Perhitungan Balok

TABEL PERENCANAAN TULANGAN BALOKPEMBANGUNAN MALL NIAS

SUMATERA BARAT

No Lantai

Type Balok B1 300 x 600Momen Tulangan Lentur Gaya Geser Tulangan Geser

Tumpuan Lapangan Tumpuan Lapangan

1 Lt.1 412.068 249.456 22 8.00 5.00 4.00 5.00 4.00 178.817 175.496 13 2.00 100.00 135.002 Lt.2 412.068 249.456 22 8.00 5.00 4.00 5.00 4.00 178.817 175.496 13 2.00 100.00 135.003 Lt.3 412.068 249.456 22 8.00 5.00 4.00 5.00 4.00 178.817 175.496 13 2.00 100.00 135.004 Lt.4 412.068 249.456 22 8.00 5.00 4.00 5.00 4.00 178.817 175.496 13 2.00 100.00 135.00

Type Balok B1 Atap 250 x 4005 Atap 154.230 63.527 22 6.00 5.00 4.00 5.00 4.00 83.069 82.773 13 2.00 70.00 80.00

No Lantai

Type Balok B2 250 x 500Momen Tulangan Lentur Gaya Geser Tulangan Geser

Tumpuan Lapangan Tumpuan Lapangan

1 Lt.1 210.612 184.835 22 6.00 4.00 4.00 5.00 4.00 153.547 153.357 13 2.00 50.00 80.002 Lt.2 210.612 184.835 22 6.00 4.00 4.00 5.00 4.00 153.547 153.357 13 2.00 50.00 80.003 Lt.3 210.612 184.835 22 6.00 4.00 4.00 5.00 4.00 153.547 153.357 13 2.00 50.00 80.004 Lt.4 210.612 184.835 22 6.00 4.00 4.00 5.00 4.00 153.547 153.357 13 2.00 50.00 80.00

Type Balok B2 Atap 200 x 3505 Atap 72.929 60.952 19 6.00 4.00 4.00 5.00 4.00 53.861 53.319 13 1 Tm ; 2 Lp 50.00 80.00

M.Tum (kN/m)

M.Lap (kN/m)

Diameter Tulangan (Ø), mm

Susut (buah)

V.Tum (kN/m)

V.Lap (kN/m)


Sengkang (kaki)Desak

(buah)Tarik (buah)

Desak (buah)

Tarik (buah)

Jarak Sengkang (s), mm


M.Tum (kN/m)

M.Lap (kN/m)


Susut (buah)

V.Tum (kN/m)

V.Lap (kN/m)


Sengkang (kaki)Desak

(buah)Tarik (buah)

Desak (buah)

Tarik (buah)



Tugas Besar

Perancangan Bangunan Teknik Sipil FST-UTY

C PERENCANAAN SLOOFI BALOK IKAT (SLOOF) S1

= 2430.33 kN

= 860 kNm

= 7.00 m

= 7.00 m

My =1,60 x 0,65

= 8601.04

= 826.923077 kNm

=

= x My =+ +

=0.1429

x 826.92310.1429 + 0.1429

= 0.500 x 826.9231= 413.4615 kNm

= 1x

10= 0.1 x 2430.33= 243.033 kN

Dimensi SloofDimensi rencana :

b = 300 mm `h = 600 mm

Ag = 180000f'c = 22.5 Mpa

< 0,1 x Ag x f'c243.033 < 0.1 x 180000 x 22.5243.033 < 405.000 kN

Pukolom

Melantai 1

Lka

Lki

Melantai 1

Msloof ki Msloof ka

Msloof 1/Lki 1/Lka

1/Lki 1/Lka 1/Lka 1/Lki

Psloof kiPukolom K1

mm2

Psloof

Jika Psloof > 0,1.Ag.f'c maka htungan tulangan sloof menggunakan perhitungan kolom penulangan 2 muka.

Jika Psloof < 0,1.Ag.f'c maka htungan tulangan sloof menggunakan perhitungan balok (tulangan tumpuan).

Tugas Besar


II HITUNGAN TULANGAN SLOOFInput Data Sloof

= 300 mmf'c = 22.5 Mpafy = 370 Mpa

= 0.85φ = 0.8ε = 0.003Es = 200000

Data Momen Sloof

= 413.462 KNm

1 Dihitung Berdasarkan Perhitungan Tulangan Tumpuan BalokRasio Tulangan (ρ)

= x600

fy 600 + fy

= x600

370 600 + 370= 0.0272

== 0.75 x 0.0272= 0.0204

ρ == 0.5 x 0.0272= 0.0136

m=

fy0.85 x f'c

=370

0.85 x 22.5= 19.3464

= ρ x f'c x (1-(0.5 x ρ x m))= 0.0135 x 22.5 x (1-(0.5 x 0.0135 x 19.3464))= 4.3668

=Mu/φ

=4.3668

= 118352398.598 mm

bsloof

β1

Msloof

ρb 0.85 x β1 x f'c

0.85x0.03x22.5

ρmax 0.75 x ρb

0.5 x ρb

Koefisien Perlawanan (R1)

R1

bd2

R1

413.462 X 106/0.8

Tugas Besar


=b

=118352398.598

300

= 628.0987

Luas tarik effektif beton (SNI 03-2847-2002)

Pusat berat tulangan baja desak : d' = 69 mmPusat berat tulangan baja tarik : z = 71.5 mm

== 699.5987 mm

Skema penampang (SNI 03-2847-2002)

Jadi, ukuran balok berdasarkan hitungan adalah :b = 300 mm

= 699.5987 mmSyarat

Diambil :

= 600 mm

dperlu bd2

mm2

hperlu dperlu + z

hperlu

hpakai < hperlu

hpakai

Tugas Besar


Tinggi efektif desak :

= h - z= 528.50 mm

Tinggi efektif tarik :

= h - d'= 531 mm

a Tulangan Desakγ = 0.50

== 0.5 x 4.3668= 2.1834239

=

== 182.95712 KNm

=Mu

-φ

=413.462

- 182.95710.8

= 333.8698 KNmLuas tulangan perlu (As')

As' =fy x (d-d')

=333.8698

370 x (528.5-69)

= 1963.7668

Diameter tulangan rencana :Luas satu tulangan

Ø (mm)6 28.2608 50.240

10 78.50013 132.66516 200.96019 283.38522 379.94025 490.625

Coba tulangan D19 dengan AsØ=283.385 mm², maka :Jumlah tulangan perlu :

n =As'

ddesak

dtarik

R2 γ x R1

M1 R2 x b x d2

2.1834239 x 300 x (528.52)

M2M1

M2

mm2

AsØ (mm2)

Tugas Besar


n =

=1963.7668

283.385= 6.9297 tulangan, ambil 7 tulangan

=

= 1983.70

b Tulangan Tarik

= γ x ρ= 0.5 x 0.0136= 0.0068

As =

=

= 3040.9868


Ø (mm)6 28.2608 50.240

10 78.50013 132.66516 200.96019 283.38522 379.94025 490.625

Coba tulangan D19 dengan AsØ=283.385 mm², maka :Jumlah tulangan perlu :n = As'

AsØ= 3040.9868

283.385= 8.0039 tulangan, ambil 11 tulangan

=

= 3117.235

AsØ

Dipakai 7D19

Luas tulangan perlu (As'perlu)

As'ada n x AsØ

mm2

ρbaru

Luas tulangan perlu (Asperlu)

(ρbaru x b x d) + As'

(0.0068 x 300 x 528.5) + 1963.7668

mm2

AsØ (mm2)

Dipakai 11D19


Asada n x AsØ

mm2

Tugas Besar


Kontrol Momen Tersediac Kontrol Momen Tersedia

= 1983.695

= 3117.235

Cc = 0.85 x f'c x a x b a = 0.85c= 4876.875 c

Cs = As' x εc x Es x c - d'

c= 1190217 x c - 69

cTs = As x fy

= 1153376.95

Cc+ Cs -Ts = 0

4877 c + 1190217 x c - 69 - 1153376.95 = 0

c

4877 36840.05 c - -82124973 = 0

7.55 c -16839.6715 = 0

Gunakan rumus ABC untuk mencari niala c :Rumus:

c = b ± - 4ac2a

= -7.5540 ± 67415.7492

= -7.5540 ± 259.64542

= 126.046 mm

= -133.60 mm

= 126.046 mmfs' = c - d' x εc x Es

c= 271.548 Mpa

= fs'= 271.548 Mpa

a = 0.85 x c

As'ada mm2

Asada mm2

c2

c2

b2

c1

c2

Dari hitungan gunakan nilai c1

< 400 Mpa…….Ok!

fs'pakai

Tugas Besar


= 107.139 mm

= = (0.85 x f'c x a x b x (d - a/2)) + As' x fs' x (d - d'))= 539.462 KNm

== 431.570 KNm > Mu = 413.462 KNm Aman!

=b x d

= 0.0125

=b x d

= 0.0197

= 0.0071

= 0.0204

0.0071 < 0.0204 Aman!

Mg- Mn-

0.8Mg- 0.8 x Mg-

ρ'ada Asp'

ρada Asp

ρada-ρ'ada

ρmax

syarat ρada-ρ'ada < ρmax


SUMATERA BARAT

No Lantai

Type Balok B1 300 x 600Momen Tulangan Lentur

Tumpuan Lapangan

1 Lt.1 412.068 249.456 22 5.00 8.00 4.00 5.002 Lt.2 412.068 249.456 22 5.00 8.00 4.00 5.003 Lt.3 412.068 249.456 22 5.00 8.00 4.00 5.004 Lt.4 412.068 249.456 22 5.00 8.00 4.00 5.00

Type Balok B1 Atap 250 x 4005 Atap 154.230 63.527 22 4.00 6.00 4.00 5.00

M.Tum (kN/m)

M.Lap (kN/m)


Desak (buah)

Tarik (buah)

Desak (buah)

Tarik (buah)


SUMATERA BARAT

Type Balok B1 300 x 600Tulangan Lentur Gaya Geser Tulangan Geser

Tumpuan Lapangan

Jarak Sengkang (s), mm Jarak Sengkang (s), mm

4.00 178.817 175.496 13 2.00 100.00 135.004.00 178.817 175.496 13 2.00 100.00 135.004.00 178.817 175.496 13 2.00 100.00 135.004.00 178.817 175.496 13 2.00 100.00 135.00

Type Balok B1 Atap 250 x 4004.00 83.069 82.773 13 2.00 70.00 80.00

Susut (buah)

V.Tum (kN/m)

V.Lap (kN/m)


Sengkang (kaki)

D PERENCANAAN PONDASI `

Data Perencanaan PondasiPu = 168359 Kgf'c = 22.5 Mpa

qc = 200SF = 10Bk = 60 cmHk = 60 cmHp = 60 cmds = 7.095 cmd = 52.905 cm

I DIMENSI PONDASIRumus Dasar :

σ =PA

Maka :

A =P x SF

qc

=(168359) x 10

200

=1683590

200

= 8417.95

Bp = A

= 8417.95

= 91.7494 cmDigunakan dimensi Pondasi :Bp = 250 cmLp = 250 cm

II CEK GAYA GESER SATU ARAH

=Bp

-Bk

- d x Hk2 2

=250

-60

- 52.9 x 602 2

= 2525.7

Vu = xqcSF

= 2525.7 x20010

Kg/cm2

cm2

A1

cm2

A1

= 50514 Kg= 505.14 kN

Vn =Vuφ

=5050.6

= 842 kN

Vc=

1 f'c x b x d6

=1 25 x 2500 x 329.056

= 1045626.873 N= 1045.627 kN

Syarat : Vn < Vc842 < 1045.6269 kN Aman!

III CEK GAYA GESER DUA ARAHApabila Hk > Bk :bc = Hk

BkApabila Hk = Bk = 1

bo = (Bk + d) x 2 + (Bp + d) x 2= (60 + 32.905) x 2 + (250 + 32.905) x 2= 451.620 cm= 45162 mm

Vu =qc

(Bp x Hp) - (Bk+d) x (Hk+d)SF

=200

(275 x 275) - (60+52.905) x (60+52.905)10

= 1237252.461 Kg= 1237.252 kN

Vn =Vuφ

=1237.252

0.6= 2062.087435 kN

= 1 +bc

2 f'c x bo x d2

= 1 +1

2 25 x 45162 x 529.052

= ### N

Vc1

= 340002.7259 kN

= 4 f'c x bo x d

= 4 25 x 250 x 32.905

= 453336967.86 N= 453336.96786 kN

Vc = 340002.726 kN

Syarat :Vn < Vc2062.087 < 340002.73 kN Aman!

