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Consider the beam as shown below. 3kN/m 30KN 10KN
A 3m C 2m D 3m B 3m ERA RB
The two equations that govern a loaded beam are:
a) Total vertical Forces Acting Downward = Total Vertical forces
acting upward
b) Total Clockwise Moment About a Turning Point = Total Anticlockwise Moment about the same turning
point.
Based on the two equations above, we are going to write two
equations.
a) Total Vertical Forces acting downward =
(3KN/m X3m) + 30KN + 10KN
Total vertical forces acting upward=RA + RB
The symbol for our equation a) is usually written
ΣFV=0: 49KN = RA + RB (1)
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.
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.
The following diagram illustrates clockwise and anticlockwise
moments about a support turning point.
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.
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.
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.
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
(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.
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.
(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.
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.
273.5KNm = RB X (3m + 2m + 3m)
RB = 34.1875KN
From the equation RA + RB = 39
RA = (49 - 34.1875)KN = 14.8125KN
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
C = 3m ; D = 5m etc
A C D B E
SHEA
R FO
RCE
AXIS
(KN
)
DISTANCE , X. (METRES)
SFD KEY POINTS1. The final vertical force at the section E = 02. The sum of the upward vertical forces = The sum of the downward vertical forces as shown by the SFD.3. ΣFV=0.
X ;Y coordinates Calculation
Point A
Y= 14.8125KN;X = 0 meters
Reaction Support A acts Upwards
Point C
Y = 5.8125KNX = 3 m
14.8125KN – (3KN/m X 3m)]
Point D
Y = 5.8125KNX = 5m
X ;Y coordinates Calculation
D Y = -24.1875KNX = 5m
(5.8125 -30 )KN 30KN acts downward
B Y = -24.1875KNX = 8 m
B Y = 10KNX = 8 m
(34.1875 – 24.1875) KN
E Y;X = 10KN ; 11m
E Y = OKNX = 11m
(10 -10)KN
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).
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.
Section Area Bending Moment
A – CTrapezium
1/2(14.8125 +5.8125) 3m= 30.9325KNm
MC =30.9325KNm
C – DRectangle
2m X 5.8125KN =11.625KNm
MD = 30.9325 + 11.625 =42.5525KNm
D – BRectangle
3m X -24.1875KN =-72.5625KNm MB=
(42.5525 – 72.5625) KNm=-30.01KNm
B – ERectangle
3m X 10KN = 30KNm ME =(30 - 30.01) = 0
BMD KEYPOINTS
1. The final bending moment at the section E = 02. The total areas (+ve bending moment) above the x axis =The total areas (-ve bending moment) below the x axis.3. ΣMA=0
BEN
DIN
G M
OM
ENT
AXIS
(KN
M)
DISTANCE , X. (METRES)
A C D B E
MC= (3m, 30.9325KNm) ; ME = (11m, 0KNm)MD = (5m,42. 5525KNm) MB = = ( 8m, 30.01KNm)
BMD KEYPOINTS4. 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.
