Unified Design Provisions

Example 1

Design axial strength of a tied reinforced concrete column — Unified Design Provisions

A 14 in × 14 in tied column uses 8-#8 bars (A_st = 6.32 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 196 in^{2}$
$A_st = 6.32 in^{2}$
$f'c = 5 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(5)(196 - 6.32) = 806.1 kips$
Steel term
$60(6.32) = 379.2 kips$
Substituting
$P_n = 0.80[806.1 + 379.2] = 948.3 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(948.3) = 616.4 kips
Steel ratio check
$\rho = 3.22% (must be 1% - 8%)$

Answer:

P_n = 948.3 kips, φP_n = 616.4 kips

Why the other options are there

1,212 kips (no 0.80 factor, gross area)
948.3 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 2

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (2)

A 14 in × 14 in tied column uses 6-#10 bars (A_st = 7.62 in²) with f′c = 6 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 196 in^{2}$
$A_st = 7.62 in^{2}$
$f'c = 6 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(6)(196 - 7.62) = 960.7 kips$
Steel term
$60(7.62) = 457.2 kips$
Substituting
$P_n = 0.80[960.7 + 457.2] = 1,134 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,134) = 737.3 kips
Steel ratio check
$\rho = 3.89% (must be 1% - 8%)$

Answer:

P_n = 1,134 kips, φP_n = 737.3 kips

Why the other options are there

1,457 kips (no 0.80 factor, gross area)
1,134 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 3

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (3)

A 20 in × 20 in tied column uses 8-#8 bars (A_st = 6.32 in²) with f′c = 6 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 400 in^{2}$
$A_st = 6.32 in^{2}$
$f'c = 6 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(6)(400 - 6.32) = 2,008 kips$
Steel term
$60(6.32) = 379.2 kips$
Substituting
$P_n = 0.80[2,008 + 379.2] = 1,910 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,910) = 1,241 kips
Steel ratio check
$\rho = 1.58% (must be 1% - 8%)$

Answer:

P_n = 1,910 kips, φP_n = 1,241 kips

Why the other options are there

2,419 kips (no 0.80 factor, gross area)
1,910 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 4

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (4)

A 18 in × 18 in tied column uses 4-#8 bars (A_st = 3.16 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 324 in^{2}$
$A_st = 3.16 in^{2}$
$f'c = 5 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(5)(324 - 3.16) = 1,364 kips$
Steel term
$60(3.16) = 189.6 kips$
Substituting
$P_n = 0.80[1,364 + 189.6] = 1,243 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,243) = 807.6 kips
Steel ratio check
$\rho = 0.98% (must be 1% - 8%)$

Answer:

P_n = 1,243 kips, φP_n = 807.6 kips

Why the other options are there

1,567 kips (no 0.80 factor, gross area)
1,243 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 5

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (5)

A 12 in × 12 in tied column uses 8-#10 bars (A_st = 10.16 in²) with f′c = 4 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 144 in^{2}$
$A_st = 10.16 in^{2}$
$f'c = 4 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(4)(144 - 10.16) = 455.1 kips$
Steel term
$60(10.16) = 609.6 kips$
Substituting
$P_n = 0.80[455.1 + 609.6] = 851.7 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(851.7) = 553.6 kips
Steel ratio check
$\rho = 7.06% (must be 1% - 8%)$

Answer:

P_n = 851.7 kips, φP_n = 553.6 kips

Why the other options are there

1,099 kips (no 0.80 factor, gross area)
851.7 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 6

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (6)

A 18 in × 18 in tied column uses 6-#9 bars (A_st = 6.00 in²) with f′c = 4 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 324 in^{2}$
$A_st = 6.00 in^{2}$
$f'c = 4 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(4)(324 - 6.00) = 1,081 kips$
Steel term
$60(6.00) = 360.0 kips$
Substituting
$P_n = 0.80[1,081 + 360.0] = 1,153 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,153) = 749.4 kips
Steel ratio check
$\rho = 1.85% (must be 1% - 8%)$

Answer:

P_n = 1,153 kips, φP_n = 749.4 kips

Why the other options are there

1,462 kips (no 0.80 factor, gross area)
1,153 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 7

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (7)

A 18 in × 18 in tied column uses 4-#9 bars (A_st = 4.00 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 324 in^{2}$
$A_st = 4.00 in^{2}$
$f'c = 5 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(5)(324 - 4.00) = 1,360 kips$
Steel term
$60(4.00) = 240.0 kips$
Substituting
$P_n = 0.80[1,360 + 240.0] = 1,280 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,280) = 832.0 kips
Steel ratio check
$\rho = 1.23% (must be 1% - 8%)$

Answer:

P_n = 1,280 kips, φP_n = 832.0 kips

Why the other options are there

1,617 kips (no 0.80 factor, gross area)
1,280 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 8

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (8)

A 16 in × 16 in tied column uses 4-#8 bars (A_st = 3.16 in²) with f′c = 6 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 256 in^{2}$
$A_st = 3.16 in^{2}$
$f'c = 6 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(6)(256 - 3.16) = 1,289 kips$
Steel term
$60(3.16) = 189.6 kips$
Substituting
$P_n = 0.80[1,289 + 189.6] = 1,183 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,183) = 769.1 kips
Steel ratio check
$\rho = 1.23% (must be 1% - 8%)$

Answer:

P_n = 1,183 kips, φP_n = 769.1 kips

Why the other options are there

1,495 kips (no 0.80 factor, gross area)
1,183 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 9

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (9)

A 16 in × 16 in tied column uses 4-#9 bars (A_st = 4.00 in²) with f′c = 6 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 256 in^{2}$
$A_st = 4.00 in^{2}$
$f'c = 6 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(6)(256 - 4.00) = 1,285 kips$
Steel term
$60(4.00) = 240.0 kips$
Substituting
$P_n = 0.80[1,285 + 240.0] = 1,220 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,220) = 793.1 kips
Steel ratio check
$\rho = 1.56% (must be 1% - 8%)$

Answer:

P_n = 1,220 kips, φP_n = 793.1 kips

Why the other options are there

1,546 kips (no 0.80 factor, gross area)
1,220 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 10

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (10)

A 14 in × 14 in tied column uses 6-#10 bars (A_st = 7.62 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

$Ag = 196 in^{2}$
$A_st = 7.62 in^{2}$
$f'c = 5 ksi$
$f_y = 60 ksi$
$\phi = 0.65, \alpha = 0.80 (tied)$

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_n = 0.80[0.85 f'c (A_g - A_st) + f_y A_st]$
Concrete term
$0.85(5)(196 - 7.62) = 800.6 kips$
Steel term
$60(7.62) = 457.2 kips$
Substituting
$P_n = 0.80[800.6 + 457.2] = 1,006 kips$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,006) = 654.1 kips
Steel ratio check
$\rho = 3.89% (must be 1% - 8%)$

Answer:

P_n = 1,006 kips, φP_n = 654.1 kips

Why the other options are there

1,290 kips (no 0.80 factor, gross area)
1,006 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Core formulas for this FE topic

Worked exam-style examples