Cheat Perhitungan Kolom

Pembuatan Diagram Interaksi Kolom InteriorData Kolom

b = 400 mmAg = 160000 mm2fc' = 3 kg/mm2fy = 39 kg/mm2tebal selimut = 30 mmdiameter sengkang = 12 mmdiameter tulangan = 25 mmAs aktual = 8*As tulangan

= 3928.5714285714 mm2As Tekan = 1964.2857142857 mm2d aktual = 345.5 mmd' aktual = 54.5 mmf = 0.7

Kondisi beban sentris

=Pn1 = 556205.35714286 kg

= 389343.75 kg

Kolom tetap diperhitungkan terhadap eksentrisitas minimume (min) = 15+(0,03*h)

= 27 mm= 10512281.25 kgmm

Kondisi seimbang

= 243.88235294118 mm

Chekfs1 =

= ((d-cb)/cb)*60= 25 kg/mm2

fs2 = 25 kg/mm2Pnb =

= 211446 kg= 148012.2 kg

Mnb == 20372822.1 kgmm= 14260975.47 kgmm

= 96.35

fPn1

fMn1

es1*Es

0,85*fc'*b*b*cb+As2*fs2-As1*fs1

fPnb0,85*fc'*b*b*cb*(0,5*h-b*cb*0,5)+As2*fy*((d-d')/2)-As1*fs1((d-d')/2)

fMnb

eb

Po=0. 85 ∗ f c '*( Ag−A st )+As∗ fy

cb=600∗d600+ fy

Kondisi beban aksial eksentrisitas besar tulangan tarik sudah leleh

Pn2 =0 = 867c+2946,42((c-54,5)/c)-2946,42((345,5-c)/c)

c = 119.40266 mmChek tegangan

fs1 = ((d-c)/c)*60= 113.61422266472 kg/mm2 Asumsi benar telah leleh

fs2 = ((c-d')/c)*60= 32.613675440731 kg/mm2 Asumsi benar telah leleh

Pn2 == 0= 0

Mn2 =Mn2 = 13625840.564784 kg/mm2

= 9538088.3953488 kg/mm2

Rekap HasilMomen Terfaktor Gaya Aksial Terfaktor

0 389343.7514260975.47 148012.2

9538088.39534883 0

PERENCANAAN KOLOM INTERIORData

0,85*fc'*b*b*c+As2*((c-d')/c)*60-As1*((d-c)/c)*60

0,85*fc'*b*b*c+As1*((c-d')/d')-As2*((d-c)/c)

fPn2

0,85*fc'*b*b*c*(0,5*h-b*c*0,5)+As2*f2*((d-d')/2)-As1*fs1((d-d')/2)

fMn2

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Diagram Interaksi Kolom Interior

Momen Terfaktor

Ga

ya

Ak

sia

l

fc' = 3 kg/mm2fy = 39 kg/mm2Pu = 6156.2 kgM1 = 12994320 kgmmM2 = 4842060 kgmmb = 400 mmb = 0.85f = 0.75

Balok : ln balok = 6200 mmbw = 300 mmhb = 500 mm

Kolom : lu kolom = 4000 mmln kolom = 3500 mm

Eksentrisitase = Mu/Pue > e min

2110.76963061629 > 27 mmDigunakan e

Menghitung Kekakuan Kolom

= 8140.6387955737 kg/mm2

= 400000000 mm4

= 0.5

= 868334804861.2 kg/mm2

Menghitung Momen Inersia balok

= 3125000000 mm4Icr = 1562500000 mm4

= 0.1058133333333

= 0.1058133333333

= 0.1058133333333

bd

Ig balok

Menghitung Faktor Kekangan ujung y yang terjadi di ujung

Ec=4700√ fc 'Igkolom=

164

b4

EI=Ec∗Ig

2. 5∗(1+βd )

