perencanaan kolom
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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
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