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force required to yield/deform a poorly-supported flat-washerHelpful Member!(7) 

electricpete (Electrical) (OP)
21 Feb 13 23:23
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)' ?

Helpful Member!  GregLocock (Automotive)
21 Feb 13 23:50
I don't think it'd be accurate but your effective length L is (Dhole-Dbolt)/2 the width of the beam =b=pi*Dbolt,

M/I=sigma/y

y=t/2

M=F*L

I=1/12*bt^3 where t is the thickness of the washer

y=t/2

sigma is the yield stress.

F is the force to initiate yield.

Cheers

Greg Locock


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Helpful Member!  israelkk (Aerospace)
22 Feb 13 1:16
Roark’s Formulas for Stress and Strain, ver 7, table 11.2, case 1 page 459
electricpete (Electrical) (OP)
22 Feb 13 1:59
Thanks for those suggestions.

Roark looks promising. I have three questions about it:
1 - Q represents shear? (I'm used to the leter V for that.
2 - I know how to convert moment and shear into bending stress and shear stress for beams, but I'm a little fuzzy for plates. Is that a straightforward conversion?
3 - My geometry has two opposing annular loads:
one at r = Dbolthead / 2
one at r = Dhole/2
...which are both between the ID/2 and OD/2

But Roark's case 1 only allows one annular force at a radius between ID/2 and OD/2 (it also allows a force at ID/2 and a force at OD/2). If I just pretended that my washer stopped at the point of application of load, that would introduce an error, right?

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electricpete (Electrical) (OP)
22 Feb 13 2:14

Quote:

2 - I know how to convert moment and shear into bending stress and shear stress for beams, but I'm a little fuzzy for plates. Is that a straightforward conversion?
Never mind -
I see a note on the top about bending stress = 6*M/t^2
I think I can figure out shear stress as a force / area = 2*pi*r*Q(r) / [2*pi*r*t] = Q/t
(I'm assuming Q is in units of force per length of circumference)

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(2B)+(2B)' ?

Helpful Member!(2)  prex (Structural)
22 Feb 13 2:57
You can use the formulae for an annular plate supported at the outer edge with a line load (see an example here) by superposing two cases: one with a line load at r = Dbolthead / 2, the other one with the same resultant and opposed direction at r = Dhole/2. The support reaction will be zero so the outer edge will in fact be free.
However I don't think, for how I understand your intent, that taking the highest stress equal to yield will be correctly representative of washer's strength. You could first take 1.5Y as the limit; this would allow for developing a full plastic hinge. Also inspecting the resulting distribution of bending moments (remember that you'll have to combine the radial and circumferential bending moments, see the curve SI, stress intensity, in the link above) will give more insight: a relatively flat distribution of SI, as should be expected, will confirm that taking the highest stress intensity with a limit of 1.5Y is a good guess.

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

electricpete (Electrical) (OP)
22 Feb 13 11:37
Superposition...great idea.

The purpose is understanding a washer which was found bent (not broken) after disassembly. I want to estimate (or at least bound) how much force it had seen

Is Sy or 1.5 Sy better for this purpose

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electricpete (Electrical) (OP)
23 Feb 13 14:45

Quote:

Is Sy or 1.5 Sy better for this purpose
Never mind - I see you mentioned we'll examine results to understand that better.

I'm trying to build a spreadhseet to implement the Roark case 1A (simply supported at outer ring).

Roark's formula uses a series of constants like F1, F2 etc.
From slide 1 attached, we see that F1 should be unitless to satisfy that equation.
From slide 2 attached, F1 does not seem to be dimensionally self-consistent.

Quote (Roark)

F1 = (1+nu*a)/2 * (b/r) * LN(r/b) + (1-nu)/4 * (r/b - b/r)
where nu is unitless poisson ratio and r, a, b all have units of length.

....the first factor (1+nu*a) seems non-sensical. How can we add unitless 1 to nu*a which has units of length.

Seems like a typo. However I have no idea how to correct the typo.

I tried removing the a from this factor and final results still do not seem correct. (I may have another error, but can't find it yet)

Does anyone know the correct expression for F1?

