Module 4 · Lesson 4.2

Deriving the moment of resistance

Three principles, no formula quoted — and a clear account of which step the code decides.

Why this matters

Almost every textbook presents the flexural design equation as something to be substituted into. It is not: it follows from three statements, two of which are pure mechanics. Deriving it once means you can rebuild it for a section shape nobody tabulated, and — more usefully — you can see exactly which step a National Annex could change.

By the end of this lesson you should be able to

State the three principles the derivation rests on
Derive M_Rd for a singly reinforced rectangular section
Identify which steps are mechanics and which are code choices
Explain what the rectangular stress block replaces, and why

What you should already know

How a section behaves, from zero load to failure (this module)
Strain compatibility (Module 1)
Equilibrium of forces and moments (Structural Analysis Fundamentals, Module 2)

Assumption

What kind of statement is each step?

This course labels every relationship, because they are not all the same kind of thing.

Mechanics — follows from equilibrium, compatibility or geometry. True regardless of any code: plane sections remain plane, C = T, M = Tz.

Material model — a chosen idealisation. Real concrete has a curved stress–strain relationship; the rectangular stress block replaces it with something simpler that gives the same resultant and roughly the same centroid. Real steel strain-hardens; the model ignores it.

Nationally determined parameter — a number the UK National Annex chooses. The partial factors γ_C and γ_S, the coefficient α_cc, the limit on x/d. These are national decisions, not scientific results, and they differ between countries.

If you can see which is which, you know what would survive a change of code and what would not.

From first principles

Moment of resistance of a singly reinforced section

We want to show: that M_Rd = A_s f_yd (d − s/2), with the stress-block depth s fixed by equilibrium.

Forget formulae for a moment and picture the section at failure. The top is being crushed; the bottom is cracked open with the bars holding it together. So there is a push in the concrete near the top and a pull in the steel near the bottom. Nothing is pushing the beam along its length, so those two must be equal and opposite. They are equal and opposite forces on parallel lines — a couple — and the moment the section resists is simply the size of that force times the distance between them. That is the whole answer: find the force, find the distance, multiply. Everything else is working out how big the force is and where it acts.

Practice

For C30/37 concrete with α_cc = 0.85 and γ_C = 1.5, what is the design compressive strength f_cd, in N/mm²?

Practice

For reinforcement with f_yk = 500 N/mm² and γ_S = 1.15, what is the yield strain, taking E_s = 200 000 N/mm²?

Practice

A section has A_s = 1473 mm², f_yd = 434.8 N/mm², b = 300 mm, d = 490 mm and η f_cd = 17.0 N/mm². Using the derived expression, what is M_Rd, in kNm?

Summary

Three principles: material models, strain compatibility, equilibrium
Strain compatibility and equilibrium are mechanics; the stress block is a chosen model
γ_C, γ_S, α_cc and the x/d limit are nationally determined
x is solved from C = T, never assumed
M_Rd = T z = A_s f_yd (d − s/2)
Always check that the steel really has yielded — the derivation assumed it

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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