kolom melintang babay

7/25/2019 Kolom MELINTANG Babay

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PERENCANAAN KOLOM

KOLOM LANTAI 2

KOLOM 40/40

Data Kolom :

b = 40 cm 400 mm

h = 60 cm 0.6 m

d' = 5 cmd = 55 cm 550 mm

L = 500 cm

P = 44905 kg

MA = 177! kgm

M" = 17!# kgm

Mb $b%ba& mat( = 1676 kgm

Data "alok :

b = 0 cm

h = 60 cm

d' = 5 cm

d = 55 cm

L = 600 cm

MD = 175! kgm

ML = 6#19 kgm

M) = 1616 kgm

*'c = 0 M+a 00 kg/cm!

*, = 00 MPa 000 kg/cm!

-c =

=

= !574 MPa !574!9.60!0

MENGHITUNG KEKAKUAN KOLOM

g kolom = 1 b h

1!

= 1 40 !16000

1!

= 7!0000 cm4

= Mom%& b%ba& mat %&ca&a

Mom%& total %&ca&a

= 654

5049

= 654

5049

= 0.7!4 1

- kolom = $-c 2 g(/!.5 = $!1019 2 !1(/!.5

1 3 0101

= 74197!5#4

1.7!4

βD

1 3 βD

4700 (

(

4700!0



= 4011774499 kg/cm!

MENGHITUNG KEKAKUAN BALOK

g balok = 1 b h

1!

1 0 !16000

1!540000cm4


Mom%& total %&ca&a

= 1.4 2 1!71

1.! 2 1!71 3 1.6 2 !5

= !59

05#0

= 0.764 1

- balok = $-c 2 g(/!.5 = $!1019 2 160000(/!.5

1 3 0#70

= 556047940#

1.764

= 15!499#16 kg/cm!

- = 4011774499 = 66171961

Lk 650

- = 15!499#16 = 5!540666

Lb 600

= = 171461555

5!540666

= .!6

= 0

"%daaka& 8omogam *akto +a&a&g %*%kt* d+%ol%h k = 095

%k k%la&g&ga& kolom

k = 1.65

Lk = 6!0

kL) = k . Lk

= 145 . 50

= 10! cm

= 0 . ;

= 0 . 40

= 1# cm

βD

1 3 βD

PERHITUNGAN

Ψata ∑$-/Lk(

∑$-/Lb(

Ψba<ah



6.666666667

= 0.00!5909091

@)la&ga& @%ka& )dah l%l%h

*' = *, = 000 kg/cm!

ab = 0#5 2 b

= 0#5 2 !5= 1.1666666667 cm

P&b = 0#5 2 *'c 2 ab 2 b

= 0#5 2 !00 2 !1!5 2 50

= 17900 kg

= 065 2 144500

= !0665 kg !4!91

%k P%&am+a&g Kolom

% = M)

P)

= 144!77.#75714

!4!91

= 59.79 cm

%' = % 3 $d > h/!(

= 59.79 3 $5>40/!(

= #4.79 cm

1>%'/d = 1 > 15!1!/5

= >0.54

1>d'/d = 1>$5/5(

= 0.909

m = *,

0#5 2 *'c

= 000

!55

= 11.765 cm

= A

b 2 d

= !!.9

40 2 5

= 0.010

P& =

=

ø. P&b

ρ = ρ'

0#5 C× b $1 (3$$1 (!3! $1 ( (E× − / − / ×× − /

0#5C!00C40C5$F0494(3$$F0494(!3!C1411#C



= 150!64 kg

= 065 2 1!1#10

= 97671 kg !4!91

G%&gka&g Lat%al

Dg)&aka& %&gka&gø

Haak G%&gka&g

1 =

= 4# 2 #

= #4 mm

! =

= 16 2 19

= 04 mm

= l%ba kolom

= 400 mm

Dg)&aka& aak %&gka&g 00 mm

TULANGAN GESER

I) = !4!91

Ic =

= !00#! kg

I& = !4!91 = 404#5

0.6

ø. P&

Haak G%&gka&g d%&ga& ø = 8 mm

4# 2 ø %&gka&g

16 2 ø kolom

@)la&ga& g%% >> ø#>00 mm

1/6 (

( .b.d



Data Kolom :

b = 40 cm

h = 60 cm

d' = 5 cmd = 55 cm

L = 400 cm

P = !4!91 kg

MA = 1096 kgm

M" = 1559 kgm

Mb $b%ba& = 045 kgm

Data "alok :

b = 0 cm

h = 60 cm

d' = 5 cm

d = 55 cm

L = 00 cm

MD = 790 kgm

ML = 1#6 kgm

M) = #4#5 kgm

*'c = 0 M+a

*, = 00 MPa

-c =

=

= !574 MPa


g kolom = 1 b

1!

