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hs05201f10

# hs05201f10 - EMSE 201 Introduction to Materials Science...

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EMSE 201 – Introduction to Materials Science & Engineering Fall 2010 Case Western Reserve University 1 of 4 Case School of Engineering Solution to Homework #5 Explain how answers were obtained. Give appropriate units for all numerical answers. 1) (C&R problem 7.4) For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 × 10 6 psi). a) (3 points) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm 2 (0.2 in. 2 ) without plastic deformation? b) (3 points) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it may be stretched without causing plastic deformation? Solution (from the publisher’s solution manual; used with permission) a) 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 F y = σ y A 0 = ( 345 × 10 6 N/m 2 )( 130 × 10 -6 m 2 ) = 44,850 N (10,000 lb f ) (3 pts) b) The maximum length to which the sample may be deformed without plastic deformation is determined from Equations 7.2 and 7.5 as l i = l 0 1 + ε ( ) = l 0 1 + σ E = (76 mm) 1 + 345 MPa 103 × 10 3 MPa = 76.25 mm (3.002 in.) (3 pts) 2) (5 points) A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elastic modulus of 110 GPa (16.0 × 106 psi). A cylindrical specimen of this alloy 15.2 mm (0.60 in.) in diameter and 380 mm (15.0 in.) long is stressed in tension and found to elongate 1.9 mm (0.075 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why (C&R problem 7.10)

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