A structural design problem uses Nominal moment capacity check. Given resistance factor (phi) = 0.8900; yield strength (Fy) = 41.0000 ksi; plastic section modulus (Zx) = 284.0 in³, determine the design moment capacity (phiMn) in kip·ft.

Start with the thinking

• The governing relation printed in this handbook section is Nominal moment capacity check.
• Everything except phiMn is given, so isolate phiMn 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 Design items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi M_n = \phi F_y Z_x$$

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

3. Step 3 — List the givens: resistance factor (phi) = 0.8900, yield strength (Fy) = 41.0000 ksi, plastic section modulus (Zx) = 284.0 in³.

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

5. Step 5 — Evaluate:

   $$phiMn = 863.6\ \text{kip·ft}$$

6. Step 6 — Check: returning phiMn = 863.6 kip·ft to

   $$\phi M_n = \phi F_y Z_x$$

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

Why the other options are there

• 1,727 — kept a factor of two that cancels in the correct rearrangement.
• 431.8 — dropped that same factor in the other direction.
• 950.0 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Design → Columns

a steel roof beam braced by purlins at 8 ft spacing Given lateral-torsional modification factor (C_b) = 1.0700; plastic moment (M_p) = 1,150 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 132.0 in^3; unbraced length (L_b) = 12.5000 ft; limiting length for yielding (L_p) = 4.1000 ft; limiting length for inelastic buckling (L_r) = 35.0000 ft, determine the nominal flexural strength (M_n) in kip-ft.

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling strength in the inelastic range.
• Everything except M_n is given, so isolate M_n 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.
• A wide-flange steel beam braced at spacing L_b between L_p and L_r fails by inelastic lateral-torsional buckling.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

2. Step 2 — Rearrange symbolically for M_n:

   $$M_{n} = C_b\left[M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

3. Step 3 — List the givens: lateral-torsional modification factor (C_b) = 1.0700, plastic moment (M_p) = 1,150 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 132.0 in^3, unbraced length (L_b) = 12.5000 ft, limiting length for yielding (L_p) = 4.1000 ft, limiting length for inelastic buckling (L_r) = 35.0000 ft.

4. Step 4 — Substitute the given values:

   $$M_{n} = C_b\left[M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

5. Step 5 — Evaluate:

   $$M_{n} = 1008\ \text{kip-ft}$$

6. Step 6 — Check: returning M_n = 1,008 kip-ft to

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

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

Why the other options are there

• 2,016 — kept a factor of two that cancels in the correct rearrangement.
• 504.0 — dropped that same factor in the other direction.
• 1,109 — rounded an intermediate value before the final step.

Reference: AISC Specification §F2.2 — Lateral-Torsional Buckling

A structural design problem uses Nominal moment capacity check. Given resistance factor (phi) = 0.8900; plastic section modulus (Zx) = 224.0 in³; design moment capacity (phiMn) = 505.0 kip·ft, determine the yield strength (Fy) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Nominal moment capacity check.
• Everything except Fy is given, so isolate Fy 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 Design items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi M_n = \phi F_y Z_x$$

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

3. Step 3 — List the givens: resistance factor (phi) = 0.8900, plastic section modulus (Zx) = 224.0 in³, design moment capacity (phiMn) = 505.0 kip·ft.

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

5. Step 5 — Evaluate:

   $$Fy = 30.3973\ \text{ksi}$$

6. Step 6 — Check: returning Fy = 30.3973 ksi to

   $$\phi M_n = \phi F_y Z_x$$

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

Why the other options are there

• 60.7945 — kept a factor of two that cancels in the correct rearrangement.
• 15.1986 — dropped that same factor in the other direction.
• 33.4370 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Design → Columns

a crane runway girder with limited lateral bracing Given nominal flexural strength (M_n) = 756.0 kip-ft; plastic moment (M_p) = 1,498 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 144.0 in^3; unbraced length (L_b) = 18.5000 ft; limiting length for yielding (L_p) = 5.1000 ft; limiting length for inelastic buckling (L_r) = 37.0000 ft, determine the lateral-torsional modification factor (C_b).

