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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.

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CHAPTER – 2 ENGINEERING CURVES

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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

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1. CONICS 2. CYCLOIDAL CURVES 3. INVOLUTE 4. SPIRAL 5. HELIX 6. SINE & COSINE CLASSIFICATION OF ENGG. CURVES

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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

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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

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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

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B - CIRCLE A - TRIANGLE CONICS C - ELLIPSE D – PARABOLA E - HYPERBOLA

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When the cutting plane contains the apex, we get a triangle as the section. TRIANGLE

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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

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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

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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

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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. HYPERBOLA Definition :- α < θ

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CONICS Definition :- 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 F V P Focus Conic Curve Directrix

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The line passing through focus & perpendicular to directrix is called as axis. The intersection of conic curve with axis is called as vertex. Axis M C F V P Focus Conic Curve Directrix Vertex

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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 C V Focus Conic Curve Directrix Vertex

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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 C F V Axis Focus Ellipse Directrix Eccentricity=PF/PM = QF/QN < 1.

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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 F1F1 A B P F2F2 O Q C D

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F1F1 A B C D P F2F2 O PF 1 + PF 2 = QF 1 + QF 2 = CF 1 +CF 2 = constant = Major Axis Q = F 1 A + F 1 B = F 2 A + F 2 B But F 1 A = F 2 B F 1 A + F 1 B = F 2 B + F 1 B = AB CF 1 +CF 2 = AB but CF 1 = CF 2 hence, CF 1 =1/2AB

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F1F1 F2F2 O A B C D Major Axis = 100 mm Minor Axis = 60 mm CF 1 = ½ AB = AO F1F1 F2F2 O A B C D Major Axis = 100 mm F 1 F 2 = 60 mm CF 1 = ½ AB = AO

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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.

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Ratio (known as eccentricity) of its distances from focus to that of directrix is constant and equal to one (1). PARABOLA 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. Definition :- Directrix Axis Vertex M C N Q F V P Focus Parabola Eccentricity = PF/PM = QF/QN = 1.

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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

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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 Axis Directrix Hyperbola M C N Q F V P Focus Vertex HYPERBOLA = QF/QN > 1.

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Nature of graph of Boyles law Shape of overhead water tanks Uses :- Shape of cooling towers etc.

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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 Circles Method Arc of Circles Method

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Normal P 2 R =A1 Tangent 1234 A B C D P1P1 P3P3 P2P2 P4P4 P4P4 P3P3 P2P2 P1P1 P 1 F2F2 P 3 P 4 P 3 P 2 P 1 90° F1F1 Rad =B1 R=B2 `R=A2 O ARC OF CIRCLES METHOD

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Axis Minor A B Major Axis 7 8 9 10 11 9 8 7 6 5 4 3 2 1 12 11 P6P6 P5P5 P4P4 P3P3 P 2` P1P1 P 12 P 11 P 10 P9P9 P8P8 P7P7 6 5 4 3 2 1 12 C 10 O CONCENTRIC CIRCLE METHOD F2F2 F1F1 D CF 1 =CF 2 =1/2 AB T N Q e = AF 1 /AQ

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N o r m a l 01 2 3 4 12 3 4 10 23 41 2 3 4AB C D Major Axis M i n o r A x i s F1F1F1F1 F2F2F2F2D i r e c t r i xEF S P P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 T a n g e n t P 1 P 1 P 2 P 2 P 3 P 3 P 4 P 4 Ø Ø R = A B / 2 P0P0P0P0 P 1 P 2 P 3 P 4 P4P4 P3P3 P2P2 P1P1 OBLONG METHOD

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B A P4P4 P0P0 D C 60 ° 6 5 4 3 2 1 0 5 4 3 2 10 1 2 3 4 5 6 5 3 2 1 0P1P1 P2P2 P3P3 Q1Q1 Q2Q2 Q3Q3 Q4Q4 Q5Q5 P6P6 Q6Q6 O 4 ELLIPSE IN PARALLELOGRAM R4R4 R3R3 R2R2 R1R1 S1S1 S2S2 S3S3 S4S4 P5P5 G H I K J Minor Axis Major Axis

