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# Units of Measurements and Conversions

08/21/2017 10:41 AM

Hello I am going to ask a seemingly elementary question but I really need some help with this.

If I have a motor that's 15HP and runs at 30RPM that equals 2626 ft-lbs of torque.

To go from ft-lbs to n.m

1ft-lb = 1.36 n.m

2626ft-lb= 3560.378Nm

This is where I have trouble and i am not sure how to solve.

How are Nm converted to Nm^2 or even Nm^3 or Nm^4

what about converting mm to mm^2, mm^3, mm^4

Are there some standard formulas to do these conversions?

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#1

### Re: units of measurements and conversions

08/21/2017 10:49 AM

The units are different, because they do not correlate to the same entity.

Nm--> Nm2 requires the number be multiplied by another (arbitrary) length over which the force (Nm) is applied. You can figure out the rest. It relates to the dimensionality of a problem, and the dimensionality of the vector space wherein the problem is defined.

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#2

### Re: units of measurements and conversions

08/21/2017 10:50 AM

You cannot convert between things which are units for different measurements. eg. mm are a unit of length whereas mm^2 is area. How can a length be an area?

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#3

### Re: units of measurements and conversions

08/21/2017 10:51 AM

It depends upon context, because <...Nm^2 or even Nm^3 or Nm^4...> are not units of torque.

<...mm^2...> is an area, <...mm^3...> is a volume and <...mm^4...> might only be found in the strangest of concepts. None of these has anything to do with torque, as torque is [force x distance perpendicular to the line drawn through the centre of rotation].

A <...motor that's [rated at] 15HP and runs at 30RPM...> might not have any mechanical load applied to it, so the torque it experiences in that circumstance might not be <...2626 ft-lbs...>; it might be zero, in which case it will be drawing considerably less than <...15HP...> from the supply.

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#4

### Re: Units of Measurements and Conversions

08/21/2017 11:12 AM

let's go back to the first example that I laid out and work from there.

a 15HP motor running at 30RPM will give out 2626ft-lbs of torque.

2626ft-lbs of torque is about equal to 3560 newtons per meter.

since it's torque let's say that I am applying the measurements to a solid cylindrical pipe.

the pipe has a diameter of 40mm and a length of 200mm.

to get the Nm^2 for an area 10mmx10mm of that solid cylindrical pipe, how would I go about calculating that?

What would happen if instead of a 10mmx10mm area, I wanted a 20mmx100mm. The calculation to get that would be the same as well?

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#5

### Re: Units of Measurements and Conversions

08/21/2017 11:37 AM

You're talking rubbish.

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#6

### Re: Units of Measurements and Conversions

08/21/2017 11:39 AM

newtons per meter newton-metres

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#7

### Re: Units of Measurements and Conversions

08/21/2017 11:40 AM

that would be 3560 Newton-m (kg-m2/sec2 ). Most definitely not Newtons/meter.

I get 3560.4 N-m, for example.

As to your "solid" pipe, are you referring to a rod shaft?

N-m is a measure of torque, not force. If that is applied to your "solid" pipe of 40 mm diameter, the torque is still the same. The force is apparently what you are really interested in, and that would be applied with 3560.4 N-m at a radius of 0.2 m, thus the force exerted by this shaft on a load would be 712.1 N, approximately speaking.

The smaller the "solid" pipe (rod or shaft), the less force would be required to stop rotation, to the extent the torque is being applied at a smaller and smaller distance.

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#8

### Re: Units of Measurements and Conversions

08/21/2017 12:10 PM

Hi James

Thank you! Yes, I made errors in what I typed but those aren't intentional. You're also right that I am looking for force.

I am talking about a 40mm rod shaft connected to a 15HP motor running at 30RPM.

like you said and I found it on wikipedia

1 Newton-metre = 1*((kilogram*meter^2)/seconds*2)

3560N-m at a radius of 20mm be calculated as

3560*(([weight of rod]*[length of rod^2]/seconds*2)

is that correct?

Also the matweb link says this material has a 490MPa yield strength, would the 15HP motor running at 30RPM create enough force to elastically strain this material?

This is the answer that I am trying to figure out but would like to understand how to calculate it.

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#9

### Re: Units of Measurements and Conversions

08/21/2017 1:39 PM

In a word, no.

