Chapter Four
Laith Batarseh
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Definition
Moment is defined as the tendency of a body lies under force to
rotate about a point not on the line of the action of that force (i.e.
there is a distance between the force and the rotation point )
Moment is a vector quantity
Description
Moment depends on two variables:
The acting force
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Moment arm
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Description
Force
Arm
Tendency to rotate
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End
Tendency for rotation
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Magnitude
D
F
Moment magnitude (M) = F.D
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End
Direction
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Solving procedures
1. Define the magnitudes of force (F) and arm (D)
2. Assume the positive direction (eg. Counter clock wise)
3. Find the magnitude of moment (M) as F.D
4. Give the moment the correct sign according to the tendency for
rotation
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Example [1]
Find the moment caused by the following forces about point O
2m
(a)
100 N
O
0.5m
100 N
(b)
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Previous
0.5m
Next
O
2m
End
Example [1]
Assume the CCW direction is the positive direction.
2m
2m
0.5m
0.5m
100 N
Branch (a)
+ Mo = F.d = -(100N)(0.5m)
Mo=-50 N.m=50N.m CW
Branch (b)
+ Mo=F.d = (100N)(2m)
Mo=200 N.m CCW
100 N
(b)
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(a)
O
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O
End
Principle Of Moments
Principle of Moments
some
times
called
Vrigonon’s
theorem
(Vrigonon
is
French
mathematician 1654-1722).
State that the moment of a force about a point equals the summation of
the moments created by the force components
In two dimensional problems: the magnitude is found as M = F.d and the
direction is found by the right hand rule
In three dimensional problems: the moment vector is found by M =rxf
and the direction is determined by the vector notation (ie. i,j and k
directions)
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Principle Of Moments
Example [1]
Find the moment caused by the following forces about point O
Next
Previous
Home
End
Principle Of Moments
Example [1]
+
Mo,1 = 100 sin(30) (10) = 500 N.m
+
Mo,2 =- 100 cos(30) (5) =- 433N.m
M = Mo,1+Mo,2=500-433=67N CCW
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Previous
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End
Principle Of Moments
Example [2]
1. Force analysis
100 cos(40)
100 sin(40)
120 sin(60)
1.2 m
0.3 m 120 cos(60)
2. Moment calculations
+ ∑ M = (100 cos(40))(1.5) –( 120 cos(60))(1.2) =43 N.m CCW
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Moment resultant
F1
d1
M1
O
M2
F2
M3
d3
F3
+ Mo = ∑Mo = M1 + M2 – M3 = F1d1+F2d2 – F3d3
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Previous
d2
End
Example [2]
Find the moment caused by the following forces about point O
100 N
2m
50 N
O
1m
60 N
30o
5m
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3m
75 N
End
Example [2]
100 N
2m
50 N
O
1m
3m
60 N
75 N
30o
5m
+  M o  1002  603  500  75 sin( 30) 5 75 cos(30) 1
 272.45 N .m  272.45 N .m CW
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o
Next
M
End
Exercise
Find the moment caused by the following forces about point O
100 N
5m
30o
O
0.3m
45o
300 N
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2m
End
Exercise
100 N 100sin (30)N
5m
O 100cos (30)N
2m
300 sin (45)N
0.3m
300 cos (45)N
300 N
+  M o  100 sin( 30) 2  300 cos( 45) 0.3  300 sin( 45) 5
 897 N .m  897 N .m CW
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Previous
o
Next
M
End
Cross product
Cross product is a mathematical operation can be done on vectors
Cross product for one time is done for two vectors
The cross product of two vectors is a vector perpendicular to the plane
of A and B
The notation of vector A cross vector B is: C = AxB where C is the
resultant vector from the cross product
the vector C can be represented as : C =CUc where Uc is a unit vector in a
direction perpendicular to the plane that contains both A and B.
The value of the scalar quantity C is given as : C=A.B.sin(ϕ)
where ϕ is the angle between A and B.
