QUESTION 1 STATICS 1.1 A bearn $$ A B C D E $$, with $$ A $$ on the left-hand side, is 10 m long and simply supported at $$ A $$ and $$ D $$. Point loads of 50 N and 30 N are at $$ B $$ and $$ E $$ respectively. $$ A B=2 \mathrm{~m} $$ and $$ D E=2 \mathrm{~m} $$ An evenly distributed load of $$ 2 \mathrm{~N} / \mathrm{m} $$ is between $$ A $$ and $$ D $$. 1.1.1 First draw the beam, then calculate the reaction forces of $$ A $$ and $$ D $$. (5) 1.1.2 Calculate the bending moments at $$ B $$ and $$ D $$ and at a point $$ C $$ halfway between B and D. 1.1.3 Draw the bending moment and shearing force diagrams with the main values indicated on the diagram. $$ 1.2 \quad y $$-axis

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To solve the problem step by step, we will first summarize the known conditions and then proceed with the calculations.

### Known Conditions:
- Length of the beam $$ L = 10 \, \text{m} $$
- Point loads:
  - $$ P_B = 50 \, \text{N} $$ at $$ B $$ (2 m from $$ A $$)
  - $$ P_E = 30 \, \text{N} $$ at $$ E $$ (8 m from $$ A $$)
- Distance from $$ A $$ to $$ B $$: $$ A B = 2 \, \text{m} $$
- Distance from $$ D $$ to $$ E $$: $$ D E = 2 \, \text{m} $$
- Uniformly distributed load (UDL): $$ w = 2 \, \text{N/m} $$ between $$ A $$ and $$ D $$

### Step 1: Calculate Reaction Forces at Supports $$ A $$ and $$ D $$

1. **Calculate the total load on the beam:**
   - Total point loads: $$ P_B + P_E = 50 \, \text{N} + 30 \, \text{N} = 80 \, \text{N} $$
   - Total UDL over the length $$ A D = 6 \, \text{m} $$:
     $$
     \text{Total UDL} = w \times \text{length} = 2 \, \text{N/m} \times 6 \, \text{m} = 12 \, \text{N}
     $$
   - Total load on the beam: $$ 80 \, \text{N} + 12 \, \text{N} = 92 \, \text{N} $$

2. **Set up equilibrium equations:**
   - Sum of vertical forces:
     $$
     R_A + R_D - 92 = 0 \quad (1)
     $$
   - Taking moments about point $$ A $$:
     $$
     \sum M_A = 0 \Rightarrow R_D \cdot 10 - 50 \cdot 2 - 30 \cdot 8 - 12 \cdot 3 = 0 \quad (2)
     $$
   - Where $$ 12 \cdot 3 $$ is the moment due to the UDL acting at its centroid (3 m from $$ A $$).

### Step 2: Solve the Equations

We will solve equations (1) and (2) simultaneously to find $$ R_A $$ and $$ R_D $$.

1. From equation (1):
   $$
   R_D = 92 - R_A \quad (3)
   $$

2. Substitute equation (3) into equation (2):
   $$
   (92 - R_A) \cdot 10 - 50 \cdot 2 - 30 \cdot 8 - 12 \cdot 3 = 0
   $$
   $$
   920 - 10R_A - 100 - 240 - 36 = 0
   $$
   $$
   10R_A = 920 - 376
   $$
   $$
   10R_A = 544 \Rightarrow R_A = 54.4 \, \text{N}
   $$

3. Substitute $$ R_A $$ back into equation (3):
   $$
   R_D = 92 - 54.4 = 37.6 \, \text{N}
   $$

### Step 3: Bending Moments at Points $$ B $$, $$ D $$, and $$ C $$

1. **Bending Moment at $$ B $$:**
   - Moment at $$ B $$ due to $$ R_A $$:
   $$
   M_B = R_A \cdot 2 - 50 \cdot 0 = 54.4 \cdot 2 = 108.8 \, \text{N m}
   $$

2. **Bending Moment at $$ D $$:**
   - Moment at $$ D $$ due to $$ R_A $$ and $$ P_B $$:
   $$
   M_D = R_A \cdot 10 - 50 \cdot 2 - 30 \cdot 8 - 12 \cdot 3
   $$
   $$
   M_D = 54.4 \cdot 10 - 100 - 240 - 36 = 544 - 376 = 168 \, \text{N m}
   $$

3. **Bending Moment at $$ C $$ (halfway between $$ B $$ and $$ D $$, 4 m from $$ A $$):**
   - Moment at $$ C $$:
   $$
   M_C = R_A \cdot 4 - 50 \cdot 2 - 12 \cdot 2
   $$
   $$
   M_C = 54.4 \cdot 4 - 100 - 24 = 217.6 - 124 = 93.6 \, \text{N m}
   $$

### Summary of Results
- Reaction forces:
  - $$ R_A = 54.4 \, \text{N} $$
  - $$ R_D = 37.6 \, \text{N} $$
- Bending moments:
  - $$ M_B = 108.8 \, \text{N m} $$
  - $$ M_D = 168 \, \text{N m} $$
  - $$ M_C = 93.6 \, \text{N m} $$

### Step 4: Draw Bending Moment and Shearing Force Diagrams

1. **Shear Force Diagram:**
   - Start at $$ R_A = 54.4 \, \text{N} $$
   - Drop by $$ 50 \, \text{N} $$ at $$ B $$
   - Drop by $$ 12 \, \text{N} $$ due to UDL from $$ A $$ to $$ D $$
   - Drop by $$ 30 \, \text{N} $$ at $$ E $$
   - End at $$ R_D = 37.6 \, \text{N} $$

2. **Bending Moment Diagram:**
   - Start at $$ 0 $$ at $$ A $$
   - Increase to $$ 108.8 \, \text{N m} $$ at $$ B $$
   - Decrease to $$ 93.6 \, \text{N m} $$ at $$ C $$
   - Increase to $$ 168 \, \text{N m} $$ at $$ D $$
   - End at $$ 0 $$ at $$ E $$

### Final Note
The diagrams can be sketched based on the calculated values, ensuring to label the key points and values accurately.
**Reaction Forces:** - $$ R_A = 54.4 \, \text{N} $$ - $$ R_D = 37.6 \, \text{N} $$ **Bending Moments:** - At $$ B $$: $$ 108.8 \, \text{N m} $$ - At $$ C $$ (midpoint between $$ B $$ and $$ D $$): $$ 93.6 \, \text{N m} $$ - At $$ D $$: $$ 168 \, \text{N m} $$ **Shearing Force Diagram:** Starts at $$ 54.4 \, \text{N} $$, drops by $$ 50 \, \text{N} $$ at $$ B $$, further drops by $$ 12 \, \text{N} $$ due to UDL, drops by $$ 30 \, \text{N} $$ at $$ E $$, and ends at $$ 37.6 \, \text{N} $$. **Bending Moment Diagram:** Starts at $$ 0 $$ at $$ A $$, increases to $$ 108.8 \, \text{N m} $$ at $$ B $$, decreases to $$ 93.6 \, \text{N m} $$ at $$ C $$, increases to $$ 168 \, \text{N m} $$ at $$ D $$, and ends at $$ 0 $$ at $$ E $$.

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