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Unformatted text preview: HW7 solutions. 1. a. elastic strain = reversible strain =E b. Plastic strain = permanent (non-reversible) strain. There is really none. c. When the stress is relieved if it goes back to the original it experienced elastic strain, if it does not go back, it experienced elastic + plastic strain. d. (answers vary) A cable on a ski lift (for example) e. (answers vary) When a paper clip is successively bent (for example): similar to the picture of Zn. f. Yield strength. g. Youngs modulus and elastic modulus 2.a. (i) This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation ( F y ). Taking the yield strength to be 345 MPa, and employment of Equation 7.1 leads to = 44,850 N (10,000 lb f ) (ii) The maximum length to which the sample may be deformed without plastic deformation is determined from Equations 7.2 and 7.5 as b. We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 1.9 mm (0.075 in.). It is first necessary to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if...
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- Spring '12