A beam is 7,5 m long. It is supported at the left end and has another support 1,5 m from the right end. From the left end, a uniformly distributed load of 60 N/m rests on the beam over a length of 3 m. Another distributed load of 20 N/m rests 3 m from the left end of the beam over a length of 4,5 m. Another support is positioned 1,5 m from the right-side end of the beam. Perform the following tasks in your groups and present them on the board to the class:
Calculate the reactions at the supports of the beam.
To calculate the reactions at the supports of the beam, we can start by drawing the free body diagram of the beam.
At the left support, there will be a vertical reaction force (R1) and a horizontal reaction force (H1). At the right support, there will be only a vertical reaction force (R2).
Next, we can write the equilibrium equations for the vertical and horizontal forces.
Vertical equilibrium:
∑Fy = R1 + R2 - (60 N/m * 3 m) - (20 N/m * 4.5 m) = 0
Horizontal equilibrium:
∑Fx = H1 = 0
Solving these equations, we can find the reactions at the supports.
From the vertical equilibrium equation:
R1 + R2 = (60 N/m * 3 m) + (20 N/m * 4.5 m)
R1 + R2 = 180 N + 90 N
R1 + R2 = 270 N
From the horizontal equilibrium equation:
H1 = 0
Finally, to find the individual reactions at each support, we can use the fact that the sum of the moments about any point on the beam is equal to zero.
Taking moments about the right support:
∑M = (R1 * 7.5 m) - (60 N/m * 3 m * 1.5 m) - (20 N/m * 4.5 m * 1.5 m) = 0
Simplifying:
7.5R1 - 270 Nm - 135 Nm = 0
7.5R1 = 405 Nm
R1 = 405 Nm / 7.5 m
R = 54 N
From the equation R1 + R2 = 270 N, we can substitute the value of R1 to solve for R2:
54 N + R2 = 270 N
R2 = 216 N
Therefore, the reactions at the supports of the beam are:
Left support: R1 = 54 N (vertical) and H1 = 0 (horizontal)
Right support: R2 = 216 N (vertical)