fabrication_calculation

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1 WORKSHOP CALCULATION

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2 PRESSURE CONVERSION 1 Kg / cm² = 14 . 223 psi ( Lb / In² ) 1 Kg / cm² = 0 . 9807 Bar. 1 PSI = 0.07031 Kg / cm² Introduction to Units ( Pressure) 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

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3 Introduction to Units ( Temperature) Temperature unit = degree centigrade or degree Fahrenheit °C = 5/9(°F- 32) If Temp. Is 100°F, Then °C=5/9( 100-32) So, °C=37.7 If Preheat Temperature Is 150 °C, Then °F=302

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4 PYTHAGORAS PRINCIPLE APPLICATION Pythagoras Principle : In Any Right Angled Triangle the Square of Sum of Adjacent Sides Is Always Equal to the Square of Hypotenuse . A B C LET US SAY  ABC is right angle triangle . AB and BC = Adjacent sides and AC = Hypotenuse. So based on pythagoras theory , AB² + BC² = AC²

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5 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 3 4 5 A B C Example : PYTHAGORAS PRINCIPLE APPLICATION

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6 TRIGONOMETRIC FUNCTIONS Trigonometric functions are used to solve the problems of different types of triangle. Let us consider  ABC is a “right angled triangle”, Angle ABC = ø , AB & BC are sides of triangle. So for this triangle. A B C ø We will see some simple formulas to solve right angle triangle which we are using in day to day work.

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7 TRIGONOMETRY COS ø = Adjacent Side Hypoteneous = BC AC TAN ø = Opposite Side Adjacent Side = AB BC SIN ø = Opposite Side Hypoteneous = AB AC C ø Hypoteneous Adjacent Side Opposite Side A B

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8 TRIGONOMETRIC FUNCTIONS Example : For triangle ABC find out value of  and . 25 mm 25 mm   A B C We Will Find Value Of  By “Tan” Formula. So , Tan  = Opposite Side / Adjacent Side = AB / BC = 25/25 =1 Tan  = 1   = Inv. Tan(1) = 45º 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

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9 TRIGONOMETRIC FUNCTIONS Example: We will Find Value Of  By “Cos” Formula. 25 mm 25 mm   A B C Cos  = Adjacent Side / Hypotenuse = AB / AC = 25 / 35.3556 = 0.7071   = Inv Cos (0.7071) = 45 º

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10 TRIGONOMETRY Example: FIND OUT ANGLE ‘ Ø ’ OF TRIANGLE ABC. OPPOSITE SIDE HYPOTENEOUS AB AC SIN ø = = = 30 50 = 0.60 ø = SIN VALUE OF 0.60 ø = 36 ° - 52’ C ø HYPOTENEOUS ADJACENT SIDE OPPOSITE SIDE A B 50 30

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11 FIND OUT SIDE ‘ ø ’ OF A TRIANGLE Example: TAN 36 ° = TAN ø = OPPOSITE SIDE ADJACENT SIDE = AB BC 20 BC BC = 20 TAN VALUE OF 36° BC = 20 0.727 BC = 27. 51 mm • • • • • • • • • OPPOSITE SIDE C ? HYPOTENEOUS ADJACENT SIDE A B 36° 20

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12 Definition : A surface covered by specific Shape is called area of that shape. i.e. area of square,circle etc. So If L = 5cm Then Area = 5 X 5 = 25cm² Area Of Square = L X L = L² 1. Square : L L Where L = Length Of Side AREA

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13 AREA Area Of Rectangle = L X B 2. Rectangle: B L Where, L = Length B = Width If L= 10 mm, And B = 6 mm Then, Area = 10 X 6 = 60mm ² Area Of Circle =  / 4 x D ² 3. Circle : D Where D= Diameter Of The Circle Same way we can find out area of quarter of circle D Area Of Half Circle =  /8 x D²

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14 AREA Hollow Circle =  x (D ² - d ²) 4 3 . Circle : WHERE D = Diameter of Greater Circle d = Diameter of Smaller Circle D d Sector Of Circle=  x D ² x Ø 4 x 360 Ø D

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15 AREA B H Area Of Triangle = ½ B x H 4. Triangle : Where B = Base Of Triangle H = Height Of Triangle 5. Cylinder : D H Surface area of Cylinder =  x D x H Where H = Height Of Cylinder D = Diameter Of Cylinder

