MOMENT
AND
COUPLES
Moment of Force
•
The turning effect of a force (torque) is known as the
moment.
•
It is the product of the force multiplied by the
perpendicular distance from the line of action of the
force to the pivot or point where the object will turn.
SMALL MOMENT
The distance from the fulcrum
to the line of action of force is
very small
LARGE MOMENT
The distance from the fulcrum
to the line of action of force is
large
(Cont…)
•
Unit: pound-feet (lb), pound-inches (lb-in), kip-feet
(kip-fit) or Newton-meter (Nm)
•
Moments taken are about a point are indicate as
being clockwise( ) or counterclockwise ( )
•
For the sake of uniformity in calculation, assume
clockwise to be +ve and counterclockwise to be -ve.
•
Moment can exspressed as 10 lb-ft ( ), + 10 lb-ft
or 10 lb.ft.
Example 3.1
Calculate the moment about point A in Figure
3.2. Notice that the perpendicular distance
can be measured to the line of action of the
force.
M=(F) (d)
= + (50) (3)
M= 150 lb-ft ( )
3’
A
Figure 3.2
Example 3.2
MO = (100 N) (2 m) = +200 Nm
MO = (50 N) (0.75 m) = 37.5 Nm
MO = (40 lb) (4 ft + 2 cos 30 ft)
= 229 lb.ft
Example 3.2
MO = (-60 lb) (1 sin 45 ft)
= -42.4 lb.ft
MO = (-7 kN) (4 m – 1 m)
= 21.0 kNm
Principle of moment
•
Sometimes refer as Varignon’s theorem
•
The moment of a force about a point is equal to the sum
of the moments of the force’s components about the
point
Fy
F
F
Fx
=
dy
dx
d
A
A
MA=Fd
=
MA=-Fy(d2)+Fx(d1)
Example 3.3
A 200 N force acts on the bracket shown in Figure. Determine the
moment of the force about point A.
Exercise 1
•
Determine the magnitude and directional sense of the
moment of the force A about point O
Exercise 2
•
Determine the magnitude and directional sense of the
moment of the force at A about point O
COUPLES
•
A couple consists of two equal , acting in
opposite directions and separated by a
perpendicular distance.
•
Example:
20’’
5 lb
Total moment
= -50 + (-50)
= -100lb.in
5 lb
•
These force could have been treated as a couple,
which consists of two forces that are:
1.
Equal
2.
Acting in opposite direction
3.
Separated by some perpendicular distance d
•
These three requirement of couple, from the
example, we have;
Couple moment = (F) (d)
= -5 (20)
= -100 lb.in
•
This is the same answer that we obtained when we
multiplied the individual forces by their distance from
the pivot.
•
Notice that when calculate moment, specified the
points or moment about which the moments were
calculated.
•
It does not matter where the moment center is located
when deal with couples.
•
A couples has the same moment about all points on a
body
MA=-(10N)(4m)-(10N)(2m)
=-40-20
=-60 N.m
=60N.m
Mb=-(10N)(11m)(10N)(5m)
=-110+50
=-60 N.m
=60N.m
Example 3.4
•
Determine the moment of the couple acting on the
member shown in Figure
Moment in 3Dimensional
Vector analysis
•
Moments in 3-D can be calculated using 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 the position vector from point O to any point
on the line of action of F.
•
•
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 = A B sin  UC
Here UC is the unit vector perpendicular to both A and B
vectors as shown (or to the plane containing the
A and B vectors).
•
The right hand rule is a useful tool 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
•
Of even more utility, the cross product can be written as
•
Each component can be determined using 2  2
determinants
•
So, using the cross product, a moment can be expressed
as
•
By expanding the above equation using 2  2
determinants, we get (sample units are N - m)
MO = (r y FZ - rZ Fy) i - (r x 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.
Example
•
The pole in Fig. Below 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.
since MA = rB x F or MA = rc x F
rB = {1i + 3j + 2k} m and
rC = {3i + 4j} m
The force has a magnitude of 60 N and a direction specified by the unit
vector uF, directed from C to B. Thus,
F = (60 N) uF = (60
 (1  3)i  (3  4)j  (2  0)k 
N) (2)2  (1)2  (2)2 


i
= {-40i – 20j + 40k} N
MA =rB x F = 1
-40
j
k
3
2
-20
40
= [3(40) – 2 (-20)]i – [1(40) – 2(-40)]j + [1(-20) – 3(-40)]k
MA = [160i -120j + 100k] Nm
Magnitude MA = (160) 2  ( 120) 2  (100) 2
= 224 N.m
•
Scalar analysis (moment at axis)
•
Recall that the moment of a force about any point A is MA= F
dA where dA is the perpendicular (or shortest) 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 N·m. However this calculation is not always
trivial and vector analysis may be preferable
Example
•
Determine the couple moment acting on the pipe
shown in Fig. 3.24a. Segment AB is directed 30 below
the x-y plan
Solution I (vector analysis)
The moment of the two couple forces can be found about any point. If
point O is considered, Fig 3.24b, we have
M = rA x (-25k) + rB x (25k)
= (8j) x (-25k) + (6 cos 30i + 8j – 6 sin 30k) x (25k)
= -200i -129.9j + 200i
= {-130j} lb.in
It is easier to take moments of the couple forces about a point lying on
the line of action of one of the forces, e.g., point A, Fig. 3.24c. In this
case the moment of the force A is zero, so that
M = rAB x (25k)
= (6 cos 30i – 6 sin 30k) x (25k)
= {-130j} lb.in
Solution II(scalar analysis)
Although this problem is shown in three dimensions,
the geometry is simple enough to use the scalar
equation M = Fd. The perpendicular distance between
the lines of action of the forces is d = 6 cos 30° = 5.20
in., Fig. 3.24d. Hence, taking moments of the forces
about either point A or B yields
M = Fd. = 25 lb (5.20 in) = 129.9 lb.in
Applying the right-hand rule, M acts in the –j direction.
