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Reinforced Concrete Design / Flexural behaviour and section analysis

Designing a section, and making the bars fit

Running the derivation backwards — and the step that turns a number into a buildable beam.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • A concrete-design reviewer must check the stated first-generation clauses, National Annex choices, detailing rules and factor sets. Numerical agreement alone does not establish code compliance.
  • A suitably qualified engineer must verify the adopted material model, standards clauses and design applicability independently.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Try it

Explore a section in bending

Walk the section up to the Klim boundary and watch the design flip from singly to doubly reinforced — and x/d cross 0.45 at the same instant. They are the same fact.

300 mm
550 mm
30 N/mm²
3 bars
25 mm
250 kNm
Section 300 by 550 millimetres with 3 bars of 25 millimetres, neutral axis at 157 millimetresneutral axis3H25b = 300h = 550d
Effective depth d
495 mm
Steel provided As
1473 mm²
K = M/(b d² fck)
0.1136
Klim
0.1673
Steel required
1311 mm²
Minimum steel
223 mm²
Neutral axis x, bars provided
157 mm
x/d, bars provided
0.317
x/d demanded by MEd
0.283
Moment of resistance MRd
276 kNm
Bars fit?
yes, 70 mm clear

Singly reinforced: K = 0.114 is below Klim = 0.167. The bars provided give MRd = 276 kNm against 250 kNm applied.

Things worth trying

  • Raise MEd until K passes Klim. Watch the x/d DEMANDED BY MEd cross 0.45 at the same instant — the two limits are one statement about ductility, written two ways.
  • Note that the other x/d, for the bars actually provided, does not move with MEd at all. It is a property of the section, not of the load.
  • Reduce h and see how fast K climbs: it goes with 1/d², so a 10% loss of depth costs 23% in K.
  • Add bars until they stop fitting. Area is necessary and not sufficient, and the fit check is where that becomes concrete.
  • Raise fck and watch how little K falls. Concrete grade is a blunt instrument in bending; depth is not.

Why this matters

Analysis asks what a section can carry. Design asks what section will carry a given moment, which is the same relationship run backwards. The part that catches people is what comes after the arithmetic: a required area of 1147 mm² is not an answer until real bars supply it, fit across the width, and leave room for concrete to pass between them.

By the end of this lesson you should be able to

  • Compute K and compare it with Klim
  • Solve for the lever arm and the required steel area
  • Select bars and check they fit
  • Apply minimum and maximum reinforcement limits

What you should already know

  • Deriving the moment of resistance (this module)
  • Effective depth and cover (Module 1)

Design turns the derivation around. The moment is known; the steel area is not.

The usual route uses a non-dimensional moment:

K = MEd / (b d² fck)

Making it non-dimensional is worth a moment's thought. It lets one number answer the question 'is this section working hard?' regardless of size or grade. Small K means a lightly stressed section; large K means one close to its ductility limit.

That limit is Klim, the largest K a singly reinforced section can take while keeping x/d inside its limit. It is not a quoted constant — it follows directly from the x/d limit:

at x/d = 0.45, s/d = λ(0.45) = 0.36, and Klim = η fcd (s/d)(1 − s/2d)/fck

For C30/37 on the current profile that gives Klim = 0.167. Below it, tension steel alone; above it, compression steel is needed — not for strength, but to keep the section ductile.

Worked example

Designing a beam section, and choosing the bars

Given

  • Design moment MEd = 220 kNm
  • Section 300 mm wide, overall depth 550 mm, C30/37 concrete, fyk = 500 N/mm²
  • Cover 30 mm, 10 mm links, assume 25 mm main bars

Find

The required steel area, a suitable bar arrangement, and whether it fits.

    Practice

    A beam has h = 550 mm, cover 30 mm, 10 mm links and 25 mm main bars in one layer. What is the effective depth, in mm?

    Practice

    For MEd = 220 kNm on a section with b = 300 mm, d = 490 mm and fck = 30 N/mm², what is K?

    Practice

    With a lever arm of 441 mm and fyd = 434.8 N/mm², what tension steel area is required for 220 kNm, in mm²?

    Summary

    • K = MEd/(b d² fck) measures how hard a section is working, independent of size
    • Klim follows from the x/d limit; it is derived, not quoted
    • Below Klim: tension steel only. Above it: compression steel, for ductility
    • The lever arm is solved, not assumed — it came out at 0.90d here
    • Effective depth is measured to the bar centroid: h − cover − link − half the bar
    • A design is not finished until the bars are shown to fit