Machinery's Handbook, 31st Edition
SAMPLE CALCULATIONS 225 The calculation is less unwieldy if carried out using millimeters in place of meters; then F = 200 kN; L = 1500 mm; A = 2830 mm 2 , and E = 200,000 N/mm 2 . Thus: = Example 3(a): Determine the size for the section of a square bar which is to be held firmly at one end and is to support a load of 3000 pounds at the other end. The bar is to be 30 inches long and is to be made from SAE 1045 medium carbon steel with a yield point of 60,000 psi. A factor of safety of 3 and a stress concentration factor of 1.3 are to be used. , 2830 200 000 200 000 1500 , . 0530mm e # # = Solution: From Equation (1) the allowable working stress s w = 60,000 ∕ 3 = 20,000 psi. The applicable equation relating working stress and nominal stress is Equation (2); hence, σ = 20,000 ∕ 1.3 = 15,400 psi. The member must be treated as a cantilever beam subject to a bending moment of 30 3 3000 or 90,000 inch-pounds. Solving Equation (11) in the Table of Simple Stresses for section modulus: Z = 90,000 ∕ 15,400 = 5.85 inch 3 . The section mod ulus for a square section with neutral axis equidistant from either side is a 3 ⁄6, where a is the dimension of the square, so . . a 351 327 3 = = inches. The bar size can be 3 5 ∕ 16 inches. Example 3(b): A similar example to that given in Example 3(a), using metric SI units is as follows. Determine the size for the section of a square bar which is to be held firmly at one end and is to support a load of 1600 kilograms at the other end. The bar is to be 1 meter long and is to be made from steel with a yield strength of 500 newtons/mm 2 . A factor of safety of 3 and a stress concentration factor of 1.3 are to be used. The calculation can be performed using millimeters throughout. Solution: From Equation (1) the allowable working stress s w = 500 N/mm 2 ⁄3 = 167 N/mm 2 . The formula relating working stress and nominal stress is Equa tion (2); hence σ = 167 ∕ 1.3 = 128 N/mm 2 . Since a mass of 1600 kg equals a weight of 1600 g newtons, where g = 9.81 meters/second 2 , the force acting on the bar is 15,700 newtons. The bending moment on the bar, which must be treated as a cantilever beam, is thus 1000 mm 3 15,700 N = 15,700,000 N·mm. Solving Equation (11) in the Table of Simple Stresses for section modulus: Z = M / σ = 15,700,000 ∕ 128 = 123,000 mm 3 . Since the section modulus for a square section with neutral axis equidistant from either side is a 3 ⁄6, where a is the dimension of the square, , . a 6 123 000 90 4 mm 3 # = = Example 4(a) : Find the working stress in a 2-inch diameter shaft through which a trans verse hole 1 ∕ 4 inch in diameter has been drilled. The shaft is subject to a torsional moment of 80,000 inch-pounds and is made from hardened steel so that the index of sensitivity q = 0.2. Solution: The polar section modulus is calculated using the equation shown in the stress concentration curve for a Round Shaft in Torsion with Transverse Hole, Fig. 7, page 214 . . c J Z 16 2 4 6 2 14inches p 3 2 3 # # = = π − = The nominal shear stress due to torsion loading is computed using Equation (16) in the Table of Simple Stresses : . , , 14 80 000 57 200 psi τ = = Referring to the previously mentioned stress concentration curve on page 214 , K t is 2.82 since d / D is 0.125. The stress concentration factor may now be calculated by means of Equation (8): K = 1 + 0.2(2.82 − 1) = 1.36. Working stress calculated with Equation (3) is s w = 1.36 3 57,200 = 77,800 psi. Example 4(b): A similar example to that given in Example 4(a) , using metric SI units, is as follows. Find the working stress in a 50 mm diameter shaft through which a transverse hole 6 mm in diameter has been drilled. The shaft is subject to a torsional moment of 8000 N·m, and has an index of sensitivity of q = 0.2. If the calculation is made in millimeters, the torsional moment is 8,000,000 N·mm.
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