6161103 4.6 moment of a couple

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


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


� 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


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


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


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


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)


� 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


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.


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


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


Example 4.11

Determine the moment of the couple acting

on the member.


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)


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


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


Example 4.12

Determine the couple moment acting on the

pipe. Segment AB is directed 30° below the

x-y plane. x-y plane.


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


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


Example 4.13

Replace the two couples acting on the pipe

column by a resultant couple moment.


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


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

Health & Medicine

6161103 4.6 moment of a couple