Download - B.tech i eg u1 engineering curves
CHAPTER – 1
ENGINEERING
CURVES
Useful by their nature & characteristics.
Laws of nature represented on graph.
Useful in engineering in understandinglaws, manufacturing of various items,designing mechanisms analysis of forces,construction of bridges, dams, watertanks etc.
USES OF ENGINEERING CURVES
Be it an arc
Be it an arch in construction in civil engineering
Be it an any spring in mechanical engineering
Be it any component of electronics and computer
engineering component
Measuring the distance
On the map
For navigation
In space technology
1. CONICS
2. CYCLOIDAL CURVES
3. INVOLUTE
4. SPIRAL
5. HELIX
6. SINE & COSINE
CLASSIFICATION OF ENGG. CURVES
It is a surface generated by moving aStraight line keeping one of its end fixed &other end makes a closed curve.
What is Cone ?
If the base/closedcurve is a polygon, weget a pyramid.
If the base/closed curve isa circle, we get a cone.
The closed curve isknown as base.
The fixed point is known as vertex or apex.
Vertex/Apex
90º
Base
If axis of cone is not
perpendicular to base, it is
called as oblique cone.
The line joins vertex/
apex to the circumference
of a cone is known as
generator.
If axes is perpendicular to base, it is called as
right circular cone.
Generator
Cone Axis
The line joins apex to the center of base is called
axis.
90º
Base
Vertex/Apex
Definition :- The section obtained by theintersection of a right circular cone by acutting plane in different position relativeto the axis of the cone are called CONICS.
CONICS
B - CIRCLE
A - TRIANGLE
CONICS
C - ELLIPSE
D – PARABOLA
E - HYPERBOLA
When the cutting plane contains theapex, we get a triangle as the section onthe surface of the cone.
TRIANGLE
When the cutting plane is perpendicular tothe axis or parallel to the base in a right conewe get circle as the section on the surface ofthe cone.
CIRCLE
Sec Plane
Circle
Definition :-When the cutting plane is inclined to the axisbut not parallel to generator or theinclination of the cutting plane(α) is greaterthan the semi cone angle(θ), we get an ellipseas the section on the surface of the cone.
ELLIPSE
α
θ
α > θ
When the cutting plane is inclined to the axisand parallel to one of the generators of the coneor the inclination of the plane(α) is equal to semicone angle(θ), we get a parabola as the section.
PARABOLA
θ
α
α = θ
When the cutting plane is parallel to theaxis or the inclination of the plane withcone axis(α) is less than semi cone angle(θ),we get a hyperbola as the section.
HYPERBOLADefinition :-
α < θα = 0
θθ
CONICSDefinition :- The locus of point moves in aplane such a way that the ratio of itsdistance from fixed point (focus) to a fixedStraight line (Directrix) is always constant.
Fixed point is called as focus.
Fixed straight line is called as directrix.
M
CF
V
P
Focus
Conic Curve
Directrix
The line passing through focus &
perpendicular to directrix is called as axis.
The intersection of conic curve with axis iscalled as vertex.
AxisM
CF
V
P
Focus
Conic Curve
Directrix
Vertex
N Q
Ratio =Distance of a point from focus
Distance of a point from directrix
= Eccentricity
= PF/PM = QF/QN = VF/VC = e
M P
F
Axis
CV
Focus
Conic Curve
Directrix
Vertex
Vertex
Ellipse is the locus of a point which moves in aplane so that the ratio of its distance from afixed point (focus) and a fixed straight line(Directrix) is a constant and less than one.
ELLIPSE
M
N Q
P
CF
V
Axis
Focus
Ellipse
Directrix
Eccentricity=PF/PM
= QF/QN
< 1.
Ellipse is the locus of a point, which moves in aplane so that the sum of its distance from twofixed points, called focal points or foci, is aconstant. The sum of distances is equal to themajor axis of the ellipse.
