Shear Force Diagrams Bending Moment Diagrams Shear Force Diagrams Calculations Bending Moment Diagrams Calculations Moments Equation Engineering Science Udl Uniformly Distributed Load Point Load Loaded Beam ( Udl and Point Load Combinations) Reaction Support Tables of BMD and SFD Calculation of BMD (Area under the SFD Curve)

1. SHEAR FORCE DIAGRAMS &
BENDING MOMENT DIAGRAMS
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2. Consider the beam as shown below.
3kN/m 30KN 10KN
A 3m C 2m D 3m B 3m E
RA RB

3. The two equations that govern a
loaded beam are:
a) Total vertical Forces Acting
Downward = Total Vertical forces
acting upward

4. b) Total Clockwise Moment About a
Turning Point = Total Anticlockwise
Moment about the same turning
point.

5. Based on the two equations above,
we are going to write two
equations.
a) Total Vertical Forces acting
downward =
(3KN/m X3m) + 30KN + 10KN

6. Total vertical forces acting upward=
RA + RB
The symbol for our equation a) is
usually written
ΣFV=0: 49KN = RA + RB (1)

7. The symbol for our equation b) is
usually written
ΣMA=0
This symbol means summation of
moments about the turning point
(support A) is zero.

8. The symbol for our equation b) is
usually written
ΣMA=0
This symbol means summation of
moments about the turning point
(support A) is zero.

9. The following diagram illustrates
clockwise and anticlockwise
moments about a support turning
point.

11. Based on the above diagram:
Let the weight of the boy on the left
hand side be A KN and the weight of
the boy on the right hand side be
B KN.
The weight of a body acts
downwards.

12. Anticlockwise moment:
AKilonewton exerts an
anticlockwise moment of A
Kilonewtons X x meters about the
support .
Assuming that the perpendicular
distance of the weight of A from the
turning point is x meters.

13. Clockwise moment:
BKN exerts an clockwise moment of
B Kilonewtons X w meters about the
support .
Assuming that the perpendicular
distance of the weight of B from the
turning point is w meters.

14. Based on the above definitions, we
are going to write an equation of
ΣMA=0 for the loaded beam.
Clockwise moments about RA
(3kN/m X 3m X (3/2)m) +
(30KN X 5m) + (10KN X 11m)
= 273.5KNm

15. (3kN/m X 3m X (3/2)m)
For the moment shown above:
3kN/m X 3m = magnitude of the
Uniformly distributed load.
(3/2)m represents the
perpendicular distance between the
turning point A and the point
through which the udl acts.

16. In order words, this moment is :
3 represents the span on which the
udl acts.
The magnitude of the udl , 3kN/m X
3m , acts through the middle of the
span.

17. (30KN X 5m)
This represents the moment of the
point load, 30KN, which acts at a
perpendicular distance of 5m from
the turning point, A.

18. For Anticlockwise moments:
RB X (3m + 2m + 3m)
RB represents the support reaction
acting vertically upwards.
(3m + 2m + 3m) represents the
distance between the support RB
and the turning point.

19. 273.5KNm = RB X (3m + 2m + 3m)
RB = 34.1875KN
From the equation RA + RB = 39
RA = (49 - 34.1875)KN = 14.8125KN

20. We shall now proceed to draw the
SFD.
Shear Force Diagram is a graph of
the vertical forces plotted on the y
axes and the horizontal
perpendicular distance of the force
on the x axis.
The point A is zero meters

21. C = 3m ; D = 5m etc
A C D B E
SHEARFORCEAXIS(KN)
DISTANCE , X. (METRES)

22. SFD KEY POINTS
1. The final vertical force at the
section E = 0
2. The sum of the upward vertical
forces = The sum of the downward
vertical forces as shown by the SFD.
3. ΣFV=0.

23. X ;Y coordinates Calculation
Point
A
Y= 14.8125KN;
X = 0 meters
Reaction
Support A
acts
Upwards
Point
C
Y = 5.8125KN
X = 3 m
14.8125KN
– (3KN/m
X 3m)]
Point
D
Y = 5.8125KN
X = 5m

24. X ;Y coordinates Calculation
D Y = -24.1875KN
X = 5m
(5.8125 -30 )KN
30KN acts
downward
B Y = -24.1875KN
X = 8 m
B Y = 10KN
X = 8 m
(34.1875 –
24.1875) KN
E Y;X = 10KN ; 11m
E Y = OKN
X = 11m
(10 -10)KN

26. BENDING MOMENT DIAGRAM
Bending Moment Diagram or Graph
is the graph of the bending moment
on the y axis plotted against the
distance, x metres, along the beam
(x axis).

27. BENDING MOMENT DIAGRAM
The values of the Bending moment
on the y axis can be evaluated by
calculating the areas under the
Shear Force diagram curve section
by section progressing from left to
right.

28. Section Area Bending
Moment
A – C
Trapezium
1/2(14.8125 +
5.8125) 3m
= 30.9325KNm
MC =
30.9325KNm
C – D
Rectangle
2m X 5.8125KN =
11.625KNm
MD =
30.9325 +
11.625 =
42.5525KNm
D – B
Rectangle
3m X -24.1875KN =
-72.5625KNm
MB=
(42.5525 –
72.5625) KNm=
-30.01KNm
B – E
Rectangle
3m X 10KN = 30KNm ME =
(30 - 30.01) = 0

29. BMD KEYPOINTS
1. The final bending moment at the
section E = 0
2. The total areas (+ve bending
moment) above the x axis =
The total areas (-ve bending
moment) below the x axis.
3. ΣMA=0

30. BENDINGMOMENTAXIS(KNM)
DISTANCE , X.
(METRES)
A C D B E
MC= (3m, 30.9325KNm) ; ME = (11m, 0KNm)
MD = (5m,42. 5525KNm) MB = = ( 8m, 30.01KNm)

31. BMD KEYPOINTS
4. The section of the Beam that
carries a uniformly distributed load
is represented as a curve on the
BMD.
5. The section of the Beam that
carries point loads are represented
by slant straight lines.