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ENT151/4
STATICS
ENT151
STATICS
ZOL BAHRI BIN RAZALI
Lecturer
School of Mechatronic Engineering
Northen Malaysia University College of Engineering (KUKUM)
(2004)
Reference and textbook : STATICS,
By Hibbeler, R.C. (2001) Prentice Hall, Inc.
Copyright © 2004 by Zol Bahri Razali. All rights reserved.
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Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
CHAPTER 1
GENERAL PRINCIPLES
1.1 Mechanics
Can be defined as that the branch of the physical science concerned with the state of rust
or motion of the bodies that are subjected to the action of forces. In general, this subject
is subdivided into three branches rigid body mechanics and fluid mechanics.
Rigid Body Mechanics
Divided into two areas: statics and dynamics.
Statics
Deals with the equilibrium of bodies that are those that are either at rest or more with the
constant velocity.
Dynamics
Is concerned with the accelerated motion of the bodies
1.2 Fundamental Concepts
a) 4 basics quantities
Length
To locate the position of a point in space and there by describe the size of a physical
system.
Time
Is conceived as a succession of events
Mass
Property manifests it self as a gravitational attraction between two bodies and provided a
quantitative measure of the resistance of matter to change in velocity.
Force
Is considered on a push or poll exerted by one body on another.
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ENT151/4
STATICS
b) Model or idealizations
Particle
Has a mass but a size that can be neglected.
Rigid body
Considered on a combination of a large number of particles in which all the particles
remain at fixed distance from one another both forever and after applying a load.
Concentrated force
Represents the effect of a loading which is assumed to act point on the body.
c) Newton's Law
First Law
A particle originally at rest a moving in a straight like with contact velocity, will remain
in this state provided the particle is not subjected to an unbalance force
F1
F2
●
(equilibrium)
F3
Second Law
A particle acted upon by an unbalance force F experience an acceleration that has same
direction on the force and magnitude that is directly proportional to the force. If F
expressed mathematically.
F = m.a
F
●
a
Third Law
The mutual force of action and reaction between to particles are equal, opposite and
collinear.
F
●●
Force of A on B
F
Force of B on A
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(action – reaction)
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School of Mechatronics, KUKUM
ENT151/4
STATICS
d) Newton's Law of Gravitational Attraction
F = G (M1 . M2 / r²)
Where : F
G
M1 ,
r
- Force of gravitation between two particles
- Universal constant of gravitation [ G = 66.73 (10‾ ¹²) m³/ (kgs²) ]
– mass of each of the two particles
– Distance between the two particles
Weight, will be the only gravitational force considered in our study of mechanics.
W = m.g
where, g - accaelaration due to gravity (9.81 m/s2) at sea
level.
1.3 SI Units
Length (m), time (s), mass (kg), force (N)
F = m.a
Thus 1 Newton is equal to a force required to give 1 kilogram of mass an acceleration of
1 m/s².
N = kg.m/s²
1kg
9.81 N
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ENT151/4
STATICS
1.4 Dimensional Homogeneity
The term of any equation want to describe a physical process must be dimensionally
homogeneous, that is, each term must be expressed in the same units.
e.g.:
e.g.:
s = vt + ½ at²
m = (m/s) (s) + ½ (m/s) (s)
m=m
where s - (m), v - (m/s), t - (s), a - (m/s)
same units – dimensionally homogenous
a = 2s/t² - 2v/t
m/s² = 2 m/s² - 2 (m/s)
s
= 2 m/s² - 2m/s²
m/s² = m/s² - dimensionally homogenous
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School of Mechatronics, KUKUM
ENT151/4
STATICS
CHAPTER 2
FORCE VECTORS
2.1 Scalars & Vectors
Scalar - a quality characterized by positive or negative members (mass, volume, length)
Vector – a quality that has both a magnitude and a direction. Vector represented by Ā
Magnitude epresented by |A|, positive quantity.
- Magnitude, length of arrow (A)
- Direction, angle between axis and the arrow’s line.
2.2
Vectors Operations
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ENT151/4
STATICS
Multiplication & division :
Addition of a vector:
parallelogram law
Substraction of a vector :
R=A−B
= A + (− B)
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ENT151/4
STATICS
Resolution of vector :
Adition of force :
FR
=
(F1 + F2) + F3
EXAMPLES
Example 2-1 (pp 22)
Use cosine law to find R, sine law to find θ.
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ENT151/4
STATICS
Note
EXAMPLES
Example 2-2 (pp 23)
Example 2-3 (pp 24)
Example 2-4 (pp 25)
2.3
Additional of Coplaner Forces
F = Fx + Fy
F’ = Fx’ + Fy’
F’ = Fx’ i + Fy’ j
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ENT151/4
STATICS
Coplanar force resultant :
Resultant :
FR =
=
=
=
where :
F1 + F2 + F3
F1x i + F1y j - F2x i + F2y j + F3x i - F3y j
( F1x - F2x + F3x ) i + ( F1y + F2y - F3y ) j
( FRx ) i + ( FRy ) j
FRx = ∑ Fx ; FRy = ∑ Fy
Resultant force, FR = √ [(FRx)2 + (FRy) 2 ]
Angle
θ
= tan -1 ( FRy / FRx )
EXAMPLES
Example 2-5 (pp 35)
Example 2-6 (pp 36)
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ENT151/4
STATICS
2.4
Position Vectors
x , y , z coordinates ;
A (x , y , z)
Position vector, is defined as a fixed vector which locates a point in space relative to
another point.
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ENT151/4
STATICS
=
=
=
=
rA + rB
rB - rA
( xB i + yB j + zB k ) - ( xA i + yA j + zA k )
(xB - xA) i + (yB - yA) j + (zB - zA) k
EXAMPLE
Example 2- 12 (pp 57)
2.5
Force Vector Directed Along a Line
F – force
r - position vector
u - unit rector = r / r
r - radius from origin
The force, F directed along the cord AB
The position vector, r directed from point A to point B on the cord.
This common direction is specified by the unit vector, u = r / r.