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P6P6 Normal P 5 P 7 P 6 P1P1 Tangent P 1 N N T T V1V1 P5P5 P 4 P4P4 P 3 P 2 F1F1 D1D1 D1D1 R1R1 b a c d e f g Q P7P7 P3P3 P2P2 Directrix R=6f` 90 ° 12 3456 7 Eccentricity = 2/3 3 R1V1R1V1 QV 1 = R1V1R1V1 V1F1V1F1 = 2 Ellipse ELLIPSE – DIRECTRIX FOCUS METHODR=1a Dist. Between directrix & focus = 50 mm 1 part = 50/(2+3)=10 mm V 1 F 1 = 2 part = 20 mm V 1 R 1 = 3 part = 30 mm < 45º S

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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.

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ARC OF CIRCLES METHOD Normal G 2 R =A1 Tangent 1234 A B G G G1G1 G3G3 G2G2 G4G4 G4G4 G3G3 G2G2 G1G1 G 1 G 3 G 4 G 3 G 2 G 1 F2F2 F1F1 R=B1 R=B2 `R=A2 O 90° directrix100140 GF1 + GF2 = MAJOR AXIS = 140 E e AF1 AF1AE e = R=70 R=70

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PROBLEM :-3 Two 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.

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O AB C 75 45 1 D 100 1 2 2 3 3 4 4 5 5 6 6 7 7 P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 P8P8 E 8 8

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PROBLEM :-5 ABCD 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.

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I3I3 C D F1F1 F2F2 P Q R S 50 I1I1 I4I4 A B I2I2 O 1 1 2 2 4 4 3 3 100

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PROBLEM :-1 Three 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.

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AB P R = 50 M N O 1 2 12 60 80 Q 1 2 1 2 P1P1 P2P2 Q2Q2 Q1Q1 R1R1 R2R2 S2S2 S1S1

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PROBLEM :-2 Draw 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.

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ELLIPSE P Q R 80mm 40mm 89mm A C B D MAJOR AXIS = PQ+QR = 129mm R=AB/2 1 2 3 12 3 12 3 O3O3O3O3 O 3 O 3 O1O1O1O1 O2O2O2O2 O 2 O 2 O 1 O 1 TANGENT NORMAL 60º 30º 231 directrix F1F1F1F1 F2F2F2F2 MAJOR AXIS MINOR AXIS S ECCENTRICITY = AP / AS ? ?

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PROBLEM :-4 Two 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.

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D C 6 6 5 4 3 2 1 0 5 4 3 21 0 1234 5 6 6 5 3 2 1 0 P2P2 P3P3 P4P4 P5P5 Q1Q1 Q2Q2 Q3Q3 Q4Q4 Q5Q5 B A O 4 ELLIPSE 100 45 75 P0P0 P1P1 P6P6 Q6Q6 G H I K J

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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.

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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.

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METHODS FOR DRAWING PARABOLA 1. Rectangle Method 2. Parabola in Parallelogram 3. Tangent Method 4. Directrix Focus Method

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2345 0 1 2 3 4 5 6 115432 6 0 1 2 3 4 5 0 V DCAB P4P4P4P4 P4P4P4P4 P5P5P5P5 P5P5P5P5 P3P3P3P3 P3P3P3P3 P2P2P2P2 P2P2P2P2 P6P6P6P6 P6P6P6P6 P1P1P1P1 P1P1P1P1 PARABOLA –RECTANGLE METHOD PARABOLA

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B0 2 0 2 6 C 6 V 5P5 30° AXD 1 2 4 5 3 1 3 4 5 4 3 1 0 5 4 3 2 1 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5P4 P3 P2 P1 P6 PARABOLA – IN PARALLELOGRAM P6

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B A O V 1 8 3 4 5 2 6 7 9 10 0 1 2 3 4 5 6 7 8 9 0 F PARABOLA TANGENT METHOD

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D D DIRECTRIX 90° 2 34 T T N N S V 1 P1P1 P2P2 PFPF P3P3 P4P4 P 1 P 2 P 3 P 4 P F AXIS RF R2 R1 R3 R4 90° R F PARABOLA DIRECTRIX FOCUS METHOD

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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.

