A steel tension coupon obeys Hooke's law up to the yield point. Given modulus of elasticity (E) = 14,900 ksi; axial strain (epsilon) = 0.0002, determine the normal stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = E \varepsilon$$

3. Step 3

   $$List the givens: modulus of elasticity (E) = 14,900 ksi, axial strain (epsilon) = 0.0002$$

4. Step 4 — Substitute the given values:

   $$\sigma = 14900 \varepsilon$$

5. Step 5 — Evaluate:

   $$\sigma = 2.9800\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 2.9800 ksi to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 5.9600 — kept a factor of two that cancels in the correct rearrangement.
• 1.4900 — dropped that same factor in the other direction.
• 3.2780 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

An aluminum tie rod's elongation is predicted using Hooke's law. Given axial strain (epsilon) = 0.0014; normal stress (sigma) = 62.7000 ksi, determine the modulus of elasticity (E) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• Everything except E is given, so isolate E 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \dfrac{\sigma}{\varepsilon}$$

3. Step 3

   $$List the givens: axial strain (epsilon) = 0.0014, normal stress (sigma) = 62.7000 ksi$$

4. Step 4 — Substitute the given values:

   $$E = \dfrac{62.7000}{\varepsilon}$$

5. Step 5 — Evaluate:

   $$E = 44786\ \text{ksi}$$

6. Step 6 — Check: returning E = 44,786 ksi to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 89,571 — kept a factor of two that cancels in the correct rearrangement.
• 22,393 — dropped that same factor in the other direction.
• 49,264 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

A structural cable's stress-strain response follows Hooke's law under service load. Given modulus of elasticity (E) = 29,300 ksi; normal stress (sigma) = 23.6000 ksi, determine the axial strain (epsilon).

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• Everything except epsilon is given, so isolate epsilon 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for epsilon:

   $$\epsilon = \dfrac{\sigma}{E}$$

3. Step 3

   $$List the givens: modulus of elasticity (E) = 29,300 ksi, normal stress (sigma) = 23.6000 ksi$$

4. Step 4 — Substitute the given values:

   $$\epsilon = \dfrac{23.6000}{29300}$$

5. Step 5 — Evaluate:

   $$\epsilon = 0.0008$$

6. Step 6 — Check: returning epsilon = 0.0008 to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 0.0016 — kept a factor of two that cancels in the correct rearrangement.
• 0.0004 — dropped that same factor in the other direction.
• 0.0009 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

A steel tension coupon obeys Hooke's law up to the yield point. Given modulus of elasticity (E) = 25,000 ksi; axial strain (epsilon) = 0.0020, determine the normal stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = E \varepsilon$$

3. Step 3

   $$List the givens: modulus of elasticity (E) = 25,000 ksi, axial strain (epsilon) = 0.0020$$

4. Step 4 — Substitute the given values:

   $$\sigma = 25000 \varepsilon$$

5. Step 5 — Evaluate:

   $$\sigma = 50.0000\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 50.0000 ksi to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 100.0 — kept a factor of two that cancels in the correct rearrangement.
• 25.0000 — dropped that same factor in the other direction.
• 55.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

An aluminum tie rod's elongation is predicted using Hooke's law. Given axial strain (epsilon) = 0.0014; normal stress (sigma) = 75.3000 ksi, determine the modulus of elasticity (E) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• Everything except E is given, so isolate E 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \dfrac{\sigma}{\varepsilon}$$

3. Step 3

   $$List the givens: axial strain (epsilon) = 0.0014, normal stress (sigma) = 75.3000 ksi$$

4. Step 4 — Substitute the given values:

   $$E = \dfrac{75.3000}{\varepsilon}$$

5. Step 5 — Evaluate:

   $$E = 53786\ \text{ksi}$$

6. Step 6 — Check: returning E = 53,786 ksi to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 107,571 — kept a factor of two that cancels in the correct rearrangement.
• 26,893 — dropped that same factor in the other direction.
• 59,164 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

A structural cable's stress-strain response follows Hooke's law under service load. Given modulus of elasticity (E) = 24,800 ksi; normal stress (sigma) = 56.4000 ksi, determine the axial strain (epsilon).

