beton dasar

STRUKTUR BETON DASAR

diketahui = fc' = 25 Mpafy = 350 Mpawl = 12 KN/m2wd = 13 KN/m2

1.Kuat perlu ( U )untuk beban merata

Wu = 1.2 Wd + 1.6 Wl

= 34.8 Kn/ m2 = 3480 Kg/m2

2. beban rencana (mu)Mu = 8808.75 = 88.0875

3.nilai P perlu diambil B = 0.85B = 0.85

Pb = 0.85 x 0.0607142857 x 0.63157895= 0.03259398496

Pmin = 0.004sehingga P perlu = 0.01422274

4.koefisien tahanan momenw = Pperlu x fy

fc'= 0.01 x 350

25= 0.20

K = fc x w (1-0.59xw)= 25 x 0.20 x 0.88252013= 4.4

dik. D = 0.8d = 158.315746832

jadi tebal plat = 158 mm

5.luas tulangan pokok (As)As = Pb x b x d

= 0.03259398496 x 1000 x 158= 5160 mm2

dari tabel di peroleh tulangan pokok = 5361.7kontrol momen lapangan

a = 88.3103529412Mn = 5361.7 x 350 x ( 158.32 - 88.31 )

2= 214233155.537= 214.233155537

Mu < Mno

110.10938 < 214.233155537

6.menghitung luas tulangan susut (As)As' = 20 % As

= 1072.34 mm2o

dari tabel di peroleh tulangan pokok = 1090.8 25 40

1.35

2.7 1 2.8 1 2.7

mutu beton fc' = 39 Mpamutu baja fy = 320 Mpabeban hidup Wl = 23 KN/m2 2300beban mati Wd = 14 KN/m2 1400

a.Mencari reaksiQ1 = 2551.5 A1 = 1.8225Q2 = 700 A2 = 0.5Q3 = 3920 A3 = 2.8Q4 = 700 A4 = 0.5Q5 = 2551.5 A5 = 1.8225

Rav = 5211.5

Rbv = 5211.5

Mx = 15406.5916667

c.Perhitungan beban hidup MLML = 29911.5

d.Menghitung momen rencana (Mu)Mu = 66346.31 = 663.4631

e.Mencari P perluPb = 0.053 B1 = 0.778

Pmax = 0.039 Pmin = 0.004375

Pperlu = 0.022

f.Mencari koefisien tahananWperlu = 0.180

K = 6.265

g.Perhitungan nilai d perlu dan lebar balok bd perlu = 594 b rencana = 300

dibulatkan 600

h.Menghitung luas tulangan pokokAs = Pperlu x b x d

= 3942o

Dari tabel diperoleh tulangan pakai As 5089.5 5 36b > 540 tidak memenuhi

i.Mencari tinggi balok hh = 658 mm tulangan begel rencana = 10 mm

kontrol = 0.028275 Pmin = 0.004375

P ada > P min

0.028 > 0.004

p max = 0.039

P ada < P max0.028 < 0.039

kontrol momen lapanganMn ada = 873.901573964 a = 126.832727

63.4163634Mn perlu = 780.544823529Mn ada > Mn perlu

873.90157 > 780.544823529

3.rencana tulangan doublediketahui mutu beton fc' = 39 Mpa

mutu baja fy = 320 Mpab = 300h = 658Mu = 663.4631lendutan jangka panjang = 5 tahun

penyelesaiana. d' = 65.8

d = 592.2

b.Konstanta keamanan KB1 = 0.778 Mn ada = 1025.45865K = 9.747 Mn perlu = 780.544824

Pb = 0.049P max = 0.037 Mn ada < Mn perlu

m = 9.653 1025.4586518 < 780.544824

c.Rasio PenulanganK = 6.3061P = 0.022 0.10359375

0.78711076310.2128892369

P > P max0.022 > 0.037 tidak menggunakan tulangan double

As = 3918.11As' = 783.62

odari tabel dapat di pakai As' 763.5 3 18

e.Perhitungan garis netrala = 228.326086957

B1 = 0.778cb = 391.304347826c = 293.47826087

Es' = 0.002Ey = 0.0016

Es' < Ey0.002 < 0.0016

f.Perhitungan Penulangan untuk daerah tarik (As)

ND = 2270.70293478NT' = 463.62

NT ada = 2734.33As = 8544.76707294 dari tabel diperoleh As

Tulangan geser dengan tulangan miring

1.jarak tulangan miringSm d-d'

400 = 0.4 m

2.sudut tulangan miringᾀ = 45

3.RumusVs perlu = 407.01754504

Y = 3791.1885736 = 3.7911886Vs Astm as tm ada Vs tm

407.01754504 799.45796 7238.7 14218.15364.07401562 715.1089 3619.35 7109.0748325.66136391 639.65933 3619.35 7109.0748291.30154692 572.17028 3619.35 7109.0748260.56695894 511.80184 3619.35 7109.0748

sin a = 0.707106781cos a = 0.707106781

odari tabel 9 32

As 7238.7

Tulangan geser dengan tulangan kombinasi

Vn = 547.53 Fy 320Vc = 140.51245496 d 450

Vs begel = 203.50877252Av = 226.2

Vs S S ada Vs adaVs 203.50877252 160.056 160 203.580

Vs1 182.03700781 178.93504 170 191.605Vs2 162.83068196 200.04092 200 162.864Vs3 145.65077346 223.6363 220 148.058Vs4 130.28347947 250.01481 250 130.291

untuk tulangan miringVs tm = 203.438 sin a = 0.7071068

y = 3.7911885736 cos a = 0.7071068sm = 0.4 400 sin a+cos a = 1.4142136

Vs As tm As tm ada Vs tmVs 181.9733 357.429 3619.35 1842.6722

Vs1 162.77369 319.718 3619.35Vs2 145.5998 285.985 3619.35Vs3 130.23788 255.811 3619.35Vs4 116.49677 228.821 3619.35

Vs ada = 1842.672188 + 203.580 > 547.53 - 140.51245= 2046.252 > 407.01755

1.45

2.9 4.15 2.9

mutu beton fc' = 22 Mpa P1 33mutu baja fy = 200 Mpa P2 33beban hidup Wl = 19 KN/m2 1900beban mati Wd = 8 KN/m2 800

a.Mencari reaksiP1 = 3300P2 = 3300Q1 = 3444.5 A1 = 4.305625

Rav = 5022.25

Rbv = 5022.25

Mx = 16946.9708333




Pmax = 0.0447 Pmin = 0.0070

Pperlu = 0.026


K = 4.45

g.Perhitungan nilai d perlu dan lebar balok b

d perlu = 736 b rencana = 300dibulatkan 740


= 5739o

Dari tabel diperoleh tulangan pakai As 6283 5 40b > 625


kontrol = 0.0283 Pmin = 0.0070

P ada > P min0.0283 > 0.0070

p max = 0.0447

P ada < P max0.0283 < 0.0447


111.996Mn perlu = 681.85Mn ada > Mn perlu789.15 > 681.85


mutu baja fy = 200 Mpab = 350h = 500Mu = 579.57315lendutan jangka panjang = 5 tahun

penyelesaiana. d' = 50

d = 450



m = 10.6952 506.23909489 < 724.4664

c.Rasio PenulanganR ada = 7.1427

P = 0.04812.m.k/fy 0.7639

P > Pmax 1-2.m.k/fy 0.23610.0481 > 0.0481 akar 0.4859 Mu/o

P = 0.0433 1-akar 0.5141 46516.251/m 0.0935 di kali dgn d20.93231

0.023258Perhitungan penulangan untuk daerah tekan as'

Mr = 218.2273 o a 0.023258Nt' = 545.568356519 dari tabel 8 22 b -20.9323As' = 3009.20218709 As' 3040.8 c 724.4664

Perhitungan letak garis netrala = 215.15625c = 253.125 maka a

Cb = 337.5es' = 0.00240740741 garis netralEy = 0.08307692308 x =

Perhitungan Penulangan untuk daerah tarik (as) Perhitungan luas tulangan tarik(As)ND 1408.1977 As =NT' 608.16

