Development of surfaces of solids

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Development of surfaces of solids

  1. 1. 1.1. SECTIONS OF SOLIDS.SECTIONS OF SOLIDS. 2.2. DEVELOPMENT.DEVELOPMENT. 3.3. INTERSECTIONS.INTERSECTIONS. ENGINEERING APPLICATIONSENGINEERING APPLICATIONS OFOF THE PRINCIPLESTHE PRINCIPLES OFOF PROJECTIONS OF SOLIDES.PROJECTIONS OF SOLIDES. STUDY CAREFULLYSTUDY CAREFULLY THE ILLUSTRATIONS GIVEN ONTHE ILLUSTRATIONS GIVEN ON NEXTNEXT SIXSIX PAGES !PAGES !
  2. 2. SECTIONING A SOLID.SECTIONING A SOLID. An object ( here a solid ) is cut byAn object ( here a solid ) is cut by some imaginary cutting planesome imaginary cutting plane to understand internal details of thatto understand internal details of that object.object. The action of cutting is calledThe action of cutting is called SECTIONINGSECTIONING a solida solid && The plane of cutting is calledThe plane of cutting is called SECTION PLANE.SECTION PLANE. wo cutting actions means section planes are recommendedwo cutting actions means section planes are recommended.. Section Plane perpendicular to Vp and inclined to Hp.Section Plane perpendicular to Vp and inclined to Hp. ( This is a definition of an Aux. Inclined Plane i.e. A.I.P.)( This is a definition of an Aux. Inclined Plane i.e. A.I.P.) NOTE:- This section plane appearsNOTE:- This section plane appears as a straight line in FV.as a straight line in FV. Section Plane perpendicular to Hp and inclined to Vp.Section Plane perpendicular to Hp and inclined to Vp. ( This is a definition of an Aux. Vertical Plane i.e. A.V.P.)( This is a definition of an Aux. Vertical Plane i.e. A.V.P.) NOTE:- This section plane appearsNOTE:- This section plane appears as a straight line in TV.as a straight line in TV. emember:-emember:- After launching a section planeAfter launching a section plane either in FV or TV, the part towards observereither in FV or TV, the part towards observer is assumed to be removed.is assumed to be removed. As far as possible the smaller part isAs far as possible the smaller part is assumed to be removed.assumed to be removed. OBSERVEROBSERVER ASSUMEASSUME UPPER PARTUPPER PART REMOVEDREMOVED SECTON PLANE SECTON PLANE IN FV. IN FV. OBSERVEROBSERVER ASSUMEASSUME LOWER PARTLOWER PART REMOVEDREMOVED SECTON PLANE SECTON PLANE IN TV. IN TV. (A)(A) (B)(B)
  3. 3. ILLUSTRATION SHOWINGILLUSTRATION SHOWING IMPORTANT TERMSIMPORTANT TERMS IN SECTIONING.IN SECTIONING. xx yy TRUE SHAPETRUE SHAPE Of SECTIONOf SECTION SECTIONSECTION PLANEPLANE SECTION LINESSECTION LINES (45(4500 to XY)to XY) Apparent ShapeApparent Shape of sectionof section SECTIONAL T.V.SECTIONAL T.V. For TVFor TV For True Shape For True Shape
  4. 4. Section PlaneSection Plane Through ApexThrough Apex Section PlaneSection Plane Through GeneratorsThrough Generators Section Plane ParallelSection Plane Parallel to end generator.to end generator. Section PlaneSection Plane Parallel to Axis.Parallel to Axis. TriangleTriangle EllipseEllipse Parabola Parabola HyperbolaHyperbola EllipseEllipse Cylinder throughCylinder through generators.generators. Sq. Pyramid throughSq. Pyramid through all slant edgesall slant edges TrapeziumTrapezium Typical Section PlanesTypical Section Planes && Typical ShapesTypical Shapes OfOf SectionsSections..
