engineering drawing is a language of all persons involved in engineering activity. engineering ideas...
TRANSCRIPT
Engineering drawing is a language of all persons involved in engineering activity. Engineering ideas are recorded by preparing drawings and execution of work is also carried out on the basis of drawings. Communication in engineering field is done by drawings. It is called as a “Language of Engineers”.
CHAPTER – 2
ENGINEERING
CURVES
Useful by their nature & characteristics. Laws of nature represented on graph.
Useful in engineering in
understanding laws,
manufacturing of various items,
designing mechanisms analysis of
forces, construction of bridges,
dams, water tanks etc.
USES OF ENGINEERING CURVES
1. CONICS
2. CYCLOIDAL CURVES3. INVOLUTE
4. SPIRAL
5. HELIX
6. SINE & COSINE
CLASSIFICATION OF ENGG. CURVES
It is a surface generated by moving a Straight line keeping one of its end fixed & other end makes a closed curve.
What is Cone ?
If the base/closed curve is a polygon, we get a pyramid.
If the base/closed curve is a circle, we get a cone.
The closed curve is known 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.
Generato
r
Cone Axis
The line joins apex to the center of base is called axis.
90º
Base
Vertex/Apex
Definition :- The section obtained by the intersection of a right circular cone by a cutting plane in different position relative to 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 the apex, we get a triangle as the section.
TRIANGLE
When the cutting plane is perpendicular to the axis or parallel to the base in a right cone we get circle the section.
CIRCLE
Sec Plane
Circle
Definition :-When the cutting plane is inclined to the axis but not parallel to generator or the inclination of the cutting plane(α) is greater than the semi cone angle(θ), we get an ellipse as the section.
ELLIPSE
α
θ
α > θ
When the cutting plane is inclined to the axis and parallel to one of the generators of the cone or the inclination of the plane(α) is equal to semi cone angle(θ), we get a parabola as the section.
PARABOLA
θ
α
α = θ
When the cutting plane is parallel to the axis or the inclination of the plane with cone 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 a plane such a way that the ratio of its distance from fixed point (focus) to a fixed Straight line (Directrix) is always constant.
Fixed point is called as focus.Fixed straight line is called as directrix.
M
C FV
P
Focu
s
Conic
CurveDirectrix
The line passing through focus & perpendicular to directrix is called as axis.The intersection of conic curve with
axis is called as vertex.
AxisM
CF
V
P
Focus
Conic
CurveDirectrix
Vertex
N Q
Ratio =Distance of a point from
focusDistance of a point from
directrix= Eccentricity= PF/PM = QF/QN = VF/VC = e
M P
F
Axis
CV
Focus
Conic
CurveDirectrix
Vertex
Vertex
Ellipse is the locus of a point which moves in a plane so that the ratio of its distance from a fixed point (focus) and a fixed straight line (Directrix) is a constant and less than one.
ELLIPSE
M
N Q
P
CF
V
Axi
s
Focu
s
Ellips
eDirectri
x
Eccentricity=PF/PM = QF/QN
< 1.
Ellipse is the locus of a point, which moves in a plane so that the sum of its distance from two fixed points, called focal points or foci, is a constant. The sum of distances is equal to the major 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
Uses :-
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 distances from focus to that of directrix is constant and equal to one (1).
PARABOLAThe parabola is the locus of a point,
which moves in a plane so that its distance from a fixed point (focus) and a fixed straight line (directrix) are always equal.
Definition :-
Directrix AxisVertex
M
C
N Q
FV
P
Focus
Parabol
a
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 a plane so that the ratio of its distances from a fixed point (focus) and a fixed straight line (directrix) is constant and grater than one.
