262_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

262_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.3.qxd 9/27/08 256 1:27 PM Page 256 CHAPTER 3 Torsion Solution 3.3-6 Socket wrench ANGLE OF TWIST TmaxL GIP From Eq. (3-15): f pd3tmax 16 From Eq. (3-12): Tmax d 8.0 mm allow L 200 mm 60 MPa G 78 GPa f MAXIMUM PERMISSIBLE TORQUE 16T From Eq. (3-12): tmax pd3 3 pd tmax Tmax 16 f a pd3t max L ba b 16 GIP pd3tmaxL(32) 16G(pd4) 2tmaxL Gd f 2(60 MPa)(200 mm) (78 GPa)(8.0 mm) f 10.03846 rad2a 3 Tmax p(8.0 mm) (60 MPa) 16 Tmax 6.03 N # m ; Problem 3.3-7 A circular tube of aluminum is subjected to torsion by torques T applied at the ends (see figure). The bar is 24 in. long, and the inside and outside diameters are 1.25 in. and 1.75 in., respectively. It is determined by measurement that the angle of twist is 4° when the torque is 6200 lb-in. Calculate the maximum shear stress max in the tube, the shear modulus of elasticity G, and the maximum shear strain max (in radians). pd4 32 IP 0.03846 rad 180 deg/radb p T 2.20° ; T 24 in. 1.25 in. 1.75 in. ...
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