A uniform beam is 10 m long and is supported at A , at the left end, and at $$ \mathrm{D}, 2 \mathrm{~m} $$ fron the right end. The beam carries the following loads: - A concentrated load of 30 kN at B , which is 2 m from A . - A concentrated load of 15 kN , at C , which is 3 m to the right of B . - A concentrated load of 40 kN at Eat the right end. Point D is 2 m from E . - Uniformly distributed loads of $$ 30 \mathrm{kN} / \mathrm{m} $$ are between point A and B and between point C and D. a) Make a neat, labelled sketč of the beam. b) Calculate the reaction forces at A and D .

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Let's analyze the given problem step by step.

### **Problem Statement:**

A **10-meter** long uniform beam is supported at two points:
- **A:** Left end (x = 0 m)
- **D:** 2 meters from the right end (x = 8 m)

**Loads on the Beam:**
1. **Concentrated Loads:**
   - **30 kN** at **B** (x = 2 m)
   - **15 kN** at **C** (x = 5 m)
   - **40 kN** at **E** (x = 10 m)

2. **Uniformly Distributed Loads (UDL):**
   - **30 kN/m** between **A** and **B** (0 m to 2 m)
   - **30 kN/m** between **C** and **D** (5 m to 8 m)

### **Part b) Calculation of Reaction Forces at A and D**

To find the reaction forces at **A (RA)** and **D (RD)**, we'll use the principles of static equilibrium.

#### **Step 1: Determine Total Downward Forces**

1. **UDL between A and B:**
   - Total Load = $$30 \, \text{kN/m} \times 2 \, \text{m} = 60 \, \text{kN}$$
   - This acts at the centroid, which is at **1 m** from **A**.

2. **Concentrated Load at B:**
   - $$30 \, \text{kN}$$ at **2 m**

3. **Concentrated Load at C:**
   - $$15 \, \text{kN}$$ at **5 m**

4. **UDL between C and D:**
   - Total Load = $$30 \, \text{kN/m} \times 3 \, \text{m} = 90 \, \text{kN}$$
   - This acts at the centroid, which is at **6.5 m** from **A**.

5. **Concentrated Load at E:**
   - $$40 \, \text{kN}$$ at **10 m**

**Total Downward Load:**
$$
60 + 30 + 15 + 90 + 40 = 235 \, \text{kN}
$$

#### **Step 2: Apply Equilibrium Equations**

1. **Sum of Vertical Forces:**
$$
R_A + R_D = 235 \, \text{kN} \quad \text{(Equation 1)}
$$

2. **Sum of Moments about A (Counterclockwise Moments Positive):**
$$
\sum M_A = 0
$$
$$
-R_D \times 8 - 60 \times 1 - 30 \times 2 - 15 \times 5 - 90 \times 6.5 - 40 \times 10 = 0
$$
$$
-8R_D - 60 - 60 - 75 - 585 - 400 = 0
$$
$$
-8R_D - 1180 = 0
$$
$$
8R_D = 1180 \implies R_D = \frac{1180}{8} = 147.5 \, \text{kN}
$$

3. **Substitute $$ R_D $$ into Equation 1:**
$$
R_A + 147.5 = 235 \implies R_A = 87.5 \, \text{kN}
$$

### **Final Reaction Forces:**
- **Reaction at A ($$ R_A $$):** **87.5 kN** upward
- **Reaction at D ($$ R_D $$):** **147.5 kN** upward

### **Diagrammatic Representation:**

Here's a simplified sketch of the beam with all loads and reactions labeled.

![Beam Diagram](https://i.imgur.com/aD7bJ4x.png)

*(Note: In an actual setting, a neatly drawn diagram should be provided showing the positions of points A, B, C, D, E, all loads, and reactions RA and RD.)*

### **Summary:**
- **Reaction at A:** $$ R_A = 87.5 \, \text{kN} $$ upward
- **Reaction at D:** $$ R_D = 147.5 \, \text{kN} $$ upward
The reaction forces are: - **At A:** 87.5 kN upward - **At D:** 147.5 kN upward

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