Chapter 4:
Force System Resultants
Engineering Mechanics: Statics
Chapter Objectives





To discuss the concept of the moment of a force and
show how to calculate it in two and three dimensions.
To provide a method for finding the moment of a force
about a specified axis.
To define the moment of a couple.
To present methods for determining the resultants of
non-concurrent force systems.
To indicate how to reduce a simple distributed loading
to a resultant force having a specified location.
Chapter Outline
 Moment
of a Force – Scalar Formation
 Cross Product
 Moment of Force – Vector Formulation
 Principle of Moments
 Moment of a Force about a Specified Axis
Chapter Outline
 Moment
of a Couple
 Equivalent System
 Resultants of a Force and Couple System
 Reduction of a Simple Distributed Loading
4.1 Moment of a Force
– Scalar Formation
Moment of a force about a point or axis – a
measure of the tendency of the force to
cause a body to rotate about the point or axis
Case 1
Consider horizontal force Fx,
which acts perpendicular to
the handle of the wrench and
is located dy from the point O
4.1 Moment of a Force
– Scalar Formation
Fx tends to turn the pipe about the z axis
 The larger the force or the distance dy, the
greater the turning effect
 Torque – tendency of
rotation caused by Fx
or simple moment (Mo) z

4.1 Moment of a Force
– Scalar Formation
Moment axis (z) is perpendicular to shaded
plane (x-y)
 Fx and dy lies on the shaded plane (x-y)
 Moment axis (z) intersects
the plane at point O

4.1 Moment of a Force
– Scalar Formation
Case 2
Apply force Fz to the wrench
 Pipe does not rotate about z axis
 Tendency to rotate about x axis
 The pipe may not actually
rotate Fz creates tendency
for rotation so moment
(Mo) x is produced
4.1 Moment of a Force
– Scalar Formation
Case 2
 Moment axis (x) is perpendicular to shaded
plane (y-z)
 Fz and dy lies on the shaded plane (y-z)
4.1 Moment of a Force
– Scalar Formation
Case 3
Apply force Fy to the wrench
 No moment is produced about point O
 Lack of tendency to rotate
as line of action passes
through O
4.1 Moment of a Force
– Scalar Formation
In General



Consider the force F and the point O which lies in
the shaded plane
The moment MO about point O,
or about an axis passing
through O and perpendicular
to the plane, is a vector quantity
Moment MO has its specified
magnitude and direction
4.1 Moment of a Force
– Scalar Formation
Magnitude

For magnitude of MO,
MO = Fd
where d = moment arm or perpendicular
distance from the axis at point O to its line
of action of the force

Units for moment is N.m
4.1 Moment of a Force
– Scalar Formation
Direction

Direction of MO is specified by
using “right hand rule”
- fingers of the right hand are
curled to follow the sense of
rotation when force rotates about
point O
4.1 Moment of a Force
– Scalar Formation
Direction
- Thumb points along the
moment axis to give the
direction and sense of the
moment vector
- Moment vector is upwards and
perpendicular to the shaded
plane
4.1 Moment of a Force
– Scalar Formation
Direction
MO is shown by a vector arrow
with a curl to distinguish it from
force vector
Example (Fig b)
 MO is represented by the
counterclockwise curl, which
indicates the action of F
4.1 Moment of a Force
– Scalar Formation
Direction
Arrowhead shows the sense of
rotation caused by F
 Using the right hand rule, the
direction and sense of the moment
vector points out of the page
 In 2D problems, moment of the
force is found about a point O

4.1 Moment of a Force
– Scalar Formation
Direction
Moment acts about an axis
perpendicular to the plane
containing F and d
 Moment axis intersects
the plane at point O

4.1 Moment of a Force
– Scalar Formation
Resultant Moment of a System of
Coplanar Forces

Resultant moment, MRo = addition of the
moments of all the forces algebraically since
all moment forces are collinear
MRo = ∑Fd
taking clockwise to be positive
4.1 Moment of a Force
– Scalar Formation
Resultant Moment of a System of
Coplanar Forces

A clockwise curl is written along the equation
to indicate that a positive moment if directed
along the
+ z axis and negative
along the – z axis
4.1 Moment of a Force
– Scalar Formation