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16 VOLUME Defination : A space covered by any object is called volume of that object. L Volume Of Sq. Block = L X L X L = L³ 1. Square block : In square block; length, width and height are equal, so L L 2. Rectangular Block : L B H Volume= L X B X H Where L = Length B = Width H = Height

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17 VOLUME 4.Prism or Triangle Block : Volume of Triangular Block = Cross Section Area of Triangle x Length ( Area of Right Angle Triangle = ½ B H ) H B L 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

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18 VOLUME 3. Cylinder : Volume of Cylinder = Cross Section Area x Length of Cylinder Volume= ¼  D² X H H D Where : D = Diameter Of The Cylinder H = Length Of Cylinder

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19 CG CALCULATION CENTRE OF GRAVITY OF D’ENDS ( CG ) ( 1 ) HEMISPHERICAL ( m ) = 0.2878  DIA ( 2 ) 2:1 ELLIPSOIDALS ( m ) = 0.1439  DIA ( 3 ) TORI - SPHERICAL ( m ) = 0.1000  DIA CG DIA m TAN LINE

FAB UNIT -1 TEST PAPER:

FAB UNIT -1 TEST PAPER 20 Q-1. 5.5 M = ____________ inches = ____________mm. Q-2. 3.4 Kg / CM² = ____________ Psi. Q-3. 900 LBS = ____________ Kgs

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21 Q-4. VOLUME Volume of shell plate = __________________ 20mm 2500 mm Dia.1200mm

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22 Q-5. AREA 100mm 300 mm 100 mm Area of above shape = ____________

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23 Q-6 PYTHAGORAS What is the value of X ? 4000 2000mm x r

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24 Q-7. PYTHAGORAS 4000 mm 5000 mm 10000 mm X What is the value of X ?

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25 Q-8. TRIGONOMETRIC FUNCTION 60º X 800 O/D =4200 mm What is the distance between two rollers “X”?

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26 q= 45 THK =60 3 Æ =60 2 18 40 X1 X2 Q-9. TRIGONOMETRIC FUNCTION Find out X1 and X2 distance.

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27 Q-10. TRIGONOMETRIC FUNCTION Find out angle between two rollers “ ” .  2150 mm 800 ID=4200 mm

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28 X 5600 mm 7800 mm Q-11. PYTHAGORAS What is the value of “X” ?

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29 WEIGHT CALCULATION Examples : Weight calculation of different items: Specific gravity for (i) C.S.= 7.86 g/cm 3 (ii) S.S.=8.00 g/cm 3 Rectangular plate Circular plate Circular plate with cutout Circular sector Shell coursce

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30 Weight of This Plate = Volume X Sp.Gravity = 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 = 70000 X 7.86 gm/cc = 546000 gms = 546 kgs WEIGHT CALCULATION Examples : 1. Rectangular plate : 200 CM 100 CM 3.5 CM

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31 WEIGHT CALCULATION Examples : 2. CIRCULAR PLATE : Weight= V X Sp. Gravity Volume V= Cross Section Area X Thk = ¼  D² X 4cm = ¼  x 300² X 4cm = 282743.33 c m³ So W = V X sp.Gravity = 282743.33 X 7.86 gms/cc = 2222362.5738 gms = 2222.362 kgs 300 cm Thk = 4cm

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32 WEIGHT CALCULATION Examples : Circular sector : R1 = 400 cm R2 = 350 cm THK = 2cm  = 120º r1 r2 Weight of Circular Plate Segment : W = Volume X Sp.Gravty. Now Volume = Cross Sec.Area X Thk =  X ( R1² - R2²) X Ø X 2 cm 360 =  X (400 ² - 350²) X 120 X 2 360 = 78539.81 c m³ Now Weight = V X Sp .Gravity = 78539.81 X 7.86 gms/cc = 617322.95 gms = 617.323 kgs

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33 WEIGHT CALCULATION Examples : Shell : W = V X Sp.Gravity V= ¼  X ( OD ² - ID ² ) X Length Here OD = 400 + 10 = 410cm ID = 400cm Length = 300cm So V = ¼  X ( 410 ² - 400² ) X 300cm = 1908517.54cm³ Now Weight W = V X Sp. Gravity = 1908517.54 X 7.86 = 15000947gms = 15000.947kgs = @ 15 Ton 300 cm 400 cm 5cm

