# Design diameter of transmission shaft

• I
tthree

## Main Question or Discussion Point

Please bear with me, I am trying to design a shaft that will be under torsion. I do not know the diameter of the shaft and I would like to be able to calculate it by the things that I do know about the system.

There's a 15HP 3 Phase motor running at 85RPM [there's a gearbox]
This will be putting out 1255Nm of torque.

I have medium carbon steal with these mechanical properties:
Hardness, Brinell 201 - 269 201 - 269
Hardness, Rockwell C 13.8 - 27.6 13.8 - 27.6
Tensile Strength, Ultimate 686 MPa 99600 psi
Tensile Strength, Yield 490 MPa 71100 psi
Elongation at Break 17 % 17 %
Reduction of Area 45 % 45 %
Modulus of Elasticity 205 GPa 29700 ksi Typical steel
Poissons Ratio 0.29 0.29 Typical steel
Machinability 55 % 55 % Based on AISI 1212 steel as 100% machinability
Shear Modulus 80.0 GPa 11600 ksi Typical steel
Impact 8.0 8.0 kg(f)/cm²

Leaving off safety factor for now, just for simplicity sake.
My question is; can I calculate the the diameter of the shaft based on the above information? What I can say so far is:
T = 1.255x10^6Nmm
G = 80GPa
Hz = 1.42
RPM = 85
If so how can I do it?

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CWatters
Homework Helper
Gold Member
Google something like... calculate size of shaft

tthree
Google something like... calculate size of shaft

I came across the above formula at: http://www.plantengineering.com/sin...ft-size/97eb6b3f6b32adcdd181341ebbdc3224.html

The shaft doesn't have a keyway so with the above formula, based on;
1255Nm of Torque
85RPM
15HP or 11.2Kw

There's two things; most of these formulas are in inches but I was looking for metrics.
Also there's a metric formula
T/J = Fs/R

With that formula:
T = 1.255x10^6Nmm
J = π R4 / 2
Fs = Modulus of rigidity or 80GPa based on the above material sheet
R = x

Solve for R?

tthree
can anyone comment on?

CWatters
Homework Helper
Gold Member
Using the formulae D = sqrt(105*P/N) I got
D = sqrt(105*15/85)
= 4.3 inches diameter

I do not know the diameter of the shaft and I would like to be able to calculate it by the things that I do know about the system.

There's a 15HP 3 Phase motor running at 85RPM [there's a gearbox]
What is the diameter of the gearbox output shaft?

Minimum diameter of the power transmission shaft will be at least this large, and usually larger to account, for instance, for right angle forces in belted systems as shown in the above equations. Another factor for long shafts is unsupported length between bearings, coupling methods, and how much misalignment is allowed. It may be necessary to increase shaft diameter to reduce flexture failure.

tthree
Using the formulae D = sqrt(105*P/N) I got
D = sqrt(105*15/85)
= 4.3 inches diameter

Where did you get this formula; why is it 105? What does that number represent?

tthree
What is the diameter of the gearbox output shaft?

Minimum diameter of the power transmission shaft will be at least this large, and usually larger to account, for instance, for right angle forces in belted systems as shown in the above equations. Another factor for long shafts is unsupported length between bearings, coupling methods, and how much misalignment is allowed. It may be necessary to increase shaft diameter to reduce flexture failure.
No gearbox yet, that will also be designed based on the information that I get here. This is why I would like to understand the process and not just use magic numbers in a formula without understanding it.

T/J = Fs/R
T = Torque = input criteria that I get from elsewhere.
J = Polar Moment of Inertia = formula based around diameter of the shaft
Fs = Modulus of rigidity = [What do I use as this, do I get it from the material sheet]
R = Radius = can be derived from the calculations.

My main question is in the above formula; does Fs come from the material sheet as the modulus of rigidity?

tthree
Is this a trivial question or is it really difficult.
Why does it seem like no one is able to help me with this question?

