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My Dad has passed me an example he is having some difficulty with, all help is greatly appreciated.
[b]The completed example (straight forward):[/b]
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[b]The Problem:[/b]
With reference to the angled joint in question; If the tensile stress and shear stress are calculated as shown and the ultimate tensile stress (UTS) of the rivet material is known (See table below), how can the factor of safety be calculated?
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Thanks again.
I think the formula is MxT/F=U
No, it's W+TF=111
conti vert pro?
You get a reserve factor for both cases - in tension it is tensile stress / UTS and in shear it will be the shear stress / the shear allowable. If you don't know what that is, a good stab might be 0.5 x UTS..... but this is a very broad approximation based on the tensile yield/shear yield relationship and will vary from material to material. Depending on the application there may be a combined tension and shear limit, but you would have to look at the relevant BS/Eurocode to see what the failue curve looks like.
Actually, it's just about converting the force into its elements across and along the rivet Shirley?
That part's already been done in the example calcualtion - the OP wants to know how to use the stresses to determine a reserve factor.
The answer is to drill it out and put the rivet in at a more sensible angle. Simple
C
You could do it with von mises stresses (google von mises stress calcs) which will give you an averaged stress taking into account shear and tensile stress. Then you need the yield stress of the material, (as von mises predicts failure to material yield not UTS).
That part's already been done in the example calculation - the OP wants to know how to use the stresses to determine a reserve factor.
Exactly. Thanks everyone so far*
* [i]almost[/i] everyone 😉
Blimey how you forget things....20 years since I did any structural stuff but I seem to think Mohr's circles is the answer? Something about tensile and shear forces combining and a Mohr's cirle givles the ultimate load strength for the given combination?? (i.e. failure in complete shear, or failure in complete tension or failure in some combination of the two fall on a graph known as a Mohr's circle???) Ask me another on Fluid dynamics.....let's see if I've fogetten that too!
von mises...maybe I mean that not Mohr's.....36 and going senile.
[url] http://www.engineersedge.com/strength_of_materials.htm [/url]
Scroll down to yielding.
S1 is tensile, S2 is zero(assume no transverse loading), S3 is shear.*
*I gave up proper engineering 2 years ago and now work on the dark side, so I may not be trusted in this.....
It takes a bunch of engineers to overcomplicate things this is a simple answer.
Lets define:
UTS = the uts from the table above (depends on which material)
TS = the tensile stress calculated in the example at the top
FOS is factor of safety.
FOS = UTS/TS
so the factor of safety in tensile using copper its 220/13.8 =15.94 so your FOS is 15.94 near as dammit 16..
Normally for shear you have a USS ultimate shear strength, but you can approxmiate USS from UTS but its material dependent.
Ok you can resolve using von mises etc but I think this is a bit advanced for what is going on here...
toys, thatwould work in a purely tensile example, but here he has combined loading so needs a more complex model. Surely what he's looking for is Factor of safety against failure point. So first needs to consider failure mode.
Determining a von Mises stress and assessing against the yield stress (factored down for material thicknesses etc) would be the 'first cut' assessment. If you are simply looking for the ultimate strength of the joint, the assessment against the UTS and USS, plus a combined case (this may not follow a yield-surface type of combanitive approach) would be acceptable as well - the strength of the connected parts woudl also have to be checked. In reality, Eurocode 3/BS5950 use the tensile and shear strength for the rivet with knock-down factors depending on the material, and then have a permissable safety factor greater than 1. This may be going slightly off-topic though.....
Edit: Toys - I agree 😉
supersessions9-2 - Membertoys, thatwould work in a purely tensile example, but here he has combined loading so needs a more complex model. Surely what he's looking for is Factor of safety against failure point. So first needs to consider failure mode.
Yeah sorry I edited probably whilst you were posting, I agree but I just don't think that level of complexity is in this question ...
Toys- fair point. Not sure what OP wants in reference to question as he seems to have the answer there.
MTT - you got a similar real life case or is this a study example?
supersessions9-2 - MemberToys- fair point. Not sure what OP wants in reference to question as he seems to have the answer there.
My assumption (always wrong) was that they didn't know the basics, the text book looking like a typical first year text. What book is it by the way?
MTT - you got a similar real life case or is this a study example?
Purely a study example, I'm inclined to take [b]Toys[/b] explanation as the most appropriate approach given the teaching level (A-Level/Foundation degree). In hindsight I should have mentioned that in the original post...
Thanks again.
What book is it by the way?
Hannah & Hillier - Applied Mechanics (third edition) 0-582-25632-1
It's principally a first year degree or HNC/D text
Ah OK, sorry if I confused things. Is this foundation Maths or Engineering?
Takes me back to my HNC days, a bit more interested in it than I was back then 😀
Ah OK, sorry if I confused things. Is this foundation Maths or Engineering?
Engineering I believe. Thanks for the responses, in reality the answer is probably much more complex (as many of you have alluded to) but in this instance simplicity is probably best. Thanks again, problem solved*
* Well... unless they all fail their exams 😉