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34 WEP CALCULATION In given figure, to find out Distance, we will use Trigonometric 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 SINGLE 'V' q= 60 100 2 3 98 A B C

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35 WEP CALCULATION Double ‘V’ q = 45 THK =60 3 Æ = 60 2 18 40 For double v also we can calculate distance by same trigonometric formula. Double v are of two types: 1. Equal v 2. 2/3 rd &1/3 rd. T joint In t joint also by tan formula we can find WEP dimensions: q= 50 40THK A B C = = AC = 20 , q = 50 , AB = ? TAN q = AB / AC AB = 20 x TAN 50 AB = 23.83

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36 WEP CALCULATION COMPOUND 'V' In such kind of compound “V”, we always do machining to take care of all calculation. As shown by dotted line, we can calculate WEP dimensions by sine or tangent formula. THK=70 Æ= 10 q= 45 R.F.= 2 R.G.= 3 56 12

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37 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 Calculation of volume , trigonometry and Weight calculation.

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38 WELD METAL WEIGHT CALCULATION Long seam weld weight = Cross section area x length of seam x density Circ. seam weld weight `= Cross section area x mean circ. of seam x density Basic fundamentals of weld metal weight Calculation 1.Single v for long seam and circseam

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39 WELD METAL WEIGHT CALCULATION 50 3  =60º 2 3 1 2 3 4 1.Crossection Area Of Joint A = A1 + A2 + A3 + A4 Now A1 = 2/3 x H x Bead Width  A1 = 2/3 x 0.3 x 6 cm² = 1.2 cm² Now A2 =A3 A2 = 1/2 x B x h = 0.5 x B x 4.7 cm² Here B= 47 Tan30º =2.713cm  A2 = 0.5 x 2.713 x 4.7 Cm² = 6.38 Cm² A3 = 6.38 Cm² A4 =0.2 * 4.7 cm² Now A = 1.2 + 6.38 + 6.38 + 0.94 cm² A = 14.9cm²

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40 WELD METAL WEIGHT CALCULATION For long seam weld weight = Cross section area x Length of seam x density = 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 = 14.9c m² X 1272.3 cm X 7.86 gms/cc = 149009gms = 149.009kgs .

TAPER CALCULATIONS:

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. 41 1:3 Taper 40 60 Thickness Difference = 60 - 40 = 20mm. X = 20 x 3 = 60mm. Instead of 1:3 taper, if 1: 5 Taper is required; X = 20 x 5 = 100 mm. x

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

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43 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 1000mm+(-1)mm Marking at 1001mm Marking at 999mm measure the length(subtract the error) Length found 1001mm Length found 999mm 1001-(+1)mm 999-(-1)mm 1000mm actual length 1000mm actual length

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Tape 01 (+1 mm error) 44 Bulb bar Measuring 1001- (+1) mm error Marking 1000+(+1) mm Actual 1000 mm

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Tape 02 (-1 mm error) 45 Bulb bar Measuring 999 - (-1) mm error Marking 1000+(-1) mm Actual 1000 mm

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

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47 CIRCUMFERENCE CALCULATION Example 1 : O/S Dia of the job is 10000mm, calculate O/S circumference. 1) 10000mm x 22/7 = 31428.571mm 2) 10000mm x 3.14 = 31400.00mm 3) 10000mm x 3.1415926 = 31415.926mm

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48 CIRCUMFERENCE CALCULATION Example 2 : Internal T-frame o/d - 9998mm Shell thickness - 34mm ,Root gap - 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 = 3.1415926 then circ. = 31626.4mm If Pie = 22/7 then circ. = 31639.14mm If Pie = 3.14 then circ. = 31610.38mm

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49 OFFSET CALCULATION Thickness difference measured from I/s or o/s on joining edges is called offset. Tolerance as per P-1402 0.1T but <= 2mm for web & <= 3mm for flange Say T = 34 mm than, Offset = 0.1 x 34mm = 3.4mm But max. 3mm allowed as mentioned above. If by mistake 0.1% T considered than, 0.1 x 34/100 = 0.034 mm offset which is wrong. offset

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50 OFFSET CALCULATION How to measure offset & kink ? Here A = D Offset = B - C Kink = ( A - B or C - D ) which ever is max. Kink is nothing but peak-in/ peak-out A B C D

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51 OFFSET CALCULATION How to measure offset& kink in case of thickness difference? Here A = D Offset = B - C Kink = ( A - B or C - D ) which ever is max. Kink is nothing but peak-in/ peak-out A B C D

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52 ORIENTATION MARKING Start orientation in following steps. Measure circumference. Check long seam orientation from drawing. Find out arc length for long seam from 0 degree. Arc length = (circ./360 ) x Orientation. Always take all digits of orientation given in drawing.

