A steel roof beam's capacity is governed by lateral-torsional buckling between brace points. Given moment gradient factor (C_b) = 1.9000; plastic moment (M_p) = 3,430 kip-in; steel yield stress (F_y) = 39.0000 ksi; elastic section modulus (S_x) = 98.0000 in^3; unbraced length (L_b) = 135.0 in; limiting length, plastic (L_p) = 68.0000 in; limiting length, inelastic (L_r) = 230.0 in, determine the nominal flexural strength (M_n) in kip-in.

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling moment.
• 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.
• Lateral-torsional buckling reduces a steel beam's nominal flexural strength when the unbraced length exceeds the plastic limit.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for M_n:

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

3. Step 3 — List the givens: moment gradient factor (C_b) = 1.9000, plastic moment (M_p) = 3,430 kip-in, steel yield stress (F_y) = 39.0000 ksi, elastic section modulus (S_x) = 98.0000 in^3, unbraced length (L_b) = 135.0 in, limiting length, plastic (L_p) = 68.0000 in, limiting length, inelastic (L_r) = 230.0 in.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$M_{n} = 5924\ \text{kip-in}$$

6. Step 6 — Check: returning M_n = 5,924 kip-in to

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

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

Why the other options are there

• 11,848 — kept a factor of two that cancels in the correct rearrangement.
• 2,962 — dropped that same factor in the other direction.
• 6,516 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Lateral-Torsional Buckling

a steel roof beam braced by purlins at 8 ft spacing Given lateral-torsional modification factor (C_b) = 1.3900; plastic moment (M_p) = 1,071 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 39.0000 in^3; unbraced length (L_b) = 12.5000 ft; limiting length for yielding (L_p) = 6.3000 ft; limiting length for inelastic buckling (L_r) = 38.5000 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) = 1.3900, plastic moment (M_p) = 1,071 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 39.0000 in^3, unbraced length (L_b) = 12.5000 ft, limiting length for yielding (L_p) = 6.3000 ft, limiting length for inelastic buckling (L_r) = 38.5000 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:

   $$M_{n} = 1232\ \text{kip-ft}$$

6. Step 6 — Check: returning M_n = 1,232 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

• 2,465 — kept a factor of two that cancels in the correct rearrangement.
• 616.2 — dropped that same factor in the other direction.
• 1,356 — rounded an intermediate value before the final step.

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

A long-span steel girder is checked for lateral-torsional buckling under gravity load. Given moment gradient factor (C_b) = 1.6000; plastic moment (M_p) = 3,630 kip-in; steel yield stress (F_y) = 54.0000 ksi; elastic section modulus (S_x) = 164.0 in^3; unbraced length (L_b) = 160.0 in; limiting length, plastic (L_p) = 96.0000 in; limiting length, inelastic (L_r) = 415.0 in, determine the nominal flexural strength (M_n) in kip-in.

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling moment.
• 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.
• Lateral-torsional buckling reduces a steel beam's nominal flexural strength when the unbraced length exceeds the plastic limit.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for M_n:

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

3. Step 3 — List the givens: moment gradient factor (C_b) = 1.6000, plastic moment (M_p) = 3,630 kip-in, steel yield stress (F_y) = 54.0000 ksi, elastic section modulus (S_x) = 164.0 in^3, unbraced length (L_b) = 160.0 in, limiting length, plastic (L_p) = 96.0000 in, limiting length, inelastic (L_r) = 415.0 in.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$M_{n} = 6633\ \text{kip-in}$$