ψ A=

∑ EILu

kolom

∑ EcIcrLn

balok

ψB=

∑ EILu

kolom

∑ EcIcrLn

balok

ψM=ψ A+ψ B

2

r = 0.3b = 120 mm

= 1.0460130529013

Chek kelangsingan Kolomk*lu/r batas kelangsingan

34.8671017633775 > 22Kolom termasuk Kolom Panjang

MetodePerbesaran Momen

= 489939.32537185 kg

= 1.673453860547

= 1.7019680185891 > 1

Mu1 = 22115917.063313Mu 2 = 8241031.2640896

Mencari eksentrisitas

2110.76963061629 > 96.35 Kondisi Keruntuhan TarikMencari Kekuatan Tumpu Kolom

= 6554.4745275125 kg

= 4915.8558956344 kg PIYEMencari Momen Kapasitas Kolom

M kap = 1,25*Mn= 17293732.221652 kgm PIYE

Perencanaan Tulangan GeserGAYA GESER

As = 3928.5714285714 mm2As' = 1964.2857142857 mm2F = 1fy = 25 kg/mm2 250

Mpr1 = 40582142.857143 kgmm

eu eb

fPn

k=20-ψm

20 √1+ψm

Pc= π2 EI(k∗lu)2

Cm=0 .6+0 . 4M 1

M 2

δs=Cm

1− ∑ Pu

φ∑ Pc

Pn=Mo

( eu−eb )+(MoPb )

M pr1=φ∗As∗1 ,25∗fy∗(d−0,5∗a )

M pr 2=φ∗As '* 1 ,25∗fy∗(d−0,5∗a)

Mpr2 = 20291071.428571 kgmm

lu = 4000 mm

= 15218.3035714286 kg

Vu = 7615.65 kg

f = 0.75

= 5711.7375 kg

> Ve OKMenentukan Jarak lo

lo < h kolom

lo < 400 mm

lo < 1/6h kolom

lo < 66.666666666667 mm

lo < 500

lo = 500 mm

PERANCANGAN TULANGAN GESER

PADA DAERAH SENDI PLASTIS

Daerah sendi plastis = 500 mm

Vc = 0 kg

Vn =

= 5711.7375 kg

Vs = Vn - Vc kg

= 5711.7375 kg

= 10 mm

fy = 25 kg/mm2 250

Av = 157.14285714286 mm2

s = 237.6372213484 mm

Syarat spasi maksimum :

s < 0,25*h

s < 100 mm

s < 0,25*b mm

s < 100 mm

Vu/f

Vu/f

Vu / f

Dipakai sengkang F

M pr 2=φ∗As '* 1 ,25∗fy∗(d−0,5∗a)

Ve=Mpr 1+Mpr 2

Ln

s=Av∗fy∗d

Vs

s < 6*db

s < 1800 mm

Maka jarak antara tulangan geser yang dipakai :

s = 100 mm

Vs aktual (Vn) = 13573.214285714 kg

< Vn OKMaka pada daerah sendi plastis, dipakai diameter D12-100

PADA DAERAH DI LUAR SENDI PLASTIS

Vn =

= 5711.7375 kg

Vc = 39894.903601003 kg

Vs = Vn - Vc

= -34183.166101003 kg

Karena Nilai Vs negatif, tidak perlu sengkang tambahanMaka pada daerah diluar sendi plastis, dipakai diameter D12-100

Vu / f

Vu / f

Vsaktual=Av∗fy∗d

s

Vc=16∗√ fc '∗b∗d

Pembuatan Diagram Interaksi Kolom Interior

*cb*0,5)+As2*fy*((d-d')/2)-As1*fs1((d-d')/2)

867c+2946,42((c-54,5)/c)-2946,42((345,5-c)/c)

Asumsi benar telah leleh


*b*c+As2*((c-d')/c)*60-As1*((d-c)/c)*60

*b*c+As1*((c-d')/d')-As2*((d-c)/c)

*c*0,5)+As2*f2*((d-d')/2)-As1*fs1((d-d')/2)

