Lab 06

Sectional views

Objects with hidden details

Sectional Views

2

Required to add information of surfaces that are represented by hidden lines in standard FV, TV, and SV.

what we need1. Cutting plane2. Part, assembly, any 3-D view.

Cross sectionSection

Section (thin parallel) lined areas are those portions that came in actual contact with cutting plane.

Visible parts behind the cutting plane are shown, but not hatched.

Hatching Pattern Continuous thin lines at convenient angle (preferably 45°) to the principal outlines.

Common Mistakes

Spacing between the hatching lines should be chosen in proportion to the size of the hatched areas, provided that the requirement for minimum spacing are maintained.

Full section view

NOTES

7

• Sectional views are always viewed in the direction defined by cutting plane arrows.

• Any hidden surface that is behind cutting plane is not included in sectional view.

Hatch lines represent location of cutting plane passing through solid material.

Full & Half section views

Example: Sectional Orthographic Views

9

AA Mistakes in dimensioning?

Representation of cutting plane?

10

Section B-B

Section A-A

Rib not sectioned

a b c

Common mistake is to omit back edge

Sectional view of Ribs

11

Ribs add strength and rigidity to an object. Usually narrow.

Pg. 167 Luzadder book

12

a) Although the spoke is in line with the front view, it can give the impression that it is a stunted spoke

b) Sectioned spoke can indicate that it is a continuous web

c) The full length of spoke is shown to indicate the structure. It should be shown along with front view to indicate the number of spokes and angles between them

Front view

Keyway

13

Revolved sections

Section rotated 90o

so that exact shape can be viewed

Cutting plane

Examples of Revolved Sections

Revolved sections examples show the shape of an object’s cross-section superimposed on a longitudinal view

Removed SectionsRemoved sections are like revolved sections but moved aside.

A

A

Section A-A

C CSection C-C

B

B

Section B-B

Offset Sections• Cutting plane lines need not be drawn as straight

lines Stepped line

16

Offset full section

Include as many features as possible without causing confusion

17

Countersunk hole

Choosing Full/Half/Broken sectional view

18

Half & Partial sectional views --- allow showing outer and inner features on the same figure.

Normal half/partial orthographic projection may include hidden lines.

Broken line is a freehand line.

Sectional view of Ribs

19

Ribs add strength and rigidity to an object. Usually narrow.

Cutting Plane Lines

20

Phantom line

Section B-B

Section A-A

ALIGNED SECTIONS• To include, in a section, certain angled elements,

cutting plane may be bent so as to pass through those features.– Plane & feature are aligned into original plane.

22

Summary• When a part is cut fully in half, the resulting view is called

a full section.• A line called the cutting-plane line shows where the object

was cut and from which direction the section is viewed.– The arrows point toward the section being viewed.

• In the section view, the areas that would have been in actual contact with the cutting plane are show with section lining.

• The visible edges of the object behind the cutting plane are generally shown because they are now visible but they are not cross-hatched.

• Section views can replace the normal top, front, side, or any other standard orthographic view.

Summary

• When a cutting plane coincides with a center line, the cutting plane line takes precedence.

• Omit hidden lines in section views.• A section-lined area is always completely bounded

by a visible outline

Labeling !!

Summary• To avoid a false impression of thickness and

solidity, ribs/webs, gear teeth, and other similar features are not hatched with section lining even though the cutting plane slices them.

• Sectional views are important for assemblies.

Sheet 7

27

PROBLEM: A room is of size L=6.5m, D=5m, H=3.5m. An electric bulb hangs 1m below the center of ceiling. A switch is placed in one of the corners of the room, 1.5m above the flooring. Draw the projections and determine real distance between the bulb & switch.

PROBLEM:- A picture frame 2 m wide and 1 m tall is resting on horizontal wall railing makes 350 inclination with wall. It is attached to a hook in the wall by two strings. The hook is 1.5 m above wall railing. Determine length of each chain and true angle between them

PROBLEM: L=6.5m, D=5m, H=3.5m. An electric bulb hangs 1m below the center of ceiling. A switch is placed in one of the corners of the room, 1.5m above the flooring. Draw the projections and determine real distance between the bulb and switch.

a

b

x y

a’

b’ b’1

6.5

3.5

5

1

1.5

Answer: a’ b’1

All Dimensions in m.

