design of fasteners

8/4/2019 Design of Fasteners

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Design of Fasteners

-V K Joshi-Assistant Professor

--Mechanical Depart-FETR


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Riveted Joint


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Riveted Joint


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Riveted JointTypes of Joint

Permanent Joint Cant be disassembled without damaging the connecting element

Welded joint

Brazed joint Soldered joint Adhesive joint Riveted joint

Detachable Joint

Can be disassembled without damaging the connecting element Threaded joint Pin joint Cotter joint Key joint


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Riveted Joint


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Riveted Joint

Rivet is a short cylindrical bar with head Head body (shank) Tail


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Riveted Joint

Rivet material strong Ductile

Mild steel Aluminum Brass

CopperUsed when Strength is not

required


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Riveted Joint

Riveting Process


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Riveted Joint


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Riveted Joint

Rivet provides Strength Rigidity Leak proof joint Can joint

dissimilar material (welding X)

Metallic and non metallic material


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Riveted Joint

Rivet could not provide Easily dissemble of parts As not leak proof as welding Weaken the parts Complicated shaped material joint Heads are inconvenient


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Riveted Joint

Type of rivet head Snap head

Structural work Pressure vessel Machine

Countersunk head Ship building


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Riveted Joint

Type of rivet head Conical head

Hand hammering

Pan head

Maximum strong Difficult to make


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Rivet JointRivetA) As per Arrangement Single Raw (Chain) Double Raw (Zigzag)B) As per column

Single Riveted Double RivetedPlateA) As per Arrangement of plate Lap

ButtB) As per number Single strap Double strap


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Rivet Joint

Lap Joint


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Rivet Joint

Lap Joint


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Rivet Joint

Butt Joint


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Rivet Joint

Butt Joint


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Riveted Joint

Quiz

Flat Countersunk Cone Pan


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Riveted Joint

Quiz


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Riveted Joint

Quiz


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Riveted Joint

Quiz


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Design of Riveted Boiler Joint


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Lozenge or Uniform strength rivet


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Type of Joint Factor of Safety

Hand Machine

Lap 4.75 4.5

Single butt 4.75 4.5

Double butt 4.25 4.0

Two plates of 7 mm thickness are connected by a double riveted lap joint of zigzag


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Given:Tensile stress t = 90 MpaShear stress s = 60 MpaCompressive stress c =120MPa

Plate thickness t = 7 mm

Note : t = 7 mm < 8mmShearing strength and crushing strength are sameEquation

Calculated d = 17.8 mmPreferred d = 19 mm (Answer 1)

cs dt d

2422

p y p j g gpattern. For 50 kN tensile loadCalculate :1) rivet diameter, 2) rivet pitch , 3) strength of rivet joint4) Number of rivets, 5) efficiency of joint, 6)width of plateAssume

t= 90 MPa

s = 60 MPa and

c=120MPa



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t= 90 MPa

s = 60 MPa and

c=120MPa

Given:Tensile stress t = 90 MpaShear stress s = 60 Mpa

Compressive stress c =120MPaPlate thickness t = 7 mmEquation

p = 73 mm (Answer 2)

st d t d p

2

42



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t= 90 MPa

s = 60 MPa and

c=120MPa

Given:Tensile stress t = 90 MpaShear stress s = 60 MpaCompressive stress c =120MPaPlate thickness t = 7 mm

From, Answer 1 and 2Pitch = 73 mmDiameter = 19 mm

Pt = 34020 N P s = 17003 N Pc = 15960 N(Answer 3)

t t t d pP ss

d P 2

4 ccdt P



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t= 90 MPa

s = 60 MPa and

c=120MPa


Load =50 kNFrom, Answer 3Strength = 15960 N

Calculated N = 3.13Preferred N = 4

(Answer 4)

SC t PPP

W N

,,min

15960

1050 3 N



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t= 90 MPa

s = 60 MPa and

c=120MPa


Load =50 kNFrom, Answer 2 and 3Pitch = 73 mmStrength = 15960 N

Calculated = 34%(Answer 5)

t

SC t

pt

PPP

,,min

90737

15960


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Try by yourself MDID, Patil R.B. , Tech Max, pp: 2-25

Two plates of 6 mm thickness are connected by a single riveted lap joint of zigzagpattern. For 50 kN tensile loadCalculate :1) rivet diameter, 2) rivet pitch , 3) strength of rivet joint4) Number of rivets, 5) efficiency of joint, 6)width of plateAssume t = 120 MPa s = 85MPa and c=185MPa

cs dt d 2

4

st d t d p

2

42

t t t d pP ss d P 2

4 ccdt P

SC t PPPW N

,,min

t

SC t

pt

PPP

,,min

b = (N-1)p+2m

Answer :1)d = 16.62 mm

d = 17 mm (Preferred)2)p = 44 mm3)P t = 19440 N

Ps = 19293.3 NPc = 18870 N

4)N = 2.65N = 3 (Preferred)

5) = 59.56% 6) b = 139 mm


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Design of EccentricallyRiveted Joint


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Design of Eccentrically Riveted Joint


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12

3 4

L1L2

L3.L4 .


