This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.
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. 33, compute the greatest tensile and compressive unit-stresses, and tell at what places they occur. (Neglect the weight.)
4,171 pounds per square inch.
11,150 " " " "
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. 30.) 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. 10. a, it being a steel I-beam whose section modulus is 20.4 inches3. ( 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 G X 12 inches. (See example for practice 1, Art. 43.)
Ans. 600 pounds per square inch.
This problem can be solved by means of equation G written in this form,
M = S I / c (6")
We substitute for S the given working strength for the material of the beam, and for 1 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 beam 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 ?
1/6 ba2=1/6 X6xl22=l44 inches3,
Hence S I/ c = 800 X 144 = 115,200 inch-pounds.
For a beam on end supports and sustaining a uniform load, the maximum bending moment equals ⅛ Wl (see Table B, page 55), W denoting; the sum of the load and weight of beam, and l the length. If W is expressed in pounds, then
1/8 Wl = 1/8 W X 12 foot-pounds = 1/8 W X 144 inch-pounds.
Hence, equating the two values of maximum bending moment and the safe resisting moment, we get
1/8 w X 144 = 115,200;
W=115,200 X 8/144 =6,400 pounds. or,
The safe load for the beam is 6,400 pounds minus the weight of the beam.
2. A steel I-beam whose section modulus is 20.4 inches3 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 16,000 pounds per square inch?
The safe resisting moment is
SI/c = 16,000 X 20.4 = 326,400 inch-pound; 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 2/3 P X 5 foot-pounds, or 2/3 P X 60 inch-pounds. Equating values of bending and resisting moments, we get
2/3 P x 60 = 326,400;
P + X 3 / 2 X 60 =8,160 pounds.
3. In 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 construct 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 dangerous section is practically always where it is under the concentrated load alone; in this case, at the load. The reactions due to the weight equal ½ X 375 = 187.5; and the reactions due to the load equal ⅓ P and ⅔ P, P denoting the value of the load. The larger reaction R: (Fig. 39) hence equals ⅔ P + 187.5. Since the weight of the beam per foot is 375 ÷ 15 = 25 pounds, the maximum bending moment (at the load) equals
( 2/3 P + 187.5) 5 - (25 X 5) 2½ = 10/3 P + 937.5 - 312.5 = 10/3 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
(10/3 P + 625) 12 = 320,400.
Solving for P, we have
10/3 p + 625 = 326,400 /12 ;
10 P + 625 X 3 =326,400 X 3/12 = 81,600; 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 R1 (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.