PROBLEM- A picture frame 2 m wide & 1 m tall is resting on horizontal wall railing makes 350

inclination with wall. It is attached to a hook in the wall by two strings.The hook is 1.5 m above the wall railing.

DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN THEM

a d

h

b c

a1

b1

a’b’

c’d’(wall railing)

(frame)

(chains)

Answers:Length of each chain= hb1True angle between chains =

(chains)

X Y

h’

1.5 m

1m

Rectangular lamina parallel to V.P.

FT

Pentagonal lamina parallel to V.P.

FT

Pentagonal lamina parallel to H.P.FV shall replace TV & TV shall replace FV

FT

31

F

T aT

o

oF

dFcFbFaF

o1

d1

b1c1

a1

Problem. A square pyramid, 40 mm base sides and axis 60 mm long, has atriangular face on the ground. Draw its projections.

1st. Angle

bT cT

dT

Ex: A line AB, 50 mm long, is inclined at 30° to the H.P. Draw its projections.

T

F

aF bF

aT bT

33

PROBLEM:- Two mangos on a tree A & B are 1.5 m and 3.0 m above ground and those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If the distance measured between them along the ground and parallel to wall is 2.6 m, Then find real distance between them by drawing their projections.

TV

A

B

0.3M THICK

a

b’

b2.6

1.5

3.0a’

1.2

PROBLEM:- Two mangos on a tree A & B are 1.5 m and 3.00 m above ground and those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If the distance measured between them along the ground and parallel to wall is 2.6 m, Then find real distance between them by drawing their projections.

1.5

B

All dimensions in m.

35

PROBLEM:-Flower A is 1.5 m & 1 m from walls Q (parallel to reference line) & P (perpendicular to reference line) respectively. Flower is 1.5 m above the ground. Orange B is 3.5m & 5.5m from walls Q & P respectively. Drawing projection, find distance between them If orange is 3.5 m above ground.

a

b

a’

b’ b’1

x y1.5

3.5

1

1.5

3.5

5.5

Wall Q

Wall P

B

Ground

All dimensions in m.

PROBLEM :- An object contains three rods OA, OB and OC whose ends A,B & C are on ground and end O is 100mm above ground. The top view of object contains three lines oa, ob & oc having length equal to 25mm, 45mm and 65mm respectively. These three lines are equally inclined and the shortest line is vertical. Draw their projections and find length of each rod.

A

O

B

C

Fv

Tv

a

b

c

o

a’b’ c’

o’

TL1TL2

c1’b1’ a1’

PROBLEM :- A top view of object (three rods OA, OB and OC whose ends A,B & C are on ground and end O is 100mm above ground) contains three lines oa, ob & oc having length equal to 25mm, 45mm and 65mm respectively. These three lines are equally inclined and the shortest line is vertical. Draw their projections and find length of each rod.

Answers:TL1 TL2 & TL3

x y

PROBLEM:- A pipeline from point A has a downward gradient 1:5 and it runs due South - East. Another Point B is 12 m from A and due East of A and in same level of A. Pipe line from B runs 150 Due East of South and meets pipeline from A at point C. Draw projections and find length of pipe line from B and it’s inclination with ground.

AB

C

1

5

12 ME

PROBLEM:- A pipe line from point A has a downward gradient 1:5 and it runs due South - East. Another Point B is 12 m from A and due East of A and in same level of A. Pipe line from B runs 150 Due East of South and meets pipe line from A at point C. Draw projections and find length of pipe line from B and it’s inclination with ground.

15

a b

c

F

T

150

450

12m

N

EAST

SOUTH

W

a’ b’

c’2c’ c’1

= Inclination of pipe line BC

PROBLEM: A person observes two objects, A & B, on the ground, from a tower, 15 M high, at the angles of depression 300 & 450 respectively. Object A is due North-West direction of observer and object B is due West direction. Draw projections of situation and find distance of objects from observer and from tower also.

W

SA

B

O

300

450

PROBLEM: A person observes two objects, A & B, on the ground, from a tower, 15 M high, at the angles of depression 300 & 450. Object A is due North-West direction of observer and object B is due West direction. Draw projections of situation and find distance of objects from observer and from tower also.

