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- 1. ENGINEERING APPLICATIONS OF THE PRINCIPLES OF PROJECTIONS OF SOLIDES.1. SECTIONS OF SOLIDS.2. DEVELOPMENT.3. INTERSECTIONS. STUDY CAREFULLY THE ILLUSTRATIONS GIVEN ON NEXT SIX PAGES !
- 2. SECTIONING A SOLID. The action of cutting is called An object ( here a solid ) is cut by SECTIONING a solid some imaginary cutting plane & to understand internal details of that The plane of cutting is called object. SECTION PLANE.Two cutting actions means section planes are recommended. OBSERVER E AN PLA) Section Plane perpendicular to Vp and inclined to Hp. ASSUME ON V. UPPER PART CT IN F SE ( This is a definition of an Aux. Inclined Plane i.e. A.I.P.) REMOVED NOTE:- This section plane appears (A) as a straight line in FV.B) Section Plane perpendicular to Hp and inclined to Vp. ( This is a definition of an Aux. Vertical Plane i.e. A.V.P.) NOTE:- This section plane appears (B) as a straight line in TV.Remember:-1. After launching a section plane ASSUME LOWER PART SE CT ON either in FV or TV, the part towards observer REMOVED IN PLA TV . NE OBSERVER is assumed to be removed.2. As far as possible the smaller part is assumed to be removed.
- 3. For TVILLUSTRATION SHOWING Fo IMPORTANT TERMS rT IN SECTIONING. ru e Sh ap e SECTION PLANE TRUE SHAPE Of SECTION x y Apparent Shape of section SECTION LINES (450 to XY) SECTIONAL T.V.
- 4. Typical Section Planes & Typical Shapes Of Sections. Section Plane Ellipse Section PlaneTriangle Through Generators Through Apex ola rab Pa Section Plane Parallel Section Plane Hyperbola to end generator. Parallel to Axis. Ellipse Trapezium Cylinder through Sq. Pyramid through generators. all slant edges
- 5. DEVELOPMENT OF SURFACES OF SOLIDS. MEANING:- ASSUME OBJECT HOLLOW AND MADE-UP OF THIN SHEET. CUT OPEN IT FROM ONE SIDE AND UNFOLD THE SHEET COMPLETELY. THEN THE SHAPE OF THAT UNFOLDED SHEET IS CALLED DEVELOPMENT OF LATERLAL SUEFACES OF THAT OBJECT OR SOLID. LATERLAL SURFACE IS THE SURFACE EXCLUDING SOLID’S TOP & BASE. ENGINEERING APLICATION: THERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BY CONVENTIONAL MANUFACTURING PROCESSES, BECAUSE OF THEIR SHAPES AND SIZES. THOSE ARE FABRICATED IN SHEET METAL INDUSTRY BY USING DEVELOPMENT TECHNIQUE. THERE IS A VAST RANGE OF SUCH OBJECTS. EXAMPLES:- Boiler Shells & chimneys, Pressure Vessels, Shovels, Trays, Boxes & Cartons, Feeding Hoppers, Large Pipe sections, Body & Parts of automotives, Ships, Aeroplanes and many more. WHAT IS To learn methods of development of surfaces of OUR OBJECTIVE different solids, their sections and frustums. IN THIS TOPIC ? 1. Development is different drawing than PROJECTIONS.But before going ahead, 2. It is a shape showing AREA, means it’s a 2-D plain drawing. note following 3. Hence all dimensions of it must be TRUE dimensions. Important points. 4. As it is representing shape of an un-folded sheet, no edges can remain hidden And hence DOTTED LINES are never shown on development. Study illustrations given on next page carefully.
