Your E factor is incorrect
(980 * 1000 ^ 3) / (48 * (70000/mm^2) * 1321) = ????
Where you get mm^2 from, I'm not sure. 80^2 maybe????
Hi All
I'm building my second router and I wanted this time to do the job properly by calculating the forces and stresses the various parts would be subjected to.
The first thing, I wanted to make a test calculation; If I have a t-slot beam of a 80x80mm profile a meter long, rested in the two ends. How much it'll deflect if I put a 100kg weight at the middle.
This is the formula:
(P * l^3) / (48 * E * I)
where:
P - the load in Newtons
l - the length of the beam in millimeters
E - Elastic modulus of the material in N/mm^2
I - moment of inertia of the profile in mm^4
P = 100kg * 9,8 = 980N
l = 1000mm
E = 70000N/mm^2 (taken from Rexroth datasheet http://www.boschrexroth.ch/country_u...ng_Catalog.pdf)
I = 1321mm^4 (taken from Rexroth datasheet http://www.boschrexroth.ch/country_u...ng_Catalog.pdf)
(980 * 1000 ^ 3) / (48 * 70000 * 1321) = 220.79mm
I can't believe an 80x80 t-slot would deflect that much!
I talked to friends who also work with t-slots, and they are also baffled and don't think it would bend nearly that much. They say 100kg is nothing for an 80x80!
I must be making and error somewhere?!
Thanks a lot in advance.
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Your E factor is incorrect
(980 * 1000 ^ 3) / (48 * (70000/mm^2) * 1321) = ????
Where you get mm^2 from, I'm not sure. 80^2 maybe????
Last edited by Jim Dawson; 09-29-2017 at 07:12 PM.
HI Jim
Thanks a lot, I'd never realize that.
I think what's needed is the crosssectional area of the profile. I got this from solidworks measurment, it's: 1814.62mm^2
I think if I put 80^2 that would mean 80x80 solid beam. 80^2=6400mm^2 which is quite a bit more.
But wait! If I divide 70000 by this number the resultant will be even smaller so the deflection would be even bigger. Shouldn't I multiply 70000 by 1814.62?
(980 * 1000 ^ 3) / (48 * 70000 * 1814.62 * 1321) = 0.12mm deflection
Now that's much more like it.
Do you agree with this calculation?
0.12mm deflection is close to what I would expect. But the math to arrive at that number does not follow the formula.
(P * l^3) / (48 * E * I)
From your numbers:
P = 100kg * 9,8 = 980N
l = 1000mm^3 = 1000000000
E = 70000N/mm^2 = 38.5755
I = 1321mm^4 = 3045168091681
so if I did the math correctly (980 * 1000000000) / (48 * 38.5755 * 3045168091681) = 1.738e-4 or about .0002 mm
Or for practical purposes, no deflection.
That kind of deflection would be fantastic.
I still have two more questions just to make sure I understand everything right.
E = 70000N / 1814.62mm^2 = 38.5755 means that if I have a solid piece of aluminum with 1814.62mm^2 area and apply 70kN force to that area I'll get 38.5755 change in form the original shape i.e. strain?
So since we want to get the linear deflection in (mm) we must "lift" 1321 to the power of 4? I thought mm^4 is the measurement unit and the number 1321 should go in the formula as it is? Why isn't then that number listed as 3.045 * 10^12mm in the datasheet. I'm not arguing, just trying to understand.
Thanks again for your explanation.
I'm just going by the formula as presented.
''we must "lift" 1321 to the power of 4? I thought mm^4 is the measurement unit''
I can not see mm^4 being a measurement unit.
Maybe look at this, it will explain the calculation much better than I can.
Beam Calculator
That calculation would be for a solid, these Extrusions are far from it, they move quite a bit, I did a FEA on 3 x 3 and @ 100Kg / 220.4Lbs. it would move more than .015"/ .38mm so you are not even in the ball park, all of these Extrusions bend and the amount of bend will depend on the Grade of Aluminum used, most are soft so they can be extruded, the length also being used, but the good side is you can get 6061 T6 which has a lot more strength and won't bend as much
Mactec54
There are a couple mistakes you are making in your calculations. If we just consider the cross section to be an 80mm square section the moment of inertia would be 3,413,333 mm^4. This is a simplified version though. As mactec54 mentioned a "T" slot section is going to be quite a bit different due to the cross section. Another method can be used to go through and add up the moment of inertia of pretty much any cross section. We would need to know the exact dimensions of the "T" slots to do this. You also can't manipulate the modulus of elasticity. It goes into the formula as 70,000N/mm^2. The N/mm^2 defines it's value, it is not a formula. To complicate things further you also have to determine if the ends of your beam are simply supported (have the ability to rotate) or if they are fixed (welded or fastened tight). This will change your initial formula. The formula you are using is for a simply supported design. The fixed end design uses the formula
P * L^3 / 192 * E * I
https://www.8020.net/tools-software
Their Tech Toolkit includes a deflection calculator. Based on their extrusions of course, though they do have an 80 x 80 profile so it should get you pretty close.