W

S

E

N

o

a

b

o’

a’1b’a’

300 450

15M

Answers:Distances of objects from observeo’a’1 & o’b’From toweroa & ob

Dimensions in m

a

a2, b2

p2

Shortest distance a2 p2

Point

Point view of line

a1

p1

b1

a’

b’

p’

pTF

b

SHORTEST distance between POINT & LINE

Auxiliary plane method. Find point view of line.Draw reference line parallel to line and obtain true length.

43

a

a

b

b

d

c

d

c

•Find T.L. of one of the lines and project its point view using auxiliary plane method

•Project the other line also in each view.

• Shortest distance between skew lines can be measured along the one line perpendicular to both. Common perpendicular !!

TF

Shortest Distance between 2 skew Lines (AB & CD)Distance between intersecting (or //) lines????

Skew (oblique) lines Lines that are not parallel & do not intersect

Distance measured alongShortest ⊥ to both.

a

a

b

b

d

c

d

c

TF

Primary auxiliary viewcTL

Required distance

a

d d, cSecondary auxiliary view

ab

d

B A

q

P

dP is ⊥ to ab

In mines, this method might be used to locate a connecting tunnel.

Piercing Point• Intersection between a line and plane

– Point – Piercing point.

45

True angle between line and plane ??

Piercing point of a line with a plane

p1

Line

Plane

Edge view of the plane

Principal line

True length of principal line

p’

p

Part of the line hidden by the plane should be shown dotted

TF

TA1

In a view showing the plane as an edge, the piercing point appears where the line intersects the edge view.

Draw auxiliary view to get EDGE VIEW.

Mistake ?

How to find angle between a line and a given plane ?

a'

a

b'

bc'

d'

cd

e'

f

e

f'

d1

c1

a1

e1

f1

Concept of Principal lines of a plane

A’

A

C’

B’

TLB

CPoint view

All the points lie on a straight line representing

the edge of the plane

Principal line

T

F

A1T

Principle lines: Lines on the boundary or within the surface, parallel to the principle planes of projection

To obtain the edge view of a plane

T

F

a’

b’

c’

a

b

c

True length

Horizontal line (parallel to top plane)

l’

lc1

a1

b1

-Draw a principle line in one principle view and project the true length linein the other principle view

-With the reference line perpendicular to the true length line, draw a primary auxiliary view of the plane, to obtain the edge view

Distances:

a1, b1, c1 from TA = a’, b’, c’ from FT

Edge view of the plane

p’

d’g’

10

50

c’, d’

d,e

F, A

a,g

c,bf,d

b’, e’

p

p1

p1’Required distance

Find the shortest distance of point P from the body diagonal AB of the cube of side 50 mm as shown

Draw an auxiliary view to get the true length of the line

Draw an auxiliary view to get the point view of the diagonal

Project the point P in these views to get the required distance

10

a1b1

a1‘, b1’

50

T

F

Draw P.A. V. such that one line (AB) shows its True Length

Draw S.A.V. view with reference line perpendicular to the True Length of the line (AB) to get the point view of the line

Draw a tertiary auxiliary view with reference line parallel to the other line in order to get its True Length

Since the secondary auxiliary view had the point view of the first line, the tertiary auxiliary view will have the True Length of the first line also.

True Angle between 2 Skew Lines (AB & CD)

x y

a’

a

c’ b’

b

d’

cd

Measure angle in view that Shows both lines in true length

TRUE LENGTH OF BOTH LINES

Point view of one line

T

F

PRIMARY AUXILIARY VIEW

SECONDARY AUXILIARY VIEW

TERTIARY AUXILIARY VIEW

True Anglebetween the two lines

Parallel

Parallel

True length

ANGLE BETWEEN TWO LINES

x y

a’

a

a1

a2 ,b2

c’ b’

b

b1

a3

b3

d’

c

c1

c2

c3

d

d1

d2

d3

Angle between two nonintersecting linesis measurable in a view that shows both lines in true shape.

