a W-shape beam web at a bolted end connection Given yield stress (F_y) = 45.0000 ksi; web area (d t_w) (A_w) = 11.4000 in^2; web shear coefficient (C_v) = 0.7200, determine the nominal shear strength (V_n) in kips.

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except V_n is given, so isolate V_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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for V_n:

   $$V_{n} = 0.6 F_y A_w C_v$$

3. Step 3 — List the givens: yield stress (F_y) = 45.0000 ksi, web area (d t_w) (A_w) = 11.4000 in^2, web shear coefficient (C_v) = 0.7200.

4. Step 4 — Substitute the given values:

   $$V_{n} = 0.6 F_y A_w C_v$$

5. Step 5 — Evaluate:

   $$V_{n} = 221.6\ \text{kips}$$

6. Step 6 — Check: returning V_n = 221.6 kips to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

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

Reference: AISC Specification §G2.1 — Shear Strength

a plate-girder web panel between stiffeners Given nominal shear strength (V_n) = 494.2 kips; yield stress (F_y) = 37.0000 ksi; web shear coefficient (C_v) = 0.7100, determine the web area (d t_w) (A_w) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except A_w is given, so isolate A_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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for A_w:

   $$A_{w} = \dfrac{V_n}{0.6 F_y C_v}$$

3. Step 3 — List the givens: nominal shear strength (V_n) = 494.2 kips, yield stress (F_y) = 37.0000 ksi, web shear coefficient (C_v) = 0.7100.

4. Step 4 — Substitute the given values:

   $$A_{w} = \dfrac{V_n}{0.6 F_y C_v}$$

5. Step 5 — Evaluate:

   A_{w} = 31.3539\ \text{in^2}

6. Step 6 — Check: returning A_w = 31.3539 in^2 to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 62.7078 — kept a factor of two that cancels in the correct rearrangement.
• 15.6769 — dropped that same factor in the other direction.
• 34.4893 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength

a coped beam web at a shear tab Given nominal shear strength (V_n) = 797.9 kips; yield stress (F_y) = 45.0000 ksi; web area (d t_w) (A_w) = 3.1000 in^2, determine the web shear coefficient (C_v).

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except C_v is given, so isolate C_v 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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for C_v:

   $$C_{v} = \dfrac{V_n}{0.6 F_y A_w}$$

3. Step 3 — List the givens: nominal shear strength (V_n) = 797.9 kips, yield stress (F_y) = 45.0000 ksi, web area (d t_w) (A_w) = 3.1000 in^2.

4. Step 4 — Substitute the given values:

   $$C_{v} = \dfrac{V_n}{0.6 F_y A_w}$$

5. Step 5 — Evaluate:

   $$C_{v} = 9.5329$$

6. Step 6 — Check: returning C_v = 9.5329 to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 19.0657 — kept a factor of two that cancels in the correct rearrangement.
• 4.7664 — dropped that same factor in the other direction.
• 10.4861 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength

a W-shape beam web at a bolted end connection Given yield stress (F_y) = 44.0000 ksi; web area (d t_w) (A_w) = 6.2000 in^2; web shear coefficient (C_v) = 0.8600, determine the nominal shear strength (V_n) in kips.

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except V_n is given, so isolate V_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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for V_n:

   $$V_{n} = 0.6 F_y A_w C_v$$

3. Step 3 — List the givens: yield stress (F_y) = 44.0000 ksi, web area (d t_w) (A_w) = 6.2000 in^2, web shear coefficient (C_v) = 0.8600.

4. Step 4 — Substitute the given values:

   $$V_{n} = 0.6 F_y A_w C_v$$

5. Step 5 — Evaluate:

   $$V_{n} = 140.8\ \text{kips}$$

6. Step 6 — Check: returning V_n = 140.8 kips to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 281.5 — kept a factor of two that cancels in the correct rearrangement.
• 70.3824 — dropped that same factor in the other direction.
• 154.8 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength

a plate-girder web panel between stiffeners Given nominal shear strength (V_n) = 276.1 kips; yield stress (F_y) = 43.0000 ksi; web shear coefficient (C_v) = 0.7100, determine the web area (d t_w) (A_w) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except A_w is given, so isolate A_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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for A_w:

   $$A_{w} = \dfrac{V_n}{0.6 F_y C_v}$$

3. Step 3 — List the givens: nominal shear strength (V_n) = 276.1 kips, yield stress (F_y) = 43.0000 ksi, web shear coefficient (C_v) = 0.7100.

4. Step 4 — Substitute the given values:

   $$A_{w} = \dfrac{V_n}{0.6 F_y C_v}$$

5. Step 5 — Evaluate:

   A_{w} = 15.0726\ \text{in^2}

6. Step 6 — Check: returning A_w = 15.0726 in^2 to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 30.1452 — kept a factor of two that cancels in the correct rearrangement.
• 7.5363 — dropped that same factor in the other direction.
• 16.5799 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength

a coped beam web at a shear tab Given nominal shear strength (V_n) = 647.4 kips; yield stress (F_y) = 38.0000 ksi; web area (d t_w) (A_w) = 17.9000 in^2, determine the web shear coefficient (C_v).

