Limits for Longitudinal Reinforcements

Structural Design · FE Reference Handbook section

Structural Design

1 formulas

10 exam-style examples

~47 min

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Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements

A beam has b = 16 in., d = 23.0 in., As = 2.00 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 16 in$
$d = 23.0 in$
$As = 2.00 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 1 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 2.00(60)/(0.85 × 4 × 16)
Evaluate
$a = 2.206 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.00(60)(23.0 − 1.103) = 2,628 kip·in
Convert and factor — Mn = 219.0 kip·ft, φMn = 0.90(219.0) = 197.1 kip·ft

Answer:

φMn ≈ 197.1 kip·ft

Why the other options are there

219.0 kip·ft (φ not applied)
230.0 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 2

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (2)

A beam has b = 14 in., d = 17.0 in., As = 3.25 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 14 in$
$d = 17.0 in$
$As = 3.25 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 2 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (2)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 3.25(60)/(0.85 × 4 × 14)
Evaluate
$a = 4.097 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 3.25(60)(17.0 − 2.048) = 2,916 kip·in
Convert and factor — Mn = 243.0 kip·ft, φMn = 0.90(243.0) = 218.7 kip·ft

Answer:

φMn ≈ 218.7 kip·ft

Why the other options are there

243.0 kip·ft (φ not applied)
276.3 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 3

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (3)

A beam has b = 12 in., d = 24.5 in., As = 1.75 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 12 in$
$d = 24.5 in$
$As = 1.75 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 3 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (3)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 1.75(60)/(0.85 × 4 × 12)
Evaluate
$a = 2.574 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 1.75(60)(24.5 − 1.287) = 2,437 kip·in
Convert and factor — Mn = 203.1 kip·ft, φMn = 0.90(203.1) = 182.8 kip·ft

Answer:

φMn ≈ 182.8 kip·ft

Why the other options are there

203.1 kip·ft (φ not applied)
214.4 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 4

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (4)

A beam has b = 11 in., d = 19.5 in., As = 1.75 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 11 in$
$d = 19.5 in$
$As = 1.75 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 4 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (4)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 1.75(60)/(0.85 × 4 × 11)
Evaluate
$a = 2.807 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 1.75(60)(19.5 − 1.404) = 1,900 kip·in
Convert and factor — Mn = 158.3 kip·ft, φMn = 0.90(158.3) = 142.5 kip·ft

Answer:

φMn ≈ 142.5 kip·ft

Why the other options are there

158.3 kip·ft (φ not applied)
170.6 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 5

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (5)

A beam has b = 11 in., d = 17.0 in., As = 3.00 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 11 in$
$d = 17.0 in$
$As = 3.00 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 5 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (5)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 3.00(60)/(0.85 × 4 × 11)
Evaluate
$a = 4.813 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 3.00(60)(17.0 − 2.406) = 2,627 kip·in
Convert and factor — Mn = 218.9 kip·ft, φMn = 0.90(218.9) = 197.0 kip·ft

Answer:

φMn ≈ 197.0 kip·ft

Why the other options are there

218.9 kip·ft (φ not applied)
255.0 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 6

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (6)

A beam has b = 10 in., d = 17.5 in., As = 1.50 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 10 in$
$d = 17.5 in$
$As = 1.50 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 6 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (6)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 1.50(60)/(0.85 × 4 × 10)
Evaluate
$a = 2.647 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 1.50(60)(17.5 − 1.324) = 1,456 kip·in
Convert and factor — Mn = 121.3 kip·ft, φMn = 0.90(121.3) = 109.2 kip·ft

Answer:

φMn ≈ 109.2 kip·ft

Why the other options are there

121.3 kip·ft (φ not applied)
131.3 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 7

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (7)

A beam has b = 15 in., d = 25.0 in., As = 3.25 in², f′c = 5 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 15 in$
$d = 25.0 in$
$As = 3.25 in^{2}$
$f'c = 5 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 7 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (7)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 3.25(60)/(0.85 × 5 × 15)
Evaluate
$a = 3.059 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 3.25(60)(25.0 − 1.529) = 4,577 kip·in
Convert and factor — Mn = 381.4 kip·ft, φMn = 0.90(381.4) = 343.3 kip·ft

Answer:

φMn ≈ 343.3 kip·ft

Why the other options are there

381.4 kip·ft (φ not applied)
406.3 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 8

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (8)

A beam has b = 16 in., d = 21.5 in., As = 2.50 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 16 in$
$d = 21.5 in$
$As = 2.50 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 8 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (8)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 2.50(60)/(0.85 × 4 × 16)
Evaluate
$a = 2.757 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.50(60)(21.5 − 1.379) = 3,018 kip·in
Convert and factor — Mn = 251.5 kip·ft, φMn = 0.90(251.5) = 226.4 kip·ft

Answer:

φMn ≈ 226.4 kip·ft

Why the other options are there

251.5 kip·ft (φ not applied)
268.8 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 9

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (9)

A beam has b = 18 in., d = 16.0 in., As = 4.25 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 18 in$
$d = 16.0 in$
$As = 4.25 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 9 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (9)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 4.25(60)/(0.85 × 4 × 18)
Evaluate
$a = 4.167 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 4.25(60)(16.0 − 2.083) = 3,549 kip·in
Convert and factor — Mn = 295.7 kip·ft, φMn = 0.90(295.7) = 266.2 kip·ft

Answer:

φMn ≈ 266.2 kip·ft

Why the other options are there

295.7 kip·ft (φ not applied)
340.0 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements

Example 10

Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (10)

A beam has b = 14 in., d = 20.5 in., As = 2.50 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 14 in$
$d = 20.5 in$
$As = 2.50 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

Find

φMn

Start with the thinking

Assume the steel yields, then confirm with the depth of the stress block.
Whitney block: C = 0.85f′c·a·b.

Figure 10 — schematic for Nominal moment of a singly reinforced beam — Limits for Longitudinal Reinforcements (10)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 2.50(60)/(0.85 × 4 × 14)
Evaluate
$a = 3.151 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.50(60)(20.5 − 1.576) = 2,839 kip·in
Convert and factor — Mn = 236.6 kip·ft, φMn = 0.90(236.6) = 212.9 kip·ft

Answer:

φMn ≈ 212.9 kip·ft

Why the other options are there

236.6 kip·ft (φ not applied)
256.3 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Limits for Longitudinal Reinforcements