It would be:

 Cross shearing from A to w1 = 01H = 10000 pounds. Cross shearing from w1 to w11 = T I = 7500 pounds. Cross shearing from w11 to w111 = V J = 5000 pounds. Cross shearing from W111 to wIv = L K = 2500 pounds. Cross shearing from WIv to wv = 0 = 0 pounds. Cross shearing from wv to wvI = MN = 2500 pounds. Cross sheering from wVl to wvII = PP1 = 5000 pounds. Cross shearing from wVll to wv111, = RR1 = 7500 pounds. Cross shearing from WvIII to B = SO = 10000 pounds.

The area of web of a 15" - 150 pounds beam (Table XX) is 7,59 square inches; the safe resistance of wrought-iron to cross-shearing per square inch being (g/f) = 8000 pounds, we need not worry any further on that score.

To find the deflection we now make the lower load line g c equal to the sum of the lengths of verticals wVlll, wvlII, wVl, etc., through parabola C E G, beginning at top g with length of right vertical wVIII. We select 2 at random, scale z j = 246" (inch scale), draw z g,zc, etc., and figure

1 Had we taken more parts, the steps in shearing figure would become smaller and smaller till they would finally assume the straight line H S, which is the real outline of shearing figure.

c1f1 g1 We now daw z o parallel c1 g1 and find it bisects g c, or greatest deflection will be at centre of beam, which we know is the case. We scale ff1 = 62" (inch scale); find from Table XX for our 15" - 150 pounds beam i = 523,5 and from Table IV for wrought-iron e = 27000000, therefore, Formula (95):

δ = 62.37,5.246.12000 = 0,486 27000000.523,5

δ = 5.20000.3003/ 384.27000000.523,5 = 0,497" or practically the same result.

The safe deflection for plastering should not exceed (28)

8 = 25.0,03 = 0,75" bo that we are perfectly safe, providing our beam is well braced sideways.

Example VI.

A wrought-iron beam, braced sideways, of 30-foot span, Figure 157, carries a uniform load of 200 pounds per foot, including weight of beam. It carries also a concentrated load wl = 10000 pounds ten feet from the right-hand support. What beam should be used?

We draw beam A B = 360" at inch scale, we divide uniform load into, say, six equal parts, each 5 feet long, or l1 = 60". The total uniform load will be u = 30.200 = 6000 pounds, therefore each part u/6 = 6000/6 =1000 pounds. We draw arrows at the centre of each uniform part, so that the end arrows will be one-half part from supports. These arrows will therefore answer for our verticals, when drawing deflection figure.

At 120" from right hand support we locate the load w1 = 10000 pounds.

We now make load line b a = 16000 pounds the total load and divide it, so that