1. STRUCTURAL MECHANICS

2. Point loads
The figure below illustrates a simply supported beam with reactions at each end and two point loadings of 50kN and 100kN.
We shall calculate the value of the end reactions. Before undertaking this calculation, we need to recap some of the basic laws associated with this type of structure;

3. Point Loads
The sum of the forces in one direction must equal the sum of the forces in the opposite direction ie the sum of downward forces equals the sum of the upward forces.
Anti-clockwise moments equal clockwise moments

4. Point Loads Moment about a point (pivot)
Moment = Force x Distance about a point
Force
Distance

5. Point Loads Moment about a point This gives a CLOCKWISE MOMENT Pivot
Force (F)
MOMENT = F x D
Distance (D)
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6. Example 1
50KN
100KN 4M 4M 3M
11M

7. Example 1 Calculate the left and right reactions
Take the left-hand reaction as the PIVOT:
Anti-clockwise moments = Clockwise moments

8. Example 1 50KN 100KN 4M 4M 3M 11M RL RR ACM = RR X 11 CM = 50 X 4

9. Example 1 Anti-clockwise moments = Clockwise moments
RR x 11 = (100 x 8) + (50 x 4)
RR x 11 = 1000
RR = 1000/11
= Kn

10. Example 1 Similarly, if we start at the right hand reaction:
Anti-clockwise moments= clockwise moments
(100 x 3) + (50 x 7) = RL x 11
650 = RL x 11
650/11 = RL = 59.09Kn

11. Example 1 RR = 90.91Kn RL = 59.09kN Check
ΣUPWARD FORCES = ΣDOWNWARD FORCES
= 150 (50+100) correct

12. Cantilever
10KN
2KN
6KN 1M 1M 1M 2M RR RL

13. Cantilever RR x 3 = (10 x 1) + (6 X 2) + (2 x 5) RR x 3 = 10 + 12 + 10
RR = 10.67kN
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14. Cantilever RR = 10.67kN Total Downward Force = 18kN Therefore,
RR + RL = 18
RL = 18 – RR
RL = 18 – 10.67
RL = 7.33kN
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15. SHEAR FORCE DIAGRAM 10KN 2KN 6KN 1M 1M 1M 2M RR RL 7.33 7.33 2.00 2.00
2.67
2.67
8.67
8.67

16. BENDING MOMENT 6KN 10KN 2KN 1M 2M A D B C E
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Find the BM values at points A, B, C, D & E
The BM value at the ends of a simply supported beam is always ZERO.
In this case BMA & BME equal 0 Knm 16

17. BENDING MOMENT 6KN 10KN 2KN 1M 2M A D B C E
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Consider all loads to the left of the point where a BM value is required.
Find the BM values at points B, C & D 17

18. BENDING MOMENT 6KN 10KN 2KN 1M 2M A D B C E
BM value at points B. Consider the following:
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BMB = +(7.33x 1) = Knm 1m
CLOCKWISE MOMENTS CONSIDERED TO BE POSITIVE.
7.33 18

19. BENDING MOMENT B C D E 10KN 6KN 1M 2M 2KN A 10Kn
BM value at points C. Consider the following:
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BMC = +(7.33x2) – (10x1) = +4.66Knm C 1m 1m
ANTI CLOCKWISE MOMENTS CONSIDERED TO BE NEGATIVE.
7.33Kn 19

20. BENDING MOMENT 6KN 10KN 2KN 1M 2M A D B C E
BM value at points D. Consider the following:
10Kn
6Kn
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BMD =
+(7.33x3) – (10x2) – (6x1) = -4.01Knm D 1m 1m 1m
7.33Kn 20

21. BENDING MOMENT DIAGRAMC D E
10KN
6KN 1M 2M
2KN A
Note!
Positive values plotted BELOW
Negative values plotted ABOVE
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-4.01 A B C D E
4.66
7.33 21

22. PROBLEM
15KN
10KN
3KN 2M 2M 1M 1M RR RL

23. 15KN 10KN 3KN 2M 2M 1M 1M (15 x 2) + (10 x 4) + (3 x 6) = RR x 5
RR = 17.6 kN
RR + RL = 28kN
RL = 28 – 17.6
RL = 10.4 kN RR
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24. SFD
15KN
10KN
3KN 2M 2M 1M 1M RR RL 3 15 3
10.4 3
-4.6
-4.6
17.6 10
-14.6
-14.6

25. BMD = +(10.4 x 5) –(15 x 3) –(10 x 1) = -3.00 kNm RRC D A E RL
BMA & E = 0
BMB = +(10.4 x 2) = kNm
BMC = +(10.4 x 4) – (15 x 2) = kNm
BMD = +(10.4 x 5) –(15 x 3) –(10 x 1) = kNm RR
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26. BMD
15KN
10KN
3KN 2M 2M 1M 1M
3.0 A B C E D
11.6
20.8