Determine the reactions at a and e. 4 kn/m 2 m 1,5 m-? 1.5 m 5 kn i

General
1

determine the reactions at a and e. 4 kn/m 2 m 1,5 m-??? 1.5 m 5 kn i

2

Determine the reactions at A and E

Answer:
To determine the reactions at points A and E for a beam subjected to the given loads, we can follow these steps:

1. Define the System and Apply Loads

Assume we have a simply supported beam with the following details:

  • Lengths of segments are: (2 \text{ m}), (1.5 \text{ m}), and (1.5 \text{ m}).
  • A uniformly distributed load (UDL) of (4 \text{ kN/m}) over (2 \text{ m}).
  • A point load of (5 \text{kN}) at the center section.
  • Supports at points A and E.

2. Calculate the Total Load Due to the UDL

The UDL of (4 \text{kN/m}) over (2 \text{ m}) creates a total load of:

F_{\text{UDL}} = 4 \text{ kN/m} \times 2 \text{ m} = 8 \text{ kN}

The UDL can be converted into a point load acting at the midpoint of the (2 \text{ m}) segment (i.e., (1 \text{ m}) from point A).

3. Define the System with Loads and Supports

Let’s represent the beam with point A as the left-end support and point E as the right-end support.
The distances from point A are:

  • UDL midpoint: (1 \text{ m})
  • Point load of (5 \text{kN}): (2 + 1.5 = 3.5 \text{ m})

4. Calculate Reactions Using Equilibrium Equations

To solve for the reactions, we use the equilibrium equations for statics:

  1. Sum of vertical forces equals zero:

    \Sigma F_y = 0 : A_y + E_y - 8 \text{ kN} - 5 \text{ kN} = 0
    A_y + E_y = 13 \text{ kN}

  2. Sum of moments about point A equals zero:

    \Sigma M_A = 0 : - (8 \text{ kN} \times 1 \text{ m}) - (5 \text{ kN} \times 3.5 \text{ m}) + E_y \times 5 \text{ m} = 0
    -8 \text{ kN}\cdot\text{m} - 17.5 \text{ kN}\cdot\text{m} + 5 \text{m} \cdot E_y = 0
    5 E_y = 25.5 \text{ kN}\cdot\text{m}
    E_y = \frac{25.5}{5} \text{ kN} = 5.1 \text{ kN}

  3. Using the vertical force balance equation to solve for (A_y):

    A_y + 5.1 \text{ kN} = 13 \text{ kN}
    A_y = 13 \text{ kN} - 5.1 \text{ kN} = 7.9 \text{ kN}

Final Answer:

  • The reaction at point A, A_y, is 7.9 \text{kN} upward.
  • The reaction at point E,E_y is 5.1 \text{kN} upward.

Thus, the reactions at A and E are \boxed{7.9 \text{kN}} and \boxed{5.1 \text{kN}} , respectively.