F = Fu = F ( r / r)
where : F is unit of force
r is unit of length
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ENT151/4
STATICS
EXAMPLES
Example 2-13 (pp 59)
2.6
Example 2-14 (pp 60)
Dot Product
The dot product of vectors A and B, written A . B is defined as the product of the
magnitudes of A and B and the cosine of the angle between their tails.
A . B = A B Cos θ
0° ≤ θ ≤ 180°
Law of operation
1. A . B = B . A
2. a (A . B ) = ( aA ) . B = A . ( aB ) = (A . B )a
3. A . ( B + D ) = ( A . B ) + ( A . D )
Cartesian Vector Formulation
i . i = (1) (1) cos 0° = 1
i . j = (1) (1) cos 90° = 0
Therefore :
i . i = 1
i . j = 0
j . j = 1
i . k = 0
k . k = 1
k . j = 0
A . B = ( Ax i + Ay j + Az k ) . ( Bx i + By j + Bz k )
= Ax Bx ( i . i ) + Ax By ( i . j ) + Ax Bz ( i . k ) +
A y Bx ( j . i ) + A y By ( j . j ) + A y Bz ( j . k ) +
Az Bx ( k . i ) + Az By ( k . j ) + Az Bz ( k . k )
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ENT151/4
STATICS
By dot operation;
A . B = Ax Bx + Ay By + Az Bz
Thus to determine the dot product of two Cartesian Vectors, multiply their corresponding
x , y , z components and sum their products algebraically. The result is a scalar.
Application Dot Product
1.
Angle formed between two vectors
A . B = A B Cos θ
If
A.B = 0;
Hence
2.
θ = cos-1 ( A . B / A B )
θ = cos-1 0
= 90°
A┴ B
The components of a vector parallel and perpendicular to a line.
A ╖ = A Cos θ
= A.u
where u is a unit vector
A┴ = A-A╖
since A = A ┴ + A ╖
To determine θ ,
θ = cos-1 ( A . u / A )
Then
A ┴ = A Sin θ
or A ┴ = √ ( A2 - A╖2 )
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ENT151/4
STATICS
EXAMPLES
Example 2-16 (pp 71)
Example 2-17 (pp 72)
EXERCISES
Exercise 2 - 6
Exercise 2 - 14
Exercise 2 - 21
Exercise 2 - 26
Exercise 2 - 31
Exercise 2 - 55
Exercise 2 - 63
Exercise 2 - 87
Exercise 2 - 88
Exercise 2 - 114
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School of Mechatronics, KUKUM
ENT151/4
STATICS
CHAPTER 3
EQUILIBRIUM OF A PARTICLE
3.1 Condition of the equilibrium
The resultant force acting on a particle to be equal to zero.
∑F = 0 ;
3.2
Newton’s Law ∑ F = m.a
where a = 0
Free Body Diagram
Account for all known and unknown forces, ∑ F which act on the particles, by draw the
particles free body diagram.
Spring :
Linear elastic spring, F = k.s
where k – stiffness (spring constant )
s - distance (from unload position)
Cable and Pulleys
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School of Mechatronics, KUKUM
ENT151/4
STATICS
EXAMPLES
Example 3-1 (pp 84)
3.3 Coplaner Force System
For equilibrium ∑ F = 0
∑ Fx i + ∑ Fy j = 0
∑ Fx = 0 ;
∑ Fy = 0
EXAMPLES
Example 3-2 (pp 87)
3.4
Example 3-3 (pp 88)
Three Dimensional Force System
∑F = 0
∑ Fx = 0 ;
∑ Fy = 0 ;
∑ Fz = 0
» ∑ Fx i + ∑ Fy j + ∑ Fz = 0
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ENT151/4
STATICS
EXAMPLES
Example 3 -5 (pp 100)
Example 3 - 7 (pp 102)
Example 3 - 6 (pp 101)
Example 3 - 8 (pp 103)
EXERCISES
Exercise 3 - 7
Exercise 3 - 11
Exercise 3 - 17
Exercise 3 - 19
Exercise 3 - 38
Exercise 3 - 41
Exercise 3 - 45
Exercise 3 - 57
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ENT151/4
STATICS
CHAPTER 4
FORCE SYSTEM RESULTANTS
4.1 Moment of a force – Scalar formulation
Moment - provides a measure of the tendency of the force to cause a body to rotate
about the point of axis.
This force tends to cause the pipe to turn about the z-axis.
Mo = Fx x d
Mo - moment of a force / torque
d - moment arm
Direction of moment - use right hand rule.
Resultant moment of a system of coplanar forces
MR = ∑F.d
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School of Mechatronics, KUKUM
ENT151/4
STATICS
The moment of any force will be +ve along the +ve z axis and –ve moment along the –ve
z axis.
EXAMPLES
Example 4-1 (pp 116)
Example 4-2 (pp 117)
Example 4-3 (pp 118)
4.2 Cross product
C = AxB;
C = AB Sin θ
C is specified by right-hand rule.
defined the direction of C.
(0o ≤ θ ≤ 180o )
AB Sin θ defined the magnitude C and unit vector uc
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ENT151/4
STATICS
Law of operation
1.
AxB ≠ BxA ;
AxB = -BxA
2.
a ( A x B ) = ( aA ) x B = A x ( aB )
3.
Ax(B+D) = (AxD)+(AxD)
=
(AxB)a
Cartesian Vector Formulation
ixj = k
ji x i = - k
jxk = i
kxj = -i
kxi = j
i x ki = - j
A x B = ( Ay Bz – Az By ) i - ( Ax Bz – Az Bx ) j + ( Ax By – Ay Bx ) k
=
│
│
│
i
j
k│
Ax Ay Az │
Bx By Bz │
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School of Mechatronics, KUKUM
ENT151/4
STATICS
4.3 Moment of a Force – vector formulation
Mo =
=
=
=
r x F
r . F Sin θ
F ( r . Sin θ )
F.d
where : r - position vector
Cartesian vector :
Mo = ( ry Fz – rz Fy ) i - ( rx Fz – rz Fx ) j + ( rx Fy – ry Fx ) k
Mo = r x F
=
│
│
│
i
rx
Fx
j
ry
Fy
k
rz
Fz
│
│
│
Resultant moment :
MRo = ∑( r x F )
EXAMPLES
Example 4-4 (pp 124)
Example 4-5 (pp 125)
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ENT151/4
STATICS
4.4
Principle of Moments
The moments of a force about a point is equal to the sum of the moments of the force’s
components about the point.
Mo = r x F1 + r x F2
= r x ( F1 + F2 )
= r x F
EXAMPLES
Example 4-6 (pp 127)
Example 4-7 (pp 128)
4.5 Moment about specified axis
To find the components of the moment along a specified axis, Ma = F . da
If the line of action of a force F is perpendicular to any specify axis, the magnitud of the
moment of F about the axis can be determine by : Ma = F . da
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School of Mechatronics, KUKUM
ENT151/4
STATICS
Where : da – the perpendicular / shortest distance between the force line to the axis.
By vector analysis : Ma = ua ( r x F )
EXAMPLES
Example 4-8 (pp 142)
4.6
Example 4-9 (pp 143)
Moment of a couple
A couple is defined as two parallel forces that have the same magnitude, have opposite
directions and are separated by a perpendicular distance, d.
Moment couple = sum of the moments of both couple forces about any arbitrary point.