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6m ROOT OF TREE BUILDING REQD.DISTANCE TOP OF TREE 3m 6m FA STONE FALLS HERE

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3m 6m ROOT OF TREE BUILDING REQD.DISTANCE GROUND TOP OF TREE 3m 6m 12 3 12 3 321 4 56 56 E FAB C D P3P3P3P3 P4P4P4P4 P2P2P2P2 P1P1P1P1 P P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 3 2 10 STONE FALLS HERE

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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.

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150 mm ABC D 1 2 3 4 5 1 2 3 45O P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5M123 4 5 1 2 3 4 5 P 1 P 1 P 2 P 2 P 3 P 3 P 4 P 4 P 5 P 5 90 mm

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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

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ground level B A 60º gunshot 80 M parabola

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ground level B A O V 1 8 3 4 5 2 6 7 9 10 0 1 2 3 4 5 6 7 8 9 0 F 60ºgunshot D D VFVE = e = 1 E

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Connect two given points A and B by a Parabolic curve, when:- 1.OA=OB=60mm and angle AOB=90° 2.OA=60mm,OB=80mm and angle AOB=110° 3.OA=OB=60mm and angle AOB=60°

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60 60 1 2 3 4 5 Parabola 5 4 3 2 1 A B 90 ° O 1.OA=OB=60mm and angle AOB=90°

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A B 80 60 5 4 3 2 1 1 2 3 4 5 ° 110 ° ParabolaO 2.OA=60mm,OB=80mm and angle AOB=110°

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54 3 2 1 5 4 3 2 1 A B O 60 60 60 ° 3.OA=OB=60mm and angle AOB=60°

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example Draw a parabola passing through three different points A, B and C such that AB = 100mm, BC=50mm and CA=80mm respectively.

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BAC 100 50 80

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1 0 0 2 0 2 6 6 5 P 5 1 2 4 5 3 1 3 4 5 4 3 5 4 3 2 1 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P 4 P 3 P 2 P 1 P 6 P 6 AB C

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METHODS FOR DRAWING HYPERBOLA 1. Rectangle Method 2. Oblique Method 3. Directrix Focus Method

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D F 12 345 5 4 3 2 P1P1 P2P2 P3P3 P4P4 P5P5 0 P6P6 P0P0 A O E X B C Y Given Point P 0 90° 6 6 Hyperbola RECTANGULAR HYPERBOLA AXIS When the asymptotes are at right angles to each other, the hyperbola is called rectangular or equilateral hyperbola ASYMPTOTES X and Y

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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.

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D F 12 345 5 4 3 2 P1P1 P2P2 P3P3 P4P4 P5P5 0 P6P6 P0P0 A O E X=20 B C Y = 50 Given Point P 0 90° 6 6 Hyperbola RECTANGULAR HYPERBOLA

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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.

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75 0 P4P4 E 6 2 1 P1P1 12 34 56 D P6P6 P5P5 P3P3 P2P2 P0P0 7 P7P7 7 C B F O Y = 30 X = 20 Given Point P 0 A

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AXIS NORMAL C V F1F1F1F1 DIRECTRIX D D 1 2 3 4 4 3 2 1 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P 1 P 1 P 2 P 2 P 3 P 3 P 4 P 4 T1T1T1T1 T2T2T2T2 N2N2N2N2 N1N1N1N1TANGENT s Directrix and focus method

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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. Superior Hypotrochoid Cycloidal Curves CycloidEpy Cycloid Hypo Cycloid Superior Trochoid Inferior Trochoid Superior Epytrochoid Inferior Epytrochoid Inferior Hypotrochoid

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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

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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. 2πr2πr Ø = 360º x r/R d Circumference of Generating Circle RdRdRdRd Rolling Circle r O Ø/2 Ø/2 P0P0 P0P0 Arc P 0 P 0 = Rd x Ø= P0P0

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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.` Directing Circle(R) P Ø Ø /2 Ø Ø = 360 x r R R T Rolling Circle Radius (r) O Vertical Hypocycloid P P

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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.

