Beams—Flexure

Structural Design · FE Reference Handbook section

Structural Design

3 formulas

10 exam-style examples

~51 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 — Beams—Flexure

A beam has b = 18 in., d = 23.5 in., As = 2.25 in², f′c = 4 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 18 in$
$d = 23.5 in$
$As = 2.25 in^{2}$
$f'c = 4 ksi$
$fy = 60 ksi$

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 1 — schematic for Nominal moment of a singly reinforced beam — Beams—Flexure

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 2.25(60)/(0.85 × 4 × 18)
Evaluate
$a = 2.206 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.25(60)(23.5 − 1.103) = 3,024 kip·in
Convert and factor — Mn = 252.0 kip·ft, φMn = 0.90(252.0) = 226.8 kip·ft

Answer:

φMn ≈ 226.8 kip·ft

Why the other options are there

252.0 kip·ft (φ not applied)
264.4 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Beams—Flexure

Example 2

Available moment φM_n of a compact steel beam — Beams—Flexure

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 38 ft simple span.

Given

$Z_x = 54 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 38 ft$

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
$w_u = 8(202.5)/(38)^{2} = 1.122 kip/ft$

Answer:

φM_n = 202.5 kip·ft, permitting w_u = 1.12 kip/ft over 38 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 → Beams—Flexure

Example 3

Nominal moment of a singly reinforced beam — Beams—Flexure (2)

A beam has b = 14 in., d = 24.5 in., As = 2.75 in², f′c = 3 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 14 in$
$d = 24.5 in$
$As = 2.75 in^{2}$
$f'c = 3 ksi$
$fy = 60 ksi$

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 — Beams—Flexure (2)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 2.75(60)/(0.85 × 3 × 14)
Evaluate
$a = 4.622 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.75(60)(24.5 − 2.311) = 3,661 kip·in
Convert and factor — Mn = 305.1 kip·ft, φMn = 0.90(305.1) = 274.6 kip·ft

Answer:

φMn ≈ 274.6 kip·ft

Why the other options are there

305.1 kip·ft (φ not applied)
336.9 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Beams—Flexure

Example 4

Available moment φM_n of a compact steel beam — Beams—Flexure (2)

A compact, fully braced A992 beam has Z_x = 234 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 22 ft simple span.

Given

$Z_x = 234 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 22 ft$

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(234) = 11,700 kip·in = 975.0 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(975.0) = 877.5 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_u = 8(877.5)/(22)^{2} = 14.504 kip/ft$

Answer:

φM_n = 877.5 kip·ft, permitting w_u = 14.50 kip/ft over 22 ft

Why the other options are there

11,700 kip·ft (inches never converted)
975.0 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Beams—Flexure

Example 5

Nominal moment of a singly reinforced beam — Beams—Flexure (3)

A beam has b = 13 in., d = 16.0 in., As = 2.25 in², f′c = 5 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 13 in$
$d = 16.0 in$
$As = 2.25 in^{2}$
$f'c = 5 ksi$
$fy = 60 ksi$

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 5 — schematic for Nominal moment of a singly reinforced beam — Beams—Flexure (3)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 2.25(60)/(0.85 × 5 × 13)
Evaluate
$a = 2.443 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.25(60)(16.0 − 1.222) = 1,995 kip·in
Convert and factor — Mn = 166.3 kip·ft, φMn = 0.90(166.3) = 149.6 kip·ft

Answer:

φMn ≈ 149.6 kip·ft

Why the other options are there

166.3 kip·ft (φ not applied)
180.0 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Beams—Flexure

Example 6

Available moment φM_n of a compact steel beam — Beams—Flexure (3)

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 39 ft simple span.

Given

$Z_x = 110 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 39 ft$

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
$w_u = 8(412.5)/(39)^{2} = 2.170 kip/ft$

Answer:

φM_n = 412.5 kip·ft, permitting w_u = 2.17 kip/ft over 39 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 → Beams—Flexure

Example 7

Nominal moment of a singly reinforced beam — Beams—Flexure (4)

A beam has b = 12 in., d = 19.5 in., As = 2.50 in², f′c = 5 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 12 in$
$d = 19.5 in$
$As = 2.50 in^{2}$
$f'c = 5 ksi$
$fy = 60 ksi$

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 7 — schematic for Nominal moment of a singly reinforced beam — Beams—Flexure (4)

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 × 12)
Evaluate
$a = 2.941 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.50(60)(19.5 − 1.471) = 2,704 kip·in
Convert and factor — Mn = 225.4 kip·ft, φMn = 0.90(225.4) = 202.8 kip·ft

Answer:

φMn ≈ 202.8 kip·ft

Why the other options are there

225.4 kip·ft (φ not applied)
243.8 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Beams—Flexure

Example 8

Available moment φM_n of a compact steel beam — Beams—Flexure (4)

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 19 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 19 ft$

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
$w_u = 8(547.5)/(19)^{2} = 12.133 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 12.13 kip/ft over 19 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 → Beams—Flexure

Example 9

Nominal moment of a singly reinforced beam — Beams—Flexure (5)

A beam has b = 14 in., d = 21.5 in., As = 2.25 in², f′c = 5 ksi, fy = 60 ksi. Find φMn (φ = 0.90).

Given

$b = 14 in$
$d = 21.5 in$
$As = 2.25 in^{2}$
$f'c = 5 ksi$
$fy = 60 ksi$

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 9 — schematic for Nominal moment of a singly reinforced beam — Beams—Flexure (5)

Step-by-step solution

Equilibrium — As·fy = 0.85f′c·a·b
Solve for a — a = As·fy/(0.85f′c·b) = 2.25(60)/(0.85 × 5 × 14)
Evaluate
$a = 2.269 in$
Nominal moment — Mn = As·fy(d − a/2)
Substituting — Mn = 2.25(60)(21.5 − 1.134) = 2,749 kip·in
Convert and factor — Mn = 229.1 kip·ft, φMn = 0.90(229.1) = 206.2 kip·ft

Answer:

φMn ≈ 206.2 kip·ft

Why the other options are there

229.1 kip·ft (φ not applied)
241.9 kip·ft (lever arm taken as d)

Reference: FE Reference Handbook — Structural Design → Beams—Flexure

Example 10

Available moment φM_n of a compact steel beam — Beams—Flexure (5)

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 31 ft simple span.

Given

$Z_x = 110 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 31 ft$

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
$w_u = 8(412.5)/(31)^{2} = 3.434 kip/ft$

Answer:

φM_n = 412.5 kip·ft, permitting w_u = 3.43 kip/ft over 31 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 → Beams—Flexure

Beams—Flexure

Core formulas for this FE topic

φMn ≥ MuNeeds technical review

Nettensilestrain:a=β1cNet tensile strain: a = \beta_{1} cNettensilestrain:a=β1​c

ft=c=aft = c = aft=c=a

Worked exam-style examples