It is simpler to take moments about a point lying on the line of action of one of the forces.
e.g
At point A:
the moment of – F about A is zero, M = r x f
M depends only upon the position vector, r directed between the forces.
M = F.d
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ENT151/4
STATICS
MR = M1 + M2
= ∑( r x F )
EXAMPLES
Example 4-10 (pp 150)
Example 4-11 (pp 151)
Example 4-12 (pp 152)
4.7 Equivalent system
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ENT151/4
STATICS
4.8 Resultant of a Force and couple system
=
M1 = r1 x F1 ;
=
M2 = r2 x F2
;
FR = F1 + F2
MRo = Mc + M1 + M2
= ∑ M c + ∑ Mo
FRx = ∑ Fx
;
FRy = ∑ Fy
EXAMPLES
Example 4-14 (pp 164)
4.9
Example 4-15 (pp 165)
Reduction of a Force and Couple System
If the force and couple moment system is reduce to a resultant system at point P, only the
force resultant will have to be applied to the body.
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ENT151/4
STATICS
EXAMPLES
Example 4-16 (pp 170)
Example 4-17 (pp 171)
Example 4-18 (pp 172)
Example 4-19 (pp 173)
4.10 Reduction of a simple Distributed Loading
FR = ∑ F
Magnitude :
FR = ∫L W (x) dx
= ∫L dA
= A
Magnitude of the resultant force is equal to the total area A under the loading diagram
W = W(x)
Location :
MRo = ∑ Mo ;
x
= ∫L x W(x) dx
∫L W(x) dx
x
= ∫L x dA
∫L dA
x . FR = ∫L x . W(x) dx
The resultant force has a line of action which passed through the centroid C (geometric
center) of the area defined by the distributed loading diagram w(x).
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ENT151/4
STATICS
EXAMPLES
Example 4-20 (pp 182)
Example 4-21 (pp 183)
Example 44-22 (pp 184)
EXERCISES
Exercise 4 - 9
Exercise 4 - 14
Exercise 4 - 22
Exercise 4 - 36
Exercise 4 - 53
Exercise 4 - 63
Exercise 4 - 70
Exercise 4 - 82
Exercise 4 - 105
Exercise 4 - 121
Exercise 4 - 145
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School of Mechatronics, KUKUM
ENT151/4
STATICS
CHAPTER 5
EQUILIBRIUM OF A RIGID BODY
5.1 Conditions
Forces :
Fi + fi = 0 ;
Moments :
ri x (Fi + fi ) = ri x Fi + ri x fi
= 0
∑Mo = ∑ ri x Fi ;
∑Fi + ∑fi = 0 ;
∑Fi = ∑F ;
∑F = 0
∑Mo = 0
The two equations of equilibrium for a rigid body :
∑F = 0
∑Mo = 0
5.1 Free Body Diagram (FBD)
To draw free body diagram :
Refer Table 5.1 below – Supports for rigid body.
5.2 Support Reactions
If a support prevents the translation of a body in a given direction, then the force is
developed on the body in that direction. If rotation is prevent, a couple moments is exert
on the body.
Types of support reactions, and forces and couple moments involved :
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ENT151/4
STATICS
EXAMPLES
Example 5-1 (pp 202)
Example 5-2 (pp 203)
Example 5-3 (pp 204)
Example 5-4 (pp 205)
Example 5-5 (pp 206)
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ENT151/4
STATICS
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STATICS
5.3 Equations of Equilibrium
∑Fx = 0
∑Fy = 0
∑Mo = 0
EXAMPLES
Example 5-6 (pp 211)
Example 5-7 (pp 212)
Example 5-8 (pp 213)
Example 5-9 (pp 214)
Example 5-10 (pp 215)
Example 5-11 (pp 216)
Example 5-12 (pp 217)
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STATICS
5.4 Two and three-force members
Two force members :
Three-force members
EXAMPLE :
Example 5-13 (pp 220)
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STATICS
5.5 Equilibrium in Three Dimensions
Free body diagram – Support reactions
A force is developed by a support that restrict the translation of the attached members
whereas a couple moment is develop when rotation of the attached member is prevented.
Magnitude of force
:
F = √ (Fx2 + Fy2 + Fz2)
The third direction angle :
Cos2 α + Cos2 β + Cos2 γ = 1
[Table 5-2 – Support for Rigid Bodies Subjected to Three Dimensional Force System]
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ENT151/4
STATICS
EXAMPLES
Example 5-14 (pp 211)
5.6
Example 5-15 (pp 212)
Equation of Equilibrium
Scalar Equations :
∑F = 0
∑M = 0
Vector Equations :
∑F = ∑Fx i + ∑Fy j + ∑Fz k
=0
∑M0 = ∑Mx i + ∑My j + ∑Mz k
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School of Mechatronics, KUKUM
ENT151/4
STATICS
Where ;
∑Fx = 0 ; ∑Fy = 0 ; ∑Fz = 0
∑Mx = 0 ; ∑My = 0 ; ∑Mz = 0
5.7
Constraints for a Rigid Body
Redundant constraints
When the body has a redundant support more supports then are necessary to hold it in
equilibrium, it becomes statically indeterminate. Statically indeterminate – there will be
more unknown loadings on the body then equations of equilibrium available for their
solution.
Eg :
Two dimensional case – five unknown but only three equation.
Three dimensional case – eight unknown but only five equation
Improper Constraints
Instability of the body because of improper constraining by the supports.
3-Dimensional – if the support reaction all intersec a common axis
2-Dimensional – the axis is perpendicular to the plane of the forces
When all the reactive forces are concurrent at the point, the body is improperly constraint.
Improper constraining leads to instability occurs when the reactive forces are all parallel.
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STATICS
Proper constraining requires
1. the line of action of the reactive forces does not intersect point on a common axis
2. the reactive force must not all be parallel on one another
EXAMPLES
Example 5-15 (pp 242)
Example 5-16 (pp 243)
Example 5-17 (pp 244)
Example 5-18 (pp 245)
Example 5-19 (pp 246)
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ENT151/4
STATICS
EXERCISES
Exercise 5 - 5
Exercise 5 - 19
Exercise 5 - 22
Exercise 5 - 24
Exercise 5 - 89
Exercise 5 - 95
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School of Mechatronics, KUKUM
ENT151/4
STATICS
CHAPTER 6
STRUCTURAL ANALYSIS
6.1
Trusses
A structure composed of slender members joined together at their end points.
Planar Truss
Lie in a single plane and are often used to support roofs and bridges.
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STATICS
Assumption for Design


