A steel beam's design strength uses resistance factors (phi) applied to nominal flexural strength. Given resistance factor (phi) = 0.8900; nominal strength (R_n) = 461.0 kip, determine the factored demand (R_u) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength (resistance factor phi).
• Everything except R_u is given, so isolate R_u 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.
• Resistance factors (phi) reduce the nominal strength of a steel member to a design strength compared against the factored demand.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n \ge R_u$$

2. Step 2 — Rearrange symbolically for R_u:

   $$R_{u} = \phi R_n$$

3. Step 3

   $$List the givens: resistance factor (phi) = 0.8900, nominal strength (R_n) = 461.0 kip$$

4. Step 4 — Substitute the given values:

   $$R_{u} = 0.8900 R_n$$

5. Step 5 — Evaluate:

   $$R_{u} = 410.3\ \text{kip}$$

6. Step 6 — Check: returning R_u = 410.3 kip to

   $$\phi R_n \ge R_u$$

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

Why the other options are there

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

Reference: FE Handbook — Structural Design: Resistance Factors (phi)

a braced-frame strut checked against its available strength Given resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 58.0000 ksi; gross area (A_g) = 22.1000 in^2, determine the design strength (\phi R_n) in kips.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength at a limit state.
• Everything except \phi R_n is given, so isolate \phi R_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.
• An A992 steel member is checked for the yielding limit state, where the available strength is the resistance factor times the nominal strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for \phi R_n:

   $$\phi R_{n} = \phi F_y A_g$$

3. Step 3 — List the givens: resistance factor for yielding (\phi) = 0.9000, specified yield stress (F_y) = 58.0000 ksi, gross area (A_g) = 22.1000 in^2.

4. Step 4 — Substitute the given values:

   $$\phi R_{n} = 0.9000 F_y A_g$$

5. Step 5 — Evaluate:

   $$\phi R_{n} = 1154\ \text{kips}$$

6. Step 6 — Check: returning \phi R_n = 1,154 kips to

   $$\phi R_n = \phi F_y A_g$$

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

Why the other options are there

• 2,307 — kept a factor of two that cancels in the correct rearrangement.
• 576.8 — dropped that same factor in the other direction.
• 1,269 — rounded an intermediate value before the final step.

Reference: AISC Specification §B3.1 — Design Basis (LRFD)

A bolted connection's capacity is checked with resistance factors (phi) for shear rupture. Given resistance factor (phi) = 0.8100; factored demand (R_u) = 71.0000 kip, determine the nominal strength (R_n) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength (resistance factor phi).
• Everything except R_n is given, so isolate R_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.
• Resistance factors (phi) reduce the nominal strength of a steel member to a design strength compared against the factored demand.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n \ge R_u$$

2. Step 2 — Rearrange symbolically for R_n:

   $$R_{n} = \dfrac{R_u}{\phi}$$

3. Step 3

   $$List the givens: resistance factor (phi) = 0.8100, factored demand (R_u) = 71.0000 kip$$

4. Step 4 — Substitute the given values:

   $$R_{n} = \dfrac{R_u}{0.8100}$$

5. Step 5 — Evaluate:

   $$R_{n} = 87.6543\ \text{kip}$$

6. Step 6 — Check: returning R_n = 87.6543 kip to

   $$\phi R_n \ge R_u$$

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

Why the other options are there

• 175.3 — kept a factor of two that cancels in the correct rearrangement.
• 43.8272 — dropped that same factor in the other direction.
• 96.4198 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Resistance Factors (phi)

a hanger rod checked at the yielding limit state Given design strength (\phi R_n) = 1,075 kips; resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 46.0000 ksi, determine the gross area (A_g) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength at a limit state.
• Everything except A_g is given, so isolate A_g 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.
• An A992 steel member is checked for the yielding limit state, where the available strength is the resistance factor times the nominal strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for A_g:

   $$A_{g} = \dfrac{\phi R_n}{\phi F_y}$$

3. Step 3 — List the givens: design strength (\phi R_n) = 1,075 kips, resistance factor for yielding (\phi) = 0.9000, specified yield stress (F_y) = 46.0000 ksi.

