689_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 689_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 08Ch08.qxd 9/18/08 11:06 AM Page 683 SECTION 8.5 Combined Loadings Solution 8.5-8 683 Torsional pendulum sx st txy L 2.0 m d 60 kg G 46.839 MPa 80 fmax (MPa) 4.0 mm M 46.839 MPa 0 sy W A sx TENSILE STRESS 80 GPa sallow A W tallow 100 MPa pd 4 50 MPa 2 12.5664 mm2 PRINCIPAL STRESSES 2 Mg (60 kg)(9.81 m/s ) 588.6 N A 23.420 ; 1123.42022 + 6400f2 max sx + sy s1, 2 ; 2 s1, 2 a sy sx 2 2 b + t2 xy ( MPa) Note that s1 is positive and s2 is negative. Therefore, the maximum in-plane shear stress is greater than the maximum out-of-plane shear stress. 23.420 ; 1123.42022 + 6400f2 max MAXIMUM ANGLE OF ROTATION BASED ON TENSILE STRESS s1 ‹ 100 MPa (100 TORQUE: T L (EQ. 3-15) 23.420)2 5316 GIp fmax 6400f2 max Tr Ip GIpfmax t a t 80 fmax L ba r b IP Units: t (EQ. 3-11) tmax Gr fmax L MPa (80 * 106 Pa)fmax fmax A 100 MPa 123.42022 + 6400f2 max fmax 0.9114 rad 52.2° 1(23.420)2 + 6400f2 max MAXIMUM ANGLE OF ROTATION BASED ON IN-PLANE SHEAR STRESS SHEAR STRESS: t sallow maximum tensile stress radians a sx 2 sy 2 b + t2 xy 1(23.420)2 + 6400f2 max tallow 50 MPa (50)2 (23.420)2 + 6400f2 max Solving, fmax 50 0.5522 rad 31.6° SHEAR STRESS GOVERNS fmax 0.552 rad 31.6° ; ...
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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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