A mechanics of materials problem uses Axial elongation. Given axial load (P) = 89.0000 kip; length (L) = 85.0000 in; area (A) = 4.4000 in²; modulus (E) = 26,000 ksi, determine the elongation (delta) in in.

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except delta is given, so isolate delta 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: axial load (P) = 89.0000 kip, length (L) = 85.0000 in, area (A) = 4.4000 in², modulus (E) = 26,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.0661\ \text{in}$$

6. Step 6 — Check: returning delta = 0.0661 in to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 0.1323 — kept a factor of two that cancels in the correct rearrangement.
• 0.0331 — dropped that same factor in the other direction.
• 0.0727 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given length (L) = 217.0 in; area (A) = 6.0000 in²; modulus (E) = 22,000 ksi; elongation (delta) = 0.9443 in, determine the axial load (P) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except P is given, so isolate P 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: length (L) = 217.0 in, area (A) = 6.0000 in², modulus (E) = 22,000 ksi, elongation (delta) = 0.9443 in.

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

5. Step 5 — Evaluate:

   $$P = 574.4\ \text{kip}$$

6. Step 6 — Check: returning P = 574.4 kip to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 1,149 — kept a factor of two that cancels in the correct rearrangement.
• 287.2 — dropped that same factor in the other direction.
• 631.9 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 65.0000 kip; length (L) = 203.0 in; modulus (E) = 27,000 ksi; elongation (delta) = 0.9724 in, determine the area (A) in in².

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except A is given, so isolate A 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: axial load (P) = 65.0000 kip, length (L) = 203.0 in, modulus (E) = 27,000 ksi, elongation (delta) = 0.9724 in.

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

5. Step 5 — Evaluate:

   $$A = 0.5026\ \text{in²}$$

6. Step 6 — Check: returning A = 0.5026 in² to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 1.0051 — kept a factor of two that cancels in the correct rearrangement.
• 0.2513 — dropped that same factor in the other direction.
• 0.5528 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 18.0000 kip; area (A) = 5.4000 in²; modulus (E) = 26,000 ksi; elongation (delta) = 0.0754 in, determine the length (L) in in.

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except L is given, so isolate L 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: axial load (P) = 18.0000 kip, area (A) = 5.4000 in², modulus (E) = 26,000 ksi, elongation (delta) = 0.0754 in.

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

5. Step 5 — Evaluate:

   $$L = 588.1\ \text{in}$$

6. Step 6 — Check: returning L = 588.1 in to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 1,176 — kept a factor of two that cancels in the correct rearrangement.
• 294.1 — dropped that same factor in the other direction.
• 646.9 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 64.0000 kip; length (L) = 97.0000 in; area (A) = 7.4000 in²; modulus (E) = 29,000 ksi, determine the elongation (delta) in in.

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except delta is given, so isolate delta 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: axial load (P) = 64.0000 kip, length (L) = 97.0000 in, area (A) = 7.4000 in², modulus (E) = 29,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.0289\ \text{in}$$

6. Step 6 — Check: returning delta = 0.0289 in to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 0.0579 — kept a factor of two that cancels in the correct rearrangement.
• 0.0145 — dropped that same factor in the other direction.
• 0.0318 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given length (L) = 88.0000 in; area (A) = 3.0000 in²; modulus (E) = 25,000 ksi; elongation (delta) = 0.6307 in, determine the axial load (P) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except P is given, so isolate P 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: length (L) = 88.0000 in, area (A) = 3.0000 in², modulus (E) = 25,000 ksi, elongation (delta) = 0.6307 in.

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

5. Step 5 — Evaluate:

   $$P = 537.5\ \text{kip}$$

6. Step 6 — Check: returning P = 537.5 kip to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 1,075 — kept a factor of two that cancels in the correct rearrangement.
• 268.8 — dropped that same factor in the other direction.
• 591.3 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 52.0000 kip; length (L) = 61.0000 in; modulus (E) = 25,000 ksi; elongation (delta) = 0.1445 in, determine the area (A) in in².

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except A is given, so isolate A 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: axial load (P) = 52.0000 kip, length (L) = 61.0000 in, modulus (E) = 25,000 ksi, elongation (delta) = 0.1445 in.

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

5. Step 5 — Evaluate:

   $$A = 0.8781\ \text{in²}$$

6. Step 6 — Check: returning A = 0.8781 in² to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 1.7561 — kept a factor of two that cancels in the correct rearrangement.
• 0.4390 — dropped that same factor in the other direction.
• 0.9659 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 40.0000 kip; area (A) = 6.7000 in²; modulus (E) = 12,000 ksi; elongation (delta) = 0.5487 in, determine the length (L) in in.

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except L is given, so isolate L 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: axial load (P) = 40.0000 kip, area (A) = 6.7000 in², modulus (E) = 12,000 ksi, elongation (delta) = 0.5487 in.

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

5. Step 5 — Evaluate:

   $$L = 1103\ \text{in}$$

6. Step 6 — Check: returning L = 1,103 in to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 2,206 — kept a factor of two that cancels in the correct rearrangement.
• 551.4 — dropped that same factor in the other direction.
• 1,213 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 80.0000 kip; length (L) = 135.0 in; area (A) = 5.8000 in²; modulus (E) = 20,000 ksi, determine the elongation (delta) in in.

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except delta is given, so isolate delta 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: axial load (P) = 80.0000 kip, length (L) = 135.0 in, area (A) = 5.8000 in², modulus (E) = 20,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.0931\ \text{in}$$

6. Step 6 — Check: returning delta = 0.0931 in to

   $$\delta = P L / (A E)$$

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

Why the other options are there

• 0.1862 — kept a factor of two that cancels in the correct rearrangement.
• 0.0466 — dropped that same factor in the other direction.
• 0.1024 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation

A mechanics of materials problem uses Axial elongation. Given length (L) = 114.0 in; area (A) = 3.3000 in²; modulus (E) = 17,000 ksi; elongation (delta) = 0.9345 in, determine the axial load (P) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Axial elongation.
• Everything except P is given, so isolate P 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:

   $$\delta = P L / (A E)$$

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

3. Step 3 — List the givens: length (L) = 114.0 in, area (A) = 3.3000 in², modulus (E) = 17,000 ksi, elongation (delta) = 0.9345 in.

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

5. Step 5 — Evaluate:

   $$P = 459.9\ \text{kip}$$

6. Step 6 — Check: returning P = 459.9 kip to

   $$\delta = P L / (A E)$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Uniaxial Loading and Deformation