TM chapter 2 moment and couple

Line of action: perpendicular to the plane containing O and the force F.. Sample Problem 1A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O.Deter

VIETNAM OIL AND GAS GROUP

PETROVIETNAM UNIVERSITY

THEORETICAL MECHANICS

CHAPTER 2: MOMENT OF A FORCE AND A COUPLE

Lecturer : Dr Vo Quoc Thang Email : thangvq@pvu.edu.vn Website : http://www.pvu.edu.vn

• Rectangular Components of the Moment of a Force

• Moment of a Force About a Given Axis

• Moment of a Couple

• Addition of Couples

• Resolution of a Force Into a Force and a Couple

• Reduction of a System of Forces

Scalar Product of Two Vectors

• Scalar product between two vectors P and Q:

01

1



i j j k k i j j k k i i

Px  Py  Pz   Qx  Qy  Qz  PxQx  PyQy  PzQz



Q i j k i j k P

2

2 z

2 y

Scalar Product of Two Vectors

• Angle between two vectors:

PQ

QPQ

PQ

PPQ

POL    PQ

z z

y y

x x

z y

x

cosP

cosP

cosP

coscos

cosP

Vector Product of Two Vectors

• Vector product V of two vectors P and Q:

1.Line of action: perpendicular to plane

Vector Product of Two Vectors

• Vector products of Cartesian unit vectors:

00

k j j k

i

i j

k j

j k

j

i

j i k k i

j i

Mixed Triple Product of Three Vectors

• Mixed triple product of three vectors:

   Scalar

 P Q S

• The six mixed triple products formed from S, P and

Q have equal magnitudes but not the same sign:

     

Q P P S Q Q P S

S

P S Q S

Q P

Q P S

x P Q P Q S P Q P QS

y y

x x

z y x

QP

QP

QP

SS

Principle of Transmissibility

Conditions of equilibrium or motion of a rigid body are not affected by transmitting a force along its line of action.

Principle of Transmissibility

• NOT always applied Ex: in determining internal forces and deformations

Moment of a Force About a Point

The moment of F applied at the point A about

a point O is defined as:

F r

M O  A / O 

1 Line of action: perpendicular to the plane

containing O and the force F

2 Magnitude:

3 Direction: right-hand rule

Fd sin

rF O

M     N m  or  lb ft 

Moment of a Force About a Point

Fd O

M  M O  Fd '  Fd sin  M O  0

• If any force F’ has the same line of

action, same magnitude and same

direction as F, does it produce the

same moment about O as F ?

F r

F r

F r

M O  1   2     i 

YES

Rectangular Components of the Moment of a Force

k j

i

k j

i

k j i

z y

x

x y

x z

y z

z y x

MM

M

yFxF

zFxF

zFyF

Fz

Fy

Fx

i F

k j

i r

F r

M

z y

x

A A

O

FF

F

zyx

Rectangular Components of the Moment of a Force

The moment of F applied at A about B:

F r

F BA

MB    A/B 

k j

i F

k j

i

r r

r

z y

x

B A

B A

B A

B A

B

/

A

FF

F

zz

yy

xx

z B

A

y B

A

x B

A B

Fz

z

Fy

y

Fx

Rectangular Components of the Moment of a Force

For two-dimensional structures:

 k k

MB  z  xA  xB Fy  yA  yB Fz

Sample Problem 1

A 100-lb vertical force is applied to the end of

a lever which is attached to a shaft at O.Determine:

Sample Problem 1

a) Moment of force applied at A about O:

12.100M

1260

cos.24d

FdM

Sample Problem 1

b) Horizontal force at A that produces the same moment:

8.20

1200F

F8.201200

FdM

8.2060

sin.24d

F 

F241200

F 

Sample Problem 1

d) To determine the point of application of a

240 lb force to produce the same moment:

in

5cos60

1200d

d2401200

Sample Problem 1

e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lbforce, none are of the same magnitude and sense, or on the same line of action None of the forces is equivalent to the 100 lb force

Sample Problem 2

The rectangular plate is supported by the brackets at A and B and by

a wire CD Knowing that the tension in the wire is 200 N, determine

the moment about A of the force exerted by the wire at C

Sample Problem 2

SOLUTION:

k j

i M

12808

.0

960

1203

.0

.03.0

A C

A

F r

MA  C A 

 N128

96120

5.0

32.024

.3

.0200r

200F

C D

C D CD

k j

i

k j

i

r λ

Moment of a Force About a Given Axis

• The moment MO of a force F applied at the point A about a point O:

F r

λ    

 O

OL M

Moment of a Force About a Given Axis

• Moment of a force about an arbitrary axis:

B BL

BL

M

The result is independent of the point B

along the given axis

• Moments of F about the coordinate axes x, y and z:

x y

z

z x

y

y z

x

yFxF

M

xFzF

M

zFyF

Sample Problem 3

a) about Ab) about the edge ABc) about the diagonal AG of the cube.d) Determine the perpendicular

distance between AG and FC

A cube is acted on by a force P at point F Determine the moment of P:

Sample Problem 3

i j k

i

M λ

i M

k j

FC λ

P

j i r

r r

P r

M

2P0

2Pa

0a

2

PFC

PP

a

A

FC

A F

A F

A F A

b) Moment of P about AB:

aP3

1M

2aP

3

13

a

aaaAGM

AG

A AG

A AG

j i

k j i M

k j i k

j i AG

λ

M λ

P3

12

Therefore, P is perpendicular to AG

Pd6

aP

6a

d 

Moment of a Couple

Two forces F and -F having the same

magnitude, parallel lines of action, and

opposite sense are said to form a couple

• Moment of the couple:

   B

/ A

B A

B A

F r

r F

r F r

M

• The moment vector of the couple is a

free vector that can be applied at any point with the same effect

Addition of Couples

• Consider two intersecting planes P1

and P2 with each containing a couple:

2

1

Pplanein

Pplanein

2 2

1 1

F r M

F r M

M     

• By Varignon’s theorem:

2 1

2 1

M M

F r F r M

Resolution of a Force Into a Force and a Couple

Force vector F applied at A can be moved to O by replacing it with

an equivalent force vector F at O and a couple vector MO=rxF

Moving F from A to a different point O’

The moments of F about O and O’ are related:

 

F s M

F s F r F s

r F

r M

Sample Problem 4

Determine the components of the single couple equivalent to the couples shown

Sample Problem 4

1) Attach equal and opposite 20 lb forces in the +x direction at A:

The three couples may be represented

by three couple vectors,

in

lb

in

lb2

in

lb3

.20M

24012

.0M

54018

.0M

z y x

lb.in.

180240

Sample Problem 4

2) Compute the sum of the moments of the four forces about D:

Only the forces at C and E contribute to the moment about D

 

k j

M M

2012

9

3018

Reduction of a System of Forces

A system of forces may be replaced by a collection of force-couple systems acting a given point O

NOTE: Two systems of forces are equivalent if they can be

R O

i

F r M

F R

Sample Problem 5

For the beam, reduce the system of forces shown to

(a) an equivalent force-couple system at A,

(b) an equivalent force couple system at B,

(c) a single force or resultant

Note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium

Sample Problem 5

j j

j j

F R

5000

0050

i

21

j i

F r M

2508

.4

1008

.2600

6.1

i i

R A

R r

M M

28801880

6008

.41880

A B

R A

R B

The couple at B is equal to the moment about

B of the force-couple system found at A:

Sample Problem 5

c) Single force or resultant:

The moment of R about A is equal to

k k

k j

i

M R

r

1880x

600

1880600

x

R A

x 

R A