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Chapter Four
Laith Batarseh
Statics
MOMENT
By
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Statics
MOMENT SCALAR
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 )
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The acting forceMoment arm
Description
Moment depends on two variables:
Moment is a vector quantity
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Statics
MOMENT SCALAR
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Description
ForceArmTendency to rotate
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Statics
MOMENT SCALAR
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Tendency for rotation
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Statics
MOMENT SCALAR
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Magnitude
FD
Moment magnitude (M) = F.D
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Statics
MOMENT SCALAR
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Direction
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Statics
MOMENT SCALAR
Solving procedures
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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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Statics
MOMENT SCALAR
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Example [1]
Find the moment caused by the following forces about point O
100 N
0.5m
2m
(b)
O
100 N
0.5m
2m
(a)
O
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Statics
MOMENT SCALAR
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Example [1]
Assume the CCW direction is the positive direction.
100 N
0.5m
2m
(b)
O
100 N
0.5m
2m
(a)
O
Branch (a) Mo = F.d = -(100N)(0.5m) Mo=-50 N.m=50N.m CWBranch (b) Mo=F.d = (100N)(2m) Mo=200 N.m CCW
+
+
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Statics
Principle Of Moments
Principle of Moments
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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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Statics
Principle Of Moments
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Example [1]
Find the moment caused by the following forces about point O
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Statics
Principle Of Moments
Example [1]
Mo,1 = 100 sin(30) (10) = 500 N.m
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Mo,2 =- 100 cos(30) (5) =- 433N.m+
M = Mo,1+Mo,2=500-433=67N CCW
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Statics
Principle Of Moments
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Example [2]
1. Force analysis
100 cos(40)1.2 m
0.3 m
100 sin(40)
120 cos(60)
120 sin(60)
2. Moment calculations
∑ M = (100 cos(40))(1.5) –( 120 cos(60))(1.2) =43 N.m CCW +
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Statics
MOMENT SCALAR
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Moment resultant
F1
F2F3
d1
d2 d3
M1O
M2
M3
Mo = ∑Mo = M1 + M2 – M3 = F1d1+F2d2 – F3d3 +
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Statics
MOMENT SCALAR
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Example [2]
Find the moment caused by the following forces about point O
2m
3m
5m
1m30o
100 N
50 N
60 N
75 NO
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MOMENT SCALAR
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Example [2]
2m
3m
5m
1m30o
100 N
50 N
60 N
75 NO
CWmNmNM
M
o
o
.45.272.45.272
1)30cos(755)30sin(750503602100
+
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MOMENT SCALAR
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Exercise
Find the moment caused by the following forces about point O
100 N
300 N
5m
2m 45o
30oO 0.3m
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Exercise
CWmNmNM
M
o
o
.897.897
5)45sin(3003.0)45cos(3002)30sin(100
+
100 N
300 N
5m
2mO 0.3m
300 sin (45)N
300 cos (45)N
100sin (30)N
100cos (30)N
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Statics
MOMENT VECTOR
Cross product
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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.The cross product is controlled by the right-hand rule
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MOMENT VECTOR
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Graphical representation
ϕ
Uc
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MOMENT VECTOR
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Cartesian vector formulation
i=jxk
j=kxi
K=ixj
y
x
z
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Statics
MOMENT VECTOR
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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
zyx
zyx
BBBAAAkji
x BA
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MOMENT VECTOR
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Moment – vector formulation
F
rθ
d
M
Ox
yMagnitude: Mo = rFsin(θ) = Fd
Direction: perpendicular to x-y plane (z-direction)
Matrix notation:
Resultant moment:MRo = ∑(rxF)
zyx
zyx
FFFrrrkji
x FrMo
Best for three dimensional
problems
Mo = rxF
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Statics
MOMENT VECTOR
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Example [1]
Find the moment caused by the following forces about point O
F = [5i + 10j + 6k]N
z
x
y3m
2m1m
O
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Statics
MOMENT VECTOR
