SME
Steel Design
Beams
Use this for “A braced W18×50 of 50 ksi steel spans 25 ft under wᵤ = 3.6 kip/ft. Is φMₚ enough, and how close must the bracing be?”
Diagram
ΣF = 0: RA + RB = 45 + 45 = 90 kip, the total load.
|M| is largest at x = 150 in: 3,380 kip·in.
The area under V between two points is the change in M.
This is an example. Type your own values to start a new problem.
φMₚ (kip·ft)
=
0.9
×
Fᵧ (ksi)
×
Zₓ (in³)
Lₚ (ft)
=
1.76
×
rᵧ (in)
×
29,000
ksi
÷
Fᵧ (ksi)
Mᵤ (kip·ft)
=
wᵤ (kip/ft)
×
(
L (ft)
)
2
÷
8
shape
W-shape
example
Mᵤ ≤ φMₚ?
Strong enough?
worked out (not typed)
Formulas
Zₓ = 101
101 = 101
rᵧ = 1.65
1.65 = 1.65
φMₚ = 0.9 × Fᵧ × Zₓ
4,545 = 0.9 × 50 × 101
Lₚ
=
1.76
×
rᵧ
×
√(
29,000
Fᵧ
)
69.94
=
1.76
×
1.65
×
√(
29,000
50
)
Mᵤ
=
wᵤ
×
L
2
8
3,375
=
0.3
×
300
2
8
Mᵤ ≤ φMₚ? = adequate when Mᵤ ≤ φMₚ
3,375 ≤ 4,545, so adequate
Formulas are worked in kip·in, in, kip/in. The step-by-step shows the conversions.
Assumptions
•
AISC 360-22 (LRFD): a compact section, braced at Lb ≤ Lₚ, reaches its plastic moment, φ = 0.90.
•
Past Lₚ lateral–torsional buckling lowers φMₙ below φMₚ: not this page.
•
Simply supported under a uniform factored load wᵤ; E = 29,000 ksi.
•
Zₓ and rᵧ are read from the W-shape table; pick “other (typed)” to type another shape’s.
Step-by-step
WE KNOW
shape: W18×50, Fᵧ = 50 ksi, wᵤ = 3.6 kip/ft, L = 25 ft
FIND
plastic section modulus (Zₓ), radius of gyration (weak axis) (rᵧ), design plastic moment (φMₚ), longest unbraced length for φMₚ (Lₚ), factored moment (Mᵤ), strong enough? (Mᵤ ≤ φMₚ?)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to kip·in, in, kip/in).
wᵤ = 3.6 kip/ft = 0.3 kip/in (1 kip/in = 12 kip/ft)
L = 25 ft = 300 in (1 ft = 12 in)
Step 2 · Find plastic section modulus (Zₓ)
Use Zₓ = 101
Read it from the W-shape table, for a W18×50.
W18×50 chosen: its row in the W-shape table
Zₓ = 101 in³
Step 3 · Find radius of gyration (weak axis): rᵧ
Use rᵧ = 1.65
Read it from the W-shape table, for a W18×50.
rᵧ = 1.65 in
Step 4 · Find design plastic moment (φMₚ)
Use φMₚ = 0.9 × Fᵧ × Zₓ
Every fiber at Fᵧ gives FᵧZₓ, in kip·in; φ = 0.90.
φMₚ = 0.9 × 50 × 101
φMₚ = 45 × 101
φMₚ = 4,545 kip·in
Step 5 · Find longest unbraced length for φMₚ (Lₚ)
Use Lₚ = 1.76 × rᵧ × √(29,000 ÷ Fᵧ)
Braced closer than Lₚ, the compression flange can’t twist sideways before the section yields.
Lₚ
=
1.76
×
1.65
×
√(
29,000
50
)
Lₚ
=
1.76
×
1.65
×
√
580
Lₚ = 1.76 × 1.65 × 24.08
Lₚ = 2.904 × 24.08
Lₚ = 69.9376 in
Step 6 · Find factored moment (Mᵤ)
Use Mᵤ = wᵤ × L² ÷ 8
A
simple
span
under
a
uniform
load
bends
most
at
midspan:
wL²
8
.
Mᵤ
=
0.3
×
300
2
8
Mᵤ
=
0.3
×
90,000
8
Mᵤ
=
27,000
8
Mᵤ = 3,375 kip·in
Step 7 · Find strong enough? (Mᵤ ≤ φMₚ?)
Use Mᵤ ≤ φMₚ? = adequate when Mᵤ ≤ φMₚ
The design strength covers the factored moment: LRFD asks φMₙ ≥ Mᵤ.
3,375 ≤ 4,545: adequate
Mᵤ ≤ φMₚ? = adequate
Step 8 · Convert the answers
Change each answer back to the unit you chose.
φMₚ = 4,545 kip·in = 378.8 kip·ft (1 kip·ft = 12 kip·in)
Lₚ = 69.94 in = 5.828 ft (1 ft = 12 in)
Mᵤ = 3,375 kip·in = 281.3 kip·ft (1 kip·ft = 12 kip·in)
Check
Put the numbers back in. Both sides should match.
101 = 101 ✓
1.65 = 1.65 ✓
4,545 = 0.9 × 50 × 101 ✓
69.94
=
1.76
×
1.65
×
√(
29,000
50
)
✓
3,375
=
0.3
×
300
2
8
✓
3,375 ≤ 4,545, so adequate ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.