# Shear and bending of a safety pin?

• WillemBouwer
In summary, WillemBouwer found that the minimum diameter of the pin for a maximum deflection of 1 mm is 18.106 mm, and that the pin will have a center of rotation within each clevis plate clearance hole. He also calculated that the pin will have a shear rupture stress level of 100%, and a normal rupture stress level of 100%. Finally, he found that the moment of inertia is wrong and that the material is EN24T.
WillemBouwer

## Homework Statement

Designing a safety pin for gate valve, picture is attached! Material: EN24 (λ=680MPa and E= 207 GPa)
Total distance between supports = 98 mm! Force applied by gate is 64705 N! Question: Minimum diameter of pin for 1) No shearing , 2) a maximum deflection of 1 mm?

## Homework Equations

For circular sections: λ=(4/3)*(V/A)
Where λ= Maximum Shear(MPa) V = applied force (N) and A = Cross-sectional area of the pin (mm^2)
Normal shear at point right next to the supports is: σ = M*c/I
Where σ= Maximum Normal Shear(MPa) M = Moment created by V (N.m) c = distance from center to outside of pin:radius (mm) I = Moment of Inertia (mm^4) = ∏*r^4/8
Maximum deflection y=(V*l^3)/(48EI) Fairly well known equation...

3.The attempt at a solution
1) Safety factor = 1.5
Hence λ=(4/3)*(V/A)
But Vmax = 64705/2 = 32352.5 N because there is a counter force at other support side
680/1.5=(4/3)*(32352.5/(∏d^2/4)
giving d = 11 mm

BUT PROLBLEM σ = M*c/I gives

680 = (32352.5*49)*r/(∏*r^4/8)
giving r = 18.106 mm, d = ±36mm which I personally think is way to big! Please help clarifing this! Sorry I'm Afrikaans not the best speller!

2) y=(V*l^3)/(48EI)
y=(64705*(0.098^3)/(48*207*10^9*((∏*r^4)/8)
r = 11 mm which gives d = 22 mm

Seeing that it is a safety pin and will have to be removed I think a 1 mm deflection over 98 mm span is not unexceptable, but now I don't know whether it will shear because of the problem listed in the 1st question... Any help will be much appreciated... Thanks

#### Attachments

• Safety pin.jpg
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WillemBouwer: Excellent work. However, I think you may have mislabeled lambda in your first paragraph. Is your steel EN24T, or what? You need to specify the heat treatment condition code, not just EN24. Is it condition T? The tensile yield strength of steel EN24T is Sty = 680 MPa. The minimum tensile ultimate strength of steel EN24T is Stu = 850 MPa. I think the minimum shear ultimate strength of steel EN24T is Ssu = 0.60*Stu = 510 MPa (?). Therefore, I think you instead should use the following criteria, where P = applied load = 64 705 N, V = shear force = 0.5*P, M = bending moment = 0.25*P*L, FSu = ultimate safety factor, and FSy = yield safety factor.

(1a) Shear rupture stress level, FSu*(4/3)(V/A)/Ssu ≤ 100 %.
(1b) Normal rupture stress level, FSu*(M*c/I)/Stu ≤ 100 %.
(1c) Normal yield stress level, FSy*(M*c/I)/Sty ≤ 100 %.
(2) Deflection criterion, [P*(L^3)/(48*E*I)]/ymax ≤ 100 %.

Also, your moment of inertia is wrong, and instead should be, I = (pi/64)*d^4. Therefore, your results for items 1 and 2 are currently wrong. Also, I think shear stress is usually called tau, not lambda, right? And sigma is normal stress (not shear stress) at midspan, not "normal shear right next to support." Try again.

Also, the pin will tend to have a center of rotation within each clevis plate clearance hole, not exactly at the inboard face of each clevis plate. Hence, your current value for beam length L is too small. Therefore, we need to know the thickness, t2, of each of the two clevis plates shown in your post 1 diagram. The two clevis plates are the two supports for your pin. You should use beam length, L = a1 + 2(pct*t2), where a1 = distance between inboard faces of clevis plates = 98 mm, t2 = clevis plate thickness, and pct = pin end center of rotation location, as a percentage of t2. For relatively thin clevis plates, you could use pct = 0.50. For relatively thick clevis plates, you might be able to use pct ≥ 0.25 (?). But we need to know, what is the thickness, t2, of each of your two clevis plates? And what is your value for FSu, and your value for FSy?