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P0P0 2 R or D 5 T T 1 2 1 2 34 6 7 8 910 11 0 12 0 3 4 5 6 7 8 9 10 11 12 P1P1 P2P2 P3P3 P4P4 P5P5 P7P7 P8P8 P9P9 P 11 P 12 C0C0 C1C1 C2C2 C3C3 C4C4 C5C5 C6C6 C7C7 C8C8 C9C9 C 10 C 11 Directing Line C 12 N N S S1S1 R P6P6 R P 10 R : Given Data : Draw cycloid for one revolution of a rolling circle having diameter as 60mm. Rolling Circle D

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C0C0C0C0 P0P0P0P0 7 8 P6P6P6P6 4 P1P1P1P11 2 3 C2C2C2C2 C3C3C3C3 P2P2P2P2 C4C4C4C4 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 C1C1C1C1 P3P3P3P3 P4P4P4P4 P5P5P5P5 P7P7P7P7 P8P8P8P87 0 C5C5C5C5 C6C6C6C6 C7C7C7C7 C8C8C8C8 123 4 5 6 D /2 π D /2 D /2 Floor WallCYCLOID5 6 7 8 Take diameter of circle = 40mm Initially distance of centre of circle from the wall 83mm (Hale circumference + D/2)

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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.

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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.

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P6P6P6P6 P4P4P4P4 r P2P2P2P2 C1C1C1C1 C0C0C0C0 C2C2C2C2 C3C3C3C3 C4C4C4C4 C5C5C5C5 C6C6C6C6 C7C7C7C7 C8C8C8C8 1 0 2 3 4 5 6 7 O RdRdRdRd Ø/2 Ø/2 P1P1P1P1 P0P0P0P0 P3P3P3P3 P5P5P5P5 P7P7P7P7 P8P8P8P8 r r Rolling Circle r R d X Ø = 2πr Ø = 360º x r/Rd Arc P 0 P 8 = Circumference of Generating Circle EPICYCLOID GIVEN: Rad. Of Gen. Circle (r) & Rad. Of dir. Circle (R d ) S º U N Ø = 360º x 25/75 = 120° = 120°

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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.

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P0P0 P1P1 Tangent P 11 r C0C0 C1C1 C2C2 C3C3 C4C4 C5C5 C6C6 C7C7 C8C8 C9C9 C 10 C 11 C 12 P 10 P8P8 0 1 2 3 4 5 6 7 8 9 10 11 12 P2P2 P3P3 P4P4 P5P5 P6P6 P9P9 P7P7 P 12 /2 = 360 x 4 12 = 360 x r R = 120° R T T N S N Normal r r Rolling Circle Radias (r) Directing Circle O Vertical Hypocycloid

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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

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C C1C1 C2C2 C3C3 C4C4 C5C5 C6C6 C7C7 C9C9 C8C8 C 10 C 11 C 12 P8P8 O 10 5 7 8 9 11 12 1 2 3 4 6 P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 P9P9 P 10 P 11 P 12 Directing Circle Rolling Circle HYPOCYCLOID

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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.

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PROB: A string is unwound from a circle of 20 mm diameter. Draw the locus of string P for unwounding the strings one turn. String is kept tight during unwound. Draw tangent & normal to the curve at any point.

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P 12 P2P2 0 02 12 6 P1P1 120910 3 4681157 12 D P3P3 P4P4 P5P5 P6P6 P7P7 P8P8 P9P9 P 10 P 11 1 2 3 4 5 7 8 9 10 11 03 04 05 06 07 08 09 010` 011 Tangent N N Normal T T.

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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.

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P R=7toP R=6toP R21 001 234 567 8 P 11 0 1 2 3 4 567 8 9 10 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 P7P7P7P7 P8P8P8P8 L= 100 mm R=1toP R=2toP R=3toP R=4toP R=5toP INVOLUTE9 ø 11 mm = 30° Then 5 mm = Ø = 30° x 5 /11 = 13.64 ° S = 2 x π x r /12

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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.

