Announcements
• Test observations
– Units
– Significant figures
– Position vectors
Moment of a Force
Today’s Objectives
• Understand and define Moment
• Determine moments of a force in 2-D
and 3-D cases
Moment of a
force
Class Activities
• Applications
• Moments in 2-D
• Moments in 3-D
• Examples
Engr221 Chapter 4
1
Applications
What is the net effect of the
two forces on the wheel shaft?
Applications - continued
What is the effect of the 30 N
force on the lug nut?
Engr221 Chapter 4
2
Moments in 2-D
The moment of a force about a point provides a measure of the
tendency for rotation (sometimes called torque).
Moments in 2-D - continued
In the 2-D case, the magnitude of the moment
is: Mo = Fd
As shown, d is the perpendicular distance from point O to the
line of action of the force F.
In 2-D, the direction of MO is either clockwise or
counter-clockwise depending on the tendency for rotation.
Engr221 Chapter 4
3
Moments in 2-D - continued
b
O
d
a
F
Fy
a
b
F
Fx
O
For example, MO = Fd and the direction is counter-clockwise.
Often it is easier to determine MO by using the components of F.
Using this approach, MO = (FY a) – (FX b). Note the different signs
on the terms. The typical sign convention for a moment in 2-D is
that counter-clockwise is considered positive. We can determine
the direction of rotation by imagining the body pinned at O and
deciding which way the body would rotate due to the force.
3-D Moments and Vector Formulation
Moments in 3-D can be calculated using a scalar (2-D) approach
but it can be difficult and time consuming. Thus, it is often easier
to use a mathematical approach called the vector cross product.
Using the vector cross product, MO = r × F
Here, r is a position vector from point O to any point on the line of
action of force F.
Engr221 Chapter 4
4
Cross Product
In general, the cross product of two vectors A and B results in
another vector C, i.e., C = A× B. The magnitude and direction of
the resulting vector can be written as
C = A × B = (AB sinθ) uC
Here, uC is the unit vector perpendicular to both the A and B
vectors as shown (or to the plane containing the A and B
vectors).
Cross Product - continued
The right hand rule is useful for determining the direction of the
vector resulting from a cross product.
For example: i × j = k
Note that a vector crossed into itself is zero, e.g., i × i = 0
Engr221 Chapter 4
5
Cross Product - continued
Of even more utility, the cross product can be written as
and each component can be determined using 2 × 2 determinants:
Moments in 3-D
So, using the cross product, a
moment can be expressed as
By expanding the above equation using the 2 × 2 determinants (see
Section 4.2 in your textbook), we get
MO = (ry Fz - rz Fy) i - (rx Fz - rz Fx) j + (rx Fy - ry Fx) k
The physical meaning of the above equation becomes evident by
considering the force components separately and using a 2-D
formulation. Units of moments are N-m or lb-ft.
Engr221 Chapter 4
6
Example A
Given: A 400 N force is
applied to the frame
and θ = 20°
Find: The moment of the
force at A
Plan:
1) Resolve the force along the x and y axes
2) Determine MA using scalar analysis
Example A - continued
+ ↑ Fy = -400 sin 20° N
+ → Fx = -400 cos 20° N
+ MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m
= 1162 N·m = 1.16 kN·m
Engr221 Chapter 4
7
Example B
Given: a = 3 in, b = 6 in, and c = 2 in
Find: Moment of F about point O
Plan:
o
1) Find rOA
2) Determine MO = rOA × F
Solution rOA = {3 i + 6 j – 0 k} in
MO =
i
3
3
j k
6 0
2 -1
= [{6(-1) – 0(2)} i – {3(-1) – 0(3)} j +
{3(2) – 6(3)} k] lb·in
= {-6 i + 3 j – 12 k} lb·in
Example C
Given: a = 3 in, b = 6 in, and c = 2 in
Find: Moment of F about point P
Plan:
1) Find rPA
2) Determine MP = rPA x F
Solution: rPA = { 3 i + 6 j - 2 k } in
MP =
Engr221 Chapter 4
i j k
3 6 -2
3 2 -1
= { -2 i - 3 j - 12 k } lb·in
8
Example D
Given: A 40 N force is
applied to the wrench
Find: The moment of the
force at O
Plan:
1) Resolve the force along the
x and y axes
2) Determine MO using scalar
analysis
Solution: + ↑ Fy = - 40 cos 20° N
+ → Fx = - 40 sin 20° N
+ MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm
= -7107 N·mm = - 7.11 N·m
Questions
1. The moment of force F about point O
is defined as MO = ___________
A) r x F
B) F x r
C) r • F
D) r * F
2. What is the moment of the 10 N force about
point A (MA)?