All loading are applied at the joints.
The members are joined together by smooth pins. Each truss member acts as a
two force member
Tension - tend to elongate the member (tensile force)
Compression - tend to shorten the member (compressive force)
Simple truss
To prevent collapse, the form of a truss must be rigid. A simple truss is constructed by
starting with a basic triangular element.
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ENT151/4
STATICS
6.2
The method of joints
Truss members are all straight two-forces members lying in the same plane, the forces
system acting at each joint is coplanar and concurrent. Moment equilibrium is
automatically satisfied at the joint or pin.
∑Fx = 0 ;
∑Fy = 0
Analysis should start at a joint at least one known force and at most two unknown.
∑Fx = 0 ;
∑Fy = 0
The member in compression “pushes” on the joint and a member in tension “pulls” on the
joint.
EXAMPLES
Example 6-1 (pp 262)
Example 6-2 (pp 263)
Example 6-3 (pp 264)
Example 6-3 (pp 265)
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STATICS
* Note : First – get all the equilibrium forces (external) through FBD.
Second – do the analysis
6.3
Zero- Force Members
Zero force members are used to increase the stability of the truss during construction and
provide support of the applied loading is changed.

If only two members from a truss joints and no external load or support reaction is
applied to the joint, the members must be zero-force member.
zero-force member
∑Fx = 0 ;

