← Design of Structures (Steel, Concrete & Masonry) · AAI Manager (Civil)

Chapter 6 of 10

Slabs, Beams and Columns

In the AAI Manager (Civil) syllabus under Design of Structures (Steel, Concrete & Masonry) · 4 parts

Singly & Doubly Reinforced Beams (Limit State of Collapse: Flexure) · One-Way & Continuous Slabs · Two-Way & Flat Slabs · Short & Long Columns

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

📑 Contents (31 sections)

Part 1 of 4

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 4

One-Way & Continuous Slabs

Last reviewed 16 Sept 2026 · 6 min read

One-way action

A slab supported on two opposite edges, or on four edges but much longer than it is wide, bends mainly in one direction — the short span.

DefinitionWhen a slab is one-way

Supported on two opposite sides only, or
Supported on all four sides with $\frac{l _{y}}{l _{x}} > 2$ (long span / short span).

Main steel runs along the short span; distribution steel runs along the long span.

Typical examples: verandah and corridor slabs, chajjas, slabs spanning between parallel beams, landings.

Effective span

For a simply supported slab: clear span + effective depth, or centre-to-centre of supports, whichever is less. For a cantilever: length to the face of support + $d /2$ . Continuous slabs: centre-to-centre of supports (with IS 456 variations for wide supports).

Loads

Dead load: self-weight (25 kN/m³ × thickness for RCC) + floor finish (typically 1.0–1.5 kN/m²) + partitions where applicable.
Imposed load (IS 875 Part 2): e.g. residential floors 2.0 kN/m², office floors 2.5–4.0 kN/m², roofs with access 1.5 kN/m².
Design load $w_{u} = 1.5 (D L + I L)$ per metre width.

Design a 1 m wide strip as a rectangular beam with $b$ = 1000 mm.

Detailing rules for slabs

Code ProvisionIS 456 — slab reinforcement

Minimum steel (each direction): 0.12% of gross area $b D$ for HYSD bars; 0.15% for mild steel bars. In slabs this rule replaces the beam minimum $0.85 b d / f_{y}$ .
Maximum bar diameter: $D /8$ (one-eighth of total slab thickness).
Maximum spacing of main bars: $3 d$ or 300 mm, whichever is less.
Maximum spacing of distribution bars: $5 d$ or 450 mm, whichever is less.
Nominal cover as per exposure (20 mm for mild exposure; 15 mm allowed for bars up to 12 mm in mild exposure).
At least 50% of the positive steel extends into the support; bars at discontinuous ends may be bent up or provided as top steel to control cracking from partial fixity (commonly 50% of mid-span steel extended over the support for 0.1 $l$ ).

Deflection control

Basic span/effective depth ratios: 20 (simply supported), 26 (continuous), 7 (cantilever) — for spans up to 10 m. Multiply by the modification factor for tension steel (IS 456 Fig. 4), which depends on the steel percentage and the service stress $f_{s} = 0.58 f_{y} \frac{A _{s t, r e q u i r e d}}{A _{s t, p r o v i d e d}}$ . Lightly reinforced slabs get a factor well above 1 (often 1.4–2.0), allowing thinner slabs.

Shear in slabs

Shear stress is usually low. Check $τ_{v} = V_{u} / (1000 d)$ against $k τ_{c}$ where $k$ = 1.30 for slabs 150 mm or thinner, falling to 1.00 at 300 mm (see Shear & Torsion). Shear reinforcement is almost never provided in ordinary slabs; the depth is increased instead.

Continuous one-way slabs — coefficient method

For slabs continuous over three or more approximately equal spans (variation within 15% of the longest), carrying uniform loads, IS 456 allows bending moments and shears from coefficients (with $l$ = effective span; for moments at supports, $l$ is the average of the two adjacent spans).

Code ProvisionIS 456 Table 12 — bending moment coefficients (multiply by

w l^{2}

)

Type of load	Near middle of end span	At middle of interior span	At support next to end support	At other interior supports
Dead load (fixed)	$+ \frac{1}{12}$	$+ \frac{1}{16}$	$- \frac{1}{10}$	$- \frac{1}{12}$
Imposed load (not fixed)	$+ \frac{1}{10}$	$+ \frac{1}{12}$	$- \frac{1}{9}$	$- \frac{1}{9}$

Code ProvisionIS 456 Table 13 — shear force coefficients (multiply by

w l

)

Type of load	At end support	At support next to end support: outer side	Inner side	At all other interior supports
Dead load	0.40	0.60	0.55	0.50
Imposed load	0.45	0.60	0.60	0.60

Imposed-load coefficients are larger because live load may act on some spans and not others (pattern loading).

