6 6 • Hence S -=- 800 X 144 = 115,200 inch..pounds.
For a beam on end supports and sustaining a uniform load, the maximum bending moment equals (see Table B, page 55), W denoting the sum of the load and weight of beam, and 1 the length. If W is expressed in pounds, then 1 1 W/ --= W X 12 foot-pounds = W x 144 inch-pounds. Hence, equating the two values of maximum bending moment and the safe resisting moment, we get W x 144 = 115,200; 115,200 X or, W = 144 8 = 6,400 pounds.
76 The safe load for the beam is 0,400 pounds minus the weight of the beam.
2. A steel I-beam whose section modulus is 20.4 inches° rests on end supports 15 feet apart. Neglecting the weight of the beam, how large a load may be placed upon it 5 feet from one.end, if the working strength is 10,000 pounds per square inch? The safe resisting moment is SI - - = 10,000 X 20.4 = 326,400 inch-pounds; c hence the bending moment must not exceed that value. The dangerous section is under the load; and if P denotes the unknown value of the load in pounds, the maximum moment (see Table B, page 55, Part I) equals P X 5 foot-pounds, or P X 60 inch pounds. Equating values of bending and resisting moments, we get 3. Iu the preceding example, it is required to take into account the weight of the beam. 375 pounds.
As we do not know the value of the safe load, we cannot con struct the shear diagram and thus determine where the dangerous section is. But in cases like this, where the distributed load (the weight) is small compared with the concentrated load, the dan gerous section is practically always where it is under the concen trated load alone; in this case, at the load. The reactions due to the weight equal X 375 = 187.5; and the reactions due to the load equal ?„- P and 1. P, P denoting the value of the load. The larger reaction R, (Fig. 39) hence equals § P + 187.5. Since 77 the weight of the beam per foot is 375 ± 15 = 25 pounds, the maximum bending moment (at the load) equals 2 ( :7T P + 187.5) 5 — (25 X 5)2 = 10 10 P 937.5 — 312.5 = P + 625.
This is in foot-pounds if P is in pounds.
The safe resisting moment is the same as in the preceding illustration, 326,400 inch-pounds; hence 10 (-- P + 625) 12 = 326,400. 3 Solving for P, we have 10 326,400 P + 625 — ; 400 X 3 10 P + 625 x 3 = 326,= 81,600; 12 10 P = 79,725; or, P = 7,972.5 pounds.
It remains to test our assumption that the dangerous section is at the load. This can be done by computing R (with P =— 7,972.5), constructing the shear diagram, and noting where the shear changes sign. It will be found that the shear changes sign at the load, thus verifying the assumption.
4. A cast-iron built-in cantilever beam projects 8 feet from the wall. Its cross-section is represented in Fig. 40, and the moment of inertia with respect to the neutral axis is 50 inches*; the working strengths in tension and compression are 2,000 and 9,000 pounds per square inch respect ively. Compute the safe uniform load which the beam can sustain, neglecting the weight of the beam.
The beam being convex up, the upper fibres are in tension and the lower in compression. The resisting moment (SI e), as determined by the compressive strength, is 78 9,000 X 50 4 = 100,000 inch-pounds; .5 and the resisting moment, as determined by the tensile strength, is 2,000 X 50 2.5 = 40.000 inch-pounds.
Hence -the safe resisting moment is the lesser of these two, or 40,000 inch-pounds. The dangerous section is at the wall (see Table B, page 55), and the valve of the maximum bending moment is WI, W denoting the load and / the length. If W is in pounds, then M= W X 8 foot-pounds = W X 96 inch-pounds. Equating bending and resisting moments, we have 1. An 8 X 8-inch timber projects 8 feet from a wall. If its working strength is 1,000 pounds per square inch, how large an end load can it safely sustain ? Ans. 890 pounds.
2. A beam 12 feet long and 8 x 16 inches in cross-section, on end supports, sustains two loads P, each 3 feet from its ends respectively. The working strength being 1,000 pounds per square inch, compute P (see Table B, page 55).