Moment of a force does not always cause rotation
Force F tends to rotate the beam clockwise about A
with moment
MA = FdA
Force F tends to rotate the beam counterclockwise
about B with moment
MB = FdB
Hence support at A prevents
the rotation
4.1 Moment of a Force
– Scalar Formation
Example 4.1
For each case, determine the moment of the
force about point O
4.1 Moment of a Force
– Scalar Formation
Solution


Line of action is extended as a dashed line to
establish moment arm d
Tendency to rotate is indicated and the orbit is
shown as a colored curl
(a) M o  (100 N )( 2m)  200 N .m(CW )
(b) M o  (50 N )(0.75m)  37.5 N .m(CW )
4.1 Moment of a Force
– Scalar Formation
Solution
(c) M o  (40 N )( 4m  2 cos 30 m)  229 N .m(CW )
(d ) M o  (60 N )(1sin 45 m)  42.4 N .m(CCW )
(e) M o  (7kN )( 4m  1m)  21.0kN.m(CCW )
4.1 Moment of a Force
– Scalar Formation
Example 4.2
Determine the moments of
the 800N force acting on the
frame about points A, B, C
and D.
4.1 Moment of a Force –
Scalar Formation
Solution
Scalar Analysis
M A  (800 N )( 2.5m)  2000 N .m(CW )
M B  (800 N )(1.5m)  1200 N .m(CW )
M C  (800 N )(0m)  0kN.m
Line of action of F passes through C
M D  (800 N )(0.5m)  400 N .m(CCW )
4.2 Cross Product
Cross product of two vectors A and B yields
C, which is written as
C=AXB
 Read as “C equals A cross B”

4.2 Cross Product
Magnitude
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
 For angle θ, 0° ≤ θ ≤ 180°
 Therefore,
C = AB sinθ

4.2 Cross Product
Direction

Vector C has a direction that is perpendicular
to the plane containing A and B such that C is
specified by the right hand rule
- Curling the fingers of the right
hand form vector A (cross) to
vector B
- Thumb points in the direction of
vector C
4.2 Cross Product

Expressing vector C when magnitude and
direction are known
C = A X B = (AB sinθ)uC
where scalar AB sinθ defines the magnitude
of vector C unit vector uC defines the
direction of vector C
4.2 Cross Product
Laws of Operations
1. Commutative law is not valid
AXB≠BXA
Rather,
AXB=-BXA

Shown by the right hand rule

Cross product A X B yields a vector opposite in
direction to C
B X A = -C
4.2 Cross Product
Laws of Operations
2. Multiplication by a Scalar
a( A X B ) = (aA) X B = A X (aB) = ( A X B )a
3. Distributive Law
AX(B+D)=(AXB)+(AXD)

Proper order of the cross product must be
maintained since they are not commutative
4.2 Cross Product
Cartesian Vector Formulation
Use C = AB sinθ on pair of Cartesian
unit vectors
Example
For i X j, (i)(j)(sin90°)
= (1)(1)(1) = 1

4.2 Cross Product
Laws of Operations
In a similar manner,
i X j = k i X k = -j i X i = 0
j X k = i j X i = -k j X j = 0
k X i = j k X j = -i k X k = 0
 Use the circle for the results.
Crossing CCW yield positive
and CW yields negative results

4.2 Cross Product
Laws of Operations
 Consider cross product of vector A and B
A X B = (Axi + Ayj + Azk) X (Bxi + Byj + Bzk)
= AxBx (i X i) + AxBy (i X j) + AxBz (i X k)
+ AyBx (j X i) + AyBy (j X j) + AyBz (j X k)
+ AzBx (k X i) +AzBy (k X j) +AzBz (k X k)
= (AyBz – AzBy)i – (AxBz - AzBx)j + (AxBy – AyBx)k
4.2 Cross Product
Laws of Operations

In determinant form,

i
 
AXB  Ax
Bx

j
Ay
By

k
Az
Bz
4.3 Moment of Force
- Vector Formulation

Moment of force F about point O can
be expressed using cross product
MO = r X F
where r represents position
vector from O to any point
lying on the line of action
of F
4.3 Moment of Force
- Vector Formulation
Magnitude


For magnitude of cross product,
MO = rF sinθ
where θ is the angle measured
between tails of r and F
Treat r as a sliding vector. Since d = r
sinθ,
MO = rF sinθ = F (rsinθ) = Fd
4.3 Moment of Force
- Vector Formulation
Direction