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The cross product is controlled by the right-hand rule
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Graphical representation
Uc
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Cartesian vector formulation
z
K=ixj
j=kxi
y
i=jxk
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x
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Cartesian vector formulation
A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
AxB=(Ay Bz -Az By )i-(Ax Bz - Az Bx )j + (Ax By - Ay Bx )k
k
Az
Bz
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j
Ay
By
Next
i
AxB  Ax
Bx
End
Moment – vector formulation
M
Mo = rxF
d
x
Magnitude: Mo = rFsin(θ) = Fd
θ
Direction: perpendicular to x-y plane
(z-direction)
i
Matrix notation: M o  rxF  rx
Fx
j
ry
Fy
r
O
y
F
k
rz
Fz
Best for three
dimensional
problems
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Resultant moment:MRo = ∑(rxF)
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Example [1]
Find the moment caused by the following forces about point O
z
3m
O
F = [5i + 10j + 6k]N
y
1m
2m
Next
x
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Example [1]
1. Formulate the position vector (r) : r = 3i+2j+1k
2. Find the moment vector (Mo) by matrix notation
i
j
k
M o  rxF  3 2 1  2i  13 j  20k
5 10 6
F = [5i + 10j + 6k]N
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r = 3i+2j+1k
End
Example [2]
Find the moment caused by the following forces about point O
z
O
y
x
F = [-5i + 5j -5k]N
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End
Example [2]
1. Formulate the position vector (r) : r= 15i + 10j +6k
2. Find the moment vector (Mo) by matrix notation
i
j
k
M o  rxF  15 10 6  80i  45 j  125k
5 5 5
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Summary
Moment is a vector can be found by cross product and
matrix notation
Matrix notation:
k
rz
Fz
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j
ry
Fy
Next
i
M o  rxF  rx
Fx
End
Principle Of Moments
Principle of Moments
some
times
called
Vrigonon’s
theorem
(Vrigonon
is
French
mathematician 1654-1722).
State that the moment of a force about a point equals the summation of
the moments created by the force components
In two dimensional problems: the magnitude is found as M = F.d and the
direction is found by the right hand rule
In three dimensional problems: the moment vector is found by M =rxf
and the direction is determined by the vector notation (ie. i,j and k
directions)
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End
Principle Of Moments
Example [1]
Find the moment caused by the following forces about point O
Next
Previous
Home
End
Principle Of Moments
Example [1]
+
Mo,1 = 100 sin(30) (10) = 500 N.m
+
Mo,2 =- 100 cos(30) (5) =- 433N.m
M = Mo,1+Mo,2=500-433=67N CCW
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Home
End
Principle Of Moments
Example [2]
The following is a gate. In which direction this gate will rotate?
100N
o
60o
120N
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Previous
0.3 m
1.2 m
Next
40
End
Principle Of Moments
Example [2]
1. Force analysis
100 cos(40)
100 sin(40)
120 sin(60)
1.2 m
0.3 m 120 cos(60)
2. Moment calculations
+ ∑ M = (100 cos(40))(1.5) –( 120 cos(60))(1.2) =43 N.m CCW
Next
Previous
Home
End
Principle Of Moments
Example [3]
Find the moment caused by the following forces about point O
Next
Previous
Home
End
Principle Of Moments
Example [2]
i
M o,1  rxF  15 10 6  30i  75k
0 5 0
i
r2= 15i + 10j -6k F2 = [-5j]N
k
j
k
M o,2  rxF  15 10  6  30i  75k
0 5 0
Mo  60i
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Next
Previous
r1= 15i + 10j +6k F1 = [5j]N
j
End
Moment of force about axis
In many real cases, the force tendency to rotate is about a specified axis.
