Yes, I can help you with that. To solve for the force in member $BD$ using the method of joints, we need to analyze the forces at each joint of the truss. Here's how to create the free-body diagram (FBD) for each point:

1. Joint A:

• At joint $A$, you have two unknown forces:
  • Force in member $AD$, at an angle of $30^\circ$ from the horizontal.
  • Reaction force at $A$, which we assume is purely vertical since the structure appears to be symmetric.
• Label these forces:
  • $F_{AD}$ acting along the direction of the member $AD$.
  • $A_y$ as the reaction force in the vertical direction at joint $A$.

2. Joint B:

• At joint $B$, the external load of $3 \, \text{kN}$ is applied horizontally.
• You also have two unknown forces:
  • Force in member $BD$.
  • Force in member $BC$.
• Label these forces:
  • $F_{BD}$ acting along the member from $B$ to $D$.
  • $F_{BC}$, along the member from $B$ to $C$.
  • External load $3 \, \text{kN}$ acting to the right.

3. Joint C:

• At joint $C$, similar to joint $A$, we assume a vertical reaction force and a force along member $CD$.
• Label these forces:
  • $F_{CD}$, acting along the member at $30^\circ$ from the horizontal.
  • $C_y$, the reaction force in the vertical direction at joint $C$.

4. Joint D:

• At joint $D$, three members meet, but since this joint is symmetrical, forces should balance out:
  • Force in member $AD$, which we already labeled as $F_{AD}$.
  • Force in member $BD$, already labeled as $F_{BD}$.
  • Force in member $CD$, labeled as $F_{CD}$.

Free-Body Diagrams (FBD) Summary:

• For each joint, you'll balance the forces using the conditions of static equilibrium:
  • Sum of forces in the $x$-direction: $\sum F_x = 0$
  • Sum of forces in the $y$-direction: $\sum F_y = 0$

Would you like me to draw these diagrams for you, or continue solving for the forces using equilibrium equations?