How to determine the maximum bending moment of a beam

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harvistar
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I have a 1.5m beam with a uniformly distributed load of 57kn/m acting over 1.2m of the beam (0.6m from the centre of the beam in both directions). I require the maximum moment and the deflection of the beam.

I know the maximum moment equation wl^2/8 and the deflection formula 5wl^4/384EI.

I have done this so far:

Take a cut at the centre:

(sum of the moments @ the cut) = 57*0.6m*0.6m/2 - Va(0.75m) => Va = 13.68

Va = Vb = 13.68kN

57 x 1.2 x 0.6 = 41.04 which does not equate to 27.36 (sum of the forces must cancel)

However, oddly enough 13.68 x 3 = 41.04kN

I am not sure where or how I am going wrong?

Any help is appreciated.
 
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harvistar said:
I have a 1.5m beam with a uniformly distributed load of 57kn/m acting over 1.2m of the beam (0.6m from the centre of the beam in both directions). I require the maximum moment and the deflection of the beam.

I know the maximum moment equation wl^2/8 and the deflection formula 5wl^4/384EI.

I have done this so far:

Take a cut at the centre:

(sum of the moments @ the cut) = 57*0.6m*0.6m/2 - Va(0.75m) => Va = 13.68

Va = Vb = 13.68kN

57 x 1.2 x 0.6 = 41.04 which does not equate to 27.36 (sum of the forces must cancel)

However, oddly enough 13.68 x 3 = 41.04kN

I am not sure where or how I am going wrong?

Any help is appreciated.
When you take a cut section at the simply supported beam mid point, you must include the bending moment that exists at that cut. Better to solve for the end reactions first by summing moments about one end then you can calculate the moment at mid point. Note also that your equations for max M and deflection are for a uniform load across the entire beam, which does not apply here.

Welcome to PF!
 
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Your formulas for max. BM and max. deflection are valid only if the beam is simply supported at both ends AND the load is distributed along the entire length of the beam.

In your case, I would recommend that you go back to first principles and construct a shear and bending moment diagram for your beam, after first solving for the reactions at the ends. BTW, I don't understand why you are cutting the beam in the center.