Perrhitungan Sambungan - Direktori File UPI

(1)

Contoh Contoh

Pehitungan

Sambungan

(2)

Soal 1

Diketahui: data spt pada gbr di bawah

kg/cm 2

1400 σ =

Hitung sambungan konstruksi tsb.

dengan: a) bout

b) paku keling c) las sudut P₁ = 1500 kg

P₂ = 2500 kg

75.75.10

plat simpul tebal 10 mm

100.100.10

P2 P1

P2

100.100.10

(3)

Penyelesaian

a) Sambungan bout:

2 2

840 1400

x 6 , 0 6

, 0

2100 1400

x 5 , 1 5

, 1

cm kg

σ

cm kg

σ σ_tp

=

= τ

Sambungan batang datar dengan plat simpul

=1cm

=1cm =1cm

100.100.10

s3

s2

s1

P2

1/2P₂

1/2P2

P2

(4)

s₁< s₂+ s₃ Æ s₁= s_min = 1 cm Bout ø ¾” Æ d = 19,7 mm Sambungan dobel

"

14 min

2 min

2 2

14 2

14

2

1 604

, 4137 0

2500 4137

4137 2100

x 1

x 97

, 1 .

5118 840

x ) 97

, 1 ( x 14 , 3 . . 2 .

2

O ambil

kg kg N

n P

kg N

N

kg cm

cm σ

s d N

kg cm

d N

tp

tp tp

gs

/

⇒

≈

=

= π τ

(5)

Sambungan batang tegak dengan plat simpul Sambungan tunggal

75.75.10 P1

1=1cm

s s_2=1cm

P1 P1

s₁= s₂= 1 cm = s_min

Bout ø ¾” Æ d = 19,7 mm

(6)

"

14 min

2 min

2 2

14 2

14

2

1 586

, 2559 0

1500 2559

4137 2100

x 1

x 97

, 1 .

2559 840

x ) 97

, 1 ( x 14 , 3 .

O ambil

kg kg N

n P

kg N

N

kg cm

cm σ

s d N

kg cm

d N

gs tp tp

gs

/

⇒

≈

=

= π τ

100 mm

10 mm a₁ t a₁ a₁ t a₁

a1

a₁ t

w

75 mm

e

75.75.10

100.100.10 100.100.10

Gambar sket sambungan bout

(7)

b) Sambungan paku keling:

2 2

1120 1400

x 8 , 0 8

, 0

2800 1400

x 2 2

cm kg

σ

cm kg

σ σ_tp

=

= τ

=1cm

=1cm =1cm

100.100.10

s3

s2

s1

P2

1/2P₂

1/2P2

P2

(8)

s₁< s₂+ s₃ Æ s₁= s_min = 1 cm pk ø 20 mm (lubang)

Sambungan dobel

mm O

ambil

kg kg N

n P

kg N

N

kg cm

cm σ

s d N

kg cm

d N

tp

tp tp

gs

20 2

1 446

, 5600 0

2500 5600

5600 2800

x 1

x 2

.

7034 0

12 1 x ) 2

( x 14 , 3 . . 2 .

2

min 2 min

2 min

2 2

14 2

14

/

⇒

≈

=

π τ

(9)

Sambungan batang tegak dengan plat simpul Sambungan tunggal

75.75.10 P1

1=1cm

s s_2=1cm

P1 P1

s₁= s₂= 1 cm = s_min pk ø 20 mm (lubang)

(10)

mm O

ambil kg

kg N

n P

kg N

N

kg cm

cm σ

s d N

kg cm

d N

gs tp tp

gs

20 2

1 426 ,

3517 0 1500

3517

5600 2800

x 1

x 2

.

3517 0

12 1 x ) 2 ( x 14 , 3 .

min 2 min

2 min

2 2

14 2

14

/

⇒

≈

=

= π τ

w

10 mm 100 mm

100 mm 100 mm 75 mm

100.100.10

e

t a1

a₁

t

t a₁ a₁ a₁

a1

100.100.10

75.75.10

Gambar sket sambungan pk

(11)

c) Sambungan las sudut

kg/cm2

40 8 σ 6 ,

0 =

τ =

e P2

100.100.10 plat simpul

tebal 10 mm

10 cm

lb la

Pb Pa

B

Dari

Dari tabeltabel profil, profil, utkutk L 100L 100.100.10 didapat.100.10 didapat:: b = 100 mm = 10 cm;

b = 100 mm = 10 cm; d = 10 mm = 1 cm; e = 2,82 cmd = 10 mm = 1 cm; e = 2,82 cm 2 baja2 baja L 100L 100.100.10.100.10 menahanmenahan gayagaya PP₂₂

1 baja1 baja L 100L 100.100.10 menahan.100.10 menahan gayagaya P = PP = P₂₂/2 /2 ÆÆ P = 2500 kg/2 = 1250 kg

P = 2500 kg/2 = 1250 kg

(12)

12 12

d = 1 cm

d = 1 cm ÆÆ a = 0,707d a = 0,707d

a = 0,707x1cm = 0,707 cm a = 0,707x1cm = 0,707 cm

bb––e = 10 cm–e = 10 cm–2,82 cm = 7,18 cm2,82 cm = 7,18 cm

cm kg

b e P P

e P b

P M

a a

B

5 , 10 352

82 , 2 x 1250

.

