Let's go through the problem step by step with a focus on solving for the final rectangular size of members AB and DE. Based on the image and instructions, the load "n" is given as $n = 21$, so the force is $21 \times 2000 = 42000 \, \text{N}$.

Step 1: Solve for the Reactions at the Supports

The given truss structure is symmetric, supported at two points:

• A hinge support at $A$, which provides reactions $A_x$ and $A_y$,
• A roller support at $E$, providing reaction $E_y$.

Loads: A vertical load of $42,000 \, \text{N}$ at joints $B$, $C$, and $D$.

Equilibrium equations:

1. Sum of Forces in $x$-direction: $\Sigma F_x = 0$ $A_x = 0 \quad (\text{since there is no horizontal load applied})$

2. Sum of Forces in $y$-direction: $\Sigma F_y = 0$ $A_y + E_y = 42,000 \times 3 = 126,000 \, \text{N}$

3. Sum of Moments about $A$: $\Sigma M_A = 0$ Taking moment about point $A$: $-42,000 \times 2 \, \text{m} - 42,000 \times 4 \, \text{m} - 42,000 \times 6 \, \text{m} + E_y \times 8 \, \text{m} = 0$ $E_y = \frac{42,000 \times (2 + 4 + 6)}{8} = \frac{42,000 \times 12}{8} = 63,000 \, \text{N}$ Now using $A_y + E_y = 126,000 \, \text{N}$: $A_y = 126,000 - 63,000 = 63,000 \, \text{N}$

Thus, the reactions are:

• $A_x = 0$,
• $A_y = 63,000 \, \text{N}$,
• $E_y = 63,000 \, \text{N}$.

Step 2: Joint Method to Solve for Member Forces

We are tasked with determining the forces in members $AB$ and $DE$. This can be done using the joint method by analyzing the forces at each joint (B and D) and