impact factor / failure deflection

[Lily]

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

 

[Relvinim]

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.

 

[Shaggy]

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

 

[Lily]

Thanks Shaggy!

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

Lily.

 

[xuhq]

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

=>1/2*dmax^2-k*dst*(h+dmax)=0

=>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.

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

Lily.