Flanged Beams — T-Beams & L-Beams — RCC Design Notes

Last reviewed 16 Sept 2026 · 5 min read

Why flanged beams

When a beam is cast monolithically with the slab it supports, part of the slab acts together with the beam in resisting bending. In sagging regions (mid-span) the slab is in compression, so the combined section behaves like a T (intermediate beam) or an L (edge beam). The large compression area means:

the neutral axis is usually shallow (often within the flange),
the lever arm is large and less tension steel is needed,
the beam can be shallower than a rectangular beam for the same moment.

In hogging regions (over interior supports of continuous beams) the slab is in tension, so the section is designed as a rectangle of width $b_{w}$ .

Effective width of flange

Compression stress is not uniform across a wide flange (shear lag). IS 456 replaces it with an effective width $b_{f}$ carrying uniform stress.

Code ProvisionIS 456 — effective flange width

Beam	$b_{f}$ (not greater than the actual flange width)
T-beam (slab on both sides)	$\frac{l _{0}}{6} + b_{w} + 6 D_{f}$
L-beam (slab on one side)	$\frac{l _{0}}{12} + b_{w} + 3 D_{f}$
Isolated T-beam	$\frac{l _{0}}{( l _{0} / b ) + 4} + b_{w}$
Isolated L-beam	$\frac{0.5 l _{0}}{( l _{0} / b ) + 4} + b_{w}$

$l_{0}$ = distance between points of zero moment (= effective span for simply supported beams; may be taken as 0.7 × effective span for continuous beams and frames); $b_{w}$ = web width; $D_{f}$ = flange (slab) thickness; $b$ = actual flange width. For T and L beams the effective width also cannot exceed the centre-to-centre spacing of beams (or half the spacing on each side plus the web).

Analysis

Case 1 — neutral axis within the flange ( $x_{u} \leq D_{f}$ )

The concrete below the flange is in tension and ignored, so the section behaves as a rectangle of width $b_{f}$ :

x_{u} = \frac{0.87 f _{y} A _{s t}}{0.36 f _{c k} b _{f}}, M_{u} = 0.87 f_{y} A_{s t} d (1 - \frac{A _{s t} f _{y}}{b _{f} d f _{c k}})

Quick check: if $M_{u} \leq 0.36 f_{c k} b_{f} D_{f} (d - 0.42 D_{f})$ , the neutral axis lies in the flange.

Case 2 — neutral axis in the web ( $x_{u} > D_{f}$ )

The compression zone consists of the web (width $b_{w}$ , depth $x_{u}$ ) plus the flange outstands ( $b_{f} - b_{w}$ ).

FormulaIS 456 Annex G — flanged sections with NA in the web

When $D_{f} / d \leq 0.2$ :

M_{u} = 0.36 f_{c k} b_{w} x_{u} (d - 0.42 x_{u}) + 0.45 f_{c k} (b_{f} - b_{w}) D_{f} (d - \frac{D _{f}}{2})

When $D_{f} / d > 0.2$ , replace $D_{f}$ by

y_{f} = 0.15 x_{u} + 0.65 D_{f} (but not more than D_{f})

$x_{u}$ from equilibrium: $0.36 f_{c k} b_{w} x_{u} + 0.45 f_{c k} (b_{f} - b_{w}) y_{f} = 0.87 f_{y} A_{s t}$ (with $y_{f} = D_{f}$ when $D_{f} / d \leq 0.2$ ).

Limiting moment: use $x_{u} = x_{u, ma x}$ .

The 0.45 $f_{c k}$ reflects that a thin flange lies in the constant-stress part of the design stress block.

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Flanged Beams — T-Beams & L-Beams

Why flanged beams

Effective width of flange

Analysis

Case 1 — neutral axis within the flange ( $x_{u} \leq D_{f}$ )

Case 2 — neutral axis in the web ( $x_{u} > D_{f}$ )

This chapter is in the syllabus of

How ready are you for RCC Design?

Why flanged beams

Effective width of flange

Analysis

Case 1 — neutral axis within the flange (xu​≤Df​)

Case 2 — neutral axis in the web (xu​>Df​)

This chapter is in the syllabus of

How ready are you for RCC Design?

Case 1 — neutral axis within the flange ( $x_{u} \leq D_{f}$ )

Case 2 — neutral axis in the web ( $x_{u} > D_{f}$ )