Part 3 of 4

Two-Way & Flat Slabs

Last reviewed 16 Sept 2026 · 6 min read

Two-way action

A slab supported on all four sides with $l_{y} / l_{x} \leq 2$ bends in both directions. Load is shared between the two spans; the short span carries more because it is stiffer (deflection varies with span⁴).

Rankine–Grashoff theory

Treat two crossing strips at the centre with equal deflection. For a uniformly loaded slab simply supported on four edges:

w_{x} = w \frac{l _{y}^{4}}{l _{x}^{4} + l _{y}^{4}}, w_{y} = w \frac{l _{x}^{4}}{l _{x}^{4} + l _{y}^{4}}

M_{x} = \frac{w _{x} l _{x}^{2}}{8}, M_{y} = \frac{w _{y} l _{y}^{2}}{8}

The theory ignores the torsional stiffness of the slab, so it overestimates moments. Marcus's correction reduces them. IS 456 Annex D gives coefficients based on yield-line and elastic analyses, which are used in practice.

Simply supported slabs — corners not held down

When corners are free to lift (no torsional restraint, no corner steel), IS 456 Table 27 gives:

M_{x} = α_{x} w l_{x}^{2}, M_{y} = α_{y} w l_{x}^{2}

(both multiplied by $l_{x}^{2}$ , the short span).

Code ProvisionIS 456 Table 27 — simply supported slabs, corners not held down

$l_{y} / l_{x}$	1.0	1.1	1.2	1.3	1.4	1.5	1.75	2.0
$α_{x}$	0.062	0.074	0.084	0.093	0.099	0.104	0.113	0.118
$α_{y}$	0.062	0.061	0.059	0.055	0.051	0.046	0.037	0.029

At least 50% of the tension steel should extend to the supports; the rest may be curtailed within $0.1 l$ of the supports.

Restrained slabs — corners held down

For slabs cast monolithically with beams, corners are prevented from lifting and edges may be continuous or discontinuous. IS 456 Table 26 lists nine cases (interior panel, one short edge discontinuous, one long edge discontinuous, two adjacent edges discontinuous, …, four edges discontinuous) with negative coefficients (at continuous edges) and positive coefficients (at mid-span) for both directions:

M = α w l_{x}^{2}

Code ProvisionIS 456 Table 26 — two cases, selected values

Case 1: Interior panel (all edges continuous)

$l_{y} / l_{x}$	1.0	1.1	1.2	1.3	1.4	1.5	1.75	2.0	Long span (all ratios)
Short span, negative	0.032	0.037	0.043	0.047	0.051	0.053	0.060	0.065	0.032
Short span, positive	0.024	0.028	0.032	0.036	0.039	0.041	0.045	0.049	0.024

Case 9: Four edges discontinuous (corners held down)

$l_{y} / l_{x}$	1.0	1.1	1.2	1.3	1.4	1.5	1.75	2.0	Long span (all ratios)
Short span, positive	0.056	0.064	0.072	0.079	0.085	0.089	0.100	0.107	0.056

The long-span coefficient does not change with $l_{y} / l_{x}$ — it depends only on the edge conditions.

Middle strips, edge strips and corner steel

Each span is divided into a middle strip (width three-quarters of the span) and two edge strips (each one-eighth of the span).
Reinforcement from the coefficients goes in the middle strips; edge strips get only minimum steel (0.12% for HYSD).
Torsion reinforcement at corners (where corners are held down):
- At a corner where both edges are discontinuous: provide top and bottom mesh, each layer in both directions, of area three-quarters of the area required for the maximum mid-span moment in the slab, over a distance $l_{x} /5$ from the edges.
- At a corner where one edge is discontinuous: half of the above.
- At a corner where both edges are continuous: none.

Loads on supporting beams

The slab load is transferred to beams by drawing 45° lines from the corners: the short beams receive triangular loads and the long beams trapezoidal loads. Equivalent UDLs for bending moment: triangular load $w l_{x} /3$ ; trapezoidal load $\frac{w l _{x}}{2} [1 - \frac{1}{3} (\frac{l _{x}}{l _{y}})^{2}]$ (per unit length of beam, with $w$ in kN/m²).

Part 4 of 4

Short & Long Columns

Last reviewed 16 Sept 2026 · 6 min read

Definitions and classification

Column — a compression member whose effective length exceeds three times its least lateral dimension.
Pedestal — a compression member whose effective length does not exceed three times its least lateral dimension (designed as plain or nominally reinforced concrete).
Short column — both $\frac{l _{e x}}{D} < 12$ and $\frac{l _{ey}}{b} < 12$ .
Slender (long) column — either ratio ≥ 12. Slender columns carry additional moments from lateral deflection.
Braced column — lateral loads resisted by walls or bracing; no significant sway.
Unbraced column — the frame itself resists sway; effective lengths are larger.