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Ga

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Ak

sia

l

Kondisi Keruntuhan Tarik

Pu = 6156.2 kg

Mu = 22115917 kgm

MPa

Pembuatan Diagram Interaksi Kolom EksteriorData Kolom


= 4910.7142857 mm2As Tekan = 2455.3571429 mm2d aktual = 527.5 mmd' aktual = 72.5 mmf = 0.85


=Pn1 = 1103256.6964 kg

= 750214.55357 kg


= 33 mm= 24757080.268 kgmm

Kondisi seimbang

= 372.35294118 mm

Chekfs1 =

= ((d-cb)/cb)*60= 25 kg/mm2


= 484245 kg= 329286.6 kg

Mnb == 68641728.75 kgmm= 54913383 kgmm

= 141.75

fPn1

fMn1

es1*Es



fMnb

eb

Po=0. 85 ∗ f c '*( Ag−A st )+As∗ fy

cb=600∗d600+ fy


Pn2 =0 = 867c+2946,42((c-54,5)/54,5)-2946,42((345,5-c)/345,5)




Pn2 == 0= 0

Mn2 =Mn2 = 30085073.123 kg/mm2

= 21059551.186 kg/mm2

Rekap HasilMomen Terfaktor Gaya Aksial Terfaktor24757080.26786 750214.553571429

54913383 329286.621059551.18592 0

PERENCANAAN KOLOM EKSTERIORData

0,85*fc'*b*b*c+As2*((c-d')/d')*60-As1*((d-c)/c)*60


fPn2


fMn2

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Momen Terfaktor

Ga

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Ak

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l

fc' = 3 kg/mm2fy = 39 kg/mm2Pu = 1623.92 kgM1 = 16379.27 kgmmM2 = 14112.37 kgmmb = 600 mmb = 0.85f = 0.75




10.0862542489778 < 27 mmDigunakan e min

Menghitung Kekakuan Kolom= 8140.6387956 kg/mm2

= 2025000000 mm4

= 0.5

= 4.39594E+12 kg/mm2


= 3125000000 mm4Icr = 1562500000 mm4

= 0.53568

= 0.53568

= 0.53568

r = 0.3b = 180 mm

bd

Ig balok



164

b4

EI=Ec∗Ig

2. 5∗(1+βd )

ψ A=

∑ EILu

kolom

∑ EcIcrLn

balok

ψB=

∑ EILu

kolom

∑ EcIcrLn

balok

ψM=ψ A+ψ B

2

= 1.2060341471




= 1865788.1708 kg

= 1.0642528505

= 1.0654893392 > 1

Mu1 = 17451.937569Mu 2 = 15036.579786


10.0862542489778 < 141.75 Kondisi Keruntuhan TekanMencari Kekuatan Tumpu Kolom

= 1011273.3673 kg

= 758455.02548 kg OK BuangetMencari Momen Kapasitas Kolom

M kap = 1,25*Mn= 12749950.372 kgm OK Buanget



Mpr1 = 78657366.071 kgmm

eu eb

fPn

k=20-ψm

20 √1+ψm

Pc= π2 EI(k∗lu)2

Cm=0 .6+0 .4M 1

M 2

δs=Cm

1− ∑ Pu

φ∑ Pc

Pn=Po

1+( PoPb−1)∗ eeb

M pr1=φ∗As∗1 ,25∗fy∗(d−0,5∗a )

M pr 2=φ∗As '* 1 ,25∗fy∗(d−0,5∗a)