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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
kjikji
x 201326105123 FrMo
F = [5i + 10j + 6k]Nr = 3i+2j+1k
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MOMENT VECTOR
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Example [2]
Find the moment caused by the following forces about point O
xy
z
O
F = [-5i + 5j -5k]N
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MOMENT VECTOR
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Example [2]
1. Formulate the position vector (r) : r= 15i + 10j +6k
2. Find the moment vector (Mo) by matrix notation
kjikji
x 1254580555
61015
FrMo
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MOMENT VECTOR
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Summary
Moment is a vector can be found by cross product and
matrix notation
Matrix notation:
zyx
zyx
FFFrrrkji
x FrMo
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Statics
Principle Of Moments
Principle of Moments
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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)
![Page 30: Statics Chapter Four Moment By Laith Batarseh Home Next Previous End.](https://reader035.fdocuments.in/reader035/viewer/2022081512/5a4d1b067f8b9ab0599887ee/html5/thumbnails/30.jpg)
Statics
Principle Of Moments
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Example [1]
Find the moment caused by the following forces about point O
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Statics
Principle Of Moments
Example [1]
Mo,1 = 100 sin(30) (10) = 500 N.m
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Mo,2 =- 100 cos(30) (5) =- 433N.m+
M = Mo,1+Mo,2=500-433=67N CCW
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Principle Of Moments
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Example [2]
The following is a gate. In which direction this gate will rotate?
100N
120N
1.2 m
0.3 m60o
40o
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Principle Of Moments
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Example [2]
1. Force analysis
100 cos(40)1.2 m
0.3 m
100 sin(40)
120 cos(60)
120 sin(60)
2. Moment calculations
∑ M = (100 cos(40))(1.5) –( 120 cos(60))(1.2) =43 N.m CCW +
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Statics
Principle Of Moments
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Example [3]
Find the moment caused by the following forces about point O
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Principle Of Moments
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Example [2]
r1= 15i + 10j +6k F1 = [5j]N kikji
x 753005061015 FrMo,1
r2= 15i + 10j -6k F2 = [-5j]N kikji
x 753005061015
FrMo,2
i60oM
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Statics
Moment of force about axis
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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)
Mo,3 =0 (about z-axis)
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Statics
Moment of force about axis
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Magnitude
Scalar analysis: M = F.d
Vector analysis:
zyx
zyx
z,ay,ax,a
a
FFFrrr
uuux.M Frua
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
To find the moment vector (Ma): Ma. ua
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Statics
Moment of force about axis
Example [1]
Mo,1 = 100 sin(30) (10) = 500 N.m
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Mo,2 =- 100 cos(30) (5) =- 433N.m+
M = Mo,1+Mo,2=500-433=67N CCW
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Statics
Moment of force about axis
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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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Statics
Moment of force about axis
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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)
m.N144105345
001M x
m.N5
4105345
010M y
m.N30
4105345
100M z
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Statics
Moment of force about axis
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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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Statics
Moment of couple
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Couple is the moment generated by two forces has the same
magnitude and opposite direction.
Definition
F
-F
d
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Statics
Moment of couple
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This analysis is considered for 2-D
problems
The magnitude of moment is found
by: M=F.d where F is the force
magnitude.
the direction of couple moment is
perpendicular to the plain that
contain the F and d and it is found by
the right hand rule.
Scalar analysis
-FF
d
M
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Statics
Moment of couple
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Example on scalar analysis
F2d1
d2
-F2-F1
F1
M1 = F1 . d1 M2 = F2 . d2
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Moment of couple
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Vector analysis
This analysis is considered for 3-D
problems
The moment vector is found by:
M=F x r where r is the position vector
directed between the forces F and –F
Note that the moment vector is
dependent on the position vector
directed between the forces F and –F
(r).