- 6. Development of lateral surfaces of different solids. (Lateral surface is the surface excluding top & base) Cylinder: A Rectangle Pyramids: (No.of triangles) Cone: (Sector of circle) SH S L L πD D H= Height D= base diameter θ Prisms: No.of Rectangles R=Base circle radius. L= Slant edge. L=Slant height. S = Edge of base R 3600 θ =L + H S S H= Height S = Edge of base Cube: Six Squares. Tetrahedron: Four Equilateral Triangles All sides equal in length
- 7. FRUSTUMSDEVELOPMENT OF DEVELOPMENT OFFRUSTUM OF CONE FRUSTUM OF SQUARE PYRAMID Base side Top side L L L1 L1 θ Rθ = L 3600 +R= Base circle radius of coneL= Slant height of cone L= Slant edge of pyramidL1 = Slant height of cut part. L1 = Slant edge of cut part. STUDY NEXT NINE PROBLEMS OF SECTIONS & DEVELOPMENT
- 8. Problem 1: A pentagonal prism , 30 mm base side & 50 mm axis Solution Steps:for sectional views: is standing on Hp on it’s base with one side of the base perpendicular to VP. Draw three views of standing prism. It is cut by a section plane inclined at 40º to the HP, through mid point of axis. Locate sec.plane in Fv as described. Draw Fv, sec.Tv & sec. Side view. Also draw true shape of section and Project points where edges are getting Development of surface of remaining solid. Cut on Tv & Sv as shown in illustration. C PE Join those points in sequence and show HA S E Section lines in it. RU B T D Y1 Make remaining part of solid dark. A A B C D E E A d ’ d” c” c’ X1 e’ e” b” b’ a’ X a” Y e DEVELOPMENT dFor True Shape: a For Development:Draw x1y1 // to sec. plane Draw development of entire solid. Name fromDraw projectors on it from c cut-open edge I.e. A. in sequence as shown.cut points. b Mark the cut points on respective edges.Mark distances of points Join them in sequence in st. lines.of Sectioned part from Tv, Make existing parts dev.dark.on above projectors fromx1y1 and join in sequence.Draw section lines in it.It is required true shape.
- 9. Q 14.11: A square pyramid, base 40 mm side and axis 65 mm long, has its base on the HP and allthe edges of the base equally inclined to the VP. It is cut by a section plane, perpendicular to theVP, inclined at 45º to the HP and bisecting the axis. Draw its sectional top view, sectional sideview and true shape of the section. Also draw its development. True shape of the section 21 A 31 o’ 1 True length 41 of slant edge 3’ D 11 4 2’ 4’ O True length of slant edge 3 1’ b’ d’ X c’ Y C a’ 45º 2 d 4 1 a c B 1 o 3 A 2 b
- 10. Q 15.17: A square pyramid, base 40 mm side and axis 65 mm long, has its base on the HP withtwo edges of the base perpendicular to the VP. It is cut by a section plane, perpendicular to theVP, inclined at 45º to the HP and bisecting the axis. Draw its sectional top view and true shape ofthe section. Also draw its development. 3 A 2 o’ 1 True length2 of slant edge 2 3 2’ 3’ D 4 1 O True length 1 4 of slant 3 1’ 4’ edge 2 X a’ d’ b’ c’ Y C a b 1 2 1 o B 3 A 4 d c
- 11. Q 14.14: A pentagonal pyramid , base 30mm side and axis 60 mm long is lying on one of its triangular faceson the HP with the axis parallel to the VP. A vertical section plane, whose HT bisects the top view of the axisand 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 ofthe pyramid. o’ C B a’ D 5 A 6’ 60 b’e’ 1’ 4 E 3 2 1 6 5’ 2’ a’ b’ e’ c’ d’ c’d’ o’X Y b1 4’ 3’ b A c c1 O 1 2 o1 a1 3 o 30 a 1 31’ 6 d1 6 4 21’ d 5 41’ e 5 e1 11’ 51’ 61’