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling strength in the inelastic range.
• Everything except C_b is given, so isolate C_b 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.
• A wide-flange steel beam braced at spacing L_b between L_p and L_r fails by inelastic lateral-torsional buckling.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

2. Step 2 — Rearrange symbolically for C_b:

   $$C_{b} = \dfrac{M_n}{M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}}$$

3. Step 3 — List the givens: nominal flexural strength (M_n) = 756.0 kip-ft, plastic moment (M_p) = 1,498 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 144.0 in^3, unbraced length (L_b) = 18.5000 ft, limiting length for yielding (L_p) = 5.1000 ft, limiting length for inelastic buckling (L_r) = 37.0000 ft.

4. Step 4 — Substitute the given values:

   $$C_{b} = \dfrac{M_n}{M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}}$$

5. Step 5 — Evaluate:

   $$C_{b} = 0.7233$$

6. Step 6 — Check: returning C_b = 0.7233 to

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

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

Why the other options are there

• 1.4467 — kept a factor of two that cancels in the correct rearrangement.
• 0.3617 — dropped that same factor in the other direction.
• 0.7957 — rounded an intermediate value before the final step.

Reference: AISC Specification §F2.2 — Lateral-Torsional Buckling

A structural design problem uses Nominal moment capacity check. Given resistance factor (phi) = 0.8900; yield strength (Fy) = 45.0000 ksi; design moment capacity (phiMn) = 1,859 kip·ft, determine the plastic section modulus (Zx) in in³.

Start with the thinking

• The governing relation printed in this handbook section is Nominal moment capacity check.
• Everything except Zx is given, so isolate Zx 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 Design items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi M_n = \phi F_y Z_x$$

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

3. Step 3 — List the givens: resistance factor (phi) = 0.8900, yield strength (Fy) = 45.0000 ksi, design moment capacity (phiMn) = 1,859 kip·ft.

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

5. Step 5 — Evaluate:

   $$Zx = 557.0\ \text{in³}$$

6. Step 6 — Check: returning Zx = 557.0 in³ to

   $$\phi M_n = \phi F_y Z_x$$

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

Why the other options are there

• 1,114 — kept a factor of two that cancels in the correct rearrangement.
• 278.5 — dropped that same factor in the other direction.
• 612.7 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Design → Columns

a W-shape floor girder braced only at third points Given lateral-torsional modification factor (C_b) = 1.1500; plastic moment (M_p) = 764.0 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 42.0000 in^3; unbraced length (L_b) = 15.0000 ft; limiting length for yielding (L_p) = 4.5000 ft; limiting length for inelastic buckling (L_r) = 42.5000 ft, determine the nominal flexural strength (M_n) in kip-ft.

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling strength in the inelastic range.
• Everything except M_n is given, so isolate M_n 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.
• A wide-flange steel beam braced at spacing L_b between L_p and L_r fails by inelastic lateral-torsional buckling.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

2. Step 2 — Rearrange symbolically for M_n:

   $$M_{n} = C_b\left[M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

3. Step 3 — List the givens: lateral-torsional modification factor (C_b) = 1.1500, plastic moment (M_p) = 764.0 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 42.0000 in^3, unbraced length (L_b) = 15.0000 ft, limiting length for yielding (L_p) = 4.5000 ft, limiting length for inelastic buckling (L_r) = 42.5000 ft.

4. Step 4 — Substitute the given values:

   $$M_{n} = C_b\left[M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

5. Step 5 — Evaluate:

   $$M_{n} = 674.8\ \text{kip-ft}$$

6. Step 6 — Check: returning M_n = 674.8 kip-ft to

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

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

Why the other options are there

• 1,350 — kept a factor of two that cancels in the correct rearrangement.
• 337.4 — dropped that same factor in the other direction.
• 742.2 — rounded an intermediate value before the final step.

Reference: AISC Specification §F2.2 — Lateral-Torsional Buckling

A structural design problem uses Nominal moment capacity check. Given resistance factor (phi) = 0.8700; yield strength (Fy) = 65.0000 ksi; plastic section modulus (Zx) = 52.0000 in³, determine the design moment capacity (phiMn) in kip·ft.

Start with the thinking

• The governing relation printed in this handbook section is Nominal moment capacity check.
• Everything except phiMn is given, so isolate phiMn 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 Design items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi M_n = \phi F_y Z_x$$

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

3. Step 3 — List the givens: resistance factor (phi) = 0.8700, yield strength (Fy) = 65.0000 ksi, plastic section modulus (Zx) = 52.0000 in³.

