A structural analysis problem uses Midspan deflection of a simply supported beam. Given uniform load (w) = 2.4000 kip/ft; span (L) = 20.0000 ft; modulus (E) = 23,000 ksi; moment of inertia (I) = 1,350 in⁴, determine the deflection (Delta) in in.

Start with the thinking

• The governing relation printed in this handbook section is Midspan deflection of a simply supported beam.
• 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.
• Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = 5 w L^4 / (384 E I)$$

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

3. Step 3 — List the givens: uniform load (w) = 2.4000 kip/ft, span (L) = 20.0000 ft, modulus (E) = 23,000 ksi, moment of inertia (I) = 1,350 in⁴.

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

5. Step 5 — Evaluate:

   $$\Delta = 0.0019\ \text{in}$$

6. Step 6 — Check: returning Delta = 0.0019 in to

   $$\Delta = 5 w L^4 / (384 E I)$$

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

Why the other options are there

• 0.0039 — kept a factor of two that cancels in the correct rearrangement.
• 0.0010 — dropped that same factor in the other direction.
• 0.0021 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

A pedestrian bridge truss's vertical deflection uses the unit load method. Given unit-load member force (n) = 0.5500; real member force (N) = 32.5000 kip; member length (L) = 294.0 in; member area (A) = 1.3500 in^2; modulus of elasticity (E) = 23,700 ksi, determine the joint deflection (Delta) in in.

Start with the thinking

• The governing relation printed in this handbook section is Truss deflection by unit load method.
• 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.
• Truss deflection by unit load method computes joint displacement from real and virtual unit-load member forces summed over all members.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = \sum \dfrac{n N L}{A E}$$

2. Step 2 — Rearrange symbolically for Delta:

   $$\Delta = \dfrac{n N L}{A E}$$

3. Step 3 — List the givens: unit-load member force (n) = 0.5500, real member force (N) = 32.5000 kip, member length (L) = 294.0 in, member area (A) = 1.3500 in^2, modulus of elasticity (E) = 23,700 ksi.

4. Step 4 — Substitute the given values:

   $$\Delta = \dfrac{0.5500 32.5000 294.0}{1.3500 23700}$$

5. Step 5 — Evaluate:

   $$\Delta = 0.1643\ \text{in}$$

6. Step 6 — Check: returning Delta = 0.1643 in to

   $$\Delta = \sum \dfrac{n N L}{A E}$$

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

Why the other options are there

• 0.3285 — kept a factor of two that cancels in the correct rearrangement.
• 0.0821 — dropped that same factor in the other direction.
• 0.1807 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Analysis: Truss Deflection by Unit Load Method

A structural analysis problem uses Midspan deflection of a simply supported beam. Given span (L) = 40.0000 ft; modulus (E) = 5,000 ksi; moment of inertia (I) = 4,800 in⁴; deflection (Delta) = 2.6230 in, determine the uniform load (w) in kip/ft.

Start with the thinking

• The governing relation printed in this handbook section is Midspan deflection of a simply supported beam.
• Everything except w is given, so isolate w 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.
• Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = 5 w L^4 / (384 E I)$$

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

3. Step 3 — List the givens: span (L) = 40.0000 ft, modulus (E) = 5,000 ksi, moment of inertia (I) = 4,800 in⁴, deflection (Delta) = 2.6230 in.

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

5. Step 5 — Evaluate:

   $$w = 157.4\ \text{kip/ft}$$

6. Step 6 — Check: returning w = 157.4 kip/ft to

   $$\Delta = 5 w L^4 / (384 E I)$$

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

Why the other options are there

• 314.8 — kept a factor of two that cancels in the correct rearrangement.
• 78.6900 — dropped that same factor in the other direction.
• 173.1 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

A steel truss chord's displacement under service load is solved by the unit load method. Given unit-load member force (n) = 0.9500; real member force (N) = 26.5000 kip; member length (L) = 186.0 in; modulus of elasticity (E) = 12,700 ksi; joint deflection (Delta) = 0.4110 in, determine the member area (A) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Truss deflection by unit load method.
• 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.
• Truss deflection by unit load method computes joint displacement from real and virtual unit-load member forces summed over all members.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = \sum \dfrac{n N L}{A E}$$

