Perencanaan kolom

Kolom Lantai 1 (salah satu ujung terjepit)

Data perencanaan kolom 18 :

Dimensi kolom : 50/50

f’c : 22,5 MPa

fy : 240 MPa

Berdasarkan perhitungan dari Staad Pro didapatkan hasil sbb :

Momen max Ma = 30497,84 kgm (kolom 18)

Mb = 30808,11 kgm (kolom 18)

Axial max Pu = 11146,54 kg ` (kolom 18)

STUDIO PERANCANGAN I

Kekakuan kolom 18

b = 50 cm

h = 50 cm

L = 550 cm

MD = 4967,55 kgm

ML = 926,24 kgm

Ec = 4700 √ f ' c = 4700 √22 ,5 = 22294 ,0575 MPa = 222940 ,575 kg /cm2

Ik=

112

b h3 = 112

x 50 x 503 = 520833 ,333 cm4

Ec x Ik = 222940,575 kg/cm2 x 520833,333 cm4 = 1,161 x1011 kg/cm2

βd=1,20D1,20D+1,60L

=1,20 x 4967,551,20 x 4967,55+1,60 x 926,24

=0 ,801

E.Ik =

Ec. I k2,50

1 + βd =

1,161 x 1011

2,501 + 0,801

=2 ,579 x1010

E . I kL

=2 ,579 x 1010

550=46883044 ,77

KEKAKUAN KOLOM 24KEKAKUAN KOLOM 24

b = 50 cmb = 50 cm

h = 50 cmh = 50 cm

L = 400 cmL = 400 cm

MMDD = 9474,34 kgm = 9474,34 kgm

MMLL = 1814,49 kgm = 1814,49 kgm

Ec = 4700 √ f ' c = 4700 √22 ,5 = 22294 ,0575 MPa = 222940 ,575 kg /cm2

Ik=

112

b h3 = 112

x 50 x 503 = 520833 ,333 cm4

Ec x Ik = 222940,575 kg/cm2 x 520833,333 cm4 = 1,161 x1011 kg/cm2

βd=1,20D1,20D+1,60L

=1,20 x 9474,341,20 x 9474,34+1,60 x 1814,49

=0 ,797

STUDIO PERANCANGAN I

E.Ik =

Ec. I k2,50

1 + βd =

1,161 x 1011

2,501 + 0,797

=2 ,584 x 1010

E . I kL

=1 ,059x 1010

400=64607679 ,47

KEKAKUAN BALOK 2KEKAKUAN BALOK 2

b = 40 cmb = 40 cm

h = 60 cmh = 60 cm

L = 750 cmL = 750 cm

MMDD = 21311,3 kgm = 21311,3 kgm

MMLL = 4199,95 kgm = 4199,95 kgm

Ec = 4700 √ f ' c = 4700 √22 ,5 = 22294 ,0575 MPa = 222940 ,575 kg /cm2

Ib=

112

b h3 = 112

x 40 x 603 = 720000 cm4

Ec x Ib = 222940,575 kg/cm2 x 720000 cm4 =1,605 x 1011 kg/cm2

βd=1,20D1,20D+1,60L

=1,20x21311,31,20x21311,3+1,60x 4199,95

=0,792

E.Ib =

Ec. I b5

1 + βd =

1,605 x1011

51 + 0,792

=1 ,79x 1010

E . I bL

=1 ,79 x 1010

750=23883928 ,57

KEKAKUAN BALOK 3KEKAKUAN BALOK 3

b = 40 cmb = 40 cm

h = 60 cmh = 60 cm

L = 150 cmL = 150 cm

MMDD = 6869,42 kgm = 6869,42 kgm

MMLL = 1459,22 kgm = 1459,22 kgm

Ec = 4700 √ f ' c = 4700 √22 ,5 = 22294 ,0575 MPa = 222940 ,575 kg /cm2

STUDIO PERANCANGAN I

Ib=

112

b h3 = 112

x 40 x 603 = 720000 cm4

Ec x Ib = 222940,575 kg/cm2 x 720000 cm4 =1,605 x 1011 kg/cm2

βd=1,20D1,20D+1,60L

=1,20x6869,421,20x6869,42+1,60x1459,22

=0 ,779

E.Ib =

Ec. I b5

1 + βd =

1,605x1011

51 + 0,779

=1 ,804 x 1010

E . I bL

=1 ,804 x1010

200=90219224 ,28

Mencari nilai k

Ψ

a =

∑E . I k

L

∑E . I b

L

=46883044 ,77+64607679 ,4723883928 ,57+90219224 ,28

=0 ,977

Ψb = 0 (terjepit penuh)