Thanks for the reply nvn, not to many people are keen on helping with this problem I see!

t2 = 5 mm therefor L = 98+2(0.5*5) = 103 mm
Never seen this before, didn't expected the rotation to make such a difference! Makes some sense though!

As for lambda, couldn't find the tau sign so I guessed you guys would be able to figure it out! Think it is a translation problem but yeah I meant normal stress(σ). Why I said it is at a point next to the support is that this is furthest away from the centre, rookie mistake as I checked my moment diagram and saw the max bending moment is indeed at midspan thanks!

Moment of inertia: rookie mistake again as I took the semi-circular equation from my textbook, is indeed (pi*r^4)/4 = (pi*d^4)/64

Material is indeed EN24T!

Fsu = 1.6
Fsy = 1.2

Newly calculated solutions:
1a) Fsu(4/3)(V/A)/Ssu ≤ 1
gives d ≥ 13.13 mm

1b) Fsu(M*c/I)/Stu ≤ 1
gives d ≥ 31.7 mm

1c) Fsy(M*c/I)/Sty ≤ 1
gives d ≥ 31.05 mm

2) (P*L^3/(48*E*I))/ymax ≤ 1
gives d ≥ 19.5 mm

So the minimum diameter that we can use for no rupture is 31.7 mm. Seeing that a 1mm deflection will be achieved if a 19.5 mm diameter pin is used the deflection will be much less than 1mm when the pin will rupture? We've done some practical tests with a 18mm pin and the pins tend to deflect to about 2 mm at midspan without rupture! This is why I thaught my calculation were wrong! Any thaughts on this... Just want to know if I understand correctly, is the minimum diameter pin we can use for this problem then 31.7 mm or what?

Thanks so much for the assistance...

WillemBouwer: Are you a student, and is this a school assignment? If not, you could have posted this thread in the mechanical engineering forum, instead of the engineering homework forum, if you wish.

There are multiple things we can investigate. But before going into that, could you provide the following information?

Could you post the exact dimensions shown in the attached diagram, for your d = 18.00 mm pin? And please include the hole tolerance values for hole diameters D1 and D2. Also, if you could indicate how the two clevis plates are supported (boundary conditions), that might be helpful, too.

It is possible your pin material is stronger than the published EN24T minimum strength requirements. Some manufacturers will, sometimes, greatly exceed the specification minimum requirements. Are you sure it is really condition T? To determine exactly what is happening, you might need an accurate stress-strain curve of your specific, actual pin material batch. Do you have an accurate stress-strain curve for your material?

You might be starting to enter the plastic zone. Are you plotting load versus deflection, when you test your 18.0 mm pin, to see if you are barely beginning to enter the yield (nonlinear) zone? Have you measured rebound, to see if your pin did not return exactly to zero deflection when the load is removed?

#### Attachments

• safety-pin-02.png
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No I am not a student, yeah sorry first post will be sure to post the next one there... 1st year of employment and I am the only mechanical engineer at the firm so no one knows what the answers are, so I truly appreciate your help...

Dimensions from diagram:
a1 = 98 mm
d = 18 mm
t2 = 5 mm
t1 = 14.5 mm
The hole is actually an upright slot like this without the strikethrough:θ (just an illustration)
The diameter of end semi circles is 20 mm and the height is 30 mm
D1 is not a dimension as the whole gate rests at the top of the pin as in the first diagram!

Clevis plates support illustration has been attached!

We are indeed entering the plastic zone, because I only saw the deflection after the pin has been removed, my question: Why did the pin not break? Though we went into the plastic deformation zone the normal rupture should be occurring before this as the minimum diameter for this condition is more than that of the yielding conditions? We use 2 pins but that I have determined:

400 mm actuator diameter at 6 bar air pressure is pressing down on the gate:
Force applied to each pin: P = ±37700 N

Lets say Safety factors have been set to 1:
Diameters calculated:
Shear rupture: 7.92 mm
Normal rupture: 22.66 mm
Yielding: 24.41 mm

Okay only just checked these, I'm not going to correct my statement in the 4th paragraph, saw now that yielding will indeed occur firstly which for me makes sense as this should not be a brittle material... Not entirely sure of the material then! May differ somewhat but that I will sort out with tests thanks!

