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) = 2.1000 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) = 2.1000 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:

   $$V_{n} = 44.2260\ \text{kips}$$

6. Step 6 — Check: returning V_n = 44.2260 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

• 88.4520 — kept a factor of two that cancels in the correct rearrangement.
• 22.1130 — dropped that same factor in the other direction.
• 48.6486 — 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) = 301.3 kips; yield stress (F_y) = 46.0000 ksi; web shear coefficient (C_v) = 0.8100, 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) = 301.3 kips, yield stress (F_y) = 46.0000 ksi, web shear coefficient (C_v) = 0.8100.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   A_{w} = 13.4774\ \text{in^2}

6. Step 6 — Check: returning A_w = 13.4774 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

• 26.9547 — kept a factor of two that cancels in the correct rearrangement.
• 6.7387 — dropped that same factor in the other direction.
• 14.8251 — 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) = 611.2 kips; yield stress (F_y) = 40.0000 ksi; web area (d t_w) (A_w) = 14.5000 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) = 611.2 kips, yield stress (F_y) = 40.0000 ksi, web area (d t_w) (A_w) = 14.5000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$C_{v} = 1.7563$$

6. Step 6 — Check: returning C_v = 1.7563 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

• 3.5126 — kept a factor of two that cancels in the correct rearrangement.
• 0.8782 — dropped that same factor in the other direction.
• 1.9320 — 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) = 42.0000 ksi; web area (d t_w) (A_w) = 15.9000 in^2; web shear coefficient (C_v) = 0.6800, 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) = 42.0000 ksi, web area (d t_w) (A_w) = 15.9000 in^2, web shear coefficient (C_v) = 0.6800.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$V_{n} = 272.5\ \text{kips}$$

6. Step 6 — Check: returning V_n = 272.5 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

• 544.9 — kept a factor of two that cancels in the correct rearrangement.
• 136.2 — dropped that same factor in the other direction.
• 299.7 — 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) = 353.4 kips; yield stress (F_y) = 45.0000 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) = 353.4 kips, yield stress (F_y) = 45.0000 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} = 14.7066\ \text{in^2}

6. Step 6 — Check: returning A_w = 14.7066 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

• 29.4132 — kept a factor of two that cancels in the correct rearrangement.
• 7.3533 — dropped that same factor in the other direction.
• 16.1773 — 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) = 160.9 kips; yield stress (F_y) = 43.0000 ksi; web area (d t_w) (A_w) = 12.3000 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) = 160.9 kips, yield stress (F_y) = 43.0000 ksi, web area (d t_w) (A_w) = 12.3000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$C_{v} = 0.5070$$

6. Step 6 — Check: returning C_v = 0.5070 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

• 1.0141 — kept a factor of two that cancels in the correct rearrangement.
• 0.2535 — dropped that same factor in the other direction.
• 0.5577 — 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) = 47.0000 ksi; web area (d t_w) (A_w) = 9.9000 in^2; web shear coefficient (C_v) = 0.6800, 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) = 47.0000 ksi, web area (d t_w) (A_w) = 9.9000 in^2, web shear coefficient (C_v) = 0.6800.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$V_{n} = 189.8\ \text{kips}$$

6. Step 6 — Check: returning V_n = 189.8 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

• 379.7 — kept a factor of two that cancels in the correct rearrangement.
• 94.9212 — dropped that same factor in the other direction.
• 208.8 — 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) = 868.1 kips; yield stress (F_y) = 44.0000 ksi; web shear coefficient (C_v) = 0.6300, 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) = 868.1 kips, yield stress (F_y) = 44.0000 ksi, web shear coefficient (C_v) = 0.6300.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   A_{w} = 52.1946\ \text{in^2}

6. Step 6 — Check: returning A_w = 52.1946 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

• 104.4 — kept a factor of two that cancels in the correct rearrangement.
• 26.0973 — dropped that same factor in the other direction.
• 57.4140 — 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) = 760.4 kips; yield stress (F_y) = 42.0000 ksi; web area (d t_w) (A_w) = 12.1000 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) = 760.4 kips, yield stress (F_y) = 42.0000 ksi, web area (d t_w) (A_w) = 12.1000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$C_{v} = 2.4938$$

6. Step 6 — Check: returning C_v = 2.4938 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

• 4.9875 — kept a factor of two that cancels in the correct rearrangement.
• 1.2469 — dropped that same factor in the other direction.
• 2.7431 — 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) = 37.0000 ksi; web area (d t_w) (A_w) = 14.9000 in^2; web shear coefficient (C_v) = 0.7900, 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) = 37.0000 ksi, web area (d t_w) (A_w) = 14.9000 in^2, web shear coefficient (C_v) = 0.7900.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$V_{n} = 261.3\ \text{kips}$$

6. Step 6 — Check: returning V_n = 261.3 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

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

Reference: AISC Specification §G2.1 — Shear Strength