NT ada 2016.3577 o 74426 >As 10081.788 dari tabel 10 36

As 10179 Perhitungan balok tulanganMn =

350 > 301a = 218.126814362 Nt = 2035.8 Kontrolh = 500 Mn >

NT ada = 2035800 861.301 >ND = 1427.64Mn = 729.998717372

Mn perlu < Mn724.4664 < 729.998717372

4.Perencanaan balok T

Mu = 579.57315b/h = 350 500

p.Balok = 9950

tebal plat = 320 mmjarak balok = 4750 mm

bE = 2487.5bE = 5470bE = 5100

724.4664375

b2-4ac 19.2552012a 0.0465163

x1 = 863.9457x2 = 36.054305= 36 mm

39 < t/B1 =

Perhitungan luas tulangan tarik(As)5240.9720417 10052.8 o

8 4010052.8

Perhitungan balok tulangan861.30103917 a =

724.4664375Mu/ф

P= 25

1.675

3.35 1.75 1.75 3.35


a.Mencari reaksiQ1 = 4690 A1 = 3.35Q2 = 4690 A2 = 3.35P1 = 2500

Rav = 5940

Rbv = 5940

Mx = 16063.25

c.Perhitungan beban hidup MLML = 28611



Pmax = 0.0423 Pmin = 0.0050

Pperlu = 0.024


K = 5.84


dibulatkan 650


= 5382o

Dari tabel diperoleh tulangan pakai As 5630.1 7 32b > 340


kontrol = 0.0247 Pmin = 0.0050

P ada > P min0.0247 > 0.0050

p max = 0.0423

P ada < P max0.0247 < 0.0423


80.287Mn perlu = 765.34

Mn ada > Mn perlu898.11 > 765.34




Wu = 1.2 Wd + 1.6 Wl= 1.2 x 13 + 1.6 x 12= 35.6 Kn/ m2 = 3560 Kg/m2



Pb = 0.85 x 0.0528378378 x 0.6185567= 0.02778071886



fc'= 0.01 x 370

23= 0.20

K = fc x w (1-0.59xw)= 23 x 0.20 x 0.88316526= 4.0

dik. D = 0.8d = 152.466336169



= 0.02778071886 x 1000 x 152= 4236 mm2


a = 101.474629156Mn = 5361.7 x 370 x ( 152.47 -

= 201812983.174= 201.812983174

Mu < Mno

93.505625 < 201.812983174


= 1072.34 mm2

dari tabel di peroleh tulangan pokok = 1047.2

1.35

2.7 1 2.8 1 2.7


a.Mencari reaksiQ1 = 2551.5 A1 = 1.8225Q2 = 700 A2 = 0.5Q3 = 3920 A3 = 2.8Q4 = 700 A4 = 0.5Q5 = 2551.5 A5 = 1.8225

Rav = 5211.5

Rbv = 5211.5

Mx = 15406.5916667



e.Mencari P perlu

Pb = 0.053 B1 = 0.778

Pmax = 0.039 Pmin = 0.004375

Pperlu = 0.022


K = 6.265


dibulatkan 600


= 3942o

Dari tabel diperoleh tulangan pakai As 5089.5 5 36b > 540 tidak memenuhi


kontrol = 0.028275 Pmin = 0.004375

P ada > P min0.028 > 0.004

p max = 0.039

P ada < P max0.028 < 0.039


63.4163634Mn perlu = 780.544823529Mn ada > Mn perlu

873.90157 > 780.544823529


mutu baja fy = 320 Mpab = 300h = 658Mu = 663.4631

lendutan jangka panjang = 5 tahun

penyelesaiana. d' = 65.8

d = 592.2



m = 9.653 1025.4586518 < 780.544824

c.Rasio PenulanganK = 6.3061P = 0.022 0.10359375

0.78711076310.2128892369

P > P max0.022 > 0.037 tidak menggunakan tulangan double

As = 3918.11As' = 783.62

odari tabel dapat di pakai As' 763.5 3 18

e.Perhitungan garis netrala = 228.326086957

B1 = 0.778cb = 391.304347826c = 293.47826087

Es' = 0.002Ey = 0.0016

Es' < Ey0.002 < 0.0016

f.Perhitungan Penulangan untuk daerah tarik (As)ND = 2270.70293478NT' = 463.62

NT ada = 2734.33As = 8544.76707294 dari tabel diperoleh As

4.1

101.47 )2




Wu = 1.2 Wd + 1.6 Wl= 1.2 x 13 + 1.6 x 12= 34.8 Kn/ m2 = 3480 Kg/m2

2. beban rencana (mu)Mu = 8808750000 = 8808.75

3.nilai P perlu diambil BB = 0.89

Pb = 0.89 x 0.0607142857 x 0.6315789474= 0.03412781955



fc'= 0.01 x 350

25= 0.21

K = fc x w (1-0.59xw)= 25 x 0.21 x 0.8777690789= 4.5

dik. D = 0.8d = 1556



= 0.03412781955 x 1000 x 1556= 53112 mm2

dari tabel di peroleh tulangan pokok = 7854kontrol momen lapangan

a = 129.36Mn = 7854 x 350 x ( 1556.279043

= 4100256607.93= 4100.25660793

Mu < Mno

11010.938 < 4100.25660793


= 1570.80 mm2


2.075

4.15 4.15


a.Mencari reaksiQ1 = 2905 A1 = 2.075Q2 = 2905 A2 = 2.075

Rav = 2905

Rbv = 2905

Mx = 6027.875



e.Mencari P perlu

Pb = 0.0434 B1 = 0.842

Pmax = 0.0325 Pmin = 0.0042

Pperlu = 0.018


K = 5.37


dibulatkan 664


= 3663

Dari tabel diperoleh tulangan pakai As 4825.8 6b > 577

i.Mencari tinggi balok hh = 720 mm tulangan begel rencana = 10

kontrol = 0.0242 Pmin = 0.0042

P ada > P min0.0242 > 0.0042

p max = 0.0325

P ada < P max0.0242 < 0.0325







d = 450



m = 9.6531 620.54516419 < 525.7394


P = 0.03942.m.k/fy 0.6163

P > Pmax 1-2.m.k/fy 0.38370.0394 > 0.0394 akar 0.6195

P = 0.0355 1-akar 0.38051/m 0.1036

Perhitungan penulangan untuk daerah tekan as'Mr = -94.8058 oNt' = -237.01438106 dari tabel 5 19As' = -826.26592664 As' 1417.5

Perhitungan letak garis netrala = 185.331521739c = 220.108695652

Cb = 293.47826087es' = 0.00231851852Ey = 0.13801916933

Perhitungan Penulangan untuk daerah tarik (as)ND 1843.122NT' 453.6

NT ada 2296.722 oAs 7177.2562 dari tabel 9 32

As 7238.7

300 > 285a = 187.308597285 Nt = 2316.384h = 500

NT ada = 2316384ND = 1862.784Mn = 845.235070957

Mn perlu < Mn525.7394 < 845.235070957

Kontrol terhadap lendutan1.Pemeriksaan tinggi maksimum

H min = 518.75 > 500

2.Kontrol pelendutan dari sisi lainEc = 29351 29500Es = 200000Fr = 4.371

3.Luas transformasin = 6.780 b 300 Mu

n As = 49075.932 d 450n AS' = 9610.169 d' 50

h 500

4.Letak garis netralx1 = 238.774071046x2 = -630.01474901

5.momen inersia penampang retak transformasiIcr = 1361323972.36 + 2189591081 + 342464635.47

= 3893379688.93 mm4

6.Momen inersia penampang utuhIg = 3125000000 mm4

7.momen yang timbul pada saat retak pertama kaliMcr = 54643732.486 Yt = 250

= 54.643732486 KNm

8.Untuk kondisi 1 (beban mati + beban hidup)Mcr/Mu = 0.12227872338

Ie = 3891974843.43

9.Untuk kondisi 2 (beban mati + 60%beban hidup)Mcr/Mu = 0.18394791265 Md 60.27875

Ie = 3888597128.47 ml 234.09

10.lendutan seketikaΔ = 243.255649718 L 8300

11.Lendutan beban mati seketikaADL = 4.69220687853

12.Lendutan beban hidup seketika kondisi 1ALL = 13.7063862313

13.Lendutan 60% beban hidup seketika kondisi 2ASL = 7.86486033816

ASL < ALL

14.Lendutan jangka panjang 3 bulan 1l = 0.0105λ = 0.65573770492

15.Beban mati + beban hidupALT = 12.0646512195

16.Beban mati + 60% beban hidupALT = 8.46953351953

17.Pemeriksaan persyaratan lendutanALL

13.706386 46.1111111111

ALL13.706386 23.0555555556

ALT12.064651 17.2916666667

ALT12.064651 34.5833333333

4500

- 129.36 )2

o32

mm

4.Perencanaan balok T tulangan geser


diketahui

Mu = 446.8785b/h = 300 500

p.Balok = 8300tebal plat = 320 mm 1.Kuat perlu ( U )

jarak balok = 4500 mm untuk beban merataWu

bE = 2075bE = 5420bE = 4800

Q1Mu/o 558.598125

68786.25di kali dgn d 30.9538125 Rav

0.034393125 Rbv

a 0.034393125 b2-4ac 29.686542 Rencana tulangan geserb -30.9538125 2a 0.0687863 Lc 558.598125 fc'

fyx1 = 881.57669 VDx2 = 18.423313 VL

maka a = 18 mm VA

garis netral 1.Perhitungan beban rencanax = 24 < t/B1 = 411.31105 VU

Perhitungan luas tulangan tarik(As) 2.Perhitungan kekuatan geser oleh betonAs = 3960.22074 6107.4 o VC

6 3668786.25 > 6107.4

VnPerhitungan balok tulangan

Mn = 851.701663 a = 28.41219 4.perhitungan kekuatan geser yang memerlukan tulangan geservs

KontrolMn > 5.kontrol

851.7016626 > 558.598125

3.Perhitungan kekuatan geser rancang (ξ = 0.6 untuk tul.geser)