FAB = 0
∑Fy = 0 ;
FAF = 0
If three members from a truss joint for which two of the members are collinear,
the third member is a zero-force member provide no–external force or support
reaction is applied to the joints.
EXAMPLES
Example 6-4 (pp 268)
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STATICS
6.4
Method of Sections
To determined the loadings acting within a body.
 if a body is in equilibrium, then any part of the body is also in equilibrium.
Each internal force as external force to the Free Body diagram (FBD)
∑Fx = 0 ;
∑Fy = 0 ;
∑Mo = 0 ;
Analysis – select the section passed through not more than three members in which the
forces are unknown.



First, find the total external forces through FBD.
Analysis for the end part to final FBC, FGC, FGF
Start with ∑MG = 0 to find FBC
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ENT151/4
STATICS
EXAMPLES
Example 6-5 (pp 276)
Example 6-6 (pp 277)
Example 6-7 (pp 278)
6.5
Space Trusses
Space truss consists of members jointed together at their ends so form a stable threedimensional structure – e.g tetrahedron (ball and socket joint).
Additional member would be redundant. Procedure of analysis : Method of joint /
method of section.
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STATICS
Method of joint – if the forces in all the members of the truss must be determined :
∑Fx = 0 ;
∑Fy = 0 ;
∑Fz = 0 ;
The analysis start at the joint at least one known and three unknown forces. Use the
Cartesian Vector analysis.
Method of section – if only a few member forces are to be determined :
When an imaginary section is passed through a truss and the truss saparated into two
parts, the force system acting on one of the parts must satisfy.
∑Fx = 0 ;
∑Mx = 0 ;
∑Fy = 0 ;
∑My = 0 ;
∑Fz = 0 ;
∑Mz = 0 ;
EXAMPLES
Example 6-8 (pp 284)
6.6
Frame and Machine
Compose of pin – connected multiforces members.
 Frame are generally stationary and are use to support loads.
 Machines contain moving parts and are designed to transmit and alter the effect of
forces.
To analysis 



Draw free body diagram
Show all the forces and couple of moments / known-unknown forces
Identify all the two-forces-member
Internal force not shown / external forces
EXAMPLES
Example 6-9 (pp 288)
Example 6-10 (pp 289)
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STATICS
Example 6-11 (pp 290)
Example 6-12 (pp 291)
Example 6-13 (pp 290)
Equations of Equilibrium
In the x – y plane :
∑Fx = 0 ;
∑Fy = 0 ;
∑Mo = 0 ;
Note : 



Draw FBD of the entire structure
The force at the connected parts of the group are internal force
Force common to two members with equal magnitude but opposite sense
Two-force mebers, equal but opposite collinear forces
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STATICS


The proper sense of the unknown forces acting
A couple moment is a free vector and can act at any point.
EXAMPLES
Example 6-14 (pp 295)
Note :
Always identify the two-force members before starting the analysis.
EXAMPLES
Example 6-15 (pp 296)
Example 6-16 (pp 297)
Example 6-17 (pp 298)
Example 6-18 (pp 299)
Example 6-19 (pp 300)
Example 6-20 (pp 301)
Example 6-21 (pp 302)
EXERCISES
Exercise 5 - 5
Exercise 5 - 19
Exercise 5 - 22
Exercise 5 - 24
Exercise 5 - 89
Exercise 5 - 95
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STATICS
CHAPTER 7
INTERNAL FORCES
7.1 Internal Forces Developed in Structure Members
The internal loadings can be determined by using the method of sections. If the internal
loadings acting on the cross section at C are to be determined, then an imaginary sections
is paased through the beam, cutting it into two segments. By doing this, the internal
loadings at the section become external on the free body diagram.
In equilibrium: – each segment is maintained provided rectangular force components Nc,
Vc, and Mc.
- equal in magnitud but opposite in directions
Direct solution – Nc : ∑Fx = 0 ;
Vc : ∑Fy = 0 ;
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Mc : ∑Mo = 0
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- where
V = shear force
N = Normal force
M = Bending Moment
My = torsional / twisting moment
All resultant force /moment acts on the center of centroid (c) of the section cross area.
Free Body Diagram
Frame and machine
- multi force members
-subjected to internal normal, shear and bending loadings.
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STATICS
Instead, we must dismember the frame and determine the reaction.
Note : We cannot apply the three equation of equilibrium to obtain these nine unknown.
Analysis





