267_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

267_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.3.qxd 9/27/08 1:27 PM Page 261 261 SECTION 3.3 Circular Bars and Tubes Problem 3.3-13 A vertical pole of solid circular cross section is twisted by c horizontal forces P 1100 lb acting at the ends of a horizontal arm AB (see figure). The distance from the outside of the pole to the line of action of each force is c 5.0 in. If the allowable shear stress in the pole is 4500 psi, what is the minimum required diameter dmin of the pole? Solution 3.3-13 B A P d Vertical pole tmax ( P(2c + d ) d 16P(2c + d) 4 pd3 pd /16 3 max)d (16P)d 32Pc 0 P 1100 lb SUBSTITUTE NUMERICAL VALUES: c 5.0 in. UNITS: Pounds, Inches allow 4500 psi Find dmin ( )(4500)d3 (16)(1100)d 1.24495d 12.4495 Solve numerically: TORSION FORMULA Tr Td tmax IP 2IP P12c + d2 d dmin IP 32(1100)(5.0) 0 or d3 T P c 0 2.496 in. 2.50 in. ; pd 4 32 Problem 3.3-14 Solve the preceding problem if the horizontal forces have magnitude P and the allowable shear stress is 30 MPa. 5.0 kN, the distance c 125 mm, ...
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