A laterally restrained beam 10m long that is simply supported at both ends support a dead uniformly distributed load of 25.2KN/m and and imposed distributed load of 12.8KN/m. The beam also carries a dead load of 33.6KN at distance 3m from both ends. Determine support reactions and draw shear force and bending moment diagram
To determine the support reactions, we can start by calculating the total load on the beam.
Dead load = 25.2 kN/m * 10 m = 252 kN
Imposed load = 12.8 kN/m * 10 m = 128 kN
Total load = dead load + imposed load = 252 kN + 128 kN = 380 kN
To balance the total load on a simply supported beam, the reactions at the two supports must be equal. Therefore, each support reaction will be 380 kN / 2 = 190 kN.
Next, let's plot the uniformly distributed load, imposed distributed load, and the dead load on the beam.
3 m 4 m 3 m
|---------|---------|---------|---------|
The dead load of 33.6 kN is located 3 m from each end. This load can be considered as a point load acting at the midpoint of each 3 m span. Therefore, each point load will be 33.6 kN / 2 = 16.8 kN.
Applying the point loads, we can now plot the shear force and bending moment diagrams. We'll start from the left end and move towards the right.
At the left support (A):
- There is a reaction of 190 kN pointing upwards.
- There is no horizontal force acting on the beam (since it is laterally restrained).
- The shear force at A is 190 kN pointing upwards.
- The bending moment at A is 0.
Moving towards the right:
- Between the left support (A) and the first point load of 16.8 kN, there is no external force acting on the beam. Hence, the shear force remains constant at 190 kN.
- The bending moment increases linearly from 0 at A to a value of -16.8 kNm (clockwise) just before the point load.
At the first point load of 16.8 kN:
- The shear force suddenly jumps to 190 kN - 16.8 kN = 173.2 kN (pointing downwards).
- The bending moment at this point is -16.8 kNm (clockwise).
Moving further right:
- Between the first point load and the second point load, there is no external force acting on the beam. So, the shear force remains constant at 173.2 kN.
- The bending moment decreases linearly from -16.8 kNm (clockwise) to -50.4 kNm (clockwise) just before the second point load.
At the second point load of 16.8 kN:
- The shear force suddenly jumps to 173.2 kN - 16.8 kN = 156.4 kN (pointing downwards).
- The bending moment at this point is -50.4 kNm (clockwise).
Moving further right:
- Between the second point load and the right support (B), there is no external force acting on the beam. Hence, the shear force remains constant at 156.4 kN.
- The bending moment decreases linearly from -50.4 kNm (clockwise) to -190 kNm (clockwise) just before the right support.
At the right support (B):
- There is a reaction of 190 kN pointing upwards.
- There is no horizontal force acting on the beam (since it is laterally restrained).
- The shear force at B is 156.4 kN pointing downwards.
- The bending moment at B is -190 kNm (clockwise).
To summarize:
Shear force diagram: (from left to right)
A: 190 kN (upwards)
First point load: 173.2 kN (downwards)
Second point load: 156.4 kN (downwards)
B: 156.4 kN (downwards)
Bending moment diagram: (from left to right)
A to first point load: 0 to -16.8 kNm (clockwise)
First point load: -16.8 kNm (clockwise)
First point load to second point load: -16.8 kNm (clockwise) to -50.4 kNm (clockwise)
Second point load: -50.4 kNm (clockwise)
Second point load to B: -50.4 kNm (clockwise) to -190 kNm (clockwise)
Note: The negative sign indicates a clockwise bending moment.
To solve this problem, we can start by calculating the support reactions at both ends of the beam. We'll then use the support reactions to determine the shear force and bending moment at various points along the beam.
Step 1: Calculate the reaction forces
Let's label the left support as A and the right support as B.
The total dead load acting on the beam is 25.2 kN/m + 33.6 kN = 58.8 kN.
The total imposed load acting on the beam is 12.8 kN/m.
Since the beam is simply supported, the reaction forces at supports A and B must balance these loads.
Let's calculate the reaction force at support A:
RA + RB = 58.8 kN + (10 m * 12.8 kN/m) - (33.6 KN * 2)
RA + RB = 58.8 kN + 128 kN - 67.2 kN
RA + RB = 119.6 kN
Since the beam is symmetrical, the reaction forces at both supports will be equal, so we can write:
2RA = 119.6 kN
RA = 59.8 kN
Now, RB can be calculated:
RB = 119.6 kN - 59.8 kN
RB = 59.8 kN
So, the support reactions are RA = 59.8 kN and RB = 59.8 kN.
Step 2: Draw the shear force diagram
To draw the shear force diagram, we need to consider the loads acting on the beam and the support reactions at each point.
At point A (left end), the shear force will be equal to the support reaction RA, which is 59.8 kN.
Moving along the beam to the right, the shear force changes due to the imposed load and the dead load. At any point where there is an imposed load, the shear force will decrease by the magnitude of the imposed load (12.8 kN/m). At any point where there is a dead load, the shear force will decrease by the magnitude of the dead load (33.6 kN).
So, the shear force diagram can be represented as follows:
Shear force (kN):
A --------------------- B
59.8 ↓ -12.8 ↓ -12.8 ↓ -33.6 ↓ -33.6 ↓
Step 3: Draw the bending moment diagram
To draw the bending moment diagram, we'll start from the left support (A) and consider the loads and support reactions acting on the beam.
At point A, the bending moment will be zero.
Moving along the beam to the right, the bending moment will change due to the loads and support reactions. At any point where there is an imposed load, the bending moment will decrease linearly. At any point where there is a dead load, the bending moment will decrease quadratically.
So, the bending moment diagram can be represented as follows:
Bending moment (kNm):
A --------------------- B
0 / ///////// ↓ -12.8x / //////// -33.6(x-3) //////// ↓
Note: The //////// indicates a quadratic decrease in the bending moment.
This completes the solution.