adi beton

Soal 1Suatu SPBU dengan data - data sebagai berikut :

- a = 1.75 m - c = 4.3 m- b = 1.7 m - h = 4.5 m- Tebal pelat atap = 12.5 cm- Mutu Beton ( f'c ) = 25 Mpa- Mutu Baja ( fy ) = 300 Mpa

Beban berdasarkan peraturan pembebanan

ta :Dimensi dan jumlah kolam dan balok

1. Penentuan beban - beban yang bekerja a. BebanMati

Berat sendiri beton disesuaikan dengan dimensib. Beban Hidup

Berat pekerja dangan peralatannya = 100 ( Peraturan Muatan Indonesia Hal. 15 )

q2. Perencanaan balok yang Melajur ke depan

D B C

L( SK SNI 03 XXX 2002 Hal 63 )

8 h

Untuk fy selain 400 Mpa, nilainya harus dikalikan dengan 0,4 +fy

jadi :700

L( 0,4 +

fy) =

1750( 0.4

300)

A8 700 8 700

= 181.25 mm 200 mm a a

b diambil = 0.6 = 120 mm 150 mm

3. Perhitungan beban - beban yang bekerja :

t plat = 0.125 mt beton = 0.2 m

3.85 m

0.2 x 0.15 = 0.03 x 2400

= 72 kg/m

0.125 x 3.85= 0.48125 x 2400= 1155 kg/m

Jadi qd total = 72 + 1155 = 1227 kg/m

Beban hidup ql = 100 x 3.85 = 385 kg/m

Kombinasi Pembebanan

= 1,2 qd + 1,6 ql= 1472.4 + 616= 2088.4 kg/m

- = 0

- 0 = 0

= 0 kg m'

- ( 2088.4 x 3.5 ) - 0 = 0RA = 7309.4 kg

Kontrol :

0 + 12791.45 - 12791.45 = 00 = 0 (Oke dehh!!!)

kg/m2

Tinggi balok ( hmin ) untuk balok kantilever :

hmin =

hmin =

hmin

Berat sendiri balok qd balok = x bj beton ( gb )

Berat sendiri plat qd plat = x bj beton (gb )

qu

4. Perhitungan reaksi perletakan akibat qu dan pu

SMA = 0

MA + (qu.a).(1/2.a) - (qu.a).(1/2.a)

MA +

MA

SVA = 0

RA - (qu.L) = 0

RA

SMC = 0

MA + RA . 2 - 2(qu.a) 2 = 0

Persamaan gaya - gaya dalam :a). Bentang AB ( 0 < x < 4.5 )

Dx = 0 Mx = 0 kg m' Nx = -7309.4 kg

b). Bentang BD ( 0 < x < 1.75 ) Dx = ( qu . x ) Nx = 0Mx = -( qu .x )(1/2 x) Dx= 2088.4 x

Mx = -1044.2

untuk : x = 1.75 mMx = -3197.8625 kg m'Dx = 3654.7 kg

c). Bentang BC ( 0 < x < 1.75 )Mx = -( qu .x)(1/2x) = 0 Dx = - (qu . x ) Nx = 0

Mx = -1044.2 0 x Dx = 0 - 2088.4 x

untuk : x = 1.75 mMx = -3197.8625 kg m'Dx = -3654.7 kg

Gambar Diagram M dan N :

3654.7 kg-3197.8625 kg m'

-3198 kg m' +- -

--3654.7 kg

5). Perencanaan tulangan utamaUntuk balok tunggal ( berdiri sendiri ), lebar Be diambil nilai terkecil dari :

- Be = L/4 = 1750 4 = 437.5 mm- Be = Bw + 16 t = 150 + 16. 125 = 2150 mm- Be = Bw + Ln = 150 + ( 1750 150 ) = 1750 mm

ta : Bw = 150d = 22

= 200 - 40 - 10 - 11= 139 mm

f'c = 25 Mpafy = 300 Mpa

Mu = 3197.8625 kg m= 31978625 N mm

t = 125 mmBe = 437.5

Sketsa Tulangan :Tul. UtamaBe

t Tul. Sengkang

Tul. Praktis

Bw

Hitung : mm

Mu=

31978625= 39973281.25f 0.8

0,85 f'c. Be. t ( d -t

)2

0.85 25 437.5 125 ( 139 -125

)2

= 88901367.1875

Syarat :

Mn perlu >39973281.25 < 88901367.1875 Balok Persegi….!!!

Mx = - MA Nx = - RA

x2

x2

( h - selimut beton - f sengkang - 1/2 f tul. Utama )

Mnperlu =

Mn a=t =

Mn a=t =

Mn a=t

2 Mu

139 + 19321 -63957250

7437.5

242.5457 mm

139 - 19321 -63957250

7437.5

35.4543 mm

r = =0.85 25 437.5 35.454329485

Bw d fy 150 139 300

r = 0.0526961581827361

1.4=

1.4= 0.00466666667

fy 300

r >0.0526962 > 0.0046667 Oke Deh!!!

As = p Bw d = 0.052696158183 150 139

As = 1098.7149

Digunakan Tulangan :

4 f 22 = 1520.5308 380.13 380.13272 f 22 760.2654

6). Analisa terhadap tulangan yang terpasang d == 200 - 40 - 10 - 22 - 12.5

0,85 f'c)

600 = 115.5 mmfy 600 + fy

= 0.850.85 25 600

300 600 + 300

= 0.0401

0,85 f'c ( Be - Bw ) t= 0.85 25

( 437.5 - 150 )125fy Bw d 300 150 115.5

= 0.1469

Bw=

1500.040138888889 + 0.1469306157 = 0.0641Be 437.5

As=

1520.53084433746= 0.0301Be d 437.5 115.5

r <

Syarat : r <0.0301 < 0.0481 Oke deh…!!!

