340_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

340_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.9-3.11.qxd 334 9/27/08 1:14 PM Page 334 CHAPTER 3 Torsion Solution 3.10-10 Rectangular tube T, t, and Lm are constants. 2T Let k t a, b Lm 2(a b) t ab k min constant constant b Am t tmin 4 8T tL2 m From the graph, we see that and the tube is square. SHEAR STRESS Lm (1 + b )2 b dimensions of the tube a b t tL2 m k thickness (constant) b T constant t is minimum when 1 ALTERNATE SOLUTION T 2tAm Am 2b(1 ) ab b2 t Lm 2(1 + b ) constant Am dt db 2 Lm bc d 2(1 + b ) b L2 m 2T (1 + b )2 c d b tL2 m (1 + b )2(1) 2T b (2)(1 + b ) c d tL2 b2 m or 2 (1 ) (1 )2 0 0 1 2 4(1 + b ) T(4)(1 + b )2 T 2tAm 2t b L2 m 2T(1 + b )2 tL2 b m ; Thus, the tube is square and is either a minimum or a maximum. From the graph, we see that is a minimum. 4 106 psi) of square cross section (see figure) with outer dimensions 2 in. 2 in. must resist a torque T 3000 lb-in. Calculate the minimum required wall thickness tmin if the allowable shear stress is 4500 psi and the allowable rate of twist is 0.01 rad/ft. Problem 3.10-11 A tubular aluminum bar (G t 2 in. 2 in. ...
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