STATICS QUESTION 1 $$ \begin{array}{l}1.1 ext { A beam } A B C D E ext {, with } A ext { on the left-hand side, is } 10 \mathrm{~m} ext { long and simply supported at } \ A ext { and } D ext {. Point loads of } 50 \mathrm{~N} ext { and } 30 \mathrm{~N} ext { are at } B ext { and } E ext { respectively. } A B=2 \mathrm{~m} ext { and } \ D E=2 \mathrm{~m} ext {. } \ ext { An evenly distributed load of } 2 \mathrm{~N} / \mathrm{m} ext { is between } A ext { and } D ext {. } \ \begin{array}{l}1.1 .1 ext { First draw the beam, then calculate the reaction forces of } A ext { and } D ext {. } \ 1.1 .2 ext { Calculate the bending moments at } B ext { and } D ext { and at a point } C ext { halfway between } \ ext { (5 and } D ext {. } \ ext { 1.1.3 Draw the bending moment and shearing force diagrams with the main values } \ ext { indicated on the diagram. }\end{array}\end{array} $$ (7)

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STATICS QUESTION 1 $$ \begin{array}{l}1.1 	ext { A beam } A B C D E 	ext {, with } A 	ext { on the left-hand side, is } 10 \mathrm{~m} 	ext { long and simply supported at } \ A 	ext { and } D 	ext {. Point loads of } 50 \mathrm{~N} 	ext { and } 30 \mathrm{~N} 	ext { are at } B 	ext { and } E 	ext { respectively. } A B=2 \mathrm{~m} 	ext { and } \ D E=2 \mathrm{~m} 	ext {. } \ 	ext { An evenly distributed load of } 2 \mathrm{~N} / \mathrm{m} 	ext { is between } A 	ext { and } D 	ext {. } \ \begin{array}{l}1.1 .1 	ext { First draw the beam, then calculate the reaction forces of } A 	ext { and } D 	ext {. } \ 1.1 .2 	ext { Calculate the bending moments at } B 	ext { and } D 	ext { and at a point } C 	ext { halfway between } \ 	ext { (5 and } D 	ext {. } \ 	ext { 1.1.3 Draw the bending moment and shearing force diagrams with the main values } \ 	ext { indicated on the diagram. }\end{array}\end{array} $$ (7)

UpStudy AI · Accepted Answer

To solve the problem step by step, we will follow the instructions provided in the question.

### Step 1: Draw the Beam
The beam $$ ABCDE $$ is 10 m long, simply supported at points $$ A $$ and $$ D $$. The loads are as follows:
- A point load of $$ 50 \, \text{N} $$ at point $$ B $$ (2 m from $$ A $$).
- A point load of $$ 30 \, \text{N} $$ at point $$ E $$ (2 m from $$ D $$).
- An evenly distributed load of $$ 2 \, \text{N/m} $$ between points $$ A $$ and $$ D $$ (which is 6 m long).

### Step 2: Calculate the Reaction Forces at Supports $$ A $$ and $$ D $$

#### Known Conditions:
- Length of beam $$ L = 10 \, \text{m} $$
- Distance $$ AB = 2 \, \text{m} $$
- Distance $$ DE = 2 \, \text{m} $$
- Distance $$ AD = 6 \, \text{m} $$ (between $$ A $$ and $$ D $$)
- Point load at $$ B = 50 \, \text{N} $$
- Point load at $$ E = 30 \, \text{N} $$
- Distributed load $$ w = 2 \, \text{N/m} $$ over $$ AD $$

#### Total Load Calculation:
1. Total load from the distributed load:
   $$
   W_d = w \times L_{AD} = 2 \, \text{N/m} \times 6 \, \text{m} = 12 \, \text{N}
   $$
2. Total vertical load on the beam:
   $$
   W_{total} = 50 \, \text{N} + 30 \, \text{N} + 12 \, \text{N} = 92 \, \text{N}
   $$

#### Moment about Point $$ A $$:
Taking moments about point $$ A $$ to find the reaction at $$ D $$ ($$ R_D $$):
$$
\sum M_A = 0 \Rightarrow R_D \cdot 10 - 50 \cdot 2 - 30 \cdot 8 - 12 \cdot 3 = 0
$$
Calculating the moment contributions:
- Moment due to $$ 50 \, \text{N} $$ at $$ B $$: $$ 50 \cdot 2 = 100 \, \text{Nm} $$
- Moment due to $$ 30 \, \text{N} $$ at $$ E $$: $$ 30 \cdot 8 = 240 \, \text{Nm} $$
- Moment due to distributed load (acting at midpoint of $$ AD $$): $$ 12 \cdot 3 = 36 \, \text{Nm} $$

