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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
EA max EAd2
EA 0 max: SOLVE QUADRATIC EQUATION FOR
2 W Weight dmax WL
b + 2L a
EA Properties of elastic cord:
E modulus of elasticity A cross-sectional area L original length
elongation of elastic cord max STATIC ELONGATION
EA IMPACT FACTOR
1 + c1 + 2EA 1/2
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
2L st dst (2.5%)(L)
EA Impact factor ; 0.025L W
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.
- Fall '11