6. Step 6 — Check: returning M_n = 6,633 kip-in to

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

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

Why the other options are there

• 13,265 — kept a factor of two that cancels in the correct rearrangement.
• 3,316 — dropped that same factor in the other direction.
• 7,296 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Lateral-Torsional Buckling

a crane runway girder with limited lateral bracing Given nominal flexural strength (M_n) = 247.5 kip-ft; plastic moment (M_p) = 361.0 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 161.0 in^3; unbraced length (L_b) = 15.0000 ft; limiting length for yielding (L_p) = 4.7000 ft; limiting length for inelastic buckling (L_r) = 45.0000 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) = 247.5 kip-ft, plastic moment (M_p) = 361.0 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 161.0 in^3, unbraced length (L_b) = 15.0000 ft, limiting length for yielding (L_p) = 4.7000 ft, limiting length for inelastic buckling (L_r) = 45.0000 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:

   $$C_{b} = 0.6367$$

6. Step 6 — Check: returning C_b = 0.6367 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

• 1.2733 — kept a factor of two that cancels in the correct rearrangement.
• 0.3183 — dropped that same factor in the other direction.
• 0.7003 — rounded an intermediate value before the final step.

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

A steel beam with widely spaced purlins is evaluated for lateral-torsional buckling. Given moment gradient factor (C_b) = 1.4500; plastic moment (M_p) = 2,740 kip-in; steel yield stress (F_y) = 54.0000 ksi; elastic section modulus (S_x) = 121.0 in^3; unbraced length (L_b) = 300.0 in; limiting length, plastic (L_p) = 55.0000 in; limiting length, inelastic (L_r) = 410.0 in, determine the nominal flexural strength (M_n) in kip-in.

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling moment.
• 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.
• Lateral-torsional buckling reduces a steel beam's nominal flexural strength when the unbraced length exceeds the plastic limit.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for M_n:

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

3. Step 3 — List the givens: moment gradient factor (C_b) = 1.4500, plastic moment (M_p) = 2,740 kip-in, steel yield stress (F_y) = 54.0000 ksi, elastic section modulus (S_x) = 121.0 in^3, unbraced length (L_b) = 300.0 in, limiting length, plastic (L_p) = 55.0000 in, limiting length, inelastic (L_r) = 410.0 in.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$M_{n} = 5808\ \text{kip-in}$$

6. Step 6 — Check: returning M_n = 5,808 kip-in to

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

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

Why the other options are there

• 11,616 — kept a factor of two that cancels in the correct rearrangement.
• 2,904 — dropped that same factor in the other direction.
• 6,389 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Lateral-Torsional Buckling

a W-shape floor girder braced only at third points Given lateral-torsional modification factor (C_b) = 1.4200; plastic moment (M_p) = 403.0 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 65.0000 in^3; unbraced length (L_b) = 19.0000 ft; limiting length for yielding (L_p) = 4.7000 ft; limiting length for inelastic buckling (L_r) = 31.5000 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) = 1.4200, plastic moment (M_p) = 403.0 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 65.0000 in^3, unbraced length (L_b) = 19.0000 ft, limiting length for yielding (L_p) = 4.7000 ft, limiting length for inelastic buckling (L_r) = 31.5000 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:

   $$M_{n} = 410.6\ \text{kip-ft}$$

6. Step 6 — Check: returning M_n = 410.6 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

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

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

A steel roof beam's capacity is governed by lateral-torsional buckling between brace points. Given moment gradient factor (C_b) = 2.0500; plastic moment (M_p) = 2,780 kip-in; steel yield stress (F_y) = 44.0000 ksi; elastic section modulus (S_x) = 146.0 in^3; unbraced length (L_b) = 165.0 in; limiting length, plastic (L_p) = 98.0000 in; limiting length, inelastic (L_r) = 470.0 in, determine the nominal flexural strength (M_n) in kip-in.

