Development of Surfaces of com

8/2/2019 Development of Surfaces of com

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1. SECTIONS OF SOLIDS.2. DEVELOPMENT.3. INTERSECTIONS.

ENGINEERING APPLICATIONSOF

THE PRINCIPLESOF

PROJECTIONS OF SOLIDES.

STUDY CAREFULLYTHE ILLUSTRATIONS GIVEN ON

NEXT SIX PAGES !


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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 appearsas 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 observeris assumed to be removed.

2. As far as possible the smaller part is

assumed to be removed.

OBSERVER

ASSUME

UPPER PART

REMOVED

OBSERVER

ASSUME

LOWER PART

REMOVED

(A)

(B)


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ILLUSTRATION SHOWINGIMPORTANT TERMS

IN SECTIONING.

x y

TRUE SHAPE

Of SECTION

SECTION

PLANE

SECTION LINES

(450 to XY)

Apparent Shape

of section

SECTIONAL T.V.

For TV


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Section PlaneThrough Apex

Section PlaneThrough Generators

Section Plane Parallelto end generator.

Section PlaneParallel to Axis.

Triangle Ellipse

Hyperbola

Ellipse

Cylinder through

generators.

Sq. Pyramid through

all slant edges

Trapezium

Typical Section Planes&

Typical ShapesOf

Sections.


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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 SOLIDS 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 its 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.


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D

H

D

SS

H

=

R

L 360

0

R=Base circle radius.L=Slant height.

L= Slant edge.

S = Edge of base

H= Height S = Edge of base

H= Height D= base diameter

Development of lateral surfaces of different solids.

(Lateral surface is the surface excluding top & base)

Prisms: No.of Rectangles

Cylinder: A RectangleCone: (Sector of circle) Pyramids: (No.of triangles)

Tetrahedron: Four Equilateral Triangles

All sides

equal in length

Cube: Six Squares.


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= RL 3600

R= Base circle radius of coneL= Slant height of cone

L1 = Slant height of cut part.

Base side

Top side

L= Slant edge of pyramid

L1 = Slant edge of cut part.

DEVELOPMENT OF

FRUSTUM OF CONE

DEVELOPMENT OF

FRUSTUM OF SQUARE PYRAMID

STUDY NEXT NINEPROBLEMS OF

SECTIONS & DEVELOPMENT

FRUSTUMS


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X Y

X1

Y1

A

B

C

E

D

a

e

d

b

c

A B C D E A

DEVELOPMENT

a

b

cd

e

Problem 1:A pentagonal prism , 30 mm base side & 50 mm axisis standing on Hp on its base with one side of the base perpendicular to VP.

It is cut by a section plane inclined at 45 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.

For True Shape:Draw x1y1 // to sec. planeDraw projectors on it fromcut points.Mark distances of pointsof Sectioned part from Tv,on above projectors fromx1y1 and join in sequence.

Draw section lines in it.It is required true shape.

For Development:Draw development of entire solid. Name fromcut-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.


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Y

h

a

b

c

d

e

g

f

X a b decg fh

o

X1

Y1

g hf ae bd c

A

B

C

D

E

F

A

G

H

SECTIONAL T.V

SECTIONAL S.V

DEVELOPMENT

Problem 2: A cone, 50 mm base diameter and 70 mm axis isstanding on its base on Hp. It cut by a section plane 450 inclinedto Hp through base end of end generator.Draw projections,sectional views, true shape of section and development of surfacesof remaining solid.

Solution Steps:for sectional views:Draw three views of standing cone.Locate sec.plane in Fv as described.Project points where generators aregetting Cut on Tv & Sv as shown inillustration.Join those points insequence and show Section lines in it.Make remaining part of solid dark.

For True Shape:Draw x1y1 // to sec. planeDraw projectors on it fromcut points.Mark distances of pointsof Sectioned part from Tv,on above projectors fromx1y1 and join in sequence.

Draw section lines in it.It is required true shape.

For Development:Draw development of entire solid.Name from cut-open edge i.e. A.in sequence as shown.Mark the cutpoints on respective edges.Join them in sequence in curvature.

Make existing parts dev.dark.


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X Yea b dcg fh

o

o

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 its projections.It is cut by a horizontal section plane through its base

center. Draw sectional TV, development of the surface of the remaining part of cone.