Eccentricity = PF/PM
AxisDirectrix
Hyperbol
aM
C
N Q
FV
P
Focu
s
Vertex
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. Directrix Focus Method
1. Arc of Circle’s Method
Nor
mal
P2’
R =
A1
Tangent
1 2 3 4A B
C
D
P1
P3
P2
P4 P4 P3
P2
P1
P1’
F2
P3’ P4’ P4’P3’
P2’
P1’
90°
F1
Rad =B1
R=B2
`R=
A2
O
ARC OF CIRCLE’S METHOD
P6
Nor
mal
P5’ P7’P6’
P1
TangentP1’
N
N
T
T
V1
P5
P4’
P4
P3’P2’
F1
D1
D1
R1
ba
cd
ef
g
Q
P7P3P2
Dir
ectr
ix
R=6f`
90°
1 2 3 4 5 6 7
Eccentricity = 2/3
3R1V1
QV1 = R1V1
V1F1 = 2
Ellipse
ELLIPSE – DIRECTRIX FOCUS METHOD
R=1a
Dist. Between directrix & focus = 50 mm
1 part = 50/(2+3)=10 mm V1F1 = 2 part = 20 mmV1R1 = 3 part = 30 mm
< 45º
S
PROBLEM :-
The distance between two coplanar fixed points is 100 mm. Trace the complete path of a point G moving in the same plane in such a way that the sum of the distance from the fixed points is always 140 mm.
Name the curve & find its eccentricity.
ARC OF CIRCLE’S METHOD
Nor
mal
G2’
R =
A1
Tangent
1 2 3 4A B
G
G’
G1
G3
G2
G4 G4 G3
G2
G1
G1’
G3’ G4’ G4’G3’
G2’
G1’
F2F1
R=B1
R=B2
`R=
A2
O
90°
90°
dir
ectr
ix
100
140
GF1 + GF2 = MAJOR AXIS = 140
E
e AF1
AEe =
R=70 R=70
PROBLEM :-
Draw an ellipse passing through A & B of an equilateral triangle of ABC of 50 mm edges with side AB as vertical and the corner C coincides with the focus of an ellipse. Assume eccentricity of the curve as 2/3. Draw tangent & normal at point A.
PROBLEM :-
Draw an ellipse passing through all the four corners A, B, C & D of a rhombus having diagonals AC=110mm and BD=70mm.
Use “Arcs of circles” Method and find its eccentricity.
METHODS FOR DRAWING PARABOLA
1. Rectangle Method
2. Parabola in Parallelogram
3. 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
0
2’
0
2
6
C
6’
V
5
P’5
30°
A X
D
1’2’
4’5’
3’
1
34
5’
4’
3’
1’
0
5
4
3
2
1
P1
P2
P3
P4
P5
P’4
P’3
P’2
P’1
P’6
PARABOLA – IN PARALLELOGRAM
P6
D
D
DIR
EC
TR
IX
90° 2 3 4T
TN
N
S
V 1
P1
P2
PF
P3
P4
P1’
P2’P3’
P4’
PF’
AXIS
RF
R2
R1
R3
R4
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 10
STONE FALLS HERE
PROBLEM:-
In a rectangle of sides 150 mm and 90 mm, inscribe two parabola such that their axis bisect each other. Find out their focus points & positions of directrix.
150 mmA
B C
D1 2 3 4 5
1
2
3
4
5
O
P1
P2
P3
P4
P5
M1’ 2’ 3’ 4’ 5’
1’
2’3’
4’5’
P1’ P
2’ P
3’ P
4’ P
5’
90 m
m
example
Draw a parabola passing through three
different points A, B and C such that AB =
100mm, BC=50mm and CA=80mm
respectively.
BA
C
100
50
80
1’
0
0
2’
0
26
6’5
P’5
1’ 2’ 4’ 5’3’134
5’
4’
3’
54
3
21
P1P
2P
3
P4
P5
P’4
P’3
P’2
P’1
P’6
P6
A B
C
METHODS FOR DRAWING HYPERBOLA
1. Rectangle Method
2. Oblique Method
3. Directrix Focus Method
D
F
1 2 3 4 5
5’4’3’
2’
P1
P2
P3P4
P5
0
P6
P0
AO EX
B
C
Y
Given Point P0
90°
6
6’
Hyperbola
RECTANGULAR HYPERBOLA
AXIS
AXIS
When the asymptotes are at right angles to each other, the hyperbola is called rectangular or equilateral hyperbola
ASYMPTOTES X and Y
Problem:-
Two fixed straight lines OA and OB are at right angle to each other. A point “P” is at a distance of 20 mm from OA and 50 mm from OB. Draw a rectangular hyperbola passing through point “P”.
