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4.6 Moment of a Couple4.6 Moment of a Couple
� Couple
- two parallel forces
- same magnitude but opposite direction
- separated by perpendicular distance d- separated by perpendicular distance d
� Resultant force = 0
� Tendency to rotate in specified direction
� Couple moment = sum of
moments of both couple
forces about any arbitrary point
4.6 Moment of a Couple4.6 Moment of a Couple
Example
Position vectors rA and rA are directed from O to
A and B, lying on the line of action of F and –FA and B, lying on the line of action of F and –F
� Couple moment about O
M = rA X (-F) + rA X (F)
� Couple moment about A
M = r X F
since moment of –F about A = 0
4.6 Moment of a Couple4.6 Moment of a Couple
� A couple moment is a free vector
- It can act at any point since Mdepends only on the position vector rdirected between forces and not directed between forces and not position vectors rA and rB, directed from O to the forces
- Unlike moment of force, it do not require a definite point or axis
4.6 Moment of a Couple4.6 Moment of a Couple
Scalar Formulation� Magnitude of couple moment
M = Fd
� Direction and sense are determined by right hand rule
� In all cases, M acts perpendicular to plane containing the forces
4.6 Moment of a Couple4.6 Moment of a Couple
Vector Formulation
� For couple moment,
M = r X FM = r X F
� If moments are taken about point A, moment of –F is zero about this point
� r is crossed with the force to which it is directed
4.6 Moment of a Couple4.6 Moment of a Couple
Equivalent Couples
� Two couples are equivalent if they produce the same momentproduce the same moment
� Since moment produced by the couple is always perpendicular to the plane containing the forces, forces of equal couples either lie on the same plane or plane parallel to one another
4.6 Moment of a Couple4.6 Moment of a Couple
Resultant Couple Moment� Couple moments are free vectors
and may be applied to any point P
and added vectoriallyand added vectorially
� For resultant moment of two
couples at point P,
MR = M1 + M2
� For more than 2 moments,
MR = ∑(r X F)
4.6 Moment of a Couple4.6 Moment of a Couple
� Frictional forces (floor) on the blades of the machine creates a moment Mc that tends to turn itto turn it
� An equal and opposite moment must be applied by the operator to prevent turning
� Couple moment Mc = Fd is applied on the handle
4.6 Moment of a Couple4.6 Moment of a Couple
Example 4.10
A couple acts on the gear teeth. Replace it
by an equivalent couple having a pair of
forces that cat through points A and B.forces that cat through points A and B.
4.6 Moment of a Couple4.6 Moment of a Couple
Solution
� Magnitude of couple
M = Fd = (40)(0.6) = 24N.m
Direction out of the page since � Direction out of the page since
forces tend to rotate CW
� M is a free vector and can
be placed anywhere
4.6 Moment of a Couple4.6 Moment of a Couple
Solution
� To preserve CCW motion,
vertical forces acting through points A and B must be directed as shownshown
� For magnitude of each force,
M = Fd
24N.m = F(0.2m)
F = 120N
4.6 Moment of a Couple4.6 Moment of a Couple
Example 4.11
Determine the moment of the couple acting
on the member.
4.6 Moment of a Couple4.6 Moment of a Couple
Solution
� Resolve each force into horizontal and vertical components
Fx = 4/5(150kN) = 120kN
F = 3/5(150kN) = 90kNx
Fy = 3/5(150kN) = 90kN
� Principle of Moment about point D,
M = 120kN(0m) – 90kN(2m)
+ 90kN(5m) + 120kN(1m)
= 390kN (CCW)
4.6 Moment of a Couple4.6 Moment of a Couple
Solution
� Principle of Moment about point A,
M = 90kN(3m) + 120kN(1m)
= 390kN (CCW)
*Note:
- Same results if take moment
about point B
- Couple can be replaced by
two couples as seen in figure
4.6 Moment of a Couple4.6 Moment of a Couple
Solution- Same results for couple replaced by two couples
- M is a free vector and acts on - M is a free vector and acts on any point on the member
-External effects such as support reactions on the member, will be the same if the member supports the couple or the couple moment
4.6 Moment of a Couple4.6 Moment of a Couple
Example 4.12
Determine the couple moment acting on the
pipe. Segment AB is directed 30° below the
x-y plane. x-y plane.
4.6 Moment of a Couple4.6 Moment of a Couple
SolutionMethod 1Take moment about point O,
M = rA X (-25k) + rB X (25k)= (8j) X (-25k) + (6cos30°i + 8j – 6sin30°k) X = (8j) X (-25k) + (6cos30°i + 8j – 6sin30°k) X (25k)= {-130j}N.cm
Take moment about point AM = rAB X (25k) = (6cos30°i – 6sin30°k) X (25k)
= {-130j}N.cm
4.6 Moment of a Couple4.6 Moment of a Couple
Solution
Method 2
Take moment about point A or B,
M = Fd = 25N(5.20cm)M = Fd = 25N(5.20cm)
= 129.9N.cm
Apply right hand rule, M acts in
the –j direction
M = {-130j}N.cm
4.6 Moment of a Couple4.6 Moment of a Couple
Example 4.13
Replace the two couples acting on the pipe
column by a resultant couple moment.
4.6 Moment of a Couple4.6 Moment of a Couple
Solution
� For couple moment developed by the forces at A and B,
M1 = Fd = 150N(0.4m) = 60N.mM1 = Fd = 150N(0.4m) = 60N.m
� By right hand rule, M1 acts in
the +i direction,
M1 = {60i} N.m
� Take moment about point D,
M2 = rDC X FC
4.6 Moment of a Couple4.6 Moment of a Couple
Solution
M2 = rDC X FC
= (0.3i) X [125(4/5)j – 125(3/5)k]
= (0.3i) X [100j – 75k]= (0.3i) X [100j – 75k]
= {22.5j + 30k} N.m
� For resultant moment,
MR = M1 + M2
= {60i + 22.5j + 30k} N.m
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