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling moment.
• 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.
• Lateral-torsional buckling reduces a steel beam's nominal flexural strength when the unbraced length exceeds the plastic limit.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for M_n:

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

3. Step 3 — List the givens: moment gradient factor (C_b) = 2.0500, plastic moment (M_p) = 2,780 kip-in, steel yield stress (F_y) = 44.0000 ksi, elastic section modulus (S_x) = 146.0 in^3, unbraced length (L_b) = 165.0 in, limiting length, plastic (L_p) = 98.0000 in, limiting length, inelastic (L_r) = 470.0 in.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$M_{n} = 6333\ \text{kip-in}$$

6. Step 6 — Check: returning M_n = 6,333 kip-in to

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

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

Why the other options are there

• 12,666 — kept a factor of two that cancels in the correct rearrangement.
• 3,166 — dropped that same factor in the other direction.
• 6,966 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Lateral-Torsional Buckling

a steel roof beam braced by purlins at 8 ft spacing Given nominal flexural strength (M_n) = 692.2 kip-ft; plastic moment (M_p) = 1,075 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 54.0000 in^3; unbraced length (L_b) = 22.0000 ft; limiting length for yielding (L_p) = 7.7000 ft; limiting length for inelastic buckling (L_r) = 45.5000 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) = 692.2 kip-ft, plastic moment (M_p) = 1,075 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 54.0000 in^3, unbraced length (L_b) = 22.0000 ft, limiting length for yielding (L_p) = 7.7000 ft, limiting length for inelastic buckling (L_r) = 45.5000 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:

   $$C_{b} = 0.9510$$

6. Step 6 — Check: returning C_b = 0.9510 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

• 1.9019 — kept a factor of two that cancels in the correct rearrangement.
• 0.4755 — dropped that same factor in the other direction.
• 1.0460 — rounded an intermediate value before the final step.

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

A long-span steel girder is checked for lateral-torsional buckling under gravity load. Given moment gradient factor (C_b) = 1.6500; plastic moment (M_p) = 5,120 kip-in; steel yield stress (F_y) = 49.0000 ksi; elastic section modulus (S_x) = 186.0 in^3; unbraced length (L_b) = 155.0 in; limiting length, plastic (L_p) = 40.0000 in; limiting length, inelastic (L_r) = 215.0 in, determine the nominal flexural strength (M_n) in kip-in.

Start with the thinking

• The governing relation printed in this handbook section is Lateral-torsional buckling moment.
• 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.
• Lateral-torsional buckling reduces a steel beam's nominal flexural strength when the unbraced length exceeds the plastic limit.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for M_n:

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

3. Step 3 — List the givens: moment gradient factor (C_b) = 1.6500, plastic moment (M_p) = 5,120 kip-in, steel yield stress (F_y) = 49.0000 ksi, elastic section modulus (S_x) = 186.0 in^3, unbraced length (L_b) = 155.0 in, limiting length, plastic (L_p) = 40.0000 in, limiting length, inelastic (L_r) = 215.0 in.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$M_{n} = 9814\ \text{kip-in}$$

6. Step 6 — Check: returning M_n = 9,814 kip-in to

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

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

Why the other options are there

• 19,628 — kept a factor of two that cancels in the correct rearrangement.
• 4,907 — dropped that same factor in the other direction.
• 10,795 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Lateral-Torsional Buckling

a crane runway girder with limited lateral bracing Given lateral-torsional modification factor (C_b) = 1.4400; plastic moment (M_p) = 418.0 kip-ft; yield stress (F_y) = 50.0000 ksi; elastic section modulus (S_x) = 231.0 in^3; unbraced length (L_b) = 10.5000 ft; limiting length for yielding (L_p) = 5.7000 ft; limiting length for inelastic buckling (L_r) = 47.5000 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) = 1.4400, plastic moment (M_p) = 418.0 kip-ft, yield stress (F_y) = 50.0000 ksi, elastic section modulus (S_x) = 231.0 in^3, unbraced length (L_b) = 10.5000 ft, limiting length for yielding (L_p) = 5.7000 ft, limiting length for inelastic buckling (L_r) = 47.5000 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:

   $$M_{n} = 644.2\ \text{kip-ft}$$

6. Step 6 — Check: returning M_n = 644.2 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

• 1,288 — kept a factor of two that cancels in the correct rearrangement.
• 322.1 — dropped that same factor in the other direction.
• 708.6 — rounded an intermediate value before the final step.

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