A 25 mm diameter steel rod, 6.0 m long, carries 120 kN in tension. With E = 200 GPa, find the stress, the strain and the elongation.

Start with the thinking

• Area first — every axial answer depends on it.
• Keep units consistent: N and mm give MPa directly.

Step-by-step solution

1. Area — A = πd²/4 = π(25)²/4 = 491 mm²

2. Stress

   $$\sigma = P/A = 120,000/491 = 244 MPa$$

3. Strain

   $$\varepsilon = \sigma/E = 244/200,000 = 0.00122$$

4. Elongation — δ = PL/(AE) = εL = 0.00122(6,000)

5. Result

   $$\delta = 7.33 mm$$

Why the other options are there

• δ = 0.733 mm (metres and millimetres mixed)
• σ = 61 MPa (diameter used as area)

Reference: FE Reference Handbook — Mechanics of Materials — Uniaxial loading

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 mechanics of materials problem uses Flexural stress. Given moment (M) = 3,370 kip·in; distance to extreme fiber (c) = 9.9000 in; moment of inertia (I) = 1,880 in⁴, determine the bending stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except sigma is given, so isolate sigma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = M c / I$$

2. Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.

3. Step 3 — List the givens: moment (M) = 3,370 kip·in, distance to extreme fiber (c) = 9.9000 in, moment of inertia (I) = 1,880 in⁴.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\sigma = 17.7463\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 17.7463 ksi to

   $$\sigma = M c / I$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 35.4926 — kept a factor of two that cancels in the correct rearrangement.
• 8.8731 — dropped that same factor in the other direction.
• 19.5209 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stress-Strain Curve for Mild Steel

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 9.8000 in; moment of inertia (I) = 1,770 in⁴; bending stress (sigma) = 16.4600 ksi, determine the moment (M) in kip·in.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except M is given, so isolate M symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = M c / I$$

2. Step 2 — Rearrange the relation so that M stands alone on the left-hand side.

3. Step 3 — List the givens: distance to extreme fiber (c) = 9.8000 in, moment of inertia (I) = 1,770 in⁴, bending stress (sigma) = 16.4600 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$M = 2973\ \text{kip·in}$$

6. Step 6 — Check: returning M = 2,973 kip·in to

   $$\sigma = M c / I$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 5,946 — kept a factor of two that cancels in the correct rearrangement.
• 1,486 — dropped that same factor in the other direction.
• 3,270 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stress-Strain Curve for Mild Steel

A mechanics of materials problem uses Flexural stress. Given moment (M) = 3,450 kip·in; moment of inertia (I) = 1,140 in⁴; bending stress (sigma) = 45.1700 ksi, determine the distance to extreme fiber (c) in in.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except c is given, so isolate c symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = M c / I$$

2. Step 2 — Rearrange the relation so that c stands alone on the left-hand side.

3. Step 3 — List the givens: moment (M) = 3,450 kip·in, moment of inertia (I) = 1,140 in⁴, bending stress (sigma) = 45.1700 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$c = 14.9257\ \text{in}$$

6. Step 6 — Check: returning c = 14.9257 in to

   $$\sigma = M c / I$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 29.8515 — kept a factor of two that cancels in the correct rearrangement.
• 7.4629 — dropped that same factor in the other direction.
• 16.4183 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stress-Strain Curve for Mild Steel

A mechanics of materials problem uses Flexural stress. Given moment (M) = 2,130 kip·in; distance to extreme fiber (c) = 13.1000 in; bending stress (sigma) = 42.6300 ksi, determine the moment of inertia (I) in in⁴.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except I is given, so isolate I symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = M c / I$$

2. Step 2 — Rearrange the relation so that I stands alone on the left-hand side.

3. Step 3 — List the givens: moment (M) = 2,130 kip·in, distance to extreme fiber (c) = 13.1000 in, bending stress (sigma) = 42.6300 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$I = 654.5\ \text{in⁴}$$

6. Step 6 — Check: returning I = 654.5 in⁴ to

   $$\sigma = M c / I$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,309 — kept a factor of two that cancels in the correct rearrangement.
• 327.3 — dropped that same factor in the other direction.
• 720.0 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stress-Strain Curve for Mild Steel

A mechanics of materials problem uses Flexural stress. Given moment (M) = 4,330 kip·in; distance to extreme fiber (c) = 14.9000 in; moment of inertia (I) = 1,750 in⁴, determine the bending stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except sigma is given, so isolate sigma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = M c / I$$

2. Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.

3. Step 3 — List the givens: moment (M) = 4,330 kip·in, distance to extreme fiber (c) = 14.9000 in, moment of inertia (I) = 1,750 in⁴.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\sigma = 36.8669\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 36.8669 ksi to

   $$\sigma = M c / I$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 73.7337 — kept a factor of two that cancels in the correct rearrangement.
• 18.4334 — dropped that same factor in the other direction.
• 40.5535 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stress-Strain Curve for Mild Steel

A mechanics of materials problem uses Flexural stress. Given distance to extreme fiber (c) = 12.2000 in; moment of inertia (I) = 1,630 in⁴; bending stress (sigma) = 55.4500 ksi, determine the moment (M) in kip·in.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except M is given, so isolate M symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = M c / I$$

2. Step 2 — Rearrange the relation so that M stands alone on the left-hand side.

3. Step 3 — List the givens: distance to extreme fiber (c) = 12.2000 in, moment of inertia (I) = 1,630 in⁴, bending stress (sigma) = 55.4500 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$M = 7408\ \text{kip·in}$$

6. Step 6 — Check: returning M = 7,408 kip·in to

   $$\sigma = M c / I$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 14,817 — kept a factor of two that cancels in the correct rearrangement.
• 3,704 — dropped that same factor in the other direction.
• 8,149 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stress-Strain Curve for Mild Steel

A mechanics of materials problem uses Flexural stress. Given moment (M) = 1,040 kip·in; moment of inertia (I) = 1,260 in⁴; bending stress (sigma) = 27.1300 ksi, determine the distance to extreme fiber (c) in in.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except c is given, so isolate c symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = M c / I$$

2. Step 2 — Rearrange the relation so that c stands alone on the left-hand side.

3. Step 3 — List the givens: moment (M) = 1,040 kip·in, moment of inertia (I) = 1,260 in⁴, bending stress (sigma) = 27.1300 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$c = 32.8690\ \text{in}$$

6. Step 6 — Check: returning c = 32.8690 in to

   $$\sigma = M c / I$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 65.7381 — kept a factor of two that cancels in the correct rearrangement.
• 16.4345 — dropped that same factor in the other direction.
• 36.1559 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stress-Strain Curve for Mild Steel

A mechanics of materials problem uses Flexural stress. Given moment (M) = 1,110 kip·in; distance to extreme fiber (c) = 5.2000 in; bending stress (sigma) = 43.3500 ksi, determine the moment of inertia (I) in in⁴.

Start with the thinking

• The governing relation printed in this handbook section is Flexural stress.
• Everything except I is given, so isolate I symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = M c / I$$

2. Step 2 — Rearrange the relation so that I stands alone on the left-hand side.

3. Step 3 — List the givens: moment (M) = 1,110 kip·in, distance to extreme fiber (c) = 5.2000 in, bending stress (sigma) = 43.3500 ksi.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$I = 133.1\ \text{in⁴}$$

6. Step 6 — Check: returning I = 133.1 in⁴ to

   $$\sigma = M c / I$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 266.3 — kept a factor of two that cancels in the correct rearrangement.
• 66.5744 — dropped that same factor in the other direction.
• 146.5 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Stress-Strain Curve for Mild Steel