ce201 statics chap4
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1/11/2013
Chapter 4 - Force System Resultants 1
STATICS: CE201
Chapter 4
Force System Resultants
Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson
Dr M. Touahmia & Dr M. Boukendakdji
Civil Engineering Department, University of Hail
(2012/2013)
4. Force System Resultants ________________________________________________________________________________________________________________________________________________
Main Goals: 1. Determine the moment of a force.
2. Define the moment of a couple.
3. Determine the resultants of force systems.
Contents: 4.1 Moment of a Force – Scalar Formulation
4.2 Cross Product
4.3 Moment of a Force – Vector Formulation
4.4 Principle of Moments
4.5 Moment of a Force about a Specified Axis
4.6 Moment of a Couple
4.7 Simplification of a Force and Couple System
4.8 Further Simplification of a Force and Couple System
4.9 Reduction of a Simple Distributed Loading
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Chapter 4 - Force System Resultants 2
4.1 Moment of a Force – Scalar Formulation
When a force is applied to a body it will produce a
tendency for the body to rotate about a point or axis that
is not on the line of action of the force.
This tendency to rotate is called “moment of a force” or
simply the moment.
3 Chapter 4 - Force System Resultant
4.1 Moment of a Force – Scalar Formulation
For example, consider a wrench used to unscrew the bolt
in the figure. If a force is applied to the handle of the
wrench it will tend to turn the bolt about point O (or the
z axis).
The magnitude of the moment is directly proportional to
the magnitude of F and the perpendicular distance or
moment arm d.
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Chapter 4 - Force System Resultants 3
4.1 Moment of a Force – Scalar Formulation
The larger the force or the longer the moment arm, the
greater the moment or turning effect.
NOTE: If force F is applied at an angle then it
will be more difficult to turn the bolt since the moment
arm will be smaller than d. If F is applied
along the wrench, its moment arm will be zero and as a
result, the moment of F about O will be zero also.
5 Chapter 4 - Force System Resultant
90
sindd
4.1 Moment of a Force – Scalar Formulation
The moment of a force F about a point O is defined
as the vector product of F and d, where d is the
perpendicular distance from O to the line of action of the
force F.
Clearly the moment is a vector and has both: magnitude
and direction.
Magnitude: The magnitude of the moment is
determined from:
Units: N.m; kN.m; N.mm
OM
FdM O
OM
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Chapter 4 - Force System Resultants 4
4.1 Moment of a Force – Scalar Formulation
Direction: The right-hand rule is used
to establish the sense of direction of the
moment .
The moment of a force will be positive
if it is directed along the +z axis and
will be negative if it is directed along
the –z axis.
OM
7 Chapter 4 - Force System Resultant
4.1 Moment of a Force – Scalar Formulation
Resultant Moment: The Resultant Moment
about point O (the z axis) can be determined by finding
the algebraic sum of the moments caused by all the
forces in the system:
As a convention, we will generally consider positive
moments (+) as counterclockwise and negative moments
(-) as clockwise.
ORM
FdMOR
332211 dFdFdFFdMOR
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Chapter 4 - Force System Resultants 5
Example 1
For each case below, determine the moment of the force
about point O.
N.m 5.37m 75.0N 50 OM
Answer
Answer
9 Chapter 4 - Force System Resultant
mN. 200m 2N 100 OM
FdM O
Example 1
kN.m 229m 30cos2 m4kN 40 o
OM
kN.m 229m 45sin1kN 60 o
OM
Answer
Answer
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Chapter 4 - Force System Resultants 6
Example 2
Determine the resultant moment of the four forces acting
on the rod about point O.
11 Chapter 4 - Force System Resultant
Solution 2
Assuming that positive moments (+) are in the
counterclockwise:
FdMOR
12 Chapter 4 - Force System Resultant
N.m 334m 30cos3 m 4 N 40
m 30sin3 N 200 N 60m 2 N 50
o
o
ROM
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Chapter 4 - Force System Resultants 7
4.2 Cross Product
The cross product of two vectors A and B yields the
vector C written:
Magnitude: The magnitude of C is defined as the
product of the magnitudes of A and B and the sine of the
angle θ between their tails:
Direction: Vector C has a direction perpendicular to the
plane containing A and B and is specified
by the right-hand rule:
defines the direction of C
BAC
Cu
13 Chapter 4 - Force System Resultant
sin ABC
CAB u BAC sin
Laws of Operation:
Commutative law: The commutative law is not valid
If the cross product is multiplied by a scalar a, it obeys
the associative law:
The vector cross product also obeys the distributive law
of addition:
ABBA
CABBA
aaaa BABABABA
DABADBA
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Chapter 4 - Force System Resultants 8
Cross products of unit vectors (i, j, k):
The direction is determined using the right hand rule.
As shown in the diagram, for this case the direction is k
and the Magnitude is:
| i j |=(1)(1)(sin90°) = (1)(1)(1)=1
so: i j = (1) k = k
and:
i j = k i k = -j i i = 0
j k = i j i = -k j j = 0
k i = j k j = -i k k = 0
Alphabetical order +
15 Chapter 4 - Force System Resultant
Cartesian Vector Formulation
The cross product of two vectors A and B expressed in
Cartesian vector form is:
16 Chapter 4 - Force System Resultant
kkjkik
kjjjij
kijiii
kjikjiBA
zzyzxz
zyyyxy
zxyxxx
zyxzyx
BABABA
BABABA
BABABA
BBBAAA
kjiBA xyyxxzzxyzzy BABABABABABA
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Chapter 4 - Force System Resultants 9
Cartesian Vector Formulation
This equation may also be
written in a more compact
determinant form as :
The three minors can be generated with the following
scheme:
Adding the results yields the expanded form of : BA
17 Chapter 4 - Force System Resultant
kjiBA xyyxxzzxyzzy BABABABABABA
4.3 Moment of a Force – Vector Formulation
The moment of a force F about point O, or about an axis
passing through O and perpendicular to the plane
containing O can be expressed using the vector cross
product:
where r is the position vector from O to any point on the
line of action of F.
FrM O
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Chapter 4 - Force System Resultants 10
4.3 Moment of a Force – Vector Formulation
Magnitude:
Direction: The direction of Mo follows the right-hand
rule as it applies to the cross product, “r cross F”.
FdrFrFMO sinsin
19 Chapter 4 - Force System Resultant
4.3 Moment of a Force – Vector Formulation
Principle of Transmissibility: We can use any position
vector r measured from point O to any point on the line
of action of the force F:
FrFrFrM 321O
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Chapter 4 - Force System Resultants 11
4.3 Moment of a Force – Vector Formulation
Resultant Moment of a System of Forces: If a body is
subjected to a system of forces, the resultant moment of
all the forces about point O is equal to the vector
addition of the moment of each force:
n
i
iiRO
1
FrM
21 Chapter 4 - Force System Resultant
Example 3
Determine the moment MO produced by the force F
about point O. Express the result as a Cartesian vector.
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Chapter 4 - Force System Resultants 12
Solution 3
As shown in the figure, either rA or rB can be used to
determine the moment about O:
F expressed as Cartesian vector:
The direction of the unit vector uAB
can be determined from the position
Vector rAB which extends from A to B.
m 12 kr A m 12 4 jir B
23 Chapter 4 - Force System Resultant
ABFuF
m 12 12 4 kjir -AB
Solution 3
The magnitude of rAB which represents the length of AB
is:
Forming the unit vector that defines
The direction of both rAB and F, we
have:
24 Chapter 4 - Force System Resultant
222m 12m 12m 4 ABr
222m 12m 12m 4
12124
kjiru
AB
AB
ABr
kN 376.1 376.1 4588.0
m 12m 12m 4
12124kN 2
222
kji
kjiuF
ABF
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Chapter 4 - Force System Resultants 13
Solution 3
The moment about O: or
, ,
OR
m 12 kr A m 12 4 jir B
FrM AO FrM BO
kN 376.1 376.1 5488.0 kjiF
25 Chapter 4 - Force System Resultant
Example 4
Two forces act on the rod shown below. Determine the
resultant moment they create about the flange at O.
Express the results as a Cartesian vector.
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Chapter 4 - Force System Resultants 14
Solution 4
Position vectors rA and rB are directed from point O to
each force:
The resultant moment about O is:
m 5 jr A m 25 4 kjir B
27 Chapter 4 - Force System Resultant
21
1
FrFrFrM
BA
n
i
iiRO
304080
254
204060
050
21
kjikji
FrFrM BARO
Solution 4
The resultant moment about O is:
The coordinate direction
angles were determined
from the unit vector for
The two forces tend to cause the rod to rotate about the
moment axis in the manner shown by the curl indicate
on the moment vector.
ORM
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Chapter 4 - Force System Resultants 15
4.3 Principle of Moments
Varignon’s Theorem: The moment of a force about a
point is equal to the sum of the moments of the
components of the force about the point.
For example, consider the moment of the force F which
has two components F1 and F2, therefore:
Then, the moment of F about O is:
2121 FrFrFFrFrM O
21 FFF
29 Chapter 4 - Force System Resultant
4.3 Principle of Moments
For two-dimensional problems (2D), we can use the
principle of moments by resolving the force into its
rectangular components and then determine the moment
using a scalar analysis:
This method is generally easier than finding the same
moment using:
FdM O
xFyFMyxO
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Chapter 4 - Force System Resultants 16
Example 3
Determine the moment of the force about point O.
31 Chapter 4 - Force System Resultant
Example 3-(I)
The moment arm d can be found from trigonometry:
The force tends to rotate clockwise about O, the moment
is directed into the page .
m 898.275sinm 3 d
kN.m 5.14m 898.2kN 5 FdM O
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Chapter 4 - Force System Resultants 17
Solution 3-(II)
We can apply the principle of moments by resolving the
force into its rectangular components and then determine
the moment using a scalar analysis:
kN.m 5.14
kN.m 5.14
m 30cos3kN 45sin5m 30sin3kN 45cos5
OM
xyyxO dFdFM
33 Chapter 4 - Force System Resultant
Solution 3-(III)
The x and y axes can be set parallel and perpendicular to
the rod’s axis. Fx produces no moment about point O
since its line of action passes through this point:
xyO dFM
kN.m 5.14 OM
kN.m 5.14
m 3kN 75sin5
OM
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Chapter 4 - Force System Resultants 18
Example 4
The force F acts at the end of the angle bracket.
Determine the moment of F about point O.
35 Chapter 4 - Force System Resultant
Solution 4-(I) Scalar Analysis
F can be resolved into its x and y components:
N.m 6.98N.m 6.98
m 4.0N 30cos400m 2.0N 30sin400
OM
xFyFM yxO
30sin400 xF
36 Chapter 4 - Force System Resultant
30cos400 yF
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Chapter 4 - Force System Resultants 19
Solution 4-(II) Vector Analysis
Using a Cartesian vector analysis, the force F and the
position vector r can be written as:
m 2.0 4.0 jir
N 4.346 0.200
N 30cos40030sin400
ji
jiF
04.3460.200
02.04.0
kji
FrMO
N.m 6.98
0.2002.04.3464.0 0 0
k
kjiM
O
37 Chapter 4 - Force System Resultant
4.6 Moment of a Couple
A couple is defined as two parallel forces that have the
same magnitude, but opposite directions, and are
separated by a perpendicular distance d.
Since the resultant force is zero, the only effect of a
couple is to produce a rotation or tendency of rotation in
a specified direction.
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Chapter 4 - Force System Resultants 20
4.6 Moment of a Couple
The moment produced by a couple is called a couple
moment. We can determine its value by finding the sum
of the moments of both couple force about any arbitrary
point.
The position vectors rA and rB are directed from point O
to points A and B lying on the line of action –F and F.
The couple moment M about O is therefore:
However, or
Therefore:
FrrFrFrM ABAB
)()(
rrr AB AB rrr
FrM
39 Chapter 4 - Force System Resultant
4.6 Moment of a Couple
Scalar Formulation: The moment of a couple, M, is
defined as having a magnitude of:
Vector Formulation: The moment of a couple can also be
expressed by the vector cross product using:
where r is directed from any point on the line of action of
one of the forces to any point on the line of action of the
other force F.
FdM
FrM
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Chapter 4 - Force System Resultants 21
4.6 Moment of a Couple
Equivalent Couples: If two couples produce a moment
with the same magnitude and direction, then these two
couples are equivalent.
The two couple moments shown in the figure are
equivalent because each couple moment has a magnitude
of 12 N.m and each is directed into the plane of the page.
N.m 12m 4.0N 30 M
N.m 12m 3.0N 40 M
41 Chapter 4 - Force System Resultant
4.6 Moment of a Couple
Resultant Couple Moment: It is simply the vector sum of
all the couple moments of the system.
For example, consider the couple moments M1 and M2
acting on the pipe. We can join their tails at any arbitrary
point and find the resultant couple moment.
FrMR
21 MMM R
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Chapter 4 - Force System Resultants 22
Example 5
Determine the resultant couple moment of the three
couples acting on the plate.
43 Chapter 4 - Force System Resultant
Solution 5
Considering counterclockwise couple moments as
positive (+), we have:
N.m 95N.m 95
m 5.0N 300m 3.0N 450m 4.0N 200
RM
332211 dFdFdFMM R
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Chapter 4 - Force System Resultants 23
Example 6
Determine the magnitude and direction of the couple
moment acting on the gear.
45 Chapter 4 - Force System Resultant
Solution 6
Resolving each force into its components. The couple
moment can be determined by summing the moments of
these force components about any point (For example
the center O or point A):
OR
N.m 9.43
2.030sin6002.030cos600
OMM
N.m 9.43
2.030sin6002.030cos600
AMM
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Chapter 4 - Force System Resultants 24
Solution 6
The same result can also be obtained using:
where d is the perpendicular distance between the lines
of action of the couple forces. However, the computation
for d is more involved.
FdM
47 Chapter 4 - Force System Resultant
4.7 Simplification of a Force and Couple System
Sometimes it is convenient to reduce a system of forces and
couple moments acting on a body to a simpler form by
replacing it with an equivalent system.
Example:
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Chapter 4 - Force System Resultants 25
System of Forces and Couple Moments:
A system of several forces and couple moments acting
on a body can be reduced to an equivalent single
resultant force acting at a point O and a resultant couple
moment.
Chapter 4 - Force System Resultant 49
System of Forces and Couple Moments:
Example: In the figure, point O is not on the line of action
of F1 and so this force can be moved to point O provided a
couple moment is added to the body.
Similarly, the couple moment should be added
to the body when we move F2 to point O.
Finally, since the couple moment M is a free vector, it can
just be moved to point O.
50 Chapter 4 - Force System Resultant
111 FrM
222 FrM
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Chapter 4 - Force System Resultants 26
System of Forces and Couple Moments:
If we sum the forces and couple moments, we obtain the
resultant force and the resultant couple moment:
Chapter 4 - Force System Resultant 51
21 FFF R 21 MMMM
OR
System of Forces and Couple Moments:
We can generalize this method of reducing a force and
couple system to an equivalent resultant forces FR acting
at point O and a resultant couple moment (MR)O by
using the following equations:
If the force system lies in the x-y plane and any couple
moments are perpendicular to this plan, then the above
equations reduce to the three scalar equations:
Chapter 4 - Force System Resultant 52
FFR
MMM OOR
xxR FF yyR FF
MMM OOR
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Chapter 4 - Force System Resultants 27
Example 7
Replace the force and couple system by an equivalent
resultant force and couple moment acting at point O.
Chapter 4 - Force System Resultant 53
Solution 7
Force Summation: The 3 kN and 5 kN forces are resolved
into their x and y components as shown below. We have:
54 Chapter 4 - Force System Resultant
kN 598.5kN 55
330coskN 3
xxR FF
kN 50.6kN 4kN 55
430sinkN 3
yyR FF
kN 58.850.6598.52222 RyRxR FFF
o
xR
yR
F
F3.49
598.5
50.6tantan 11
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Chapter 4 - Force System Resultants 28
Solution 7
Moment Summation: The moments of 3 kN and 5 kN
about point O will be determined using their x and y
components. We have:
55 Chapter 4 - Force System Resultant
kN 46.2kN.m 46.2m 2.0kN 4
m 5.0kN 55
4m 1.0kN 5
5
3
m 1.030coskN 3m 2.030sinkN 3
OOR MM
Example 8
Replace the force and couple system acting on the
member by an equivalent resultant force and couple
moment acting at point O.
Chapter 4 - Force System Resultant 56
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Chapter 4 - Force System Resultants 29
Solution 8
Force Summation: Since the couple forces of 200 N are
equal but opposite, they produce zero resultant force.
The 500 N force is resolved into its x and y components:
Chapter 4 - Force System Resultant 57
N 300kN 5005
3
xxR FF
N 3507505
4N 500
yyRFF
N 462N 350N 3002222 yRxRR FFF
4.49
N 300
N 350tantan 11
xR
yR
F
F
Solution 8
Moment Summation: Since the couple moment is a free
vector, it can act at any point on the member, then:
Chapter 4 - Force System Resultant 58
N.m 5.37N.m 5.37
m 1N 200m 25.1N 750
m 15
3N 500N 5.2
5
4N 500
cOOR MMM
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Chapter 4 - Force System Resultants 30
4.8 Reduction of a Simple Distributed Loading
Sometimes, a body may be subjected to a loading that is
distributed over its surface.
For example:
The pressure of the wind on the face of a sign.
The pressure of water within a tank.
The weight of sand on the floor of a storage container.
The pressure exerted at each point on the surface
indicates the intensity of the loading. It is measured
using Pascals Pa (N/m2).
Chapter 4 - Force System Resultant 59
Uniform Loadings Along a Single Axis:
The most common type of distributed
loading encountered in engineering practice
is generally uniform along a single axis.
Example: The beam (or plate) that has a
constant width and is subjected to a pressure
loading that varies only along the x axis.
This loading can be described by the
function:
We can replace this coplanar parallel force
system with a single equivalent resultant
force FR acting at a specific location on the
beam.
60 Chapter 4 - Force System Resultant
2N/m )(xpp
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Chapter 4 - Force System Resultants 31
Magnitude of Resultant Force:
The magnitude of FR is equivalent to the
sum of all the forces in the system:
Integration must be used since there is an
infinite number of parallel forces dF acting
on the beam.
Since dF is acting on an element of
length dx, and w(x) is a force per unit
length, Then:
Therefore, the magnitude of the
resultant force is equal to the total area A
under the loading diagram.
AL
R AdAdxxwF
61 Chapter 4 - Force System Resultant
FFR
dAdxxwdF
Location of Resultant Force:
The location of the line of action of FR
can be determined by equating the moments
of the force resultant and the parallel force
distribution about point O.
dF produces a moment of:
Then for the entire length:
x
L
R dxxxwFx
A
A
L
L
dA
xdA
dxxw
dxxxwx
62 Chapter 4 - Force System Resultant
dxxxwxdF
OOR MM
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Chapter 4 - Force System Resultants 32
Example 9
Determine the magnitude and location of the equivalent
resultant force acting on the shaft.
Chapter 4 - Force System Resultant 63
Solution 9
Since w = w(x) is given, this problem will be solved by
integration.
The differential element has an area:
Chapter 4 - Force System Resultant 64
N 1603
0
3
260
0
m2
36060x
33m2
0
2
xdxdAF
A
R
m 5.1
N 160
4
0
4
260
N 160
0
m 2
460
N 160
60444m 2
0
2
x
dxxx
dA
xdA
x
A
A
dxxwdxdA 260
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Chapter 4 - Force System Resultants 33
Example 10
The granular material exerts the distributed loading on
the beam. Determine the magnitude and location of the
equivalent resultant of this load.
Chapter 4 - Force System Resultant 65
Solution 10
The area of the loading diagram is a trapezoid. It can be
divided into a rectangular and triangular loading.
Chapter 4 - Force System Resultant 66
kN 225kN/m 50m 92
11 F
kN 450kN/m 50m 92 F
m 3m 93
11 x
m 5.4m 92
12 x
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Chapter 4 - Force System Resultants 34
Solution 10
The two parallel forces F1 and F2 can be reduced to a
single resultant force FR.
The magnitude of FR is:
We can find the location of F with reference to point A:
Chapter 4 - Force System Resultant 67
kN 675450225 RF
FFR
AAR MM
4505.42253675 x
m 4x