Direction and sense of MO are determined by
right-hand rule
- Extend r to the dashed position
- Curl fingers from r towards F
- Direction of MO is the same
as the direction of the thumb
4.3 Moment of Force
- Vector Formulation
Direction
*Note:
- “curl” of the fingers indicates the sense of
rotation
- Maintain proper order of r
and F since cross product
is not commutative
4.3 Moment of Force
- Vector Formulation
Principle of Transmissibility
For force F applied at any point A,
moment created about O is MO = rA x
F
 F has the properties of a sliding vector
and
therefore act at any point
along its line of action and
still create the same
moment about O

4.3 Moment of Force
- Vector Formulation
Cartesian Vector Formulation

For force expressed in Cartesian
form,



i

 
M O  r XF  rx
j
ry
k
rz
Fx
Fy
Fz
where rx, ry, rz represent the x, y, z
components of the position vector
and Fx, Fy, Fz represent that of the
force vector
4.3 Moment of Force
- Vector Formulation
Cartesian Vector Formulation



With the determinant expended,
MO = (ryFz – rzFy)i – (rxFz - rzFx)j + (rxFy – yFx)k
MO is always perpendicular to
the plane containing r and F
Computation of moment by cross
product is better than scalar for
3D problems
4.3 Moment of Force
- Vector Formulation
Cartesian Vector Formulation

Resultant moment of forces about point
O can be determined by vector addition
MRo = ∑(r x F)
4.3 Moment of Force
- Vector Formulation



Moment of force F about point
A, pulling on cable BC at any
point along its line of action,
will remain constant
Given the perpendicular
distance from A to cable is rd
MA = rdF
In 3D problems,
MA = rBC x F
4.3 Moment of Force
- Vector Formulation
Example 4.4
The pole is subjected to a 60N force that is
directed from C to B. Determine the magnitude
of the moment created by this force about the
support at A.
4.3 Moment of Force
- Vector Formulation
Solution
 Either one of the two position vectors can be
used for the solution, since MA = rB x F or MA
= rC x F
 Position vectors are represented as
rB = {1i + 3j + 2k} m and
rC = {3i + 4j} m
 Force F has magnitude 60N
and is directed from C to B
4.3 Moment of Force
- Vector Formulation
Solution


F  (60 N )u F



 (1  3)i  93  4) j  92  0)k 
 (60 N ) 

2
2
2

(2)  (1)  (2)




  40i  20 j  40k N


Substitute into
formulation

 determinant
i



M A  rB XF  1
j
3
k
2
 40  20 40



 [3(40)  2(20)]i  [1(40)  2(40)] j  [1(20)  3(40)]k


4.3 Moment of Force
- Vector Formulation
Solution
Or  

i

M A  rC XF  3

k
0

j
4
 40  20 40



 [4(40)  0(20)]i  [3(40)  0(40)] j  [3(20)  4(40)]k

Substitute
into determinant
formulation






M A  160i  120 j  100k N .m
For magnitude,

M A  (160) 2  (120) 2  (100) 2
 224 N .m

4.4 Principles of Moments
Also known as Varignon’s Theorem
“Moment of a force about a point is equal to
the sum of the moments of the forces’
components about the point”
 For F = F1 + F2,
MO = r X F1 + r X F2
= r X (F1 + F2)
=rXF

4.4 Principles of Moments


The guy cable exerts a
force F on the pole and
creates a moment about
the base at A
MA = Fd
If the force is replaced by
Fx and Fy at point B where
the cable acts on the pole,
the sum of moment about
point A yields the same
resultant moment
4.4 Principles of Moments


Fy create zero moment
about A
MA = Fxh
Apply principle of
transmissibility and slide
the force where line of
action intersects the
ground at C, Fx create zero
moment about A
MA = F yb
4.4 Principles of Moments
Example 4.6
The force F acts at the end of the angle
bracket. Determine the moment of the force
about point O.
4.4 Principles of Moments
Solution
Method 1
MO = 400sin30°N(0.2m)-400cos30°N(0.4m)
= -98.6N.m
= 98.6N.m (CCW)
As a Cartesian vector,
MO = {-98.6k}N.m
4.4 Principles of Moments
Solution
Method 2:
 Express as Cartesian vector
r = {0.4i – 0.2j}N
F = {400sin30°i – 400cos30°j}N
= {200.0i – 346.4j}N



