Chapter 5 of 12

Shear, Bond & Development Length

In the NTPC NGEL Engineer Civil syllabus under RCC Design · 2 parts

Shear & Torsion in RCC Members · Bond, Anchorage & Development Length

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📑 Contents (18 sections)

Part 1 of 2

Shear & Torsion in RCC Members

Last reviewed 16 Sept 2026 · 7 min read

How RCC beams fail in shear

Near supports, shear force is high. Shear and bending combine to produce principal tensile stresses inclined at about 45°. When these exceed concrete's tensile strength, diagonal tension cracks form. Without web reinforcement, a beam can fail suddenly by:

diagonal tension (inclined crack running to the compression face),
shear compression (crushing of the compression zone above the crack),
shear bond / splitting along the tension bars.

Shear failures are brittle, so IS 456 always provides a minimum of shear reinforcement in beams.

Shear is resisted by: concrete in the compression zone, aggregate interlock across cracks, dowel action of longitudinal bars, and stirrups or bent-up bars crossing the cracks.

Nominal shear stress

τ_{v} = \frac{V _{u}}{b d}

For members of varying depth: $τ_{v} = \frac{V _{u} \pm \frac{M _{u}}{d} tan β}{b d}$ — the negative sign when the bending moment increases numerically in the same direction as the depth increases.

Design shear strength of concrete, τc

$τ_{c}$ depends on concrete grade and the tension steel percentage $p_{t} = 100 A_{s t} / (b d)$ at the section (bars continuing at least $d$ beyond it).

Code ProvisionIS 456 Table 19 — τc (N/mm²), selected values

$p_{t}$ (%)	M20	M25	M30
≤ 0.15	0.28	0.29	0.29
0.25	0.36	0.36	0.37
0.50	0.48	0.49	0.50
0.75	0.56	0.57	0.59
1.00	0.62	0.64	0.66
1.25	0.67	0.70	0.71
1.50	0.72	0.74	0.76
2.00	0.79	0.82	0.84
≥ 3.00	0.82	0.92	0.96

Interpolate linearly between values.

Code ProvisionIS 456 Table 20 — maximum shear stress τc,max (N/mm²)

M15	M20	M25	M30	M35	M40 and above
2.5	2.8	3.1	3.5	3.7	4.0

If $τ_{v} > τ_{c, ma x}$ , the section must be enlarged — shear reinforcement cannot help.

Design procedure for shear

$τ_{v} = V_{u} / b d$ at the critical section.
If $τ_{v} > τ_{c, ma x}$ → revise section.
If $τ_{v} \leq τ_{c}$ → provide minimum shear reinforcement (except in minor members such as lintels, where $τ_{v} < 0.5 τ_{c}$ ).
If $τ_{c} < τ_{v} \leq τ_{c, ma x}$ → design shear reinforcement for $V_{u s} = V_{u} - τ_{c} b d$ .

FormulaShear reinforcement (IS 456 cl. 40.4)

Vertical stirrups:

V_{u s} = \frac{0.87 f _{y} A _{s v} d}{s _{v}} \Rightarrow s_{v} = \frac{0.87 f _{y} A _{s v} d}{V _{u s}}

Inclined stirrups at angle $α$ : $V_{u s} = \frac{0.87 f _{y} A _{s v} d}{s _{v}} (sin α + cos α)$

Single bent-up bar (or group at one section): $V_{u s} = 0.87 f_{y} A_{s v} sin α$ — bent-up bars may resist not more than half the total shear reinforcement demand.

$A_{s v}$ = total area of stirrup legs (a 2-legged 8 mm stirrup gives $2 \times 50.3 = 100.5$ mm²).

Code ProvisionMinimum shear reinforcement and spacing

Minimum: $\frac{A _{s v}}{b s _{v}} \geq \frac{0.4}{0.87 f _{y}}$ → $s_{v} \leq \frac{0.87 f _{y} A _{s v}}{0.4 b}$
Maximum spacing of vertical stirrups: 0.75d or 300 mm, whichever is less.
For inclined stirrups at 45°: spacing not more than $d$ (and 300 mm).
Where $τ_{v} > τ_{c}$ , stirrups must also satisfy the design spacing above.

