To determine the force in member BD of the truss structure shown in the image, we can use the method of joints or the method of sections. Since we are focusing on member BD and there are external loads applied at joint D, the method of joints might be more appropriate in this case.

Let's walk through the steps to solve for the force in member BD:

Step 1: Analyze Support Reactions

There are reactions at joints $A$ and $C$ since the structure is supported at those points. To start, we need to find the reaction forces at supports $A$ and $C$. Let's assume the structure is statically determinate.

1. Sum of moments about point A: $\sum M_A = 0$ Taking moments about point $A$, we consider the vertical and horizontal forces. The applied load at $D$ (120 kN vertically and 60 kN horizontally) will generate moments. The moment arms are determined by the distances to point $A$.

   The equation for the moment about point $A$ is: $120 \text{kN} \times 8 \text{m} + 60 \text{kN} \times 6 \text{m} - R_C \times 8 \text{m} = 0$ Solving for $R_C$ (reaction at $C$): $R_C = \frac{120 \times 8 + 60 \times 6}{8} = 135 \text{kN}$

2. Sum of forces in the vertical direction: $\sum F_y = 0 \implies R_A + R_C = 120 \text{kN}$ Substituting $R_C = 135 \text{kN}$: $R_A = 120 \text{kN} - 135 \text{kN} = -15 \text{kN}$ This means $R_A$ is 15 kN acting downwards.

Step 2: Analyze Joint D

Now that we have the reaction forces, we can move to joint $D$. At this joint, we have two applied loads: 60 kN to the right and 120 kN downward. We assume the forces in the members $BD$ and $CD$ are acting along their respective axes.

1. Sum of forces in the vertical direction at D: $F_{BD} \sin \theta + F_{CD} \sin \theta = 120 \text{kN}$ Here, $\theta$ is the angle between members $BD$ and $CD$ and the horizontal axis. From the geometry, we can determine $\theta$.

   The height difference is 2 m, and the horizontal distance between points $B$ and $D$ is 4 m. So, $\tan \theta = \frac{2}{4} = 0.5$, and $\sin \theta = \frac{1}{\sqrt{5}}$.

2. Sum of forces in the horizontal direction at D: $F_{BD} \cos \theta - 60 \text{kN} = 0$ Solving for $F_{BD}$: $F_{BD} = \frac{60}{\cos \theta}$ Since $\cos \theta = \frac{2}{\sqrt{5}}$, we get: $F_{BD} = \frac{60}{\frac{2}{\sqrt{5}}} = 60 \times \frac{\sqrt{5}}{2}$

Step 3: Final Calculation

$F_{BD} = 60 \times \frac{\sqrt{5}}{2} \approx 67.08 \text{kN}$

Thus, the force in member BD is approximately 67.08 kN.

Would you like further details or clarification on any part of the solution? Here are five related questions to explore this topic further:

1. How would you calculate the forces in members $AB$ and $BC$ using the method of joints?
2. What would change if the load at point $D$ was increased?
3. Can you verify the force in member $CD$ using the same method?
4. How do you check if a truss structure is statically determinate?
5. What would be the effect on support reactions if joint $C$ were moved horizontally?

Tip: In truss analysis, it's essential to resolve forces into horizontal and vertical components based on geometry, which helps solve complex structures systematically.