As£

1.4Bw d fy

0.0878 > 0.004666666666667 Oke deh!!!

w =As fy

=1098.71489811005 300

= 0.2609Be d f'c 437.5 115.5 25

1,18 w d£ t

41.8361 < 125 Analisa sbg balok biasa/persegi

a =As fy

=1520.5308 300

= 49.0659 mm0,85 f'c Be 0.85 25 437.5

As fy ( d -a

)2

= 1520.53084433746 300 ( 115.5 -49.0659

)2

= 41495468.6135787

aaktual = d ± d2 -0,85 f'c f Be

aaktual =

aaktual =

aaktual =

aaktual =

0,85 f'c Be aaktual

rmin =

rmin

mm2

mm2 mm2

mm2

( h - selimut beton - f sengkang - f tul. Utama - 1/2 x)

rb = b1 (

rf =

rb = ( rb + rf )

r =

rmaks

75 % rb

b1

Mnada =

Tingkat kekekonomisan Desain :

= x 100%

=39973281.25

x 100%41495468.6135787

= 96.3317 %

7). Perencanaan Tulangan GeserDesain Tulangan UtamaData :

Bw = 150 mmd = 115.5 mmf'c = 25 Mpafy = 300 MpaVu = 3654.7 kg

= 36547 N

Diagram Geser :Asumsi Kolom 25 25

x = 1 2 250 = 125-3654.7 kg

Vud=

VuL-(x+d) L

Vud Vud=

3654.701509.5 1750

115.5 mm Vud = 3152.4398 kgVud = 31524.398 N

1750 mm

Hitung :

Vn =Vud

=31524.398

= 42032.5307 Nf 0.75

Vc =1

f'c Bw d =1

25 150 115.5 = 14437.5000 N6 6

Pada posisi sepanjang Balok terdapat : 2

f'c Bw d < ( Vn - Vc )3

57750.0000 > 27595.0307 Penampang Cukup!!!

Vud >31524.398 > 8662.5 Tul. Geser struktural

Av =( Vn - Vc ) s d

57.75 mmfy d 2Nilai s max = 600 mm Nilai s = 50 mm

Av =42032.531 - 14437.5 50

300 115.5

Av = 39.81966907166910.5 39.81966907 = 19.909834535835

Digunakan : f 8 -100 mm = 50.286

Sketsa Tulangan :437.5 mm

125

200 mm

Mnperlu

Mnada

f Vc

s £

mm2

1 f =mm2

4 f 20 mm

f 8-100 mm

2 f 20 mm

8). Perencanaan Balok UtamaPelat

0.12 m

E C D F

4.5 m

Kolom

A B

1.7 m 4.3 m 1.7 m

a). Penentuan Dimensi BalokUntuk Bentang C - D

-Untuk Komponen Struktur Balok Kedua Ujung Menerus

LSK SNI 03 XXX 2002 Hal. 6321


700Jadi :

L21

=4300

21

= 169.65986 mm = 200 mm

Lebar Balok :

0.6 = 133.333 mm

Jadi dimensi balok :b = 133.333 mmh = 200 mm

Untuk Bentang E - C = F - D

-Untuk komponen struktur balok kantilever

LSK SNI 03 XXX 2002 Hal. 638


700Jadi:

L8

=1700

8

= 176.07143 mm = 200 mm

Lebar Balok :

0.6 = 150 mm 133.33Jadi dimensi Balok =

b = 150 mmh = 200 mm

Dari perhitungan diatas maka digunakan :b = 150 mmh = 200 mm

Tebal Balok Minimum ( hmin )

hmin =

hmin =

hmin

Tebal balok Minimum ( hmin )

hmin =

hmin =

hmin

- Perhitungan Yang Bekerja Pada Balok

Tebal plat = 0.125 mTebal Beton = 0.2 m

3.5 m

7.7 m

- Berat Sendiri balok qd balok = 0.2 x 0.15 x

= 0.03 x 2400 = 72 kg/m

- Berat Sendiri pelat qd pelat = 0.125 x 3.5 x

= 0.4375 x 2400 = 1050 kg/m

Jadi, qd total = 1122 kg/m

Beban hidup

ql = 100 x 3.5 = 350 kg/m

Kombinasi pembebanan :qu = 1,2 qd +1,6 ql

= 1346.4 + 560= 1906.4 kg/m

qu

E B C F

4.5 m

A D

1.7 4.3 1.7

Menghitung Momen Primer ( FEM )

0

0

-112

= -11906.40 18.4912

= -2937.444667 kg m

2937.444667 kg m

0

0Menghitung Momen dari balok yang melajur ke samping

1=

11906.4 2.89

2 2= 2754.748 kg m

qu

B C

h

A D

L2

Menentukan perbandingan kekakuan batang

1

I2 h=

1 12 15 8000 4.5=

45000= 0.3215

L I1 1 12 25 15625 4.3 139974

1

Beban mati ( qd )

bj beton ( gc )

kg/m3

bj beton ( gc )

kg/m3

b). Perhitungan reaksi perletakan akibat qu

M' AB =

M' BA =

M' BC = qu L2

M' CB =

M' CD =

M' DC =

M BE = M CF = qu L2

kAB =

kBC =

KCD =

Menurunkan persamaan Slope Deflectionrumus :

o

= =o

= =

= -2937.44466666667 + 0.3215

= 2937.44466666667 + 0.3215o

= =o

= =

Syarat sambungan :

= 0

+ -2937.44466667 + 0.643 + 0.3215 + 2754.748 = 0

2.64298 + 0.3215 = 182.6966666667 …………. ( 1 )

= 0

2937.444666667 + 0.3215 + 0.643 + - 2754.748 = 0

0.3215 + 2.643 = -182.696666667 …………….( 2 )

Eliminasi Persamaan (1) dan (2)

2.643 + 0.3215 = 182.696666667 x 0.3215

0.3215 + 2.6429767 = -182.69666667 x 2.6430

0.1033548 = 58.7349

6.9853 = -482.8630

-6.8820 = 541.5979

= -78.6981

2.643 + 0.3215 = 182.696666667

= 78.6981

= 78.6981 kg.m= 157.3962 kg.m

= -2912.1442 kg.m

= 2912.1442 kg.m

= -157.3962 kg.m

= -78.6981 kg.m

Kontrol :

157.3962 + -2912.1442 + 2754.748 = 0

0 = 0 Oke deh…!!!