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53 ORIENTATION MARKING Example : O/s circ. = 25300mm L/s orientation = 75.162 degree Find out arc length for 75.167 Arc length for l/s = ( 25300/360 ) x 75.1 = 5277.86mm = ( 25300/360 ) x 75.16 = 5282.07mm = ( 25300/360 ) x 75.167 = 5282.56mm

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54 TOLERANCES Always read the drawing carefully to interpret tolerance correctly. (1) Pre-tilt of web : For 101 mm to 150 mm frame height -- 0.025H but  3mm Example: H = 120mm then, pre tilt = 0.025 x 120 = 3mm

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55 TOLERANCES How to check Pre tilt of web :[ X-Y ] = pre tilt X Y

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56 TOLERANCES (2) Flange pre tilt : <= 3mm [ X-Y ] = Pre tilt X Y

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57 TOLERANCES (4) Out of circularity (OOC) : 0.2 % R ( R-theoretical radius of PRB ) Example : R = 4000mm OOC = 0.2 x 4000/100 = 8mm (5) Flange position w.r.t web : (Flange unbalance) :+/- 1mm [ X - Y ] = 2mm X Y

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58 l l = ARC / LENGTH a = AREA OF SEGMENT c = CHORD LENGTH q = ANGLE r = RADIUS h = HEIGHT BETWEEN CHORD TO ARC ( 2 ) a = 1/2 [ rl - c ( r - h ) ] ( 3 ) h = r - 1/2 ( 4 ) r = c 2 + 4 h 2 8 h ( 1 ) c = 2 h ( 2 r - h )  ( 5 ) l = 0.01745  r  q ( 6 ) q = 57. 296  l r ( 7 ) h = r [ 1 - COS ( q / 2 ) ] 4 r 2 - C 2  q r C h a Example:

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59 CHORD LENGTH Example : Web segment size - 60 0 Inside radius R - 4000mm Sine 30 = CB/4000mm 1/2 chord length CB = 0.5 x 4000mm = 2000mm Full chord length = 4000mm A B 60 R C

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60 If you are marking 3500mm length with tape error as (+2mm), what will be the actual dimensions you will mark? Add ERROR. I.e. 3500mm +2 CALCULATION PRACTICE 1

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61 3200mm 24mm What is plate length required ? If 2

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62 6600mm 55 0 44mm 0 0 What is circumferential distance for marking centre of nozzle ? 3

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63 What is kink & offset. 60mm 32mm 21 24 54 55 I/S 4

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64 X1 X2 60 0 4000mm 3500mm S2 S1 Find out arc length & chord length. 5

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65 440mm X1 X2 What is the maximum allowable difference between X1 & X2 6

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66 330mm 600mm Y1 Y2 What is the maximum deviation allowed on Y1 & Y2 7

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67 16mm 200 94 90 Is the above stiffener acceptable 8

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68 27mm 18mm 42mm 9mm Section 1 Section 2 Outside Circumference of section 1 = 25240mm out side Circumference of section 2 = 25295mm 8000mm ID Theoretical surface 9

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69 7900mm 27mm 27mm OUT OF CIRCULARITY Maximum Out of circularity allowable= _______mm 10

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70 PYTHAGORAS PRINCIPLE APPLICATION T.L A B C D E Trimming height calculation in hemispherical D’end For matching OD / ID of D’end to shell OD / ID we have to do actual Marking on D’end for trimming height We can find out trimming height by Pythagoras theory As shown in figure, we can have Following dimension before Marking trimming AB = Radius of D’end. Based on act Circumference at that end AC = CD = D’end I/S Radius as per DRG. from T.L BC = Straight face or height from T.L TO D’end. edge ED = D’end radius calculated from its matching part’s Circumference BE = Trimming height req to maintain for req circumference of Matching part circumference

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71 PYTHAGORAS PRINCIPLE APPLICATION T.L A B C D E Example : AB = 1500mm AC = CD = 1510mm BC = 173.5mm ED = 1495mm BE = ? Based on Pythagoras theory In triangle  CED CE² + ED² = CD²  CE ² = CD² - ED² = 1510 ² - 1495²  CE = 212.3mm Now CE = CB + BE  BE = CE - CB = 212.3 - 173.5 = 38.8mm