Derivation: M = rB x F + rA x –F = (rB - rA) x F But: (rB - rA) = r Then: M = rx F
F-F
O
r
rArB
A
B
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Statics
Moment of couple
Example [1]
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Find the resultant moment couple produced by the following forces. All dimensions in m
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Moment of couple
Example [1]
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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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Statics
Moment of couple
Example [2]
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Find the moment couple produced by the following force. All dimensions in m
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Statics
Moment of couple
Example [2]
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1. F = 150 j 2. r = 3k 3. M = r x F = 150 j x 3k = {450N.m} i
M
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Statics
Moment of couple
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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)
Equivalent couples
F1
-F1
d1 d2
F2
-F2
=F1.d1 = F2.d2
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Statics
Moment of couple
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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.
Resultant couple
M1 M2
M1
M2
MR
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Statics
Moment of couple
Example [1]
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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.
150 N
150 N
100N
100N
200N
200N
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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)
Statics
Moment of couple
Example [1]
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Statics
Moment of couple
Example [2]
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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
3m
6m
0.6m
45o 100 N
100 N200 N
45o
200 N
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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)
Statics
Moment of couple
Example [2]
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Statics
Moment of couple
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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
3-D problems
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Statics
Moment of couple
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3-D problems
M1 M2
x
z
y
θ
Example:
The figure show an object subjected to
two moments: M1 and M2. as you can
see M2 has an angle θ from the y-axis.
The moment M1 can be represented as:
M1= M1i.
While the moment
M2 can be represented as:
M2 = M2 (0i + cos(θ) j + sin(θ) k)
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Statics
Simplification of a force and couple system
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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.
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
Equivalent couples
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Statics
Simplification of a force and couple system
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Simplification conditions
B A BF F-F F
Note: the acting force can be transport from one position to another on its line of action (i.e. force vector)
A B A B
F
M = F.d
F
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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Statics
Simplification of a force and couple system
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Simplification conditions
F1
O
F2
r1r2
M
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.
Fig.a
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Statics
Simplification of a force and couple system
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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.
Fig.b
O
F2 M
M1 = r1 x F1M2 = r2 x F2
F1
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Statics
Simplification of a force and couple system
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Simplification conditions
•Then all forces and moments are summed using the following formulas:
FR = ∑F MR,O = ∑MO + ∑M
Fig.c
O
FRMR,o
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Statics
Simplification of a force and couple system
Example [1]
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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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Solution:
First, calculate the resultant force FR
∑Fx = 300 – 100 = 200 N
∑Fy = 200 + 150 = 350 N
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.
Statics
Simplification of a force and couple system
Example [1]
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NFR 403350200 22 o60
200350tan 1
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Statics
Simplification of a force and couple system
Example [2]: couple resultant
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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
3m
6m
0.6m
45o 100 N
100 N200 N
45o
200 N399 N.m
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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)
Note that the resultant force (FR ) = 0 which is true for all the coupled forces systems
Statics
Simplification of a force and couple system
Example [2]
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Statics
Simplification of a force and couple system
Example [3]
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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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Solution:
First, calculate the resultant force FR
Finally, calculate the resultant moment MR using cross product
Statics
Simplification of a force and couple system
Example [3]
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kk ]210[8050180 NFR
mN
kjikjikjix
.}22515{
500005.02
800005.025.1
180000025.1
ji
FrMo
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Statics
Simplification of a force and couple system
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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
Special cases:
F1 F2
F4F3
OFR
O=
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Statics
Simplification of a force and couple system
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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.
Special cases:
F1 F2
F4F3
OFR
OM
=
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Statics
Simplification of a force and couple system
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Parallel forces system:
F2
F1
F3
F4 O
zFR = ∑F
O
zMR,O
b
aFR = ∑F O
z
b
a
d
A reverse process can be done to transform the single force – moment system into a single force with moment arm from the rotation point
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Statics
Simplification of a force and couple system
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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.
Analysis procedures
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Statics
Simplification of a force and couple system
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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.
Special cases:
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Statics
Simplification of a force and couple system
Example [1]
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Replace the following forces-moment system to a single force system.