= 1 40

1!

= 7!0000 cm4

= Mom%& b%ba&

Mom%& tota

= 159!0

!097

= 159!0

!097

= 0.760

- kolom = $-c 2 g(/!.5 =

=

βD

1 3 βD

4700$ (

4700!0



=


g balok = 1 b

1!

1 0

1!540000cm4

= Mom%& b%ba&

Mom%& tota

= 1.4 2 1!71

1.! 2 1!71 3 1.6 2 !5

= 506

6766

= 0.7#4

- balok = $-c 2 g(/!.5 =

=

=

- = 4!115#7707 =

Lk 400

- = 11640746 =

Lb 00

βD

1 3 βD



KolomLA8JG8J

!6#75. 0.09!055 0.907945

1.101## 1.111111

= !!.9 cm!

= !!.9 cm!

0.0015



0.0015

kg (OK)

011$0#



kg (OKE)

= #

!00#!kg



400 mm

0.6m

550 mm

00kg/cm!

000 kg/cm!

!574!9.60!0!74!#

h

!16000

at %&ca&a

l %&ca&a

1

$!1019 2 !1(/!.5

1 3 0101

74197!5#.#99

1.760



4!115#7707 kg/cm!

h

!16000

at %&ca&a

l %&ca&a

1

$!1019 2 160000(/!.5

1 3 0#70

556047940.79!45

1.7#4

11640746 kg/cm!

105!#9594

10##01!



PERENCANAAN KOLOM

KOLOM LANTAI 2

KOLOM 40/40

Data Kolom :

b = 40 cm 400 mm

h = 60 cm 0.6 m

d' = 5 cmd = 55 cm 550 mm

L = 400 cm

P = 516 kg

MA = 15#00 kgm

M" = !9!4 kgm

Mb $b%ba& = 117! kgm

Data "alok :

b = 0 cm

h = 50 cm

d' = 5 cm

d = 45 cm

L = 1000 cm

MD = 175! kgm

ML = 6#19 kgm

M) = 1616 kgm

*'c = 0 M+a 00 kg/cm!

*, = 00 MPa 000 kg/cm!

-c =

=

= !574 MPa !574!9.60!0


g kolom = 1 b h

1!

= 1 40 !16000

1!

= 7!0000 cm4


Mom%& total %&ca&a

= ##7

5049

= ##7

5049

= 0.770 1

- kolom = $-c 2 g(/!.5 = $!1019 2 !1(/!.5

1 3 0101

= 74197!5#4

1.770

βD

1 3 βD

4700$ (

4700!0



= 41#90!7!!1 kg/cm!


g balok = 1 b h

1!

1 0 1!5000

1!1!500cm4


Mom%& total %&ca&a

= 1.4 2 1!71

1.! 2 1!71 3 1.6 2 !5

= !59

05#0

= 0.764 1

- balok = $-c 2 g(/!.5 = $!1019 2 160000(/!.5

1 3 0#70

= !17#700!5

1.764

= 1#!4!#691 kg/cm!

- = 41#90!7!!1 = 6444657

Lk 650

- = 1#!4!#691 = 1#!4!#7

Lb 1000

= = 16976167

1#!4!#7

= 9.0

= 0

"%daaka& 8omogam *akto +a&a&g %*%kt* d+%ol%h k = 095

%k k%la&g&ga& kolom

k = 1.65

Lk = 6!5

kL) = k . Lk

= 145 . 50

= 101.!5 cm

= 0 . ;

= 0 . 40

= 1# cm

βD

1 3 βD

PERHITUNGAN

Ψata ∑$-/Lk(

∑$-/Lb(

Ψba<ah



kL) = 101.!5

1#

= 57.!916666667 !! KolomLA8

PERBESARAN MOMEN

Pc =$kL)(!

=

$!.5(!

= 41770!11!.6!1

106476.56!5

= #9076 kg !5!#99.5

= 1 1.!56541

1>P)/065 2 Pc

= 1

1>!!57!/065 2 14551#60!

