fondasi bujur sagkar
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Tugas FondasiTRANSCRIPT
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Pondasi Tapak Beton Type Bujursangkar
Project : Ruko Permata Juanda SurabayaClient : Mr Bunawan
Engineer Checked Approved
Name Lutfi AriDate 4/19/2023 19:23
Data Struktur : Pu
Mu1 Dimensi Kolom b = 400 mm b
h = 400 mm Muka TanahType Kolom αs = 30
2 Dimensi Pondasi B = 2.40 m haL = 2.40 m B = Lht = 0.40 m
3 Mutu Beton fc' = 20 Mpa ht4 Mutu Baja fy' = 320 Mpa5 Besi tulangan D = 19
(dipakai) L6 BJ Beton γc = 24 KN/m3
LData Tanah :
6 Daya dukung tanah = 105 Kpa7 Berat tanah γt = 17.20 KN/m38 Tebal tanah diatas ha = 1.20 m
pondasi hB
Data Beban :B = L
9 Beban P ultimate Pult = 384 KN10 Beban M ultimate Mult = 18.25 KNm
b
ResultCek Status Pondasi :
Tegangan yang terjadi pada tanah = 104.82767 < 105 OK!= 88.98566 < 105 OK!
Kontrol gaya geser 1 arah Ø.Vc = 422.617Vu = 168.620 < 422.617 OK!
Kontrol gaya geser 2 arah Ø.Vc = 1007.237Vu = 508.641 < 1007.237 OK!
Penulangan Pondasi Arah X = Y D 19 - 219
Author : Lutfi Andrian, WSWEB :E-mail :CR 2011
σt
σmaksσmin
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SKEMA HITUNGAN PONDASI
Data : tebal fondasi (ht, d, ds), mutu bahanMutu Bahan (Fc, Fy)Beban (Pu, Mu)
Penentuan Dimensi Pondasi
Pult Mu,x Mu,yσ = + + + q ≤
B x L 1/6 B x 1/6 B x
Menghitung tegangan tanah didasar fondasi
Pult Mu,x Mu,yσmaks = + + + q ≤
B x L 1/6 B x 1/6 B x
Pult Mu,x Mu,yσmin = - - + q ≤
B x L 1/6 B x 1/6 B x
Kontrol gaya geser 1 arah
σmaks + σaVu = a x B
2Dimana
L - a x σmaks - σminσa = σmin +
Lfc'
Vc = x B x d6
Daya dukung tanah (σt)
σt
L2 L2
σt
L2 L2
σt
L2 L2
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Syarat :
Vu ≤ Ø Vc dengan Ø = 0.75
Kontrol gaya geser 2 arah
σmaks +
Vu = B 2 - b + d h + d x2
Dipilih nilai Vc terkecil dari :
2 fc' bo dVc1 = 1 + x
6
αs d fc' bo dVc2 = 2 + x
bo 12
1Vc3 = x fc' bo d
3Syarat :
Vu ≤ Ø Vc dengan Ø = 0.75
Hitung Penulangan Fondasi
1) ds = 75 + D + D/2dimanad = h - ds
2) σx = σmin + B - x x σmaks - σmin
σmaks - σx
3) Mu = 0.5 σx x 2 + x 23
Mu4) K = , b = 1000
βc
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Ø x b x d 2
382.5 x 0.85 x 600 + fy - 225 x β15) Kmaks =
600 + fy 2
2 x K6) a' = 1 - 1 - d
0.85 x fc'
0.85 x fc' x a x b7) As(1) =
fy
8) fc' ≤ 31.36 Mpamaka
1.4 x b x dAs(2) ≥ Pilih "As" yang
fy terbesar
fc' > 31.36 Mpamaka
fc' x b x dAs(3) =
4 fy
9) Jarak tulangan
0.25 x phi x D 2s = S
As terpilih Pilih "s" yang
s ≤ 2 x ht terkecil
s ≤ 450
Cek Kuat Dukung Fondasi
Pu = Ø x 0.85 x fc' x A ..... ( Ø = 0,7 )
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Syarat : Pult ≤ Pu
Referensi : Asroni, A., Buku Kolom Fondasi & Balok T Beton Bertulang,
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Notasi :
σ = Tegangan yang terjadi pada dasar fondasi ...........σt = Daya dukung tanah ...........
Pult = Beban axial terfaktor pada kolom...........B,L = Panjang dan lebar fondasi
Mu,x = Momen terfaktor kolom searah sumbu X ...........Mu,y = Momen terfaktor kolom searah sumbu Y ...........
q = Beban terbagi rata akibat beban sendiri fondasiditambah beban tanah diatas fondasi ...........
ht = Tebal fondasi ≥ 150 mm
(lihat pasal 17.7 SNI 03-2847-2002)ha = Tebal tanah diatas fondasi ...........γc = Berat per volume beton ...........γt = Berat per volume tanah ...........Vu = Gaya geser pons terfaktor...........Vc = Gaya geser yang sanggup ditahan oleh beton...........Ø =βc = Rasio dari sisi panjang terhadap sisi pendek pada
kolom, daerah beban terpusat / atau daerah reaksi bo = Keliling penampang krisis dari fondasi ...........
