Shear and Development Length

SHEAR AND DEVELOPMENT LENGTH

Shear in homogeneous beams: Shear in a beam is indused due to the change of bending

moment along the span.

b

cN A

Section xx stress shear stress diagram distribution.

For homogeneous beams, shear stress distribution across the section, = (V*A y)/(b*Ixx)

V= shear at the section

A= area out side the section when shear stress is required = b*c

y = distance from the centroid of A to the center of section. = (D/2)-(c/2)

Ixx = M.I. of beam = bD³/12

For rectangular homogeneous sections, theshear stress distribution is parabolic.the

max.shaer stress at the center,

(V/b)* (b*D/2)*(D/4) max =

(bD³/12)

= (3/2)*(V/bD)

M C A M+δM

P P+δP yc b δyY z a

D B C/S δx

consider two sections AB&CD of a beam δx a part.let M and M+δM be the moments

acting on the sections respectively.

To find shear stress at a distance ‘a’ from the N.A (sec.C/S) when breadth is ‘z’, take

b*dy elementry area at a distance y from N.A. above ‘a’

Stress at the section = (M/I)*y

Elementary force on area (b*dy) = (M/I)*y*(b*dy)

Total force above distance ‘a’ from N.A on section CD

yc

= P = (M/I)*y*(b*dy)

a yc = (M/I) (y*b*dy)

a

parallally total force above ‘a’ on section AB. yc = P+δP = (M+δM) (y*b*dy)

Ia

yc

δP = (δM/I) (b*y*dy).a

total shear force on area z*δx = δP yc

, shear stress = δP/(z*δx) = (1/z*δx)*(δM/I) (b*y*dy)

a yc = (δM/δx)*(1/z*I) (b*y*dy)

ayc

= (F/I*z) (b*y*dy). [ since (δM/δx) = F ]

ayc

_ (b*y*dy) = moment of area above ‘a’ about N.A if the area above ‘a’ is ‘A’ and y is the distance of its .a

yc _ (b*y*dy) = A. ya

_therefore = (F/I*z)*A* y _for a particular section, (F/I) is a constant.therefore varies as (a* y )*1

at extreme fibres intensity of shear stress will be zero.

When a beam is loaded with transverse loads, the B.M varies from section to

section. Shearing stresses in the beam are caused by this variation of B.M along the span.

C A w KN/m

b

xN A

d

D Bδx cross section

C A C C+δC M M+δM

V V+δx N A z

T T+δTD B δx δx

short length δx forces of δx shear stress distribution

consider ashort length δx of an RCC beam as shown in fig. M and V be the moment and

shear at section CD, and M+δM and V+δV the moment and shear at section AB.taking

the moment of forces about side AB,

M+V*δx – w(δx)²/2 – (M-δM) = 0 as the section is in equilibrium.

w (δx)²/2 is a very small and can be neglected.

Therefore V*δx – δM = 0

Or V*δx = δM

The M.R = δc*z [where z is lever arm]

Or δT*z

External moment = internal M.R

Therefore δM = δT*z = δc*z = V*δx

δc = (V*δx/z) or δT = (V*δx/z)

at any level, the difference in longitudinalforces induce horizontal shear stresses. At the

N.A, the difference in forces above it is δc, which causes the shear stress on the area

(b*δx) thus at N.A, = [δc/(b*δx)] = (Vδx/z)*(1/b*δx) = (V/b*z)

= (V/b*jd)

the shear stress distribution in RCC beam is zero at top and parabolic up to N.A at

the value is maximum and is equal to (V/b*jd).

Below the N.A as the concrete is cosidered ineffective in tension, the change in

longitudinal forces remains constant and is equal to δC or δT.

Shear stress, = (δT/b*δx) = (V*δx/z)*(1/b*δx) = (V/b*z)= (v/b*jd)

At C.G of bars the compression forces causing stress δc are neutralised by equal and

opposite force δT and hence, shear stress drops to zero.

Diagonal tension and diagonal comprission

A B A B

σ2f f f

σ1

D C D

small element along principal stresses the length of beam

cosider a small element along the length of the beam. This is subjected to shear stress

and tensile stress f .

the principal stresses on this element are

σ1 = (f/2)±√(f/2)²+² and the inclination of principal plane is

tan 2 = 2/f

the major principal stress is tensile and is equal to

σ1 = (f/2)+ √(f/2)²+² and is known as diagonal tension.