Angle between 2 planes

TFa’

e’

d’ c’

b’

f’

ab

cd

e f

• Obtain an auxiliary view such that the reference line is perpendicular to the True Length of the line of intersection of the planes

• In this case, the intersection line is parallel to both principle planes and hence is in True Length in both front and top views

• Both planes will be seen as edge views in the auxiliary view.

• The angle between the edge views is the angle between the planes

x

Line of intersection of the 2 planes (here it is True Length)

TF

PRIMARY AUXILIARY VIEW

x1

y

y1a’

e’

d’ c’

b’

f’

ab

cd

e f

b1, a1c1, d1

f1, e1

Auxiliary Views

Third angle projection method

Auxiliary Views

First angle projection method

56

T Y

a

c

b

c’

b’

a’

10

15

15

cA1

bA1aA1

aA2

bA2

cA2

A2

A1

Problem: TV is a triangle abc. ab is 50 mm long, angle cab is 30° and angle cba is 65°. a’b’c’ is a FV. a’ is 25 mm, b’ is 40 mm and c’ is 10 mm above Hp respectively. Draw projections of that figure and find it’s true shape.

300 650

50 mm

DISTANCES FOR NEW FV come from PREVIOUS FV AND FOR NEW TV, DISTANCES OF PREVIOUS TV are accounted.

F

c’1

b’1

a’1

A1A2

b1

c1

a1

c’

T

F

a’

b’

b

ca

10

20

15

15

1’

140

5025

Problem Fv & Tv of a triangular plate are shown. Determine it’s true shape.

DISTANCES FOR NEW FV come from PREVIOUS FV

AND FOR NEW TV, DISTANCES from PREVIOUS TV are

accounted.

Angle between a line and a plane

p1

Line

Plane

Edge view of the plane

Principal line

True length of principal line

p’

p

Part of the line hidden by the plane should be shown dotted

TF

TA1

a'

a

b'

bc'

d'

cd

e'

f

e

f'

d1

c1

a1

e1

f1 A1 A2

e2

c2

a2f2

d2

A2A3

59

Projection of point P on A.I.P. (⊥ to VP and inclined at β to HP)

• Draw FA such that it makes an angle β with FT

• Project PA on AIP by drawing a line PF PA such that PF PA is ⊥ to FA and O PT = O’ PA.

Helix

• Space curve– FV & TV specify helix completely. – Widely employed on screw threads, helical

springs, conical spring, screw conveyors, staircases, etc.

Draw a helix of one convolution, upon a cylinder. Given 80 mm pitch and 50 mm diameter of a cylinder.

1

2

3

4

5

6

7

8

P

P1

P

P2

P3

P4

P5

P6

P7

P8

1

2

3

4

5

6

7

HELIX (UPON A CYLINDER)

PROBLEM: Draw a helix of one convolution, upon a cylinder. Given 80 mm pitch and 50 mm diameter of a cylinder.

Pitch: Axial advance during one complete revolution .

FT

P

1

2

3

4

5

6

7

PP1

P2

P3

P4

P5

P6

P7

P8

P1

P2

P3

P4

P5P6

P7

P8

X Y

HELIX (UPON A CONE)

PROBLEM: Draw a helix of one convolution, upon a cone, diameter of base 70 mm, axis 90 mm and 90 mm pitch.

7 6 5 4 3 2 1P

1

2

3

4

5

6

7

P2

P6

P1

P3

P5

P7

P4 O

SPIRALProblem: Draw a spiral of one convolution. Take distance PO 40 mm.

IMPORTANT APPROACH FOR CONSTRUCTION!FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENTAND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS.

16 13 10 8 7 6 5 4 3 2 1 P

1,9

2,10

3,11

4,12

5,13

6,14

7,15

8,16

P1

P2

P3

P4

P5

P6

P7

P8

P9P10

P11

P12

P13 P14

P15

SPIRAL of two convolutionsProblem: Point P is 80 mm from point O. It starts moving

towards O and reaches it in two revolutions around. It Draw locus of point P (To draw a Spiral of TWO convolutions).

Problem: A link 60 mm long, swings on a point Ofrom its vertical position of the rest to the left through60° and returns to its initial position at uniformvelocity. During that period a point P moves atuniform speed along the center line of the link from Oat reaches the end of link. Draw the locus of P.

