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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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slide1

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

slide2

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)

slide3

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.

slide4

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

slide5

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.

slide6

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

slide7

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.

slide8

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

slide9

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

slide10

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)

slide11

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

slide12

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.

slide13

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’

slide14

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’

slide16

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

slide17

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

slide18

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

slide19

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

slide20

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

slide21

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

slide22

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’

slide23

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

slide24

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

slide25

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