Fixed Ended Beam applied to couple moment

[socrates_1]

Homework Statement

Hi, i am trying to understand how to find the reactions (Ra & Rb) and moments (Ma & Mb) of the beam shown in the sketch below. Not an exact number but in terms of Mo(couple moment) and L(length).

Homework Equations

I have the answers which are Ra = -Rb = 3Mo /2L
AND Ma = -Mb = Mo/4

The Attempt at a Solution

taking moments about B: Ma - Ra(L) + Mo - Mb = 0
After that i broke the beam into 2 pieces (in the middle of the beam)
took the elastic equation for both sides where:
Mx = Ra(x) - Ma (left side)
Mx =Rb (x) - Mb (right side)
Intergrated both equations twice and found C1 and C2,and set these 2 equations equal for x= L/2 (as the delfection in the middle is 0)

But i can't get this result

Have i done something wrong?

 

[SteamKing]

It would probably be easier if you set the deflections and slopes at A and B equal to zero, and solved for the reactions. Is the deflection at x = L/2 equal to zero? Maybe, but the deflections at A and B must be zero, as should the slopes.

 

[socrates_1]

It would probably be easier if you set the deflections and slopes at A and B equal to zero, and solved for the reactions. Is the deflection at x = L/2 equal to zero? Maybe, but the deflections at A and B must be zero, as should the slopes.

Thank you for your response.I just tried to solve it like this but I couldn't get the result.
Any other suggestions?

 

[socrates_1]

any suggestions??

 

[SteamKing]

Show your detailed calculation. There may be a mistake in your algebra.

 

[socrates_1]

Show your detailed calculation. There may be a mistake in your algebra.

For Ra:
φEI= (Ra (x^2 ))/2- Ma (x)+ C1 (eqn 1)

where φ=slope, x=distance, C1 = (Ra (x^3))/6 + Ma (x^2) vEI= (Ra (x^3))/6 - (Ma (x^2))/2 + C1(x) + C2 (eqn 2)

Where v=delfection,x=distance, C1 = (Ra (x^3))/6 + Ma (x^2), C2=0By setting (eqn 1) equal to (eqn 2) I cannot get the result I want because the result for Ra must be in terms of Ra and Mo.
What u think?

 

[SteamKing]

Remember, at x = 0, the slope = 0 and the deflection = 0.
At x = L, the slope = 0 and the deflection = 0.

You've got to use these four conditions in order to solve for the unknown reactions and fixed end moments.