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Unformatted text preview: Sec_2.8-2.12.qxd 9/25/08 11:44 AM Page 217 SECTION 2.8 Impact Loading Problem 2.8-8 A cable with a restrainer at the bottom hangs
vertically from its upper end (see figure). The cable has an
effective cross-sectional area A 40 mm2 and an effective
modulus of elasticity E 130 GPa. A slider of mass M 35 kg
drops from a height h 1.0 m onto the restrainer.
If the allowable stress in the cable under an impact load is
500 MPa, what is the minimum permissible length L of the cable? Cable Slider
Restrainer Probs. 2.8-8, 2.8-2, 2.8-9 Solution 2.8-8 Slider on a cable STATIC STRESS
A 343.4 N
40 mm2 8.585 MPa MINIMUM LENGTH Lmin
59): smax Eq. (2 sst c 1 + a 1 + 2hE 1/2
Square both sides and solve for L:
L Lmin 2Ehsst
2sst) ; SUBSTITUTE NUMERICAL VALUES:
(35 kg)(9.81 m/s2) W Mg A 40 mm2 h 1.0 m E 130 GPa allow max 343.4 N Find minimum length Lmin 500 MPa Lmin 2(130 GPa) (1.0 m) (8.585 MPa)
(500 MPa) [500 MPa
9.25 mm ; 217 ...
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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