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• Everything except epsilon is given, so isolate epsilon 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for epsilon:

   $$\epsilon = \dfrac{\sigma}{E}$$

3. Step 3

   $$List the givens: modulus of elasticity (E) = 24,800 ksi, normal stress (sigma) = 56.4000 ksi$$

4. Step 4 — Substitute the given values:

   $$\epsilon = \dfrac{56.4000}{24800}$$

5. Step 5 — Evaluate:

   $$\epsilon = 0.0023$$

6. Step 6 — Check: returning epsilon = 0.0023 to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 0.0045 — kept a factor of two that cancels in the correct rearrangement.
• 0.0011 — dropped that same factor in the other direction.
• 0.0025 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

A steel tension coupon obeys Hooke's law up to the yield point. Given modulus of elasticity (E) = 19,100 ksi; axial strain (epsilon) = 0.0016, determine the normal stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = E \varepsilon$$

3. Step 3

   $$List the givens: modulus of elasticity (E) = 19,100 ksi, axial strain (epsilon) = 0.0016$$

4. Step 4 — Substitute the given values:

   $$\sigma = 19100 \varepsilon$$

5. Step 5 — Evaluate:

   $$\sigma = 30.5600\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 30.5600 ksi to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 61.1200 — kept a factor of two that cancels in the correct rearrangement.
• 15.2800 — dropped that same factor in the other direction.
• 33.6160 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

An aluminum tie rod's elongation is predicted using Hooke's law. Given axial strain (epsilon) = 0.0008; normal stress (sigma) = 80.0000 ksi, determine the modulus of elasticity (E) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• Everything except E is given, so isolate E 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \dfrac{\sigma}{\varepsilon}$$

3. Step 3

   $$List the givens: axial strain (epsilon) = 0.0008, normal stress (sigma) = 80.0000 ksi$$

4. Step 4 — Substitute the given values:

   $$E = \dfrac{80.0000}{\varepsilon}$$

5. Step 5 — Evaluate:

   $$E = 100000\ \text{ksi}$$

6. Step 6 — Check: returning E = 100,000 ksi to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 200,000 — kept a factor of two that cancels in the correct rearrangement.
• 50,000 — dropped that same factor in the other direction.
• 110,000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

A structural cable's stress-strain response follows Hooke's law under service load. Given modulus of elasticity (E) = 29,000 ksi; normal stress (sigma) = 54.9000 ksi, determine the axial strain (epsilon).

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• Everything except epsilon is given, so isolate epsilon 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for epsilon:

   $$\epsilon = \dfrac{\sigma}{E}$$

3. Step 3

   $$List the givens: modulus of elasticity (E) = 29,000 ksi, normal stress (sigma) = 54.9000 ksi$$

4. Step 4 — Substitute the given values:

   $$\epsilon = \dfrac{54.9000}{29000}$$

5. Step 5 — Evaluate:

   $$\epsilon = 0.0019$$

6. Step 6 — Check: returning epsilon = 0.0019 to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 0.0038 — kept a factor of two that cancels in the correct rearrangement.
• 0.0009 — dropped that same factor in the other direction.
• 0.0021 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law

A steel tension coupon obeys Hooke's law up to the yield point. Given modulus of elasticity (E) = 13,000 ksi; axial strain (epsilon) = 0.0019, determine the normal stress (sigma) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Hooke's law (axial).
• 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.
• Hooke's law states that stress is proportional to strain within the elastic limit of the material.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\sigma = E \varepsilon$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = E \varepsilon$$

3. Step 3

   $$List the givens: modulus of elasticity (E) = 13,000 ksi, axial strain (epsilon) = 0.0019$$

4. Step 4 — Substitute the given values:

   $$\sigma = 13000 \varepsilon$$

5. Step 5 — Evaluate:

   $$\sigma = 24.7000\ \text{ksi}$$

6. Step 6 — Check: returning sigma = 24.7000 ksi to

   $$\sigma = E \varepsilon$$

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

Why the other options are there

• 49.4000 — kept a factor of two that cancels in the correct rearrangement.
• 12.3500 — dropped that same factor in the other direction.
• 27.1700 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Hooke's Law