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A steel plate of 8 mm thickness subjected to a force of 5 kN and fixed to verticalchannel shown in figure with four identical rivets. Tensile yield strength of 220 N/mm 2,factor of safety id 2, determine the diameter of rivet, Assume that the yield strength incompression is 30% more than yield strength in tension.

Given:Yield tensile stress S yt = 220 N/mm 2

Yield Compressive stress S yc =1.3 x Syt Plate thickness t = 8 mm

Load W =500 kNNumber of rivet n = 4Factor of safety Nf = 2


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Yield Compressive stress S yc =1.3 x Syt Plate thickness t = 8 mmLoad W =500 kNNumber of rivet n = 4Factor of safety Nf = 2Calculation

L32 =

50

50L3

22

505071.70=

L1 = L2 = L3 = L4 = 70.71 mm


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Yield Compressive stress S yc =1.3 x Syt Plate thickness t = 8 mmLoad W =5000 N

Number of rivet n = 4Factor of safety Nf = 2Calculation

W x e = w x l1

2+ w x l2

2 + w x l3

2 + w x l3

2 Fs = w x l

Fs1 = w x l1Fs2 = w x l2Fs3 = w x l3F

s4= w x l

4

W x e = FS1xl1+ FS2xl2+ FS3xl3+ FS4xl4

24

23

22

21 llll

eW w

2

75.704

2505000w = 62.5 N/mm

l l f h k b d f f k d f d l


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50

50L3

50

50tan 1 =45 0

Secondary Shear ForceFs1 = w x l1=62.5x70.75=4419.38NFs2 = w x l2=62.5x70.75=4419.38N Fs3 = w x l3=62.5x70.75=4419.38N Fs4 = w x l4=62.5x70.75=4419.38N

A l l f 8 hi k bj d f f 5 kN d fi d i l


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Primary shear force =F p=n

W

Primary shear force =F p= 1250 N



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Resultant shear force =F R

cos2 112

s ps p R F F F F F

45cos1250238.44191250 2 RF

= 5376.42 N



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Number of rivet n = 4Factor of safety N f = 2Calculation

Rivet Size 55

5.0

f

yt

f

ye

N

S

N

S N/mm 2

1433.1

f

yt

f

ye

N

S

N

S N/mm 2

55

4

42.5376

2d A

F R

d = 11.16 mm (calculated)

d = 13.5 mm(preferred)

A t l l t f 8 thi k bj t d t f f 5 kN d fi d t ti l


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Number of rivet n = 4Factor of safety N f = 2Calculation

Rivet checking in crushing

14378.4985.13

42.5376

dt F R N/mm 2

A bracket plate of 12 5 mm thickness is riveted to column by 6 rivets of equal


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A bracket plate of 12.5 mm thickness is riveted to column by 6 rivets of equalsize as shown in figure. It carries a load of 50 kN at the distance of 150 mmfrom the center of column. If the permissible shear stress for the rivet are 75N/mm2 and 140 N/mm2 respectively, determine the diameter of rivet.



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Given :t = 12.5 mm n = 6W = 50000 N e = 150 mm all= 75 N/mm 2 all= 140 N/mm 2

W x e = FS1xl1+ FS2xl2+ FS3xl3+ FS4xl4 + FS5xl5+ FS6xl6

26

25

24

23

22

21 llllll

eW w

Fs = w x l

Primary shear force =F p=

n

W

cos211

2

2 s ps p R F F F F F

44 SP R F F F

2

4

,max,max 4242

d

F F

A

F F R R R R



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Given :t = 12.5 mm n = 6W = 50000 N e = 150 mm all= 75 N/mm 2 all= 140 N/mm 2

Answer

l1= l2 = l5=l6=90.138 mml3= l4 = 50 mmFp = 8333.33 Nw = 200 N/mmFs2 = Fs6 = 18027.87 NFs4 = 10000 N

FR4 =18333.33 NFR2 = 23687.88 N = 20.05 mmc = 88.14 N/ mm 2


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A steel plate is subjected to a force of 8 kN is fixed to a channel by means of three identical rivets as shown I figure. The rivets are made of 15C8 (S yt = 250N/mm 2), If the required FOS is 3, determine the size of Rivets.