IV PERHITUNGAN TULANGAN

X = Bp - Bk2 2

= 250 - 602 2

= 95 cm

P = X x Hp (qc/SF)= 108 x 60 (200/10)= 114000 Kg= 1140 kN

Mu = P x 0.5X= 1290 x (0.5 x 1.08)= 542 kN/m

Mn = Muφ

= 542 0.8

= 676.875 kN/m

Mn = C x ZMn = 0.85 x f'c x b x a (d-a/2)

676875000 = 0.85 x 22.5 x 1000 a 529 - a2

Vc2

Diantara Vc1 dan Vc2 diambil nilai yang paling kecil :

676875000 = 19125 a 529 - a2

676875000 = 10118081.25 a -9562.5

Didapat persamaan :

1 -1058.10 a + 70784.314 = 0

Gunakan Rumus ABC :

a = b ± - 4ac2a

= 1058.10 ± 836438.3552

= 1058.10 ± 914.57002

= 986.335 mm

= 71.765 mm

= 71.765 mm

fy = 400 MpaMaka :T = CAs.fy = 0.85 x f'c x b x aAs.400 = 0.85 x 22.5 x 1000 x 71.765As = 1372505.258

400

= 3431.26

Jarak tulangan pokok :D = 22 mm

= 379.940

s = AØ x 1000As

= 379.940 x 10003431.26

= 111 mm ≈ 100 mm

Syarat :s ≤ 2 h

100 ≤ 2 x 600100 ≤ 1200 mm

a2

a2

b2

c1

c2


mm2

AØ mm2

Jarak tulangan susut :Menggunakan tulangan polosD = 16 mm

= 200.960

Asst = 0.002 x b x h= 0.002 x 1000 x 600= 1200

s = AØ x 1000100

= 200.960 x 10001200.00

= 167 mm ≈ 150 mm

Digunakan :Jarak (s) tulangan pokok : D22-100Jarak (s) tulangan susut : D16-150

AØ mm2

PERENCANAAN BALOK 1

A PERENCANAAN BALOK 1I TULANGAN LENTUR BALOK

Input Data Balokb 300 mmf'c 22.5 Mpa Data Momen Balok B1 dari SAP2000

fy 370 Mpa = 268.948 KNm

0.85 = 221.876 KNmφ 0.8ε 0.003

Es 200000

1 Tulangan TumpuanRasio Tulangan (ρ)

=x

600fy 600 + fy

=x

600370 600 + 370

= 0.0272

= ρ == 0,75 x 0,0272 = 0,5 x 0,0272 = 0.0204 = 0.0136

m = fy0,85 x f'c

= 3700,85 x 25

= 19.3464

= ρ x fy x (1-(0,5 x ρ x m))= 0.0136 x 370 x (1-(0,5 x 0,0136 x 19,3464))= 4.3668

Mu TumB1

β1 Mu LapB1

ρb 0,85 x β1 x f'c

0,85 x 0,85 x 22,5

ρmax 0,75 x ρb 0,5 x ρb


R1

= Mu/φ

=4.3668

= 76985736.126 mm

=b

= 76985736.126300

= 506.5759

Luas tarik effektif beton (SNI 03-2847-2002) Skema penampang (SNI 03-2847-2002)

Pusat berat tulangan baja desak : d' = 50 mmPusat berat tulangan baja tarik : z = 109 mm

== 615.5759 mm


= 615.5759 mm

bd2

R1

268,948 x106/0.8

dperlu bd2

mm2

hperlu dperlu + z

hperlu

Syarat

Diambil :

= 600 mm

Tinggi efektif desak : Tinggi efektif tarik :

= h - z = h - d'= 491 mm = 550.5 mm


== 0.5 x 4,3668= 2.1834

= Mu -

= φ

= = 268.948 - 157.9146= 157.914609 KNm 0.8

= 178.3 KNm

Luas tulangan perlu (As')

As' =fy x (d-d')

= 178270391.27489370 x (491-50)

= 1091.31Diameter tulangan rencana :Luas satu tulangan

Ø (mm) Coba tulangan D19 dengan AsØ=283,385 mm², maka :6 28.260 Jumlah tulangan perlu :8 50.240 n = As'

10 78.500 AsØ13 132.665 = 1091.3116 200.960 379.94019 283.385 = 3.8510 tulangan, ambil 4 tulangan22 379.94025 490.625

hpakai < hperlu

hpakai

ddesak dtarik

R2 γ x R1

M2 M1

M1 R2 x b x d2

2,1834 x 300 x (4912)

M2

mm2

AsØ (mm2)

=

= 1133.54

b Tulangan Tarik

= γ x ρ= 0.5 x 0.0135= 0.0068

As =

=


Ø (mm) Coba tulangan D19 dengan AsØ=283,385 mm², maka :6 28.260 Jumlah tulangan perlu :8 50.240 n = As'


=

= 2267.080

Dipakai 4D19


As'ada n x AsØ

mm2

ρbaru



(0.0068 x 300 x 550,5) + 1091,31

mm2

AsØ (mm2)

Dipakai 8D19


Asada n x AsØ

mm2

c

= 1133.540

= 2267.080

Cc = 0.85 x f'c x a x b a = 0.85c= 4876.875 c


c= 680124

x c - 40

cTs = As x fy

= 838819.6

Cc+ Cs -Ts = 04876.875 c + 680124

x c - 40 - 838819.6 = 0

c

4876.875 -158695.6 c - -33666138 = 0

-32.5404 c -6903.2194 = 0


c = b ± - 4ac2a

= 32.5404 ± 28671.7572

= 32.5404 ± 169.32742

= 100.934 mm

= -68.393 mm

= 100.93 mmfs' = c - d' x εc x Es

c= 100,93 - 50 x 0.003 x 200000

100.93= 305.748 Mpa < 370 Mpa Ok!

= fs'= 305.748 Mpa

a = 0.85 x c= 0.85 x 100,93= 85.794 mm

Kontrol Momen Tersedia1

Kontrol Momen Tersedia Mg-

As'ada mm2

Asada mm2

c2

c2

b2

c1

c2


fs'pakai


= 0.8 x 373,589= 298.871 KNm > Mu = 268.948 KNm Aman!

=b x d

= 2267.080300 x 491

= 0.0077

= = 0.0077

b x d = 0.0204

= 1133.540300 x 491 0.0077 < 0.0204 Aman!

= 0.0154

d

= 1133.540

= 2267.080d' = 50 mmz = 109 mm

= 550.50 mm

Cc = 0.85 x f'c x a x b a = 0.85c= 4876.875 c

Cs= As' x εc x Es x

c - zc

= 1360248 x c - 59c

Ts = As x fy= 566,77 x 370= 419409.8

Cc+ Cs -Ts = 04876.875 c + 1360248 x c - 59 - 419409.8 = 0

c

4876.875 940838.2 c -148267032 = 0

192.9183 c -30402.0571 = 0

Mg- Mn-

0.8Mg-

ρ'ada Asada'

ρada Asada ρada-ρ'ada

ρmax


Kontrol Momen Tersedia Mg+2

As'ada mm2

Asada mm2

dtarik

c2 +

c2 +


c = b ± - 4ac2a

= -192.9183 ± 158825.6802

= -192.9183 ± 398.52942

= 102.806 mm

= -295.724 mm

= 102.81 mmfs' = c - d' x εc x Es

c= 102,81 - 50 x 0.003 x 200000

102.81= 311.11 Mpa < 370 Mpa Ok!

= fs'= 311.105 Mpa

a = 0.85 x c= 0.85 x 102,81= 87.385 mm


= 0.8 x 607,454= 485.963 KNm

2Mu Lap = 221.876 KNm m = fy

0.85 x f'c

= = 370= 0.0204 0.85 x 22,5

= 19.3464

b = 300 mmh = 600 mm

= h - z= 491 mm

=x d

=x 491

= 303.7113402062 mm

b2

c1

c2


fs'pakai

Mg+ Mn-

0.8Mg+

Tulangan Lapangan Balok3

ρmax 0.75 x ρb

ddesak

cb εc x Esεc x Es x fy

0.003 x 2000000.003 x 200000 x 370

=== 193.62 mm

Mu = 0.85 x f'c x a x b x (d - a/2)φ

277.3446 = 5737.50 a x (491 - a/2)

277.3446 = 2817112.5 a -2868.75Didapat persamaan :

2868.75 -2817112.5 a 277344625 = 0

-982 a 96677.865 = 0


a = b ± - 4ac2a

= 982 ± 577612.542

= 982 ± 760.0082

= 871.004 mm

= 110.996 mm

= 110.996 mmTulangan yang dignakan yaitu Tulangan sebelah

Cc = Ts0.85 x f'c x a x b = As x fy0.85 x 22,5x 83,860 x 300 = As x 370636838.8324 = As 370

As = 1721.186

amax 0.75 x β x cb

0.75 x 0.8 x 303,7113402

a2

a2

a2

b2

a1

a2

Dari hitungan gunakan nilai a2 < amax = 193,62 mm

mm2

Perencanaan tulangan lapangan balok sama dengan perencanaan tulangan tumpuan balok. Momen yang digunakan adalah momen pada daerah daerah lapangan balok yang terbesar.


Ø (mm)6 28.260 Jumlah tulangan perlu :8 50.240 n = As'

10 78.50013 132.665 = 1721.18616 200.960 283.38519 283.385 = 6.0737 tulangan, ambil 7 tulangan22 379.940

25 490.625

= 1983.695

Kontrol momen tersediaCc = Ts

0.85 x f'c x a x b =Maka :

a =0.85 x f'c x b

= 127.925 mmMn = 0.85 x f'c x a x b x (d - a/2)

= 313.431659 KNm

= 250.745327 KNm > Mu = 221.876 KNm Aman!

AsØ (mm2) Coba tulangan D19 dengan AsØ= 283,385 mm², maka :

AsØ

Dipakai 7D19

Luas tulangan perlu (Asada)

Asada mm2

Asada x fy

Asada x fy

φMn

Perencanaan tulangan lapangan balok sama dengan perencanaan tulangan tumpuan balok. Momen yang digunakan adalah momen pada daerah daerah lapangan balok yang terbesar. Ukuran balok pada perencanaan tulangan lapangan mengikuti ukuran balok padaperencanaan tulangan tumpuan. diameter tulangan yang digunakan harus sama dengan diameter pada tulangan tumpuan.

Jumlah maksimum tulangan dalam satu barisb - d'

=251

= 5.693 ≈ 6 buah tulangan25 + D 44

Tulangan Susutfy = 370 Mpa

Tabel. Luas minimum tulangan susut (pasal 9.12)Mutu Baja (fy) Asst

0,0020.b.h > 0,0014.b.hBJTD-40 0,0018.b.h > 0,0014.b.h

≥BJTD-41 0,0018.b.h (400/fy) > 0,0014.b.h

Asst D10 =

= 324Diameter tulangan rencana :

Luas satu tulangan

Ø (mm) Jumlah tulangan susut :6 28.260 n = Asst D10

8 50.24010 78.500 = 32413 132.665 78.50016 200.960 = 4.1274 tulangan, ambil 4 tulangan19 283.385

22 379.940

25 490.625 = 314.000

Momen Kapasitas Balok

Tarik

=

=

= 283.385

== 8 x 283,385

= 2267.080

Desak

=

=

= 283.385

== 4x 283,385

≤BJTD-30

0.0018 x b x hbalok

mm2

Coba tulangan D10 dengan AsØ=78.50 mm², maka :

AsØ (mm2)

AsØ

Dipakai 4D10

Luas tulangan perlu (Aspakai)

Aspakai mm2

Momen kapasitas negatif (Mpr- )5

A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

Asada n x AsØ

mm2

A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

As'ada n x AsØ

= 1133.540 mm2

Skema penampang dalam keadaan lentur batas (SNI 03-2847-2002)

Cc = 0.85 x f'c x a x b= 4876.875 c

Cs =As' x εc x Es x

c - d'c

=680124.00 x

c - 40c

Ts = As x 1.25fy= 1048524.5

Cc+ Cs -Ts = 0

4876.875 c + 680124x c - 40

- 1048524.5 = 0c

4876.875 -368400.5 c -33666138 = 0

-75.5403 c -6903.2194 = 0

c2

c2


c = b ± - 4ac2a

c = 75.5403 ± 33319.2112

= 75.5403 ± 182.53552

= 129.038 mm

= -53.498 mm

= 129.038 mmfs' = c - d'

x εc x Esc

= 129,038 - 50x 0.003 x 200000

129.038= 369.8351 Mpa < 370 MPa Ok!

= fs'= 369.835 Mpa

a = 0.85 x c= 0.85 x 129,038= 109.682 mm

= (0.85 x f'c x a x b x (d - a/2)) + As' x fs' x (d - d'))= 459.562 KNm

= 367.650 KNm > Mu = 268.948 KNm Aman!

Tarik

=

=

= 283.385

== 4 x 283,385

= 1133.540Desak

=

=

= 283.385

b2

c1

c2


fs'pakai

MPR-

0,8MPR-

Momen kapasitas positif (Mpr+ )

A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

As'ada n x AsØ

mm2

A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

== 8 x 283,385

= 2267.080Cc = 0.85 x f'c x a x b

= 4876.875 c


c

= 170031 x c - 40

cTs = As x 1,25fy

= 1133,54 x 1,25(370)= 524262.25

Cc+ Cs -Ts = 0

4876.875 c + 170031 x c - 40

- 524262.25 = 0c

4876.875 -354231.25 c -18533379 = 0

-72.6349 c -3800.2571 = 0


c = b ± - 4ac2a

= 72.6349 ± 20476.8552

= 72.6349 ± 143.09742

= 107.866 mm

= -35.231 mm

= 107.866 mmfs' = c - d' x εc x Es

c= 107,866 - 50 x 0.003 x 200000

107.866= 324.659 MPa < 370 MPa Ok!