0 .

.

0

=

−

= Σ

kg kg

kg P

P P

P

P P

P H

b

a b

b a

5 , 897 5

, 352 1250

0 0

=

−

=

−

=

− +

= Σ

(13)

cm 4 atas

las panjang

4 4

715 ,

2

) 707

, 0 x 3 ( 594

, 0 3

594 ,

707 0 ,

0 42 , 0

. 707 ,

0 42

, 0

. 707 ,

0 .

42 , 840 0

5 , 352

840

min 2

2

2 2

ambil 2

=

⇒

=

<

=

+

= +

=

⇒

=

cm l

ambil cm

l cm l

cm cm

a l

l

cm cm l cm

l cm cm

l cm l

a F

cm cm kg

kg F P

cm F kg

P

br a br

a

an br

a an

an an an

gsa gsa a

a gsa

a a

τ

τ τ

τ

(14)

14 14

cm 4

bawah las

panjang

4 4

632 ,

3

) 707

, 0 x 3 ( 511

, 1 3

511 ,

707 1 ,

0

068 ,

1

. 707

, 0 068

, 1

. 707

, 0 .

068 ,

840 1

5 , 897

840

min 2

2

2 2

ambil 2

=

⇒

=

<

=

+

= +

=

⇒

=

cm l

ambil cm

l cm l

cm cm

a l

l

cm cm l cm

l cm cm

l cm l

a F

cm cm kg

kg F P

cm F kg

P

br b br

b

n b br

b n b

n b

n b n

b b

gs

b b

gs

b b

gs b b

τ

τ τ

τ

(15)

Sambungan batang tegak dengan plat simpul

e

Pki Pka B

lka lki

plat simpul tebal 10 mm b

P1

kg/cm2

40 8 σ 6 ,

0 =

τ =

1 baja1 baja L 75L 75.75.10 Æ.75.10 Æ PP₁₁=1500 kg=1500 kg

Dari

Dari tabeltabel profil, profil, utkutk L 75.75.10 L 75.75.10 didapatdidapat:: b = 75 mm = 7,5 cm;

b = 75 mm = 7,5 cm;

d = 10 mm = 1 cm; e = 2,21 cm d = 10 mm = 1 cm; e = 2,21 cm

(16)

16 16

d = 1 cm

d = 1 cm ÆÆ a = 0,707d a = 0,707d

a = 0,707x1cm = 0,707 cm a = 0,707x1cm = 0,707 cm

bb––e = 7,5 cm–e = 7,5 cm–2,21 cm = 5,29 cm2,21 cm = 5,29 cm

cm kg

b

e b

P P

e b

P b

P M

ki ki

B

5 1058 ,

7

29 , 5 x 1500

) .(

0 )

.(

.

0

1 1

=

− =

=

−

= Σ

kg kg

kg P

P P

P

P P

P H

ka

ki ka

442 1058

1500

0 0

1

=

−

=

−

=

− +

= Σ

(17)

cm 4 kanan las

panjang

4 4

865 ,

2

) 707

, 0 x 3 ( 744

, 0 3

744 ,

707 0 ,

0

526 ,

0

. 707 ,

0 526

, 0

. 707 ,

0 .

526 ,

840 0 442

840

min 2

2

2 2

ambil 2

=

⇒

=

<

=

+

= +

=

⇒

=

cm l

ambil cm

l cm l

cm cm

a l

l

cm cm l cm

l cm cm

l cm l

a F

cm cm kg

P kg F

cm F kg

P

br ka br

ka

n ka br

ka n ka

n ka

n ka n

ka ka

gs

ka ka

gs

ka ka

gs ka ka

τ

τ τ

τ

(18)

18 18

cm 4

kiri las

panjang

4 4

903 ,

3

) 707

, 0 x 3 ( 782

, 1 3

782 ,

707 1 ,

0 26 , 1

. 707

, 0 26

, 1

. 707

, 0 .