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(2B)+(2B)' ?

electricpete (Electrical) (OP)
23 Feb 13 14:47

Quote (electricpete)

attached

HERE is the file that was supposed to be attached.

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desertfox (Mechanical)
23 Feb 13 17:03
Hi electricpete

Try this ebsite for bending formula of circular plates ith holes

http://www.roymech.co.uk/Useful_Tables/Mechanics/P...
electricpete (Electrical) (OP)
24 Feb 13 0:34
Thanks DesertFox. That has loadings at the edges of the washer, but I need loading at an intermediate point.

Here are some more details about the configuration.
(powerpoint attached).

There is a mechanically-soft electrically-insulating washer between the steel washer and the bracket that I didn't mention (I was trying to look at the simplest form of the question first). I assume it wouldn’t make a big different in this analysis, but I’m open to comment, including which way is the direction of error due to neglecting this insulating piece. I’m thinking the effect of the soft washer is to make the bracket hole act slightly smaller which actually helps the washer resist bending. Therefore I’m thinking the effect of neglecting this piece would cause us calculate force (to bend washer) lower than actual required force to bend washer?

Slide 1 shows the parts:
* Steel Flat washer – identified on drawing as “SAE 5/8 NARROW”. Original material is unknown to me.. can we tell material from this description? Washer is not available for inspection (pictures are old). Dimensions of washer are 21/32” (ID), 1+5/16” (OD), 3/32” (thickness).
* Allen head bolt (not hex head as I said before).
* Insulating sleeves. They don’t play any role in the problem, but they give an idea of how big the hole in the bracket must be since the sleeve fits inside the hole of the bracket (between bolt and bracket hole). I don’t have exact diameter of the hole in the bracket yet, but I’m definitely going to be able to get that measurement shortly.

The remainder of the slides show the same parts from a sister unit, where the washer deformation is much more pronounced. On slides 3 & 4, notice the steep-sided pocket in the washer face formed by the allen head, where the bottom of the pocket almost looks like a right angle to the steep walls of the pocket. On slide 5 (opposite side of washer), the deformation is more gradual. It’s not just a bend but as if the material flowed to allow the bolt head to sink in (this location is <120F.... the piece was installed approx 20 years). What do we make of this appearance...does it tell us anything?

Having a rough idea of dimensions, is thin-plate approximation going to do reasonably well here?

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electricpete (Electrical) (OP)
24 Feb 13 13:39
Another question: based on appearance of the washer (attached to previous post) would you guess that the bending occurred during initial bolt tightening (this bolt goes into a tapped hole in another piece, so the bolt turns during tightening) or after initial installation when additional axial load was applied between these pieces. I do see some circumferential marks on slide 3 which suggested they occurred during tightening, but that overall appearance of steep step in one side and gradual step on the other seems to me more reminiscent of creep. What do you think?

(I don't know what torque was used when originally assembled, and the axial force seen by the pieces during operation is also somewhat unknown).

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Helpful Member!  Tmoose (Mechanical)
24 Feb 13 17:39
The way the SHCS appears to have sunk embedded in the "steel" washer suggest the under head contact or bearing stress is WAY over the washer material yield strength >>OR<< the head contact is reduced by a washer with excessively large ID . Holokrome recommends " The bearing stress under the head at these preloads
will be approximately 80,000 psi, so indentation should not occur when the parts clamped are of steel or cast iron with a hardness equal to or
in excess of Rockwell B 85. With softer materials, washers may be required under the heads of the screws to avoid indentation."
http://www.holo-krome.com/pdf/Consolidated_Tech_Ma...

A lowly grade 2 washer ought to have hardness 70>100 hB.
http://www.selfdrillers.com/brochures/TDS005-Grade...

Do new OEM washers have the curious rounded/eroded/pitted appearance?