- 12. Q 15.26: draw the projections of a cone resting on the ground on its base and show on them, the shortest pathby which a point P, starting from a point on the circumference of the base and moving around the cone willreturn to the same point. Base of cone 65 mm diameter ; axis 75 mm long. 1 12 Where r is radius of base circle 11 and L is slant height 10 9 8 θ=143º 7 O’ θ=r/L X 360 º 6 θ=32.5/81.74 X 360º 5 L 4 = 143º 3 X 2 2’ 1’ 12’ 3’ 11’ 4’ 10’ 5’ 9’ 6’ 8’ 7’ Y 4 3 5 2 6 1 r 7 12 8 11 9 10
- 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 asection plane perpendicular to the V.P., inclined at 45º to the H.P. and bisecting the axis. Draw the projectionsof the truncated cone and develop its lateral surface. 1 12 A Where r is radius of base circle 11 and L is slant height B 10 C 9 D 8 E θ=129º F θ=r/L X 360 º 7 G H I 6 θ=25/69.64 X 360º J g’ 5 K e’f’ i’ h’ L d’ j’ = 129.2º 4 c’ L k’ b’l’ 3 A a’ X 2 2’ 1’ 12’ 3’ 11’ 4’ 10’ 5’ 9’ 6’ 8’ 7’ Y 4 3 5 2 c d 6 b e f 1 a gr 7 h l i 12 j 8 k 11 9 10
- 14. Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largest TO DRAW PRINCIPALcircle.If the semicircle is development of a cone and inscribed circle is some VIEWS FROM GIVENcurve on it, then draw the projections of cone showing that curve. DEVELOPMENT. E D F o’ R=Base circle radius. 4 L=Slant height. 3 5 R 3600 C G θ = L + 1’ L 7’ 2 6 B H 6’ 2’ 1 3’ 5’ 4’ θ 7 X a’ h’ b’ c’ g’ d’f’ e’ Y A A O g L 6 h f Solution Steps: 5 Draw 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 7 o Semicircle being dev.of a cone it’s radius is slant height of cone.( L ) 4 e Then using above formula find R of base of cone. Using this data 1 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’ b 3 d and name 1’ Similarly locate all points on Fv. Then project all on Tv 2 on respective generators and join by smooth curve. c
- 15. Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largest circle.If thesemicircle is development of a cone and inscribed circle is some curve on it, then draw theprojections of cone showing that curve. L 1 A 2 12 A 1’ R=Base circle radius. θ L=Slant height. 3 11 B L R 3600 θ = L + 4 10 C K R= θXL/360º D 5 9 J R= 180ºX50/360º 8 E 6 I X Y F 7 R= 25mm G H a’ b’ c’ d’ e’ f’ g’ ….l’ k’ j’ i’ h’ j k i Solution Steps: l Draw semicircle of given diameter, divide it in 12 Parts and inscribe in it h 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 ) a g Then using above formula find R of base of cone. Using this data draw Fv & Tv of cone and form 12 generators and name. b f 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 Tv c e on respective generators and join by smooth curve. d
- 16. Q.15.11: A right circular cylinder, base 50 mm diameter and axis 60 mm long, is standing on HP on itsbase. It has a square hole of size 25 in it. The axis of the hole bisects the axis of the cylinder and isperpendicular to the VP. The faces of the square hole are equally inclined with the HP. Draw itsprojections and develop lateral surface of the cylinder. 2’ 3’ 4’ 5’ 6’ 1’ 12’ 11’ 10’ 9’ 8’ 7’ b’ B B a’ c’ A C C A d’ D D X a c Y 1 2 3 4 5 6 7 8 9 10 11 12 1 b4d 3 5 a c 2 6 1 7 12 8 a c 11 9 10 b d
- 17. Q.15.21: A frustum of square pyramid has its base 50 mm side, top 25 mm side and axis 75 mm. Drawthe development of its lateral surface. Also draw the projections of the frustum (when its axis is verticaland a side of its base is parallel to the VP), showing the line joining the mid point of a top edge of oneface with the mid point of the bottom edge of the opposite face, by the shortest distance. o’ A A1 D True P length of slant edge C R B D1 a’ b’ d’ p’ c’ A r’ S C1 s’ 75 Q B1 a1 ’ b1 ’ X d1’ q’ c1’ Y d1 c1 A1 p r d c 50 25 o a b s q a1 b1
