# Breaking solid shaft in torsion purposefully

• thissong
In summary: So maybe the best engineering solution would be a click torque wrench design that permanently disengages after the first "click"...Conventional practice is to use a shear pin. More predictable and reliable because of the simpler nature of the shearing action and in any event where shear stress cannot be avoided.
thissong
I am trying to design a drive shaft that breaks when i apply a specific torque to protect fro over torquing. I made a few samples, and now i am trying to compare the results to my equations. My problem is, none of the equations seem to match the actual results. Your help would be appreciated:

I made a bar that was 0.078in. in diameter, necked it down to 0.067in. diameter (.0335in. radius)

In testing, I found that the bar shears/breaks at around 9.6 in-lbs.

τ = T*r/J = 9.6in-lbs*.0335in/((π/2)*.0335in^4) = 164ksi

Now, my material states that I have an tensile strength of 170 ksi and yield strength of 140 ksi, and i thought, "Great! 3.5% error on my yield strength", but then i remembered the 0.577*σ rule. (making shear stress only .577*170 = 98ksi). I'd expect this to break at 5.7in-lbs using that conversion. Can any explain why my test results don't match up to my theoretical values?

(I considered using a stress concentration factor, but that would only make my answer more puzzling because it would increase the discrepancy where τ(max) = τ(nom)*K and K>1)Thank you

Joe591
thissong said:
I am trying to design a drive shaft that breaks when i apply a specific torque to protect fro over torquing. I made a few samples, and now i am trying to compare the results to my equations. My problem is, none of the equations seem to match the actual results. Your help would be appreciated:

I made a bar that was 0.078in. in diameter, necked it down to 0.067in. diameter (.0335in. radius)

In testing, I found that the bar shears/breaks at around 9.6 in-lbs.

τ = T*r/J = 9.6in-lbs*.0335in/((π/2)*.0335in^4) = 164ksi

Now, my material states that I have an tensile strength of 170 ksi and yield strength of 140 ksi, and i thought, "Great! 3.5% error on my yield strength", but then i remembered the 0.577*σ rule. (making shear stress only .577*170 = 98ksi). I'd expect this to break at 5.7in-lbs using that conversion. Can any explain why my test results don't match up to my theoretical values?

(I considered using a stress concentration factor, but that would only make my answer more puzzling because it would increase the discrepancy where τ(max) = τ(nom)*K and K>1)Thank you
Welcome to the PF.

Couldn't you just use a mechanism design similar to a click-type ratcheting torque wrench? It has the advantage of being non-destructive...

berkeman said:
Welcome to the PF.

Couldn't you just use a mechanism design similar to a click-type ratcheting torque wrench? It has the advantage of being non-destructive...
Thank you for the welcome. I've used PF for years, but finally came across an issue that i didn't see a solution to already.

I could use a ratchet, but that wouldn't work for this application. I'm really interested in just leaving the part behind if that torque is overloaded.

Joe591
thissong said:
I am trying to design a drive shaft that breaks when i apply a specific torque to protect fro over torquing. I made a few samples, and now i am trying to compare the results to my equations. My problem is, none of the equations seem to match the actual results. Your help would be appreciated:

I made a bar that was 0.078in. in diameter, necked it down to 0.067in. diameter (.0335in. radius)

In testing, I found that the bar shears/breaks at around 9.6 in-lbs.

τ = T*r/J = 9.6in-lbs*.0335in/((π/2)*.0335in^4) = 164ksi

Now, my material states that I have an tensile strength of 170 ksi and yield strength of 140 ksi, and i thought, "Great! 3.5% error on my yield strength", but then i remembered the 0.577*σ rule. (making shear stress only .577*170 = 98ksi). I'd expect this to break at 5.7in-lbs using that conversion. Can any explain why my test results don't match up to my theoretical values?

(I considered using a stress concentration factor, but that would only make my answer more puzzling because it would increase the discrepancy where τ(max) = τ(nom)*K and K>1)Thank you
In my experience, designing a structure (like a shaft) to fail at a precise value of shear is very difficult.

