Calculating torque in a shaft

[jamesy_81]

Hi, I'm stuck on this progress checker question. I've come up with the answer of 6,041875 Nmm or 6,042 NM but this is wrong but I've been told my calcualators are correct. find my progress below. thanks for any help


Calculate the power that can be safely transferred by 100mm diameter shaft rotating at 150 RPM.

Assume the material is medium carbon steel.

Ultimate shear stress = 580N/mm2 ÷ 6 for dead load = 96.67N/mm2 [ζ] ]

Torque T = Jζ/R or 2Jζ/D

Where J = Polar second moment of area = ПD4/32 for a solid shaft.

Now T = 2ПD4ζ
32D

= ПD3ζ
16

= П*1003*96.67
16

 

[Mech_Engineer]

The answer you came up with, 6.042 N-m, is not a unit of power. Do you know how to calculate power tranmitted in the shaft, given the calculated torque and angular velocity?

 

[jamesy_81]

Do you know how to calculate power tranmitted in the shaft, given the calculated torque and angular velocity?

Thanks for the reply. no i dont? is that what i need to be doing? and what units should the answer be in?
Thanks

 

[Mech_Engineer]

The basic equation you're looking for is:

$$P_{max} = T_{max} * \omega$$

Where P is in units of Watts, T is N*m, and Omega is rad/s. This will give you the max power for the shaft based on its max torque it can carry.

 

[jamesy_81]

Is the max torque i got earlyer of 6,041875 Nmm or 6,042 NM right? if so do i times that by 15.7 rps. as 150 rpm is 15.7 rps?

 

[DickL]

If you are after the max. safe power that can be transmitted, then you probably should not use the ultimate shear stress. When the ultimate stress is reached it is pretty certain the shaft will have failed. I would use the yield stress and then apply a factor of safety to reduce the stress to a level that would ensure no failure resulted. In most machines, when a member yields (permanent deformation) that is considered a failure. For an exam or class problem using a SF of 1 maybe reasonable. In an actual design the SF might range for 1.2 to 3 or higher, depending on the consequences of a failure.

 

[spanky489]

Is the max torque i got earlyer of 6,041875 Nmm or 6,042 NM right? if so do i times that by 15.7 rps. as 150 rpm is 15.7 rps?

the torque on the edge of your shaft is T=F*r , where F is the force transmitted through various mechanical mates unless your shaft is coming out of the power source directly.
r is the radius=d/2. Power of your torque is P=T*w where w is the angular velocity like mech_engineer said.

Now in order to get 6.042 Nm you need a force and radius, you have only 1, so there is no way that your torque is correct based on your data. if you wanted to get your result your force would have to be 120.84 N, since 120.84 N * 100mm/2 = 6.042 Nm

your best shot would be to give the forum some more information on your problem and we can help you out.

also 150 rpm is not 15.7 rps but 2.5 rps, 1 minute has 60 seconds so you just divide 150 by 60.

 

[DickL]

The maximum safe torque can be calculated using the formula you've listed, but use the maximum safe stress rather and the ultimate stress. The maximum safe power is then computed from P=T*angular velocity (rad/s).

The angular velocity is rpm / 60 sec/min * 2PI rad/rev

 

[nvn]

jamesy_81: Homework questions go in the engineering homework forum, not here. We are not allowed to give you the relevant equations nor tell you how to solve your homework problem. You must list relevant equations yourself. And you must use the homework template. Read the PF rules.

Your formula in post 1 is correct. But do not forget to convert torque T in post 1 from N*mm to N*m, before you multiply T by angular velocity w.

For zeta in post 1, use shear ultimate strength (580 MPa) divided by 6, as your given problem in post 1 states. You are correct; zeta = 96.67 MPa.

 

[Mech_Engineer]

Is the max torque i got earlyer of 6,041875 Nmm or 6,042 NM right? if so do i times that by 15.7 rps. as 150 rpm is 15.7 rps?

Correct, although "rps" can be misconstrued as "revolutions per second" instead of "radians per second." I would recommend stating the units of your angular speed as 15.7 rad/s to avoid ambiguity.

Everything said and done, a torque of 6.042 N-m and an angular speed of 15.7 rad/s gives you a power throughput of 95 watts.

 

[nvn]

Is the max torque I got earlier of 6,041875 N*mm or 6,042 N*m right?

No, that is wrong. Try again.