2 X 6 x 14 = 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 com puted 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 P, and the formula for greatest shearing unit stress becomes 3 IP 50 = 2 X 2X 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 woodMi beam 4 X 11 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 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 con sidered 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 L„ and L, and to two equal end pulls P and P. The reactions R, and R, are due to the trans verse 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 (L„ L„ R, and RO 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; c, and the distances from the neutral axis to the highest and the low. est fibre in the section; and S, and the corresponding unit-fibre stresses due to the transverse loads. Then The stress due to the end pulls is a simple tension, and it equals P; this is sometimes called the direct stress. Let denote the unit-tension due to P, and A the area of the cross-section; then Both systems of loads to the left of a section between L, and are represented in Fig. 43, L; 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 flex ure. Let S, denote the total (resultant) unit-stress on the upper fibre, and that on the lower fibre, due to all the forces acting on the beam. In combining the stresses there are two cases to con sider: (1) The flexural compressive unit-stress on the upper fibre is greater than the direct unit-stress; that is, S, is greater than S.. The resultant stress on the upper fibre is S, = S, — S. (compressive); and that on the lower fibre is = S. (tensile).
The combined stress is as represented in Fig. 43, c, part tensile and part compressive.