Question:-
A wooden beam of kgf rectangular cross-section 10 cm deep by 5 cm wide carries maximum shear force of 2000 kgf. Shear stress at neutral axis of the beam section is
Option (A)
zero
Option (B)
40 kgf/cm2
Option(C)
60 kgf/cm2
Option(D)
80 kgf/cm2
Correct Option:
(C)
Question Solution:
Distribution of shear stress over a Rectangular section

A= b{(d/2)-y}
`y = y + 1/2{(d/2)-y}= (y/2)+y/4

t = (F/2l){(d2/4)-y2}
When y = 0; t maximum. {1=(bd3/12)}

= (3/2) tavg
tmax = (3/2)x(2000/(10x5)) = 60 kg.f/cm2
question-answer-faq-5148