# Failure analysis of a 2 OD solid 6063 aluminum round bar

1. Dec 5, 2008

### EMCSQ

Failure analysis of a 2" OD solid 6063 aluminum round bar

I have a 6 foot piece of 2" OD solid 6063 aluminum bar that is supported at both ends. Hanging from the center is a 5000 pound weight. What techniques and equations would be used to determine if the aluminum bar will successfully bear this load?

This setup is going to be used as an anchor point for fall arrest protection on a communications tower.

Any help anyone can provide would be appreciated.

2. Dec 8, 2008

### minger

Re: Failure analysis of a 2" OD solid 6063 aluminum round bar

Well, the static case is quite simple. Easiest way would be to simple look up the load case in Roark. Find your max stress (should be at the center). Compare that to the strength of your material. Factor in your safety factor.

However, if the load is going to essentially be dropped onto it, then there are additional factors one needs to account for. I won't get into it in detail, but there are factors that you can use to account for sudden loading. There is a section also in Roark (while you're there) that gives the factors.

Make sure that you account for any possible decrease in strength in the material being that this is essentially used as human safety. Corrosion, heat, anything, PLUS use a beefy safety factor.

3. Dec 8, 2008

### stewartcs

Re: Failure analysis of a 2" OD solid 6063 aluminum round bar

I agree with the advice minger gave and will stress the importance of accounting for the additional load due to the fall arrest requirement. Additionally, you might want to consider the fatigue life of it since it will probably be in service for a while based on the application.

CS

4. Dec 15, 2008

### EMCSQ

Re: Failure analysis of a 2" OD solid 6063 aluminum round bar

Ok thanks for the assistance. I appreciate it.

5. Dec 22, 2008

### nvn

EMCSQ: That type of requirement is written such that the given load, P, already includes the ultimate factor of safety, dynamic amplification factor, and any fatigue factor. All you need to do is apply the given load, and ensure the bending stress does not exceed the material strength. When you perform an elastic analysis, you immediately see your bending stress exceeds the material tensile yield strength, Sty = 214 MPa. Therefore, you know the beam is yielding, and an inelastic analysis is required. To perform a simplistic plastic analysis, you can divide the elastic bending stress by an ultimate plastic shape factor, sf. In your particular case, sf = 1.822. Therefore, for your given problem, the bending stress is sigma = 8*P*L/(pi*d^3) = 8(22 241 N)(1829 mm)/[pi*(50.8 mm)^3] = 790.16 MPa. The ultimate bending stress level is therefore R = sigma/(sf*Sty) = (790.16 MPa)/[1.822(214 MPa)] = 202.7 % > 100 %. Ensure R does not exceed 100 %. Therefore, the above indicates you need to, e.g., increase the beam section modulus, and/or decrease the beam length.

6. Jan 13, 2009

### stewartcs

Re: Failure analysis of a 2" OD solid 6063 aluminum round bar

What failure?

CS