Sambungan Paku Keling dan Baut

Dedison Gasni

Riveted, bolted and welded connections

Ordinary bolt (or rivet)

INTRODUCTION

3

Friction type connections High strength bolt

Bearing type connections

Lab joints

main plate

TYPE OF RIVETED AND BOLTED JOINT:

DEFENITION

4

Butt jointscover plate

main plate

Repeating

group

Boiler joints (pressure joints)Repeating

group

Inner outer

in calculation: bolt holeD D

5

Inner

cover

plate

outer

cover

plate

1 (or 1.5 mm)16hole bolt

D D

= +

in calculation:

Structural joints

Multiple rows

6

long pitch

back pitch

short

pitch

intermediate

pitch

7

diagonal

pitch

Failure of bolted or riveted joints

Efficiency:

compares the strength of the joint with that of solids plates

9

P0P0

Efficiency = P1/P0P1P1

(1) Shearing (at connector)

10

(2) Tearing (of plate) t

(3) Bearing (of plate) b

11

(4) Unlikely to occur

if the distance from the edge of

the plate to the connector is

large enough (1.75-2 times the

diameter of the connector)

3.25

D

12

in calculation: 11/16 in.rivet holeD D =

2

2

2( )4

2 (11/16) 88004

6533.5 lb.

dpi

pi

=

=

=

(1) Rivet capacity (shearing of rivet and bearing failure)

s sP A=Shearing:

(1) Rivet capacity (shearing of rivet and bearing failure) (cont) b = 19 ksi

2( )7 112( ) 19000

16 1611429.7 lb.

ttd =

=

=

(2) Tearing capacity (tension failure of plate) t = 11 ksi

bearing:b b bP A =

13

3.25

tP

11 7(3.25 ) 1100016 16

12332 lb.

=

=

tearing:

t t tP A=

Hence the strength of

the repeating section is

6533.5 lb per 3.25 in.

3.25

6533.5 0.418 41.8 %3.25 (7 /16) 11000= = =

6533.5 lb. per 3.25P =Internal pressure

14

2

4DF ppi=

6533.5 lb. per 3.25sP =

2 6533.56 lb4 3.25D p Dpi pi=

2(15 12) 6533.56(15 12)4 3.25

ppi pi =

134 psi.p =

(1) Preliminary calculation:

load that can transmitted by one rivet (shear or bearing)

(2) Possible method of failure

15

16

300 60 MPa5650 130 MPa5

400 80 MPa5

b

t

= =

= =

= =

d = 20.5 mm

17

18

P-19.8 kN

19.8 kN

Row 1 (has 1-rivet)

19.8 kN

single shear

main plate

cover plate

P-37.3 kN

37.3 kN

Row 1 (has 1-rivet)main plate

cover plate

crash in the main plate

37.3 kN

37.3 kN

P

19

P-26.7 kN

26.7 kN

Hence strength of row-1 is

19.8 kN

main plate

cover plate

crash in the cover plate

26.7 kN

26.7 kN

P

Row 2 (has 2-rivets)

double shear

2 x 19.8 kN39.6 kN

Shear strength at row-2 is

20

19.8 kN

59.4 kN

19.8 kN

P-99 kN

2 39.6 79.2 kN =2 x 19.8 kN

2 x 19.8 kN

2 x 19.8 kN

One possible

Row 2 (has 2-rivets)

crushing at main plate 37.3 kN One possible bearing strength at row-2 is

2 37.3 74.6 kN =

19.8 kN

P-94.4 kN2 x 37.3 kN

37.3 kN

57.1 kN

37.3 kN

21

crushing at cover plate 53.4 kN

4 26.7 106.8 kN =

19.8 kN

P-126.6 kN4 x 26.7 kN

73.2 kN

2 x 26.7 kN

One possible bearing strength at row-2 is2 x 26.7 kN

Hence

strength of

row 2 is

74.6 kN

Row 3 (has 2-rivets)