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

5. Step 5 — Evaluate:

   $$phiMn = 245.0\ \text{kip·ft}$$

6. Step 6 — Check: returning phiMn = 245.0 kip·ft to

   $$\phi M_n = \phi F_y Z_x$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Structural Design → Columns

a steel roof beam braced by purlins at 8 ft spacing Given nominal flexural strength (M_n) = 958.1 kip-ft; plastic moment (M_p) = 910.0 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 216.0 in^3; unbraced length (L_b) = 9.0000 ft; limiting length for yielding (L_p) = 7.6000 ft; limiting length for inelastic buckling (L_r) = 49.5000 ft, determine the lateral-torsional modification factor (C_b).

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling strength in the inelastic range.
• Everything except C_b is given, so isolate C_b 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.
• A wide-flange steel beam braced at spacing L_b between L_p and L_r fails by inelastic lateral-torsional buckling.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

2. Step 2 — Rearrange symbolically for C_b:

   $$C_{b} = \dfrac{M_n}{M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}}$$

3. Step 3 — List the givens: nominal flexural strength (M_n) = 958.1 kip-ft, plastic moment (M_p) = 910.0 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 216.0 in^3, unbraced length (L_b) = 9.0000 ft, limiting length for yielding (L_p) = 7.6000 ft, limiting length for inelastic buckling (L_r) = 49.5000 ft.

4. Step 4 — Substitute the given values:

   $$C_{b} = \dfrac{M_n}{M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}}$$

5. Step 5 — Evaluate:

   $$C_{b} = 1.0638$$

6. Step 6 — Check: returning C_b = 1.0638 to

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

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

Why the other options are there

• 2.1276 — kept a factor of two that cancels in the correct rearrangement.
• 0.5319 — dropped that same factor in the other direction.
• 1.1702 — rounded an intermediate value before the final step.

Reference: AISC Specification §F2.2 — Lateral-Torsional Buckling

A structural design problem uses Nominal moment capacity check. Given resistance factor (phi) = 0.8800; plastic section modulus (Zx) = 20.0000 in³; design moment capacity (phiMn) = 563.0 kip·ft, determine the yield strength (Fy) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Nominal moment capacity check.
• Everything except Fy is given, so isolate Fy 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 Design items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi M_n = \phi F_y Z_x$$

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

3. Step 3 — List the givens: resistance factor (phi) = 0.8800, plastic section modulus (Zx) = 20.0000 in³, design moment capacity (phiMn) = 563.0 kip·ft.

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

5. Step 5 — Evaluate:

   $$Fy = 383.9\ \text{ksi}$$

6. Step 6 — Check: returning Fy = 383.9 ksi to

   $$\phi M_n = \phi F_y Z_x$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Structural Design → Columns

a crane runway girder with limited lateral bracing Given lateral-torsional modification factor (C_b) = 1.5200; plastic moment (M_p) = 858.0 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 294.0 in^3; unbraced length (L_b) = 20.0000 ft; limiting length for yielding (L_p) = 6.3000 ft; limiting length for inelastic buckling (L_r) = 48.0000 ft, determine the nominal flexural strength (M_n) in kip-ft.

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling strength in the inelastic range.
• Everything except M_n is given, so isolate M_n 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.
• A wide-flange steel beam braced at spacing L_b between L_p and L_r fails by inelastic lateral-torsional buckling.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

2. Step 2 — Rearrange symbolically for M_n:

   $$M_{n} = C_b\left[M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

3. Step 3 — List the givens: lateral-torsional modification factor (C_b) = 1.5200, plastic moment (M_p) = 858.0 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 294.0 in^3, unbraced length (L_b) = 20.0000 ft, limiting length for yielding (L_p) = 6.3000 ft, limiting length for inelastic buckling (L_r) = 48.0000 ft.

4. Step 4 — Substitute the given values:

   $$M_{n} = C_b\left[M_p - \left(M_p - \dfrac{0.7 F_y S_x}{12}\right)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

5. Step 5 — Evaluate:

   $$M_{n} = 1304\ \text{kip-ft}$$

6. Step 6 — Check: returning M_n = 1,304 kip-ft to

   $$M_n = C_b\left[M_p - (M_p - 0.7 F_y S_x)\dfrac{L_b - L_p}{L_r - L_p}\right]$$

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

Why the other options are there

• 2,608 — kept a factor of two that cancels in the correct rearrangement.
• 652.0 — dropped that same factor in the other direction.
• 1,434 — rounded an intermediate value before the final step.

Reference: AISC Specification §F2.2 — Lateral-Torsional Buckling