ME13A: CHAPTER THREE

STATICS OF RIGID BODIES

3.1 INTRODUCTION

A rigid body is one which does not suffer deformation. It can be continuous or

connected members.

F2 F3

F1

P2

P1

F3 F1 F2

Continuous Member Connected Members

INTRODUCTION CONTD.

The forces acting on rigid bodies can be internal or external. F1, F2 and F3

which are applied by an external force on the rigid body are called external forces.

P1 and P2 which are forces internal to

the rigid body are called internal forces.

INTRODUCTION CONCLUDED.

The external forces are completely responsible for the bulk motion of the rigid body.

As far as this bulk motion is concerned, the internal forces are in equilibrium.

3.2 PRINCIPLE OF TRANSMISSIVITY OF FORCES

External forces generally cause translation i.e. linear motion and/or rotation (motion about a pivot) of the rigid body.

PRINCIPLE OF TRANSMISSIVITY OF FORCES

Principle of transmissivity states that the condition of rest or motion of a rigid body is unaffected if a force, F acting on a point A is moved to act at a new point, B provided that the point B lies on the same line of action of that force.

A B

F F

3.3 CROSS OR VECTOR PRODUCT OF TWO VECTORS

The moment of a force will be formulated using Cartesian vectors in the next section.

It is necessary to first expand our knowledge of vector algebra and introduce the cross-product method of vector multiplication.

Cross Products of ForcesP and Q

The cross product of two vectors , P and Q yields the vector, V

which is written:

V = P x Q ( i.e V = P cross Q).

3.3.1 Magnitude: The magnitude of V is the product of the magnitudes

of P and Q and sine of the angle between their tails ( 0 < < 180o).

Thus : V = P Q Sin

3.3.2 Direction

Vector, V has a direction that is perpendicular to the plane containing P and Q such that the direction of V is specified by the right hand rule i.e. curling the fingers of the right hand from vector P (cross) to vector Q, the thumb then points in the direction of V.

Direction of Cross Product

The sense of V is such that a person located at the tip of V will observe as

counterclockwise the rotation through that brings the vector P in line with vector Q.

Knowing the magnitude and direction of V:

V = P x Q = (P Q sin ) v

Where: the scalar PQ sin defines the magnitude of V and the unit vector, v

defines the direction of V.

V = P x Q V

Q PQ sin = V

Q

P P

3.3.3. Laws of Operation

1. The cummutative law does not apply i.e. P x Q Q x P

Rather: P x Q = - Q x P 2. Multiplication by a scalar a ( P x Q) = (a P) x Q = P x ( a

Q) = ( P x Q ) a 3. The distributive law: P x (Q + S ) = ( P x Q ) + ( P x

S ) 4. The associative property does not apply

to vector products (P x Q ) x S P ( Q x S )

Cartesian (Rectangular) Vector

Formulation

y

j x

k = (i x j) i

z

To find i x j, the magnitude of the resultant is:

i x j = i . j . sin 90o = 1 x 1 x 1 = 1.

Using the right hand rule, the resultant vector points in the k direction.

Thus i x j = 1 k. But: j x i = - k since the 90o rotation that

brings j into i is observed as counterclockwise by a person located

at the tip of - k.

Cartesian Vector Formulation Contd.

Likewise: i x j = k i x k = - j i x i = 0 j

j x k = i j x i = - k j x j = 0 k i

k x i = j k x j = - i k x k = 0

Rules: If a circle is constructed as shown, then the vector product of two unit vectors in

a counterclockwise fashion around the circle yields the positive third unit vector eg.

k x i = j. Moving clockwise, a negative unit vector is obtained e.g. i x k = - j.

k i

j

-ve

Cross Product of Two Vectors P and Q

Consider the cross product of two vectors P and Q expressed in

Cartesian vector form:

V = P x Q = ( Px i + Py j + Pz k) x ( Qx i + Qy j + Qz k)

= Px Qx (i x i) + Px Qy (i x j) + Px Q z ( i x k)

+ Py Qx ( j x i) + Py Qy (j x j) + Py Q z (j x k)

+ Pz Q x (k x i) + Pz Qy (k x j) + Pz Qz (k x k)

= ( Px Qy ) k + Px Qz (-j) + Py Qx (-k) + Py Qz (i) + Pz Qx (j) + Pz Qy (-i)

V = (Py Q z - Pz Qy) i + (Pz Qx - Px Qz) j + (Px Qy - Py Qx) k

Equation For Product, V in Determinant Form

The equation for V may be written in a more compact determinant form as:

  V = P x Q = i j k

Px Py Pz

Qx Qy Qz

3.4 MOMENT OF A FORCE ABOUT A POINT

The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.

