# Design Shear Pin for Gate Valve at 763 N.m Torque

• WillemBouwer
In summary, Willem Bouwer suggests that the shear stress to be used when fitting a shear pin between the gearbox spindle and hand wheel is either average direct shear stress or transverse shear stress. He also suggests that the pin should rupture when a certain torque is reached. The calculations show that a shear pin between the gearbox spindle and hand wheel would need to have a diameter of 1.45 mm for average direct shear stress and 2.9 mm for transverse shear stress to prevent too much torque from being transmitted.
WillemBouwer
Background: To prevent too much torque having to be transmitted through the gearbox on a gate valve, a shear pin is to be fitted between the hand wheel and the gearbox spindle. When a certain torque is reached the pin should rupture and no torque transmitted. See attachment for reference diagram.

Solution needed: What diameter pin to use at certain torque figures?

Calculations: There are 2 kinds of shear stress I am considering.
Average direct shear stress and Transverse shear stress.

Inputs: Output torque of the gearbox: To = 763 N.m (15% SF included)
Gearbox ratio: rg = 12
Gearbox efficiency: η = 85%
Spindle diameter: ∅D = 0.05 m (Note the spindle would not be threaded as in the picture)
Material: SS 304 thus Su = 505 MPa and Ssu = 0.6*505 = 303MPa

Average direct shear stress:
Ssu = V/A
with Ssu the ultimate shear stress in MPa
V = Normal load on pin due to torque in N
A = Cross sectional area of pin in m^2

V = [(To/(rg*η))/(D/2)]/2 because we have double shear
V = 1496.08 N

A = ∏p^2 where p is the radius of the pin

Thus 303MPa = 1496.08N/∏p^2
so p = 1.25 mm and ∅P = 2.5 mm

Transverse shear stress:
Ssu = VQ/It = (4/3)*(V/A)
303MPa = (4/3)*(1496.08/∏p^2)
so p = 1.45 mm and ∅P = 2.9 mm≈ 3 mm

The reason I think the Transverse shear stress should be used is because the pin would not be rupturing linearly, the hand wheel hole diameter would also be larger than the spindle diameter leaving a gap which would cause a bend hence not rupturing the pin due to normal shearing...

Due to the fact that we use an M6 bolt and machine it down to get to the right diameter, it is very difficult to keep this peace from deforming as you machine. The solution is to machine it down to a 4mm diameter and drill a hole through the pin! To check the diameter of the hole I used the following equations:

Ssu=VQ/(It)
with V= T/r/2;
Q= y*A;
y=(4/3π)*(So²+SoSi+Si²)/(So+Si);
A= π*(So²-Si²);
I=¼π*(So4-Si4);
t=(So-Si)

for a input of: So (Outside diameter) = 4mm
Si (inside diameter) = 1mm

it gives: Ssu = 297 MPa

which is more or less the Ssu of the material...

I also realize that the material is strain hardening which could cause the material to get even harder during normal working torque so this could affect the Ssu of the material, but I am not o worried about this.

Any thoughts or feedback on this problem would be appreciated! Am I over thinking it?

HARSLE
WillemBouwer: I guess using peak shear stress, (4/3)*V/A, might give an absolute minimum shear rupture load for the shear pin. It perhaps would be an acceptable starting point. Then, through experimentation, you might find that the shear rupture load is higher, perhaps closer to when V/A = Ssu, or even slightly higher (?). Or even higher than that, because you included a safety factor of 1.15. For now, I will go along with your assumption to use peak shear stress.

I could not understand your last V formula, V = (T/r)/2. I thought V was simply, V = 1496.08 N, already given, above.

I could not understand your hollow shear pin formulas. They currently seem wrong to me. I currently think the correct formula for peak shear stress on a round, hollow, non-split shear pin (i.e., a non-split, round tube) is as follows.

tau = V*Q/(I*t) = [16/(3*pi)](V)(d2^3 - d1^3)/[(d2^4 - d1^4)(d2 - d1)],

where tau = shear pin peak shear stress (MPa), pi = 3.14159..., V = applied shear force (N), d2 = shear pin outside diameter (mm), and d1 = shear pin inside diameter (mm).

Therefore, compute the right-hand side of the above equation, iteratively, until it equals, e.g., Ssu. If Ssu = 303 MPa, and d2 = 4.0 mm, then the above formula gives, d1 = 2.0526 mm.

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I agree with nvn, I found myself sitting here trying to figure out where you got those formulas from. I also thought the direct shear stress would just be 1496.08N and for the transverse, I would just use the bending shear stress formula too.

The V calculated in the first paragraph above is the correct one but is the same as the V = (T/r)/2 one, it is defined in my spreadsheets as:

T = Input torque of gearbox acting on pin(N.m)
T = To/(rg*η)
and
r = radius of shaft (m)
r = (D/2)

So you can see that it is the same, sorry for not defining this better!

