TABLE P4.4 Stress, MPa Strain, 10 3 m/m Stress, MPa Strain, 10 3 m/m 0.0 399.9 5.2 55.2 0.6 427.5 5.8 117.2 1.5 441.3 6.2 186.2 2.4 448.2 6.5 241.3 3.2 462.0 7.3 296.5 4.0 468.9 8.1 344.8 4.6 482.7 9.7

1. The tensile strength vs. temperature curves of aluminum are plotted below. Explain the meaning of these two curves. 400 Ultimate strength 300 Yield strength 200 100 -200 -150 -100 -50 50 100 150 200 250 Temperature, °C Strength, kPa

A steel specimen is tested in tension. The specimen is 25 mm wide by 12.5 mm thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 160 kN and fractured at 214 kN. a. Determine the tensile stress at yield and at fracture. b. If the original gauge length was 100 mm, estimate the gauge length when the specimen is stressed to 1/2 the yield stress.

A steel component is subjected to alternate cyclical loading. The steel follows Basquin's law for high cycle fatigue, o, x N = C, (where the stress amplitude is in MPa). Ignore the geometric detail and assume that Marin's modifying factors are all equal to 1. You are given the minimum stress ain = -213 MPa, the maximum stress omax = 213 MPa. The material data are Tensile strength oUTS = 539 MPa, Basquin's constant c, = 875 MPa, Basquin's exponent a = 0.085. a) Calculate the stress ratio R, the stress amplitude o, in MPa and the mean stress am in MPa. The answers are acceptable with a tolerance of 0.01 for R and of 1 MPa the stresses. R: MPa MPа b) Calculate the corresponding life, in 10° cycles, (tolerance of 0.1 106 cycles) N :