A mechanics of materials problem uses Axial elongation. Given axial load (P) = 85.0000 kip; length (L) = 108.0 in; area (A) = 4.1000 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) = 85.0000 kip, length (L) = 108.0 in, area (A) = 4.1000 in², modulus (E) = 29,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.0772\ \text{in}$$

6. Step 6 — Check: returning delta = 0.0772 in to

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

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

Why the other options are there

• 0.1544 — kept a factor of two that cancels in the correct rearrangement.
• 0.0386 — dropped that same factor in the other direction.
• 0.0849 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given length (L) = 129.0 in; area (A) = 5.9000 in²; modulus (E) = 17,000 ksi; elongation (delta) = 0.9995 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) = 129.0 in, area (A) = 5.9000 in², modulus (E) = 17,000 ksi, elongation (delta) = 0.9995 in.

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

5. Step 5 — Evaluate:

   $$P = 777.1\ \text{kip}$$

6. Step 6 — Check: returning P = 777.1 kip to

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

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

Why the other options are there

• 1,554 — kept a factor of two that cancels in the correct rearrangement.
• 388.6 — dropped that same factor in the other direction.
• 854.8 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 37.0000 kip; length (L) = 99.0000 in; modulus (E) = 26,000 ksi; elongation (delta) = 0.1275 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) = 37.0000 kip, length (L) = 99.0000 in, modulus (E) = 26,000 ksi, elongation (delta) = 0.1275 in.

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

5. Step 5 — Evaluate:

   $$A = 1.1050\ \text{in²}$$

6. Step 6 — Check: returning A = 1.1050 in² to

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

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

Why the other options are there

• 2.2100 — kept a factor of two that cancels in the correct rearrangement.
• 0.5525 — dropped that same factor in the other direction.
• 1.2155 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 26.0000 kip; area (A) = 2.1000 in²; modulus (E) = 20,000 ksi; elongation (delta) = 0.5204 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) = 26.0000 kip, area (A) = 2.1000 in², modulus (E) = 20,000 ksi, elongation (delta) = 0.5204 in.

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

5. Step 5 — Evaluate:

   $$L = 840.6\ \text{in}$$

6. Step 6 — Check: returning L = 840.6 in to

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

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

Why the other options are there

• 1,681 — kept a factor of two that cancels in the correct rearrangement.
• 420.3 — dropped that same factor in the other direction.
• 924.7 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 80.0000 kip; length (L) = 191.0 in; area (A) = 7.2000 in²; modulus (E) = 27,000 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) = 191.0 in, area (A) = 7.2000 in², modulus (E) = 27,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.0786\ \text{in}$$

6. Step 6 — Check: returning delta = 0.0786 in to

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

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

Why the other options are there

• 0.1572 — kept a factor of two that cancels in the correct rearrangement.
• 0.0393 — dropped that same factor in the other direction.
• 0.0865 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given length (L) = 181.0 in; area (A) = 6.9000 in²; modulus (E) = 27,000 ksi; elongation (delta) = 0.2371 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) = 181.0 in, area (A) = 6.9000 in², modulus (E) = 27,000 ksi, elongation (delta) = 0.2371 in.

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

5. Step 5 — Evaluate:

   $$P = 244.0\ \text{kip}$$

6. Step 6 — Check: returning P = 244.0 kip to

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

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

Why the other options are there

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

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 51.0000 kip; length (L) = 82.0000 in; modulus (E) = 10,000 ksi; elongation (delta) = 0.2376 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) = 51.0000 kip, length (L) = 82.0000 in, modulus (E) = 10,000 ksi, elongation (delta) = 0.2376 in.

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

5. Step 5 — Evaluate:

   $$A = 1.7601\ \text{in²}$$

6. Step 6 — Check: returning A = 1.7601 in² to

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

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

Why the other options are there

• 3.5202 — kept a factor of two that cancels in the correct rearrangement.
• 0.8801 — dropped that same factor in the other direction.
• 1.9361 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 46.0000 kip; area (A) = 3.8000 in²; modulus (E) = 21,000 ksi; elongation (delta) = 0.0527 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) = 46.0000 kip, area (A) = 3.8000 in², modulus (E) = 21,000 ksi, elongation (delta) = 0.0527 in.

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

5. Step 5 — Evaluate:

   $$L = 91.4230\ \text{in}$$

6. Step 6 — Check: returning L = 91.4230 in to

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

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

Why the other options are there

• 182.8 — kept a factor of two that cancels in the correct rearrangement.
• 45.7115 — dropped that same factor in the other direction.
• 100.6 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given axial load (P) = 95.0000 kip; length (L) = 224.0 in; area (A) = 9.3000 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) = 95.0000 kip, length (L) = 224.0 in, area (A) = 9.3000 in², modulus (E) = 21,000 ksi.

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

5. Step 5 — Evaluate:

   $$\delta = 0.1090\ \text{in}$$

6. Step 6 — Check: returning delta = 0.1090 in to

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

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

Why the other options are there

• 0.2179 — kept a factor of two that cancels in the correct rearrangement.
• 0.0545 — dropped that same factor in the other direction.
• 0.1199 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

A mechanics of materials problem uses Axial elongation. Given length (L) = 34.0000 in; area (A) = 8.2000 in²; modulus (E) = 13,000 ksi; elongation (delta) = 0.7367 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) = 34.0000 in, area (A) = 8.2000 in², modulus (E) = 13,000 ksi, elongation (delta) = 0.7367 in.

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

5. Step 5 — Evaluate:

   $$P = 2310\ \text{kip}$$

6. Step 6 — Check: returning P = 2,310 kip to

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

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

Why the other options are there

• 4,620 — kept a factor of two that cancels in the correct rearrangement.
• 1,155 — dropped that same factor in the other direction.
• 2,541 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations