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03-02ChapGere.0031 - 32 d 1 4 5 0.643398 in 4 Steel sleeve...

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Problem 3.8-13 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d 1 1.6 in. for the brass core and d 2 2.0 in. for the steel sleeve. The shear moduli of elasticity are G b 5400 ksi for the brass and G s 12,000 ksi for the steel. Assuming that the allowable shear stresses in the brass and steel are b 4500 psi and s 7500 psi, respectively, determine the maximum per- missible torque T max that may be applied to the shaft. ( Hint: Use Eqs. 3-44a and b to find the torques.) Solution 3.8-13 Composite shaft shrink fit 234 CHAPTER 3 Torsion d 1 1.6 in. d 2 2.0 in. G B 5,400 psi G S 12,000 psi Allowable stresses: B 4500 psi S 7500 psi B RASS CORE ( ONLY ) G B I PB 3.47435 10 6 lb-in. 2 S TEEL SLEEVE ( ONLY ) G S I PS 11.1288 10 6 lb-in. 2 I PS
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Unformatted text preview: 32 d 1 4 5 0.643398 in. 4 Steel sleeve S Brass core B d 1 d 2 T B T B T ORQUES Total torque: T 5 T B 1 T S Eq. (3-44a): 5 0.237918 T Eq. (3-44b): 5 0.762082 T T 5 T B 1 T S (CHECK) A LLOWABLE TORQUE T BASED UPON BRASS CORE Substitute numerical values: T 5 15.21 k-in. A LLOWABLE TORQUE T BASED UPON STEEL SLEEVE Substitute numerical values: T 5 9.13 k-in. S TEEL SLEEVE GOVERNS T max 5 9.13 k-in. T S 5 0.762082 T 5 2(7500 psi)(0.927398 in. 4 ) 2.0 in. t S 5 T S ( d 2 / 2) I PS T S 5 2 t S I PS d 2 5 2(4500 psi)(0.643398 in. 4 ) 1.6 in. T B 5 0.237918 T t B 5 T B ( d 1 / 2) I PB T B 5 2 t B I PB d 1 T S 5 T ¢ G S I PS G B I PB 1 G S I PS ≤ T B 5 T ¢ G B I PB G B I PB 1 G S I PS ≤ T S T S A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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