Topic :- Projection of solids

Subject :- Engineering graphics (2110013)

CO (evening) -: Prepared Bye :-

PATEL ADARSH (36)PATEL AVNI (37)PATEL JANKI (38)PATEL KRIMA (39)PATEL MOSAM (40)PATEL NIDHI (41)PATEL UJJAVAL (42)PATEL YASH (43)

Guided Faculty :-

PROF. BHAVIK BHAGATPROF. ROHIT DUDHAT PROF. NITIN VIBHAKARPROF. AMIT PATEL

SARVAJANIK COLLEGE OF ENGG. & TECH.

Definition of Solid :-

A solid is a three dimensional object having length, breadth and thickness. It is completely bounded by a surface or surfaces which may be curved or plane.

The shape of the solid is described by drawing its two orthographic views usually on the two principle planes i.e. H.P. & V.P.

Solids may be divided into two main groups

(A) Polyhedra

(B) Solids of revolution

A Polyhedra is defined as a solid bounded by planes called faces which meet in straight lines called edges.

There are seven regular Polyhedra which may be defined as stated below;

(3) Tetrahedron(4) Cube or Hexahedron

(5) Octahedron:(6) Dodecahedron:

(7) Icosahedron:

(1) Prism(2) Pyramid

(A) Polyhedra

It is a polyhedra having two equal and similar faces called its ends or bases, parallel to each other and joined by other faces which are rectangles.

(1) Prism

Edge

Axis

Faces

Square Prism

Pentagonal Prism

Hexagonal Prism

Triangular Prism

This is a polyhedra having plane surface as a base and a number of triangular faces meeting at a point called the Vertex or Apex.

(2) Pyramid

Edge

Base

Axis

Square Pyrami

dTriangular

Pyramid

Pentagonal Pyramid

Hexagonal Pyramid

(B) Solids of Revolutions:

When a solid is generated by revolutions of a plane figure about a fixed line (Axis) then such solids are named as solids of revolution.

Solids of revolutions may be of following types

(1) Cylinder

(2) Cone

(3) Sphere

(4) Ellipsoid

(5) Paraboloid

(6) Hyperboloid

(1) Cylinder

A right regular cylinder is a solid generated by the revolution of a rectangle about its vertical side which remains fixed.

RectangleAxis

Base

(2) Cone

A right circular cone is a solid generated by the revolution of a right angle triangle about its vertical side which remains fixed.

Right angle triangle

Axis

Base

Generators

(3) Right Solid

A solid is said to be a Right Solid if its axis is perpendicular to its base.

AxisBase

A solid is said to be a Oblique Solid if its axis is inclined at an angle other than 90° to its base.

Axis

Base

(4) Oblique Solid

(5) Frustum of Solid

When a Pyramid or a Cone is cut by a Plane parallel to its base, thus removing the top portion, the remaining lower portion is called its frustum.

FRUSTUM OF A PYRAMID

CUTTING PLANE PARALLEL TO BASE

c’,3’b’,2’

A:- Axis perpendicular to V.P. and hence parallel to both H.P. & P.P.

f’,6’

a

e’,5’

d’,4’a’,1’

b,f c,e d

43,52,61X Y

H

A hexagonal prism side of base 30 mm & axis length 60 mm is resting on one of its base edge on HP such that its axis inclined at 45⁰ to H.P. and the side on which it is resting is inclined at 30⁰ to V.P. draw the projection.

Problem :- 1

6030

45⁰

30⁰

a,1

b,2

c,3

d,4e,5

f,6

o,s

1 ‘,6 ‘ 5 ‘,2 ‘ 4 ‘,3 ‘

c ‘,d ‘b ‘ e ‘,s’a ‘,f ‘

1 ‘,6 ‘

5 ‘,2 ‘

4 ‘,3 ‘

c ‘,d ‘

b ‘,e ‘,s’

a ‘,f ‘

31

61

21

41

1151

31

61

21

11

51

41

c1’ d1’

b1’ e1’

e1

b1

a1 c1

d1f1

e1

b1

a1

c1

d1

f1

11’

21’

31’ 41’

61’

51’

a1’f1’

Problem : 2

A cube of 50 mm long edges is so placed on HP on one corner that a body diagonal is parallel to HP and perpendicular to VP Draw it’s projections.

Steps :1.Assuming standing on HP, begin with TV,a square with all sidesequally inclined to XY. Project FV and name all points of FV & TV.2.Draw a body-diagonal joining c’ with 1’( This can become // to xy)3.From 3’ drop a perpendicular on this and name it p’4.Draw 2nd Fv in which 3’p’ line is vertical means c’-1’ diagonal must be horizontal. .Now as usual project TV.. 6.In final TV draw same diagonal is perpendicular to VP as said in problem.Then as usual project final FV.