A

B

C

D

E

F

A

G

H

O

a1

h1

g1

f1

e1

d1

c1

b1

o1

SECTIONAL T.V

DEVELOPMENT

(SHOWING TRUE SHAPE OF SECTION)

HORIZONTAL

SECTION PLANE

h

a

b

c

d

e

g

f

O

Follow similar solution steps for Sec.views - True shape Development as per previous problem!


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A.V.P300 inclined to VpThrough mid-point of axis.

X Y1

2

3 4

5

6

78

b fa ec d

a

b

c

d

e

f

a1

d1b1

e1

c1

f1

X1

Y1

AS SECTION PLANE IS IN T.V.,

CUT OPEN FROM BOUNDRY EDGE C1 FOR DEVELOPMENT.

C D E F A B C

DEVELOPMENT

SECTIONAL F.V.

Problem 4: A hexagonal prism. 30 mm base side &55 mm axis is lying on Hp on its rect.face with axis

// to Vp. It is cut by a section plane normal to Hp and300 inclined to Vp bisecting axis.Draw sec. Views, true shape & development.

Use similar steps for sec.views & true shape.

NOTE: for development, always cut open object fromFrom an edge in the boundary of the view in whichsec.plane appears as a line.Here it is Tv and in boundary, there is c1 edge.Henceit is opened from c and named C,D,E,F,A,B,C.

Note the steps to locatePoints 1, 2 , 5, 6 in sec.Fv:Those are transferred to1st TV, then to 1st Fv and

Then on 2nd Fv.


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1

2

3

4

5

6

7

7

1

5

4

3

2

6

7

1

6

5

4

32

a

b

c

d

e

f

g

4

4 5

3

6

2

7

1

A

B

C

D

E

A

F

G

O

O

de cf gb aX Y

X1

Y1

F.V.

SECTIONAL

TOP VIEW.

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.)

Note:

Fv & TV 8f two solidssandwichedSection lines style in both:Development ofhalf cone & half pyramid:


13/25

o

h

a

b

c

d

g

f

o e

a b cg df ehX Y

= RL 3600


A

B

C

D E F

G

H

AO

1

3

2

4

7

6

5

L

1

2

3

4

5

6

7

1

2

3 45

6

7

Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largestcircle.If the semicircle is development of a cone and inscribed circle is somecurve on it, then draw the projections of cone showing that curve.

Solution Steps:Draw semicircle of given diameter, divide it in 8 Parts and inscribe in ita largest circle as shown.Name intersecting points 1, 2, 3 etc.Semicircle being dev.of a cone its radius is slant height of cone.( L )Then using above formula find R of base of cone. Using this datadraw Fv & Tv of cone and form 8 generators and name.Take o -1 distance from dev.,mark on TL i.e.oa on Fv & bring on oband name 1 Similarly locate all points on Fv. Then project all on Tv

on respective generators and join by smooth curve.

TO DRAW PRINCIPALVIEWS FROM GIVEN

DEVELOPMENT.


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o

h

a

b

c

d

g

f

o e

a b cg df ehX Y

= RL 3600


A

B

C

D E F

G

H

AO

1

3

2

4

7

6

5

L

1

2

3

4

5

6

7

1

2

3 45

6

7

Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largestcircle.If the semicircle is development of a cone and inscribed circle is somecurve on it, then draw the projections of cone showing that curve.

Solution Steps:Draw semicircle of given diameter, divide it in 8 Parts and inscribe in ita largest circle as shown.Name intersecting points 1, 2, 3 etc.Semicircle being dev.of a cone its radius is slant height of cone.( L )Then using above formula find R of base of cone. Using this datadraw Fv & Tv of cone and form 8 generators and name.Take o -1 distance from dev.,mark on TL i.e.oa on Fv & bring on oband name 1 Similarly locate all points on Fv. Then project all on Tv

on respective generators and join by smooth curve.


DEVELOPMENT.


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h

a

b

c

d

g

f

e

o

a b dc g fh eX Y

A

B

C

DE

F

G

H

AO L

= RL 3600


1

2 3

4

56

7

12

3

4

5

67

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.


DEVELOPMENT.

Solution Steps:Similar to previousProblem:


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a b c d

o

e

a

b

c

d

o e

X Y

A

B

C

D

E

A

O

2

3

4

1

Problem 8: A half cone of 50 mm base diameter, 70 mm axis, is standing on its half base on HP with its flat face

parallel and nearer to VP.An inextensible string is wound round its surface from one point of base circle and

brought back to the same point.If the string is ofshortest length, find it and show it on the projections of the cone.

