Transcript Applied Mechanics

Engineering Mechanics
Chapter 3 :
Moments and Couples
1
Objectives
Understand the Principle of Transmissibility
of a Force;
Determine the Moment of a Force System
from Basic Definition and/or with the
application of Varignon’s Theorem;
Understand the Properties of a Couple;
Determine a Force-Couple Equivalent to
Replace a System of Forces.
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Introduction
A force will cause motion along its
direction.
A force may also cause rotation about a
fixed point some distance away.
This rotation or turning effect of a force
is called MOMENT.
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Principle of Transmissibility
If we move (TRANSMIT) the Force P from point A to
point B which lies on the Line of Action of Force P, the
same effect would be expected.
P
Point of
B
Application
A
P
Line of action
Principle of transmissibility states that a force acting on
a rigid body at different points along the force’s line of
action will produce the same effect on the body.
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Moment of A Force
The moment M of a force F about a
fixed point A is defined as the product of
the magnitude of force F and the
perpendicular distance d from point A
to the line of action of force F.
MA = F x d
d
Where force F is in newtons, N
A
And distance d is in meters, m
Thus moment MA is in newton-meter, Nm.
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F
Moment Sign Convention
Anti-clockwise : + VE
(Counter-clockwise)
Clockwise :
- VE
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Example 3.1
Calculate the moment of the 500 N
force about the point A as shown in the
diagram.
500 N
A
1.5 m
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Example 3.1
Solution
500 N
A
1.5 m
Since the perpendicular distance from the force
to the axis point A is 1.5 m, from
MA = F x d
MA = - 500 x 1.5
= - 750 Nm
= 750 Nm
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Example 3.2
Calculate the moment about point A
caused by the 500 N force as shown
in the diagram.
500 N
A
1.5 m
60o
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A
Example 3.2
Solution:
A
600
600
d
B
500 N
60o
500 N
1.5 m
1.5 m
MA = F x d
= -500 x AB
= -500 x 1.5 sin 60
Line of action
= -649.5 Nm
= 649.5 Nm
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Addition of Moments of Coplanar Forces
Coplanar Forces refers to forces acting
on the Same Plane ( i.e. 2-D ).
For a system of several coplanar forces,
the combined turning effect of the
forces can be determined by adding
algebraically the moment caused by
each individual force, taking into
account their sense of rotation,
i.e +ve for anti-clockwise and
–ve for clockwise rotation.
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Example 3.3
Determine the resulting moment about
point A of the system of forces on bar
ABC as shown in the diagram.
500 N
800 N
0.5m
1.5 m
60 0
A
B
C
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500
Example 3.3
Solution:
1.5 m
600
D
0.5 m
60 0
500 N
1.5 m
0.5 m
A
800 N
A
B
C
800 N
AD = (1.5 + 0.5) sin 60
C
= 2.0 sin 60
Line of Action = 1.732 m
MA = (-500 x 1.5) + (-800 x
1.732)
= - 750 – 1385.6
= - 2135.6 Nm
= 2135.6 Nm
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Varignon’s Theorem
Varignon’s Theorem states that “ the
moment of a force about any point is
equal to the sum of the moments of its
components about the same point”.
To calculate the moment of any force
with a slope or at an angle to the x or yaxis, resolve the force into the Fx and
the Fy components, and calculate the
sum of the moment of these two force
components about the same point.
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Example 3.4
Re-Calculate Example 3.3 Above
Employing Varignon’s Theorem.
500 N
800 N
0.5m
1.5 m
60 0
A
B
C
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Example 3.4
Solution
500 N
800 N
0.5m
1.5 m
60 0
A
500 N
1.5 m
A
Fy
0.5m
B
C
800 N
600
MA
= (-500 x 1.5) + (- 800 sin 60 x 2)
= -2135.6 Nm
Fy = 800 sin 600
B
C
600
Fx = 800 cos
Fx
= 2135.6 Nm
600
* Same answer as in example 3.3
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Example 3.5
Determine the resulting moment about
point A for the system of forces acting
on the plate ABCD as shown in the
diagram.
15 cm
20 N
45o
D
C
8cm
30o
35 N
A
B
10.38 cm
80o
45 N
Example 3.5
Solution
35 N
15 cm
45o
D
8cm
30o
A
B
10.38 cm
Fx
800
45 N
Fy
Fx
300
Fy
20 N
C
80o
45 N
45 N force:
Fx = 45 cos 80 = 7.814 N
Fy = 45 sin 80 = 44.32 N
35 N force:
Fx = 35 cos 60 = 17.5 N
Fy = 35 sin 60 = 30.31 N
35 N
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Example 3.5
Solution
60o
D
0.15 m
45o
20 N
C
0.08m
A
B
35 N
20 N
450
Fx
0.1038 m
Fy
20 N force:
80o
45 N
Fx = 20 cos 45 = 14.14 N
Fy = 20 sin 45 = 14.14 N
MA = (- 45 sin 80 x 0.1038) + (- 35 cos 60 x 0.08)
+ (20 cos 45 x 0.08) + (20 sin 45 x 0.15)
= - 4.591 – 1.4 + 1.131 + 2.121
= - 2.74 Nm = 2.74 Nm
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3.6
F
Couples
d
F

