Impact failure calculation
Impact failure calculation
(OP)
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.
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.





RE: Impact failure calculation
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
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RE: Impact failure calculation
-handleman, CSWP (The new, easy test)
RE: Impact failure calculation
RE: Impact failure calculation
RE: Impact failure calculation
TTFN
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RE: Impact failure calculation
Cheers
Greg Locock
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RE: Impact failure calculation
.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.
RE: Impact failure calculation
RE: Impact failure calculation
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
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