**** Is there an OEM spec available for bolt installation? I know you said there is no record of how they actually were installed. I can picture a well meaning novice millwright continuing to crank down on the bolt because it did not "feel" tight yet.
My hunch is Compressing any kind of remotely resilient soft insulating washer, or crappy low yield strength metallic washer would limit installation torque to some value way below the nominal recommendation for a SHCS.
desertfox (Mechanical)
24 Feb 13 20:22
Hi electricpete

well the insulating washer does complicate the situation somewhat in so much as the bolt is being tightened it will give and reduce the preload and most plastics will flow under load.
with the amount of deformation seen I'm not sure how accurate the bending calculation will be.
I think because the joint has insulating washers there must be a preload figure given for the bolt particularly if this is a bought out piece of kit, maybe you can contact the supplier and see if there is a set preload.
Screwman1 (Mechanical)
25 Feb 13 11:33
You should not be using a SHC head design in this joint. the bearing surface area of the head is designed for a hardened steel washer are die set and you have neither of these. The washer shows gross embedment from tightening and the displaced material from that at least started your deformation. You have obviously yielded the washer material during the assembly process from the compressive force created by the undersized fastener head. Just calculate your compressive stress at that bearing surface and you'll see what is happening.
If you switch to a larger bearing surface head design (hex flange) your problem should go away.
electricpete (Electrical) (OP)
25 Feb 13 15:27
Thanks, I appreciate the replies and insights.

I should clarify that we did not experience an undesirable loosening or loss of preload (at least that's not the reason I'm interested in these washers).

I'm going to provide some more info on the overall problem I'm working on (not because I want you to solve my overall problem.... I can't possibly give you all the background info), but to help understand the specific question I'm asking.

We experienced a lower bearing failure on a vertical motor. Inspection of the failed bearing suggests upthrust load when we only expect radial load in this application. On this motor, the upper bearing takes downthrust and the upper bearing takes any upthrust applied external to the motor. Upthrust may originate externally to the motor or a result of endplay/thermal considerations.

Slide 2 shows the lower bearing area. The hardware we've been talking about are labeled in the solid red boxes ("Bolts" and "Steel & Insulating Washers"). These parts are inverted from the orientation of my original attachment (the bolt head is actually down and the applied force on the shaft/inner cap would be up, adding tension to the bolt). The washers I'm showing you are from sister motors to the one that experienced the bearing failure.

I'm sure you'll have questions/suggestions based on what I posted to help me solve my overall problem, but I'm not asking for that here.

The specific question I'm asking here and now: based on the appearance of the washers I posted above, can we make an educated guess as to whether the appearance of those washers suggests that particular deformation occurred during original tightening or from external/thermal loading during subsequent operation.

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(2B)+(2B)' ?

electricpete (Electrical) (OP)
25 Feb 13 15:31
Correction:

Quote (electricpete corrected)

. On this motor, the upper bearing takes downthrust and the upper lower bearing takes any upthrust applied external to the motor

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IRstuff (Aerospace)
25 Feb 13 15:38
What's the material of the "insulating washer?" Was there something unusual about the hexheads? I'm trying to scale stuff in Visio, and it looks like the head contact area is along its circumference from the first picture, so there ought not be any force on the washer strong enough to deform it. Do you have the hexhead from the failed washer?

TTFN
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desertfox (Mechanical)
25 Feb 13 20:35
Hi electricpete
I think the washer distortion was caused during tightening, my reason is that if the distortion was caused during operation then the joint would be separating and the original bolt preload would have been insufficient but you clearly stated that
the joint didn't work loose.
electricpete (Electrical) (OP)
26 Feb 13 22:37
Thanks.

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 1.63”
* Steel flatwasher dimensions (as stated before): 21/32” (ID), 1+5/16” (OD), 3/32” (thickness). Washer material - still working on.

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(2B)+(2B)' ?

electricpete (Electrical) (OP)
26 Feb 13 22:38
Whoops, I just noticed that would make the insulated washer smaller than the hole on which it sits. Let me double check that...

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(2B)+(2B)' ?

Helpful Member!  desertfox (Mechanical)
27 Feb 13 16:08
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 (Electrical) (OP)
27 Feb 13 17:31
Here is correction to previous info:

Quote (electricpete, corrrected)


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 1.63” 1.0"
* Steel flatwasher dimensions (as stated before): 21/32” (ID), 1+5/16” (OD), 3/32” (thickness). Washer material - still working on.

Quote:

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)' ?

Helpful Member!  Screwman1 (Mechanical)
28 Feb 13 11:44
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 (Electrical) (OP)
20 May 13 17:21
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 (Mechanical)
20 May 13 18:07
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%20n...

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