Singly-Reinforced Beams

Structural Design · FE Reference Handbook section

Structural Design

2 formulas

10 exam-style examples

~49 min

All Structural Design lectures

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 — Singly-Reinforced Beams

A beam has b = 15 in., d = 23.0 in., As = 2.50 in², f′c = 5 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 15 in$
$d = 23.0 in$
$As = 2.50 in^{2}$
$f'c = 5 ksi$
$fy = 60 ksi$

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 1 — schematic for Nominal moment of a singly reinforced beam — Singly-Reinforced Beams

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 × 5 × 15)
Evaluate
$a = 2.353 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.50(60)(23.0 − 1.176) = 3,274 kip·in
Convert and factor — Mn = 272.8 kip·ft, φMn = 0.90(272.8) = 245.5 kip·ft

Answer:

φMn ≈ 245.5 kip·ft

Why the other options are there

272.8 kip·ft (φ not applied)
287.5 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 2

Available moment φM_n of a compact steel beam — Singly-Reinforced Beams

A compact, fully braced A992 beam has Z_x = 110 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 18 ft simple span.

Given

$Z_x = 110 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 18 ft$

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_p = F_y Z_x$
Substituting — M_p = 50(110) = 5,500 kip·in = 458.3 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(458.3) = 412.5 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_u = 8(412.5)/(18)^{2} = 10.185 kip/ft$

Answer:

φM_n = 412.5 kip·ft, permitting w_u = 10.19 kip/ft over 18 ft

Why the other options are there

5,500 kip·ft (inches never converted)
458.3 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 3

Nominal moment of a singly reinforced beam — Singly-Reinforced Beams (2)

A beam has b = 14 in., d = 25.5 in., As = 1.75 in², f′c = 3 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 14 in$
$d = 25.5 in$
$As = 1.75 in^{2}$
$f'c = 3 ksi$
$fy = 60 ksi$

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 — Singly-Reinforced Beams (2)

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 × 3 × 14)
Evaluate
$a = 2.941 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 1.75(60)(25.5 − 1.471) = 2,523 kip·in
Convert and factor — Mn = 210.3 kip·ft, φMn = 0.90(210.3) = 189.2 kip·ft

Answer:

φMn ≈ 189.2 kip·ft

Why the other options are there

210.3 kip·ft (φ not applied)
223.1 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 4

Available moment φM_n of a compact steel beam — Singly-Reinforced Beams (2)

A compact, fully braced A992 beam has Z_x = 54 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 29 ft simple span.

Given

$Z_x = 54 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 29 ft$

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_p = F_y Z_x$
Substituting — M_p = 50(54) = 2,700 kip·in = 225.0 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(225.0) = 202.5 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_u = 8(202.5)/(29)^{2} = 1.926 kip/ft$

Answer:

φM_n = 202.5 kip·ft, permitting w_u = 1.93 kip/ft over 29 ft

Why the other options are there

2,700 kip·ft (inches never converted)
225.0 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 5

Nominal moment of a singly reinforced beam — Singly-Reinforced Beams (3)

A beam has b = 12 in., d = 22.5 in., As = 4.00 in², f′c = 3 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 12 in$
$d = 22.5 in$
$As = 4.00 in^{2}$
$f'c = 3 ksi$
$fy = 60 ksi$

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 — Singly-Reinforced Beams (3)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 4.00(60)/(0.85 × 3 × 12)
Evaluate
$a = 7.843 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 4.00(60)(22.5 − 3.922) = 4,459 kip·in
Convert and factor — Mn = 371.6 kip·ft, φMn = 0.90(371.6) = 334.4 kip·ft

Answer:

φMn ≈ 334.4 kip·ft

Why the other options are there

371.6 kip·ft (φ not applied)
450.0 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 6

Available moment φM_n of a compact steel beam — Singly-Reinforced Beams (3)

A compact, fully braced A992 beam has Z_x = 146 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 28 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 28 ft$

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_p = F_y Z_x$
Substituting — M_p = 50(146) = 7,300 kip·in = 608.3 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(608.3) = 547.5 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_u = 8(547.5)/(28)^{2} = 5.587 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 5.59 kip/ft over 28 ft

Why the other options are there

7,300 kip·ft (inches never converted)
608.3 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 7

Nominal moment of a singly reinforced beam — Singly-Reinforced Beams (4)

A beam has b = 17 in., d = 15.5 in., As = 3.50 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 17 in$
$d = 15.5 in$
$As = 3.50 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

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 7 — schematic for Nominal moment of a singly reinforced beam — Singly-Reinforced Beams (4)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 3.50(60)/(0.85 × 4 × 17)
Evaluate
$a = 3.633 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 3.50(60)(15.5 − 1.817) = 2,874 kip·in
Convert and factor — Mn = 239.5 kip·ft, φMn = 0.90(239.5) = 215.5 kip·ft

Answer:

φMn ≈ 215.5 kip·ft

Why the other options are there

239.5 kip·ft (φ not applied)
271.3 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 8

Available moment φM_n of a compact steel beam — Singly-Reinforced Beams (4)

A compact, fully braced A992 beam has Z_x = 77 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 29 ft simple span.

Given

$Z_x = 77 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 29 ft$

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_p = F_y Z_x$
Substituting — M_p = 50(77) = 3,850 kip·in = 320.8 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(320.8) = 288.8 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_u = 8(288.8)/(29)^{2} = 2.747 kip/ft$

Answer:

φM_n = 288.8 kip·ft, permitting w_u = 2.75 kip/ft over 29 ft

Why the other options are there

3,850 kip·ft (inches never converted)
320.8 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 9

Nominal moment of a singly reinforced beam — Singly-Reinforced Beams (5)

A beam has b = 17 in., d = 19.5 in., As = 3.50 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 17 in$
$d = 19.5 in$
$As = 3.50 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

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 9 — schematic for Nominal moment of a singly reinforced beam — Singly-Reinforced Beams (5)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 3.50(60)/(0.85 × 4 × 17)
Evaluate
$a = 3.633 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 3.50(60)(19.5 − 1.817) = 3,714 kip·in
Convert and factor — Mn = 309.5 kip·ft, φMn = 0.90(309.5) = 278.5 kip·ft

Answer:

φMn ≈ 278.5 kip·ft

Why the other options are there

309.5 kip·ft (φ not applied)
341.3 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams

Example 10

Available moment φM_n of a compact steel beam — Singly-Reinforced Beams (5)

A compact, fully braced A992 beam has Z_x = 146 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 36 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 36 ft$

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_p = F_y Z_x$
Substituting — M_p = 50(146) = 7,300 kip·in = 608.3 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(608.3) = 547.5 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_u = 8(547.5)/(36)^{2} = 3.380 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 3.38 kip/ft over 36 ft

Why the other options are there

7,300 kip·ft (inches never converted)
608.3 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Singly-Reinforced Beams