To determine the normal force, shear force, and moment acting at point $C$, we need to analyze the forces and moments on the beam segment. Let's proceed with the following steps:

Problem Breakdown

1. Beam Details:

   • A point load of $8 \ \text{kN}$ acts vertically down at point $A$.
   • A couple moment of $18 \ \text{kN} \cdot \text{m}$ acts clockwise at point $A$.
   • A uniformly distributed load (UDL) of $5 \ \text{kN/m}$ spans from point $C$ to point $B$ over a length of $0.5 \ \text{m}$.
   • Dimensions:
     • $AC = 1.25 \ \text{m}$
     • $CB = 0.5 \ \text{m}$

2. Support Reactions: Since the beam's reactions are not provided, we assume points $A$ and $B$ are supported to resist loads (e.g., fixed or simply supported beam). For simplicity in this case, assume there are no additional supports outside the reactions shown.

Solution Steps

Step 1: Calculate Reactions at Supports

Let's assume that point $B$ has a reaction, although we are primarily focused on point $C$ (we'll return to calculating exact reactions if detailed constraints are provided).

Step 2: Analyze the Forces and Moment at Point $C$

We’ll take a section cut at point $C$ to the left and analyze the segment from $A$ to $C$.

1. Normal Force at $C$:

   • Since there is no horizontal force applied to the beam in this setup, the normal force at $C$ should be 0 kN.

2. Shear Force at $C$:

   • Calculate the shear force at $C$ by summing vertical forces in the segment $A$-$C$.
   • Forces acting in the $A$-$C$ segment:
     • Downward point load at $A$: $-8 \ \text{kN}$.
   • Shear Force at $C$: $V_C = -8 \ \text{kN}$
   • So, the shear force $V_C$ at point $C$ is $-8 \ \text{kN}$ (downward).

3. Moment at $C$:

   • Calculate the moment at $C$ by taking moments about point $C$.

   • Forces and moments to consider:

     • The $8 \ \text{kN}$ load at $A$ creates a moment about $C$. $\text{Moment due to } 8 \ \text{kN load} = 8 \ \text{kN} \times 1.25 \ \text{m} = 10 \ \text{kN} \cdot \text{m}$ This is counterclockwise about point $C$.
     • The $18 \ \text{kN} \cdot \text{m}$ couple moment at $A$.
       • This is already a moment, and it is clockwise.

   • Sum of moments about $C$: $M_C = 18 \ \text{kN} \cdot \text{m} - 10 \ \text{kN} \cdot \text{m} = 8 \ \text{kN} \cdot \text{m}$

     • The result is a clockwise moment at point $C$.

   So, the moment $M_C$ at point $C$ is $8 \ \text{kN} \cdot \text{m}$ clockwise.

Summary of Results

• Normal Force at $C$: $0 \ \text{kN}$
• Shear Force at $C$: $-8 \ \text{kN}$ (downward)
• Moment at $C$: $8 \ \text{kN} \cdot \text{m}$ (clockwise)

Would you like further details on any of these calculations or clarification on beam theory concepts?

Additional Questions to Explore:

1. What would change if there were a horizontal load acting on the beam?
2. How would the analysis differ if point $B$ were a pinned support?
3. What is the effect of a concentrated moment at a different location, such as point $B$?
4. How would the results change if the uniformly distributed load covered the entire beam?
5. How does the location of the section cut affect the shear force and moment calculations?

Tip:

When analyzing forces and moments, always start by isolating segments and using free-body diagrams for clarity in determining unknowns.