Mu/ф

dengan menggunakan rumus perbandingan segi tiga

446.8785 Y

X + Y

1/2 Ya 150b 58686.101695 -58686.1017 Penulangan geser dengan begelc -22564677.97

b2-4ac 130318.32301 -130318.323 S12a 300

banyaknya begel bag.1n

KontrolVs begel

247.28755

begel bag.2Vs1S2

banyak begel bag.2n

KontrolVs begel

123.64377

digunakan ф 12

tulangan geser


1.jarak tulangan miring

Sm45004.5

2.sudut tulangan miringSTRUKTUR BETON DASAR ᾀ =

= fc' = 25 Mpa 3.Rumusfy = 350 Mpa Vs perlu =wl = 12 KN/m2 Y =wd = 13 KN/m2


= 1.2 Wd + 1.6 Wl= 1.2 x 13 + 1.6 x 12= 34.8 Kn/ m2 = 3480 Kg/m2

sin a == 156.6 cos a =

= 78.3 dari tabel 9= 78.3 As 7238.7

Rencana tulangan geser Tulangan geser dengan tulangan kombinasi= 8300 8.3= 39 Vn == 320 Vc == 29.05 Vs begel == 7470 74.7 Av == 78.3

1.Perhitungan beban rencana Vs= 232.68 Kn Vs1

Vs22.Perhitungan kekuatan geser oleh beton Vs3

= 140.51245 Vs4

untuk tulangan miring= 387.8 Vs tm =

y =4.perhitungan kekuatan geser yang memerlukan tulangan geser sm =

= 247.28755Vs

Vs 37.101008Vs1 31.493071

387.8 70.256227 Vs2 26.732791Vs3 22.692043Vs4 19.262067


1.Bila Vn < 0.5 Vc

2.Bila 0.5 Vc < Vs < 1/6 √fc' bw.d

70.256227 247.28755 382.42646Vs ada =

=140.51245 247.28755 281.02491

4281.02491 247.28755 562.04982

5247.28755 562.04982

dengan menggunakan rumus perbandingan segi tiga4150

= 2646.3211 X = 1503.6789

= 4150

= 1323.1605 1/2 Vc = 70.256227

Penulangan geser dengan begel= 226.2 mm

= 132

banyaknya begel bag.1= 10 buah

Vs perlu S1 ada = 131.72034247.28755

begel bag.2= 123.64377= 263

banyak begel bag.2= 5

Vs perlu S2 ada = 263.44068123.64377

3.Vc < Vs < 1/3 √fc'.bw.d

digunakan ф 12

d-d'400 = 0.4 m

2.sudut tulangan miring45

247.287545042646.3210725 = 2.6463211

Vs Astm as tm ada Vs tm247.28754504 485.71861 7238.7 14218.15209.90923178 399.80671 3619.35 7109.0748178.18077162 349.98009 3619.35 7109.0748151.24817097 297.07946 3619.35 7109.0748128.38652012 252.17494 3619.35 7109.0748

0.7071067810.707106781

o32


387.8 Fy 320140.51245496 d 450123.64377252

226.2

Vs S S ada Vs ada123.64377252 263.441 160 203.580104.95461589 310.35129 170 191.60589.090385812 365.61521 200 162.86475.624085485 430.71992 220 148.05864.193260061 507.41776 250 130.291

43.708 sin a = 0.70710682.6463210725 cos a = 0.7071068

0.4 400 sin a+cos a = 1.4142136

As tm As tm ada Vs tm72.873 3619.35 1842.672261.858 3619.3552.508 3619.3544.571 3619.3537.834 3619.35

1842.672188 + 203.580 > 387.8 - 140.512452046.252 > 247.28755

Perencanaan kolom persegidik Fc' 39

Fy 320

Mu 446.8785Mu' 148.9595Vu 232.68Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450l

Penyelesaian1.Menentukan momen dan gaya aksial rencana (Mn,Vn,Vk)

Mn = 248.2658333Vu = 387.8Vk = 1350 = 13.5Vn = 646.3333333

2.Mencari jarak eksentritase = 0.124360912 = 124.36091

Pemeriksaan luas tulanganl = 3075 Ag = 150000

As = As' = 1537.5daari tabel diperoleh luas tulangan 5 o 1571 mm2

20As tot = 3142 10 o

203142 > 3075

l = 0.010473333 1/2 l = 0.01025

0.010473333 > 0.01025

3.Pemeriksaan Mnb terhadap beban seimbang Pnbcb = 293.4782609ab = 247.1086957Es' = 0.00239298B1 = 0.842

fy/Es0.002393 > 0.0016

fs' = 497.7777778 > 320

Pemeriksaan Nilai Pnb dan mnbND = 2457495.978NT' = 450641.35NT = 502720

Pnb = 2405417.328

= 2405.417328 kN




Wu = 1.2 Wd + 1.6 Wl= 1.2 x 14 + 1.6 x 11= 34.4 Kn/ m2 = 3440 Kg/m2

Q1 = 16168Q2 = 3784

2. beban rencana (mu)Rav = 7641.19148936 Mx = -2081.2Rbv = 12310.8085106 x = 2.2212765957


Pb = 0.85 x 0.065 x 0.6382978723= 0.03526595745



fc'= 0.02 x 340

26= 0.20

K = fc x w (1-0.59xw)= 26 x 0.20 x 0.882081475= 4.6

dik. D = 0.8d = 189



= 0.03526595745 x 1000 x 189= 6679 mm2


a = 116.963076923Mn = 7602.6 x 340 x ( 189.3890398

= 338380705.66= 338.38070566

Mu < Mno

164.40755 < 338.38070566


= 1520.52 mm2 o10 50


0.45

1 0.9 1 3.3


a.Mencari reaksiQ1 = 800 A1 = 0.5Q2 = 1440 A2 = 0.9Q3 = 800 A3 = 0.5Q4 = 4356 A4 = 2.7225Q5 = 800 A5 = 0.5Q6 = 1440 A6 = 0.9Q7 = 800 A7 = 0.5

Rav = 5218

Rbv = 5218

Mx = 13120



e.Mencari P perlu

Pb = 0.0557 B1 = 0.770

Pmax = 0.0418 Pmin = 0.0045

Pperlu = 0.023


K = 6.41


dibulatkan 580


= 4026



kontrol = 0.0277 Pmin = 0.0045

P ada > P min0.0277 > 0.0045

p max = 0.0418

P ada < P max0.0277 < 0.0418







d = 360



m = 9.1176 407.51703905 < 631.5525


P = 0.04182.m.k/fy 0.6166

P > Pmax 1-2.m.k/fy 0.38340.0418 > 0.0418 akar 0.6192

P = 0.0376 1-akar 0.38081/m 0.1097

Perhitungan penulangan untuk daerah tekan as'Mr = 224.0355 oNt' = 700.110815459 dari tabel 5 29As' = 2536.63338934 As' 3302.5


Cb = 237.362637363es' = 0.00232592593Ey = 0.13328025478 0.00155



As 8043

300 > 285a = 144.074019608 Nt = 2493.33h = 400

NT ada = 2493330ND = 1469.555Mn = 750.785452058

Mn perlu < Mn631.5525 < 750.785452058


H min = 443.75 > 400



n As = 53620.000 d 360n AS' = 22016.667 d' 40

h 400



= 2729702016.15 mm4



= 35.4175097939 KNm


Ie = 2729312863.82


Ie = 2728679327.72 ml 217.37625





ASL < ALL




13.144867 < 50.5555555556

ALL13.144867 < 25.2777777778

ALT14.521018 < 18.9583333333

ALT14.521018 < 37.9166666667

4.7 1.1

Mu = 13152.6041= 131.526041

o22 50

- 116.96307692 )2

1 0.9 1

1.650.55

o32

mm



diketahui

Mu = 505.242b/h = 300 500



bE = 2275bE = 5420bE = 4800

Q1Mu/o 631.5525 Q277350

di kali dgn d 27.846 Rav0.038675 Rbv


fyx1 = 696.55648 VDx2 = 23.443519 VL

maka a = 23 mm VA



6 3679845.16129 > 6107.4




658.4147852 > 631.5525


Mu/ф


505.242 Y

X + Y


b2-4ac 133533.61129 -133533.611 S12a 300


KontrolVs begel372.6712

begel bag.2Vs1S2

banyak begel bag.2n

KontrolVs begel186.3356

digunakan ф 13

tulangan geser



Sm4700 11004.7 1.1




= 1.2 Wd + 1.6 Wl= 34.4 Kn/ m2 = 3440 Kg/m2

sin a == 16168 cos a == 3784

= 76.411915 dari tabel 10= 123.10809 As 8043

Rencana tulangan geser Tulangan geser dengan tulangan kombinasi= 9100 9.1= 40 Vn == 310 Vc == 52.18 Vs begel == 9555 95.55 Av == 76.411915



