section and development [compatibility mode]

8/13/2019 Section and Development [Compatibility Mode]

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ILLUSTRATION SHOWINGILLUSTRATION SHOWING

IMPORTANT TERMSIMPORTANT TERMS

IN SECTIONING.IN SECTIONING.

xx yy

TRUE SHAPETRUE SHAPE

Of SECTIONOf SECTION

SECTIONSECTION

PLANEPLANE

SECTION LINESSECTION LINES

(45(4500 to XY)to XY)

Apparent ShapeApparent Shape

of sectionof section

SECTIONAL T.V.SECTIONAL T.V.

For TVFor TV

Section PlaneSection Plane

Through ApexThrough Apex


Through GeneratorsThrough Generators

Section Plane ParallelSection Plane Parallel

to end generator.to end generator.


Parallel to Axis.Parallel to Axis.

TriangleTriangle EllipseEllipse

HyperbolaHyperbola

EllipseEllipse

Cylinder throughCylinder through

generators.generators.

Sq. Pyramid throughSq. Pyramid through

all slant edgesall slant edges

TrapeziumTrapezium

Typical Section PlanesTypical Section Planes

&&

Typical ShapesTypical ShapesOfOf

SectionsSections..



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DEVELOPMENT OF SURFACES OF SOLIDS.DEVELOPMENT OF SURFACES OF SOLIDS.

MEANING: MEANING:--

ASSUME OBJECT HOLLOW AND MADEASSUME OBJECT HOLLOW AND MADE--UP OF THIN SHEET. CUT OPEN IT FROM ONE SIDE ANDUP OF THIN SHEET. CUT OPEN IT FROM ONE SIDE AND

UNFOLD THE SHEET COMPLETELY. THEN THEUNFOLD THE SHEET COMPLETELY. THEN THE SHAPE OF THAT UNFOLDED SHEET IS CALLEDSHAPE OF THAT UNFOLDED SHEET IS CALLED

DEVELOPMENT OF LATERLAL SUEFACESDEVELOPMENT OF LATERLAL SUEFACES OF THAT OBJECT OR SOLID.OF THAT OBJECT OR SOLID.

LATERLAL SURFACELATERLAL SURFACE IS THE SURFACE EXCLUDING SOLID’S TOP & BASE.IS THE SURFACE EXCLUDING SOLID’S TOP & BASE.

ENGINEERING APLICATION ENGINEERING APLICATION ::

THERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BYTHERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BY

CONVENTIONAL MANUFACTURING PROCESSES, BECAUSE OF THEIR SHAPES AND SIZES.CONVENTIONAL MANUFACTURING PROCESSES, BECAUSE OF THEIR SHAPES AND SIZES.

THOSE ARE FABRICATED IN SHEET METAL INDUSTRY BY USINGTHOSE ARE FABRICATED IN SHEET METAL INDUSTRY BY USING

DEVELOPMENT TECHNIQUE. THERE IS A VAST RANGE OF SUCH OBJECTS.DEVELOPMENT TECHNIQUE. THERE IS A VAST RANGE OF SUCH OBJECTS.

EXAMPLES: EXAMPLES:--

Boiler Shells & chimneys, Pressure Vessels, Shovels, Trays, Boxes & Cartons, Feeding Hoppers,Boiler Shells & chimneys, Pressure Vessels, Shovels, Trays, Boxes & Cartons, Feeding Hoppers,

Large Pipe sections, Body & Parts of automotives, Ships, Aeroplanes and many more.Large Pipe sections, Body & Parts of automotives, Ships, Aeroplanes and many more.

WHAT ISWHAT IS

OUR OBJECTIVEOUR OBJECTIVE

IN THIS TOPIC ?IN THIS TOPIC ?

To learn methods of development of surfaces of To learn methods of development of surfaces of

different solids, their sections and frustumsdifferent solids, their sections and frustums..

1. Development is different drawing than PROJECTIONS.1. Development is different drawing than PROJECTIONS.

2. It is a shape showing AREA, means it’s a 22. It is a shape showing AREA, means it’s a 2--D plain drawing.D plain drawing.