Just want to know if the calculation approach that we are following are indeed accurate! Can you post this at mechanical engineering as I don't have a clue how to, thanks a bunch!

#### Attachments

• SKG clevis pin illustration.jpg
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WillemBouwer: What do you mean by, "We use 2 pins but that I have determined:"? Could you reword that sentence?

Are you sure the load applied to your 18 mm pin is truly P = 64 705 N?

You said your pin is entering the plastic range. Do you want to allow yielding of the 18 mm pin? Or do you want to prevent yielding?

Let me know about the above questions, and then I will review the calculation approach.

We use 2 pins to secure the gate, one at either side of the centerline! But my previous posts only calculated for 1 pin! In the last post: I calculated the diameters when using 2 pins!

The load applied is at 6 bar and 400 mm actuator! This gives a load of 75400 N! Because 2 pins are used the net force P on a single pin is 37700 N! Right?

I do not want yielding... Only stated that there was yielding as the pin stayed in the deformed position! But yeah I just wanted to check if there is indeed yielding, up to what point will the pin yield before it ruptures! Hope this makes sense...

WillemBouwer: But what applied load was applied at the midspan of the 18 mm pin you tested? Was it 64 705 N? Or was it 37 700 N? Why did you say 64 705 N, earlier?

Does the gate (14.5 mm thick) have a circular edge, which presses on the pins? Or is it a flat edge? If it is a circular gate, are you sure you did the vector mechanics correctly, to obtain the force applied to each pin? If a circle (a disk) presses on the two pins, the total force on the pins will be greater than the total axial applied force. Do you have a free-body diagram (FBD) showing the system, the 400 mm actuator rod, and how the force is applied to each of the two pins?

WillemBouwer: Under static loading, the extreme fiber of your pin will begin to yield long before your pin ruptures. Because of your ultimate and yield factors of safety, and because of your pin material and ultimate plastic shape factor, bending ultimate stress does not govern for your pin. Bending yield stress governs over bending ultimate stress. You can currently ignore bending ultimate stress. Therefore, to analyze your specific pin, you only need to compute the bending yield stress level (criterion 1c in post 2).

You stated you do not want your pin to begin yielding, and you stated you want to use a yield factor of safety of FSy = 1.20. Therefore, you only need to use criterion 1c in post 2, to design your pin.

Because of the thickness of your center plate (t1), and because of your clevis plate thickness (t2), you can currently use a beam effective length of L = 95.8 mm, instead of 103 mm (because t1 slightly reduces the moment arm).

I am currently assuming your applied load is P = 37 700 N (?), until I see your responses to my questions in post 8. Therefore, using P = 37 700 N, L = 95.8 mm, and FSy = 1.20, then criterion 1c in post 2 currently gives, d = 25.32 mm. Therefore, the required pin diameter is currently 25.32 mm.

You greatly overstressed your 18 mm pin. After the test, you observed a large amount of permanent deformation. Although you did not apply enough load to cause the pin to rupture, the pin was greatly overstressed, nonetheless.

nvn: In my first post I merely put in some known so that I can know what the correct way of getting to a solution was, as I did not want tyou to do the problem for me, but I realized that some things may depend on the exact criteria and in post 7 I clarified this! So it is indeed a load of 37700 N,thanks! Okay so there must have been something wrong with the testing equipment or air pressure reading as this would have caused overstressing! The gate has a flat edge at the end! How did you get to L = 95.8 mm? I get the same answer as you, 25.32 mm for no yielding! So this will mean that the pin will not deform plastically but only in the elastic region and will return to its normal position after the load has been taken off...?

Question: If I do allow for yielding of 1mm say, with an initial test, will the pin deflect more or less another 1 mm on the next test?

WillemBouwer said:
How did you get to L = 95.8 mm?

WillemBouwer: Your applied load, P, is equivalent to two loads, 0.5*P, centered on the beam, and separated by a distance 0.5*t1. Therefore, the maximum moment is M = (0.5*L2)(0.5*P) = 0.25*P*L2, where L2 = beam effective length for moment = L - 0.5*t1 = 103 - 0.5*14.5 = 95.8 mm.