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A steel plate is subjected to a force of 8 kN is fixed to a channel by means of three identical rivets as shown I figure. The rivets are made of 15C8 (S yt = 250N/mm 2), If the required FOS is 3, determine the size of Rivets.

GivenW = 8000 N n = 3Syt = 250 N/mm2 N f = 3e = 300 mm


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A steel plate is subjected to a force of 8 kN is fixed to a channel by means of three identical rivets as shown I figure. The rivets are made of 15C8 (S yt = 250N/mm 2), If the required FOS is 3, determine the size of Rivets.GivenW = 8000 N n = 3Syt = 250 N/mm2 N f = 3e = 300 mm

n

W F P

23

21 ll

eW w

Fs = w x l

cos21

22

s ps p R F F F F F

f

yt all N

S5.0

2

4

1

d

F Rall

l l b d f f k f d h l b f


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A steel plate is subjected to a force of 8 kN is fixed to a channel by means of three identical rivets as shown I figure. The rivets are made of 15C8 (S yt = 250N/mm 2), If the required FOS is 3, determine the size of Rivets.GivenW = 8000 N n = 3Syt = 250 N/mm2 N f = 3e = 300 mm

N F P 67.2666

mm N w / 33.213

Fs = 16000 N

N F R 7.16220

2 / 66.41 mm N all

mmd 26.22


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QuizC =100

e=400P=50kN


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Quiz


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Welded Joint

Welding is a process of Joining two metallic parts together by heating to a plastic state with or without the application of pressure and filler metal

Semi molten


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Welded Joint

Permanent joint The heat obtained by

burning of gas (gas welding) an electric arc (electric arc welding).

More speedthen gaswelding


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Welded Joint

Alternative method for casting or forging Replacement for bolted and riveted joints.

It is also used as a repair medium Reunite metal at a crack, To build up broken part

such as gear tooth or worn surface such as a bearing surface.


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Welded Joint


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Welded Joint

Joint Type Sub type

WeldedJoint

Butt

Square

VSingle

Double

U SingleDouble

Fillet

Parallel

TransverseSingle

Double

Other

Corner

Edge

T


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Welded Joint


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Welded Joint

Butt Joint


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Welded Joint

Other Joint


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Welded Joint

Symbols


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Welded Joint

Symbols


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Welded Joint

Symbols


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Welded Joint

Symbols


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Welded Joint

Welded Joint


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Welded Joint

hlP

t = average tensile stress in weld , N/mm 2 P = tensile force on the weld, Nh = weld throat thickness

= plate thicknessl = length of the weld

Tensile Stress

Butt Weld


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Welded Joint

lhhP

t 21

Double Butt Weld

t = average tensile stress in weld , N/mm 2

P = tensile force on the weld, Nh 1= weld throat thickness at toph 2= weld throat thickness at bottoml = length of the weld

Tensile Stress


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Welded Joint

hlP

= average tensile stress in weld , N/mm 2 P = tensile force on the weld, Nl = length of the weld

Machined offreinforcementto avoid stressconcentration

Shear Stress

Welded Joint


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Welded Joint

tlP

lhP

045cos

lhP

707.0

Fillet Weld

Two plates joined by fillet welds,


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are subjected to tensile load 200 N.If the allowable shear stress for theweld material is 85Pa.

Given :l1 = 125 mml2 = 100 mml3 = 125 mmP = 200 x 10 2 N/ mm 2

= 85 x10 6 Pa = 85 N/ mm 2 t lll

P

321

t 12510012510200

853

t = 6.722 mm

045cos

t h

h= 10 mm

Two plates joined byunsymmetrical fillet welds,


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are subjected to axial force P.Lengths of weld should besuch a way that the

sum of resisting momentsAbout the axis of forceis zero .

For top weld,Moment = force x distance

= stress x area x distance= x t x l a x a

For bottom weld,Moment = force x distance

= stress x area x distance

= x t x l b x bFor balance condition, x t x l a x a = x t x l b x b

And, l = l a + lb

lbaa

l

lba

bl

b

a

A 200x150x10 mm steel angle is to bewelded to a steel plate by fillet welds alongh d f 200 l h l i


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the edges of 200 mm leg. The angle issubjected to static load 200 kN. Find thelength of weld for allowable shear stress is75 MPa.

