Calculation help

[MMagis]

2-1/2" round bar? That will be almost 60 lbs of steel... do you mean pipe? How about adding a center support and using 1 or 1-1/4 bar?

It needs to be a solid bar.

 

[LD1]

I think most forklifts do have a center support to. those that have the tube style forks like that.

No reason to have the left fork go right of the center support and vice versa

 

[Mendonsy]

2-1/2" round bar? That will be almost 60 lbs of steel... do you mean pipe? How about adding a center support and using 1 or 1-1/4 bar?

That's what I did when I built my forks. I've had all that my 3pt could carry on the forks (about 1800#) and nothing bent.

 

[Mysfyt]

The center support is where the calculation is aimed at. I want to add a pair of cylinders to move the forks in and out with my third function and I figured it would be best to get the structural components established so I can place them accordingly. I agree with LD1 about the outward force vs the downward, but I was trying to keep it simple as far as the deflection calculations go. What I'm getting at is, if there was 1000 lbs on the forks as they are in the the image, would that be the maximum deflection on the bar? Or, let's say we just hung that weight by a chain in the center, what would the deflection be?

 

[LD1]

To get the moment of inertia, its (pi x r^4)/4

to get deflection its (W x l^3)/(48EI)

Where:
W is weight in pounds
l is length in inches
E is modulus of elasticity. (steel is 30,000,000)
And I is the moment of inertia calculated in the first equation

So: The radius of 2.5" is 1.25. so (pi x 1.25^4)/4 = 1.917

(1000# x 44"^3)/(48 x 30,000,000 x 1.917) = 0.030" deflection

 

[Egon]

LD1 is correct, that bar is subjected to as much horizontal force as it is vertical, if not more. You also need to define the width of the center support, in this case the two fork brackets, as well as how the two ends are supported including the span of the end supports like jb1390 noted. It's much more complicated than a simple deflection calculation.

It seems reality is more than a simple equation?

 

[SPYDERLK]

How is the end of the bar supported? Into a rigid structure (concrete wall or similar), or supported from underneath? It will make a pretty significant difference in the result.

Also the yield strength of the material (100 ksi), won't make a noticeable difference in the deflection. It will, however, affect the load which the bar will carry and still be able to return to its original shape when unloaded.

,,,,,,,,,,,,,,:thumbsup: ... [The amt of deflection the bar can withstand elastically.]

 

[Mysfyt]

To get the moment of inertia, its (pi x r^4)/4

to get deflection its (W x l^3)/(48EI)

Where:
W is weight in pounds
l is length in inches
E is modulus of elasticity. (steel is 30,000,000)
And I is the moment of inertia calculated in the first equation

So: The radius of 2.5" is 1.25. so (pi x 1.25^4)/4 = 1.917

(1000# x 44"^3)/(48 x 30,000,000 x 1.917) = 0.030" deflection

Thanks! Looks like I can proceed.