2. Step 2 — Rearrange symbolically for A:

   $$A = \dfrac{n N L}{\Delta E}$$

3. Step 3 — List the givens: unit-load member force (n) = 0.9500, real member force (N) = 26.5000 kip, member length (L) = 186.0 in, modulus of elasticity (E) = 12,700 ksi, joint deflection (Delta) = 0.4110 in.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{0.9500 26.5000 186.0}{0.4110 12700}$$

5. Step 5 — Evaluate:

   A = 0.8971\ \text{in^2}

6. Step 6 — Check: returning A = 0.8971 in^2 to

   $$\Delta = \sum \dfrac{n N L}{A E}$$

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

Why the other options are there

• 1.7942 — kept a factor of two that cancels in the correct rearrangement.
• 0.4485 — dropped that same factor in the other direction.
• 0.9868 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Analysis: Truss Deflection by Unit Load Method

A structural analysis problem uses Midspan deflection of a simply supported beam. Given uniform load (w) = 1.3000 kip/ft; span (L) = 16.0000 ft; modulus (E) = 9,000 ksi; deflection (Delta) = 1.4780 in, determine the moment of inertia (I) in in⁴.

Start with the thinking

• The governing relation printed in this handbook section is Midspan deflection of a simply supported beam.
• 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.
• Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = 5 w L^4 / (384 E I)$$

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

3. Step 3 — List the givens: uniform load (w) = 1.3000 kip/ft, span (L) = 16.0000 ft, modulus (E) = 9,000 ksi, deflection (Delta) = 1.4780 in.

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

5. Step 5 — Evaluate:

   $$I = 1.0008\ \text{in⁴}$$

6. Step 6 — Check: returning I = 1.0008 in⁴ to

   $$\Delta = 5 w L^4 / (384 E I)$$

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

Why the other options are there

• 2.0015 — kept a factor of two that cancels in the correct rearrangement.
• 0.5004 — dropped that same factor in the other direction.
• 1.1008 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

A roof truss joint deflection is found using truss deflection by unit load method. Given unit-load member force (n) = 0.5000; real member force (N) = 37.5000 kip; member area (A) = 6.0000 in^2; modulus of elasticity (E) = 21,300 ksi; joint deflection (Delta) = 0.0350 in, determine the member length (L) in in.

Start with the thinking

• The governing relation printed in this handbook section is Truss deflection by unit load method.
• 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.
• Truss deflection by unit load method computes joint displacement from real and virtual unit-load member forces summed over all members.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = \sum \dfrac{n N L}{A E}$$

2. Step 2 — Rearrange symbolically for L:

   $$L = \dfrac{\Delta A E}{n N}$$

3. Step 3 — List the givens: unit-load member force (n) = 0.5000, real member force (N) = 37.5000 kip, member area (A) = 6.0000 in^2, modulus of elasticity (E) = 21,300 ksi, joint deflection (Delta) = 0.0350 in.

4. Step 4 — Substitute the given values:

   $$L = \dfrac{0.0350 6.0000 21300}{0.5000 37.5000}$$

5. Step 5 — Evaluate:

   $$L = 238.6\ \text{in}$$

6. Step 6 — Check: returning L = 238.6 in to

   $$\Delta = \sum \dfrac{n N L}{A E}$$

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

Why the other options are there

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

Reference: FE Handbook — Structural Analysis: Truss Deflection by Unit Load Method

A structural analysis problem uses Midspan deflection of a simply supported beam. Given uniform load (w) = 1.9000 kip/ft; span (L) = 26.0000 ft; modulus (E) = 9,000 ksi; moment of inertia (I) = 2,150 in⁴, determine the deflection (Delta) in in.

Start with the thinking

• The governing relation printed in this handbook section is Midspan deflection of a simply supported beam.
• 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.
• Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = 5 w L^4 / (384 E I)$$

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

3. Step 3 — List the givens: uniform load (w) = 1.9000 kip/ft, span (L) = 26.0000 ft, modulus (E) = 9,000 ksi, moment of inertia (I) = 2,150 in⁴.