26 , 840 1

1058

840

min 2

2

2 2

ambil 2

=

⇒

=

<

=

+

= +

=

⇒

=

cm l

ambil cm

l cm l

cm cm

a l

l

cm cm l cm

l cm cm

l cm l

a F

cm cm kg

kg F P

cm F kg

P

br ki br

ki

n ki br

ki n ki

n ki

n ki n

ki ki

gs ki ki gs

ki ki

gs ki ki

τ

τ τ

τ

(19)

Gambar

Gambar sket sket sambungan sambungan las las

100.100.10

100.100.10 75.75.10

75 mm

lb lb

l_ka l_ki

b

l_a l_a

e

b

10 mm

100 mm

100 mm 100 mm

(20)

Soal 2

Diketahui: data spt pada gbr di bawah

kg/cm 2

1400 σ =

A₁ = - 4000 kg A₂ = - 2000 kg D₁ = - 2000 kg T₁ = + 2000 kg

A₁ dan A₂ merupakan batang menrus

T1

??100.100.10

L 75.7 5.10

L 75.75.10

60°

D1 A1

A2

Hitung sambungan konstruksi tsb.

dengan: a) paku keling b) las sudut

(21)

Penyelesaian

a) Sambungan paku keling:

2 2

1120 1400

x 8 , 0 8

, 0

2800 1400

x 2 2

cm kg

σ

cm kg

σ σ_tp

=

= τ

Sambungan batang D₁ dengan plat simpul

D L 75.7

5.10

D1

D1 s

2=1cm 1=1cm

s

= - 2000 kg

s₁= s₂ Æ s₁= s_min = 1 cm pk ø 20 mm (lubang)

Sambungan tunggal

(22)

mm O

ambil kg

kg N

n P

kg N

N

kg cm

cm σ

s d N

kg cm

d N

gs tp tp

gs

20 2

1 567 , 3517 0

2000 3517

5600 2800

x 1

x 2

.

3517 0

12 1 x ) 2 ( x 14 , 3 .

min 2 min

2 min

2 4 2

2 1 14

/

⇒

≈

=

= π τ

Sambungan batang T₁ dengan plat simpul

L 75.75.10

T1 T1

T1

1=1cm

s

s₂

s₁= s₂ Æ s₁= s_min = 1 cm

pk ø 20 mm (lubang) Sambungan tunggal

mm O

ambil kg

kg N

n P

kg N

N kg

N

kg N

gs tp

gs

20 2

1 567 , 3517 0

2000

3517 5600

3517

min 2

min

/

⇒

≈

=

⇒

=

(23)

Sambungan batang A₁/A₂ dengan plat simpul P = A₁ – A₂

= -4000kg+2000 kg

= - 2000 kg

s₁= s₂ Æ s₁= s_min = 1 cm pk ø 20 mm (lubang)

Sambungan dobel

1 2P

3=1cm 2=1cm

s

s s₁^=1cm

A1

+ P A2

P

100.100.10

mm O

ambil kg

kg N

n P

kg N

N

kg cm

cm σ

s d N

kg cm

d N

tp tp tp

gs

20 2

1 357 , 5600 0

2000 5600

5600 2800

x 1 x 2

.

7034 0

12 1 x ) 2 ( x 14 , 3 . . 2 .

2

2 min

2 2

14 2

14

/

⇒

≈

=

= π τ

(24)

Gambar

Gambar sketsket sambungansambungan pakupaku kelingkeling

75 mm

1

a t t

a

1 1

1

a t a a

a t

L 75.7 5.10

L 75.75.10

100.100.10

(25)

b) Sambungan las sudut

kg/cm2

40 8 σ 6 ,

0 =

τ =

Sambungan batang D₁ dengan plat simpul

L 75.7

5.10

D1 P_ka

P_ki _ka

ka ki

b

e b-e

DD₁₁= -= -2000 kg2000 kg L L 7575.75.10 .75.10

Dari

Dari tabeltabel profil, profil, utkutk L 75.75.10 L 75.75.10 didapatdidapat:: b = 75 mm = 7,5 cm;

b = 75 mm = 7,5 cm;

d = 10 mm = 1 cm; e = 2,21 cm d = 10 mm = 1 cm; e = 2,21 cm

(26)

d = 1 cm

d = 1 cm ÆÆ a = 0,707d a = 0,707d a = 0,707x1cm = 0,707 cm a = 0,707x1cm = 0,707 cm

b–b–e = 7,5 cme = 7,5 cm–2,21 cm = 5,29 cm–2,21 cm = 5,29 cm

cm kg

b

e b

P P

e b

P b

P M

ka ka

B

5 1410 ,

7

29 , 5 x 2000

) .(

0 )

.(

.

0

1 1

=

− =

=

−

= Σ

kg kg

kg P

P P

P

P P

P H

ki

ka ki

ki ka

590 1410

2000

0 0

1

=

−

=

−

=

− +

= Σ

(27)

cm cm l cm

l cm cm

l cm l

a F

cm cm kg

P kg F

cm F kg

P

n ka

n ka n

ka ka

gs

ka ka

gs

ka ka

gs ka ka

375 , 707 2

, 0

679 , 1

. 707

, 0 679

, 1

. 707

, 0 .

