Title: ENGINEERING GRAPHICS By R.Nathan Assistant Professor Department of Mechanical Engineering
1ENGINEERING GRAPHICSBy R.NathanAssistant
ProfessorDepartment of Mechanical Engineering
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
I - SEMESTER
20.09.2012
2UNIT 1BASICS OF ENGINEERING GRAPHICS
- Engineering Graphics Lecture Notes
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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18.09.2012
3- Engineering Graphics
Lecture Notes
PENCIL GRADES
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
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4- Engineering Graphics
Lecture Notes
Types of Lines
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5- Engineering Graphics
Lecture Notes
Types of Lines
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6- Engineering Graphics
Lecture Notes
Example for Lines
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7- Engineering Graphics
Lecture Notes
Pencil Holding Technique
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
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8- Engineering Graphics
Lecture Notes
ALPHABETS FOR TITLE BLOCK
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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9- Engineering Graphics
Lecture Notes
ALPHABETS FOR TITLE BLOCK (FOR LEFT HAND USERS)
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
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10- Engineering Graphics
Lecture Notes
LIST OF DRAWING EQUIPMENTS MATERIALS
- DRAWING BOARD
- MINI DRAFTER
- DRAWING SHEET
- PENCILS (HB, 2H, 4H, 2B, 4B etc)
- NON DUST RUBBER
- SCALES (150 mm 300 mm)
- INSTRUMENT BOX
- SET SQUARES (30, 45, 60 etc..)
- DRAWING BOARD CLIPS, CLAMPS, PINS, CELLO TAPE
- PROTECTOR or PROCIRCLE or CIRCLE MASTER
- PENCIL SHARPENER
- SMALL PAPER KNIFE or STITCHING THREAD to cut the
sheetS - A HANDKERCHIEF or HAND TOWEL
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
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11- Engineering Graphics
Lecture Notes
DRAWING BOARD
ERASER
SCALES
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12- Engineering Graphics
Lecture Notes
COMPASS
BOW COMPASS
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DIVIDER
BOW DIVIDER
SET SQUARES
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14- Engineering Graphics
Lecture Notes
DRAWING CLIPS, CLAMPS,
PROTRACTOR
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15- Engineering Graphics
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PAPER KNIFE or STITCHING THREAD
CIRCLE MASTER or PRO-CIRCLE
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16- Engineering Graphics
Lecture Notes
ROLLER
MINI DRAFTER
INSTRUMENT BOX
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17- Engineering Graphics
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HOW TO KEEP DRAWING SHEET CLEAN
- ALWAYS CLEAN YOUR HANDS INSTRUMENTS BEFORE
FIXING THE DRAWING SHEET - HANDS SHOULD BE FREQUENTLY WIPED WITH CLEAN
HANDKERCHIEF - THE SET SQUARES MAY BE LIGHTLY MOVE WITH THE
FINGER NAILS WHILE USING - CLEAN PENCIL SMULGE WITH A CLEAN CLOTH
- THE RUBBER PPOWER OF PENCIL IS REMOVED AWAY FROM
THE DRAWING SHEET - DRAWING SHEET SHOULD BE ERASED WITH A SOFT
RUBBER ONLY - REMOVE RUBBER DUST WITH A CLEAN PIECE OF CLOTH or
HANDKERCHIEF - DO NOT KEEP ANYTHING or ARTICLE ON THE DRAWING
SHEET - AVOID UNNECCESARY RUBBING OF LINES
- DO NOT TOUCH DRAWING SHEET WITH DIRECT HANDS
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18- Engineering Graphics
Lecture Notes
DIMENSIONING
ARROWS
SYMBOLS
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TYPES OF DIMENSIONING
1. CHAIN DIMENSIONING
2. DIMENSIONING FROM A COMMON FEATURE
3. PARALLEL DIMENSIONING
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TYPES OF DIMENSIONING
4. SUPER IMPOSED RUNNING DIMENSIONING
5. COMBINED DIMENSIONING
6. PROGRESSIVE DIMENSIONING
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SYSTEM OF PLACING DIMENSION
1. ALIGNED SYSTEM
2. UNIDIRECTIONAL SYSTEM
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Lecture Notes
TYPES OF DIMENSIONING
DIMENSIONING IN THE HATCHING
DIMENSIONING ANGLES
DIMENSIONING CIRCLES
DIMENSIONING ARCS
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
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Lecture Notes
CORRECT INCORRECT METHODS OF DIMENSIONING
CORRECT METHOD
INCORRECT METHOD
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Lecture Notes
TYPES OF TRIANGLES
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CONSTRUCTION OF TRIANGLES
1. Construct an equilateral triangle given the
length of the side 50 mm
- Solution-
- Method 1 (using Compass)
- Draw a line AB of 50 mm length
- With centres A B and radius equal to 50 mm draw
arcs intersecting each other at C - Draw lines joining C with A and B.
