Workshop Calculation

WORKSHOP CALCULATION 1

Module : Fabrication calculation Faculty : Course co-ordinators ( DRM, CPP, SIS ) Demonstrator: Workmen from Shops Duration : Maximum 16 Hours Participants : Max. 06 / Module (Workmen from shops) No. 1 2 3 4 5 6 7

Topics p Introduction to SRMs. Pre test Classroom training Demonstration P Practical ti l Skill test Post test and feed back

Time 30 Min. 30 Min. 6.0 Hrs. 3.0 Hrs. 30H 3.0 Hrs. 2.0 Hrs. 1 0 Hrs 1.0 Hrs. 2

U it : 1 M Unit Module d l : Workshop W k h Calculation C l l ti Topics

Time

 Introduction t oduct o and a d induction duct o test 10 0e examples a p es

1.0 0 hr

 Units of length, Area, Volume, Weight, Temperature p and Pressure

1.0 hr

 Pythagoras theorem and demonstration

0.5 hr

 Trigonometric functions & demo. 0.5 hr  Practice examples = 10

1 0 hr 1.0 3

MODULE : WORKSHOP CALCULATION UNIT : 2 Weight calculation and weld deposition  weight i ht with ith demonstration d t ti

2h 2hours

 WEP calculation, 1:3 and 1:5 taper

1 hour

calculation  Practice examples = 10 nos nos.

1 hour

4

MODULE : WORKSHOP CALCULATION UNIT : 3 Measure tape error correction and circumference calculation = with demonstration (1 hour) O i t ti marking Orientation ki ( 0.5 0 5 hour h ) Offset and kink, web and flange tilt, flange unbalance calculation l l ti (1 hour) h ) Arc length and chord length calculation for web layout= with demonstration ( 0.5 hour ) Practice examples = 10 nos. (1 hour) 5

MODULE : WORKSHOP CALCULATION UNIT : 4 Tank rotator location calculation and sling angle for handling a job calculation ( 0.5hour ) M hi i allowance Machining ll calculation l l ti for f overlay l

and d

machining allowance for bracket calculation (0.5 hour) Marking PCD and holes for flange calculation = with demonstration ( 0.5 hour) Practice examples = 5 Nos. (0.5hour) Test => > theory = 10 questions Practical= 4 questions ( 2 hours )

6

U it : 1 M Unit Module d l : Workshop W k h Calculation C l l ti Topics

Time

 Introduction t oduct o and a d induction duct o test 10 0e examples a p es

1.0 0 hr

 Units of length, Area, Volume, Weight , Temperature p and Pressure

1.0 hr

 Pythagoras theorem and demonstration

0.5 hr

 Trigonometric functions demonstration

0.5 hr

 Practice 10 examples

1.0 hr

7

Introduction to Units ( Pressure) PRESSURE CONVERSION 1 Kg / cm² = 14 . 223 psi ( Lb / In² ) 1 Kg / cm² = 0 . 9807

Bar Bar.

1 PSI = 0.07031 Kg / cm²

Introduction to Units (Length) 1m = 100 cm 1cm = 10 mm 1m = 1000 mm 1in. = 25.4 mm

Introduction to Units (Weight) 1 kg = 2.204 lbs 8

Introduction to Units ( Temperature) Temperature unit = degree centigrade or degree Fahrenheit

°C = 5/9(°F 5/9(°F- 32) If Temp. Is 100°F, Then °C=5/9( 100-32) °C=37.7

So,

If Preheat Temperature Is 150 °C, Then °F=302

9

PYTHAGORAS PRINCIPLE APPLICATION A

Pythagoras Principle : In Any Right Angled Triangle the Square of Sum of Adjacent Sides Is Always Equal to th Square the S off Hypotenuse H t . B

C

LET US SAY  ABC is right g angle g triangle g . AB and BC = Adjacent sides and AC = Hypotenuse. So based on pythagoras theory ,

