The free body diagram is
As a first approximation, I shall assume that the beam's weight can be neglected. We shall examine the validity of this assumption later.
The shear at A is 180 kN. dV/dx equals the negative of the distributed load. But there is no distributed load and so V is constant from A to B (180 kN). At B, the 90 kN force drives the shear down by 90 kN to 180 - 90 = 90 kN. At C, the external 90 kN force again drives the shear down and the shear between C and D is zero. From D to E, the shear is -90 kN.
The bending moment is zero at the ends. The change in the bending moment equals the areaunder the shear curve. Since the area under the shear curve from A to B is (180 kN)(0.6 m) = 108 kN-m, the moment at B is 108 kN-m. The area under the shear curve from B to C is (90)(0.6) = 54 kN-m, and so the moment at C equals 108 + 54 = 162 kN-m. Since the area under the shear curve from C to D is zero, there is no change in the moment from C to D. Finally, the area under the shear curve from D to E is (- 90)(1.8) = -162 kN-m which causes the moment to fall to zero, as I deduced it must be at the beginning of this paragraph.
Therefore, the maximum moment is 162 kN-m. Since the normal stress is given by
From Appendix C, we may choose:
Let's examine our assumption that the beam's mass is negligible. This beam has a weight of 2.1 kN. Disregarding the fact that this is really a distributed load, I'll consider it a concentrated load acting at the center. If included, it would cause a one percent difference in the shear values, which is within the engineering precision we agreed upon at the beginning of the course. So it is not necessary to include the weight in this case. But that will not always be so.