A square tied column in a parking structure is sized using design column strength, tied columns provisions. Given resistance factor (phi) = 0.6400; concrete strength (f_c) = 3.2000 ksi; gross column area (A_g) = 325.0 in^2; steel reinforcement area (A_st) = 23.0000 in^2; steel yield strength (f_y) = 61.0000 ksi, determine the design strength of tied column (P_n_phi) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for P_n_phi:

   $$P_{nphi} = 0.80\,\phi\left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6400, concrete strength (f_c) = 3.2000 ksi, gross column area (A_g) = 325.0 in^2, steel reinforcement area (A_st) = 23.0000 in^2, steel yield strength (f_y) = 61.0000 ksi.

4. Step 4 — Substitute the given values:

   $$P_{nphi} = 0.80\,0.6400\left[0.85 f_c'(A_g - 23.0000) + f_y 23.0000\right]$$

5. Step 5 — Evaluate:

   $$P_{nphi} = 1139\ \text{kip}$$

6. Step 6 — Check: returning P_n_phi = 1,139 kip to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

• 2,278 — kept a factor of two that cancels in the correct rearrangement.
• 569.5 — dropped that same factor in the other direction.
• 1,253 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

An interior building column's design column strength, tied columns capacity is checked against factored axial load. Given resistance factor (phi) = 0.6000; concrete strength (f_c) = 3.9000 ksi; steel reinforcement area (A_st) = 2.0000 in^2; steel yield strength (f_y) = 59.0000 ksi; design strength of tied column (P_n_phi) = 791.0 kip, determine the gross column area (A_g) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for A_g:

   $$A_{g} = \dfrac{P_n}{0.80\,\phi\,0.85 f_c'} + A_{st}\left(1-\dfrac{f_y}{0.85f_c'}\right)+A_{st}$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6000, concrete strength (f_c) = 3.9000 ksi, steel reinforcement area (A_st) = 2.0000 in^2, steel yield strength (f_y) = 59.0000 ksi, design strength of tied column (P_n_phi) = 791.0 kip.

4. Step 4 — Substitute the given values:

   $$A_{g} = \dfrac{P_n}{0.80\,0.6000\,0.85 f_c'} + 2.0000\left(1-\dfrac{f_y}{0.85f_c'}\right)+2.0000$$

5. Step 5 — Evaluate:

   A_{g} = 465.5\ \text{in^2}

6. Step 6 — Check: returning A_g = 465.5 in^2 to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

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

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

A tied column supporting a mezzanine is evaluated for design column strength, tied columns. Given resistance factor (phi) = 0.6200; concrete strength (f_c) = 3.6000 ksi; gross column area (A_g) = 200.0 in^2; steel reinforcement area (A_st) = 22.0000 in^2; steel yield strength (f_y) = 67.0000 ksi, determine the design strength of tied column (P_n_phi) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for P_n_phi:

   $$P_{nphi} = 0.80\,\phi\left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6200, concrete strength (f_c) = 3.6000 ksi, gross column area (A_g) = 200.0 in^2, steel reinforcement area (A_st) = 22.0000 in^2, steel yield strength (f_y) = 67.0000 ksi.

4. Step 4 — Substitute the given values:

   $$P_{nphi} = 0.80\,0.6200\left[0.85 f_c'(A_g - 22.0000) + f_y 22.0000\right]$$

5. Step 5 — Evaluate:

   $$P_{nphi} = 1001\ \text{kip}$$

6. Step 6 — Check: returning P_n_phi = 1,001 kip to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

• 2,003 — kept a factor of two that cancels in the correct rearrangement.
• 500.6 — dropped that same factor in the other direction.
• 1,101 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

A square tied column in a parking structure is sized using design column strength, tied columns provisions. Given resistance factor (phi) = 0.6400; concrete strength (f_c) = 3.7000 ksi; steel reinforcement area (A_st) = 3.5000 in^2; steel yield strength (f_y) = 43.0000 ksi; design strength of tied column (P_n_phi) = 553.0 kip, determine the gross column area (A_g) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for A_g:

   $$A_{g} = \dfrac{P_n}{0.80\,\phi\,0.85 f_c'} + A_{st}\left(1-\dfrac{f_y}{0.85f_c'}\right)+A_{st}$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6400, concrete strength (f_c) = 3.7000 ksi, steel reinforcement area (A_st) = 3.5000 in^2, steel yield strength (f_y) = 43.0000 ksi, design strength of tied column (P_n_phi) = 553.0 kip.