Most standards organizations dealing with structural design tend to use relatively high factors of safety (compared with tensile loading) when assessing the shear strength of a material. AISC typically places the max shear stress for steel at 0.4 * sy, while tensile strength can be 0.6 * sy minimum.

thissong said:
I could use a ratchet, but that wouldn't work for this application. I'm really interested in just leaving the part behind if that torque is overloaded.
SteamKing said:
In my experience, designing a structure (like a shaft) to fail at a precise value of shear is very difficult.
So maybe the best engineering solution would be a click torque wrench design that permanently disengages after the first "click"...

Conventional practice is to use a shear pin . More predictable and reliable because of the simpler nature of the shearing action and in any case you can always find a suitable pin configuration by trial and error without damaging the parent components .

There are many other possibilities .

For example - two face to face wave cams held together by an axial spring . This is re-useable - it will slip above a certain torque but continue to drive when torque is reduced again .

Another one is the friction drive - a shaft light press fit in a tube . Works well with plastic components .

Tell us the actual problem ?

Nidum said:
Tell us the actual problem ?

The design is already set for its application. I am trying to better understand how to make it work reliably. The actual problem is that both FEA and hand calculations don't line up with the actual part.

jack action said:

Looks like this may actually make my problem worse. If the actual stress is higher than what I calculate, shouldn't the bar break even sooner?

SteamKing said:
AISC typically places the max shear stress for steel at 0.4 * sy, while tensile strength can be 0.6 * sy minimum.

So are you saying that there is a factor of safety placed on my certificate of material properties?

thissong said:
So are you saying that there is a factor of safety placed on my certificate of material properties?
No, what I'm saying is that your material properties are based on a tensile test, not a shear test. To use those test results for calculating the max. allowable shear stress, the figure 0.4 sy is supposed to account for going from a tensile loading to a shear loading and any other unknown factors. The test figures on your material data sheets are the actual stress results from the tension test, but no one normally designs to the yield strength of the material.

For design in bending, where there are tensile and compressive stresses, you can design to 0.6 sy, where sy is the min. yield strength of the material. That extra 0.2 sy on the allowable stress is used as a cushion to guard against the vagaries of shear loading.

With your shaft, you want it to fail when a certain torque figure is encountered. Stressing a bar axially to yield doesn't necessarily produce failure; steel has a different figure called the rupture stress, which is the stress measured just before the test piece breaks.

thissong said:
The design is already set for its application. I am trying to better understand how to make it work reliably. The actual problem is that both FEA and hand calculations don't line up with the actual part.

What is the actual configuration of the shaft and surrounding parts - can you supply a drawing ?

For my test I put the fixed end in a drill press and the rotating end in a torque gauge and twisted the shaft (by hand, with the drill press off) until it came apart. I measure at almost 9.6 in-lbs every time i repeated the test. Obviously the drive shaft shears off at the necked down area. Wrapping my head around why the shaft withstood a higher shear than expected is why i am reaching out to you all.

Thank you for your help so far. I've been reading around in my old machine design books and doing other research, but haven;t come up with much yet. I appreciate the outside perspectives.

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I'll give you a clue :

A simple beam does not fail when stress in outer fibres reaches yield stress . In fact you can usually load a beam a good bit more before actual failure occurs . Think about why this is and how you can apply similar thinking to torsional failure situations .

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Nidum said:
I'll give you a clue :

Are you suggesting that the failure is not happening out the outside radius? or possible there is some necking occurring which diminishes the radius? I think this is a possibility, however, to get a number that is relatively close to what I'd expect, the radius would need to change from 0.0335in to 0.0200 in. (which isn't impossible).

Is it possible that i am incurring some cold work into the material as well during my machining operation? it IS a 300 series steel I am working with.

thissong said:
Looks like this may actually make my problem worse. If the actual stress is higher than what I calculate, shouldn't the bar break even sooner?
Yes, your number would be higher because of the smaller «true» area, but if you look at the diagram carefully, you'll notice that the «true» stress at point #3 (rupture) is also higher than the given ultimate strength. All in all, it may even out.