= 113.842 Vs4



= 372.6712Vs

Vs 170.33461Vs1 154.69556

486.51319 56.920998 Vs2 140.49239Vs3 127.59327Vs4 115.87846


1.Bila Vn < 0.5 Vc

2.Bila 0.5 Vc < Vs < 1/6 √fc' bw.d

56.920998 372.6712 346.41016Vs ada =

=113.842 372.6712 227.68399

4227.68399 372.6712 455.36798

5372.6712 455.36798


= 3485.3196 X = 1064.6804

= 4550

= 1742.6598 1/2 Vc = 56.920998


= 79


Vs perlu S1 ada = 79.476601372.6712

begel bag.2= 186.3356= 159


Vs perlu S2 ada = 158.9532186.3356

3.Vc < Vs < 1/3 √fc'.bw.d

digunakan ф 13

d-d'320 = 0.32 m


372.671195723485.319556 = 3.4853196


0.707106781 444733.930.707106781

o32


486.51319149 Fy 310113.84199577 d 360186.33559786

265.4

Vs S S ada Vs ada186.34 158.953 160 185.117169.23 175.02268 180 164.548153.69 192.71671 200 148.093139.58 212.19953 215 137.761126.76 233.65197 235 126.037

187.555 sin a = 0.70710683.485319556 cos a = 0.7071068

0.32 320 sin a+cos a = 1.4142136


1983.4318691 + 185.117 > 486.51319 - 113.8422168.548 > 372.6712

Perencanaan kolom persegi Perencanaan kolom persegidik Fc' 40 dik

Fy 310

Mu 505.242Mu' 168.414Vu 291.9079149Lk 3.75 3750b 300 0.3h 400 0.4d' 40d 360l 2.05


Mn = 280.69Vu = 291.9079149Vk = 1080 = 10.8Vn = 486.5131915



As = As' = 1230daari tabel diperoleh luas tulangan 5 o 1571 mm2

20As tot = 3142 10 o

203142 > 2460

l = 0.013091667 1/2 l = 0.01025

0.013091667 > 0.01025

3.Pemeriksaan Mnb terhadap beban seimbang Pnbcb = 237.3626374ab = 182.7692308 Ec = 0.003Es' = 0.002343434 Es = 200000B1 = 0.77

fy/Es0.0023434 > 0.00155

fs' = 498.8888889 > 310fs = 310 = 310

Pemeriksaan Nilai Pnb dan mnbND = 1864246.154NT' = 433596NT = 487010.00

Pnb = 1810832.154= 1810.832154 kN > 291.90791

Mnb = 202485813 + 98181333 + 98181333= 398848479.7= 398.8484797

eb = 0.220257012 = 220.257012eb < e

220.25701 < 564.41294e' = 380.2570121m = 9.117647059l = 0.01025

e'/d = 1.056269478d'/d = 0.111111111

1-e'/d -0.0562691-d'/d 0.8888889

Pn = 1304307.4 0.16931005= 1304.3074 0.41147302

0.35520354φ begel 12φ tul.memanjang 20

a 576b 320

Perencanaan kolom persegiFc' 40Fy 310

Mu 505.242Mu' 168.414Vu 291.90791Lk 3.75 3750b 300 0.3h 400 0.4d' 40d 360

Ag 120000

π 3.1428571r = 195.40168418

D = 390.80336837 = 400

Mn = 210.5175Vn = 486.51319149

e = 0.2539858225 = 260Vk + Vn = 828.85531915

h = 320

b = 392.85714286 = 400

Menghitung luas tulangand' = 32d = 288

oAs = As' = 1180.8 3 25

1472.7As tot = 2945.4 6 25

2945.4

P = 0.0127838542Ds = 336

d-d' = 224 = 230

Cb = 189.89010989 = 190

ab = 146.3Fs' = 498.94736842

Fs' > Fy498.94737 > 310

Pnb = 2267942.7895= 2267.9427895

1/2 h 1601/2 ab 73.15

Mnb = 172803708 + 58436736 + 58436736= 289677180= 289.67718

eb = 0.1277268463= 127.72684626

e eb0.2539858 > 0.1277268463

D = 400Ds = 336Pg = 0.0234293182m = 9.1176470588

Karena e > b maka kolom akan mengalami keruntuhan tarikPn = 0.85 x fc' x D2 = 5440000

= 0.1500486304= 0.0717764053 = 0.4709831= 0.1725

Pnb = 1623747.8056= 1623.7478056

Kontrol Pn > Vn

1623.7478 > 486.51319149

Merencanakan tulangan spiraldigunakan tul.spiral = 10 = 78.5

Dc = 336Ac = 88704Ag = 125714.28571

Persentase tul.spiralPs min = 0.0242264648

jarak tulangan spiral d 10 78.5s = 37.426451162 = 37




Wu = 1.2 Wd + 1.6 Wl= 36 Kn/ m2 = 3600 Kg/m2

Q1 = 2880Q2 = 17640Q3 = 2880

2. beban rencana (mu)Rav = 11700 Mu = 96.525Mx = -1800 x2 + 8820 x - 1152


Pb = 0.85 x 0.06 x 0.6382978723= 0.03255319149



fc'= 0.01 x 340

24= 0.20

K = fc x w (1-0.59xw)= 24 x 0.20 x 0.8807577571= 4.3

dik. D = 0.8d = 168.055213532



= 0.03255319149 x 1000 x 168= 5471 mm2


a = 113.096666667Mn = 6785.8 x 340 x ( 168.0552135

= 257265551.801= 257.265551801

Mu < Mno

120.65625 < 257.265551801


= 1357.16 mm2


P = 30

1.7

3.4 1.1 1.1 3.4


a.Mencari reaksiQ1 = 4760Q2 = 4760P = 3000

Rav = 6260

Rbv = 6260

Mx = 14842



e.Mencari P perlu

Pb = 0.0344 B1 = 0.898

Pmax = 0.0258 Pmin = 0.0041

Pperlu = 0.015


K = 4.45


dibulatkan 710


= 3186



kontrol = 0.0227 Pmin = 0.0041

P ada > P min0.0227 > 0.0041

p max = 0.0258

P ada < P max0.0227 < 0.0258







d = 450



m = 16.6667 418.25055894 < 846.8700


P = 0.02582.m.k/fy 0.6750

P > Pmax 1-2.m.k/fy 0.32500.0258 > 0.0258 akar 0.5701

P = 0.0232 1-akar 0.42991/m 0.0600



Cb = 287.234042553es' = 0.0023037037Ey = 0.14758842444



As 7539.6

300 > 317a = 282.505555556 Nt = 2563.464h = 500

NT ada = 2563464ND = 1728.934Mn = 867.615569906

Mn perlu < Mn846.8700 < 867.615569906


H min = 562.5 > 500



n As = 54833.455 d 450n AS' = 17850.909 d' 50

h 500



= 4443123711.07 mm4



= 42.8660704987 KNm


Ie = 4442789841.38


Ie = 4442171519 ml 312.12





ASL < ALL





17.232655 < 50

ALL17.232655 < 25

ALT16.659804 < 18.75

ALT16.659804 < 37.5

0.8 4.9 0.8

x 2.45

diameter 36jarak 150

- 113.09666667 )2

diameter 19 mmjarak 200 mm

3.4

o32

mm


0.8


diketahui

Mu = 677.496b/h = 300 500



bE = 2250bE = 5420bE = 4800

Q1Mu/o 846.8745900



fyx1 = 856.939 VDx2 = 43.061004 VL

maka a = 43 mm VA



6 3643200 > 6107.4




887.4614081 > 846.87


Mu/ф


677.496 Y

X + Y

1/2 Ya 150b 72684.364 -72684.3636 Penulangan geser dengan begelc -25567600

b2-4ac 143609.11084 -143609.111 S12a 300


KontrolVs begel

449.97296

begel bag.2Vs1S2

banyak begel bag.2n

KontrolVs begel

224.98648

digunakan ф 12

tulangan geser



Sm4900 0.84.9




= 1.2 Wd + 1.6 Wl= 1.2 x 13 + 1.6 x 12= 36 Kn/ m2 = 3600 Kg/m2

sin a == 176.4 cos a =

= 88.2 dari tabel 6= 88.2 As 7539.6

Rencana tulangan geser Tulangan geser dengan tulangan kombinasi= 9000 9= 24 Vn == 340 Vc == 62.6 Vs begel == 10800 108 Av == 88.2



= 110.22704 Vs4



= 449.97296Vs

Vs 194.98612Vs1 173.40828

560.2 55.113519 Vs2 154.21833Vs3 137.152Vs4 121.97428


1.Bila Vn < 0.5 Vc

2.Bila 0.5 Vc < Vs < 1/6 √fc' bw.d

55.113519 449.97296 300Vs ada =

=110.22704 449.97296 220.45408

4220.45408 449.97296 440.90815

5449.97296 440.90815


= 3614.5632 X = 885.43676

= 4500

= 1807.2816 1/2 Vc = 55.113519


= 77


Vs perlu S1 ada = 76.912621449.97296

begel bag.2= 224.98648= 154


Vs perlu S2 ada = 153.82524224.98648

3.Vc < Vs < 1/3 √fc'.bw.d

digunakan ф 12

d-d'400 = 0.4 m


449.972961573614.5632401 = 3.6145632


0.7071067810.707106781

o40


560.2 Fy 340110.22703843 d 450224.98648079

226.2

Vs S S ada Vs ada224.98648079 153.825 150 230.724200.08870301 172.96629 170 203.580177.94619895 194.48912 190 182.151158.25406055 218.69012 200 173.043140.74112192 245.90254 210 164.803