= fs'= 324.659 MPa

a = 0.85 x c= 0.85 x 107,866= 91.686 mm


= 249.255 KNm

Asada n x AsØ

mm2

c2+

c2 -

b2

c1

c2


fs'pakai

MPR+

0,8MPR+

II TULANGAN GESER BALOKInput Data Balok

459.562 kNm =

311.568 kNm = 7,00 - 0,60f'c 22.5 Mpa = 6.40 mfy 370 MPab 300 mm Data Geser Balok B1 dari SAP2000

h 600 mm = 205.416 kN

541 mm = -205.416 kN

7.00 m

0.60 mPerhitungan1 Gaya geser akibat beban gempa

Ve = (Mpr -) + (Mpr +) Ve = 120.4892 kNLnetto

= 459.562 + 311.568 A B

6.4= 120.489166 kN

2 Gaya geser akibat beban gravitasi 205.416 kN

= 205.416 kN Vg

= -205.416 kNA B

205.416 kN

3 Menghitung gaya geser ultimate (Vu)Vu = Ve + Vg Vu= 325.91 kN

= 325.905 kN

y = 255.48

x = 2h-84.927 kN

x = 1200 mm

MPR - Lnetto LB1- Lkolom

MPR+

Vmax B1

ddesak Vmin B1

LB1

Lkolom

Vmax B1

Vmin B1

a. Daerah dalam sendi plastisVc = 1/6 x f'c x b x d

= 1/6 x 22,5 x 300 x 541= 128.31 kN

Cek Jika Ve > 0,5 Vu Vc = 0120.4892 > 162.952583 Vc = 0

Vs = Vu- Vc

= 325.905- 0

0.75= 434.540221 kNm

Kontrol = 2/3 x f'c x b x d= 2/3 x 22,5x 300 x 541= 513.24 kNm

Syarat Vs< Kontrol434.5402 < 513.24

Jarak Begel (Sengkang)Jml. kaki = 2

= 10 mm

=

=

= 78.50

S =Vs

= 2 x 370 x 541 x 78,50434540.221461566

= 31426690434540.221

= 72.3217 mm ≈ 70 mm

0.25d = 135 mm8db = 152 mm24ds = 240 mm300 mm

digunakan P10 - 70b. Daerah luar sendi plastis

Vc = 1/6 x f'c x b x d= 1/6 x 22,5 x 300 x 541= 128 N

Cek Jika Ve > 0,5 Vu Vc = 0

φgeser

…Ok!

Dtulangan

AP10 0.25 x π x D2

0.25 x 3.4 x 102

mm2

Jml kaki x fy x d x AP10

Smax Daerah dalam sendi plastis

Maka dari perhitungan jarak sengkang dan syarat Smax diambil yang terkecil,

120.4892 > 162.952583 Vc = 0

Vs = y - Vc0.75

= 255.48 - 00.75

= 340.64 kNm

Kontrol = 2/3 x f'c x b x d= 2/3 x 22,5x 300 x 541= 513.237664 Nmm

Syarat Vs< Kontrol340.64 < 513.24

Jarak Begel (Sengakang)Jml. kaki = 2

= 10 mm

=

= 78.5

S =Vs

= 2 x 370 x 541 x 78,5340640.00

= 31426690340640.00

= 92.258 mm ≈ 90 mm

0.5d = 271 mm600 mm

digunakan P10 - 90

Jadi hasil dari perhitungan diatas didapat hasil sebagai berikut :Tul. TumpuanDesak 4D19Tarik 8D19Tul Lapangan7D19Tul. Susut4D10Tul. Geserdalam sendi plastis P10 - 70Luar sendi plastis P10 - 90

…Ok!

Dtulangan

AP13 0.25 x π x D2

mm2


Syarat Smax daerah luar sendi plastis


PERENCANAAN BALOK 2

A PERENCANAAN BALOKI TULANGAN LENTUR BALOK

Input Data Balokb 250 mmf'c 22.5 Mpa Data Momen Balok dari SAP2000

fy 370 Mpa = 163.132 KNm

0.85 = 155.3512 KNmφ 0.8ε 0.003

Es 200000


=x

600fy 600 + fy

=x

600370 600 + 370

= 0.0272

= ρ == 0,75 x 0,0272 = 0,5 x 0,0272 = 0.0204 = 0.0136

m=

fy0,85 x f'c

=370

0,85 x 22,5= 19.3464

= ρ x fy x (1-(0,5 x ρ x m))= 0.0136 x 370 x (1-(0,5 x 0,0136 x 19,3464))= 4.3668

= Mu/φ

=4.3668

= 46696153.553 mm

=b

= 46696153.553300

= 432.1859

Mu TumB2

β1 Mu LapB2

ρb 0,85 x β1 x f'c

0,85 x 0,85 x 22,5



R1

bd2

R1

163,132 x106/0.8

dperlu bd2

mm2



== 491.1859 mm


= 491.1859 mm

Syarat

Diambil :

= 500 mm


= h - z = h - d'= 441 mm = 460.5 mm


== 0.5 x 4,3668= 2.1834

= = Mu -

= φ= 106.158618 KNm = 163.132 - 106.1586

0.8

= 97.7564 KNm


As' =fy x (d-d')

= 97756381.7366102370 x (441-40)

= 658.05

hperlu dperlu + z

hperlu

hpakai < hperlu

hpakai

ddesak dtarik

R2 γ x R1

M1 R2 x b x d2 M2 M1

2,1834x 250 x (4412)

M2

mm2


Ø (mm) Coba tulangan D19 dengan AsØ=283,385 mm², maka :6 28.260 Jumlah tulangan perlu :8 50.240 n = As'10 78.500 AsØ13 132.665 = 658.0516 200.960 283.38519 283.385 = 2.3221 tulangan, ambil 2 tulangan22 379.94025 490.625

=

= 566.77

b Tulangan Tarik

= γ x ρ= 0.5 x 0.0136= 0.0068

As =

=


Ø (mm) Coba tulangan D19 dengan AsØ=283,385 mm², maka :6 28.260 Jumlah tulangan perlu :8 50.24010 78.500 n = As'13 132.665 AsØ16 200.960 = 1407.10919 283.385 283.38522 379.940 = 4.9654 tulangan, ambil 5 tulangan25 490.625

=

= 1416.925

c

= 566.770

= 1416.925

Cc = 0.85 x f'c x a x b a = 0.85c= 4064.0625 c

AsØ (mm2)

Dipakai 2D19


As'ada n x AsØ

mm2

ρbaru



(0.0068 x 250 x 441) + 658,05

mm2

AsØ (mm2)

Dipakai 5D19


Asada n x AsØ

mm2



As'ada mm2

Asada mm2


c= 340062

x c - 40

cTs = As x fy

= 524262.25

Cc+ Cs -Ts = 04064.0625 c + 340062

x c - 40 - 524262.25 = 0

c

4064.0625 -184200.25 c - -13432449 = 0

-45.3242 c -3305.1778 = 0


c = b ± - 4ac2a

= 45.3242 ± 15274.9912

= 45.3242 ± 123.59202

= 84.458 mm

= -39.134 mm

= 84.46 mm

fs' = c - d' x εc x Esc

= 89,91 - 40 x 0.003 x 20000084.46

= 319.387 Mpa < 370 Mpa Ok!

= fs'= 319.387 Mpa

a = 0.85 x c= 0.85 x 89,91= 71.789 mm


= 0.8 x 316,067= 169.383 KNm > Mu = 163.132 KNm Aman!

=b x d

= 1416.925250 x 441

= 0.0051

= = 0.0077

b x d = 0.0204= 566.770

c2

c2

b2

c1

c2


fs'pakai

Mg- Mn-

0.8Mg-

ρ'ada Asada'


ρmax

250 x 441= 0.0129 0.0077 < 0.0204 Aman!


d

= 566.770

= 1416.925d' = 40 mmz = 59 mm

= 460.50 mm

Cc = 0.85 x f'c x a x b a = 0.85c= 4064.0625 c


c - zc

= 850155 x c - 59c

Ts = As x fy= 566,77 x 370= 209704.9

Cc+ Cs -Ts = 04064.0625 c + 850155 x c - 59 - 209704.9 = 0

c

4064.0625 640450.1 c -50159145 = 0

157.5886 c -12342.1195 = 0Gunakan rumus ABC untuk mencari niala c :Rumus:

c = b ± - 4ac2a

= -157.5886 ± 74202.6592

= -157.5886 ± 272.40162

= 57.407 mm

= -214.995 mm

= 57.41 mmfs' = c - d' x εc x Es

c= 55,64 - 40 x 0.003 x 200000

55.64= 187.15 Mpa < 370 Mpa Ok!

= fs'= 187.155 Mpa

a = 0.85 x c= 0.85 x 57,407= 48.796 mm


= 0.8 x 213,387= 170.709 KNm


As'ada mm2

Asada mm2

dtarik

c2 +

c2 +

b2

c1

c2


fs'pakai

Mg+ Mn-

0.8Mg+

2Mu Lap = 155.3512 KNm m

=fy

0.85 x f'c

==

370= 0.0204 0.85 x 22,5

= 19.3464

b = 250 mmh = 500 mm

= h - z= 441 mm

=x d

=x 441

= 272.783505 mm

=== 173.90 mm

Mu = 0.85 x f'c x a x b x (d - a/2)φ

194.189 = 4781.25 a x (441 - a/2)

194.189 = 2108531.25 a -2390.63Didapat persamaan :

2390.63 -2108531.25 a 194189000 = 0

-882 a 81229.386 = 0


a = b ± - 4ac2a

= 882 ± 453006.45752

= 882 ± 673.0582

= 777.529 mm

= 104.471 mm


Cc = Ts0.85 x f'c x a x b = As x fy0.85 x 22,5x 104,471x 250 = As x 370


ρmax 0.75 x ρb

ddesak


0.003 x 2000000.003 x 200000 x 370

amax 0.75 x β x cb

0.75 x 0.8 x 272,7835052

a2

a2

a2

b2

a1

a2


Perencanaan tulangan lapangan balok sama dengan perencanaan tulangan tumpuan balok. Momen yang digunakan adalah momen pada daerah daerah lapangan balok yang terbesar.

499503.0589 = As 370

As = 1350.008 mm2


Ø (mm)6 28.260 Jumlah tulangan perlu :8 50.240 n

=As'

10 78.50013 132.665

=1350.008

16 200.960 283.38519 283.385 = 4.7639 tulangan, ambil 5 tulangan22 379.940

25 490.625

= 1416.925



a =0.85 x f'c x b

= 109.650 mmMn = 0.85 x f'c x a x b x (d - a/2)

= 202.4570738 KNm

= 161.965659 KNm > Mu = 155.351 KNm Aman!


=211

= 4.784 ≈ 5 buah tulangan25 + D 44



0,0020.b.h > 0,0014.b.hBJTD-40 0,0018.b.h > 0,0014.b.h

≥BJTD-41 0,0018.b.h (400/fy) > 0,0014.b.h

Asst D10 =

= 225


AsØ

Dipakai 5D19


Asada mm2

Asada x fy

Asada x fy

φMn

≤BJTD-30

0.0018 x b x hbalok

mm2



Ø (mm)6 28.260 Jumlah tulangan susut :8 50.240 n = Asst D10

10 78.50013 132.665 = 22516 200.960 78.50019 283.385 = 2.8662 tulangan, ambil 3 tulangan22 379.94025 490.625

= 235.500


Tarik

=

=

= 283.385

== 5 x 379.940

= 1416.925

Desak

=

=

= 283.385

== 2 x 379.940

= 566.770


Cc = 0.85 x f'c x a x b= 4064.0625 c


c - d'c

= 340062.00 x c - 40c

Ts = As x 1.25fy= 655327.8125

Cc+ Cs -Ts = 04064.0625 c + 340062 x c - 40 - 655327.8125 = 0

c

4064.0625 -315265.8125 c -13432449 = 0

AsØ (mm2) Coba tulangan D10 dengan AsØ=78.50 mm², maka :

AsØ

Dipakai 3D10


Aspakai mm2


A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

Asada n x AsØ

mm2

A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

As'ada n x AsØ

mm2

c2

-77.5741 c -3305.1778 = 0


c = b ± - 4ac2a

c = 77.5741 ± 19238.4452

= 77.5741 ± 138.70272

= 108.138 mm

= -30.564 mm

= 108.138 mmfs' = c - d' x εc x Es

c= 34,887 - 40 x 0.003 x 200000

108.138= 380.8364 Mpa < 400 MPa Ok!

= fs'= 380.836 Mpa

a = 0.85 x c= 0.85 x 770,038= 91.918 mm


= 208.220 KNm > Mu = 163.132 KNm Aman!

