Tw Pondasi

Diketahui dari data soal yang didapat : P = 5.0 ton DL = 36% P = LL = 64% P = MB = ML = 5.00 tm tm

50

1 .8 3 .2

ton ton 300

2

f'c

=

225

kg/cm

fy FK

= =

1910 3

kg/cm

2

B

Perhitungan untuk mendapatkan q c

qc diambil dari kedalaman tanah 0,20 s/d 2,60 meter ; maka didapatkan nilai qc : qc

 q n

=

364 15

=

=

2

24.26667

kg/cm

Perhitungan untuk mendapatkan q ult & qa

qult

5 + 0,34 qc

=

=

13.25066667

Maka dari perhitungan tersebut didapatkan qa

2

kg/cm



=

qult

=

13.250667



=

4.416889

kg/cm

=

44.16889 t/m

FK 3 2

2

Perhitungan dimensi pondasi telapak

e

=

M/P

;

e b

=

MB / P

=

5

=

 1.000

el

5

=

ML / P

m

= =

 1.000

q max =

P 2

B

(

1

±

(

1

±

6 e b

±

B

6 el

)

B

q min

q max

=

5 2

B

6.000 B

6.000

±

B

)

Dengan mencoba-coba nilai B : q mak < qa 2

B

B

q max

q min

q rata-rata

Rasio

0.85 0.90 0.95 1.00 1.50 2.00 2.50 3.00

0.72 0.81 0.90 1.00 2.25 4.00 6.25 9.00

104.62 88.48 75.52 65.00 20.00 8.75 4.64 2.78

-90.78 -76.13 -64.44 -55.00 -15.56 -6.25 -3.04 -1.67

6.92 6.17 5.54 5.00 2.22 1.25 0.80 0.56

0.07 0.07 0.07 0.08 0.11 0.14 0.17 0.20

Sehingga dipergunakan ukuran pondasi

1.50

x

1.50

150 cm

5

Perhitungan kedalaman pondasi efektif

Pult

=

1, 1,4DL

+

1,7LL

qult

=

q

=

(Pult/B ) (1+6e b/B+6el/B)

=

7.96

ton

2

=

=

7960

88.4 t/m

2

Untuk bujur sangkar digunakan Balok aksi dua arah dengan f Maka didapatkan Vc

=

1/2

1,060.f.(f c')

Dengan Dimensi kolom sebesar

40

= x

13.52

40 cm

2

2

=

kg 8.84

2

kg/cm

= 0.85 2 kg/cm

maka 2

Dengan menggunakan menggunakan rumus : d (vc+ q/4)+d(vc+ q/2) w = (B -w ) q/4

15.726

2

d

+

190.404

Dengan menggunakan rumus abc maka akan didapatkan nilai d

d

d

=

= = d positif d negatif Maka d efektif adalah

2

= = =

1/2

- b ± (b - 4ac ) 2a

= =

46212.22

-

190.404

±

1  2

2943208 2x 15.726 190.404 ± 1715.578 2 x 15.726 190.404 ± 1715.578 31.45 48.492 cm -60.599 cm 48.492 cm  diambil 48.0 cm

Karena d efektif yang dihasilkan > 305 mm maka tidak perlu menghitung d sesungguhnya.


Pult

=

1, 1,4DL

+

1,7LL

qult

=

q

=


=

7.96

ton

2

=

=

7960

88.4 t/m

2


=

1/2

1,060.f.(f c')


40

= x

13.52

40 cm

2

2

=

kg 8.84

2

kg/cm

= 0.85 2 kg/cm

maka 2

Dengan menggunakan menggunakan rumus : d (vc+ q/4)+d(vc+ q/2) w = (B -w ) q/4

15.726

2

d

+

190.404


d

d

=


2

= = =

1/2

- b ± (b - 4ac ) 2a

= =

46212.22

-

190.404

±

1  2

2943208 2x 15.726 190.404 ± 1715.578 2 x 15.726 190.404 ± 1715.578 31.45 48.492 cm -60.599 cm 48.492 cm  diambil 48.0 cm


Perhitungan baja / lebar satuan

Menghitung keadaan kritis (I)

