Failure Load Calculations

[var10]

Hi Everyone,

I am trying to validate the design by verifying that the design will withstand the applied load. Just want to know what will be right to use interms of figuring out failure load.

Yield stress or tensile stress? The ultimate yield only gives you the Force where it starts to deform but ultimate tensile will give me the force at which it shall fail. I have got both yield and tensile of all the materials except for the screws (which is most likely to fail before others - hoping to get it from the supplier today) and I am approximately through some tests figuring out the Force applied area (its not easy! especially with screws).

My question is in product design environment what stress do you use to work out failure? I will have some safety factors on the loads later.

Regards,
V

 

[var10]

I guess for Aluminium I can use Proof Stress instead of the Yield stress? Any thoughts?

 

[desertfox]

Hi var10

In general I would use proof stress or yield stress as the failure stress, however I would use safety factors on the yield or proof stress so that my components never get into that stress region at least for static stress, if you have alternating stresses due to dynamic loads, then fatigue comes into play and that's a whole new ball game.
Might help if you identify what the component that you are concerned with actually is, or provide a sketch please.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[var10]

thanks desertfox. The load is just around 50kg applied and in one case there could be this shear load on 2 screws. All other parts are very well within the allowable range.

Worst case will be screws fail due to shear. Since there is no perfect way to calculate shear, I am assuming a percentage of tensile stress for shear. Correct me if I am wrong.

When it comes to dynamic loads yes it gets a bit trickier and based on some tests I only have very little dynamic load around 10N on carbon steel and aluminium materials so I am going to neglect it. Even the moment will have no effect.

 

[TLHS]

At that load, your materials are cheap.

If it's a one off design, overdesign by a large factor. The cost of the materials are less than the cost of you thinking about it.

If it's going to be repeated, make your best estimate (likely using yield stresses, because that's where you'll start seeing indications of failure... plus some safety factor) and then test so you can make sure you're as efficient as you can get.

 

[desertfox]

hi var10

Yes its correct to use a % of the tensile stress although I usually take a % of the yield stress because that's the figure you consider to be the failure point, my rule of thumb for shear is 50% of the yield stress.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[var10]

thanks guys.

@ TLHS - It will be used more than thousand times if not more in its life cycle. I am pretty sure they are well equipped for the load all except screws which are the limiting factors. Yeah I was also leaning towar

ds using Yield Stress as it is a bit more safer and you identify failu​

re earlier.

With screws though I am not a 100 percent sure on how to calculate the applied area for stress. I am just using the standard tensile area (At) to calculate the shear area (As). As except for one screw all others only experience Shear forces. Any advice on how to proceed with these calculations? Because the table does not consider the length of the screw. What happens if you have a very long screw? I do realise that the load is distributed on the first 5 or 6 threads from the head but that is for tensile condition not for shear.

 

[var10]

http://www.roymech.co.uk/Useful_Tables/Screws/Thread_Calcs.html

Not completely sure if this Tensile area is applicable for any length of the screw. Help and advice please!

 

[desertfox]

Hi var10
This is where a sketch of your situation might be useful, the link you have sent deals with shearing or threads during the bolt tightening phase only. I thought you meant shearing of bolts at ninety degrees to the bolt axis which can be found on this site. Roymech

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[var10]

Hi desert fox. Based on the table in that site, the allowable shear load on an M4 screw with medium carbon steel is about 150 kg. Can you achieve this on hand tightening or during tightening? I thought they did mean thats static loading scenarios. Can you please share the link where you think I will get the right information.

Thanks again mate!

 

[desertfox]

Hi var10
See below, now an grade 8.8 bolt as a proof stress of about 640 N/mm^2 so at say 50% it's shear stress is 320N/mm^2 and so if you multiply this by the minimum tensile area of the required bolt it will provide the shear load you're looking for.

http://www.roymech.co.uk/Useful_Tables/Screws/Bolted_Joint.html

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[BUGGAR]

I just did a test on 0.125 inch 6061 T6 aluminum with a #8-32 Flat head screw and got over 1000 pounds to failure. My gage didn't read any higher. Total deflection was 0.10 inches. Calculations per Aluminum Association predicted a safe load of about 700 pounds with a safety factor of 1.5 Bearing on the screw head was the failure mode.

 

[var10]

thanks guys. That was the table that I used too. Cheers!