Before the member is ‘cut’, determined the all members support reaction.
Keep all the forces, moments at exact locations.
Draw FBD of the segment that has the least number of loads on it (x,y,z components)
Coplaner system only N, V & M
Moments should summed at section passing through centre
EXAMPLES
Example 7-1 (pp 327)
Example 7-2 (pp 328)
Example 7-3 (pp 329)
Example 7-4 (pp 330)
Example 7-5 (pp 331)
Example 7-6 (pp 332)
7.2
Shear and moment equations and diagram
Beams – to support loadings applied perpendicular to their axis. In general, beams are
long, straight bars, constant cross-sectional areas.
Simply supported beam – is pinned at one end and roller support at the other.
Centilever beam – is fixed at one end and free at the other.
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V
- internal shear force / shear diagram
M
- bending moment / bending moment diagram.
The variation of V and M as functions of the position x along the beam’s axis can be
obtain by using the method of sections. It is necessary to section the beam at an arbitrary
distance x from one end rather then at a specified point.
If the result are plotted, the graphical variation of V and M as functions of x are termed
the shear diagram and bending moment diagram. Because of the internal shear and
bending moment functions will be discontinous, the functions must be determined for
each segment of the beam located between any two discontinuities of loading.
Procedure for analysis






Determined all the reactive forces and couple moment acting on the beam, and
resolve all the forces into components acting perpendicular and parallel to the
beam axis.
Specify separate coordinates and having an origin at the beam’s left-end
Section the beam perpendicular to the axis.
Shear V-summing forces perpendicular to the axis
Moment M-slimming moments about section end.
Plot the shear diagram (V vs x) and moment diagram (M vs x)
EXAMPLES
Example 7-7 (pp 342)
Example 7-8 (pp 343)
7.3 Relations Betwen Distributed Load, Shear and Moment
--- TO BE CONTINUE --51
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STATICS
EXERCISES
Exercise 7 - 2
Exercise 7 - 4
Exercise 7 - 10
Exercise 7 - 33
Exercise 7 - 49
Exercise 7 - 59
Exercise 7 - 86
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STATICS
CHAPTER 8
FRICTION
8.1
Characteristic of dry friction
Friction
A force of resistance acting on a body which prevents slipping of the body relative to a
second body or surface with which it is in contact.
∆ F - frictional force
∆ N - normal force
For equilibrium;
and
∆N
W
P
∆F
Free body diagram ;
F - acts tangent to the contacting surface
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STATICS
- opposite direction of P
N - directed upward
- balance the block weight (W)
Impending motion
When P slowly increase, F correspondingly increase until maximum value, Fs - limiting
static frictional force.
When the value reached, the block unstable equilibrium. Increase P, the block begin to
move. Fs directly propotional normal force, Ns.
Fs = μs N ;
where μs - coefficient of static friction.
When the block verge of sliding, N & F create Rs and the angle ø called angle of static
friction.
Øs = tan -1 (Fs / N)
= tan -1 (μs N / N)
= tan -1 μs
Tabular Value Ms
μs
- coefficient of static friction – dimensionless
- depends on the characteristic of two surfaces.
Motion
When P > Fs, frictional force drops slingtly to smaller value Fk - kinetic frictional force.
P > Fk – block not in equilibrium – begin to slide with increasing speed.
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Fk directly propotional to the magnitude of the N.
Fk = μk N
where μk – coefficient of kinetic friction.
= 25 % smaller than μs
Øk = tan -1 (Fk / N)
= tan -1 (μk N / N)
= tan -1 μk
Øs ≥ Øk
Note : slipping about to occur,
8.2
Fs = μs N ; slipping is occuring,
Fk = μk N
Problems involving dry friction
Equilibrium: The total number of unknowns to be equal to the total number of available
equilibrium equations.
F ≤ μs N
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STATICS
From the equations of equilibrium, FA, NA, Fc and Nc are determined. The bar remain in
equilibrium provide
FA ≤ 0.3 NA , Fc ≤ 0.5 Nc
Impending motion at all points :
Total number of unknowns
= total number of available equilibrium equations
+ total number of available frictional.
If motion is impending, Fs ≤ μs N
If the body is shipping, Fk ≤ μk N
Impending motion at some point
Total number of unknowns
< number of available equilibrium equation
+ the total number of frictional equation.
Unknowns : NA, FA, Nc, Fc, Bx, By, P
Equations : six equilibrium equation (three of each member) one static frictional
equation
Conditions :
1.
Slipping at A, no slipping at C
FA = 0.3 NA ,
Fc ≤ 0.5 Nc
2.
No slipping at A, slipping at C
Fc = 0.5 Nc ,
FA ≤ 0.3 NA ; depend on which P is smaller.
Equilibrium Versus Frictional Equestions
F (equilibrium force) < μs N
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Procedure for analysis

Draw free-body diagrams – always show frictional force as unknowns.