  5. 5. DEVELOPMENT OF SURFACES OF SOLIDS.DEVELOPMENT OF SURFACES OF SOLIDS. MEANING:-MEANING:- ASSUME OBJECT HOLLOW AND MADE-UP OF THIN SHEET. CUT OPEN IT FROM ONE SIDE ANDASSUME OBJECT HOLLOW AND MADE-UP OF THIN SHEET. CUT OPEN IT FROM ONE SIDE AND UNFOLD THE SHEET COMPLETELY. THEN THEUNFOLD THE SHEET COMPLETELY. THEN THE SHAPE OF THAT UNFOLDED SHEET IS CALLEDSHAPE OF THAT UNFOLDED SHEET IS CALLED DEVELOPMENT OF LATERLAL SUEFACESDEVELOPMENT OF LATERLAL SUEFACES OF THAT OBJECT OR SOLID.OF THAT OBJECT OR SOLID. LATERLAL SURFACELATERLAL SURFACE IS THE SURFACE EXCLUDING SOLID’S TOP & BASE.IS THE SURFACE EXCLUDING SOLID’S TOP & BASE. ENGINEERING APLICATIONENGINEERING APLICATION:: THERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BYTHERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BY CONVENTIONAL MANUFACTURING PROCESSES, BECAUSE OF THEIR SHAPES AND SIZES.CONVENTIONAL MANUFACTURING PROCESSES, BECAUSE OF THEIR SHAPES AND SIZES. THOSE ARE FABRICATED IN SHEET METAL INDUSTRY BY USINGTHOSE ARE FABRICATED IN SHEET METAL INDUSTRY BY USING DEVELOPMENT TECHNIQUE. THERE IS A VAST RANGE OF SUCH OBJECTS.DEVELOPMENT TECHNIQUE. THERE IS A VAST RANGE OF SUCH OBJECTS. EXAMPLES:-EXAMPLES:- Boiler Shells & chimneys, Pressure Vessels, Shovels, Trays, Boxes & Cartons, Feeding Hoppers,Boiler Shells & chimneys, Pressure Vessels, Shovels, Trays, Boxes & Cartons, Feeding Hoppers, Large Pipe sections, Body & Parts of automotives, Ships, Aeroplanes and many more.Large Pipe sections, Body & Parts of automotives, Ships, Aeroplanes and many more. WHAT ISWHAT IS OUR OBJECTIVEOUR OBJECTIVE IN THIS TOPIC ?IN THIS TOPIC ? To learn methods of development of surfaces ofTo learn methods of development of surfaces of different solids, their sections and frustumsdifferent solids, their sections and frustums.. 1. Development is different drawing than PROJECTIONS.1. Development is different drawing than PROJECTIONS. 2. It is a shape showing AREA, means it’s a 2-D plain drawing.2. It is a shape showing AREA, means it’s a 2-D plain drawing. 3. Hence all dimensions of it must be TRUE dimensions.3. Hence all dimensions of it must be TRUE dimensions. 4. As it is representing shape of an un-folded sheet, no edges can remain hidden4. As it is representing shape of an un-folded sheet, no edges can remain hidden And hence DOTTED LINES are never shown on development.And hence DOTTED LINES are never shown on development. But before going ahead,But before going ahead, note followingnote following ImportantImportant pointspoints.. Study illustrations given on next page carefully.Study illustrations given on next page carefully.
  6. 6. πD H D SS H L θ θ = R L + 3600 R=Base circle radius. L=Slant height. L= Slant edge. S = Edge of base L S S H= Height S = Edge of base H= Height D= base diameter Development of lateral surfaces of different solids. (Lateral surface is the surface excluding top & base) Prisms: No.of Rectangles Cylinder: A Rectangle Cone: (Sector of circle) Pyramids: (No.of triangles) Tetrahedron: Four Equilateral Triangles All sides equal in length Cube: Six Squares.
