Originally Posted by noelex 77
Can someone calculate the size of stainless steel
rod that would have the same yield strength as a 16mm diameter aluminium rod
The aluminium is 6061 T6 (yield strength from Wikipedia 35,000 psi)
My guess is 16 mm diameter rod in aluminium would be similar to about 12mm stainless steel
Ok, for the sake of learning
etc i've gone through a step by step thought process below.
You've been using 6061 T6 and it works.
The length of the rod is no more than 2x the diameter
You're planning on replacing it with AISI316L (preferred) or equivalent.
The rod is supported on both ends.
Firstly, don't use ultimate tensile (UTS) in your calculations as you're not wanting to know when it breaks, but you're wanting it to operate in its elastic region and not go plastic. For this always use the 0.2% offset yield (sometime referred to as "yield strength"). Most PEs will usually use a material factor too, somewhere in the region of 15%, this is to take into account any material variations in the stock that you have. This way you keep the actual stresses within a workable repeatable limit without causing any permenant damage to the material.
So, if your rod length is 2 diameters or less then you don't need to think about bending moments.. only shear strength.
Your allowable stresses for each would work
Aluminum 6061-T6: 275MPa/1.15 = 239MPa
Stainless AISI316L: 355Mpa/1.15 = 308MPa
The allowable shear stress, i.e. trying to snap the pin in a clean break across its section, from Von Mises, would be:
Aluminum 6061-T6: 275MPa/(1.15*root(3)) = 138MPa
Stainless AISI316L: 355Mpa/(1.15*root(3))= 178MPa
Remember that 1 MPa is equivalent to 1 Newton per square mm.
You mentioned that you aluminum rod is 16mm diameter. Remember that the rod's supported on both ends, so you have 2 shear (or break) regions through the section of the rod.
The maximum allowable shear load for the Aluminum would be 2*x-sectional-area*allowable-shear-stress:
2*((16)^2)*PI/4)*138MPa = 17664PI = 55493N = 5657Kg
Knowing that the rod can take 5657Kg in shear, we can then look to see what happens if it's made from AISI316L - so we reverse:
55493 N /178MPa = 311.76 sq.mm.
311.76 / 2 areas = 155.9 sq.mm.
root(155.9*4/PI) = 14mm diameter rod.
If you looked at the problem as a simple scaling of the two yields you'd end up with an answer that didn't take into account that area changes as a square and not as a linear.. in that respect you'd get an answer of 12mm which would be incorrect...
Remember that a 16mm rod has 77% more cross-sectional area than a 12mm rod, eventhough it only has 33.3% more diameter!
If it turns out that the rod you're looking at is longer than 2 diameters i'd be happy to show you the appropriate strength calculation if you give me a sketch of the loading scenario, supports and geometry of the rod etc.
If you want to have a stab as this yourself:
You calculate the shear stress.
You calculate the stress due to bending moments (remember the correct moment of inertia of a rod calc and sectional W-value).
You then combine the two using Von-mises which is Tot-stress=root(((bending-stress)^2)+(3*(bending-stress)^2)))
As long as your total stress comes out below your allowable stress (in this case 308MPa) then you're in good shape..
Hope this helps.. good luck!