Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 8)
8.
Two shafts 'A' and 'B' transmit the same power. The speed of shaft 'A' is 250 r.p.m. and that of shaft 'B' is 300 r.p.m. The shaft 'B' has the greater diameter.
Discussion:
52 comments Page 1 of 6.
Paresh said:
1 decade ago
The torque relation can be obtained from above relation, but not the diameter because, if dia is smaller (as per answer, since torque is less) than even a larger diameter (than dia of shaft A) will be able to sustain that amount of torque.
Though the ' Minimum diameter ' of B will be less than that of A.
Though the ' Minimum diameter ' of B will be less than that of A.
Kishan P said:
8 years ago
As Rpm increasing, torque in reduces. As torque reduce the maximum shear stress at the outer most fibre will reduce.
As it's having less shear stress, it required smaller diameter.
Hence when Rpm increasing, the diameter will be smaller When compared to a member having lower Rpm.
As it's having less shear stress, it required smaller diameter.
Hence when Rpm increasing, the diameter will be smaller When compared to a member having lower Rpm.
Barath said:
1 decade ago
The power transmitted by the shaft is directly proportional to speed where as the torque is inversely proportional so torque reduces with speed. While designing we take smaller dia for the shaft which has to transmit lesser torque. Remember the formula P = 2*3.14*N*T/60000 KW.
Govind.T said:
1 decade ago
There is direct relation.
Power(P) = Torque(t)*Angular velocity(w).
(shear stress(s) = 16t/(3.14*d^3).
(w = 2*3.14*n/60).
= s*pi (3.14)*d^3*2*pi*n/60.
Power is directly proportional to d^3 and r.p.m(n).
To get same power for less r.p.m shaft it should have more diameter.
Power(P) = Torque(t)*Angular velocity(w).
(shear stress(s) = 16t/(3.14*d^3).
(w = 2*3.14*n/60).
= s*pi (3.14)*d^3*2*pi*n/60.
Power is directly proportional to d^3 and r.p.m(n).
To get same power for less r.p.m shaft it should have more diameter.
Debashish said:
9 years ago
When power is constant then torque is inversely proportional to speed so when speed increase torque is decreased. Again torque is directly proportional to radius from eqn T = f * r. So less torque has less diameter also.
VISHNUVARDAN CHADA said:
1 decade ago
Power = 2*pi*N*T/60. as power is same in both the cases (N*T)1=(N*T)2. But T=F*R.
=> (N*F*R)1 = (N*F*R)2.
For same force N*R = CONSTANT.
Hence 250d1 = 350d2.
=> d2 = 250d1/350.
So d2 is smaller than d1.
=> (N*F*R)1 = (N*F*R)2.
For same force N*R = CONSTANT.
Hence 250d1 = 350d2.
=> d2 = 250d1/350.
So d2 is smaller than d1.
Naga Raju said:
9 years ago
P=2πNT/60.
For same power of two shafts NT=CON,
N=1/T>>>N=1/F. R>>> N1/R.
Here N1<N2 hence R1>R2.
But here it is given B has larger diameter.
Hence it is false statement.
For same power of two shafts NT=CON,
N=1/T>>>N=1/F. R>>> N1/R.
Here N1<N2 hence R1>R2.
But here it is given B has larger diameter.
Hence it is false statement.
Prasenjit said:
1 decade ago
Power transmission does not depend on diameter of shaft. Its depend on the torque. Torque is equal to force*radius of shaft. We don't know the force. So we can't say which dia is more than other.
Ankibulla said:
1 decade ago
Torque is inversely proportional to the speed (RPM) And Again Torque is proportional to the diameter of the shaft.
So which A shaft is less speed, torque is greater and then diameter is greater.
So which A shaft is less speed, torque is greater and then diameter is greater.
Vikram Korpale said:
1 decade ago
Speed of Shaft and dia. are inversely proportional to each other.
N1*d1 = N2*d2 = Constant.
Therefore, (d2/d1) = (N1/N2).
If N1<N2 then (N1/N2)<1, hence (d2/d1)<1.
Therefore d2<d1.
N1*d1 = N2*d2 = Constant.
Therefore, (d2/d1) = (N1/N2).
If N1<N2 then (N1/N2)<1, hence (d2/d1)<1.
Therefore d2<d1.
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