An A36 plate 6 in. × ½ in. has two ⅞ in. bolt holes in the critical section. With Fy = 36 ksi, Fu = 58 ksi and U = 1.0, what governs the design tensile strength?

Start with the thinking

• Two limit states: gross-section yielding (φ = 0.90) and net-section rupture (φ = 0.75).
• Hole diameter includes the 1/16 in. damage allowance.

Step-by-step solution

1. Gross yielding — φPn = 0.90 Fy Ag = 0.90(36)(3.00) = 97.2 kips

2. Net area

   $$An = 3.00 - 2(1.00)(0.5) = 2.00 in^{2}$$

3. Effective net

   $$Ae = U An = 1.0(2.00) = 2.00 in^{2}$$

4. Rupture — φPn = 0.75 Fu Ae = 0.75(58)(2.00) = 87.0 kips

5. Governing — the smaller value, 87.0 kips (net-section rupture)

Why the other options are there

• 97.2 kips (yielding taken as governing)
• 104 kips (holes ignored)

Reference: FE Reference Handbook — Structural Design — Steel tension members

A steel connection is checked against multiple limit states and available strengths. Given resistance factor (phi) = 0.7800; nominal strength (limit state) (R_n) = 175.0 kip, determine the required strength (R_u) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Available strength (LRFD limit states).
• 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.
• Limit states and available strengths compare the governing nominal strength, reduced by phi, against the required factored strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = R_u$$

2. Step 2 — Rearrange symbolically for R_u:

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

3. Step 3

   $$List the givens: resistance factor (phi) = 0.7800, nominal strength (limit state) (R_n) = 175.0 kip$$

4. Step 4 — Substitute the given values:

   $$R_{u} = 0.7800 R_n$$

5. Step 5 — Evaluate:

   $$R_{u} = 136.5\ \text{kip}$$

6. Step 6 — Check: returning R_u = 136.5 kip to

   $$\phi R_n = R_u$$

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

Why the other options are there

• 273.0 — kept a factor of two that cancels in the correct rearrangement.
• 68.2500 — dropped that same factor in the other direction.
• 150.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Limit States and Available Strengths

a braced-frame strut checked against its available strength Given resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 61.0000 ksi; gross area (A_g) = 20.6000 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) = 61.0000 ksi, gross area (A_g) = 20.6000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$\phi R_{n} = 1131\ \text{kips}$$

6. Step 6 — Check: returning \phi R_n = 1,131 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

• 2,262 — kept a factor of two that cancels in the correct rearrangement.
• 565.5 — dropped that same factor in the other direction.
• 1,244 — rounded an intermediate value before the final step.

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

A beam-column's governing limit states and available strengths are compared for design. Given resistance factor (phi) = 0.7600; required strength (R_u) = 596.0 kip, determine the nominal strength (limit state) (R_n) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Available strength (LRFD limit states).
• 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.
• Limit states and available strengths compare the governing nominal strength, reduced by phi, against the required factored strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = R_u$$

2. Step 2 — Rearrange symbolically for R_n:

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

3. Step 3

   $$List the givens: resistance factor (phi) = 0.7600, required strength (R_u) = 596.0 kip$$

4. Step 4 — Substitute the given values:

   $$R_{n} = \dfrac{R_u}{0.7600}$$

5. Step 5 — Evaluate:

   $$R_{n} = 784.2\ \text{kip}$$

6. Step 6 — Check: returning R_n = 784.2 kip to

   $$\phi R_n = R_u$$

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

Why the other options are there

• 1,568 — kept a factor of two that cancels in the correct rearrangement.
• 392.1 — dropped that same factor in the other direction.
• 862.6 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Limit States and Available Strengths

a hanger rod checked at the yielding limit state Given design strength (\phi R_n) = 300.5 kips; resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 45.0000 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) = 300.5 kips, resistance factor for yielding (\phi) = 0.9000, specified yield stress (F_y) = 45.0000 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} = 7.4198\ \text{in^2}

6. Step 6 — Check: returning A_g = 7.4198 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

• 14.8395 — kept a factor of two that cancels in the correct rearrangement.
• 3.7099 — dropped that same factor in the other direction.
• 8.1617 — rounded an intermediate value before the final step.

