250 APPLIED MECHANICS 
dead load, the total equivalent uniform dead load upon a cross girder is 
6°4 + 52°2 = 58°6 tons. 
Using Unwin’s formula (p. 223), assuming a ratio of depth to span 
of =4;, and a constant of 1500 for steel plate web girders; the span 
being about 16 feet, and the stress limited to 7 tons per square inch ; the 
weight of each cross girder is 1:1 tons. Add 10 per cent. for gussets and 
fastenings at the ends. Hence the dead weight carried by the main 
girders, exclusive of their own weight, is— 
Tons. 
Bridge floor . ‘ ‘ . : . ; : : - 128-2 
21 cross girders. : : 4 : . 254 
Overhead wind bracing, ‘etc. fn ‘say . ; - : : i 5:0 
Total . f . 158°6 
The total equivalent uniform load on the two main girders is therefore _ 
159 + 256 =415 tons, exclusive of their own weight. Applying Unwin’s — 
formula, the span being 150 feet, depth at centre 12} feet, and takin 
the safe working stress in compression at 54 tons per square inch, <a 
the constant at 1900 for steel open web girders of the type under con- — 
sideration, the weight of the -two main girders is 87 tons. 
Hence the total dead load on the main girders is 159 + 87 = 246 tons. 
Unit Load Stresses.—Find the stress in each member of the main 
girders with unit load at each panel point, preferably both graphically — 
and analytically. Tabulate on the stress sheet. 
Dead Load Stresses.—Tabulate also the stress in each member due — 
to the actual dead loads at the panel points. ‘This stress is 6-2 times the — 
unit load stress, since the total dead load is 246 tons, and there are 2 x 20 
panels, 
Maximum Live Load Stresses in the Booms.—Find the maximum 
stresses in the boom members due to the live load. These will occur 
when the bridge is fully covered, and are obtained from the equivalent 
uniform load, which is 256 tons. The corresponding load at each panel 
point is 6°4 tons, and the stresses in the boom members are therefore 6-4 
times the unit load stresses ; they can therefore now be tabulated. 
Maximum Live Load Stresses in the Web Members.—The maximum 
live load stresses in the web members must be obtained from the maxi- 
mum shear force diagram (not 6°4 times the unit load stresses). Tabulate 
these stresses both for the front and back of the travelling load, giving to 
each its correct sign. 
Wind Load Stresses.—Calculate and tabulate the wind load stresses 
under both conditions (a) and (b), Art. 224, p. 246. 
Under condition (a) the exposed area is— 
Sq. Ft. 
Twice the face area of the upper flange . , 4 . 450 
The face area of the lower flange and floor system . ; . 420 
Three times the face area of the verticals : : : . 480 
Three times the face area of the diagonals. , P - 530 
Total . : . 1880 
The distributed wind load at 56 lbs. per square foot is therefore — 
47 tons. 
