Chapter 4 of 13

Design of beams

In the UPSC ESE Civil syllabus under RCC Design · 2 parts

Singly & Doubly Reinforced Beams (Limit State of Collapse: Flexure) · Flanged Beams — T-Beams & L-Beams

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (11 sections)

Part 1 of 2

Singly & Doubly Reinforced Beams (Limit State of Collapse: Flexure)

Last reviewed 16 Sept 2026 · 5 min read

Singly reinforced rectangular beam

A singly reinforced beam has tension steel only (hanger bars in the compression zone hold stirrups but are ignored in strength).

Notation: $b$ = width, $d$ = effective depth (compression face to centroid of tension steel), $D$ = overall depth, $A_{s t}$ = tension steel area, $x_{u}$ = depth of neutral axis.

Depth of neutral axis

Equating compression and tension ( $C = T$ ):

0.36 f_{c k} b x_{u} = 0.87 f_{y} A_{s t} \Rightarrow x_{u} = \frac{0.87 f _{y} A _{s t}}{0.36 f _{c k} b}

Check $x_{u} \leq x_{u, ma x}$ ( $0.53 d$ , $0.48 d$ , $0.46 d$ for Fe 250, 415, 500).

Moment of resistance

FormulaMoment of resistance (IS 456 Annex G)

If $x_{u} \leq x_{u, ma x}$ (under-reinforced):

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

which equals $0.87 f_{y} A_{s t} (d - 0.42 x_{u})$ .

If $x_{u} > x_{u, ma x}$ : $M_{u} = M_{u, l im} = 0.36 \frac{x _{u, ma x}}{d} (1 - 0.42 \frac{x _{u, ma x}}{d}) b d^{2} f_{c k}$ .

Steel required for a given moment

Solving the quadratic above:

A_{s t} = \frac{0.5 f _{c k}}{f _{y}} [1 - 1 - \frac{4.6 M _{u}}{f _{c k} b d ^{2}}] b d

Minimum depth for a given moment (balanced)

d_{min} = \frac{M _{u}}{k f _{c k} b}, k = 0.149, 0.138, 0.133 (Fe 250, 415, 500)

Adopt a depth larger than $d_{min}$ so that the section is under-reinforced.

Detailing rules for beams

Code ProvisionIS 456 — reinforcement limits in beams

Minimum tension steel: $\frac{A _{s}}{b d} = \frac{0.85}{f _{y}}$ (0.34% for Fe 250, 0.205% for Fe 415, 0.17% for Fe 500).
Maximum tension steel: 4% of $b D$ . Maximum compression steel: 4% of $b D$ .
Horizontal clear distance between bars: not less than the largest bar diameter or the nominal maximum aggregate size + 5 mm.
Maximum clear distance between tension bars (no redistribution): 215 mm (Fe 250), 180 mm (Fe 415), 150 mm (Fe 500) — crack control.
Side-face reinforcement: where the depth of the web exceeds 750 mm, provide 0.1% of the web area distributed equally on both faces, spaced not more than 300 mm or the web thickness, whichever is less.
Vertical distance between layers: not less than 15 mm, two-thirds of the aggregate size, or the maximum bar diameter.

Effective span (simply supported beam)

Clear span + effective depth, or centre-to-centre of supports — whichever is less. For a cantilever: length to the face of the support plus half the effective depth (except where it forms the end of a continuous beam).

Deflection control by span/depth ratio

Basic $l / d$ : 7 (cantilever), 20 (simply supported), 26 (continuous) for spans up to 10 m. For spans over 10 m (except cantilevers), multiply by $10/ span$ . Further modification factors depend on tension steel percentage and its stress, and compression steel; flanged beams get a reduction.

Doubly reinforced beams

A doubly reinforced beam has steel in both tension and compression zones. It is used when:

the moment exceeds $M_{u, l im}$ and the depth cannot be increased (headroom, architectural limits);
the section is subject to reversal of moments (wind, earthquake, continuous beams at supports);
compression steel is needed to reduce long-term deflection (it reduces creep) or to improve ductility.

Design procedure

$M_{u, l im}$ for the section; if $M_{u} \leq M_{u, l im}$ , design as singly reinforced.
$A_{s t 1}$ for $M_{u, l im}$ : $A_{s t 1} = \frac{M _{u, l im}}{0.87 f _{y} ( d - 0.42 x _{u, ma x} )}$ .
Extra moment $M_{u 2} = M_{u} - M_{u, l im}$ .
Strain in compression steel: $ε_{sc} = 0.0035 (1 - \frac{d ^{'}}{x _{u, ma x}})$ . Read $f_{sc}$ from the design stress–strain curve (it is $0.87 f_{y}$ if $ε_{sc}$ exceeds the yield strain). Concrete stress at that level $f_{cc} = 0.446 f_{c k}$ if $ε_{sc} \geq 0.002$ .
$A_{sc} = \frac{M _{u 2}}{( f _{sc} - f _{cc} ) ( d - d ^{'} )}$ .
$A_{s t 2} = \frac{A _{sc} ( f _{sc} - f _{cc} )}{0.87 f _{y}}$ ; total $A_{s t} = A_{s t 1} + A_{s t 2}$ .

FormulaMoment of resistance of a doubly reinforced section

M_{u} = 0.36 f_{c k} b x_{u} (d - 0.42 x_{u}) + (f_{sc} - f_{cc}) A_{sc} (d - d^{'})

with $x_{u}$ from $0.36 f_{c k} b x_{u} + (f_{sc} - f_{cc}) A_{sc} = 0.87 f_{y} A_{s t}$ .

Exam TipCompression bars need ties

Compression bars must be restrained against buckling by closed stirrups (links), as in columns. Otherwise they can push out the cover.

Part 2 of 2

Flanged Beams — T-Beams & L-Beams

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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