SHEAR DESIGN.doc

7/27/2019 SHEAR DESIGN.doc

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SHEAR REINFORCEMENT DESIGN

Concrete beam can fail due to excessive shear as shown below.

Shear Failure

Shear failure in reinforced concrete beams is complex andcan occur in several ways. A typical failure mode for a simply

supported beam is illustrated in the Figure below , whichalso shows how reinforcement can assist in resisting theshear.

HJ ROSLAN BIN KOLOPPUSAT PENGAJIAN DIPLOMA (PPD)

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Typical failure mode due to shear for a simply supportedbeam

Shear reinforcement is either in the form of :(1) Stirrups (Links)(2) Inclined bars (used together with stirrups)

In EC2 the method of shear design is known as The Variable StrutInclination Method.

In a reinforced concrete beam with vertical links, shear forces areconsidered to be carried by the links in tension acting with diagonalconcrete struts in compression, as shown below:


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Beam carrying shear: links in tension and concrete incompression

The angle of concrete strut varies depending on the shear forceapplied. Usually designer choose any value of , between 22 and 45 o

(i.e cot between 1 and 2.5) depending on the applied shear. It canbe noticed that if the value assumed for is small then therequirement for shear reinforcement is also small. Thus, to be

economical use the smallest value of as possible as shown below.


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Fig: Choice of

EC2 allows the designer to vary the angle of the strut to obtain themost economical solution. However an angle of 22 (the minimum

allowed in EC2) will give practical designs in most cases, and thisapproach is adopted in this manual.

Note that for a rectangular beam the breadth bw used in the shearcalculation is equal to the overall breadth b . For T and L beams bw isthe breadth of the web, as shown in below:

Fig: b and bw for rectangular, L and T beams in shear

The procedure for checking the shear resistance of a concrete beaminvolves:(a) verifying that the concrete has suffi cient capacity at the

face of the support.


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(b) reinforcement design is based on the shear force at adistance equal to one effective depth from the face of thesupport as shown in the Figure below:

(c) the requirements for minimum reinforcement should be

checked and a suitable arrangement of links should bechosen.

Figure: Critical sections for a beam carrying shear


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Design Procedures 1 (For = 220)

1. Find VEd1 = shear force at the face of the support.

2. Find vEd1=shear stress at the face of the support = VEd1 /(0.9 bwd).

3. Find the concrete strut capacity vRd, from Table below .

Table: Rd concrete strut capacities for calculations of shearin beams

4. IfvEd1 is greater than vRd then see the design procedure 2for use of other values of up to 45 . In some cases it maybe necessary to use a larger beam or a higher class ofconcrete.

5. IfvEd1 is not greater than vRd then use = 22 and follow

steps 6 11.

6. Find VEd2 = shear force at a distance dfrom the face of thesupport.

7. Find vEd2 = shear stress at a distance dfrom the face ofthe

support = VEd2 /(0.9 bwd).

8. Calculate the area of shear reinforcement required:Asw/s = 0.4vEd2b w / 0.87 fyk .

Since fyk is always 500 N/mm2 , this givesAsw/s = 0.00092 v

Ed2bw .


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9. Find the minimumAsw/s from Table below .

Table: Minimum shear reinforcement in beams

10. Consider the following limits to the spacing of the linksalong the

beam:Minimum spacing 75 mm

Maximum spacing 0.75 dbut not more than 600 mm.

Economy will be achieved by having as few links aspossible, so in step 11 it is best to choose a spacing sclose to the maximum permitted value.

11. Choose a link sizeAsw and link spacing s so thatAsw / s isnot less than the values from steps 8 and 9. Tablebelow may be used for single links, and otherarrangements using multiple links are shown in Figure3.13 .

Table 3.22 : Area of shear linksAsw/s (mm2 /mm) for various

link sizes and spacings (based on two legs per link)


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Figure 3.13 : Examples of shear reinforcement in the form oflinks.

This procedure determines the shear reinforcement neededat the support. In many beams it is possible to use less shear

reinforcement towards the centre of the span where theshear force is lower, and this can often be achieved by usingthe same size of links but increasing the links spacing s .

This module does not cover the calculations required to dothis, and the chosen link arrangement should be used for thefull length of the beam.

EXAMPLE

A simply supported reinforced concrete beam with aneffective span of 7.0 m is 500 mm deep overall by 250 mmwide (see Figure below). It supports the followingcharacteristic loads:

Permanent dead loads: 12.0 kN/m plus beam self-weightVariable imposed loads: 11.0 kN/m.

The concrete is grade C40/50, and 25-mm cover is requiredto all reinforcement. Determine the shear reinforcement

required.


