Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 3 (Q.No. 46)
46.
Keeping breadth constant, depth of a cantilever of length l of uniform strength loaded with uniformly distributed load w varies from zero at the free end and
at the fixed end
at the fixed endDiscussion:
3 comments Page 1 of 1.
B.Singh said:
3 years ago
Considering Cantilever beam with udl.
Then, M =(WL^2) / (2) ------------------ at fixed end.
From Bending Eqn: M/I=F/y,
M= (F*I)/(y),
We know, Section modulus (Z) = I/y.
M = F * Z.
{ F = M/Z } ----------------------> (i)
I = (bd^3)/(12),
y = d/2,
Z = (bd^2)/(6).
Putting value in F = M/Z.
F = [(WL^2)/(2)] / [(bd^2)/(6)],
F = [(WL^2)/(2)] x [(6)/(bd^2)].
F = (3WL^2)/(bd^2).
d^2 = (3WL^2) / (bF)
d = √ (3WL^2)/(bF).
d = L x √[(3W) / (bF)].
Thank you.
Then, M =(WL^2) / (2) ------------------ at fixed end.
From Bending Eqn: M/I=F/y,
M= (F*I)/(y),
We know, Section modulus (Z) = I/y.
M = F * Z.
{ F = M/Z } ----------------------> (i)
I = (bd^3)/(12),
y = d/2,
Z = (bd^2)/(6).
Putting value in F = M/Z.
F = [(WL^2)/(2)] / [(bd^2)/(6)],
F = [(WL^2)/(2)] x [(6)/(bd^2)].
F = (3WL^2)/(bd^2).
d^2 = (3WL^2) / (bF)
d = √ (3WL^2)/(bF).
d = L x √[(3W) / (bF)].
Thank you.
Parmar said:
5 years ago
Thanks @Abdullah.
Abdullah. said:
6 years ago
The question asks to depth at the fixed end.
M/I=F/Y.
F*(bd^2/6)=wl^2.
d=Sqrt(3wl^2/bf).
wl=W.
d=Sqrt(3wl/bf).
May Almighty make us knowledgeable.
M/I=F/Y.
F*(bd^2/6)=wl^2.
d=Sqrt(3wl^2/bf).
wl=W.
d=Sqrt(3wl/bf).
May Almighty make us knowledgeable.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers