How do you compute the force to deform a flat plate?

  • #1
I am a trying to work out a design issue that you may be able to help with or maybe point me in the right direction.

I am designing a 17-4 stainless steel plate to be used to support a trust bearing. The dimensions are 3.333 inches in diameter with a 1.20 inch center bore and .097 inches thick.

My question is how much load can this plate take before it begins to deform?

If you could point me in the right direction in the form of an equation or a table that can be used I would greatly appreciate it.

Thanks for any help....
 

Answers and Replies

  • #2
cronxeh
Gold Member
974
10
do you have a sketch of how you want it connected
 
  • #3
minger
Science Advisor
1,496
2
Well...it seems theoreticaly, ANY force will deform it. How much deformation are you willing to take? I actually had a VERY similar problem not too long ago that I posted here. I was trying to calculate stresses on a flat plate. It seems your best bet is to find someone on here, someone you know, or (it would be a good investment) go out and buy the Roark Formulas for Stress and Strain. That book is very helpful. At the moment another engineer has our department's copy, so I can't look it up for you.

However, the book has many many situations of loadings, for all types of supportings. You look up your loading and support situation, then it lists forumlas for stress, strain and max deflections as a function of load and aspect ratio (ratio of length of plate to width).

p.s. It also has formulas for things like circular plates and the like, so it is really useful for anyone who calculates things like that.
 
  • #4
FredGarvin
Science Advisor
5,067
9
Questions, questions, questions...

Is there any reason why you are using a precipitation hardened steel? What condition heat treat will the final part be in?

What kind of thrust loads are you expecting to see at the bearing? Also, what kind of mounting are you going to have. That will dictate the boundary conditions of the analysis.
 
  • #5
attachment

I have attached a zipped file of the design.
 

Attachments

  • bearingPlate.ZIP
    12.6 KB · Views: 150
  • #6
Thanks for the suggestion.
I will get the book and do some homework.
 
  • #7
The choice of stainless 17-4 seal was a suggestion by a machinist I know.
I was not planning on heat treating the final part (once machined).

The bearing plate will be mounted inside a dodge caravan steering knuckle. The plate will provide a thrust surface to sandwich a 3-piece roller bearing assembly and the drive shaft’s machined surface.

The reason for such a bizarre arrangement is that I am converting my front wheel drive to a rear wheel drive and the front wheels have about .030 play since the drive shafts are no longer connected to the transaxle.

The basic idea is to install a thrust bearing so the detached/standalone driveshaft has something to press against when the hub nut is tightened down.

As far as thrust loads: the hub nut needs to be torqued to 180 ft lbs and the lateral forces on the wheel during a turn should not exceed half of the cars weight (about 1500 lbs).
 
  • #8
FredGarvin
Science Advisor
5,067
9
You should just go with 304 or 316 then, don't waste your money on 17-4 if you're not going to heat treat it.

Does the thrust bearing sit inside the Ø2.833 ID?
 
  • #9
Reply: FredGarvin

Yes, the bearing assembly will be housed inside the steering knuckle;
in-between the machined surface of the driveshaft and the bearing plate I am trying to design (see the attached images of the shafts and the steering knuckle).

Thanks.....
 

Attachments

  • 003.jpg
    003.jpg
    8 KB · Views: 370
  • 011.jpg
    011.jpg
    15.9 KB · Views: 363
  • #10
FredGarvin
Science Advisor
5,067
9
I am trying to get an idea of how the plate will be loaded and constrained. Obviously, if the entire plate is sandwiched between the two objects, it won't deform. The way I am seeing it now is that the very outer edge of the plate is going to be constrained all the way around the perimeter and the face of the thrust bearing is what is applying the load. Is this correct (I'm not really a car guy so I'm not familiar with the mechanism)?
 
  • #11
Q_Goest
Science Advisor
Homework Helper
Gold Member
3,001
39
Tony,
If you sketched a cross sectional view of the bearing, thrust plate, threaded bolt, nut, etc... it would be a whole lot easier to evaluate this.

One other question, I assume there's a nut as shown on the first pic above that bears on the thrust plate and bearing, and that's the one you're torquing to 180 ft lb. What thread size is that? (diameter and pitch)
 
  • #12
Reply: FredGarvin

Fred,

The plate has an inner cavity 2.883 in diameter and 0.040 in depth. This area is on the bottom of the plate which forms a pocket for the rotating part of the hub assembly – which has the 0.030 lateral play. Due to this cavity I have concerns about deformation of the plate under heavy loads (like those experienced when turning a sharp corner).

However, you are correct when you say the outer edge of the plate is constrained all the way around the perimeter and the face of the thrust bearing is applying the load.

I have included more pictures and a drawing of the whole assembly (minus some parts for clarity).

Photo 1: Steering Knuckle
The knuckle mounts on the ball joint stud that is attached to the lower control arm. Also attaches the tie-rod end which is part of the steering linkage. The knuckle pivots on the ball joint as the steering linkage asserts input from the steering wheel.

Photo 2: Drive Shaft inserted indo Steering Knuckle
The Hub Nut is torqued to 180 ft-lbs. Note the machined surface on the drive shaft – this will be used as a surface for the roller thrust bearing.

Photo 3: Hub Assembly press fit into an adapter plate which holds the brake caliper
Note the outer surface of the hub assembly does not rotate. However, the inner surface does. In fact this is the surface that has the 0.030 lateral play that I am trying to constrain from moving back-and-forth. Also note the bearing plate will sit on top of this assembly resting on the non-rotating outer surface of the hub.

see next post for more photos and drawing

Regards,
Tony….
 

Attachments

  • 1.jpg
    1.jpg
    17.6 KB · Views: 332
  • 2.jpg
    2.jpg
    19.2 KB · Views: 358
  • 3.jpg
    3.jpg
    20.6 KB · Views: 343
  • #13
cont. previous post

Continuation,

Photo 4: Steering Knuckle with out drive shaft mounted onto brake caliper adapter plate
Note the 0.325 space between the hub assembly and the bottom of the lubrication seal. This is the space the roller bearing assembly and bearing plate will need to fit into.

Photo 5: Drive Shaft inserted into Steering Knuckle – side view
Just for perspective

Drawing: The whole assembly

Regards,
Tony….
 

Attachments

  • 4.jpg
    4.jpg
    15.9 KB · Views: 347
  • 5.jpg
    5.jpg
    15.8 KB · Views: 322
  • HubAssembly.jpg
    HubAssembly.jpg
    7.2 KB · Views: 362
  • #14
Q_Goest,

I have enclosed a cross sectional view in the prior posting (good idea, thanks).

You are correct in the assumption that the hub nut is what bears down on the thrust plate and bearing. I don’t have a thread gauge but I measured the thread diameter to be 0.845 inches and counted 17 threads per inch.

Regards,
Tony….
 
  • #15
Q_Goest
Science Advisor
Homework Helper
Gold Member
3,001
39
When the nut is tightened to 180 ft lb of torque, you'll get roughly 15,000 pounds of force pulling on the driveshaft. From the pictures, it would look as if that force is compressing this stainless washer you're putting in. It looks like the drive shaft is pressing against a thrust bearing which is pressing against this washer, which is pressing against the outer hub assembly. Is that correct?

Is the thrust bearing fully contained and have it's own races? I'd assume it does but it almost looks like the balls of this bearing are riding on this washer as you've drawn it.

The stresses seem way too high in this washer if the above is true, but I can't be sure without knowing what the thrust bearing OD and ID is. If the OD of the thrust bearing (where it bears on the washer) is at least 2.880" then it might be alright (that diameter almost eliminates any bending in the plate*) but you'll need the 17-4ph steel and preferably a hardened version of it. But then if the race is not properly supported, it will be damaged anyway, so that's probably not a good design. This plate seems to be supporting the race on the bearing, so it needs to be relatively thick. Can you make the plate thicker? (say at least 0.250", preferably more) You might be able to machine down the drive shaft to add some thickness for the plate. It looks like the drive shaft has quite a bit of added meat there.

*I used Roark's case 1, table 11.2 for this.
 
  • #16
Q_Goest,

You’re right on the mark with your assessment of the drive shaft pressing against the thrust bearing which is pressing against the washer (bearing plate), which is pressing against the outer hub assembly.