  7. 7. L L θ θ = R L + 3600 R= Base circle radius of cone L= Slant height of cone L1 = Slant height of cut part. Base side Top side L1 L1 L= Slant edge of pyramid L1 = Slant edge of cut part. DEVELOPMENT OF FRUSTUM OF CONE DEVELOPMENT OF FRUSTUM OF SQUARE PYRAMID STUDY NEXTSTUDY NEXT NINENINE PROBLEMS OFPROBLEMS OF SECTIONS & DEVELOPMENTSECTIONS & DEVELOPMENT FRUSTUMSFRUSTUMS
  8. 8. X Y X1 Y1 a’ b’ e’ c’ d’ A B C E D a e d b c TRUE SHAPE A B C D E A DEVELOPMENT a” b” c”d” e” Problem 1:Problem 1: A pentagonal prism , 30 mm base side & 50 mm axisA pentagonal prism , 30 mm base side & 50 mm axis is standing on Hp on it’s base with one side of the base perpendicular to VP.is standing on Hp on it’s base with one side of the base perpendicular to VP. It is cut by a section plane inclined at 40It is cut by a section plane inclined at 40ºº to the HP, through mid point of axis.to the HP, through mid point of axis. Draw Fv, sec.Tv & sec. Side view. Also draw true shape of section andDraw Fv, sec.Tv & sec. Side view. Also draw true shape of section and Development of surface of remaining solid.Development of surface of remaining solid. Solution Steps:Solution Steps:for sectional views:for sectional views: Draw three views of standing prism.Draw three views of standing prism. Locate sec.plane in Fv as described.Locate sec.plane in Fv as described. Project points where edges are gettingProject points where edges are getting Cut on Tv & Sv as shown in illustration.Cut on Tv & Sv as shown in illustration. Join those points in sequence and showJoin those points in sequence and show Section lines in it.Section lines in it. Make remaining part of solid dark.Make remaining part of solid dark. For True Shape:For True Shape: Draw xDraw x11yy11 // to sec. plane// to sec. plane Draw projectors on it fromDraw projectors on it from cut points.cut points. Mark distances of pointsMark distances of points of Sectioned part from Tv,of Sectioned part from Tv, on above projectors fromon above projectors from xx11yy11 and join in sequence.and join in sequence. Draw section lines in it.Draw section lines in it. It is required true shape.It is required true shape. For Development:For Development: Draw development of entire solid. Name fromDraw development of entire solid. Name from cut-open edge I.e. A. in sequence as shown.cut-open edge I.e. A. in sequence as shown. Mark the cut points on respective edges.Mark the cut points on respective edges. Join them in sequence in st. lines.Join them in sequence in st. lines. Make existing parts dev.dark.Make existing parts dev.dark.
  9. 9. Q 14.11: A square pyramid, base 40 mm side and axis 65 mm long, has its base on the HP and all the edges of the base equally inclined to the VP. It is cut by a section plane, perpendicular to the VP, inclined at 45º to the HP and bisecting the axis. Draw its sectional top view, sectional side view and true shape of the section. Also draw its development. A B C D A O 1 1 2 3 4 X 45º a b c d o a’ b’ c’ d’ o’ 1 2 3 4 1’ 2’ 3’ 4’ 11 41 21 31 True shape of the section Y True length of slant edge True length of slant edge
  10. 10. Q 15.17: A square pyramid, base 40 mm side and axis 65 mm long, has its base on the HP with two edges of the base perpendicular to the VP. It is cut by a section plane, perpendicular to the VP, inclined at 45º to the HP and bisecting the axis. Draw its sectional top view and true shape of the section. Also draw its development. X o’ Y A B C D A O a b cd o a’ d’ b’ c’ 1 2 3 4 1’ 4’ 2’ 3’ 2 3 1 2 True length of slant edge 1 4 1 1 4 2 3 2 3 True length of slant edge