2912.1442 + -157.3962 - 2754.748 = 00 = 0 Oke deh…!!!

qu = 1906.4 kg/m

B C

4.5 m

Ha A D Hd

Va Vd

4.3 m

Reaksi Perletakan

4.3 - 1 2 1906.4 18.49 = 0

4.3 = 17624.668

= 4098.76 kg

= 52.4654 kg4.5

Mxy = MF XY + kxy . ( 2fx fy )

MAB 0 + 1 . ( 2qA + qB ) qB

MBA 0 + 1 . ( 2qB + qA ) 2qB

MBC ( 2qB + qC )

MCB ( 2qC + qB )

MCD 0 + 1 . ( 2qC + qD ) 2qC

MDC 0 + 1 . ( 2qD + qC ) qC

SMB

MBA + MBC + MBE = 0

2qB qB qC

qB qC

SMC

MCB + MCD - MCF = 0

qB qC 2qC

qB qC

qB qC

qB qC

qC

qC

qC

qC

Subtitusi niLai qC ke persamaan (1), maka:

qB qC

qB

Subtitusi nilai qB dan qC ke persamaan Slope Deflection, maka didapat :

MAB

MBA

MBC

MCB

MCD

MDC

MBA + MBC + MBE = 0

MCB + MCD - MCF = 0

SMD = 0

VA

VA

VA

SH = 0

HA = - HD =MAB + MBA

Menentukan Momen MaxBentang BC ( 0 < x < 4.3 )

Mx =

= 4098.76 x - 953.2 + -2912.1442

= -953.2 + 4098.76 x - 2912.1442Dx = -1906.4 x + 4098.76

-1906.4 x + 4098.76 = 0

x = 2.15 m = 1494.0228 kg m

Gambar Diagram Momen dan Geser

2912.1442 kg m' 2912.14415056 kg m'

2754.7480 kg m' 2754.748 kg m'

-157.3962 kg m' -157.3962 kg m'1494.0228 kg m'

78.6981 kg m' 78.6980752792 kg m'

4098.76 kg3240.88 kg

3240.8800 kg 4098.76 kg

52.4654 kg 52.4654 kg

9). Perencanaan Tulangan pada Tumpuan

Untuk Balok T dari suatu sistem lantai, lebar Be diambil dari nilai terkecil dari :- Be = L/4 = 4300 4 = 1075 mm- Be = Bw + 16 t = 150 + 16. 125 = 2150 mm- Be = Bw + Ln = 150 + ( 4300 150 ) = 4300 mm

Data :Bw = 150 mm

d = 19= 200 - 40 - 10 - 9.5= 140.5 mm


Mu = 2754.7480 kg m= 27547480 N mm

t = 125 mmBe = 1075 mm

Hitung :Mu

=27547480

= 34434350f 0.8

0,85 f'c. Be. t ( d -t

)2

0.85 25 1075 125 ( 140.5 -125

)2

= 222726562.5

Syarat :

Mn perlu >34434350 < 222726562.5 Balok Persegi….!!!

2 Mu

140.5 + 19740.3 -55094960

18275

VA . x - 1/2 qu . x2 + MBC

x2

x2

Mmax pada saat D=0

Mmax


Mnperlu =

Mn a=t =

Mn a=t =

Mn a=t

aaktual = d ± d2 - 0,85 f'c f Be

aaktual =

3014.8

269.8270 mm 16725

140.5 - 19740.3 -55094960

18275

11.1730 mm

r = =0.85 25 1075 11.1730

Bw d fy 150 140.5 300

r = 0.0404

1.4=

1.4= 0.00466666667fy 300

r >0.040369 > 0.0046667 Oke Deh!!!

As = = 0.040368964667 150 140.5

As = 850.7759304


4 f 19 = 1134.1149 283.52872 f 19 = 567.0575

10). Analisa terhadap tulangan yang terpasang d = ###= 200 - 40 - 10 - 19 - 12.5

0,85 f'c)

600 = 118.5 mmfy 600 + fy

= 0.850.85 25 600 0.071 0.6667

300 600 + 300

= 0.0401388888888889

0,85 f'c ( Be - Bw ) t= 0.85 25

( 1075 - 150 )125fy Bw d 300 150 118.5

= 0.460765353961557

Bw=

1500.040138888889 + 0.460765354 = 0.069893615281Be 1075

As=

1134.11494794592= 0.00890287467723219Be d 1075 118.5

Syarat : r <0.00890287467723 < 0.052420211461093 Oke deh…!!!

As£

1.4Bw d fy

0.06380393518683 > 0.004666666666667 Oke deh!!!

w =As fy

=1134.11494794592 300

= 0.106834496126786Be d f'c 1075 118.5 25

1,18 w d£ t


a =As fy

=1134.1149479459 300

= 14.893985636499 mm0,85 f'c Be 0.85 25 1075

As fy ( d -a

)2

= 1134.11494794592 300 ( 118.5 -14.8939856365

)2

= 37784062.6377505


= x 100%

=34434350

x 100%37784062.6377505

= 91.1346

aaktual =

aaktual =

aaktual =

0,85 f'c Be aaktual

rmin =

rmin

r Bw d

mm2

mm2


rb = b1 (

rf =

rb = ( rb + rf )

r =

75 % rb

b1

Mnada =

Mnperlu

Mnada

`11). Perencanaan Tulangan pada Lapangan

Untuk Balok T dari suatu sistem lantai, lebar Be diambil dari nilai terkecil dari :- Be = L/4 = 4300 4 = 1075 mm- Be = Bw + 16 t = 150 + 16. 125 = 2150 mm- Be = Bw + Ln = 150 + ( 4300 150 ) = 4300 mm

Data :Bw = 150 mm

d = 12= 200 - 40 - 10 - 6= 144 mm


Mu = 1494.0228 kg m= 14940228.4944168 N mm

t = 125 mmBe = 1075 mm

Hitung :Mu

=14940228.494417

= 18675285.618f 0.8

0,85 f'c. Be. t ( d -t

)2

0.85 25 1075 125 ( 144 -125

)2

= 232720703.125

Syarat :

Mn perlu >18675285.6180211 < 232720703.125 Balok Persegi….!!!

2 Mu

144 + 20736 -29880456.9888337

182751635

282.206202733218 mm 19101

144 - 20736 -29880456.9888337

18275

5.79379726678246 mm

r = =0.85 25 1075 5.7937972668

Bw d fy 150 144 300

r = 0.020424700048312

1.4=

1.4= 0.00466666667fy 300

r >0.0204247 > 0.0046667 Oke Deh!!!

As = = 0.020424700048 150 144

As = 441.173521


4 f 12 = 452.3893 113.09732 f 12 226.19472 f 12 = 226.1947


Mnperlu =

Mn a=t =

Mn a=t =

Mn a=t

aaktual = d ± d2 - 0,85 f'c f Be

aaktual =

aaktual =

aaktual =

aaktual =

0,85 f'c Be aaktual

rmin =

rmin

r Bw d

mm2

mm2

12). Analisa terhadap tulangan yang terpasang d = ###= 200 - 40 - 10 - 20

0,85 f'c)

600 = 130 mmfy 600 + fy

= 0.850.85 25 600

300 600 + 300

= 0.0401388888888889

0,85 f'c ( Be - Bw ) t= 0.85 25

( 1075 - 150 )125fy Bw d 300 150 144

= 0.379171489197531

Bw=

1500.040138888889 + 0.3791714892 = 0.058508424849Be 1075

As=

452.38934211693= 0.0029224117707812Be d 1075 144

Syarat : r <0.00292241177078 < 0.043881318636951 Oke deh…!!!