Imagine hanging a weight (mass times acceleration = weight) from a rope wound on said shaft of 40 mm diameter. The work done in rotating the shaft is the force (weight) falling through a distance (extension of the rope) as it turns the shaft while falling.

The maximum speed of the falling weight is limited by gravitational acceleration of 9.80 m/sec2 so it depends on how long a time the weight falls in gravity, any shaft resistance to rotation, etc.

When you think of a rotating shaft: consider the rotation direction, and the torque required to rotate it is the Force acting upon the attachment to the shaft at its diameter, but the torque is the vector product of the tangential force with the radius of the shaft. Plain and simple, no?

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#12

### Re: Units of Measurements and Conversions

08/21/2017 5:07 PM

So, to restate your newly restated quest:

Will a solid round bar driveshaft 40mm diameter withstand the torque from my as yet undefined 15 hp motor?

What about couplings/slider - pillow blocks - U-Joint/direction changes - etc

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#15

### Re: Units of Measurements and Conversions

08/22/2017 9:11 AM

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#16

### Re: Units of Measurements and Conversions

08/22/2017 3:43 PM

30rpm is a bit slow for a motor! Presumably there is a gearbox too.

You say it's driving a solid pipe (whatever that is).

Whether the shaft is solid or hollow, you need to calculate the 2nd moment of area in torsion, that's in length4. To help you, for solid it's pi*r4/2. Divide the torque by that, then multiply by the radius. That gives you the stress, you can check the units work out right. As your allowable stress (which btw isn't the same as yield stress) looks like being in MPa = N/mm2, you're probably best using mm and N*mm. Unless you use Mathcad, which does it all for you.

The shaft length doesn't affect the stress, only the (torsional) deflection.

Let us know how you get on and I'll check your answer.

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#17

### Re: Units of Measurements and Conversions

08/22/2017 3:56 PM

Buddy, I hate to say it, but I think your flight just zoomed over the head of OP's tower just a tad fast.

To the OP I say this. 2π x Torque x speed = power (just make sure your units are in rotations per second). 2Π x Torque x speed x time of operation = energy (example: set up xyz used 5 gallons of gas in three hours, with 200 ft-lb and 10 rps. 5 gallons gas being the amount of energy consumed in driving the system, not the energy output.

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#19

### Re: Units of Measurements and Conversions

08/23/2017 5:29 AM

Maybe, but if OP just does what I said he'll get the stress, which appears to be what he wants. Perhaps he won't understand the calculation.

I think he'd have more trouble trying to understand your #7 and #9

Not to mention the latest puzzle you set him. Converting CV of gas in BTU/lb (if that's how it's expressed in USA, over here we use MJ/kg) to ft.lbf/gal would be a challenge for a start!

OP (or somebody) has already got the right relationship between speed, torque and power.

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#23

### Re: Units of Measurements and Conversions

08/23/2017 10:20 AM

I am not sure at all he understand what he is wanting or how to get there. If stress is what we are talking about I confess, that I am getting stressed over this thread, and need to bug out, before I get torqued. Off. Over and out.

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#25

### Re: Units of Measurements and Conversions

08/23/2017 10:29 AM

Thank you for not being insulting.

You are correct in that there is a gearbox as well. The gearbox ratio can be changed and so before I make the gearbox I would like to be sure that the materials that I have on hand will be able to withstand the forces.

The info you typed didn't go over my head I actually wrote some calculations in this reply: http://cr4.globalspec.com/comment/1219177/Re-Units-of-Measurements-and-Conversions

after the calculation: 1255*0.02/((pi/2)*0.02^4) I get about 99 MPa does that mean a solid pipe with a diameter of 40mm would be okay under 1255/m2 of force?

Doubling the radius of the pipe to 80mm then the formula becomes

1255*0.04/((pi/2)*0.04^4) and I get about 125MPa, is that correct?

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#26

### Re: Units of Measurements and Conversions

08/23/2017 10:30 AM

What is a <...solid pipe...>, please? If it were truly a <...pipe...>, then what is the wall thickness, please?

Of what material is the object made?

Note that <.../m2...> is not a unit of force.

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#20

### Re: Units of Measurements and Conversions

08/23/2017 5:58 AM

As you appear to know the relation between speed, torque and power, an approach is to work out motor power required to produce the same torque but at typical motor speed, say 1500rpm. Then check the shaft diameter of a motor with that rated power (from a catalogue). It's a lot bigger than 40mm!

