## Punching shear central column C

At C2, applied shear force, VEd = 1204.8 kN Check at perimeter of column:

P = factor dealing with eccentricity; recommended value 1.15 VEd = applied shear force U = control perimeter under consideration.

For punching shear adjacent to interior columns u0 = 2(cx + cy) = 1600 mm d = mean effective depth = (260 + 240) / 2 = 250 mm

= 1.15 x 1204.8 x 103 / 1600 x 250 = 3.46 MPa where v = 0.6(1 - fck / 250) = 0.528 fcd = accXfck / = 1.0 x 1.0 x 30 / 1.5 = 20 = 0.5 x 0.528 x 20 = 5.28 MPa Check shear stress at basic perimeter u1 (2d from face of column):

P, VEd, d as before u1 = control perimeter under consideration.

For punching shear at 2d from interior columns u1 = 2(cx + cy) + 2nx 2d = 4741 mm vEd = 1.15 x 1204.8 x 103 / 4741 x 250 = 1.17 MPa vRd,c = 0.18 / x k x (100^fCk)0333 where

OK <Concise EC2 Table 15.7ttttt> <6.4.2>

Column C2 is taken to be an internal column. In the case of a penultimate column, an additional elastic reaction factor should have been considered.

twt At the perimeter of the column, vRd max assumes the strut angle is 45°, i.e, that cot 6 = 1.0. Where cot 6= < 1.0, vRdmax is available from Concise EC2[10] Table 15.7.

P = (PxPy)05 = (0.0085 x 0.0048)05 = 0.0064 where px, py = areas of bonded steel in a width of the column plus 3d each side of column*****

<Concise EC2 Table 15,

.'. Punching shear reinforcement required

Perimeter required such that punching shear links are no longer required: <Exp. (6.54)>

uout = 1204.8 x 1.15 x 103 / (250 x 0.61) = 9085 mm

Length of column faces = 4 x 400 = 1600 mm

Radius to uout = (9085 - 1600) / 2n = 1191 mm from face of column

Perimeters of shear reinforcement may stop 1191 - 1.5 x 250 = 816 m from face of column <6.4.5(4) & NA>

Shear reinforcement (assuming rectangular arrangement of links)

sr,max = 250 x 0.75 = 187, say = 175 mm <9.4.3(1)>

Inside 2d control perimeter, stmax = 250 x 1.5 = 375, say 350 mm <9.4.3(2)>

Outside basic perimeter stmax = 250 x 2.0 = 500 mm Assuming vertical reinforcement at the basic control perimeter, u1, 2d from the column:

4w > (vEd - 0.75vRd,c) sr u1 / 1.5fywd.ef) Where fy ywd,ef

: effective design strength of reinforcement = (250 + 0.25d) < fyd = 312 MPa For perimeter u1

Asw = (1.17 - 0.75 x 0.61) x 175 x 4741 / (1.5 x 312) = 1263 mm2 per perimeter

Asw.min > 0.08fCk°5(sr x st) / (1.5 fyk sin a + cos a) Where

<Exp. (6.52)> <6.4.5(1)> <Exp. (9.11)>

= area of a link leg a = angle between main reinforcement and shear reinforcement;

for vertical reinforcement sin a= 1.0 Aew,min > 0.08 x 3005 (175 x 350) / (1.5 x 500) = 36 mm2

Try H8 legs of links in perimeters at 175 mm cc Asw / u1 > 1250 / 4741 = 0.26 mm2 / mm

Using H8 max. spacing = min[50 / 0.2; 1.5d] <9.4.3>

= min[192; 375] = 192 mm cc Use H8 legs of links at 175 mm cc around perimeters******

Rd,c ttttt The values used here for px, ply ignore the fact that the reinforcement is concentrated over the support. Considering the concentration would have given a higher value of VRdc at the expense of further calculation to determine px, py at 3d from the side of the column.

Clause 6.4.5 provides Expression (6.52), which by substituting vEd for vRdc, allows calculation of the area of required shear reinforcement, Asw, for the basic control perimeter, u,. This should be considered as the required density of shear reinforcement. The sssss „ ^ for various values of d and pl is available from Concise EC2[ ] Table 15.6.

which are also at 175 mm centres Check 26 H8 legs of links (1250 mm2) in perimeter u, 2d from column face 1st perimeter

1st perimeter to be > 0.3d but < 0.5d from face of column. Say 0.4d = 100 mm from face of column

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