02-02ChapGere.0006

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Problem 2.3-14 A bar ABC revolves in a horizontal plane about a vertical axis at the midpoint C (see figure). The bar, which has length 2 L and cross-sectional area A , revolves at constant angular speed v . Each half of the bar ( AC and BC ) has weight W 1 and supports a weight W 2 at its end. Derive the following formula for the elongation of one-half of the bar (that is, the elongation of either AC or BC ): d 5 } 3 L g 2 E v A 2 } ( W 1 1 3 W 2 ) in which E is the modulus of elasticity of the material of the bar and g is the acceleration of gravity. Solution 2.3-14 Rotating bar SECTION 2.3 Changes in Lengths under Nonuniform Conditions 85 AC B v LL W 2 W 1 W 1 W 2 C B v L W 1 W 2 x F(x) dx d j j v 5 angular speed A 5 cross-sectional area E 5 modulus of elasticity g 5 acceleration of gravity F ( x ) 5 axial force in bar at distance x from point C Consider an element of length dx at distance x from point C . To find the force F ( x ) acting on this element, we must find the inertia force of the part of the bar from distance x to distance L , plus the inertia force of the
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## This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Tech.

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