Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 8 (Q.No. 37)
37.
A simply supported beam with a gradually varying load from zero at B and w per unit length at A is shown in the below figure. The shear force at B is equal to
Discussion:
9 comments Page 1 of 1.
John hirpesa said:
2 years ago
Best explanation, thanks.
RC... said:
6 years ago
@ M.S.R.
It is SSB, it is never SF Zero at the support.
BM is zero at the support.
It is SSB, it is never SF Zero at the support.
BM is zero at the support.
M.S.R said:
7 years ago
@All.
The answer given is Reaction force at B but in the question, they have asked about shear force which will be zero as there is no force on the right of B.
The answer given is Reaction force at B but in the question, they have asked about shear force which will be zero as there is no force on the right of B.
Anand said:
8 years ago
Very nice explanation @Mulkesh.
Shadab khan said:
8 years ago
Ra + Rb= wl/2,
The moment about A point.
Rb * l= wl/2 * l/3,
Rb = wl/6.
The moment about A point.
Rb * l= wl/2 * l/3,
Rb = wl/6.
Mulkesh mishra said:
8 years ago
First of all, its centroid it at l/3.
And we know r(a)+r(b)-w*(l/3)=0,
And r(a)=r(b),
So r(b)=wl/6.
And we know r(a)+r(b)-w*(l/3)=0,
And r(a)=r(b),
So r(b)=wl/6.
Sweta said:
8 years ago
@Josh.
You have taken the wrong length, 2l/3 from B and l/3 from A.
So the answer should be B.
You have taken the wrong length, 2l/3 from B and l/3 from A.
So the answer should be B.
Josh said:
9 years ago
L/3 from B and 2l/3 from A so the load is wl then taking moment about B l/3 * wl/2 = A * l
= wl/6.
= wl/6.
Shah said:
9 years ago
I think the answer is option D. If it is wrong then please correct me.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers