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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.
2.
3.
Determine the moment of a force.
Define the moment of a couple.
Determine the resultants of force systems.
Contents:
4.1
4.2
4.3
4.4
4.5
4.6
4.7
Moment of a Force – Scalar Formulation
Cross Product
Moment of a Force – Vector Formulation
Principle of Moments
Moment of a Force about a Specified Axis
Moment of a Couple
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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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.
Chapter 4 - Force System Resultant
4.1
3
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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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   90  then it
will be more difficult to turn the bolt since the moment
arm d   d sin  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.
Chapter 4 - Force System Resultant
4.1
5
Moment of a Force – Scalar Formulation

The moment M O 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
determined from:
M O is
M O  Fd

Units: N.m; kN.m; N.mm
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4.1

Moment of a Force – Scalar Formulation
Direction: The right-hand rule is used
to establish the sense of direction of the
moment M O .

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.
Chapter 4 - Force System Resultant
4.1

7
Moment of a Force – Scalar Formulation
Resultant Moment: The Resultant Moment M R O
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:
M R O   Fd

As a convention, we will generally consider positive
moments (+) as counterclockwise and negative moments
(-) as clockwise.
M R O   Fd  F1d1  F2 d 2  F3 d 3
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Example 1

For each case below, determine the moment of the force
about point O.
Answer
M O  Fd
M O  100 N2 m  200 N.m
Answer
M O  50 N0.75 m  37.5 N.m
Chapter 4 - Force System Resultant
9
Example 1
Answer


M O  40 kN 4m  2 cos 30 o m  229 kN.m
Answer


M O  60 kN 1sin 45 o m  229 kN.m
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Example 2

Determine the resultant moment of the four forces acting
on the rod about point O.
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Solution 2

Assuming that positive moments (+) are in the
counterclockwise:
M R O   Fd

M RO  50 N 2 m   60 N 0  20 N 3 sin 30o m



 40 N 4 m  3 cos 30o m  334 N.m
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4.2
Cross Product

The cross product of two vectors A and B yields the
vector C written: C  A  B

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: C  AB sin 

Direction: Vector C has a direction perpendicular to the
plane containing A and B and is specified
by the right-hand rule:
C  A  B   AB sin   uC
u C defines the direction of C
Chapter 4 - Force System Resultant
13
Laws of Operation:

Commutative law: The commutative law is not valid
AB  B A
A  B  B  A  C

If the cross product is multiplied by a scalar a, it obeys
the associative law: aA  B  aA  B  A  aB  A  Ba

The vector cross product also obeys the distributive law
of addition:
A  B  D  A  B  A  D
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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  +
Chapter 4 - Force System Resultant
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Cartesian Vector Formulation

The cross product of two vectors A and B expressed in
Cartesian vector form is:
A  B  Ax i  Ay j  Az k   B x i  B y j  B z k 
 Ax B x i  i   Ax B y i  j  Ax B z i  k 
 Ay B x j  i   Ay B y j  j  Ay B z j  k 

 Az B x k  i   Az B y k  j  Az B z k  k 
A  B  Ay Bz  Az B y i   Ax Bz  Az Bx j  Ax B y  Ay Bx k
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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 A  B :
A  B  Ay Bz  Az B y i   Ax Bz  Az Bx j  Ax B y  Ay Bx k
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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:
MO  r  F
where r is the position vector from O to any point on the
line of action of F.
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4.3 Moment of a Force – Vector Formulation


Magnitude: M O  rF sin   F r sin    Fd
Direction: The direction of Mo follows the right-hand
rule as it applies to the cross product, “r cross F”.
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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:
M O  r1  F  r2  F  r3  F
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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
M RO   ri  Fi 
i 1
Chapter 4 - Force System Resultant
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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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Solution 3

As shown in the figure, either rA or rB can be used to
determine the moment about O:
rA  12 k m

rB  4 i  12 j m
F expressed as Cartesian vector:
F  Fu AB

The direction of the unit vector uAB
can be determined from the position
Vector rAB which extends from A to B.
rAB  4 i  12 j-12 k m
Chapter 4 - Force System Resultant
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Solution 3

The magnitude of rAB which represents the length of AB
is:
2
2
2
rAB  4 m  12 m   12 m

Forming the unit vector that defines
The direction of both rAB and F, we
have:
u AB 
rAB

rAB
4i  12 j  12k
4 m  12 m2   12 m2
2

4i  12 j  12k
F  Fu AB  2 kN 
2
 4 m   12 m 2   12 m 2

 0.4588 i  1.376 j  1.376 k kN




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Solution 3

The moment about O: M O  rA  F
or
M O  rB  F
rA  12 k m, rB  4 i  12 j m , F  0.5488 i  1.376 j  1.376 k kN
OR
Chapter 4 - Force System Resultant
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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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Solution 4

Position vectors rA and rB are directed from point O to
each force:
rA  5 j m

rB  4 i  5 j  2 k m
The resultant moment about O is:
n
M RO   ri  Fi  rA  F1   rB  F2 
i 1
M RO  rA  F1   rB  F2 
i
j k
i
j
k
 0
5 0  4 5 2
 60 40 20 80 40  30
Chapter 4 - Force System Resultant
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Solution 4

The resultant moment about O is:

The coordinate direction
angles were determined
from the unit vector for M R
O

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.
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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:
F  F1  F2

Then, the moment of F about O is:
M O  r  F  r  F1  F2   r  F1  r  F2
Chapter 4 - Force System Resultant
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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:
M  F y  F x
O

x
y
This method is generally easier than finding the same
moment using:
M O  Fd
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Example 3

Determine the moment of the force about point O.
Chapter 4 - Force System Resultant
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Example 3-(I)

The moment arm d can be found from trigonometry:
d  3 msin 75  2.898 m
M O  Fd  5 kN2.898 m  14.5 kN.m

The force tends to rotate clockwise about O, the moment
is directed into the page .
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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:
M O   Fx d y  Fy d x


 

M O   5 cos 45 kN 3 sin 30 m  5 sin 45 kN 3 cos 30 m

 14.5 kN.m
 14.5 kN.m
Chapter 4 - Force System Resultant
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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:
M O   Fy d x


M O   5 sin 75 kN 3 m 
 14.5 kN.m
M O  14.5 kN.m
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Example 4

The force F acts at the end of the angle bracket.
Determine the moment of F about point O.
Chapter 4 - Force System Resultant
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Solution 4-(I) Scalar Analysis

F can be resolved into its x and y components:
Fx  400 sin 30  
Fy  400 cos 30 
M O  Fx y  Fy x




M O  400 sin 30  N 0.2 m  400 cos 30  N 0.4 m
 98.6 N.m  98.6 N.m
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Solution 4-(II) Vector Analysis

Using a Cartesian vector analysis, the force F and the
position vector r can be written as:


F  400 sin 30  i  400 cos 30  j N
 200.0 i  346.4 j N
r  0.4 i  0.2 j m
i
M O  r  F  0.4
j
 0.2
k
0
200.0  346.4 0
M O  0 i  0 j  0.4 346.4   0.2200.0 k
  98.6 k N.m
Chapter 4 - Force System Resultant
4.6
37
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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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:
M  (r  F)  (r  F)  r  r   F
B
A
B

However, rB  rA  r

Therefore:
or
A
r  rB  rA
M rF
Chapter 4 - Force System Resultant
4.6

39
Moment of a Couple
Scalar Formulation: The moment of a couple, M, is
defined as having a magnitude of:
M  Fd

Vector Formulation: The moment of a couple can also be
expressed by the vector cross product using:
M rF
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.
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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.
M  30 N0.4 m  12 N.m
M  40 N0.3 m  12 N.m

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.
Chapter 4 - Force System Resultant
4.6

41
Moment of a Couple
Resultant Couple Moment: It is simply the vector sum of
all the couple moments of the system.
M R   r  F 

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.
M R  M1  M 2
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Example 5

Determine the resultant couple moment of the three
couples acting on the plate.
Chapter 4 - Force System Resultant
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Solution 5

Considering counterclockwise couple moments as
positive (+), we have:
M R   M   F1 d1  F2 d 2  F3 d 3
M R  200 N 0.4 m  450 N 0.3 m  300 N 0.5 m
 95 N.m  95 N.m
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Example 6

Determine the magnitude and direction of the couple
moment acting on the gear.
Chapter 4 - Force System Resultant
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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):
M  MO







 600 cos 30  0.2  600 sin 30  0.2
 43.9 N.m

OR
M  MA

 600 cos 30  0.2  600 sin 30  0.2
 43.9 N.m
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Solution 6

The same result can also be obtained using:
M  Fd
where d is the perpendicular distance between the lines
of action of the couple forces. However, the computation
for d is more involved.
Chapter 4 - Force System Resultant
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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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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
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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 M1  r1  F1 is added to the body.
Similarly, the couple moment M 2  r2  F2 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.
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System of Forces and Couple Moments:

If we sum the forces and couple moments, we obtain the
resultant force and the resultant couple moment:
M R O  M  M1  M 2
FR  F1  F2
Chapter 4 - Force System Resultant
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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:
FR   F
M R O   M O   M

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:
FR x   Fx
FR  y   Fy
M R O   M O   M
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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
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Solution 7
Force Summation: The 3 kN and 5 kN forces are resolved
into their x and y components as shown below. We have:
FR  x   Fx  3 kN cos 30    3 5 kN   5.598 kN
5
FR  y   Fy  3 kN sin 30
FR 
FRx 2  FRy 2

 

4
  5 kN   4 kN  6.50 kN
5
5.5982  6.502
 8.58 kN
 FR y 
6.50 
o
  tan 1 
  49.3

 5.598 
 FR x 
  tan 1 
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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:
M R O   M O  3 kN  sin 30  0.2 m   3 kN  cos 30  0.1 m 
 3
4
  5 kN 0.1 m    5 kN 0.5 m 
5
 
5
 4 kN 0.2 m   2.46 kN.m  2.46 kN
Chapter 4 - Force System Resultant
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Example 8

Replace the force and couple system acting on the
member by an equivalent resultant force and couple
moment acting at point O.
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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:
FR  x   Fx
F 
R
y
FR 
 3
  500 kN   300 N
5
 4
  F  500 N    750  350 N
5
y
FR x 2  FR 2 y

300 N2  350 N2
 462 N
 FR y 
350 N 

  tan 1 
  49.4



F
 300 N 
 R x
  tan 1 
Chapter 4 - Force System Resultant
57
Solution 8

Moment Summation: Since the couple moment is a free
vector, it can act at any point on the member, then:
M R O   M O   M c  500 N  4 2.5 N   500 N  3 1 m
5
5
 
 
 750 N 1.25 m   200 N 1 m 
 37.5 N.m  37.5 N.m
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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
2
function: p  p(x) N/m
We can replace this coplanar parallel force
system with a single equivalent resultant
force FR acting at a specific location on the
beam.
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Magnitude of Resultant Force:
The magnitude of FR is equivalent to the
sum of all the forces in the system: FR   F

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:
dF  wx dx  dA

FR   wx dx   dA  A
L
A
Therefore, the magnitude of the
resultant force is equal to the total area A
under the loading diagram.

Chapter 4 - Force System Resultant
61
Location of Resultant Force:
The location x 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: xdF  xwx dx

Then for the entire length: M R O   M O
 xFR   xwx dx
L
x
 xwx dx   xdA
 wx dx  dA
L
L
A
A
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Example 9

Determine the magnitude and location of the equivalent
resultant force acting on the shaft.
Chapter 4 - Force System Resultant
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Solution 9


Since w = w(x) is given, this problem will be solved by
integration.
The differential element has an area: dA  wdx  60x 2 dx
FR   dA 
A
 x3
2

60x
dx

60
0
 3
2m
 xdA  x60 x dx
2m
2
x
A
 dA

0
160 N
 2m
 23 0 

 60    160 N
 0
 3 3
 x4  2 m
 24 04
60 
60 
4
 4  0
 4


160 N
160 N



 1.5 m
A
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Example 10

The granular material exerts the distributed loading on
the beam. Determine the magnitude and location of the
equivalent resultant of this load.
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Solution 10

The area of the loading diagram is a trapezoid. It can be
divided into a rectangular and triangular loading.
F1 
1
9 m50 kN/m  225 kN
2
F2  9 m50 kN/m  450 kN
x1 
1
9 m  3 m
3
x2 
1
9 m  4.5 m
2
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Solution 10

The two parallel forces F1 and F2 can be reduced to a
single resultant force FR.

The magnitude of FR is:
FR   F
FR  225  450  675 kN

We can find the location of F with reference to point A:
M R  A   M A
x675  3225  4.5450
x  4m
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