03-02ChapGere.0029

03-02ChapGere.0029 - 2.25 in. G 5 11.6 3 10 6 psi P OLAR...

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Problem 3.8-11 A solid steel bar of diameter d 1 5 1.50 in. is enclosed by a steel tube of outer diameter d 3 5 2.25 in. and inner diameter d 2 5 1.75 in. (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B . The composite bar, which has length L 5 30.0 in., is twisted by a torque T 5 5000 lb-in. acting on the end plate. (a) Determine the maximum shear stresses t 1 and t 2 in the bar and tube, respectively. (b) Determine the angle of rotation f (in degrees) of the end plate, assuming that the shear modulus of the steel is G 5 11.6 3 10 6 psi. (c) Determine the torsional stiffness k T of the composite bar. ( Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.) Solution 3.8-11 Bar enclosed in a tube 232 CHAPTER 3 Torsion d 1 5 1.50 in. d 2 5 1.75 in. d 3 5
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Unformatted text preview: 2.25 in. G 5 11.6 3 10 6 psi P OLAR MOMENTS OF INERTIA Tube: I P 2 5 p 32 d 3 4 2 d 2 4 5 1.595340 in. 4 Bar: I P 1 5 p 32 d 1 4 5 0.497010 in. 4 T ORQUES IN THE BAR (1) AND TUBE (2) FROM E QS . (3-44 A AND B ) (a) M AXIMUM SHEAR STRESSES (b) A NGLE OF ROTATION OF END PLATE (c) T ORSIONAL STIFFNESS k T 5 T f 5 809 k-in. f 5 0.354 8 f 5 T 1 L GI P 1 5 T 2 L GI P 2 5 0.00618015 rad Tube: t 2 5 T 2 ( d 3 / 2) I P 2 5 2690 psi Bar: t 1 5 T 1 ( d 1 / 2) I P 1 5 1790 psi Tube: T 2 5 T I P 2 I P 1 1 I P 2 5 3812.32 lb-in. Bar: T 1 5 T I P 1 I P 1 1 I P 2 5 1187.68 lb-in. A Tube (2) T = 5000 lb-in. End plate B Bar (1) L = 30.0 in. d 3 d 2 d 1 A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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