Available Moment φ Mn (foot-kips)

Structural Design · FE Reference Handbook section

Structural Design

3 formulas

10 exam-style examples

~51 min

All Structural Design lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

AISC APPROXIMATE VALUES OF EFFECTIVE LENGTH FACTOR, K
END CONDITION CODE ROTATION FIXED AND TRANSLATION FIXED
ROTATION FREE AND TRANSLATION FIXED
ROTATION FIXED AND TRANSLATION FREE
ROTATION FREE AND TRANSLATION FREE
FOR COLUMN ENDS SUPPORTED BY, BUT NOT RIGIDLY CONNECTED TO, A FOOTING OR FOUNDATION, G IS

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 — Available Moment φ Mn (foot-kips)

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 → Available Moment φ Mn (foot-kips)

Example 2

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (2)

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 27 ft simple span.

Given

$Z_x = 110 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 27 ft$

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
$w_u = 8(412.5)/(27)^{2} = 4.527 kip/ft$

Answer:

φM_n = 412.5 kip·ft, permitting w_u = 4.53 kip/ft over 27 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 → Available Moment φ Mn (foot-kips)

Example 3

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (3)

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 31 ft simple span.

Given

$Z_x = 54 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 31 ft$

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(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
$w_u = 8(202.5)/(31)^{2} = 1.686 kip/ft$

Answer:

φM_n = 202.5 kip·ft, permitting w_u = 1.69 kip/ft over 31 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 → Available Moment φ Mn (foot-kips)

Example 4

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (4)

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 24 ft simple span.

Given

$Z_x = 54 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 24 ft$

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(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
$w_u = 8(202.5)/(24)^{2} = 2.813 kip/ft$

Answer:

φM_n = 202.5 kip·ft, permitting w_u = 2.81 kip/ft over 24 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 → Available Moment φ Mn (foot-kips)

Example 5

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (5)

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 36 ft simple span.

Given

$Z_x = 77 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 36 ft$

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(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
$w_u = 8(288.8)/(36)^{2} = 1.782 kip/ft$

Answer:

φM_n = 288.8 kip·ft, permitting w_u = 1.78 kip/ft over 36 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 → Available Moment φ Mn (foot-kips)

Example 6

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (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 27 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 27 ft$

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
$w_u = 8(547.5)/(27)^{2} = 6.008 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 6.01 kip/ft over 27 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 → Available Moment φ Mn (foot-kips)

Example 7

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (7)

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 33 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 33 ft$

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
$w_u = 8(547.5)/(33)^{2} = 4.022 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 4.02 kip/ft over 33 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 → Available Moment φ Mn (foot-kips)

Example 8

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (8)

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 23 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 23 ft$

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
$w_u = 8(547.5)/(23)^{2} = 8.280 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 8.28 kip/ft over 23 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 → Available Moment φ Mn (foot-kips)

Example 9

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (9)

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 35 ft simple span.

Given

$Z_x = 146 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 35 ft$

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
$w_u = 8(547.5)/(35)^{2} = 3.576 kip/ft$

Answer:

φM_n = 547.5 kip·ft, permitting w_u = 3.58 kip/ft over 35 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 → Available Moment φ Mn (foot-kips)

Example 10

Available moment φM_n of a compact steel beam — Available Moment φ Mn (foot-kips) (10)

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 33 ft simple span.

Given

$Z_x = 54 in^{3}$
$F_y = 50 ksi$
$\phi_b = 0.90$
$L = 33 ft$

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
$w_u = 8(202.5)/(33)^{2} = 1.488 kip/ft$

Answer:

φM_n = 202.5 kip·ft, permitting w_u = 1.49 kip/ft over 33 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 → Available Moment φ Mn (foot-kips)