A) 10 Nm
B) 30 Nm
C) 13 Nm
D) (10/3) Nm
E) 7 Nm
Engr221 Chapter 4
F = 10 N
•A
d=3m
9
Question
If a force of magnitude F can be applied in four different 2-D
configurations (P, Q, R, & S), select the cases resulting in the
maximum and minimum torque values on the nut.
A) (Q, P)
B) (R, S)
S
C) (P, R)
D) (Q, S)
R
P
Q
Example Problem
Determine the direction θ for 0° ≤ θ ≤ 180° of the force F so
that F produces (a) the maximum moment about point A and
(b) the minimum moment about point A. Calculate the
moment in each case.
Maximum
θ= 56.3º MA = 1.44KN-M ccw
Mimimum
θ = 146º MA = 0 N-M
Engr221 Chapter 4
10
Example Problem
The two boys push on the gate with forces of FA = 30 lb and
FB = 50 lb as shown. Determine the moment of each force
about point C. Which way will the gate rotate, clockwise or
counterclockwise? Neglect the thickness of the gate.
MCA = 162 lb-ft cw
MCB = 260 lb-ft ccw
Example Problem
Strut AB of the 1m diameter hatch door exerts a force of
450 N on point B. Determine the moment of this force
about point O.
MO = {373i – 100j +173k} N-M
Engr221 Chapter 4
11
Textbook Problem 4.40
The force F = {600i + 300j – 600k} N acts at the end of the
beam. Determine the moment of the force about point A.
MO = {-840i + 360j - 660k} N-M
Summary
• Understand and define Moment
• Determine moments of a force in 2-D and 3-D cases
Engr221 Chapter 4
12
Announcements
Moment About an Axis
Today’s Objectives
Be able to determine the moment of a force about an axis using
• Scalar analysis
• Vector analysis
Class Activities
• Applications
• Scalar analysis
• Vector analysis
• Examples
Engr221 Chapter 4
13
Applications
With the force F, a person is
creating the moment MA.
What portion of MA is used in
turning the socket?
The force F is creating the
moment MO. How much of
MO acts to unscrew the
pipe?
Scalar Analysis
Recall that the moment of a force about any point A is MA = FdA
where dA is the perpendicular distance from the point to the force’s
line of action. This concept can be extended to find the moment of
a force about an axis.
In the figure above, the moment about the y-axis would be My =
20(0.3) = 6 Nm. This calculation is not always trivial and vector
analysis is usually preferable.
Engr221 Chapter 4
14
Vector Analysis
Our goal is to find the
moment of F (the
tendency to rotate the
body) about the axis a’a.
First, compute the moment of F about any arbitrary point O that
lies on the a’a axis using the cross product:
MO = rOA × F
Now, find the component of MO along the axis a’a using the dot
product:
Ma’a = ua • MO
Vector Analysis - continued
Ma’a can also be obtained as
The above equation is called the
triple scalar product.
