Let's analyze and solve the given problem step by step.

Problem Overview:

We are tasked with calculating the internal bending moment at point $C$, located just to the right of the 20 kN vertical force $F$. The given data includes:

• A distributed load $MD = 5 \, \text{kN}\cdot \text{m}$ at point $D$,
• A downward force $F = 20 \, \text{kN}$ acting at point $C$,
• Beam spans:
  • $AB = 2 \, \text{m}$,
  • $BD = 8 \, \text{m}$.

Hint:

The distance from $F$ to point $C$ is $0.00 \, \text{m}$, meaning the force is acting directly at $C$.

Step 1: Calculate Support Reactions

We calculate the reactions at supports $A$ (vertical force $R_A$) and $D$ (vertical force $R_D$).

Moment About $A$:

Using the moment equilibrium about $A$: $\sum M_A = 0 \implies R_D \cdot 10 - 20 \cdot 2 - 5 = 0$ $10 R_D = 40 + 5 \implies R_D = \frac{45}{10} = 4.5 \, \text{kN}$

Vertical Force Equilibrium:

Using the vertical force balance: $\sum F_y = 0 \implies R_A + R_D - 20 = 0$ $R_A + 4.5 = 20 \implies R_A = 20 - 4.5 = 15.5 \, \text{kN}$

Step 2: Shear Force and Bending Moment at $C$

Now, we analyze the bending moment just to the right of $C$.

Shear Force to the Left of $C$:

The shear force at $C^-$ (left side of $C$): $V_C^- = R_A - 20 = 15.5 - 20 = -4.5 \, \text{kN}$

Bending Moment at $C^+$:

The bending moment at $C^+$ is calculated by summing moments from $A$ to $C$: $M_C^+ = R_A \cdot 2 - 20 \cdot 0 = 15.5 \cdot 2 = 31.00 \, \text{kN}\cdot \text{m}$

Final Answer:

The internal bending moment just to the right of $C$ is: $\boxed{31.00 \, \text{kN}\cdot \text{m}}$

Would you like a more detailed explanation of any step or visual aids?

Related Questions:

1. How do reaction forces affect bending moment calculations?
2. What are the key differences between shear force and bending moment?
3. Why is the bending moment zero at a free end of a beam?
4. What happens to the bending moment when distributed loads are involved?
5. How is equilibrium used in beam analysis?

Tip:

Always ensure units are consistent when solving for reactions, shear forces, and bending moments!