As£

1.4Bw d fy

0.02094395102393 > 0.004666666666667 Oke deh!!!

w =As fy

=452.38934211693 300

= 0.0350689412493744Be d f'c 1075 144 25

1,18 w d£ t


a =As fy

=452.38934211693 300

= 5.94109122342343 mm0,85 f'c Be 0.85 25 1075

As fy ( d -a

)2

= 452.38934211693 300 ( 144 -5.94109122342

)2

= 53991573.32 19140066.6269482


= x 100%

=18675285.6180211

x 100%19140066.6269482

= 97.5717

13). Perencanaan Tulangan GeserDesain Tulangan GeserData :

Bw = 150 mmd = 144 mmf'c = 25 Mpafy = 300 MpaVu = 4098.76 kg

Diagram Geser : 25 25

Vu x = 1 2 250 = 125

Vud Vud=

VuL-(x+d) L

Vud=

4098.81881 2150

144 mm Vud = 3585.9384 kg2150 mm Vud = 35859.384 N


rb = b1 (

rf =

rb = ( rb + rf )

r =

75 % rb

b1

Mnada =

Mnperlu

Mnada

C817

david: danger

E817

david: danger

Hitung :

Vn =Vud

=35859.384

= 47812.512 Nf 0.75

Vc =1

f'c Bw d =1

25 150 144 = 18000 N6 6

Pada posisi sepanjang Balok terdapat : 2

f'c Bw d < ( Vn - Vc )3

72000 > 29812.512 Oke dehh!!!

Vud >35859.384 > 10800 Tul. Geser struktural

Av =( Vn - Vc ) s d

72 mmfy d 2Nilai s max = 600 mm Nilai s = 50 mm

Av =47812.512 - 18000 50

300 144

Av = 34.50522222222220.5 34.50522222 = 17.252611111111

Digunakan : f 8 -100 mm = 50.24

.Sketsa Tulangan pada Tumpuan

1075 mm

125

150 mm

Sketsa Tulangan pada Lapangan

1075 mm

125

150 mm

14). Perencanaan KolomData - Data Perencanaan

- Dimensi Kolomb = 250 mmh = 250 mmd = 144 mm

= 2400- Mutu Beton (f'c) = 25 Mpa- Mutu Baja (fy) = 300 Mpa

- Beban Mati1.Beton = 0.2 x 0.15 = 252 Kg2.Beton = 0.2 x 0.15 = 277.2 Kg

Pelat = = 4042.5 kg

Kolom = = 675 kg

Σ = 5246.7 kgKombinasi Pembebanan

1,2 Pd + 1,6 Pl= 6296.04 + 2156= 8452.04 kg= 84520.4 N

- Beban HidupBerat seorang pekerja dan peralatannya = 100 . (B+0.5 C) . 2A = 1347 kg

- Momen yang bekerja =Mu = -157.3962 kg m' = -1573961.50558 N mm

Perhitungan kekakuan lenturEc = 4700 f'cEc = 4700 25Ec = 23500.0000 Mpa

1,2 Pd =

6296.041,2 Pd 1,6 Pl 6296.04 2156

0.7449

- Untuk Balok0.2 Ec Ib1 + b

f Vc

s £

mm2

1 f =mm2

4 f 20 mm

f 8- 100 mm

2 f 13 mm

2 f 10 mm

f 8- 100 mm

2 f 20 mm

gc kg/m3

x bj beton ( gb ) (2A)x bj beton ( gc ) (B+0.5C)

t . gc . (B+0.5C) 2A

b. h . gc . H

PU =

b =

b =

Eib =

=0.2 23500 1 12 150 200 3

1 + 0.744913653981761

= 269354302390.548

- Untuk Kolom0.4 Ec Ik1 + b

=0.4 23500 1 12 250 250 3

1 + 0.744913653981761

= 1.75361E+12

Eik =

Menentukan Faktor tekukBagian atas kolom :

Lb = 2 A + B + 0.5 C

= 7.35

1.75361E+12 4.52.69354E+11 7.35

10.6337

Bagian Bawah Kolom :0 ( Terjepit sempurna ) 0.1538

0.6154Dari nomogram faktor panjang efektif kolom diperoleh :

k = 1.46 > 1 ( Ok )

Menentukan radius Girasi ( r )IA

1 12 250 250 3

250 250

72.1688 mm

atau :0.289 h0.289 25072.25 mm

Di ambil r = 72.25 mm = 0.07225 m

Kontrol kelangsingan kolom

kLu

< 22r

1.464.5

< 220.0722590.934256 > 22 Kolom Langsing

Menentukan faktor reduksi kekuatanAgr = 250 250 = 62500 mmSyarat :

8452.04 = 0.1 25 625008452.04 < 156250 f =0,8

j =EIk / Lk

EIb / Lb

j =

j =

j =

r =

r =

r =

r =r =r =

Pu < 0,1 f'c Agr maka f = 0,8Pu > 0,1 f'c Agr maka f = 0,65

Rumus :1

> 11 -

ΣPu

Pc =

Pc =3.14 2 1753.61

1.46 4.5 2

Pc = 400.5541708

1> 1

1 -Σ Pu

1

184.5204

0.8 400.554

1.358254272

Mc = d MuMc = 1.358254272 -1573961.50558316Mc = -2137839.939 N mm

Menentukan Eksentrisitas ( e )

e =McPu

e =-2137839.939

84520.4

e = -25.29377451

15 + 0.03 h

15 + 0.03 250

22.5

> e gunakan e_min

e=

22.5= 0.09h 250

Sumbu Vertikal :

k =Pu

k =84520.4

0.85 25 0.8 250 250

k = 0.08Sumbu Horizontal :

ke

= 0.08 0.09h= 0.01

Syarat :Mu > Pu

-1573962 < 84520.4 TuL. 2 sisi

Faktor Pembesar Momen (d)

ds =

f Pc

p2 Eik( k Lu )2

ds =

f Pc

ds =

ds =

e min =

e min =

e min =

e min

Penentuan Harga r dengan menggunakan grafik

0,85 f'c f b h

Dari Tabel diperoleh rasio tulangan :r = 0.014 0.8

Syarat :

