02-07ChapGere.0007

02-07ChapGere.0007 - 166 A-PDF Axially Loaded : Purchase...

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Unformatted text preview: 166 A-PDF Axially Loaded : Purchase CHAPTER 2 Split DEMOMembers from www.A-PDF.com to remove the watermark Nonlinear Behavior (Changes in Lengths of Bars) Problem 2.11-1 A bar AB of length L and weight density hangs vertically under its own weight (see figure). The stress-strain relation for the material is given by the Ramberg-Osgood equation (Eq. 2-71): 0 m 0 A E E Derive the following formula L2 2E 0 L (m L 1)E L 0 m for the elongation of the bar. B Solution 2.11-1 Bar hanging under its own weight STRAIN AT DISTANCE x Let A cross-sectional area axial force at distance x Ax N A gx 0 L e s E s0 E dx Let N L N x ELONGATION OF BAR L s m s0 gx E s0 E L s e dx 0 gL2 2E s0 L gL m (m 1)E s0 gx dx E s0 E 0 gx m s0 Q.E.D. gx m dx s0 Problem 2.11-2 A prismatic bar of length L 1.8 m and cross-sectional area A 480 mm2 is loaded by forces P1 30 kN and P2 60 kN (see figure). The bar is constructed of magnesium alloy having a stress-strain curve described by the following Ramberg-Osgood equation: 45,000 1 618 10 A 2L -- 3 B P1 C L -- 3 P2 170 ( MPa) in which has units of megapascals. (a) Calculate the displacement C of the end of the bar when the load P1 acts alone. (b) Calculate the displacement when the load P2 acts alone. (c) Calculate the displacement when both loads act simultaneously. Solution 2.11-2 A 2L -- 3 Axially loaded bar B P1 C L -- 3 P2 L P1 1.8 m A 480 mm2 60 kN 30 kN P2 s 45,000 Ramberg Osgood Equation: e Find displacement at end of bar. 1 s 10 (s 618 170 MPa) ...
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