= 1.!57 1 OK-

Mc =

= 1.1 2 651

= 19#5.4007!#79!

PENULANGAN KOLOM

?ao @)la&ga& : At = 4 4

A = A' = 1

A = 0.01 2 b 2 d

A = 0.01 2 40 2 5

= !!

Dcoba

@)la&ga& @ak = 5D>19 A =

@)la&ga& @%ka& = 5D>19 A =

%k K%adaa& mba&g

= *, = 000 = 0.0015

- !B106

b = = 000 2 50

%c 3 %, 000 3 0001!

= 0.165

0.0045

= 6.666666667 cm

= = 000 $!5 > 5(

b !5

= 0.095

π!. -k

π!. 16!#74!##

δ

δ . M)

ε,

εc 2 d

ε' εc $b > d'(



6.666666667

= 0.00!5909091 0.0015

@)la&ga& @%ka& )dah l%l%h

*' = *, = 000 kg/cm!

ab = 0#5 2 b

= 0#5 2 !5= 1.1666666667 cm

P&b = 0#5 2 *'c 2 ab 2 b

= 0#5 2 !00 2 !1!5 2 50

= 17900 kg

= 065 2 144500

= !0665 kg 516 kg

%k P%&am+a&g Kolom

% = M)

P)

= 19#54.007!#79!

516

= #.451 cm

%' = % 3 $d > h/!(

= #.451 3 $5>40/!(

= 6.451 cm

1>%'/d = 1 > 15!1!/5

= >0.154

1>d'/d = 1>$5/5(

= 0.909

m = *,

0#5 2 *'c

= 000

!55

= 11.765 cm

= A

b 2 d

= !!.9

40 2 5

= 0.010

P& =

=

ø. P&b

ρ = ρ'

0#5× C b× $1 − / (3$$1 − / (!3!××$1 − / ( (E

0#5C!00C40C5$F0494(3$$F0494(!3!C1411#C0011$0#



= 196565 kg

= 065 2 1!1#10

= 1!7767 kg 516 kg

G%&gka&g Lat%al

Dg)&aka& %&gka&gø =

Haak G%&gka&g

1 =

= 4# 2 #

= #4 mm

! =

= 16 2 19

= 04 mm

= l%ba kolom

= 400 mm

Dg)&aka& aak %&gka&g 00 mm

TULANGAN GESER

I) = 516

Ic =

= !00#! kg

I& = 516 = #6055 !00#!

0.6

ø. P&

Haak G%&gka&g d%&ga& ø = 8 mm

4# 2 ø %&gka&g

16 2 ø kolom

@)la&ga& g%% >> ø#>00 mm

1/6$ (.b.d



Data Kolom :

b = 40 cm 400

h = 60 cm 0.6

d' = 5 cmd = 55 cm 550

L = 400 cm

P = 516 kg

MA = !9!4 kgm

M" = 15#00 kgm

Mb $b%ba& = 117! kgm

Data "alok :

b = 0 cm

h = 50 cm

d' = 5 cm

d = 45 cm

L = 400 cm

MD = 790 kgm

ML = 1#6 kgm

M) = #4#5 kgm

*'c = 0 M+a 00

*, = 00 MPa 000

-c =

=

= !574 MPa !574!9.60!0!74!#


g kolom = 1 b h

1!

= 1 40 !16000

1!

= 7!0000 cm4


Mom%& total %&ca&a

= 159!0

!097

= 159!0

!097

= 0.760 1

- kolom = $-c 2 g(/!.5 = $!1019 2 !1

1 3 0101

= 74197!5#.#99

1.760

βD

1 3 βD

4700$ (

4700!0



= 4!115#7707


g balok = 1 b h

1!

1 0 1!5000

1!1!500cm4


Mom%& total %&ca&a

= 1.4 2 1!71

1.! 2 1!71 3 1.6 2 !5

= 506

6766

= 0.7#4 1

- balok = $-c 2 g(/!.5 = $!1019 2 160000

1 3 0#70

= !17#700!5.4!#5

1.7#4

= 1#047490

- = 4!115#7707 = 105!#9594

Lk 400

- = 1#047490 = 450#6#6

Lb 400

βD

1 3 βD



G8J

0.!04164 0.795#6

1.111111

!!.9cm!

!!.9cm!



(OK)



(OKE)

#

kg



mm

m

mm

kg/cm!

kg/cm!

(/!.5



kg/cm!

(/!.5

kg/cm!

kolom melintang babay

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