Example :
Moment components:
Mo,1 = (100)(10) (about y-axis)
Mo,2 = (100)(15) (about x-axis)
(about z-axis)
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Mo,3 =0
End
Moment of force about axis
Magnitude
Scalar analysis: M = F.d
Vector analysis:
u a ,x
M a  u a .rxF   rx
Fx
ua ,y
ry
Fy
u a ,z
rz
Fz
Where: ua is a unit vector defining the direction of a-axis and given as
ua = ua,x i + ua,y j + ua,z k
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To find the moment vector (Ma): Ma. ua
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Moment of force about axis
Example [1]
+
Mo,1 = 100 sin(30) (10) = 500 N.m
+
Mo,2 =- 100 cos(30) (5) =- 433N.m
M = Mo,1+Mo,2=500-433=67N CCW
Next
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End
Moment of force about axis
Example [2]
Determine the magnitude of the moments of the force F about the x, y, and z axes. Solve
the problem
using a Cartesian vector approach
using a scalar approach.
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Moment of force about axis
Example [2]
using a Cartesian vector approach
rAB = {5i + 4j -3k} m
For the axes: x, y and z the unit vectors are i, j and k respectively.
Mx = i . (rAB x F)
My = j . (rAB x F)
Mz = k . (rAB x F)
0
M x  5 4  3  14 N .m
5 10  4
0
1
0
M y  5 4  3  5 N .m
5 10  4
0
0
1
M z  5 4  3  30 N .m
5 10  4
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0
Next
1
End
Moment of force about axis
Example [2]
using a scalar approach
Mx = ∑Mx = 10(3) – 4(4) = 14 N.m
My = ∑My = -5(3) + 4(5) = 5 N.m
Mz = ∑Mz = -5(4) + 10(5) = 30N.m
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Moment of couple
Definition
Couple is the moment generated by two forces has the same
magnitude and opposite direction.
F
d
-F
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Moment of couple
Scalar analysis
This analysis is considered for 2-D
M
problems
The magnitude of moment is found
by: M=F.d where F is the force
magnitude.
F
-F
d
the direction of couple moment is
perpendicular to the plain that
contain the F and d and it is found by
Next
the right hand rule.
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Moment of couple
Example on scalar analysis
d1
-F2
d2
-F1
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Previous
M1 = F1 . d1
M2 = F2 . d2
F2
Next
F1
End
Moment of couple
Vector analysis
This analysis is considered for 3-D
problems
Derivation:
M = rB x F + rA x –F = (rB - rA) x F
But: (rB - rA) = r Then: M = rx F
The moment vector is found by:
M=F x r where r is the position vector
B
directed between the forces F and –F
r
Note that the moment vector is
dependent on the position vector
F
-F
A
directed between the forces F and –F
rA
O
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rB
Next
(r).
End
Moment of couple
Example [1]
Find the resultant moment couple produced by the following forces. All
dimensions in m
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End
Moment of couple
Example [1]
Solution:
1. By traditional moment analysis
+
∑Mo = -(150)(7)-(150)(7) – (200)(9)-(200)(9) = -5700 N.m
2.By cpouple moment analysis
∑Mc = -(150)(14) – (200)(18) = -5700 N.m
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+
End
Moment of couple
Example [2]
Find the moment couple produced by the following force. All
dimensions in m
Next
Previous
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End
Moment of couple
Example [2]
1. F = 150 j
2. r = 3k
3. M = r x F = 150 j x 3k = {450N.m} i
Next
M
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End
Moment of couple
Equivalent couples
in many of life applications, an equivalent couple is required to solve
some technical problems such as space and size.
Equivalent couples are the couples that have the same magnitude and
same direction
As you can see, the relation between the forces and the arm distances in
equivalent coupels is reverse (for example, as we reduce the moment
arm, the required force for equivalent couple increases)
F1
F1.d1 = F2.d2
=
d2
-F1
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-F2
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d1
F2
End
Moment of couple
Resultant couple
Resultant couple is the vectorial summation of the couples act on the
body as you can see.
In simple situation as shown in the figure, the parallelogram is used to
sum the moments and in more complicated cases or three dimensional
problems, the Cartesian notation is used.
