../ ~ A solid has three dimensions, viz. length, breadth and thickness. To represent a solid on a flat surface having only length and breadth, at least two orthographic views are necessary. Sometimes, additional views projected on auxiliary planes become necessary to make the description of a solid complete. This chapter deals with the following topics: 1. Types of solids. 2. Projections of solids in simple positions. (a) Axis perpendicular to the H.P. (b) Axis perpendicular to the V.P. (c) Axis parallel to both the H.P. and the V.P. 3. Projections of solids with axes inclined to one of the reference planes and parallel to the other. (a) Axis inclined to the V.P. and parallel to the H.P. (b) Axis inclined to the H.P. and parallel to the V.P. 4. Projections of solids with axes inclined to both the H.P. and the V.P. 5. Projections of spheres. YA; 1 This book is accompanied by a computer CD, which contains an audiovisual animation presented for better visualization and understanding of the subject. Readers are to refer Presentation module 28 for the types of Solids may be divided into two main groups: (1) Polyhedra (2) Solids of revolution. (1) When all A polyhedron is defined as a solid bounded by planes called faces. faces are equal and regular, the polyhedron is said to be regular. There are seven regular polyhedra which may be defined as stated below: (i) Tetrahedron 3-·1): It has four equal faces, each an equilateral triangle. (ii) Cube or hexahedron ·1 3-2): It has six faces, all equal squares.
[Ch. 13 Engineering Drawing (iii) Octahedron (fig. 13-3): It has eight equal equilateral triangles as faces. Cube Octahedron FIG. 'J 3-2 Tetrahedron FIG. 13-1 FIG. 13-3 (iv) Dodecahedron (fig. 13-4): It has twelve equal and regular pentagons as faces. (v) Icosahedron (fig. 13-5): It has twenty faces, all equal equilateral triangles. Dodecahedron FIG. 'l 3-4 Icosahedron FIG. '13-5 (vi) Prism: This is a polyhedron having two equal and similar faces called its ends or bases, parallel to each other and joined by other faces which are parallelograms. The imaginary line joining the centres of the bases is called the axis. A right and regular prism (fig. 13-6) has its axis perpendicular to the bases. All its faces are equal rectangles. i i ~i ~i i i -----~Triangular Square Pentagonal Hexagonal Prisms FIG. 13-6 (vii) Pyramid: This is a polyhedron having a plane figure as a base and a number of triangular faces meeting at a point called the vertex or apex. The imaginary line joining the apex with the centre of the base is its axis. A right and regular pyramid (fig. 13-7) has its axis perpendicular to the base which is a regular plane figure. Its faces are all equal isosceles triangles.
Art. 13-1] 273 Oblique prisms and pyramids have their axes inclined to their bases. Prisms and pyramids are named according to the shape of their bases, as triangular, square, pentagonal, hexagonal etc. APEX TRIANGULAR SQUARE PENTAGONAL HEXAGONAL Pyramids FIG. 13-7 (2) Solids of revolution: (i) Cylinder (fig. 13-8): A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides which remains fixed. It has two equal circular bases. The line joining the centres of the bases is the axis. It is perpendicular to the bases. (ii) Cone (fig. 13-9): A right circular cone is a solid generated by the revolution of a right-angled triangle about one of its perpendicular sides which is fixed. It has one circular base. Its axis joins the apex with the centre of the base to which it is perpendicular. Straight lines drawn from the apex to the circumference of the base-circle are all equal and are called generators of the cone. The length of the generator is the slant height of the cone. Cylinder FIG. 13-8 Cone FIG. 13-9 Sphere FIG. 13-10 (iii) Sphere (fig. 'l 3-10): A sphere is a solid generated by the revolution of a semi-circle about its diameter as the axis. The mid-point of the diameter is the centre of the sphere. All points on the surface of the sphere are equidistant from its centre. Oblique cylinders and cones have their axes inclined to their bases. (iv) Frustum: When a pyramid or a cone is cut by a plane parallel to its base, thus removing the top portion, the remaining portion is called its frustum (fig. 13-11).
