Beams

Structural Design · FE Reference Handbook section

Structural Design

1 formulas

10 exam-style examples

~47 min

All Structural Design lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

For doubly symmetric compact I-shaped members bent about their major axis, the design flexural strength φbMn is determined

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

Available moment φM_n of a compact steel beam — Beams

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 38 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 38 ft$

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(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
$w_u = 8(547.5)/(38)^{2} = 3.033 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 3.03 kip/ft over 38 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

Example 2

Available moment φM_n of a compact steel beam — 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 26 ft simple span.

Given

$Z_x = 54 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 26 ft$

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
$w_u = 8(202.5)/(26)^{2} = 2.396 kip/ft$

Answer:

φM_n = 202.5 kip·ft, permitting w_u = 2.40 kip/ft over 26 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

Example 3

Available moment φM_n of a compact steel beam — Beams (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

Example 4

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

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 29 ft simple span.

Given

$Z_x = 234 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 29 ft$

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(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
$w_u = 8(877.5)/(29)^{2} = 8.347 kip/ft$

Answer:

φM_n = 877.5 kip·ft, permitting w_u = 8.35 kip/ft over 29 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

Example 5

Available moment φM_n of a compact steel beam — Beams (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 24 ft simple span.

Given

$Z_x = 110 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 24 ft$

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
$w_u = 8(412.5)/(24)^{2} = 5.729 kip/ft$

Answer:

φM_n = 412.5 kip·ft, permitting w_u = 5.73 kip/ft over 24 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

Example 6

Available moment φM_n of a compact steel beam — Beams (6)

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 22 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 22 ft$

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(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
$w_u = 8(547.5)/(22)^{2} = 9.050 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 9.05 kip/ft over 22 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

Example 7

Available moment φM_n of a compact steel beam — Beams (7)

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 40 ft simple span.

Given

$Z_x = 77 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 40 ft$

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
$w_u = 8(288.8)/(40)^{2} = 1.444 kip/ft$

Answer:

φM_n = 288.8 kip·ft, permitting w_u = 1.44 kip/ft over 40 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 → Beams

Example 8

Available moment φM_n of a compact steel beam — Beams (8)

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 40 ft simple span.

Given

$Z_x = 234 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 40 ft$

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
$w_u = 8(877.5)/(40)^{2} = 4.388 kip/ft$

Answer:

φM_n = 877.5 kip·ft, permitting w_u = 4.39 kip/ft over 40 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

Example 9

Available moment φM_n of a compact steel beam — Beams (9)

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 20 ft simple span.

Given

$Z_x = 110 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 20 ft$

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
$w_u = 8(412.5)/(20)^{2} = 8.250 kip/ft$

Answer:

φM_n = 412.5 kip·ft, permitting w_u = 8.25 kip/ft over 20 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

Example 10

Available moment φM_n of a compact steel beam — Beams (10)

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 38 ft simple span.

Given

$Z_x = 77 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 38 ft$

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(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
$w_u = 8(288.8)/(38)^{2} = 1.600 kip/ft$

Answer:

φM_n = 288.8 kip·ft, permitting w_u = 1.60 kip/ft over 38 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 → Beams

Beams

Handbook notes for this section

Core formulas for this FE topic

with φb = 0.90 and Ω = 1.67 as follows:Needs technical review

Worked exam-style examples