CHAPTER THREE
STATICS OF RIGID
BODIES
3.1 INTRODUCTION
A
rigid body is one which does not suffer deformation.
It can be continuous
connected members.
F2
F3
F1
P2
P1
F3
Continuous Member
F1
Connected Members
F2
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.
F
A
F
B
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 = PxQ
V
PQ sin  = V
Q


P
Q
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
k = (i x j)
x
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 xi =0
j xk=i
j x i = -k
j xj =0
jj
kk
ii
-ve
kxi =j
k x j =-i
kx 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.
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
Px
Qx
j
Py
Qy
k
Pz
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 an
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
Moment, M0 = + F d (Counter-clockwise , +ve)
F
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.
Dot product is used to determine the angle formed by two vectors or inters
lines. The angle between two vectors P and Q is required.
P = Px i + Py j + Pz k
Q = Qx i + Qy j + Qz k
Recall that: P. Q = P Q cos  ............. (1)
P. Q = Px Qx + Py Qy + Pz Qz ...... (2)
Equating (1) and (2): PQ cos  = Px Qx + Py Qy + Pz Qz
Px Qx  Py Qy  Pz Qz
i. e. cos 
PQ
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
Px Qx  Py Qy  Pz Qz
P Q cos 
PQ
( 100 x  600)  (300 x 300)  ( 600 x  600)
  cos
678.2 x 900
1
 cos1 0.8355  33.3o
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) .
or Ma =  a .( rA/O x F) = .
i
j
k
rA/Ox
rA/Oy
rA/Oz
Fx
Fy
Fz
 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 COUPLE
3.3.4 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
F = 5 kN
5 kN
A
5 kN
M = 25 kN m 5 kN
B
B
5 kN
5m
(a)
(b)
(c)
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