O, P

M

N

66

Necessity of Auxiliary view

Ex: A line AB, 50 mm long, is inclined at 30° to the H.P. and its top view makes an angle of 60° with the V.P. Draw its projections.

T

F

aF bF

aTbT

68

Draw A.V. of a plane ABC (A(50,10,30), B(10,40,0), C(10,30,50)) on a plane which is ⊥ to frontal plane and inclined at an angle of 45o to top plane. Draw another A.V. on a plane which is perpendicular to the top plane and inclined at an angle 60o to the frontal plane. USE III rd ANGLE

T

F

X

Y

Z

1020304050

1020304050

60o

45oO

b’c’

a’b

c

aa1’

c1’

b1’

a1

c1

b1

x1

y1

x2

y2

2010 50

Distance of a1 from FA = distance of a from OZ

Distance of b1 from FA = distance of b from OZ

Distance of c1 from FA = distance of c from OZ

Distance of a1’ from TA = distance of a’ from OZ

Distance of b1’ from TA = distance of b’ from OZ

Distance of c1’ from TA = distance of c’ from OZ

h

a

bc

d

e

g

f

F a’ b’ d’ e’c’ g’ f’h’

o’

450

a1

h1 f1

e1

d1

c1

b1

g1

o11

Problem: A right circular cone, 40 mm base diameter and 60 mm long axis is resting on Hp on one point of base circle such that it’s axis makes 450 inclination with Hp. Draw it’s projections in I angle projection method.

T

o

Projection of Solids using Auxiliary Plane Method

• Projections of solids, whose axes are inclined to H.P./V.P. can be drawn using Primary Auxiliary Plane method.

• In this method, instead of changing the position of views w.r.t. fold line, we change the position of fold line.

Secondary auxiliary view of a cube

Direction of view is perpendicular to the fold line

45o

e’,h’ f’, g’

a’, d’

a, e

d, h c, g

b, f

f1, a1

g1, d1

e1

h1

b1

c1

T

F

Distances:

e1, f1, a1, b1 from TA = e, f, a, b from FT

h1, g1, d1, c1 from TA = h, g, d, c from FT

P.A.V.

b’, c’

S.A.V. of a cube

Fold line

45o

60o

e’,h’ f’, g’

a’, d’ b’, c’

a, e

d, h c, g

b, f

f1, a1

g1, d1

e1

h1

b1

c1

Distancese’, f’, a’, b’ from TA1 = e2, f2, a2, b2 from A1A2h’, g’, d’, c’ from TA1 = h2, g2, d2, c2 from A1A2

a2

f2

d2

g2

c2

e2b2

h2

TF

SECONDARY AUXILIARY VIEW

F

F

AIP // to slant edgeShowing true lengthi.e. a’- 1’

a’ b’ e’ c’ d’

1’ 2’5’ 3’4’

Aux.Tv

1

23

45

a

b

d

c

e

1 2

34

5

b1

c1d1

e1

a1

Problem: A frustum of regular pentagonal pyramid is standing on it’s larger base on Hp with one base side perpendicular to Vp. Draw it’s Fv & Tv. Project it’s Aux. Tv on an AIP parallel to one of the slant edges showing TL. Base side is 50 mm long, top side is 30 mm long and 50 mm is height of frustum.

1. Attendance & Marks http://web.iitd.ac.in/~hirani/marks & attendance.pdf 2. Lab sheets (http://web.iitd.ac.in/~hirani/mel110-lab-sheets.pdf)

T

A1

b

FT

a

d

co

d’ c’b’a’

o’

Problem: A square pyramid 30 mm base side and 50 mm long axis is resting on it’s apex on Hp, such that it’s one slant edge is vertical and a triangular face through it is perpendicular to Vp. Draw it’s 1st angle projections.

b

Hidden lines in Projections of solid !

b

FT

a

d

co

d’ c’b’a’

o’

b1

c1a1

d1

o1

Problem: A square pyramid 30 mm base side and 50 mm long axis is resting on it’s apex on Hp, such that it’s one slant edge is vertical and a triangular face through it is perpendicular to Vp. Draw it’s 1st

angle projections.b

Auxiliary Views

Projection lines should be thin continuous lines.

Mistake ???