D
F
1 2 3 4 5
5’4’3’
2’
P1
P2
P3P4
P5
0
P6
P0
AO EX=20
B
C
Y =
50
Given Point P0
90°
6
6’
Hyperbola
RECTANGULAR HYPERBOLA
PROBLEM:-
Two straight lines OA and OB are at 75° to each other. A point P is at a distance of 20 mm from OA and 30 mm from OB. Draw a hyperbola passing through the point “P”.
75 0
P4
E
6’
2’1’ P1
1 2 3 4 5 6 D
P6P5
P3P2
P0
7’P7
7C
B F
O
Y =
30
X = 20
Given Point P0
A
AXIS
NORMAL
C V F1
DIR
EC
TR
IXD
D
1 2 3 4
4’
3’
2’
1’P1
P2
P3
P4
P1’
P2’
P3’
P4’
T1
T2
N2
N1
TAN
GE
NT
s
Directrix and focus method
CYCLOIDAL GROUP OF CURVESWhen one curve rolls over another curve without slipping or sliding, the path Of any point of the rolling curve is called as ROULETTE.When rolling curve is a circle and the curve on which it rolls is a straight line Or a circle, we get CYCLOIDAL GROUP OF CURVES.
Cycloidal Curves
Cycloid Epy Cycloid Hypo Cycloid
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
R d
Rolling Circle
r
O
Ø/2
Ø/2
P0 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
Hypocycloid
P P
P0
2R or D 5
T
T
1
2
1 2 3 4 6 7 8 9 10 110 120
3
4
567
8
9
1011 12
P1
P2
P3
P4
P5 P7
P8
P9
P11
P12
C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10C11
Directing Line
C12
N
N
S
S1
R
P6
R
P10
R
: Given Data :
Draw cycloid for one revolution of a rolling circle having diameter as 60mm.
Rolling Circle
D
C0
P0
78
P64
P1
1
2
3
C2 C3
P2 C4
Problem 1:A circle of diameter D rolls without
slip on a horizontal surface (floor) by Half revolution and then it rolls up a vertical surface (wall) by another half revolution. Initially the point P is at the Bottom of circle touching the floor. Draw the path of the point P.
5
6 C1
P3
P4
P5
P7
P8
70
C 5
C 6
C 7
C 8
1 2 3 4
56
D/2
πD
/2
πD/2 D/2Floor
Wal
l
CYCLOID
56
7
8
Take diameter of circle = 40mmInitially distance of centre of
circle from the wall 83mm (Hale circumference + D/2)
Problem : 2
A 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
R d
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) S
ºU
N
Ø = 360º x 25/75
= 120°
Problem :3
A 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
Tangent
P11
r
C0
C1
C2
C3
C4C5
C6 C7 C8 C9
C10
C11
C12
P10P8
0
1 2 3
4
5
67
89
10
11
12P2 P3
P4P5 P6
P9P7
P12
/2
/2
=360 x 412
= 360 x rR
= 120°
R
T
T
N
S
N
Norm
alr
r
Rolling Circle
Radias (r)
DirectingCircle
O
Vertical
Hypocycloid
Problem :
Show by means of drawing that when the diameter of rolling circle is half the diameter of directing circle, the hypocycloid is a straight line
C
C1
C2
C3
C4C5
C6 C7
C9
C8
C10
C11
C12P8O
10
5
7
89
11
12
1
2 34
6P1
P2
P3 P4P5 P6 P7 P9
P10
P11
P12
Directing Circle
Rolling Circle
HYPOCYCLOID
INVOLUTE
DEFINITION :- If a straight line is rolled
round a circle or a polygon without
slipping or sliding, points on line will
trace out INVOLUTES.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.
PROB:
A string is unwound from a circle of 20 mm diameter. Draw the locus of string P for unwounding the string’s one turn. String is kept tight during unwound. Draw tangent & normal to the curve at any point.
P12
P2
002 12
6
P11 20 9 103 4 6 8 115 7 12
DP3
P4
P5
P6
P7
P8
P9
P10
P11
123
45 7
8910
11
03
04
05
06
07
08
09
010`
011
Tangent
N
N
Norm
alT
T.