  11. 11. Q 14.14: A pentagonal pyramid , base 30mm side and axis 60 mm long is lying on one of its triangular faces on the HP with the axis parallel to the VP. A vertical section plane, whose HT bisects the top view of the axis and makes an angle of 30º with the reference line, cuts the pyramid removing its top part. Draw the top view, sectional front view and true shape of the section and development of the surface of the remaining portion of the pyramid. YX a’ b’ e’ c’ d’ a b c d e o o’ 60 c’d’ o’ a’ b’e’ 30 a1 b1 c1 d1 e1 o1 1’ 2’ 3’4’ 5’ 6’ 1 2 3 4 5 6 31’ 41’ 21’ 11’ 61’ 51’ O A B C D E A 1 2 3 4 5 6 1 5 6
  12. 12. X Y 1 2 3 4 5 6 7 8 9 10 11 12 Q 15.26: draw the projections of a cone resting on the ground on its base and show on them, the shortest path by which a point P, starting from a point on the circumference of the base and moving around the cone will return to the same point. Base of cone 65 mm diameter ; axis 75 mm long. 1’ 2’ 12’ 3’ 11’ 4’ 10’ 5’ 9’ 6’ 8’ 7’ 2 3 4 5 6 7 8 9 10 11 12 1 θ=143º O’ θ=r/L X 360 º θ=32.5/81.74 X 360º = 143º Where r is radius of base circle and L is slant height r L
  13. 13. Q 14.24: A right circular cone, base 25 mm radius and height 65 mm rests on its base on H.P. It is cut by a section plane perpendicular to the V.P., inclined at 45º to the H.P. and bisecting the axis. Draw the projections of the truncated cone and develop its lateral surface. X Y 1 2 3 4 5 6 7 8 9 10 11 12 1’ 2’ 12’ 3’ 11’ 4’ 10’ 5’ 9’ 6’ 8’ 7’ 2 3 4 5 6 7 8 9 10 11 12 1 a b c k d e f g h il j a’ f’ b’ c’ k’ d’ e’ g’ h’i’ l’ j’ A C D E B A F G H I J K L θ=129º θ=r/L X 360 º θ=25/69.64 X 360º = 129.2º Where r is radius of base circle and L is slant height r L
  14. 14. o’ h a b c d g f o e a’ b’ c’ g’ d’f’ e’h’X Y θ = R L + 3600 R=Base circle radius. L=Slant height. θ A B C D E F G H A O 1 3 2 4 7 6 5 L 11 22 33 44 55 66 77 1’1’ 2’2’ 3’3’ 4’4’5’5’ 6’6’ 7’7’ Problem 6:Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largestDraw a semicircle 0f 100 mm diameter and inscribe in it a largest circle.If the semicircle is development of a cone and inscribed circle is somecircle.If the semicircle is development of a cone and inscribed circle is some curve on it, then draw the projections of cone showing that curve.curve on it, then draw the projections of cone showing that curve. Solution Steps:Solution Steps: Draw semicircle of given diameter, divide it in 8 Parts and inscribe in itDraw semicircle of given diameter, divide it in 8 Parts and inscribe in it a largest circle as shown.Name intersecting points 1, 2, 3 etc.a largest circle as shown.Name intersecting points 1, 2, 3 etc. Semicircle being dev.of a cone it’s radius is slant height of cone.( L )Semicircle being dev.of a cone it’s radius is slant height of cone.( L ) Then using above formula find R of base of cone. Using this dataThen using above formula find R of base of cone. Using this data draw Fv & Tv of cone and form 8 generators and name.draw Fv & Tv of cone and form 8 generators and name. Take o -1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’Take o -1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’ and name 1’ Similarly locate all points on Fv. Then project all on Tvand name 1’ Similarly locate all points on Fv. Then project all on Tv on respective generators and join by smooth curve.on respective generators and join by smooth curve. LL TO DRAW PRINCIPALTO DRAW PRINCIPAL VIEWS FROM GIVENVIEWS FROM GIVEN DEVELOPMENT.DEVELOPMENT.
