A steel tension brace is checked for yielding of the gross cross-sectional area. Given resistance factor (phi) = 0.9400; steel yield stress (F_y) = 48.0000 ksi; gross area (A_g) = 8.5000 in^2, determine the design tensile yield strength (P_n_phi) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Tension yielding of gross section.
• Everything except P_n_phi is given, so isolate P_n_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.
• Yielding of the gross section limits a steel tension member's design strength to the yield stress times the gross area.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for P_n_phi:

   $$P_{nphi} = \phi F_y A_g$$

3. Step 3 — List the givens: resistance factor (phi) = 0.9400, steel yield stress (F_y) = 48.0000 ksi, gross area (A_g) = 8.5000 in^2.

4. Step 4 — Substitute the given values:

   $$P_{nphi} = 0.9400 F_y A_g$$

5. Step 5 — Evaluate:

   $$P_{nphi} = 383.5\ \text{kip}$$

6. Step 6 — Check: returning P_n_phi = 383.5 kip to

   $$\phi P_n = \phi F_y A_g$$

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

Why the other options are there

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

Reference: FE Handbook — Structural Design: Yielding

a braced-frame strut checked against its available strength Given resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 50.0000 ksi; gross area (A_g) = 15.6000 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) = 50.0000 ksi, gross area (A_g) = 15.6000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$\phi R_{n} = 702.0\ \text{kips}$$

6. Step 6 — Check: returning \phi R_n = 702.0 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

• 1,404 — kept a factor of two that cancels in the correct rearrangement.
• 351.0 — dropped that same factor in the other direction.
• 772.2 — rounded an intermediate value before the final step.

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

A hanger rod's tensile design strength is governed by yielding at the gross section. Given resistance factor (phi) = 0.9200; gross area (A_g) = 7.4000 in^2; design tensile yield strength (P_n_phi) = 1,289 kip, determine the steel yield stress (F_y) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Tension yielding of gross section.
• 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.
• Yielding of the gross section limits a steel tension member's design strength to the yield stress times the gross area.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for F_y:

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

3. Step 3 — List the givens: resistance factor (phi) = 0.9200, gross area (A_g) = 7.4000 in^2, design tensile yield strength (P_n_phi) = 1,289 kip.

4. Step 4 — Substitute the given values:

   $$F_{y} = \dfrac{P_n}{0.9200 A_g}$$

5. Step 5 — Evaluate:

   $$F_{y} = 189.3\ \text{ksi}$$

6. Step 6 — Check: returning F_y = 189.3 ksi to

   $$\phi P_n = \phi F_y A_g$$

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

Why the other options are there

• 378.7 — kept a factor of two that cancels in the correct rearrangement.
• 94.6680 — dropped that same factor in the other direction.
• 208.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Yielding

a hanger rod checked at the yielding limit state Given design strength (\phi R_n) = 824.8 kips; resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 51.0000 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) = 824.8 kips, resistance factor for yielding (\phi) = 0.9000, specified yield stress (F_y) = 51.0000 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} = 17.9695\ \text{in^2}

6. Step 6 — Check: returning A_g = 17.9695 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

• 35.9390 — kept a factor of two that cancels in the correct rearrangement.
• 8.9847 — dropped that same factor in the other direction.
• 19.7664 — rounded an intermediate value before the final step.

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

A steel tie member in a truss is checked against yielding of the gross section. Given resistance factor (phi) = 0.9400; steel yield stress (F_y) = 58.0000 ksi; design tensile yield strength (P_n_phi) = 1,558 kip, determine the gross area (A_g) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Tension yielding of gross section.
• 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.
• Yielding of the gross section limits a steel tension member's design strength to the yield stress times the gross area.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for A_g:

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

3. Step 3 — List the givens: resistance factor (phi) = 0.9400, steel yield stress (F_y) = 58.0000 ksi, design tensile yield strength (P_n_phi) = 1,558 kip.