The minor principal stress is compression and is equal to

σ2 = (f/2)- √(f/2)² + ² and is known as diagonal compression.

Design for shear by working stress method

v= v/(bd)

B code

N A d Elastic theory

= V/(bd)

Indian standard load recommonds to use normal shear stress given by the expression

v, nominal shear stress = V/bd

Where V = S.F at the section due to design loads.

b = breadth of beam, for flanged beams breadth of web

c, permissible shear stress in concrete in beams with out shear reinforcement is given in

table 23 of IS:456-2000.

The permissible shear stress depends upon two factors

(1) grade of concrete.

(2) Percentage of reinforcement (100As/bd)

c,max, maximum shear stress.

When shear reinforcement is provided the nominal shear stress v in beams shall

not exceed c,max given in table 24 of IS:456-2000.

Design of shear reinforcement.

V = S.F at the section due to design loads

Vc = shear resistance of concrete = cbd

Vs = shear to be resisted by steel

V = Vc+Vs or

Vs = V-Vc

= V - cbd

when v exceeds c, shear reinforcement shall be provide.

Shear reinforcement shall be provided in any of the following forms.

(A) Vertical stirrups.

(d-dי)=d holding bars

dי

d-dי 45

Tension reinforcement

In the absence of shear reinforcement, the beam fails in diagonal tension, the inclination

of the tension cracks being at 45 to the axis of the beam and extended upto horizantal

distance equal to d-dי d.

There fore number of vertical stirrups resisting shear force = d/Sv

Asv = area of each stirrup.

For 2 legged vertical stirrups, area of each stirrup = area of 2- legs.

σsv = permissible tensile stress in shear reinforcement

there fore shear resistance of stirrups = shear resistance of each stirrup*no.of stirrups.

Vs = (Asv*σsv)d/Sv

Or Sv = Asv*σsv*d/Vs

(B) Inclined stirrups or series of bent up bars

(P) A simply supported beam 200mm wide and 380mmm eff.depth is loaded with a

UDL of 20KN/m as shown in fig. The span of the beam is 3.5m. Chech the beam for

shear reinforcement if the tension steel provided is 4 bars of 20mm dia.(mild steel)

and the concrete used is M20. 20KN/m

x

c.s.for shear380

x

b=200 d

S.F. at critical section

S.F. at support

Shear Force diagram

For simply supported beams critical section for S.F is located at a distance ‘d’

from the face of support (ref: cl.22.6.2.1 of IS:456-2000)

Therefore S.F at the C/S = (wl/2) – (wd)

V = (20*3.5/2)-(20*0.38)

V = 27.4KN.

c,nominal shear stress = (V/bd) = (27.4*10³/200*380) = 0.36N/mm².

c,permissible shear stress:

% of tension steel = 100As/bd.

= (100*4*314)/(200*380)

= 1.65%

for 1.65% steel and M20 grade concrete,

c = 0.46N/mm² (table 23 of IS:456-2000)

c>v therefore no shear reinforcement is required.

However, provideing minimum shear reinforcement as per (cl.26.5.1.6 of IS:456-2000)

(Asv/b*sv) (0.4/0.87fy)

providing 6mmΦ 2-legged vertical stirrups, spacing required.

Sv ≤ [(Asv*0.87fy)/(0.4*b)] = [(2*28.3*0.87*250)/(0.4*200)]

= 153.74mm

maximum spacing of shear reinforcement = 0.75*d = 0.75*380 = 285mm

Therefore provide 6mmΦ 2-legged vertical stirrups @ 150mm c/c through out the span.

(2) A reinforcement concrete rectangular beam 400mm wide and 700mm overall

depth is simply supported over a span of 6m. It carries a concentrated load of

150KN at 1.0m from one end in addition to a super imposed load of 18KN/m. The

section is reinforced with 6 nos of 20mmΦ bars of grade Fe415 using M20 concrete,

calculatethe shear reinforcement if

(a) Only vertical stirrups are used.

(b) Two bars are bent up at 45 at the same cross section.

