types of shear failure · = shear force in the compression zone (20 – 40%) v a = interlocking...

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V

V

CASE 1: av/d 6

av

d

V CASE 2: 2 av/d 6

av

d

V

av

V

CASE 3: av/d 2

d

Types of Shear Failure

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Shear force is transmitted through the crack member by a combination of the uncracked concrete in compression

zone, Vcz, the dowelling action of the flexural reinforcement, Vd and aggregate interlocking across

tension cracks, Va

V

Vcz = Shear force in the compression zone

(20 – 40%)

Va = Interlocking between aggregates (35 – 50%)

Vd = Dowel action (35 – 50%)

Concrete compression

Steel tension

d

Shear Resistance

EC2 Shear Design

EC 2 introduces the strut inclination method

for shear capacity checks. In this method the

shear is resisted by concrete struts acting in

compression and shear reinforcement acting in

tension.

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EC2: Cl. 6.2.3

EC2: Cl. 6.2.3

z cot

VEd

Longitudinal steel in tension

Concrete strut in compression Vertical shear steel in tension

z =

0.9

d

d

bw

z cos

z

fcdbwz cos VRd, max

VEd

sin

VEdcot

VEd

Assumed truss model for the strut inclination method

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(1) Diagonal Compressive Strut

VRd, max = [fcd (bw z cos θ)] sin θ

= fcd bw z cos θ sin θ

In EC2 this equation is modified by the inclusion of a strength

reduction factor for concrete cracked in shear v1 and the introduction

of coefficient taking account of the state of the stress in compression

chord αcw thus,

VRd, max = αcwv1fcd bw z / (cot θ + tan θ)

= αcwv1(fck/1.5) bw (0.9d) / (cot θ + tan θ)

= αcwv10.6fck bwd / (cot θ + tan θ)

)tan(cot

)250/1(36.0 ckwckmaxRd,

fdbfV

It is set by the EC2 to limit the θ value between 22 to 45.

The recommended value for αcw and v1 are given in Cl. 6.2.3

EC2. For the purpose of this module the following values are

used: αcw = 1.0 and v1 = 0.6 (1 – fck/250), hence

(1) Diagonal Compressive Strut

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(2) Vertical Shear Reinforcement

The shear resistance of the link is given by

VEd = VRd, s = fywdAsw

= (fyk / 1.15) Asw

= 0.87fykAsw

If the links are spaced at a distance s apart, then

the shear resistance of the link is increased

proportionately and is given by;

VEd = VRd, s = 0.87fykAsw (z cot θ / s)

= 0.87fykAsw (0.9d cot θ / s)

= 0.78fykAswd (cot θ / s)


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Thus, rearranging them gives;

cot78.0 yk

Edsw

df

V

s

A


EC2 (Cl. 9.2.2) specifies a minimum value for Asw/s

such that;

yk

ckwsw08.0

f

fb

s

A

EC2 (Cl. 9.2.2) also specifies that the maximum

spacing of vertical link, s should not exceed 0.75d.


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(3) Additional Longitudinal Force

The longitudinal tensile force, Ftd is caused by the

horizontal component required to balance the

compressive force in the inclined concrete strut;

Longitudinal force = (VEd/sin ) cos

= VEd cot

Assumed half of this force is carried by the reinforcement

in the tension zone of the beam, then the additional tensile

force provided in the tensile zone is given by;

∆Ftd = 0.5VEd cot θ

Shear Design Procedure in EC2

Start

Determine shear

force, VEd

Determine VRd, max for

cot = 1.0 ( = 45) and

cot = 2.5 ( = 22)

If VEd VRd, max

cot = 1.0

If VRd, max cot θ = 2.5

VEd VRd, max cot θ = 1.0

Red

esig

n s

ection

If VEd VRd, max

cot = 2.5

A B

)tan(cot

)250/1(36.0 ckwckmaxRd,

fdbfV

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Calculate shear

reinforcement

using cot = 2.5

A

df

V

df

V

s

A

yk

Ed

yk

Edsw 513.0

cot78.0

C


Calculate

B

)250/1(18.0sin5.0

ckck

Ed1

fdfb

V

w

Calculate shear

links cot78.0 yk

Edsw

df

V

s

A

C

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Calculate additional

longitudinal tensile

force caused by shear

C

End

Calculate minimum links

required by EC2: Cl.