The shaft isn't sized purely for torque, but then your shaft might also get other loads you haven't thought of.

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#21

### Re: Units of Measurements and Conversions

08/23/2017 6:30 AM

Better for the motor to stall and the circuit protective device(s) to operate than the shaft to bend and break, perhaps?

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#28

### Re: Units of Measurements and Conversions

08/23/2017 10:45 AM

I would prefer for the gear to stall than to break the shaft. This is why I am trying to calculate the maximum shear force that the shaft can bear.

For this project the limitation is the material with a 450MPa yield strength. I would prefer to adjust the gearbox gear ratio so that the maximum output force isn't too close to the total yield strength for the reason you say; there might be other stresses as well.

If these calculations below are accurate;

after the calculation: 1255*0.02/((pi/2)*0.02^4) I get about 99 MPa does that mean a solid pipe with a diameter of 40mm would be okay under 1255/m2 of force?

Doubling the radius of the pipe to 80mm then the formula becomes

1255*0.04/((pi/2)*0.04^4) and I get about 125MPa

does that mean that the 40mm solid shaft is still able to withstand about 390MPa worth of force?

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#10

### Re: Units of Measurements and Conversions

08/21/2017 1:52 PM

What examples does your textbook have?...

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#11

### Re: Units of Measurements and Conversions

08/21/2017 2:06 PM
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#13

### Re: Units of Measurements and Conversions

08/21/2017 9:55 PM

thanks for the book, I will have to pick that up.

1Nm = 1((kg*m^2)/sec^2) to get Nm^2 would I just multiply that by the area of some surface?

For example the area of a circle X = PI*r^2 so to get the Nm^2

Would it be correct to do this

Nm^2 = (pi*r^2)*(1((kg*m^2)/sec^2))

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#14

### Re: Units of Measurements and Conversions

08/21/2017 10:34 PM
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#18

### Re: Units of Measurements and Conversions

08/22/2017 8:17 PM

Its sad when people without even the most minimal background in science suddenly jump in an try to become engineers.

Then, when they realize that they are way beyond their background they ask for help.

Would such a person decide that they are now a "doctor" begin to operate on a patient and then, as the blood flowed, ask for help.?????

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#27

### Re: Units of Measurements and Conversions

08/23/2017 10:44 AM

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#22

### Re: Units of Measurements and Conversions

08/23/2017 10:20 AM

moment of inertia

max shear stress

t = T * r / J

15hp motor at 85RPM = approximately 926ft-lb or 1255/m2 of force

with the above info, I tried calculating the max stress of a solid metal shaft, the diameter of that shaft is 40mm.

radius in meters is 0.02 applying this to the formula:
1255*0.02/((pi/2)*0.02^4) = 99869727

That's 99,869,727 Pascals or 99 MPa

The material has a tensile strength of 450MPa, does that mean that under 1255/m2 force, this material would not fail elastically?

Does that mean that a metal shaft that's 40mm in diameter can wit

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#24

### Re: Units of Measurements and Conversions

08/23/2017 10:26 AM

See #21⇑.

Please note that <...ft-lb or.../m2...> are not units of force.

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#29

### Re: Units of Measurements and Conversions

08/23/2017 10:48 AM

The calculation is OK, but you mean 1255N*m torque, not 1255/m2 of force. m2 isn't force anyway.

And you've moved the goalposts to 85rpm.

Whether 99MPa is acceptable you need to decide, but don't forget allowable shear stress is quite a bit lower than allowable tensile, which is itself well below yield strength. I still suggest you try the approach in #20 (with updated figures).

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#30

### Re: Units of Measurements and Conversions

08/23/2017 11:04 AM

Thank you, and my apologies; I did use the wrong words it's not 1255/m2 but like you said 1255n-m torque.

I've seen the shaft of the 1200HP motor, it's about 60mm in diameter but it is machined with a 5mm key so the total diameter is 65mm and the key is 5mm.

That motor will connect to a gearbox that reduces the RPM, I think I started with 30 to stress the maximum I force that I am willing to go; For this application, that much force is way too much.

I can understand that there are unseen stresses that could act on this shaft that I want to make; that's why I am trying to figure out if my calculations are correct that would help me adjust different parts of the shaft, gearbox, etc.

