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
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.
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)
B.Singh said:
3 years ago
I think the answer is 0.025.
Dheeraj said:
5 years ago
The right answer is 0.025.
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)].
$=0.025.
T/I=Fs/Ymax=G$/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)].
$=0.025.
Manoj paridwal said:
7 years ago
Yes, @Pinki.
The right answer is .025.
The right answer is .025.
Pinki said:
7 years ago
According to me, the Right answer is 0.025.
Rohit kumar said:
8 years ago
By Torsion Equation:
T/Ip = Fs/ymax = G$/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$/L So 50/100 = 0.8 * 100000/4000 (All diamension in MM) = 0.0025 Answer.
T/Ip = Fs/ymax = G$/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.
Vishal patil said:
9 years ago
Explain it?
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