=

( B-w ) / 2 150 2

= Maka Mult

=

40

55

cm

2

( qult . I ) / 2 8.84

=

x 2

3025

=

Dengan menggunakan persamaan dimana b lebar satuan

=

0.85 x 1623.5

13377.22

=

77928

Dengan menggunakan rumus abc Maka akan didapatkan nilai As

1910 As

x

As

-

As

As ( 48

=

=

0 .2

= =



9.99 As )

2

2

2

1/2

- b ± (b - 4ac ) 2a

=

=

d positif d negatif

( 48 48 4.99

8106.889 As

=

2

As

Mu = f As f y.( d - a/2 )

= =

0.2 cm 0 .2 x

9.99

1 cm

13377.22 13377.22

= =

13377.22 kgcm

=

Dengan menggunakan persamaan

Maka As Total As

=

± 6506563801 2 x 8106.889 77928 ± 80663.27 2 x 8106.889 77928 ± 80663.27 16213.78 0.169 cm -9.781 cm -

77928

1  2

 cm lebar B

x

150

=

2

cm

25.31

Lindungan beton minimum 7 cm Dipakai lindungan beton

9 cm

maka Befektif

= = =

Dipakai ba baja tu tulangan di diameter

B - 2 x lind lindung ungan an beton beton yang dipa dipaka kaii 150 132

2 cm

x

19 mm

maka luas penampangnya adalah

Juml Jumlaah tul tulan ang gan yang ang dipe diperrlukan ukan ( n )

Jarak tulangan = B efektif / ( n-1 )

2

= =

1/4 .p. diameter 1/4 . 22/7 . 3.61

= = = = =

2.836 cm As / luas luas pena penamp mpan ang g 25.31  2.836 8.921 dipakai 132 9 - 1 16.66 cm dipakai

=

2

9 buah

17 cm

9

Jadi dipakai total

= =

tulangan yang diperlukan x luas penampang 9 x 2.836

= 25.305 Sehingga As sesungguhnya = = =

2

cm Total As / B 25.305 0.169



150

2

cm / cm lebar satuan

Rasio baja sesungguhnya

0.004

= As / b x d = 0.169  1 x 48 = 0.004 > 0.002 OKE < Rasio baja ijin ( Pd ) yaitu 0,049 ( pada lampiran tabel ) OKE

Kedalaman pondasi ( Dc )

Maka Dc dibuat

59 cm

= = =

d ( yang dipakai ) + 1 diameter + lindungan beton 48 + 1.9 + 9 58.9 cm = 58.9 cm

dan d yang dipakai

Lindungan beton di bagian bawah

= = =

48.0 cm

Dc yang dibuat - d - diameter tulangan yang dipakai 59 - 48.0 1.9 9.1 cm

Sketsa gambar dimensi telapak pondasi dan pembesian telapak pondasi :

20  19

20  19

B= 180 cm

sketsa penulangan telapak

Dc = 81cm 180 cm

sketsa dimensi telapak

Memeriksa dukungan kolom.

A1 A2

= =

2

w

2

1600 cm

=

( w + 4d )

2

= =

sehingga Y

=

( A2 / A1 )

f c = 0,85 . f . f c' . y = =

=

40

x

325.1

kg/cm

0.85

=

1/2

) 53824  1600 5.8 > 2 maka Y yang digunakan adala 225

x

2

< fc maka digunakan angker minimum

0.005 .A1 0.005 x

=

2

1600

cm

8

16 mm 2

1/4 .p. diameter 1/4 . 22/7 . 2.56

=

= =

x

= =

= =

Angker yang diperlukan ( n )

2

48 )

1600 2

maka luas penampangnya

Jadi Aa yang dipakai

(

kg/cm

Dipakai baja tulangan diameter

x

2



Luas angker minimum ( Aa )

4

2

= =

0.85

4.975

+ cm

53824

1/2

7960

f a = Pult / A1 =

(

= = =

2

cm

2.011

Aa / Luas penampang 8  2.011 3.977 dipakai

4 batang

luas penampang x Angker yang diperlukan 2.011 x 4 8.05

2

cm

Sketsa gambar kolom dan pembesian angker :

