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Help Calculating Shear Stress 2

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RohitGogna

Mechanical
Nov 9, 2015
17
Hello,

I am trying to calculate the shear stress on the bolt in the attached picture. This is just so I can confirm on paper that a 0.25" cap screw bolt is strong enough to withstand the applied forces at the pivot. Can someone please help guide me on how to calculate the shear stress at this point? The reason as to why I am trying to calculate the shear stress is because this bolt at this location has had several repetitive failures in the past where it has been sheared.

I believe I have all the variables I require to solve this problem but am not sure exactly how to approach it.

I originally believed I should have used the equation:
max tau = (16*F)/(pi*d^3) but soon learned that this equation is used for rotating shafts. In the attached image the rotating arm is at its lower limit and is unable to rotate further.

Any help would be greatly appreciated.
 
 http://files.engineering.com/getfile.aspx?folder=14a2494a-76a2-4881-8a64-f180b36d81f7&file=Calcuate_Shear_Stress.jpg
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Could you post pictures of the failed bolt fracture surfaces?

Is the bolt in single shear or double shear, and is the body/shank full diameter ( no threads ) in the shear planes?

What is the bolt made of?
 
The bolt is in single shear.
Currently the threads are in the plane of the shear, however in the near future we will be custom fitting the bolt to avoid having any threads in the shear plane.
The bolt is made of Grade 8 Steel.
The thickness of the rotating arm shown at its lower limit in the attached picture above is 0.375"

I do not have any pictures of the bolt at the moment however can post some tomorrow if required.

I would like to prove on paper though that the 0.25" bolt is strong enough to withstand the shear force.
 
What does that cylinder do? what is the expected rotation at the pivot? Is it a free falling 3 kg object and does it bounce off after impact?
 
Typically the cylinder is used to rotate the arm up and down to allow some cores to roll onto it for the next stage of manufacturing, however there are a few procedures where the extra manufacturing steps are not required. During these instances the pivot arms are left down and the cores drop onto the arms and skip to the final stage of manufacturing via a conveyor belt (located directly underneath the pivot arm).

Yes it is a free falling object.

The bounce produced from the object after it impacts is very tiny. I would estimate the bounce to be between 10-15 millimeters.

The arm is expected to rotate 50 degrees in the typical scenario, however in the instance I am concerned about, the arm is fixed in the lower limit and the cores are dropping directly onto the arm.
 
RohitGogna:
You really have to give us much more info. on what this thing is, how it works and what it does; including enough dimensions, thicknesses, part sizes, and few more views of the drawing, etc. etc. You don’t have to give away any national secrets, but we really can’t see it from here. We can imagine about how it works, but we really can’t analyze it without much more info.
 
Hi,

You can simplified the model in single line model to eliminate all other confusion.
Then use kinematic or what you have learn in stress analysis.
It's quite simple actually, but my method may not give exact value. you can try 3D analysis.
See my attachment (i did not add the mass of the equipment in the line drawing).
Your screw is almost like M6. That is quite small actually, compared to other components.
You must know that due to dynamic loading also, it may cause the failure even though you are using high grade steel considering the size of your bolt.
 
 http://files.engineering.com/getfile.aspx?folder=bd908b19-fc90-49a9-90d2-29a4e2bd2ac2&file=Engr-Tips-1.jpg
I have attached some images of diferent views with some more dimensions. The material of the arm is 1/2 H.R. Steel.

Pivot Arm Dimensions (force acting on this component)

Section View

Picture of Sheared Capscrew

Picture Showing Diagram for Calculations

Attempt to Calculate Shear Stress at 1/4" grade 8 bolt
I attempted to calculate it the Conantar had suggested using the single line model and determined that the impact force is 334 Newtons using conservation of energy.

And the two Reaction forces are 601.77 Newtons and 268 Newtons from left to right as shown on Conantars' image.
From this I was able to determine the largest Shear force acting on arm would be 334 Newtons.
334 Newtons = V

For the Moment of Inertia I took the cross sectional area at the point where the bolt I am concerned about keeps breaking
[(0.375in)*(4.3in)^3]/12 = 2.48in^4 = 1.034x10^-6 m^4 = I

(not sure what the technical term for this value is if someone could let me know would be greatly appreciated)
Q = (2.15.in/2)*[(0.375in)*(2.15in)] = 0.866in^3 = 1.426x10^-5 m^3

The thickness of the plate is 0.375in = 0.009525 m = b

From the equation: shear stress = (V*Q)/(I*b) = [(334 N)*(1.426x10^-5 m^3)]/[(1.034x10^-6 m^4)*(0.009525 m)] = 480803.55 N/m^2 = 69.73 PSI

Is the method I used to calculate this correct? I feel as though I have mised something crtical.
 
Hi RG,

Looks like a 1/4-20 fastener. About the weakest fastener for it's size ( deep threads ).

So the steel arm rotates on the non-rotating brass/bronze bushing?
How tight do you tighten the bolt/screw?
 
RohitGogna:
At the very least, that should be a shoulder screw or shoulder bolt, not a cap screw with the threads in the shear plain. Have all the screws failed the same way and at the same length location? It actually looks like the failure location is some distance into the .5" supporting plate. Is that correct?
 