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72 TRIGONOMETRIC FUNCTIONS Tank rotator rollers dist. Calculation   A B C D As shown in figure we can find out Two things : 1. Angle  between two rollers 2. Dist. Between two roller for specific diameter of shell . We will check it one by one. For safe working, angle  Should be between 45- 60º

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73 TRIGONOMETRIC FUNCTIONS Tank rotator rollers dist calculation 1. Angle  between 2 roller: As shown in figure BC = Half of the dist between two rollers AD = Shell o/s radius DC = Roller radius So we can get above dimensions from DRG and Actual dist from tank rotator Now as per sine formula Sin /2 = BC/ AC AC = AD + DC ( Shell OD + Roller DIA ) Sin /2 = BC / (AD +DC) Now If We Take BC = 1500 mm, AD = 2000mm AND DC = 400 mm Then Sin  /2 = 1500 / (2000 + 400 ) = 1500 / 2400 = 0.625 Sin  /2 = 0.625   /2 = INV Sin 0.625 = 38.68º   = 2  38.68º = 77.36º  A B D C

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74 TRIGONOMETRIC FUNCTIONS Tank rotator rollers dist calculation : 2.Roller dist. By deciding angle Between two roller If We Keep Roller Angle = 75º AD = Shell O/s Radius = 3000mm DC = Roller Radius = 400mm CE = Dist. Between Two Roller = CH + BE = 2  CH (CH = CE) Now By Sine Law Sin  /2 = BC/AC BC = Sin /2  AC  BC = Sin37.5 º  3400 ( = 75 º  /2 = 37.5 º , AC = AD + DC = 3000 + 400)  BC = 0.6087  3400 = 2069.78 mm  Dist.Between Roller CE = 2  BC = 2  2069.78 = 4139.56mm  A B D C E

PCD & HOLE MARKING CALCULATIONS:

PCD & HOLE MARKING CALCULATIONS For Example, consider a flange 14”-1500# with P.C.D.=600 mm & No. of Holes N = 12. Mark P.C.D. = 600 mm. Angular distance y = 360 / N = 360/12 = 30 degrees. Chord length between holes = 2 x PCD x Sin ( y/2 ) 2 = 2 x 600 x Sin (30/2) 2 = 2 x 600 x 0.2588 = 155.28 mm. 2 75 ‘N’ Holes P.C.D. y

SLING ANGLE CALCULATION.:

SLING ANGLE CALCULATION. 76 Hook 5000 4000

SLING ANGLE CALCULATION.:

SLING ANGLE CALCULATION. 77 5000 2000 ø

CALCULATIONS:

CALCULATIONS 78 Sin Ø = x/y x = 2000 & y = 5000 Ø = 23.5 0 2 Ø = 23.5 X 2 = 47 0

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79 M/CING ALLOWANCES Add 3 mm (min.) on all dimensions to provide for m/cing allowances. Example of O/Lay on Gasket face of Flange: 2106 dia.(min.) 8 (min.) 5 1894 dia.(max.) 1900 dia.

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80 MACHINING ALLOWANCE CALCULATION 710 650 600 30 30 Machining Allowance 700

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81 FABRICATION UNIT 1 THEORY EXAMINATION 5.5 M 7.28 M 1. AREA of given figure = ___________mm² 2. 3500 psi = ________________kgs/cm² 3. Weight of shell = __________________kgs 5500mm 2200mm

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82 FABRICATION UNIT 1 THEORY EXAMINATION x 60° 4 mm 34 mm 4.What is the value of x ? 5. Dimension of taper = ___________mm 1:3 Taper 56mm 28mm 6. If OD of shell is 2800 mm then Circumference of shell at 37.7° = ___________mm

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83 FABRICATION UNIT 1 THEORY EXAMINATION 6000mm x 10000 mm 7. Value of x = ____________mm 50 51 53 55 8.Value of Kink = _______ Offset = _______

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84 FABRICATION UNIT 1 THEORY EXAMINATION 9. Is handling safe ? Why ? __________ 3000mm 5400 mm 

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85 FABRICATION UNIT 1 THEORY EXAMINATION 10. FIND THE WEIGHT OF WELD METAL 60° 75° 4mm 90mm 56mm