**1. Introduction**

In this question, we are required to draw shear force and bending moment diagrams for the simply supported beam shown in Figure 4. We also need to determine the internal normal force, shear force, and moment at points C and D. Point D is located just to the left of the 10kN concentrated load.

**2. Beam and Loading**

The beam shown in Figure 4 is a simply supported beam with a length of L units. It is subjected to various loads, including a concentrated load of 10kN at point D.

**3. Shear Force Diagram**

To draw the shear force diagram, we need to consider the different loads acting on the beam and their effects on the shear force.

- At point A (left support), the shear force is zero as there are no loads to the left of it.
- Moving to point B, the shear force remains zero as there are no loads between points A and B.
- At point C, there is a concentrated load of 20kN acting downwards. This load creates a sudden drop in the shear force from zero to -20kN.
- Moving towards point D, the shear force remains constant at -20kN as there are no loads between points C and D.
- At point D, there is a concentrated load of 10kN acting downwards. This load creates an additional sudden drop in the shear force from -20kN to -30kN.

Therefore, the shear force diagram for the beam is as follows:

```
Shear Force (kN)
|
|
-30 | D
|
-20 | C
|
-10 |
|
0 | B
|
|
|
|
| A
+-----------------------------------------
```

**4. Bending Moment Diagram**

To draw the bending moment diagram, we need to consider the changes in the shear force along the beam and their effects on the bending moment.

- Starting at point A, the bending moment is zero as there are no loads to the left of it.
- Moving towards point B, the bending moment remains zero as there are no loads between points A and B.
- At point C, the sudden drop in the shear force from zero to -20kN creates a positive bending moment. The magnitude of the bending moment at C can be calculated using the equation M = V * d, where V is the shear force and d is the distance from the point of interest (C in this case) to the point where the shear force changes (B in this case). Let's assume the distance d is 1 unit. Therefore, the bending moment at C is M = -20kN * 1 = -20kNm.
- Moving towards point D, the bending moment remains constant at -20kNm as there are no changes in the shear force between points C and D.
- At point D, the sudden drop in the shear force from -20kN to -30kN creates an additional positive bending moment. Let's assume the distance d is 1 unit. Therefore, the bending moment at D is M = -30kN * 1 = -30kNm.

Therefore, the bending moment diagram for the beam is as follows:

```
Bending Moment (kNm)
|
|
-30 |