Question

A beam ABCDE, with A on the left is 7 m long and simply supported at B and E.
The lengths of the various portions are as follows: AB = 1,5 m, BC = 1,5m ,
CD = 1m and DE = 3m. There is a uniformly distributed load of 15kN /m between
B and 2m to the right of B. Concentrated loads of 20 kN act at A and D with one of 50 kN at C.
1.1.1 Make a neat, labelled drawing of the beam described above.
1.1.2 Calculate the reaction forces at the supports by taking moments.

Answer 1

Let's break down the problem step by step.

### 1.1.1 Draw the Beam

1. Draw a horizontal line representing the beam.
2. Label the supports at points B and E (these will be the locations where the beam is supported).
3. Mark lengths along the beam according to the given distances:
- From A to B is 1.5 m
- From B to C is 1.5 m
- From C to D is 1 m
- From D to E is 3 m

4. Add the following loads:
- A concentrated load of 20 kN acting downward at point A (1.5 m from the left).
- A concentrated load of 50 kN acting downward at point C (3 m from the left).
- A concentrated load of 20 kN acting downward at point D (6 m from the left).
- A uniformly distributed load (UDL) of 15 kN/m acting from point B (1.5 m) to a point 2 m to the right of B (i.e., from B to a point located at 3.5 m from A).

The diagram would look something like this:

```
A B C D E
|------|----------|--------------|----------|
20kN | UDL | 50 kN | 20 kN |
|<----1.5m-->|<---1m--->|<---3m--->| (total length = 7m)
|<---3.5m--->|
```

### 1.1.2 Calculate the Reaction Forces

To find the reaction forces at supports B and E (RB and RE), we will use the equations of equilibrium, specifically taking moments about E.

#### Load Calculations

**Step 1: Calculate the total loads.**
1. UDL load between B and point 3.5 m from A is 1.5 m long:
- Total UDL = \(15 \, \text{kN/m} \times 1.5 \, \text{m} = 22.5 \, \text{kN}\)

2. Concentrated loads:
- 20 kN at A
- 50 kN at C (3.0 m from A)
- 20 kN at D (6.0 m from A)

**Total Load (W):**
\[
W = 20 \, \text{kN} + 22.5 \, \text{kN} + 50 \, \text{kN} + 20 \, \text{kN} = 112.5 \, \text{kN}
\]

#### Moment Calculation about E

Let \( R_B \) = Reaction at Support B
Let \( R_E \) = Reaction at Support E

Taking moments about E (counter-clockwise positive), we have:

\[
R_B \times 7 \, \text{m} - 20 \, \text{kN} \times 6.0 \, \text{m} - 50 \, \text{kN} \times 3.0 \, \text{m} - 20 \, \text{kN} \times 1.0 \, \text{m} - 22.5 \, \text{kN} \times 2.5 \, \text{m} = 0
\]

Calculating moments:
\[
R_B \times 7 \, \text{m} - (120 + 150 + 20 + 56.25) = 0
\]
\[
R_B \times 7 \, \text{m} - 346.25 = 0
\]
\[
R_B \times 7 = 346.25
\]
\[
R_B = \frac{346.25}{7} = 49.464 \, \text{kN} \, \text{(approx. 49.46 kN)}
\]

#### Summation of Vertical Forces

Using vertical force equilibrium (\( R_B + R_E = W \)):
\[
R_B + R_E = 112.5 \, \text{kN}
\]

Substituting \( R_B \):
\[
49.464 + R_E = 112.5
\]
\[
R_E = 112.5 - 49.464 = 63.036 \, \text{kN} \, \text{(approx. 63.04 kN)}
\]

### Summary of Reactions
- Reaction at support B (\( R_B \)): **49.46 kN (up)**
- Reaction at support E (\( R_E \)): **63.04 kN (up)**