αs = Suatu konstanta yang digunakan untuk menghitung Vc , yang nilai tergantung pada letak fondasi,40 = Kolom dalam30 = Kolom tepi
20 = Kolom sudutσmin = Tegangan tanah minimum ...........
σmaks = Tegangan tanah maksimum ...........σa = Tegangan tanah sejarak "a" dari tepi fondasi ..........σx = Tegangan tanah sejarak "x" dari tepi fondasi ..........a = Jarak dari area bidang yang menerima tekanan
keatas dari tanah (area keruntuhan)fc' = Kuat tekan beton ...........fy = Mutu beton ...........ds = decking
D = Diameter tulangan ...........a' = Tinggi blok tegangan beton teka persegi ekivalen
Mu = Momen yang terjadi padafondasiK, Kmaks = Faktor momen pikul
s = Jarak tulangan ...........As = Luas besi tulangan yang dibutuhkan ...........
Faktor keamanan = 0,7 (lihat SNI)
minimal = 75 mm (berhubungan dengan tanah)
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σmin
B
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fc'
Pilih "As" yang
Pilih "s" yang
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(Kpa) atau KN/m2(Kpa) atau KN/m2KNmKNmKNm
(KN/m2)mm
mmKN/m3KN/m3KNKN
mm
(Kpa) atau KN/m2(Kpa) atau KN/m2(Kpa) atau KN/m2(Kpa) atau KN/m2mm
MpaMpamm
mmmm
mmmm2
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Job No. Sheet No1
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner Lutfi DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Analisa
q = Berat Fondasi + Berat Tanah= ht x γc + ha x γt= 0.40 x 24 + 1.20 x 17.20= 30.24 KN/m2
Cek Fondasi Terhadap Tegangan Izin Tanah
Tegangan yang terjadi pada tanah
Pult Mult= + + q ≤
B x L 1/6 B x
384 18.25= + + 30.24 ≤ 105
2.40 x 2.40 1/6 2.40 x 2.40 2
= 66.666667 + 7.921 + 30.24
= 104.828 ≤ 105 Save!
Pult Mult= - + q ≤
B x L 1/6 B x
384 18.25= - + 30.24 ≤ 105
2.40 x 2.40 1/6 2.40 x 2.40 2
= 66.666667 - 7.921 + 30.24
= 88.986 ≤ 105 Save!
Kontrol Tegangan Geser 1 Arah
0.40ds = 75 + D/2
= 75 + 9.50= 85.00 mm
d = ht - ds= 400 - 85.00
0.40 = 315.00
L = 2.40 a = B/2 - b/2 - d= 1200 - 200 - 315.00
2.40 = 685.00 mm= 0.685 m
σa = σmin + B - a x σmaks - σmin b
= 88.986 + 2.40 - 0.685 x 104.828 - 88.986 2.402.40
0.4 = 100.306 KN/m2
Gaya tekan ke atas dari tanah (Vu)
Vu = a x B x σmaks + σa 20.4
= 0.685 x 2.40 x 104.828 + 100.306 2
= 168.620 KN
σmaks σt
L2
σmin σt
L2
Program Hitung Pondasi Tapak
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Job No. Sheet No1
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner Lutfi DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Program Hitung Pondasi Tapak
0.40 0.315 0.685
fc'Ø.Vc = Ø x x B x d
620
= 0.75 x x 2.40 x 315.006
0.315= 422.617 KN
2.40Ø.Vc = 422.617 > Vu = 168.620 ......... Save!
σmin σa σmaks88.986 100.306 104.828
Kontrol Tegangan Geser 2 Arah (Geser Pons)
Dimensi Kolom, b = 400h = 400
b + d = 400.00 + 315.00 = 715.00 mm = 0.715 mh + d = 400 + 315.00 = 715.00 mm = 0.715 m
Gaya Tekan Ke Atas (Geser Pons)
σmaks + σmin
Vu = B 2 - b + d x h + d x2
104.828 + 88.986
= 2.40 2 - 0.715 x 0.715 x2
= 508.64129
hk 400= = = 1.000
Bk 400
bo = 2 x bk + d + hk + d
bo = 2 x 400 + 315.00 + 400 + 315.00
= 2860 mm
Gaya geser yang ditahan oleh beton
2 fc' bo dVc1 = 1 + x
0.158 6
0.40 2 20 2860 315.00Vc1 = 1 + x
0.158 1.000 6
Vc1 = 2014473.641 N= 2014.474 KN
0.843 0.158 0.40 0.158 0.8425αs d fc' bo d
Vc2 = 2 + xbo 12
30 315.00 20 2860 315.00Vc2 = 2 + x
0.315 2860 12
0.085 Vc2 = 1445114.832 N2.40 = 1445.115 KN
Gaya geser yang dapat ditahan oleh beton (Ø.Vc)
βc
βc
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Job No. Sheet No1
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner Lutfi DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Program Hitung Pondasi Tapak
1Vc3 = x fc' bo d
31
σmin Vc3 = x 20 2860 315.0088.986 σmaks 3
104.828= 1342982.4273 N= 1342.982 KN
Jadi
Vc1 = 2014.474Vc2 = 1445.115 Diambil yang terkecilVc3 = 1342.982
Vc = 1342.982 KNØ.Vc = 0.75 x 1342.982
= 1007.237 KN
Ø.Vc = 1007.237 > Vu = 508.641 ......... Save!
