Transcript unit – 1 – basics

ENGINEERING GRAPHICS
By
R.Nathan
Assistant Professor
Department of Mechanical Engineering
I - SEMESTER
C.R.ENGINEERING COLLEGE
Alagarkovil, Madurai - 625301
20.09.2012
Engineering Graphics
Lecture Notes
UNIT – 1
BASICS OF ENGINEERING GRAPHICS
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Alagarkovil, Madurai - 625301
18.09.2012
Engineering Graphics
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PENCIL GRADES
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Types of Lines
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Engineering Graphics
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Types of Lines
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Example for Lines
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Pencil Holding Technique
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ALPHABETS FOR TITLE BLOCK
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ALPHABETS FOR TITLE BLOCK (FOR LEFT HAND USERS)
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LIST OF DRAWING EQUIPMENTS & MATERIALS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
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
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Engineering Graphics
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DRAWING BOARD
ERASER
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SCALES
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COMPASS
BOW COMPASS
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DIVIDER
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BOW DIVIDER
SET SQUARES
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DRAWING CLIPS, CLAMPS,
PROTRACTOR
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PAPER KNIFE or STITCHING THREAD
CIRCLE MASTER or PRO-CIRCLE
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ROLLER
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MINI DRAFTER
INSTRUMENT BOX
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HOW TO KEEP DRAWING SHEET CLEAN
1. ALWAYS CLEAN YOUR HANDS & INSTRUMENTS BEFORE FIXING THE DRAWING SHEET
2. HANDS SHOULD BE FREQUENTLY WIPED WITH CLEAN HANDKERCHIEF
3. THE SET SQUARES MAY BE LIGHTLY MOVE WITH THE FINGER NAILS WHILE USING
4. CLEAN PENCIL SMULGE WITH A CLEAN CLOTH
5. THE RUBBER PPOWER OF PENCIL IS REMOVED AWAY FROM THE DRAWING SHEET
6. DRAWING SHEET SHOULD BE ERASED WITH A SOFT RUBBER ONLY
7. REMOVE RUBBER DUST WITH A CLEAN PIECE OF CLOTH or HANDKERCHIEF
8. DO NOT KEEP ANYTHING or ARTICLE ON THE DRAWING SHEET
9. AVOID UNNECCESARY RUBBING OF LINES
10. DO NOT TOUCH DRAWING SHEET WITH DIRECT HANDS
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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
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6. PROGRESSIVE DIMENSIONING
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SYSTEM OF PLACING DIMENSION
1. ALIGNED SYSTEM
2. UNIDIRECTIONAL SYSTEM
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TYPES OF DIMENSIONING
DIMENSIONING IN
THE HATCHING
DIMENSIONING ANGLES
DIMENSIONING CIRCLES
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DIMENSIONING ARCS
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CORRECT & INCORRECT METHODS OF DIMENSIONING
CORRECT METHOD
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INCORRECT METHOD
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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)
1. Draw a line AB of 50 mm length
2. With centres A & B and radius equal to 50
mm draw arcs intersecting each other at C
3. Draw lines joining C with A and B.
4. 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)
1. Draw a line AB of 50 mm length
2. Draw a line through A, making 60° angle with AB
3. Similarly through B, draw a line making the same
angle with AB
4. Intersecting point is C
5. Now, ABC is the required equilateral triangle
I - SEMESTER
C.R.ENGINEERING COLLEGE
Alagarkovil, Madurai - 625301
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Engineering Graphics
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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:1.
2.
3.
4.
5.
6.
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
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CONSTRUCTION OF SQUARE
1. Construct a square ABCD with AB = 60 mm
Solution:Steps Involved:-
1. Draw a line segment AB of 60 mm
2. At A, draw a perpendicular at B.
3. With A as centre, 60 mm as radius draw an arc
and to intersect the ┴ to get the point D
4. With B as centre, 60 mm as radius draw an
arc and to intersect the ┴ through B, to get the
point C.
5. Now ABCD, is the perfect square
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Alagarkovil, Madurai - 625301
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CONSTRUCTION OF RECTANGLE
1. Construct a rectangle PQRS when PG = 70 mm, QR = 50 mm
Solution:Steps Involved:-
1. Draw a line segment PQ of 70 mm long
2. At P and Q erect perpendiculars
3. With P as centre, 50 mm as radius draw an
arc, to cut the ┴ at the point S
4. With Q as centre, 50 mm as radius draw an
arc, to cut the ┴ at the point R
5. Join R with S
6. Now PQRS is the required rectangle
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CONSTRUCTION OF RHOMBUS
1. 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:1. Draw a line segment AB of 40 mm length
2. At B, draw a line BX at an angle of 105°
3. B as centre 40 mm as radius draw an arc to
cut the line BX. Intersection point is C
4. With A and C as centres, 40 mm as radius
draw arcs, intersecting point is D
5. Join A with D and C with D
6. Thus ABCD is the required rhombus
I - SEMESTER
C.R.ENGINEERING COLLEGE
Alagarkovil, Madurai - 625301
18.09.2012
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CONSTRUCTION OF RHOMBUS
1. 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:1. Draw a line segment QS of 60 mm length
2. Draw a bisector of QS which passes through
the point O
3. With O as centre, ½ PR (23 mm) as radius
draw arcs above and below to cut the ┴
bisector. Intersection points are P & R
4. Draw lines joining P with Q & S.
5. Thus PQRS is the required rhombus
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C.R.ENGINEERING COLLEGE
Alagarkovil, Madurai - 625301
18.09.2012
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CONSTRUCTION OF QUADRILATERAL
1. 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:1. Draw AB of 45 mm length
2. With B as centre, 55 mm as radius draw an arc
3. With A as centre, 70 mm as radius draw an arc
to cut the previous arc at the point C
4. With C as centre, 40 mm as radius draw an arc
5. With A as centre, 60 mm as radius cut the
previous arc to get the intersection point D
6. Draw lines joining D with C and A
7. Thus ABCD is the required quadrilateral
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C.R.ENGINEERING COLLEGE
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CONSTRUCTION OF TRAPEZIUM
1. 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:1. Draw a line segment AB of 45 mm, difference
between parallel sides = 20 mm (20 + 25=45 mm)
2. With A as centre and 30 mm as radius draw an arc
3. With E as centre 35 mm as radius draw an arc to
cut the previous arc (ie; ED║BC)
4. Intersecting point is D. D as centre 25 mm as
radius draw an arc
5. With B as centre, 35 mm as radius cut the previous
arc. Intersection point is C
6. Draw lines joining A with D, D with C and C with B
7. Thus ABCD is the required trapezium
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CONSTRUCTION OF POLYGONS
HEPTAGON
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OCTAGON
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NONAGON
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CONSTRUCTION OF PENTAGON
1. Construct a regular pentagon given the lengh of its side as 40 mm
Solution:1. Draw a line segment AB of 40 mm length
2. At B, draw a perpendicular BL such that BK = AB,
join A with K
3. With B as centre, AB as radius draw an arc (ie; arc
AK)
4. Draw a perpendicular bisector of AB
5. Mark the point of intersection of perpendicular with
line AK as 4 and arc AK as 6
6. Mark a point 5 which is the midpoint of 4-6
obtained by bisecting it
7. With 5 as centre and 5A as radius draw a circle
8. With B as centre and radius = AB, draw an arc to
cut the circle at C
9. In a similar way, set off the measurement of side
along the circle to get the points D and E
10. Draw lines joining B with C, C with D, D with E & E
with A
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18.09.2012