# hmwk3b - ME541 Fatigue of Materials Homework 3 Solutions...

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- 1 - ME541: Fatigue of Materials Homework 3 Solutions 2.1) It is assumed that the strains is uniform throughout, therefore the relations between true stress/strain and engineering stress/strain are applied such that: s = s(1+e) and e = ln(1+e). For s =62.2 ksi and e=0.0098 in/in respectively, s = 62.8 ksi and e = 0.0098. For s =90.8 ksi and e=0.0898in/in respectively, s = 98.95 ksi and e = 0.0860. At the lower stress/strain values, presumably in the elastic region, the true stress/strain and engineering stress/strain are nearly identical, but as the stress/strain increases into the plastic region, the true stress become greater than the engineering stress and the true strain becomes less than the engineering strain. 2.2) For a 50 mm initial diameter and a 250 mm initial gage length. Structural steel at room temperature. Elastic modulus, GPa 210.084 e s A L d P ?e ?s E elastic elastic 0 0 elastic0 elastic0 = = = = Upper yield, MPa A P S 0 upper upper y 4 . 188 = = Lower yield, MPa A P S 0 lower lower y 1 . 168 = = 0.2% yield, MPa 168.1 A P S S 0 lower lower y yp = Ultimate tensile strength, MPa 269.9 A P S 0 max uts = = Engineering stress and strain at necking, m/m 0.20 e MPa, 269.9 S u uts = =

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- 2 - True stress and strain at necking, m/m 0.18 e MPa, 323.9 s u u = = Strain hardening exponent, 0.18 e n u = Percentage Reduction in Area, % RA = 64 d d d * 100 2 0 2 f 2 0 = - True fracture ductility, m/m d d ln A A ln e or, m/m 0.23 L L ln e 2 f 2 0 0 f f 0 f f 02 . 1 = = = = = True fracture strength, MPa 424.4 s f = Strength coefficient, MPa 441.8 e s K n u u = = 2.11) For the material, the values of E = 30,000 ksi, n’ = 0.202 and K’ = 74.6 ksi are given. The cyclic stress-strain curve is determined from 0 to 0.02 in/in strain
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## This note was uploaded on 11/28/2010 for the course ENES enes100 taught by Professor Staff during the Spring '10 term at Maryland.

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hmwk3b - ME541 Fatigue of Materials Homework 3 Solutions...

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