Mark
Last edited by wendtmk; 10-03-2017 at 11:19 AM. Reason: Gotta learn how to type my own name...
Yes that calculator is quite good, I checked it to ways 40 series 8080 extrusion with fixed Ends it will deflect .002" or .05mm most are not built as fixed ends, with the same material it would be.008" or .2mm
This was with a 1m length or 1000mm with 100Kg / 220lbs of 40 series 80x80
So my FEA that I did for the smaller profile was very close to the same
Mactec54
The deflection will be the same for any grade of Aluminum, 6061/7075/2024 etc etc. The E (youngs modulus) is similar for all Aluminums (with very small variations) The difference being that the aluminums (and steels) with higher yields will deflect further before yielding.
You might want to learn some fundementals about stress/strain before you do any FEA if you think different aluminums will have different deflections.
What you said is correct, the E ( young's modulus ) is similar, the material grades that you list are not what most extrusions are made from, 6063 is the most common, material used, and is like butter in it's T1 state when extruded
Get on the same page, we are only talking about what deflection to expect, 6063 will show a greater defection than 6061 of the same profile of the same temper and same loading, there is not a lot of difference, but enough for one grade to stand out better than the other, we have tested them physically as well as FEA and almost always have a very similar outcome, the outcome is not the same for the different grades even though the modulus does not change the rate of deflection does change
This is very basic engineering, and the link to the extrusions software is all most here need to find out what to expect for deflection with what ever extrusion they select
https://www.8020.net/tools-software
Mactec54
Your talking out of your arse, the Youngs modulus for 6061 and 6063 is the same (68.9GPa), therefore the deflection is the same upto the yield point regardless of the heat treatment.
Not in the real world, do you get paid to be an engineer, or just play one on day time TV?
And who is talking out the A______ deflection and yield go hand in hand, are you smoking something check the chart below
As I said you are saying the same thing as what I am, I guess you are trying to make your self look like something, or do you feel robbed because of your small size man hood, what ever it is you need some attention, below is a list of the different grades that in it's self tells everyone the TV is on and working well
I already stated the Young's Modulus does not change, is that all you can come up with I guess you can't read
The 2 results between 6063 and 6061 are very different, numbers don't lie, You can pretend though, the chart below gives you those numbers to work with
6063 Yield Strength 8,000 PSI as Extruded
6061 Yield Strength 14,000 PSI as Extruded
So Mr. wizard calculate the difference between the 2 material Grades and see who is correct with the deflection being different between the 2 Grades, a high school kid could do this
Mactec54
Wow wow! I don't want a flame war in my thread. Chill out guys. I'm sure you both know your stuff, but it seems to me you two talk about the same stuff in different terms, which creates the confusion IMO. No need to use harsh words to make a point!
Mactec54
Deflection has nothing to do with yield strength!! It has everything to do with Young's modulus!!
The Youngs modulus is the same (with VERY minor variations) for any heat treatment for 6061/6063 from T0 to any T6(XXX) condition. Therefore the deflection will be the same for any heat treatment/tensile strength.
The difference being that the T0 condition materials will yield and retain a permanent set at lower deflections than a T6(XXX) material. But up until yield the deflection will be the same.
What's so hard to understand
The formula for deflection is D = L^3 X W / 48EI
in that formula E is the value of young's modulus.
Where in that formula is the value for ultimate/yield strength of the material??
You are a bit late to the party, this whole exercise was to inform the builders or users of aluminum extrusion, that they have a choice of what grade of aluminum to buy, 6061 T6 or 6063 T5, it would be very hard to find 6063 T6, I would suggest that you get the 2 extrusions of the 2 grades and have it tested
Mactec54