9.2.2(5) and s 0.75d

Flanged

beam?

Yes

No

D

Example 1

Design the required shear reinforcement from the beam section shown

below. Take fyk = 500 N/mm2 and fck = 30 N/mm2.

100 kN/m 225 mm

d = 500 mm

2H16

3H25

(1473 mm2)

8 m

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Example 1: Solution

Maximum shear force;

VEd = wL/2 = 100 8.0 /2 = 400 kN

Concrete strut capacity;

VRd, max = 0.36bwdfck(1 – fck/250) / (cot + tan )

= 0.36 225 500 30 (1 – 30/250)

(cot + tan )

for = 22, cot = 2.5 VRd, max = 371 kN

= 45, cot = 1.0 VRd, max = 535 kN

VRd, max cot = 2.5 (371 kN) VEd (400 kN) VRd, max cot = 1.0 (535 kN)

Therefore, 22

Example 1: Solution

= 0.5sin-1 [VEd / 0.18bwdfck(1 – fck/250)]

= 0.5sin-1 400 103

0.18 225 500 30 (1 – 30/250)

= 24.2

tan = 0.45 ; cot = 2.22

Shear links;

Asw/s = VEd / 0.78fykdcot

= 400 103 / (0.78 500 500 2.22)

= 0.923

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Example 1: Solution

Try link: H10 Asw = 157 mm2

Spacing, s = 157/0.923

= 170 mm smax = 0.75d (375 mm)

Provide H10-150 mm

Minimum links;

Asw/s = 0.08fck1/2bw / fyk

= 0.08 (30)1/2 225 / 500

= 0.197


Spacing, s = 157/0.197

= 797 mm smax = 0.75d (375 mm)

Provide H10-350 mm

Example 1: Solution

400

400

217

217

H10-350 H10-150 H10-150

1.83 m 1.83 m 4.33 m

Vmin = (Asw/s)(0.78d fykcot )

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Example 1: Solution

Additional longitudinal reinforcement;

Additional tensile force, Ftd = 0.5VEd cot

= 0.5 400 2.22

= 445 kN

Additional tension reinforcement, As = Ftd / 0.87fyk

= 445 103 / 0.87 500

= 1022 mm2

Provide 3H25 ( 1473 mm2)

Flanged Beam

Shear between the web and flanged of a flanged section

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Flanged Beam

).(

dEd

xh

Fv

f

f

wf

f

d

2/)(x

)2/( b

bb

hd

MF

The longitudinal shear stress, vEd at the web-flange interface is determine

according to;

M = the change in moment over the distance ∆x

where:

x = half the distance between the sections with zero moment and that

where maximum moment occurs. Where point loads occur, x should

not exceed the distance between the loads.

Flanged Beam

The permitted range of the values of cot f are recommended as follows:

1.0 cot f 2.0 for compression flanges (45 f 26.5)

1.0 cot f 1.25 for tension flanges (45 f 38.6)

The concrete strut capacity of the flange is given by;

vRd = v fcd sin f cos f

= 0.6 (1 – fck/250) (fck/1.5) sin f cos f

= 0.4fck (1 – fck/250)

(cot f + tan f)

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Flanged Beam: Design Procedure

D

Calculate longitudinal

design shear stress, vEd

vEd 0.27fctk 0.4fctd = 0.4(fctk/1.5) = 0.27fctk

No

Yes

Check the stresses in the

incline strut. Compare vEd

with vRd, max

).(

dEd

xh

Fv

f

vEd vRd, max Use min

Yes No E F

F

Flanged Beam: Design Procedure

E

Calculate

and cot f

Calculate transverse

shear reinforcement

)250/1(2.0sin5.0

ckck

Ed1

ff

v

fyk

Ed

f

sf

cot87.0 f

hv

s

A f

Calculate minimum

transverse steel area

Calculate additional

longitudinal tensile

force caused by shear

End

F

where b = 1000 mm

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Example 2

Design the required shear reinforcement from the beam section shown

below. Take fyk = 500 N/mm2 and fck = 25 N/mm2.