In this equation,
ua represents the unit vector along the a’a axis
r is the position vector from any point on the a’a axis to
any point on the line of action of the force (A)
F is the force vector
Engr221 Chapter 4
15
Example A
Given: A force is applied to the
tool to open a gas valve
A
Find: The magnitude of the
moment of this force
about the z axis
B
Plan:
1) Find Mz = u • (r × F)
2) Note that u = 1k
3) The vector r is the position vector from A to B
4) Force F is already given in Cartesian vector form
Example A - continued
A
u = 1k
rAB = {0.25 sin 30° i + 0.25 cos30° j} m
B
= {0.125 i + 0.2165 j} m
F = {-60 i + 20 j + 15 k} N
Mz = u • (rAB × F)
0
0
1
0.125
0.2165
Mz = with
Created
the Trial
Edition of0 SmartDraw 5.
-60
20
15
= 1{0.125(20) – 0.2165(-60)} Nm
= 15.5 Nm
Engr221 Chapter 4
16
Example B
Given: A force of 80 lb acts
along the edge DB
Find:
The magnitude of the
moment of this force
about the axis AC
Plan:
1) Find MAC = uAC • (rAB × FDB)
2) Find uAC = rAC /rAC
3) Find FDB = 80(uDB) lb = 80(rDB/rDB) lb
4) Complete the triple scalar product
Example B - continued
rAB = { 20 j } ft
rAC = { 13 i + 16 j } ft
rDB = { -5 i + 10 j – 15 k } ft
uAC = (13 i + 16 j) / (13 2 + 16 2) ½ ft
= 0.6306 i + 0.7761 j
FDB = 80 {rDB / (5 2 + 10 2 + 15 2) ½ } lb
= {-21.38 i + 42.76 j – 64.14 k } lb
Engr221 Chapter 4
17
Example B - continued
Now find the triple product, MAC = uAC • ( rAB × FDB )
MAC =
0.6306 0.7706
0
0
20
0
-21.38 42.76 -64.14
ft•lb
MAC = 0.6306 {20 (-64.14) – 0 – 0.7706 (0 – 0)} lb·ft
= -809 lb·ft
IMPORTANT: The negative sign indicates that the
sense of MAC is opposite to that of uAC
Questions
1. For finding the moment of the
force F about the x-axis, the
position vector in the triple
scalar product should be ___
A) rAC
B) rBA
C) rAB
D) rBC
2. If r = {1i + 2j} m and F = {10i + 20j + 30k} N, then the
moment of F about the y-axis is ____ m
A) 10
B) -30
C) -40
D) None of the above
Engr221 Chapter 4
18
Questions
1. When determining the moment of a force about a specified
axis, the axis must be along _____________
A) the x axis
B) the y axis C) the z axis
D) any line in 3-D space E) any line in the x-y plane
2. The triple scalar product u • ( r × F ) results in:
A) a scalar quantity ( + or - )
B) a vector quantity
C) zero
D) a unit vector
E) an imaginary number
Question
The force F is acting along
DC. Using the triple
product to determine the
moment of F about the bar
BA, you could use any of
the following position
vectors except ______
A) rBC
B) rAD
C) rAC
D) rDB
E) rBD
Engr221 Chapter 4
19
Summary
• Be able to determine the moment of a force about an axis
using
– Scalar analysis
– Vector analysis
Announcements
Engr221 Chapter 4
20
Moment of a Couple
Today’s Objectives
• Define a couple
• Determine the moment of a couple
Class activities
• Applications
• Moment of a Couple
• Examples
Applications
A torque or moment of 12 Nm is required to rotate the wheel.
Which one of the two grips of the wheel above will require less
force to rotate the wheel?
Engr221 Chapter 4
21
Applications - continued
The crossbar lug wrench is being used to loosen a lug
net. What is the effect of changing dimensions a, b, or c
on the force that must be applied?
Moment of a Couple
A couple is defined as two
parallel forces with the same
magnitude but opposite in
direction, separated by a
perpendicular distance d.
The moment of a couple is defined as:
M = Fd (using a scalar analysis) or as
M = r × F (using a vector analysis).
Here, r is any position vector from the line of action of –F
to the line of action of F.
Engr221 Chapter 4
22
Moment of a Couple - continued
The net external effect of a couple is that
the net force equals zero and the magnitude
of the net moment equals Fd.