Dari nomogram (SNI 03-2847-2002 hal.78) untuk struktur bergoyang, didapat nilai k = 0,563

STUDIO PERANCANGAN I

Kontrol kelangsingan :

[ K . lur ]≥22

, kolom langsing (SNI 03-2847-2002 hal. 79)

[ K . lur ]≤22

, kolom pendek

[ 0 ,56 x 5500,3 x60 ]=17 ,1875≥22

kolom Langsing tidak perlu pembesaran

momen

Perencanaan Tulangan Kolom

Pu = 11146,54 kg

Mu = 30808,11 kgm

Agr = 50 x 50 = 2500 cm2

d’ = 50 mm

Φ sengkang = 10 mm

Φ tul. utama = 22 mm

d = 500-10-50-1/2 x 22 = 429 mm

ρmin = 1%

ρmax = 8% (SNI 03-2847-2002 hal. 75)

Digunakan 10 - D22 (3801 mm2) pada 2 sisi :

ρ = Asb . d

= 38,0150 . 42,9

= 0,0177 ⇒ 0,01 < 0,0177< 0,08 ⇒ OK

tulangan tarik = 5 – D22 = 19.01 cm2

tulangan tekan = 5 – D22 = 19,01 cm2

Kontrol !

Kontrol terhadap keadaan seimbang

cb = 600d600 + fy

=600 . (429 )600 + 240

=306,429 mm

a = 0,85 x cb = 0,85 x 306,429 = 260,464 mm = 26,0464cm

STUDIO PERANCANGAN I

ɛs’ =

cb−d 'cb

0,003=306 ,429−50306 ,429

0 ,003=0 ,00251> fyEs

fs’ = Es x εs’ = 200000 x 0,00251 = 502 MPa

Pnb = 0,65[0,85 x fc’ x a x b + As’ x fs’ – As x fy]

= 0,65[0,85 x 22,5 x 260,464 x 500 + 3801 x 502 - 3801 x 240]

= 3486549 N

= 348654,9 kg > Pu = 11146,54 kg……………..OK!

Kontrol penampang kolom

e=MuPu

=30808,11 . 103

11146,54 =2763 ,917mm

m= fy0 ,85 . f ' c

=2400 ,85 . 22 ,5

=12 ,549

e '=e+(d−h2 )=2763 ,917+(329−500

2 )=2842 ,917mm

1− e 'd

=1−2842 ,917429

=−5 ,627

1−d 'd

=1−50429

=0 ,883

ρ= Asb .d

=3801500 . 429

=0 ,0177

Pn=0 ,85 . f ' c .b .d [(1− e 'd )+√(1− e '

d )2

+2 .m . ρ(1−d 'd ).]

=0 ,85 .22 ,5 .500 .429 [ (−5 ,627 )+√(−5 ,627 )2+2.12 ,549 . 0 ,0177 (0 ,883 ) ]

= 1425463347 N

= 142546334,7 kg

STUDIO PERANCANGAN I

φ Pn=0,65 x 142546334,7 = 92655,117 kg > Pu = 11398,72 kg...........OK!

Penulangan Geser Kolom

Vu maksimum = 11398,72

Pemeriksaan kebutuhan tulangan geser

Syarat kebutuhan tulangan geser :

Vn > Vc

Didapat :

Vc = 16

. √ fc ' . bw . d = 16

. √22 ,5 . 500 . 429 = 169577 ,14 N=16957 ,714 kg

= 0,6 ( Faktor reduksi untuk geser )

Vn = Vuφ

= 11398,720,6

=18997 ,867 kg

Vc = 16957,714 kg < Vn = 18997,867 kg

Sehingga digunakan sengkang praktis 10 – 200

STUDIO PERANCANGAN I

STUDIO PERANCANGAN I