PONDASI

1. Diketahui :

C = 4 t/m^2 SF = 327 qa = 32 t/m^21.7 t/m^2 2.6 t/m^2

Diasumsikan: a = 0.6 mb = 0.6 mc = 0.8 md = 0.2 md' = 0.5 m

Penyelesaian :

Df = Dw + z + d'z = Df - (Dw+d')

= 4,5 - ( 3 + 0,5 )= 1

Dimensi pondasi :

qu = qa * SF= 32 * 3 = 96 t/m^2w1 =

= ( 5,5 - 0,5 )* 0,6 * 0,6 * 2,4= 4.32

w2 == ( 0,5 - 0,2 ) * 0,8 * 0,8 * 2,4= 0.4608

w3 == 0,2 * B^2 * 2,4= 0,48 B^2

w4 == 2 * ( 0,5 *( 0,5 - 0,2 ) * ( 0,5 * B - 0,5 * 0,8 ) * B * ( 2,4 - 1)= 0,21 * B^2 - 0,42 * B

= = sat =

( h - d' ) * a * b *beton

( d' - d ) * c * c *beton

d * B * B * beton

2 * ( 0,5 *(d' - d)*( 0,5 *B - 0,5 *c )) * B * '

w5 = = 2 * ( 0,5 * B - 0,5 * 0,6 ) * 3 * B * 1,7= 4,1 *B ^2 - 1,23 * B

w6 == 2 * ( 0,5 * B - 0,5 *0,6 ) * 1 * B * ( 2,6 - 1 )= 0,16 * B^2 - 0,96 * B

w7 == 2 * ( 0,5 * ( 0,5 - 0,2 ) * ( 0,5 * B - 0,5 * 0,8 )) * B * ( 2,6 - 1 )= 0,12 * B^2 - 0,96 * B

w1 + w2 + w3 + w4 + w5 + w6 + w7= 4,32+0,4608+(0,48*B^2)+(0,21*B^2-0,42*B)+(4,1*B^2-1,23*B)+(0,16*B^2-0,96*B)+(0,12*B^2-0,96*B)= 4,728 + 5,07 * B^2 - 3,57 * b

Q = 4,728 + 5,07 * B^2 - 3,57 * b + 605,07 * B^2 - 3,57 * B + 64,728

qu = 96 t/m^2

Q

Mx * xA Iy

96 =5,07 * B^2 - 3,57 * B + 64,728

35 + ( 25 * 5,5 ) * B/2

B^2 1/12 *B * B ^3

96 = 5,07 - 3.57

+64.728

1035

B B^2 B^3x B^3

96*B ^3 =

96*B ^3 - 5,07*B^3 + 3,57*B^2 - 64,728

90,93*B^3 + 3,57*B^2 - 64,728*B

Cara coba-coba :B B^3 B^2 90,93*B^3 + 3,57*B^2 - 64,728*B

2.34127 12.83379 5.48155 1035.000

Sehingga B = 2.34127 2.3 m

B = L = 2.3 2.4 m

2 * ( 0,5 *B - 0,5 * a ) * Dw * B * b

2 * ( 0,5 * B - 0,5 * a ) * z * B * '

2 * ( 0,5 * ( d' - d ) * ( 0,5 * B - 0,5 * c )) * B * '

wtotal =

wtotal + P

max =

max =

5,07*B^3 - 3,57*B^2 + 64,728*B 1035

1035

1035

m

m

2.

Df

M . A . T d

A B

a l b

LDiketahui :

80 ton 2.4 t/m^380 ton 1.75 t/m^34 t/m l = 6 m

35 t/m 2.6 t/m^3M.A.T = 2 m 35 t/m^2

5 m 0.35 m0.4 m 0.35 m0.4 m

Letak resultan kolom :R = P1 + P2

= 80 + 80= 160 TON

Misal : a = 1.5 mb = 3.5 mL = 11 m

P1 * a + P2 * ( a + l ) + M1 + M2 - R * x = 080*1,5 + 80 * ( 1,5 + 6 ) + 40 + 35 - 160 * ro = 0

795 = 160 * xro = 4.96875 m

Diasumsikan tebal pondasi d = 1.4 mBerat tiap luasan pondasi = ( 1,4 * 2,4 ) + (( 5 - 1,4) * 2,4 )

= 12 t/m^2Tekanan tanah izin yang tersedia untuk menahan beban :

qa - berat pondasi - tanah urug= 35 - 12 - (2 * (1,75))+((5-2)*(2,6-1)= 24.3 t/m^2

P1 P2

M1 M2

C1 C2

C1' C2'

P1 = beton =P2 = b =M1 =M2 = sat =

qa =Df = C1' =C1 = C2' =C2 =

= 0

beban =

Luas pondsasi yang diperlukan :

