Question

At full plasticity, the stress in web is given by

(a) fw = fy√[1+(τw/τy)^2 ].

(b) fw = fy√[(τw/τy)^2 ].

(c) fw = fy√[1-(τw/τy)^2 ].

(d) fw = fy√[1+2(τw/τy)^2 ].

I have been asked this question in an interview for job.

My doubt is from Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity in chapter Plastic and Local Buckling Behaviour of Steel of Design of Steel Structures

Answer 1

The correct choice is (c) fw = fy√[1-(τw/τy)^2 ].

The explanation: When full plasticity is produced, the stress in web will be equal to fw, fw =√(fy^2 -3τw^2 ) = fy√[1-(w/τy)^2 ]  where τy is shear strengths when yield is in pure shear.