Moment of a Force About a Point Consider a force, F acting at point A

and a point O which lie on the same plane. The position of A is defined by the position vector, r, which joins the reference point O with A

Moment of a Force About a Point

Moment (Mo) of F about O is defined as the vector product of r and F: i.e. Mo = r x F

Direction of Mo

Using the right hand rule, the direction of Mo can be found. Curling the fingers of the right hand to follow the sense of rotation, the thumb points along the moment axis so the direction and sense of the moment vector is upward and perpendicular to the plane containing r and F.

Finally, denoting by the angle between the lines of action of the position vector, r and

the force, F, the magnitude of the moment of F about O is:

Mo = r F sin = F d

Where: d is the perpendicular distance from O to the line of action of F.

In two dimensions:

F

F

F

Moment, M0 = + F d (Counter-clockwise , +ve) Moment, M0 = + F d clockwise , +ve)

3.5 VARIGNON’S THEOREM It states that the moment of a force about a point is equal

to the sum of the moments of its components about the same point.

Consider the Force, F with two components: F = F1 + F2.

Using the distributive law of vectors ( see section 3.3.3)

Mo = r x F1 + r x F2 = r x (F1 + F2) = r x F

y

F1 Moment of F about 0

F Mo = r x F

F2

r

o x

3.6 RECTANGULAR COMPONENTS OF THE MOMENT OF A FORCE (CARTESIAN VECTOR FORMULATION)

 

Convention

Alternative Method

Fx = 100 cos 25o = 90.63 N Fy = 100 sin 25o = 42.26 N MB = - (0.204 x 90.63) + (42.26 x 0.0951)

= - 14.47 N m = 14.47 N m Clockwise

Note

Cross or vector product has a distinct advantage over the scalar formulation when solving problems in three dimensions. This is because, it is easier to establish the position vector, r to the force rather than determine the moment-arm distance, d perpendicular to the line of the force.

For two dimensional problems, it is easier to use the Scalar Formulation

3.7 SCALAR OR DOT PRODUCT OF TWO VECTORS

Scalar product of two vectors, P and Q P.Q = P Q cos ..... magnitude

Where: 0 < < 180o The dot product is often referred to as

scalar product of vectors, since the result is a scalar, not a vector.

Q

P

3.7.1. Laws of Operation

(i) Commutative law: P. Q = Q. P

(ii) Multiplication by a scalar: a (P. Q) = (a P). Q = P. (a Q) = (P. Q) a

(iii) Distributive law: P. ( Q1 + Q2) =

P Q1 + P Q2

3.7.2 Cartesian Vector Formulation

3.7.3 Applications of Dot Product

1 . D o t p r o d u c t i s u s e d t o d e t e r m i n e t h e a n g l e f o r m e d b y t w o v e c t o r s o r i n t e r s e c t i n g

l i n e s . T h e a n g l e b e t w e e n t w o v e c t o r s P a n d Q i s r e q u i r e d .

P = P x i + P y j + P z k Q = Q x i + Q y j + Q z k

R e c a l l t h a t : P . Q = P Q c o s . . . . . . . . . . . . . ( 1 )

P . Q = P x Q x + P y Q y + P z Q z . . . . . . ( 2 )

E q u a t i n g ( 1 ) a n d ( 2 ) : P Q c o s = P x Q x + P y Q y + P z Q z

i eP x Q x P y Q y P z Q z

P Q. . c o s

Solution Contd.

AC = - 100 mm i + 300 mm j - 600 mm k, AC = 678.2 mm

BC = - 600 mm i + 300 mm j - 600 mm k, BC = 900 mm

Let AC be P and BC be Q

P QPx Qx Py Qy Pz Qz

P Q

x x x

xo

cos

cos( ) ( ) ( )

.

cos . .

1

1

100 600 300 300 600 600

678 2 900

0 8355 33 3

Solution Using Cosine Law

3 Using triangle ABC and cosine rule:

C

678.2 900

A B

500

5002 = 678.22 + 9002 - 2 x 678.2 x 900 cos

250000 = 1269955.2 - 1220760 cos

cos = (12.7 - 2.5)/ 12.207 = 0.8356 = 33.3o.