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

A bolted splice is evaluated for limit states and available strengths in bearing and shear. Given nominal strength (limit state) (R_n) = 372.0 kip; required strength (R_u) = 500.0 kip, determine the resistance factor (phi).

Start with the thinking

• The governing relation printed in this handbook section is Available strength (LRFD limit states).
• 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.
• Limit states and available strengths compare the governing nominal strength, reduced by phi, against the required factored strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = R_u$$

2. Step 2 — Rearrange symbolically for phi:

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

3. Step 3

   $$List the givens: nominal strength (limit state) (R_n) = 372.0 kip, required strength (R_u) = 500.0 kip$$

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$\phi = 1.3441$$

6. Step 6 — Check: returning phi = 1.3441 to

   $$\phi R_n = R_u$$

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

Why the other options are there

• 2.6882 — kept a factor of two that cancels in the correct rearrangement.
• 0.6720 — dropped that same factor in the other direction.
• 1.4785 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Limit States and Available Strengths

an A992 tension chord checked for gross-section yielding Given design strength (\phi R_n) = 101.1 kips; resistance factor for yielding (\phi) = 0.9000; gross area (A_g) = 24.1000 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) = 101.1 kips, resistance factor for yielding (\phi) = 0.9000, gross area (A_g) = 24.1000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$F_{y} = 4.6611\ \text{ksi}$$

6. Step 6 — Check: returning F_y = 4.6611 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

• 9.3223 — kept a factor of two that cancels in the correct rearrangement.
• 2.3306 — dropped that same factor in the other direction.
• 5.1272 — rounded an intermediate value before the final step.

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

A steel connection is checked against multiple limit states and available strengths. Given resistance factor (phi) = 0.9500; nominal strength (limit state) (R_n) = 889.0 kip, determine the required strength (R_u) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Available strength (LRFD limit states).
• 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.
• Limit states and available strengths compare the governing nominal strength, reduced by phi, against the required factored strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = R_u$$

2. Step 2 — Rearrange symbolically for R_u:

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

3. Step 3

   $$List the givens: resistance factor (phi) = 0.9500, nominal strength (limit state) (R_n) = 889.0 kip$$

4. Step 4 — Substitute the given values:

   $$R_{u} = 0.9500 R_n$$

5. Step 5 — Evaluate:

   $$R_{u} = 844.6\ \text{kip}$$

6. Step 6 — Check: returning R_u = 844.6 kip to

   $$\phi R_n = R_u$$

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

Why the other options are there

• 1,689 — kept a factor of two that cancels in the correct rearrangement.
• 422.3 — dropped that same factor in the other direction.
• 929.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Limit States and Available Strengths

a braced-frame strut checked against its available strength Given resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 48.0000 ksi; gross area (A_g) = 2.2000 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) = 48.0000 ksi, gross area (A_g) = 2.2000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$\phi R_{n} = 95.0400\ \text{kips}$$

6. Step 6 — Check: returning \phi R_n = 95.0400 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

• 190.1 — kept a factor of two that cancels in the correct rearrangement.
• 47.5200 — dropped that same factor in the other direction.
• 104.5 — rounded an intermediate value before the final step.

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

A beam-column's governing limit states and available strengths are compared for design. Given resistance factor (phi) = 0.7700; required strength (R_u) = 165.0 kip, determine the nominal strength (limit state) (R_n) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Available strength (LRFD limit states).
• 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.
• Limit states and available strengths compare the governing nominal strength, reduced by phi, against the required factored strength.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi R_n = R_u$$

2. Step 2 — Rearrange symbolically for R_n:

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

3. Step 3

   $$List the givens: resistance factor (phi) = 0.7700, required strength (R_u) = 165.0 kip$$

4. Step 4 — Substitute the given values:

   $$R_{n} = \dfrac{R_u}{0.7700}$$

5. Step 5 — Evaluate:

   $$R_{n} = 214.3\ \text{kip}$$

6. Step 6 — Check: returning R_n = 214.3 kip to

   $$\phi R_n = R_u$$

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

Why the other options are there

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

Reference: FE Handbook — Structural Design: Limit States and Available Strengths