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683_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 683_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 677 SECTION 8.5 Combined Loadings Solution 8.5-2 Gondola on a ski lift b W 677 180 mm sallow 12 kN 2 100 MPa (tension) pd 4 2 50 MPa S pd 32 W M + A S pst 3 or a bd 4W d 4(6 kN) dmin 32 Wb 2 + 8b pd 3 0 13,089.97 1 m2 1.44 m d 1.44 Solve numerically: d 4W pd p(100 Mpa) psallow 4W 13,090 d3 3 MAXIMUM TENSILE STRESS st tallow pst 4W 8b Find dmin A SUBSTITUTE NUMERICAL VALUES: 6 kN 0 (d meters) 0.04845 m 48.4 mm ; MAXIMUM SHEAR STRESS st (uniaxial stress) 2 Since tallow is one-half of sallow, the minimum diameter for shear is the same as for tension. tmax Problem 8.5-3 The hollow drill pipe for an oil well (see figure) is 6.2 in. in outer diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the pipe (due to the weight of the pipe) is 62 k and the torque (due to drilling) is 185 k-in. Determine the maximum tensile, compressive, and shear stresses in the drill pipe. ...
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