Batang Tarik(1)

7/23/2019 Batang Tarik(1)

1/30

PERENCANAAN BATANG TARIK

Jumlah Baut = 4 buah

Diameter Baut = 0.250 in

Luas Bruto = 2.48Tebal Plat = 0.5 in

= 15 kips

= 35 kips

= 58 ksi

= 3 ksi

= kips

Limit State of Yielding

Gross Area

= 2.4

Nominal Strengt!

= 3 ! 2.48

= ".2 #i$s

%esign Strengt!

= 0." ! 8".28

= &.'(2 #i$s

)ra*t+re Control

)ra*t+re of T!e Net Se*tion

= 0.3#5

= 2.48 $ 0.3#5

= 2.,&(

=

= 2.,&(

Nominal Strengt!

in2

LL

LD

%u

%&

Pu

'( in

2

Pn

tP

n

'holes in

2

'n

in2

'e

'n

in2


2/30

= !

= 58 ! 2.105

= ,22.&" #i$s

%esign Strengt!

= 0.#5 ! 122.0"

= ",.(-( #i$s

C!e*# )ra*t+re and Yielding Points

Desi(n )tren(th = &.'(2 #i$s *

Pn

%u

'e

tP

n


3/30

/01LA BA0T TERKENA GAYA = 2

= 1.4 ! 35

= 1.2 ! 35

= -- #i$s

+ =, '-

Pu

Pu

Pu

Pn


4/30

Yielding Points


5/30

PE1ASANGAN BA0T

Baut 1 Baut 2

)b ! )b & )b ! )b &

#.5 $2.5 12.5 $2.5

Baut 3 Baut 4

)b ! )b & )b ! )b &

#.5 $#.5 12.5 $#.5

Be3an 0ltimate

= 4" kips

/ 1. ! 15 = kips

Ce# Keamanan

'

0 5 10 15

-30

-25

-20

-15

-10

-5

0

5

10

15

20

PEMASANGAN


6/30


7/30

BIR0 1E

)b ! )b & )b !

0 20

1 15 1"2 14 18

3 13 1#

4 12 1

5 11 15

10 14

# " 13

8 8 12

" # 11

10 10

11 5 "

12 4 8

13 3 #

14 2

15 1 (

15 0 5

15 $1 5

15 $2 5

15 $3 5

15 $4 5

15 $5 5

15 $ 5

15 $# 5

15 $8 5

15 $" 5

14 $10 (

13 $11 5

12 $12 11 $13 #

10 $14 8

" $15 "

8 $1 10

# $1# 11

$18 12

5 $1" 13

4 $20 14

3 $21 15

2 $22 1

20 25

AUT

Gusset

Plat

Baut 1

Baut 2

Baut 3

Baut 4


8/30

1 $23 1#

0 $24 18

$25 1"

20


9/30

A

)b &

0

00

0

0

0

0

0

0

0

0

0

0

0

0

0

0

$1

$2

$3

$4

$5

$

$#

$8

$"

$10

$10

$10$10

$10

$10

$10

$10

$10

$10

$10

$10

$10


10/30

$10

$10

$10

$10


11/30

KR%INAT BA0T

. B'T in. in.

1 0 0

2 3 0

3 0 3

4 3 3

5 0

3

# 0 "

Jumlah Baut #

x ,.2(,42(

'.(,42(,

PE1ASANGAN BA0T

B'67) 1 2 buah

B'67) 2 2 buah

B'67) 3 2 buah

B'67) 4 1 buah

J-L' # buah

= 4 B'67) 9767

= 3 B'67) 9''

/ARAK BA0T TERA%AP S01B0 0TA1A

= 0 in.

= 3 in.

= 3 in.

= 3 in.

J-L' B'67)B'T

J-L' B'67)B'T

&1

&2

&3

&4


12/30

= 0 in.

= 3 in.

= 0 in.

= 0 in.

= 2 ! 0

= 3.85#14285#1 in.

! = 4 ! 0

= 1.285#14285# in.

N. BA0T 5 6in.7 Y 6in.7 8 6in.7 9 6in.7

1 0 0 1.28 3.85#14285#1

2 3 0 1.#14 3.85#14285#1

3 0 3 1.28 0.85#14285#1

4 3 3 1.#14 0.85#14285#1

5 0 1.28 2.14285#142"

3 1.#14 2.14285#142"

# 0 " 1.28 5.14285#142"

EKSENTRISITAS

e = 3 / 5

= .#14285#143 in.

