A steel member is fully restrained between rigid abutments and heated 40 °C. With α = 11.7 × 10⁻⁶ /°C and E = 200 GPa, what stress develops?

Start with the thinking

• Full restraint means the free thermal strain is cancelled by an equal mechanical strain.
• Length cancels out — the answer does not depend on the span.

Step-by-step solution

1. Free thermal strain — ε_T = αΔT = 11.7×10⁻⁶(40) = 4.68×10⁻⁴

2. Restraint condition

   $$\varepsilon_mech = -\varepsilon_T$$

3. Stress

   $$\sigma = E\varepsilon = 200,000(4.68\times10^{-4})$$

4. Result

   $$\sigma = 93.6 MPa compression$$

Why the other options are there

• 0 MPa (free expansion assumed)
• 93.6 MPa tension (sign of restraint reversed)

Reference: FE Reference Handbook — Mechanics of Materials — Thermal deformations

A steel rod of area 4.50 in² and length 274 in. carries 47 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 47/4.50 = 10.44 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 47(274)/(4.50 \times 29,000)$$

5. Evaluate

   $$\delta = 0.0987 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.000360 = \sigma/E = 0.000360 ✓$$

Why the other options are there

• 10,444 ksi (psi/ksi confusion)
• 2,862 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain

A steel rod of area 2.50 in² and length 237 in. carries 105 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 105/2.50 = 42.00 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 105(237)/(2.50 \times 29,000)$$

5. Evaluate

   $$\delta = 0.3432 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.001448 = \sigma/E = 0.001448 ✓$$

Why the other options are there

• 42,000 ksi (psi/ksi confusion)
• 9,954 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain

A steel rod of area 2.25 in² and length 256 in. carries 93 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 93/2.25 = 41.33 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 93(256)/(2.25 \times 29,000)$$

5. Evaluate

   $$\delta = 0.3649 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.001425 = \sigma/E = 0.001425 ✓$$

Why the other options are there

• 41,333 ksi (psi/ksi confusion)
• 10,581 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain

A steel rod of area 3.50 in² and length 134 in. carries 62 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 62/3.50 = 17.71 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 62(134)/(3.50 \times 29,000)$$

5. Evaluate

   $$\delta = 0.0819 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.000611 = \sigma/E = 0.000611 ✓$$

Why the other options are there

• 17,714 ksi (psi/ksi confusion)
• 2,374 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain

A steel rod of area 5.25 in² and length 154 in. carries 55 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 55/5.25 = 10.48 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 55(154)/(5.25 \times 29,000)$$

5. Evaluate

   $$\delta = 0.0556 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.000361 = \sigma/E = 0.000361 ✓$$

Why the other options are there

• 10,476 ksi (psi/ksi confusion)
• 1,613 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain

A steel rod of area 4.50 in² and length 143 in. carries 47 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 47/4.50 = 10.44 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 47(143)/(4.50 \times 29,000)$$

5. Evaluate

   $$\delta = 0.0515 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.000360 = \sigma/E = 0.000360 ✓$$

Why the other options are there

• 10,444 ksi (psi/ksi confusion)
• 1,494 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain

A steel rod of area 6.75 in² and length 77 in. carries 47 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 47/6.75 = 6.96 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 47(77)/(6.75 \times 29,000)$$

5. Evaluate

   $$\delta = 0.0185 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.000240 = \sigma/E = 0.000240 ✓$$

Why the other options are there

• 6,963 ksi (psi/ksi confusion)
• 536.1 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain

A steel rod of area 2.00 in² and length 197 in. carries 116 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 116/2.00 = 58.00 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 116(197)/(2.00 \times 29,000)$$

5. Evaluate

   $$\delta = 0.3940 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.002000 = \sigma/E = 0.002000 ✓$$

Why the other options are there

• 58,000 ksi (psi/ksi confusion)
• 11,426 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain

A steel rod of area 7.50 in² and length 169 in. carries 116 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Start with the thinking

• Stress is load over area; deformation adds length and stiffness.
• Keep kips and inches so ksi comes out directly.

Step-by-step solution

1. Stress

   $$\sigma = P/A$$

2. Substituting

   $$\sigma = 116/7.50 = 15.47 ksi$$

3. Elongation

   $$\delta = PL/(AE)$$

4. Substituting

   $$\delta = 116(169)/(7.50 \times 29,000)$$

5. Evaluate

   $$\delta = 0.0901 in$$

6. Strain check

   $$\varepsilon = \delta/L = 0.000533 = \sigma/E = 0.000533 ✓$$

Why the other options are there

• 15,467 ksi (psi/ksi confusion)
• 2,614 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering Strain