force required to yield/deform a poorly-supported flat-washer 7

[electricpete]

I'm trying to figure out the force required to deform a washer which is captured between a bolt-head of known dimensions and a hole of known dimensions.

If all the dimensions shown are known and washer material properties are known, can we estimate the force required to deform the washer into a cupped shape be estimated (by means other than FE) ?

I'm willing to use simplifying approximations:
* bolthead acts circular rather than hex
* just find the force to get to yield, don't complictate it with plastic deformation
* others?

Any suggestions/references for how to approach this problem?



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(2B)+(2B)' ?

 

[electricpete]

Whoops, I just noticed that would make the insulated washer smaller than the hole on which it sits. Let me double check that...

=====================================
(2B)+(2B)' ?

 

[desertfox]

Hi electricpete
I was thinking about your problem and how you just wanted a fairly simple way of calculating the force that caused the distortion.
Firstly trying to calculate forces after the material as yielded is a field on its own, so calculate the load on the steel washer by using the washer material yield stress (when you get it) and multiply it by the area under the bolt head, this will give you a ball park minimum force that the washer must have experienced at the point of yielding.

 

[electricpete]

Here is correction to previous info:

Here is some info:
* Socket Head Cap Screw: 5/8-11 (5/8 diameter, 11 tpi). Head diameter is 0.925”
* Insulating washer - Thickness 0.252”, Outer Diameter 1.472”, Original material (as in the photos) is unknown. Replacement material used during recent refurbishment is Electrical Grade Fiberglass (GP03) with plain backing, smooth finish, furnished originally as square sheet 1/4" thickness. McMaster-Carr Part Number 8549K66. Tensile Strength: Excellent [sic].
* hole in the lower bracket onto which the insulated washer seats has diameter of [COLOR=red yellow] 1.63” 1.0"[/color]
* Steel flatwasher dimensions (as stated before): 21/32” (ID), 1+5/16” (OD), 3/32” (thickness). Washer material - still working on.

Hi electricpete
I was thinking about your problem and how you just wanted a fairly simple way of calculating the force that caused the distortion.
Firstly trying to calculate forces after the material as yielded is a field on its own, so calculate the load on the steel washer by using the washer material yield stress (when you get it) and multiply it by the area under the bolt head, this will give you a ball park minimum force that the washer must have experienced at the point of yielding.

Let's give it a try:

washer ID = 21/32 = 0.65625"
bolt head OD = 0.925”
Area = pi*(0.925^2 - 0.65625^2)/4 = 0.333764 in^2
F = Area * Sy = 0.333764 in^2* Sy
I am thinking grade 2 or grade 5.... What is Sy?

With some googling, I am unable to find yield strength of SAE flatwasher. There's plenty of info on bolts, but not on flatwashers (which makes sense because we normally care about hardness, not yield strength for flatwashers). Does an SAE grade 5 flatwasher have the same yield strength as an SAE grade 5 bolt?


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(2B)+(2B)' ?

 

[Screwman1]

Unless the washer was specified as a hardened washer (which is not common), then it will be unhardened, in an as rolled condition. The yield will probably be in the range of 35 - 65ksi. This is a big range, but this type of washer is not very well controlled.

 

[electricpete]

The hardening applies to the surface characteristics, as I understand.

Is there any significant difference in yield strength of a hardened washer vs unhardened washer?

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(2B)+(2B)' ?

 

[desertfox]

Hi electric piete

Yes I think there will be a difference in strength of a hardened washer versus unhardenedband.
Have a look at this article:-

http://www.ajaxfast.com.au/downloads/Technical noteFunction of a Washer.pdf