Tarik

=

=

= 283.385

== 2 x 379.940

= 566.770Desak

=

=

= 379.940

== 5x 379.940

= 1899.700Cc = 0.85 x f'c x a x b

= 4064.0625 c


c - d'c

= 227964x

c - 40

c2

b2

c1

c2


fs'pakai

MPR-

0,8MPR-


A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

As'ada n x AsØ

mm2

A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

Asada n x AsØ

mm2

x c

Ts = As x 1,25fy= 566,77 x 1,25(400)= 262131.125

Cc+ Cs -Ts = 04064.0625 c + 227964 x c - 40 - 262131.125 = 0

c

4064.0625 -34167.125 c -13449876 = 0

-8.4071 c -3309.4658 = 0


c = b ± - 4ac2a

= 8.4071 ± 13308.5432

= 8.4071 ± 115.36272

= 61.885 mm

= -53.478 mm

= 61.885 mmfs' = c - d' x εc x Es

c= 154.942 - 55 x 0.003 x 200000

61.885= 27.970 MPa < 400 MPa Ok!

= fs'= 27.970 MPa

a = 0.85 x c= 0.85 x 61,885= 52.602 mm


= 90.941 KNm

c2+

c2 -

b2

c1

c2


fs'pakai

MPR+

0,8MPR+

II TULANGAN GESER BALOK

Input Data Balok =

260.276 kNm = 4 - 0,60

113.677 kNm = 6.40 mf'c 22.5 Mpafy 370 MPa Data Geser Balok B2 dari SAP2000

b 250 mm = 161.946 kN

h 500 mm = -161.946 kN

441 mm

7.00 m

0.60 m

Perhitungan1 Gaya geser akibat beban gempa


= 260.276 + 113.677 A B6.4

= 58.43005852 kN


= 161.946 kN Vg

= -161.946 kNA B

161.946 kN3 Menghitung gaya geser ultimate (Vu)

Vu = Ve + Vg Vu= 220.38 kN= 220.376 kN

y = 174.11

x = 2h-103.516 kN

x = 1000 mm

Lnetto LB1- Lkolom

MPR -

MPR+

Vmax B2

Vmin B2

ddesak

LB2

Lkolom

Vmax B2

Vmin B2


= 1/6 x 22,5 x 250 x441= 87.16 kN


Vs = Vu - Vc

= 220.376 - 00.75

= 293.8347447 kNm

Kontrol = 2/3 x f'c x b x d= 2/3 x 22,5x 250 x441= 348.64 kNm

Syarat Vs < Kontrol293.83474 < 348.64


= 10 mm

=

=

= 78.50

S =Vs

= 2 x 370 x 441 x 78,5293834.74469859

= 25617690293834.7447

= 87.1840 mm ≈ 80 mm

φgeser

…Ok!

Dtulangan

AP10 0.25 x π x D2

0.25 x 3.4 x 102

mm2


0.25d = 110 mm8db = 152 mm24ds = 240 mm300 mm

digunakan P10-80b. Daerah luar sendi plastis

Vc = 1/6 x f'c x b x d= 1/6 x 22,5 x 250 x 441= 87 N


Vs = y - Vc0.75

= 174.11 - 00.75

= 232.1466667 kNm

Kontrol = 2/3 x f'c x b x d= 2/3 x 22,5x 250 x441= 348.641112 Nmm



= 10 mm

=

= 78.5

S =Vs

= 2 x 370 x441 x 78,5232146.67

= 25617690232146.67

= 110.351 mm ≈ 100 mm

0.5d = 221 mm600 mm

digunakan P10-100

Jadi hasil dari perhitungan diatas didapat hasil sebagai berikut :Tul. TumpuanDesak 2D19Tarik 5D19Tul Lapangan5D19Tul. Susut3D10Tul. Geserdalam sendi plastis digunakan P10-80



…Ok!

Dtulangan

AP10 0.25 x π x D2

mm2




Luar sendi plastis digunakan P10-100

PERENCANAAN BALOK 3


Input Data Balokb 200 mm Data Momen Balok dari SAP2000

f'c 22.5 Mpa = 45.26704 KNm

fy 370 Mpa = 26.72676 KNm

0.85φ 0.8ε 0.003

Es 200000


=x

600fy 600 + fy

=x

600370 600 + 370

= 0.0272

= ρ == 0,75 x 0,0272 = 0,5 x 0,0272 = 0.0204 = 0.0136

m = fy0,85 x f'c

= 3700,85 x 22.5

= 19.3464

= ρ x fy x (1-(0,5 x ρ x m))= 0.0136 x 370 x (1-(0,5 x 0,0136 x 19,3464))= 4.3668

= Mu/φ

=4.3668

= 12957584.353 mm

=b

= 12957584.353200

= 254.5347

Mu TumB3

Mu LapB3

β1

ρb 0,85 x β1 x f'c

0,85 x 0,85 x 22,5



R1

bd2

R1

45.26704 x106/0.8

dperlu bd2

mm2


Pusat berat tulangan baja desak d' = 40 mmPusat berat tulangan baja tarik z = 59 mm

== 313.5347 mm


= 313.5347 mmSyarat

Diambil :

= 300 mm


= h - z = h - d'= 241 mm = 260.50 mm


== 0.5 x4,3668= 2.1834

= = Mu-

= φ= 25.36308927805 KNm = 45.267

- 25.36310.8

= 31 KNmLuas tulangan perlu (As')

As' =fy x (d-d')

= 31220710.7219484370 x (241-40)


Ø (mm)6 28.260 Jumlah tulangan perlu :8 50.240 n = As'10 78.500 AsØ13 132.665 = 418.7616 200.960 283.38519 283.385 = 1.4777 tulangan, ambil 2 tulangan22 379.94025 490.625

hperlu dperlu + z

hperlu

hpakai < hperlu

hpakai

ddesak dtarik

R2 γ x R1

M1 R2 x b x d2 M2M1

2,1834 x 200 x (2412)

M2

mm2

AsØ (mm2) Coba tulangan D19 dengan AsØ=283,385 mm², maka :

=

= 566.77

b Tulangan Tarik

= γ x ρ= 0.5 x 0.0136= 0.0068

As =

=


Ø (mm) Coba tulangan D19 dengan AsØ=283,385 mm², maka :6 28.260 Jumlah tulangan perlu :8 50.240 n = As'10 78.500 AsØ13 132.665 = 746.24116 200.960 283.38519 283.385 = 2.6333 tulangan, ambil 4 tulangan22 379.94025 490.625

=

= 1133.540

c

= 566.770

= 1133.540

Cc = 0.85 x f'c x a x b a = 0.85c= 3251.25 c


c= 340062

x c - 40

cTs = As x fy

= 419409.8

Dipakai 2D19


As'ada n x AsØ

mm2

ρbaru



(0.0068 x 200 x 241) + 418.76

mm2

AsØ (mm2)

Dipakai 4D19


Asada n x AsØ

mm2



As'ada mm2

Asada mm2

Cc+ Cs -Ts = 03251.25 c + 340062

x c - 40 - 419409.8 = 0

c

3251.25 -79347.8 c - -13432449 = 0

-24.4053 c -4131.4722 = 0Gunakan rumus ABC untuk mencari niala c :Rumus:

c = b ± - 4ac2a

= 24.4053 ± 17121.5092

= 24.4053 ± 130.84922

= 77.627 mm

= -53.222 mm

= 77.63 mm

fs' =c - d'

x εc x Esc

=77.63 - 40

x 0.003 x 20000077.63

= 294.695 Mpa < 370 Mpa Ok!

= fs'= 294.695 Mpa

a = 0.85 x c= 0.85 x 77.63= 65.983 mm


= 0.8 x 316,067= 68.923 KNm > Mu = 45.267 KNm Aman!

=b x d

=1133.540200 x 241

= 0.0118

== 0.0118

b x d = 0.0204

=566.770

200 x 241 0.0118 < 0.0204 Aman!= 0.0235

d

= 566.770

= 1133.540d' = 40 mmz = 59 mm

= 260.50 mm

Cc = 0.85 x f'c x a x b a = 0.85c= 3251.25 c

Cs = As' x εc x Es x c - z

c= 680124 x c - 59

cTs = As x fy

= 566,77 x 370= 209704.9

c2

c2

b2

c1

c2


fs'pakai

Mg- Mn-

0.8Mg-

ρ'adaAsada'

ρadaAsada ρada-ρ'ada

ρmax



As'ada mm2

Asada mm2

dtarik

Cc+ Cs -Ts = 03251.25 c + 680124 x c - 59 - 209704.9 = 0

c

3251.25 470419.1 c -40127316 = 0

144.6887 c -12342.1195 = 0


c = b ± - 4ac

2a

= -144.6887 ± 70303.2952

=-144.6887 ± 265.1477

2

= 60.229 mm

= -204.918 mm

= 60.23 mm

fs' =c - d'

x εc x Esc

=60.23 - 40

x 0.003 x 20000060.23

= 206.51 Mpa < 370 Mpa Ok!

= fs'= 206.505 Mpa

a = 0.85 x c= 0.85 x 60.23= 51.195 mm


= 0.8 x 108,877= 78.185 KNm

2

Mu Lap = 26.72676 KNm m = fy0.85 x f'c

= = 370= 0.0204 0.85 x 22,5

= 19.3464b = 200 mmh = 300 mm

= h - z= 241 mm

= x d

=x 241

= 149.0721649485 mm

=== 95.03 mm

Mu = 0.85 x f'c x a x b x (d - a/2)φ

33.40845 = 3825.00 a x (241 - a/2)

33.40845 = 921825 a -1912.50Didapat persamaan :

1912.50 -921825 a 33408450 = 0

-482 a 17468.471 = 0

c2 +

c2 +

b2

c1

c2


fs'pakai

Mg+ Mn-

0.8Mg+


ρmax 0.75 x ρb

ddesak

cbεc x Es

εc x Es x fy0.003 x 200000

0.003 x 200000 x 370

amax 0.75 x β x cb

0.75 x 0.8 x 149.072165

a2

a2

a2


a = b ± - 4ac2a

= 482 ± 162450.122

= 482 ± 403.0512

= 442.526 mm

= 39.474 mm


Cc = Ts0.85 x f'c x a x b = As x fy0.85 x 22,5x 39,474 x 200 = As x 370150989.9410 = As 370

As = 408.081Diameter tulangan rencana :Luas satu tulangan


10 78.50013 132.665 = 408.08116 200.960 283.38519 283.385 = 1.4400 tulangan, ambil 2 tulangan22 379.94025 490.625

= 850.155Kontrol momen tersediaCc = Ts


a =0.85 x f'c x b

= 82.237 mmMn = 0.85 x f'c x a x b x (d - a/2)

= 62.87416103117 KNm

= 50.29932882494 KNm > Mu = 26.727 KNm Aman!


=161

= 3.648 ≈4 buah tulangan25 + D 44

b2

a1

a2


mm2


AsØ

Dipakai 2D19


Asada mm2

Asada x fy

Asada x fy

φMn


Secara teoritis, perencanaan ini tidak memerlukan tulangan desak karena a < amax , tetapi untuk dapat memasang sengkang, maka digunakan 4buah tulangan dengan diameter yang sama dengan diamater tulangan tarik


Tabel. Luas minimum tulangan susut (pasal 9.12)

Mutu Baja (fy) Asst0,0020.b.h > 0,0014.b.h

BJTD-40 0,0018.b.h > 0,0014.b.h≥BJTD-41 0,0018.b.h (400/fy) > 0,0014.b.h

Asst D10 =

= 108Diameter tulangan rencana :Luas satu tulangan

Ø (mm)6 28.260 Jumlah tulangan susut :8 50.240 n = Asst D10


= 157


Tarik

=

=

= 283.385

== 4 x 283,385

= 1133.540

Desak

=

=

= 283.385

== 2 x 283,385

= 566.770


Cc = 0.85 x f'c x a x b= 3251.25 c


c= 340062.00 x c - 40

cTs = As x 1.25fy

= 524262.25Cc+ Cs -Ts = 0

3251.25 c + 340062 x c - 40 - 524262.25 = 0

c

3251.25 -184200.25 c -13432449 = 0

≤BJTD-30

0.0018 x b x hbalok

mm2


AsØ

Dipakai 2D10


Aspakai mm2


A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

Asada n x AsØ

mm2

A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

As'ada n x AsØ

mm2

c2

-56.6552 c -4131.4722 = 0c2


c= b ± - 4ac

2a

c =56.6552 ± 19735.702

2= 56.6552 ± 140.4838

2

= 98.570 mm

= -41.914 mm

= 98.570 mm

fs' =c - d'

x εc x Esc

=98.570 - 40

x 0.003 x 20000098.570

= 359.5605 Mpa < 400 MPa Ok!

= fs'= 359.561 Mpa

a = 0.85 x c= 0.85 x 98.570= 83.784 mm


= 83.898 KNm > Mu = 45.267 KNm Aman!