(c) Two groups of bars are bent up at 45, each group consisting of 2-bars each

group separated by 600mm (Design in L.S.M)

150KN

wd w = 18KN/m

400mm

1.0m

L = 6m

665mm

0.99m

199 KN 174.44 KN 6-20ΦAst = 6*314

99.16 KN

Assuming the given span as clear span shear force at the critical section (at ‘d’

from the face of support )

V = (W*5/6)+(w*l/2) – (w*d).

Where d = 700-35 = 665mm

Self weight of beam = 0.4*0.7*1*25 = 7KN/m

Therefore V = (150*5/6)+(25*6/2) – (25*0.665).

= 182.7KN

Vu = 1.5V = 275KN.

v,nominal shear stress = (Vu/bd) = (275*10³/400*665)

= 1.03/mm²

c,permissible shear stress:

% of steel = (100As/bd) = (100*314*6)/(400*665)

= 0.71

for 0.71% steel and M20 grade concrete,

c = 0.54N/mm² (table19)

c,max. = 2.2N/mm² (table 24 of IS:456-2000)

c < v < c,max. therefore provide reinforcement

(A) Only vertical stirrups are used.

Vus, shear to be resisted by stirrups = Vu-Vc

Vc, shear resistance of concrete = c.b.d

= 0.54*400*665

= 143.6KN

Vus = 275-143.6 = 131.4KN

Using 10mmΦ 2-legged vertical stirrups,spacing required

Sv = (Asv*0.87fy*d)/Vus

= (2*78.5*0.87*415*665)/131400

= 286.9mm.

Maximum spacing as per minimum shear reinforcement requirement

Sv = (Asv*0.87*fy)/(0.4*b)

= (2*78.5*0.87*415)/(0.4*400)

= 354mm.

However maximum spacing shall not exceed 0.75d = 0.75*665 = 499mm

Therefore provide 10mmΦ 2- legged vertical stirrups @280mm c/c for a length of 1m

from the respective ends. At the centre minimum shear reinforcement of 10mmΦ 2-

legged @350mm c/c shall be provided.

400mm

2-20mmΦ

700mm d= 665mm4-20mmΦ

6-20mmΦ

6-20mmΦv, nominal shear stress = 0.69N/mm² (as before)

Assume that bars 2 and 5 are bent up at 45.

When 2 bars are bent up, remaining bars are 4 nos and they continued to the support.

Therefore % of tension reinforcement = (100*As/bd)

= (100*4*314/400*665)

= 0.47%.

therefore for 0.47% steel & M20 grade concrete,

c, permissible shear stress = 0.47N/mm².

0.47<1.03<2.8

c < v < c,max

Therefore provide shear reinforcement.

Vuc, shear resistance of concrete = c.b.d

= 0.47*400*665

= 125KN

Vus, shear to be resisted by stirrups = Vu-Vc

= 275-125

= 150KN.

Actual shear strength of 2 bent up bars.

= Asv*0.87fy*sin

= 2*314*0.87*415*sin45

= 160.3KN.

However, maximum contribution of shear possible by bent up bars = 50% of Vus

= 0.5*150

= 75KN

Net shear to be resisted by vertical stirrups = 0.5*150 = 75KN.

Using 8mmΦ 2-legged vertical stirrups, spacing

Sv = (Asv*0.87fy*d)/(0.5*Vus)

= (2*50*0.87*415*665)/(75*10³)

= 320mm

Maximum spacing from minimum shear reinforcement,

Sv = (Asv*0.87fy)/(0.4b)

= (2*50*0.87*415)/(0.4*400)

= 225.6mm

Maximum spacing = 0.75d

= 499mm

Therefore provide 8mmΦ 2-legged stirrups @225mm c/c through out the span in

addititional to bent up bars near the supports.

(C) When two groups(2-bars of 20mmΦ) of bars are bent at 45 and each group is

separated by 600mm.

600 600

d

Sv

a b tan = d/a tan = d/b a = (d/tan)= d.cot b = d/tan = d.cot

a+b = d.cot + d.cot

= d(cot +d.cot)

n, no of inclined bars or stirrups = d(cot +d.cot)/Sv

vertical component of tensile resistance in inclined bars crossing diagonal crack.

Vs = Asv*σsv*sin*n.

