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SECTIONS OF SOLIDS. DEVELOPMENT. INTERSECTIONS.

ENGINEERING APPLICATIONS OF THE PRINCIPLES OF PROJECTIONS OF SOLIDES. SECTIONS OF SOLIDS. DEVELOPMENT. INTERSECTIONS. STUDY CAREFULLY THE ILLUSTRATIONS GIVEN ON NEXT SIX PAGES !. Laxmi Institute of Technology, sarigam. OBSERVER. ASSUME UPPER PART REMOVED . SECTON PLANE

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SECTIONS OF SOLIDS. DEVELOPMENT. INTERSECTIONS.

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  1. ENGINEERING APPLICATIONS OF THE PRINCIPLES OF PROJECTIONS OF SOLIDES. • SECTIONS OF SOLIDS. • DEVELOPMENT. • INTERSECTIONS. STUDY CAREFULLY THE ILLUSTRATIONS GIVEN ON NEXT SIX PAGES ! Laxmi Institute of Technology, sarigam

  2. OBSERVER ASSUME UPPER PART REMOVED SECTON PLANE IN FV. ASSUME LOWER PART REMOVED SECTON PLANE IN TV. OBSERVER SECTIONING A SOLID. An object ( here a solid ) is cut by some imaginary cutting plane to understand internal details of that object. The action of cutting is called SECTIONING a solid & The plane of cutting is called SECTION PLANE. Two cutting actions means section planes are recommended. A) Section Plane perpendicular to Vp and inclined to Hp. ( This is a definition of an Aux. Inclined Plane i.e. A.I.P.) NOTE:- This section plane appears 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 as a straight line in TV. Remember:- 1. After launching a section plane either in FV or TV, the part towards observer is assumed to be removed. 2. As far as possible the smaller part is assumed to be removed. (A) (B)

  3. For TV ILLUSTRATION SHOWING IMPORTANT TERMS IN SECTIONING. For True Shape 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 Through Generators Ellipse Triangle Section Plane Through Apex Parabola Section Plane Parallel to Axis. Section Plane Parallel to end generator. Hyperbola Ellipse Trapezium Cylinder through generators. Sq. Pyramid through 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 OUR OBJECTIVE IN THIS TOPIC ? To learn methods of development of surfaces of different solids, their sections and frustums. 1. Development is different drawing than PROJECTIONS. 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. 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. But before going ahead, note following Important points. Study illustrations given on next page carefully.

  6. Development of lateral surfaces of different solids. (Lateral surface is the surface excluding top & base) =  H L D R + 3600 D L R=Base circle radius. L=Slant height. H S S S S Cylinder: A Rectangle Pyramids: (No.of triangles) Cone: (Sector of circle) L  H= Height D= base diameter Prisms: No.of Rectangles L= Slant edge. S = Edge of base H= Height S = Edge of base Cube: Six Squares. Tetrahedron: Four Equilateral Triangles All sides equal in length

  7. FRUSTUMS =  R + 3600 L L L L1 L1 STUDY NEXT NINE PROBLEMS OF SECTIONS & DEVELOPMENT DEVELOPMENT OF FRUSTUM OF CONE DEVELOPMENT OF FRUSTUM OF SQUARE PYRAMID Base side Top side  R= Base circle radius of cone L= Slant height of cone L1 = Slant height of cut part. L= Slant edge of pyramid L1 = Slant edge of cut part.

  8. Problem 1:A 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. It is cut by a section plane inclined at 45º to the HP, through mid point of axis. Draw Fv, sec.Tv & sec. Side view. Also draw true shape of section and Development of surface of remaining solid. Solution Steps:for sectional views: Draw three views of standing prism. Locate sec.plane in Fv as described. Project points where edges are getting Cut on Tv & Sv as shown in illustration. Join those points in sequence and show Section lines in it. Make remaining part of solid dark. C B D Y1 A A B C D E A E d” c” X1 b” e” a” e d For True Shape: Draw x1y1 // to sec. plane Draw projectors on it from cut points. Mark distances of points of Sectioned part from Tv, on above projectors from x1y1 and join in sequence. Draw section lines in it. It is required true shape. a For Development: Draw development of entire solid. Name from cut-open edge I.e. A. in sequence as shown. Mark the cut points on respective edges. Join them in sequence in st. lines. Make existing parts dev.dark. c b TRUE SHAPE a’ b’ e’ c’ d’ X Y DEVELOPMENT