Since the moment of a couple depends
only on the distance between the forces,
the moment of a couple is a free vector. It
can be moved anywhere on the body and
have the same external effect on the body.
Moments due to couples can be added
using the same rules as adding any vectors.
Example A – Scalar Approach
Given: Two couples act on the
beam and d = 8 ft
Find: The resultant couple
Plan:
1) Resolve the forces in x and y directions so they can
be treated easily as couples
2) Determine the net moment due to the two couples
Engr221 Chapter 4
23
Example A - continued
The x and y components of the top
60 lb force are:
(4/5)(60 lb) = 48 lb vertically up
(3/5)(60 lb) = 36 lb to the left
Similarly, for the top 40 lb force:
(40 lb) (sin 30°) up
(40 lb) (cos 30°) to the left
The net moment is:
+ ΣM = -(48)(4) + (40)(cos 30º)(8) lb·ft
= -192.0 + 277.1 = 85.1 lb·ft
Example B – Vector Approach
Given: A force couple acting on the
rod
Find: The couple moment acting
on the rod in Cartesian
vector notation
A
B
Plan:
1) Use M = r × F to find the couple moment
2) Set r = rAB and F = {14i – 8j – 6k} N
3) Calculate the cross product to find M
Engr221 Chapter 4
24
Example B - continued
rAB = {0.8i + 1.5j – 1k} m
F = {14i – 8j – 6k} N
A
B
M = rAB × F
i
j k
0.8
1.5
Nm 5.
Created with the Trial Edition-1of SmartDraw
14 -8 -6
=
= {i (-9 – (8)) – j (-4.8 – (-14)) + k (-6.4 – 14(1.5))} Nm
= {-17 i – 9.2 j – 27.4 k} Nm
Example C – Scalar Approach
Given: Two couples act on the beam.
The resultant couple is zero.
Find: The magnitudes of the forces
P and F and the distance d.
Plan:
1) Use the definition of a couple to find P and F
2) Resolve the 300N force in x and y directions
3) Determine the net moment
4) Equate the net moment to zero to find d
Engr221 Chapter 4
25
Example C - continued
From the definition of a
couple:
P = 500 N and
F = 300 N
Resolve the 300 N force into vertical and horizontal
components. The vertical component is (300 cos 30º) N and
the horizontal component is (300 sin 30º) N
It was given that the net moment equals zero. So
+ ΣM = - (500)(2) + (300 cos 30º)(d) - (300 sin 30º)(0.2) = 0
Solving this equation for d yields
d = (1000 + 60 sin 30º) / (300 cos 30º) = 3.96 m
Example D – Vector Approach
Given: F = {25 k} N and - F = {25 k} N
Find: The couple moment acting
on the pipe assembly, using
Cartesian vector notation.
Plan:
1) Use M = r × F to find the couple moment
2) Set r = rAB and F = {25 k} N
3) Calculate the cross product to find M
Engr221 Chapter 4
26
Example D - continued
rAB = { - 350 i – 200 j } mm
= { - 0.35 i – 0.2 j } m
F = {25 k} N
M = rAB × F
=
i
-0.35
0
j
k
-0.2 0
0
25
Nm
= { i (-5 – 0 ) – j (-8.75 – 0) + k (0) } Nm
= { -5 i + 8.75 j } Nm
Questions
1. A couple is applied to the beam as shown. Its moment equals
_____ Nm
50 N
A) 50
B) 60
1m
2m
5
C) 80
D) 100
3
4
2. You can determine the couple
moment as M = r × F
If F = {-20k} lb, then r is
A) rBC
B) rAB
C) rCB
D) rAC
Engr221 Chapter 4
27
Questions
1. In statics, a couple is defined as __________ separated by a
perpendicular distance.