= [Asv*σsv*d*(cot+cot)*sin] / Sv

Assuming that the crack is located at = 45, the above relation simplifies to.

Vs = [Asv*σsv* d*(cot+cot)] / Sv

c,permissible shear stress:

% of steel = [(2*314)/(400*665)]*100

= 0.24%

for 0.24% steel & M20 concrete, c = 0.28N/mm²

0.35<1.03<2.8

Therefore design shear reinforcement

Vc, shear to be resisted by steel = 182.72-58.52

= 124.2KN

Actual shear resistance of series of bent up bars

= [Asv*σsv*d*(sin +cos)] / Sv

= [2*314*230*665*(sin 45 +cos45)] / 600

= 226.7KN.

However maximum contribution of bent up bars = 0.5*124.2

= 62.1KN.

Therefore Net shear to be resisted by vertical stirrups = 62.1KN.

Spacing of 8mmΦ 2-legged vertical stirrups

Sv = (Asv*σsv*d)/(0.5*Vs) = (2*50*230*665)/(0.5*124.2*10³)

= 246.3mm.

minimum shear reinforcement = 226mm

maximum spacing = 499mm

Therefore provide 8mmΦ 2-legged vertical stirrups @225mm c/c.

L.S.M

Vu = factored shear force or design shear force = 1.5V

v = nominal shear stress = Vu/bd

c = permissible shear stress is related to % of tension steel& grade of concrete(table:19)

c,max. = refer table 20 of IS:456-2000

Vuc = shear resistance of concrete = c.b.d.

Vus = shear to be resisted by steel = Vu-c.b.d

For vertical stirrups:

V'us = (0.87*fy*Asv*d)/Sv

For bent up bars(single group):

Vus = 0.87*fy*Asv*sin.

For series of bent up bars or inclined stirrups:

V'us = [0.87*fy*Asv*d*( sin +cos)]/Sv.

However V'us ≤0.5Vus

Minimum shear reinforcement:

Sv = (0.87*fy*Asv)/(0.4*b)

(Problem)

a rectangular beam 350mm wide and 500mm effective consist of 4-25mmΦ bars out

of which two bars are to be bent up at 1m from support with 45 inclination. Design

the shear reinforcement the beam is subjected to shear force of linear variation with

200KN at the support and zero at mid span, which is 3m from support. Use M25

concrete and Fe415 steel. Adopt L.S.M

Maximum shear force at the face of support = 200KN

Shear force at the critical section located at ‘d’ from the face of support

Vu = 200(2.5/3) = 166.7KN.

v, nominal shear stress = (Vu/bd) = (166.7*10³/350*500)

= 0.95N/mm²

% of tension steel after bending up 2-25mmΦ bars is (remaining bars consists of 2-

25mmΦ)

= 100*As/(bd) = (100*2*491)/(350*500) = 0.56

for 0.56% steel & M25 concrete

c,permissible shear stress = 0.51N/mm²

c,max. for M25 concrete = 3.1N/mm²

c <v < c,max.

Design shear reinforcement.

350 c.s for shear

2-25mmΦ

d = 500

2-25mmΦ 4-25mmΦ

Ast = 4-25mmΦ d

3m

1m

2.5

200KN Vu = 200*2.5/3

3m

Shear to be resisted by steel, Vus = Vu – Vuc

= Vu - c.b.d

= 166.7 – (0.51*350*500/1000)

= 77.5KN.

shear resistance of a single group of bent up bars , V'us = 0.87*fy*Asv*sin

= 0.87*415*2*491*sin45

= 250.7KN.< 0.5Vus

shear to be resisted by vertical stirrups = 0.5*Vus

= 0.5*77.5

= 38.75KN.

providing 8mmΦ 2-legged vertical stirrups spacing = (0.87*fy*Asv*d)/(0.5*Vus)

= (0.87*415*2*50*500)/38.5*10³

= 465.8mm

Spacing of minimum shear reinforcement = (0.87*fy*asv)/(0.4*b)

= (0.87*415*2*50)/(.04*350)

= 258mm

Maximum spacing = 0.75*d = 0.75*500

= 375mm

provide 8mmΦ 2-legged vertical stirrups @ 250mm c/c through out.

DEVELOPMENT LENGTH

BOND: The grip of the reinforcement and concrete due to adhesion or bearing is termed

as bond.

