G-02 — steel beam design

pick the shape. check the math.

beam design tool

choose a W-shape. see if it survives.

Pick a section, set the demands, and watch the flexure and shear checks update in real time.

material
Fy
section
shape
Zx54.0
d13.7
tw0.255
demands
Mu180
Vu40
—
flexure utilization (Mu / φbMn)
φbMn = — kip-ft  |  φvVn = — kips
W-shape cross-section
utilization — flexure & shear
live equation
set values above to see the live calculation
Mu = Maximum bending demand from load combinations
φMn = φFyZx for compact, fully braced beams
Zx = Section property controlling flexural capacity
Vu = Maximum shear demand from load combinations
Aw = Area of the web resisting shear: d × tw
explained
A steel beam must pass two checks: flexure (bending) and shear. For flexure, the beam must have enough plastic section modulus Zx so that φMp = φFyZx exceeds the factored moment demand. For shear, the web area must resist the factored shear. Both must pass — the beam is only as strong as its weakest check.
key concepts
overviewFlexure and shear checks for compact W-shapes

A steel beam must satisfy both flexure (φbMn ≥ Mu) and shear (φvVn ≥ Vu). For compact sections with full lateral bracing, Mn = Mp = Fy·Zx. Shear capacity for most W-shapes: Vn = 0.6·Fy·Aw where Aw = d·tw. The beam passes when BOTH checks are satisfied.

the two checks every beam needsWhy both flexure and shear must pass independently

Every steel beam must pass both flexure (φbMn ≥ Mu) and shear (φvVn ≥ Vu). Flexure usually governs for longer spans, where bending moment builds up over distance. Shear governs near supports or for short, heavily loaded beams where the internal vertical forces are large relative to the moment. A beam that passes one check can still fail the other — both must be satisfied independently.

what makes a section “compact”?Width-to-thickness ratios and full plastic moment

AISC classifies sections by their flange and web width-to-thickness ratios. A compact section can develop the full plastic moment Mp = Fy × Zx before any local buckling occurs. Most standard W-shapes are compact for Fy = 50 ksi (A992 steel). If a section is not compact, the available moment must be reduced below Mp. The interactive tool below assumes a compact, fully braced section — the simplest and most common case in practice.

the flexure equationComputing the plastic moment capacity

For a compact, fully braced beam: φbMn = 0.90 × Fy × Zx. Here Fy is the yield strength (50 ksi for A992 steel) and Zx is the plastic section modulus, which you look up in the AISC Steel Construction Manual for your chosen W-shape. The result is in kip-inches — divide by 12 to convert to kip-feet. This is the maximum moment the beam can resist before forming a full plastic hinge.

the shear equationWeb shear yielding and the resistance factor

For most W-shapes where h/tw ≤ 2.24√(E/Fy): φvVn = 1.00 × 0.6 × Fy × Aw, where Aw = d × tw (total depth times web thickness). The web carries nearly all the shear in a W-shape — the flanges contribute very little. The resistance factor φv = 1.0 because web shear yielding is well-understood and ductile: it gives ample warning before failure, so no additional reduction is needed.