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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
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= 4011774499 kg/cm!
MENGHITUNG KEKAKUAN BALOK
g balok = 1 b h
1!
1 0 !16000
1!540000cm4
= Mom%& b%ba& mat %&ca&a
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
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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
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= 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
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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
MENGHITUNG KEKAKUAN KOLOM
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
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=
MENGHITUNG KEKAKUAN BALOK
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
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KolomLA8JG8J
!6#75. 0.09!055 0.907945
1.101## 1.111111
= !!.9 cm!
= !!.9 cm!
0.0015
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0.0015
kg (OK)
011$0#
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kg (OKE)
= #
!00#!kg
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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
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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!
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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 = 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
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
= ##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
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= 41#90!7!!1 kg/cm!
MENGHITUNG KEKAKUAN BALOK
g balok = 1 b h
1!
1 0 1!5000
1!1!500cm4
= Mom%& b%ba& mat %&ca&a
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(
∑$-/Lb(
Ψba<ah
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kL) = 101.!5
1#
= 57.!916666667 !! KolomLA8
PERBESARAN MOMEN
Pc =$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'(
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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$0#
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= 196565 kg
= 065 2 1!1#10
= 1!7767 kg 516 kg
G%&gka&g Lat%al
Dg)&aka& %&gka&gø =
Haak G%&gka&g
1 =
= 4# 2 #
= #4 mm
! =
= 16 2 19
= 04 mm
= l%ba kolom
= 400 mm
Dg)&aka& aak %&gka&g 00 mm
TULANGAN GESER
I) = 516
Ic =
= !00#! kg
I& = 516 = #6055 !00#!
0.6
ø. P&
Haak G%&gka&g d%&ga& ø = 8 mm
4# 2 ø %&gka&g
16 2 ø kolom
@)la&ga& g%% >> ø#>00 mm
1/6$ (.b.d
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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!#
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
= 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
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= 4!115#7707
MENGHITUNG KEKAKUAN BALOK
g balok = 1 b h
1!
1 0 1!5000
1!1!500cm4
= Mom%& b%ba& mat %&ca&a
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
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G8J
0.!04164 0.795#6
1.111111
!!.9cm!
!!.9cm!
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(OK)
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(OKE)
#
kg
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mm
m
mm
kg/cm!
kg/cm!
(/!.5
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kg/cm!
(/!.5
kg/cm!
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