- 18. Q: A square prism of 40 mm edge of the base and 65 mm height stands on its base on the HPwith vertical faces inclined at 45º with the VP. A horizontal hole of 40 mm diameter is drilledcentrally through the prism such that the hole passes through the opposite vertical edges of theprism, draw the development of the surfaces of the prism. a’ b’d’ c’ 4’ 4 4 3’ 5’ 3 5 5 3 2’ 6’ 2 6 6 2 1’ 7’ 1 7 7 1 12’ 8’ 12 8 8 12 11’ 9’ 11 9 9 11 10’ 10 10 X a’ b’d’ c’ 1 2 3 Y 3 4 5 6 7 C 7 6 5 4 11 2 1 4 b10 A 12 11 10 9 8 8 9 10 12 A 5 B D 3 11 9 2 6 12 8 1 7 a c 1 7 2 6 12 8 3 5 11 40 9 d 4 10
- 19. Problem 2: A cone, 50 mm base diameter and 70 mm axis is Solution Steps:for sectional views:standing on it’s base on Hp. It cut by a section plane 450 inclined Draw three views of standing cone. to Hp through base end of end generator.Draw projections, Locate sec.plane in Fv as described.sectional views, true shape of section and development of surfaces Project points where generators areof remaining solid. getting Cut on Tv & Sv as shown in illustration.Join those points in ON C TI sequence and show Section lines in it. SE Make remaining part of solid dark. OF E Y1 A H AP SECTIONAL S.V S UE o’ E N B AN IO TR PL CT SE DEVELOPMENT C D X1 E X a’ h’b’ c’ g’ f’ d’ e’ Y g” h”f” a”e” b”d” c” F gFor True Shape: h G fDraw x1y1 // to sec. planeDraw projectors on it from For Development: Hcut points. a e Draw development of entire solid.Mark distances of points Name from cut-open edge i.e. A. Aof Sectioned part from Tv, in sequence as shown.Mark the cuton above projectors from b d points on respective edges.x1y1 and join in sequence. c Join them in sequence inDraw section lines in it. SECTIONAL T.V curvature. Make existing partsIt is required true shape. dev.dark.
- 20. Problem 9: A particle which is initially on base circle of a cone, standingon 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.Take base circle diameter 50 mm and axis 70 mm long. It’s a construction of curve o’ Helix of one turn on cone: 7’ DEVELOPMENT Draw Fv & Tv & dev.as usual HELIX CURVE 6’ On all form generators & name. A Construction of curve Helix:: 5’ 4’ B Show 8 generators on both views 3’ Divide axis also in same parts. 2’ 1 Draw horizontal lines from those 1’ C points on both end generators. 2 1’ is a point where first horizontalX Y a’ h’b’ c’ g g’ f’ d’ e’ D Line & gen. b’o’ intersect. 3 2’ is a point where second horiz. h f O 4 E Line & gen. c’o’ intersect. 7 6 6 5 In this way locate all points on Fv. 5 7 F Project all on Tv.Join in curvature. a e O 4 For Development: Then taking each points true 3 G Distance From resp.generator b 1 d 2 from apex, Mark on development c H & join. A
- 21. Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largest TO DRAW PRINCIPALcircle.If the semicircle is development of a cone and inscribed circle is some VIEWS FROM GIVENcurve on it, then draw the projections of cone showing that curve. DEVELOPMENT. E D F o’ R=Base circle radius. 4 L=Slant height. 3 5 R 3600 C G θ = L + 1’ L 7’ 2 6 B H 6’ 2’ 1 3’ 5’ 4’ θ 7 X a’ h’ b’ c’ g’ d’f’ e’ Y A A O g L 6 h f Solution Steps: 5 Draw 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 7 o Semicircle being dev.of a cone it’s radius is slant height of cone.( L ) 4 e Then using above formula find R of base of cone. Using this data 1 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’ b 3 d and name 1’ Similarly locate all points on Fv. Then project all on Tv 2 on respective generators and join by smooth curve. c