  9. Problem 2: A cone, 50 mm base diameter and 70 mm axis is standing on it’s base on Hp. It cut by a section plane 450 inclined to Hp through base end of end generator.Draw projections, sectional views, true shape of section and development of surfaces of remaining solid. Solution Steps:for sectional views: Draw three views of standing cone. Locate sec.plane in Fv as described. Project points where generators are getting Cut on Tv & Sv as shown in illustration.Join those points in sequence and show Section lines in it. Make remaining part of solid dark. Y1 X1 e’ a’ h’ b’ c’ g’ f’ d’ g h f For True Shape: Draw x1y1 // to sec. plane Draw projectors on it from cut points. Mark distances of points of Sectioned part from Tv, on above projectors from x1y1 and join in sequence. Draw section lines in it. It is required true shape. a e For Development: Draw development of entire solid. Name from cut-open edge i.e. A. in sequence as shown.Mark the cut points on respective edges. Join them in sequence in curvature. Make existing parts dev.dark. b d c TRUE SHAPE OF SECTION A SECTIONAL S.V B o’ SECTION PLANE DEVELOPMENT C D E X Y g” h”f” a”e” b”d” c” F G H A SECTIONAL T.V

  10. 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 a’ e’ c’g’ h’b’ d’f’ D a’ h’ b’ c’ g’ f’ e’ d’ E g1 g h f1 h1 F f G a e1 a1 o1 e O H b1 d1 b d A c c1 Follow similar solution steps for Sec.views - True shape – Development as per previous problem! DEVELOPMENT o’ HORIZONTAL SECTION PLANE Y X o’ O SECTIONAL T.V (SHOWING TRUE SHAPE OF SECTION)

  11. Problem 4: A hexagonal prism. 30 mm base side & 55 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 300 inclined to Vp bisecting axis. Draw sec. Views, true shape & development. c’ f’ Note the steps to locate Points 1, 2 , 5, 6 in sec.Fv: Those are transferred to 1st TV, then to 1st Fv and Then on 2nd Fv. e’ c’ a’ d’ f’ b’ e’ a’ d’ b’ 8 3 7 4 6 5 1 2 Use similar steps for sec.views & true shape. NOTE: for development, always cut open object from From an edge in the boundary of the view in which sec.plane appears as a line. Here 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. 5 6 2 1 X Y 4 3 8 7 f f1 e1 a1 e a d1 b1 d b c1 c X1 B A C F E D C Y1 SECTIONAL F.V. A.V.P300 inclined to Vp Through mid-point of axis. 1,2 3,8 4,7 5,6 AS SECTION PLANE IS IN T.V., CUT OPEN FROM BOUNDRY EDGE C1 FOR DEVELOPMENT. TRUE SHAPE OF SECTION DEVELOPMENT

  12. 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.) 3 2 4 Y1 5 1 Note: Fv & TV 8f two solids sandwiched Section lines style in both: Development of half cone & half pyramid: 6 A B 7 4’ C 3’ 5’ X1 2 4 2’ 3 6’ D 1 1’ E 7’ X Y O a’ c’f’ g’b’ d’e’ f 7 F g 6 e 5 G 4 5 6 7 4 A a 1 3 d 2 b c TRUE SHAPE O’ DEVELOPMENT F.V. SECTIONAL TOP VIEW.

  13. TO DRAW PRINCIPAL VIEWS FROM GIVEN DEVELOPMENT. 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. =  4 E 5 3 R + 3600 D F L R=Base circle radius. L=Slant height. 1’ C G L 6 2 7’ 6’ 2’ B H 1 7  5’ 3’ 4’ X Y A g L A O 6 Solution Steps: 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. 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 data 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’ and name 1’ Similarly locate all points on Fv. Then project all on Tv on respective generators and join by smooth curve. h f 5 a 7 o e 4 1 g’ a’ b d h’ b’ c’ d’f’ e’ 3 2 c o’

  14. TO DRAW PRINCIPAL VIEWS FROM GIVEN DEVELOPMENT. 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. =  4 E 5 3 R + 3600 D F L R=Base circle radius. L=Slant height. 1’ C G L 6 2 7’ 6’ 2’ B H 1 7  5’ 3’ 4’ X Y A g L A O 6 Solution Steps: 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. 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 data 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’ and name 1’ Similarly locate all points on Fv. Then project all on Tv on respective generators and join by smooth curve. h f 5 a 7 o e 4 1 g’ a’ b d h’ b’ c’ d’f’ e’ 3 2 c o’