ELLIPSE
F1
A B
P
F2
O
Q
C
D
F1
A B
C
D
P
F2
O
PF1 + PF2 = QF1 + QF2 = CF1 +CF2 = constant
= Major Axis
Q
= F1A + F1B = F2A + F2B
But F1A = F2B
F1A + F1B = F2B + F1B = AB
CF1 +CF2 = AB
but CF1 = CF2
hence, CF1=1/2AB
F1 F2
OA B
C
D
Major Axis = 100 mm
Minor Axis = 60 mm
CF1 = ½ AB = AO
F1 F2
OA B
C
D
Major Axis = 100 mm
F1F2 = 60 mm
CF1 = ½ AB = AO
APPLICATION :-
Shape of a man-hole.
Flanges of pipes, glands and stuffing boxes.
Shape of tank in a tanker.
Shape used in bridges and arches.
Monuments.
Path of earth around the sun.
Shape of trays etc.
Ratio (known as eccentricity) of its distancesfrom focus to that of directrix is constant andequal to one (1).
PARABOLA
The parabola is the locus of a point, which movesin a plane so that its distance from a fixed point(focus) and a fixed straight line (directrix) arealways equal.
Definition :-
Directrix AxisVertex
M
C
N Q
FV
P
Focus
Parabola
Eccentricity = PF/PM
= QF/QN
= 1.
Motor car head lamp reflector.
Sound reflector and detector.
Shape of cooling towers.
Path of particle thrown at any angle with earth,etc.
Uses :-
Bridges and arches construction
Home
It is the locus of a point which moves in aplane so that the ratio of its distances froma fixed point (focus) and a fixed straightline (directrix) is constant and grater thanone.
Eccentricity = PF/PM
AxisDirectrix
HyperbolaM
C
NQ
FV
P
FocusVertex
HYPERBOLA
= QF/QN
> 1.
Nature of graph of Boyle’s law
Shape of overhead water tanks
Uses :-
Shape of cooling towers etc.
METHODS FOR DRAWING ELLIPSE
2. Concentric Circle Method
3. Loop Method
4. Oblong Method
5. Ellipse in Parallelogram
6. Trammel Method
7. Parallel Ellipse
8. Directrix Focus Method
1. Arc of Circle’s Method
P2’
1 2 3 4A B
C
D
P1
P3
P2
P4 P4P3
P2
P1
P1’
F2
P3’P4’ P4’
P3’
P2’
P1’
90°
F1 O
ARC OF CIRCLE’S METHOD
A BMajor Axis 7
8
910
11
9
8
7
6
54
3
2
1
12
11
P6
P5P4
P3
P2`
P1
P12
P11
P10P9
P8
P7
6
54
3
2
1
12C10
O
CONCENTRIC
CIRCLE
METHOD
F2F1
D
CF1=CF2=1/2 AB
T
N
Q
e = AF1/AQ
0
1
2
3
4
1 2 3 4 1’0’
2’3’4’
1’
2’
3’
4’
A B
C
D
Major Axis
Min
or
Axis
F1 F2
Dir
ectr
ix
E
F
S
P
P1
P2
P3
P4
P1’
P2’
P3’P4’
P0
P1’’
P2’’
P3’’
P4’’P4
P3
P2
P1
OBLONG METHOD
BA
P4
P0
D
C
60°
6
543
21
0
5 4 3 2 1 0 1 2 3 4 5 6
5
3210P1P2
P3
Q1
Q2Q3Q4
Q5
P6 Q6O
4
ELLIPSE IN PARALLELOGRAM
R4
R3
R2R1
S1
S2
S3
S4
P5
G
H
I
K
J
P6
P5’ P7’P6’
P1
P1’
T
T
V1
P5
P4’
P4
P3’P2’
F1
D1
D1
R1
ba
cd
ef
g
Q
P7P3P2
Directr
ix
90°
1 2 3 4 5 6 7
Eccentricity = 2/3
3R1V1
QV1=
R1V1
V1F1=
2
Ellipse
ELLIPSE – DIRECTRIX FOCUS METHOD
Dist. Between directrix
& focus = 50 mm
1 part = 50/(2+3)=10
mm V1F1 = 2 part = 20 mm
V1R1 = 3 part = 30 mm
< 45º
S
METHODS FOR DRAWING PARABOLA
1. Rectangle Method
2. Parabola in Parallelogram
3. Tangent Method
4. Directrix Focus Method