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

5. Step 5 — Evaluate:

   $$\Delta = 0.0070\ \text{in}$$

6. Step 6 — Check: returning Delta = 0.0070 in to

   $$\Delta = 5 w L^4 / (384 E I)$$

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

Why the other options are there

• 0.0140 — kept a factor of two that cancels in the correct rearrangement.
• 0.0035 — dropped that same factor in the other direction.
• 0.0077 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

A pedestrian bridge truss's vertical deflection uses the unit load method. Given unit-load member force (n) = 1.9500; real member force (N) = 5.0000 kip; member length (L) = 242.0 in; member area (A) = 2.3000 in^2; modulus of elasticity (E) = 27,800 ksi, determine the joint deflection (Delta) in in.

Start with the thinking

• The governing relation printed in this handbook section is Truss deflection by unit load method.
• 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.
• Truss deflection by unit load method computes joint displacement from real and virtual unit-load member forces summed over all members.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = \sum \dfrac{n N L}{A E}$$

2. Step 2 — Rearrange symbolically for Delta:

   $$\Delta = \dfrac{n N L}{A E}$$

3. Step 3 — List the givens: unit-load member force (n) = 1.9500, real member force (N) = 5.0000 kip, member length (L) = 242.0 in, member area (A) = 2.3000 in^2, modulus of elasticity (E) = 27,800 ksi.

4. Step 4 — Substitute the given values:

   $$\Delta = \dfrac{1.9500 5.0000 242.0}{2.3000 27800}$$

5. Step 5 — Evaluate:

   $$\Delta = 0.0369\ \text{in}$$

6. Step 6 — Check: returning Delta = 0.0369 in to

   $$\Delta = \sum \dfrac{n N L}{A E}$$

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

Why the other options are there

• 0.0738 — kept a factor of two that cancels in the correct rearrangement.
• 0.0185 — dropped that same factor in the other direction.
• 0.0406 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Analysis: Truss Deflection by Unit Load Method

A structural analysis problem uses Midspan deflection of a simply supported beam. Given span (L) = 21.0000 ft; modulus (E) = 19,000 ksi; moment of inertia (I) = 3,350 in⁴; deflection (Delta) = 0.7140 in, determine the uniform load (w) in kip/ft.

Start with the thinking

• The governing relation printed in this handbook section is Midspan deflection of a simply supported beam.
• Everything except w is given, so isolate w 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.
• Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = 5 w L^4 / (384 E I)$$

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

3. Step 3 — List the givens: span (L) = 21.0000 ft, modulus (E) = 19,000 ksi, moment of inertia (I) = 3,350 in⁴, deflection (Delta) = 0.7140 in.

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

5. Step 5 — Evaluate:

   $$w = 1496\ \text{kip/ft}$$

6. Step 6 — Check: returning w = 1,496 kip/ft to

   $$\Delta = 5 w L^4 / (384 E I)$$

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

Why the other options are there

• 2,991 — kept a factor of two that cancels in the correct rearrangement.
• 747.8 — dropped that same factor in the other direction.
• 1,645 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

A steel truss chord's displacement under service load is solved by the unit load method. Given unit-load member force (n) = 1.8500; real member force (N) = 11.0000 kip; member length (L) = 136.0 in; modulus of elasticity (E) = 12,800 ksi; joint deflection (Delta) = 1.5080 in, determine the member area (A) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Truss deflection by unit load method.
• 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.
• Truss deflection by unit load method computes joint displacement from real and virtual unit-load member forces summed over all members.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\Delta = \sum \dfrac{n N L}{A E}$$

2. Step 2 — Rearrange symbolically for A:

   $$A = \dfrac{n N L}{\Delta E}$$

3. Step 3 — List the givens: unit-load member force (n) = 1.8500, real member force (N) = 11.0000 kip, member length (L) = 136.0 in, modulus of elasticity (E) = 12,800 ksi, joint deflection (Delta) = 1.5080 in.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{1.8500 11.0000 136.0}{1.5080 12800}$$

5. Step 5 — Evaluate:

   A = 0.1434\ \text{in^2}

6. Step 6 — Check: returning A = 0.1434 in^2 to

   $$\Delta = \sum \dfrac{n N L}{A E}$$

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

Why the other options are there

• 0.2868 — kept a factor of two that cancels in the correct rearrangement.
• 0.0717 — dropped that same factor in the other direction.
• 0.1577 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Analysis: Truss Deflection by Unit Load Method