ecommended values for β (see Concise EC2)

ProjectExercise 9)

Part of structurePunching Shear design for two way spanning slab

Job ref

Sheet no1 / 3

Drawing ref.see Ref

Calc by

KWDate

29/11/2010Checkb

Date

Ref Calculations Output

Exercise 9:An internal 400 mm square column supports a 300mm thick slab.

The ULS reaction is 1042kN

The cover was determined to 30mm.

Characteristic concrete strength f = 30N /mm2 ck

The flexural design requires H20 @ 125mm c/c (Asl= 2513mm2) in y direction (T1)and H25@150mm (Asl= 3272mm2) c/c in z direction (T2)

‐ Calculate applied shear force VEd,eff = β x VEd

whereV Ed = applied shear forceβ = factor dealing with eccentricity

R

Therefore β = 1.15

For internal rectangular columns with No load eccentricity.

For columns with loading eccentric to one axis.(See calculation of b Concise EC2 (8.2.3))

Ved,eff = 1.15 x 1042 = 1199 kN

ProjectExercise 9)

Part of structurePunching Shear design for two way spanning slab

Job ref

Sheet no2 / 3

Drawing ref.see Ref

Calc by

KWDate

29/11/2010Checkb

Date

Ref Calculations Output

6.6N

2.4.2.4 (1)

BS EN1992

6.47

9.2.2(5), Exp. (9.4)

Exp. (9.5N) & NA

BS EN1992

6.47

- Mean Value of effective depth:

dy = 300 - 30 -20 - 25/2 = 237.5 mm dz = 300 - 30 -20 / 2 = 260 mm

deff = (237.5 + 260 ) / 2 = 249 mm

- Shear stress at face of column (check at column perimeter uo)

check if shear reinforcement is required @ u0

Adjacent to the column the punching shear resistance is limited toa maximum of:

uo = 400 x 4 = 1600 mm

Shear stress ved = β x Ved / (uo x d ) = 1199 x 10^3 / ( 1600 x 249 ) =

- Shear resistance of concrete strut:

= 0.5 x 0.53 x 20 =

= 0.6 [1-30/250] = 0.53

fck = 30N /mm2 fcd = αcc fck / γc = 1 x 30/1.5 = 20 N/mm2

3.01 N/mm^2 < 5.30 N/mm^2

- Shear resistance of longitudinal reinforcement:

Determine V Rd,c

With No axial force (σcp= 0) and C Rd,c is 0,18/ γ c = 0,18/ 1,5 = 0.12

= 0.12 x 1.90 (100 x 0.006 x 30 )^1/3

Where:

= 1 + (200/249)1/2 = 1.90 < 2

= (0.005 x 0.006)^0.5 = 0.006 < 0.02

= 2513 mm2/ ((400+249 x 6) x 249) = 0.005

= 3272mm2/ ((400+249 x 6) x 249) = 0.007

But a minimum of:

With No axial force (σcp= 0)

= 0.035 x 30^(1/2) x 1.90 ^(3/2)

Therefore v = 1.4 N/mm2< 3.01 N/mm^2Rd,c

3.01 N/mm^2

5.30 N/mm^2

Ok

1.40 N/mm^2

Ok

0.50 N/mm^2

ProjectExercise 9)

Part of structurePunching Shear design for two way spanning slab

Job ref

Sheet no3 / 3

Drawing ref.see Ref

Calc by

KWDate

29/11/2010Checkb

Date

Ref Calculations Output

EC2 (6.52)

- Calculate shear reinforcement between u0 and u1:

From:

Where:

vRd,cs = vEd = 3.01 N/mm^2

vRd,c = 1.40 N/mm^2

d = deff = 249 mm

cos α = 1

fywd,ef = 250 + 0.25deff = 250 + 0.25 x 249mm= 312 N/mm2

fywd,ef = 312 N/mm2 < 500 / γs = 500 / 1,15 = 435 N/mm2

Therefore use fywd,ef = 312 N/mm2

uradius = 2 x 249mm x 2 x π = 3129 mm

u1 = 3129 mm + 4 x 400mm = 4729 mm

sr < 0.75 deff = 0.75 x 249mm = 187 mm use say sr =

Therefore resolve equation as follows:

Asw > (vEd - 0.75 vRd,c) sr u1 / 1.5 fywd,ef

Asw > ((3.01N/mm2 - 0.75 x 1.4N/mm2) x 125mm x 4729mm) / (1.5 x 312.25N/mm2) =

Asw,min > 0.08 fck (sr x st) /(1.5fyk sin α + cos α)

where:

st = (transversal spacing)

st = 1.5 x d (if inside u1) st = 2.0 x d (if outside u1)

therefore st < 1.5 x 249mm = 373.5 mm2 use

Asw,min > 0.08 x 30^0.5 (125 x 300) /(1.5 x 500 x 1 + 0) = 21.9 mm2

Therefore use :

Asw > 2474 mm2 use : H20 @ 125mm c/c = 2513 mm2

Note: To choose the reinforcement you vary the spacing to find a sensible diameter.

- check if shear reinforcement is required @ u1:

Shear stress ved = β x Ved / (u1 x d ) = 1199 x 10^3 / ( 4729 x 249 ) =

ved = 1.02 N/mm^2 < 1.40 N/mm^2 = vRd,c

Ok

125 mm

2474 mm2

300 mm

1.02 N/mm^2

Therefore shear reinforcement not required @ u1 provide Asw beween u0 and u1 as above