Mpr2 = 39328683.036 kgmm

lu = 4000 mm

= 29496.512277 kg

Vu = 7615.65 kg

f = 0.75

= 5711.7375 kg

> Ve OKMenentukan Jarak lo

lo < h kolom

lo < 400 mm

lo < 1/6h kolom

lo < 66.666666667 mm

lo < 500

lo = 500 mm




Vc = 0 kg

Vn =

= 5711.7375 kg

Vs = Vn - Vc kg

= 5711.7375 kg

= 10 mm

fy = 25 kg/mm2 250

Av = 157.14285714 mm2

s = 362.81804417 mm


s < 0,25*h

s < 150 mm

s < 0,25*b mm

s < 150 mm

s < 6*db

Vu/f

Vu/f

Vu / f

Dipakai sengkang F

Ve=Mpr 1+Mpr 2

Ln

s=Av∗fy∗d

Vs

s < 1800 mm


s = 100 mm

Vs aktual (Vn) = 20723.214286 kg



Vn =

= 5711.7375 kg

Vc = 91365.680099 kg

Vs = Vn - Vc

= -85653.9426 kg


Vu / f

Vu / f


s

Vc=16∗√ fc '∗b∗d

Pembuatan Diagram Interaksi Kolom Eksterior

*cb*0,5)+As2*fy*((d-d')/2)-As1*fs1((d-d')/2)

867c+2946,42((c-54,5)/54,5)-2946,42((345,5-c)/345,5)



PERENCANAAN KOLOM EKSTERIOR

*b*c+As2*((c-d')/d')*60-As1*((d-c)/c)*60

*b*c+As1*((c-d')/d')-As2*((d-c)/c)

*c*0,5)+As2*f2*((d-d')/2)-As1*fs1((d-d')/2)

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Ga

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Ak

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l

Kondisi Keruntuhan Tekan

Pu = 1623.92 kg

Mu = 17451.94 kgm

MPa

Pembuatan Diagram Interaksi Kolom TanggaData Kolom


= 2036.5714286 mm2As Tekan = 1018.2857143 mm2d aktual = 249 mmd' aktual = 51 mmf = 0.7


=Pn1 = 306329.65714 kg

= 214430.76 kg


= 24 mm= 5146338.24 kgmm

Kondisi seimbang

= 175.76470588 mm

Chekfs1 =

= ((d-cb)/cb)*60= 25 kg/mm2


= 114291 kg= 80003.7 kg

Mnb == 8606112.3 kgmm= 6024278.61 kgmm

= 75.3

fPn1

fMn1

es1*Es



fMnb

eb

Po=0. 85 ∗ f c '*( Ag−A st )+As∗ fy

cb=600∗d600+ fy


Pn2 =0 = 650,25c+117857((c-54,5)/c)-117857((245,5-c)/c)




Pn2 == 0= 0

Mn2 =Mn2 = 6944119.351 kg/mm2

= 4860883.5457 kg/mm2

Rekap HasilMomen TerfaktGaya Aksial Terfaktor

5146338.24 214430.766024278.61 80003.7

4860883.5457 0

PERENCANAAN KOLOM TANGGAData

0,85*fc'*b*b*c+As2*((c-d')/c)*60-As1*((d-c)/c)*60


fPn2


fMn2

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Momen Terfaktor

Ga

ya

Ak

sia

l

fc' = 3 kg/mm2fy = 25 kg/mm2Pu = 1481.1 kgM1 = 7028.23 kgmmM2 = 2111.23 kgmmb = 300 mmb = 0.85f = 0.75




4.74527715886841 < 27 mmDigunakan e min

Menghitung Kekakuan Kolom= 8140.6387956 kg/mm2

= 126562500 mm4

= 0.5

= 2.74747E+11 kg/mm2


= 3125000000 mm4Icr = 1562500000 mm4

= 0.01593

= 0.01593

= 0.01593

r = 0.3b = 90 mm

bd

Ig balok



164

b4

EI=Ec∗Ig

2. 5∗(1+βd )

ψ A=

∑ EILu

kolom

∑ EcIcrLn

balok

ψB=

∑ EILu

kolom

∑ EcIcrLn

balok

ψM=ψ A+ψ B

2

= 1.0071307105




= 167220.64072 kg

= 1.9315896421

= 1.9546734498 > 1

Mu1 = 13737.89458Mu 2 = 4126.7652275


4.74527715886841 < 75.3 Kondisi Keruntuhan TekanMencari Kekuatan Tumpu Kolom

= 276999.0218 kg

= 207749.26635 kg OK BuangetMencari Momen Kapasitas Kolom

M kap = 1,25*Mn= 1643046.414 kgm OK Buanget



Mpr1 = 14896247.143 kgmm

eu eb

fPn

k=20-ψm

20 √1+ψm

Pc= π2 EI(k∗lu)2

Cm=0 .6+0 . 4M 1

M 2

δs=Cm

1− ∑ Pu

φ∑ Pc

Pn=Po

1+( PoPb−1)∗ eeb

M pr1=φ∗As∗1 ,25∗fy∗(d−0,5∗a )

