The stressstrain diagram for an aluminum alloy specimen | StudySoup

Textbook Solutions for Mechanics of Materials

Chapter 3 Problem 3-10

Question

The stress–strain diagram for an aluminum alloy specimen having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. Determine approximately 3 the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support.

Solution

Step 1 of 3

Calculate the modulus of elasticity from the stress-strain diagram.

\(\mathrm{E}_{\text {approx }}=\frac{\sigma_{@ \varepsilon=0.005}-\sigma_{@ \varepsilon=0}}{\varepsilon_{@ \varepsilon=0.05-} \varepsilon_{@ \varepsilon=0}}\)

Here, \(E_{@ \varepsilon=0.005}\) is the stress of the alloy when strain is 0.005 in./in., \(E_{@ \varepsilon=0}\) is stress of the alloy when strain is 0 in./in., \(\varepsilon_{@ \varepsilon=0.005}\) is strain of the alloy when strain is 0.005 in./in., and \(\varepsilon_{@ \varepsilon=0}\) is strain of the alloy when strain is 0 in./in..

Substitute 50 ksi for \(E_{@ \varepsilon=0.005}\), \(0\ \mathrm{ksi}\) ksi for \(E_{@ \varepsilon=0}\), 0.005 in./in. for \(\varepsilon_{@ \varepsilon=0.005}\), and 0 in./in. for \(\varepsilon_{@ \varepsilon=0}\).

\(\begin{aligned}E_{\text{approx }}&=\frac{50-0}{0.005-0}\\ &=10\times10^3\mathrm{\ ksi}\end{aligned}\)

Therefore, the modulus of elasticity is \(10 \times 10^{3}\ \mathrm{ksi}\)

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full solution

Title Mechanics of Materials 10 
Author Russell C. Hibbeler
ISBN 9780134319650

The stressstrain diagram for an aluminum alloy specimen

Chapter 3 textbook questions

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