Thus,
M = {130j} lb.in
Resultant A force and couple
system
•
When a rigid body is subjected to a system of forces and couple
moments
•
The external effects on the body by replacing the system by an
equivalent single resultant force acting at a specified point O and a
resultant couple moment
•
Point O is not on the line of action of the forces, an equivalent effect
is produced if the forces are moved to point O and the corresponding
couple moments M1=r1xF1 and M2=r2xF2 are applied to body
AN EQUIVALENT SYSTEM (Section 4.7)
=
•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).
MOVING A FORCE OFF OF 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.
FINDING THE 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.
Example 3.5
•
Replace the forces acting on the brace shown in Figure
by an equivalent resultant and couple moment acting
at point A.
+ FRx = Fx; FRx = -100 N - 400 cos 45 = - 382.8 N =
382.8 N
+ FRy = Fy; FRy = -600 N - 400 sin 45 = - 882.8 N =
882.8 N
FR has a magnitude of
FR  (FR x )2  (FRy )2  (382.8)2  (882.8)2  962 N
and a direction of
 FRy 
882.8
  tan 1 
θ  tan 1 
  66.6
F
382.8


 Rx 
•
The resultant couple moment MRA is determined by
summing the moments of the forces about point A.
Assuming that positive moments act clockwise, we
have
+ MRA = MA
MRA = 100 N (0) + 600 N (0.4m) + (400 sin 45) (0.8 m)
+ (400 cos 45) (0.3 m)
= 551 Nm
Example (Equivalent resultant
force and couple moment)
A structural member is subjected to a couple moment M and
forces F1 and F2 as shown in Fig. below. Replace this system by an
equivalent resultant force and couple moment acting at its base,
point O.
The three-dimensional aspects of the problem can be simplified by
using a Cartesian vector analysis. Expressing the forces and couple
moment as Cartesian vectors, we have
F1 = {-800k)N
F2 = (300 N)uCB = (300 N) (rcb/rcb)
= 300 [-0.15i+0.1j/ (0.15)2 + (0.1)2] = {-249.6i + 166.4j}N
M = -500 (4/5)j + 500 (3/5)k = {-400j + 300k) Nm
Force Summation
FR = F; FR = F1 + F2 = -800k – 249.6i + 166.4j
= {-249.6i + 166.4j – 800k} N
Moment Summation
MRO = MC + MO
MRO = M + rC x F1 + rB x F2
MRO = (-400j + 300k) + (1k) x (-800k)
+
i
j
k
-0.15
0.1 1
- 249.6 166.4 0
= (-400j + 300k) + (0) + (-166.4i – 249.6j)
= {-166i -650j + 300k} Nm
Exercise 3:
Replace the three forces shown with an equivalent force-couple
system at A.
F1
F2
F3
To find the equivalent set of
forces at A.
3
  tan    36.87o
4
Rx   Fx
1
 400 N cos 180o   750 N cos  36.87 o   100 N cos  90o 
 200 N
Ry   Fy
 400 N sin 180o   750 N sin  36.87o   100 N sin  90o 
 550 N
Find the moments about point A.
Using the line of action for the force at B. The force can be
moved along the line of action until it reaches perpendicular
distance from A
M1  FB d
 100 N  360 mm 
 36000 N-mm
Find the moments about point A.
The force at O can be broken up into its two components in the x and
y direction
Fx  750 N cos  36.87 o 
 600 N
Fy  750 N sin  36.87 o 
 450 N
Using the line of action for each component, their moment contribution can
be determined.
Find the moments about point A.
Using the line of action for Fx component d is 160 mm.
M 2  FOx d
 600 N 160 mm 
 96000 N-mm
Fy component is 0 since in line with A.
MB   Mi
 M1  M 2  M 3
 36000 N-mm k  96000 N-mm k  0 N-mm k
 132000 N-mm k
The final result is
R = 585 N at 70.0o
M = 132 Nm 
M = 132 Nm