M pr 2=φ∗As '* 1 ,25∗fy∗(d−0,5∗a)

Mpr2 = 7448123.5714 kgmm

lu = 4000 mm

= 5586.0926786 kg

Vu = 7615.65 kg

f = 0.75

= 5711.7375 kg

< Ve EmbuhMenentukan Jarak lo

lo < h kolom

lo < 400 mm

lo < 1/6h kolom

lo < 66.666666667 mm

lo < 500

lo = 500 mm




Vc = 0 kg

Vn =

= 5711.7375 kg

Vs = Vn - Vc kg

= 5711.7375 kg

= 10 mm

fy = 25 kg/mm2 250

Av = 157.14285714 mm2

s = 171.26387298 mm


s < 0,25*h

s < 1000 mm

s < 0,25*b mm

s < 75 mm

s < 6*db

Vu/f

Vu/f

Vu / f

Dipakai sengkang F

Ve=Mpr 1+Mpr 2

Ln

s=Av∗fy∗d

Vs

s < 1800 mm


s = 75 mm

Vs aktual (Vn) = 13042.857143 kg



Vn =

= 5711.7375 kg

Vc = 21564.032554 kg

Vs = Vn - Vc

= -15852.29505 kg


Vu / f

Vu / f


s

Vc=16∗√ fc '∗b∗d

Pembuatan Diagram Interaksi Kolom Tangga

*cb*0,5)+As2*fy*((d-d')/2)-As1*fs1((d-d')/2)

650,25c+117857((c-54,5)/c)-117857((245,5-c)/c) 650.2561097.14



*b*c+As2*((c-d')/c)*60-As1*((d-c)/c)*60

*b*c+As1*((c-d')/d')-As2*((d-c)/c)

*c*0,5)+As2*f2*((d-d')/2)-As1*fs1((d-d')/2)

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Momen Terfaktor

Ga

ya

Ak

sia

l

Kondisi Keruntuhan Tekan

Pu = 1481.1 kg

Mu = 13737.89 kgm

MPa

PERANCANGAN JOINT

Datafc' = 30

= 24 kg/m2 240 MPa= 39 kg/m2 390 MPa= 0.85= 12 mm

Keterangan :l'ka, l'ki = bentang bersih baloklka, lki = bentang p.k.p balokhka, hkb = tinggi kolom (tinggi antar lantai)

Pertemuan Balok Kolom EksteriorPERHITUNGAN GAYA-GAYA DALAMUkuran balok

fy f < 12mmfy f > 12mmb1f tulangan

V kolom

0,7 Mkap, ki

Cki

Tki

l'ka

Cka

Tka

l'ki

hka

hkb

V kolomhc

bKolom b

ZkaZki

b = 300 mmh = 500 mmd = 409 mmc = 64 mmb = 0.85a = 54.23 mmUkuran kolomb kol = 400 mmZka = Zki = d-0,5a = 381.88515406 mmhka = 4000 mmhkb = 4000 mmlka = 3600 mmlki = 3600 mmlka' = 3200 mmlki' = 3200 mmMkap ka =Mkap ki =

Vkol = kg

Tka = kg

Tki = kg

Vj,h = kg

KONTROL TEGANGAN GESER HORISONTAL

Vkol=0 . 7∗( lkilki'

∗Mkapki+lkalka '

∗Mkapka)12

(hka+hkb )

Tka=0 .7∗MkapkaZka

Tki=0 . 7∗MkapkiZki

Vj ,h=Cki+Tka−Vkol

0,7 Mkap, kaCka

Cheat Perhitungan Kolom

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