However, you cannot use L2 for deflection. For deflection, you can use L = 103 mm, but then deflection will have an error of 0.73 %, due to the thickness of t1. To avoid the 0.73 % error, you can use y = P*(L3^3)/(48*E*I), where L3 = beam effective length for deflection = 102.75 mm. Therefore, if P = 37 700 N, E = 207 000 MPa, and d = 25.32 mm, then

M = 0.25*P*L2 = 0.25(37 700)(95.8) = 902 915 N*mm,
y = P*(L3^3)/(48*E*I) = (37 700)(102.75^3)/{48(207 000)[(pi/64)*25.32^4]} = 0.2040 mm.
WillemBouwer said:
I get the same answer as you, 25.32 mm, for no yielding! So this will mean that the pin will not deform plastically but only in the elastic region, and will return to its normal position after the load has been taken off?

Yes, that is correct.
WillemBouwer said:
If I do allow for yielding of 1 mm say, with an initial test, will the pin deflect more or less another 1 mm on the next test?
The above deflection equation applies only in the elastic (linear) range. If you deflect a d = 25.32 mm pin 1 mm, you will exceed the minimum tensile yield strength specified in the specification. (Keep in mind, some manufacturers will, sometimes, greatly exceed the specification minimum requirements.) You will probably be entering the inelastic (plastic, nonlinear) range. Probably most of the 1 mm deflection will rebound, when you remove the applied load. But some of it will probably remain, if you greatly exceeded the tensile yield strength. If, e.g., 0.25 mm remains, then I think the maximum deflection on the next test might be roughly 1.25 mm (?).

WillemBouwer: I forgot to mention in post 9, you can change criterion 1b in post 2 to FSu*(M*c/I)/(psfu*Stu) ≤ 100 %, where psfu = plastic shape factor based on ultimate (not yield) strength. For your particular pin and material, psfu = 1.56. To clarify, items 1b and 1c are now,

(1b) Normal rupture stress level, FSu*(M*c/I)/(psfu*Stu) ≤ 100 %,
(1c) Normal yield stress level, FSy*(M*c/I)/Sty ≤ 100 %.

nvn: Yes this is only for static loading as the pins will be used once every 2 months so maintenance may be done!

Thanks for all the help! Manufactured some new pins this weekend and the calculations seem to work fine for the different line pressures and actuator diameters...

Happy holidays!

WillemBouwer: Keep in mind, FSy = 1.20 is rather low, and is generally used only where low weight is a particularly important consideration. For normal-weight, ground-based applications, FSy = 1.50, and FSu = 2.0, is perhaps more common. Or even FSy = 1.70, in applications where robustness is more important, and weight is not such a critical issue. Therefore, unless you have a good reason for choosing FSy = 1.20, you might consider increasing FSy to 1.30, 1.40, or 1.50, and increasing FSu to, say, 1.90, or something like that. Nonetheless, I would probably recommend you review and compare what FSy value is typically used in your industry for similar applications, and ensure you are being consistent with general practices in your industry and with typical design code requirements. Good luck.

## 1. What is shear and bending in the context of a safety pin?

Shear and bending refer to two types of mechanical stress that can occur in a safety pin. Shear stress is the force that causes two parts of the pin to slide past each other, while bending stress is the force that causes the pin to bend or deform.

## 2. What factors can cause shear and bending stress in a safety pin?

Shear and bending stress can be caused by various factors, including the weight of the object attached to the pin, the force applied to the pin during use, and the material and design of the pin itself.

## 3. How can I determine the maximum shear and bending stress that a safety pin can withstand?

The maximum shear and bending stress that a safety pin can withstand depends on its material properties and design. You can consult the manufacturer's specifications or perform a stress analysis to determine these values.

## 4. What are some ways to prevent shear and bending failure in a safety pin?

To prevent shear and bending failure in a safety pin, you can use a pin with a larger diameter or a stronger material, avoid overloading the pin, and regularly inspect the pin for any signs of wear or damage.

## 5. Can a safety pin fail due to shear or bending stress?

Yes, if the shear or bending stress exceeds the pin's maximum capacity, it can fail and break. This can result in the pin not being able to fulfill its intended function, such as holding an object in place or securing a garment.

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