Transcript Static: PowerPoint Notes

Principle of Engineering
ENG2301
 Mechanics
Section
 Textbook:
A Foundation Course in Statics and Dynamics
 Addison Wesley Longman 1997

Syllabus Overview
 A Statics

B Dynamics
Units
 force
Newton
(N)
 stress
Newton per metre squared (N/m2 )

or Pascal, 1 Pa = 1 N/m2 (Pa)
 pressure Newton per metre squared (N/m2 )

or bar, 1 bar = 1x105 N/m2 (bar)
 moment, torque, couple

Newton . Metre
(Nm)
Units
 Most
commonly used prefixes
 micro
x 10-6

 milli
x 10-3
m
 kilo
x 103
k
 mega
x 106
M
 giga
x 109
G
 * Note Capitals and lower case letters are
important
Scalars and Vector
 Two
kind of quantities:
 Scalar
 Vector
 Scalar quantities have magnitude but no
directional properties
 can be handled by ordinary algebra, e.g.
c= a+b, c= 8 if a=3, b= 5
 e.g. time, mass, speed and energy etc. etc....
Vector
 Associated
with directions and
magnitude
 e.g. Force, displacement,
acceleration and velocity
 Can be represented by a straight
line with arrowhead and the
magnitude is shown by the
length
l

Vector Addition and Subtraction
 By
Triangle or Parallelogram laws
Addition

V = V1 + V2
V is called the resultant vector
V2
V1
V
1
V1
V1
V
V2
2
(a)
V2
(b)
(c)
Vector Addition and Subtraction
Subtraction
V’ = V1 - V2
can be regarded as V’ = V1 + (- V2)
 - V2 is drawn in the opposite direction
V’ is the resultant vector

V1
-V2
V1
1
V'
2
-V 2
(a)
V1
V'
-V 2
(b)
(c)
Vector Addition and Subtraction
Adding

more than two vectors
V’ = V1 + V2 +V3 +V4
V3
V2
V2
V1
V4
V1+V2
V3
V1
V1+V2+V3
V4
V=V 1+V2+V3 +V4
(a)
(b)
Resolution of Vectors
 Any
vector can be resolved into
components
 Commonly resolve into two components
y
perpendicular to each other
 V = Vx + Vy
Vy
V
 Vx = V cos 

 Vy = V sin 
o
Vx
2
2
 magnitude V = Vx + Vy )
  = tan-1 (Vy /Vx )
x
Force and Newton’s First Law
 First
Law - If the resultant force acting on a
particle is zero, the particle will remain at rest
(if originally at rest), or will move with constant
speed in a straight line (if originally in motion).
 State of Equilibrium - Equilibrium exists when
all the forces on a particle are in balance . The
velocity of a particle does not change , if the
particle is in Equilibrium .
Interpretation of First Law
 A body
is in Equilibrium if it moves with
constant velocity. A body at rest is a special
case of constant velocity i.e. v = 0 = constant.
 For a body to be in Equilibrium the resultant
force (meaning the vector addition of all the
forces) acting on the body must be zero.
 A Force can be defined as 'that which tends to
cause a particle to accelerate', assuming that
the force is not in Equilibrium with other
forces acting on the body.
Force
 A force
cannot be seen, only the effect of a
force on a body may be seen.
 Force Units: S.I. Unit ,Newton, (N) or
(kN)
 Force is a vector quantity. It has both
magnitude and direction.
Force Vectors
 Polar
and Rectangular Coordinates
F = Fx  Fy ;
Fx = F.cos ;
F = Fx  Fy
2
fy
Fy  F.sin  ;
2
 Fy 
  tan  
 Fx 
Fy
F

Fx
-1
fx
Example 1
 Calculate
the components in rectangular
coordinates of the 600 N force.
 Solution
Fx  600. cos 35  491 N
Fy  600. sin 35  344 N

