MOMENT%20AND%20COUPLES

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MOMENT AND COUPLES
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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.

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SMALL MOMENTThe distance from the fulcrum to the
line of action of force is very small
LARGE MOMENTThe distance from the fulcrum to the
line of action of force is large
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(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.

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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 ( )
• 
  
• 
  Figure 3.2

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A
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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
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Example 3.2
MO (-60 lb) (1 sin 45? ft) -42.4
lb.ft
MO (-7 kN) (4 m 1 m) 21.0 kNm
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Principle of moment

• Sometimes refer as Varignons theorem
• The moment of a force about a point is equal to
  the sum of the moments of the forces components
  about the point
• MAFd MA-Fy(d2)Fx(d1)

A
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Example 3.3

• A 200 N force acts on the bracket shown in
  Figure. Determine the moment of the force about
  point A.

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Exercise 1

• Determine the magnitude and directional sense of
  the moment of the force A about point O

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Exercise 2

• Determine the magnitude and directional sense of
  the moment of the force at A about point O

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COUPLES

• A couple consists of two equal , acting in
  opposite directions and separated by a
  perpendicular distance.
• Example

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5 lb
Total moment -50 (-50) -100lb.in
5 lb
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• These force could have been treated as a couple,
  which consists of two forces that are
• Equal
• Acting in opposite direction
• 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
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• 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

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MA-(10N)(4m)-(10N)(2m) -40-20 -60
N.m 60N.m
Mb-(10N)(11m)-(10N)(5m) -11050
-60 N.m 60N.m
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Example 3.4

• Determine the moment of the couple acting on the
  member shown in Figure

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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.

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• 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 theA and B vectors).

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• 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

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• Of even more utility, the cross product can be
  written as
• Each component can be determined using 2 ? 2
  determinants

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• 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.

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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.

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• 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 N)
• -40i 20j 40k N

3(40) 2 (-20)i 1(40) 2(-40)j
1(-20) 3(-40)k
MA 160i -120j 100k Nm Magnitude MA
224 N.m
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• 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 forces
  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. However this
  calculation is not always trivial and vector
  analysis may be preferable

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Example

• Determine the couple moment acting on the pipe
  shown in Fig. 3.24a. Segment AB is directed 30?
  below the x-y plan

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• 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

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• 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

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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 M1r1xF1 and M2r2xF2 are applied
  to body

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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.

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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).
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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.
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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.

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Example 3.5

• Replace the forces acting on the brace shown in
  Figure by an equivalent resultant and couple
  moment acting at point A.

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• 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
• and a direction of

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• 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

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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.

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• 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 -0.15i0.1j/? (0.15)2 (0.1)2
  -249.6i 166.4jN
• 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

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Moment Summation MRO ?MC ?MO
MRO M rC x F1 rB x F2
i j k MRO (-400j 300k)
(1k) x (-800k) -0.15 0.1
1 - 249.6 166.4 0
(-400j 300k) (0) (-166.4i 249.6j)
-166i -650j 300k Nm
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Exercise 3
Replace the three forces shown with an
equivalent force-couple system at A.
F1
F2
F3
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To find the equivalent set of forces at A.
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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
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Find the moments about point A.
The force at O can be broken up into its two
components in the x and y direction
Using the line of action for each component,
their moment contribution can be determined.
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Find the moments about point A.
Using the line of action for Fx component d is
160 mm.
Fy component is 0 since in line with A.
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The final result is
R 585 N at 70.0o
M 132 Nm ?
M 132 Nm