219.249 sin a = 0.70710683.6145632401 cos a = 0.7071068

0.4 400 sin a+cos a = 1.4142136

As tm As tm ada Vs tm360.460 3769.8 2039.2231320.570 3769.8 sda285.095 3769.8 sda253.545 3769.8 sda225.487 3769.8 sda

2039.2231244 + 230.724 > 560.2 - 110.227042269.947 > 449.97296


Fy 340

Mu 677.496Mu' 225.832Vu 336.12Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450l 1.85 nilai ditentukan sama bapa'


Mn = 282.29Vu = 336.12Vk = 1350 = 13.5Vn = 560.2




20As tot = 3142 10 o

203142 > 2775

l = 0.010473333 1/2 l = 0.01025

0.010473333 > 0.01025


fy/Es0.0024185 > 0.0017

fs' = 495.5555556 > 340fs = 340 = 340

Pemeriksaan Nilai Pnb dan mnbND = 1578569.362NT' = 502091.6NT = 534140.00

Pnb = 1546520.962= 1546.520962 kN > 336.12

Mnb = 191057272.6 + 137516667 + 137516667= 466090606= 466.090606

e = 0.301380077 = 301.380077e' = 101.3800766m = 16.66666667l = 0.00925

e'/d = 0.225289059d'/d = 0.111111111

1-e'/d 0.77471091-d'/d 0.8888889

Pn = 4708582.392 0.87425112= 4708.582392 0.93501397


a 576b 320

Mu 677.496Mu' 225.832Vu 336.12Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450

Ag 150000

π 3.1428571r = 218.46572438

D = 436.93144875 = 440

Mn = 282.29Vn = 560.2

e = 0.4920515949 = 700Vk + Vn = 573.7

h = 352

b = 432.14285714 = 440

Menghitung luas tulangand' = 35.2d = 316.8

oAs = As' = 1289.376 3 25

1472.7As tot = 2578.752 6 25

2945.4

P = 0.0105651687Ds = 369.6

d-d' = 246.4 = 250

Cb = 202.21276596 = 210

ab = 188.58Fs' = 499.42857143

Fs' > Fy499.42857 > 340

Pnb = 1927484.5371= 1927.4845371

1/2 h 1761/2 ab 94.29

Mnb = 138310033.28 + 70501094.4 + 70501094.40= 279312222.08= 279.31222208

eb = 0.1449102271= 144.91022714

e eb0.4920516 > 0.1449102271

D = 440Ds = 369.6Pg = 0.0193630729m = 16.666666667


= 2.2444730952= 0.1084332081 = 1.5339186104= 0.9722727273

Pnb = 2218186.7167= 2218.1867167

Kontrol Pn > Vn

2218.1867 > 560.2


Dc = 369.6Ac = 107331.84Ag = 152114.28571






Wu = 1.2 Wd + 1.6 Wl= 51.2 Kn/ m2 = 5120 Kg/m2



Pb = 0.41 x 0.0959677419 x 0.6593406593= 0.02594292804



fc'= 0.01 x 310

35= 0.11

K = fc x w (1-0.59xw)= 35 x 0.11 x 0.9373611264= 3.5

dik. D = 0.8d = 222.850950529



= 0.02594292804 x 1000 x 223= 5781 mm2


a = 45.885210084Mn = 4403.5 x 310 x ( 222.8509505

= 272891883.799= 272.891883799

Mu < Mno

172.98 < 272.891883799

6.menghitung luas tulangan susut (As) dari tabel di peroleh tulangan pokokAs' = 20 % As

= 880.70 mm2P = 31 P =

1.15

2.3 5.2 2.3


a.Mencari reaksiQ1 = 7280p1 = 3100p2 = 3100

Rav = 6740

Rbv = 6740

Mx = 20234


d.Menghitung momen rencana (Mu)Mu = 74220 = 742.2

e.Mencari P perlu

Pb = 0.0444 B1 = 0.818

Pmax = 0.0333 Pmin = 0.0041

Pperlu = 0.019


K = 5.66


dibulatkan 664


= 3725



kontrol = 0.0242 Pmin = 0.0041

P ada > P min0.0242 > 0.0041

p max = 0.0333

P ada < P max0.0242 < 0.0333







d = 450



m = 9.6531 620.54516419 < 927.7500


P = 0.03942.m.k/fy 0.6163

P > Pmax 1-2.m.k/fy 0.38370.0394 > 0.0394 akar 0.6195

P = 0.0355 1-akar 0.38051/m 0.1036



Cb = 293.47826087es' = 0.00231851852Ey = 0.13801916933



As 7238.7

300 > 285a = 187.308597285 Nt = 2316.384h = 500

NT ada = 2316384ND = 1862.784Mn = 845.235070957

Mn perlu < Mn927.7500 < 845.235070957


H min = 612.5 > 500



n As = 49075.932 d 450n AS' = 9610.169 d' 50

h 500



= 3893379688.93 mm4



= 54.643732486 KNm


Ie = 3893073045.19


Ie = 3892594210.72 ml 312.12





ASL < ALL





22.856548 54.4444444444

ALL22.856548 27.2222222222

ALT29.386517 20.4166666667

ALT29.386517 40.8333333333

1.25 4.65 1.25

= 3.575= 18304

diameter 29jarak 150

- 45.885210084 )2

= 1570.8 diameter 20 mmjarak 200 mm

31

2.3

2.3 2.6

o32

mm


Mu = 742.2b/h = 300 500

p.Balok = 9800tebal plat = 320 mm

jarak balok = 4500 mm

bE = 2450bE = 5420bE = 4800

Mu/o 927.7581217.5

di kali dgn d 36.5478750.04060875

a 0.04060875 b2-4ac 34.424528b -36.547875 2a 0.0812175c 927.75

x1 = 873.85604x2 = 26.143965

maka a = 26 mm

garis netralx = 34 < t/B1 = 411.31105

Perhitungan luas tulangan tarik(As)As = 6635.46084 6107.4 o

6 3681217.5 > 6107.4

Perhitungan balok tulanganMn = 855.951245 a = 24.063385

KontrolMn >

855.9512448 > 927.75Mu/ф

742.2

a 150b 58686.101695 -58686.1017c -22564677.97

b2-4ac 130318.32301 -130318.3232a 300

tulangan geser



Sm46504.65

2.sudut tulangan miringSTRUKTUR BETON DASAR ᾀ

diketahui = fc' = 35 Mpa 3.Rumusfy = 310 Mpa Vs perluwl = 23 KN/m2 Ywd = 12 KN/m2


Wu = 1.2 Wd + 1.6 Wl= 1.2 x 13 + 1.6 x 12= 51.2 Kn/ m2 = 5120 Kg/m2

sin aQ1 = 238.08 cos a

Rav = 119.04 dari tabelRbv = 119.04 As

Rencana tulangan geser Tulangan geser dengan tulangan kombinasiL = 9800 9.8

fc' = 39 Vnfy = 320 VcVD = 67.4 Vs begelVL = 11760 117.6 AvVA = 119.04

1.Perhitungan beban rencanaVU = 388.08 Kn

2.Perhitungan kekuatan geser oleh betonVC = 140.51245

untuk tulangan miringVn = 646.8 Vs tm

y4.perhitungan kekuatan geser yang memerlukan tulangan geser sm

vs = 506.28755

5.kontrol VsVs1

646.8 70.256227 Vs2Vs3Vs4


1.Bila Vn < 0.5 Vc

2.Bila 0.5 Vc < Vs < 1/6 √fc' bw.d

70.256227 506.28755 382.42646Vs ada

140.51245 506.28755 281.02491

4281.02491 506.28755 562.04982

5506.28755 562.04982


Y = 3835.5117 X = 1064.4883

X + Y = 4900

1/2 Y = 1917.7559 1/2 Vc = 70.256227


S1 = 64

banyaknya begel bag.1n = 30 buah

KontrolVs begel Vs perlu S1 ada = 64.336562

506.28755 506.28755

begel bag.2Vs1 = 253.14377S2 = 129

banyak begel bag.2n = 15


253.14377 253.14377

3.Vc < Vs < 1/3 √fc'.bw.d

digunakan ф 12

d-d'400 = 0.4 m

2.sudut tulangan miring= 45

= 506.28754504= 3835.5117048 = 3.8355117


= 0.707106781= 0.707106781

o9 32

7238.7


= 646.8 Fy 320= 140.51245496 d 450= 253.14377252= 226.2

Vs S S ada Vs adaVs 253.14377252 128.673 160 203.580

Vs1 226.74377252 143.65466 180 180.960Vs2 203.09699055 160.38052 200 162.864Vs3 181.91629747 179.05378 230 141.621Vs4 162.94450842 199.90118 260 125.280

untuk tulangan miring= 302.708 sin a = 0.7071068= 3.8355117048 cos a = 0.7071068= 0.4 400 sin a+cos a = 1.4142136

Vs As tm As tm ada Vs tm271.13861 532.567 3619.35 1842.6722242.86195 477.026 3619.35217.53423 427.278 3619.35194.8479 382.717 3619.35