Tarik

=

=

= 283.385

== 2x 283,385

= 566.770Desak

=

=

= 283.385

== 4 x 379.940

= 1133.540Cc = 0.85 x f'c x a x b

= 3251.25 c


c - d'c

= 170031x

c - 40c

Ts = As x 1,25fy= 1133,54 x 1,25(370)= 262131.125

Cc+ Cs -Ts = 03251.25 c + 170031 x c - 40 - 262131.125 = 0

c

3251.25 -92100.125 c -10031829 = 0

-28.3276 c -3085.5299 = 0


c = b ± - 4ac

2a

=28.3276 ± 13144.573

2

b2

c1

c2


fs'pakai

MPR-

0,8MPR-


A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

As'ada n x AsØ

mm2

A1D19 0.25 x π xd2

0.25 x 3.14 x192

mm2

Asada n x AsØ

mm2

c2+

c2 -

b2

=28.3276 ± 114.6498

2

= 71.489 mm

= -43.161 mm

= 71.489 mm

fs' =c - d'

x εc x Esc

=71.489 - 59

x 0.003 x 20000071.489

= 268.479 MPa < 400 MPa Ok!

= fs'= 268.479 MPa

a = 0.85 x c= 0.85 x 71.489= 60.765 mm


= 56.240 KNm


104.872 kNm =

70.300 kNm = 4,00 - 0,60f'c 22.5 Mpa = 3.40 mfy 370 MPab 200 mm Data Geser Balok dari SAP2000

h 300 mm = 45 kN

241 mm = -45 kN

4 m

0.60 m



= 104.872 + 70.300 A B3.4

= 51.52123552018 kN


= 45.115 kN Vg

= -45.115 kNA B


Vu = Ve + Vg Vu= 96.64 kN= 96.636 kN

6.406 kN y = 83.1

x = 2h

x = 600 mma. Daerah dalam sendi plastis

Vc = 1/6 x f'c x b x d= 1/6 x 22,5 x 200 x 241= 38.11 kN


Vs =Vu

- Vc

=96.636

- 0

c1

c2


fs'pakai

MPR+

0,8MPR+


MPR+

Vmax B3

ddesak Vmin B3

LB3

Lkolom

Vmax B3

Vmin B3

φgeser

=0.75

- 0

= 128.8483140269 kNm

Kontrol = 2/3 x f'c x b x d= 2/3 x 22,5x 200 x 241= 152.42 kNm



= 10 mm

=

=

= 78.50

S =Vs

=2 x 370 x 241 x 78.50

128848.314026908

=13999690

128848.3140269= 108.6525 mm ≈ 100 mm

0.25d = 60 mm8db = 152 mm24ds = 240 mm300 mm


Vc = 1/6 x f'c x b x d= 1/6 x 22,5 x 200 x 241= 38 N


Vs =y

- Vc0.75

=83.10

- 00.75

= 110.8 kNm




= 10 mm

=

= 78.5

S =Vs

= 2 x 370 x241 x 78.5110800.00

= 13999690110800.00

= 126.351 mm ≈ 120 mm

0.5d = 121 mm600 mm

digunakan P10-120

…Ok!

Dtulangan

AP10 0.25 x π x D2

0.25 x 3.14 x 102

mm2




…Ok!

Dtulangan

AP10 0.25 x π x D2

mm2




PERENCANAAN BALOK 4



fy 370 Mpa = 63.512 KNm

0.85 = 52.825 KNmφ 0.8ε 0.003Es 200000


=x

600fy 600 + fy

=x

600370 600 + 370

= 0.0272

= ρ == 0,75 x 0,0272 = 0,5 x 0,0272 = 0.0204 = 0.0136

m = fy0,85 x f'c

= 3700,85 x 22,5

= 19.3464

= ρ x fy x (1-(0,5 x ρ x m))= 0.0136 x 370 x (1-(0,5 x 0,0136 x 19,3464))= 4.3668

Mu TumB4

β1 Mu LapB4

ρb 0,85 x β1 x f'c

0,85 x 0,85 x 22,5



R1

A B C D E F G H I J K L M N O P Q R S1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

= Mu/φ

=4.3668

= 18180146.034 mm

=b

= 18180146.034200

= 301.4975



== 357.4975 mm


= 357.4975 mmSyarat

bd2

R1

63.511946 x106/0.8

dperlu bd2

mm2

hperlu dperlu + z

hperlu

A B C D E F G H I J K L M N O P Q R S

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

Diambil :

= 400 mm


= h - z = h - d'= 344 mm = 362.00 mm


== 0.5 x4,3668= 2.1834

= = Mu -

= φ= 51.675531 KNm = 63.512 - 51.6755

0.8= 27.71440121 KNm


As' =fy x (d-d')

= 27714401.2108088370 x (344-38)




=

= 401.92

hpakai < hperlu

hpakai

ddesak dtarik

R2 γ x R1

M1 R2 x b x d2 M2 M1

2,1834 x 200 x (3442)

M2

mm2


Dipakai 2D16


As'ada n x AsØ

mm2


74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

b Tulangan Tarik

= γ x ρ= 0.5 x 0.0136= 0.0068

As =

=


Ø (mm) Coba tulangan D16 dengan AsØ= 200,960 mm², maka :6 28.260 Jumlah tulangan perlu :8 50.240 n = As'


=

= 803.840

ρbaru



(0.0068 x 200 x 344) + 256,48

mm2

AsØ (mm2)

Dipakai 4D16


Asada n x AsØ

mm2


119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

142

143

144

c

= 401.920

= 803.840

Cc = 0.85 x f'c x a x b a = 0.85c= 3251.25 c


c= 241152

x c - 38

cTs = As x fy

= 297420.8

Cc+ Cs -Ts = 03251.25 c + 241152

x c - 38 - 297420.8 = 0

c

3251.25 -56268.8 c - -9163776 = 0

-17.3068 c -2818.5393 = 0


c = b ± - 4ac2a

= 17.3068 ± 11573.6832

= 17.3068 ± 107.58112

= 62.444 mm

= -45.137 mm

= 62.44 mmfs' = c - d' x εc x Es

c= 87.16 - 38 x 0.003 x 200000

62.44= 234.872 Mpa < 370 Mpa Ok!

= fs'= 234.872 Mpa

a = 0.85 x c= 0.85 x 87.16= 53.077 mm



As'ada mm2

Asada mm2

c2

c2

b2

c1

c2


fs'pakai


145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188

189

190


= 0.8 x 66,197= 74.670 KNm > Mu = 63.512 KNm Aman!

=b x d

= 803.840200 x 344

= 0.0058

= = 0.0058

b x d = 0.0204= 401.920

200 x 344= 0.0117 0.0058 < 0.0204 Aman!

d

= 401.920

= 803.840d' = 38 mmz = 56 mm

= 362.00 mm

Cc = 0.85 x f'c x a x b a = 0.85c= 3251.25 c


c - zc

= 482304 x c - 56c

Ts = As x fy= 401,92 x 370= 148710.4

Cc+ Cs -Ts = 03251.25 c + 482304 x c - 56 - 148710.4 = 0

c

3251.25 333593.6 c -27009024 = 0

102.6047 c -8307.2738 = 0

Mg- Mn-

0.8Mg-

ρ'ada Asada'


ρmax



As'ada mm2

Asada mm2

dtarik

c2 +

c2 +


191

192

193

194

195

196

197

198

199

200

201

202

203

204

205

206

207

208

209

210

211

212

213

214

215

216

217

218

219

220

221

222

223

224

225

226

227

228

229

230

231


c = b ± - 4ac2a

= -102.6047 ± 43756.8242

= -102.6047 ± 209.18132

= 53.288 mm

= -155.893 mm

= 53.288 mmfs' = c - d' x εc x Es

c= 53.90 - 38 x 0.003 x 200000

53.29= 172.14 Mpa < 370 Mpa Ok!

= fs'= 172.139 Mpa

a = 0.85 x c= 0.85 x 53.90= 45.295 mm


= 0.8 x 126,351= 82.901 KNm

2Mu Lap = 52.82515 KNm m = fy

0.85 x f'c

= = 370= 0.0204 0.85 x 22,5

= 19.3464b = 200 mmh = 400 mm

= h - z= 344 mm

b2

c1

c2


fs'pakai

Mg+ Mn-

0.8Mg+


ρmax 0.75 x ρb

ddesak


233

234

235

236

237

238

239

240

241

242

243

244

245

246

247

248

249

250

251

252

253

254

255

256

257

258

259

260

261

262

263

264

265

266

267

268

269

270

271

=x d

=x 344

= 212.78351 mm

=== 135.65 mm

Mu = 0.85 x f'c x a x b x (d - a/2)φ

66.03144 = 3825.00 a x (344 - a/2)


1912.50 -1315800 a 66031437.5 = 0

-688 a 34526.242 = 0


a = b ± - 4ac2a

= 688 ± 335239.03272

= 688 ± 578.9982

= 633.499 mm

= 54.501 mm


Cc = Ts0.85 x f'c x a x b = As x fy0.85 x 22,5x 55.072 x 200 = As x 370208465.7491 = As 370

As = 563.421


0.003 x 2000000.003 x 200000 x 370

amax 0.75 x β x cb

0.75 x 0.8 x 210,9278

a2

a2

a2

b2

a1

a2


mm2



272

273

274

275

276

277

278

279

280

281

282

283

284

285

286

287

288

289

290

291

292

293

294

295

296

297

298

299

300

301

302

303

304

305

306

307

308

309

310

311

312

313

314

315

316

317



10 78.50013 132.665 = 563.42116 200.960 200.96019 283.385 = 2.8036 tulangan, ambil 3 tulangan22 379.94025 490.625

= 602.880



a =0.85 x f'c x b

= 58.318 mmMn = 0.85 x f'c x a x b x (d - a/2)

= 70.230218 KNm

= 56.184175 KNm > Mu = 52.825 KNm Aman!


AsØ

Dipakai 3D16


Asada mm2

Asada x fy

Asada x fy

φMn

Perencanaan ini tidak memerlukan tulangan desak karena a < amax , tetapi untuk dapat memasang sengkang, maka digunakan 4buah tulangan dengan diameter yang sama dengan diamater tulangan tarik


319

320

321

322

323

324

325

326

327

328

329

330

331

332

333

334

335

336

337

338

339

340

341

342

343

344

345

346

347

348

349

350

351

352

353

354

355

356

357


=162

= 3.682 ≈ 4 buah tulangan25 + D 44



0,0020.b.h > 0,0014.b.hBJTD-40 0,0018.b.h > 0,0014.b.h

≥BJTD-41 0,0018.b.h (400/fy) > 0,0014.b.h

Asst D10 =


Ø (mm)6 28.260 Jumlah tulangan susut :8 50.240 n =Asst D10


= 157Momen Kapasitas Balok

Tarik

=

=

= 200.960

== 4 x 200,96

= 803.840

≤BJTD-30

0.0018 x b x hbalok

mm2


AsØ

Dipakai 2D10


Aspakai mm2


A1D16 0.25 x π xd2

0.25 x 3.14 x162

mm2

Asada n x AsØ

mm2


359

360

361

362

363

364

365

366

367

368

369

370

371

372

373

374

375

376

377

378

379

380

381

382

383

384

385

386

387

388

389

390

391

392

393

394

395

Desak

=

=

= 200.96

== 2 x 200,96

= 401.920


Cc = 0.85 x f'c x a x b= 3251.25 c


c - d'c

= 241152.00 x c - 38c

Ts = As x 1.25fy= 371776

Cc+ Cs -Ts = 03251.25 c + 241152 x c - 38 - 371776 = 0

c

3251.25 -130624 c -9163776 = 0

-40.1765 c -2818.5393 = 0


c = b ± - 4ac2a

c = 40.1765 ± 12888.3122

= 40.1765 ± 113.52672

= 76.852 mm

= -36.675 mm

A1D16 0.25 x π xd2

0.25 x 3.14 x162

mm2

As'ada n x AsØ

mm2

c2

c2

b2

c1

c2


397

398

399

400

401

402

403

404

405

406

407

408

409

410

411

412

413

414

415

416

417

418

419

420

421

422

423

424

425

426

427

428

429

430

431

432

433

434

435

436

437

438

439

440

441

442

443

444

= 76.852 mmfs' = c - d' x εc x Es

c= 76.852 - 38 x 0.003 x 200000

76.852= 303.3244 Mpa < 400 MPa Ok!

= fs'= 303.324 Mpa

a = 0.85 x c= 0.85 x 76.852= 65.324 mm


= 92.078 KNm > Mu = 63.512 KNm Aman!

