Presentation on theme: "Engineering drawing is a language of all persons involved in engineering activity. Engineering ideas are recorded by preparing drawings and execution."— Presentation transcript:
1 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”.
3 USES OF 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.
4 CLASSIFICATION OF ENGG. CURVES 1. CONICS2. CYCLOIDAL CURVES3. INVOLUTE4. SPIRAL5. HELIX6. SINE & COSINE
5 What is Cone ?It is a surface generated by moving a Straight line keeping one of its end fixed & other end makes a closed curve.The fixed point is known as vertex or apex.Vertex/ApexThe closed curve is known as base.90ºIf the base/closed curve is a circle, we get a cone.there is no objectionIf the base/closed curve is a polygon, we get a pyramid.Base
6 The line joins apex to the center of base is called axis. If axes is perpendicular to base, it is called as right circular cone.If axis of cone is not perpendicular to base, it is called as oblique cone.BaseVertex/ApexCone AxisGeneratorThe line joins vertex/ apex to the circumference of a cone is known as generator.90º
7 CONICSDefinition :- 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.
9 TRIANGLEWhen the cutting plane contains the apex, we get a triangle as the section.
10 CIRCLEWhen the cutting plane is perpendicular to the axis or parallel to the base in a right cone we get circle the section.Sec PlaneCircle
11 α > θ ELLIPSE 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.αθα > θ
12 PARABOLAWhen 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.α = θθα
13 α = 0 α < θ HYPERBOLA Definition :- 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.α = 0θα < θ
14 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.Conic CurveMPDirectrixFCVFocusFixed straight line is called as directrix.Fixed point is called as focus.
15 The intersection of conic curve with axis is called as vertex. The line passing through focus & perpendicular to directrix is called as axis.The intersection of conic curve with axis is called as vertex.MCFVPFocusConic CurveDirectrixAxisVertex
16 = Eccentricity Distance of a point from focus Ratio = AxisCVFocusConic CurveDirectrixVertexMPNQDistance of a point from focusRatio =Distance of a point from directrix= Eccentricity= PF/PM = QF/QN = VF/VC = e
17 ELLIPSEEllipse 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.EllipsePMAxisDirectrixVertexFCVFocusEccentricity=PF/PM = QF/QN< 1.NQ
18 ELLIPSEEllipse 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.F1ABPF2OQCD
20 Major Axis = 100 mm Minor Axis = 60 mm CF1 = ½ AB = AO DMajor Axis = 100 mmMinor Axis = 60 mmOABF1F2CF1 = ½ AB = AOCDMajor Axis = 100 mmF1F2 = 60 mmOABCF1 = ½ AB = AOF1F2
21 Shape of tank in a tanker. Uses :-Shape of a man-hole.Shape of tank in a tanker.Flanges of pipes, glands and stuffing boxes.Shape used in bridges and arches.Monuments.Path of earth around the sun.Shape of trays etc.
22 PARABOLA Definition :- The 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.Ratio (known as eccentricity) of its distances from focus to that of directrix is constant and equal to one (1).MParabolaPDirectrixAxisVertexFCVEccentricity = PF/PM = QF/QN= 1.FocusNQ
23 Uses :- Motor car head lamp reflector. Sound reflector and detector. Bridges and arches constructionShape of cooling towers.Path of particle thrown at any angle with earth, etc.Home
24 HYPERBOLAIt 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.AxisDirectrixHyperbolaMCNQFVPFocusVertexEccentricity = PF/PM= QF/QN> 1.
