Civil Engineering - Theory of Structures - Discussion

Discussion Forum : Theory of Structures - Section 2 (Q.No. 4)
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
0.001
0.002
0.0025
0.003
0.005
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Ramjan Ali said:   2 years ago
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.
(1)

Vishal patil said:   9 years ago
Explain it?

Rohit kumar said:   8 years ago
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:   7 years ago
According to me, the Right answer is 0.025.

Manoj paridwal said:   7 years ago
Yes, @Pinki.

The right answer is .025.

Asay said:   7 years ago
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:   4 years ago
The right answer is 0.025.

B.Singh said:   2 years ago
I think the answer is 0.025.

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