Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 2 (Q.No. 34)
34.
The maximum shear stress in a solid shaft of circular cross-section having diameter d subjected to a torque T is τ. If the torque is increased by four times and the diameter of the shaft is increased by two times, the maximum shear stress in the shaft will be
Discussion:
4 comments Page 1 of 1.
Keshav Kaushik said:
7 years ago
T/IP = t/r = Gπ/L.
Ip= πd^4/32.
t= shear stress.
Now, new Torque = 4T and new diameter = 2d.
So new Ip= π(2d)^4/32.
t= (4T/new Ip)*r.
Also,
Here r is new i.e 2d/2 = d,
So r=d.
Old r is d/2,
Upon solving you get 8T/πd^3.
Also upon solving the initial case you get
16T/πd^3.
Clearly, the new conditions have shear stress equal to half of the one in the first case.
Ip= πd^4/32.
t= shear stress.
Now, new Torque = 4T and new diameter = 2d.
So new Ip= π(2d)^4/32.
t= (4T/new Ip)*r.
Also,
Here r is new i.e 2d/2 = d,
So r=d.
Old r is d/2,
Upon solving you get 8T/πd^3.
Also upon solving the initial case you get
16T/πd^3.
Clearly, the new conditions have shear stress equal to half of the one in the first case.
Praveen said:
8 years ago
Shr.stres=16T/πD^3,
= 16(4T)/π(2D)^3,
= 1/2.
= 16(4T)/π(2D)^3,
= 1/2.
Anand said:
9 years ago
T/Ip = t/r, Ip = Pie * D^4/32 now according to question T = 4T and D = 2D (r = 2r). Put these values in equation we have 4T * 2r/(Pie * (2D)^4P) = t therefore t = T/2.
Sathya said:
10 years ago
How it will come?
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