  15. TO DRAW PRINCIPAL VIEWS FROM GIVEN DEVELOPMENT. Problem 7:Draw a semicircle 0f 100 mm diameter and inscribe in it a largest rhombus.If the semicircle is development of a cone and rhombus is some curve on it, then draw the projections of cone showing that curve. =  Solution Steps: Similar to previous Problem: E R + 3600 D F L R=Base circle radius. L=Slant height. C G B H 6’ 2’ 3’ 5’ a’ h’ b’ c’ g’ f’ d’ e’  Y X 1’ 7’ 4’ A g A O L 7 6 h f 5 a e 4 b 3 d 2 1 c o’ 1 2 3 4 5 6 7

  16. 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 face parallel 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 of shortest length, find it and show it on the projections of the cone. TO DRAW A CURVE ON PRINCIPAL VIEWS FROM DEVELOPMENT. o’ A Concept: A string wound from a point up to the same Point, of shortest length Must appear st. line on it’s Development. Solution steps: Hence draw development, Name it as usual and join A to A This is shortest Length of that string. Further steps are as usual. On dev. Name the points of Intersections of this line with Different generators.Bring Those on Fv & Tv and join by smooth curves. Draw 4’ a’ part of string dotted As it is on back side of cone. B C 1 D 3’ 2’ 2 4’ 1’ 3 E 4 X Y a’ c’ d’ e’ O b’ e a o 4 3 1 2 b d c A

  17. Problem 9: A 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. 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 Helix of one turn on cone: Draw Fv & Tv & dev.as usual On all form generators & name. Construction of curve Helix:: Show 8 generators on both views Divide axis also in same parts. Draw horizontal lines from those points on both end generators. 1’ is a point where first horizontal Line & gen. b’o’ intersect. 2’ is a point where second horiz. Line & gen. c’o’ intersect. In this way locate all points on Fv. Project all on Tv.Join in curvature. For Development: Then taking each points true Distance From resp.generator from apex, Mark on development & join. 7’ HELIX CURVE 6’ A 5’ 4’ B 3’ 1 2’ C 1’ 2 a’ h’ b’ c’ g’ f’ e’ d’ D g 3 O 4 h E f 7 5 6 6 5 7 F a e O 4 G 3 1 b d 2 H c A o’ DEVELOPMENT X Y

  18. 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 ofn cone 65 mm diameter ; axis 75 mm long. 4 3 5 2 6 1 7 12 8 11 9 10 1 12 11 10 9 8 θ=103º 7 6 5 4 3 Y 2 2 12 3 11 4 10 5 9 6 8 X 1 7

  19. Q e: A right circular cone base 30 mm side and height 50 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. 1 4 12 3 5 A 11 2 6 B 10 C 9 1 7 D 8 E θ=103º 12 8 F 7 G H 11 9 I 6 10 J f g 5 e h i K d j 4 c k L a l b 3 A 2 c d b e f g a h l i j k Y 2 12 3 11 4 10 5 9 6 8 X 1 7

  20. 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 D C B A 21 31 o’ 1 41 3’ 11 4 2’ 4’ O 3 1’ b’ d’ Y X c’ a’ 45º 2 d 4 1 1 a o c 3 2 b

  21. 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. a’ 60 b’e’ c’d’ a’ b’ e’ c’ d’ o’ b1 b c1 c o1 a1 o 30 a d1 d e1 e C o’ B D 5 A 6’ 1’ 4 E 3 2 1 6 5’ 2’ Y X 4’ 3’ A O 1 2 3 1 31’ 6 6 4 21’ 41’ 5 5 11’ 51’ 61’

  22. Q 14.11: 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. A D C B A 3 o’ 2 1 True length of slant edge 2 2 3 2’ 3’ 4 1 O True length of slant edge 3 1 4 1’ 4’ 2 Y X a’ d’ b’ c’ b a 1 2 1 o 3 4 c d

  23. 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. 4 3 5 2 6 1 7 b’ 12 8 a’ c’ 11 9 10 d’ 4’ 10’ 3’ 11’ 6’ 8’ 2’ 12’ 5’ 9’ 7’ 1’ B B A C C A D D X Y a c 1 3 5 9 2 4 6 7 10 1 8 11 12 b d a c c a b d

  24. 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. o’ A A1 D True length of slant edge P C R D1 B a’d’ b’c’ p’ A r’ S C1 s’ 75 B1 Q q’ a1’d1’ b1’c1’ X Y A1 d1 c1 p r d c 50 o 25 a b s b1 a1 q

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