40

40

sketsa kolom dan tulangan angker

4  16

2

5 m

As )

cm

P

=

115 ton

DL

=

36% P

=

41.4 ton

LL

=

64% P

=

73.6 ton

MB=ML

=

qc

=

35.4

+

25.3

2 qc

=

qult

=

2

30.3 kg/cm

10 tm 2

F'c

=

200 kg/cm

Fy FK

= =

1910 kg/cm 3

2

5

+

0.34 qc 2

15.3 kg/cm

= Maka didapatkan q a =

qult



FK 2

5.11 kg/cm

= Langkah 1 : Menghitung dimensi pondasi telapak

e

=

M  P

;

e b

MB  P

= =

el

0.087 meter

=

ML  P

=

0.087 meter

q max P

=

2

B

(

1

±

(

1

+

6 e b

6 el

±

B

)

B

q min

q max

=

115 2

B

Dengan mencoba-coba nilai B : q 2

0.522

+

B <

max

qa

0.522

)

B(m)

B

q max

q min

q rata-rata

Rasio

1.00 1.10 1.20 1.30 1.40 1.50 1.60

1.00 1.21 1.44 1.69 1.96 2.25 2.56

235.00 185.20 149.31 122.67 102.41 86.67 74.22

-5.00 4.88 10.42 13.43 14.94 15.56 15.63

115.00 95.04 79.86 68.05 58.67 51.11 44.92

0.49 0.51 0.53 0.55 0.57 0.59 0.61

sehingga dipakai ukuran pondasi

1.10

x

)

B

1.10

=

110 cm

Langkah 2 : Menghitung Kedalaman Pondasi Efektif (d)

Pult qult

=

1,4DL

=

q

+ =

1,7LL

=

183.08 ton

2


= =

183080 2

295 t/m

Untuk bujur sangkar digunakan Balok aksi dua arah dengan f Maka didapatkan

=

1/2

1,060.f.(f c')

Dengan dimensi kolom sebesar

40

= x

kg 2

29.48 kg/cm

= =

0.85

2

12.74 kg/cm

40 cm

maka ;

2

2

2

Dengan persamaan ( 4 - 11a) diperoleh rumus : d (vc+ q/4)+d(vc+ q/2) w = (B -w ) q/4 2

20.11 d

+

602.4 d

=

77394.70


d

= = = d positif d negatif

2

- b ± (b - 4ac ) 2a

=

= =

± ± x ± 40.23 48.83855 cm -78.790141 cm -

602.4 602.4 2 602.4

1/2

6589460.1 2566.9944 20.1 2566.9944

1  2

maka d efektif adalah

48.84

2

2

Dengan persamaan ( 4 - 11c) diperoleh rumus : 4d +4dw = B q / vc 2

4d

+

160 d

=

27998.032


d

= = d positif d negatif Maka di pakai d efektif =50.00 cm

2

- b ± (b - 4ac ) 2a

=

= =

± x ± 8 66.020394 cm -106.02039 cm -

160 2 160

1/2

473568.52 4 688.16315

1  2

maka d sesungguhnya adalah

karena d efektif > 305 mm

Langkah 3 : Menghitung baja / lebar satuan

Menghitung keadaan kritis (I) =

2

(B-w)/2

=

110

=

2

Maka M ult ( qult . I )/2 40

= 29.48

x 1225 2 = 18058.76 kgcm

35 cm

Dari persamaan ( 4 - 5 )

= 11.24 As

dimana b lebar satuan

=

1 cm

Dari persamaan ( 4 - 6 ) Mu = f As f y.( d - a/2 ) 18058.764 18058.764

= =

0.85 x 1623.5

18058.764

= -

9120.25

2

As

1910 As

x (

81175

As

-

9120.25

81175

As

=

18058.764


As

Maka As = Total As = =

( -

50

2

50 - 11.24  2 5.6176471 As ) 2

1/2

- b ± (b - 4ac ) 2a

=

± 2 x 81175 ± 18240.5 0.2171683 -81179.667

= =

-

81175

7248182398 9120.25 85136.258

2

0.2 cm / cm lebar 0.2 x B 0.2

x

110

=

Lindungan beton minimum 7 cm Dipakai lindungan beton 9 cm,

2

23.89 cm

maka B efektif = B-2 x lindungan beton yang dipakai = 110 = 92


16 mm

2 cm

x

9

2

maka luas penampangnya adala 1/4 . p .diameter = 1/4 . 22/7 . 1.6 2

= 2.0114286 cm Jumlah tulangan yang diperlukan (n) As / luas penampang = 23.89  2.0114286 = 11.87638978 dipakai 12 buah Jarak tulangan = B efektif / (n-1= 92 12 -1 = 8.363636364 cm dipakai 8 cm