Nidum
Gold Member
Somewhere in between .

The way to deal with this problem is not to try to solve for the diameter analytically but rather to pick a diameter based on conventional practice , your experience , other similar equipment installations or any other useful source of information and then determinine whether stress levels , distortion (twist) and possible other consideration make your choice of diameter an acceptable one for the application . It either is acceptable in which case job done or it is not acceptable in which case you can make a more well informed selection of a new diameter and try again

Normally this process converges very rapidly . Starting from best guess you will usually get the answer required after only one or two iterations .

The use of this method ties in well with the general design process for the shaft in terms of matching size to other components , practical sizes for support bearings , available sizes of shafting and a host of other detail considerations .

To make progress in this particular case :

Make your best estimate of a suitable shaft diameter . If necessary just pick a diameter that seems to be in good proportion to rest of machinery installation .

Post a reasonable sketch of the shaft arrangement including what it is connected to each end and any bearings . Show basic installation dimensions .

Mention the class of equipment involved . I'm sure that you will appreciate that requirements for an adequate shaft to drive a rock crusher are very different to the requirements for an adequate shaft for a church clock or for an aeroengine .

Last edited:
Asymptotic
There are well established shaft design procedures taught in most ME machine design courses. You design against a presumed failure mode, where the presumption is guided by experience. The design should consider stresses due to both shear and bending, as well as the possibility of stress reversals (fatigue). Check out any well recognized machine design text, such as Shigley or Spotts for more details.

Asymptotic
tthree
There are well established shaft design procedures taught in most ME machine design courses. You design against a presumed failure mode, where the presumption is guided by experience. The design should consider stresses due to both shear and bending, as well as the possibility of stress reversals (fatigue). Check out any well recognized machine design text, such as Shigley or Spotts for more details.
I specifically asked to keep it simple and then u reply with this, if you do not understand my obstacle, cannot help or do not want to provide some assistance at least be respectful enough to not reply.

@tthree -- There was nothing at all disrespectful in my answer to you. You initially said that you wanted to "design" a shaft. To design means to go through the steps required to get the final result correct based on physics and mathematics, rather than to simply guess at something. I simply summarized the basic requirements and told you where to go for more information; where is the disrespect in that? I offered you real information, serious guidance, and direction, and then you snap back as you did.

In many problems, simple and correct are mutually exclusive. Which do you want?

tthree
Please bear with me, I am trying to design a shaft that will be under torsion. I do not know the diameter of the shaft and I would like to be able to calculate it by the things that I do know about the system.

There's a 15HP 3 Phase motor running at 85RPM [there's a gearbox]
This will be putting out 1255Nm of torque.

I have medium carbon steal with these mechanical properties:
Hardness, Brinell 201 - 269 201 - 269
Hardness, Rockwell C 13.8 - 27.6 13.8 - 27.6
Tensile Strength, Ultimate 686 MPa 99600 psi
Tensile Strength, Yield 490 MPa 71100 psi
Elongation at Break 17 % 17 %
Reduction of Area 45 % 45 %
Modulus of Elasticity 205 GPa 29700 ksi Typical steel
Poissons Ratio 0.29 0.29 Typical steel
Machinability 55 % 55 % Based on AISI 1212 steel as 100% machinability
Shear Modulus 80.0 GPa 11600 ksi Typical steel
Impact 8.0 8.0 kg(f)/cm²

Leaving off safety factor for now, just for simplicity sake.
My question is; can I calculate the the diameter of the shaft based on the above information? What I can say so far is:
T = 1.255x10^6Nmm
G = 80GPa
Hz = 1.42
RPM = 85
If so how can I do it?
The main part of the question was
Leaving off safety factor for now, just for simplicity sake.
My question is; can I calculate the the diameter of the shaft based on the above information? What I can say so far is:
T = 1.255x10^6Nmm
G = 80GPa
Hz = 1.42
RPM = 85
If so how can I do it?
For a solid shaft there's the formula for strength criteria
T/J = Fs/R

I can get J [2nd polar moment] as well as the T [Torque] and R can be rewritten as D/2

What I am asking is: in this formula T/J = Fs/R; is Fs the modulus of rigidity of the material?