For moment,
i
j
k



M O  r XF  0.4
 0.2 0
200.0  346.4 0

  98.6k N .m


4.6 Moment of a Couple




Couple
- two parallel forces
- same magnitude but opposite direction
- separated by perpendicular distance d
Resultant force = 0
Tendency to rotate in specified direction
Couple moment = sum of
moments of both couple
forces about any arbitrary point
4.6 Moment of a Couple
Example
Position vectors rA and rA are directed from O
to
A and B, lying on the line of action of F and –F

Couple moment about O
M = rA X (-F) + rA X (F)

Couple moment about A
M=rXF
since moment of –F about A = 0
4.6 Moment of a Couple
A couple moment is a free vector
- It can act at any point since M
depends only on the position vector r
directed between forces and not
position vectors rA and rB, directed
from O to the forces
- Unlike moment of force, it do not
require a definite point or axis

4.6 Moment of a Couple
Scalar Formulation
Magnitude of couple moment
M = Fd
 Direction and sense are
determined by right hand
rule
 In all cases, M acts
perpendicular to plane
containing the forces

4.6 Moment of a Couple
Vector Formulation
For couple moment,
M=rXF
 If moments are taken about point A,
moment of –F is zero about this point
 r is crossed with the force to which it is
directed

4.6 Moment of a Couple
Equivalent Couples
Two couples are equivalent if they
produce the same moment
 Since moment produced by the couple
is always perpendicular to the plane
containing the forces, forces of equal
couples either lie on the same plane or
plane parallel to one another

4.6 Moment of a Couple
Resultant Couple Moment
Couple moments are free vectors
and may be applied to any point P
and added vectorially
 For resultant moment of two
couples at point P,
MR = M1 + M2
 For more than 2 moments,
MR = ∑(r X F)

4.6 Moment of a Couple



Frictional forces (floor)
on the blades of the
machine creates a
moment Mc that tends
to turn it
An equal and opposite
moment must be
applied by the operator
to prevent turning
Couple moment Mc =
Fd is applied on the
handle
4.6 Moment of a Couple
Example 4.10
A couple acts on the gear teeth. Replace it
by an equivalent couple having a pair of
forces that cat through points A and B.
4.6 Moment of a Couple
Solution
 Magnitude of couple
M = Fd = (40)(0.6) = 24N.m
 Direction out of the page since
forces tend to rotate CW
 M is a free vector and can
be placed anywhere
4.6 Moment of a Couple
Solution

To preserve CCW motion,
vertical forces acting through
points A and B must be directed as
shown

For magnitude of each force,
M = Fd
24N.m = F(0.2m)
F = 120N
4.7 Equivalent System




A force has the effect of both translating and
rotating a body
The extent of the effect depends on how and
where the force is applied
We can simplify a system of forces and
moments into a single resultant and moment
acting at a specified point O
A system of forces and moments is then
equivalent to the single resultant force and
moment acting at a specified point O
4.7 Equivalent System
Point O is on the Line of Action


Consider body subjected to force F applied to
point A
Apply force to point O without altering external
effects on body
- Apply equal but opposite forces F and –F at O
4.7 Equivalent System
Point O is on the Line of Action
- Two forces indicated by the slash across them can
be cancelled, leaving force at point O
- An equivalent system has be maintained between
each of the diagrams, shown by the equal signs
4.7 Equivalent System
Point O is on the Line of Action
- Force has been simply transmitted along its line
of action from point A to point O
- External effects remain unchanged after force is
moved
- Internal effects depend on location of F
4.7 Equivalent System
Point O is Not on the Line of Action
F is to be moved to point ) without altering the
external effects on the body
 Apply equal and opposite forces at point O
 The two forces indicated by a slash across
them, form a couple that has a moment
perpendicular to F

4.7 Equivalent System
Point O is Not on the Line of Action
The moment is defined by cross product
M=rXF
 Couple moment is free vector and can be
applied to any point P on the body

4.8 Resultants of a Force
and Couple System



Consider a rigid body
Since O does not lies on the line of
action, an equivalent effect is
produced if the forces are moved to
point O and the corresponding
moments are
M1 = r1 X F1 and M2 = r2 X F2
For resultant forces and moments,
FR = F1 + F2 and MR = M1 + M2
4.8 Resultants of a Force
and Couple System




Equivalency is maintained thus each
force and couple system cause the
same external effects
Both magnitude and direction of FR
do not depend on the location of
point O
MRo depends on location of point O
since M1 and M2 are determined
using position vectors r1 and r2
MRo is a free vector and can acts on
any point on the body
4.8 Resultants of a Force
and Couple System