Vadar2012, in my opinion, transverse shear stress = bending shear sress!

The reason why is used transverse is because it is the peak shear stress, and I do not want the pin to break before the required torque was reached to close the valve! However I also do not want it to be to strong and not break before damaging the gearbox!

The adaptor fitting over the shaft through which the pin will be positioned has in inner diameter of say 30.2 mm when the shaft is 30mm. Does this mean I should rather use direct shear stress?

The equation I used, seems to have dissappeared from all my textbooks, I will try to find it but it is beyond me where I got that at the moment, because I"'ve done the spreadsheet a while back!
I will use yours as I do not have a reference to my equation at the moment!

The following is a direct example of one of our valves:

Output torque (To) = 119 N.m
Gearbox ratio (gr) = 4
Gearbox effieciency (η) = 85%
Spindle diameter (D) = 20 mm
Maximum shear stress = 303 MPa
Maximum gearbox ouput = 593 N.m

So by itteration with an outside diameter of do = 6mm we get di = 4.631, so by choosing a suitable drill say 4.5 mm, we get Ssu = 279 MPa and safety factor of 1.085! Now to be clear I want the pin to transver the required 119 N.m but it should rupture before coming close to the maximum allowable torque of 593 N.m!

By looking at the safety factor of this pin does this give me a torque transfer of 129 N.m before rupture which is still well below 593 N.m?

If we look at direct shear Ssu = V/A for do = 6mm adn di = 4.5mm we get Ssu = 2.14 with a max output of 254 N.m which is still below 593 N.m! I show this to illustrate that it makes a big difference by using direct shear stress or transverse shear stress! Which one would then be better suited for the application?

Any remarks? Thanks

WillemBouwer: I now noticed, your formulas at the bottom of post 1 work correctly, and give the same answer as my formula in post 2. Perhaps you made a mistake on your calculator. When you compute it correctly, your formulas at the bottom of post 1 give Si = 2.0526 mm, not 1 mm. Try it again.

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WillemBouwer said:
I show this to illustrate that it makes a big difference by using [average] shear stress or [peak] shear stress. Which one would then be better suited for the application?
WillemBouwer: If I had to guess, I currently would guess that the non-split, round, tubular shear pin will break at n*Ssu = V/A, where Ssu = shear pin material shear ultimate strength, A = 0.25*pi*(d2^2 - d1^2), and n = 1.15 or 1.20. This guess uses the nominal, average shear stress formula (V/A).

If you test your shear pin(s) in the future, I hope you will post here the shear load at which your shear pin breaks.

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WillemBouwer said:

Vadar2012, in my opinion, transverse shear stress = bending shear sress!

Gah, that was a bad typo. I have bending on the brain at the moment.

nvn: yes I wanted to say that I am pretty sure about those formulas, I think I might have derived them from first principles perhaps! I checked my calculations and somehow I manged to forget a bracket in my equation on excell, bad mistake to make, haha!

I will indeed be doing some tests and post it here, but I am also a bit worried as some of these pins have already been installed and going to customers, I guess I'll have to wait and see!

Just a final question. In the above example I use a M8 bolt and machine(machining it also heat hardens the material) it down to a 6 mm OD (difficult to turn less then 5 mm OD, pin bends), then I drill a 4.5 mm hole to give the ID. For the formula in post 1 the Ssu is then 238MPa with safety factor of 1.27. So this gives a max torque output of 968 N.m. For the suggested formula in your last post you then have Ssu as 100 MPa with a safety factor of 3 which gives a max output torque of 2294 N.m!

The problem is the shear pin should break before 1055 N.m because that is the maximum output torque of the gearbox, or else the shear pin would have had no purpose at all!

For the formula you suggested the ID should be 5.5 to give Ssu as 276 MPa and transfer a maximum of 837 N.m! But for this ID the peak stress could be as high as 662 MPa which will suggest the pin will rupture at 349 N.m which will also be of no use because the valve will not be closing!

Which way do I go?

I believe your answer will be that I should test the material and see what the breaking torque of each pin is,haha...

WillemBouwer said:
Output torque (To) = 119 N*m
Gearbox ratio (rg) = 4
Gearbox efficiency (eta) = 85%
Spindle diameter (D) = 20 mm
Maximum gearbox output torque = 593 N*m

I want the pin to transfer the required 119 N*m, but it should rupture before coming close to the maximum allowable torque of 593 N*m.
WillemBouwer said:
The shear pin should break before 1055 N*m, because that is the maximum output torque of the gearbox.
WillemBouwer: What is the required valve closing output torque (N*m)? What is the maximum allowable gear box output torque (N*m)? The above quotes give different values, which currently confused me.