AB² + BC AB BC² = AC² AC 10

PYTHAGORAS PRINCIPLE APPLICATION Example :

A

5 3

B

4

C

Proof of P. theory in triangle ABC AB = 3 , BC = 4 and AC = 5 SO AC² = AB² + BC² = 3² + 4 ² = 25 so, AC = 5 11

TRIGONOMETRIC FUNCTIONS A

Trigonometric go o et c functions u ct o s are a e used to so solve e the problems of different types of triangle. ø B

C

We will see some simple formulas to solve right angle triangle which we are using in day to day work. work

Let us consider  ABC is a “right angled triangle”, A l ABC = ø , AB & BC are sides Angle id off ttriangle. i l So for this triangle.

12

TRIGONOMETRY A

Hypoteneous

SIN

ø = Opposite Side = AB AC Hypoteneous

pp Opposite Side

COS ø =

ø B

Adjacent Side

Adjacent Side = BC AC Hypoteneous yp

C

TAN

Opposite it Side Sid = AB ø = O BC Adjacent Side

13

TRIGONOMETRIC FUNCTIONS Example : For triangle ABC find out value of  and . A

Tan  = Opposite Side / Adjacent Side = AB / BC = 25/25 =1 Tan  = 1   = Inv. Tan(1) = 45º

25 mm

W Will Fi We Find d Value V l Of  By B “T “Tan”” Formula. F l S , So

B



 25 mm

Now, We Will Find AC By Using “Sin” Formula. Sin  = Opposite Side /Hypotenuse = AB / AC  AC = AB / Sin  = 25 / Sin45 =25 / 0.7071 = 35.3556mm

14

C

TRIGONOMETRIC FUNCTIONS Example: We will Find Value Of  By “Cos” Formula. A



25 mm

 B

25 mm

C

Cos  = Adjacent Side / Hypotenuse = AB / AC = 25 / 35.3556 = 0.7071 0 7071   = Inv Cos (0.7071) = 45º

15

TRIGONOMETRY Example:

FIND OUT ANGLE ‘ Ø ’ OF TRIANGLE ABC.

A

SIN

AB AC

ø = OPPOSITE SIDE = HYPOTENEOUS

OPPOSITE SIDE

HYPOTENEOUS 50

= 0.60

30

ø = SIN VALUE OF 0.60 ø = 36° - 52’

ø B

ADJACENT SIDE

= 30 50

C

16

FIND OUT SIDE ‘ø ’ OF A TRIANGLE Example: TAN ø = OPPOSITE SIDE = AB ADJACENT SIDE BC

A

OPPOSITE SIDE

TAN 36° =

HYPOTENEOUS

•

20 36° B

? ADJACENT SIDE

C

20 BC

• • BC =

20 TAN VALUE OF 36°

•

20 0.727

• • BC = •

• • BC =

27. 51 mm

17

AREA Definition :

A surface covered by specific Shape is called area of that shape. i.e. area of square,circle etc.

1 S 1. Square :

Area Of Square = L X L = L² L Where L = Length Of Side L

So If L Th Area Then A

= 5cm = 5 X 5 = 25 25cm²² 18

AREA 2. Rectangle:

Area Of Rectangle = L X B Where,

3. Circle :

L B If L= 10 mm, And B

= Length = Width = 6 mm

Then, Area

= 10 X 6 = 60mm²

Area Of Circle =

 / 4 x D²

B L

D

Wh Where D= D Diameter Di Of The Th Circle Ci l

Area Of Half Circle = /8 x D² D

Same way we can find out area of quarter of circle19

AREA 3 . Circle :

Hollow Circle =

 x (D² - d²) d

4 WHERE D = Diameter of Greater Circle d = Diameter of Smaller Circle

Sector Of Circle=

xD²xØ 4 x 360

D

Ø

D 20

AREA 4. Triangle g :

H

Area Of Triangle = ½ B x H Where B H

= Base Of Triangle = Height Of Triangle

B

5. Cylinder : Surface area of Cylinder y =xDxH Where H D

H

D

= Height Of Cylinder = Diameter Of Cylinder 21

VOLUME Defination : A space covered by any object is called volume of that object.