Enhanced shear strength near supports

Where a support reaction causes compression in the region and the section is within $2 d$ of the support face, IS 456 allows the design shear strength to be increased to

τ_{c}^{'} = \frac{2 d τ _{c}}{a _{v}} (not exceeding τ_{c, ma x})

where $a_{v}$ is the distance of the section from the support face. The critical section for shear in beams with such supports is commonly taken at distance $d$ from the support face.

Part 2 of 2

Bond, Anchorage & Development Length

Last reviewed 16 Sept 2026 · 7 min read

Bond

Bond is the grip between reinforcement and the surrounding concrete. It lets the tensile force in a bar be transferred to concrete so the two act together. Without bond, bars would slip and the beam would behave like unreinforced concrete.

Bond is developed by:

Adhesion — chemical gum between cement paste and steel.
Friction — due to shrinkage of concrete gripping the bar.
Mechanical interlock — ribs of deformed (HYSD/TMT) bars bearing on concrete. This is by far the most important for modern bars.

Type of bond	Where it acts
Flexural (local) bond	Along a beam where the bar force changes with bending moment; stress $\propto$ rate of change of moment (shear)
Anchorage (development) bond	At bar ends, cut-off points and laps, where the full bar force must be transferred over a length

IS 456:2000 designs primarily for anchorage bond through development length; adequate development length also takes care of local bond.

Design bond stress

Code ProvisionIS 456 — design bond stress τbd for plain bars in tension (N/mm²)

M20	M25	M30	M35	M40 and above
1.2	1.4	1.5	1.7	1.9

Deformed bars (IS 1786): increase these values by 60%.
Bars in compression: increase the tension values by 25%.

Development length

The development length $L_{d}$ is the embedment needed on each side of a section to develop the design stress in the bar.

Equating the bar force to bond resistance:

\frac{π ϕ ^{2}}{4} σ_{s} = π ϕ L_{d} τ_{b d} \Rightarrow L_{d} = \frac{ϕ σ _{s}}{4 τ _{b d}}

with $σ_{s} = 0.87 f_{y}$ at the limit state of collapse.

FormulaDevelopment length values (commonly used)

Bar and concrete	Tension	Compression
Fe 250 plain, M20	$\frac{217.5 ϕ}{4 \times 1.2} = 45.3 ϕ$	$36.3 ϕ$
Fe 415 deformed, M20	$\frac{361 ϕ}{4 \times 1.92} = 47.0 ϕ$	$37.6 ϕ$
Fe 415 deformed, M25	$40.3 ϕ$	$32.2 ϕ$
Fe 500 deformed, M20	$56.6 ϕ$	$45.3 ϕ$
Fe 500 deformed, M25	$48.5 ϕ$	$38.8 ϕ$

Bundled bars: $L_{d}$ of each bar is increased by 10% for two bars in contact, 20% for three and 33% for four.

Anchorage

Where the straight length available is short, bars are anchored with hooks and bends.

Anchorage value of a standard U-hook (180°): 16φ; of a 90° bend: 8φ — counted toward $L_{d}$ . The anchorage value of a bend is 4φ for each 45° of bend, up to 16φ.
Deformed bars in tension may be anchored with straight lengths or bends; hooks are generally used for plain bars.
Bars in compression: hooks and bends are not counted in anchorage (they are ineffective and may cause spalling); only the straight length counts.
Stirrups and links are anchored by a 135° or 180° hook around a longitudinal bar, or a 90° bend with sufficient extension.
Minimum radius of bends for bars: IS 456 specifies bend radii to avoid crushing the concrete inside the bend; bearing stresses inside bends must be checked when bars are heavily stressed.

Check at simple supports and points of inflexion

At a simple support (and at points of inflexion), the positive-moment tension bars must satisfy:

FormulaIS 456 — development length at supports and inflexion points

L_{d} \leq \frac{1.3 M _{1}}{V} + L_{0}

$M_{1}$ = moment of resistance of the section, assuming all bars at the section are stressed to $0.87 f_{y}$ .
$V$ = shear force at the section due to design loads.
$L_{0}$ = sum of the anchorage beyond the centre of the support and the equivalent anchorage value of any hook or bend; at a point of inflexion, $L_{0}$ = the greater of $d$ and $12 ϕ$ .
The factor 1.3 applies only where the reaction confines the bar ends with compression (e.g. a beam bearing on a wall or column); otherwise use 1.0.

At least one-third of the positive moment steel in simple members (one-fourth in continuous members) must extend along the same face into the support to a length of $L_{d} /3$ .

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