Discussion :: Theory of Structures - Section 2 (Q.No.4)
|Vishal Patil said: (Nov 11, 2015)|
|Rohit Kumar said: (Jul 6, 2016)|
|By Torsion Equation:
T/Ip = Fs/ymax = G$/L.
Where T = Torsion
Ip = Polar moment of inertia
Ymax = Max distance from nuetral axis
G = MODULUS of rigidity
$ = Angle of twist in Radian
Fs = Max shear force.
Hence Fs/Ymax = G$/L So 50/100 = 0.8 * 100000/4000 (All diamension in MM) = 0.0025 Answer.
|Pinki said: (Jun 23, 2017)|
|According to me, the Right answer is 0.025.|
|Manoj Paridwal said: (Sep 11, 2017)|
The right answer is .025.
|Asay said: (Nov 21, 2017)|
|By TORSION EQUATION.
====> $=Fs.L/G.Ymax ----> (1).
Fs=Maximum Shear Stress=50 N/mm2.
G=0.8 x 105 N/mm2.
$=Twisting Moment=? (Required)
Now , Putting the Values in eqn (1).
|Dheeraj said: (Feb 24, 2020)|
|The right answer is 0.025.|
|B.Singh said: (Apr 4, 2022)|
|I think the answer is 0.025.|
|Ramjan Ali said: (Oct 15, 2022)|
Shear stress/radius = G * angle of twist/L.
From the above equation after solving we will get the value of the angle of twist i.e.
θ = 0.0025.
I have converted the value of Dia to Radius (R=100mm).
So, option (C) is correct.
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