Fy
600N
35

Fx
Example 2
 A force
vector has the components 600 kN
and 300 kN in the x and y directions
respectively, calculate the components in
polar coordinates.
 Solution
F  300  600  670.8 kN
2
2
300
  tan
 26.56
600
1
Resultant Force
 Parallelogram
Method
Fy
Force F1
F2
R
Force F2
Particle
Note
Resultant R
F1
means "equivalent to"
Fx
Resultant Force
 Algebraic
Method
Rx  F1, x  F2, x  F1 cos1  F2 cos 2
F2,y
F2
2
F1,x
F2,x
Ry  F1, y  F2, y   F1 sin 1  F2 sin  2
R  Rx  R y
2
1
 Ry 
  tan  
 Rx 
1
F1,y
F1
2
Resultant Force
 Triangle
of Forces Method
Fy
Fy
R
F2
F1
R
F2
F1
Fx
Order is not important
Fx
Example 3
 Find
the magnitude and direction of the
resultant (i.e. in polar coordinates) of the
two forces shown in the diagram,
 a) Using the Parallelogram Method
 b) Using the Triangle of Forces Method
 c) Using the algebraic calculation method
 Solution
Example 3 (Solution)
6kN
4kN
15
30
Rx  6 cos 30  4 cos 15  1.332 kN
Ry  6 sin 30  4 sin 15  4.035 kN
R  (1.332) 2  (4.035) 2  4.249 kN
 4.035

  tan
 71.73
 1.332
1
Or -108.260 from +ve
x axis
Equilibrium of Concurrent forces
F1
Equilibrant E
Equilibrant E
Force
F2
Resultant R
Equilibrant E are equal and
opposite to Resultant R
R
Fy
E
F2
F1
E = -R
Fx
Conditions for Equilibrium
 Coplanar:
all forces being in the same plane
(e.g.only x-y plane, no forces in z direction)
 Concurrent: all forces acting at the same
point (particle)
i=n
i=n
 F  0; or  F
i
i=0
i=n
x, i
i=0
For three forces
acting on a
particle
 0; and  Fy, i  0;
i=0
Fx,1  Fx,2  Fx,3  0;
Fy,1  Fy,2  Fy,3  0;
Some Definitions
 Particle
is a material body whose linear
dimensions are small enough to be
irrelevant
 Rigid Body is a body that does not deform
(change shape) as a result of the forces
acting on it .
Polygon of Forces
 Equilibrium
under multiple forces
F2
F3
F2
F5
F1
F3
F5
F1
F4
F4
Forces acting
on particle
Rigid body under
concurrent forces
Resultant and Equilibrant
F1
F1
F2
F2
F4
F4
Fy
F3
F3
y
F5
R
Fx
x
Resultant = - Equilibrant
R
= - F5
Example 4
 The
diagram shows three forces acting on a
particle .
 Find the equilibrant by drawing the polygon
of forces.
45 kN
84 kN
20
30
50
122 kN
Newton’s Third Law
 The
forces of action and reaction between
bodies in contact have the same magnitude,
but opposite in direction.
Hammer
BANG!
Solid Surface
Action and Reaction
Free Body Diagram of Hammer
Hammer
Free Body
Boundary
Force on Hammer caused by Surface
Force on Surface caused by Hammer
Solid Surface
Free Body diagram of Surface
Free Body Diagram
 Free
Body Diagram - used to describe the
system of forces acting on a body when
considered in isolation
R
R
R
mg
R
Free Body Diagram
Free Body
W
Boundary
Wr Reaction
W
Wa Action
System of Particles or Bodies
Two or more bodies or particles connected
together are referred to as a system of bodies
or particles.
External Force
External forces are all the forces acting on a
body defined as a free body or free system of
bodies, including the actions due to other
bodies and the reactions due to supports.
Transmissibility of Force
F
F
F
Load and Reaction
 Loads
are forces that are applied to bodies
or systems of bodies.
 Reactions at points supporting bodies are a
consequence of the loads applied to a body
and the equilibrium of a body.
Tensile and Compressive Forces
 Push
on the body which is called a
compressive force
 Pull
on a body which is called a tensile
force
Reaction
Action
Compressive Force
Reaction
Action
Tensile Force
Procedure for drawing a free body diagram
 Step
1: Imagine the particle to be isolated or
cut “free from its surroundings. Draw or
sketch its outlined shape.
 Step
2: Indicate on this sketch all the forces
that act on the particle. These forces can be
applied surface forces, reaction forces
and/or force of attraction.
Procedure for drawing a free body diagram
Fp
Supported
pulley
Pulling
Force
Fs
Spring
F1
Pulling
Force
F1
Spring
Mass
Mass
Fs
mg
Procedure for drawing a free body diagram
 Step
3: The forces that known should be
labeled with their proper magnitudes and
directions. Letters are used to represent the
magnitudes and directions of forces that
unknown.
Example 5
Ceiling Support
Ceiling Support
Action Load = 10 N
Ring
Reaction Force = 10 N
Ring
Weight = 10 N
Weight = 10 N
Gravity Load =10 N
Free Body boundary
Example 6
tow rope
Tug No.1
25
Barge
60
Ring
.
tow rope
Tug No.2
Example 6 (Solution)