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P 13 P 11 3 13 14 15 P0P0P0P0 P 12 O 7 10 1 2 3 4 5 6 8 9 11 12 1 2 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 P7P7P7P7 P8P8P8P8 P9P9P9P9 P 10 P 14 P L=160 mm R=21mm 6 45 78 9 1011 12 13 14 15 ø

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PROBLEM:- Draw an involute of a pantagon having side as 20 mm.

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P5P5 R=01 R=2 01 P0P0 P1P1 P2P2 P3P3 P4P4 R=3 01 R=4 01 R=5 01 2 3 4 5 1 T T N N S INVOLUTE OF A POLYGON Given : Side of a polygon 0

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PROBLEM:- Draw an involute of a square having side as 20 mm.

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P2P2 1 2 3 0 4 P0P0 P1P1 P3P3 P4P4 N N S R=3 01 R=4 01 R=2 01 R=01 INVOLUTE OF A SQUARE

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PROBLEM:- Draw an involute of a string unwound from the given figure from point C in anticlockwise direction. 60° ABC R21 30°

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R21 60° ABC 30° X X+A1 X X+A2 X+A3 X+A5 X+A4 X+AB R =X+AB X+66+BC 1 2 3 45 C0C0 C1C1 C2C2 C3C3 C4C4 C5C5 C6C6 C7C7 C8C8

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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:-

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A6A6 B6B6 5 A B C B1B1 A1A1 B2B2 A2A2 B3B3 A3A3 B4B4 A4A4 B5B5 A5A5 1 2 3 4 5 O 1 2 3 4 6 INVOLUTE

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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.

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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

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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.

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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:

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P 10 1 2 3 4 5 6 7 8 9 10 11 12 0 8 7 0 12 3 4 569 11 12 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 P7P7P7P7 P8P8P8P8 P9P9P9P9 P 11 P 12 o

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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

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31 26 5 8 4 79 10 11 2 1 3 4 5 6 7 8 9 10 11 12 P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 P8P8 P9P9 P 10 P 11 P 12 O N N T T S R min R max Diff. in length of any two radius vectors Angle between them in radians Constant of the curve = = OP – OP 3 Π/2 100 – 90 = Π/2 = 6.37 mm

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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. O AB 40 25

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BAO 1 2 3 4 5 6 7 8 9 10 11 P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 P8P8 P9P9 P 10 P 11 P 12 1 2121 3131 4141 5151 6161 7171 8181 9191 10 1 11 1 4025

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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.

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A Initial Position of point P POPOPOPO P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 P7P7P7P7 P8P8P8P8 213 4 5 6 7 O12 3 4 5 6 7 8 2/3 X 360° = 240°120º

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A0A0A0A0 Linear Travel of point P PP P on A AA AB = 96 =16x (6 div.) AB B PB A P P 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. P 1 P 1 AB A1A1A1A1 A2A2A2A2 A3A3A3A3 A4A4A4A4 A5A5A5A5 A6A6A6A6 P0P0P0P0 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 P 2 P 2 P 3 P 3 P 4 P 4 P 5 P 5 P 6 P 6 96 Link AB = 96 C T a n g e n t Angular Swing of link AB = 180° + 90° = == = 270 ° =45 °X 6 div. ARCHIMEDIAN SPIRAL DN O R M A L MN

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Arch.Spiral Curve Constant B BB BC = Linear Travel ÷Angular Swing in Radians = 96 ÷ (270º×π /180º) =20.363636 mm / radian

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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.

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θ o 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1819 20 21 22 2324 23 13 22 24 1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 18 19 20 21 17 P3P3 P9P9 P 15

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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.

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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.

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Rolling circle P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P0P0P0P0 P6P6P6P6 P7P7P7P7 P8P8P8P8 P5P5P5P5 P9P9P9P9 P 10 1 2 3 4 5 6 7 8 9 10 1 23 4 5 6 7 8 9 10 0 C0C0C0C0 C1C1C1C1 C2C2C2C2 C3C3C3C3 C4C4C4C4 Directing line Length of directing line = 3 D/2 540 = 360 + 180 540 = 360 + 180 540 = D + D/2 Total length for 540 rotation = 3 D/2 C5C5C5C5 C6C6C6C6 C7C7C7C7 C8C8C8C8 C9C9C9C9 C 10 S normal

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