A 0.640 in diameter bar of original length 4 in carries 8,000 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

   $$A_{0} = (\pi/4)(0.640)^{2} = 0.32170 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 8000/0.32170 = 24,868 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.07/4 = 0.01750$$

6. Formula

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

7. Substituting

   $$\sigma_true = 24,868(1 + 0.01750) = 25,303 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.01750) = 0.01735$$

Why the other options are there

• σ_true = 24,440 psi (divided instead of multiplied)
• ε_true = 0.0175 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.750 in diameter bar of original length 8 in carries 20,000 lb and stretches 0.35 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

   $$A_{0} = (\pi/4)(0.750)^{2} = 0.44179 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 20000/0.44179 = 45,271 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.35/8 = 0.04375$$

6. Formula

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

7. Substituting

   $$\sigma_true = 45,271(1 + 0.04375) = 47,251 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.04375) = 0.04282$$

Why the other options are there

• σ_true = 43,373 psi (divided instead of multiplied)
• ε_true = 0.0438 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.445 in diameter bar of original length 4 in carries 18,000 lb and stretches 0.30 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

   $$A_{0} = (\pi/4)(0.445)^{2} = 0.15553 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 18000/0.15553 = 115,734 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.30/4 = 0.07500$$

6. Formula

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

7. Substituting

   $$\sigma_true = 115,734(1 + 0.07500) = 124,415 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.07500) = 0.07232$$

Why the other options are there

• σ_true = 107,660 psi (divided instead of multiplied)
• ε_true = 0.0750 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.475 in diameter bar of original length 5 in carries 9,000 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

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

2. Formula

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

3. Substituting

   $$\sigma_eng = 9000/0.17721 = 50,789 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.12/5 = 0.02400$$

6. Formula

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

7. Substituting

   $$\sigma_true = 50,789(1 + 0.02400) = 52,007 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.02400) = 0.02372$$

Why the other options are there

• σ_true = 49,598 psi (divided instead of multiplied)
• ε_true = 0.0240 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.570 in diameter bar of original length 7 in carries 11,000 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

   $$A_{0} = (\pi/4)(0.570)^{2} = 0.25518 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 11000/0.25518 = 43,108 psi$$

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

   $$\sigma_true = 43,108(1 + 0.02857) = 44,339 psi$$

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 = 41,910 psi (divided instead of multiplied)
• ε_true = 0.0286 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.420 in diameter bar of original length 8 in carries 14,000 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

   $$A_{0} = (\pi/4)(0.420)^{2} = 0.13854 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 14000/0.13854 = 101,051 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.12/8 = 0.01500$$

6. Formula

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

7. Substituting

   $$\sigma_true = 101,051(1 + 0.01500) = 102,567 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.01500) = 0.01489$$

Why the other options are there

• σ_true = 99,557 psi (divided instead of multiplied)
• ε_true = 0.0150 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.685 in diameter bar of original length 3 in carries 7,000 lb and stretches 0.05 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

   $$A_{0} = (\pi/4)(0.685)^{2} = 0.36853 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 7000/0.36853 = 18,994 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.05/3 = 0.01667$$

6. Formula

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

7. Substituting

   $$\sigma_true = 18,994(1 + 0.01667) = 19,311 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.01667) = 0.01653$$

Why the other options are there

• σ_true = 18,683 psi (divided instead of multiplied)
• ε_true = 0.0167 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.605 in diameter bar of original length 2 in carries 16,000 lb and stretches 0.17 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

   $$A_{0} = (\pi/4)(0.605)^{2} = 0.28748 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 16000/0.28748 = 55,657 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.17/2 = 0.08500$$

6. Formula

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

7. Substituting

   $$\sigma_true = 55,657(1 + 0.08500) = 60,388 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.08500) = 0.08158$$

Why the other options are there

• σ_true = 51,297 psi (divided instead of multiplied)
• ε_true = 0.0850 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.440 in diameter bar of original length 2 in carries 24,000 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

   $$A_{0} = (\pi/4)(0.440)^{2} = 0.15205 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 24000/0.15205 = 157,840 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.13/2 = 0.06500$$

6. Formula

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

7. Substituting

   $$\sigma_true = 157,840(1 + 0.06500) = 168,099 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.06500) = 0.06297$$

Why the other options are there

• σ_true = 148,206 psi (divided instead of multiplied)
• ε_true = 0.0650 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress

A 0.420 in diameter bar of original length 8 in carries 10,000 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

   $$A_{0} = (\pi/4)(0.420)^{2} = 0.13854 in^{2}$$

2. Formula

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

3. Substituting

   $$\sigma_eng = 10000/0.13854 = 72,179 psi$$

4. Formula — ε_eng = ΔL/L₀

5. Substituting

   $$\varepsilon_eng = 0.25/8 = 0.03125$$

6. Formula

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

7. Substituting

   $$\sigma_true = 72,179(1 + 0.03125) = 74,435 psi$$

8. Formula

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

9. Substituting

   $$\varepsilon_true = \ln (1.03125) = 0.03077$$

Why the other options are there

• σ_true = 69,992 psi (divided instead of multiplied)
• ε_true = 0.0313 (no logarithm)

Reference: FE Reference Handbook — Materials Science → Engineering stress