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.

If the working strength in shear is taken equal to one-wentieth the working fibre strength, then it can be shown that,

1. For a beam on end supports loaded in the middle, the safe load depends on the shearing or fibre strength according as the ratio of length to depth (I ÷ a) is less or greater than 10.

2. For a beam on end supports uniformly loaded, the safe load depends on the shearing or fibre strength according as I ÷ a is less or greater than 20.

Examples. 1. It is required to design a timber beam to sustain loads as represented in Fig. 11, the working fibre strength being 550 pounds and the working shearing strength 50 pounds per square inch.

The maximum bending moment (see example for practice 3, Art. 43; and example for practice 2, Art. 44) equals practically 7,000 foot-pounds or, 7,000 X 12 = 84,000 inch-pounds. Hence, according to equation 6'",

I/C = 84,000/550 =152.7 inches3.

Since for a rectangle

I/C =1/6 ba2,

1/6 ba2 = 152.7, or ba2 = 916.2.

Now, if we let b = 4, then a2 = 229; or, a = 15.1 (practically 16 inches.

If, again, we let b = 6, then a2 = 152.7; or a = 12.4 (practically 14) inches.

Either of these sizes will answer so far as fibre stress is concerned, but there is more "timber " in the second.

The maximum external shear in the beam equals 1,556 pounds, neglecting the weight of the beam (see example 3, Art. 37; and example 2, Art. 38). Therefore, for a 4 X 16-inch beam,

*See "Materials of Construction." - Johnson. Page 55.

Greatest shearing unit-stress = 3/2 X 1,556 / 4 X 16

= 36.5 pounds per square inch; and for a 6 X 14-inch beam, it equals

3/2 X 1,556 / 6 X14 = 27.7 pounds per square inch.

Since these values are less than the working strength in shear, either size of beam is safe as regards shear.

If it is desired to allow for weight of beam, one of the sizes should be selected. First, its weight should be computed, then the new reactions, and then the unit-fibre stress may be computed as in Art. 64, and the greatest shearing unit-stress as in the foregoing. If these values are within the working values, then the size is large enough to sustain safely the load and the weight of the beam.

2. What is the safe load for a white pine beam 9 feet long and 2 X 12 inches in cross-section, if the beam rests on end supports and the load is at the middle of the beam, the working fibre strength being 1,000 pounds and the shearing strength 50 pounds per square inch.

The ratio of the length to the depth is less than 10; hence the safe load depends on the shearing strength of the material Calling the load P, the maximum external shear (see Table B, page 55) equals ½ P, and the formula for greatest shearing unit stress becomes

50 = 3/2 X (½ p/2 X 12); or P = 1,600 pounds-

1. What size of wooden beam can safely sustain loads as in Fig. 12, with shearing and fibre working strength equal to 50 and 1,000 pounds per square inch respectively?

Ans. 6 X 12 inches

2. What is the safe load for, a wooden beam 4 X 14 inches, and 18 feet long, if the beam rests on end supports and the load is uniformly distributed, with working strengths as in example 1?

Ans. 3,730 pounds

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73. Kinds of Loads and Beams. We shall now discuss the strength of beams under longitudinal forces (acting parallel to the beam) and transverse loads. The longitudinal forces are supposed to be applied at the ends of the beams and along the axis* of the beam in each case. We consider only beams resting on end supports.

The transverse forces produce bending or flexure, and the longitudinal or end forces, if pulls, produce tension in the beam; if pushes, they produce compression. Hence the cases to be considered may be called "Combined Flexure and Tension" and "Combined Flexure and Compression."

74. Flexure and Tension. Let Fig. 43, a, represent a beam subjected to the transverse loads L1, L2 and L3, and to two equal end pulls P and P. The reactions R1 and R2 are due to the transverse loads and can be computed by the methods of moments just as though there were no end pulls. To find the stresses at any cross-section, we determine those due to the transverse forces (L1, L2, L3, R1 and R2 ) and those due to the longitudinal; then combine these stresses to get the total effect of all the applied forces.

The stress due to the transverse forces consists of a shearing stress and a fibre stress; it will be called the flexural stress. The fibre stress is compressive above and tensile below. Let M denote the value of the bending moment at the section considered; c1 and c2 the distances from the neutral axis to the highest and the lowest fibre in the section; and Sj and S2 the corresponding unit-fibre stresses due to the transverse loads. Then

St = Mc1/I; and S2 =Mc2 / I.

The stress due to the end pulls is a simple tension, and it equals P; this is sometimes called the direct stress. Let S0 denote the unit-tension due to P, and A the area of the cross-section; then

S0 = P/A

Both systems of loads to the left of a section between L1 and

* Note. By " axis of a beam " is meant the line through the centers of gravity of all the cross-sections.

L2 are represented in Fig. 43, b; also the stresses caused by them at that section. Clearly the effect of the end pulls is to increase the tensile stress (on the lower fibres) and to decrease the compressive stress (on the. upper fibres) due to the flexure. Let Sc denote the total (resultant) unit-stress on the upper fibre, and St that on the lower fibre,, due to all the forces acting on the beam. In combining the stresses there are two cases to consider:

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