Module 6 · Lesson 6.2

Bending, shear, and what happens when they combine

One interaction in this subject falls out of mechanics. Every other one is fitted — and knowing which is which changes how much you trust them.

Why this matters

Real members rarely carry one action. A beam carries moment and shear; a column carries axial load and moment; a portal rafter carries all three at once. The expressions that combine them look similar to each other and have quite different pedigrees — one of them is exact and the rest are fitted to test data. A designer who cannot tell them apart will trust the wrong ones.

By the end of this lesson you should be able to

Choose the right section modulus from the class
Derive the plastic shear resistance and identify the calibrated part
Say when shear reduces bending, and by roughly how much
Derive the rectangle's N–M interaction and compare it with an I section's

Bending: the class decides the modulus

This is where Module 5 pays off.

M_c,Rd = W f_y / γ_M0

and W is:

W_pl for Class 1 and 2 — yielding spreads through the depth;
W_el for Class 3 — only the extreme fibre reaches yield;
W_eff for Class 4 — part of the section is discounted first.

For an I section in major-axis bending the shape factor W_pl/W_el is about 1.13, so dropping from Class 2 to Class 3 costs about 12% of the bending resistance. Nothing about the steel changed. The reserve is physically there and cannot be claimed, because local buckling arrives before the section can use it.

Shear: one part derived, one part chosen

V_pl,Rd = A_v (f_y/√3) / γ_M0

The f_y/√3 is derived. Von Mises says yielding starts when the distortion energy reaches its uniaxial value; work that through for a pure shear state and τ_y = f_y/√3 = 0.577 f_y falls out exactly. No code chooses it.

The shear area A_v is chosen. Real shear stress in an I section is far from uniform — it is roughly parabolic through the web and small in the flanges. A_v is the area which, carrying a uniform τ_y, gives the same total force as the real distribution. That is a practical simplification, and a good one, but it is a simplification.

And the plastic expression may not apply at all

A slender web buckles in shear before it yields, in a diagonal mode. Above about h_w/t_w = 72ε the plastic resistance above is simply unreachable and a quite different check is needed. Nothing in the plastic expression hints at this, which is why the slenderness has to be checked separately.

Worked example

A plate girder web: does it yield or buckle in shear?

Given

Welded girder: web 1200 mm deep, 8 mm thick; flanges 300 × 25
S355, so ε = 0.814
No intermediate stiffeners

Find

Whether the plastic shear resistance may be used.

Assumptions

Unstiffened web, so the threshold takes its lower value
The threshold is code-calibrated and is not verified in this course

Bending and shear together

Below half the shear resistance, shear is ignored entirely. Not because it has no effect, but because the effect is small enough to absorb — and that threshold is calibrated.

Above it, the yield strength in the shear area is reduced by (1 − ρ), where

ρ = (2V_Ed/V_pl,Rd − 1)²

The squared form is fitted. What is physically real is the mechanism: the web is already using part of its yield capacity to carry shear, so less remains for the longitudinal stress that carries moment.

How much this costs depends on how much of W_pl the web contributes, and in an I section that is not much — the flanges, being far from the neutral axis, do most of the bending work. For a 600 × 200 × 10 × 15 section the moment resistance falls by:

V_Ed/V_pl,Rd	ρ	Moment lost
0.6	0.04	1.3%
0.7	0.16	5.3%
0.8	0.36	12.0%
0.9	0.64	21.3%
1.0	1.00	33.3%

So the common shorthand that 'shear barely affects bending' is true up to about 0.7 and stops being true after that. At high shear the penalty is substantial, and since ρ grows as a square it accelerates fast.

The practical consequence is that moment–shear interaction rarely governs a beam in mid-span, where shear is low and moment is high, and can matter at an internal support, where both are large at once. It is worse again for a section whose web does carry a large share of the bending — a deep plate girder with small flanges.

Try it

Interaction surfaces, derived and fitted

One of these curves falls out of mechanics and the other is fitted to test data. They are visibly different shapes — which is the evidence that the second could not have come from the first.