M2
M1
MR
Next
M2
Home
Previous
M1
End
Moment of couple
Example [1]
The member shown in the figure is subjected to three coupling
forces : 150, 200 and 100 N. the moment arms are shown.
If All dimensions in m, find the resultant couple moment produced
by the following forces.
200N
150 N
100N
150 N
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Next
100N
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200N
End
Moment of couple
Example [1]
Solution:
First, the Moment direction: +
Second, the calculate the coupled moments
M1 = -(200)(13) = -2600 N.m
M2 = -(150)(7) = -1050 N.m
M3 = - (100)(8) = -800 N.m
Finally, calculate the moment sum (resultant)
MT = -2600-1050 -800=-4450 N.m
MT = 4450 N.m (clockwise)
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Previous
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End
Moment of couple
Example [2]
The member shown in the figure is subjected to two coupling forces : 200
and 100 N. the moment arms are shown.
Find the resultant couple moment produced by the following forces.
3m
6m
200 N
45o
100 N
0.6m
100 N
Previous
3m
200 N
Home
Next
45o
End
Moment of couple
Example [2]
Solution:
First, the Moment direction: +
Second, the calculate the coupled moments
M1 = -(100)(0.6) = -60 N.m
M2,x = (200 cos(45))(0.6) = 85 N.m
M2,y = -(200 sin(45))(3) = -424 N.m
Finally, calculate the moment sum (resultant)
MT = --60+85-242=-399 N.m
MT =399 N.m (clockwise)
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Moment of couple
3-D problems
For three dimensional problems, it is better to use the Cartesian
notation (I, j and k) to represent the monuments.
The moment in a direction not one of the principle axes (x, y or z)
can be represented as: M = Mu. Where u is a unit vector in the
direction of the moment M.
The resultant moment can be found finally by vector addition for
the moments vectors
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Moment of couple
3-D problems
z
Example:
The figure show an object subjected to
two moments: M1 and M2. as you can
The moment M1 can be represented as:
M1= M1i.
While the moment
M1
x
θ
y
M2
M2 can be represented as:
Next
M2 = M2 (0i + cos(θ) j + sin(θ) k)
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see M2 has an angle θ from the y-axis.
End
Simplification of a force and couple system
Equivalent couples
In many of situation where there a group of forces and moments acting
on an object, it is seem more convenient to reduce the large number of
forces and moments to one force and one moment.
Physical meaning: replacing a system of forces and moments by a
system of one force and one moment.
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Previous
Condition: the external effects produced by the forces and moments on
the body for the original system are the same of the single force and
moment in the new simplified system
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Simplification of a force and couple system
Simplification conditions
F
-F
F
F
Note: the acting force can be transport from one position to another on
its line of action (i.e. force vector)
F
F
M = F.d
Note: the force acts on a member can be transport from one
position to another on a line perpendicular force vector by
adding the moment generated by the original force (i.e. M=F.d)
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Simplification of a force and couple system
Simplification conditions
Assume an object as shown in Fig.a is subjected to two forces ( F1 and
F2) and one moment M. The forces F1 and F2 has a position vectors r1
and r2 respectively from the rotation point O to the line of action for
each force.
M
F2
F1
r1
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Fig.a
Next
O
Previous
r2
End
Simplification of a force and couple system
Simplification conditions
To convert the previous system into one force-moment system we must:
•first move each force to the point of rotation O. this step include adding
the moments produced by both forces (M1 and M2 respectively )at the
rotation point as shown in the Fig.b.
F2
F1
Fig.b
M1 = r1 x F1
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O
Next
M2 = r2 x F2
M
End
Simplification of a force and couple system
Simplification conditions
•Then all forces and moments are summed using the following
formulas:
FR = ∑F
MR,O = ∑MO + ∑M
MR,o
Home
Fig.c
Next
O
Previous
FR
End
Simplification of a force and couple system
Example [1]
Figure below shows a plate a group of forces (100, 150, 200 and 300
N)
If All dimensions in m, Simplify the following force system to single
force and moment system about point O
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Previous
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End
Simplification of a force and couple system
Example [1]
Solution:
First, calculate the resultant force FR
∑Fx = 300 – 100 = 200 N
∑Fy = 200 + 150 = 350 N
FR  2002  3502  403N
 350 
o
  tan 1 
  60
 200 
Second, calculate the resultant moment MR
MR =-(200)(13) – (300)(4) –(150)(5) – (100)(4) = 4950 N.m
Finally, represent the new force and moment on the original system as
shown.