4. Step 4 — Substitute the given values:

   $$A_{g} = \dfrac{0.9000 R_n}{0.9000 F_y}$$

5. Step 5 — Evaluate:

   A_{g} = 25.9614\ \text{in^2}

6. Step 6 — Check: returning A_g = 25.9614 in^2 to

   $$\phi R_n = \phi F_y A_g$$

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

Why the other options are there

• 51.9227 — kept a factor of two that cancels in the correct rearrangement.
• 12.9807 — dropped that same factor in the other direction.
• 28.5575 — rounded an intermediate value before the final step.

Reference: AISC Specification §B3.1 — Design Basis (LRFD)

A steel column's available strength is found using resistance factors (phi) on its nominal strength. Given nominal strength (R_n) = 515.0 kip; factored demand (R_u) = 421.0 kip, determine the resistance factor (phi).

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength (resistance factor phi).
• Everything except phi is given, so isolate phi 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.
• Resistance factors (phi) reduce the nominal strength of a steel member to a design strength compared against the factored demand.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n \ge R_u$$

2. Step 2 — Rearrange symbolically for phi:

   $$\phi = \dfrac{R_u}{R_n}$$

3. Step 3

   $$List the givens: nominal strength (R_n) = 515.0 kip, factored demand (R_u) = 421.0 kip$$

4. Step 4 — Substitute the given values:

   $$\phi = \dfrac{R_u}{R_n}$$

5. Step 5 — Evaluate:

   $$\phi = 0.8175$$

6. Step 6 — Check: returning phi = 0.8175 to

   $$\phi R_n \ge R_u$$

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

Why the other options are there

• 1.6350 — kept a factor of two that cancels in the correct rearrangement.
• 0.4087 — dropped that same factor in the other direction.
• 0.8992 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Resistance Factors (phi)

an A992 tension chord checked for gross-section yielding Given design strength (\phi R_n) = 51.4000 kips; resistance factor for yielding (\phi) = 0.9000; gross area (A_g) = 9.6000 in^2, determine the specified yield stress (F_y) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength at a limit state.
• Everything except F_y is given, so isolate F_y 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.
• An A992 steel member is checked for the yielding limit state, where the available strength is the resistance factor times the nominal strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for F_y:

   $$F_{y} = \dfrac{\phi R_n}{\phi A_g}$$

3. Step 3 — List the givens: design strength (\phi R_n) = 51.4000 kips, resistance factor for yielding (\phi) = 0.9000, gross area (A_g) = 9.6000 in^2.

4. Step 4 — Substitute the given values:

   $$F_{y} = \dfrac{0.9000 R_n}{0.9000 A_g}$$

5. Step 5 — Evaluate:

   $$F_{y} = 5.9491\ \text{ksi}$$

6. Step 6 — Check: returning F_y = 5.9491 ksi to

   $$\phi R_n = \phi F_y A_g$$

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

Why the other options are there

• 11.8981 — kept a factor of two that cancels in the correct rearrangement.
• 2.9745 — dropped that same factor in the other direction.
• 6.5440 — rounded an intermediate value before the final step.

Reference: AISC Specification §B3.1 — Design Basis (LRFD)

A steel beam's design strength uses resistance factors (phi) applied to nominal flexural strength. Given resistance factor (phi) = 0.7400; nominal strength (R_n) = 776.0 kip, determine the factored demand (R_u) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength (resistance factor phi).
• Everything except R_u is given, so isolate R_u 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.
• Resistance factors (phi) reduce the nominal strength of a steel member to a design strength compared against the factored demand.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n \ge R_u$$

2. Step 2 — Rearrange symbolically for R_u:

   $$R_{u} = \phi R_n$$

3. Step 3

   $$List the givens: resistance factor (phi) = 0.7400, nominal strength (R_n) = 776.0 kip$$

