# 7.59 From The Stress Strain Data For Poly (Methyl Methacphiet?

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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.

(b) From Equation 8.4, the critical resolved shear stress is just

8.18 The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single crystal of Fe pulled in tension.

Solution In order to determine the maximum possible yield strength for a single crystal of Fe pulled in tension, we simply employ Equation 8.5 as

8.23 (a) From the plot of yield strength versus (grain diameter)–1/2 for a 70 Cu–30 Zn cartridge brass in Figure 8.15, determine values for the constants σ0 and ky in Equation 8.7. (b) Now predict the yield strength of this alloy when the average grain diameter is 1.0 10-3 mm.

Solution (a) Perhaps the easiest way to solve for 0 and ky in Equation 8.7 is to pick two values each of y and d-1/2 from Figure 8.15, and then solve two simultaneous equations, which may be created. 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

For copper, the critical resolved shear stress is 2.10 MPa (305 psi) at a dislocation density of 105 mm-2. If it is known that the value of A for copper is 6.35 10-3 MPa-mm (0.92 psi-mm), compute the crss at a dislocation density of 107 mm-2.

Solution We are asked in this problem to compute the critical resolved shear stress at a dislocation density of 107 mm-2. It is first necessary to compute the value of the constant 0 (in the equation provided in the problem statement) from the one set of data. Rearranging this expression such that 0is the dependent variable, and incorporating appropriate values given in the problem statement

Now, the critical resolved shear stress may be determined at a dislocation density of 107 mm-2

*…show more content…*(b) From Equation 8.4, the critical resolved shear stress is just

8.18 The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single crystal of Fe pulled in tension.

Solution In order to determine the maximum possible yield strength for a single crystal of Fe pulled in tension, we simply employ Equation 8.5 as

8.23 (a) From the plot of yield strength versus (grain diameter)–1/2 for a 70 Cu–30 Zn cartridge brass in Figure 8.15, determine values for the constants σ0 and ky in Equation 8.7. (b) Now predict the yield strength of this alloy when the average grain diameter is 1.0 10-3 mm.

Solution (a) Perhaps the easiest way to solve for 0 and ky in Equation 8.7 is to pick two values each of y and d-1/2 from Figure 8.15, and then solve two simultaneous equations, which may be created. 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

*…show more content…*For copper, the critical resolved shear stress is 2.10 MPa (305 psi) at a dislocation density of 105 mm-2. If it is known that the value of A for copper is 6.35 10-3 MPa-mm (0.92 psi-mm), compute the crss at a dislocation density of 107 mm-2.

Solution We are asked in this problem to compute the critical resolved shear stress at a dislocation density of 107 mm-2. It is first necessary to compute the value of the constant 0 (in the equation provided in the problem statement) from the one set of data. Rearranging this expression such that 0is the dependent variable, and incorporating appropriate values given in the problem statement

Now, the critical resolved shear stress may be determined at a dislocation density of 107 mm-2