Simplifying any force and couple
system,
FR = ∑F
MR = ∑MC + ∑MO
If the force system lies on the x-y
plane and any couple moments are
perpendicular to this plane,
FRx = ∑Fx
FRy = ∑Fy
MRo = ∑MC + ∑MO
4.8 Resultants of a Force
and Couple System
Procedure for Analysis

When applying the following equations,
FR = ∑F
MR = ∑MC + ∑MO
FRx = ∑Fx
FRy = ∑Fy
MRo = ∑MC + ∑MO
Establish the coordinate axes with the origin
located at the point O and the axes having a
selected orientation
4.8 Resultants of a Force and Couple
System
Procedure for Analysis
Force Summation
 For coplanar force system, resolve each force
into x and y components
 If the component is directed along the positive x
or y axis, it represent a positive scalar
 If the component is directed along the negative x
or y axis, it represent a negative scalar
 In 3D problems, represent forces as Cartesian
vector before force summation
4.8 Resultants of a Force and Couple
System
Procedure for Analysis
Moment Summation
 For moment of coplanar force system about
point O, use Principle of Moment
 Determine the moments of each components
rather than of the force itself
 In 3D problems, use vector cross product to
determine moment of each force
 Position vectors extend from point O to any point
on the line of action of each force
4.8 Resultants of a Force and Couple
System
Example 4.14
Replace the forces acting on the brace by an
equivalent resultant force and couple moment
acting at point A.
4.8 Resultants of a Force and Couple
System
Solution
Force Summation
 For x and y components of resultant force,
  FRx  Fx ;
FRx  100 N  400 cos 45 N
 382.8 N  382.8 N 
  FRy  Fy ;
FRy  600 N  400 sin 45 N
 882.8 N  882.8 N 
4.8 Resultants of a Force and Couple
System
Solution
 For magnitude of resultant force
FR  ( FRx ) 2  ( FRy ) 2  (382.8) 2  (882.8) 2
 962 N

For direction of resultant force
 FRy 
1  882.8 


  tan 
 tan 


 382.8 
 FRx 
 66.6
1
4.8 Resultants of a Force and Couple
System
Solution
Moment Summation
 Summation of moments about point A,
M RA  M A ;
M RA  100 N (0)  600 N (0.4m)  (400 sin 45 N )(0.8m)
 (400 cos 45 N )(0.3m)
 551N .m  551N .m(CW )

When MRA and FR act on point A,
they will produce the same
external effect or reactions at the support
4.9 Reduction of a Simple
Distributed Loading
Large surface area of a body may be
subjected to distributed loadings such as
those caused by wind, fluids, or weight of
material supported over body’s surface
 Intensity of these loadings at each point
on the surface is defined as the pressure p
 Pressure is measured in pascals (Pa)
1 Pa = 1N/m2

4.9 Reduction of a Simple
Distributed Loading



Most common case of distributed pressure loading is
uniform loading along one axis of a flat rectangular
body
Direction of the intensity of the pressure load is
indicated by arrows shown on the load-intensity
diagram
Entire loading on the plate is a
system of parallel forces,
infinite in number, each
acting on a separate
differential area of the plate
4.9 Reduction of a Simple
Distributed Loading




Loading function p = p(x) Pa, is a function of x
since pressure is uniform along the y axis
Multiply the loading function by the width
w = p(x)N/m2]a m = w(x) N/m
Loading function is a measure
of load distribution along line
y = 0, which is in the symmetry
of the loading
Measured as force per unit
length rather than per unit area
4.9 Reduction of a Simple
Distributed Loading
Load-intensity diagram for w = w(x) can
be represented by a system of coplanar
parallel
 This system of forces can be simplified into
a single resultant force FR
and its location can be
specified

4.9 Reduction of a Simple
Distributed Loading
Magnitude of Resultant Force



FR = ∑F
Integration is used for infinite number of parallel
forces dF acting along the plate
For entire plate length,
  FR  F ; FR   w( x)dx   dA  A
L

A
Magnitude of resultant force is equal to the total
area A under the loading diagram w = w(x)
4.9 Reduction of a Simple
Distributed Loading
Location of Resultant Force