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except C_v is given, so isolate C_v 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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for C_v:

   $$C_{v} = \dfrac{V_n}{0.6 F_y A_w}$$

3. Step 3 — List the givens: nominal shear strength (V_n) = 647.4 kips, yield stress (F_y) = 38.0000 ksi, web area (d t_w) (A_w) = 17.9000 in^2.

4. Step 4 — Substitute the given values:

   $$C_{v} = \dfrac{V_n}{0.6 F_y A_w}$$

5. Step 5 — Evaluate:

   $$C_{v} = 1.5863$$

6. Step 6 — Check: returning C_v = 1.5863 to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 3.1726 — kept a factor of two that cancels in the correct rearrangement.
• 0.7931 — dropped that same factor in the other direction.
• 1.7449 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength

a W-shape beam web at a bolted end connection Given yield stress (F_y) = 45.0000 ksi; web area (d t_w) (A_w) = 2.2000 in^2; web shear coefficient (C_v) = 0.7800, determine the nominal shear strength (V_n) in kips.

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except V_n is given, so isolate V_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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for V_n:

   $$V_{n} = 0.6 F_y A_w C_v$$

3. Step 3 — List the givens: yield stress (F_y) = 45.0000 ksi, web area (d t_w) (A_w) = 2.2000 in^2, web shear coefficient (C_v) = 0.7800.

4. Step 4 — Substitute the given values:

   $$V_{n} = 0.6 F_y A_w C_v$$

5. Step 5 — Evaluate:

   $$V_{n} = 46.3320\ \text{kips}$$

6. Step 6 — Check: returning V_n = 46.3320 kips to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 92.6640 — kept a factor of two that cancels in the correct rearrangement.
• 23.1660 — dropped that same factor in the other direction.
• 50.9652 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength

a plate-girder web panel between stiffeners Given nominal shear strength (V_n) = 589.6 kips; yield stress (F_y) = 46.0000 ksi; web shear coefficient (C_v) = 0.8900, determine the web area (d t_w) (A_w) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except A_w is given, so isolate A_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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for A_w:

   $$A_{w} = \dfrac{V_n}{0.6 F_y C_v}$$

3. Step 3 — List the givens: nominal shear strength (V_n) = 589.6 kips, yield stress (F_y) = 46.0000 ksi, web shear coefficient (C_v) = 0.8900.

4. Step 4 — Substitute the given values:

   $$A_{w} = \dfrac{V_n}{0.6 F_y C_v}$$

5. Step 5 — Evaluate:

   A_{w} = 24.0026\ \text{in^2}

6. Step 6 — Check: returning A_w = 24.0026 in^2 to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 48.0052 — kept a factor of two that cancels in the correct rearrangement.
• 12.0013 — dropped that same factor in the other direction.
• 26.4029 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength

a coped beam web at a shear tab Given nominal shear strength (V_n) = 110.6 kips; yield stress (F_y) = 44.0000 ksi; web area (d t_w) (A_w) = 5.8000 in^2, determine the web shear coefficient (C_v).

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except C_v is given, so isolate C_v 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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for C_v:

   $$C_{v} = \dfrac{V_n}{0.6 F_y A_w}$$

3. Step 3 — List the givens: nominal shear strength (V_n) = 110.6 kips, yield stress (F_y) = 44.0000 ksi, web area (d t_w) (A_w) = 5.8000 in^2.

4. Step 4 — Substitute the given values:

   $$C_{v} = \dfrac{V_n}{0.6 F_y A_w}$$

5. Step 5 — Evaluate:

   $$C_{v} = 0.7223$$

6. Step 6 — Check: returning C_v = 0.7223 to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 1.4446 — kept a factor of two that cancels in the correct rearrangement.
• 0.3612 — dropped that same factor in the other direction.
• 0.7945 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength

a W-shape beam web at a bolted end connection Given yield stress (F_y) = 40.0000 ksi; web area (d t_w) (A_w) = 3.0000 in^2; web shear coefficient (C_v) = 0.8900, determine the nominal shear strength (V_n) in kips.

Start with the thinking

• The governing relation printed in this handbook section is Shear strength of a steel web.
• Everything except V_n is given, so isolate V_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.
• The web of a rolled steel beam is checked for the shear limit state at the support.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V_n = 0.6 F_y A_w C_v$$

2. Step 2 — Rearrange symbolically for V_n:

   $$V_{n} = 0.6 F_y A_w C_v$$

3. Step 3 — List the givens: yield stress (F_y) = 40.0000 ksi, web area (d t_w) (A_w) = 3.0000 in^2, web shear coefficient (C_v) = 0.8900.

4. Step 4 — Substitute the given values:

   $$V_{n} = 0.6 F_y A_w C_v$$

5. Step 5 — Evaluate:

   $$V_{n} = 64.0800\ \text{kips}$$

6. Step 6 — Check: returning V_n = 64.0800 kips to

   $$V_n = 0.6 F_y A_w C_v$$

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

Why the other options are there

• 128.2 — kept a factor of two that cancels in the correct rearrangement.
• 32.0400 — dropped that same factor in the other direction.
• 70.4880 — rounded an intermediate value before the final step.

Reference: AISC Specification §G2.1 — Shear Strength