Engineering Drawing (v) [Ch. 13 Truncated: When a solid is cut by a plane inclined to the base it is said to be truncated. In this book mostly right and regular solids are dealt with. Hence, when a solid is named without any qualification, it should be understood as being right and regular. Frustums FIG. 13-11 A solid in simple position may have its axis perpendicular to one reference plane or parallel to both. When the axis is perpendicular to one reference plane, it is parallel to the other. Also, when the axis of a solid is perpendicular to a plane, its base will be parallel to that plane. We have already seen that when a plane is parallel to a reference plane, its projection on that plane shows its true shape and size. Therefore, the projection of a solid on the plane to which its axis is perpendicular, will show the true shape and size of its base. Hence, when the axis is perpendicular to the ground, i.e. to the H.P., the top view should be drawn first and the front view projected from it. When the axis is perpendicular to the V.P., beginning should be made with the front view. The top view should then be projected from it. When the axis is parallel to both the H.P. and the V.P., neither the top view nor the front view will show the actual shape of the base. In this case, the projection of the solid on an auxiliary plane perpendicular to both the planes, viz. the side view must be drawn first. The front view and the top view are then projected from the side view. The projections in such cases may also be drawn in two stages. (1) Axis perpendicular to the H.P.: Problem 13-1. (fig. 13-12): Draw the projections of a triangular prism, base 40 mm side and axis 50 mm long, resting on one of its bases on the H.P. with a vertical face perpendicular to the V.P. (i) As the axis is perpendicular to the ground i.e. the H.P. begin with the top view. It will be an equilateral triangle of sides 40 mm long, with one of its a'r.-:----,c' b' FIG. 13-12
Ari. 13-2] 275 sides perpendicular to xy. Name the corners as shown, thus completing the top view. The corners d, e and fare hidden and coincide with the top corners a, b and c respectively. (ii) Project the front view, which will be a rectangle. Name the corners. The line b'e' coincides with a'd'. Problem 13-2. (fig. 13-13): Draw the projections of a pentagonal pyramid, base 30 mm edge and axis 50 mm long, having its base on the H.P. and an edge of the base parallel to the V.P. Also dravv its side view. X1 FIG. 13-13 (i) Assume the side DE which is nearer the V.P., to be parallel to the V.P. as shown in the pictorial view. In the top view, draw a regular pentagon abcde with ed parallel to and nearer xy. Locate its centre o and join it with the corners to indicate the slant edges. (iii) Through o, project the axis in the front view and mark the apex o', 50 mm (ii) above xy. Project all the corners of the base on xy. Draw lines o'a', o'b' and o'c' to show the visible edges. Show o'd' and o'e' for the hidden edges as dashed lines. (iv) For the side view looking from the left, draw a new reference line x 1y 1 perpendicular to xy and to the right of the front view. Project the side view on it, horizontally from the front view as shown. The respective distances of all the points in the side view from x 1y 1 , should be equal to their distances in the top view from xy. This is done systematically as explained below: (v) From each point in the top view, draw horizontal lines upto x 1y1 . Then draw lines inclined at 45° to x 1y1 (or xy) as shown. Or, with q, the point of intersection between xy and x 1y1 as centre, draw quarter circles. Project up all the points to intersect the corresponding horizontal lines from the front view and complete the side view as shown in the figure. Lines o 1d 1 and o 1 c 1 coincide with o 1 e1 and o 1 a1 respectively. Problem 13-3. (fig. 13-14): Draw the projections of (i) a cylinder, base 40 mm diameter and axis 50 mm Jong, and (ii) a cone, base 40 mm diameter and axis 50 mm long, resting on the H.P. on their respective bases.
Engineering Drawing [Ch. 13 (i) Draw a circle of 40 mm diameter in the top view and project the front view which will be a rectangle [fig. 13-14(ii)]. (ii) Draw the top view [fig. 13-14(iii)]. Through the centre o, project the apex o', 50 mm above xy. Complete the triangle in the front view as shown. (i) (iii) (ii) FIG. 13-14 In the pictorial view [fig. 13-14(i)], the cone is shown as contained by the cylinder. 13-4. (fig. 13-15): A cube of 50 mm long edges is resting on the H.P. with its vertical faces equally inclined to the V.P. Draw its projections. b' d' a' I c' I C X h' e' f' d g' y b FIG. 13-15 Begin with the top view. (i) Draw a square abed with a side making 45° angle with xy. (ii) Project up the front view. The line d' h' will coincide with b' f'. Problem 13-5. (fig. 13-16): Draw the projections of a hexagonal pyramid, base 30 mm side and axis 60 mm long, having its base on the H.P. and one of the edges of the base inclined at 45° to the V.P.