174.52749 342.804 3619.35

= 1842.672188 + 203.580 > 646.8 - 140.51245= 2046.252 > 506.28755


Fy 320

Mu 742.2Mu' 247.4Vu 388.08Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450l


Mn = 412.3333333Vu = 388.08Vk = 1350 = 13.5Vn = 646.8




20As tot = 3142 10 o

203142 > 3075

l = 0.010473333 1/2 l = 0.01025

0.010473333 > 0.01025


fy/Es0.0023752 > 0.0016

fs' = 497.7777778 > 320fs = 320 = 320


Pnb = 2335369.937= 2335.369937 kN > 388.08

Mnb = 310290464.7 + 153066667 + 153066667= 616423798.1= 616.4237981

e = 0.263951243 = 263.951243e' = 63.95124314m = 9.653092006l = 0.01025

e'/d = 0.142113874d'/d = 0.111111111

1-e'/d 0.85788611-d'/d 0.8888889

Pn = 8112754.026 0.91186939= 8112.754026 0.95491853


a 576b 320


Fy 320

Mu 742.2Mu' 247.4Vu 388.08Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450

Ag 150000

π 3.1428571r = 218.46572438

D = 436.93144875 = 440

Mn = 412.33333333Vn = 646.8

e = 0.6244636276 = 700Vk + Vn = 660.3

h = 352

b = 432.14285714 = 440


oAs = As' = 1428.768 3 25

1472.7As tot = 2857.536 6 25

2945.4

P = 0.0105651687Ds = 369.6

d-d' = 246.4 = 250

Cb = 206.60869565 = 210

ab = 171.78Fs' = 499.42857143

Fs' > Fy499.42857 > 320

Pnb = 2769827.5371= 2769.8275371

1/2 h 1761/2 ab 85.89

Mnb = 225778091.34 + 66353971.2 + 66353971.20= 358486033.74= 358.48603374

eb = 0.1294253988= 129.42539885

e eb0.6244636 > 0.1294253988

D = 440Ds = 369.6Pg = 0.0193630729m = 9.653092006


= 2.2444730952= 0.0628029441 = 1.5189720337= 0.9722727273

Pnb = 3508628.677= 3508.628677

Kontrol Pn > Vn

3508.6287 > 646.8


Dc = 369.6Ac = 107331.84Ag = 152114.28571






Wu = 1.2 Wd + 1.6 Wl= 1.2 x 9 + 1.6 x 15= 34.8 Kn/ m2 = 3480 Kg/m2

Q1 = 2610Q2 = 14964

2. beban rencana (mu)Rav = 10319.6162791 Mx = 0Rbv = 7254.38372093 x = 2.9654069767


Pb = 0.85 x 0.0575806452 x 0.6593406593= 0.03227047146



fc'= 0.01 x 310

21= 0.21

K = fc x w (1-0.59xw)= 21 x 0.21 x 0.8749356685= 3.9

dik. D = 0.8d = 286



= 0.03227047146 x 1000 x 286= 9221 mm2


a = 170.5Mn = 9817.5 x 310 x ( 285.754222

= 610219561.944= 610.219561944

Mu < Mno

318.02567 < 610.219561944

6.menghitung luas tulangan susut (As)As' = 20 % As = 2261.9

= 1963.50 mm2 o12 50

dari tabel di peroleh tulangan pokokp 30

0.875

1 1.75 1 1.25 1.25


a.Mencari reaksiQ1 = 700 A1 = 0.5Q2 = 2450 A2 = 1.75Q3 = 700 A3 = 0.5Q4 = 700 A4 = 0.5Q5 = 2450 A5 = 1.75Q6 = 700 A6 = 0.5P = 3000

Rav = 5350

Rbv = 5350

Mx = 13781.25



e.Mencari P perlu

Pb = 0.0417 B1 = 0.79

Pmax = 0.0313 Pmin = 0.0038

Pperlu = 0.018


K = 5.82


dibulatkan 664


= 3495



kontrol = 0.0185 Pmin = 0.0038

P ada > P min0.0185 > 0.0038

p max = 0.0313

P ada < P max0.0185 < 0.0313







d = 360



m = 11.7647 367.45576077 < 608.6719


P = 0.03132.m.k/fy 0.6010

P > Pmax 1-2.m.k/fy 0.39900.0313 > 0.0313 akar 0.6316

P = 0.0282 1-akar 0.36841/m 0.0850



Cb = 222.680412371es' = 0.00228148148Ey = 0.16217532468



As 8043

300 > 285a = 185.901960784 Nt = 2975.91h = 400

NT ada = 2975910ND = 1753.985Mn = 859.415974657

Mn perlu < Mn608.6719 < 859.415974657


H min = 500 > 400



n As = 53620.000 d 360n AS' = 22016.667 d' 40

h 400



= 2729702016.15 mm4



= 34.0634701697 KNm


Ie = 2729315283.44


Ie = 2728731406.3 ml 200.9765625





ASL < ALL




13.193361 < 55.5555555556

ALL13.193361 < 27.7777777778

ALT17.042978 < 20.8333333333

ALT17.042978 < 41.6666666667

0.75 4.3

Mu = 25442.054= 254.42054

o50 200

- 170.5 )2

1 1.75 1

o28

mm


0.75750


diketahui

Mu = 486.9375b/h = 300 500



bE = 2500bE = 5420bE = 4800

Q1Mu/o 608.671875 Q278625



fyx1 = 697.81221 VDx2 = 22.187788 VL

maka a = 22 mm VA



6 3668000 > 6107.4




781.0324474 > 608.671875


Mu/ф


486.9375 Y

X + Y


b2-4ac 133533.61129 -133533.611 S12a 300


KontrolVs begel

17476.871

begel bag.2Vs1S2

banyak begel bag.2n

KontrolVs begel

8738.4354

digunakan ф 12

tulangan geser



Sm4.3

43002.sudut tulangan miring

STRUKTUR BETON DASAR ᾀ =



= 1.2 Wd + 1.6 Wl= 34.8 Kn/ m2 = 3480 Kg/m2

sin a == 2610 cos a == 14964

= 10319.616 dari tabel 10= 7254.3837 As 8043

Rencana tulangan geser Tulangan geser dengan tulangan kombinasi= 10000 10= 37 Vn == 370 Vc == 53.5 Vs begel == 10500 105 Av == 10319.616



= 109.48973 Vs4



= 17476.871Vs

Vs 16177.335Vs1 15135.499

17586.36 54.744863 Vs2 14160.758Vs3 13248.792Vs4 12395.557


1.Bila Vn < 0.5 Vc

2.Bila 0.5 Vc < Vs < 1/6 √fc' bw.d

54.744863 17476.871 333.16662Vs ada =

=109.48973 17476.871 218.97945

4218.97945 17476.871 437.9589

517476.871 437.9589


= 4968.8708 X = 31.12916

= 5000

= 2484.4354 1/2 Vc = 54.744863


= 2


Vs perlu S1 ada = 1.723983717476.871

begel bag.2= 8738.4354= 3


Vs perlu S2 ada = 3.44796748738.4354

3.Vc < Vs < 1/3 √fc'.bw.d

digunakan ф 12

d-d'320 = 0.32 m


17476.870744968.8708401 = 4.9688708

Vs Astm as tm ada Vs tm17476.87074 29688.922 8043 13663.08716351.34367 27776.927 4021.5 6831.5434

15298.301612 25988.067 4021.5 6831.543414313.076463 24314.411 4021.5 6831.543413391.300748 22748.54 4021.5 6831.5434

0.7071067810.707106781

o32


17586.360465 Fy 370109.48972555 d 3608738.4353698

226.2

Vs S S ada Vs ada8738.4353698 3.448 162 185.9878175.6718349 3.6853045 200 150.6497649.150806 3.9389784 240 125.541