4. Step 4 — Substitute the given values:

   $$A_{g} = \dfrac{P_n}{0.9400 F_y}$$

5. Step 5 — Evaluate:

   A_{g} = 28.5767\ \text{in^2}

6. Step 6 — Check: returning A_g = 28.5767 in^2 to

   $$\phi P_n = \phi F_y A_g$$

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

Why the other options are there

• 57.1533 — kept a factor of two that cancels in the correct rearrangement.
• 14.2883 — dropped that same factor in the other direction.
• 31.4343 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Yielding

an A992 tension chord checked for gross-section yielding Given design strength (\phi R_n) = 595.7 kips; resistance factor for yielding (\phi) = 0.9000; gross area (A_g) = 19.2000 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) = 595.7 kips, resistance factor for yielding (\phi) = 0.9000, gross area (A_g) = 19.2000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$F_{y} = 34.4734\ \text{ksi}$$

6. Step 6 — Check: returning F_y = 34.4734 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

• 68.9468 — kept a factor of two that cancels in the correct rearrangement.
• 17.2367 — dropped that same factor in the other direction.
• 37.9207 — rounded an intermediate value before the final step.

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

A steel tension brace is checked for yielding of the gross cross-sectional area. Given resistance factor (phi) = 0.9400; steel yield stress (F_y) = 41.0000 ksi; gross area (A_g) = 6.0000 in^2, determine the design tensile yield strength (P_n_phi) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Tension yielding of gross section.
• Everything except P_n_phi is given, so isolate P_n_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.
• Yielding of the gross section limits a steel tension member's design strength to the yield stress times the gross area.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for P_n_phi:

   $$P_{nphi} = \phi F_y A_g$$

3. Step 3 — List the givens: resistance factor (phi) = 0.9400, steel yield stress (F_y) = 41.0000 ksi, gross area (A_g) = 6.0000 in^2.

4. Step 4 — Substitute the given values:

   $$P_{nphi} = 0.9400 F_y A_g$$

5. Step 5 — Evaluate:

   $$P_{nphi} = 231.2\ \text{kip}$$

6. Step 6 — Check: returning P_n_phi = 231.2 kip to

   $$\phi P_n = \phi F_y A_g$$

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

Why the other options are there

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

Reference: FE Handbook — Structural Design: Yielding

a braced-frame strut checked against its available strength Given resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 63.0000 ksi; gross area (A_g) = 9.2000 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) = 63.0000 ksi, gross area (A_g) = 9.2000 in^2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$\phi R_{n} = 521.6\ \text{kips}$$

6. Step 6 — Check: returning \phi R_n = 521.6 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

• 1,043 — kept a factor of two that cancels in the correct rearrangement.
• 260.8 — dropped that same factor in the other direction.
• 573.8 — rounded an intermediate value before the final step.

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

A hanger rod's tensile design strength is governed by yielding at the gross section. Given resistance factor (phi) = 0.9000; gross area (A_g) = 28.5000 in^2; design tensile yield strength (P_n_phi) = 1,549 kip, determine the steel yield stress (F_y) in ksi.

Start with the thinking

• The governing relation printed in this handbook section is Tension yielding of gross section.
• 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.
• Yielding of the gross section limits a steel tension member's design strength to the yield stress times the gross area.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = \phi F_y A_g$$

2. Step 2 — Rearrange symbolically for F_y:

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

3. Step 3 — List the givens: resistance factor (phi) = 0.9000, gross area (A_g) = 28.5000 in^2, design tensile yield strength (P_n_phi) = 1,549 kip.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$F_{y} = 60.3899\ \text{ksi}$$

6. Step 6 — Check: returning F_y = 60.3899 ksi to

   $$\phi P_n = \phi F_y A_g$$

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

Why the other options are there

• 120.8 — kept a factor of two that cancels in the correct rearrangement.
• 30.1949 — dropped that same factor in the other direction.
• 66.4288 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Yielding

a hanger rod checked at the yielding limit state Given design strength (\phi R_n) = 193.8 kips; resistance factor for yielding (\phi) = 0.9000; specified yield stress (F_y) = 49.0000 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) = 193.8 kips, resistance factor for yielding (\phi) = 0.9000, specified yield stress (F_y) = 49.0000 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} = 4.3946\ \text{in^2}

6. Step 6 — Check: returning A_g = 4.3946 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

• 8.7891 — kept a factor of two that cancels in the correct rearrangement.
• 2.1973 — dropped that same factor in the other direction.
• 4.8340 — rounded an intermediate value before the final step.

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