309_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

309_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.8.qxd 9/27/08 1:13 PM Page 303 303 SECTION 3.8 Statically Indeterminate Torsional Members Solution 3.8-2 Circular bar with fixed ends (a) ANGLE OF TWIST AT SECTIONS B AND C φB φ max 0 fB From Eqs. (3-46a and b): fAB L/A TAx GIP TA T0 LB L dfB dx T0 (L GIPL TB T0 LA L dfB dx 0; L APPLY THE ABOVE FORMULAS TO THE GIVEN BAR: L 4 or x x L/2 T0 (L GIPL 2x)(x) 4x) 4x 0 ; (b) MAXIMUM ANGLE OF TWIST fmax T0(L L TA x) T0x L T0 (L L 2x) TD (fB)max (fB)x T0L 8GIP L 4 ; TA Problem 3.8-3 A solid circular shaft AB of diameter d is fixed against rotation at both ends (see figure). A circular disk is attached to the shaft at the location shown. What is the largest permissible angle of rotation max of the disk if the allowable shear stress in the shaft is allow? (Assume that a b. Also, use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.) Disk A d B a b Solution 3.8-3 Shaft fixed at both ends Assume that a torque T0 acts at the disk. The reactive torques can be obtained from Eqs. (3-46a and b): T0b T0a TB L L Since a b, the larger torque (and hence the larger stress) is in the right hand segment. TA L a a b b ...
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