228_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

228_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.8-2.12.qxd 222 9/25/08 11:44 AM Page 222 CHAPTER 2 Axially Loaded Members Problem 2.8-13 A weight W rests on top of a wall and is attached to one end of a very flexible cord having cross-sectional area A and modulus of elasticity E (see figure). The other end of the cord is attached securely to the wall. The weight is then pushed off the wall and falls freely the full length of the cord. W W (a) Derive a formula for the impact factor. (b) Evaluate the impact factor if the weight, when hanging statically, elongates the band by 2.5% of its original length. Solution 2.8-13 Weight falling off a wall CONSERVATION OF ENERGY P.E. U or W(L + dmax) d2 max 2WL d EA max EAd2 max 2L 2WL2 EA 0 max: SOLVE QUADRATIC EQUATION FOR 2 W Weight dmax WL WL WL 1/2 + ca b + 2L a bd EA EA EA Properties of elastic cord: E modulus of elasticity A cross-sectional area L original length elongation of elastic cord max STATIC ELONGATION dst WL EA IMPACT FACTOR 1 + c1 + 2EA 1/2 d W P.E. potential energy of weight before fall (with respect to lowest position) dmax dst P.E. W(L NUMERICAL VALUES Let U U max) strain energy of cord at lowest position EAd2 max 2L st dst (2.5%)(L) WL EA Impact factor ; 0.025L W EA EA W 40 1 + [1 + 2(40)]1/2 10 0.025 ; ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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