Axial ratio n = N/N_pl0.30

Section area A11,700 mm²

Flange width b200 mm

Flange thickness t_f15 mm

Shear ratio V/V_pl0.40

Axial ratio n: 0.30
Rectangle: m = 1 − n²: 0.910
I section: m: 0.925
Web area ratio a: 0.487
Difference: 1.5 points
A linear interaction would give: 0.700
Shear ratio V/V_pl: 0.40
Interaction parameter ρ: 0.000
Moment lost to shear: 0.0 %

At n = 0.30 a rectangle keeps 91% of its moment and the I section 93%, against 70% for the linear interaction almost everyone assumes. V_Ed is 40% of the shear resistance, at or below the 0.5 threshold, so no reduction applies. Shear is not absent — it is judged small enough to absorb, and that threshold is calibrated.

Things worth trying

Set n to 0.2. The rectangle keeps 96% of its moment. The linear interaction would have said 80% — a difference big enough to change a section size, and it is the derived curve that is right.
Follow the rectangle's curve from n = 0 to n = 1. It is almost flat at the start and almost vertical at the end. The axial force is carried by material near the neutral axis, which contributes least to bending — so it costs little until there is nothing left.
Compare the two curves at low n. The I section sits ABOVE the rectangle, because its axial force goes into the web, which was doing very little bending. The shapes differ because the material is distributed differently.
Thicken the flanges. The web area ratio a falls and the I-section curve moves — the fitted expression is tracking the section's proportions, which is what a calibrated formula does well within its range and badly outside it.
Raise the shear ratio past 0.5 and watch ρ appear. Below that threshold shear is ignored entirely; above it the reduction grows as a SQUARE. Take it from 0.6 to 1.0 and watch the moment loss accelerate — small at moderate shear, and substantial by the time the web is fully used.
Note the labels on the right. One curve is DERIVED and one is CALIBRATED, and the difference matters when you are designing to a utilisation of 0.99 on an unusual section.

The one interaction you can derive

Almost every interaction expression in steel design is fitted to test data. The rectangle under N and M is the exception, and it is worth doing because it shows where the shape of an interaction curve comes from.

Put the plastic neutral axis a distance y either side of the centroid.

The axial force is carried by the central band of depth 2y, which is in compression throughout:

N = 2y b f_y, and N_pl = h b f_y ⟹ n = N/N_pl = 2y/h

The moment is carried by the two outer blocks, which form a couple:

M = f_y b (h²/4 − y²), and M_pl = f_y b h²/4

Divide the second pair by M_pl and substitute:

m = 1 − (2y/h)² = 1 − n²

An exact parabola, with no calibration anywhere.

Two things follow, and both are useful intuition:

The curve is flat at low axial load. At n = 0.2 the section still has 96% of its moment capacity. A small axial force barely matters.
The penalty accelerates. Going from n = 0.8 to n = 0.9 costs more moment capacity than going from 0 to 0.4.

Why an I section is different

The rectangle's parabola does not carry over, because an I section's material is not spread evenly. Axial load is taken first by the web, which was contributing little to bending anyway. So the moment capacity is almost unaffected until the web is fully used, and then falls quickly.

The design expression reproduces that behaviour. It does not derive it.

Predict first

A rectangular section carries 20% of its squash load. What fraction of its plastic moment remains?

Practice

A rectangular section carries 40% of its squash load. What fraction of its plastic moment remains?

Practice

A web is 1150 mm deep and 10 mm thick, in S355 (ε = 0.814). Is a shear buckling check required? Give the threshold 72ε.

Practice

A beam carries V_Ed = 0.75 V_pl,Rd. What is the interaction parameter ρ?

Check yourself

Why is the moment–shear interaction usually a small effect for an I-section beam?

Summary

M_c,Rd = W f_y/γ_M0, with W chosen by the class — Class 2 to 3 costs about 12%
V_pl,Rd = A_v (f_y/√3)/γ_M0: the √3 is derived, the shear AREA is a simplification
Above about h_w/t_w = 72ε the plastic shear expression does not apply at all
A formula does not know its own range — the slenderness check establishes it
Shear reduces bending only above 0.5 V_pl — about 5% at V/V_pl = 0.7, but 33% at full shear
For a rectangle m = 1 − n² EXACTLY — flat at low n, steep at high n
An I section is flatter still at low n, because the web takes the axial load
Know which interactions are derived and which are fitted; the margin you allow should differ

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint