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 2 of 6.
Nilesh said:
9 years ago
Hihow can we calculate stress and strain?
If you know the voltage of motor and rpm and torque. Using this formula if the shaft is made up of ms steel and it is 380gpa. Please tell me.
If you know the voltage of motor and rpm and torque. Using this formula if the shaft is made up of ms steel and it is 380gpa. Please tell me.
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.
Rahul Kumar said:
10 years ago
As per the chemistry rule, we all know that mobility of any particle (i.e. speed is inversely proportional to its size) Here B's mobility is more so it must have a less diameter.
(1)
Univ said:
9 years ago
The speed(rpm) is inversely proportional to Diameter of shaft.
we know
P = (2 * 3.14 * N * T)/60 and also for shaft T = F * radius.
From this, we can observe the answer.
we know
P = (2 * 3.14 * N * T)/60 and also for shaft T = F * radius.
From this, we can observe the answer.
Nikhil Agrawal said:
9 years ago
Speed is inversely proportional to the torque.
So, as the speed of 2nd shaft is more than the 1st shaft so its diameter will be than that of 1st with speed less than 2nd.
So, as the speed of 2nd shaft is more than the 1st shaft so its diameter will be than that of 1st with speed less than 2nd.
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
Kartik Uttarkar said:
8 years ago
If the RPM is increasing then diameter value should be small but not big. And vice-versa if the RPM value is decreasing then it should have bigger diameter.
(1)
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?
Harshal Dongre said:
7 years ago
P=(2*π*N*T/60).
So for same power, NT=Constant.
Therefore for less rpm more torque is transferred, so greater the torque greater is the diameter.
So for same power, NT=Constant.
Therefore for less rpm more torque is transferred, so greater the torque greater is the diameter.
(7)
OM PRAKASH said:
1 decade ago
(2*pi*n*t)/60 = (2*pi*N*T)/60.
T in the shaft = (pi*tou*d3)/16.
300*d3 = 250*D3.
d3 = (250/300) *D3.
d = 0.94*D.
Then d is smaller than D.
T in the shaft = (pi*tou*d3)/16.
300*d3 = 250*D3.
d3 = (250/300) *D3.
d = 0.94*D.
Then d is smaller than D.
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