0.0112ambil yang terbesar

Menentukan Luas tulangan :As =

= 0.25 0.0112 250 250

= 175

Digunakan Tulangan : 2 f 12 = 226.1947

Total 8 f 12 = 904.7787

15). Menentukan tulangan geser Data : b = 250 mm

h = 250 mmd = 144 mm


Vud = 52.4654 kgVud = 524.6538 N

Hitung :

Vn =Vud

=524.65384

= 699.538f 0.75

Vc =1

f'c b d 1Nu

6 14 Ag

Vc = 0.1666667 35 250 144 140987.6875000

Vc = 33833.7182 N

Periksa :Vud >

524.653835194385 < 20300.2308966626 Tulangan Geser Minumum

Rumus :

Av =b s s = 72 mm3 fy s = 50 mm

Av =250 50

3 300

Av = 13.8888890.5 13.88888889 = 6.9444444444444

Digunakan :

8 100 mm = 50.24

Sketsa Tulangan :

100 mm250 mm

h

b250 mm

b =r = r b

0.01 ≤ r ≤ 0.06r =r =

1/4 r Agr

mm2

mm2

mm3

f Vc

Dimana s £ d/2 , 600 mm

1 f =

f mm2

8 f 12 mm

f 8

2). Diketahui suatu balok seperti gambar di bawah ini :

C A D E B

L1 L2 L3 L1

Catatan : 4 model pembebanan Data Data :

L1 = 3.15 m P1 = 3.2 tonL2 = 4.35 m P2 = 3.75 tonL3 = 3.25 mfy = 400 Mpaf'c = 32 Mpa

= 1.03 Ton/mDiminta :

a. Tentukan dimensi penampang balok yang memberi nilai ekonomisb. Tentukan tulangan balokc. Tentukan Volume Beton dan tulangan yang digunakan

1). Penentuan Dimensi Balok- Untuk komponen struktur Dua tumpuan sederhana

L18.5

=1075018.5

= 581.0811 mm = 600 mm

b = 2/3 = 2/3 600 = 400 mm = 400 mm

- Untuk komponen struktur Kantilever

L8

=3150

8

= 393.75 mm = 400 mm

b = 2/3 = 2/3 400 = 266.7 mm = 300 mm

2). Penentuan Beban yang bekerja- Beban Mati

0.4 0.6 2400 = 576 kg/m = 0.576 t/m

- Beban Hidup

1.03 t/m

1.2 + 1.6

= 1.2 0.576 + 1.6 1.03

= 2.3392 t/m

qdu = 0.6912 t/m

Kondisi 1 3.20 ton 3.75 ton

qu = = 2.3392 t/m

C A D E B

3.15 4.35 3.25 3.15Reaksi Perletakan

Va 10.75 - 13.90 ( 0.50 13.90 ) - 6.40 - 3.15 = 010.75 Va - 225.98 - 20.48 - 11.8125 = 010.75 Va - 258.27 = 0

Va = 24.0252

qLL

P1 P2

qLL

hmin =

hmin

hmin =

hmin

qDL = b.h.γb

qDL =

qLL =

Maka : qu = qDL qLL

åMB = 0

qu P1 P2

- Vb 10.75 + ( 10.75 ) ( 0.5 10.75 ) + 4.35 + 7.60 - ( 3.15 ) ( 0.5 3.15 ) = 0- Vb 10.75 + 135.2 + 13.92 + 28.5 - 11.60536 = 0- Vb 10.75 + 165.9765 = 0

Vb = 15.43968

Kontrol

Va + Vb = Q + P1 + P224.025201488 + 15.43968 = 32.51488 + 3.2 + 3.75

39.46488 = 39.46488 Oke !!!

Gaya - Gaya DalamBentang CA ( 0 £ x £ 3.15 )

C A

Mx =1 x Mx Dx Nx2 0 0 0 0

1 -1.1696 -2.3392 0

Mx =1

2.33922 -4.6784 -4.6784 0

2 3 -10.5264 -7.0176 03.15 -11.6054 -7.36848 0

= -1.1696

Dx = -2.3392 x

Bentang AD ( ( 0 £ x £ 4.35 )

A DVa

Mx = Va x - Da x - MA -1 x Mx Dx Nx2 0 -11.6054 16.6567 0

1 3.8818 14.3175 0

= 24.0252 x -7.36848 x -11.6054 - 0.5 2.3392 2 17.0297 11.9783 04.35 38.7196 6.4812 0

= 16.6567 x - 11.6054 - 1.1696

Dx = 16.6567 - 2.3392 x

Bentang DE ( 0 £ x £ 3.25 )P1

D EVd x Mx Dx Nx

0 38.7196 3.2812 0

Mx = 1 40.8312 0.9420 0

= 38.72 + 6.48 x - 3.2 x - 0.5 2.3392 2 40.6036 -1.3972 0

= 38.7196 + 3.2812 x - 1.17 3.25 37.0296 -4.3212 0Dx = 3.2812 - 2.3392 xMomen Maks = Dx = 0

3.281201 - 2.339 x = 0x = 1.4027 mEmenUhi

Mmaks = 41.0209

åMA = 0

qu P1 P2 qu

qu x2

x2

x2

qu x2

x2

x2

Md + Dd x - P1 x - 0,5 qu x2

x2

x2

Bentang DE (EB ( 0 £ x £ 3.15 )P2

E BVd

Mx = x Mx Dx Nx

= 37.03 - 4.32 x - 3.75 x - 0.5 2.3392 0 37.0296 -8.0712 0

= 37.0296 - 8.0712 x - 1.17 3.15 0.0000 -15.4397 0Dx = -8.07 - 2.3392 x

Gambar M dan D


qu = 2.339 t/m

C A D E B

3.15 4.35 3.25 3.15

-11.60536 tm 1.4027

0 - - 0

+

41.0209 tm

16.6567 t 6.4812

3.2812

4.3212 t

7.3685 t 8.0712 t15.4397 t


2.3392 t/m

0.6912

C A D E B

3.15 4.35 3.25 3.15

Reaksi Perletakan

Va 10.75 - ( 0.6912 3.15 ) ( 0.5 3.15 + 10.75 ) - ( 2.3392 4.35 ) ( 0.5 4.35 + 6.4 ) -( 2.3392 3.25 ) ( 0.5 3.25 + 3.15 ) - 3.20 6.4 - ( 2.3392 3.15 ) ( 0.5 3.15 ) - 3.75 3.15 = 0

