QUESTION 1
STATICS
1.1 A beam ABCDE, with on the left-hand side, is 10 m long and simply supported at and . Point loads of 50 N and 30 N are at and respectively. and . An evenly distributed load of is between and .
1.1.1 First draw the beam, then calculate the reaction forces of and .
(5)
1.1.2 Calculate the bending moments at and and at a point halfway between and .
(3)
1.1,3 Draw the bending moment and shearing force diagrams with the main values indicated on the diagram.
(7)

Ask by Thornton Dawson. in South Africa

Mar 15,2025

Question

QUESTION 1
STATICS
1.1 A beam ABCDE, with on the left-hand side, is 10 m long and simply supported at and . Point loads of 50 N and 30 N are at and respectively. and . An evenly distributed load of is between and .
1.1.1 First draw the beam, then calculate the reaction forces of and .
(5)
1.1.2 Calculate the bending moments at and and at a point halfway between and .
(3)
1.1,3 Draw the bending moment and shearing force diagrams with the main values indicated on the diagram.
(7)

Ask by Thornton Dawson. in South Africa

Mar 15,2025

UpStudy AI · Accepted Answer

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:
  - $$ F_B = 50 \, \text{N} $$ at $$ B $$ (2 m from $$ A $$)
  - $$ F_E = 30 \, \text{N} $$ 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 $$ D $$ (which is 6 m long)

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

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

2. **Set up the equilibrium equations:**
   - Sum of vertical forces: $$ R_A + R_D - W = 0 $$
   - Taking moments about point $$ A $$:
     $$
     \sum M_A = 0 \Rightarrow R_D \times 8 - 50 \times 2 - 30 \times 8 - 12 \times 3 = 0
     $$

3. **Calculate $$ R_D $$:**
   - $$ R_D \times 8 = 100 + 240 + 36 $$
   - $$ R_D \times 8 = 376 $$
   - $$ R_D = \frac{376}{8} = 47 \, \text{N} $$

4. **Calculate $$ R_A $$:**
   - From the vertical forces equation:
   $$
   R_A + 47 - 92 = 0 \Rightarrow R_A = 92 - 47 = 45 \, \text{N}
   $$

### Step 2: Calculate the Bending Moments at Points $$ B $$, $$ D $$, and $$ C $$

1. **Bending Moment at $$ B $$:**
   - Moment at $$ B $$ due to $$ R_A $$ and $$ F_E $$:
   $$
   M_B = R_A \times 2 - F_E \times (8 - 2) = 45 \times 2 - 30 \times 6
   $$
   $$
   M_B = 90 - 180 = -90 \, \text{N m}
   $$

2. **Bending Moment at $$ D $$:**
   - Moment at $$ D $$ due to $$ R_A $$ and $$ F_B $$:
   $$
   M_D = R_A \times 6 - F_B \times (6) - \text{UDL moment}
   $$
   - UDL moment (acting at the center of the UDL):
   $$
   \text{UDL moment} = 12 \times 3 = 36 \, \text{N m}
   $$
   $$
   M_D = 45 \times 6 - 50 \times 4 - 36 = 270 - 200 - 36 = 34 \, \text{N m}
   $$

3. **Bending Moment at $$ C $$ (halfway between $$ B $$ and $$ D $$):**
   - Distance from $$ A $$ to $$ C $$ is $$ 4 \, \text{m} $$:
   $$
   M_C = R_A \times 4 - F_B \times (4 - 2) - \text{UDL moment}
   $$
   $$
   M_C = 45 \times 4 - 50 \times 2 - 12 \times 2 = 180 - 100 - 24 = 56 \, \text{N m}
   $$

### Step 3: Draw the Bending Moment and Shearing Force Diagrams

1. **Shear Force Diagram:**
   - From $$ A $$ to $$ B $$: $$ V = R_A = 45 \, \text{N} $$
   - At $$ B $$: $$ V = 45 - 50 = -5 \, \text{N} $$
   - From $$ B $$ to $$ D $$: $$ V = -5 \, \text{N} $$ (constant)
   - At $$ D $$: $$ V = -5 + 30 = 25 \, \text{N} $$
   - From $$ D $$ to $$ E $$: $$ V = 25 - 12 = 13 \, \text{N} $$
   - At $$ E $$: $$ V = 13 - 30 = -17 \, \text{N} $$

2. **Bending Moment Diagram:**
   - At $$ A $$: $$ M = 0 $$
   - At $$ B $$: $$ M = -90 \, \text{N m} $$
   - At $$ C $$: $$ M = 56 \, \text{N m} $$
   - At $$ D $$: $$ M = 34 \, \text{N m} $$
   - At $$ E $$: $$ M = 0 $$

### Summary of Results:
- Reaction Forces: $$ R_A = 45 \, \text{N}, R_D = 47 \, \text{N} $$
- Bending Moments: $$ M_B = -90 \, \text{N m}, M_C = 56 \, \text{N m}, M_D = 34 \, \text{N m} $$

### Diagrams:
- The diagrams can be drawn based on the calculated values, indicating the shear forces and bending moments at the respective points along the beam.

If you need further assistance with the diagrams or any specific part, please let me know!
**Reaction Forces:** - $$ R_A = 45 \, \text{N} $$ - $$ R_D = 47 \, \text{N} $$ **Bending Moments:** - At $$ B $$: $$ -90 \, \text{N m} $$ - At $$ C $$: $$ 56 \, \text{N m} $$ - At $$ D $$: $$ 34 \, \text{N m} $$ **Shear Force Diagram:** - From $$ A $$ to $$ B $$: $$ 45 \, \text{N} $$ - At $$ B $$: $$ -5 \, \text{N} $$ - From $$ B $$ to $$ D $$: $$ -5 \, \text{N} $$ - At $$ D $$: $$ 25 \, \text{N} $$ - From $$ D $$ to $$ E $$: $$ 13 \, \text{N} $$ - At $$ E $$: $$ -17 \, \text{N} $$ **Bending Moment Diagram:** - At $$ A $$: $$ 0 \, \text{N m} $$ - At $$ B $$: $$ -90 \, \text{N m} $$ - At $$ C $$: $$ 56 \, \text{N m} $$ - At $$ D $$: $$ 34 \, \text{N m} $$ - At $$ E $$: $$ 0 \, \text{N m} $$ *Diagrams can be drawn based on these values to visualize the shear forces and bending moments along the beam.*

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