# impact factor / failure deflection



## Lily (Oct 5, 2010)

Hi everyone,

The problem is: Design requierements specify that a cell phone's plastic shell must survive a 6ft fall onto concrete. Static compression tests indicate shell fracture at F=1250 lbf and 0.005 deflection. If a factor of safety 6 is used what is most nearly the maximum allowable phone weight.

My solution : I used the formula's on page 58-19 from MERM to find detalv =(2*g*h)^0.5 = 19.66 ft/s^2, then natural frequency = delta v / deflection = 47184 rad/sec, then acceleration=natural frequency X delta v=927637.44 ft/sec^2, then 6 x F=m*a/gc to find m. m=0.26 lbm , I find the same m using m*deltav^2/gc+mgh/gc=6 x F x deflection ,

This is the same solution presented in problem 73 of SMS MD.

Lindeburg solution: impact factor = 1+(1+2*h/deflection)^0.5, F/safety factor = weight X impact factor, m=1.22 lbm

1) where Linderburg formulas came from?

2) why both methods don't give the same solution

Thanks,

Lily


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## Relvinim (Oct 6, 2010)

I am not familiar with the Sample exam you are taking but I have a 2nd edition Lindburg PE Sample Exam and my feelings were you need a PhD in Engineering to solve some of the problems. I stopped looking at it because non of the problems could be solved in 6 minutes.


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## Shaggy (Oct 7, 2010)

That impact factor formula comes from The Fundamentals of Mechanical Design by Juvinall. It is a text that is comparable (mostly) to Shigley. Here is a good link that has similar info.

See Issue No 1 (06/2008)

http://www.brushwellman.com/TechnicalTidbits.aspx


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## Lily (Oct 7, 2010)

Thanks Shaggy!

Yes, Shigley has some similar formulas, but they are not written the same way as in the solution.

Lily.


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## xuhq (Oct 24, 2010)

It was applied twice in the sample exam, 104 and 115. I was also wondering and checked

one of my books to understand the formula.

Let d be the deflection, dmax the max deflection, and dst the static deflection,

m*g/gc*(dmax+h)=1/2*k*dmax^2

m*g/gc=k*dst

=&gt;1/2*dmax^2-k*dst*(h+dmax)=0

=&gt;dmax=dst+(dst^2+2h*dst)^0.5

The impact factor fi=Fi/Fst=dmax/dst=1+(1+2h/dst)^0.5

Also, fi=1 for static load

=2 for suddenly applied load

=3 for suddenly applied and reversed

Some assumptions are needed for the above derivations.



Lily said:


> Thanks Shaggy!
> Yes, Shigley has some similar formulas, but they are not written the same way as in the solution.
> 
> Lily.


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