a' = U mn/2 dkl = 120000 mn/2x3 1/2x9000x50=0.038095 mn.

Now, when m = n = 25, we have the middle point; then -

a' = 0.038095 m n = 0.038095 x 25 x 25 = 23.81;

or, the area of the bottom flange at mid-length is 23.81 inches.

When m = 20, then n == 30, and -

a' = 0.038095 x20 x 30 = 22.86;

or, the required area, at 5 feet either way from the middle, is 22 7/8 inches.

When m = 15, then n = 35, and -

a' = 0.038095 x 15 x 35 = 20.0;

or, at 10 feet either way from the middle, the required area is 20 inches.

When m = 10, then n as 40, and -

a' = 0.038095 x 10 x 40 = 15 . 24;

or, at 15 feet either way from the middle, the required area is 15 1/4 inches.

When m as 5, then n = 45, and -

a = 0.038095 x 5x45 = 8.57;

or, at 20 feet each side of the middle, the required area is 8 5/8 inches.

The area of cross-section found in every case is that of the uncut fibres; to this is to be added as much as will be cut by the rivets. This is usually about one sixth of the area given by the rule. The top flange is to be made equal in area to the bottom flange. The flanges are unvarying in width from end to end, the variation of area being obtained by varying the thickness of the flanges, and this being attained by building the flange in lamina, or plates; but these should not be less than a quarter of an inch thick. There should be added to the length of the girder, in the clear, about one tenth of its length for supports on the walls: thus, a girder 30 feet long requires 3 feet added for supports, or 18 inches on each wall.

186. - Tubular Iron Girders: Shearing Strain. - The top and bottom flanges are provided of sufficient size to resist the transverse strain; the two upright plates, technically termed the web, need, therefore, to be thick enough to resist only the shearing strain. This, upon a beam uniformly loaded, is at the middle theoretically nothing, but from thence it increases regularly towards each support, where it equals half the whole weight. For example, the girder of Art. 185, 50 feet long between supports, carries 120,000 pounds uniformly distributed over its length. In this case the shearing strain at the wall at each end is the half of 120,000 pounds, or 60,000 pounds; at 5 feet from the wall it is 5/25 or 1/5 less, or 48,000 pounds; at 10 feet from the wall it is 2/5 less, or 36,000 pounds; at 15 feet it is 24,000; at 20 feet it is 12,000; and at 25 feet or the middle, it is nothing.

187. - Tubular Iron Girders: Thickness of Wen. - The equation appropriate for this is -

t = G/dk; (84.)

in which t is the thickness of the web (equal to the sum of the thicknesses of the two side plates), d is the height of the plate (t and d both in inches), G is the shearing strain, and k' is the effective resistance of wrought iron to shearing per inch of cross-section. This may be put in words as follows;

Rule LXII. - Divide the shearing strain by the product of the depth in inches into the value of k', and the quotient will be the thickness of the web, or of the two side plates taken together.

Example. - What is the required thickness of web in a girder 50 feet between bearings, side plates 38 inches high between top and bottom flanges, and to carry 120,000 pounds, uniformly distributed? Here, putting the shearing resistance of the plates at 7000 pounds per inch, we have -

t = G/dk = G/38x7000 = G/266000.

The shearing strain at the supports, as in last article, is 60000; therefore, we have for this point -

t = 60000/266000 = 0.225.

When G = 48000, then -

t = 48000/266000=0.18; and when G = 36000, then -

t = 36000/266000 = 0.135.

Those nearer the middle of the girder are still less than these; and these are all below the practicable thickness, which is half an inch for the two plates. The plates ought not in practice ever to be made less than a quarter of an inch thick.