Art. 13-2] 277 (i) In the top view, draw a line af 30 mm long and inclined at 45° to xy. Construct a regular hexagon on af. Mark its centre o and complete the top view by drawing lines joining it with the corners. (ii) Project up the front view as described in problem 13-2, showing the line o'e' and o'f for hidden edges as dashed lines. o' I a d C FIG. 13-16 FIG. 13-17 Problem 13-6. (fig. 13-17): A tetrahedron of 5 cm long edges is resting on the f-1.P. on one of its faces, with an edge of that face parallel to the V.P. Draw its projections and measure the distance of its apex from the ground. All the four faces of the tetrahedron are equal equilateral triangles of 5 crn side. (i) Draw an equilateral triangle abc in the top view with one side, say ac, parallel to xy. Locate its centre o and join it with the corners. (ii) In the front view, the corners a', b' and c' will be in xy. The apex o' will lie on the projector through o so that its true distance from the corners of the base is equal to 5 cm. (iii) To locate o', make oa (or ob or oc) parallel to xy. Project a1 to a' 1 on xy. With a' 1 as centre and radius equal to 5 cm cut the projector through o in o'. Draw lines o'a', o'b' and o'c' to complete the front view. o'b' will be the distance of the apex from the ground. (2) Axis perpendicular to the V.P.: Problem 13-7. (fig. 13-18): A hexagonal prism has one of its rectangular faces parallel to the H.P. Its axis is perpendicular to the V.P. and 3.5 cm above the ground. Draw its projections when the nearer end is 2 cm in front of the V.P. Side of base 2.5 cm long; axis 5 cm long. (i) Begin with the front view. Construct a regular hexagon of 2.5 cm long sides with its centre 3.5 cm above xy and one side parallel to it. (ii) Project down the top view, keeping the line for nearer end, viz. 1-4, 2 cm below xy.
[Ch. 13 Engineering Drawing o' x........i.---1--1----1---1--y 2 P 6 5 e f a 3 b O C d FIG. 13-18 Problem 13-8. (fig. 13-19): A square pyramid, base 40 mm side and axis 65 mm long, has its base in the V.P. One edge of the base is inclined at 30° to the H.P. and a corner contained by that edge is on the H.P. Draw its projections. (i) Draw a square in the front view with the corner d' in xy and the side d'c' inclined at 30° to it. Locate the centre o' and join it with the corners of the square. (ii) Project down all the corners in xy (because the base is in the V.P.). Mark the apex o on a projector through o'. Draw lines for the slant edges and complete the top view. b' 0 FIG. 13--19 (3) Axis to the H.P. and the V.P.: Problem 13-9. (fig. 13-20): A triangular prism base 40 mm side and height 65 mm is resting on the H.P. on one of its rectangular faces with the axis parallel to the V.P. Draw its projections. 1
Projections of Solids As the axis is parallel to both the planes, begin with the side view. e' (i) Draw an equilateral triangle representing the side view, with one side in xy. (ii) Project the front view horizontally from this triangle. (iii) Project down the top view from the front view and the side view, as shown. p' a1 279 b' ·-·-·-·-· o' c' a' ~----~~-t-------~Y X-'-1 This problem can also be solved in two stages as explained in the next article. d' f' ~1--------....,c eP a d FIG. 13-20 Draw the projections of the following solids, situated in their respective positions, taking a side of the base 40 mm long or the diameter of the base 50 mm long and the axis 65 mm long. 1. 2. 3. 4. 5. 6. 7. 8. A hexagonal pyramid, base on the H.P. and a side of the base parallel to and 25 mm in front of the V.P. A square prism, base on the H.P., a side of the base inclined at 30° to the V.P. and the axis 50 mm in front of the V.P. A triangular pyramid, base on the H.P. and an edge of the base inclined at 45° to the V.P.; the apex 40 mm in front of the V.P. A cylinder, axis perpendicular to the V.P. and 40 mm above the H.P., one end 20 mm in front of the V.P. A pentagonal prism, a rectangular face parallel to and 10 mm above the H.P., axis perpendicular to the V.P. and one base in the V.P. A square pyramid, all edges of the base equally inclined to the H.P. and the axis parallel to and 50 mm away from both the H.P. and the V.P. A cone, apex in the H.P. axis vertical and 40 mm in front of the V.P. A pentagonal pyramid, base in the V.P. and an edge of the base in the H.P. ~4 ~~ When a solid has its axis inclined to one plane and parallel to the other, its projections are drawn in two stages. (1) In the initial stage, the solid is assumed to be in simple position, i.e. its axis perpendicular to one of the planes. If the axis is to be inclined to the ground, i.e. the H.P., it is assumed to be perpendicular to the H.P. in the initial stage. Similarly, if the axis is to be inclined to the V.P., it is kept perpendicular to the V.P. in the initial stage.
![Art. 13-3]](./files/e7d84ad503d6411115eca825d4904418/small_images/pg_9-P.jpg "Art. 13-3]")
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