bearing allowable 2

[midsidenode]

A pin bears on the inside dia of a hole. The hole's cross section is made of two dissimilar types of aluminum sheets with different bearing allowables. can anyone provide a technique for calculating one bearing allowable value? thanks.

 

[rb1957]

you could assume that each sheet works to their respective bearing allowables ... this means that both sheets share the load when they are both effective, and only the stronger sheet is effective above the lower allowable.

 

[midsidenode]

thanks rb

 

[swearingen]

Just make sure the two materials have very similar elasticity moduli. If they are different, even by a small amount, the stiffer of the two materials may end up taking almost all of the load.


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[rb1957]

sure, you could apply the "rule of mixtures" (factor one sheet thickness by the ratio of Ec) ... but will it make any difference in th elong run ?

the stiffer sheet will react more of the applied load untill the lower bearing allowable stress is reached. the load reacted by this sheet is fixed (Fbr*A), all that would change is the applied load when this is reached (but who cares ?). the load reacted by the stronger sheet is also fixed, so the full load is fixed. the assumption is that the weaker sheet doesn't fail in bearing untill the stronger sheet does, not 100% correct but near enough, no?

 

[midsidenode]

Fortunately, both sheets are of aluminum (just different types of aluminum). One sheet is twice as thick as the other. Do both sheets share the load evenly until the weaker allowable is reached?

 

[rb1957]

they'd strain equally, sharing load in proportion to thickness

 

[swearingen]

You're absolutely correct, rb, as long as you can stomach the strain required to mobilize both.

Since both are aluminum, it really shouldn't be an issue.


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