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SOLUTIONS

Beam width b= bw = 250 mm

Beam overall height h = 500 mm

Effective span L = 7.0 m

F= 1.35 G k + 1.50 Q k, = 1.35 x 105.9 + 1.50 x 77.0 =258.5 kN

Concrete strength fck = 40 N/mm2

Effective depth d= 500 25 10 12.5 = 452.5 mm

The figure below shows part of the Shear Force diagram. Thevalues in the diagram are calculated below:


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Reaction = F/2 = 258.5/2 = 129.3 kN

Width of support =150 mm, so distance from centre ofsupport

to face of support = 0.15/2 = 0.075 m

VEd1 = 129.3 258.5 x 0.075/7.0 = 126.5 kN = 126.5 x 103 N

vEd1 = VEd1 /(0.9 bwd) = 126.5 x 103 /(0.9 x 250 x 452.5)

= 1.24 N/mm2

From Table with fck = 40 N/mm2 ; vRd = 4.63 N/mm

2

Checkv

Ed1 is not more thanv

R d OK

Effective depth d= 0.4525 m, soVE d2 = 126.5 258.5 x 0.4525/7.0 = 109.8

kN

vEd2 = VEd2 /(0.9 b wd)= 109.8 x 103 /(0.9 x 250 x 452.5)

= 1.08 N/mm2

Asw /

s= 0.00092

vEd2

bw = 0.00092 x 1.08 x 250 = 0.248mm2 /mm

From Table , with fck = 40 N/mm2 ,

Min.A sw / s = 0.0010 b w = 0.0010 x 250 = 0.25 mm2 /mm

Max . link spacing = 0.75 d= 0.75 x 452.5 = 339 mm

From Table, withA sw/s not less than 0.25 mm2 /mm,

by interpolation.Use H8 links at 325 mm centres

A sw/s = 0.309 mm2 /mm(alternatively, 2x x d2/4 /325 = 0.25, d=7.19 mm (Use H8)


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DESIGN PROCEDURE 2 ( 220 220 and

therefore must be calculated from the followingexpression:

1sin5.0

= {}

)250

1(18.0 ckck

w

Ef

ff

db

V

450 --------(2)


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Or0

)45max(,

145sin5.0

=

Rd

Ef

V

V

If this calculation gives a value of > 450 then the beam

should be re-sized or a higher class concrete could beused.

3. The shear link required can be calculated using theexpression:

cot78.0yk

Edsw

df

V

s

A= -------(3)

where Asw is the cross-sectional area of the legs of the links

(2x4

2

for single stirrups)

For a predominantly Uniformly Distributed Load the shear VEDshould be calculated at a distance d from the face of thesupport and the shear reinforcement should continue to theface of the support.

The shear resistance for the links actually specified is

Vmin = cot78.0 yksw dfxs

A-------(4)

4. Calculate the minimum links required by EC2 from:

yk

wcksw

f

bf

s

A 5.0min, 08.0= ----------(5)

5. Calculate the additional longitudinal tensile force caused bythe shear.

cot5.0 Edtd VF = ------(6)


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EXAMPLE

The beam shown below spans 8.0 m on 300 mm wide supports. Itcarries a uniformly distributed ultimate load of 200 kN/m. Check if theshear reinforcement in the form of vertical links shown is sufficient.Given fck = 30 N/mm

2)

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SOLUTION

Total ultimate load on beam = 200 x 8.0 = 1600 kNSupport Reaction = 1600/2 = 800 kNShear, VEf at face of support = 800 200x0.3/2 = 770 kNShear, VEd distance d from face of support= 770 200x0.65 = 640 kN

STEP 1

vEd1 = VEd1 /(0.9 bwd) = 770x10

3

/(0.9x350x650) = 3.76N/mm2

STEP 2

From Table, for fck = 30 N/mm2 vRd = 3.64 N/mm

2


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Since vEd1 > vRd proceed to Procedure 2

Check the crushing strength VRd,max of the concrete diagonal strut atthe face of the beam support.

Using equation 1 with =220

ckckwRd ffdbV )250/1(124.0)22max(, =

= 0.124 x 350 x 650(1 - 30/250)30= 745 KN (VEf= 770 kN)

Therefore, 220 <


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s = 175 mm ( max spacing = 0.75d = 0.75 x 650 = 487 mm ) OK

29.1175

226==

s

ASW

The shear resistance of the links can be calculated using equation 3:

cot78.0, ykSW

sRd dfxs

AV =

= 1.29 x 0.78 x 650 x 500 x 2.39 x 10 -3 = 781 kN

Design shear, VEd distance d from the face of the support = 640 kN

(


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Q2. An RC beam section is shown below, and resists an ultimate factoredshear of 200 kN at the face of support. Use EC2 to determine the shearreinforcement to be provided.

d = 550

425

b = 300

Given, fck =40 N/mm2, fy = 500 N/mm

2


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DL=2.25 kN/m; IL=200kN/m

d=400

200

225

300


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ASSIGNMENT

Rajah di bawah menunjukkan pelan lantai tingkat kedua sebuah bangunan.

Data berikut diberi,

Ketebalan papak konkrit = 125 mmBeban kenaan di atas papak = 3.0 kN/m2

Berat konkrit = 25 kN/m3

fyk = 500 N/mm2

fck = 30 N/mm2

Berat dinding bata di atas rasuk = 2.0 kN/m

Kemasan lantai = 1.5 kN/m2

Reinforcement Cover = 25 mm

Dengan mengandaikan rasuk sekunder 2/B-C sebagai tupang mudah,

(i) Kirakan beban rekabentuk pada rasuk.(ii) Kirakan daya ricih rekabentuk untuk rasuk.

(iii) Rekabentuk tetulang ricih bagi rasuk tersebut.


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125

450

300

All dimensions in mm

Beam D/1-3

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