  15. 15. Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largest circle.If the semicircle is development of a cone and inscribed circle is some curve on it, then draw the projections of cone showing that curve. Solution Steps:Solution Steps: Draw semicircle of given diameter, divide it in 12 Parts and inscribe in itDraw semicircle of given diameter, divide it in 12 Parts and inscribe in it a largest circle as shown.Name intersecting points 1, 2, 3 etc.a largest circle as shown.Name intersecting points 1, 2, 3 etc. Semicircle being dev.of a cone it’s radius is slant height of cone.( L )Semicircle being dev.of a cone it’s radius is slant height of cone.( L ) Then using above formula find R of base of cone. Using this dataThen using above formula find R of base of cone. Using this data draw Fv & Tv of cone and form 12 generators and name.draw Fv & Tv of cone and form 12 generators and name. Take o -1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’Take o -1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’ and name 1’ Similarly locate all points on Fv. Then project all on Tvand name 1’ Similarly locate all points on Fv. Then project all on Tv on respective generators and join by smooth curve.on respective generators and join by smooth curve. A B C D E F G H I J K L A 1 2 3 4 5 6 7 8 9 10 11 12 θ θ = R L + 3600 R=Base circle radius. L=Slant height. L X Y a b c d e f g h i j k l R= θXL/360º R= 180ºX50/360º R= 25mma’ b’ c’ d’ e’ f’ g’ ….l’ k’ j’ i’ h’ 1’
  16. 16. Q.15.11: A right circular cylinder, base 50 mm diameter and axis 60 mm long, is standing on HP on its base. It has a square hole of size 25 in it. The axis of the hole bisects the axis of the cylinder and is perpendicular to the VP. The faces of the square hole are equally inclined with the HP. Draw its projections and develop lateral surface of the cylinder. Y 1 2 3 4 5 6 7 8 9 10 11 12 X 1’ 2’ 12’ 3’ 11’ 4’ 10’ 5’ 9’ 6’ 8’ 7’ a’ b’ c’ d’ 1 2 3 4 5 6 7 8 9 10 11 12 1 a a b d b d c c A B D C C B D A a c
  17. 17. Q.15.21: A frustum of square pyramid has its base 50 mm side, top 25 mm side and axis 75 mm. Draw the development of its lateral surface. Also draw the projections of the frustum (when its axis is vertical and a side of its base is parallel to the VP), showing the line joining the mid point of a top edge of one face with the mid point of the bottom edge of the opposite face, by the shortest distance. YX 50 25 75 a b cd a1 b1 c1 d1 a’ d’ b’ c’ a1 ’ d1’ b1 ’ c1’ o o’ True length of slant edge A1 B1 C1 D1 A1 A B C D A P Q R S p’ p q’ q r’ r s’ s
  18. 18. Q: A square prism of 40 mm edge of the base and 65 mm height stands on its base on the HP with vertical faces inclined at 45º with the VP. A horizontal hole of 40 mm diameter is drilled centrally through the prism such that the hole passes through the opposite vertical edges of the prism, draw the development of the surfaces of the prism. YX a b c d a’ b’d’ c’ a’ b’d’ c’ 1’ 2’ 3’ 4’ 5’ 6’ 7’ 8’ 9’ 10’ 11’ 12’ 1 1 2 12 2 12 3 11 3 11 4 10 4 10 5 9 5 9 6 8 6 8 1 2 12 3 11 4 10A B C 7 7 5 9 6 8 7 6 8 5 9 4 10 7 12 12 3 11 A 1 2 12 11 3 10 4 9 5 8 6 7 7 6 8 9 11 3 12 1 5 4 2 10 D 40