10.75 Va - 26.83498 - 87.255084 - 36.30146 - 20.48 - 11.6054 - 11.81 = 010.75 Va - 194.289376 = 0

Va = 18.07343 ton

- Vb 10.75 + ( 2.339 3.15 ) ( 0.5 3.15 + 8 ) + 3.75 7.6 + ( 2.339 3.25 ) ( 0.5 3.25 + 4.35 ) + 3.2 4.35 +( 2.3392 4.35 ) ( 0.5 4.35 ) - ( 0.691 3.15 ) ( 0.5 3.15 )

- 10.75 Vb + 67.6058 + 28.5 + 45.42434 + 13.92 + 22.1318 - 3.429216 = 0- 10.75 Vb + 174.1527 = 0

Vb = 16.2002496744 ton

Kontrol Va + Vb = ( 0.691 3.15 ) + ( 2.339 10.75 ) + 3.2 + 3.7518.073430326 + 16.20025 = 2.17728 + 25.1464 + 3.2 + 3.75

34.27368 = 34.27368 Oke !!

Gaya Gaya Dalam

Bentang CA ( 0 £ x £ 3.15 )

C A

Mx =12 x Mx Dx Nx

0 0 0 0

Mx =1

0.69123.15 -3.4292 -2.1773 0

2

= -0.3456

Dx = -0.6912 x

Me - De x - P2 x - 0,5 qu x2 + Vd x

x2

x2

åMB = 0

åMA = 0

qd x2

x2

x2

Bentang AD ( ( 0 £ x £ 4.35 )

A DVa


4.35 43.58728 5.72063 0

= 18.0734 x -2.17728 x -3.42922 - 0.5 2.3392

= 15.8962 x - 3.42922 - 1.17

Dx = 15.8962 - 2.339 x

Bentang DE ( ( 0 £ x £ 3.25 )P1

D EVd

Mx = x Mx Dx Nx

= 43.5873 + 5.721 x - 3.2 x - 0.5 2.3392 0 43.5873 2.5206 0

= 43.5873 + 2.521 x - 1.17 3.25 39.4254 -5.0818 0Dx = 2.52063 - 2.339 xMomen Maksimum = Mx/Dx = 0

2.52063 - 2.339 x = 0x = 1.0776 memenuHi

Mmaks = 44.94535

Bentang EB ( 0 £ x £ 3.15 )P2

E BVd

Mx = x Mx Dx Nx

= 39.4254 - 5.082 x - 3.75 x - 0.5 2.3392 0 39.42543 -8.83177 0

= 39.4254 - 8.832 x - 1.17 3.15 0 -16.2002 0Dx = -8.8318 - 2.339 x

Kondisi 23.20 ton 3.75 ton

qu = 2.3392 t/m

qdu = 0.6912

C A D E B

3.15 4.35 3.25 3.15

1.078 m3.4292 tm

0

44.9453 tm15.8962 t

5.7206 t

2.5206 t

2.1773 t 5.0818 t

16.2002 t

Kondisi 3

3.2 ton 3.75 tonqu = 2.3392 t/m

qdu = 0.6912 t/m

C A D E B

3.15 4.35 3.25 3.15

qu x2

x2

x2

Mad + Dad x - P1 x - 0,5 qu x2

x2

x2

Mde - Dde x - P2 x - 0,5 qu x2

x2

x2

Reaksi Perletakan

Va 10.75 - ( 2.3392 3 ) ( 0.5 3.15 + 10.75 ) - ( 0.6912 4.35 ) ( 0.5 4.35 + 6.4 ) -( 0.6912 3.25 ) ( 0.5 3.25 + 3.15 ) - 3.2 6.4 - ( 0.691 3.15 ) ( 0.5 3.15 ) - 3.75 3.15 = 0

10.75 Va - 90.81652 - 25.78262 - 10.72656 - 20.48 - 3.42922 - 11.81 = 010.75 Va - 163.047416 = 0

Va = 15.1672 ton

- Vb 11 + ( 0.691 3.15 ) ( 0.5 3.15 + 8 ) + 4 7.6 + ( 0.691 3.25 ) ( 0.5 3.25 + 4.35 ) + 3.2 4.35 +( 0.6912 4.35 ) ( 0.5 4.35 ) - ( 2.339 3.15 ) ( 0.5 3.15 ) = 0

- 10.75 Vb + 19.97654 + 28.5 + 13.42224 + 13.92 + 6.53962 - 11.60536 = 0- 10.75 Vb + 70.75304 = 0

Vb = 6.581678512 ton

KontrolVa + Vb = ( 2.3392 3.15 ) + 0.691 10.75 + 3.2 + 3.75

15.16720149 + 6.5816785 = 7.36848 + 7.4304 + 3.2 + 3.7521.74888 = 21.7489 Ok !!

Menghitung Gaya Gaya DalamBentang CA ( 0 £ x £ 3.15 )

C A

Mx =12

Mx =1

2.3392x Mx Dx Nx

2 0 0 0 03.15 -11.6054 -7.36848 0

= -1.1696

Dx = -2.3392 x

Bentang AD ( ( 0 £ x £ 4.35 )

A DVa


4.35 15.7795 4.7920 0

= 15.1672 x -7.36848 x -11.6054 - 0.5 0.6912

= 7.7987 x - 11.6054 - 0.3456

Dx = 7.7987 - 0.6912 x

Bentang DE ( 0 £ x £ 3.25 )P1

D EVd

åMB = 0

åMA = 0

qu x2

x2

x2

qd x2

x2

x2

Mx = x Mx Dx Nx

= 15.779 + 4.792 x - 3.2 x - 0.5 0.6912 0 15.7795 1.5920 0


1.592 - 0.691 x = 0x = 2.30324 memenuHi

Mmaks = 17.3031

Bentang EB ( 0 £ x £ 3.15 )P2

E BVd

Mx = x Mx Dx Nx

= 17.303 + -0.6544 x - 3.75 x - 0.5 0.6912 0 17.3031 -4.4044 0

= 17.303 + -4.4044 x - 0.346 3.15 0.0000 -6.5817 0Dx = -4.404 - 0.6912 x

Kondisi 3

3.2 ton 3.75 ton2.339 t/m

0.6912

C A D E B

3.15 4.35 3.25 3.15

-11.60536 tm

0

17.30307 tm

7.799 t

-0.6544

4.4044-7.36848 t

6.5817 t

Kondisi 4

qu = 2.3392 t/m 3.2 ton 3.75 ton

qdu = 0.6912 t/m

C A D E B

3.15 4.35 3.25 3.15

Reaksi Perletakan

Va 10.75 - ( 2.3392 13.90 ) ( 0.5 13.90 ) - ( 0.6912 13.90 ) ( 0.5 13.90 ) - 3.2 6.4 - 3.75 3.1510.75 Va - 225.9784 - 66.77338 - 20.48 - 11.8125 = 010.75 Va - 325.044292 = 0

