69. The Resisting Shear. The shearing stress on a cross section of a loaded beam is not a uniform stress; that is, it is not uniformly distributed over the section. In fact the intensity or unit-stress is actually zero on the highest and lowest fibres of a cross-section, and is greatest, in such beams as are used in prac tice, on fibres at the neutral axis. In the following article we explain how to find the maximum value in two cases--cases which are practically important.
7o. Second Beam Formula. Let denote the average value of the unit-shearing stress on a cross-section of a loaded beam, and A the area of the cross-section. Then the value of the whole shearing stress on the section is : Resisting shear = A.
Since the resisting shear and the external shear at any section of a beam are equal (see Art. 59), A „= (7) This is called the ".second beam formula " It is used to investi gate and to design for shear in beams.
In beams uniform in cross-section, A is constant, and is greatest in the section for which AT is greatest. Hence the great est unit-shearing stress in a loaded beam is at the neutral axis of the section at which the external shear is a maximum. There is a formula for computing this maximum value in any case, but it is not simple, and we give a simpler method for computing the value in the two practically important cases: 1. In wooden beams (rectangular or square in cross-section), the greatest unit-shearing stress in a section is 50 per cent larger than the average value 2. In I-beams, and in others with a thin vertical web, the greatest unit-shearing stress in a section practically equals S., as given by equation 7, if the area of the web is substituted for A.
85, Examples. 1. What is the greatest value of the unit shearing stress in a wooden beam 12 feet long and 6 X 12 inches in cross-section when resting on end supports and sustaining a uni form load of 6,400 pounds ? (This is the safe load as determined by working fibre stress; see example 1, Art. 65.) The maximum external shear equals one-half the load (see Table B. page 55), and comes on the sections near the supports. Since A = 6 X 12 = 72 square inches; 3,200 = 7244 pounds per square inch, and the greatest unit-shearing stress equals 3 3 = 44 = 66 pounds per square inch.
Apparently this is very insignificant; but it is not negligible, as is explained in the next article.
2. A steel I-beam resting on end supports 15 feet apart sustains a load of 8,000 pounds 5 feet from one end. The weight of the beam is 375 pounds, and the area of its web section is 3.2 square inches. (This is the beam and load described in examples 2 and 3, Art. 65.) What is the greatest unit-shearing stress ? The maximum external shear occurs near the support where the reaction is the greater, and its value equals that reaction. Calling that reaction R, and taking moments about the other end of the beam, we have 1 R x 15 — 375 x 7 7- — x 10 = 0; therefore 15 It = 80,000 + 2,812.5 = 82,812.5; or, R = 5,520.8 pounds.
Hence Ss 3.2 20.8 = 1,725 pounds per square inch.
1. A wooden beam 10 feet long and 2 X 10 inches in cross section sustains a middle load of 1,000 pounds. Neglecting the weight of the beam. compute the value of the greatest unit-shearing stress.
Ans. 37.5 'pounds per square inch.
813 2. Solve the preceding example taking into account the weight of the beam, 60 pounds.
Ans. 40 pounds per square inch.
3. A wooden beam 12 feet long and 4 X 12 inches in cross section sustains a load of 3,000 pounds 4 feet from one end. Neglecting the weight of the beam, compute the value of the greatest shearing unit-stress.
Ans. 62.5 pounds per square inch.
71. Horizontal Shear. It can be proved that there is a shearing stress on every horizontal section of a loaded beam. An experimental explanation will have to suffice here. Imagine a pile of six boards of equal length supported so that they do not bend. If the intermediate supports are removed, they will bend and their ends will not be flush but somewhat as represented in Fig. 41. This indicates that the boards slid over each other during the bending, and hence there was a rubbing and a frictional re sistance exerted by the boards upon each other. Now, when a solid beam is being bent, there is an exactly similar tendency for the horizontal layers to slide over each other; and, instead of a frictional resistance, there exists shearing stress on all horizontal sections of the beam.