  19. 19. Y h a b c d e g f X a’ b’ d’ e’c’ g’ f’h’ o’ X1 Y1 g” h”f” a”e” b”d” c” A B C D E F A G H SECTIONAL T.V SECTIONAL S.V TRUE SH APE OF SECTIO N DEVELOPMENT SECTION PLANE Problem 2:Problem 2: A cone, 50 mm base diameter and 70 mm axis isA cone, 50 mm base diameter and 70 mm axis is standing on it’s base on Hp. It cut by a section plane 45standing on it’s base on Hp. It cut by a section plane 4500 inclinedinclined to Hp through base end of end generator.Draw projections,to Hp through base end of end generator.Draw projections, sectional views, true shape of section and development of surfacessectional views, true shape of section and development of surfaces of remaining solid.of remaining solid. Solution Steps:Solution Steps:for sectional views:for sectional views: Draw three views of standing cone.Draw three views of standing cone. Locate sec.plane in Fv as described.Locate sec.plane in Fv as described. Project points where generators areProject points where generators are getting Cut on Tv & Sv as shown ingetting Cut on Tv & Sv as shown in illustration.Join those points inillustration.Join those points in sequence and show Section lines in it.sequence and show Section lines in it. Make remaining part of solid dark.Make remaining part of solid dark. For True Shape:For True Shape: Draw xDraw x11yy11 // to sec. plane// to sec. plane Draw projectors on it fromDraw projectors on it from cut points.cut points. Mark distances of pointsMark distances of points of Sectioned part from Tv,of Sectioned part from Tv, on above projectors fromon above projectors from xx11yy11 and join in sequence.and join in sequence. Draw section lines in it.Draw section lines in it. It is required true shape.It is required true shape. For Development:For Development: Draw development of entire solid.Draw development of entire solid. Name from cut-open edge i.e. A.Name from cut-open edge i.e. A. in sequence as shown.Mark the cutin sequence as shown.Mark the cut points on respective edges.points on respective edges. Join them in sequence inJoin them in sequence in curvature. Make existing partscurvature. Make existing parts dev.dark.dev.dark.
  20. 20. X Y e’a’ b’ d’c’ g’ f’h’ o’ h a b c d e g f O DEVELOPMENT A B C D E F A G H OO 11 22 33 44 66 55 77 1’1’ 2’2’ 3’3’ 4’4’ 5’5’ 6’6’ 7’7’ 11 22 33 44 5566 77 HELIX CURVEHELIX CURVE Problem 9:Problem 9: A particle which is initially on base circle of a cone, standingA particle which is initially on base circle of a cone, standing on Hp, moves upwards and reaches apex in one complete turn around the cone.on Hp, moves upwards and reaches apex in one complete turn around the cone. Draw it’s path on projections of cone as well as on it’s development.Draw it’s path on projections of cone as well as on it’s development. Take base circle diameter 50 mm and axis 70 mm long.Take base circle diameter 50 mm and axis 70 mm long. It’s a construction of curveIt’s a construction of curve Helix of one turn on coneHelix of one turn on cone:: Draw Fv & Tv & dev.as usualDraw Fv & Tv & dev.as usual On all form generators & name.On all form generators & name. Construction of curve Helix::Construction of curve Helix:: Show 8 generators on both viewsShow 8 generators on both views Divide axis also in same parts.Divide axis also in same parts. Draw horizontal lines from thoseDraw horizontal lines from those points on both end generators.points on both end generators. 1’ is a point where first horizontal1’ is a point where first horizontal Line & gen. b’o’ intersect.Line & gen. b’o’ intersect. 2’ is a point where second horiz.2’ is a point where second horiz. Line & gen. c’o’ intersect.Line & gen. c’o’ intersect. In this way locate all points on Fv.In this way locate all points on Fv. Project all on Tv.Join in curvature.Project all on Tv.Join in curvature. For Development:For Development: Then taking each points trueThen taking each points true Distance From resp.generatorDistance From resp.generator from apex, Mark on developmentfrom apex, Mark on development & join.& join.
