engineering mechanics statics - rami zakaria...scalar approach example given: two couples act on the...
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6. Moment of a Couple
Engineering Mechanics
Statics
Dr. Rami Zakaria
Mechanical Systems Engineering_2016
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We need a moment (or torque)
of (12 N · m) to rotate the
wheel. Notice that one of the
two grips of the wheel above
require less force to rotate the
wheel.
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).
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Moment of a Couple
The moment of a couple is defined as :
MO = F d (using a scalar analysis)
or as:
MO = 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.
We can add Couple Moments together using the same rules as adding any vectors.
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Scalar approach Example
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.
From the definition of a couple: P = 2 kN, F = 4 kN
Determine the net moment
+ M = (2)(0.3) – (4)(d)
It was given that the net moment equals zero. So:
+ M = (2)(0.3) – (4)(d) = 0
Now solve this equation. d = (0.6) N·m / (4) N = 0.15 m
Solution:
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Classroom
Exercise
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1) Use M = r F to find the couple moment.
2) Set r = rAB and F = {35 k} N .
3) Calculate the cross product to find M.
Given: A 35 N force couple
acting on the rod.
Find: The couple moment
acting on the rod in
Cartesian vector notation.
Plan:
Vector approach Example
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= {(– 8.75 – 0) i – (0 – 0) j – (0 – 0) k} N·m
= {– 8.75 i + 0 j + 0 k} N·m
M = rAB F
= N·m
i j k 0 – 0.25 0.1443
0 0 35
rAB = { 0 i – (0.25) j + (0.25 tan 30°) k} m
rAB = {– 0.25 j + 0.1443 k} m
F = {0 i + 0 j + 35 k} N
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SIMPLIFICATION OF FORCE AND COUPLE 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.
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MOVING A FORCE ON ITS LINE OF ACTION
Moving a force from A to B, when both points are on the
vector’s 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).
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MOVING A FORCE OFF OF ITS LINE OF ACTION
When a force is moved, but not along its line of action, there
is a change in its external effect!
Essentially, moving a force from point A to B requires
creating an additional couple moment. So moving a force
means you have to “add” a new couple.
Since this new couple moment is a “free vector”, it can be
applied at any point on the body.
B
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SIMPLIFICATION 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.
WR = W1 + W2
(MR)o = W1 d1 + W2 d2
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PROBLEM SOLVING
1) Sum all the x and y components of the two forces to find FRA.
2) Find and sum all the moments resulting from moving each force to
A and add them to the 45 kN m free moment to find the resultant
MRA .
Given: A 2-D force and couple
system as shown.
Find: The equivalent resultant
force and couple
moment acting at A.
Plan:
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+ Fx = (5/13) 26 – 30 sin 30°
= – 5 kN
+ Fy = – (12/13) 26 – 30 cos 30°
= – 49.98 kN
Summing the force components:
Solution:
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+ MRA = {30 sin 30° (0.3m) – 30 cos 30° (2m) – (5/13) 26 (0.3m)
– (12/13) 26 (6m) – 45 } = – 239 kN m
Now find the magnitude and direction of the resultant.
FRA = ( 5 2 + 49.98 2 )1/2 = 50.2 kN and = tan-1 (49.98/5)
= 84.3 °
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Classroom
Exercise
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FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
In three special cases, concurrent, coplanar, and parallel systems of
forces, the system can always be reduced to a single force.
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.
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EXAMPLE
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 component to A.
3) Shift FRA to a distance d such that
Given: A 2-D force system
with geometry as shown.
Find: The equivalent resultant
force and couple
moment acting at A and
then the equivalent
single force location
measured from A.
Plan:
Ry
RAx
F
Md
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Solution:
+ FRx = 150 (3/5) + 50 –100 (4/5)
= 60 lb
+ FRy = 150 (4/5) + 100 (3/5)
= 180 lb
+ MRA = 100 (4/5) 1 – 100 (3/5) 6
– 150(4/5) 3 = – 640 lb·ft
FR = ( 602 + 1802 )1/2 = 190 lb
= tan-1 ( 180/60) = 71.6 °
The resultant force FR can be located at a distance d measured from A.
dx = MRA/FRy = 640 / 180 = 3.56 ft.
FR
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EXAMPLE (simplification of a system)
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
Given: The slab is subjected to three
parallel forces.
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:
RZ
RxO
RZ
RyO
F
My
F
Mx ,
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FRO = {100 k – 500 k – 400 k} = – 800 k N
MRO = (3 i) (100 k) + (4 i + 4 j) (-500 k)
+ (4 j) (-400 k)
= {–300 j + 2000 j – 2000 i – 1600 i}
= { – 3600 i + 1700 j }N·m
The location of the single equivalent resultant force is given as,
x = + |MRyo| / |FRzo| = (1700) / (800) = 2.13 m
y = + |MRxo| / |FRzo| = (3600) / (800) = 4.5 m
Solution:
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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:
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3. A couple is applied to the beam as shown. Its moment equals _____ N·m.
A) 50 B) 60
C) 80 D) 100
3
4
5 1m 2m
50 N
4. 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
5. 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
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DISTRIBUTED LOADING
In such cases, w(x) is the load intensity
distribution function and it has units of
force per length.
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.
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MAGNITUDE OF RESULTANT FORCE
The net force on the beam is given by
+
Here A is the area under the loading
curve w(x).
Consider an element of length dx.
The force magnitude dF acting on it is
given as
dxxwdF )(
LL
R AdxxwdFF )(
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LOCATION OF THE RESULTANT FORCE
The total moment about point O is given as
Assuming that FR acts at , it will produce the
moment about point O as
The force dF will produce a moment of (x)(dF)
about point O.
x
LL
Ro dxxxwxdFM )(
L
RRo dxxwxFxM )()()(
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Comparing the last two equations,
we get
You will learn more detail later, but
FR acts through a point “C,” which is
called the geometric center or
centroid of the area under the loading
curve w(x).
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Example: Determine the magnitude
and location of the equivalent resultent
force (FR) acting on the shaft
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Area Cx
b.h b/2
b.h/2 (a +b)/3
π.a2 a
π.a2 /2 a
Note: for simples shapes we can find the shift easily using the geometric centroid
(we will learn more about the centroid later!)
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Examples
The rectangular load:
ftx
lbFR
52
10
400010400
The triangular loading:
mx
NFR
43
66
180066002
1
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Classroom
Exercise
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