259_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

259_Mechanics - Sec_3.3.qxd 1:27 PM Page 253 253 SECTION 3.3 Circular Bars and Tubes Solution 3.3-2 Torsion of a drill bit(b RATE OF TWIST From

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Unformatted text preview: Sec_3.3.qxd 9/27/08 1:27 PM Page 253 253 SECTION 3.3 Circular Bars and Tubes Solution 3.3-2 Torsion of a drill bit (b) RATE OF TWIST From Eq. (3-14): u d 4.0 mm T 0.3 N m G 75 GPa (a) MAXIMUM SHEAR STRESS u From Eq. (3-12): tmax tmax tmax 16T u pd3 T GIP 0.3 N # m p (75 GPa) a b (4.0 mm)4 32 0.1592 rad/m 9.12°/m ; 16(0.3 N # m) p(4.0 mm)3 23.8 MPa ; Problem 3.3-3 While removing a wheel to change a tire, a driver applies forces P 25 lb at the ends of two of the arms of a lug wrench (see figure). The wrench is made of steel with shear modulus of elasticity G 11.4 106 psi. Each arm of the wrench is 9.0 in. long and has a solid circular cross section of diameter d 0.5 in. (a) Determine the maximum shear stress in the arm that is turning the lug nut (arm A). (b) Determine the angle of twist (in degrees) of this same arm. P 9.0 in. A 9.0 in. d = 0.5 in. P = 25 lb ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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