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End
Simplification of a force and couple system
Example [2]: couple resultant
The system shown in the figure is subjected to two coupling forces : 200
and 100 N. the moment arms are as shown.
Simplify the following force system to single force and moment system
about point O
3m
6m
399 N.m
200 N
45o
100 N
0.6m
Previous
3m
200 N
Home
100 N
Next
45o
End
Simplification of a force and couple system
Example [2]
Solution:
First, the Moment direction: +
Second, calculate the coupled moments
M1 = -(100)(0.6) = -60 N.m
M2,x = (200 cos(45))(0.6) = 85 N.m
M2,y = -(200 sin(45))(3) = -424 N.m
Finally, calculate the moment sum (resultant)
MT = -60+85-242=-399 N.m
MT =399 N.m (clockwise)
Next
Note that the resultant force (FR ) = 0 which is true for all the
coupled forces systems
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End
Simplification of a force and couple system
Example [3]
The three forces act on the pipe
assembly. If F1 = 50 N and F2 = 80 N,
replace this force system by an
equivalent resultant force and couple
moment acting at O. Express the
results in Cartesian vector form..
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Previous
Home
End
Simplification of a force and couple system
Example [3]
Solution:
First, calculate the resultant force FR
FR  180  50  80k  [210N ]k
Finally, calculate the resultant moment MR using cross product
i
j
k
i
j
k
i
j
k
M o   rxF  1.25 0
0  1.25 0.5
0  2 0.5 0
0
0  180
0
0  80 0 0 50
 {15i  225 j}N .m
Next
Previous
Home
End
Simplification of a force and couple system
Special cases:
Concurrent forces: forces that’s lines of action intersect at a common
point
Concurrent forces are simply summed to find FR and as seen the moment
is zero due to the passing of forces lines of action through the rotation
point
F1
F2
F3
F4
Home
Previous
O
FR
Next
=
O
End
Simplification of a force and couple system
Special cases:
Coplanar forces: forces share the same plane
Coplanar forces produce moments about the point of rotation and are
summed to find FR . All the moments produced by the acting forces are
summed to find the equivalent moment M.
F1
F2
=
O
M
O
FR
F4
Next
F3
Previous
Home
End
Simplification of a force and couple system
Parallel forces system:
b
O
z F z= ∑F
O
R FR = ∑F
z
F3
MO
dR,O
a
a
F1
Home
Previous
F4
b
A reverse process can be
done to transform the
single force – moment
system into a single force
with moment arm from
the rotation point
Next
F2
End
Simplification of a force and couple system
Analysis procedures
Establish the coordinate system (x, y and z axes). It is preferred to put
the origin of this system at the rotation point.
Force summation
find the resultant force by summing the acting forces. You may resolve
the forces to their rectangular components.
Moment summation
The resultant moment is the summation of the moments acting on the
body and the moments produced by the acting forces.
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End
Simplification of a force and couple system
Special cases:
In three dimensional systems, we can find an equivalent force
and moment. However, in general cases the moments and
force are not perpendicular to each other. Because of that, it
become impossible to reduce the system to single force with
moment arm from the rotation point.
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Previous
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End
Simplification of a force and couple system
Example [1]
Replace the following forces-moment system to a single force
system.
