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297_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 297_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.5-3.7.qxd 9/27/08 1:10 PM Page 291 SECTION 3.5 Pure Shear 291 Problem 3.5-6 A solid circular bar of steel (G 78 GPa) transmits a torque T 360 N m. The allowable stresses in tension, compression, and shear are 90 MPa, 70 MPa, and 40 MPa, respectively. Also, the allowable tensile strain is 220 10 6. Determine the minimum required diameter d of the bar. Solution 3.5-6 Solid circular bar of steel T 360 N m G DIAMETER BASED UPON ALLOWABLE TENSILE STRAIN 78 GPa ALLOWABLE STRESSES Tension: 90 MPa Compression: 70 MPa Shear: 40 MPa Allowable tensile strain: max 220 10 max tmax 6 DIAMETER BASED UPON ALLOWABLE STRESS The maximum tensile, compressive, and shear stresses in a bar in pure torsion are numerically equal. Therefore, the lowest allowable stress (shear stress) governs. allow tmax 40 MPa 16T 3 pd 45.837 d 16T pt allow 0.0358 m d max; 16T 3 pd G max d3 16T ptmax 10 6 16(360 N # m) 6 53.423 * 10 0.0377 m 6 ) m3 37.7 mm dmin 37.7 mm ; p(40 MPa) m3 35.8 mm surface of a circular tube (see figure) is 880 10 6 when the torque T 750 lb-in. The tube is made of copper alloy with G 6.2 106 psi. Strain gage T 45° Solution 3.5-7 Circular tube with strain gage T Strain gage at 45°: 750 lb-in. G max 880 10 6.2 6 106 psi T = 750 lb-in. d 2 = 0.8 in. If the outside diameter d2 of the tube is 0.8 in., what is the inside diameter d1? 0.80 in. max 16T 2pGâmax 2p(78 GPa)(220 * 10 d 2G 16(360 N # m) d3 Problem 3.5-7 The normal strain in the 45° direction on the d2 max TENSILE STRAIN GOVERNS d3 3 2 MAXIMUM SHEAR STRAIN max 2 max ...
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