Determine the number of unknowns.

Apply the frictional equations
EXAMPLES
Example 8-1 (pp 385)
Example 8-2 (pp 386)
Example 8-3 (pp 387)
Example 8-4 (pp 388)
Example 8-5 (pp 389)
Example 8-5 (pp 390)
EXERCISES
Exercise 8 - 5
Exercise 8 - 11
Exercise 8 - 39
Exercise 8 - 43
Exercise 8 - 58
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STATICS
CHAPTER 9
CENTER OF GRAVITY AND CENTROID
9.1
Center of gravity / mass of the particle
Center of gravity – G - locates the resultant weight of a system of particles.
WR = ∑ W ;
Sum of moments of all weights = moments of the resultant weight.
∑ Mall = MR
WR = x1 W1 + x2 W2 + x3 W3
=
cordinate o gravity (c.g)
cordinate =>
Hence,
= ∑xW
∑ W
WR = y1 W1 + y2 W2 + y3 W3
=
∑yW
∑ W
= ∑zW
∑ W
- cordinate of the center of gravity
x , y , z - cordinate of the each particles.
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Center of mass ; W = m.g
= ∑xW
∑ W
Hence,
=
= ∑xm.g
∑ m.g
= ∑xm
∑ m
∑ym
∑ m
= ∑zm
∑ m
9.2
Center of Gravity / Mass and Centroid of Body
Center of gravity (c.g) :
Arbitrary particle (x
x , y , z ) , weight dW
= ∫ x dW ;
∫ dW
=
If γ - specific weight, dW = γ dv ;
∫ y dW
∫ dW
= ∫v x γ dv
∫v γ dv
=
= ∫ z dW
∫ dW
∫v y γ dv
∫v γ dv
= ∫v z γ dv
∫v γ dv
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STATICS
Center of mass :
Density , ρ
= mass per unit volume ;
γ = ρ.g
= ∫v x ρg dv
∫v ρg dv
=
∫v y ρg dv
∫v ρg dv
= ∫v x ρ dv
∫v ρ dv
= ∫v y ρ dv
∫v ρ dv
= ∫v z ρg dv
∫v ρg dv
= ∫v z ρ dv
∫v ρ dv
Centroid :
Centroid, C -- the geometric center of an object and depends only the body’s geometric.
Volume
= ∫v x dv
∫v dv
=
∫v y dv
∫v dv
= ∫v z dv
∫v dv
=
∫A y dA
∫A dA
= ∫A z dA
∫A dA
=
∫L y d L
∫L d L
= ∫L z d L
∫L d L
Area
= ∫A x dA
∫A dA
Line
= ∫L x dL
∫L d L
Symetry
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EXAMPLES
Example 9-1 (pp 441)
Example 9- 2 (pp 442)
Example 9-3 (pp 443)
Example 9- 4 (pp 444)
Example 9-5 (pp 445)
Example 9- 6 (pp 446)
Example 9-7 (pp 447)
Example 9- 8 (pp 448)
9.3 Composite Bodies
Composite body - consists of a series of connected simpler shaped bodies.
 Devide into its composite parts
 Provide weight and location of c.g of each parts
 Determine the c.g for entire body
Hence,
= ∑xW
∑ W
=
where
- c.g composite body
x - c.g each composite part
∑W - total weight of the body
∑yW
∑ W
= ∑zW
∑ W
When the body has a constant density or specific weight, center of gravity coincide with
the centroid of body.
EXAMPLES
Example 9-9 (pp 461)
Example 9- 10 (pp 462)
Example 9-11 (pp 463)
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EXERCISES
Exercise 9 - 46
Exercise 9 - 61
Exercise 9 - 70
Exercise 9 - 75
TO BE CONTINUE
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CHAPTER 10
MOMENT OF INERTIA
10.1 Moment of Inertia for an Areas
∫ x 2 dA ; Moment of inertia for areas
The moment of inertia of the different planar area dA about x and y axes are :
dIx = y2 dA
and
dIy = x2 dA
Hence,
Ix = ∫A y2 dA ;
Iy = ∫A x 2 dA
Second moment of dA about the pole O or z axis, refered to as the polar moment of
inertia,
dJo = r2 dA ; where r is the perpendicular distance from the pole
(z axis)
Polar moment of inertia, Jo = ∫A r2 dA
= Ix + Iy
Where : r2 = x2 + y2 ;
Jo, Ix, Iy always positive
10.2 Parallel-axis Theorem for an Area
When the moment of inertia for an area is known about an axis passing through its
centroid, determine the moment of inertia for an area about a corresponding parallel axis
using the Parallel-axis Theorem.
Figure below shown, a differential element dA is located at an arbitrary distance y’ from
the centroidal x’ axis, whereas the fixed distance between the parallel x and x’ axes is
defined as dy . Since the moment of inertia of dA about the x axis is
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dIx = ( y’ + dy ) 2 dA
For entire area;
Ix = ∫A ( y’ + dy ) 2 dA
= ∫A y’2 dA + 2 dy ∫A y’ dA + dy2 ∫A dA
The first integral :
moment of inertia for an area about centroidal axis Ix’
The second integral : zero, since the x’ axis passed through the area’s centroid
The third integral : the total area
Therefore ;
Ix = ¯Ix’ + A dy2 ;
For polar moment of inertia
Iy = ¯Iy’ + A dx2
Jo = ¯Jc + Ad2
The three equations state that :
The moment of inertia for an area about an axis is equal to the moment of inertia
of the area about a parallel axis passing through the area’s centroid plus the product of
the area and the square of the perpendicular distance between the axes.
10.2 Radius of Gyration of an Area
The radius of gyration of a planar area has units of length and is a quantity that is often
used for the design of columns in structural mechanics. Provided the areas and moments
of inertia are known, the radii of gyration are:
kx = √(Ix / A) ; ky =
Where
√(Iy / A) ;
ko = √(Jo / A)
Ix = kx2 A
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10.4 Moment of Inertia for an Areas by Integration
Procedure :