25 Uses :- Nature of graph of Boyle’s law Shape of overhead water tanks Shape of cooling towers etc.
26 METHODS FOR DRAWING ELLIPSE 1. Arc of Circle’s Method2. Concentric Circle Method3. Loop Method4. Oblong Method5. Ellipse in Parallelogram6. Trammel Method7. Parallel Ellipse8. Directrix Focus Method
28 CONCENTRIC CIRCLE METHOD 10119CP10128P11P910N11AxisMinor9P12P8T128P7Major AxisQ1P17ABF11O7F26P6P2`253P34P526P4De = AF1/AQCF1=CF2=1/2 AB354
29 Directrix 1 2 3 4 4’ 3’ 2’ 1’ C E F Normal A F1 F2 B Tangent D OBLONG METHODP4CP1’P2’P3’P4’EF44’P3Normal33’DirectrixP2Minor Axis2SR=AB/22’P1Ø11’P0Major Axis0’A1F12344’3’2’F21’BTangentP1’’P2’’P3’’P4’’P4P3P2P1PD
30 ELLIPSE IN PARALLELOGRAM DCHP1Q1P01P21Q2K2P32Q33Q4P434Q545P5Minor AxisP65BAQ6O654321123456Major AxisS4R4GJS3R360°S2R2S1IR1
31 ELLIPSE – DIRECTRIX FOCUS METHOD gf < 45ºeEllipseD1dcEccentricity = 2/3bP5P6P7P4aP3P23R1V1QV1=V1F12QP1DirectrixR=6f`R=1a1234567R1V1F1Dist. Between directrix & focus = 50 mm90°P1’TTangentP2’N1 part = 50/(2+3)=10 mmP3’P4’P5’P6’P7’SV1F1 = 2 part = 20 mmTV1R1 = 3 part = 30 mmNormal
32 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.
33 R=B1 R =A1 F2 F1 A B 1 2 3 4 `R=A2 R=B2 Tangent Normal ARC OF CIRCLE’S METHODGG’G4G4AF1AEe =eG3G3directrixG2G2G1R=70G1R=70R=B1R =A1F2F1EOAB14090°1001234`R=A2R=B2G1’G1’TangentG2’G2’90°NormalG3’G3’G4’G4’GF1 + GF2 = MAJOR AXIS = 140
34 PROBLEM :-3Two points A & B are 100 mm apart. A point C is 75 mm from A and 45 mm from B. Draw an ellipse passing through points A, B, and C so that AB is a major axis.
36 PROBLEM :-5ABCD is a rectangle of 100mm x 60mm. Draw an ellipse passing through all the four corners A, B, C and D of the rectangle considering mid – points of the smaller sides as focal points.Use “Concentric circles” method and find its eccentricity.
38 PROBLEM :-1Three points A, B & P while lying along a horizontal line in order have AB = 60 mm and AP = 80 mm, while A & B are fixed points and P starts moving such a way that AP + BP remains always constant and when they form isosceles triangle, AP = BP = 50 mm. Draw the path traced out by the point P from the commencement of its motion back to its initial position and name the path of P.
40 PROBLEM :-2Draw an ellipse passing through 60º corner Q of a 30º - 60º set square having smallest side PQ vertical & 40 mm long while the foci of the ellipse coincide with corners P & R of the set square.Use “OBLONG METHOD”. Find its eccentricity.
41 ELLIPSE C Q S A B P D directrix O3 O3’ 3 2 3 1 O2 60º O2’ 2 O1 O1’ 1 TANGENTdirectrixNORMALO3O3’QR=AB/23231MINOR AXISO260ºO2’2ELLIPSE40mmO189mmO1’1F1F2S80mmMAJOR AXIS30ºABP1’’2’’3’’?R1’2’3’DMAJOR AXIS = PQ+QR = 129mmECCENTRICITY = AP / AS
42 PROBLEM :-4Two points A & B are 100 mm apart. A point C is 75 mm from A and 45 mm from B. Draw an ellipse passing through points A, B, and C so that AB is not a major axis.
43 ELLIPSE H 75 45 K 100 J G I C P1 P2 1 P0 Q1 1 P3 2 Q2 2 Q3 3 P4 3 Q4 4 HP145P21P0Q11P32Q22Q3K3P43Q444P5Q5556OQ6P6BA6100654321123456JGID
44 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.
45 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.
46 METHODS FOR DRAWING PARABOLA 1. Rectangle Method2. Parabola in Parallelogram3. Tangent Method4. Directrix Focus Method
51 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.