7156.5382313 4.2101137 280 107.6076695.6503741 4.4999124 350 86.085

17290.884 sin a = 0.70710684.9688708401 cos a = 0.7071068

0.32 320 sin a+cos a = 1.4142136


2367.3219082 + 185.987 > 17586.36 - 109.489732553.309 > 17476.871

Perencanaan kolom persegi Perencanaan kolom bundardik Fc' 37 dik

Fy 370

Mu 486.9375Mu' 162.3125Vu 10551.81628Lk 3.75 3750b 300 0.3h 400 0.4d' 40d 360l


Mn = 270.5208333Vu = 10551.81628Vk = 1080 = 10.8Vn = 17586.36047



As = As' = 1230daari tabel diperoleh luas tulangan 5 o 1571 mm2

20As tot = 3142 10 o

203142 > 2460

l = 0.013091667 1/2 l = 0.01025

0.013091667 > 0.01025


fy/Es0.0023213 > 0.00185

fs' = 492.2222222 > 370fs = 370 = 370


Pnb = 1618777.864= 1618.777864 kN > 10551.816

Mnb = 186162657.8 + 96869333 + 96869333= 379901324.4= 379.9013244

e = 0.234684037 = 234.684037e' = 74.68403713m = 11.76470588l = 0.01025

e'/d = 0.207455659d'/d = 0.111111111

1-e'/d 0.79254431-d'/d 0.8888889

Pn = 5809630.877 0.84250562= 5809.630877 0.91788105


a 576b 320

Perencanaan kolom bundarFc' 37Fy 370

Mu 486.9375Mu' 162.3125Vu 10551.816Lk 3.75 3750b 300 0.3h 400 0.4d' 40d 360

Ag 120000

π 3.1428571r = 195.40168418

D = 390.80336837 = 440

Mn = 202.890625Vn = 17586.360465

e = 0.006917842 = 300Vk + Vn = 29328.600775

h = 352

b = 432.14285714 = 440


oAs = As' = 1428.768 3 25

1472.7As tot = 2857.536 6 25

2945.4

P = 0.0105651687Ds = 369.6

d-d' = 246.4 = 250

Cb = 195.95876289 = 210

ab = 166.74Fs' = 499.42857143

Fs' > Fy499.42857 > 370

Pnb = 2497957.5771= 2497.9575771

1/2 h 1761/2 ab 83.37

Mnb = 213729656.36 + 76721779 + 76721779= 367173214.76= 367.17321476

eb = 0.1469893717= 146.98937168

e eb0.0069178 > 0.1469893717

D = 440Ds = 369.6Pg = 0.0193630729m = 11.764705882


= 0.1805606118= 0.0765410881 = 0.507052= 0.1995454545

Pnb = 1872321.0684= 1872.3210684

Kontrol Pn > Vn

1872.3211 > 17586.360465


Dc = 369.6Ac = 107331.84Ag = 152114.28571




diketahui = fc' = 27 Mpafy = 370 Mpawl = 9.5 KN/m2wd = 4.5 KN/m2


Wu = 1.2 Wd + 1.6 Wl= 1.2 x 4.5 + 1.6 x 9.5= 20.6 Kn/ m2 = 2060 Kg/m2

Q1 = 9682Q2 = 2266



Pb = 0.85 x 0.062027027 x 0.618556701= 0.03261214823



fc'= 0.01 x 370

27= 0.19

K = fc x w (1-0.59xw)= 27 x 0.19 x 0.8858254821= 4.6

dik. D = 0.8d = 146



= 0.03261214823 x 1000 x 146= 4756 mm2


a = 122.569150327Mn = 7602.6 x 370 x ( 145.8480795

= 237873924.19= 237.87392419

Mu < Mno

98.453359 < 237.87392419


= 1520.52 mm2 o10 50


0.8

1 1.6 1 1.2



Rav = 5292

Rbv = 5292

Mx = 19098



e.Mencari P perlu

Pb = 0.0417 B1 = 0.794

Pmax = 0.0313 Pmin = 0.0038

Pperlu = 0.018


K = 5.82


dibulatkan 580


= 3053



kontrol = 0.0277 Pmin = 0.0038

P ada > P min0.0277 > 0.0038

p max = 0.0313

P ada < P max0.0277 < 0.0313




mutu baja fy = 370 Mpab = 300h = 500Mu = 497.212 1

lendutan jangka panjang = 0.25 tahun 0.25


d = 450



m = 11.7647 574.1496262 < 621.5150


P = 0.03132.m.k/fy 0.6010

P > Pmax 1-2.m.k/fy 0.39900.0313 > 0.0313 akar 0.6316

P = 0.0282 1-akar 0.36841/m 0.0850



Cb = 278.350515464es' = 0.00228148148Ey = 0.16217532468 0.00185



As 4825.5

300 > 285a = 173.619607843 Nt = 1785.435h = 500

NT ada = 1785435ND = 1638.101Mn = 653.875823386

Mn perlu < Mn621.5150 < 653.875823386


H min = 525 > 500


3.Luas transformasi

n = 6.993 b 300 Mun As = 33744.755 d 450n AS' = 2784.615 d' 50

h 500



= 2930365342.06 mm4



= 53.2241721401 KNm


Ie = 2930604080.37


Ie = 2930728356.26 ml 211.68





ASL < ALL




19.604633 < 46.6666666667

ALL19.604633 < 23.3333333333

ALT30.772088 < 17.5

ALT30.772088 < 35

4.7 1.1

Mu = 7876.26872= 78.7626872

o22 50

- 122.56915033 )2

1 1.6 1

0.60.2

k 497.2120.8 1875 202500

o32

mm


12

3

Mu = 497.212b/h = 300 650

p.Balok = 7500tebal plat = 170 mm

jarak balok = 5000 mm

bE = 1875bE = 3020bE = 5300

Mu/o 621.51558968.75

di kali dgn d 26.53593750.029484375

a 0.029484375 b2-4ac 25.116848b -26.5359375 2a 0.0589688c 621.515

x1 = 875.93489x2 = 24.065111

maka a = 24 mm

garis netralx = 30 < t/B1 = 214.10579

Perhitungan luas tulangan tarik(As)As = 3835.37701 3936.2 o

6 2927093.75 > 3936.2

Perhitungan balok tulanganMn = 637.39249 a = 24.697725

KontrolMn >

637.3924904 > 621.515

k 497212000 1.63691193

Mu/ф

303750000

p 0.085 0.19735629 0.0167752845

m 370 11.764705931.45

as 0.0167752845 1875 450 14154.146

497.212

a 150b 36529.371 -36529.371c -15324370.63

b2-4ac 102610.99988 -1026112a 300

208977.84

tulangan geser



Sm4700 11004.7 1.1

2.sudut tulangan miringSTRUKTUR BETON DASAR ᾀ

diketahui = fc' = 27 Mpa 3.Rumusfy = 370 Mpa Vs perluwl = 9.5 KN/m2 Ywd = 4.5 KN/m2


Wu = 1.2 Wd + 1.6 Wl= 20.6 Kn/ m2 = 2060 Kg/m2

sin aQ1 = 9682 cos aQ2 = 2266

Rav = 45.75829787234 dari tabelRbv = 73.72170212766 As

Rencana tulangan geser Tulangan geser dengan tulangan kombinasiL = 7500 7.5

fc' = 37 Vnfy = 370 VcVD = 52.92 Vs begelVL = 9000 90 AvVA =

1.Perhitungan beban rencanaVU = 207.504 Kn

2.Perhitungan kekuatan geser oleh betonVC = 136.86215693171

untuk tulangan miringVn = 345.84 Vs tm

y4.perhitungan kekuatan geser yang memerlukan tulangan geser sm

vs = 208.97784306829

5.kontrol VsVs1

345.84 68.431078 Vs2Vs3Vs4


1.Bila Vn < 0.5 Vc

2.Bila 0.5 Vc < Vs < 1/6 √fc' bw.d

68.431078 208.97784 372.49161 Vs ada

136.86216 208.97784 273.72431

4273.72431 208.97784 547.44863

5208.97784 547.44863


Y = 2265.981122791 X = 1484.0189

X + Y = 3750

1/2 Y = 1132.990561396 1/2 Vc = 68.431078

Penulangan geser dengan begel= 157 mm

S1 = 125

banyaknya begel bag.1n = 9 buah


208.977843 208.97784306829

begel bag.2Vs1 = 104.4889215341S2 = 250

banyak begel bag.2n = 5

KontrolVs begel Vs perlu S2 ada = 250.17485104.49 104.49

3.Vc < Vs < 1/3 √fc'.bw.d

digunakan ф 10

d-d'400 = 0.4 m

2.sudut tulangan miring= 45

= 208.97784307= 2265.9811228 = 2.2659811


= 0.707106781= 0.707106781

2840180.7263o

6 32 345.84 208.977844825.5


= 345.84 Fy 370= 136.86215693 d 450= 104.48892153= 157

Vs S S ada Vs adaVs 104.49 250.175 250 104.562

Vs1 86.04 303.80344 300 87.135Vs2 70.86 368.92809 370 70.650Vs3 58.35 448.01316 450 58.090Vs4 48.05 544.05125 550 47.528

untuk tulangan miring= 104.416 sin a = 0.7071068= 2.2659811228 cos a = 0.7071068= 0.4 400 sin a+cos a = 1.4142136

Vs As tm As tm ada Vs tm85.983943 146.066 2412.75 1420.304870.805716 120.282 2412.7558.306809 99.049 2412.7548.014259 81.564 2412.7539.538591 67.166 2412.75

= 1420.304845 + 104.562 > 345.84 - 136.86216= 1524.867 > 208.97784

37.5


Fy 370

Mu 497.212Mu' 248.606Vu 207.504Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450l 2.05


Mn = 414.3433333Vu = 207.504Vk = 1350 = 13.5Vn = 345.84




20As tot = 3142 10 o

203142 > 3075

l = 0.010473333 1/2 l = 0.01025

0.010473333 > 0.01025


fy/Es0.0023213 > 0.00185

fs' = 492.2222222 > 370fs = 370 = 370


Pnb = 2035824.318= 2035.824318 kN > 207.504

Mnb = 290879152.8 + 151358333 + 151358333= 593595819.4= 593.5958194

eb = 0.291575169 = 291.575169eb < e

291.57517 < 1153.0677e' = 491.5751689m = 11.76470588l = 0.01025

e'/d = 1.092389264d'/d = 0.111111111

1-e'/d -0.0923891-d'/d 0.8888889

Pn = 1612320.758 0.22291486= 1612.320758 0.4721386


a 576b 320


Fy 370

Mu 497.212Mu' 248.606Vu 207.504Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450

