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 4 of 6.
Chetan said:
1 decade ago
According to the equation of power (p) = (pi*d*n)/60, The speed of the shaft is directly proportional to the shaft diameter. Then how it can be possible?
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.
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.
Vrushang said:
1 decade ago
P = F*V .
P = F*omega*r.
So if both shaft transmit same power then omega*r=constant so,
for same power omega increase radius decreases.
P = F*omega*r.
So if both shaft transmit same power then omega*r=constant so,
for same power omega increase radius decreases.
ANUP said:
1 decade ago
P = 2(PI)NT/60.
HERE P1 = P2.
SO, 2(PI)N1T1/60 = 2(PI)N2T2/60
SINCE N1 < N2.
T1 > T2 TO MAINTAIN SAME POWER.
T = SHEAR STRESS * (PI) * D3 / 16.
SO, D1 > D2
HERE P1 = P2.
SO, 2(PI)N1T1/60 = 2(PI)N2T2/60
SINCE N1 < N2.
T1 > T2 TO MAINTAIN SAME POWER.
T = SHEAR STRESS * (PI) * D3 / 16.
SO, D1 > D2
Pari.siva said:
1 decade ago
Power is produced same,
We imagine,
Both pulleys are connected to same shaft.
And speed is inversely proposition to diameter in case greater speed pulley is the smaller diameter.
We imagine,
Both pulleys are connected to same shaft.
And speed is inversely proposition to diameter in case greater speed pulley is the smaller diameter.
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.
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.
Swapnil said:
1 decade ago
Angular speed is inversely proportional to Torque.
And Torque is directly proportional to radius.
And Torque is directly proportional to radius.
Vikash kumar said:
1 decade ago
The shaft having more speed will transmit less torque at same power level and for transmitting less torque less diameter shaft is required.
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