Machinery's Handbook, 31st Edition
226 SAMPLE CALCULATIONS 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: , 24 544 2500 22 044 mm − = The nominal shear stress due to torsion loading is computed using Equation (16) in the Table of Simple Stresses : , c J Z = = π 16 50 6 6 50 # p 3 2 3 # − = mm 363 megapascals 2 = Referring to the previously mentioned stress concentration curve on page 214, K t is 2.85, since a / d = 6 ∕ 50 = 0.12. The stress concentration factor may now be calculated by means of Equation (8): K = 1 + 0.2(2.85 − 1) = 1.37. From Equation (3), working stress s w = 1.37 3 363 = 497 N/mm 2 = 497 MPa. Example 5(a) : For Case 3 in the Tables of Combined Stresses , calculate the least factor of safety for a 5052-H32 aluminum beam that is 10 inches long, 1 inch wide, and 2 inches high. Yield strengths are 23,000 psi tension; 21,000 psi compression; 13,000 psi shear. The stress concentration factor is 1.5; F y is 600 lbs; F x 500 lbs. Solution: From Tables of Combined Stresses , Case 3: 1 2 1 2 b # # # σ ′ =− + =− a k The other formulas for Case 3 give σ a ′ = 8750 psi (in tension); τ a ′ = 4375 psi, and τ b ′ = 4625 psi. Using Equation (4) for the nominal compressive stress of 9250 psi: S w = 1.5 3 9250 = 13,900 psi. From Equation (1) f s = 21,000 ∕ 13,900 = 1.51. Applying Equations (1), (4) and (5) in appropriate fashion to the other calculated nominal stress values for tension and shear will show that the factor of safety of 1.51, governed by the compressive stress at b on the beam, is minimum. Example 5(b): What maximum F can be applied in Case 3 if the aluminum beam is 200 mm long; 20 mm wide; 40 mm high; θ = 30 ° ; f s = 2, governing for compression, K = 1.5, and S m = 144 N/mm 2 for compression. Solution: From Equation (1) S w = − 144 N/mm 2 . Therefore, from Equation (4), σ b ′ = − 72⁄1.5 = − 48 N/mm 2 . Since F x = F cos 30 ° = 0.866 F , and F y = F sin 30 ° = 0.5 F : . . F F F N 48 20 40 1 0866 40 6 200 05 2420 # # # − = − + = a k , 22 000 8 000 000 363 , , = N τ = 6 10 600 500 9250psi (in compression) Stresses and Deflections in a Loaded Ring.— For thin rings, that is, rings in which the dimension d shown in the accompanying diagram is small compared with D , the maxi- mum stress in the ring is due primarily to bending moments produced by the forces P . The maximum stress due to bending is: (26) For a ring of circular cross section where d is the diame ter of the bar from which the ring is made, (27) The increase in the vertical diameter of the ring due to load P is: S I PDd 4 π = . . S d PD or P D Sd 1621 0617 3 3 = = P D P d
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