223_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

223_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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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 L h Restrainer Probs. 2.8-8, 2.8-2, 2.8-9 Solution 2.8-8 Slider on a cable STATIC STRESS sst W A 343.4 N 40 mm2 8.585 MPa MINIMUM LENGTH Lmin 59): smax Eq. (2 sst c 1 + a 1 + 2hE 1/2 bd Lsst or smax 2hE 1/2 1 a1 + b sst Lsst Square both sides and solve for L: L Lmin 2Ehsst smax(smax 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 2(8.585 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.

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