With regard to the bearing: it is a three part assembly consisting of two washers, matched to the bearing, and a roller thrust bearing with race (Timken p/n NTA-1828 and TRA-1828 – 1.750 OD, 1.125Bore, and overall thickness of 0.1381 Width). The specs on the bearing assembly are 13,700 lbs static load and 3,040 lbs dynamic. The limiting speed is 9,800 RPM. I plan on using BF-Goodrich 245/50 WR 16 tires with a diameter of 25.6 inches. Accordingly, the bearing should only be rotating 787.8 RPM’s @ 60 MPG’s and twice that at 120 (not that I would actually drive at that speed). My sketch neglected this level of detail.

You make a good point about removing some of the drive shaft material to gain clearance and beefing up the bearing plate to 0.250 or more.

I have the Roark’s on order and it should arrive late next week.

Regards,
Tony….
 
  • #17
Q_Goest
Science Advisor
Homework Helper
Gold Member
3,001
39
Tony, if the thrust bearing is only rated for 3040 lb dynamic, then something seems way out of place here. The bearing can't take anything like 15000 lb. Per my calculations, you'll hardly make it out your driveway before the bearing starts screaming like a cop siren... Here's Timken's standard equation:
http://www.timken.com/products/bearings/fundamen/calculate.asp#Bearing Life Equation
and here's a calculator similar to one I have:
http://www.ahrinternational.com/bearing_life_calculator.htm
Both my calculator and this one on the net give 0.1 hours of life for a bearing rated at 3040 lb dynamic, RPM is 788 and a load of 15,000 lb. Note that when using the calculator on the internet, you'll want to select "Cylindrical Roller Bearing" because the bearing elements are rollers as opposed to balls.

You don't want any more than about 1000 lb of load on that size bearing, preferably less. That's a torque of only 12 ft lb, which seems very light to me. If you use that bearing and 12 ft lb of torque, you can get away with a 0.150" thick plate.

The best answer is to get the biggest one that will fit the space, then use the calculator on the internet to find out what load is good for it, then determine torque. I can't find a bolt torque calculator on the net, but I'm sure there must be one out there.

If the bearing OD is at least 2.88", then you're probably ok with the .250" thick plate, but you may even need to go thicker if the bearing OD is smaller. For the bearing you have, if you were to actually torque it to 180 ft lb, you'll need the plate to be roughly .500" thick. And don't forget to look at the material on the drive shaft if you cut some of that away.
 
Last edited:
  • #18
Q_Goest,

Thanks for pointing out the clamping force problem. I was totally unaware; I thought that the torque on the hub nut was the total force clamping down on the bearing assembly.

So my next step was to compute this force myself.
I searched many a web sites and finally found a Clamp Load equation.

Using my values:
T = 180 (ft-lb)
D = 0.853 (drive shaft dia)
K = .3 (non-plated fastener)

W = (12 * 180) / (.3 * .853)
W = 8,440.8 lbs

Our results are a little more than a bit off. How did you compute the clamping force of 15,000 lbs?


I found the following at:
http://www.eng-tips.com/viewthread.cfm?qid=118494&page=1

Handbooks typically state:

T = (K*D*W)/12

T=torque
D= dia.
W = Clamp load
K=0.09 to 0.3 depending on factors affecting friction.
0.3 for non-plated fastener, 0.12 if anti-seize compound is used.

Regards,
Tony…..
 
  • #19
Q_Goest
Science Advisor
Homework Helper
Gold Member
3,001
39
I used a K of 0.17

The typical fastener will be between 0.15 and 0.20, so for lack of a better value, I generally use 0.17 as an in between point since it is so highly variable depending on the use of lubricant and other factors. It's inexact, but it's close enough for most work like this.
 

Related Threads on How do you compute the force to deform a flat plate?

Replies
1
Views
5K
  • Last Post
Replies
10
Views
13K
Replies
13
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
2
Views
1K
Replies
1
Views
25K
Replies
14
Views
23K
Replies
2
Views
5K
  • Last Post
2
Replies
27
Views
5K
Top