A) two forces in the same direction
B) two forces of equal magnitude
C) two forces of equal magnitude acting in the same direction
D) two forces of equal magnitude acting in opposite directions
2. The moment of a couple is called a _________ vector.
A) free
B) spin
C) romantic
D) sliding
Questions
1. F1 and F2 form a couple. The moment
of the couple is given by ____
A) r1 × F1
B) r2 × F1
C) F2 × r1
D) r2 × F2
F1
r1
r2
F2
2. If three couples act on a body, the overall result is that
A) the net force is not equal to 0
B) the net force and net moment are equal to 0
C) the net moment equals 0 but the net force is not necessarily
equal to 0
D) the net force equals 0 but the net moment is not necessarily
equal to 0
Engr221 Chapter 4
28
Textbook Problem 4-79
Express the moment of the couple acting on the pipe
assembly in Cartesian vector form. Solve the problem
(a) using Eq. 4-13, and (b) summing the moment of
each force about point O. Take F = {25k} N.
MC = (-5i + 8.75j) Nm
Summary
• Define a couple
• Determine the moment of a couple
Engr221 Chapter 4
29
Announcements
Equivalent Force and Couple Systems
Today’s Objectives
• Determine the effect of moving a force
• Find an equivalent force-couple system for a
system of forces and couples
Class Activities
• Applications
• Equivalent systems
• System reduction
• Examples
Engr221 Chapter 4
30
Applications
What is the resultant effect on the person’s hand
when the force is applied in four different ways?
Applications - continued
Several forces and a couple
moment are acting on this
vertical section of an I-beam.
||
??
Can you replace them with just
one force and one couple
moment at point O that will
have the same external effect?
If yes, how will you do that?
Engr221 Chapter 4
31
An Equivalent Force System
=
When a number of forces and couple moments are acting on a
body, it is easier to understand their overall effect on the body if
they are combined into a single force and couple moment having
the same external effect.
The two force and couple systems are called equivalent systems
since they have the same external effect on the body.
Moving a Force On its Line of Action
Moving a force from A to O, when both points are on the
vectors’ line of action, does not change the external effect.
Hence, a force vector is called a sliding vector. (But the
internal effect of the force on the body does depend on where
the force is applied).
Engr221 Chapter 4
32
Moving a Force Off its Line of Action
Moving a force from point A to O (as shown above) requires
creating an additional couple moment. Since this new couple
moment is a “free” vector, it can be applied at any point P on the
body.
Resultant of a Force and Couple System
When several forces and couple moments
act on a body, you can move each force
and its associated couple moment to a
common point O.
Now you can add all the forces and
couple moments together and find one
resultant force-couple moment pair.
Engr221 Chapter 4
33
Resultant of a Force and Couple System
If the force system lies in the x-y plane (the 2-D case), then the
reduced equivalent system can be obtained using the following
three scalar equations:
Force-Moment Reduction
=
=
If FR and MRO are perpendicular to each other, then the system
can be further reduced to a single force, FR , by simply moving
FR from O to P.
In three special cases, concurrent, coplanar, and parallel
systems of forces, the system can always be reduced to a single
force.
Engr221 Chapter 4
34
Example A
Given: A 2-D force and couple
system as shown.
Find:
The equivalent resultant
force and couple moment
acting at A, and then the
equivalent single force
location along the beam AB.
Plan:
1) Sum all the x and y components of the forces to find FRA
2) Find and sum all the moments resulting from moving each
force to A
3) Shift FRA to a distance d such that d = MRA/FRy
Example A - continued
+ → ΣFRx = 25 + 35 sin 30° = 42.5 lb
+ ↓ ΣFRy = 20 + 35 cos 30° = 50.31 lb
FR
+ MRA
= 35 cos30° (2) + 20(6) – 25(3)
= 105.6 lb·ft
FR = ( 42.52 + 50.312 )1/2 = 65.9 lb
θ = tan-1 (50.31/42.5) = 49.8 °
The equivalent single force FR can be located on the
beam AB at a distance d measured from point A.
d = MRA/FRy = 105.6/50.31 = 2.10 ft
Engr221 Chapter 4
35
Example B
Given: The building slab has four
columns. F1 and F2 = 0.