What I am asking is: in this formula T/J = Fs/R; is Fs the modulus of rigidity of the material?
T/J = Fs/R
T = Torque = input criteria that I get from elsewhere.
J = Polar Moment of Inertia = formula based around diameter of the shaft
Fs = Modulus of rigidity = [What do I use as this, do I get it from the material sheet]
R = Radius = can be derived from the calculations.

My main question is in the above formula; does Fs come from the material sheet as the modulus of rigidity?
Yes. In your spec sheet it is referred to as shear modulus.

If you "design" the shaft diameter based on that information alone, you may have a satisfactory shaft but quite likely will not. Reality is not quite that simple.

Asymptotic
jack action
Gold Member
What I am asking is: in this formula T/J = Fs/R; is Fs the modulus of rigidity of the material?
Fs in your equation is the shear yield strength, which you do not have in your mechanical properties. As a rough guide, shear yield strength (SYS) for steels is 58% of tensile yield strength (TYS), so 284 MPa (41240 psi) in your case.

For your polar moment equation, you have a sample calculation on Wikipedia to see how to use it to find the shaft diameter.

This is good for a shaft in pure shear, but if there are perpendicular or axial loads on the shaft, it gets a lot more complicated. You have to consider the Von Mises equations, which can get nasty. Note how the «Pure shear» case has ##\sigma_v = \sqrt{3}\lvert\sigma_{12}\lvert## for the Von Mises equation. ##\sigma_v## can be seen as the tensile stress, ##\lvert\sigma_{12}\lvert## as the shear stress and therefore ##\lvert\sigma_{12}\lvert = 0.58\sigma_v## (this is where the 58% presented earlier comes from).

If you have a perpendicular load on the rotating shaft (like an axle for example), you have to also consider the fatigue strength.

If you have an axial component, you have to consider buckling as well.

And then you have to consider stuff like environment (corrosion), notches, etc. Of course, you can wrap these in a large safety factor, but without any other designs to base your criteria upon, it is very difficult to define one.

People are afraid to give you clear answers because designing a shaft can get complicated very quickly. And no one would want to misguide you.

Asymptotic and tthree
You may want to select a gearbox first before worrying too much about the shaft.
No gearbox yet, that will also be designed based on the information that I get here. This is why I would like to understand the process and not just use magic numbers in a formula without understanding it.
Is it right angle, a parallel shaft, or something more esoteric?

You've mentioned a 15 HP, 3 phase motor with 85 RPM and 1255 Nm at the gearbox output shaft, but didn't state the input speed. 85 RPM at 1750 RPM (typical full load speed for a 4 pole motor at 60 Hz) is a 20.59:1 ratio. Gearboxes with 20.0:1 ratios abound, but 20.59:1 may be a special creature.

Usually a gearbox is selected first, and by pouring over transmission catalogs you'll learn whether a single or double reduction box is required, and a lot about service factor and other aspects that also pertain to shafting.

tthree
Fs in your equation is the shear yield strength, which you do not have in your mechanical properties. As a rough guide, shear yield strength (SYS) for steels is 58% of tensile yield strength (TYS), so 284 MPa (41240 psi) in your case.

For your polar moment equation, you have a sample calculation on Wikipedia to see how to use it to find the shaft diameter.

This is good for a shaft in pure shear, but if there are perpendicular or axial loads on the shaft, it gets a lot more complicated. You have to consider the Von Mises equations, which can get nasty. Note how the «Pure shear» case has ##\sigma_v = \sqrt{3}\lvert\sigma_{12}\lvert## for the Von Mises equation. ##\sigma_v## can be seen as the tensile stress, ##\lvert\sigma_{12}\lvert## as the shear stress and therefore ##\lvert\sigma_{12}\lvert = 0.58\sigma_v## (this is where the 58% presented earlier comes from).

