The stress-strain diagram of reinforcement steel having a cross-sectional diameter of 12 mm diameter and 100 mm gage length is determined after its tensile strength test as follows. Based on the stress-strain diagram determine the followings properties of the material (Poisson’s ratio of the material is 0.32) a) Resilience b) Shear modulus c) Bulk modulus

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 :

Draw the stress-strain diagram for any ductile and brittle material and explain the following points: (a) Modulus of elasticity (b) Yield strength (c) Ultimate Strength (d) Ductility (e) Toughness

The composite bar is firmly attached to unyielding supports. The bar is stress-free at 50°F. Compute the stress in each material after the 80-kip force is applied and the temperature is increased to 100°F. Use a = 6.5 x106/°F for steel and a = 12.8 x106/°F for aluminum. Aluminum A = 2 in.² E = 10 x 106 psi 15 in. Steel A = 3 in.2 E = 29 × 106 psi 80 kips 10 in.