Va = 30.2367 ton

- Vb 10.75 - ( 3.15 2.3392 ) ( 0.5 3.15 ) - ( 0.6912 3.15 ) ( 0.5 3.15 ) + 3.2 4.35 + 3.75 7.6 +( 2.3392 10.75 ) ( 0.5 10.75) + ( 0.691 10.75 ) ( 0.5 10.75 ) = 0

- 10.75 Vb - 11.60536 - 3.429216 + 13.92 + 28.5 + 135.162 + 39.9384 = 0- 10.75 Vb + 202.4857 = 0

Vb = 18.83588167 ton

KontrolVa + Vb = ( 2.3392 13.9 ) + ( 0.6912 13.9 ) + 3.2 + 3.75

30.23667833 + 18.835882 = 32.5149 + 9.60768 + 3.2 + 3.7549.07256 = 49.07256 Ok !!

Mad + Dad x - P1 x - 0,5 qd x2

x2

x2

Mde + Dde x - P2 x - 0,5 qd x2

x2

x2

åMB = 0

åMA = 0

Menghitung Gaya Gaya DalamBentang CA ( 0 £ x £ 3.15 )

C A

Mx =1

-1

2 2

Mx =1

2.33921

0.6912x Mx Dx Nx

2 2 0 0 0 03.15 -15.0346 -9.5458 0

= -1.5152

Dx = -3.0304 x

Bentang AD ( ( 0 £ x £ 4.35 )

A DVa

Mx = Va x - Da x - MA -1

-1 x Mx Dx Nx

2 2 0 -15.0346 20.6909 04.35 46.2996 7.5087 0

= 30.2367 x -9.54576 x -15.0346 - 0.5 2.3392 - 0.5 0.691

= 20.6909 x - 15.0346 - 1.5152

Dx = 20.69092 - 3.0304 x

Bentang DE ( 0 £ x £ 3.25 )P1

D EVd

Mx = x Mx Dx Nx

= 46.3 + 7.50868 x - 3.2 x - 0.5 3.0304 0 46.2996 4.3087 0


4.30868 - 3.03 x = 0x = 1.42182 memenuHi

Mmaks = 49.36263

Bentang EB ( 0 £ x £ 3.15 )P2

E BVd

Mx = x Mx Dx Nx

= 44.298 + -5.5401 x - 3.75 x - 0.5 3.0304 0 44.2985 -9.2901 0

= 44.298 + -9.2901 x - 1.515 3.15 0.0000 -18.8359 0Dx = -9.29 - 3.0304 x

Kondisi 4

2.339 t/m 3.2 ton 3.75 ton

0.6912

C A D E B

3.15 4.35 3.25 3.15

-15.03457 tm

0

49.36263 tm

qu x2 qd x2

x2 x2

x2

qu x2 qd x2

x2 x2

x2

Mad + Dad x - P1 x - 0,5 qu x2 - 0,5 qd x2

x2

x2

Mde + Dde x - P2 x - 0,5 qu x2 - qd x2

x2

x2

20.6909 t

7.5087 t

4.3087 t

5.5401 t

9.2901 t-9.54576 t

18.8359 t

3). Rekapitulasi hasil peninjauan tiap kondisi* Kombinasi momen lentur ( Mu )

KondisiMomen tiap titik ( ton/m )

C A D Mmaks E BI 0 -11.6054 38.7196 41.0209 37.0296 0II 0 -3.4292 43.5873 44.9453 39.4254 0III 0 -11.6054 15.7795 17.3031 17.3031 0IV 0 -15.0346 46.2996 49.3626 44.2985 0

* Kombinasi gaya geser ( D )

Bentang C - A A - D D - E E - BTitik C A A D D E E B

Kondisi I 0 -7.3685 16.6567 6.4812 3.2812 -4.3212 -8.0712 -15.4397Kondisi II 0 -2.1773 15.8962 5.7206 2.5206 -5.0818 -8.8318 -16.2002Kondisi III 0 -7.3685 7.7987 4.7920 1.5920 -0.6544 -4.4044 -6.5817Kondisi IV 0 -9.5458 20.6909 7.5087 4.3087 -5.5401 -9.2901 -18.8359

Dari tabel kombinasi diperoleh :

= -15.0346 tm

= 49.3626 tmGeser Maksimum = 20.6909 ton

4). Penulangan pada balok* Perencanaa penulanagan pada tumpuan

Data : Mu = 15.0346 tm= 150345720 Nmm

b = 400 mmd = h - selimut beton - - 1/2 tul. Utama

= 600 - 40 - 8 - 12= 546 mm

f'c = 32 MPafy = 400 MPa

Hitung :

1.4=

1.4=

0.0035fy 400

0.85

0.85 - 8f'c - 30 untuk 30 < fc < 55 Mpa

1000

0.65

0.75 0.85 (0.85 f'c

)600

fy 600 + fy

= 0.75 0.85 (0.85 32

)600

400 600 + 400

= 0.0260

=600

d600 + fy

= 0.85600

546 = 278.46 mm600 + 400

= 75%

= 75% 278.46 = 208.845 mm

= 0.85 f'c b d -2

= 0.85 32 400 mm 208.845 546 mm -208.845

= 1,003,367,232.502

Momen Tumpuan ( Mu ) -

Momen Lapangan ( Mu )+

f sengkang

rmin =

b1 =

untuk 0 £ fc £ 30 Mpa

untuk fc ³ 55 Mpa

rmaks =

ab b1

amax ab

Mnada amaxamax

=Muf

=150,345,720

= 187,932,1500.8

Syarat :Mn ada > Mn perlu

1,003,367,233 > 187,932,150 tulangan tunggal…!!

m =fy

0.85 f'c

=400

= 14.7060.85 32

Rn =Mu

f b

=150,345,720 Nmm

= 1.57600.80 400 298,116

r =1

1 - 1 -2 m Rn

m fy

=1

1 - 1 -2 14.71 1.57599852

= 0.004114.71 400

Syarat :

r <0.0041 < 0.0260 OK…!!