MR = ∑MO
Location x of the line of action of FR can be
determined by equating the moments of the
force resultant and the force distribution about
point O
dF produces a moment of
xdF = x w(x) dx about O
For the entire plate,
M Ro  M O ; xFR   xw( x)dx
L
4.9 Reduction of a Simple
Distributed Loading
Location of Resultant Force

Solving,
x
 xw( x)dx  xdA
L
 w( x)dx
L


A
 dA
A
Resultant force has a line of action which passes
through the centroid C (geometric center) of the
area defined by the distributed loading diagram
w(x)
4.9 Reduction of a Simple
Distributed Loading
Location of Resultant Force


Consider 3D pressure loading p(x), the resultant
force has a magnitude equal to the volume under
the distributed-loading curve p = p(x) and a line of
action which passes through the centroid (geometric
center) of this volume
Distribution diagram can be
in any form of shapes such
as rectangle, triangle etc
4.9 Reduction of a Simple
Distributed Loading


Beam supporting this stack of lumber is subjected
to a uniform distributed loading, and so the loadintensity diagram has a rectangular shape
If the load-intensity is wo, resultant is determined
from the are of the rectangle
FR = wob
4.9 Reduction of a Simple
Distributed Loading



Line of action passes through the centroid or
center of the rectangle,x = a + b/2
Resultant is equivalent to the distributed load
Both loadings produce same “external” effects or
support reactions on the beam
4.9 Reduction of a Simple
Distributed Loading
Example 4.20
Determine the magnitude and location of the
equivalent resultant force acting on the shaft
4.9 Reduction of a Simple
Distributed Loading
Solution
For the colored differential area element,
dA  wdx  60 x 2dx
For resultant force
FR  F ;
2
FR   dA   60 x 2dx
A
0
3 2
x
 23 03 
 60   60  
 3 0
3 3
 160 N
4.9 Reduction of a Simple
Distributed Loading
Solution
For location of line of action,
2
2
 x4 
 24 04 
2
 xdA  x(60 x )dx 60 4  60 4  4 


0 
xA
0

160
160
160
dA

A
 1.5m
Checking,
ab 2m(240 N / m)

 160
3
3
3
3
x  a  (2m)  1.5m
4
4
A
4.9 Reduction of a Simple
Distributed Loading
Example 4.21
A distributed loading of p = 800x Pa acts
over the top surface of the beam. Determine
the magnitude and location of the equivalent
force.
4.9 Reduction of a Simple
Distributed Loading
Solution
 Loading function of p = 800x Pa indicates that
the load intensity varies uniformly from p = 0 at
x = 0 to p = 7200Pa at x = 9m
 For loading,
w = (800x N/m2)(0.2m) = (160x) N/m
 Magnitude of resultant force
= area under the triangle
FR = ½(9m)(1440N/m)
= 6480 N = 6.48 kN
4.9 Reduction of a Simple
Distributed Loading
Solution
 Resultant force acts through the centroid of
the volume of the loading diagram p = p(x)
 FR intersects the x-y plane at point (6m, 0)
 Magnitude of resultant force
= volume under the triangle
FR = V = ½(7200N/m2)(0.2m)
= 6.48 kN
4.9 Reduction of a Simple
Distributed Loading
Example 4.22
The granular material exerts the distributed
loading on the beam. Determine the magnitude
and location of the equivalent resultant of this
load
4.9 Reduction of a Simple
Distributed Loading
Solution
 Area of loading diagram is trapezoid
 Magnitude of each force = associated area
F1 = ½(9m)(50kN/m) = 225kN
F2 = ½(9m)(100kN/m) = 450kN
 Line of these parallel forces act
through the centroid of associated
areas and insect beams at
1
1
x1  (9m)  3m, x2  (9m)  4.5m
3
2
4.9 Reduction of a Simple
Distributed Loading
Solution
 Two parallel Forces F1 and F2 can be reduced to a
single resultant force FR
 For magnitude of resultant force,
  FR  F ;
FR  225  450 x  675kN

For location of resultant force,
M Ro  M O ;
x (675)  3(225)  4.5(450)
x  4m
4.9 Reduction of a Simple
Distributed Loading
Solution
*Note:
 Trapezoidal area can be divided into two triangular
areas,
F1 = ½(9m)(100kN/m) = 450kN
F2 = ½(9m)(50kN/m) = 225kN
1
1
x1  (9m)  3m, x2  (9m)  3m
3
3