Question

The moment capacity of plastic section for V > 0.6Vd is given by

(a) Mdv = Md – β(Md – Mfd)

(b) Mdv = Md + β(Md – Mfd)

(c) Mdv = Md – β(Md + Mfd)

(d) Mdv = Md + β(Md + Mfd)

I have been asked this question in an international level competition.

I'm obligated to ask this question of Design Strength of Laterally Supported Beams in portion Design of Beams of Design of Steel Structures

Answer 1

Correct answer is (a) Mdv = Md – β(Md – Mfd)

The best I can explain: The moment capacity of plastic or compact section for V > 0.6Vd is given by Mdv = Md – β(Md – Mfd), where Md = plastic design moment of whole section disregarding high shear force effect but considering web buckling effect, Mfd = plastic design strength of area of cross section excluding shear area, considering partial safety factor γm0, β is constant.