1. Square block : In square block;

length,

width and height are equal, so

L L

Volume Of Sq. Block = L X L X L = L³

L

2. Rectangular Block : Volume= L X B X H Where L = Length B = Width H = Height

H L 22

B

VOLUME

H

4.Prism or Triangle Block : B

Volume of Triangular Block = Cross Section Area of Triangle x Length

L

( Area of Right Angle Triangle = ½ B H ) Volume = ½ B H X L

Where B = Base of R.A.Triangle H = Height of R.A.Triangle L = Length of R.A.Triangle R A Triangle 23

VOLUME 3. Cylinder : Volume of Cylinder = Cross Section Area x Length of

Volume= ¼D² X H Where : D = Diameter Of The Cylinder H = Length Of Cylinder

D

Cylinder

H

24

CG CALCULATION CG m

TAN LINE

DIA

CENTRE OF GRAVITY OF D D’ENDS ENDS ( CG ) (1)

HEMISPHERICAL

( m ) = 0.2878  DIA

(2)

2 1 ELLIPSOIDALS 2:1

( m ) = 0.1439 0 1439  DIA

(3)

TORI - SPHERICAL

( m ) = 0.1000  DIA

25

FAB UNIT -1 TEST PAPER Q-1. 5.5 M

= ____________ inches = ____________mm.

Q-2 3 Q-2. 3.4 4 Kg / CM CM² = ____________ Psi Psi.

Q-3. 900 LBS

= ____________ Kgs

26

Q-4. VOLUME

Dia.120 00mm

20mm

2500 mm

Volume of shell plate = __________________

27

Q-5. AREA

100mm

300 mm

100 mm

A Area off above b shape h = ____________

28

Q-6 PYTHAGORAS 2000mm

x

r

Wh t iis th What the value l off X ? 29

Q-7. PYTHAGORAS 4000 mm

5000 0 mm

10000 mm

What is the value of X ? 30

Q-8. TRIGONOMETRIC FUNCTION

O/D =4200 mm 60º 60

800

X

What is the distance between two rollers “X”?

31

Q-9. TRIGONOMETRIC FUNCTION

X2 q= 45 THK =6 60

40 2 18

Æ =60

X1

3

Find out X1 and X2 distance. 32

Q-10. TRIGONOMETRIC FUNCTION

ID=4200 mm



800

21 0 mm 2150

Find out angle between two rollers “”. 33

Q-11. PYTHAGORAS

7800 mm X

What is the value of “X” ? 34

MODULE : WORKSHOP CALCULATION UNIT : 2

Weight calculation and weld deposition weight = with demonstration

2hours

WEP calculation calculation, 1:3 and 1:5 taper calculation Practice examples = 10 nos.

1 hour 1 hour

35

WEIGHT CALCULATION Examples : Weight calculation of different items: • • • • •

Rectangular R t l plate l t Circular plate Circular p plate with cutout Circular sector Shell coursce

Specific gravity for (i) C.S.= 7.86 g/cm3 (ii) S.S.=8.00 g/cm3 36

WEIGHT CALCULATION Examples : 1. Rectangular plate : Weight of This Plate 3.5 CM = Volume X Sp.Gravity 200 CM = L X B X H X 7.86gm / CC Here L = 200cm, B = Width = 100cm And H = Thk = 3.5 cm So Volume = 200 X 100 X 3.5 cm³ = 70000 cm³ Now Weight Of Plate = Volume X Sp .Gravity gm/cc = 70000 X 7.86 g = 546000 gms 37 = 546 kgs