Tmoose,
Yes it is a 1/4"-20 fastener.
The arm rotates around the bushing which is fixed.
We try to tighten up the bolt as much possible without creating issues with the rotation. There is no specific torque we tighten the bolt to.

dhengr,
Yes all of the bolts have broken in the same manner, more or less in the same location. We will be changing the bolt so that the threads are no longer in the shear plane to avoid stress concentrations. All of them are broken approx 0.5" from the supporting plane.

If we are to change the bolt do we need to use a shoulder bolt? or can we just use a 1/4" bolt that has a threaded and unthreaded section and thread the bolt to the length we require to achieve the perfect unthreaded length (0.625"). This way no threads are in the shear plane still. I just wanted to avoid having to make the hole bigger as there isn’t to much excess material around the pivot hole to work with and shoulder bolts typically have a shoulder 1 size bigger than the threaded section.

Also is the method I used to calculate the shear stress above correct for the situation?

Also thank you to everyone for all there help so far it is greatly appreciated.
 
"All of them are broken approx 0.5" from the supporting plane. "
Is that inside the bushing?

First, I'd modify the bushing shoulder length and spacer thickness so the bolt/screw could be tightened fully and still have 0.010" or so clearance to the arm.
A loose fastener would be subjected to bending.
A "sufficiently tight" fastener would make the larger diameter bushing resist the bending, leaving essentially pure tension in the screw/bolt.
Might help to make a shallow spotface in the arm .002" larger than the diameter of the bushing.

After that, On principal I'd also modify a bolt/screw so no threads were not in the "shear plane."
 
Bending failure. Not shear. It's a fatigue failure starting at the thread root.

Not sure why you say the bolt is failing .5 inch from the supporting plane unless your section diagram is wrong.

The brass is too soft to prevent bending. Its elastic modulus is too low, so the bolt is subject to bending. Using a through bolt with a close fitting hole in the support may help as it won't have a stress riser for the fatigue crack to start.
 
While reviewing the cross section, I believe that torsional stress is the problem. With a new set of your equipment, pay close attention to the nut and bolt head and determine if either or both are rotating while the arm is rotating during the impact of the 3 kg or while the piston is activated. If the bolt rotates, even a little bit while the bolt head is stationary then torsion would part of or the problem.
 
Hi 3DDave,

"The brass is too soft to prevent bending. Its elastic modulus is too low, so the bolt is subject to bending."

I believe the OD of the bronze bushing is 0.5".
The bolt OD is 0.25"

I believe the Area moment of inertia of the bronze bushing is almost 16 times greater than the bolt. ( 2X bolt OD ^4 )

The modulus of elasticity of steel is only about 2X that of bronze.

I think if the bronze bushing has a flat ( or slightly concave ) face and is clamped hard (enough) against the steel arm, the bronze will be about 8 times stiffer in bending than the bolt was, so it will bear the brunt of the load.

Whether the bronze face will yield under impact is another question.

reards,

Dan T
 
I would like to confirm all the dimensions Tmoose mentioned in the above post are correct. Yes the bushing also has a flat surface on the end with rounded corners

Also I'm not sure if this is relevant but from all the failures that have occurred so far, there has been no damage to the brass bushing.
 
For in-plane motion, the bushing can slide side-to-side, allowing the bolt to bend. The moment of inertia of the bushing makes no difference. The resistance to that bending is a compression under the flange of the bolt embedding into the bushing until the bushing moves far enough to contact the bolt. This is similar to the failure mode seen on Junkers testing of bolted joints.

You are right that in compression or pure bending the bulk of the bushing is much stiffer than the bolt, if the bushing was uniformly loaded, but my suspicion is that it's the in-plane motion that's the problem.

To add to the description of the problem: because the cylinder is free to rotate, the mechanism is more like an over-center linkage where the loads on the pivot can be amplified. Because it's an impact load the situation is even more complicated.
 
I don't think bending is the only problem as the brass bushing and spacer wear down . Shaving from the brass bushing and spacer will also occur and will cause torsion on the bolt and ultimately with some bending will shear at the designated area shown in the section view.
 
Hi 3DDave,

OP's description of assembly procedure was "We try to tighten up the bolt as much possible without creating issues with the rotation."

That description strongly suggested to me that, at least sometimes, the bushing was too short ( or the spacer too thick) to define the endplay/clearance, and the bolt was left loose to allow the arm to pivot, and like you said, leaving the bolt on it's own to resist the applied force's induced bending and shear.

My April 1 approach would be to eliminate any in plane motion AND tension the bolt fully.
First defence is a good bushing/plate interface clamped together with 2800 lbs force.
Adding a close fitting shallow counterbore in the plate (as recommended in my April 1 post) would provide a bunch of insurance that the bushing could not slide, even if the impact did occasionally overcome the friction from bolt/screw clamping.
In order to allow proper tensioning of the bolt, the dimensions of the bronze bushing and spacer need to be controlled to provide axial clearance with the arm. There may be some additional machining costs

As far as the Junker's test, the few I've looked at included a defined amount of sliding motion.
With very few exceptions, if a bolted joint I had designed slid or separated in service, I failed.

regards,

Dan T

 
Thank you very much for all your help everybody. I really do appreciate all the time and effort everyone has spent helping me.

I believe I know what I need to do from here to resolve this situation.

Best Regards,
RohitGogna
 
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