@thissong

As load is increased stress in outer fibres increases until it reaches yield stress .

What happens when load is increased further ?

Apart from the above issue, you state "Now, my material states that I have an tensile strength of 170 ksi and yield strength of 140 ksi". What is the reference that is giving you those values? I am asking this because most standard material specs for purchased materials are statements of minimum material properties and generally the actual properties of the purchased materials exceed those quoted minimum values.

Dullard
As well as everything already mentioned there is the problem of ensuring that the shaft is subject to a pure torque . Any significant bending forces could change the value of the breaking torque considerably . Made worse in this particular situation because the shaft is so small in diameter and has a built in stress raiser .

There are really many problems in trying to get an exact figure for the breaking torque of this shaft . The best you will ever manage is to find upper and lower bounds for the value .

I actually reread some of your posts and it seems that you are using the actual strength from the material cert delivered with the bar stock... Yeah, I'm reasonably sure that is the actual strength of the bar stock...
The only thing I can think in this case is that the torque that you are using might be slightly off or as the outer fibres of the shaft start to yield the fibres slightly deeper are forced to take more load until it partially yields as well and so forth until the shaft finally breaks. So when the shaft starts to yield more of it is at a higher stress and it ends up being able to take a higher torque. In normal shaft torsion theory you assume that the stress varies linearly with radius. After yielding starts this isn't true anymore though. The classical theory starts to break down at that point. I think this is more likely to be the cause of the error you are seeing. There is a name for it. Plasticity or something like that- I forget now...

Its the same mechanism that causes ductile metals to deal better with stress raisers than brittle materials. When there is a stress raiser in a ductile metal it yields partially, causing stresses to redistribute and ''average out''. I think the same thing is happening with the shaft. It ends with a higher percentage of the shaft being at higher stress and thus it's taking higher torque.
Maybe a hollow shaft will cause your shaft material to reach the same stress throughout faster and cause more uniform and predictable failure. Otherwise I'm stumped...

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Slightly off topic but putting an inclusion in the shaft will probably do what you want

When the 2,000 MW Didcot Power Station in the UK was being commissioned in the 60s/70s staff were investigating vibration in a 500 MW generator when the shaft driving it broke. If I remember correctly the shaft was about 30" in diameter.

Examination of the break showed an inclusion about the size and shape of a teaspoon which had acted as a stress raiser.

As an aside I think your test result and calculation seem quite reasonable in the circumstances.

etotheipi

## 1. What is the purpose of breaking a solid shaft in torsion purposefully?

The purpose of breaking a solid shaft in torsion purposefully is to study the behavior and strength of materials under extreme conditions. By intentionally breaking the shaft, scientists can gather data on the maximum stress and strain the material can withstand before failure.

## 2. How is the solid shaft broken in torsion purposefully?

The solid shaft is usually broken by applying a twisting force, also known as torque, to the shaft. This torque is gradually increased until the shaft reaches its breaking point and fractures.

## 3. What factors affect the breaking of a solid shaft in torsion purposefully?

There are several factors that can affect the breaking of a solid shaft in torsion purposefully, such as the material properties of the shaft, the amount of torque applied, and the speed at which the torque is applied. Additionally, the shape and size of the shaft can also play a role in its breaking behavior.

## 4. What are the potential risks involved in breaking a solid shaft in torsion purposefully?

Breaking a solid shaft in torsion purposefully can be a dangerous process and should only be done by trained professionals in a controlled environment. The high levels of torque and force involved can cause the shaft to break suddenly and potentially injure those nearby. It is important to take all necessary safety precautions and follow proper procedures when conducting this type of experiment.

## 5. How is the data collected and analyzed after breaking a solid shaft in torsion purposefully?

Various instruments, such as strain gauges and sensors, are used to collect data on the torque, stress, and strain applied to the shaft during the breaking process. This data is then analyzed to determine the maximum strength of the material and its failure point. This information can be used to improve the design and performance of materials in various applications.

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