2345
0
1
2
3
4
5
6 1 1 5432 6
0
1
2
3
4
5
0
VD C
A B
P4
P4
P5
P5
P3
P3
P2
P2
P6
P6
P1
P1
PARABOLA –RECTANGLE METHOD
PARABOLA
B
2’
0
6
C
P’5
30°
A X
D5’
4’
3’
1’
0
5
4
3
2
1
P1
P2
P3
P4
P5
P’4
P’3P’2P’1
P’6
PARABOLA – IN PARALLELOGRAM
P6
BA O
V
1
8
3
4
5
2
6
7
9
10
0
1
2
3
4
5
6
7
8
9
1
0
0
F
PARABOLA
TANGENT METHOD
D
D
90° 2 3 4T
TN
N
S
V 1
P1
P2
PF
P3
P4
P1’
P2’P3’
P4’
PF’
AXIS
90°
R F
PARABOLA DIRECTRIX FOCUS METHOD
PROBLEM:-
A stone is thrown from a building 6 m
high. It just crosses the top of a palm
tree 12 m high. Trace the path of the
projectile if the horizontal distance
between the building and the palm
tree is 3 m. Also find the distance of
the point from the building where the
stone falls on the ground.
6m
ROOT OF TREE
BUILDING
REQD.DISTANCE
TOP OF TREE
3m
6m
F
A
STONE FALLS HERE
3m
6m
ROOT OF TREE
BUILDING
REQD.DISTANCE
GROUND
TOP OF TREE
3m
6m
1
2
3
1
2
3
321 4 5 6
5
6
EF
A B
CD
P3
P4
P2
P1
P
P1
P2
P3
P4
P5
P6
3 2 1
0
STONE FALLS HERE
EXAMPLE
A shot is discharge from the ground level at an angle 60 to the horizontal at a point 80m away from the point of discharge. Draw the path trace by the shot. Use a scale 1:100
ground level BA
60º
gun shot
80 M
parabola
ground level BA O
V
1
8
3
4
5
2
6
7
9
10
0
1
2
3
4
5
6
7
8
9
1
0
0
F
60º
gun shot
D D
VF
VE = e = 1
E
CYCLOIDAL GROUP OF CURVES
SuperiorHypotrochoid
Cycloidal Curves
Cycloid Epy Cycloid Hypo Cycloid
SuperiorTrochoid
InferiorTrochoid
SuperiorEpytrochoi
d
InferiorEpytrochoi
d
InferiorHypotrochoi
d
Rolling Circle or Generator
CYCLOID:-
Cycloid is a locus of a point on the circumference of
a rolling circle(generator), which rolls without slipping or sliding along a fixed straight line or a directing line or a director.
C
P P
P
R
C
Directing Line or Director
EPICYCLOID:-
Epicycloid is a locus of a point(P) on the circumference
of a rolling circle(generator), which rolls without slipping or
sliding OUTSIDE another circle called Directing Circle.
2πr
Ø = 360º x r/Rd
Circumference of
Generating Circle
Rolling
Circle
r
O
Ø/2 Ø/2P0 P0
Arc P0P0 =
Rd x Ø =
P0
HYPOCYCLOID:-
Hypocycloid is a locus of a point(P) on the circumference of a
rolling circle(generator), which rolls without slipping or sliding
INSIDE another circle called Directing Circle.`
DirectingCircle(R)
P
Ø /2 Ø /2
Ø =360 x r
RR
T
Rolling CircleRadius (r)
O
Vertical
Hypocycloi
d
P P
If the point is inside the circumference of the circle,
it is called inferior trochoid.
If the point is outside the circumference of the
circle, it is called superior trochoid.
What is TROCHOID ?
DEFINITION :- It is a locus of a point
inside/outside the circumference of a rolling
circle, which rolls without slipping or sliding
along a fixed straight line or a fixed circle.