3. Hence all dimensions of it must be TRUE dimensions.3. Hence all dimensions of it must be TRUE dimensions.

4. As it is representing shape of an un4. As it is representing shape of an un--folded sheet, no edges can remain hiddenfolded sheet, no edges can remain hidden

And hence DOTTED LINES are never shown on development.And hence DOTTED LINES are never shown on development.

But before going ahead, But before going ahead,

note followingnote following

Important Important points points..

Study illustrations given on next page carefully.Study illustrations given on next page carefully.

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.

X Y45º

a

b

c

d

o

a’b’

c’d’

o’

1

2

3

4

1’

2’

3’

4’11

41

21 31

X1

Y1

d” a”c” b”

o”

3”

2”4”

1”



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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.

X Y

a

b

c

d

e

o

a’ b’e’c’d’

o’

6 0

3 0

c’d’ o’

a’

b’e’

a1

b1

c1

d1

e1

o1

1’

2’

3’4’

5’

6’

1

2

3

4

5

6 31’

41’

21’

11’

61’

51’

Q 14.6: A Hexagonal prism has a face on the H.P. and the axis parallel to the V.P. It is cut by a vertical section

plane the H.T. of which makes an angle of 45 with XY and which cuts the axis at a point 20 mm from one of

its ends. Draw its sectional front view and the true shape of the section. Side of base 25 mm long height

65mm.

X Y

a

b

c

d

e

f

a’ b’ c’d’e’f’

2 5

6 5

a’ b’ c’d’e’f’

a’

b’

c’d’

e’

f’a’

b’

c’d’

e’

f’

d1

a1

b1

c1

e1

f 1d1

a1

b1

c1

e1

f 120

1’

2’

3’4’

5’

6’ 7’

1 23

4

5

6

7

X1

Y1

31’

41’

21’

11’

71’

61’51’



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X Y

1

2

34

5

6

7

8

910

11

12

Q 14.24: A Cone base 75 mm diameter and axis 80 mm long is resting 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 cutting the axis at a point 35 mm from the

apex. Draw the front view, sectional top view, sectional side view and true shape of the section.

12

12

3

11

4

105

9

6

8 7

o

o’

3 5

a

b

k

c d

l

e

f

g

h

i

j

a’

b’

k’c’

d’

l’

e’f’

g’

h’i’

j’

X1

Y1

4” 5” 6” 7” 8” 9”10”

11”12”1”2”3”

o”

a”

b”

c”

d”

e”

f”g”

h”

i”

j”

k”

l”

D

H

D

SS

H

= R L

3600

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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Y

h

a

b

c

d

e

g

f

X a’ b’ d’ e’c’ g’ f’h’

o’

X1

Y1

g” h”f” a”e” b”d” c”

A

B

C

D

E

F

A

G

H

SECTIONAL T.V

SECTIONAL S.V

DEVELOPMENT

Problem 2:Problem 2: A cone, 50 mm base diameter and 70 mm axis is A cone, 50 mm base diameter and 70 mm axis is

standing on it’s base on Hp. It cut by a section plane 45standing on it’s base on Hp. It cut by a section plane 4500 inclinedinclined

to Hp through base end of end generator.Draw projections,to Hp through base end of end generator.Draw projections,

sectional views, true shape of section and development of surfacessectional views, true shape of section and development of surfaces

of remaining solid.of remaining solid.

Solution Steps:Solution Steps:for sectional views:for sectional views:

Draw three views of standing cone.Draw three views of standing cone.

Locate sec.plane in Fv as described.Locate sec.plane in Fv as described.

Project points where generators areProject points where generators are

getting Cut on Tv & Sv as shown ingetting Cut on Tv & Sv as shown in

illustration.Join those points inillustration.Join those points insequence and show Section lines in it.sequence and show Section lines in it.Make remaining part of solid dark.Make remaining part of solid dark.

For True Shape:For True Shape:

Draw xDraw x11yy11 // to sec. plane// to sec. plane

Draw projectors on it fromDraw projectors on it fromcut points.cut points.

Mark distances of pointsMark distances of points

of Sectioned part from T v,of Sectioned part from T v,

on above projectors fromon above projectors fromxx11yy11 and join in sequence.and join in sequence.

