T

'cote

Using high yield reinforcement

Therefore A

Eqn 4.44

Use 4T12 bars

Reinforcement will also be required in the bottom flange to cater for flexure of the flange acting as a continuous nib.

Ak = (1500 - 107) X (250 - 107) = 199.2 x 103 mm2 Therefore

106 x 106

cd k

Again by reference to Figure 3.1, cote should fall within the limits of 0.67 to 1.5. Similarly use cote = 1.5

As the web is subject to shear and torsion, the combined effects should now be checked to satisfy the condition

Sd vr

2vf JA.

2 x 0.385 x 20 x 107 x 199.2 x 103 15 + <iï>

355 kN

Therefore

Rd1 1

Where the entire section is used to resist normal shear, each sub-section should be checked to satisfy the above interaction condition.

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