Unit IIIDesign of Structures - I

Design of Beams

The flexural member — the stress block, Mu,lim and the tension steel.

≈ 40 min + worked example

A beam resists bending: concrete takes the compression at the top, steel takes the tension at the bottom. The whole of limit-state flexure follows from one picture — the stress block — and one threshold, the limiting moment Mu,lim, that decides whether the section needs steel in tension only, or in compression too.

Learning objectives

By the end of this lesson, you will be able to — mapped to the course outcomes for Design of Structures I:

CO3 · Understand

Describe the limit-state stress block and its 0.36 (force) and 0.42 (lever-arm) constants.

CO3 · Apply

Compute the limiting moment Mu,lim = K fck b d² for Fe250/415/500.

CO3 · Analyse

Decide singly vs doubly reinforced by comparing Mu with Mu,lim.

CO6 · Apply

Find the tension steel Ast for a rectangular beam.

The core idea

The stress block and Mu,lim

At collapse the compressive resultant is C = 0.36 fck b xu at 0.42 xu from the top; the tension is T = 0.87 fy Ast. Capping the neutral axis for ductility fixes the limiting moment Mu,lim = K fck b d².^{[1, 3]}

DiagramThe limit-state stress block of a singly reinforced beam with strain 0.0035, compressive force 0.36 fck b xu and tension 0.87 fy Ast

DiagramA singly reinforced beam section versus a doubly reinforced section with compression steel

The limit-state stress block

At collapse the extreme concrete fibre strains to 0.0035; the compressive resultant is C = 0.36 fck b xu acting at 0.42 xu from the top, so the lever arm is z = d − 0.42 xu, balanced by the steel tension T = 0.87 fy Ast.^{[1, 3]}

PhotoBeam and floor formwork (shuttering) supporting the reinforcement before the concrete pour.WTF Formwork - http://www.wallties.com · CC BY-SA 3.0 · via Wikimedia Commons

The moments

Simple and continuous beams

A simply-supported beam under a uniform load has a sagging parabola of moment wL²/8; a continuous beam adds hogging moments over the interior supports — both designed by the same flexure rules.

DiagramBending moment diagrams for a simply supported beam and a continuous beam

Live calculator

Design a beam section

A 300 × 500 mm M20 beam with Fe415 has Mu,lim ≈ 207 kN·m. Apply 150 kN·m and it is singly reinforced with about 958 mm² of steel; push past 207 and the calculator flags that you need compression steel.

Beam design · rectangular, flexure

Width b300 mmEffective depth d500 mmDesign moment Mu150 kN·mConcrete grade

Steel grade

Mu,lim = K·fck·b·d² (K = 0.138 Fe415); if Mu ≤ Mu,lim design singly, with Ast = 0.5(fck/fy)[1 − √(1 − 4.6 Mu/fck b d²)] b d.

0.0 kN·m

Limiting moment Mu,lim

0 mm²

Tension steel Ast

Applied moment is within the singly-reinforced limit.

At a glance

Singly vs doubly reinforced

Aspect	Singly	Doubly / grade
Decision	Mu ≤ Mu,lim: singly reinforced	Mu > Mu,lim: doubly reinforced
Failure mode	Under-reinforced: ductile (steel yields)	Over-reinforced: brittle (concrete crushes) — disallowed
K in Mu,lim	Fe415: K = 0.138	Fe500: K = 0.133 · Fe250: K = 0.149
xu,max / d	Fe415: 0.48	Fe500: 0.46 · Fe250: 0.53
Steel added	Singly: tension only	Doubly: tension + compression

Vocabulary

Key terms

Neutral axis (xu)

Depth where bending stress is zero; concrete above in compression, steel below in tension.

Stress-block 0.36 / 0.42

Compressive force C = 0.36 fck b xu, acting at 0.42 xu from the top fibre.

xu,max / d

Ductility limit on the neutral axis: 0.53 (Fe250), 0.48 (Fe415), 0.46 (Fe500).

Mu,lim = K fck b d²

Limiting moment of a singly-reinforced section; K = 0.149 / 0.138 / 0.133.

Under- vs over-reinforced

Under-reinforced (steel yields first) is ductile and required; over-reinforced (concrete crushes first) is brittle and disallowed.

Singly / doubly reinforced

Singly = tension steel only (Mu ≤ Mu,lim); doubly adds compression steel (Mu > Mu,lim).

Lever arm z

Distance between the compressive and tensile resultants, z = d − 0.42 xu.

Minimum tension steel

Ast/bd = 0.85/fy (IS 456 cl. 26.5.1).

Apply it

Worked example

For b = 300 mm, d = 500 mm, M20, Fe415: Mu,lim = 0.138 × 20 × 300 × 500² = 207 kN·m. An applied moment of 150 kN·m is below this, so the section is singly reinforced, with Ast = 0.5(20/415)[1 − √(1 − 4.6 × 150×10⁶ /(20 × 300 × 500²))] × 300 × 500 ≈ 958 mm². Verify in the calculator, then raise Mu above 207 to see it call for compression steel.

Check your understanding

Self-assessment

1. In the IS 456 stress block, the compressive resultant acts at what depth from the top fibre?

2. For an M20–Fe415 beam, Mu,lim is approximately —

3. When the applied moment exceeds Mu,lim, the correct action is to —

In a nutshell

Recap

At collapse the stress block gives C = 0.36 fck b xu at 0.42 xu, balanced by T = 0.87 fy Ast.

Ductility caps the neutral axis (xu,max/d) and so the moment a singly-reinforced section can carry: Mu,lim = K fck b d².

If Mu ≤ Mu,lim design singly reinforced; if greater, add compression steel (doubly reinforced).

The tension steel follows Ast = 0.5 (fck/fy)[1 − √(1 − 4.6 Mu/(fck b d²))] b d, within the min/max limits.

The evidence

References & further reading

[1]IS 456:2000 — Plain and Reinforced Concrete, Code of Practice. Bureau of Indian Standards. (cl. 38, Annex G; cl. 26.5.1.)
[2]SP 16:1980 — Design Aids for Reinforced Concrete to IS 456. Bureau of Indian Standards.
[3]S.U. Pillai & Devdas Menon, Reinforced Concrete Design (3rd ed.). McGraw-Hill Education, 2009.