If you have a perpendicular load on the rotating shaft (like an axle for example), you have to also consider the fatigue strength.

If you have an axial component, you have to consider buckling as well.

And then you have to consider stuff like environment (corrosion), notches, etc. Of course, you can wrap these in a large safety factor, but without any other designs to base your criteria upon, it is very difficult to define one.

People are afraid to give you clear answers because designing a shaft can get complicated very quickly. And no one would want to misguide you.
Thanks for the reply with a lot of references to other issues that can occur. I will eventually tackle those issues as well but for now if I can't figure out where that "Fs" value comes from then I have a lot bigger issues.

.58 * TYS at 284MPa vs 80GPa
converting GPa to MPa to MPa I get 80000
Using 80000 vs 284 will give wildly varying results.

Isn't that a bit strange?

tthree
You may want to select a gearbox first before worrying too much about the shaft.

Is it right angle, a parallel shaft, or something more esoteric?

You've mentioned a 15 HP, 3 phase motor with 85 RPM and 1255 Nm at the gearbox output shaft, but didn't state the input speed. 85 RPM at 1750 RPM (typical full load speed for a 4 pole motor at 60 Hz) is a 20.59:1 ratio. Gearboxes with 20.0:1 ratios abound, but 20.59:1 may be a special creature.

Usually a gearbox is selected first, and by pouring over transmission catalogs you'll learn whether a single or double reduction box is required, and a lot about service factor and other aspects that also pertain to shafting.
It's a 3 phase 15HP motor running at 1200RPM. I am also going to have to design and build the gearbox, not only that but I want to have a worm gearbox that's mounted directly on the motor output shaft and then have the output from the gearbox perpendicular to to motor input shaft.

There will also be hexagonal sections to the main transmission shaft. So as you can see, I cannot go and select a gearbox I have to build one.

It will be 2x 45teeth gears and 2x 12teeth gears. These gears will need to be mounted in the gearbox as well, that means I will need to know how to calculate a solid shaft to mount those gears to the gearbox.

If I cannot figure out what "Fs" is referring to in that calculation then I have many more serious problems. I wanted to keep it simple and figure out what "Fs" is referring to and find some documentation on it so that I can proceed to designing this machine.

It would take quite a bit of time to have all those pieces explained and I am sure very few would be willing to do the work for me, which I am not asking for.

I wanted to keep things simple and try to understand where does the value "Fs" come from in the formula T/J = Fs/R so that I can reason about different parts of the design.

Can you explain where does the "Fs" come from, it can't be some magic number that's just guessed; what's the logic behind it?

I am also going to have to design and build the gearbox
OK, so 1200 RPM into a right angle worm gearbox. My advice is for you to read up on gearbox design, AGMA literature, and go through manufacturer's catalogs, and consult with one or two of them before deciding to build a gearbox from scratch. Even if you plan on using off-the-shelf gears, putting together a functional, reliable gearbox is far more complex and demanding a task than fabricating a piece of shafting.

tthree
OK, so 1200 RPM into a right angle worm gearbox. My advice is for you to read up on gearbox design, AGMA literature, and go through manufacturer's catalogs, and consult with one or two of them before deciding to build a gearbox from scratch. Even if you plan on using off-the-shelf gears, putting together a functional, reliable gearbox is far more complex and demanding a task than fabricating a piece of shafting.
I understand that and if I can't even get straight answers on how large to make the shaft because I can't sort out a simple equation the problems will be a lot worse. For instance what's the diameter of the shaft holding the gears to the gearbox?

So I am still back to the original question. Yes I understand there needs to be factor of safety, shock, fatigue, etc.. but those can be integrated later.

jack action
Gold Member
Let's go back to the fundamental concepts for defining a material.