Tarik

=

=

= 200.96

== 2x 200,96

= 401.920Desak

=

=

= 200.960

== 4 x 200,96

= 803.840Cc = 0.85 x f'c x a x b

= 3251.25 c


fs'pakai

MPR-

0,8MPR-


A1D16 0.25 x π xd2

0.25 x 3.14 x162

mm2

As'ada n x AsØ

mm2

A1D16 0.25 x π xd2

0.25 x 3.14 x162

mm2

Asada n x AsØ

mm2


445

446

447

448

449

450

451

452

453

454

455

456

457

458

459

460

461

462

463

464

465

466

467

468

469

470

471

472

473

474

475

476

477

478

479

480


c - d'c

= 120576x

c - 38c

Ts = As x 1,25fy= 803,84 x 1,25(370)= 185888

Cc+ Cs -Ts = 03251.25 c + 120576 x c - 38 - 185888 = 0

c

3251.25 -65312 c -6752256 = 0

-20.0883 c -2076.8185 = 0


c = b ± - 4ac2a

= 20.0883 ± 8710.8132

= 20.0883 ± 93.33172

= 56.710 mm

= -36.622 mm

= 56.710 mmfs' = c - d' x εc x Es

c= 56.710 - 38 x 0.003 x 200000

56.710= 197.955 MPa < 400 MPa Ok!

= fs'= 197.955 MPa

a = 0.85 x c= 0.85 x 56.710= 48.204 mm


= 60.152 KNm

c2+

c2 -

b2

c1

c2


fs'pakai

MPR+

0,8MPR+


482

483

484

485

486

487

488

489

490

491

492

493

494

495

496

497

498

499

500

501

502

503

504

505

506

507

508

509

510

511

512

513

514

515

516

517

518

519

520

521


115.097 kNm =

75.190 kNm = 7,00 - 0,60f'c 22.5 Mpa = 6.40 mfy 370 MPab 200 mmh 400 mm Data Geser Balok dari SAP2000

341 mm = 43.771 kN

7 m = -43.769 kN

0.6 m



= 115.097 + 75.190 A B6.4

= 29.732407 kN


= 66.678 kN Vg

= -66.678 kNA B


Vu = Ve + Vg Vu= 96.41 kN= 96.410 kN

y = 81.17

x = 2h

-36.946 kN x = 800 mma. Daerah dalam sendi plastis

Vc = 1/6 x f'c x b x d= 1/6 x 22,5 x 200 x344= 53.92 kN


MPR+

ddesak Vmax B3

LB4 Vmin B3

Lkolom

Vmax B4

Vmin B4


523

524

525

526

527

528

529

530

531

532

533

534

535

536

537

538

539

540

541

542

543

544

545

546

547

548

549

550

551

552

553

554

555

556

557

558

559

560


Vs = Vu - Vc

= 96.410 - 00.75

= 128.54721 kNm


Syarat Vs < Kontrol128.5472 < 215.67 Ok!


= 10 mm

=

=

= 78.50

S =Vs

= 2 x 370 x 344 x 78,5128547.209979995

= 19808690128547.21

= 154.0966 mm ≈ 150 mm

0.25d = 85 mm8db = 128 mm24ds = 240 mm300 mm


Vc = 1/6 x f'c x b x d= 1/6 x 22,5 x 200 x 344= 54 N


Vs = y - Vc0.75

φgeser

Dtulangan

AP10 0.25 x π x D2

0.25 x 3.14 x 102

mm2





562

563

564

565

566

567

568

569

570

571

572

573

574

575

576

577

578

579

580

581

582

583

584

585

586

587

588

589

590

591

592

593

594

595

596

597

598

599

600

601

602

603

604

605

606

= 81.17 - 00.75

= 108.22667 kNm




= 10 mm

=

= 78.5

S =Vs

= 2 x 370 x344 x 78.5108226.67

= 19808690108226.67

= 183.030 mm ≈ 180 mm

0.5d = 171 mm600 mm

digunakan P10-180

Jadi hasil dari perhitungan diatas didapat hasil sebagai berikut :Tul. TumpuanDesak Dipakai 2D16Tarik Dipakai 4D16Tul LapanganDipakai 3D16Tul. SusutDipakai 2D10Tul. Geserdalam sendi plastis digunakan P10-150Luar sendi plastis digunakan P10-180

…Ok!

Dtulangan

AP10 0.25 x π x D2

mm2





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PERENCANAAN BALOK 5



fy 370 Mpa = 25.6052 KNm

0.85 = 9.0919 KNmφ 0.8ε 0.003Es 200000


=x

600fy 600 + fy

=x

600370 600 + 370

= 0.0272

= ρ =

= 0,75 x 0,0272 = 0,5 x 0,0272

= 0.0204 = 0.0136

m = fy0,85 x f'c

= 370

0,85 x 22,5= 19.3464

= ρ x fy x (1-(0,5 x ρ x m))= 0.0136 x 370 x (1-(0,5 x 0,0136 x 19,3464))= 4.3668

Mu TumB5

β1 Mu LapB5

ρb 0,85 x β1 x f'c

0,85 x 0,85 x 22,5



R1


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34

35

36

37

38

= Mu/φ

=4.3668

= 7329428.628 mm

=b

= 7329428.628200

= 191.4344



== 247.4344 mm


= 247.4344 mmSyarat

bd2

R1

25,6052 x106/0.8

dperlu bd2

mm2

hperlu dperlu + z

hperlu


39

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71

72

Diambil :

= 300 mm


= h - z = h - d'= 244 mm = 262.00 mm


== 0.5 x4,3668= 2.1834

= Mu -

= φ

= = 25.605 - 25.9985= 25.998465647 KNm 0.8

= 6.00803435 KNmLuas tulangan perlu (As')

As' =fy x (d-d')

= 6008034.35274737370 x (244 - 38)




=

= 401.92

b Tulangan Tarik

= γ x ρ= 0.5 x 0.0135= 0.0068

As =

=

= 410.382

hpakai < hperlu

hpakai

ddesak dtarik

R2 γ x R1

M2 M1

M1 R2 x b x d2

2,1834 x 200 x (2412)

M2

mm2

AsØ (mm2)

Dipakai 2D16


As'ada n x AsØ

mm2

ρbaru



(0.0068 x 200 x 244) + 78,82

mm2


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=

= 602.880

AsØ (mm2)

Dipakai 3D16


Asada n x AsØ

mm2


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129

130

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132

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139

140

c

= 401.920

= 602.880

Cc = 0.85 x f'c x a x b a = 0.85c= 3251.25 c


c= 241152

x c - 38

cTs = As x fy

= 223065.6

Cc+ Cs -Ts = 03251.25 c + 241152

x c - 38 - 223065.6 = 0

c

3251.25 18086.4 c - -9163776 = 0

5.5629 c -2818.5393 = 0


c = b ± - 4ac2a

= -5.5629 ± 11305.1032

= -5.5629 ± 106.32552

= 50.381 mm

= -55.944 mm

= 50.38 mmfs' = c - d' x εc x Es

c= 50,38 - 38 x 0.003 x 200000

50.38= 147.451 Mpa < 370 Mpa Ok!

= fs'= 147.451 Mpa

a = 0.85 x c= 0.85 x 50,38= 42.824 mm



As'ada mm2

Asada mm2

c2

c2

b2

c1

c2


fs'pakai


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185

186


= 0.8 x 47,898= 38.935 KNm> Mu = 25.605 KNm Aman!

=b x d

= 602.880200 x 244

= 0.0082

= = 0.0041

b x d = 0.0204

= 401.920200 x 244 0.0041 < 0.0204 Aman!

= 0.0124

d

= 401.920

= 602.880d' = 38 mmz = 56 mm

= 262.00 mm

Cc = 0.85 x f'c x a x b a = 0.85c= 3251.25 c


c - zc

= 361728 x c - 56c

Ts = As x fy= 401,92 x 370= 148710.4

Cc+ Cs -Ts = 03251.25 c + 361728 x c - 56 - 148710.4 = 0

c

3251.25 213017.6 c -20256768 = 0

65.5187 c -6230.4554 = 0

Mg- Mn-

0.8Mg-

ρ'ada Asada'


ρmax



As'ada mm2

Asada mm2

dtarik

c2 +

c2 +


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c = b ± - 4ac2a

= -65.5187 ± 29214.5192

= -65.5187 ± 170.92262

= 52.702 mm

= -118.221 mm

= 52.70 mmfs' = c - d' x εc x Es

c= 52,70 - 38 x 0.003 x 200000

52.70= 167.38 Mpa < 370 Mpa Ok!

= fs'= 167.378 Mpa

a = 0.85 x c= 0.85 x 52,70= 44.797 mm


= 0.8 x 63,659= 50.927 KNm

2Mu Lap = 9.0919 KNm m

=fy

0.85 x f'c

==

370= 0.0204 0.85 x 22,5

= 19.3464b = 200 mmh = 300 mm

= h - z= 244 mm

b2

c1

c2


fs'pakai

Mg+ Mn-

0.8Mg+


ρmax 0.75 x ρb

ddesak


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=x d

=x 244

= 150.92783505 mm

=== 96.22 mm

Mu = 0.85 x f'c x a x b x (d - a/2)φ

11.364875 = 3825.00 a x (244 - a/2)


1912.50 -933300 a 11364875 = 0

-488 a 5942.418 = 0


a = b ± - 4ac2a

= 488 ± 214374.32682

= 488 ± 463.0062

= 475.503 mm

= 12.497 mm


Cc = Ts0.85 x f'c x a x b = As x fy0.85 x 22,5x 12,497 x 200 = As x 370

47801.4984 = As 370

As = 129.193


0.003 x 2000000.003 x 200000 x 370

amax 0.75 x β x cb

0.75 x 0.8 x 150,9278351

a2

a2

a2

b2

a1

a2


mm2



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Ø (mm)

6 28.2608 50.240 Jumlah tulangan perlu :

10 78.500 n = As'

13 132.66516 200.960 = 129.19319 283.385 200.96022 379.940 = 0.6429 tulangan, ambil 2 tulangan25 490.625

= 401.920



a =0.85 x f'c x b

= 38.879 mmMn = 0.85 x f'c x a x b x (d - a/2)

= 33.394516284 KNm

= 26.715613027 KNm > Mu = 9.092 KNm Aman!

AsØ (mm2)

Coba tulangan D16 dengan AsØ= 200,96 mm², maka :

AsØ

Dipakai 2D16


Asada mm2

Asada x fy

Asada x fy

φMn

Perencanaan ini tidak memerlukan tulangan desak karena a < amax , tetapi untuk dapat memasang sengkang, maka digunakan 4buah tulangan dengan diameter yang sama dengan diamater tulangan tarik


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354


=162

= 3.951 ≈ 4 buah tulangan25 + D 41



0,0020.b.h > 0,0014.b.hBJTD-40 0,0018.b.h > 0,0014.b.h

≥BJTD-41 0,0018.b.h (400/fy) > 0,0014.b.h

Asst D10 =


Ø (mm)6 28.260 Jumlah tulangan susut :8 50.240 n =Asst D10


= 157

≤BJTD-30

0.0018 x b x hbalok

mm2


AsØ

Dipakai 2D10


Aspakai mm2


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Tarik

=

=

= 200.960

== 3 x 200,96

= 602.880

Desak

=

=

= 200.96

== 2 x 200,96

= 401.920


Cc = 0.85 x f'c x a x b= 3251.25 c


c - d'c

= 241152.00 x c - 38c

Ts = As x 1.25fy= 278832

Cc+ Cs -Ts = 03251.25 c + 241152 x c - 40 - 278832 = 0

c

3251.25 -37680 c -9163776 = 0

-11.5894 c -2818.5393 = 0


A1D16 0.25 x π xd2

0.25 x 3.14 x162

mm2

Asada n x AsØ

mm2

A1D16 0.25 x π xd2

0.25 x 3.14 x162

mm2

As'ada n x AsØ

mm2

c2

c2


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c = b ± - 4ac2a

c = 11.5894 ± 11408.4712

= 11.5894 ± 106.81042

= 59.200 mm

= -47.611 mm

= 59.200 mmfs' = c - d' x εc x Es

c= 59,2 - 83 x 0.003 x 200000

59.200= 214.8643 Mpa < 400 MPa Ok!

= fs'= 214.864 Mpa

a = 0.85 x c= 0.85 x 59,2= 50.320 mm


= 47.929 KNm > Mu = 25.605 KNm Aman!

Tarik

=

=

= 200.96

== 2x 200,96

= 401.920Desak

=

=

= 200.960

b2

c1

c2


fs'pakai

MPR-

0,8MPR-


A1D16 0.25 x π xd2

0.25 x 3.14 x162

mm2

As'ada n x AsØ

mm2

A1D16 0.25 x π xd2

0.25 x 3.14 x162

mm2


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474

== 3 x 200,96

= 602.880Cc = 0.85 x f'c x a x b

= 3251.25 c


c - d'c

= 120576x

c - 38c

Ts = As x 1,25fy= 401,920 x 1,25(370)= 185888

Cc+ Cs -Ts = 03251.25 c + 120576 x c - 38 - 185888 = 0

c

3251.25 -65312 c -6752256 = 0

-20.0883 c -2076.8185 = 0


c = b ± - 4ac2a

= 20.0883 ± 8710.8132

= 20.0883 ± 93.33172

= 56.710 mm

= -36.622 mm

= 56.710 mmfs' = c - d' x εc x Es

c= 56,71 - 38 x 0.003 x 200000

56.710= 197.955 MPa < 400 MPa Ok!

