A beam 6 m long is simply supported at the ends,and carries a uniform distributed load of 1500kg/m (including its own weight ) and three concentrated loads of 1000,2000 and 3000 kg acting respectively at the left quarter point,centre point and right quarter point.
Draw the shear force ans bending moment diagram and determine the maximum bending moment.
get end forces up first
beam kg = 6*1500 = 9000 kg at center
forces F1 up at left and F2 up at right
moments about left to get F2
F2*6 = 1000*1.5 + 11000*3 + 3000*4.5
so
F2 = 48,000/6 = 8,000 up at right end
total mass = (1+11+3)10^3 = 15,000 kg
so
F1 = 15,000- 8000 = 7,000 up at left
shear in left 1/4 where 0<x<1.5
= 7000 - 1500x
integrate to get
BM in left 1/4 = 7000 x - 7,500 x^2
from 1.5<x<3
shear = 7,000 - 1500 x - 1000 = 6,000-1500x
BM, integrate again
7,000 x - 7,500 x^2 - 1000 (x-1.5)
= 6000 x - 7500 x^2 + 1500
check to see if BM is continuous at x = 1.5
BM at x = 1.5 from left
= 7000 (1.5) - 7,500 (2.25)
= -6375
BM at x = 1.5 from left of center 1.5<x<3
= 6000(1.5) -7500(2.25) + 1500
= 9000 - 16875 + 1500
= - 6375 whew, good
continue on to the right hald the same way
I am not sure I have designed the sign conventions the same as your class, if not you can remedy that
By the way I did all this as if a kg was a force
If you need Newtons and Newton meters, multiply final answers by 9.81
To draw the shear force and bending moment diagrams, we need to follow these steps:
1. Calculate the reactions at the supports:
- Since the beam is simply supported, the reactions at the supports are equal.
- To calculate the reactions, we need to consider the total weight of the beam plus the distributed load and the concentrated loads.
- The total distributed load is given as 1500 kg/m, so the total distributed load over the 6 m span is 1500 kg/m * 6 m = 9000 kg.
- The concentrated loads are given as 1000 kg, 2000 kg, and 3000 kg.
- Thus, the total weight is 9000 kg + 1000 kg + 2000 kg + 3000 kg = 15000 kg.
- Since the beam is simply supported at the ends, each support will carry half of the total weight, which is 15000 kg / 2 = 7500 kg.
- Therefore, the reaction at each support is 7500 kg.
2. Determine the positions of the concentrated loads on the beam:
- The beam is 6 m long.
- The left quarter point is at 1.5 m (one-fourth of 6 m).
- The center point is at 3 m (half of 6 m).
- The right quarter point is at 4.5 m (three-fourths of 6 m).
3. Draw the shear force diagram (SFD):
- Start the shear force diagram at one end of the beam and label the vertical axis as "Shear Force (SF)."
- The SFD will show the variation of shear force along the length of the beam.
- Since the beam is simply supported and the reactions are equal, the initial shear force at the left end is 7500 kg.
- At the left quarter point (1.5 m), the shear force will decrease by the magnitude of the concentrated load at that point, which is 1000 kg.
- At the center point (3 m), the shear force will further decrease by the magnitude of the concentrated load at that point, which is 2000 kg.
- At the right quarter point (4.5 m), the shear force will decrease by the magnitude of the concentrated load at that point, which is 3000 kg.
- Finally, the shear force at the right end of the beam will be zero since it is simply supported.
- Plot these points on the SFD and connect them with straight lines.
4. Draw the bending moment diagram (BMD):
- Start the bending moment diagram below the shear force diagram and label the vertical axis as "Bending Moment (BM)."
- The BMD will show the variation of bending moment along the length of the beam.
- Since the beam is simply supported, the bending moment is zero at both ends of the beam.
- At each concentrated load, the bending moment will change abruptly.
- The bending moment will be maximum between the concentrated loads.
- Calculate the bending moment at each point by considering the loads and distances from the supports.
- At the left end, the bending moment is zero.
- At the left quarter point (1.5 m), the bending moment will be equal to the reaction at the left support (7500 kg) multiplied by the distance from the support (1.5 m).
- At the center point (3 m), the bending moment will be increased by the magnitude of the concentrated load at that point (2000 kg) multiplied by the distance from the left quarter point (1.5 m).
- At the right quarter point (4.5 m), the bending moment will be increased by the magnitude of the concentrated load at that point (3000 kg) multiplied by the distance from the left quarter point (3 m).
- Finally, the bending moment at the right end will be zero.
- Plot these points on the BMD and connect them with a smooth curve.
5. Determine the maximum bending moment:
- The maximum bending moment occurs at the point of the beam with the highest value on the bending moment diagram.
- In this case, the maximum bending moment occurs at the center point (3 m) on the BMD.
- Read the value of the bending moment at that point from the BMD, which will give you the maximum bending moment.
Note: It's important to note that I have explained the general process of drawing shear force and bending moment diagrams. To obtain the exact values and accurately draw the diagrams, calculations should be performed using appropriate units and considering the beam's dimensions and loads.