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I'm not sure if you're asking for just the equation for the reactions, or how to calculate the FEM of a cantilever via the Moment Distribution method (aka Hardy Cross method), so I'll anser both.

If you assume clockwise moments are negative...

If your cantilevered span has a length "L", and a uniform load "W" acting downward along the entirelength of the span, then the fixed end moment is (W*L2)/8

If the fixed end is to the left, the equation is positive (the load is acting clockwise, so the reacting FEM is counter-clockwise, therefore positive.) If the fixed end is on the right, the equation becomes -(W*L2)/8

If your uniform load W is acting downward, and is acting along length "d" (which is NOT the entire span), its easiest to convert the uniform load W to a point load "P", where P = W*d

Then your FEM = [P*a*b*(2*L-a)]/(2*L2)

where:

L = entire length of span

a = distance P is acting from the fixed end,

b = distance P is acting from the cantilevered end (also, b = L - a)

OR, if W is centered along the span (therefore Pis also centered), then the equation becomes (3*P*L)/16

When going through the distributive process, the cantilevered end will ALWAYS have a final moment of zero. At the fixed point, the cantilevered side has a distribution factor of zero, while the other side has a D.F. of one. So, if you have length ABCD, where A is a cantelever, and B, C and D are assumed fixed, FEMAB = 0, D.F.BA = 0, D.F.BC = 1, and D.F.CB = D.F.CD = D.F.DC = 0.5

So what ends up happening is that the final FEM for the cantilever ("AB") is zero, and the opposite end of the span ("BA") remains as its initial value. The reason for this is when you sum up the reactions at the point (in this case B) supporting the cantilever and flip the sign, the distribution factors tell you to distribute everything to the side of the support opposite the cantilever ( -(FEMBA + FEMBC) is added to FEMBC). So you can go through the Cross Method as normal, but you don't touch AB or BA, and BC is adjusted only the once that I just mentioned (you don't carry over the 1/2 reactions from CB).

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Q: How do you calculate fixed end moment for cantilever beam with uniformly distributed load?
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