resultant R of the two forces in tow ropes No.1 &
No. 2 from the components in the x and y
directions:
Rx  10 cos 25  30 cos 60  24.06 kN
Ry  10 sin 25  30 sin 60  21.75 kN
tow rope
Tug No.1
25
Barge
60
Ring
.
tow rope
Tug No.2
Example 6 (Solution)
R  24.06  21.75  32.43 kN
2
2
 21.75

  tan
 42.1
24.06
1
Equilibrant E = - R
E = 32.43 kN
42.1
42.1
R = 32.43 kN
Example 6 (Solution)
tow rope
Tug No.1
25
Barge
60
Ring
.
Tug No.2
tow rope
Resultant R is the sum of the actions
of the tow ropes on the barge
E=-R
Equilibrant E is the reaction of the barge to the ropes
Moment and Couple
 Moment
of Force
 Moment M of the force
F about the point O is
defined as:
M=Fd
where d is the
perpendicular distance
from O to F
 Moment is directional
M
o
d
F
Moment and Couple
F
A
r

d
B
M = F.d
= F.r.cos 
Moment = Force x Perpendicular Distance
Resultant of A System
of Forces
arbitrary body
subjected to a number of
forces F1, F2 & F3.
 Resultant
R = F1 + F2 + F3
 Components
Rx = F1x + F2x + F3x
Ry = F1y + F2y + F3y
d3
O
F3
d2
 An
d
R
F2
d1
F1
Rx
F 3y
F3
R
Ry
F2
F1
F 1x
F 1y
F 3x
F 2y
F 2x
Resultant of A System
of Forces
d3
O
F3
d2
 Resultant
moment Mo
= Sum of Moments
 Mo = F1 d1 + F2 d2 + F3 d3
 Mo = R d
d
R
F2
d1
F1
Rx
F 3y
F3
R
Ry
F2
F1
F 1x
F 1y
F 3x
F 2y
F 2x
Couple
 For
a Couple
O
 R =F = 0
 But Mo  0
l
 Mo = F(d+l) - Fl = Fd
 Moment of couple is the
same about every point in
its plane
F
d
F
Example 7
 Calculate
the total (resultant) moment on
the body.
300 mm
15 N
30 N
170 mm
30 N
A
100 mm 15 N
50 mm
Example 7 (Solution)
 Taking
moments about the corner A
M  30  0.17  15  0.3  30  0.05  15  0.1
 30  0.120  15  0.2  6.6 Nm
 Note
that the forces form two couples or
pure moments 3.6 Nm and 3.0 Nm
(resultant force =0, moment is the same
about any point).
Equilibrium of Moments
 The
sum of all the moments is zero when
the body is in moment equilibrium.
or
n
M  0
i 1
 If
i
M 1  M 2  .......  0
the body is in equilibrium the sum of the
moments of all the forces on acting on a
rigid body is the same for all points on the
body. It does not matter at which point on
a rigid body you choose for taking
moments about
Example 8
 Calculate
the resultant moment and the
equilibrant moment.
0.5 m
10 N
B
15 N
1.0 m 1.25 m
A
5N
2.5 m
3.0 m
Example 8 (Solution)
Take moment about A
M R  10  0.5  15 1.0  5  2.5  22.5 Nm
Take moment about B
M R  10  2.5  15  0.25  5  0.5  18.75 Nm
0.5 m
10 N
B
15 N
1.0 m 1.25 m
A
5N
2.5 m
3.0 m
Example 8 (Solution)
 Note
that the body is not in vertical and
horizontal equilibrium.
 There is no unique value for the resultant
moment.
 The value depends on where the resultant
force acts, ie., depends on the perpendicular
distance between the resultant force and the
point for taking moment.
 Therefore, the moments about A and B are
different.
Example 9
 Cantilever
Wall
beam
Built in End
or Fixed End
Point Load of weight W
d
Beam
W
Find the reaction force and
moment at the built in end
Example 9 (Solution)
 Taking
moment about A
 M  M  W .d  0
A
M A  W .d
   Fy  V  W  0
V W
W
Moment Reaction MA
C
A
d
Reaction V
Free Body Diagram of Beam
B
General Equations of Equilibrium of a Plane
(Two Dimensional) Rigid Body
(Non-concurrent forces)
i=n
i=n
 F  0; and  F  0;  M  0
x, i
i=0
n
y, i
i =0
For complete equilibrium, all 3
equations must be satisfied
Fx,1  Fx,2  Fx,3  ...  0;
Fy,1  Fy,2  Fy,3  ...  0;
M 1  M 2  .......  0
i 1
i
Types of Beam Supports
Simply
supported
beam
A beam simply supported on points or knife-edges
A side view or elevation of a simply supported beam
A diagrammatic view of a simply supported beam
Types of Beam Supports
Force perpendicular