As )

As

=

= As positif As negatrif

As

1  2

Jadi dipakai total As

= =


=

2

cm

24.13714286

Sehingga As sesungguhnya

total As = 24.137143 =


Kedalaman pondasi = ( Dc ) = = Maka Dc dibuat

= = =

As/bd 0.2194 0.004 > <

 

B 110

2

cm / cm lebar

0.2194

 1 x 50 0.002 (OK) Pd (Tabel 4.3 diperoleh 0,049) (OK)

d ( yang dipakai ) + 1 diameter + lindungan beton 50 + 1.6 + 9 60.60 cm

60 cm

dan

=

50

;

4

x

lindungan beton bagian bawah Dc yang dibuat - d - tulan = 60 50 = 8.4 cm

Langkah 4 : Memeriksa dukungan kolom 2

A1 = w A2 =

2

1600 cm

= 2

(w+4d)

sehingga Y

=

(

40

=

57600 1/2

(A2/A1)

=

f c = 0,85 . f . f c' .

0.85 =

f a = Pult / A1 = = Luas angker (Aa)

+ 2

cm

= =

x

2

50 )

(

57600 >

6

0.85

x

200

x

 2

1/2

1600 ) maka Y yang digunakan adalah

2

2

289 183080

kg/cm

114.425

kg/cm



1600 2

=

0,005 . A1

=

8



2

cm

16 mm

maka luas penampangnya =

2

1/4 . p .diameter = 1/4 . 22/7 . 1.6 2

= 2.0114286 cm

 2.0114286 Angker yang diperlukan (n) = Aa / Luas penampang = 8 = 3.98 dipakai 4 batang Jadi Aa yang dipakai = luas penampang . Angker yang diperlukan = 2.011428571 x 4 =

8.05

2

cm

2 cm

cm

66.0 cm

gan yang dipakai 1.6

Diketahui dari data soal yang didapat : P = 4.0 ton DL = 36% P = LL = 64% P = MB = ML = 10.00 tm

50

1.44 2.56

ton ton 300

2

f'c

=

275

kg/cm

fy FK

= =

1910 3

kg/cm

2

B



 q n

=

364 15

=

=

2

24.26667

kg/cm


qult

5 + 0,34 qc

=

=

13.25066667


2

kg/cm



=

qult

=

13.250667



=

4.416889

kg/cm

=

44.16889 t/m

FK 3 2

2


e

=

M/P

;

e b

=

MB / P

=

 2.500

10

=

el

4

=

ML / P

m

= =

 2.500

10

q max =

P 2

B

(

1

±

(

1

±

6 e b

±

B

6 el

)

B

q min

q max

=

4 2

B

15.000 B

15.000

±

B

)


B

B

q max

q min

q rata-rata

Rasio

0.85 0.90 0.95 1.00 1.50 2.00 2.50 3.00

0.72 0.81 0.90 1.00 2.25 4.00 6.25 9.00

200.94 169.55 144.39 124.00 37.33 16.00 8.32 4.89

-189.86 -159.67 -135.53 -116.00 -33.78 -14.00 -7.04 -4.00

5.54 4.94 4.43 4.00 1.78 1.00 0.64 0.44

0.03 0.03 0.03 0.03 0.05 0.06 0.08 0.09


1.50

x

1.50

150 cm


Pult

=

1,4DL

+

1,7LL

qult

=

q

=


=

6.37

ton

2

=

=

6368

165 t/m

2


=

1/2

1,060.f.(f c')


40

= x

14.94

40 cm

2

2

=

kg 16.51

2

kg/cm

= 0.85 2 kg/cm

maka 2

Dengan menggunakan rumus : d (vc+ q/4)+d(vc+ q/2) w = (B -w ) q/4

19.069

2

d

+

345.134


d

d

=


2

= = =

1/2

- b ± (b - 4ac ) 2a

= =

86262.81

-

345.134

±

6698832 2x 19.069 345.134 ± 2588.21 2 x 19.069 345.134 ± 2588.21 38.14 58.815 cm -76.915 cm 58.815 cm  diambil