Stiffness

A spring is modeled by the equation ##F=kx##, where ##F## is the force applied to the spring, ##x## is the spring displacement under that force and ##k## is the spring constant, i.e. what defines the spring such that ##k = \frac{F}{x}## is always true, no matter the force or the displacement.

A solid rod can be viewed as a very stiff spring than can deform under a force. With materials, we prefer to look at stress ##\sigma## and strain ##\epsilon## such that ##E = \frac{\sigma}{\epsilon}##, where ##E## is the elastic modulus.

Because the stress is defined as ##\tau = \frac{F}{A}## and the strain is defined as ##\epsilon = \frac{x}{L}##, where ##A## is the cross-sectional area of the rod and ##L## is the original length of the rod. Therefore:
$$E = \frac{\sigma}{\epsilon} = \frac{\frac{F}{A}}{\frac{x}{L}} = \frac{F}{x}\frac{L}{A} = k\frac{L}{A}$$
Since ##L## and ##A## are constants, then the elastic modulus is comparable to a spring constant, i.e. how stiff the material is.

In torsion, there are similar relations where:

##T = K\theta##: ##T## is the torque, ##K## is the torsional spring constant and ##\theta## is the angular displacement;
##G = \frac{\tau}{\gamma}##: ##G## is the shear modulus, ##\tau## is the shear stress and ##\gamma## is the shear strain;
##\tau = \frac{F}{A}= \frac{T}{r A}##: ##\tau## is the shear stress and ##r## is the rod radius;
##\gamma= \frac{x}{L} = \frac{\theta r}{L}##

And thus:
$$G = K \frac{L}{r^2 A}$$
Like for the «linear» case, since ##L##, ##A## and ##r## are constants, then the shear modulus is comparable to a torsional spring constant, i.e. how stiff the material is in torsion.

All of this to demonstrate to you that the elastic modulus (modulus of elasticity in OP) and shear modulus do not give anything as to when material will fail; it just tells how the material will deform.

Stress

I already introduced stress definitions, where the stress is defined by a force divided by the cross-sectional area of the rod. Note that with tensile stress, the force is perpendicular to the area, and with shear stress the force is in the same plane as the area.

The tensile stress is usually noted ##\sigma## and the shear stress ##\tau##. But there are somewhat related, because stresses can be measured in the 3 dimensional planes, so you may see ##\sigma_x, \sigma_y, \sigma_z, \tau_{xy}, \tau_{yz}, \tau_{xz}## (sometimes with ##1, 2, 3## instead of ##x, y, z##); With this notation, you may see ##\sigma## and ##\tau## use interchangeably, depending on the author.

Stress is something you can measure, like pressure for example.

Strength

The strength of a material is the stress limit that the material can endure before failing. There is usually a value for tensile strength and one for shear strength. For both of these strengths, there are also a yield strength and an ultimate strength. The ultimate strength is when the material breaks apart. The yield strength is when the material deforms to a point where when the force or torque is removed, the material doesn't go back to it's original shape, i.e. it doesn't react like a spring anymore because there is permanent plastic deformation.

When you want to evaluate when a material will fail, you always use the yield tensile strength and/or yield shear strength to determine when permanent damage occurs (a part should always go back to its original shape when the load is removed).

Conclusion

All the equations you will encounter will have the tensile stress or shear stress as a variable (again, Fs in your equation is the shear stress). But if you want to evaluate the maximum value the material can handle (for example, the maximum torque in your equation) then you must use the corresponding material yield strength (tensile or shear) for the stress variable in the equation.

So - in your equation - if the torque is greater than the one calculated with the yield shear strength, then you know the shear stress will exceed the material yield shear strength and thus, the part will fail.

All of this can be confusing because all of these values have units of pressure. But there are not the same. I suggest you go through some wikipedia articles to study those concepts a little further to better understand them.

