Discussion :: Theory of Structures - Section 1 (Q.No.19)
|Ferd said: (Apr 27, 2016)|
|It should be :(2wa^2)/(9 * square root of 3).|
|Prem Khandeep said: (Feb 13, 2017)|
|According to me, It is wl^2÷9 √3EI.|
|Muzzamil said: (Jul 25, 2017)|
|It is 2wl/9*√ 3.|
|Lokesh Sahu said: (Aug 5, 2017)|
|Let reaction loads are RA & RB @ point A & B.
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
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)
|Nami said: (Apr 2, 2018)|
Really good explanation.
|Ankit said: (Feb 22, 2019)|
|Well said, Thanks, @Lokesh.|
|Sandeep Kshetri said: (Mar 31, 2019)|
|How the reaction at support b is negative? it would be positive.|
|Vaibhav Tatawar said: (Nov 19, 2020)|
|Thanks for explaining @Llokesh Sahu.|
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