At the left end of the beam the moment will equal zero. The 220# force will cause the moment to increase at a constant rate of 2200#/ft. The value of the moment diagram at the point four feet from the left end can be found using the area method. Its va lue will be equal to the area of the shear diagram to that point or (2200#)(4ft) = 8800#ft.
Continuing to the right, the shear diagram drops to -200#. The shear diagram is constant, so the moment diagram will once again have a constant slope. This time the slope will be negative because the shear diagram is negative. Using the area method, th e value of the moment diagram at 6 ft will be 8800#ft - (200#)(2ft) = 8400#ft.
From here, the moment diagram must close to zero because no concetrated moment acts on the right end. The shear diagram for this portion has a constant negative slope with an increasing ordinate. therfore, the moment diagram will follow an x^2 curve wit h a slope approaching the vertical.
The closure of the moment diagram can and should be checked using the area method. The area of this part of the shear diagram is a composite area consisting of a rectangle and a triangle. The shear diagram is also negative, so the area will be subtracte d from the last know moment value of 8400#ft.
8400#ft - (200#)(6ft) - 1/2(400#/ft)(6ft)(6ft) = 8400#ft - 1200#ft -7200#ft = 0The moment diagram will close to zero.
Note that a beam such as this can experience posative and negative shear and still maintain a moment which remains posative along the entire length of the beam!