1  2

58.0 cm




=

( B-w ) / 2 150 2

= Maka Mult

=

40

55

cm

2

( qult . I ) / 2 16.51

=

x 2

3025

=


=

0.85 x 1623.5

24970.81

=

94163


1910 As

x

As

-

As

As ( 58

=

=

0.3

= =



8.17 As )

2

2

2

1/2

- b ± (b - 4ac ) 2a

=

=

d positif d negatif

( 58 4.09

6632.909 As

=

2

As

Mu = f As f y.( d - a/2 )

= =

0.3 cm 0.3 x

8.17

1 cm

24970.81 24970.81

= =

24970.81 kgcm

=


Maka As Total As

=

± 9529187147 2 x 6632.909 94163 ± 97617.56 2 x 6632.909 94163 ± 97617.56 13265.82 0.260 cm -14.457 cm -

94163

1  2

 cm lebar B

x

150

=

2

cm

39.06

Lindungan beton minimum 7 cm Dipakai lindungan beton

9 cm

maka Befektif

= = =


B - 2 x lindungan beton yang dipakai 150 132

2 cm

x

19 mm


Jumlah tulangan yang diperlukan ( n )


2

= =

1/4 .p. diameter 1/4 . 22/7 . 3.61

= = = = =

2.836 cm As / luas penampang 39.06  2.836 13.77 dipakai 14 buah 132 14 - 1 10.34 cm dipakai 10 cm

=

2

9

Jadi dipakai total

= =


= 39.062 Sehingga As sesungguhnya = = =

2

cm Total As / B 39.062 0.260



150

2



0.004

= As / b x d = 0.260  1 x 58 = 0.004 > 0.002 OKE < Rasio baja ijin ( Pd ) yaitu 0,060 ( pada lampiran tabel ) OKE


Maka Dc dibuat

69 cm

= = =

d ( yang dipakai ) + 1 diameter + lindungan beton 58 + 1.9 + 9 68.9 cm = 68.9 cm

dan d yang dipakai


= = =

58.0 cm



20  19

20  19

B= 180 cm


Dc = 81cm 180 cm



A1 A2

= =

2

w

2

1600 cm

=

( w + 4d )

2

= =

sehingga Y

=

( A2 / A1 )

f c = 0,85 . f . f c' . y = =

=

40

x

397.4

kg/cm

0.85

=

1/2

) 73984  1600 6.8 > 2 maka Y yang digunakan adala 275

x

2


0.005 .A1 0.005 x

=

2

1600

cm

8

16 mm 2

1/4 .p. diameter 1/4 . 22/7 . 2.56

=

= =

x

= =

= =


2

58 )

1600 2

maka luas penampangnya


(

kg/cm


x

2




4

2

= =

0.85

3.98

+ cm

73984

1/2

6368

f a = Pult / A1 =

(

= = =

2

cm

2.011


4 batang


2

cm


40

40


4  16

2

4 m

As )

cm

Diketahui dari data soal yang didapat : P = 115 ton DL = 36% P = 41.4 LL = 64% P = 73.6 MB = ML = 10 tm

50

ton ton 250

2

f'c

=

200

kg/cm

fy FK

= =

1910 3

kg/cm

2

B



 q n

=

319 13

=

=

2

24.53846

kg/cm


qult

5 + 0,34 qc

=

=

13.34307692


2

kg/cm



=

qult

=

13.343077



=

4.447692

kg/cm

=

44.47692 t/m

FK 3 2

2


e

=

M/P

;

e b

=

MB / P

=

 0.087

10

=

el

115

=

ML / P

m

= =

 0.087

10

q max =

P 2

B

(

1

±

(

1

±

6 e b

±

B

6 el

)

B

q min

q max

=

115 2

B

0.522 B

0.522

±

B

)


