A mechanics of materials problem uses Axial elongation. Given axial load (P) = 44.0000 kip; length (L) = 199.0 in; area (A) = 4.2000 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) = 44.0000 kip, length (L) = 199.0 in, area (A) = 4.2000 in², modulus (E) = 26,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.0802\ \text{in}$$

6. Step 6 — Check: returning delta = 0.0802 in to

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

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

Why the other options are there

• 0.1604 — kept a factor of two that cancels in the correct rearrangement.
• 0.0401 — dropped that same factor in the other direction.
• 0.0882 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given length (L) = 103.0 in; area (A) = 4.5000 in²; modulus (E) = 19,000 ksi; elongation (delta) = 0.4663 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) = 103.0 in, area (A) = 4.5000 in², modulus (E) = 19,000 ksi, elongation (delta) = 0.4663 in.

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

5. Step 5 — Evaluate:

   $$P = 387.1\ \text{kip}$$

6. Step 6 — Check: returning P = 387.1 kip to

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

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

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 12.0000 kip; length (L) = 125.0 in; modulus (E) = 10,000 ksi; elongation (delta) = 0.3448 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) = 12.0000 kip, length (L) = 125.0 in, modulus (E) = 10,000 ksi, elongation (delta) = 0.3448 in.

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

5. Step 5 — Evaluate:

   $$A = 0.4350\ \text{in²}$$

6. Step 6 — Check: returning A = 0.4350 in² to

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

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

Why the other options are there

• 0.8701 — kept a factor of two that cancels in the correct rearrangement.
• 0.2175 — dropped that same factor in the other direction.
• 0.4785 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 12.0000 kip; area (A) = 2.5000 in²; modulus (E) = 15,000 ksi; elongation (delta) = 0.2215 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) = 12.0000 kip, area (A) = 2.5000 in², modulus (E) = 15,000 ksi, elongation (delta) = 0.2215 in.

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

5. Step 5 — Evaluate:

   $$L = 692.2\ \text{in}$$

6. Step 6 — Check: returning L = 692.2 in to

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

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

Why the other options are there

• 1,384 — kept a factor of two that cancels in the correct rearrangement.
• 346.1 — dropped that same factor in the other direction.
• 761.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 81.0000 kip; length (L) = 44.0000 in; area (A) = 6.1000 in²; modulus (E) = 30,000 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) = 81.0000 kip, length (L) = 44.0000 in, area (A) = 6.1000 in², modulus (E) = 30,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.0195\ \text{in}$$

6. Step 6 — Check: returning delta = 0.0195 in to

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

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

Why the other options are there

• 0.0390 — kept a factor of two that cancels in the correct rearrangement.
• 0.0097 — dropped that same factor in the other direction.
• 0.0214 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given length (L) = 167.0 in; area (A) = 6.2000 in²; modulus (E) = 14,000 ksi; elongation (delta) = 0.6120 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) = 167.0 in, area (A) = 6.2000 in², modulus (E) = 14,000 ksi, elongation (delta) = 0.6120 in.

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

5. Step 5 — Evaluate:

   $$P = 318.1\ \text{kip}$$

6. Step 6 — Check: returning P = 318.1 kip to

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

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

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 34.0000 kip; length (L) = 197.0 in; modulus (E) = 27,000 ksi; elongation (delta) = 0.1059 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) = 34.0000 kip, length (L) = 197.0 in, modulus (E) = 27,000 ksi, elongation (delta) = 0.1059 in.

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

5. Step 5 — Evaluate:

   $$A = 2.3425\ \text{in²}$$

6. Step 6 — Check: returning A = 2.3425 in² to

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

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

Why the other options are there

• 4.6851 — kept a factor of two that cancels in the correct rearrangement.
• 1.1713 — dropped that same factor in the other direction.
• 2.5768 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 77.0000 kip; area (A) = 7.6000 in²; modulus (E) = 18,000 ksi; elongation (delta) = 0.9217 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) = 77.0000 kip, area (A) = 7.6000 in², modulus (E) = 18,000 ksi, elongation (delta) = 0.9217 in.

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

5. Step 5 — Evaluate:

   $$L = 1638\ \text{in}$$

6. Step 6 — Check: returning L = 1,638 in to

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

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

Why the other options are there

• 3,275 — kept a factor of two that cancels in the correct rearrangement.
• 818.8 — dropped that same factor in the other direction.
• 1,801 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 23.0000 kip; length (L) = 153.0 in; area (A) = 6.0000 in²; modulus (E) = 21,000 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) = 23.0000 kip, length (L) = 153.0 in, area (A) = 6.0000 in², modulus (E) = 21,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.0279\ \text{in}$$

6. Step 6 — Check: returning delta = 0.0279 in to

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

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

Why the other options are there

• 0.0559 — kept a factor of two that cancels in the correct rearrangement.
• 0.0140 — dropped that same factor in the other direction.
• 0.0307 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

A mechanics of materials problem uses Axial elongation. Given length (L) = 48.0000 in; area (A) = 1.5000 in²; modulus (E) = 22,000 ksi; elongation (delta) = 0.4420 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) = 48.0000 in, area (A) = 1.5000 in², modulus (E) = 22,000 ksi, elongation (delta) = 0.4420 in.

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

5. Step 5 — Evaluate:

   $$P = 303.9\ \text{kip}$$

6. Step 6 — Check: returning P = 303.9 kip to

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

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

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation