Info

1000 x 185

Figure 4.18

Eqn 4.50 Figure 4.21

Note:

The amount of tensile reinforcement in two perpendicular directions > 0.5%. 43.4.1(9) Assume p/x + p = 2 (0.0072) > 0.005 OK

Therefore

!/Rdi = 0.34 x 1.415 x (1.2 + 40 x 0.0072) x 185 = 133 N/mm . = 174 N/mm > vn..

Sd Rd1

Therefore shear reinforcement required such that i/Rd, > vSd 4.3.4.3(3)

Check that applied shear does not exceed the maximum section capacity vRd2 = 2.0 vRd1 = 2.0 x 133 = 266 > 174 N/mm OK NAD

Table 3

Shear stress around column perimeter = - = 2.0 N/mnrr

1200 x 185

Design shear reinforcement using EC2 Eqn 4.58 since vSd/i/Rd1 = 174/133 < 1.6 NAD 6.4(d)

Using type 2 deformed high yield bars as links

Therefore

0 0

Post a comment