Bending moment:

Since we cannot have sagging or hogging concept for the columns AB and CD we shall continue with the convention of drawing the BM on the tension side as in the beam BC.

For member AB, the positive end moment of + 1.992 kNm causes a bending moment producing tension on the right hand face; while at end B the positive end moment of + 3.996 kNm produce a tension on the opposite, i.e. left hand face. Thus, they are plotted on opposite sides of line AB and joined by a straight line. Similarly, for column CD. The compression side is marked 'C' the opposite side of the member will be in tension. The points of contraflexure (where the bending moments change the sign) are the points of zero bending moments and can be easily calculated.

For vertical members AB:

AH = 1.992/(1.992 + 3.996) × AC = 1.992/5.998  × 3 = 1.0 m

For horizontal member BC, BF is acquired by putting the moment equation to zero.

Mx = - 3.996 + 6 . x - 2 . x( x/2)

For point of contraflexure M_x = 0, when x = 0.764 m and 5.236 m (from the above quadratic equation).

The maximum positive BM in the beam is at the point where, dM x/dx = 0 , or 6 - 2x  = 0 or x = 3 m (i.e. at the centre of the BM).

The maximum positive BM

wl²/8  - FEM  = 2 × 6²/8 - 3.996 = + 5.004 kN m

 (b) For the beam member BC, you are already familiar with the method. For the vertical members AB, CD the shear force is equal to the horizontal reaction ± any lateral force acting on them. As there are no lateral forces, a members are subjected to constant SF of 1.996 kN.