Substituting in equation 6', we find that 78,240 S = 96 = 815 pounds per square inch.
The weight of the beam therefore increases the unit-stress pro duced by the load at the dangerous section by 15 pounds per square 3. A T-bar (see Fig. 38) 8 feet long and supported at each end, bears a uniform load of 1200 pounds. The moment of inertia of its cross-section with respect to the neu tral axis being 2.42 inches', compute the maximum tensile and compressive unit-stresses in the beam rig. as.
Evidently the dangerous section is in the middle, and the value of the maximum bending moment (see Table B, page 55, Part I) is Wi, W and 1 denoting the load and length respectively. Here W/ = X 200 X 8 = 1,200 foot-pounds, 8 8 or 1,200 X 12 = 14,400 inch-pounds.
The section modulus equals 2.42 .÷ 2.28 = 1.06; hence 14,400 S — 1.06 = 13,585 pounds per square inch.
This is the unit-fibre stress on the lowest fibre at the middle sec tion, and hence is tensile. On the highest fibre at the middle section the unit-stress is compressive, and equals (see page 62): c' 0.72 = c 2.28 —S = X = 4,290 pounds per square inch.
74 1. A beam 12 feet long and 6 X 12 inches in cross-section rests on end supports, and sustains a load of 3,000 pounds in the middle. Compute the greatest tensile and compressive unit stresses in the beam, neglecting the weight of the beam.
Ans. 750 pounds per square inch.
2. Solve the preceding example taking into account the weight of the beam, 300 pounds Ans. 787.5 pounds per square inch.
3. Suppose that a built-in cantilever projects 5 feet from the wall and sustains an end load of 250 pounds. The cross-section of the cantilever being represented in Fig. 38, compute the greatest tensile and compressive unit-stresses, and tell at what places they occur. (Neglect the weight.) Tensile, 4,469 pounds per square inch. .
Ans. Compressive, 14,150 66 " 4. Compute the greatest tensile and compressive unit-stresses in the beam of Fig. 18, a, due to the loads and the weight of beam (400 pounds). (A moment diagram is represented in Fig. 18, b; for description see. example 2, Art. 44, p. 39.) The section of
the beam is a rectangle 8 X 12 inches.
Ans. 580 pounds per square inch.
5. Compute the greatest tensile and compressive unit-stresses in the cantilever beam of Fig. 19, a, it being a steel I-beam whose section modt;lus is 20.4 inches'. (A bending moment diagram for it is represented in Fig. 19, b; for description, see Ex. 3, Art. 44.) .Ans. 11,470 pounds per square inch.
6. Compute the greatest tensile and compressive unit-stresses in the beam of Fig. 10, neglecting its weight, the cross-sections being rectangular 6 X 12 inches. (See example for practice 1, Art. 43.) Ans. 600 pounds per square inch.
65. Second Application,. The dimensions and the work ing strengths of a beam are given, and it is required to determine its safe load (the manner of application being given).
This problem can be solved by means of equation 6 written in this form, We substitute for S the given working strength for the ma terial of the beam, and for I and c their values as computed from the given dimensions of the cross-section; then reduce, thus obtaining the value of the safe resisting moment of the beam, which equals the greatest safe bending moment that the beam can stand. We next compute the value of the maximum bending moment in terms of the unknown load; equate this to the value of the resisting moment previously found; and solve for the unknown load.
In cast iron, the tensile and compressive strengths are very different; and the smaller (the tensile) should always be used if the neutral surface of the beam is midway between the top and bottom of the, beam; but if it is unequally distant from the top and bottom, proceed as in example 4, following. • Examples. 1. A wooden Wain 12 feet long and 6 X 12 inches in cross-section rests on end supports. If its working strength is 800 pounds per square inch, how large a load uniformly distributed can it sustain I The section modulus is Oa', b and a denoting the base and altitude of the section (see Table A, page 54); and here 1 1 baa = X 6 X 12' = 144 inches'.