Draw section lines in it.Draw section lines in it.

It is required true shape.It is required true shape.

For Development:For Development:

Draw development of entire solid.Draw development of entire solid.

Name from cutName from cut--open edge i.e. A.open edge i.e. A.

in sequence as shown.Mark the cutin sequence as shown.Mark the cut

points on respective edges.points on respective edges.

Join them in sequence in curvature.Join them in sequence in curvature.

Make existing parts dev.dark.Make existing parts dev.dark.

X Ye’a’ b’ d’c’ g’ f’h’

o’

o’

Problem 3: A cone 40mm diameter and 50 mm axis is resti ng o n one generator on Hp( l ying on Hp)

which is // to Vp.. Draw it’s pro jections.It is cut by a horizontal section plane through it ’s base

center. Draw sectional TV, development of th e surface of t he remaining part of cone.

A

B

C

D

E

F

A

G

H

O

a1

h1

g1

f 1

e1

d 1

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.viewsFollow similar solution steps for Sec.views -- True shapeTrue shape – – Development as p er previous probl em!Development as per previous problem!



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A.V.P300 inclined to Vp

Through mid-point of axis.

X Y1

2

3 4

5

6

78

b’ f’a’ e’c’ d’

a

b

c

d

e

f

a1

d 1 b1

e1

c1

f 1

X1

Y1

AS SECTION PLANE IS IN T.V.,

CUT OPEN FROM BOUNDRY EDGE C 1 FOR DEVELOPMENT.

C D E F A B C

DEVELOPMENT

SECTIONAL F.V.

Problem 4:Problem 4: A hexagonal prism. 30 mm base side & A hexagonal prism. 30 mm base side &

55 mm axis is lying on Hp on it’s rect.face with axis55 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// to Vp. It is cut by a section plane normal to Hp and

303000 inclined to Vp bisecting axis.inclined to Vp bisecting axis.

Draw sec. Views, true shape & development.Draw sec. Views, true shape & development.

Use similar steps for sec.views & t rue shape.Use similar steps for sec.views & t rue shape.NOTE:NOTE: for development, always cut open object fromfor development, always cut open object from

From an edge in the boundary of the view in whichFrom an edge in the boundary of the view in which

sec.plane appears as a line.sec.plane appears as a line.

Here it is Tv and in boundary, there is c1 edge.HenceHere it is Tv and in boundary, there is c1 edge.Henceit is opened from c and named C,D,E,F,A,B,C.it is opened from c and named C,D,E,F,A,B,C.

NoteNote the steps to locatethe steps to locate

Points 1, 2 , 5, 6 in sec.Fv:Points 1, 2 , 5, 6 in sec.Fv:

Those are transferred toThose are transferred to

11stst TV, then to 1TV, then to 1stst Fv andFv and

Then on 2Then on 2ndnd Fv.Fv.

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’

d’e’ c’f’ g’b’ a’X 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 45 0 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:Note:Fv & TV 8f two solidsFv & TV 8f two solids

sandwichedsandwiched

Section lines style in both:Section lines style in both:

Development ofDevelopment of

half cone & half pyramid:half cone & half pyramid:



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o’

h

a

b

c

d

g

f

o e

a’ b’ c’ g’ d’f’ e’h’X Y

= R L

3600


A

B

C

DE

F

G

H

AO

1

3

2

4

7

6

5

L

11

22

33

44

55

66

77

1’1’

2’2’

3’3’ 4’4’5’5’

6’6’

7’7’

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

Solution Steps:Solution Steps:

Draw semicircle of given diameter, divide it in 8 Parts and inscribe in itDraw 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.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 )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 dataThen using above formula find R of base of cone. Using this data

draw Fv & Tv of cone and form 8 generators and name.draw Fv & Tv of cone and form 8 generators and name.

Take oTake o --1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’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 Tvand name 1’ Similarly locate all points on Fv. Then project all on Tv

on respective generators and join by smooth curve.on respective generators and join by smooth curve.