P0
2R or D
51
2
1 2 3 4 6 7 8 9 10 110 120
3
4
56
7
8
9
10
11 12P1
P2
P3
P4
P5 P7
P8
P9
P11
P12
C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10C11
Directing Line
C12
N
S1
R
P6
R
P10
R
: Given Data :
Draw cycloid for one revolution of a rolling circle having
diameter as 60mm.
Rolling
Circle
D
Problem : 2A circle of 25 mm radius rolls on the circumference of another circle of 150 mm diameter and outside it. Draw the locus of the point P on the circumference of the rolling circle for one complete revolution of it. Name the curve & draw tangent and normal to the curve at a point 115 mm from the centre of the bigger circle.
First Step : Find out the included angle by
using the equation
360º x r / R = 360 x 25/75 = 120º.
Second step: Draw a vertical line & draw two
lines at 60º on either sides.
Third step : at a distance of 75 mm from O,
draw a part of the circle taking radius = 75 mm.
Fourth step : From the circle, mark point C
outside the circle at distance of 25 mm & draw
a circle taking the centre as point C.
P6
P4
rP2
C1
C0
C2
C3C4 C5
C6
C7
C8
1
0
23
4
5
6 7
O
Ø/2 Ø/2
P1P0
P3 P5
P7P8
r rRolling
Circle
r
Rd X Ø = 2πrØ = 360º x r/Rd
Arc P0P8 = Circumference of
Generating Circle
EPICYCLOIDGIVEN:
Rad. Of Gen. Circle (r)
& Rad. Of dir. Circle (Rd) Sº
U
N
Ø = 360º x 25/75
= 120°
Problem :3A circle of 80 mm diameter rolls on the circumference of another circle of 120 mm radius and inside it. Draw the locus of the point P on the circumference of the rolling circle for one complete revolution of it. Name the curve & draw tangent and normal to the curve at a point 100 mm from the centre of the bigger circle.
P0 P1 P11
C0
C1
C2
C3
C4
C5C6 C7 C8
C9
C10
C1
1
C12
P1
0P8
0
1 23
4
5
67
89
10
1
1
1
2
P2
P3
P4P5 P6
P9P7
P12
/2
/2
=360 x 412
= 360 x rR
=120°
R
T
T
N
S
N
Rolling Circle
Radias (r)
DirectingCircle
O
Vertical
Hypocycloi
d
INVOLUTEDEFINITION :- If a straight line is rollesround a circle or a polygon without slipping or sliding, the locus of points on the straight line is an INVOLUTES to the curve.
OR
Uses :- Gears profile
Involute of a circle is a curve traced out by a point on a tights string unwound or wound from or on the surface of the circle.
PROBLEM:A string is unwound from a
circle of 20 mm diameter. Draw the
locus of string P for unwinding the
string’s one turn. String is kept tight
during unwound. Draw tangent &
normal to the curve at any point.
P12
P2
012
6
P11 20 9 103 4 6 8 115 7 12
DP3
P4
P5
P6
P7
P8
P9
P10
P11
123
45
78
910
11N
.
PROBLEM:-
Trace the path of end point of a thread
when it is wound round a circle, the length of
which is less than the circumference of the
circle.
Say Radius of a circle = 21 mm &
Length of the thread = 100 mm
Circumference of the circle = 2 π r
= 2 x π x 21 = 132 mm
So, the length of the string is less than
circumference of the circle.
P
R=6toP
0
0 1 2 3 4 5 6 7 8 P
11 01
2
3
456
7
8
9
10
P1
P2
P3
P4
P5
P6
P7
P8
L= 100 mm
R=
3to
P
INVOLUTE
9
ø
11 mm = 30°Then 5 mm =
Ø = 30° x 5 /11 = 13.64 °
S = 2 x π x r /12
PROBLEM:-
Trace the path of end point of a thread
when it is wound round a circle, the length of
which is more than the circumference of the
circle.
Say Radius of a circle = 21 mm &
Length of the thread = 160 mm
Circumference of the circle = 2 π r = 2 x π x 21 = 132 mm
So, the length of the string is more than
circumference of the circle.