crushing at main plate crushing at cover plate

2 x 19.8 kN

2 x 19.8 kN

double shear

22

Hence

strength of

row 3 is also

74.6 kN2 x 37.3 kN37.3 kN

37.3 kN

crushing at main plate

4 x 26.7 kN

2 x 26.7 kN

2 x 26.7 kN

crushing at cover plate

Total rivet capacity

19.8 kN74.6 kN 74.6 kN

23

24

Tearing at row 1 of main plate Tearing at row 2

of main plate

19.8 kN 74.6 kN 74.6 kN

Row 1 Row 2 Row 3 80 MPa14 mm180 mm20.5 mm

t

t

pd

=

=

=

=

main plate1P 180 mmp =

t 1 ( ) tP p d t= (180 20.5) 14 80= 178.6 kN=

Tearing capacity of main plate

25

main plate

19.8 kN

2P80 MPat =

main plate

19.8 kN

37.3 kN

37.3 kN3P

80 MPat =

2 ( 2 ) 19.8 kNtP p d t= +175.5 kN=

3 ( 2 ) 19.8 74.6 kNtP p d t= + +250.1 kN=

Tearing capacity of cover plate

P

Row 1 Row 3Row 2

26

P

Row 1 Row 3Row 2

19.8 kN

PRow 1 Row 3Row 2

19.8 kN

outer cover plate

19.8 kN

19.8 kN

19.8 kN

9.9 kN

9.9 kN

79.2 kN

27

79.2 111.2190.4 kN

cP = +=

9.9 kN

80 10 (180 2 20.5)111.2 kN

=

=

80 MPat =

inner cover plate

180 mmp =

10 mmt =20.5 mmd =

( 2 )tP t p d =

Rivets failure: 169.0 kN

Tearing of main plate: 175.5 kN

Tearing of cover plate: 190.4 kN

28

Assume each rivet carries a load proportional to its resisting shear area

1/212

Totally 9 shear areas:

12-5 STRESSES IN BEARING-TYPE CONNECTIONS

29

1/21

2

2 the average load transmitted by 1 shear

area is 36 / 9 4 kipssP = =

the average shearing stress is

Row1Row2 Row3

3/ 8 in

22

4=5.80 ksi 15/ 4 ( )

4 16

sPd

pipi

= =

15 in

16d =

30

4 kips3/ 8 in

1/ 2 in 4 kips

4 kips

3/ 8 in

3/ 8 in

1/ 2 in

the maximum average bearing stress is

8=17.07 ksi 1 15( )

2 16

bb

Ptd

= =

15 116 2

36 10.2 ksi(8 )t = =

36 4

=

main plate (Row 2)

36 kips 2 kips

main plate (Row 1)

36 kips

31

15 116 22(8 )

10.45 ksi

t

=

=

2 kips

main plate (Row 3)

36 kips 8 kips

8 kips

2 kips

2 kips

15 116 22

36 20(8 )5.23 ksi

t

=

=

Upper cover plate

16 kips4 kips

4 kips

2 kips

2 kips

4 kips

Lower cover plate

20 kips4 kips

4 kips

2 kips 2 kips4 kips

32

4 kips2 kips 2 kips

Lower cover plate

20 kips4 kips

4 kips

2 kips

2 kips

2 kips

2 kips

4 kips

33

34

Lab joints () Butt joints ()

shearing capacity of 1-rivet

2( / 4)sP d pi=295( 19 / 4) Npi=

26.935 kN=bearing capacity of 1-rivet

1.5 mmd d= + 20.5 mm=

35

( )b bP td=220(8 19) N= 33.44 kN=upper plate

1P250 mmL =

t

9 26.935 kN242.42 kN

rivetP = =

Total rivets capacity

1.5 mmhole rivetd d= + 20.5 mm=

1 ( )t holeP t L d= 140 8 (250 20.5) N= 257 kN=

Row1

1.5 mmhole rivetd d= +

2 26.94 kN ( 2 )t holeP t L d = 140 8(250 2 20.5) N=

upper plate

2P250 mmL =

tRow2

Hence the joint capacity

is 242.42 kN

Efficiency =

242.42 kN 86.6%140 8 250 N

=

36

140 8(250 2 20.5) N=

2 261 kNP =

250 mmL =26.94 kN

3 3 26.94 kN ( 3 )t holeP t L d = 140 8(250 3 20.5) N=

Row3

t

3P 26.94 kN26.94 kN

26.94 kN3 291.9 kNP =

Beban Eksentrik

Example 1

Example 2

40

41

42

43

44

(1.5,4)

(1.5,0)

(4.5,4)

(-4.5,0)

(-4.5,4)

(4.5,0)(-1.5,0)

(-1.5,4)

28.8 2.4 kips12dx

P = =

38.4 3.2 kips12dy

P = =

45

(1.5,-4)(-4.5,-4)

(4.5,0)

(4.5,-4)(-1.5,-4)

38.4 4.5 kips.inT = 172.8 kips.in=

Twisting moment

46

47

48

49

Type of Joints

Lap Joint (single Joint) But Joint

Example 1

Example 3