= fs'= 197.955 MPa

a = 0.85 x c= 0.85 x 56,71= 48.204 mm


= 42.219 KNm

Asada n x AsØ

mm2

c2+

c2 -

b2

c1

c2


fs'pakai

MPR+

0,8MPR+


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477

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II TULANGAN GESER BALOK

Input Data Balok =

59.911 kNm = 4,00 - 0,60

52.774 kNm = 3.40 mf'c 22.5 Mpafy 370 MPa Data Geser Balok dari SAP2000

b 200 mm = 32.188 kN

h 300 mm = -32.188 kN

241 mm

4 m

0.6 m



= 59.911 + 52.774 A B3.4

= 33.142644434 kN


= 66.678 kN Vg

= -66.678 kNA B


Vu = Ve + Vg Vu= 99.82 kN= 99.821 kN

y = 79.82

x = 2h-33.535 kN

x = 600 mm

Lnetto LB5- Lkolom

MPR -

MPR+

Vmax B3

Vmin B3

ddesak

LB5

Lkolom

Vmax B4

Vmin B4


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555


= 1/6 x 22,5 x 200 x241= 38.11 kN


Vs = Vu - Vc

= 99.821 - 00.75

= 133.09419258 kNm




= 10 mm

=

=

= 78.50

S =Vs

= 2 x 370 x 241 x 78,5133094.19257932

= 13999690133094.19258

= 105.1863 mm ≈ 100 mm

0.25d = 60 mm8db = 128 mm24ds = 240 mm300 mm

b. Daerah luar sendi plastisVc = 1/6 x f'c x b x d

= 1/6 x 22,5 x 200 x 241= 38 N


Vs = y - Vc0.75

= 79.82 - 00.75

= 106.4267 kNm

φgeser

Dtulangan

AP10 0.25 x π x D2

0.25 x 3.14 x 102

mm2



Maka dari perhitungan jarak sengkang dan syarat Smax diambil yang terkecil, digunakan P10-100


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= 10 mm

=

= 78.5

S =Vs

= 2 x 370 x241 x 78,5106426.67

= 13999690106426.67

= 131.543 mm ≈ 130 mm

0.5d = 121 mm600 mm

digunakan P10-130

Jadi hasil dari perhitungan diatas didapat hasil sebagai berikut :Tul. TumpuanDesak Dipakai 2D16Tarik Dipakai 3D16Tul LapanganDipakai 2D16Tul. SusutDipakai 2D10Tul. Geserdalam sendi plastis digunakan P10-100Luar sendi plastis digunakan P10-130

Dtulangan

AP10 0.25 x π x D2

mm2





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Tugas BesarPerancangan Bangunan Teknik Sipil

FST-UTY

PLAT LANTAI 7 X 3.5 MPERHITUNGAN BEBAN PLAT LANTAI

Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung (PPIUG)1983a. Beban Mati (Qd)

Komponen Platvolume berat jenis hasil perkalian

1.Beton (12cm) 0.13 2400 3.12 kN/m²2.Spesi (2 cm) 2 21 0.42 kN/m²3.Pasir (3cm) 0.03 1800 0.54 kN/m²4.Keramik (0,5 cm) 0.5 24 0.12 kN/m²5Plafon+penggantung 0.18 kN/m²6.ME 0.25 kN/m²

Qd = 4.63 kN/m²

b.Beban Hidup (Ql)Beban hidup pada lantai gedung kampus adalah

Ql = 2.5 kN/m²

c. Beban Ultimate (qu)qu = 1.2 Qd+1.6Ql

= 9.556 kN/m²Perhitungan

Analisis dimensi platLy = 7 mLx = 3.5 m

= 2 dua arahTabel untuk menentukan momen plat (PBI - 1971)

Ly/Lx 2 Analisis tebal minimum platClx 41 S = Lx x 1000Cly 12 S = 3500 mmCtx 83Cty 57 h = S

28h = 125 mm

jadi dipakai 130 mm

Syarat tebal plat ≥ 120mmDipakai tebal plat

h = 125 mmjadi dipakai = 130 mm

f'c (Mpa) = 22.5 Mpafy (Mpa) = 240 MPa karena tulangan polos


FST-UTY

d.PERHITUNGAN MOMEN0.001 x qU x Clx x Lx² = 4.799501 KNm

0.001 X qU X Cly X Lx² = 1.404732 KNm

0.001 X qU X Ctx X Lx² = 9.716063 KNm

0.001 X qU X Cty X Lx² = 6.672477 KNm

1. Penulangan Arah Xa.Tulangan lapangan

4.799501 KNm

Rencana diameter tulangan (Ø) mm = 10 mmSelimut beton (Pb) = 25 mm

ds x = Pb + 0,5xØ= 30 mm

d = h plat - ds= 100 mm

Faktor momen pikul (K)interpolasi k maks untuk mencari nilai k maks dari f'c =22.5 dan fy = 370 data dari tabel faktor momen pikul maksimal dari Mpa

f'c20 5.3828

22.5 x25 6.7295

5=

1.31672.5 6.7295 -x

5x = 3.36075 - 33.64755x = -30.28675

x = 6.05735

K = Mu

= 0.599937625 Mpa < 6.057 Mpa aman

a = (1- √1 - (2 x K / 0.85 x f'c)) x d= 3.187737097 mm

Mlx ⁺=

Mly ⁺=

Mtx ¯=

Mty ¯=

Mlx ⁺=

Ø x b x d


FST-UTY

Perhitungan Tulangan PokokAs = (0.85 x f'c x a x b)/fy

= 254.0227999

Luas tulangan minimal platTulangan pokok (Pasal 12.5.1. SNI 04-2847-2002)f'c < 31.36Mpa

= (1.4 x b x d)/fy= 583.3333333

dipilih yang terbesar, jadi As pakai := 583.3333333

Perhitungan Jarak Tulangan (S)

Jarak maksimal tulangan (Pokok)Plat 1 arah : S ≤ 3.h dan S ≤ 450mm (Pasal 12.5.4)Plat 2 arah : S ≤ 2.h dan S ≤ 450mm (Pasal 15.3.2)

S = (1/4 x π x D² x S)/As,u= 134.5714286 mm

untuk mempermudah proses pemasangan tul. S pakai :S pakai = 100 mm < 2.h

= 100 mm < 260 aman

Luas tulangan = (1/4 x π x D² x S)/S= 785 > As,u 583.3 aman

di gunakan P10-140b.Perhitungan tulangan tumpuan

= 9.716063 KNmRencana diameter tulangan (Ø) = 10 mm

K = Mu

= 1.214507875 Mpa < 6.057 Mpa aman

a = (1- √1 - (2 x K / 0.85 x f'c)) x d= 6.565924817 mm


= 523.2221339


As,u = (1.4 x b x d)/fy= 583.3333333

mm²

As ≥(1.4xbxd)/fy

As,umm²

As,u mm²

Mtx¯

Ø x b x d

mm²

As,u ≥(1.4xbxd)/fy

mm²


FST-UTY

dipilih yang terbesar, jadi As pakai :As,u = 583.3333333

Perhitungan Jarak Tulangan (S)S = (1/4 x π x D² x S)/As,u

= 134.5714286 mmuntuk mempermudah proses pemasangan tul. S pakai :

S pakai = 100 mm < 2.h 260 amanLuas tulangan = (1/4 x π x D² x S)/S

= 785 > As,u 583.3 amandi gunaka P10-140

b.Perhitungan tulangan bagi (tul.susut)Rencana diameter tulangan (Ø) = 8 mm

Luas tulangan minimal platTulangan bagi/tul.susut (Pasal 9.12.2.1. SNI 04-2847-2002)fy < 300Mpafy = 400Mpafy > 400Mpa

Jarak maksimal tulangan (Pokok)S ≤ 5.h dan S ≤ 450mm (Pasal 9.12.2.2)

Asb = 20% x As,u= 116.6666667

Asb = 0.002.b.h= 260

dipilih yang terbesar, maka Asb pakai :Asb,u = 260

Perhitungan Jarak Tulangan (S)S = (1/4 x π x D² x S)/Asb,u


S = 150 mm < 5.h 650150 mm < 650 mm aman

Luas tulangan = (1/4 x π x D² x S)/S= 334.9333333 > Asb,u 260= 334.9333333 mm² > 260 mm² aman

Dari perhitungan diatas, maka dipakai tul.pokok :D 10 - 100

Dari perhitungan diatas, maka dipakai tul.susut :D 8 - 150

2. Penulangan Arah Ya.Tulangan lapangan

= 1.404732 KNm

Rencana diameter tulangan (Ø) = 10 mmSelimut beton (Pb) = 20 mm

ds y = Pb + Ø+0,5xØ= 35 mm

d = h plat - ds= 130 - 35= 95 mm

mm²

Asb ≥ 0.0020.b.hAsb ≥ 0.0018.b.hAsb ≥ 0.0018.b.h.(400/fy)

mm²

mm²

mm²

mm²

Mly⁺


FST-UTY

K = Mu

= 0.194561219 Mpa < 6.057a = (1- √1 - (2 x K / 0.85 x f'c)) x d

= 0.971414442 mm


= 77.40958838Luas tulangan minimal platTulangan pokok (Pasal 12.5.1. SNI 04-2847-2002)f'c < 31.36Mpa

As,u = (1.4 x b x d)/fy= 554.1666667





= 785 > As,u 554.167 aman

b.Perhitungan tulangan tumpuan= 6.672477 KNm

Rencana diameter tulangan (Ø) = 10

K = Mu

= 0.924165789 Mpa < 6.057a = (1- √1 - (2 x K / 0.85 x f'c)) x d

= 4.707249547 mmPerhitungan Tulangan Pokok

As = (0.85 x f'c x a x b)/fy= 375.1089483

Luas tulangan minimal platTulangan pokok (Pasal 12.5.1. SNI 04-2847-2002)

f'c < 31.36MpaAs,u = (1.4 x b x d)/fy

= 554.1666667



= 141.6541353 mm

Ø x b x d²

mm²


mm²

mm²

mm²

Mty¯

Ø x b x d

mm²


mm²

mm²


FST-UTY

untuk mempermudah proses pemasangan tul. S pakai :S pakai = 100 mm < 2.h 260 aman

Luas tulangan = (1/4 x π x D² x S)/S= 785 > As,u 554.167 aman

b.Perhitungan tulangan bagi (tul.susut)Rencana diameter tulangan (Ø) = 8 mmLuas tulangan minimal platTulangan bagi/tul.susut (Pasal 9.12.2.1. SNI 04-2847-2002)fy < 300Mpafy = 400Mpafy > 400MpaJarak maksimal tulangan (Pokok)S ≤ 5.h dan S ≤ 450mm (Pasal 9.12.2.2)

Asb = 20% x As,u= 20% x 554.1666667= 110.8333333

Asb = 0.002.b.h= 260



= 193.2307692 mm

untuk mempermudah proses pemasangan tul. S pakai :S = 150 mm < 5.h 650 aman

Luas tulangan = (1/4 x π x D² x S)/S= 334.9333333 > Asb,u 260 aman

Dari perhitungan diatas, maka dipakai tul.pokok :D 10 - 100

Dari perhitungan diatas, maka dipakai tul.susut :D 8 - 150

mm²


mm²

mm²

mm²

mm²


FST-UTY

PLAT LANTAI 7 X 3.5 MPERHITUNGAN BEBAN PLAT LANTAI

Berdasarkan Peraturan Pembebanan Indonesia Untuk Gedung (PPIUG)1983a. Beban Mati (Qd)

Komponen Platvolume berat jenis

Beton (12cm) 0.13 2400 3.12 kN/m²Spesi (2 cm) 2 21 0.42 kN/m²Pasir (3cm) 0.03 1800 0.54 kN/m²Keramik (0,5 cm) 0.5 24 0.12 kN/m²Plafon+penggantung 0.18 kN/m²MEE 0.25 kN/m²

Qd = 4.63 kN/m²

b. Beban Hidup (Ql)Beban hidup pada lantai gedung kampus adalah

Ql = 2.5 kN/m²


= 9.556 kN/m²perhitunganAnalisis dimensi plat

Ly = 4 mLx = 3.5 m

= 1.14285714 dua arah

Tabel untuk menentukan momen plat (PBI - 1971)Ly/Lx 1.142857143

Clx 32 Analisis tebal minimum platCly 18.7 S = Lx x 1000Ctx 70.3 3500 mmCty 57

h = s28

125 mm


jadi h dipakai h = 125 mm= 130 mm

f'c (Mpa) = 22.5 Mpafy (Mpa) = 240 MPa karena tulangan polos


FST-UTY

d. PERHITUNGAN MOMEN0.001 X qU X Clx X Lx²

= 3.745952 KNm0.001 X qU X Cly X Lx²

= 2.1890407 KNm0.001 X qU X Ctx X Lx²

= 8.2293883 KNm0.001 X qU X Cty X Lx²

= 6.672477 KNm

1. Penulangan Arah Xa. Tulangan lapangan

3.745952 KNmRencana diameter tulangan (Ø) mm = 10Selimut beton (Pb) = 20 mm

ds x = Pb + 0,5xØ ds y = Pb + Ø+0,5xØ= 25 = 35

d = h plat - ds d = h plat - ds= 105 mm = 95 mm

Faktor momen pikul (K)K = Mu

= 0.42471 < K maksa = (1- √1 - (2 x K / 0.85 x f'c)) x d

= 2.35823 mm

Mlx ⁺=

Mly ⁺=

Mtx ¯=

Mty ¯=

Mlx ⁺=

Ø x b x d


FST-UTY



= (1.4 x b x d)/fy= 612.5


Perhitungan Jarak Tulangan (S)Jarak maksimal tulangan (Pokok)Plat 1 arah : S ≤ 3.h dan S ≤ 450mm (Pasal 12.5.4)Plat 2 arah : S ≤ 2.h dan S ≤ 450mm (Pasal 15.3.2)