5m
10 kN 7 kN
30o
2m
8m
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Solution:
First, calculate the resultant force F
∑Fx = 10 + 7cos(30) = 16.06kN
∑Fy = - 7 sin(30) = 3.5 kN
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
Statics
Simplification of a force and couple system
Example [1]
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kNFR 43.165.306.16 22 o3.12
06.165.3tan 1
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Statics
Simplification of a force and couple system
Example [2]
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Replace the following forces-moment system to a single force system.
3m3m3m
x
z
y
200N
300 N600 N
4m
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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
Statics
Simplification of a force and couple system
Example [2]
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Solution:
Finally, you can represent the new force on the original system.
Statics
Simplification of a force and couple system
Example [2]
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z
y700 N
4.29m4.72 m
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Statics
Distributed Loads
Uniform loading along a single axis
x
w
L
w = w(x)
x
w
L
dF = dA
x
dxw = w(x)
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).
The force function could be linear or none linear
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
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Statics
Distributed Loads
Magnitude of resultant force
x
FRw
w = w(x)
'x
Let us first assume a distributed force w(x) acting on the member as shown in the fig. 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:
AdAdxxwF
AR .
As you can see from the above equation that the magnitude of FR equal the area under the curve w(x)
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Statics
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:
x
FRw
w = w(x)
'x
A
A
L
L
dA
dAx
dxxw
dxxxwx
.
.
.'
For this stage, the location of the centroid for the given shape will be given
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Statics
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
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Statics
Distributed Loads
Example 1:
x
w200 N/m
2m(a)
x
w 200 N/m
3m(b)
Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figures.
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Statics
Distributed Loads
Example 1:
x
w200 N/m
2m
Solution: part a
FR = area under the loading diagram
FR = (200 N/m) (2m) = 400 N
(x’) at the center of load rectangle
x‘ = 1m
x
w
1m
400 NYou can note that the centroid of a rectangular area is its geometric center
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Statics
Distributed Loads
Example 1:
Solution: part b
Similar to part aFR = area under the loading diagram
FR = (1/2)(200 N/m) (3m) = 300 N
(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.
x
w 200 N/m
3m
x
w
2m
300 N
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Statics
Distributed Loads
Example 2:
Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figure.
x
w 250 N/m
w = (30)x2 N/m
2.5m
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Statics
Distributed Loads
Example 2:
Solution: magnitude
x
w156.25 N
NF
xdxxdAF
R
AR
25.15630
35.230
330.30
33
5.2
0
35.2
0
2
x
w 250 N/m
w = (30)x2 N/m
2.5m
Try to solve it by your self and verify the solution
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Statics
Distributed Loads
Example 2:
Solution: location
x
w156.25 N
1.875 m
x
w 250 N/m
w = (30)x2 N/m
2.5m
m
x
x
dxxx
dA
dAxx
A
A
875.125.156
430
25.156
.30.
5.2
0
4
5.2
0
2
Try to solve it by your self and verify the solution
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Statics
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
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Statics
Distributed Loads
Example 1:
Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figure.
x
w400N/m
300N/m
2m 4m 2m
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Statics
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
x
w
2/3 m
400N
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Statics
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
x
w
4m
1600N
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Statics
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
x
w
7m
600N
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Statics
Distributed Loads
Example 1:
Solution: representing all forces
w
2/3m 4m
7m
600N1600N400N
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Statics
Distributed Loads
Example 1:
Solution: representing all forces
w
x’=4.2m
2600NFR =∑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
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
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Statics
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
200 N/m 400 N/m
60o 45o
5m
5m 10m
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Statics
Distributed Loads
Example 2:
1250 N
1000 N
4000 N
60o 45o
5m
5m 10m
∑Fx = -(4000)(cos45) + (1000)(cos30) = -2198 N∑Fy = -1250 – (4000)(sin45) – (1000)(sin30) = -4578 NFR = {(-2198)2 + (-4578)2}1/2 = 5078 N Ө = tan-1(-4578/-2198) = 64o
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Statics
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
O
1250 N
1000 N
4000 N
60o 45o
5m
5m 10m(4000)(cos45)
(4000)(sin45)
(1000)(cos30)
(1000)(sin30)