I have many different valves, hence different torque figures that I have calculated. The first post is the example of one of the valves which I am a bit worried about!

Output torque (To) = 763 N*m
Gearbox ratio (rg) = 12
Gearbox efficiency (eta) = 85%
Spindle diameter (D) = 50 mm
Maximum gearbox output torque = 1055 N*m

And it is this one that I referred to in my last post, to illustrate how the 2 different formulas can give a very different outcome, so which one to use is my question?

WillemBouwer: Is the applied shear force (V) on the shear pin reversing? In other words, is the valve opening output torque applied to this same shear pin, in the opposite direction, and is the valve opening output torque equal to the valve closing output torque (763 N*m)? If so, how many times (cycles) is the valve opened and closed in its life span, using this shear pin?

I would probably currently use the following equations. Parameter k2 is the peak shear stress factor; k3 reduces this to "pct" percentage of its effect.

Stu = shear pin material tensile ultimate strength.
Sty = shear pin material tensile yield strength.
Ssu = shear pin material shear ultimate strength = 0.60*Stu.
Ssy = shear pin material shear yield strength = 0.577*Sty = 0.577*205 = 118.3 MPa.
k2 = (4/3)(d2^2 + d2*d1 + d1^2)/(d2^2 + d1^2).
k3 = 1.3333 + pct*(k2 - 1.3333).
pct = 0.70.
nn = 1.15.
tau1 = k3*(V/A).
tau2 = V/A.
Rsy = tau1/Ssy.
Rsu = tau2/(nn*Ssu).

Ensure Rsy is less than or equal to 1.0. If Rsy > 1.0, it indicates the shear pin is overstressed, with respect to shear yield stress, during valve closing. Ensure Rsu is greater than or equal to 1.0. If Rsu ≥ 1.0, it indicates the shear pin encounters shear rupture.

I currently could not make it work. I.e., when the shear pin is strong enough to avoid shear yielding during valve closing, then it currently appears the shear pin is far too strong to encounter shear rupture at the gear box maximum allowable output torque.

nvn, sorry for the delayed reply, been a bit busy with some other work!

It is indeed a very tricky situation to get the correct sizing, for the bigger torque figures I will definitely built a testing unit to check the breaking torque and also repeatable cycles without breakage as to see what the effect of the yielding of the pin would have.

For the smaller valves it works out well however:
DN100 gate valve: Spindle torque needed: To=28 N.m
gearbox ratio: 4
Spindle diameter = 20mm
Maximum gearbox output torque = 288 N.m
Choosing outside diameter of pin as 6mm and inside diameter as 5mm.

For these inputs and using the equations in your last post you get:

Rsy = 0.722 and Rsu =0.154 so this equates to no yielding or rupture for the desired spindle torque.

Now if something goes wrong and the torque spikes to 200 N.m

Rsy = 5.16 and Rsu = 1.1 which relates to the rupturing of the pin before the maximum gearbox torque could be reached!

I would definitely post my test results here so that you could have a look at it!

Thanks for the assistance...

## 1. How do you determine the appropriate shear pin strength for a gate valve at a specific torque?

To determine the appropriate shear pin strength for a gate valve, you will need to consider the maximum torque that the valve will experience during operation. This torque can be calculated by considering the applied force and the distance from the point of application to the axis of rotation. Once you have determined the maximum torque, you can select a shear pin with a strength that is slightly higher than this value to ensure the valve will remain stable during operation.

## 2. What material is commonly used for shear pins in gate valves?

The most common material used for shear pins in gate valves is stainless steel. This is because stainless steel has a high strength and is resistant to corrosion, making it suitable for use in various environmental conditions.

## 3. How often do shear pins need to be replaced in gate valves?

The frequency of shear pin replacement will depend on the amount of torque and stress the valve experiences during operation. Generally, it is recommended to replace shear pins every 6 months to 1 year to ensure the valve remains in optimal working condition. However, if the valve experiences high levels of torque or is used frequently, more frequent replacement may be necessary.

## 4. Can shear pins be customized for specific torque requirements?

Yes, shear pins can be customized for specific torque requirements. This can be done by selecting a material with a higher or lower strength, or by adjusting the diameter or length of the shear pin. It is important to consult with a professional to ensure the customized shear pin meets the necessary safety and functionality requirements.

## 5. What happens if a shear pin fails in a gate valve?

If a shear pin fails in a gate valve, the valve will no longer be able to withstand the specified torque and may become unstable or malfunction. This can result in potential damage to the valve or other equipment, and may also pose safety hazards. It is important to regularly inspect and replace shear pins to prevent failure and ensure the proper functioning of the gate valve.

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