60

C 3a 1

B 2

D 4

3’ 2’4’ 1’

C’B’D’ A’

3’

2’4’

1’C’

B’D’

A’

1

4d

c 3 a

b2

c

b

1

4

d 3

2

3’

2’

B’

A’

D’

4’

C’ 1’P’

Problem : 3

A pentagonal prism side of base 50 mm & axis height 80 mm is held on a corner of its base on HP such that the opposite rectangular face appears to be a square in plan draw the projection of prism when the vertical edge containing that corner is inclined at 30⁰ to VP. find the inclination of the prism with HP. The top pentagonal faces of the prism is towards the observer

Problem : 4

A Hexagonal pyramid, side of base 30 mm and axis length 70 mm is resting on the H.P. on one of its base edges in such way that its apex is 55 mm above H.P. Draw the Projection of pyramid when the base edge which is on the H.P. is parallel to V.P. Keep apex of the solid on V.P.

• Initial position: Step 1:Since an edge of the base of the pyramid is on H . P.,

initially we will keep the base of the pyramid on H.P. with a side of base perpendicular to V.P.(i.e. XY)

Step 2:Draw the hexagon of side 30 mm in the T.V. with side ab perpendicular to XY line. Projection the F.V.

• Intermediate position: Step 1:Apex of the pyramid must be 55 mm above H.P.Hence

redraw the F.V. of initial stage keeping d’e’ on the XY line and apex o’ 55 mm above XY line.

Step 2:Project various points of the F.V. vertically down and project the points of initial stage T.V. parallel to XY line and to the right to locate same point in second stage T.V.

Step 3:Join the points in proper sequence following the rules of visibility and non-visibility.

Final position: Step 1:Apex of the pyramid is kept on V.P. of side 30

mm in elevation, base edge ef must be parallel to V.P. redraw the T.V. of intermediate stage with base edge ef parallel to XY line.

Step 2:Project the points from final stage T.V. vertically up and project the points from intermediate stage F.V. parallel to the XY line and to the right to locate the points in the final stage F.V.

Step 3: Join the points in proper sequence following the rules of visibility and non-visibility.

Problem:5 A triangular pyramid side of base 30 mm & axis height 60 mm is freely suspended from midpoint of one of its slant edge. Draw the projection of pyramid when axis is inclined at 20˚ to VP.

Step 1: Pyramid is freely suspended from mid-point ‘P’ of slant edge .We will assume the base of the pyramid on HP in such a way that slant edge oc should be parallel to VP.

Step 2: Draw an equilateral triangle of side 30 mm in the TV with TV of slant edge parallel to XY line. Project the FV locate the C.G. of the pyramid on the axis o’ s’ as g’. 15 mm above the base also locate point P’ on the mid-point of slant edge o’ c’ , joint p’ to g’.

Step 3: Redraw the FV of initial stage by keeping p’g’ perpendicular to XY line. Care should be taken that no part of the FV of intermediate stage is touching the XY line as pyramid is freely suspended.

Step 4: project various point from FV vertically down and project points of initial stage TV parallel to the XY Line and to the right to locate same points in second stage TV.

Step 5: Joint the point in proper sequence following the rules the visibility and non- visibility.

Step 6: Axis of the pyramid os is making an angle of 20˚ with the VP but it is seen in apparent length in the TV Hence we will calculate βos Redraw the TV of intermediate stage with plan of axis os making βos with the XY line.

Step 7: project the points from final stage TV vertically up and project the points intermediate stage FV ???parallel??? to XY line and to the right to locate the points in the final stage FV.

Step 8 : Joint the point in proper sequence following the rules the visibility and non- visibility.

c’

c

o’

a’(b’)

p’

g’

s’

a

bg,po,s

c’

o’

g’

p’

a’(b’)s’

c₁

a₁

b₁

o₁g₁,p₁s₁

c₁’

a₁’c₁

a₁

b₁

o₁

b₁’

o₁’

X Y

Problem No : 9A right cylinder of 50 mm diameter & 100 mm length. Draw the projection when it’s axis is inclined at angle of 50 to H.P. And 17 to V.P.

100

R=25

50

17

Ɓ

Q-9

Locus of oo

Problem No : 10A frustum of a cone, having bottom base diameter 70 mm top base diameter and axis is 50 mm long is resting on one of its generator on the ground. Its axis is inclined at 45˚ to VP. Draw its projections.