Rd3 cd

The VSI method gives A

Eqn 4.27

Re-arranging gives sw S


In the above equation the contribution of the concrete, V to the shear resistance of the section is not taken into account. °

With cote = 1.5 which is the maximum value permitted in the NAD, reductions in shear reinforcement will only occur when

Putting ^ = ^Rd3 gives l/Sd > 3Vcd lf ^sd > 3^cd, then the VSI method will allow a reduction in shear reinforcement.

If this inequality is not satisfied, use of the variable strut inclination method will produce an uneconomic amount of shear reinforcement. In this case the standard method should be used.

For elements with vertical shear reinforcement, l/Rd2 is given by b zv f .

Putting \/Sd = VRd2 and re-arranging gives

bwzvfcd cote + tane

Figure 3.1 shows cote plotted against 1/(cote + tane) together with the EC2 and NAD limits for cote. Hence for a given VSd, the limits for cote can be found.

Increasing the value of cote will reduce the shear reinforcement required but increase the force in the tension reinforcement.

In this example, cote will be chosen to minimize the shear reinforcement. Position 1 - at d from support

From above

0 0

Post a comment