Find: The equivalent resultant
force and couple moment
at the origin O. Also find
the location (x,y) of the
single equivalent resultant
force.
Plan:
1) Find FRO = ∑Fi = FRzo k
2) Find MRO = ∑ (ri × Fi) = MRxO i + MRyO j
3) The location of the single equivalent resultant force is given
as x = MRyO/FRzO and y = MRxO/FRzO
Example B - continued
FRO = {-50 k – 20 k} = {-70 k} kN
MRO = (10 i) × (-20 k) +
(4 i + 3 j) × (-50 k)
= {200 j + 200 j – 150 i} kN·m
= {-150 i + 400 j } kN·m
The location of the single equivalent resultant force is given as:
x = MRyo/FRzo = 400/(70) = 5.71 m
y = MRxo/Frzo = (150)/(70) = 2.14 m
Engr221 Chapter 4
36
Example C
Given: A 2-D force and couple
system as shown
Find: The equivalent resultant
force and couple moment
acting at A
Plan:
1) Sum all the x and y components of the forces to find FRA
2) Find and sum all the moments resulting from moving each
force to A and add them to the 500 lb-ft free moment to find
the resultant MRA
Example C - continued
Summing the force
components:
+ → ΣFx = (4/5) 150 lb + 50 lb sin 30° = 145 lb
+ ↑ ΣFy = (3/5) 150 lb + 50 lb cos 30° = 133 lb
Now find the magnitude and direction of the resultant:
FRA = ( 145 2 + 133.3 2 )1/2 = 197 lb and θ = tan-1 (133.3/145)
= 42.6 °
+ MRA = { (4/5)(150)(2) – 50 cos30° (3) + 50 sin30° (6) + 500 }
= 760 lb·ft
Engr221 Chapter 4
37
Example D
Given: Handle forces F1 and F2 are
applied to the electric drill
Find: An equivalent resultant
force and couple moment at
point O
Plan:
a) Find FRO = Σ Fi
b) Find MRO = Σ ( ri × Fi )
Fi are the individual forces in Cartesian vector form
ri are the position vectors from the point O to any point on the
line of action of Fi
Example D - continued
F1 = {6 i – 3 j – 10 k} N
F2 = {0 i + 2 j – 4 k} N
FRO = {6 i – 1 j – 14 k} N
r1 = {0.15 i + 0.3 k} m
r2 = {-0.25 j + 0.3 k} m
MRO = r1 × F1 + r2 × F2
i
j
MRO = { 0.15 0
6
-3
k
0.3
-10
+
i
j
k
0 - 0.25 0.3
0
2
-4
} Nm
= {0.9 i + 3.3 j – 0.45 k + 0.4 i + 0 j + 0 k} Nm
= {1.3 i + 3.3 j – 0.45 k} Nm
Engr221 Chapter 4
38
Questions
1. A general system of forces and couple moments acting on a
rigid body can be reduced to a ___
A) single force
B) single moment
C) single force and two moments
D) single force and a single moment
2. The original force and couple system and an equivalent
force-couple system have the same _____ effect on a body.
A) internal
B) external
C) internal and external D) microscopic
Questions
•Z
S
1. The forces on the pole can be reduced to
a single force and a single moment at
point ____
A) P
B) Q
C) R
D) S
E) Any of these points
•
R
•
Q
•P
Y
X
2. Consider two couples acting on a body. The simplest possible
equivalent system at any arbitrary point on the body will have:
A) one force and one couple moment
B) one force
C) one couple moment
D) two couple moments
Engr221 Chapter 4
39
Summary
• Determine the effect of moving a force
• Find an equivalent force-couple system for a system of
forces and couples
Announcements
Engr221 Chapter 4
40
Distributed Loading Reduction
Today’s Objective
• Determine an equivalent force for a
distributed load
Class Activities
• Applications
• Equivalent force
• Examples
Applications
A distributed load on the
beam exists due to the
weight of the lumber.
Engr221 Chapter 4
Is it possible to reduce this
force system to a single
force that will have the
same external effect? If
yes, how?