- 22. Problem 7:Draw a semicircle 0f 100 mm diameter and inscribe in it a largest TO DRAW PRINCIPALrhombus.If the semicircle is development of a cone and rhombus is some curve VIEWS FROM GIVENon it, then draw the projections of cone showing that curve. DEVELOPMENT. Solution Steps: o’ Similar to previous Problem: E D F 4 C 3 5 G 2 6 1 7 2’ 6’ B H 3’ 5’ a’ h’ b’ c’ g’ X 1’ 7’ f’ d’ e’ 4’ Y θ g A A O L 7 6 h f 5 R=Base circle radius. L=Slant height. R 3600 a 4 e θ = L + b 3 d 2 1 c
- 23. Note the steps to locate Problem 4: A hexagonal prism. 30 mm base side & Points 1, 2 , 5, 6 in sec.Fv: 55 mm axis is lying on Hp on it’s rect.face with axis Those are transferred to // to Vp. It is cut by a section plane normal to Hp and 1st TV, then to 1st Fv and 300 inclined to Vp bisecting axis. a’ b’ c’ f’ d’ e’ Then on 2nd Fv. Draw sec. Views, true shape & development. 3 SECTIONAL F.V. Use similar steps for sec.views & true shape. a’ b’ 4 NOTE: for development, always cut open object from 2 5 From an edge in the boundary of the view in which sec.plane appears as a line. c’ f’ Here it is Tv and in boundary, there is c1 edge.Hence 1 6 it is opened from c and named C,D,E,F,A,B,C.X Y d’ e’ 8 7 f 1 ,2 f1 3,8 e a1 e1 a A.V.P300 inclined to Vp 4,7 Through mid-point of axis. b d b1 d1 5,6 AS SECTION PLANE IS IN T.V., c c1 CUT OPEN FROM BOUNDRY EDGE C1 FOR DEVELOPMENT. X1 8 1 C D E F A B C 2 7 Y1 3 6 TR UE SH AP E OF SE 5 CT 4 IO N DEVELOPMENT
- 24. Problem 5:A solid composed of a half-cone and half- hexagonal pyramid is PE shown in figure.It is cut by a section plane 450 inclined to Hp, passing through S HA 3 UE 2 mid-point of axis.Draw F.v., sectional T.v.,true shape of section and TR 4 development of remaining part of the solid. Y1 ( take radius of cone and each side of hexagon 30mm long and axis 70mm.) 51 O’ Note: 6 A Fv & TV 8f two solids T sandwiched EN Section lines style in both: PM 7 B Development of LO 4’ VE half cone & half pyramid: DE 3’ X1 5’ C 2’ 6’ F.V. 2 4 3 1 D 1’ 7’ X Y d’e’ c’f’ g’b’ a’ O f E g 7 6 e 5 SECTIONAL 7 4 5 6 4 TOP VIEW. a F 1 3 d 2 b G c A
- 25. 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 faceparallel and nearer to VP.An inextensible string is wound round it’s surface from one point of base circle andbrought back to the same point.If the string is of shortest length, find it and show it on the projections of the cone.TO DRAW A CURVE ON PRINCIPAL VIEWSFROM DEVELOPMENT. Concept: A string wound from a point up to the same o’ Point, of shortest length A Must appear st. line on it’s Development. B Solution steps: Hence draw development, 1 C Name it as usual and join A to A This is shortest 2’ 3’ 4’ D Length of that string. 1’ 2 Further steps are as usual. 3 On dev. Name the points of Intersections of this line withX Y 4 E a’ b’ c’ d’ e’ Different generators.Bring O Those on Fv & Tv and join a o e 4 by smooth curves. Draw 4’ a’ part of string dotted 3 As it is on back side of cone. 1 2 b d c A
- 26. 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.Follow similar solution steps for Sec.views - True shape – Development as per previous problem! o’ DEVELOPMENT A a’ B ’b h HORIZONTAL ’ c’ SECTION PLANE C g’ d’f’ D X a’ h’b’ c’ g’ f’ d’ e’ o’ Y e’ g O E g1 h f f1 h1 F a e e1 a1 o1 G O H b d d1 b1 A c c1 SECTIONAL T.V (SHOWING TRUE SHAPE OF SECTION)

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