1 2

3

4

12 3 4

TO DRAW A CURVE ONPRINCIPAL VIEWS

FROM DEVELOPMENT. Concept: A string woundfrom a point up to the samePoint, of shortest lengthMust appear st. line on itsDevelopment.Solution steps:Hence draw development,

Name it as usual and joinA to A This is shortestLength of that string.Further steps are as usual.On dev. Name the points ofIntersections of this line withDifferent generators.BringThose on Fv & Tv and joinby smooth curves.Draw 4 a part of string dotted

As it is on back side of cone.


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X Yea b dcg fh

o

h

a

b

c

d

e

g

f

O

DEVELOPMENT

A

B

C

D

E

F

A

G

H

O

12

3

4

6 57

1

2

3

4

5

6

7

1

2

3

4

567

HELIX CURVE

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 its path on projections of cone as well as on its development.

Take base circle diameter 50 mm and axis 70 mm long.

Its a construction of curve

Helix of one turn on cone:Draw Fv & Tv & dev.as usualOn all form generators & name.Construction of curve Helix::Show 8 generators on both viewsDivide axis also in same parts.Draw horizontal lines from those

points on both end generators.1 is a point where first horizontalLine & gen. bo intersect.2 is a point where second horiz.Line & gen. co intersect.In this way locate all points on Fv.Project all on Tv.Join in curvature.For Development:

Then taking each points trueDistance From resp.generatorfrom apex, Mark on development& join.

Q 15 26 d th j ti f ti th d it b d h th th h t t th


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X Y

1

2

34

5

6

7

8

910

11

12

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.

12

12311

410

59

68 7

2

3

4

5

6

7

8

9

10

11

12

1

=103

Q A ight i l b 30 mm id d h ight 50 mm t it b H P It i t b ti


20/25

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 45 to the H.P. and bisecting the axis. Draw the projections of the

truncated cone and develop its lateral surface.

X Y

1

2

34

5

6

7

8

910

11

12

12

12311

410

59

68 7

2

3

4

5

6

7

8

9

10

1112

1

a

b

c

k

d

ef

g

h

ilj

a

f

b

ck

d

e

g

hi

l

j

A

C

D

E

B

A

F

G

H

I

J

K

L

=103

Q 14 11 A id b 40 id d i 65 l h it b th HP d ll


21/25

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 45 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.

X45

a

b

c

d

o

ab

cd

o

1

2

3

4

1

2

3

411

41

21 31

Y

A

B

C

D

A

O

1

1

2

3

4

Q 14 14: A pentagonal pyramid base 30mm side and axis 60 mm long is lying on one of its triangular faces


22/25

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 30 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.

YXa b e c d

a

b

c

d

e

o

o

60

cd o

a

be

30 a1

b1

c1

d1

e1

o1

1

2

34

5

6

1

2

3

4

5

631

41

21

11

61

51

O

A

BC

D

E

A

12

3

4

5

6

1

5

6

Q 14 11: A square pyramid base 40 mm side and axis 65 mm long has its base on the HP with


23/25

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 45 to the HP and bisecting the axis. Draw its sectional top view and true shape of

the section. Also draw its development.

X

o

Y

A

B

C

D

A

O

a b

cd

o

a d b c

1

2

3

4

1 4

2 3

2

3

1

2True length

of slantedge

1 4

1

1

4

2 3

2

3True length

of slantedge

Q 15 11: A right circular cylinder base 50 mm diameter and axis 60 mm long is standing on HP on its


24/25

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.

Y

1

2

34

5

6

7

8

9

10

11

12

X

1

2

12

3

11

4

105

9

6

87

a

b

c

d

1 2 3 4 5 6 7 8 9 10 11 12 1

a

a

bd

bd

c

c

A

B

D

C C

B

D

A

a c

Q 15 21: A frustum of square pyramid has its base 50 mm side top 25 mm side and axis 75 mm Draw


25/25

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.

YX

50 25

75

a b

cd

c1d1

a

db

c

a1d1

b1c1

o

o

Truelength of

slantedge

A1

B1

C1

D1

A1

A

B

C

D

A

P

Q

R

S

p

p

q

r

r

s

s

Documents

Development of Surfaces of com