# Double Shear stress on a bolt

• ABPL
In summary, the conversation discusses a problem with a bolt being used in equipment and the concern that it may not be able to withstand the forces it will face. The speaker is trying to solve the problem mathematically as they cannot perform an actual test. They are specifically trying to find the shear stress on the bolt. The conversation also mentions the dimensions and forces involved in the setup, as well as the approach taken to calculate the shear stress. Finally, there is a warning to thoroughly analyze the situation for safety before conducting any tests.
ABPL
Hello.

I know what follows seems like a HW problem, but it's an actual problem I'm trying to solve with some equipment I will be using.

I am concerned that a bolt I am using will be unable to withstand the forces that it will face, and so I am trying to solve the problem out mathematically as I cannot really do an actual test with the equipment.

I need to find the shear stress that is being put on a bolt.

The bolt is 0.5 inches in Diameter, and joins 3 bars. Each bar has a thickness of 0.375 inches. There is a 1.125 inch gab between one of the plates and the other two. The force on the lone bar is 25,000 lbs. The opposite force on each of the paired bars is unknown, but all the forces balance out to zero.

Please see the attached diagram I've illustrated.

http://postimg.org/image/50ouq73sd/

What I've tried:

Since the bar with F3 is not centered, this problem is unlike any previous ones I remember doing in class, or can find examples of online. None the less, I took the approach of assuming that force was inversely linearly proportional to the distance along the bolt. So F2 has most of the load (75%) of it, and F1 has 25% of the load from F3. I don't know if this is correct (If the relationship is a simple linear one), but it seems like it should be.

Then I tried to calculate the shear stress. Here I confused myself as I found I perhaps did not need to do the previous step at all - even though it struck me as necessary to know each force acting on the bolt. Following the equation for Shear Force of tau = F/A, I ended up just using the force F3 (25k lbs) and an area of 0.3927 in^2 (The x-sectional area of the bolt, times two, since there are two plates attached).

This gave me an answer of 15,900 lbs/in^2... I have no clue if I did it right, or if that answer is reasonable for me.

Any obvious fault in my work? Anyone know what I am missing here, and how to find the shear stress of this bolt? Did I *gasp* do it right?

ABPL said:
Hello.

I know what follows seems like a HW problem, but it's an actual problem I'm trying to solve with some equipment I will be using.

I am concerned that a bolt I am using will be unable to withstand the forces that it will face, and so I am trying to solve the problem out mathematically as I cannot really do an actual test with the equipment.

I need to find the shear stress that is being put on a bolt.

The bolt is 0.5 inches in Diameter, and joins 3 bars. Each bar has a thickness of 0.375 inches. There is a 1.125 inch gab between one of the plates and the other two. The force on the lone bar is 25,000 lbs. The opposite force on each of the paired bars is unknown, but all the forces balance out to zero.

Please see the attached diagram I've illustrated.

http://postimg.org/image/50ouq73sd/

What I've tried:

Since the bar with F3 is not centered, this problem is unlike any previous ones I remember doing in class, or can find examples of online. None the less, I took the approach of assuming that force was inversely linearly proportional to the distance along the bolt. So F2 has most of the load (75%) of it, and F1 has 25% of the load from F3. I don't know if this is correct (If the relationship is a simple linear one), but it seems like it should be.

Then I tried to calculate the shear stress. Here I confused myself as I found I perhaps did not need to do the previous step at all - even though it struck me as necessary to know each force acting on the bolt. Following the equation for Shear Force of tau = F/A, I ended up just using the force F3 (25k lbs) and an area of 0.3927 in^2 (The x-sectional area of the bolt, times two, since there are two plates attached).

This gave me an answer of 15,900 lbs/in^2... I have no clue if I did it right, or if that answer is reasonable for me.

Any obvious fault in my work? Anyone know what I am missing here, and how to find the shear stress of this bolt? Did I *gasp* do it right?

It's not just shear stress which you have to be worried about here. Because F2 is acting at a distance away from F1 and F3, there is going to be bending set up in the shaft of the bolt. In other words, you must analyze the shaft of the bolt as a beam and not just something which resists shear.

It's not clear in all the details of what this setup represents, but I would urge you not to proceed with any actual testing until you have analyzed this situation thoroughly for safety.

You have a potentially complicated situation here. The actual loading on the bold depends considerably on the details of the fit of the bolt in the holes.

1. Steamking is correct in saying that this needs to be analyzed as a beam.

2. Is the bolt a snug fit in the outer plates, or is it a loose fit? If it is a snug fit, you have something approximating a fixed end (no rotation), whereas, if it is a loose fit, you have something like a simple support at the inner edge.

3. The localized bearing stresses can be analyzed by Hertz contact stress theory.

4. Is the bolt a snug fit or a loose fit in the center member? This will also establish the boundary conditions applicable there.

5. You need to worry about both the possible breakage of the bolt and the tear-out of the supporting member.

As mentioned by Steamking, but sure to analyze this situation well from a safety standpoint before actually testing it.

I would agree with the above, but thought I'd add a few more thoughts.