TO DRAW PRINCIPALTO DRAW PRINCIPAL

VIEWS FROM GIVENVIEWS FROM GIVEN

DEVELOPMENT.DEVELOPMENT.

h

a

b

c

d

g

f

e

o’

a’ b’ d’c’ g’ f’h’ e’X Y

A

B

C

DE

F

G

H

AO L

= R L

3600


1’1’

2’2’ 3’3’

4’4’

5’5’6’6’

7’7’

1122

33

44

55

6677

Problem 7:Problem 7:Draw a semicircle 0f 100 mm diameter and inscribe in it a largestDraw a semicircle 0f 100 mm diameter and inscribe in it a largest

rhombusrhombus.If the semicircle is development of a cone and rhombus is some curve.If the semicircle is development of a cone and rhombus is some curve

on it, then draw the projections of cone showing that curve.on it, then draw the projections of cone showing that curve.

TO DRAW PRINCIPALTO DRAW PRINCIPAL

VIEWS FROM GIVENVIEWS FROM GIVEN

DEVELOPMENT.DEVELOPMENT.

Solution Steps:Solution Steps:

Similar to previousSimilar to previous

Problem:Problem:



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a’a’ b’ b’ c’c’ d’d’

o’o’

e’e’

aa

b b

cc

d d

oo ee

XX YY

AA

BB

CC

DD

EE

AA

OO

22

33

44

11

Problem 8: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 faceA 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 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 brought back to the same point.If the string is of shortest length shortest length, find it and show it on the projections of the cone., find it and show it on the projections of the cone.

11 22

33

44

1’1’2’2’ 3’3’ 4’4’

TO DRAW A CURVE ONTO DRAW A CURVE ONPRINCIPAL VIEWSPRINCIPAL VIEWS

FROM DEVELOPMENT.FROM DEVELOPMENT. Concept:Concept: A string wound A string wound

from a point up to the samefrom a point up to the same

Point, of shortest lengthPoint, of shortest length

Must appear st. line on it’sMust appear st. line on it’s

Development.Development.Solution steps:Solution steps:

Hence draw development,Hence draw development,

Name it as usual and joinName it as usual and join

A to A This is shortest A to A This is shortest

Length of that string.Length of that string.

Further steps are as usual.Further steps are as usual.

On dev. Name the points of On dev. Name the points of

Intersections of this line withIntersections of this line with

Different generators.BringDifferent generators.Bring

Those on Fv & Tv and joinThose on Fv & Tv and joinby smooth curves.by smooth curves.

Draw 4’ a’ part of string dottedDraw 4’ a’ part of string dotted As it is on back side of cone. As it is on back side of cone.

X Ye’a’ b’ d’c’ g’ f’h’

o’

h

a

b

c

d

e

g

f

O

DEVELOPMENT

A

B

C

D

E

F

A

G

H

OO

1122

33

44

66 5577

1’1’

2’2’

3’3’

4’4’

5’5’

6’6’

7’7’

11

22

33

44

556677

HELIX CURVEHELIX CURVE

Problem 9:Problem 9: A particle which is initially on base circle of a cone, standingA 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.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.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.Take base circle diameter 50 mm and axis 70 mm long.

It’s a construction of curveIt’s a construction of curve

Helix of one turn on coneHelix of one turn on cone::Draw Fv & Tv & dev.as usualDraw Fv & Tv & dev.as usual

On all form generators & name.On all form generators & name.

Construction of curve Helix::Construction of curve Helix::

Show 8 generators on both viewsShow 8 generators on both views

Divide axis also in same parts.Divide axis also in same parts.

Draw horizontal lines from thoseDraw horizontal lines from those

points on both end generators.points on both end generators.

1’ is a point where first horizontal1’ is a point where first horizontal

Line & gen. b’o’ intersect.Line & gen. b’o’ intersect.

2’ is a point where second horiz.2’ is a point where second horiz.Line & gen. c’o’ intersect.Line & gen. c’o’ intersect.

In this way locate all points on Fv.In this way locate all points on Fv.

Project all on Tv.Join in curvature.Project all on Tv.Join in curvature.

For Development:For Development:

Then taking each points trueThen taking each points trueDistance From resp.generatorDistance From resp.generator

from apex, Mark on developmentfrom apex, Mark on development

& join.& join.

section and development [compatibility mode]

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