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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.

1 answer

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)