Last edited:
Asymptotic
tthree
Let's go back to the fundamental concepts for defining a material.

Stiffness

A spring is modeled by the equation ##F=kx##, where ##F## is the force applied to the spring, ##x## is the spring displacement under that force and ##k## is the spring constant, i.e. what defines the spring such that ##k = \frac{F}{x}## is always true, no matter the force or the displacement.

A solid rod can be viewed as a very stiff spring than can deform under a force. With materials, we prefer to look at stress ##\sigma## and strain ##\epsilon## such that ##E = \frac{\sigma}{\epsilon}##, where ##E## is the elastic modulus.

Because the stress is defined as ##\tau = \frac{F}{A}## and the strain is defined as ##\epsilon = \frac{x}{L}##, where ##A## is the cross-sectional area of the rod and ##L## is the original length of the rod. Therefore:
$$E = \frac{\sigma}{\epsilon} = \frac{\frac{F}{A}}{\frac{x}{L}} = \frac{F}{x}\frac{L}{A} = k\frac{L}{A}$$
Since ##L## and ##A## are constants, then the elastic modulus is comparable to a spring constant, i.e. how stiff the material is.

In torsion, there are similar relations where:

##T = K\theta##: ##T## is the torque, ##K## is the torsional spring constant and ##\theta## is the angular displacement;
##G = \frac{\tau}{\gamma}##: ##G## is the shear modulus, ##\tau## is the shear stress and ##\gamma## is the shear strain;
##\tau = \frac{F}{A}= \frac{T}{r A}##: ##\tau## is the shear stress and ##r## is the rod radius;
##\gamma= \frac{x}{L} = \frac{\theta r}{L}##

And thus:
$$G = K \frac{L}{r^2 A}$$
Like for the «linear» case, since ##L##, ##A## and ##r## are constants, then the shear modulus is comparable to a torsional spring constant, i.e. how stiff the material is in torsion.

All of this to demonstrate to you that the elastic modulus (modulus of elasticity in OP) and shear modulus do not give anything as to when material will fail; it just tells how the material will deform.

Stress

I already introduced stress definitions, where the stress is defined by a force divided by the cross-sectional area of the rod. Note that with tensile stress, the force is perpendicular to the area, and with shear stress the force is in the same plane as the area.

The tensile stress is usually noted ##\sigma## and the shear stress ##\tau##. But there are somewhat related, because stresses can be measured in the 3 dimensional planes, so you may see ##\sigma_x, \sigma_y, \sigma_z, \tau_{xy}, \tau_{yz}, \tau_{xz}## (sometimes with ##1, 2, 3## instead of ##x, y, z##); With this notation, you may see ##\sigma## and ##\tau## use interchangeably, depending on the author.

Stress is something you can measure, like pressure for example.

Strength

The strength of a material is the stress limit that the material can endure before failing. There is usually a value for tensile strength and one for shear strength. For both of these strengths, there are also a yield strength and an ultimate strength. The ultimate strength is when the material breaks apart. The yield strength is when the material deforms to a point where when the force or torque is removed, the material doesn't go back to it's original shape, i.e. it doesn't react like a spring anymore because there is permanent plastic deformation.

When you want to evaluate when a material will fail, you always use the yield tensile strength and/or yield shear strength to determine when permanent damage occurs (a part should always go back to its original shape when the load is removed).

Conclusion

All the equations you will encounter will have the tensile stress or shear stress as a variable (again, Fs in your equation is the shear stress). But if you want to evaluate the maximum value the material can handle (for example, the maximum torque in your equation) then you must use the corresponding material yield strength (tensile or shear) for the stress variable in the equation.

So - in your equation - if the torque is greater than the one calculated with the yield shear strength, then you know the shear stress will exceed the material yield shear strength and thus, the part will fail.

All of this can be confusing because all of these values have units of pressure. But there are not the same. I suggest you go through some wikipedia articles to study those concepts a little further to better understand them.