# Civil Engineering - Theory of Structures - Discussion

Discussion Forum : Theory of Structures - Section 1 (Q.No. 19)

19.

A simply supported beam carries varying load from zero at one end and

*w*at the other end. If the length of the beam is*a*, the maximum bending moment will beDiscussion:

8 comments Page 1 of 1.
Vaibhav Tatawar said:
4 years ago

Thanks for explaining @Llokesh Sahu.

Sandeep kshetri said:
6 years ago

How the reaction at support b is negative? it would be positive.

Ankit said:
6 years ago

Well said, Thanks, @Lokesh.

Nami said:
7 years ago

Thanks @Lokesh.

Really good explanation.

Really good explanation.

Lokesh Sahu said:
7 years ago

Let reaction loads are RA & RB @ point A & B.

RA+RB=wl/2.

To calculating we found RA= wl/6 & RB= -wl/3,

let take a section X-X @ a distance of x from A.

UVL will be @ X-X = wx/l,

& load on triangle ACX = wx2 /2l,

& find the SF @ X-X,

Fx= RA - load on triangle ACX,

Fx=( wl/6) - (wx2/2l) -------> (1)

Fx at A will be. X=0

Fx = wl/6

Fx at B will be. X=l

Fx=-wl/3.

Moment will be max, where Fx=0,

So, x= l/(3)1/2.

Now find the moment at X-X.

Mx = (RA*x) " (load on triangle ACX * x/3)load of the triangle will be act ta 1/3rd distance from A.

Mx= (wlx/6) " (wx3/6l)

Moment at x= l/(3)1/2

Mx= (wl*( l/(3)1/2)/6) - (w*( l/(3)1/2)3/6l)

Mx= wl2/9(3)1/2.

RA+RB=wl/2.

To calculating we found RA= wl/6 & RB= -wl/3,

let take a section X-X @ a distance of x from A.

UVL will be @ X-X = wx/l,

& load on triangle ACX = wx2 /2l,

& find the SF @ X-X,

Fx= RA - load on triangle ACX,

Fx=( wl/6) - (wx2/2l) -------> (1)

Fx at A will be. X=0

Fx = wl/6

Fx at B will be. X=l

Fx=-wl/3.

Moment will be max, where Fx=0,

So, x= l/(3)1/2.

Now find the moment at X-X.

Mx = (RA*x) " (load on triangle ACX * x/3)load of the triangle will be act ta 1/3rd distance from A.

Mx= (wlx/6) " (wx3/6l)

Moment at x= l/(3)1/2

Mx= (wl*( l/(3)1/2)/6) - (w*( l/(3)1/2)3/6l)

Mx= wl2/9(3)1/2.

(1)

Muzzamil said:
7 years ago

It is 2wl/9*√ 3.

Prem khandeep said:
8 years ago

According to me, It is wl^2÷9 √3EI.

Ferd said:
9 years ago

It should be :(2wa^2)/(9 * square root of 3).

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