to surface only.
Pin Joint supported by a roller
Ry
Both parallel and
Rx
perpendicular forces
Pin Joint Fixed to the ground
Ry
Fix Support
Two components of force and
Rx
M
Ry
a moment
Types of Supports and
Connections
Simply
supported
beam
A beam simply supported on points or knife-edges
A side view or elevation of a simply supported beam
A diagrammatic view of a simply supported beam
Types Wof Loading on Beams
Point load
a)

W
Diagrammatic representation
W / unit length
Distrubuted load (uniform)
b)
W / unit length
Diagrammatic representation
W / unit length
Another way of representing
Uniformly distributed loads
Types of Loading on Beams

Non-uniformly distributed loads
W
Types of Loading on Beams

C
A
B
W
Diagrammatic view of simply
supported beam with a
b
a
concentrated load W
W
B
A
RA
C
Free body diagram replacing
R B loads and supports by forces
Types of Loading on Beams
A

w N/m
a
b
B
Diagrammatic view of simply
supported beam with uniformly
distributed load w
F= w(b-a)
(a+b)/2
Free body diagram replacing
loads and supports by forces
Example 10
 Find
the reactions at the supports for the
beam shown in the diagram.
20 kN
5 kN/m
3.5m
3.5m
7m
Example 10(Solution)
W  5000  7  35 kN
 RB 14  35 10.5  20  3.5  0
20 kN
RB  31.25 kN
RA  RB  35  20
 RA  23.75 kN
35 kN
RB
RA
3.5m
7m
3.5m
Example 11
 Express
F
F in terms of m, a and b.
a
b
W
Example 11(Solution)
F
b
a
W
R
 F = 0; + ; R - F - W = 0
 M  Wb  Fa  0
b
F W
a
Ratio a/b is called
Mechanical Advantage
Example 12
 Find
reaction forces at supports A and B.
350N
A
B
0.15m
o
30
0.30 m
0.40 m
200N
Example 12 (Solution)
 Consider
the sum of vertical forces and
horizontal forces are zero, and since RAx = 0
as point B can only take up vertical force.
 RAx = 200 x sin30o = 100 N
 RAy


+ RBy = 350 + 200 cos30o
= 350 + 173.2 N
= 523.2 N
Example 12 (Solution)
 Taking
moment about A,
 RBy x 0.3 = 350 x 0.15 + 200 cos30o x 0.4

RBy x 0.3 = 121.8 N

RBy = 405.9 N
 Therefore the reaction at point B is 405.9N
upward.

RAy = 523.2 - 405.9 = 117.3 N
Example 12 (Solution)
 The
reaction at point A is 117.3 N upward
and 100 N to the left.
 Resultant at A is:

RA = (117.32 + 1002)

= 154.1 N
117.3

angle  =
49.55o

100
Example 13
 Find
reaction forces at supports A and B.
55 kN
A
750 mm
B
600 mm
Example 13 (Solution)
   Fy  VA  55  0
55 kN
HA



 Fx  H A  H B  0
VA
750 mm
 M H  0.75  55  0.6  0
HB
A
600 mm
VA  55 kN
H A  44 kN H B  44 kN
Example 14
20
o
T
75 N
O
250 mm
100 mm
Example 14 (Solution)
T
75 N
O
R

Rx
Ry
Example 14 (Solution)
 Fy
= 0:Ry - 75 = 0
Ry = 75 N
 Fx = 0:T - Rx = 0T = Rx
 MO = 0:
75 x 250 - T x 100 x cos 200
=0



Therefore Rx = T = 200 N
R = (Rx2 + Ry2) = 214 N
 = tan-1 (Ry/Rx) = 20o 33’
Example 14 (Solution)
 Reaction
R is that exerted by the frame on
the bellcrank, which is equal and opposite
to that on the chassis.
R

R = 214 N
75 N
20o 33'
T = 200 N
Example 14 (Graphical Solution)
 Having
determined the line of reaction R, a
scaled force polygon can be drawn.
 By measurement, T = 200 N, R = 214 N
R = 214 N
75 N
20o 33'
T = 200 N