Classification by loading: axially loaded, uniaxially eccentric, biaxially eccentric. By ties: tied (lateral ties) and spiral / helically reinforced.

Effective length

Code ProvisionIS 456 Table 28 — effective length of compression members (selected)

End condition	Theoretical	Recommended design value
Effectively held in position and restrained against rotation at both ends	$0.50 l$	$0.65 l$
Held in position at both ends, restrained against rotation at one end	$0.70 l$	$0.80 l$
Held in position at both ends, not restrained against rotation	$1.00 l$	$1.00 l$
Held in position and restrained against rotation at one end, other end restrained against rotation but not held in position	$1.00 l$	$1.20 l$
Held in position and restrained against rotation at one end, other end free	$2.00 l$	$2.00 l$

$l$ = unsupported length.

Minimum eccentricity

No column is perfectly axially loaded. IS 456 requires every column to be designed for a minimum eccentricity:

e_{min} = \frac{l}{500} + \frac{D}{30} \geq 20 mm

( $l$ = unsupported length, $D$ = lateral dimension in the direction considered.)

Axially loaded short columns

When $e_{min}$ does not exceed $0.05 D$ , the IS 456 simplified formula (which already includes the effect of minimum eccentricity) applies:

FormulaShort axially loaded column (IS 456 cl. 39.3)

P_{u} = 0.4 f_{c k} A_{c} + 0.67 f_{y} A_{sc}

$A_{c}$ = area of concrete ( $A_{g} - A_{sc}$ ); $A_{sc}$ = area of longitudinal steel.

The coefficients 0.4 and 0.67 are lower than the pure axial values ( $0.45 f_{c k}$ and $0.75 f_{y}$ ) to allow for $e_{min}$ .

Helically reinforced columns

Closely spaced helical (spiral) reinforcement confines the core, improving ductility and strength. IS 456 allows the load capacity to be taken as 1.05 times that of a tied column when

\frac{volume of helical reinforcement}{volume of core} \geq 0.36 (\frac{A _{g}}{A _{k}} - 1) \frac{f _{c k}}{f _{y}}

$A_{k}$ = area of core measured to the outside of the helix.

Detailing requirements

Code ProvisionIS 456 — longitudinal reinforcement in columns

Minimum 0.8% and maximum 6% of gross area (4% is the practical maximum recommended to allow proper placing and compaction; laps count in this).
Minimum number of bars: 4 in rectangular columns, 6 in circular columns.
Minimum bar diameter 12 mm.
Spacing of longitudinal bars along the periphery not more than 300 mm.
Nominal cover to longitudinal bars ≥ 40 mm or bar diameter (see Concrete & Reinforcement).

Code ProvisionIS 456 — transverse reinforcement (ties)

Diameter of lateral ties: not less than one-fourth of the largest longitudinal bar diameter, and not less than 6 mm.
Pitch of ties: not more than the least of
- the least lateral dimension of the column,
- 16 times the smallest longitudinal bar diameter,
- 300 mm.
Every corner bar and alternate bars must be held by the corner of a tie with an included angle not more than 135°; a bar more than 75 mm from a restrained bar needs its own tie.

Helical reinforcement: pitch not more than 75 mm or one-sixth of the core diameter, and not less than 25 mm or three times the helix bar diameter.

Columns with uniaxial bending

For a short column with $P_{u}$ and $M_{u}$ (the larger of the applied moment and $P_{u} e_{min}$ ), design uses interaction diagrams (SP 16 charts) built from strain compatibility: for each neutral-axis position, compute the axial force and moment capacity. Parameters:

\frac{P _{u}}{f _{c k} b D}, \frac{M _{u}}{f _{c k} b D ^{2}} \to \frac{p}{f _{c k}} from the chart

The interaction curve shows that a moderate axial compression can increase moment capacity (up to the balanced point) — compression delays tension yielding.

Columns with biaxial bending

FormulaIS 456 interaction formula (cl. 39.6)

(\frac{M _{ux}}{M _{ux 1}})^{α_{n}} + (\frac{M _{u y}}{M _{u y 1}})^{α_{n}} \leq 1.0

$M_{ux 1}$ , $M_{u y 1}$ = uniaxial moment capacities under $P_{u}$ ; $α_{n}$ depends on $P_{u} / P_{u z}$ : 1.0 for $P_{u} / P_{u z} \leq 0.2$ , 2.0 for $P_{u} / P_{u z} \geq 0.8$ , linear in between.

P_{u z} = 0.45 f_{c k} A_{c} + 0.75 f_{y} A_{sc}

Downloaded from civilgyantests.com · Not for resale. Verify every code clause and date against the official source.

Finished reading? Test yourself.

A timed chapter test from the AAI Manager (Civil) series, on exactly this chapter.

Practice this chapter →