4. Step 4 — Substitute the given values:

   $$R_{u} = 0.7400 R_n$$

5. Step 5 — Evaluate:

   $$R_{u} = 574.2\ \text{kip}$$

6. Step 6 — Check: returning R_u = 574.2 kip to

   $$\phi R_n \ge R_u$$

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

Why the other options are there

• 1,148 — kept a factor of two that cancels in the correct rearrangement.
• 287.1 — dropped that same factor in the other direction.
• 631.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Resistance Factors (phi)

a braced-frame strut checked against its available strength Given resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 39.0000 ksi; gross area (A_g) = 8.3000 in^2, determine the design strength (\phi R_n) in kips.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength at a limit state.
• Everything except \phi R_n is given, so isolate \phi R_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.
• An A992 steel member is checked for the yielding limit state, where the available strength is the resistance factor times the nominal strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for \phi R_n:

   $$\phi R_{n} = \phi F_y A_g$$

3. Step 3 — List the givens: resistance factor for yielding (\phi) = 0.9000, specified yield stress (F_y) = 39.0000 ksi, gross area (A_g) = 8.3000 in^2.

4. Step 4 — Substitute the given values:

   $$\phi R_{n} = 0.9000 F_y A_g$$

5. Step 5 — Evaluate:

   $$\phi R_{n} = 291.3\ \text{kips}$$

6. Step 6 — Check: returning \phi R_n = 291.3 kips to

   $$\phi R_n = \phi F_y A_g$$

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

Why the other options are there

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

Reference: AISC Specification §B3.1 — Design Basis (LRFD)

A bolted connection's capacity is checked with resistance factors (phi) for shear rupture. Given resistance factor (phi) = 0.8600; factored demand (R_u) = 490.0 kip, determine the nominal strength (R_n) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength (resistance factor phi).
• Everything except R_n is given, so isolate R_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.
• Resistance factors (phi) reduce the nominal strength of a steel member to a design strength compared against the factored demand.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n \ge R_u$$

2. Step 2 — Rearrange symbolically for R_n:

   $$R_{n} = \dfrac{R_u}{\phi}$$

3. Step 3

   $$List the givens: resistance factor (phi) = 0.8600, factored demand (R_u) = 490.0 kip$$

4. Step 4 — Substitute the given values:

   $$R_{n} = \dfrac{R_u}{0.8600}$$

5. Step 5 — Evaluate:

   $$R_{n} = 569.8\ \text{kip}$$

6. Step 6 — Check: returning R_n = 569.8 kip to

   $$\phi R_n \ge R_u$$

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

Why the other options are there

• 1,140 — kept a factor of two that cancels in the correct rearrangement.
• 284.9 — dropped that same factor in the other direction.
• 626.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Resistance Factors (phi)

a hanger rod checked at the yielding limit state Given design strength (\phi R_n) = 1,223 kips; resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 61.0000 ksi, determine the gross area (A_g) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is LRFD design strength at a limit state.
• Everything except A_g is given, so isolate A_g 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.
• An A992 steel member is checked for the yielding limit state, where the available strength is the resistance factor times the nominal strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for A_g:

   $$A_{g} = \dfrac{\phi R_n}{\phi F_y}$$

3. Step 3 — List the givens: design strength (\phi R_n) = 1,223 kips, resistance factor for yielding (\phi) = 0.9000, specified yield stress (F_y) = 61.0000 ksi.

4. Step 4 — Substitute the given values:

   $$A_{g} = \dfrac{0.9000 R_n}{0.9000 F_y}$$

5. Step 5 — Evaluate:

   A_{g} = 22.2732\ \text{in^2}

6. Step 6 — Check: returning A_g = 22.2732 in^2 to

   $$\phi R_n = \phi F_y A_g$$

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

Why the other options are there

• 44.5464 — kept a factor of two that cancels in the correct rearrangement.
• 11.1366 — dropped that same factor in the other direction.
• 24.5005 — rounded an intermediate value before the final step.

Reference: AISC Specification §B3.1 — Design Basis (LRFD)