Drawing bending moment diagram for a beam with constant moment

[frozen7]

Can anyone draw me a correct Bending moment diagram and shear force diagram for this beam?
i have drawn the shear force diagram for this case but I do not know how to draw the bending moment disgram for this case because this question involve a constant moment. Normally the rule for bending moment is the value of bending moment at certain point is the area under shear force diagram from the starting point to that certain point. Can this rule applied on this question as well? I felt the bending moment diagram is quite strange if follow this rule for this question. Can anyone help me?
Thanks.

 

[Pyrrhus]

Hello, the only thing a couple or moment does for the bending diagram is create a "discontinuity" (more like a derivative doesn't exist at said point), just like the loads do on the shear diagram. If the moment is clockwise it goes up (adds to the area [method]), and if its counterclockwise it goes down (substract to the area [method]).

 

[frozen7]

Does it mean when there is a constant moment at point b, then we should either add or substract that value of constant moment in that point (either draw a straight line goes up or goes down)?

 

[Pyrrhus]

Does it mean when there is a constant moment at point b, then we should either add or substract that value of constant moment in that point (either draw a straight line goes up or goes down)?

exactly what i said above.

 

[frozen7]

One more thing have to be confirmed.
For uniformly distributed load, Mx = (Ay)(x) - (wx^2)/2 (where x is the distance from staring point to x, Ay is the reaction force at the starting point and w is the force per unit length and Mx is the bending moment)
Let say if there is a constant moment which is 50N/m at point a ,then Mx become Mx = (Ay)(x) - (wx^2)/2 +50 ??

 

[Pyrrhus]

One more thing have to be confirmed.
For uniformly distributed load, Mx = (Ay)(x) - (wx^2)/2 (where x is the distance from staring point to x, Ay is the reaction force at the starting point and w is the force per unit length and Mx is the bending moment)
Let say if there is a constant moment which is 50N/m at point a ,then Mx become Mx = (Ay)(x) - (wx^2)/2 +50 ??

Yes +50 if its clockwise.