(Fy). Taking the yield strength to be 275 MPa, and employment of Equation 6.1 leads toР= 89,375 N (20,000 lbf)Р?(b) The maximum length to which the sample may be deformed without plastic deformation is determined from Equations 6.2 and 6.5 asР?6.8 РA cylindrical rod of copper (E = 110 GPa, 16 ´ 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6660 N (1500 lbf). If the length of the rod is 380 mm (15.0 in.), what must be the diameter to allow an elongation of 0.50 mm (0.020 in.)?Р SolutionР?This problem asks us pute the diameter of a cylindrical specimen of copper in order to allow an elongation of 0.50 mm. Employing Equations 6.1, 6.2, and 6.5, assuming that deformation is entirely elasticРOr, solving for d0Р= 7.65 ´ 10-3 m = 7.65 mm (0.30 in.)Р?6.9
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