52 A F 6m BUILDING 6m 3m TOP OF TREE STONE FALLS HERE ROOT OF TREE REQD.DISTANCE
53 TOP OF TREEDCP6mP1P1P2P3P4123P2123P3BUILDINGFABP43216m3m32145656P5STONE FALLS HEREROOT OF TREEEREQD.DISTANCE3mGROUNDP6
54 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.
55 A B C O 90 mm M 150 mm D 1’ 2’ 3’ 4’ 5’ P1 P2 P3 P4 P5 P1’ P2’ P3’ P4’
56 EXAMPLEA 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
67 RECTANGULAR HYPERBOLA When the asymptotes are at right angles to each other, the hyperbola is called rectangular or equilateral hyperbolaBF6’P6Given Point P0AXIS612345CDP0P1P22’P33’HyperbolaP44’P5Y5’ASYMPTOTES X and YAOXE90°AXIS
68 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”.
69 RECTANGULAR HYPERBOLA F6’P6Given Point P0612345CDP0P1P22’P33’HyperbolaP44’P5Y = 505’AO90°X=20E
70 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”.
71 Oblique Hyperbola B F X = 20 P7 7’ Given Point P0 7 1 2 3 4 5 6 C D P0 1’P12’P2Y = 30P3P4P5P66’OAE750
72 Directrix and focus method 4’DIRECTRIXDP43’T2P32’P2N2N11’sP1NORMALTANGENTAXISCV1F1234T1P1’P2’P3’P4’
73 CYCLOIDAL GROUP OF CURVES When 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 CurvesCycloidEpy CycloidHypo CycloidSuperiorHypotrochoidInferiorTrochoidSuperiorTrochoidInferiorHypotrochoidInferiorEpytrochoidSuperiorEpytrochoid
74 Rolling Circle or Generator Directing Line or Director 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.Rolling Circle or GeneratorPRCCPPDirecting Line or Director
75 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.OØ/2P0Rolling CircleDirecting CirclerP0P0Rd x Ø=2πrRdCircumference of Generating CircleArc P0P0Ø = 360º x r/Rd=
76 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)PØ /2Ø =360 x rRTRolling CircleRadius (r)OVerticalHypocycloid
77 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.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.
78 : Given Data :Draw cycloid for one revolution of a rolling circle having diameter as 60mm.DRolling CircleTN6P65P57P7SRP4P848RP9P39C0C1C2C3C4C5C6C7C8C9C10C11C123S1RP22P1010P1P1112111P0P12123456789101112Directing Line2R or D
79 Take diameter of circle = 40mm 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.C88D/2πD/2C7Take diameter of circle = 40mmInitially distance of centre of circle from the wall 83mm (Hale circumference + D/2)7WallP8C6P7P6P564P48CYCLOIDC5753P35P226C0C1C2C3C46P1715P0πD/2D/2Floor1234
80 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.
81 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.
82 S U N Ø = 360º x 25/75 = 120° Rd X Ø = 2πr Ø = 360º x r/Rd EPICYCLOIDGIVEN:Rad. Of Gen. Circle (r)& Rad. Of dir. Circle (Rd)P4SP3P5rUrRolling CircleC4C3C5P2C2C6P632NC1C74r1Directing CircleP1rC0C85P0P8P767Ø/2Ø/2OØ = 360º x 25/75 = 120°Arc P0P8 = Circumference of Generating CircleRdRd X Ø = 2πrØ = 360º x r/Rd
83 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.
84 Hypocycloid r = Directing Circle Vertical Rolling Circle Radias (r) 2rNormal1r3P0P12P1rP2P114P1012P3TP8P9P5P6P7P4STangent11510N6 /2798 =360 x 412360 x rR120°ORHypocycloid
85 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
86 10 5 7 8 9 11 12 1 2 3 4 6 C O Directing Circle Rolling Circle P2P11P3P4P7P9HYPOCYCLOIDP1CP5OP8P10P12P6
87 INVOLUTEDEFINITION :- If a straight line is rolled round a circle or a polygon without slipping or sliding, points on line will trace out INVOLUTES.ORInvolute 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.Uses :- Gears profile
88 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.
