The following true stresses produce the corresponding true plastic strains for a brass alloy: $$\begin{aligned} &\text { True }\\\ &\text {Stress}\\\ &\begin{array}{cc} (p s i) & \text {True Strain} \\ \hline 60,000 & 0.15 \\ 70,000 & 0.25 \end{array} \end{aligned}$$ What true stress is necessary to produce a true plastic strain of \(0.21 ?\)

Short Answer

Expert verified
Answer: The true stress necessary to produce a true plastic strain of 0.21 is 66,000 psi.

Step by step solution

01

Calculate the slope and intercept of the linear relationship between true stress and true plastic strain

We can write the linear relationship between true stress (Y) and true plastic strain (X) as: $$Y = mX + b$$ Where m is the slope and b is the intercept. Using point 1: \((X_{1}, Y_{1}) = (0.15, 60000)\) and point 2: \((X_{2}, Y_{2}) = (0.25, 70000)\), we can calculate the slope and intercept as follows: Slope: $$m = \frac{Y_{2} - Y_{1}}{X_{2} - X_{1}} = \frac{70000 - 60000}{0.25 - 0.15} = \frac{10000}{0.1} = 100000$$ Next, we can substitute the slope and point 1's coordinates to find the intercept: $$60000 = 100000\cdot0.15 + b$$ $$b = 60000 - 100000\cdot0.15 = 45000$$ So the linear relationship looks like this: $$Y = 100000X + 45000$$
02

Use the linear relationship to find the true stress for a true plastic strain of 0.21

Now that we have our linear relationship between true stress and true plastic strain, we can plug in the desired true plastic strain of 0.21 to find the corresponding true stress. $$Y = 100000 \cdot 0.21 + 45000 = 21000 + 45000 = 66000$$ The true stress necessary to produce a true plastic strain of 0.21 is 66,000 psi.

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