WEIGHT CALCULATION Examples : 2. CIRCULAR PLATE : g V X Sp. Gravity y Weight=

300 cm

Volume V= Cross Section Area X Thk = ¼D² X 4cm

Thk = 4cm

= ¼ x 300² X 4cm = 282743.33 cm³ So W = V X sp.Gravity sp Gravity = 282743.33 X 7.86 gms/cc = 2222362.5738 gms = 2222.362 kgs 38

WEIGHT CALCULATION Examples : Circular sector : Weight of Circular Plate Segment : W = Volume X Sp.Gravty. p y Now Volume = Cross Sec.Area X Thk =  X ( R1² - R2²) X Ø X 2 cm 360 =  X (400² - 350²) X 120 X 2 360 = 78539.81 cm³ Now Weight = V X Sp .Gravity = 78539.81 X 7.86 g gms/cc = 617322.95 gms = 617.323 kgs

r1 r2

R1 = 400 cm R2 = 350 cm THK = 2cm  = 120º

39

WEIGHT CALCULATION Examples : Shell : W = V X Sp.Gravity V= ¼  X ( OD² - ID² ) X Length H Here OD = 400 + 10 = 410 410cm ID = 400cm Length = 300cm So V = ¼ X ( 410² - 400² ) X 300cm = 1908517.54cm³ 1908517.54cm Now Weight W = V X Sp. Gravity = 1908517.54 X 7.86 = 15000947gms = 15000.947kgs = @ 15 Ton

40

WEP CALCULATION SINGLE 'V' A

In

B

given

figure,

to

find

out

Distance, we will use

C

2 3

100

9 98

q= 60

Trigonometric formula. formula Tan Q / 2 = AB / BC Here AB = ?, BC = 98, Q / 2 = 30º  Tan 30 = AB / 98  AB = Tan30  98 = 0.577  98 = 56.54 mm 41

WEP CALCULATION Double ‘V’ V For double v also we can calculate distance by 40 2 18

THK =60

q = 45

Æ = 60 3

same trigonometric formula. formula

Double v are

two types: 1. Equal v 2. 2/3 rd &1/3 rd.

T joint • In I t joint j i t also l by b tan t formula f l we can find WEP dimensions: 40THK

=

= B

q= 50

C

A

AC = 20 , q = 50 , AB = ? TAN q = AB / AC AB = 20 x TAN 50 AB = 23.83 42

of

WEP CALCULATION COMPOUND 'V' Æ=

10 56

q=

THK=70

45 12 R.F.=

R.G.=

2

3

In such kind of compound “V”, we always do machining to take care of all calculation. y dotted line,, we can calculate WEP As shown by dimensions by sine or tangent formula. 43

WELD METAL WEIGHT CALCULATION Weld metal weight = Cross section area of particular WEP x length / circumference of seam x density

Basically weld metal weight calculation involves C Calculation off volume, trigonometry and Weight calculation. 44

WELD METAL WEIGHT CALCULATION Basic fundamentals of weld metal weight g Calculation 1.Single v for long seam and circseam • Long seam weld weight = Cross section area x length of seam x density • Circ. seam weld weight `= Cross C section ti area x mean circ. i off seam x density d it

45

WELD METAL WEIGHT CALCULATION 3

1

Now A1 = 2/3 x H x Bead Width  A1 = 2/3 x 0.3 x 6 cm² = 1.2 cm²

 =60º 2 4

50 0

3

Now A2 =A3

3 2 1.Crossection Area Of Joint A = A1 + A2 + A3 + A4

A2 = 1/2 x B x h = 0.5 x B x 4.7 cm² H Here B B= 47 T Tan30º 30º =2.713cm 2 713  A2 = 0.5 x 2.713 x 4.7 Cm² = 6.38 Cm² A3 = 6.38 6 38 Cm² Cm