B

B

q max

q min

q rata-rata

Rasio

1.00 1.50 1.60 1.70 1.80 1.90 2.00 2.10

1.00 2.25 2.56 2.89 3.24 3.61 4.00 4.41

235.00 86.67 74.22 64.22 56.07 49.35 43.75 39.03

-5.00 15.56 15.63 15.37 14.92 14.36 13.75 13.12

115.00 51.11 44.92 39.79 35.49 31.86 28.75 26.08

0.49 0.59 0.61 0.62 0.63 0.65 0.66 0.67


2.00

x

2.00

200 cm


Pult

=

1,4DL

+

1,7LL

qult

=

q

=


=

183.08

2

ton =

=

183080

124 t/m

2

Untuk bujur sangkar digunakan Balok aksi dua arah d engan f Maka didapatkan Vc

=

1/2

1,060.f.(f c')


40

= x

= = 2 kg/cm

12.74

40 cm

2

kg 12.38

2

kg/cm

0.85

maka 2

2

Dengan menggunakan rumus : d (4vc+ q)+d(2vc+ q)(w+w) = (B -w ) q

63.350

2

d

+


3029.308 d d

=


2

= = =

1/2

- b ± (b - 4ac ) 2a

= =

475477.33

1  2

3029.308 ± 129663575 2x 63.350 3029.308 ± 11386.991 2 x 63.350 3029.308 ± 11386.991 126.70 65.964 cm -113.782 cm 65.964 cm  diambil 66.0 cm -




=

( B-w ) / 2 200 2

= Maka Mult

=

40

80

cm

( qult . I2 ) / 2 12.38

=

x 2

6400

=


=

0.85 x 1623.5

39623.11

=

107092


1910 As

x

As

-

As

As ( 66

= = d positif d negatif

=

0.4

= =

( 66 5.62



11.24 As )

2

2

9120.25 As 2

1/2

- b ± (b - 4ac ) 2a

= =

2

As

Mu = f As f y.( d - a/2 )

= =

0.4 cm 0.4 x

11.24

1 cm

39623.11 39623.11

= =

39623.11 kgcm

=


Maka As Total As

=

± 12914254841 2 x 9120.25 107092 ± 113640.9 2 x 9120.25 107092 ± 113640.9 18240.50 0.359 cm -12.101 cm -

107092

1  2

 cm lebar B

x

200

=

2

cm

71.80

Lindungan beton minimum 7 cm Dipakai lindungan beto

9 cm

maka Befektif

= = =


B - 2 x lindungan beton yang dipakai 200 182

2 cm

x

19 mm


Jumlah tulangan yang diperlukan ( n


2

= =

1/4 .p. diameter 1/4 . 22/7 . 3.61

= = = = =

2.836 cm As / luas penampang 71.80  2.836 25.31 dipakai 25 buah 182 25 - 1 7.485 cm dipakai 7.5 cm

=

2

9

Jadi dipakai total

= =


= 71.803 Sehingga As sesungguhnya = =

2

cm Total As / B 71.803

=

0.359



200

2



0.005

= As / b x d = 0.359  1 x 66 = 0.005 > 0,002 ( OK ) < Rasio baja ijin ( Pd ) yaitu 0,043 ( pada lampiran tabel ) ( OK )


Maka Dc dibuat

80 cm

= = =

d ( yang dipakai ) + 1 diameter + lindungan beton 65.96 + 1.9 + 9 76.86 cm

dan d yang dipakai


= = =

66.0 cm



25  19 - 7,5 cm

25  19 - 7,5 cm

B= 200cm


Dc = 80cm B = 200 cm



A1 A2

= =

2

w

= 2

( w + 4d )

2

1600 cm =

(

=

( A2 / A1 )

f c = 0,85 . f . f c' . y =

1/2

289

183080

f a = Pult / A1 =

=

114.425


Dipakai baja tulangan diameter maka luas penampangnya

x

= = =

2

66 )

1/2

0.85

x

200

x

2

2

kg/cm



1600 2

kg/cm


= =

0.005 .A1 0.005 x

=

2

1600

cm

8

16 mm 2

1/4 .p. diameter 1/4 . 22/7 . 2.56

=


x

) ( 92328.11  1600 7.6 > 2 maka Y yang digunakan adalah

= =

= =


4

2

0.85

=

+

92328.112 cm

= sehingga Y

40

= = =

2

cm

2.011


4 batang


2

cm


40

4  16

40


2

115 m

As )

Tw Pondasi

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