679 , 840 1

1410

840

2 2

ambil 2

=

⇒

=

τ

τ τ

τ

cm 5 kanan las

panjang

5 496

, 4

) 707

, 0 x 3 ( 375

, 2 3

=

⇒

=

+

= +

=

cm l

ambil cm

l

cm cm

a l

l

br ka br

ka

n ka br

ka

(28)

cm cm l cm

l cm cm

l cm l

a F

cm cm kg

kg F P

cm F kg

P

n ki

n ki n

ki ki

gs

ki ki

gs

ki ki

gs ki ki

993 , 707 0

, 0

702 , 0

. 707

, 0 702

, 0

. 707

, 0 .

702 , 840 0

590

840

2 2

ambil 2

=

⇒

=

τ

τ τ

τ

cm 4 kiri las

panjang

4 4

114 , 3

) 707

, 0 x 3 ( 993

, 0 3

min

=

⇒

=

<

=

+

= +

=

cm l

ambil cm

l cm l

cm cm

a l

l

br ki br

ki

n ki br

ki

Batang

Batang TT₁₁= 2000 kg; = 2000 kg; L L 75.75.10 75.75.10 ÆÆ analog dgnanalog dgn perhit

perhit. . laslas batangbatang DD₁₁; ; didapatdidapat :: Panjang

Panjang laslas kirikiri = 4 cm= 4 cm Panjang

Panjang laslas kakankakan = 5 cm= 5 cm

(29)

2 baja2 baja L L 100.100.10100.100.10 menahanmenahan gayagaya PP

1 1 bajabaja L L 100100.100.10 .100.10 menahanmenahan gayagaya PP₁₁ = = P/2 P/2 ÆÆ PP₁₁ = = ---- 2000 kg/2 = 1000 kg2000 kg/2 = 1000 kg

kg/cm2

40 8 σ 6 ,

0 =

τ =

Sambungan batang A₁ dan A₂ dengan plat simpul

100.100.10

s = 10 mm Pb P_a

A2

7,18cm 2,82cm

A1

B

P = A₁ – A₂

= -4000kg+2000 kg

= - 2000 kg

Dari

Dari tabeltabel profil, profil, utkutk L 100.100.10 L 100.100.10 didapatdidapat:: b = 100 mm = 10 cm;

b = 100 mm = 10 cm; d = 10 mm = 1 cm; e = 2,82 cmd = 10 mm = 1 cm; e = 2,82 cm

(30)

d = 1 cm

d = 1 cm ÆÆ a = 0,707d a = 0,707d a = 0,707x1cm = 0,707 cm a = 0,707x1cm = 0,707 cm

b–b–e = 10 cme = 10 cm–2,82 cm = 7,18 cm–2,82 cm = 7,18 cm

cm kg

b e b

P P

e b

P b

P M

a a

B

10 718

18 , 7 x 1000

) .(

0 )

.(

.

0

1

=

− =

=

−

= Σ

kg kg

kg P

P P

P

P P

P H

b

a b

b a

282 718

1000

0 0

1

=

−

=

−

=

− +

= Σ

(31)

secukupnya dipasang

cm 4 atas

las panjang

4 4

33 , 3

) 707

, 0 x 3 ( 209

, 1 3

209 ,

707 1 ,

0

855 ,

0

. 707 ,

0 855

, 0

. 707 ,

0 .

855 ,

840 0 718

840

min 2

2

2 2

ambil 2

⇒

=

⇒

=

<

=

+

= +

=

⇒

=

cm l

ambil cm

l cm l

cm cm

a l

l

cm cm l cm

l cm cm

l cm l

a F

cm cm kg

kg F P

cm F kg

P

br a br

a

an br

a an

an an an

gsa gsa a

a gsa

a a

τ

τ τ

τ

(32)

secukupnya dipasang

cm 4 bawah las

panjang

4 4

596 ,

2

) 707

, 0 x 3 ( 475

, 0 3

475 ,

707 0 ,

0

336 ,

0

. 707 ,

0 336

, 0

. 707 ,

0 .

336 ,

840 0 282

840

min 2

2

2 2

ambil 2

⇒

=

⇒

=

<

=

+

= +

=

⇒

=

cm l

ambil cm

l cm l

cm cm

a l

l

cm cm l cm

l cm cm

l cm l

a F

cm cm kg

kg F P

cm F kg

P

br b br

b

n b br

b n b

n b

n b n

b b

gs b b gs

b b

gs b b

τ

τ τ

τ

(33)

Gambar

Gambar sketsket sambungansambungan laslas sudutsudut

2,21 cm

5,29 cm

75 mm

l L 75.7

5.10

L 75.75.10

100.100.10

ka

lki ki

ka

2,82 cm

7,18 cm

2,21 cm