Chapter 1 of 13

Limit state design for bending

In the SSC JE Civil syllabus under RCC Design · 2 parts

Limit State Method — Principles, Loads & Partial Safety Factors · Singly & Doubly Reinforced Beams (Limit State of Collapse: Flexure)

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📑 Contents (15 sections)

Part 1 of 2

Limit State Method — Principles, Loads & Partial Safety Factors

Last reviewed 16 Sept 2026 · 6 min read

What a limit state is

A limit state is a condition beyond which a structure (or part of it) no longer fulfils its design purpose. Limit state design makes the structure safe against all relevant limit states with an acceptable probability, instead of only keeping elastic stresses below permissible values (as working stress design does).

Group	Limit states
Limit state of collapse (strength, safety)	Flexure, compression, shear, torsion, bond; overall stability (overturning, sliding); buckling
Limit state of serviceability (fitness for use)	Deflection, cracking, vibration, durability, fire resistance

The structure is designed for the collapse limit states and checked for serviceability.

Characteristic values

Characteristic strength of a material — the value below which not more than 5% of test results are expected to fall ( $f_{c k}$ for concrete, $f_{y}$ for steel).
Characteristic load — the load that has only a 95% probability of not being exceeded during the life of the structure. In practice, loads from IS 875 (dead, imposed, wind) and IS 1893 (earthquake) are used as characteristic loads.

Partial safety factors

For materials ( $γ_{m}$ )

Design strength = \frac{characteristic strength}{γ _{m}}

Material	Limit state of collapse	Limit state of serviceability
Concrete	1.5	1.0
Steel	1.15	1.0

Concrete gets the larger factor because its strength is more variable (site mixing, compaction, curing).

For loads ( $γ_{f}$ )

Design load F_{d} = γ_{f} \times characteristic load

Code ProvisionIS 456 Table 18 — partial safety factors for loads

Load combination	Collapse: DL	Collapse: IL	Collapse: WL/EL	Serviceability: DL	Serviceability: IL	Serviceability: WL/EL
DL + IL	1.5	1.5	—	1.0	1.0	—
DL + WL (or EL)	1.5 or 0.9*	—	1.5	1.0	—	1.0
DL + IL + WL (or EL)	1.2	1.2	1.2	1.0	0.8	0.8

*0.9 is used for dead load where it helps stability (overturning, uplift, stress reversal).

Assumptions for flexure (limit state of collapse)

Plane sections remain plane — strain varies linearly with depth.
Maximum strain in concrete at the outermost compression fibre is 0.0035 in bending.
The design stress–strain curve of concrete is parabolic–rectangular, with maximum design stress $0.67 f_{c k} /1.5 = 0.446 f_{c k}$ .
Tensile strength of concrete is ignored.
Stresses in reinforcement follow the design stress–strain curve of steel, with $γ_{m}$ = 1.15 (design yield $0.87 f_{y}$ ).
The maximum strain in the tension steel at failure must not be less than $\frac{f _{y}}{1.15 E _{s}} + 0.002$ (that is, $\frac{0.87 f _{y}}{E _{s}} + 0.002$ ), so that steel yields before concrete crushes (ductile failure).

The stress block

Integrating the parabolic–rectangular stress diagram over the compression depth $x_{u}$ of a rectangular section of width $b$ :

FormulaCompressive force and its position

C = 0.36 f_{c k} b x_{u}, acting at 0.42 x_{u} from the compression face

Tension: $T = 0.87 f_{y} A_{s t}$ (when steel yields).

Lever arm $= d - 0.42 x_{u}$ ; moment of resistance $M_{u} = 0.36 f_{c k} b x_{u} (d - 0.42 x_{u}) = 0.87 f_{y} A_{s t} (d - 0.42 x_{u})$ .

Where the numbers come from: the average stress of the block is $0.8095 \times 0.446 f_{c k} = 0.36 f_{c k}$ , and its centroid lies at $0.416 x_{u} \approx 0.42 x_{u}$ .

Part 2 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.

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