4. Step 4 — Substitute the given values:

   $$A_{g} = \dfrac{P_n}{0.80\,0.6400\,0.85 f_c'} + 3.5000\left(1-\dfrac{f_y}{0.85f_c'}\right)+3.5000$$

5. Step 5 — Evaluate:

   A_{g} = 302.6\ \text{in^2}

6. Step 6 — Check: returning A_g = 302.6 in^2 to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

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

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

An interior building column's design column strength, tied columns capacity is checked against factored axial load. Given resistance factor (phi) = 0.6200; concrete strength (f_c) = 4.2000 ksi; gross column area (A_g) = 205.0 in^2; steel reinforcement area (A_st) = 18.0000 in^2; steel yield strength (f_y) = 50.0000 ksi, determine the design strength of tied column (P_n_phi) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for P_n_phi:

   $$P_{nphi} = 0.80\,\phi\left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6200, concrete strength (f_c) = 4.2000 ksi, gross column area (A_g) = 205.0 in^2, steel reinforcement area (A_st) = 18.0000 in^2, steel yield strength (f_y) = 50.0000 ksi.

4. Step 4 — Substitute the given values:

   $$P_{nphi} = 0.80\,0.6200\left[0.85 f_c'(A_g - 18.0000) + f_y 18.0000\right]$$

5. Step 5 — Evaluate:

   $$P_{nphi} = 777.5\ \text{kip}$$

6. Step 6 — Check: returning P_n_phi = 777.5 kip to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

• 1,555 — kept a factor of two that cancels in the correct rearrangement.
• 388.8 — dropped that same factor in the other direction.
• 855.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

A tied column supporting a mezzanine is evaluated for design column strength, tied columns. Given resistance factor (phi) = 0.6500; concrete strength (f_c) = 5.5000 ksi; steel reinforcement area (A_st) = 21.0000 in^2; steel yield strength (f_y) = 53.0000 ksi; design strength of tied column (P_n_phi) = 328.0 kip, determine the gross column area (A_g) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for A_g:

   $$A_{g} = \dfrac{P_n}{0.80\,\phi\,0.85 f_c'} + A_{st}\left(1-\dfrac{f_y}{0.85f_c'}\right)+A_{st}$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6500, concrete strength (f_c) = 5.5000 ksi, steel reinforcement area (A_st) = 21.0000 in^2, steel yield strength (f_y) = 53.0000 ksi, design strength of tied column (P_n_phi) = 328.0 kip.

4. Step 4 — Substitute the given values:

   $$A_{g} = \dfrac{P_n}{0.80\,0.6500\,0.85 f_c'} + 21.0000\left(1-\dfrac{f_y}{0.85f_c'}\right)+21.0000$$

5. Step 5 — Evaluate:

   A_{g} = -61.1510\ \text{in^2}

6. Step 6 — Check: returning A_g = -61.1510 in^2 to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

• -122.3 — kept a factor of two that cancels in the correct rearrangement.
• -30.5755 — dropped that same factor in the other direction.
• -67.2661 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

A square tied column in a parking structure is sized using design column strength, tied columns provisions. Given resistance factor (phi) = 0.6100; concrete strength (f_c) = 4.9000 ksi; gross column area (A_g) = 300.0 in^2; steel reinforcement area (A_st) = 2.0000 in^2; steel yield strength (f_y) = 71.0000 ksi, determine the design strength of tied column (P_n_phi) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for P_n_phi:

   $$P_{nphi} = 0.80\,\phi\left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6100, concrete strength (f_c) = 4.9000 ksi, gross column area (A_g) = 300.0 in^2, steel reinforcement area (A_st) = 2.0000 in^2, steel yield strength (f_y) = 71.0000 ksi.