I'd suggest modeling the 3 points of contact as pinned joints (ie: unable to resist a moment). I say that for a few reasons... it's the most conservative assumption, the hole could wallow out over time and the contacting support may be relatively free to bend even if it is a press fit. I might consider it a fixed joint if I was absolutely sure it really was fixed and couldn't get loose over time.

Next, do a freebody diagram of the bolt. Assume the forces F1, F2, F3, are acting at the center of those .375" wide members. I wouldn't worry about bending between F1 and F3 since they're so close, but if there's any possibility that those points can move, I'd suggest looking at how much they can move and consider the worst possible case. I'd look at bending between F2 and F3.

The other things you need to look out for regard allowable stresses and applicable codes. If this is for lifting for example, certain ASME codes apply in the US. Also, sheer stress is not the same as tensile stress when it comes to stress allowables. Depending on material, the allowable shear stress will be much less than the stress allowable per ASME code. It isn't hard to see the shear stresses in this case are going to be tremendously high.

For the most part, I agree with Q_Goest, although I think I would put F1 at the inner edge of the left plate and F3 at the inner edge of the right plate. If the bolt bends, these are the effective load points rather than the member center points.

If it were me, I would not feel comfortable unless the system were analyzed using finite element stress analysis. The geometry just seems too complicated to take chances.

Chet

Being not myself a mechanical engineer, my intuition is to use something like this

clamp it between your elements F1 and F2 and let the swivel handle mis-alignment that's what it is for.
Now you've got rid of bending in your bolt and replaced it with shear.

I'll leave it to the genuine mechanicals to advise on whether that's a good idea.

Shown is a "top link end" for a tractor, about \$5 at farm supply stores.
They come in several sizes. This one is cat1 having a 3/4 inch hole.
I haven't found load ratings for the agricultural equipment parts. 25000 pounds is substantial so be real careful with this project

Here's a page from a manufacturer of aircraft hardware with way better information:

Keeping in mind that ordinary steel oughtn't be loaded near(maybe 2/3?) its yield point of about 36,000 psi you need to make sure your 25,000 lb pull is not imposed across less than one square inch of steel.
How much cross sectional area is left in those 3/8 bars after drilling your half inch hole?
A 3/8 inch square is only 9/64 square inch so should not be subjected to more than 9/64 X 2/3 X 36,000 = 3375 lbs of pull.

And from that manufacturer's site , for 25,000 lbs of pull i'd pick his 3/4 inch bore rod end.

I hope this ramble makes clear why there are so many raised eyebrows among previous posters.

The sketch looks to me weak at the boltholes.
Don't get hurt.

old jim, just an old electrical guy

Last edited by a moderator:
Chet proposes an FEA on this part, and that is a good idea,
1) provided the project has money/justifies the expense of such an analysis, and
2) provided that there is a really competent engineer to make the FEA.

The FEA is not one bit better than the model it uses. Just what will the model say about the exact way the load is transferred from one body to the next? In a real sense, this is the hard part, and FEA does nothing at all to make it any easier.

If there FEA is not justified (can't meet both 1 and 2 above), then make it extremely stout! Keep any stress you can calculate very, very low, and then cross your fingers.

Q_Goest
Jim Hardy has suggested the use of a ball joint connection for the center element. This will have the effect of centering the load in that element, but it will not eliminate bending in the bolt (the bold is still a beam with lateral loads). This really does not solve very much since the location of the load transfer between the central load element and the bolt is not really the issue here.

The issue is two-fold:
1) potential failure of the bolt:
2) potential failure of the load application elements (end breaking off or tear-out).

I stand by my previous advice to make it extremely stout if you cannot analyze it properly. This is common practice in the oil field, off-shore, and many other heavy service areas.

Sadly, Guys, like a lot of posters, it appears the OP briefly surfaced with his one post and then disappeared back into the interwebs (never to be seen again?)

## 1. What is double shear stress on a bolt?

The double shear stress on a bolt refers to the stress that occurs when a bolt is subjected to two equal and opposite forces in different planes, perpendicular to the axis of the bolt. This happens when the bolt is used to connect two or more parts that are loaded in different directions.

## 2. How is double shear stress calculated on a bolt?

The double shear stress on a bolt can be calculated by dividing the total force applied on the bolt by the cross-sectional area of the bolt that is experiencing the shear load. This value is then multiplied by a factor of 2 to account for the double shear.

## 3. What factors can affect double shear stress on a bolt?

The magnitude and direction of the applied forces, the material properties of the bolt and the connected parts, and the geometry of the bolt and its connections can all affect the double shear stress on a bolt.

## 4. Why is double shear stress important to consider in bolt design?

Double shear stress is important to consider in bolt design because it can significantly increase the stress and potential failure of the bolt. Failure of a bolt due to double shear stress can lead to catastrophic consequences, making it crucial for engineers to accurately calculate and account for this stress in their designs.

## 5. How can double shear stress on a bolt be minimized?

Double shear stress on a bolt can be minimized by using bolts with larger cross-sectional areas, increasing the number of bolts used, or using alternative connection methods such as welding. Properly designing the connection geometry and considering the direction and magnitude of the applied forces can also help minimize double shear stress on a bolt.

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