Substituting these values:
$$
R_D \cdot 10 - 100 - 240 - 36 = 0
$$
$$
R_D \cdot 10 = 376 \Rightarrow R_D = \frac{376}{10} = 37.6 \, \text{N}
$$

#### Vertical Force Equilibrium:
Using vertical force equilibrium to find $$ R_A $$:
$$
\sum F_y = 0 \Rightarrow R_A + R_D - W_{total} = 0
$$
Substituting $$ R_D $$:
$$
R_A + 37.6 - 92 = 0 \Rightarrow R_A = 92 - 37.6 = 54.4 \, \text{N}
$$

### Summary of Reaction Forces:
- $$ R_A = 54.4 \, \text{N} $$
- $$ R_D = 37.6 \, \text{N} $$

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

#### Bending Moment at Point $$ B $$:
$$
M_B = R_A \cdot AB - 50 \cdot 0 = 54.4 \cdot 2 = 108.8 \, \text{Nm}
$$

#### Bending Moment at Point $$ C $$ (5 m from $$ A $$):
$$
M_C = R_A \cdot AC - 50 \cdot (AC - AB) - 12 \cdot \left(\frac{AC}{2}\right)
$$
Where $$ AC = 5 \, \text{m} $$:
$$
M_C = 54.4 \cdot 5 - 50 \cdot 3 - 12 \cdot 2.5
$$
Calculating:
$$
M_C = 272 - 150 - 30 = 92 \, \text{Nm}
$$

#### Bending Moment at Point $$ D $$:
$$
M_D = 0 \, \text{Nm} \quad (\text{as it is a support})
$$

### Summary of Bending Moments:
- $$ M_B = 108.8 \, \text{Nm} $$
- $$ M_C = 92 \, \text{Nm} $$
- $$ M_D = 0 \, \text{Nm} $$

### Step 4: Draw Bending Moment and Shearing Force Diagrams
1. **Shear Force Diagram**:
   - From $$ A $$ to $$ B $$: $$ +54.4 \, \text{N} $$
   - At $$ B $$: $$ +54.4 - 50 = +4.4 \, \text{N} $$
   - From $$ B $$ to $$ C $$: $$ +4.4 \, \text{N} $$
   - At $$ C $$: $$ +4.4 - 12 = -7.6 \, \text{N} $$
   - From $$ C $$ to $$ D $$: $$ -7.6 \, \text{N} $$
   - At $$ D $$: $$ -7.6 + 37.6 = 0 $$

2. **Bending Moment Diagram**:
   - At $$ A $$: $$ 0 $$
   - At $$ B $$: $$ 108.8 \, \text{Nm} $$
   - At $$ C $$: $$ 92 \, \text{Nm} $$
   - At $$ D $$: $$ 0 $$

### Final Diagrams
- **Shear Force Diagram**: Starts at $$ +54.4 \, \text{N} $$, drops to $$ +4.4 \, \text{N} $$ at $$ B $$, drops to $$ -7.6 \, \text{N} $$ at $$ C $$, and returns to $$ 0 $$ at $$ D \
**Reaction Forces:** - $$ R_A = 54.4 \, \text{N} $$ - $$ R_D = 37.6 \, \text{N} $$ **Bending Moments:** - $$ M_B = 108.8 \, \text{Nm} $$ - $$ M_C = 92 \, \text{Nm} $$ - $$ M_D = 0 \, \text{Nm} $$ **Shear Force Diagram:** - From $$ A $$ to $$ B $$: $$ +54.4 \, \text{N} $$ - At $$ B $$: $$ +4.4 \, \text{N} $$ - From $$ B $$ to $$ C $$: $$ +4.4 \, \text{N} $$ - At $$ C $$: $$ -7.6 \, \text{N} $$ - From $$ C $$ to $$ D $$: $$ -7.6 \, \text{N} $$ - At $$ D $$: $$ 0 \, \text{N} $$ **Bending Moment Diagram:** - At $$ A $$: $$ 0 \, \text{Nm} $$ - At $$ B $$: $$ 108.8 \, \text{Nm} $$ - At $$ C $$: $$ 92 \, \text{Nm} $$ - At $$ D $$: $$ 0 \, \text{Nm} $$

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