E-01 — stress

same load, smaller section, bigger problem.

σ = P ÷ A
stress visualizer

apply a force. watch the stress change.

Drag the section handles or use the sliders — same force through a smaller section means higher stress.

P 100 kips
b 4 in
d 8 in
stress (ksi)
A = 32 in²
live equation
A = b × d = 4 × 8 = 32 in² · σ = P/A = 100/32 = 3.13 ksi 8.7% of Fy = 36 ksi — well within elastic range
σ = Internal force per unit area
P = Force along the member axis
A = Cross-sectional area of the member
Fy = Stress at which steel begins permanent deformation
explained
Stress is force per unit area — it tells you how hard the material is working. A larger section spreads the same force over more area, reducing stress. The material yields when stress reaches Fy.
key concepts
overview Stress is a condition, not a property

Stress isn't a material property — it's a condition. The same 100-kip load through a 10"×10" column puts every square inch of steel to relatively gentle work. Squeeze that column down to 2"×2" and the same load is packed into 4 square inches instead of 100. The material doesn't know the load changed. It just knows how hard it's being pushed per square inch. That's stress. When that number exceeds what the material can hold, it yields.

what stress really means Force per unit area controls member sizing

Stress is force per unit area: σ = F / A. It's what the material actually "feels." A 100-kip force on a 10 in² section creates 10 ksi of stress. The same force on a 1 in² section creates 100 ksi — enough to fracture steel. This is why member sizing is fundamentally about controlling stress. Bigger sections spread the same force over more area, keeping stress below the material's capacity.

normal vs. shear Perpendicular and parallel stress act together in bending

Normal stress (σ) acts perpendicular to the cross-section — tension pulls fibers apart, compression pushes them together. Shear stress (τ) acts parallel to the cross-section — it tries to slide one face past the other. In bending, both exist simultaneously: normal stress from the bending moment (maximum at extreme fibers) and shear stress from the transverse force (maximum at the neutral axis). Design must check both.

the full picture Axial, shear, and bending stress compared

Axial stress is uniform — every square inch of the section carries exactly the same load. That uniformity is what makes σ = P/A so clean. But when loads act differently, the distribution stops being uniform, and a new equation takes over.

P
this module
Axial (Normal)
σ = P / A
Load along the member axis. Identical stress at every point across the section — perfectly uniform.
V
see bc-01
Shear
τ = VQ / Ib
Force parallel to the cut face. Maximum at the neutral axis, zero at the extreme edges — parabolic.
0 –C +T
see bc-03
Bending
σ = Mc / I
A moment compresses one side, tensions the other. Zero at the neutral axis, max at the extreme fibers — linear.

Stress tells you how hard the material is working. Next: strain tells you how much it moves.

E-02: Strain & Deformation →