  21. 21. o’ h a b c d g f o e a’ b’ c’ g’ d’f’ e’h’X Y θ = R L + 3600 R=Base circle radius. L=Slant height. θ A B C D E F G H A O 1 3 2 4 7 6 5 L 11 22 33 44 55 66 77 1’1’ 2’2’ 3’3’ 4’4’5’5’ 6’6’ 7’7’ Problem 6:Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largestDraw a semicircle 0f 100 mm diameter and inscribe in it a largest circle.If the semicircle is development of a cone and inscribed circle is somecircle.If the semicircle is development of a cone and inscribed circle is some curve on it, then draw the projections of cone showing that curve.curve on it, then draw the projections of cone showing that curve. Solution Steps:Solution Steps: Draw semicircle of given diameter, divide it in 8 Parts and inscribe in itDraw semicircle of given diameter, divide it in 8 Parts and inscribe in it a largest circle as shown.Name intersecting points 1, 2, 3 etc.a largest circle as shown.Name intersecting points 1, 2, 3 etc. Semicircle being dev.of a cone it’s radius is slant height of cone.( L )Semicircle being dev.of a cone it’s radius is slant height of cone.( L ) Then using above formula find R of base of cone. Using this dataThen using above formula find R of base of cone. Using this data draw Fv & Tv of cone and form 8 generators and name.draw Fv & Tv of cone and form 8 generators and name. Take o -1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’Take o -1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’ and name 1’ Similarly locate all points on Fv. Then project all on Tvand name 1’ Similarly locate all points on Fv. Then project all on Tv on respective generators and join by smooth curve.on respective generators and join by smooth curve. LL TO DRAW PRINCIPALTO DRAW PRINCIPAL VIEWS FROM GIVENVIEWS FROM GIVEN DEVELOPMENT.DEVELOPMENT.
  22. 22. h a b c d g f e o’ a’ b’ d’c’ g’ f’h’ e’ X Y A B C D E F G H A O L 1 2 3 4 5 6 7 θ = R L + 3600 R=Base circle radius. L=Slant height. θ1’1’ 2’2’ 3’3’ 4’4’ 5’5’ 6’6’ 7’7’ 11 22 33 44 55 6677 Problem 7:Problem 7:Draw a semicircle 0f 100 mm diameter and inscribe in it a largestDraw a semicircle 0f 100 mm diameter and inscribe in it a largest rhombusrhombus.If the semicircle is development of a cone and rhombus is some curve.If the semicircle is development of a cone and rhombus is some curve on it, then draw the projections of cone showing that curve.on it, then draw the projections of cone showing that curve. TO DRAW PRINCIPALTO DRAW PRINCIPAL VIEWS FROM GIVENVIEWS FROM GIVEN DEVELOPMENT.DEVELOPMENT. Solution Steps:Solution Steps: Similar to previousSimilar to previous Problem:Problem:
  23. 23. A.V.P300 inclined to Vp Through mid-point of axis. X Y 1,2 3,8 4,7 5,6 1 2 3 4 5 6 78 2 1 8 7 6 54 3 b’ f’a’ e’c’ d’ a b c d e f b’f’a’e’c’d’ a1 d1b1 e1 c1 f1 X1 Y1 AS SECTION PLANE IS IN T.V., CUT OPEN FROM BOUNDRY EDGE C1 FOR DEVELOPMENT. TRUE SHAPE OF SECTION C D E F A B C DEVELOPMENT SECTIONAL F.V. Problem 4:Problem 4: A hexagonal prism. 30 mm base side &A hexagonal prism. 30 mm base side & 55 mm axis is lying on Hp on it’s rect.face with axis55 mm axis is lying on Hp on it’s rect.face with axis // to Vp. It is cut by a section plane normal to Hp and// to Vp. It is cut by a section plane normal to Hp and 303000 inclined to Vp bisecting axis.inclined to Vp bisecting axis. Draw sec. Views, true shape & development.Draw sec. Views, true shape & development. Use similar steps for sec.views & true shape.Use similar steps for sec.views & true shape. NOTE:NOTE: for development, always cut open object fromfor development, always cut open object from From an edge in the boundary of the view in whichFrom an edge in the boundary of the view in which sec.plane appears as a line.sec.plane appears as a line. Here it is Tv and in boundary, there is c1 edge.HenceHere it is Tv and in boundary, there is c1 edge.Hence it is opened from c and named C,D,E,F,A,B,C.it is opened from c and named C,D,E,F,A,B,C. NoteNote the steps to locatethe steps to locate Points 1, 2 , 5, 6 in sec.Fv:Points 1, 2 , 5, 6 in sec.Fv: Those are transferred toThose are transferred to 11stst TV, then to 1TV, then to 1stst Fv andFv and Then on 2Then on 2ndnd Fv.Fv.