- Now, ABC is the required equilateral triangle
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CONSTRUCTION OF TRIANGLES
2. Construct an equilateral triangle given the
length of the side 50 mm
- Solution-
- Method 2(using Set Squares)
- Draw a line AB of 50 mm length
- Draw a line through A, making 60 angle with AB
- Similarly through B, draw a line making the same
angle with AB - Intersecting point is C
- Now, ABC is the required equilateral triangle
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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27- Engineering Graphics
Lecture Notes
CONSTRUCTION OF TRIANGLES
3. Construct a triangle given the altitude 55
mm and two base angles 40 and 65
- Solution-
- Let ?A and ?B are the given base angles and CD be
the altitude - Steps Involved-
- Draw a base line of any convenient length
- Draw a - at a point D
- Make CD equal to the given altitude 55 mm
- Through C, draw a line EF to AB
- Make ?ECA 40, ?FCB 65
- Thus ABC is the required triangle
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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28- Engineering Graphics
Lecture Notes
CONSTRUCTION OF SQUARE
- Construct a square ABCD with AB 60 mm
- Solution-
- Steps Involved-
- Draw a line segment AB of 60 mm
- At A, draw a perpendicular at B.
- With A as centre, 60 mm as radius draw an arc and
to intersect the - to get the point D - With B as centre, 60 mm as radius draw an arc and
to intersect the - through B, to get the point C. - Now ABCD, is the perfect square
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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29- Engineering Graphics
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CONSTRUCTION OF RECTANGLE
- Construct a rectangle PQRS when PG 70 mm, QR
50 mm
- Solution-
- Steps Involved-
- Draw a line segment PQ of 70 mm long
- At P and Q erect perpendiculars
- With P as centre, 50 mm as radius draw an arc,
to cut the - at the point S - With Q as centre, 50 mm as radius draw an arc, to
cut the - at the point R - Join R with S
- Now PQRS is the required rectangle
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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30- Engineering Graphics
Lecture Notes
CONSTRUCTION OF RHOMBUS
- Construct a rhombus ABCD, having its side equal
to 40 mm and base angle at B of 105
- Solution-
- Let us use the properties of rhombus here.
- All sides of a rhombus are equal
- Steps Involved-
- Draw a line segment AB of 40 mm length
- At B, draw a line BX at an angle of 105
- B as centre 40 mm as radius draw an arc to cut
the line BX. Intersection point is C - With A and C as centres, 40 mm as radius draw
arcs, intersecting point is D - Join A with D and C with D
- Thus ABCD is the required rhombus
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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Lecture Notes
CONSTRUCTION OF RHOMBUS
- Construct a rhombus PQRS with diagonals 46 mm and
QS 60 mm
- Solution-
- We know that, the diagonals of a rhombus bisect
each other at right angles - Steps Involved-
- Draw a line segment QS of 60 mm length
- Draw a bisector of QS which passes through the
point O - With O as centre, ½ PR (23 mm) as radius draw
arcs above and below to cut the - bisector.
Intersection points are P R - Draw lines joining P with Q S.
- Thus PQRS is the required rhombus
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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32- Engineering Graphics
Lecture Notes
CONSTRUCTION OF QUADRILATERAL
- Construct a Quadrilateral with AB 45 mm, BC
55 mm, CD 40 mm, AD 60 mm, AC 70 mm.
- Solution-
- From our earlier classes, we learnt that, to draw
a quadrilateral, minimum five dimensions are
required. Let u now draw this quadrilateral. - Steps Involved-
- Draw AB of 45 mm length
- With B as centre, 55 mm as radius draw an arc
- With A as centre, 70 mm as radius draw an arc to
cut the previous arc at the point C - With C as centre, 40 mm as radius draw an arc
- With A as centre, 60 mm as radius cut the
previous arc to get the intersection point D - Draw lines joining D with C and A
- Thus ABCD is the required quadrilateral
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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33- Engineering Graphics
Lecture Notes
CONSTRUCTION OF TRAPEZIUM
- Construct a trapezium ABCD, having its sides AD
30 mm, DC 25 mm, CB 35 mm and the difference
of parallel sides is 20 mm
- Solution-
- Steps Involved-
- Draw a line segment AB of 45 mm, difference
between parallel sides 20 mm (20 2545 mm) - With A as centre and 30 mm as radius draw an arc
- With E as centre 35 mm as radius draw an arc to
cut the previous arc (ie EDBC) - Intersecting point is D. D as centre 25 mm as
radius draw an arc - With B as centre, 35 mm as radius cut the
previous arc. Intersection point is C - Draw lines joining A with D, D with C and C with
B - Thus ABCD is the required trapezium
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
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Lecture Notes
CONSTRUCTION OF POLYGONS
HEPTAGON
OCTAGON
NONAGON
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CONSTRUCTION OF PENTAGON
- Construct a regular pentagon given the lengh of
its side as 40 mm
- Solution-
- Draw a line segment AB of 40 mm length
- At B, draw a perpendicular BL such that BK AB,
join A with K - With B as centre, AB as radius draw an arc (ie
arc AK) - Draw a perpendicular bisector of AB
- Mark the point of intersection of perpendicular
with line AK as 4 and arc AK as 6 - Mark a point 5 which is the midpoint of 4-6
obtained by bisecting it - With 5 as centre and 5A as radius draw a circle
- With B as centre and radius AB, draw an arc to
cut the circle at C - In a similar way, set off the measurement of side
along the circle to get the points D and E - Draw lines joining B with C, C with D, D with E
E with A
C.R.ENGINEERING COLLEGE Alagarkovil, Madurai -
625301
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18.09.2012