If a single integration is performed to determine the moment of inertia of an area
about an axis, it will be necessary to specify the differential element dA.
Most often the element will be rectangular, such that it will have a finite length
and differential width.
The element should be located so that it intersects the boundary of the area at the
arbitrary point (x,y), by two possible ways.
Case 1 : The length of the element can be oriented parallel to the axis. Direct
application Iy = ∫A x 2 dA can be made since the element has an infinitisimel
thickness dx, therefore all parts of the element lie at the same moment-arm
distance x from y axis.
Case 2 : The length of the element can be oriented perpendicular to the axis. For
this case, first , calculate the moment of inertia of the elemen about the horizontal
axis passing through the element’s centroid then determine the moment of inertia
of the element about the x axis by using the parallel-axis theorem. Integration of
this result will yield Ix.
EXAMPLES
Example 10-1 (pp 501)
Example 10-2 (pp 502)
Example 10-3 (pp 503)
Example 10-4 (pp 504)
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10.5 Moment of Inertia for Composite Areas
A composite area consists of a series of connected “simpler” parts or shape, such as
semicircles, rectangulars and triangles. Provide the moment of inertia of aech af these
parts is known or can be determined about a common axis, then the moment inertia of
the composit area equal the algebric sum of the moments of inertia of all its parts.
Procedure :



Using a sketch, divide the area into its parts and indicate the perpendicular
distance from the centroid of each part to the reference axis.
The moment of inertia of aech part should be determined about its centroidal axis,
which is parallel to the reference axis. For calculation, use the table given.
If the centroidal axis does not coincide with the reference axis, the parallel-axis
theorem I = ¯I + A d2 should be used to determine the moment of inertia of
the part about the reference axis.
Summations:


The moment of inertia of the entire area about the reference axis is determine by
summing the result of its components parts.
If the composite parts has a hole, its moment of inertia is found by substracting
the moment of inertia for the hole from the moment of inertia of the entire part
including the hole.
EXAMPLES
Example 10-5 (pp 509)
Example 10-6 (pp 510)
10.6 Mass Moment of Inertia
The mass moment of inertia of a body is a property that measures the resistance of the
body to angular acceleration. Mass moment of inertia as the integral of the second
moment about an axis of all the elements of mass dm which compose of the body.
For the figure below, the body’s moment of inertia about z axis is
I = ∫m r 2 dm
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The moment arm, r is the perpendicular distance from the axis to the arbitrary element
dm.
If the body consists of material having a variable density, ρ = ρ (x , y , z), the elemental
mass dm of the body may be eexpressed in terms of its density and volume as
dm = ρ dV
subsituting dm into
I = ∫m r 2 dm
the body’s moment of inertia is then computed using volume elements for integration
I = ∫v ρ dV
In the special case of ρ being constant
I = ρ ∫v dV
When the element volume choosen for integration has differential size in all three
directions, eg: dV = dx dy dz;
the moment inertia of the body must be determine using ‘triple integration’.
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STATICS
EXAMPLES
Example 10-11 (pp 535)
Example 10-12 (pp 536)
Example 10-13 (pp 539)
Example 10-14 (pp 540)
EXERCISES
Exercise 10 - 2
Exercise 10 - 13
Exercise 10 - 28
Exercise 10 - 27
Exercise 10 - 34
Exercise 10 - 49
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