Civil Engineering - Strength of Materials - Discussion


For a simply supported beam of length L, the bending moment M is described as M = a (x - x3/L2), 0 ≤ x < L; where a is a constant. The shear force will be zero at

[A]. the supports
[B]. x = L/2
[C]. x = L/3
[D]. x = L/3

Answer: Option C


No answer description available for this question.

Kashi said: (Sep 24, 2013)  
Maximum bending moment occurs at zero shear force*.

d(M)/dx = 0 (for maximum bending moment).

on differentiating above equation you get option C.

Maneesha said: (Aug 15, 2017)  
Give me the correct explanation.

Ebeyehu said: (Aug 21, 2017)  
The derivation of the moment is always shear having this idea we can derivate and we obtain dm/dx =a (1-3x*x/l*l) say this equation is zero and we will get 1-3x*x/l*l =0 from this x=l/3thepower of half.

Venkatesh said: (Oct 17, 2017)  
Shear force is the partial derivative of bending moment so to obtain shear force the given bending moment is differentiated with respect to x and submit the given options so that you should get zero.

Hussein said: (May 27, 2019)  
1-(3x2/l2) = 00.

Anjali said: (Dec 24, 2019)  
Shear force always zero at maximum bending moment.
For max BM,dm/dx =0.
So 1-3x^2/l^2 =0.
Hence, 1-3x^2=0.

Shreyas A said: (Oct 31, 2020)  
We know that,

Rate of change of Bending moment, dM/dx = Shear force.
Rate of change of Shear force, dF/dx = Rate of loading.

So, they gave B.moment as an equation. To get Shear force then differentiate that wrt x and we will get,

dM/dx = a( 1 - 3x^2/l^2).
Now to get max shear force, dM/dx = 0.
so, a( 1 - 3x^2/l^2) = 0.
l^2 = 3x^2.
x = l/√2.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.