Civil Engineering - Theory of Structures - Discussion

4. 

If a solid shaft (diameter 20 cm, length 400 cm, N = 0.8 x 105 N/mm2) when subjected to a twisting moment, produces maximum shear stress of 50 N/mm2, the angle of twist in radians, is

[A]. 0.001
[B]. 0.002
[C]. 0.0025
[D]. 0.003
[E]. 0.005

Answer: Option C

Explanation:

No answer description available for this question.

Vishal Patil said: (Nov 11, 2015)  
Explain it?

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)  
Yes, @Pinki.

The right answer is .025.

Asay said: (Nov 21, 2017)  
By TORSION EQUATION.

T/I=Fs/Ymax=G$/L.
====> Fs/Ymax=G$/L.
====> $=Fs.L/G.Ymax ----> (1).

Here,
Fs=Maximum Shear Stress=50 N/mm2.
And,
Ymax=D/2=20/2=10cm=100mm
And,
G=0.8 x 105 N/mm2.
$=Twisting Moment=? (Required)
L=400cm=4000mm.
Now , Putting the Values in eqn (1).
$=50*4000/[0.8*100000*(100)].
$=0.025.

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)  
We know,

Shear stress/radius = G * angle of twist/L.
So, 50/100=(0.8*10^5)*theta/4000.

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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