Impact failure calculation 1

[markarnoldTN]

Mechanism: Rod assembly made of 304 stainless with diameter of 4.6mm impacts onto housing of hard anodized 6061 aluminum. The aluminum housing has a hole of 4.1mm concentric to the rod assembly. Thus the annular area of contact is 4.67mm^2 (0.00724in^2). Static stress would be 1472 psi based on a driving force of 7.79 lbf, which is acceptable.

Problem: During testing, the repeated impact of the rod assembly deformed the aluminum housing. The estimated speed of the rod assembly is 12.3 in/s. Mass of the rod assembly is 0.0131 lb. We have changed the hole diameter to 3.9 mm but have changed nothing else. Obviously this will reduce the static stress due to the increased area. However, I would like to determine the increase in impact resistance, if any, based on this. Also, we have increased the thickness of the had anodizing from 20 microns to 30 microns, but I doubt this will really help. Any help would be appreciated.

 

[IRstuff]

I get 3.4 mm^2 area originally, and 4.7 mm^2 for the modification. That's about a 30% change in contact area, assuming that the rod actually impacts uniformly, which seems a bit farfetched.

Why does this have to be metal on metal impact? Whacking an aluminum structure with a steel hammer just doesn't seem to be a good idea. Even if you could get it to work in the short run, over the life of the system, there will be deformation, given enough impacts.

Every contact, no matter how small, causes changes. Even cotton thread can cut steel, given enough contact time and abrasion.

TTFN

FAQ731-376

 

[handleman]

Hi, Mark, and welcome to Eng-Tips! Working on a new cylinder? When will Matt be showing it off to us? ;-)

-handleman, CSWP (The new, easy test)

 

[markarnoldTN]

IRstuff, the goal is repeatable positioning after contact. A compliant dampening material could not do this. Obviously stainless steel on aluminum is not ideal but are necessary for the production process used. The rod assembly is guided concentrically to the hole in the aluminum housing. The rod assembly indented the aluminum to almost .040" within 2 million cycles. The goal is to change no more than .007" within 200 million cycles.

 

[israelkk]

Can you use a stronger insert in the 6061? Something like AL 7050?

 

[IRstuff]

For that matter, why not a steel insert?

TTFN

FAQ731-376

 

[GregLocock]

What is the important failure mechanism? burring of the hole? loss of accuracy in the positioning requirement? or what?

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[zekeman]

I would first write the energy equation allowing the KE of the moving mass to be absorbed elasically by the impacted material, namely
.5*V^2/2g=.5k*x^2=.5* Fm^2/k
where k is the elastic equivalent of the impacted area and x is the elastic deformation;you get k by using the impact area,a, and the modulus of elesticity of both the cylinder and the aluminum. Worst case it by assuming only the deformation in the aluminum using E of aluminum in compression and/or shear . Using compression, only

e=s/E
F=s*a=E*e*a
x=l*e
k=F/x=E*e*a/(l*e)=Ea/l
where
Fm= F max corresponding to emax
l= thickhness of aluminum
e= strain in compression
E =modulus Al in compression
Now, from the above equation, you choose Fm so that Fm=smax*a
where smax is less than the yield.
The RHS=.5*(smax*a)^2/Ea/l=.5*smax^2*a*l/E
from which you get the minimum area, a, to assure an elastic collision.

 

[markarnoldTN]

Thank you Zekeman, that is exactly what I was looking for.

 

[IRstuff]

Again, note that elasticity is a macro phenomenon.

Hitting something 200 million times requires both elasticity and wear resistance. If you lose a mere 1 Angstrom of aluminum per impact, after 200 million impacts, you'd have almost 0.8 inches of lost material

TTFN

FAQ731-376