2m
5m
8m
10 kN
30o
7 kN
Next
Previous
Home
End
Simplification of a force and couple system
Example [1]
Solution:
First, calculate the resultant force F
16.062   3.52
∑Fx = 10 + 7cos(30) = 16.06kN
FR 
∑Fy = - 7 sin(30) = 3.5 kN
  tan 1 
 16.43kN
  3.5 
o
  12.3
 16.06 
Second, calculate the resultant moment MR
MR =-(7sin30)(5) -(7cos30)(8)-(10)(8-2) = 126 kN.m
Finally, you can represent the new force and moment on the original
system.
Xo= MR/FR = 126/16.43 = 7.67 m from the base point
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Previous
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End
Simplification of a force and couple system
Example [2]
Replace the following forces-moment system to a single force
system.
z
200N
3m
300 N
3m
3m
y
600 N
4m
Next
x
Previous
Home
End
Simplification of a force and couple system
Example [2]
Solution:
First, calculate the resultant force F
FR = 300 + 600 – 200 = (700k) N
Second, calculate the resultant moment Mx and My
Mx =(300)(0) – (200)(3) + (600)(6) = 3000 N.m = FRy → y = 4.29m
My =-(300)(3) + (200)(0) - (600)(4) = 3300N.m = FRx→ x = 4.71m
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Simplification of a force and couple system
Example [2]
Solution:
Finally, you can represent the new force on the original system.
z
700 N
y
4.72 m
4.29m
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x
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Distributed Loads
Uniform loading along a single axis
w
w = w(x)
x
The force function could be linear or
none linear
L
w
dF = dA
dx
w = w(x)
x
x
L
Distributed force is a force acting on a
line or surface of the rigid body. The
value of this force (w) is represented
by a function in terms of dimensions.
For example: w(x).
We can represent the distributed force
by a single. To do that we first take an
infinitesimal
segment
of
the
distributed force (dF) which equal the
infinitesimal segment of the area
under the force function as you can
see on the screen
Distributed Loads
Magnitude of resultant force
Let us first assume a distributed force w(x)
acting on the member as shown in the fig.
w
FR
Now, assume there is an equivalent force
called FR for the distrusted force w(x) and it
is located at a distance equal x’
The magnitude of FR can obtained by
integrating the function w(x) over the
distance x:
FR   wx .dx   dA  A
A
As you can see from the above equation
that the magnitude of FR equal the area
under the curve w(x)
w = w(x)
x
x'
Distributed Loads
Location of resultant force
The location of the resultant force (d) can
be found using the principle of centroid
(will be discussed later) as:
xw x .dx

x' 
 wx .dx
L
L

w
FR
w = w(x)
 x.dA
x
A
 dA
A
For this stage, the location of the centroid
for the given shape will be given
x'
Distributed Loads
Analysis procedures
To analyze the distributed forces you have to follow the
following procedures:
distributed load is defined as function w = w(x) with
unit of N/m or lbf/ft.
The effect of distributed load is simplified by single
concentrated force acts at certain point in the body
The resultant force equals the area under the loading
diagram and acts on the centroid of this area
Distributed Loads
Example 1:
Determine the magnitude and the location of the equivalent resultant
force acting on the beam shown in the figures.
w
200 N/m
w
200 N/m
x
2m
(a)
x
3m
(b)
Distributed Loads
Example 1:
Solution: part a
w
200 N/m
FR = area under the loading diagram
FR = (200 N/m) (2m) = 400 N
x
(x’) at the center of load rectangle
2m
x‘ = 1m
You can note that the centroid of a
rectangular area is its geometric
center
w
400 N
x
1m
Distributed Loads
Example 1:
Solution: part b
200 N/m
w
Similar to part a
FR = area under the loading diagram
x
FR = (1/2)(200 N/m) (3m) = 300 N
3m
(x’) at the centroid of triangle load
x‘ = (2/3)(3) = 2m
You can note that the centroid of a
triangular area is located at a
distance equal 1/3 of its height fro its
base.
w
300 N
x
2m
Distributed Loads
Example 2:
Determine the magnitude and the location of the equivalent resultant
force acting on the beam shown in the figure.
