moment and couples. moment of force the turning effect of a force (torque) is known as the moment....

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 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

(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 ( )

Figure 3.2

3’

A

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

MA=Fd = MA=-Fy(d2)+Fx(d1)

AA

F

d

F

Fx

Fy

dx

dy=

Example 3.3A 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

5 lb Total moment

= -50 + (-50)

= -100lb.in

• These force could have been treated as a couple, which consists of two forces that are:1. Equal2. Acting in opposite direction3. 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 theA 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 N)

= {-40i – 20j + 40k} Ni j k

MA =rB x F = 1 3 2

-40 -20 40

= [3(40) – 2 (-20)]i – [1(40) – 2(-40)]j + [1(-20) – 3(-40)]k

222 (2)1)(2)(

0)(24)(33)(1 kji

MA = [160i -120j + 100k] NmMagnitude MA =

= 224 N.m

222 (100)120)((160)

• 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 30i + 8j – 6 sin 30k) 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 30i – 6 sin 30k) 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

and a direction of 962 N 222Ry

2xRR (882.8)(382.8))(F)(FF

66.6

382.8882.8

tanF

Ftanθ 1

Rx

Ry1

• 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

i j kMRO = (-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

Exercise 3: Replace the three forces shown with an equivalent Replace the three forces shown with an equivalent force-couple system at A.force-couple system at A.

FF11

FF22

FF33

To find the equivalent set of forces at A.

x x

o o o400 N cos 180 750 N cos 36.87 100 N cos 90

200 N

R F

1 o3tan 36.87

4

y y

o o o400 Nsin 180 750 Nsin 36.87 100 Nsin 90

550 N

R F

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

1 B

100 N 360 mm

36000 N-mm

M F d

OOOOOOOOOOOOOOOOOOOOOOOOOOOO

Find the moments about point A.

The force at O can be broken up into its two components in the x and y direction

ox

oy

750 N cos 36.87

600 N

750 Nsin 36.87

450 N

F

F

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.

Fy component is 0 since in line with A.

2 Ox

600 N 160 mm

96000 N-mm

M F d

OOOOOOOOOOOOOOOOOOOOOOOOOOOO

B i

1 2 3

36000 N-mm 96000 N-mm 0 N-mm

132000 N-mm

M M

M M M

k k k

k

OOOOOOOOOOOOOOOOOOOOOOOOOOOO

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

The final result is

R = 585 N at 70.0o

M = 132 Nm

M = 132 Nm