The problem is to determine the reactions at supports A and D for a simply supported beam subjected to two point loads. The beam has a length of $1.5 + 3 + 1.5 = 6$ m. There are two point loads of $10$ kN each, applied at points B and C, which are located at $1.5$ m and $4.5$ m from support A, respectively. The given reactions at the supports are $R_1 = 10$ kN and $R_2 = 10$ kN. We will verify if these are correct.

1. Problem Description

The problem is to determine the reactions at supports A and D for a simply supported beam subjected to two point loads. The beam has a length of $1.5 + 3 + 1.5 = 6$ m. There are two point loads of $10$ kN each, applied at points B and C, which are located at $1.5$ m and $4.5$ m from support A, respectively. The given reactions at the supports are $R_1 = 10$ kN and $R_2 = 10$ kN. We will verify if these are correct.

2. Solution Steps

Let $R_1$ be the vertical reaction at support A and $R_2$ be the vertical reaction at support D. The vertical loads are $10$ kN downwards at B and $10$ kN downwards at C.

For vertical equilibrium, the sum of the vertical forces must be zero.

$R_1 + R_2 - 10 \text{ kN} - 10 \text{ kN} = 0$

$R_1 + R_2 = 20 \text{ kN}$

To determine the reactions $R_1$ and $R_2$, we can take the moment about point A.

$\sum M_A = 0$

$R_2 \times 6 \text{ m} - 10 \text{ kN} \times 1.5 \text{ m} - 10 \text{ kN} \times 4.5 \text{ m} = 0$

$6 R_2 = 15 + 45 = 60$

$R_2 = \frac{60}{6} = 10 \text{ kN}$

Substituting $R_2$ into the vertical equilibrium equation:

$R_1 + 10 \text{ kN} = 20 \text{ kN}$

$R_1 = 10 \text{ kN}$

Thus, $R_1 = 10 \text{ kN}$ and $R_2 = 10 \text{ kN}$.

3. Final Answer

$R_1 = 10 \text{ kN}$

$R_2 = 10 \text{ kN}$