250 N/m
w
w = (30)x2 N/m
x
2.5m
Distributed Loads
Example 2:
x
FR   dA   30 x .dx  30
A
 3
0
2.5
250 N/m
w
Solution: magnitude
3
2
2.5
w = (30)x2 N/m


0
x
 2.53 03 
FR  30
   156.25 N
3
 3
2.5m
156.25 N
Try to solve it by your self and verify
the solution
w
x
Distributed Loads
Example 2:
 x30 x .dx
2.5
x.dA

x

 dA
A
w = (30)x2 N/m
2
x
0
156.25
A
2.5m
2.5
x 
30 
 4 0
x
 1.875m
156.25
4
250 N/m
w
Solution: location
Try to solve it by your self and verify
the solution
156.25 N
w
x
1.875 m
Distributed Loads
Combined distributed loads
In this lecture we will learn how find the resultant force
from a combined distributed forces
If you have a several disturbed loads , you can find the
resultant force for the whole combination by finding the
resultant force from each distributed force and then sum
the resultant forces to obtain one equivalent force
Distributed Loads
Example 1:
Determine the magnitude and the location of the equivalent resultant
force acting on the beam shown in the figure.
400N/m
w
300N/m
x
2m
4m
2m
Distributed Loads
Example 1:
Solution: first load
FR = area under the loading diagram
FR = (0.5)(400 N/m) (2m) = 400 N
(x’) at the centroid of triangle load
x‘ = (1/3)(2) = 2/3m
w
400N
x
2/3 m
Distributed Loads
Example 1:
Solution: second load
FR = area under the loading diagram
FR = (400 N/m) (4m) = 1600 N
(x’) at the centroid of load area
x‘ = 2+(0.5)(4)= 4m
w
1600N
x
4m
Distributed Loads
Example 1:
Solution: third load
FR = area under the loading diagram
FR = (300 N/m) (2m) = 600 N
(x’) at the centroid of load area
x‘ = 2+4+(0.5)(2)= 7m
w
600N
x
7m
Distributed Loads
Example 1:
Solution: representing all forces
w
1600N
400N
2/3m
4m
7m
600N
Distributed Loads
Example 1:
Solution: representing all forces
FR =∑F = 400 + 1600 + 600 = 2600 N
To find the location of this force we
must use the technique of
simplifying of a force and couple
moment you learn previously
w
2600N
x’=4.2m
MR = FR * x’ → x’ = MR/FR
MR =(400)(2/3) + (1600)(4) + (600)(7) = 10867 N.m
x‘ = 10867/2600 = 4.2 m
Try to solve the same problem if the second load (400N/m) acts on the lower
surface of the body. Be aware to the sum of forces and the direction of resultant
moment
Distributed Loads
Example 2:
Determine the magnitude and the location of the equivalent resultant
force acting on the beam shown in the figure.
250 N/m
400 N/m
200 N/m
5m
5m
60o
10m
45o
Distributed Loads
Example 2:
1250 N
4000 N
1000 N
5m
5m
60o
10m
∑Fx = -(4000)(cos45) + (1000)(cos30) = -2198 N
∑Fy = -1250 – (4000)(sin45) – (1000)(sin30) = -4578 N
FR = {(-2198)2 + (-4578)2}1/2 = 5078 N
Ө = tan-1(-4578/-2198) = 64o
45o
Distributed Loads
Example 2:
to find the location of the resultant force, we can take the moment about a
certain point in the body. Let us take point O.
∑Mo = -(1000)(cos30)(0.5)(5sin60) – (1000)(sin30)(0.5)(5cos60)
– (1250)(2.5+5cos60) – (4000)(sin45)(5cos60+5+(0.5)10cos45)
+ (4000)(cos45)(0.5)(10sin45) = -29963 N.m
x‘ = Mo/FR = 29963 /5078 = 5.9 m from point O
1250 N
(4000)(sin45)
4000 N
1000 N (1000)(sin30)
5m
5m
(1000)(cos30)
O
60o
10m
(4000)(cos45)
45o