P13
P11
313
14
15
P0
P12
O
7
10 123
456
89
1112 1 2
P1
P2
P3P4
P5
P6
P7
P8 P9 P10
P14P
L=160 mm
R=21mm
64 5 7 8 9 10 11 12 131415
ø
PROBLEM:-
Draw an involute of a pantagon having
side as 20 mm.
P5
P0
P1
P2
P3
P4
23
4 5
1
INVOLUTE
OF A POLYGON
Given :
Side of a polygon 0
PROBLEM:-
Draw an involute of a square
having side as 20 mm.
P2
1
2 3
04
P0
P1
P3
P4
INVOLUTE OF A SQUARE
PROBLEM:-
Draw an involute of a string unwound
from the given figure from point C in
anticlockwise direction.
60°
A
B
C
30°
60°
A
B
C
30°
X
X+
A3
X+AB
12
3
4
5
C0
C1
C2
C3
C4
C5
C6
C7
C8
SPIRALS
If a line rotates in a plane about one of its
ends and if at the same time, a point moves
along the line continuously in one direction,
the curves traced out by the moving point is
called a SPIRAL.
The point about which the line rotates is
called a POLE.
The line joining any point on the curve with
the pole is called the RADIUS VECTOR.
The angle between the radius vector and the line
in its initial position is called the VECTORIAL
ANGLE.
Each complete revolution of the curve is
termed as CONVOLUTION.
Spiral
Arche Median Spiral for Clock
Semicircle Quarter
CircleLogarithmic
ARCHEMEDIAN SPIRAL
It is a curve traced out by a point
moving in such a way that its
movement towards or away from the
pole is uniform with the increase of
vectorial angle from the starting line.
USES :-
Teeth profile of Helical gears.
Profiles of cams etc.
To construct an Archemedian Spiral of
one convolutions, given the radial
movement of the point P during one
convolution as 60 mm and the initial
position of P is the farthest point on the
line or free end of the line.
Greatest radius = 60 mm &
Shortest radius = 00 mm ( at centre or at pole)
PROBLEM:
P10
1
2
3
4
5
6
7
89
10
11
120
8 7 012345691112
P1
P2P3
P4
P5
P6
P7P8
P9
P11
P12
o
To construct an Archemedian
Spiral of one convolutions, given
the greatest & shortest(least)
radii.
Say Greatest radius = 100 mm &
Shortest radius = 60 mm
To construct an Archemedian
Spiral of one convolutions, given
the largest radius vector &
smallest radius vector.
OR
3 1
26
5
8 4
79
1
011
2
1
34
5
6
7
89
10
11
12
P1
P2P3
P4
P5
P6
P7
P8 P9
P1
0
P1
1
P12
O
Diff. in length of any two radius vectors
Angle between them in radiansConstant of the curve =
=
OP – OP3
Π/2
100 – 90=
Π/2
= 6.37 mm
PROBLEM:-
A link OA, 100 mm long rotates about O in
clockwise direction. A point P on the link,
initially at A, moves and reaches the other end
O, while the link has rotated thorough 2/3 rd of
the revolution. Assuming the movement of the
link and the point to be uniform, trace the path
of the point P.
AInitial Position of point PPO
P1
P2
P3
P4P5
P6
P7
P8
21
34567O
1
2
3
4
56
7
8
2/3 X 360°
= 240°
120º
PROBLEM :
A monkey at 20 m slides down from a
rope. It swings 30° either sides of
rope initially at vertical position. The
monkey initially at top reaches at
bottom, when the rope swings about
two complete oscillations. Draw the
path of the monkey sliding down
assuming motion of the monkey and
the rope as uniform.
θ
o
012
3
45 6
78
9
1011
121314
15
16 17 18 1920
21
222324
23
13
22
24
12
34
56
78
910
1112
1415
16
1819
2021
17
P3
P9
P15
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A text book of engineering graphics- Prof. P.J
SHAH
Engineering Drawing-N.D.Bhatt
Engineering Drawing-P.S.Gill
Thank You