_ 
Z DESIGN OF STRUCTURES 229 
+15=2°53 tons; they are in double shear, and, as has been shown, 
his worth 3°55 tons. But the load is not equally divided over the 
sts, and therefore a margin is desirable. If the area of the cross 
tion of the stiffening be worked out, it will be found that the direct 
‘stress is small. Taking only the end plate 12 inches x 2 inch, and the 
‘two end angles 4 inches x 4 inches x $ inch into account, it is 3-2 tons 
er square inch. 
the depth of the web. rat each passes 7 rivets, and if a section 
4 inches x 3 inches x $ inch be used, the direct stress will be less than 
1 ton per square inch. 
g bsitend of Inertia and Moment of Resistance of Cross Section.— 
Subtracting the rivet holes from both flanges and allowing for the middle 
1 foot 8 inches of the web only at 33 per cent. efficiency, that is, the worth 
of the riveted joint (the joint is weakest in compression between the 
rivets and plates), the moment of inertia of the central cross section is 
_ found to be 9740 (inches)*. - Since the distance of the extreme fibres 
from the neutral axis is 15°81 inches, the modulus of the section is 
9740 + 15°81 =616 (inches)*. Hence the maximum stress at the central 
cross section is 4440+ 616 =7-2 tons per square inch, which exceeds the 
limit. The explanation of this is, that in so deep a flange the variation of 
stress between the outer and inner fibres is considerable, a point often 
overlooked when the approximate method is applied, and this shows the 
necessity of calculating the moment of resistance everywhere. It will be 
_ found necessary to increase the thickness of the outer plate from ;', inch 
-to}inch. The moment of inertia of the central cross section will then 
become 10,060 (inches)*, and the moment of resistance will increase 
to 634 (inches)*®, which will reduce the maximum stress to 7 tons 
per square inch. 
_ The moment of resistance diagram is plotted on the base line of the 
_ bending moment diagram (Fig. 331), and it will be seen that the former 
lies entirely outside the latter. 
__ The moment of inertia diagram is also plotted on the base line of the 
_ bending moment diagram (Fig. 331). 
Deflection and Camber.—Assuming that M +I is constant through- 
out the girder (a rough approximation) and equal to aaah’ then, 
; A ML? 4440 x 450? ‘ : oes 
_ deflection = SEI ~8x9000 x 100607 1} inches. This deflection is yy 
_ of the span, which may be considered as reasonable. 
If 2 inch of camber be allowed per 10 feet of span the total camber 
required i is 13 inches, agreeing fairly well with the estimated deflection. 
Actual Weight of Girder.—The weight of the girder as designed 
‘is 4tons Ocwt. 2qrs. 13 lbs., showing that the estimated weight is 
sufficiently accurate. 
Exercises XIV. 
_ 1. A steel plate web Faber with parallel booms, 100 feet long, is to support 
a dead load of # ton, and a rolli load of 1} tons per foot-run. Select a suitable 
h, and assuming suitable working stresses, design the centre section and the 
dinal section of the booms, taking 30 feet as the maximum length of 
