._ xvii., 1887), and another by L.M. Hoskins (_Proc. Am. Soc. C.E._ xxv., 1899).

28. _Eddy's Method._--Another method of investigating the maximum shear at a section due to any distribution of a travelling load has been given by Prof. H.T. Eddy (_Trans. Am. Soc. C.E._ xxii., 1890). Let hk (fig. 56) represent in magnitude and position a load W, at x from the left abutment, on a girder AB of span l. Lay off kf, hg, horizontal and equal to l. Join f and g to h and k. Draw verticals at A, B, and join no. Obviously no is horizontal and equal to l. Also mn/mf = hk/kf or mn-W(l-x)/l, which is the reaction at A due to the load at C, and is the shear at any point of AC. Similarly, po is the reaction at B and shear at any point of CB. The shaded rectangles represent the distribution of shear due to the load at C, while no may be termed the datum line of shear. Let the load move to D, so that its distance from the left abutment is x+a. Draw a vertical at D, intersecting fh, kg, in s and q. Then qr/ro = hk/hg or ro = W(l-x-a)/l, which is the reaction at A and shear at any point of AD, for the new position of the load. Similarly, rs = W(x+a)/l is the shear on DB. The distribution of shear is given by the partially shaded rectangles. For the application of this method to a series of loads Prof. Eddy's paper must be referred to.

29. _Economic Span._--In the case of a bridge of many spans, there is a length of span which makes the cost of the bridge least. The cost of abutments and bridge flooring is practically independent of the length of span adopted. Let P be the cost of one pier; C the cost of the main girders for one span, erected; n the number of spans; l the length of one span, and L the length of the bridge between abutments. Then, n = L/l nearly. Cost of piers (n-1)P. Cost of main girders nG. The cost of a pier will not vary materially with the span adopted. It depends mainly on the character of the foundations and height at which the bridge is carried. The cost of the main girders for one span will vary nearly as the square of the span for any given type of girder and intensity of live load. That is, G = al², where a is a constant. Hence the total cost of that part of the bridge which varies with the span adopted is--

C = (n-i)P+nal²
= LP/l-P+Lal.

Differentiating and equating to zero, the cost is least when

dC LP
-- = - -- + La = 0,
dl l²

/*
P = al² = G;

that is, when the cost of one pier is equal to the cost erected of the main girders of one span. Sir Guilford Molesworth puts this in a convenient but less exact form. Let G be the cost of superstructure of a 100-ft. span erected, and P the cost of one pier with its protection. Then the economic span is l = 100[root]P/[root]G.

30. _Limiting Span._--If the weight of the main girders of a bridge, per ft. run in tons, is--

w_3 = (w_1+w_2)lr/(K-lr)

according to a formula already given, then w_3 becomes infinite if k-lr = 0, or if

l = K/r,

[v.04 p.0554] where l is the span in feet and r is the ratio of span to depth of girder at centre. Taking K for steel girders as 7200 to 9000,

Limiting Span in Ft.
r = 12 l = 600 to 750
= 10 = 720 to 900
= 8 = 900 to 1120

In a three-span bridge continuous girders are lighter than discontinuous ones by about 45% for the dead load and 15% for the live load, if no allowance is made for ambiguity due to uncertainty as to the level of the supports. The cantilever and suspended girder types are as economical and free from uncertainty as to the stresses. In long-span bridges the cantilever system permits erection by building out, which is economical and sometimes necessary. It is, however, unstable unless rigidly fixed at the piers. In the Forth bridge stability is obtained partly by the great excess of dead over live load, partly by the great width of the river piers. The majority of bridges not of great span have girders with parallel booms. This involves the fewest difficulties of workmanship and perhaps permits the closest approximation of actual to theoretical dimensions of the parts. In spans over 200 ft. it is economical to have one horizontal boom and one polygonal (approximately parabolic) boom. The hog-backed girder is a compromise between the two types, avoiding some difficulties of construction near the ends of the girder.

Most braced girders may be considered as built up of two simple forms of truss, the king-post truss (fig. 61, a), or the queen-post truss (fig. 61, b). These may be used in either the upright or the inverted position. A _multiple truss_ consists of a number of simple trusses, e.g. Bollman truss. Some timber bridges consist of queen-post trusses in the upright position, as shown diagrammatically in fig. 62, where the circles indicate points at which the flooring girders transmit load to the main girders. _Compound_ trusses consist of simple trusses used as primary, secondary and tertiary trusses, the secondary supported on the primary, and the tertiary on the secondary. Thus, the Fink truss consists of king-post trusses; the Pratt truss (fig. 63) and the Whipple truss (fig. 64) of queen-post trusses alternately upright and inverted.

A combination bridge is built partly of timber, partly of steel, the compression members being generally of timber and the tension members of steel. On the Pacific coast, where excellent timber is obtainable and steel works are distant, combination bridges are still largely used (Ottewell, _Trans. Am. Soc. C.E._ xxvii. p. 467). The combination bridge at Roseburgh, Oregon, is a cantilever bridge, The shore arms are 147 ft. span, the river arms 105 ft., and the suspended girder 80 ft., the total distance between anchor piers being 584 ft. The floor beams, floor and railing are of timber. The compression members are of timber, except the struts and bottom chord panels next the river piers, which are of steel. The tension members are of iron and the pins of steel. The chord blocks and post shoes are of cast-iron.

33. _Graphic Method of finding the Stresses in Braced Structures._--Fig. 65 shows a common form of bridge truss known as a _Warren girder_, with lines indicating external forces applied to the joints; half the load carried between the two lower joints next the piers on either side is directly carried by the abutments. The sum of the two upward vertical reactions must clearly be equal to the sum of the loads. The lines in the diagram represent the directions of a series of forces which must all be in equilibrium; these lines may, for an object to be explained in the next paragraph, be conveniently named by the letters in the spaces which they separate instead of by the method usually employed in geometry. Thus we shall call the first inclined line on the left hand the line AG, the line representing the first force on the top left-hand joint AB, the first horizontal member at the top left hand the line BH, &c; similarly each point requires at least three letters to denote it; the top first left-hand joint may be called ABHG, being the point where these four spaces meet. In this method of lettering, every enclosed space must be designated by a letter; all external forces must be represented by lines _outside_ the frame, and each space between any two forces must receive a distinctive letter; this method of lettering was first proposed by O. Henrici and R. H. Bow (_Economics of Construction_), and is convenient in applying the theory of reciprocal figures to the computation of stresses on frames.

34. _Reciprocal Figures._--J. Clerk Maxwell gave (_Phil. Mag. 1864_) the following definition of reciprocal figures:--"Two plane figures are reciprocal when they consist of an equal number of lines so that corresponding lines in the two figures are parallel, and corresponding lines which converge to a point in one figure form a closed polygon in the other."

Let a frame (without redundant members), and the external forces which keep it in equilibrium, be represented by a diagram constituting one of these two plane figures, then the lines in the other plane figure or the reciprocal will represent in direction and magnitude the forces between the joints of the frame, and, consequently, the stress on each member, as will now be explained.

Reciprocal figures are easily drawn by following definite rules, and afford therefore a simple method of computing the stresses on members of a frame.

The external forces on a frame or bridge in equilibrium under those forces may, by a well-known proposition in statics, be represented by a closed polygon, each side of which is parallel to one force, and represents the force in magnitude as well as in direction. The sides of the polygon may be arranged in any order, provided care is taken so to draw them that in passing round the polygon in one direction this direction may for each side correspond to the direction of the force which it represents.