A structural analysis problem uses Midspan deflection of a simply supported beam. Given uniform load (w) = 1.8000 kip/ft; span (L) = 36.0000 ft; modulus (E) = 11,000 ksi; moment of inertia (I) = 900.0 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) = 1.8000 kip/ft, span (L) = 36.0000 ft, modulus (E) = 11,000 ksi, moment of inertia (I) = 900.0 in⁴.

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

5. Step 5 — Evaluate:

   $$\Delta = 0.0477\ \text{in}$$

6. Step 6 — Check: returning Delta = 0.0477 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

• 0.0954 — kept a factor of two that cancels in the correct rearrangement.
• 0.0239 — dropped that same factor in the other direction.
• 0.0525 — rounded an intermediate value before the final step.

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

A pedestrian bridge truss's vertical deflection uses the unit load method. Given unit-load member force (n) = 1.4000; real member force (N) = 26.5000 kip; member length (L) = 117.0 in; member area (A) = 5.5500 in^2; modulus of elasticity (E) = 12,900 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) = 1.4000, real member force (N) = 26.5000 kip, member length (L) = 117.0 in, member area (A) = 5.5500 in^2, modulus of elasticity (E) = 12,900 ksi.

4. Step 4 — Substitute the given values:

   $$\Delta = \dfrac{1.4000 26.5000 117.0}{5.5500 12900}$$

5. Step 5 — Evaluate:

   $$\Delta = 0.0606\ \text{in}$$

6. Step 6 — Check: returning Delta = 0.0606 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

• 0.1213 — kept a factor of two that cancels in the correct rearrangement.
• 0.0303 — dropped that same factor in the other direction.
• 0.0667 — 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) = 39.0000 ft; modulus (E) = 21,000 ksi; moment of inertia (I) = 2,550 in⁴; deflection (Delta) = 2.4600 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) = 39.0000 ft, modulus (E) = 21,000 ksi, moment of inertia (I) = 2,550 in⁴, deflection (Delta) = 2.4600 in.

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

5. Step 5 — Evaluate:

   $$w = 364.4\ \text{kip/ft}$$

6. Step 6 — Check: returning w = 364.4 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

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

Reference: FE Reference Handbook — Structural Analysis → Truss 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) = 0.2500; real member force (N) = 55.5000 kip; member length (L) = 135.0 in; modulus of elasticity (E) = 16,600 ksi; joint deflection (Delta) = 0.8500 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) = 0.2500, real member force (N) = 55.5000 kip, member length (L) = 135.0 in, modulus of elasticity (E) = 16,600 ksi, joint deflection (Delta) = 0.8500 in.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{0.2500 55.5000 135.0}{0.8500 16600}$$

5. Step 5 — Evaluate:

   A = 0.1328\ \text{in^2}

6. Step 6 — Check: returning A = 0.1328 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

• 0.2655 — kept a factor of two that cancels in the correct rearrangement.
• 0.0664 — dropped that same factor in the other direction.
• 0.1460 — 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) = 3.0000 kip/ft; span (L) = 17.0000 ft; modulus (E) = 21,000 ksi; deflection (Delta) = 0.7340 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) = 3.0000 kip/ft, span (L) = 17.0000 ft, modulus (E) = 21,000 ksi, deflection (Delta) = 0.7340 in.

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

5. Step 5 — Evaluate:

   $$I = 2.5399\ \text{in⁴}$$

6. Step 6 — Check: returning I = 2.5399 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

• 5.0799 — kept a factor of two that cancels in the correct rearrangement.
• 1.2700 — dropped that same factor in the other direction.
• 2.7939 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Truss 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) = 39.5000 kip; member area (A) = 1.6000 in^2; modulus of elasticity (E) = 23,500 ksi; joint deflection (Delta) = 0.7290 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) = 39.5000 kip, member area (A) = 1.6000 in^2, modulus of elasticity (E) = 23,500 ksi, joint deflection (Delta) = 0.7290 in.

4. Step 4 — Substitute the given values:

   $$L = \dfrac{0.7290 1.6000 23500}{0.5000 39.5000}$$

5. Step 5 — Evaluate:

   $$L = 1388\ \text{in}$$

6. Step 6 — Check: returning L = 1,388 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

• 2,776 — kept a factor of two that cancels in the correct rearrangement.
• 693.9 — dropped that same factor in the other direction.
• 1,527 — 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) = 1.8000 kip/ft; span (L) = 38.0000 ft; modulus (E) = 20,000 ksi; moment of inertia (I) = 3,000 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) = 1.8000 kip/ft, span (L) = 38.0000 ft, modulus (E) = 20,000 ksi, moment of inertia (I) = 3,000 in⁴.

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

5. Step 5 — Evaluate:

   $$\Delta = 0.0098\ \text{in}$$

6. Step 6 — Check: returning Delta = 0.0098 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

• 0.0195 — kept a factor of two that cancels in the correct rearrangement.
• 0.0049 — dropped that same factor in the other direction.
• 0.0108 — rounded an intermediate value before the final step.

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

A pedestrian bridge truss's vertical deflection uses the unit load method. Given unit-load member force (n) = 1.7000; real member force (N) = 46.0000 kip; member length (L) = 125.0 in; member area (A) = 5.2500 in^2; modulus of elasticity (E) = 14,000 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) = 1.7000, real member force (N) = 46.0000 kip, member length (L) = 125.0 in, member area (A) = 5.2500 in^2, modulus of elasticity (E) = 14,000 ksi.

4. Step 4 — Substitute the given values:

   $$\Delta = \dfrac{1.7000 46.0000 125.0}{5.2500 14000}$$

5. Step 5 — Evaluate:

   $$\Delta = 0.1330\ \text{in}$$

6. Step 6 — Check: returning Delta = 0.1330 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

• 0.2660 — kept a factor of two that cancels in the correct rearrangement.
• 0.0665 — dropped that same factor in the other direction.
• 0.1463 — 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) = 36.0000 ft; modulus (E) = 23,000 ksi; moment of inertia (I) = 1,300 in⁴; deflection (Delta) = 1.4360 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) = 36.0000 ft, modulus (E) = 23,000 ksi, moment of inertia (I) = 1,300 in⁴, deflection (Delta) = 1.4360 in.

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

5. Step 5 — Evaluate:

   $$w = 163.6\ \text{kip/ft}$$

6. Step 6 — Check: returning w = 163.6 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

• 327.2 — kept a factor of two that cancels in the correct rearrangement.
• 81.8023 — dropped that same factor in the other direction.
• 180.0 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Truss 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) = 0.4000; real member force (N) = 24.0000 kip; member length (L) = 225.0 in; modulus of elasticity (E) = 17,700 ksi; joint deflection (Delta) = 1.2520 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) = 0.4000, real member force (N) = 24.0000 kip, member length (L) = 225.0 in, modulus of elasticity (E) = 17,700 ksi, joint deflection (Delta) = 1.2520 in.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{0.4000 24.0000 225.0}{1.2520 17700}$$

5. Step 5 — Evaluate:

   A = 0.0975\ \text{in^2}

6. Step 6 — Check: returning A = 0.0975 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

• 0.1949 — kept a factor of two that cancels in the correct rearrangement.
• 0.0487 — dropped that same factor in the other direction.
• 0.1072 — rounded an intermediate value before the final step.

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