ec2 shear design

7/22/2019 EC2 Shear Design

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Sheet 1 of 6

Shear Design to EC2

30 N/mm2

1.5 #

20 N/mm2

500 N/mm2

1.15 #

750 mm

750 mm

50 mm

32 mm

10452 mm2

12 mm

672 mm2

2.19 N/mm2

0.0 N/mm2 (+) Compression, (-) Tension

Calculate Shear Capacity of Concrete, νRd,c

νRd,c = VRd,c/(bw*d) = [CRd,c*k*(100*ρ1*f ck) + k1*σcp]

Where;

k = 1 + (200/d) = 1 + √(200/672) = 1.55 (≤2)

ρ1 = (≤0.02)

0.18/1.5 = 0.12

k1 = 0.15

σcp = Applied axial stress = 0 (≤0.2*fcd = 4)

σcp = 0 N/mm2

νRd,c = [0.12*1.55*(100*0.02*30)^1/3 + 0.15*0] = 0.73 N/mm2

Calculate Minimum Concrete Shear Capacity, (νRd,c)min

(νRd,c)min = (VRd,c/bw*d)min = νmin + k1*σcp

where, νmin = 0.035*k3/2

*fck1/2

(νRd,c)min = [0.035*(0.15)^3/2*(30) 1̂/2]+[0.15* 0] = 0.37 N/mm2

Therefore, νRd,c = 0.73 N/mm2

0.87

60

νRd,c < νEd → Designed Shear Links Required

Applied Shear Stress, νEd:

Co-existent Axial Stress, σcp:

Section Properties

Design Cylinder Strength of Concrete, fcd:

Thickness/Depth of Element/Section t:

Width/Breadth of Element/Section bw:

Cover to Links :

Main Bar Diameter, Øs

Area of Tension Steel Provided, As1:

CRd,c = 0.18/γc =

Ratio of tensile steel = 10452/(750*672) = As1/(bw*d) =

Link Diameter, Øw:

Effective Depth, d:

Loading

Yield Strength of Reinforcement, fyk:

Material Safety Factor for Reinforcement, γs:

Chracteristic Cylinder Strength of Concrete, fck:

Material Safety Factor for Concrete, γc:

Calculated By:

Checked By:

Material Properties

EC2 Clause

Job No.:Div/Dept.:

Date:

Date:

Project:

Calculations for:

6.2.2(1)

0.0200



Sheet 2 of 6

Calculated By:

Checked By:

Job No.:Div/Dept.:

Date:

Date:

Project:

Calculations for:

Calculate Angle of Diagonal Compression Strut, θ

θ = 0.5*Sin-1{VEd/[0.18*bw*d*(1-fck/250)*fck]} = 0.5*Sin-1{νEd/[0.18*(1-fck/250)*fck]}

θ = 0.5 * arcsin{2.19/[0.18*(1-30/250)*30]} = 13.7 Degrees

But, 21.80 ≤ θ ≤ 450

So, θ = 21.8 Deg. & 2.5

0.4

Calculate Shear Links

Asw/s = VEd/(z*fywd*Cotθ) = VEd/(0.78*d*fyk*Cotθ) = νEd*bw*d/(0.78*d*fyk*Cotθ) (z=0.9d & fyk = fywd/γs)

Asw/s = 2.19*750*672/(0.78*672*500*2.5)

Asw/s = 1.68

Check Minimum Shear Links Requirement:

(Asw/s)min = 0.08*f ck0.5

*bw/f yk = 0.08*[(30)^0.5]*750/500 = 0.66

Therefore, Asw/s = 1.68

300 mm

250 mm

12 mm

113 mm2

504 mm2

4.46 #

3 #

339 mm2

0.67 FAIL

Shear Resistance Provided by Links:

νRd,s = VRd,s/(bw*d) = (Asw,prov/s)*z*f ywd*Cotθ = 0.78*(Asw,prov/s)*f yk*Cotθ/bw

νRd,s = 0.78*339*500*2.5/(750*300) = 1.469 N/mm2

Check Maximum Resistance to be Provided by Shear Links:

Max Shear Resistance Provided By Links:

νRd,max = VRd,max/(bw*d) = [αcw*bw*z*ν1*f cd/(Cotθ + Tanθ)]/(bw*d) = 0.9*αcw*ν1*f cd/(Cotθ + Tanθ)

6.2.3(3)

Tanθ =

Vertical

90

Type of shear reinforcement :