4. Step 4 — Substitute the given values:

   $$P_{nphi} = 0.80\,0.6100\left[0.85 f_c'(A_g - 2.0000) + f_y 2.0000\right]$$

5. Step 5 — Evaluate:

   $$P_{nphi} = 675.0\ \text{kip}$$

6. Step 6 — Check: returning P_n_phi = 675.0 kip to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

• 1,350 — kept a factor of two that cancels in the correct rearrangement.
• 337.5 — dropped that same factor in the other direction.
• 742.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

An interior building column's design column strength, tied columns capacity is checked against factored axial load. Given resistance factor (phi) = 0.6200; concrete strength (f_c) = 6.0000 ksi; steel reinforcement area (A_st) = 6.5000 in^2; steel yield strength (f_y) = 67.0000 ksi; design strength of tied column (P_n_phi) = 2,120 kip, determine the gross column area (A_g) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for A_g:

   $$A_{g} = \dfrac{P_n}{0.80\,\phi\,0.85 f_c'} + A_{st}\left(1-\dfrac{f_y}{0.85f_c'}\right)+A_{st}$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6200, concrete strength (f_c) = 6.0000 ksi, steel reinforcement area (A_st) = 6.5000 in^2, steel yield strength (f_y) = 67.0000 ksi, design strength of tied column (P_n_phi) = 2,120 kip.

4. Step 4 — Substitute the given values:

   $$A_{g} = \dfrac{P_n}{0.80\,0.6200\,0.85 f_c'} + 6.5000\left(1-\dfrac{f_y}{0.85f_c'}\right)+6.5000$$

5. Step 5 — Evaluate:

   A_{g} = 765.7\ \text{in^2}

6. Step 6 — Check: returning A_g = 765.7 in^2 to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

• 1,531 — kept a factor of two that cancels in the correct rearrangement.
• 382.8 — dropped that same factor in the other direction.
• 842.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

A tied column supporting a mezzanine is evaluated for design column strength, tied columns. Given resistance factor (phi) = 0.6000; concrete strength (f_c) = 3.8000 ksi; gross column area (A_g) = 555.0 in^2; steel reinforcement area (A_st) = 3.0000 in^2; steel yield strength (f_y) = 73.0000 ksi, determine the design strength of tied column (P_n_phi) in kip.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for P_n_phi:

   $$P_{nphi} = 0.80\,\phi\left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6000, concrete strength (f_c) = 3.8000 ksi, gross column area (A_g) = 555.0 in^2, steel reinforcement area (A_st) = 3.0000 in^2, steel yield strength (f_y) = 73.0000 ksi.

4. Step 4 — Substitute the given values:

   $$P_{nphi} = 0.80\,0.6000\left[0.85 f_c'(A_g - 3.0000) + f_y 3.0000\right]$$

5. Step 5 — Evaluate:

   $$P_{nphi} = 960.9\ \text{kip}$$

6. Step 6 — Check: returning P_n_phi = 960.9 kip to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

• 1,922 — kept a factor of two that cancels in the correct rearrangement.
• 480.5 — dropped that same factor in the other direction.
• 1,057 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns

A square tied column in a parking structure is sized using design column strength, tied columns provisions. Given resistance factor (phi) = 0.6500; concrete strength (f_c) = 3.8000 ksi; steel reinforcement area (A_st) = 22.5000 in^2; steel yield strength (f_y) = 43.0000 ksi; design strength of tied column (P_n_phi) = 2,773 kip, determine the gross column area (A_g) in in^2.

Start with the thinking

• The governing relation printed in this handbook section is Design strength of tied columns.
• 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.
• Design column strength, tied columns, combines the concrete and reinforcement contributions with a 0.80 reduction for accidental eccentricity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

2. Step 2 — Rearrange symbolically for A_g:

   $$A_{g} = \dfrac{P_n}{0.80\,\phi\,0.85 f_c'} + A_{st}\left(1-\dfrac{f_y}{0.85f_c'}\right)+A_{st}$$

3. Step 3 — List the givens: resistance factor (phi) = 0.6500, concrete strength (f_c) = 3.8000 ksi, steel reinforcement area (A_st) = 22.5000 in^2, steel yield strength (f_y) = 43.0000 ksi, design strength of tied column (P_n_phi) = 2,773 kip.

4. Step 4 — Substitute the given values:

   $$A_{g} = \dfrac{P_n}{0.80\,0.6500\,0.85 f_c'} + 22.5000\left(1-\dfrac{f_y}{0.85f_c'}\right)+22.5000$$

5. Step 5 — Evaluate:

   A_{g} = 1396\ \text{in^2}

6. Step 6 — Check: returning A_g = 1,396 in^2 to

   $$\phi P_n = 0.80 \phi \left[0.85 f_c'(A_g - A_{st}) + f_y A_{st}\right]$$

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

Why the other options are there

• 2,793 — kept a factor of two that cancels in the correct rearrangement.
• 698.2 — dropped that same factor in the other direction.
• 1,536 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Design Column Strength, Tied Columns