90 kN/m

9 m

250 mm

d =

53

0 m

m

3H20

d’ =

45

mm

2H12

600 mm

110 mm

Example 2: Solution

Maximum shear force;

VEd = wL/2 = 90 9.0 /2 = 405 kN

Concrete strut capacity;

VRd, max = 0.36bwdfck(1 – fck/250) / (cot + tan )

= 0.36 250 500 25 (1 – 25/250)

(cot + tan )

for = 22, cot = 2.5 VRd, max = 373 kN

= 45, cot = 1.0 VRd, max = 537 kN

VRd, max cot = 2.5 (373 kN) VEd (405 kN) VRd, max cot = 1.0 (537 kN)

Therefore, 22

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Example 2: Solution

= 0.5sin-1 [VEd / 0.18bwdfck(1 – fck/250)]

= 0.5sin-1 400 103

0.18 250 500 25 (1 – 25/250)

= 24.5

tan = 0.46 ; cot = 2.19

Shear links;

Asw/s = VEd / 0.78fykdcot

= 400 103 / (0.78 500 530 2.19)

= 0.893

Example 2: Solution


Spacing, s = 157/0.893

= 176 mm smax = 0.75d (398 mm)

Provide H10-150 mm

Minimum links;

Asw/s = 0.08fck1/2bw / fyk

= 0.08 (25)1/2 250 / 500

= 0.200


Spacing, s = 157/0.200

= 786 mm smax = 0.75d (398 mm)

Provide H10-375 mm

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Example 2: Solution

405

405

214

214

H10-375 H10-150 H10-150

2.12 m 2.12 m 4.76 m

Vmin = (Asw/s)(0.78d fykcot )

Example 2: Solution

Transverse steel in the flange;

x = 0.5(L/2) = 9000/4 = 2250 mm

Change of moment over distance x from zero moment:

M = (wL/2)(L/4) – (wL/4)(L/8) = 683.44 kNm

Change in longitudinal force;

= 683.44 103 (600 – 250)

(530 – 55) (2 600)

= 420 kN

f

wf

f

d

2/)(x

)2/( b

bb

hd

MF

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Example 2: Solution

Longitudinal shear stress;

vEd = Ftd / (hf x)

= 420 103 / (110 2250)

= 1.70 N/mm2

Since vEd (1.70 N/mm2) 0.27fctk = 0.27 1.80 = 0.49 N/mm2

Transverse steel reinforcement is required

Example 2: Solution

Concrete strut capacity in the flange;

vRd, max = 0.4fck (1 – fck/250) / (cot + tan )

= 0.4 25 (1 – 25/250)

(cot + tan )

for = 27, cot = 2.0 vRd, min = 3.59 N/mm2

= 45, cot = 1.0 vRd, max = 4.50 N/mm2

vEd (1.70 N/mm2) vRd, min cot = 2.0 (3.59 N/mm2)

and

vEd (1.70 N/mm2) VRd, max cot = 1.0 (4.50 N/mm2)

Therefore, use = 27 ; tan = 0.50 ; cot = 2.0

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Example 2: Solution

Transverse shear reinforcement;

Asf / sf = vEdhf / 0.87fykcot

= 1.70 110 / (0.87 500 2.0)

= 2.0

Try H10: Asf = 79 mm2

Spacing, sf = 79/0.21 = 367 mm

Minimum transverse steel area;

As, min = 0.26 (fctm/fyk) bhf

= 0.26 (2.60/500) bhf

= 0.0013bhf 0.0013bhf

= 0.0013 1000 110 = 147 mm2

Provide H10 – 300 (As = 262 mm2/m)

Example 2: Solution

Additional longitudinal reinforcement;

Additional tensile force, Ftd = 0.5VEd cot

= 0.5 405 2.19

= 444 kN

Additional tension reinforcement, As = Ftd / 0.87fyk

= 444 103 / 0.87 500

= 1021 mm2

Provide 3H25 ( 1473 mm2)

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types of shear failure · = shear force in the compression zone (20 – 40%) v a = interlocking...

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