x axis load (deflection)

[Zygoat]

hay guys i am designing a new machine.

newbie diy router build.

i am trying to calculate the deflection of the x axis rails, worst case scenario. weight at the center only. supported at each end.

my beam will be made of steel. 96" long 1/4" wide and 3" tall.

obviously the force will be acting on the hight or the 3" portion as above.

the equation i have is

(w*1^3)/(48*e*i)

w is weight in lbs. (70lbs. weight of my gantry fully loaded.)
1 is the length of the beam (96")
e is the modulus of elasticity. (3.04579X10^7) i am using plane old carbon steel for the calculations i do not know for sure what alloy it will be (tho it will be cold rolled).
i is moment of inertia.(20776.160)

when i plug in i get .000002" this does not seem right to me. heres why.

1. when i preform the calculation in solidworks it comes out to ~ .010" and this is for a uniform distribution. my formula above is for weight concentrated at the center. so it should deflect MORE not less.

2. i am pulling the modulus of elasticity and the moment of inertia from solidworks. i am not 100% sure that these are correct. can someone confirm these. particularly the moment of inertia. i am pretty confident about the elasticity.

if anybody can run over these numbers or sees any mistakes please let me know.

[TomB]

The formula you have is the correct one for beam deflection where the deflection is limited to planar motion. Your beam is not limited in that sense. Just as a thought experiment imagine your beam being loaded while supported laterally on both sites, i.e. like it was clamped loosely between the jaws of an eight foot wide vise. In that case the beam would deflect in one plane and your formula would be correct. But as the jaws of the vise were opened the beam would no longer stay flat. The top would deflect out of plane in one direction while the bottom went the other direction. At that point the deflection would be the result of both your deflection formula and twisting geometry. Neat formulas to estimate deflection in these three dimensional cases are not (to my knowledge) available and analysis of such cases involves the use of finite element analysis. Unless you knew what that was before you asked your question I expect you will find that analysis methodology to be too dense for your current math capability.

Although I’ve never used the tool there is something on the web called Beam Boy. It might be easy to find but unless you understand the assumptions that go into formulas you could end up automating the same analysis mistake that is shown in your post, i.e. applying a formula to the entirely wrong case. In trying to make an ‘un-engineered’ machine you are probably going to have to ‘over kill’ the size of supporting sections. If the 3” by ¼” rail was replaced by a 2”by 2” angle bolted to a 2” by 4” channel with some lateral supports then you might get something adequate to the need, but even then you would probably end up having to prototype and modify.

Tom

[stangtjk]

Your moment of inertia is way off. For a rectangular cross section I=bh^3/12

[Zygoat]

ok, i had re worked the moment of inertia manually. i thought i had posted that but i guess not.

i am coming up with I = 31104

and the deflection is .0067"

for some reason when i worked it out before (using the correct moment of inertia, i was still getting crazy answers.) i must have done something wrong mathematically or did not catch something my calculator was doing that it should not have been.

oh well all is good now. thanks guys.


ps. i feel that this formula is valid for my case because the beam is constrained in the x and y directions by being mounted to the bed. i know it will still deflect some, but i think this is certainly close enough.


thanks again guys.

[stangtjk]

I think you're using the wrong dimensions for I. Its calculated using the cross section, so in this case your base (b) is the thickness (1/4") and the height (h) is 3" so your moment of inertia is 0.5625in^4. That works out to a deflection of 0.076"

[Zygoat]

ok. i worked it again, and my answers coincide with yours. i assumed b was the area of the base. i got it now. (deflection is .075309")

[stangtjk]

Keep in mind this is purely vertical loading. If you calculate deflection from a horizontal load your deflection is going to be considerably more because the moment of inertia becomes much lower as b would then be 3" and h would be 1/4".

[Zygoat]

oh yea, i ran those #'s for the y axis. because it will see the loads in the "skinny" direction. steps have been taken to firm that up.