Given :P = 200 x10 3 N = 75 N / mm 2h = 10 mm

lba

al

l

ba

bl

b

a

t lP

2

hl

P

1075

210200 3l

mml 378

mma

a

mmb

b

A A y A y A

b

11.139

88.60200

88.60

101501010200

510150105101020021

2211

mml

l

lll

mml

l

b

b

ab

a

a

937.262

06.115378

06.115

378200

88.60

190

10

10

150


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Eccentric Loading

d


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Eccentric Loading

Primary shear stress

Secondary shear stress

J = IXX+IYY

Moment of Inertia about YY axis I YY= MOI about axis parallel to CG + A 1H12

Moment of Inertia about XX axis I xx= MOI about axis parallel to CG + A 1H12

AP

d

J r T

r

22

113

212

1 lt llt I XX

21

13

1212

1 X

lt lt l I YY

Eccentric Loading


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A1 = l1t= 150 t

A2 = l2t= 150t

A = A1+A2= 150t+150t= 300t

902

180

752

150

Y

X

43

2

2

2

13

2

211

3

102430

9015022

901500

2

1801500

12

1

212

1

mmt

t I

t I

t l I

lt llt I

XX

XX

XX

XX

3

3

3

21

3

2

11

31

105.562

15012

122

0150

12

1

015012

1

212

1

t

t I

t I

t lt I

X l

t lt l I

YY

YY

YY

YY

J = I XX+IYY=2430t x 10 3 + 562.5t x 10 3

= 2992.5t x 103

mm4

For shear stress = 80 N/ mm 2

Eccentric Loading For shear stress = 80 N/ mm 2


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2

3

/ 667.66

300

1020

mm N t

t

AP

d

d

J = I XX+IYY=2430t x 10 3 + 562.5t x 10 3= 2992.5t x 10 3 mm 4

2

3

/ 6.156

105.2992

15.1174000

mm N t

J Tr

r

r

75

90r

mm

r

15.117

7590 22

T = P x e= 20 x 10 3 x 200= 4000 x 10 3 N - mm

0

1

194.50

75

90tan

t

t t t t

r d r d

755.205

194.50cos6.156667.66

26.156667.66

cos2

022

22

mmt t

t

357.2

80755.205

80755.205

X A X A X Ax 332211

A = A1+A2 +A3


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For shear stress= 110 N/ mm 2

J = I XX+IYY

xle

l y

A A A x

1

2

321

300

2

2

13

1 212

1Y t lt l I XX

21

13

1

2

2 22 X

lt lt l X t l I YY

AP

2

221

2

l

X lr

T =P x e

X l

l

1

2

21tan

J Tr

t

cos222t d t d t

mm N t

mmr

mm N T

mm N t

mmt J

mmt I

mmt I e

mmY

mm X

tmm A

t

d

YY

XX

55.719

/ 20.617

49.59

144.232

102.50143

/ 428.171

1086.18

1052678.1

10333.1786.417

200

14.32

700

2

0

3

2

46

46

4

2

6

t=6.54 mmh=10 mm


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Welded Joint


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Bending

Moment

Solid rectangular bar of cross section 50 mmx 75 mm is welded to a support by fillet weldsubjected to a static load of 14 kN at a


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22

13

2212

12

lt lt l I XX

subjected to a static load of 14 kN at adistance of 150 mm from the plane of welds is 70 N/mm 2 .

23

2

755075

12

12 t t I XX

4310937.210 mmt I XX

A = A1+A

2+A

3+A

4=2(l1t+l2t)=2(50t+75t)= 250t mm 2

2 / 56

250

1014 3

mm N t

t

AP

d

M= P x e=14 x 10 3 x 150=2100 x 10 3 N-mm

mml

y 5.372

75

22

Solid rectangular bar of cross section 50 mmx 75 mm is welded to a support by fillet weldsubjected to a static load of 14 kN at a


91/129

subjected to a static load of 14 kN at adistance of 150 mm from the plane of welds is 70 N/mm 2 .

XX

b

I

y M .

3

3

10937.210

5.37102100

t b

2 / 33.373

mm N t b

22

d b

22

5633.373 t t

t

51.377

7051.377

t

t 70

51.377

ht 270

51.3772

h=7.62mm8mm t=h/2=5.65 mm

Ixx=210.937tx103

mm4

Y=37.5mmM =2100x10 3 N-mm

2 / 56

mm N t d


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A = A1+A2+A3

t llt l I XX 3

2

2

21

12

1

22

AP

d

XX b I

y M .

