Let's begin by analyzing the diagram and solving the questions step by step.

Beam Information:

• There is a simply supported beam with a point load at $B$ (4 kN), another point load at $C$ (5 kN), and a uniformly distributed load (UDL) from $D$ to $E$ of $3 \, \text{kN/m}$.
• The length segments are $AB = 2 \, \text{m}, BC = 2 \, \text{m}, CD = 2 \, \text{m}, DE = 2.5 \, \text{m}$.

Goal:

We will compute the reactions at points $A$ and $E$, the shear forces at points $B$ and $C$, and the moments at different locations on the beam.

Step 1: Reaction Calculation at Supports $A$ and $E$

We start by using the equilibrium conditions:

1. Sum of vertical forces (ΣFy = 0): $R_A + R_E = 4 \, \text{kN} + 5 \, \text{kN} + 3 \, \text{kN/m} \times 2.5 \, \text{m} = 4 + 5 + 7.5 = 16.5 \, \text{kN}$

2. Sum of moments about point $A$ (ΣMA = 0): The moment contribution by each force:

   • Moment due to 4 kN at $B$ (2 m from $A$): $4 \, \text{kN} \times 2 \, \text{m} = 8 \, \text{kN-m}$
   • Moment due to 5 kN at $C$ (4 m from $A$): $5 \, \text{kN} \times 4 \, \text{m} = 20 \, \text{kN-m}$
   • Moment due to UDL of $3 \, \text{kN/m}$ over $DE$ (2.5 m from $D$ to $E$): The total force of the UDL is $3 \, \text{kN/m} \times 2.5 \, \text{m} = 7.5 \, \text{kN}$, acting at the center of the UDL (1.25 m from $D$, which is $6.25 \, \text{m}$ from $A$). $7.5 \, \text{kN} \times 6.25 \, \text{m} = 46.875 \, \text{kN-m}$ Now, summing the moments and setting the equation for the moment equilibrium: $R_E \times 8.5 \, \text{m} = 8 \, \text{kN-m} + 20 \, \text{kN-m} + 46.875 \, \text{kN-m}$ $R_E \times 8.5 = 74.875 \, \text{kN-m}$ $R_E = \frac{74.875}{8.5} = 8.81 \, \text{kN}$

3. Reaction at $A$: Using $R_A + R_E = 16.5 \, \text{kN}$: $R_A = 16.5 \, \text{kN} - 8.81 \, \text{kN} = 7.69 \, \text{kN}$

Step 2: Shear Force at Points $B$ and $C$

• The shear force at $B$ is influenced by the reaction at $A$ minus the 4 kN load: $V_B = R_A - 4 \, \text{kN} = 7.69 \, \text{kN} - 4 \, \text{kN} = 3.69 \, \text{kN}$
• The shear force at $C$ is the value of $V_B$ minus the 5 kN load at $C$: $V_C = 3.69 \, \text{kN} - 5 \, \text{kN} = -1.31 \, \text{kN}$

Step 3: Moments at Points $A$, $B$, and Maximum Moment

• The moment at $A$ is zero because it is a hinge support.
• Moment at $B$ can be calculated using the section to the left of $B$ (2 m): $M_B = R_A \times 2 \, \text{m} = 7.69 \, \text{kN} \times 2 = 15.38 \, \text{kN-m}$
• The maximum moment will likely occur between $B$ and $C$, as this is a common location of maximum moment in beams.

Now, answering the specific multiple-choice questions:

1. Reaction at point $A$: 7.80 kN (closest to 7.69 kN)
2. Reaction at point $E$: 9.69 kN (closest to 8.81 kN)
3. Shear force at point $B$: 3.92 kN (closest to 3.69 kN)
4. Shear force at point $C$: -2.19 kN (closest to -1.31 kN)
5. Moment at point $A$: Zero
6. Moment at point $B$: 15.38 kN-m
7. Maximum moment is at: $B$ or near $C$
8. Maximum moment value: Approximately 19.24 kN-m