Balance Beam formula

[walp25]

Hi folks,
Sorry to have to ask a seemingly simple question. Ive been out of practice for some time, and the "googling" route is not providing me any clarity.

I need to calculate how much weight to put on one side of an asymmetric beam. This is somewhat easy to figure if the loading was balanced, and i had the weight of all materials including the beam.

However Im working with evenly distributed loads on either side of the fulcrum.
my example is thus:
Beam is 20ft long
Fulcrum is at 8ft from left end
I need to figure how much weight (distributed) say over 2ft from extreme left to apply to the beam to create equilibrium across the fulcrum.

I know there is an easy solution to this, but I just cannot put my finger on it.

Can someone advise me on a formula or location of one?

Thanks,
Sam

 

[delagina]

summation of moment @ fulcrum = 0
W= weight of beam
W1= weight of distributed load 2ft from left

W(12)(12/2) = W(8)(8/2) + W1(2)(7)

 

[rb1957]

i'm guessing it's slightly more complicated than that ...

LL = span on the left of the fulcrum = 12'
LR = span on the right, 8'
W = weight (per ft) of the beam
WL = weight added on the left side
XL = distance from the fulcrum (of WL)
WR = weight added to the right
XR = where this is
if you're applying a distributed load, apply a concentrated load in the middle of the distributed span (clear as mud)

LHS moments = W*LL*LL/2+WL*XL
RHS moments = W*LR*LR/2+WR*XR

 

[walp25]

delagina & rb1957,
Thanks for your responses.....It didnt even occur to me I could turn the dist. load into a point load. This will help me figure this out. I know that the moment needs to "null" out at the fulcrum to be in equilibrium, I was just having a bear of a time dealing with the loading on the beam (being distributed).
Thanks again for your responses, this made my afternoon.
-Sam

 

[JAE]

What exactly is this for?

 

[BAretired]

So, now that you know how to do it, what is the answer? What is the uniform load required to balance the beam? No hinting from the audience.

BA

 

[csd72]

be careful of changing distributed loads to point loads when the distributed load goes past the hinge point or support, treat the load on each side of this as a separate point load.

 

[BAretired]

delagina and rb1957 showed a nice way of solving the problem. Another way would be to say that the first 8' of the 12' portion is already balanced, so the unbalanced moment is 4*w*(8+4/2) or 40w where w is the uniform load of the beam.

If the additional load on the end of the 8' span is w₁, then 2*w₁*(8-2/2) or 14w₁ is its moment about the fulcrum.

The beam is balanced when those two expressions are equal.

BA