Check for punching shear column B

As the beam is wide and shallow it should be checked for punching shear.

At B, applied shear force, VFd = 569.1 + 517.9 = 1087.0 kN Check at perimeter of 400 x 400 mm column

P = factor dealing with eccentricity; recommended value 1.15 VFd = applied shear force

U = control perimeter under consideration. For punching shear adjacent to interior columns u0 = 2(cx + cy) = 1600 mm d = mean d = (245 + 226) / 2 = 235 mm vFd = 1.15 x 1087.0 x 103 / 1600 x 235 = 3.32 MPa

VRd,max = 0.5vfcd where v = 0.6(1 - fck / 250) = 0.516 fcd = accAfck / Yc = 1.0 x 1.0 x 35 / 1.5 = 23.3 vRd,max = 0.5 x 0.516 x 23.3 = 6.02 MPa

Check shear stress at basic perimeter u1 (2.0d from face of column)

P, VFd, d as before u1 = control perimeter under consideration. For punching shear at 2d from interior columns = 2(cx + cy) + 2nx 2d = 1600 + 2nx 2 x 235 = 4553 mm vfa = 1.15 x 1087.0 x 103 / 4553 x 235 = 1.17 MPa Vrac = 0.18 / Yc x k x (100 Mk)0333 where

Yc = 1.5 k = 1 + (200/d)05 < 2 k = 1 +(200 / 235)05 = 1.92 PI = (Px Py)05 where

PX = areas of bonded steel in a width of the column

<Fxp. (6.32)> <6.4.5(3) Note> <Fxp. (6.6) & NA>

OK <Concise FC2 Table 15.

In this case, at the perimeter of the column, it is assumed that the strut angle is 45°, i.e. that cot 0 = 1.0. In other cases, where cot 0 < 1.0, vRdmax is available from Concise FC2 Table 15.7.

plus 3d each side of column. = 6874 / (2000 x 226) = 0.0152 ply = 741 / (900 x 245) = 0.0036 p = (0.0152 x 0.0036)05 = 0.0074 fck = 35

^Rdc = 0.18 / 1.5 x 1.92 x (100 x 0.0074 x 35)0333 = 0.68 MPa <Conclse EC2 Table 15.6 >

.'. punching shear reinforcement required Shear reinforcement (assuming rectangular arrangement of links) At the basic control perimeter, u1, 2d from the column:

Asw ^ (^ - 0.75^) s U1 / 1.5fyWd.ef) <Exp. (6.52)>

fywdef = effective design strength of reinforcement

= (250 + 0.25d) < fyd = 309 MPa <6.4.5(1)>

For perimeter u1

Asw = (1.17 - 0.75 x 0.68) x 175 x 4553 / (1.5 x 309) = 1135 mm2 per perimeter

Try 15 no. H10 (1177 mm2) OK by inspection Figure 4.21

Shear links and punching shear perimeter u1

Figure 4.21

Shear links and punching shear perimeter u1

Perimeter at which no punching shear links are required:

Length of column faces = 4 x 400 = 1600 mm

Radius to uout = (7823 - 1600) / 2n= 990 mm from face of column i.e. in ribs, therefore beam shear governs vRdc for various values of d and p, is available from Concise EC2 Table 15.6 Greener Homes for You

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Responses

• quinto
Why punching shear check at 2d from face of column?
2 years ago