Development length: A length of reinforcement embedded in concrete so that it can

develop the stress by bond is termed as development length.

The bond between concrete and steel is responsible for the transfer of axial

force from a reinforcing bar to the surrounding concrete, there by providing strain

compatibility and composite action of concrete and steel.If this bond is adequate, slipping

of reinforcing bar will occur, destroying full composite action.

It is through the action of bond resistance that the axial stress in a reinforcing bar

can under go variation from point to point along its length had the bond been absent, the

stress at all points on a straight bar would be constant, as a string or a straight cable.

Bond resistance in R.C.C. is achieved through the following mechanism.

(1) Chemical action: Due to a gum-like property in the products of hydration formed

during the making of concrete.

(2) Frictional resistance: Due to surface roughness of the reinforcement and the grip

extended by the concrete shrinkage.

(3) Mechanical interlock: Due to surface protrusions or ribs provided in deformed

bars.

Bond stress( bd):

Bond resistance is achieved by the development of tangential stress (shear)

components along the inter face (contact surface) between the reinforcing bar and the

surrounding concrete. The stress so developed at the inter face is called bond stress, and

is expressed in terms of the tangential face per unit nominal surface area of the

reinforcing bar.

bd

T

Development length:

Pull out test:To find development length, consider a pull out test as shown in fig.

Diameter of rod = Φ

Embedment length = Ld

Pull = P

Average bond stress in concrete = bd

Working stress in bar = σs

Let the axial pull ‘P’ is increased gradully pull P is noted when the bar is puled out.

The bond resistance of concrete and the strength of bar can be equated.

bd*ΠΦ*Ld = (σs*ΠΦ²)/4

Ld = (σs*Φ)/(4*bd) = (Φσs)/(4*bd) [cl.26.2.1 of IS:456]

Where Ld is known as development length.

In working stress designs (ref. Table 21)

σs = σst = 140N/mm² for mild steel

= 230 N/mm² for HYSD bars.

In limit state method (ref.26.2.1.1)

σs =0.87fy

Flexural bond stress:

(1) (2)

M M+M

T T+T

x

At certain locations in a beam, high bond stress may arise due to large variations of

bending moment over a short distance, i.e. high shear force.These bond stress are called

flexural bond stresses and must be checked at the face of a simply support and at the

points of inflection with in continuous spans.At these locations, tensile capacity to be

developed is usually small but the rate of change of tensile stresses in the bars is high.

Let the increase in tension between sectio(1) and (2) x meters apart be equal to T

T = M/jd

Let bd be the local bond stress and o the parimeter of steel bars provided.

Equating the forces

bd * o * dx = T = dM /jd

bd = (dM/dx) * 1/(o*jd) = V/(o * jd) (1)

parallally for a group of ‘n’ bars the development length Ld required ,

n*Π*Φ*Ld*bd = Ast *σs

Ld = (Ast*σs)/( o*bd)

Or bd = (Ast*σs)/( o*Ld) (2)

Equating bd from equations (1) and (2)

V/(o*jd) = (Ast*σs)/( o*Ld)

Ld = (Ast*σs*jd)/V = M1/V (3)

Where M1 = M.R.with respect to tension steel alone at the section under consideration

V = shear force at the same section

If the permissible bond stress bd is not be exceeded, the ratio M1/V be must be equal to

or greater than Ld. As an additional factor of safety, the code has added, the anchorage

length Lo to the right hand side of Eq (3).,i.e.

Ld ≤ (M1/V)+Lo.

Where Lo = sum of the anchorage beyond the centre of support.

Anchorage bond stress:

(ref. development length)

anchorage value of hooks:

26.2.2.1

(a) Deformed bars may be used without end anchorages provided development length

required is satisfied.

(b) (1) Bends: The anchorage value of bend shall be taken as 4 times of the diameter

of the bar for each 45 bend subjected to a max. of 16Φ.

(2) Hooks: The anchorage value of standard U- type hook shall be equal to 16Φ.

4Φ-minBEND

nΦ8Φ

Φ (n+1)Φ

4Φ-min

nΦ STANDARD U-HOOK

16Φ Φ

(n+1)Φ

Min. value of ‘n’ for mild steel is 2 and for other steels is 4.