  24. 24. 1’ 2’ 3’ 4’ 5’ 6’ 7’ 7 1 5 4 3 2 6 7 1 6 5 4 3 2 a b c d e f g 4 4 5 3 6 2 7 1 A B C D E A F G O O’ d’e’ c’f’ g’b’ a’ X Y X1 Y1 TRUE SHAPE F.V. SECTIONAL TOP VIEW. DEVELOPMENT Problem 5:A solid composed of a half-cone and half- hexagonal pyramid is shown in figure.It is cut by a section plane 450 inclined to Hp, passing through mid-point of axis.Draw F.v., sectional T.v.,true shape of section and development of remaining part of the solid. ( take radius of cone and each side of hexagon 30mm long and axis 70mm.) Note:Note: Fv & TV 8f two solidsFv & TV 8f two solids sandwichedsandwiched Section lines style in both:Section lines style in both: Development ofDevelopment of half cone & half pyramid:half cone & half pyramid:
  25. 25. a’a’ b’b’ c’c’ d’d’ o’o’ e’e’ aa bb cc dd oo ee XX YY AA BB CC DD EE AA OO 22 33 44 11 Problem 8:Problem 8: A half cone of 50 mm base diameter, 70 mm axis, is standing on it’s half base on HP with it’s flat faceA half cone of 50 mm base diameter, 70 mm axis, is standing on it’s half base on HP with it’s flat face parallel and nearer to VP.An inextensible string is wound round it’s surface from one point of base circle andparallel and nearer to VP.An inextensible string is wound round it’s surface from one point of base circle and brought back to the same point.If the string is ofbrought back to the same point.If the string is of shortest lengthshortest length, find it and show it on the projections of the cone., find it and show it on the projections of the cone. 11 22 33 44 1’1’ 2’2’ 3’3’ 4’4’ TO DRAW A CURVE ONTO DRAW A CURVE ON PRINCIPAL VIEWSPRINCIPAL VIEWS FROM DEVELOPMENT.FROM DEVELOPMENT. Concept:Concept: A string woundA string wound from a point up to the samefrom a point up to the same Point, of shortest lengthPoint, of shortest length Must appear st. line on it’sMust appear st. line on it’s Development.Development. Solution steps:Solution steps: Hence draw development,Hence draw development, Name it as usual and joinName it as usual and join A to A This is shortestA to A This is shortest Length of that string.Length of that string. Further steps are as usual.Further steps are as usual. On dev. Name the points ofOn dev. Name the points of Intersections of this line withIntersections of this line with Different generators.BringDifferent generators.Bring Those on Fv & Tv and joinThose on Fv & Tv and join by smooth curves.by smooth curves. Draw 4’ a’ part of string dottedDraw 4’ a’ part of string dotted As it is on back side of cone.As it is on back side of cone.
  26. 26. X Ye’a’ b’ d’c’ g’ f’h’ a’ h’b’ e’ c’g’ d’f’ o’ o’ Problem 3: A cone 40mm diameter and 50 mm axis is resting on one generator on Hp( lying on Hp) which is // to Vp.. Draw it’s projections.It is cut by a horizontal section plane through it’s base center. Draw sectional TV, development of the surface of the remaining part of cone. A B C D E F A G H O a1 h1 g1 f1 e1 d1 c1 b1 o1 SECTIONAL T.V DEVELOPMENT (SHOWING TRUE SHAPE OF SECTION) HORIZONTAL SECTION PLANE h a b c d e g f O Follow similar solution steps for Sec.views - True shape – Development as per previous problem!Follow similar solution steps for Sec.views - True shape – Development as per previous problem!

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