S = (1/4 x π x D² x S)/As,u= 128.163 mm


Luas tulangan = (1/4 x π x D² x S)/S= 560.714 > As,u 612.5 aman

b. Perhitungan tulangan tumpuan= 8.22939 KNm

Rencana diameter tulangan (Ø) mm = 10

K = Mu

= 0.93304 Mpa < Kmaksa = (1- √1 - (2 x K / 0.85 x f'c)) x d


As = (0.85 x f'c x a x b)/fy= 418.679


As,u = (1.4 x b x d)/fy= 612.5



= 128.163 mm

mm²

As ≥(1.4xbxd)/fyAs,u

mm²

As,u mm²

Mtx¯

Ø x b x d

mm²


mm²

mm²


FST-UTY


Luas tulangan = (1/4 x π x D² x S)/S= 560.714 > As,u 612.5 td aman

b. Perhitungan tulangan bagi (tul.susut)Rencana diameter tulangan (Ø) mm = 8Luas tulangan minimal platTulangan bagi/tul.susut (Pasal 9.12.2.1. SNI 04-2847-2002)fy < 300Mpafy = 400Mpafy > 400MpaJarak maksimal tulangan (Pokok)S ≤ 5.h dan S ≤ 450mm (Pasal 9.12.2.2)

Asb = 20% x As,u= 122.5

Asb = 0.002.b.h= 260




S = 200 mm < 5.h 650 aman

Luas tulangan = (1/4 x π x D² x S)/S= 251.2 > Asb,u 260 td aman

Dari perhitungan diatas, maka dipakai tul.pokok : D 10 - 140Dari perhitungan diatas, maka dipakai tul.susut : D 8 - 200

2. Penulangan Arah Ya. Tulangan lapangan

= 2.18904 KNmRencana diameter tulangan (Ø) mm = 10Selimut beton (Pb) = 20 mm

ds y = Pb + Ø+0,5xØ= 35 mm


K = Mu

= 0.30319 Mpa < Kmaks


mm²

mm²

mm²

mm²

Mly⁺

Ø x b x d²


FST-UTY

a = (1- √1 - (2 x K / 0.85 x f'c)) x d= 1.51818 mm



As,u = (1.4 x b x d)/fy= 554.167





= 523.333 > As,u 554.167td aman



K = Mu



As = (0.85 x f'c x a x b)/fy= 375.109


As,u = (1.4 x b x d)/fy= 554.167



= 141.654 mm

mm²


mm²

mm²

mm²

Mty¯

Ø x b x d

mm²


mm²

mm²


FST-UTY


Luas tulangan = (1/4 x π x D² x S)/S= 523.333 > As,u 554.167td aman


Asb = 20% x As,u= 110.833

Asb = 0.002.b.h= 260




S = 200 mm < 5.h 650 aman

Luas tulangan = (1/4 x π x D² x S)/S= 251.2 > Asb,u 260 td aman


mm²


mm²

mm²

mm²

mm²


FST-UTY

PERHITUNGAN PLAT DAG ( AREA 6 m x 3 m )KELOMPOK III

1.1. Beban matiPlat lantai (12 cm) = T x ϒb

= 2.88 Kn/m3Pasir urug (3cm) = T x ϒ

= 0.054 Kn/m3Keramik (1cm) = T x ϒ

= 0.24 Kn/m3Spesi (2cm) = T x ϒ

= 0.42 Kn/m3Plafon & penggantung = 0.18 Kn/m3

1.2. Beban hidupfungsi bangunan ruang kuliah

qL = 2.5 Kn/m2

PerhitunganAnalisis dimensi plat

Ly = 7 mLx = 3.5 m

= 2 dua arahTabel untuk menentukan momen plat (PBI - 1971)

Ly/Lx 2Clx 41Cly 12Ctx 83Cty 57

Analisis tebal minimum platS = 3500 mmh = 125 mm


h = 125.000 mm= 100 mm

f'c (Mpa) = 22.5 Mpafy (Mpa) = 240 MPa


FST-UTY

PERHITUNGAN BEBAN PLAT LANTAIBerdasarkan Peraturan Pembebanan Indonesia Untuk Gedung (PPIUG)1983a. Beban Mati (Qd)

Komponen PlatBeton (10cm) = 0.1 x 2400 = 2.4 kN/m²Spesi (1,5 cm) = 2 x 21 = 0.42 kN/m²Plafon+penggantung = = 0.18 kN/m²water proofing = 0.5 kN/m²

Qd = 3.5 kN/m²


Ql = 2.5 kN/m²beban hujan (qR) = 0.5 kN/m²


= 8.2 kN/m²


= 4.11845 KNm0.001 X qU X Cly X Lx²

= 1.2054 KNm0.001 X qU X Ctx X Lx²

= 8.33735 KNm0.001 X qU X Cty X Lx²

= 5.72565 KNm



ds x = Pb + 0,5xØ ds y = Pb + Ø+0,5xØ= 25 = 35

d = h plat - ds d = h plat - ds= 75 mm = 65 mm


= 0.9152 < K maksa = (1- √1 - (2 x K / 0.85 x f'c)) x d


Mlx ⁺=

Mly ⁺=

Mtx ¯=

Mty ¯=

Mlx ⁺=

Ø x b x d


FST-UTY

As = (0.85 x f'c x a x b)/fy= 293.2


= (1.4 x b x d)/fy= 437.5



S = (1/4 x π x D² x S)/As,u= 179.43 mm < 2.h 200 aman

untuk mempermudah proses pemasangan tul. S pakai :S pakai = 170 mm




K = Mu



As = (0.85 x f'c x a x b)/fy= 610.13


As,u = (1.4 x b x d)/fy= 437.5



= 179.43 mm < 2.h 200 aman

mm²


mm²

As,u mm²

Mtx¯

Ø x b x d

mm²


mm²

mm²


FST-UTY

untuk mempermudah proses pemasangan tul. S pakai :S pakai = 175 mm



Asb = 20% x As,u= 87.5

Asb = 0.002.b.h= 200



= 251.2 mm < 5.h 500 amanuntuk mempermudah proses pemasangan tul. S pakai :

S = 250 mm




= 1.2054 KNmRencana diameter tulangan (Ø) mm = 10Selimut beton (Pb) = 20 mm

ds y = Pb + Ø+0,5xØ= 35 mm


K = Mu

= 0.3566 Mpa < Kmaks


mm²

mm²

mm²

mm²

Mly⁺

Ø x b x d²


FST-UTY

a = (1- √1 - (2 x K / 0.85 x f'c)) x d= 1.2236 mm



As,u = (1.4 x b x d)/fy= 379.17





= 523.33 > As,u 379.167 aman



K = Mu



As = (0.85 x f'c x a x b)/fy= 481.13


As,u = (1.4 x b x d)/fy= 379.17



= 207.03 mm

mm²


mm²

mm²

mm²

Mty¯

Ø x b x d

mm²


mm²

mm²


FST-UTY




Asb = 20% x As,u= 75.833

Asb = 0.002.b.h= 200



= 251.2 mm < 5.h 500 amanuntuk mempermudah proses pemasangan tul. S pakai :

S = 200 mm



mm²


mm²

mm²

mm²

mm²


FST-UTY

PERHITUNGAN PLAT DAG ( AREA 4 m x 3 m )KELOMPOK III

1.1. Beban matiPlat lantai (12 cm) = T x ϒb

= 2.88 Kn/m3Pasir urug (3cm) = T x ϒ

= 0.054 Kn/m3Keramik (1cm) = T x ϒ

= 0.24 Kn/m3Spesi (2cm) = T x ϒ

= 0.42 Kn/m3Plafon & penggantun = 0.18 Kn/m3

1.2. Beban hidupfungsi bangunan ruang kuliah

qL = 2.5 Kn/m2

PerhitunganAnalisis dimensi plat

Ly = 4 mLx = 3.5 m

= 1.1429 dua arahTabel untuk menentukan momen plat (PBI - 1971)

Ly/Lx 1.142857143Clx 32Cly 18.7Ctx 70.3Cty 57

Analisis tebal minimum platS = 3500 mmh = 125 mm


h = 125.000 mm= 100 mm

f'c (Mpa) = 22.5 Mpafy (Mpa) = 240 MPa karena dipakai tulangan polos


FST-UTY

PERHITUNGAN BEBAN PLAT LANTAIBerdasarkan Peraturan Pembebanan Indonesia Untuk Gedung (PPIUG)1983a. Beban Mati (Qd)

Komponen PlatBeton (12cm) = 0.1 x 2400 = 2.4 kN/m²Spesi (1,5 cm) = 1.5 x 21 = 0.315 kN/m²Plafon+penggantu = = 0.18 kN/m²MEE = 0.25 kN/m²

Qd = 3.145 kN/m²


Ql = 2.5 kN/m²beban hujan (qR) = 0.5 kN/m²


= 7.774 kN/m²


= 3.047408 KNm0.001 X qU X Cly X Lx²

= 1.78082905 KNm0.001 X qU X Ctx X Lx²

= 6.69477445 KNm0.001 X qU X Cty X Lx²

= 5.4281955 KNm



ds x = Pb + 0,5xØ= 25



= 0.6772 < K maksa = (1- √1 - (2 x K / 0.85 x f'c)) x d


As = (0.85 x f'c x a x b)/fy

Mlx ⁺=

Mly ⁺=

Mtx ¯=

Mty ¯=

Mlx ⁺=

Ø x b x d


FST-UTY

= 215.51 mm²


FST-UTY


= (1.4 x b x d)/fy= 437.5



S = (1/4 x π x D² x S)/As,u= 179.43 mm





K = Mu



As = (0.85 x f'c x a x b)/fy= 484.56


As,u = (1.4 x b x d)/fy= 437.5



= 179.43 mm


mm²

As,u mm²

Mtx¯

Ø x b x d

mm²


mm²

mm²


FST-UTY



b. Perhitungan tulangan bagi (tul.susut)Rencana diameter tulangan (Ø) = 8Luas tulangan minimal platTulangan bagi/tul.susut (Pasal 9.12.2.1. SNI 04-2847-2002)fy < 300Mpafy = 400Mpafy > 400MpaJarak maksimal tulangan (Pokok)S ≤ 5.h dan S ≤ 450mm (Pasal 9.12.2.2)

Asb = 20% x As,u= 87.5

Asb = 0.002.b.h= 200




S = 250 mm < 5.h 500 aman


Dari perhitungan diatas, maka dipakai tul.pokok D 10 - 170Dari perhitungan diatas, maka dipakai tul.susut : D 8 - 250


= 1.7808 KNmRencana diameter tulangan (Ø) = 10Selimut beton (Pb) = 20 mm

ds y = Pb + Ø+0,5xØ= 35 mm


K = Mu

= 0.5269 Mpa < Kmaks


mm²

mm²

mm²

mm²

Mly⁺

Ø x b x d²


FST-UTY

a = (1- √1 - (2 x K / 0.85 x f'c)) x d= 1.816 mm



As,u = (1.4 x b x d)/fy= 379.17



= 207 mmuntuk mempermudah proses pemasangan tul. S pakai :


= 436.11 > As,u 379.167 aman



K = Mu



As = (0.85 x f'c x a x b)/fy= 454.93


As,u = (1.4 x b x d)/fy= 379.17



= 207 mm

mm²


mm²

mm²

mm²

Mty¯

Ø x b x d

mm²


mm²

mm²


FST-UTY



b. Perhitungan tulangan bagi (tul.susut)Rencana diameter tulangan (Ø) = 8Luas tulangan minimal platTulangan bagi/tul.susut (Pasal 9.12.2.1. SNI 04-2847-2002)fy < 300Mpafy = 400Mpafy > 400MpaJarak maksimal tulangan (Pokok)S ≤ 5.h dan S ≤ 450mm (Pasal 9.12.2.2)

Asb = 20% x As,u= 75.833

Asb = 0.002.b.h= 200




S = 200 mm < 5.h 500 aman


Dari perhitungan diatas, maka dipakai tul.pokok D 10 - 180Dari perhitungan diatas, maka dipakai tul.susut : D 8 - 200

mm²


mm²

mm²

mm²

mm²

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