Angle of shear reinf to long. axis, α:

Link Details:

6.2.3(3)

Cotθ =

9.2.2(5)

Longitudinal spacing, s:

6.2.3(2)

No of legs provided:

Area Provided, Asw,prov:

Asw,prov/Asw:

Horizontal spacing btwn legs, sh:

Link diameter, Øw:

Bar area, Aw:

Area required, Asw:

No of legs required:



Sheet 3 of 6

Calculated By:

Checked By:

Job No.:Div/Dept.:

Date:

Date:

Project:

Calculations for:

Calculate Design Stress of Shear Reinforcement, σw:

σw = vEd*bw*d*s/(Asw,prov*z*Cotθ)

σw = 2.19*750*672*300/(339*0.9*672*2.5)σw = 646 N/mm

2

σw > 0.8*fyk & fck ≤ 60 ==========> ν1 = ν*(1-0.5*Cosα)

Calculate Strength Reduction Factor for Concrete Cracked in Shear, ν1:

ν = 0.6*[1 - f ck/250] = 0.53

ν1 = ν*(1-0.5*Cosα)

ν1 = 0.53*(1-0.5*Cos90) = 0.53

Calculate Coefficient to Account for Stress in Compr Chord, αcw:

σcp = 0.0 N/mm2

=======> No Pre-Stress!

αcw = 1.00

0.9*1*0.53*20/(2.5+0.4)

3.29 N/mm2

Calculate Additional Longitudinal Force:

Additional longitudinal force = (VEd/Sinθ)*Cosθ = (νEd*bw*d/Sinθ)*Cosθ = νEd*bw*d*Cotθ

Longitudinal force to be resisted by tension zone steel, ΔFtd = 0.5*νEd*bw*d*Cotθ

ΔFtd = 0.5*2.19*750*672*2.5 =

mm

mm2

mm2

#

mm

mm

1380 kN

6.2.2(6)

6.2.3(3)

188

156

νRd,max =

νRd,max =

Clear spacing between bars:

Shear capacity provided by links below max allowable… OK!

3174

No of bars required:

Horiz. c/c spacing of bars, s:

4

Bar area, As:

32

804

Equivalent Tensile Steel:

Area required, As,req:

Bar diameter, Øw:



Shear Design to EC2

30 N/mm2

1.5 #

20 N/mm2

500 N/mm2

1.15 #

750 mm

750 mm

50 mm

32 mm

10452 mm2

12 mm

672 mm2

2.19 N/mm2

0.0 N/mm2 (+) Compression, (-) Tension

Calculate Shear Capacity of Concrete, νRd,c

νRd,c = VRd,c/(bw*d) = [CRd,c*k*(100*ρ1*fck)1/3 + k1*σcp]

Where;

k = 1 + (200/d) = 1 + √(200/672) = 1.55 (≤2)

ρ1 = (≤0.02)

0.18/1.5 = 0.12

k1 = 0.15

σcp = Applied axial stress = 0 (≤0.2*fcd = 4) σcp = 0 N/mm

2

νRd,c = [0.12*1.55*(100*0.02*30)^1/3 + 0.15*0] = 0.73 N/mm2

Calculate Minimum Concrete Shear Capacity, (νRd,c)min

(νRd,c)min = (VRd,c/bw*d)min = νmin + k1*σcpwhere, νmin = 0.035*k3/2*fck1/2(νRd,c)min [0.035*(0.15) 3̂/2*(30)^1/2]+[0.15* 0] = 0.37 N/mm

2

Therefore, νRd,c 0.73 N/mm2

0.87

60

νRd, < νEd → Designed Shear Links Required

Project:

KOC New Water Centre

Calculations for: Div/Dept.: Greenfield Job No.: EF1920

Apr-13

Checked By: Date:

Material Safety Factor for Reinforcement, γs:

Vehicle Access Bridge - Shear DesignCalculated By: SdeS Date:

Material PropertiesChracteristic Cylinder Strength of Concrete, fck:

Material Safety Factor for Concrete, γc:

Design Cylinder Strength of Concrete, fcd:

Yield Strength of Reinforcement, fyk:

EC2 Clause

Section Properties

Thickness/Depth of Element/Section t:

Width/Breadth of Element/Section bw:

Cover to Links :

Main Bar Diameter, Øs

Area of Tension Steel Provided, As1:

Link Diameter, Øw:

Effective Depth, d:

Loading

Applied Shear Stress, νEd:

Co-existent Axial Stress, σcp:

6.2.2(1)

Ratio of tensile steel = As1/(bw*d) = 10452/(750*672) = 0.0200

CRd,c = 0.18/γc =



Project:



Apr-13

Checked By: Date:Vehicle Access Bridge - Shear Design

Calculated By: SdeS Date:

Calculate Angle of Diagonal Compression Strut, θ

θ = 0.5*Sin-1{VEd/[0.18*bw*d*(1-fck/250)*fck]} = 0.5*Sin-1{νEd/[0.18*(1-fck/250)*fck]}

θ = 0.5 * arcsin{2.19/[0.18*(1-30/250)*30]} = 13.7 Degrees

But, 21.80 ≤ θ ≤ 450

So, θ = 21.8 Deg. & 2.5

0.4

Calculate Shear Links

Asw/s = VEd/(z*fywd*Cotθ) = VEd/(0.78*d*fyk*Cotθ) = νEd*bw*d/(0.78*d*fyk*Cotθ) (z=0.9d & fyk = fywd/γs)

Asw/s = 2.19*750*672/(0.78*672*500*2.5)

Asw/s = 1.68

Check Minimum Shear Links Requirement:

(Asw/s)min 0.08*fck0.5*bw/fyk 0.08*[(30)^0.5]*750/500 = 0.66

Therefore, Asw/s = 1.68

250 mm

213 mm

12 mm

113 mm2

420 mm2

3.72 #

4 #

452 mm2

1.08 OK

Shear Resistance Provided by Links:

νRd,s = VRd,s/(bw*d) = (Asw,prov/s)*z*fywd*Cotθ = 0.78*(Asw,prov/s)*fyk*Cotθ/bw

νRd,s = 0.78*452*500*2.5/(750*250) = 2.350 N/mm2

Check Maximum Resistance to be Provided by Shear Links:

Max Shear Resistance Provided By Links:

νRd,max = VRd,max/(bw*d) = [αcw*bw*z*ν1*fcd/(Cotθ + Tanθ)]/(bw*d) = 0.9*αcw*ν1*fcd/(Cotθ + Tanθ)

Cotθ =

Tanθ =

9.2.2(5)

6.2.3(2)

6.2.3(3)

Type of shear reinforcement : Vertical

Angle of shear reinf to long. axis, α: 90

6.2.3(3)

Link Details:

Longitudinal spacing, s:

Horizontal spacing btwn legs, sh:

Link diameter, Øw:

Bar area, Aw:

Area required, Asw:

No of legs required:

No of legs provided:

Area Provided, Asw,prov:

Asw,prov/Asw:



Project:



Apr-13

Checked By: Date:Vehicle Access Bridge - Shear Design

Calculated By: SdeS Date:

Calculate Design Stress of Shear Reinforcement, σw:σw = vEd*bw*d*s/(Asw,prov*z*Cotθ)σw = 2.19*750*672*250/(452*0.9*672*2.5)

σw = 404 N/mm2

σw > 0.8*fyk & fck ≤ 60 ==========> ν1 = ν*(1-0.5*Cosα)

Calculate Strength Reduction Factor for Concrete Cracked in Shear, ν1:ν = 0.6*[1 - fck/250] = 0.53

ν1 = ν*(1-0.5*Cosα)

ν1 = 0.53*(1-0.5*Cos90) = 0.53

Calculate Coefficient to Account for Stress in Compr Chord, αcw:σcp = 0.0 N/mm2 =======> No Pre-Stress!

αcw = 1.00

0.9*1*0.53*20/(2.5+0.4)

3.29 N/mm2

Calculate Additional Longitudinal Force:

Additional longitudinal force = (VEd/Sinθ)*Cosθ = (νEd*bw*d/Sinθ)*Cosθ = νEd*bw*d*Cotθ

Longitudinal force to be resisted by tension zone steel, ΔFtd = 0.5*νEd*bw*d*Cotθ

ΔFtd = 0.5*2.19*750*672*2.5 =

mm

mm2

mm2

#

mm

mm

6.2.2(6)

6.2.3(3)

νRd,max =

νRd,max =

Shear capacity provided by links below max allowable… OK!

Equivalent Tensile Steel:

Bar diameter, Øw: 32Bar area, As: 804

1380 kN

Clear spacing between bars: 156

Area required, As,req: 3174

No of bars required: 4

Horiz. c/c spacing of bars, s: 188

ec2 shear design

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