A structure, along with the loads applied on it, is shown in the figure. Self-weight of all the members is negligible and all the pin joints are friction-less. AE is a single member that contains pin C. Likewise, BE is a single member that contains pin D. Members GI and FH are overlapping rigid members. The magnitude of the force carried by member CI is ________ kN (in integer).

The correct answer is 18.

Step 1: Understand the Geometry of the Members
We are given that:
- $AE$ is a single straight member pinned at the wall support $A$ and the joint $E$, and it contains the pin joint $C$.
- $BE$ is a single horizontal member pinned at the wall support $B$ and joint $E$, containing the pin joint $D$.
- The vertical distance between the wall supports $A$ and $B$ is $3m$.
- The horizontal distance $BD=2m$ and $DE=2m$, so the total horizontal length $BE=4m$.

Let $\theta$ be the angle that the member $AE$ makes with the horizontal member $BE$. From the right-angled triangle formed by the wall and the member $AE$ (triangle $AEB$):
$$\tan \theta =\frac{AB}{BE}=\frac{3}{4}=0.75$$

Let $C^{\text{'}}$ be the projection of the joint $C$ on the vertical wall $AB$. Since $C$ is directly above $D$, the horizontal distance from the wall to $C$ is $C{C}^{\text{'}}=BD=2m$.
Using triangle $AC{C}^{\text{'}}$:
$$\tan \theta =\frac{A{C}^{\text{'}}}{C{C}^{\text{'}}}=\frac{A{C}^{\text{'}}}{2}$$
Substituting $\tan \theta =0.75$:
$$A{C}^{\text{'}}=2\times 0.75=1.5m$$
Therefore, the height of joint $C$ (and the horizontal member $CI$) above the bottom member $BE$ is:
$$h=AB-A{C}^{\text{'}}=3-1.5=1.5m$$

Step 2: Method of Sections
To find the force in member $CI$, we pass a section cut through the structure. The cut divides the truss into left and right portions by cutting through:
1. Member $CI$
2. Member $AE$ (between joints $C$ and $E$)
3. Member $BE$ (between joints $D$ and $E$)

Consider the equilibrium of the right-hand portion of the cut structure.
Because the members $AE$ and $BE$ are straight members pinned at joint $E$, the internal forces transmitted at their cut sections must have lines of action that pass directly through the pin joint $E$. Therefore, taking the sum of moments about point $E$ will eliminate the contribution of the forces in members $AE$ and $BE$.

Step 3: Moment Equilibrium Equation about Joint E
Taking counter-clockwise moments as positive, the equations of equilibrium for the right-hand section about joint $E$ gives:
$$\sum M_E=0$$
The forces contributing to the moment about $E$ are:
- The force $F_{CI}$ in member $CI$ acting horizontally at a vertical distance of $1.5m$ (producing a counter-clockwise moment).
- The horizontal external load of $2kN$ at joint $H$ acting at a vertical distance of $1.5m$ (producing a clockwise moment).
- The vertical external load of $4kN$ at joint $G$ acting at a horizontal distance of $3m+3m=6m$ from $E$ (producing a clockwise moment).

Setting up the moment equation:
$$F_{CI}\times 1.5-(2\times 1.5)-(4\times 6)=0$$
Simplify and solve for $F_{CI}$:
$$1.5F_{CI}=3+24$$
$$1.5F_{CI}=27$$
$$F_{CI}=\frac{27}{1.5}=18kN$$

Thus, the magnitude of the force carried by member $CI$ is 18 kN.