PERIT0NGAN 11EN

- = P ! e

= 53 ! .#14285#143= 355.85#1428 in.$kips

!1

!2

!3

!4

!2

!2/&2


13/30

PERIT0NGAN TEGANGAN BA0T AKIBAT BEBAN

=53

= #.5#14285#14#

= 0.0000 = 0#

PERIT0NGAN TEGANGAN BA0T A

PERIT0NGAN TEGANGAN TI

N. BA0T

1 1.808035#1 5.502#85#1 1.808035#14

2 1.808035#1 #.413"04#2 1.808035#14

3 3.#08452381 5.502#85#1 3.#08452381

4 3.#08452381 #.413"04#2 3.#084523815 ".2#1130"52 5.502#85#1 ".2#1130"52

".2#1130"52 #.413"04#2 ".2#1130"52

# 22.2410#142" 5.502#85#1 22.2410#1428

CECK )

Diameter Baut = 0.8#5 in.

Tebal Pro:il = 0.455 in.

= 58

= 2 in.

s = 3 in.

3; = 2.25 in.

1.5; = 1.3125 in.


14/30


15/30

BEBAN

P = 53 kips

)u;ut 9emirin(an = "0

Psin = 53 kips

P*os = 3.244340$015 * 0.0000

= 5 in.

Jarak Baut ke Tepi = 2 in.

o

Posisi Beban Terha;ap BautTer;ekat

0 3

0

3

6

9

12

PEMASANGAN BAUT


16/30

PERIT0NGAN TITIK BERAT

/ 2 ! 3 / 2

#

/ 3 ! 3#

1.5301 14.8##551

2."38## 14.8##551

1.5301 0.#34"4

2."38## 0.#34"4

1.5301 4.5"183#

2."38## 4.5"183#

1.5301 2.448"80

15.4285#1 .85#143

2.2(,4

$ 1.285#14285#

82 6in27 92 6in27

&2


17/30

kips

kips

IBAT 11EN

P BA0T

P 6#i$s.7

13.131"428 21.22"4#1801

14."8511"04# 22.4232"1#8#

13.131"428 13.448581052

14."8511"04# 15.43#"02"1113.131"428 1.0#2324"3

14."8511"04# 1#.1"1140

13.131"428 25.82840"0

P 6Kritis7 2(.2 BA0T N

R BEARING STRENGT

= 48

+ 3 =, 9

P96#i$s.7

%A


18/30

+ 2 =, 9

kips

! 55.41"

#i$s

kips , 25.828408"5" kips =, K

R SEAR STRENGT

48 ! 0.01525

+ 25.828408"5" kips =, NT K


19/30

kips

6

Baut 1

Baut 2Baut 3

Baut 4

Baut 5

Baut 6

Baut 7


20/30

! / 1 ! "


21/30


22/30


23/30


24/30


25/30


26/30


27/30

S01B0 0TA1

Ba+t 8 9 8 6Bar+7 9 6

1 0 0 $1.28 $

2 3 0 1.#14 $

3 0 3 $1.28 $

4 3 3 1.#14 $

5 0 $1.28 2

3 1.#14 2

# 0 " $1.28 5

x = 1.28 in. e

= 3.85# in.

= 1.5#1 in. $1.5#1

x = 1.285#14285# in.

= 3.85#14285#1 in.

P =6r

/ e

=1#10.285

1.5#10 / .#14285#14

= 2&-.42'- kips

%& = 6& $ P

= 20.424 $ 20.423

= 0.000 * 0.000

CECK 1A5I101 )ACT

ltimate )tren(ht = #4 kips

-a!imum )tren(th = 21.525 kips =,

-a!imum %a?tore; Loa; = 20.424 !21.525

#4

= 0.410350" kips


28/30

BERA%A %I TITIK BERAT SA1B0NGAN

ar+7 r ; R Rr

.85# 3.88 0.255 #0.##4 2#3.#31

.85# 5.0# 0.334 #2.553 3#.5"8

.85# 0."03 0.00 4#.4" 43.04

.85# 3.3"5 0.224 ".53 23.188

.143 2.12 0.143 3.31 13#.555

.143 3."22 0.25" #0.8"1 2#8.01

.143 5.151 0.340 #2.31 3#4.10#


29/30

R9

5.221

4#.045

15.050

#.310

8.3"8

5".3##

4.023

20.424

0 3

0

3

6

9

12

PEMASANGAN BAU

3

Baut 1

Baut 2

Baut 3

Baut 4

Baut 5

Baut 6

Baut 7


30/30

6

Baut 1

Baut 2

Baut 3

Baut 4

Baut 5

Baut 6

Baut 7

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Batang Tarik(1)