F
d

F
A couple consists of a pair of 2 forces which
has the following properties :- Equal magnitude and opposite in
direction
- Act along parallel lines of action
- Separated by a perpendicular distance d.
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What a couple does?
A couple causes a body to rotate only
without translational motion since the
two forces ‘cancels’ out each other
giving zero resultant.
A couple acting in a system of forces
will only contribute to the resulting
moment but not to the resulting force.
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Magnitude of a Couple
Consider a light bar acted upon by a couple
as shown :F
d1
A
d
d2
B
F
The moment of the couple about A is
MA
= + ( F x d2 ) - ( F x d1 )
MA
= F x ( d2 - d1 )
MA
=Fxd
What is the total moment of the couple about
point B ?
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From above, we see that :
The couple’s moment about any pivot
point is equal to F x d
A couple has the same moment about
all points on a body.
F
F
F
F
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Example 3.6
A light bracket ABC is subjected to two
forces and two couples as shown.
Determine the moment at (a) point A
and (b) point B.
2 KN
30o
80 Nm
120 Nm
1.5 m
A
2.5 m
B
3 KN
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Example 3.6
Solution:
2 KN
30o
80 Nm
120 Nm
1.5 m
A
2.5 m
B
3 KN
a)  MA = (2000 cos 60 x 2.5) – (2000 sin 60 x 1.5) + ( 3000 x 0)
+ 120 – 80
= 2500 – 2598 + 0 +120 - 80
= - 58 Nm = 58 Nm
(b)  MB = (2000 cos 60 x 0) – (2000 sin 60 x 1.5) + (3000 x 0)
+ 120 – 80
= -2558 Nm
= 2558 Nm
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3.7 Force Couple Equivalent
The Force-Couple Equivalent concept will
enable us to transfer a force to another
location outside its line of action.
Consider a force F acting at a point B on a
rigid body as shown in diagram (a) below.
How do we transfer the force F from point B
to point A?
d
A
F
B
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F
F
d
A
B
F
B
F
B
MA = F x d
F
d
A
A
Force Couple Equivalent
The above shows how a
force can be replaced by a
force-couple equivalent.
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Example 3.7
(Single Force System)
Determine the force-couple equivalent at
point A for the single force of 20 kN acting
at point C on the bracket ABC.
20 KN
40o
C
1.2 m
A
2.3 m
B
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Example 3.7
solution
Fy
20 KN
40o
C
Fx
1.2 m
A
2.3 m
B
 MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2)
= 11.18 kNm
Answer:
11.18 kNm 20 kN
40 0
A
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Example 3.8 (Multiple force system)
Another 30 kN horizontal force is added to
Example 3.7 at point B. Determine the forcecouple equivalent at point A.
20 KN
40o
C
1.2 m
A
30 KN
2.3 m
B
30
Example 3.8
Solution
Fy
20 KN
40o
C
Fx
1.2 m
A
2.3 m
400
Fy
Fx
Therefore
And
B
30 KN
Rx =  Fx = 20 cos 40 + 30
= 45.32 kN
Ry =  Fy = 20 sin 40
= 12.86 kN
R =  (45.322 + 12.862)
= 47.11 kN
tan  = Ry = 12.86 = 15.84 0
Rx
45.32
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Example 3.8
Solution
Fy
20 KN
40o
C
Fx
1.2 m
A
2.3 m
B
30 KN
 MA = (20 sin 40 x 2.3) – (20 cos 40 x 1.2) + (30 x 0)
= 11.18 kNm
Answer:
47.11 kN
11.18 kNm
15.84 0
End of Chapter 3
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