1255n-m of torque on a 40mm solid shaft produces 99MPa of shear stress;

doubling the diameter of the solid shaft to 80mm

1255*0.04/((pi/2)*0.04^4) and I get about 125MPa,

both of those values are well below the 450 yield strength of this material so it shouldn't fail, is that correct?

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#31

### Re: Units of Measurements and Conversions

08/23/2017 11:13 AM

What size is the output connection of the gearbox?

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#33

### Re: Units of Measurements and Conversions

08/23/2017 12:07 PM

the gearbox has a 65mm diameter tap; its 60mm with a 5mm key. But that gearbox tap is different than the steel that I will be using. I will be connecting my shaft to their gearbox with a coupler; the gearbox has a 1/14 gear ratio; the input motor has a 65mm tap as well with 5mm key, that motor is 15hp 1200rpm.

The gearbox and motor will output 926ft-lb or 1255n-m of torque.

I am trying to make sure that the shaft that I am trying to design; rated at 450 yield stress will be able to withstand those torques.

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#38

### Re: Units of Measurements and Conversions

08/24/2017 2:47 AM

...and what is it driving that presents all that torque at that speed?

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#32

### Re: Units of Measurements and Conversions

08/23/2017 11:51 AM

Where does 1200HP come from? On original figures, at 1500rpm it is 15*1500/30 = 750HP, 560kW. And rated torque 3591N*m checks with 2626ft.lb = 3560N. My data says 400 frame size, 100mm dia. shaft.

New rpm gives 15*1500/85 = 265HP, 200kW. Rated torque 1285N*m = 950ft.lb vs your 926ft.lb, 315 frame size, 80mm dia. shaft.

Doubling dia from 40 to 80mm gives increased stress? You need to look at that again.

I'm not going to decide for you what the correct dia is. I don't know the full story, and I'm not a rotating machinery man. Read my earlier posts.

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#34

### Re: Units of Measurements and Conversions

08/23/2017 12:15 PM

In the first post I wrote:

[ Hello I am going to ask a seemingly elementary question but I really need some help with this.

If I have a motor that's 15HP and runs at 30RPM that equals 2626 ft-lbs of torque.

To go from ft-lbs to n.m

1ft-lb = 1.36 n.m

2626ft-lb= 3560.378Nm ]

it's a 15HP motor, it runs at 1200RPM, there's a gearbox that has a gear ratio of between 1/14 - 1/40.

The first post I was using the 1/40 gear ratio. that's where I got the 2626ft-lbs.

I was asking to figure out if my shaft at 40mm and 450 yield strength would be able to avoid failure if the motor & gearbox puts out 926ft-lbs or 1255n-m of torque.

That's why I am asking are the calculations below correct.

1255n-m of torque on a 40mm solid shaft produces 99MPa of shear stress;

doubling the diameter of the solid shaft to 80mm

1255*0.04/((pi/2)*0.04^4) and I get about 125MPa,

both of those values are well below the 450 yield strength of this material so it shouldn't fail, is that correct?

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#35

### Re: Units of Measurements and Conversions

08/23/2017 12:19 PM

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#36

### Re: Units of Measurements and Conversions

08/23/2017 12:36 PM

Where does 1200HP come from? There is no 1200HP; If I wrote that I made a horrible mistake; it's a 15 HP, 1200RPM motor.

Doubling dia from 40 to 80mm gives increased stress?

1255*0.04/((pi/2)*0.04^4) = 99MPa

1255*0.04/((pi/2)*0.04^4) = 12.48MPa

No that's not doubling; the stress is decreasing

I'm not going to decide for you what the correct dia is. I don't know the full story, and I'm not a rotating machinery man. Read my earlier posts.

I wasn't asking you to figure out the diameter for me;

I though if I wrote the calculations and the steps that I took on a engineering/ physics forum; people would say no that's wrong because x, y, z or maybe you're on the right path look at this or that formula or theory.

If the calculations that are used are correct and valid, then I can figure out the diameter by myself, I was just asking is there anything that I could be overlooking.

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#37

### Re: Units of Measurements and Conversions

08/23/2017 1:55 PM

1200HP is in your #30, 2nd para.

In #34 you said 99 and 125MPa.

The calcs are OK if you use the right inputs. It's my formula! But you've asked several times whether the stress is acceptable. That's what I'm not prepared to say. If you're happy to select a working stress you've got all you need, IMO.

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