Substituting the value of D in the above equation, viz.:

D= (p + q) { } + 8 +8 + + +++}},

and arranging it according to the rule, to ascertain the effect of the moving load:

0 = Y. sin - p { } + { + } + ÷ + § − (1 − §) − (1 − }) }

− q ( + + + + 1) + q { (1 − q) + (1 − }) }.

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 λ be the length of a bay, and h the height of the girder, take moments first round G and then round F, thus:

0 = Xh+ (p + q) { (} + } + { + 1 + §). 3λ + (§. 3 x − λ)

0=Zh+(p + q) { (} + 2 + 3 + + + §). 2x + (}. 2λ − λ) }.

From which it is evident that these bars are in the greatest state of stress when the bridge is fully loaded.

Lastly, to find the stress in the adjacent vertical to the right, take a section y8 (Fig. 52). The point about which to take moments is at infinity, and

0 = − V. ∞ - D. ∞ + (2 + 2) ∞ + (p + 2 ) ∞ + (p +?)∞ .

The only difference between this equation and the one previously obtained for Y.sin is that -V is written instead of Y.sin . Hence it is obvious that

The values Y.sin and -V are therefore identical; hence by first finding V, Y can be ascertained by dividing by (—sin 4). This is expressed in the following rule: the resolved 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 RIGHT-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).

Since Vo and D are the only vertical forces acting at A

(Fig. 54):

0

Vo 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:

2

Further, since Z, is the only horizontal force at A:

Two vertical forces act at S (Fig. 53), viz., the load at this point (the greatest value of which is 3000 kilos.) and Vg. Therefore,

And since X, is the only horizontal force at S:

X, = 0.

Calculation of X1, Z2, V1, Y1.

(Section a B. Fig. 55.)

Taking moments for the part of the girder shown in Fig. 55, about the point C:

The equation of moments about the point D is

0 = − Z2 x 2 + (1000 + 5000) (} +

...

+7) 2.

1

The diagonal Y1 makes an angle of 45° with the horizontal. Therefore, the resolved part of Y1 vertically is Y1. sin 45o,

Taking moments about F:

Z3 × 2 +(1000 + 5000) { († + ..... + g) 4 + (} . 4 — 2)}
Z2 (max.) = + 36000 kilos.

By leaving out of this equation, first the negative, then the positive members, multiplied by 5000:

Then multiplying these values by 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 = X, x 2 + (1000 + 5000) { (+ ... + §) 6 + (§. 6 − 2) + (7.6 −4)}

X, (min.) =

4

0=Z, x 2 + (1000+ 5000) { ( + ... + §) 6 + (§. 62) + (3.64)} Z1 (max.) = + 45000 kilos.

0 = X, × 2 + (1000 + 5000) { (} + . +1) 8+ (3.8 - 2)

...

0=Z, x 2 + (1000 + 5000) { ( + ... + 1) . 8 + (§.82)

0 =