89 P9 T P8 P10 09 N 08 010` P7 07 P11 Normal 011 Tangent 06 P6 5 6 4 7 8 053.92P51004111NP12120302129103468115712P1P4P2P3D
90 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 mmCircumference of the circle = 2 π r= 2 x π x 21 = 132 mmSo, the length of the string is less than circumference of the circle.
91 11 mm = 30° Then 5 mm = Ø = 30° x 5 /11 = 13.64 ° P3 P4 P2 INVOLUTE R=3toPP2INVOLUTER=4toPR=2toPP5R=5toPP1P6R=6toP1112345678910R=1toPR=7toPP7øP8PR2112345678PL= 100 mm9S = 2 x π x r /12
92 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 mmCircumference of the circle = 2 π r= 2 x π x 21 = 132 mmSo, the length of the string is more than circumference of the circle.
94 PROBLEM:-Draw an involute of a pantagon having side as 20 mm.
95 INVOLUTE OF A POLYGON Given : Side of a polygon P3 R=301 P2 R=201 45P1P4P0NR=501TSP5
96 PROBLEM:-Draw an involute of a square having side as 20 mm.
97 INVOLUTE OF A SQUARE R=301 P1 R=01 P0 R=201 P4 1 4 2 3 P2 R=401 S N R=201P41423P2R=401R=301NSP3
98 PROBLEM:-Draw an involute of a string unwound from the given figure from point C in anticlockwise direction.60°ABCR2130°
99 C8 C7 C6 C0 C5 C1 C2 C4 C3 B 5 C 4 3 2 X 1 A X+66+BC R =X+AB X+AB 60° 30°X+AB12345C6X+A5XX+A4X+A3C0XX+A2X+A1C5C1C2C4C3
100 PROBLEM:-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.
101 INVOLUTE B A6 6 A5 B1 5 4 A4 B2 3 A3 B3 2 3 2 4 1 A2 1 5 B4 A C A1 B5
102 SPIRALSIf 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.
103 Spiral for Clock Semicircle Quarter Circle 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.SpiralArche MedianSpiral for Clock Semicircle Quarter CircleLogarithmic
104 ARCHEMEDIAN SPIRALIt 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.
105 PROBLEM: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 = mm &Shortest radius = mm ( at centre or at pole)
107 To construct an Archemedian Spiral of one convolutions, given the greatest & shortest(least) radii. To construct an Archemedian Spiral of one convolutions, given the largest radius vector & smallest radius vector.Say Greatest radius = 100 mm &Shortest radius = mm
108 Diff. in length of any two radius vectors Constant of the curve =Angle between them in radians32OP – OP34P3P2=TP4Π/215P1100 – 90=Π/2P5R minN1086426P1212=6.37 mmP6O1197531SP11P7P10117R maxP8P98109
109 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.2540BAO
111 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.
112 Initial Position of point P 2/3 X 360°123456782134567= 240°P1P2120ºP8OP3P7P6P4P5
113 Angular Swing of link AB = 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.Angular Swing of link AB =AA0180° + 90°P696= 270 °P5=45 °X 6 div.A1P4P3ARCHIMEDIAN SPIRALP2P1P1’A6CP0P2’A2BP6’NORMALP3’MNP5’P4’A3Link AB = 96DA5Linear Travel of point P on AB= 96 =16x (6 div.)TangentA4
114 Arch.Spiral Curve Constant BC = Linear Travel ÷Angular Swing in Radians= 96 ÷ (270º×π /180º)=mm / radian
115 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.
117 Problem : 2Draw 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.
118 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= mm i.e. 1.5 x x 60 mm360Third step : take the point P at the top of the circle.
119 S normal Length of directing line = 3D/2 540 = 360 + 180 Rolling circleP0P8P18P7P9719SC0C3C4C8C9P106102C1C2P2P6C5C6C7C10P33P554P4Directing line12345678910Length of directing line = 3D/2540 = 360 + 180540 = D + D/2Total length for 540 rotation = 3D/2