A 0.505 in diameter bar of original length 6 in carries 17,000 lb and stretches 0.08 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.505)^{2} = 0.20030 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 17000/0.20030 = 84,874 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.08/6 = 0.01333$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 84,874(1 + 0.01333) = 86,006 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.01333) = 0.01325$$

Why the other options are there

• σ_true = 83,758 psi (divided instead of multiplied)
• ε_true = 0.0133 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.480 in diameter bar of original length 5 in carries 10,000 lb and stretches 0.09 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.480)^{2} = 0.18096 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 10000/0.18096 = 55,262 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.09/5 = 0.01800$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 55,262(1 + 0.01800) = 56,257 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.01800) = 0.01784$$

Why the other options are there

• σ_true = 54,285 psi (divided instead of multiplied)
• ε_true = 0.0180 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.475 in diameter bar of original length 8 in carries 7,000 lb and stretches 0.15 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.475)^{2} = 0.17721 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 7000/0.17721 = 39,502 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.15/8 = 0.01875$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 39,502(1 + 0.01875) = 40,243 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.01875) = 0.01858$$

Why the other options are there

• σ_true = 38,775 psi (divided instead of multiplied)
• ε_true = 0.0188 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.635 in diameter bar of original length 4 in carries 4,000 lb and stretches 0.13 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.635)^{2} = 0.31669 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 4000/0.31669 = 12,631 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.13/4 = 0.03250$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 12,631(1 + 0.03250) = 13,041 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.03250) = 0.03198$$

Why the other options are there

• σ_true = 12,233 psi (divided instead of multiplied)
• ε_true = 0.0325 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.505 in diameter bar of original length 6 in carries 5,000 lb and stretches 0.25 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.505)^{2} = 0.20030 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 5000/0.20030 = 24,963 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.25/6 = 0.04167$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 24,963(1 + 0.04167) = 26,003 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.04167) = 0.04082$$

Why the other options are there

• σ_true = 23,965 psi (divided instead of multiplied)
• ε_true = 0.0417 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.595 in diameter bar of original length 7 in carries 21,000 lb and stretches 0.20 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.595)^{2} = 0.27805 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 21000/0.27805 = 75,526 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.20/7 = 0.02857$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 75,526(1 + 0.02857) = 77,684 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.02857) = 0.02817$$

Why the other options are there

• σ_true = 73,428 psi (divided instead of multiplied)
• ε_true = 0.0286 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.545 in diameter bar of original length 6 in carries 21,000 lb and stretches 0.07 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.545)^{2} = 0.23328 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 21000/0.23328 = 90,019 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.07/6 = 0.01167$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 90,019(1 + 0.01167) = 91,070 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.01167) = 0.01160$$

Why the other options are there

• σ_true = 88,981 psi (divided instead of multiplied)
• ε_true = 0.0117 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.650 in diameter bar of original length 5 in carries 5,000 lb and stretches 0.09 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.650)^{2} = 0.33183 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 5000/0.33183 = 15,068 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.09/5 = 0.01800$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 15,068(1 + 0.01800) = 15,339 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.01800) = 0.01784$$

Why the other options are there

• σ_true = 14,801 psi (divided instead of multiplied)
• ε_true = 0.0180 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.560 in diameter bar of original length 3 in carries 15,000 lb and stretches 0.12 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.560)^{2} = 0.24630 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 15000/0.24630 = 60,901 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.12/3 = 0.04000$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 60,901(1 + 0.04000) = 63,337 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.04000) = 0.03922$$

Why the other options are there

• σ_true = 58,559 psi (divided instead of multiplied)
• ε_true = 0.0400 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain

A 0.720 in diameter bar of original length 7 in carries 10,000 lb and stretches 0.28 in. Compute the engineering stress and engineering strain, then convert both to true stress and true strain.

Start with the thinking

• Engineering stress uses the original area; true stress uses the instantaneous area.
• Below necking, constant volume gives σ_true = σ_eng(1 + ε) and ε_true = ln(1 + ε).

Step-by-step solution

1. Area

   $$A_{0} = (\pi/4)(0.720)^{2} = 0.40715 in^{2}$$

2. Formula

   $$\sigma_eng = P/A_{0}$$

3. Substituting

   $$\sigma_eng = 10000/0.40715 = 24,561 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.28/7 = 0.04000$$

6. Formula

   $$\sigma_true = \sigma_eng(1 + \varepsilon_eng)$$

7. Substituting

   $$\sigma_true = 24,561(1 + 0.04000) = 25,543 psi$$

8. Formula

   $$\varepsilon_true = \ln (1 + \varepsilon_eng)$$

9. Substituting

   $$\varepsilon_true = \ln (1.04000) = 0.03922$$

Why the other options are there

• σ_true = 23,616 psi (divided instead of multiplied)
• ε_true = 0.0400 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering strain