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=7toPR=6toP
R21
0
0 1 2 3 4 5 6 7 8 P11 0
1
2
3
456
7
8
9
10
P1
P2
P3
P4
P5
P6
P7
P8
L= 100 mm
R=1toP
R=2
toPR
=3
toPR
=4toPR=5toP
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
45689
1112 1 2
P1
P2
P3P4
P5
P6
P7
P8 P9 P10
P14P
L=160
mm
R=21mm
64 5 7 8 9 101112131415
ø
PROBLEM:-
Draw an involute of a pantagon having side as 20 mm.
P5
R=01
R=201
P0
P1
P2
P3
P4
R=3
01
R=401
R=501
23
4 51
T
TN
NS
INVOLUTEOF 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
N
NS
R=3
01
R=401
R=201
R=0
1
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
R2130°
R21
60°
A
B
C
30°X
X+A1
XX+
A2
X+
A3
X+A5
X+
A4
X+AB
R
=X+AB
X+
66+
BC
12
3
45
C0
C1
C2C3
C4
C5
C6
C7
C8
A stick of length equal to the circumference of a
semicircle, is initially tangent to the semicircle
on the right of it. This stick now rolls over the
circumference of a semicircle without sliding till
it becomes tangent on the left side of the
semicircle. Draw the loci of two end point of this
stick. Name the curve. Take R= 42mm.
PROBLEM:-
A6
B6
5
A
B
C
B1
A1
B2
A2
B3A3
B4
A4
B5
A5
12 3
4
5
O
1
2
3
4
6
INVOLUTE
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 Circle Logarithmic
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
23
4
5
6
7
89
10
11
120
8 7 012345691112
P1
P2P3
P4
P5
P6
P7P8
P9
P11P12
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
1011
2
1
34
5
6
7
89
10
11
12
P1
P2P3
P4
P5
P6
P7
P8 P9
P10
P11
P12
O
N
N
T
T
S
R m
in
R max
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 slotted link, shown in fig rotates in the horizontal plane about a fixed point O, while a block is free to slide in the slot. If the center point P, of the block moves from A to B during one revolution of the link, draw the locus of point P. OAB
40 25
B A O1234567891011
P1
P2
P3 P4
P5
P6
P7
P8
P9P10
P11
P12
11
21
31
41
51
61
71
81
91
101
111
40 25
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
A0
Linear Travel of point P on AB
= 96 =16x (6 div.)
EXAMPLE: A link AB, 96mm long initially is vertically upward w.r.t. its pinned end B, swings in clockwise direction for 180° and returns back in anticlockwise direction for 90°, during which a point P, slides from pole B to end A. Draw the locus of point P and name it. Draw tangent and normal at any point on the path of P.
P1’
A
B
A1
A2
A3
A4
A5
A6P0
P1
P2
P3
P4
P5
P6
P2’
P3’
P4’P5’
P6’
9 6
Link AB = 96
C
Tangent
Angular Swing of link AB = 180° + 90°
= 270 ° =45 °X 6 div.
ARCHIMEDIAN SPIRAL
D
NO
RM
AL
M
N
Arch.Spiral Curve Constant BC
= Linear Travel ÷Angular Swing in Radians
= 96 ÷ (270º×π /180º)
=20.363636 mm / radian
Problem : 2
Draw a cycloid for a rolling circle, 60
mm diameter rolling along a straight
line without slipping for 540°
revolution. Take initial position of the
tracing point at the highest point on
the rolling circle. Draw tangent &
normal to the curve at a point 35 mm
above the directing line.
First Step : Draw a circle having diameter of 60 mm.
Second step: Draw a straight line tangential to the circle from bottom horizontally equal to
(540 x ) x 60 mm= 282.6 mm i.e. 1.5 x x 60 mm 360
Third step : take the point P at the top of the circle.
Rolling circle
P1
P2
P3
P4
P0
P6
P7
P8
P5
P9
P10
1
2
34
5
6
7 89
10
1 2 3 4 5 6 7 8 9 100
C0 C1 C2C3 C4
Directing line
Length of directing line = 3D/2
540 = 360 + 180
540 = D + D/2 Total length for 540 rotation = 3D/2
C5 C6 C7C8 C9 C10
S
norm
al