# Home Worksss

1065 WordsJun 3, 20135 Pages
7.59 From the stress–strain data for poly(methyl methacrylate) shown in Figure 7.24, determine the modulus of elasticity and tensile strength at room temperature [20°C (68°F)], and compare these values with those given in Tables 7.1 and 7.2. Solution From Figure 7.24, the elastic modulus is the slope in the elastic linear region of the 20C curve, which is The value range cited in Table 7.1 is 2.24 to 3.24 GPa (325,000 to 470,000 psi). Thus, the plotted value is a little on the high side. The tensile strength corresponds to the stress at which the curve ends, which is 52 MPa (7500 psi). This value lies within the range cited in Table 7.2—48.3 to 72.4 MPa (7000 to 10,500 psi). 7.66 On the basis of the curves in Figure…show more content…
For example d-1/2 (mm) -1/2 y (MPa) 4 75 12 175 The two equations are thus Solution of these equations yield the values of 0 = 25 MPa (3630 psi) (b) When d = 1.0 10-3 mm, d-1/2 = 31.6 mm-1/2, and, using Equation 8.7, 8.28 Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16 and 11 mm, respectively. The second specimen, with an initial radius of 12 mm, must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation. Solution In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed to the same percent cold work. For the first specimen, using Equation 8.8, the percent cold work is equal to For the second specimen, the deformed radius is computed using the above equation and solving for rd as For a percent cold work of 52.7% and an r0 value of 12 mm, rd is equal to 8.32 Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress τcrss is a function of the dislocation density ρD as where τ0 and A are constants. For copper, the critical resolved shear stress is 2.10 MPa (305 psi) at a