## Chapter 6

## Q. 6.5

True Stress and True Strain

Compare engineering stress and strain with true stress and strain for the aluminum alloy in Example 6-1 at (a) the maximum load and (b) fracture. The diameter at maximum load is 0.4905 in. and at fracture is 0.398 in.

## Step-by-Step

## Verified Solution

(a) At the maximum load,

Engineering stress S = \frac{F}{A_{0}}=\frac{8000 \ lb}{(\pi /4)(0.505 \ in.)^2}=39,941 \ psi

Engineering strain e = \frac{\Delta l}{l_{0}}=\frac{2.120 \ - \ 2.000}{2.000}=0.060 \ in./in.

True stress = σ = S(1 + e) = 39,941 psi (1 + 0.060) = 42,337 psi

True strain = ln (1 + e) = ln (1 + 0.060) = 0.058 in./in.

The maximum load is the last point at which the expressions used here for true stress and true strain apply. Note that the same answers are obtained for true stress and strain if the instantaneous dimensions are used:

\sigma=\frac{F}{A}=\frac{8000 \ lb}{(\pi /4)(0.4905 \ in.)^2}=42,337 \ psi

\varepsilon =\ln (\frac {A_{0}}{A})=\ln \left[\frac{(\pi /4)(0.505 \ in.)^2}{(\pi /4)(0.4905 \ in.)^2} \right]=0.058 \ in./in.

Up until the point of necking in a tensile test, the engineering stress is less than the corresponding true stress, and the engineering strain is greater than the corresponding true strain.

(b) At fracture,

S=\frac{F}{A_{0}}=\frac{7600 \ lb}{(\pi /4)(0.505 \ in.)^2}=37,944 \ psi

e=\frac{\Delta l}{l_{0}}=\frac{2.205 \ – \ 2.000}{2.000}=0.103 \ in./in.

\sigma =\frac{F}{A_{f}}=\frac{7600 \ lb}{(\pi /4)(0.398 \ in.)^2}=61,088 \ psi

\varepsilon = \ln (\frac{A_{0}}{A_{f}} ) = \ln \left[\frac{(\pi /4)(0.505 \ in.)^2}{(\pi /4)(0.398 \ in.)^2} \right]= 0.476 \ in./in.

It was necessary to use the instantaneous dimensions to calculate the true stress and strain, since failure occurs past the point of necking. After necking, the true strain is greater than the corresponding engineering strain.