Syarat :

r ³0.0041 > 0.0035 OK ...

As = r b d

= 0.0041 400 mm 546 mm = 886.9825

Digunakan tulangan : 2 f 25 = 981.7477

* Analisa terhadap tulangan yang terpasang

Jarak antar tulangan:

x =b - 2.d' - 2 Ø tul. Sengkang - n.Ø tul. Utama

n - 1

=400 - 80 - 20 - 50

1= 250 > ###mm Oke …!!!

d = h - selimut beton - - 1/2 tul. Utama= 600 - 40 - 8 - 25= 539.5 mm

r =A 2 Æ 25

b d

=981.7477

= 0.0045400 mm 539.5

=1.4fy

=1.4

= 0.0035400

Syarat :

r ³0.0045 > 0.0035 OK ...

=0.85 f'c 600

fy 600 + fy

= 0.850.85 32 600

= 0.0347400 600 + 400

= 75%

= 75% 0.0347 = 0.0260

Mnperlu

d2

rmax

rmin

mm2

mm2

f sengkang

rmin

rmin

rb b1

rmax rb

Syarat :

r <0.0045 < 0.0260 OK…!!

a =A 2 Æ 25 fy

0.85 f'c b

=981.7477 400

= 36.09370.85 32 400

= 0.85 f'c b a d -a2

= 0.85 32 400 36.09367 539.5 -36.0937

= 204,774,1802

=Muf

=150,345,720.00

= 187,932,150.000.8

Tingkat ekonomis =Mn perlu

x 100%Mn ada

=187,932,150

x 100%204,774,180

= 91.78 %

* Perencanaan penulangan lapangan

Data : Mu = 49.3626 tm= 493626296.685 Nmm

b = 400 mmd = h - selimut beton - - 1/2 tul. Utama

= 600 - 40 - 8 - 12= 546 mm

f'c = 32 MPafy = 400 MPa

Hitung :

1.4=

1.4= 0.0035fy 400

0.85

0.85 - 8f'c - 30 untuk 30 < fc < 55 Mpa

1000

0.65

0.75 0.85 (0.85 f'c

)600

fy 600 + fy

= 0.75 0.85 (0.85 32

)600

400 600 + 400

= 0.02601

=600

d600 + fy

= 0.85600

546 = 278.46 mm600 + 400

= 75%

= 75% 278.46 = 208.845 mm

= 0.85 f'c b d -2

= 0.85 32 400 mm 208.845 546 mm -208.845

= 1,003,367,232.502

=Muf

=493,626,297

= 617,032,8710.8

rmax

Mnada

Mnperlu

f sengkang

rmin =

b1 =

untuk 0 £ fc £ 30 Mpa

untuk fc ³ 55 Mpa

rmaks =

ab b1

amax ab

Mnada amaxamax

Mnperlu

Syarat :Mn ada > Mn perlu

1,003,367,233 > 617,032,871 tulangan tunggal…!!

m =fy

0.85 f'c

=400

= 14.7060.85 32

Rn =Mu

f b

=493,626,297 Nmm

= 5.174436048850.80 400 298,116

r =1

1 - 1 -2 m Rn

m fy

=1

1 - 1 -2 14.71 5.1744

= 0.01447718483314.71 400

Syarat :

r <0.0144772 < 0.0260100 OK…!!

Syarat :

r ³0.0144772 > 0.0035000 OK ...

As = r b d

= 0.0145 400 mm 546 mm = 3161.8172

Digunakan tulangan : 8 f 25 = 3926.990817

* Analisa terhadap tulangan yang terpasang

Jarak antar tulangan:


n - 1

=400 - 80 - 20 - ###

7= 14.2857 > ###mm Tul. SUSUN


n - 1

=400 - 80 - 20 - ###

3= 66.6667 > ###mm Oke …!!!

d = h - selimut beton - - 1/2 tul. Utama= 600 - 40 - 8 - 25= 539.5 mm

r =A 8 Æ 25

b d

=3926.9908

= 0.0181973624512847400 mm 539.5

=1.4fy

=1.4

= 0.0035400

Syarat :

r ³0.0181974 > 0.0035000 OK ...

=0.85 f'c 600

fy 600 + fy

= 0.850.85 32 600

= 0.03468400 600 + 400

= 75%

= 75% 0.03468 = 0.02601

Syarat :

r <0.0181974 < 0.0260100 OK…!!

d2

rmax

rmin

mm2

mm2

f sengkang

rmin

rmin

rb b1

rmax rb

rmax

a =A 8 Æ 25 fy

0.85 f'c b

=3926.9908 400

= 144.3746623892370.85 32 400

= 0.85 f'c b a d -a2

= 0.85 32 400 144.3746624 539.5 -144.3747

= 734,053,0242

=Muf

=493,626,296.68

= 617,032,870.860.8

Tingkat ekonomis =Mn perlu

x 100%Mn ada

=617,032,871

x 100%734,053,024

= 84.06 %

5). Perencanaan tulangan geserData : Vu = 20.6909 ton

= 206909 Nb = 400 mmd = 546 mmfy = 400 Mpaf'c = 32 Mpa

546 mmVud = 64812 +

( 4350 -546 ) ( 136090 - 64812 )4350

Vud

Vud = 64812 +348978178

64812 4350

136090 N Vud = 145036.883 N4350 mm

Hitung :

Vn =Vudf

=145036.883282448

= 193382.5110433 N0.75

Vc =1

fc bw d6

=1

32 400 546 = 205,909.49 N6

pada posisi sepanjang balok terdapat :

2fc bw d > Vn - Vc3

823637.978726091 > -12526.9836383 penampang cukup

periksa :Vud >

145,036.88 > 123,545.70 tulangan geser stRukturaL

s £d

, = 6002

diambil jarak sengkang s £d

=546

= 273 mm 250 mm2 2

Av =b s

=400 250

3 fy 3 400

Av =100000

= 83.3333331200

0.5 83.33333 = 41.6666667

Digunakan tulangan geser f 8 = 50.26548

Sketsa tulangan

Penulangan pada tumpuan Penulangan pada lapangan

60

0 m

m

60

0 m

m

2 f 25 2 Æ 25

Æ 8 - 250 mm Æ 8 - 250 mm

2 f 25 8 Æ 25

400 mm 400 mm

Mnada

Mnperlu

f Vc

smax

mm2

1 f = mm2

mm2

adi beton

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