Now A = 1.2 + 6.38 + 6.38 + 0.94 cm² A = 14.9cm²

A4 =0.2 * 4.7 cm² 46

WELD METAL WEIGHT CALCULATION For long seam weld weight = Cross C section ti area x Length L th off seam x density d it = 14.9cm² x 100cm x 7.86gm/cm³ = 11711.4gms = 11.712kgs for 1 mtr long seam For circ. seam = Cross section area x Mean circ. x Density For Circ. seam having OD = 4000 mm and Thk. = 50 mm Weld Weight = 14.9cm² X 1272.3 cm X 7.86 gms/cc = 149009gms = 149.009kgs. 47

TAPER CALCULATIONS Whenever a Butt joint is to be made between two plates of different thickness, a taper is generally provided on thicker plate to avoid mainly stress concentration.

1:3 Taper p 40

x 60

Thickness Difference = 60 - 40 = 20mm. X = 20 x 3 = 60mm. Instead of 1:3 taper, taper if 1: 5 Taper is required; X = 20 x 5 = 100 mm.

48

MODULE : WORKSHOP CALCULATION UNIT : 3 Measure tape error correction and circumference calculation = with demonstration (1 hour) O i t ti marking Orientation ki ( 0.5 0 5 hour h ) Offset and kink, web and flange tilt, flange unbalance b l calculation l l ti (1 hour) h ) Arc length and chord length calculation for web layout= with demonstration ( 0.5 hour ) Practice examples = 10 nos. (1 hour) 49

USE OF CALIBRATION TAPE How to refer calibration report? p Consider total error for calculation. Standard error & relative error are for calibration purpose only. H How tto use calibration lib ti report? t? Marking - Add the error. (Mad) Measuring - Subtract the error (Mes) During calculation, always put error value in brackets. 50

USE OF CALIBRATION TAPE. Example: Cut 1meter long bulbar Tape-01

Tape 02

Total error at 1m (+1)

Total error at 1m (-1)

Marking of 1 m (add the error) 1000mm+(+1)mm M ki att 1001mm Marking 1001

1000mm+(-1)mm M ki att 999mm Marking 999

measure the length(subtract the error) Length found f 1001mm

Length found f 999mm

1001-(+1)mm

999-(-1)mm

1000mm actual length

1000mm actual length 51

Tape 01 (+1 mm error) Bulb bar

Marking 1000+(+1) mm Measuring 1001- (+1) mm error Actual 1000 mm

52

Tape 02 (-1 mm error) Bulb bar

Marking 1000+( 1000+(-1) 1) mm Measuring 999 - (-1) mm error Actual 1000 mm

53

CIRCUMFERENCE CALCULATION Circumference = Pie x Diameter of job If I/D is known and O/S circ. circ Is required then, then Circumference = Pie x ( I/D + 2 x thick ) Here Pie value is very important important. Which is the correct value of pie? 22/7 3.14 3 1415926 (Direct 3.1415926 (Di t from f calculator/ l l t / computer) t )

54

CIRCUMFERENCE CALCULATION

Example 1 : O/S Dia of the job is 10000mm, calculate O/S circumference. circumference 1) 10000mm x 22/7

= 31428.571mm

2) 10000mm 10000 x 3.14 3 14

= 31400.00mm 31400 00

3) 10000mm x 3.1415926 = 31415.926mm

55

CIRCUMFERENCE CALCULATION Example 2 : Internal T-frame o/d - 9998mm Shell thickness - 34mm ,Root Root gap - 0.5mm 0 5mm Calculate shell o/s circumference. Shell o/d

= T - fr o/d 9998mm + root gap (0.5mm x 2) + thickness (34 x 2mm) = 10067mm

Circumference = Pie x 10067mm If pie i = 3.1415926 3 1415926

th then circ. i = 31626 31626.4mm 4

If Pie = 22/7

then circ. = 31639.14mm

If Pie = 3.14

then circ. = 31610.38mm 56

OFFSET CALCULATION Thickness difference measured from I/s or o/s on joining edges is called offset.

offset

Tolerance as per P P-1402 1402 0.1T but