EM 319 - Review for 1st Midterm Exam

EM 319 - Review for 1st Midterm Exam - Review 1 Fall 08 1 A...

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Unformatted text preview: Review 1 Fall 08 1. A solid circular bar is fixed against rotation at the ends and loaded by a distributed torque t ( x ) that varies linearly in intensity from zero at end A to t o at end B (see figure). Assume that the bar has shear modulus G and diameter d . Find: (a) The reaction torques at two supports. (b) The vale and position of maximum shear stress. (c) The value and position of maximum angle of twist. Solution: FBD EQM L t T T T C A o 2 1 + = = ∑ (1) internal torque L x t T x T T A AB 2 ) ( : 2 o − = = ∑ Load/displacement relations ( ) 6 1 2 1 ) ( 2 o 2 o L t L T GI dx L x t T I G I G dx x T A P L A P L P AB AB − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = ∫ ∫ φ A B t o L t ( x ) T A T C t o T A A T AB ( x ) x t o x/L Compatibility = AB φ ( ) 6 1 2 o = − L t L T GI A P or L t T A o 6 1 = (2) From (1) (a) L t T B o 3 1 = . (b) To find maximum shear stress, have to find maximum internal torque first. L t T o max 3 1 = at point B ( x = L ) 3 o 4 o max max 3 16 32 2 ) 3 ( d L t d d L t I r T P π π τ = = = Maximum shear stress occurs at point B ( x = L ), outer surface of the bar....
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This note was uploaded on 09/18/2011 for the course EM 319 taught by Professor Staff during the Spring '11 term at Oklahoma State.

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EM 319 - Review for 1st Midterm Exam - Review 1 Fall 08 1 A...

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