Hitungan Penulangan Fondasi
0.40 ds = 75 + 19 + 9.5= 103.500 mm≈ 0.105 m
1.001.20 d = ht - ds
= 0.40 - 0.105= 0.295 m
0.40 0.295 = 295 mm0.105
B h2.40 x = -
2 2
2400 400x = -
2 2σmin σx σmaks x = 1000 mm88.986 98.227 104.828 = 1 m
σx = σmin + B - x x σmaks - σmin B
σx = 88.986 + 2.40 - 1 x 104.828 - 88.986 2.40`
= 98.227 KN/m3
σmaks - σx
Mu = 0.5 σx x 2 + x 23
104.828 - 98.227
= 0.5 98.227 1 2 + 1 23
= 51.314 KNm
MuK =
Ø x b x d 2
51313696.952=
0.8 x 1000 x 295 2
= 0.737 Mpa
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Job No. Sheet No1
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner Lutfi DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Program Hitung Pondasi Tapak
382.5 x 0.85 x 600 + fy - 225 x β1 fc'Kmaks =
600 + fy 2
382.5 x 0.85 x 600 + 320 - 225 x 0.85 20Kmaks =
600 + 320 2
= 5.599 Mpa
K < KmaksK = 0.737 < Kmaks = 5.599 ......... OK!
2 x Ka' = 1 - 1 - d
0.85 x fc'
2 x 0.737= 1 - 1 - 295
0.85 x 20
= 13.080 mm
0.85 x fc' x a' x bAs(1) =
fy
0.85 x 20 x 13.080 x 1000=
320= 694.877 mm2
Jikafc' ≤ 31.36 Mpa
maka1.4 x b x d
As ≥ ........... (R.1)fy ....... SNI 03-2847-2002 (Pasal 12.5.1)
Jikafc' > 31.36 Mpa
makafc' x b x d
As = ........... (R.2)4 fy ....... SNI 03-2847-2002 (Pasal 12.5.1)
fc' = 20 < 31.36maka yang dipakai adalah pers ..................... (R.1)
1.4 x b x d fc' x b x dAs(2) = ....... (R.1) As(2') = ....... (R.2)
fy 4 fy
1.4 x 1000 x 295 0 x 0 x 0= =
320 4 0
= 1290.625 mm2 ........... As dipakai = #DIV/0! mm2 ........... As tidak Dipakai!
Dipilih yang terbesar dari As(1) dan As(2)......... Sehingga,
As(1) = 694.877 mm2As = 1290.625 mm2
As(2) = 1290.625 mm2
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Job No. Sheet No1
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner Lutfi DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Program Hitung Pondasi Tapak
Jarak tulangan,
0.25 x phi x D 2 x Ss =
As
0.25 x 3.14 x 19 2 x 1000=
1290.625
= 219.572 mm
s ≤ 2 x ht≤ 2 x 400≤ 800
s ≤ 450 mm
Dipilih (s) yang terkecil = 219.572 mm
Jadi dipakai tulangan = D 19 - 219
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Job No. Sheet No2
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Cek Panjang Penyaluran Tegangan Tulangan
9 x fy x α β x γ x λ= db
c + Ktr10 fc'
db
9 x 320 x 1 1 x 0.8 x 1= 19
75 + 010 20
19
9 x 320 x 1 1 x 0.8 x 1= 19
10 20 2.5
= 391.544 mm > 300 mm≈ 392 mm
B BkPanjang Tersedia (λt) = - - 75
2 2
2400 400= - - 75
2 2
= 925
>925 > 392 ....... OK!
Kuat Dukung Fondasi
Pu = Ø x 0.85 x fc' x A
= 0.7 x 0.85 x 20 x 160000
= 1904000 N
= 1904 KN
Pu,k = 384 KN
Pu = 1904 > Pu,k = 384 ....... OK!
Panjang penyaluran tegangan (λd)
Panjang Tersedia (λt) Panjang penyaluran tegangan (λd)
Program Hitung Pondasi Tapak
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Job No. Sheet No2
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Program Hitung Pondasi Tapak
SKSNI 03-2847-2002
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Job No. Sheet No2
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Program Hitung Pondasi Tapak
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Job No. Sheet No3
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
2.40
2.40
D 19 - 219
DETAIL PONDASINon Scale
400
D 19 - 219
400 295
1052400
POTONGAN 1-1Non Scale
SNI 03-2847-2002 (Pasal 17.4.3)
SNI 03-2847-2002 (Pasal 17.7)
SNI 03-2847-2002 (Pasal 9.7.1)
Program Hitung Pondasi Tapak
1
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Job No. Sheet No3
PartJob Title Ref
Ruko Permata Juanda Surabaya Enginner DateClient. Mr Bunawan File Date/time 4/19/2023 19:23
Program Hitung Pondasi Tapak