Quiz (Open note, book, calc)
Problem 4

116 (pg. 181)
Replace the loading acting on the beam
by a single resultant force. Specify where
the force acts measured from
B
.
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.
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
+
F
R
=
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
+ M
RO
=
L
x dF =
L
x w(x) dx
Assuming that F
R
acts at , it will produce
the moment about point O as
+ M
RO
= ( ) (F
R
) =
L
w(x) dx
x
x
x
Comparing the last two equations,
we get
LOCATION OF THE RESULTANT FORCE
(continued)
You will learn later that F
R
acts
through a point “C,” which is
called the geometric center or
centroid of the area under the
loading curve w(x).
In a triangular loading ,
F
R
= (0.5) (6000) (6) = 1,800 N and = 6
–
(1/3) 6 = 4 m.
Please note
that the centroid in a right triangle is at a distance one third the width of
the triangle as
measured from its base
.
EXAMPLES
Until you learn more about centroids, we will consider only rectangular and
triangular loading diagrams whose centroids are well defined and shown on the
inside back cover of your textbook.
In a rectangular loading, F
R
= 400
10 = 4,000 lb and = 5 ft.
x
x
CONCEPT QUIZ
1. What is the location of F
R
, i.e., the
distance d?
A) 2 m
B) 3 m
C) 4 m
D) 5 m
E) 6 m
2. If F
1
= 1 N, x
1
= 1 m, F
2
= 2 N
and x
2
= 2 m, what is the location
of F
R
, i.e., the distance x.
A) 1 m B) 1.33 m C) 1.5 m
D) 1.67 m E) 2 m
F
R
x
F
2
F
1
x
1
x
2
F
R
B
A
d
B
A
3 m
3 m
GROUP PROBLEM SOLVING
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 F
R
and for each of these two distributed loads.
3) Determine the overall F
R
and for the three point loadings.
GROUP PROBLEM SOLVING
(continued)
For the combined loading of the three forces,
F
R
= 1.5 kN + 3 kN + 1.5 kN = 6 kN
+ M
RA
= (1.5) (1.5) + 3 (1) + (1.5) 4 = 11.25 kN
• m
Now, F
R
= 11.25 kN
• m
Hence, = (11.25) / (6) = 1.88 m from A.
x
x
For the rectangular loading of height
0.5 kN/m and width 3 m,
F
R1
= 0.5 kN/m
3 m = 1.5 kN
= 1.5 m from A
x
1
For the triangular loading of height 2 kN/m and width 3 m,
F
R2
= (0.5) (2 kN/m) (3 m) = 3 kN
and its line of action is at = 1 m from A
x
2
Group Problem Solving
4

154
Replace the distributed loading with an equivalent
resultant force, and specify its location on the
beam measured from point A
EQUILIBRIUM OF A RIGID BODY & FREE

BODY
DIAGRAMS
Today’s Objectives
:
Students will be able to:
a) Identify support reactions, and,
b) Draw a free

body diagram.
In

Class Activities
:
•
Check Homework
•
Reading Quiz
•
Applications
•
Support Reactions
•
Free
–
Body Diagram
•
Concept Quiz
•
Group Problem Solving
•
Attention Quiz
READING QUIZ
1. If a support prevents translation of a body, then the support exerts a
___________ on the body.
1) couple moment
2) force
3) Both A and B.
4) None of the above
2. Internal forces are _________ shown on the free body diagram of a
whole body.
A) always
B) often
C) rarely
D) never
APPLICATIONS
A 200 kg platform is suspended off an oil rig. How do we
determine the force reactions at the joints and the forces in the
cables?
How are the idealized model and the free body diagram used to
do this? Which diagram above is the idealized model?
APPLICATIONS
(continued)
A steel beam is used to support
roof joists.
How can we determine the
support reactions at A & B?
Again, how can we make use of an idealized model and a free
body diagram to answer this question?
CONDITIONS FOR RIGID

BODY EQUILIBRIUM
(Section
5.1)
In contrast to the forces on a particle, the
forces on a rigid

body are not usually
concurrent and may cause rotation of the
body (due to the moments created by the
forces).
For a rigid body to be in equilibrium, the
net force as well as the net moment
about any arbitrary point O must be
equal to zero.
F
= 0 and
M
O
= 0
Forces on a rigid body
Forces on a particle
THE PROCESS OF SOLVING RIGID BODY
EQUILIBRIUM PROBLEMS
For analyzing an actual physical system, first we need to create an
idealized model
.
Then we need to draw a
free

body diagram showing all the external
(active and reactive) forces
.
Finally, we need to
apply the equations of equilibrium
to solve for
any unknowns.
FREE

BODY DIAGRAMS
(Section 5.2)
1.
Draw an outlined shape
.
Imagine the body to be isolated
or cut “free” from its constraints and draw its outlined
shape.
2.
Show all the external forces and couple moments.
These
typically
include: a) applied loads, b) support reactions,
and, c) the weight of the body.
Idealized model
Free

body diagram
FREE

BODY DIAGRAMS
(continued)
3.
Label loads and dimensions
:
All known forces and
couple moments should be labeled with their
magnitudes and directions. For the unknown forces
and couple moments, use letters like A
x
, A
y
, M
A
, etc..
Indicate any necessary dimensions.
Idealized model
Free

body diagram
SUPPORT REACTIONS IN 2

D
As a general rule, if a
support prevents translation
of a body in a
given direction, then
a force is developed
on the body in the
opposite direction. Similarly,
if
rotation is prevented, a couple
moment
is exerted on the body.
A few examples are shown above. Other support reactions
are given in your textbook (in Table 5

1, pg. 205).
EXAMPLE
Given
:
An operator applies 20 lb to the
foot pedal. A spring with
k = 20 lb/in is stretched 1.5 in.
Draw
:
A free

body diagram of the foot
pedal.
The idealized model
The free

body diagram
CONCEPT QUIZ
1. The beam and the cable (with a frictionless pulley at D) support an 80 kg load at C.
In a FBD of only the beam, there are how many unknowns?
1) 2 forces and 1 couple moment
2) 3 forces and 1 couple moment
3) 3 forces
4) 4 forces
CONCEPT QUIZ
2. If the directions of the force and the couple moments are reversed, then
what will happen to the beam?
A)
The beam will lift from A.
B)
The beam will lift at B.
C)
The beam will be restrained.
D)
The beam will break.
GROUP PROBLEM SOLVING
Draw a FBD of the bar, which
has smooth points of contact at
A, B, and C.
Draw a FBD of the 5000 lb
dumpster (D). It is supported
by a pin at A and the hydraulic
cylinder BC (treat as a short
link).
GROUP PROBLEM SOLVING
(continued)
Homework
Due Next Tuesday
5

5 (pg. 211)
5

7 (pg. 212)
5

19 (pg. 227)
5

35 (pg. 230)
5

37 (pg. 231)
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