41
Applications - continued
The sandbags on the beam create a distributed load.
How can we determine a single equivalent resultant
force and its location?
Distributed Loading
In many situations a surface area
of a body is subjected to a
distributed load. Such forces are
caused by winds, fluids, or the
weight of items on the body’s
surface.
We will analyze the most common
case of a distributed pressure
loading. This is a uniform load
along one axis of a flat rectangular
body.
In such cases, w is a function of x
and has units of force per length.
Engr221 Chapter 4
42
Magnitude of Resultant Force
Consider an element of length dx.
The force magnitude dF acting on it is
given as
dF = w(x) dx
The net force on the beam is given by
+ ↓ FR = ∫L dF = ∫L w(x) dx = A
Here, A is the area under the
loading curve w(x).
Location of the Resultant Force
The force dF will produce a moment of
(x)(dF) about point O.
The total moment about point O is
given as
+ MRO = ∫L x dF = ∫L x w(x) dx
Assuming that FR acts at x , it will
produce the moment about point O as
+ MRO = x (FR) = ∫L x w(x) dx
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Location of the Resultant Force - continued
Comparing the last two equations,
we get
In other words, FR acts through a
point “C,” which is called the
geometric center or centroid of the
area under the loading curve w(x).
Examples
We will usually consider only rectangular and triangular loading
diagrams whose centroids are well defined and shown on the inside
back cover of your textbook.
In the rectangular loading, FR = 400 × 10 = 4,000 lb, x = 5 ft
In the triangular loading,
FR = (0.5) (600) (6) = 1,800N and x = 6 – (1/3) 6 = 4 m
Note that the centroid in a right triangle is at a distance one
third the width of the triangle as measured from its base.
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Example A
Given: The loading on the beam as
shown
Find: The equivalent force and its
location from point A
Plan:
1) Consider the trapezoidal loading as two separate loads (one
rectangular and one triangular)
2) Find FR and x for each of these two distributed loads
3) Determine the overall FR and
for the three point loadings
x
Example A - continued
For the rectangular loading of height
0.5 kN/m and width 3m,
FR1 = 0.5 kN/m × 3 m = 1.5 kN
x1 = 1.5m from A
For the triangular loading of height 2 kN/m and width 3m,
FR2 = (0.5) (2 kN/m) (3 m) = 3 kN
and its line of action is at 1m from A
For the combined loading of the three forces,
FR = 1.5 kN + 3 kN + 1.5 kN = 6 kN
+ MRA = (1.5) (1.5) + 3 (1) + (1.5) 4 = 11.25 kNm
MRA = FR(x) x
Hence, x = x
(11.25) / (6) = 1.88m from A
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Questions
FR
100 N/m
12 m
x
1. FR = ____________
A) 12 N B) 100 N
C) 600 N D) 1200 N
2. x = __________
A) 3 m
B) 4 m
C) 6 m
D) 8 m
Questions
w
Distributed load curve
FR
1. The resultant force (FR) due to a
distributed load is equivalent to
the _____ under the distributed
x
loading curve, w = w(x).
A) centroid
B) arc length
C) area
D) volume
2. The line of action of the distributed load’s equivalent force passes
through the ______ of the distributed load.
A) centroid
B) mid-point
C) left edge
D) right edge
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Questions
1. What is the location of FR, i.e., the
distance d?
B) 3m
C) 4m
B A) 2m
D) 5m
E) 6m
FR
A
BA
3m 3m
F1
x2
F2
x1
d
x
FR
2. If F1 = 1N, x1 = 1m, F2 = 2N and
x2 = 2m, what is the location of
FR, i.e., the distance x?
A) 1m
B) 1.33m C) 1.5m
D) 1.67m E) 2m
Textbook Problem 4-156
Simplify the distributed loading to an equivalent
resultant force and specify the magnitude and
location of the force, measured from A.
FR = 1.25 kN
xA = 8.00 m
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Summary
• Determine an equivalent force for a distributed load
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