Ag 150000

π 3.1428571r = 218.46572438

D = 436.93144875 = 400

Mn = 310.7575Vn = 345.84

e = 0.5188804475 = 260Vk + Vn = 598.9

h = 320

b = 392.85714286 = 400


oAs = As' = 1180.8 3 25

1472.7As tot = 2945.4 6 25

2945.4

P = 0.0127838542Ds = 336

d-d' = 224 = 230

Cb = 178.1443299 = 190

ab = 150.86Fs' = 498.94736842

Fs' > Fy498.94737 > 370

Pnb = 2087719.5895= 2087.7195895

1/2 h 1601/2 ab 75.43

Mnb = 160498535.92 + 69747072 + 69747072= 299992679.92= 299.99267992

eb = 0.1436939527= 143.69395269

e eb0.5188804 > 0.1436939527

D = 400Ds = 336Pg = 0.0234293182m = 11.764705882


= 0.1500486304= 0.0926147166 = 0.4926087= 0.1725

Pnb = 1610787.0584= 1610.7870584

Kontrol Pn > Vn

1610.7871 > 345.84


Dc = 336Ac = 88704Ag = 125714.28571




diketahui = fc' = 27 Mpafy = 370 Mpawl = 9.5 KN/m2wd = 4.5 KN/m2


Wu = 1.2 Wd + 1.6 Wl= 1.2 x 4.5 + 1.6 x 9.5= 20.6 Kn/ m2 = 2060 Kg/m2

Q1 = 9682Q2 = 2266



Pb = 0.85 x 0.062027027 x 0.618556701= 0.03261214823



fc'= 0.01 x 370

27= 0.19

K = fc x w (1-0.59xw)= 27 x 0.19 x 0.8858254821= 4.6

dik. D = 0.8d = 146



= 0.03261214823 x 1000 x 146= 4756 mm2


a = 122.569150327Mn = 7602.6 x 370 x ( 145.8480795

= 237873924.19= 237.87392419

Mu < Mno

98.453359 < 237.87392419


= 1520.52 mm2 o10 50


0.8

1 1.6 1 1.2



Rav = 5292

Rbv = 5292

Mx = 19098



e.Mencari P perlu

Pb = 0.0417 B1 = 0.794

Pmax = 0.0313 Pmin = 0.0038

Pperlu = 0.018


K = 5.82


dibulatkan 580


= 3053



kontrol = 0.0277 Pmin = 0.0038

P ada > P min0.0277 > 0.0038

p max = 0.0313

P ada < P max0.0277 < 0.0313




mutu baja fy = 370 Mpab = 300h = 500Mu = 497.212 1

lendutan jangka panjang = 0.25 tahun 0.25


d = 450



m = 11.7647 574.1496262 < 621.5150


P = 0.03132.m.k/fy 0.6010

P > Pmax 1-2.m.k/fy 0.39900.0313 > 0.0313 akar 0.6316

P = 0.0282 1-akar 0.36841/m 0.0850



Cb = 278.350515464es' = 0.00228148148Ey = 0.16217532468 0.00185



As 4825.5

300 > 285a = 173.619607843 Nt = 1785.435h = 500

NT ada = 1785435ND = 1638.101Mn = 653.875823386

Mn perlu < Mn621.5150 < 653.875823386


H min = 525 > 500


3.Luas transformasi

n = 6.993 b 300 Mun As = 33744.755 d 450n AS' = 2784.615 d' 50

h 500



= 2930365342.06 mm4



= 53.2241721401 KNm


Ie = 2930604080.37


Ie = 2930728356.26 ml 211.68





ASL < ALL




19.604633 < 46.6666666667

ALL19.604633 < 23.3333333333

ALT30.772088 < 17.5

ALT30.772088 < 35

4.7 1.1

Mu = 7876.26872= 78.7626872

o22 50

- 122.56915033 )2

1 1.6 1

0.60.2

k 497.2120.8 1875 202500

o32

mm


12

3


diketahui

Mu = 497.212b/h = 300 650



bE = 1875bE = 3020bE = 5300

Q1Mu/o 621.515 Q2

58968.75di kali dgn d 26.5359375 Rav

0.029484375 Rbv


fyx1 = 875.93489 VDx2 = 24.065111 VL

maka a = 24 mm VA



6 2927093.75 > 3936.2




637.3924904 > 621.515


Mu/ф


497.212 Y

X + Y


b2-4ac 102610.99988 -102611 S12a 300


KontrolVs begel

285.24167

begel bag.2Vs1S2

banyak begel bag.2n

KontrolVs begel

142.62084

digunakan ф 13

tulangan geser



Sm4700 11004.7 1.1


= fc' = 27 Mpa 3.Rumusfy = 370 Mpa Vs perlu =wl = 9.5 KN/m2 Y =wd = 4.5 KN/m2


= 1.2 Wd + 1.6 Wl= 20.6 Kn/ m2 = 2060 Kg/m2

sin a == 9682 cos a == 2266

= 45.758298 dari tabel 6= 73.721702 As 4825.5

Rencana tulangan geser Tulangan geser dengan tulangan kombinasi= 7500 7.5= 37 Vn == 370 Vc == 52.92 Vs begel == 9000 90 Av == 45.758298



= 136.86216 Vs4



= 285.24167Vs

Vs 7.7829607Vs1 6.6860737

422.10383 68.431078 Vs2 5.7437759Vs3 4.9342803Vs4 4.2388705


1.Bila Vn < 0.5 Vc

2.Bila 0.5 Vc < Vs < 1/6 √fc' bw.d

68.431078 285.24167 372.49161Vs ada =

=136.86216 285.24167 273.72431

4273.72431 285.24167 547.44863

5285.24167 547.44863


= 2838.1998 X = 1361.8002

= 4200

= 1419.0999 1/2 Vc = 68.431078


= 155


Vs perlu S1 ada = 154.91811285.24167

begel bag.2= 142.62084= 310


Vs perlu S2 ada = 309.83621142.62084

3.Vc < Vs < 1/3 √fc'.bw.d

digunakan ф 13

d-d'400 = 0.4 m


285.241672862838.1998491 = 2.8381998


0.7071067810.707106781

o32


422.10382979 Fy 370136.86215693 d 450142.62083643

265.4

Vs S S ada Vs ada142.62 309.836 160 276.182122.52 360.66654 180 245.495105.25 419.83585 200 220.94590.42 488.71222 215 205.53177.68 568.88812 235 188.039

9.060 sin a = 0.70710682.8381998491 cos a = 0.7071068

0.4 400 sin a+cos a = 1.4142136


1420.304845 + 276.182 > 422.10383 - 136.862161696.487 > 285.24167


Fy 370

Mu 497.212Mu' 165.7373333Vu 253.2622979Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450l 2.05


Mn = 276.2288889Vu = 253.2622979Vk = 1350 = 13.5Vn = 422.1038298




20As tot = 3142 10 o

203142 > 3075

l = 0.010473333 1/2 l = 0.01025

0.010473333 > 0.01025


fy/Es0.0023213 > 0.00185

fs' = 492.2222222 > 370fs = 370 = 370


Pnb = 2035824.318= 2035.824318 kN > 253.2623

Mnb = 290879152.8 + 151358333 + 151358333= 593595819.4= 593.5958194

eb = 0.291575169 = 291.575169eb < e

291.57517 < 634.1287e' = 491.5751689m = 11.76470588l = 0.01025

e'/d = 1.092389264d'/d = 0.111111111

1-e'/d -0.0923891-d'/d 0.8888889

Pn = 1612320.758 0.22291486= 1612.320758 0.4721386


a 576b 320

Mu 497.212Mu' 165.73733Vu 253.2623Lk 3.75 3750b 300 0.3h 500 0.5d' 50d 450

Ag 150000

π 3.1428571r = 218.46572438

D = 436.93144875 = 400

Mn = 207.17166667Vn = 422.10382979

e = 0.2853579135 = 260Vk + Vn = 726.00638298

h = 320

b = 392.85714286 = 400


oAs = As' = 1180.8 3 25

1472.7As tot = 2945.4 6 25

2945.4

P = 0.0127838542Ds = 336

d-d' = 224 = 230

Cb = 178.1443299 = 190

ab = 150.86Fs' = 498.94736842

Fs' > Fy498.94737 > 370

Pnb = 2087719.5895= 2087.7195895

1/2 h 1601/2 ab 75.43

Mnb = 160498535.92 + 69747072 + 69747072= 299992679.92= 299.99267992

eb = 0.1436939527= 143.69395269

e eb0.2853579 > 0.1436939527

D = 400Ds = 336Pg = 0.0234293182m = 11.764705882


= 0.1500486304= 0.0926147166 = 0.4926087= 0.1725

Pnb = 1610787.0584= 1610.7870584

Kontrol Pn > Vn

1610.7871 > 422.10382979


Dc = 336Ac = 88704Ag = 125714.28571



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