M= P x e

22

d b

22l y

A= 250t mm 2y=75mmh=7mmt =4.95 mm

2 / 80

mm N

t d

2 / 44.444

mm N t b

2 / 587.451

mm N

t

ht 2


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, prove


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8

33 t d t r I XX

2

66.5

d h M

2

d r y

XX I My

8

23t d

d M

2

4

2 hd

M

2

66.5

d h M

Q i


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Quiz

Find CG and MOI of

21

1

l X t l A 11

Y


96/129

321

332211

A A A X A X A X A

X

2

02

13

2

l X

X t l A

t l A

13

22

t lt lt l

lt lt l

lt l

X 121

112

11

20

2

21

21

2 lll

X

t ll

t lt l

X 21

21

21

2

22

X

1 0Y t l A 11Y


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321

332211

A A AY AY AY A

Y

23

22 2

lY

lY

t l A

t l A

13

22

t lt lt l

lt ll

t lt l

Y 121

212

21 20

22lY

t ll

t lll

Y 21

21

22

2

2

X

Y

22lY 21

21

2 lll X


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XIYY1= Moment of inertia of weld 1

Parallel vertical axisThrough part 1 st C.G.+ A1H12

IYY1=3

12

1bd 2

11 H A

IYY1=3

112

1tl

2

11 2

X l

t l

Y

22lY 21

21

2 lll X


99/129


Parallel vertical axisThrough part 2 nd C.G.+ A2H22

IYY2=3

12

1bd 2

22 H A

IYY2=3

212

1t l

IYY2= 22 X t l

22 X t l

Y

22lY 21

21

2 lll X


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Parallel vertical axisThrough part 3 rd C.G.+ A3H32

IYY3=3

12

1bd 2

33 H A

IYY3=3

112

1tl

2

11 2

X l

t l

Y


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X

IYY1= 31121

tl 2

11 2

X l

t l

IYY2= 22 X t lIYY3=

3

112

1tl

2

11 2

X l

t l

IYY = IYY1+ IYY2 + IYY3

2

2

2

11

31 212

12 X t l X

lt llt I YY

Y

22lY 21

21

2 lll X


102/129

XIxx1= Moment of inertia of weld 1

Parallel horizontal axisThrough part 1 st C.G.+ A1H12

Ixx1=3

12

1bd 2

11 H A

Ixx1= 13

12

1lt

2

11 2

l

t l

2

11 2

l

t lIxx1=

Y

22lY 21

21

2 lll X


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XIxx2= Moment of inertia of weld 2

Parallel horizontal axisThrough part 2 nd C.G.+ A2H22

Ixx2=3

12

1bd 2

22 H A

Ixx2=3

212

1tl 21 0t l

Ixx2=3

212

1tl


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Y


105/129

X

Ixx = Ixx1+ Ixx2 + Ixx3

2

11

2

lt lIxx1=

Ixx2=3

212

1tl

2

11 2

l

t lIxx3=

t ll

t l I XX 3

2

2

21 12

1

22


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C.G. Inertia

21

21

2 lll

X

22lY

t ll

t l I XX 3

2

2

21 12

1

22

222

11

31 212

12 X t l X lt llt I YY


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Cotter Joint


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Cotter Joint


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1. Diameter of rod =d2. Diameter of spigot =d 13. Thickness of cotter =t4. Outside diameter of socket =D 15. Distance between slot end and spigot end =a

6. Diameter of socket collar = D 2 7. Thickness of socket collar =c8. Diameter of spigot collar =d 29. Thickness of spigot collar =t 110. Width of Cotter =b11. Failure of cotter under bending


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Diameter of rod =dSpigot

1. Diameter of spigot =d 12. Diameter of spigot collar =d 23. Thickness of spigot collar =t 14. Distance between slot end and spigot end =aSocket1. Outside diameter of socket =D 1

2. Diameter of socket collar =D 2 3. Thickness of socket collar =cCotter1. Width of Cotter =b2. Thickness of cotter =tFailure of cotter under bending


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115/129

Diameter of Spigot d 1and Thickness of cotter t


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t d d

Pt

12

14

t=0.3d

t d P

c1

Outside diameter of socket D 1


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t d Dd D

Pt

112

12

14

Distance from the end of slot to the end of spigota


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ad P

12

diameter of socket collar D 2


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t d DP

c12

thickness of socket collarC


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cd DP

122

Diameter of spigot collard


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d 2

21214 d DPc


122/129

Width of cotter b


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bt P

2

Failure of cotter under bending


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yy

b I x M

462112 d d DP M

12

3tb I yy

2

b x

2

6

tb M

b

2

12

4

2

tbd DP

b


125/129


126/129

FETR, Bardoli


127/129

,

FETR, Bardoli


128/129

,

FETR, Bardoli


129/129

,

design of fasteners

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