Moment of a Force About A Specified Axis

First, compute moment of force, F about any point O that lies

on the aa’ axis.

Moment of F about O, Mo = rA/O x F .................... (1)

Where rA/O = OA. Mo acts along the moment axis bb’ which is

perpendicular to the plane containing rA/O and F.

The component or projection of Mo onto the

aa’ axis is then represented by Ma.

The magnitude of Ma is given by the dot product:

Ma = Mo cos = Mo . a ....... (2)

Where a defines the direction of the aa’ axis.

Combining the two steps ie. Equations (1) and (2):

Ma = (rA/O x F) . a

Moment of a Force About A Specified Axis Concluded

Since dot product is commutative, Ma = a .( rA/O x F)

In vector algebra, this combination of dot and cross products

yielding the scalar, Ma is called the triple scalar product.

The triple scalar product can be written as:

Ma = (ax i + ay j + az k) . i j k

rA/Ox rA/Oy rA/Oz

Fx Fy Fz

or Ma = a .( rA/O x F) = . ax ay j az

rA/Ox rA/Oy rA/Oz

Fx Fy Fz

Solution Concluded

BG = - 400 mm i + 740 mm j - 320 mm k , BG = 900 mm

TBG = TBG. BG/BG = 1125 N x 1/900mm x BG

= - 500 N i + 925 N j - 400 N k

rB/A = AB = 400 mm i

Unit vector, AD, along AD = AD/AD = (800 mm i - 600 mm k)/1000 mm = 0.8 i - 0.6 k MAD = AD . ( rB/A x TBG) = 0.8 0 -0.6

400 mm 0 0

-500 N 925 N - 400

= - 0.6 ( 925 N x 400 mm) = - 222000 Nmm = - 222 Nm

3.9 MOMENT OF A COUPLE3.3.4 Introduction: A couple is defined as two parallel forces

which have the same magnitude, opposite directions and are

separated by a perpendicular distance, d

-F

F

Since the resultant force of the two forces of the couple is zero,

the only effect of a couple is to produce a rotation or tendency of

a rotation in a specified direction.

-F B r A F

rB rA

O

To derive the moment of the couple, consider two vectors , rA and rB

from O to points A and B lying on the line of action of F and - F.

The moment of the couple about O is:

M = rA x (F) + rB x (-F) = (rA - rB) x F

By triangle law of vector addition, rB + r = rA or r = rA - rB

So: M = r x F

This result shows that a couple moment is a free vector i.e

can act at any point, since M depends only on the position

vector directed between the forces and not position vectors

rA and rB from O to the forces.

3.9.2. Scalar Formulation

Moment of a couple, M has magnitude:

M = F x d = r F sin F is the magnitude of one of the

forces; d is the perpendicular distance or

moment arm between the forces.

Scalar Formulation Concluded

The direction and sense of the couple moment are determined by the right hand rule, where the thumb indicates the direction when the fingers are curled with the sense of rotation caused by the two forces.

The moment acts perpendicular to the plane containing the two forces.

3.9.3 Vector Formulation

The moment of a couple can be expressed using the cross product.

M = r x F

 

3.9.4 Equivalent Couples

Two couples are equivalent if they produce the same moment.

Since the moment produced by a couple is always perpendicular to the plane containing the couple forces, it is necessary that the forces of equal couples lie either in the same plane or in planes that are parallel to one another.

Resolution of a Force into a Force at B and a couple

A Force, F acting at point A on a rigid body can be resolved to the same force acting on another point B and in the same direction as the original force plus a couple M equal to r x F i.e. moment of F about B

i.e Force in (a) equal to that in (b) equal to that in C

A

F = 5 kN 5 kN 5 kN

5 kNB

5 m

5 kNM = 25 kN m

(a)(b) (c)

B

Example

A force and couple act as shown on a square plate of side a = 625 mm. Knowing that P = 300 N, Q = 200 N and = 50o, replace the given force and couple by a single force applied at a point located (a) on line AB (b) on line AC. In each case, determine the distance for A to the point of application of the force.

k (143.63 – 192.8 x’) = 125 N m k , x’ = 0.0966 m = 96.6 mm = AE