A = =160

= 6.584 t/m^224.3

Lebar pondasi A dan B

A =B + A

* L2

6.584 =B + A

* 112B + A = 1.197 m

Titik berat dari luasan pondasi trapesium dengan kolom yang berjarak x, sehingga diperoleh :

ro =L 2 * A + B 3 A + B

4.96875 =11 A + 1,1973 1.197

4.96875 =11 * A + 13,167

3.5914717.84512 = 11 * A + 13,1674.678118 = 11 * A

A = 0.425283 m

B = ( B + A ) - A= 1,197 - 0,425= 0.772 m

Maka luas dasar pondasi :

A =B + A

* L2

A =0,772 + 0,425

* 112= 6.584362 m^2

Tekanan dasar pondasi :

Q

Mx * xA Iy

1.197

160 * 4,9686.584362 1/12*0,772*11^3

= 0.181794 9.285906

9.468-9.104

Gambar tekanan pada dasar pondasi :

-9.104

-

+

9.468

Pbeban

=

=

max = qaminx = qa

3. Diketahui :

121.7 t/m^3 1.6 t/m^325 273 t/m^2 3.5 t/m^2

1.8 t/m^3 1.6 t/m^328 252.5 t/m^2 4 t/m^2

Koefisien tekanan tanah aktit ( Ka ) :

Ka =

= 0.978

= 0.978

= 0.442

= 0.978

= 0.978

= 0.678

= 0.978

= 0.978

= 0.698

Koefisien tekanan tanah pasif ( Kp ) :

Kp =1

=1

= 1.4330.698

= = = = =C1 = C3 = = = = =C2 = C4 =

cos cos cos^2 - cos^2 cos cos^2 - cos^2

Ka1 = cos 12cos 12 cos^2 12 - cos^2 25cos 12 cos^2 12 - cos^2 25

0,978 0,957 - 0,8210,978 0,957 - 0,821

0,9780,3690,9780,369


0,978 0,957 - 0,7800,978 0,957 - 0,780

0,9780,1770,9780,177


0,978 0,957 - 0,7940,978 0,957 - 0,794

0,9780,1630,9780,163

Ka3

SELURUH KONSTRUKSI

a = 10.5 1.7 H1 = 3 c1 = 3b = 7.5 1.8 H2 = 3 c2 = 2.5c = 10 1.6 H3-TEBAL = 1 c3 = 3.5d = 9 1.6 TEBAL = 1 c4 = 4

H5 = tan 12 * d 1.917 vi1 = 25 Ka1 = 0.442q 2.5 vi2 = 15 Ka2 = 0.678

vi3 = 30 Ka3 = 0.698vi4 = 13 Kp = 1.433

PERHITUNGAN AKIBAT MOMEN VERTIKAL ( BEBAN DINDING PENAHAN + TANAH )

No Rumus Lengan Mv

1 ( a + b + c + d ) * tebal * 2,4 88.800 18.500 1642.8002 0,5 * b * ( H1 + H2 + ( H3-tebal ) * 2,4 63.000 0.900 56.7003 c * ( H1 + H2 + ( H3-tebal ) * 2,4 168.000 1.250 210.0004 6.300 0.400 2.5205 0.321 0.086 0.0286 14.665 32.500 476.6147 45.900 28.750 1319.6258 48.600 28.750 1397.2509 5.400 28.750 155.250

440.986 5260.787PERHITUNGAN GAYA HORISONTAL

No RUMUS Pa LENGAN

1 q * Ka1 * ( H1 + H5 ) 6.630 6.485 4.959 32.1562 q * Ka2 * H2 5.085 4.974 6.639 33.0223 q * Ka3 * H3 3.490 3.414 3.500 11.9484 9.083 8.885 6.639 58.9865 27.866 27.257 3.500 95.4016 28.688 28.061 1.000 28.0617 5.492 5.372 3.000 16.1158 11.308 11.061 1.000 11.0619 0.838 0.819 0.667 0.546

10 4.500 4.402 0.667 2.934100.729 290.231

No RUMUS Pp LENGAN

1 0.6 0.587 0.667 0.3912 6.4 6.260 0.667 4.173

6.847 4.565107.577 294.795

1 =2 =3 =4 =

Gaya berat (W)

a * ( H3-tebal ) * (3-1)0,5 * 1,071 * ( H3-tebal ) * ( 3-1 )0,5 * H5 * d * 1d * H1 * 1d * H2 * 2d * ( H3-tebal ) * ( 3-1 )

V = MV =

PaH (Pa*COS12)

MH (PaH*LENGAN)

0,5 * 1 * ( H1 + H5)^2 * Ka11 * ( H1 + H5 )^2 * Ka21 * ( H1 + H5 )^2 * Ka30,5 * 2 * H2^2 * Ka22 * H2^2 * Ka30,5 1 ) * H3^2 * Ka30,5 * air * H3^2

PaH (Pa*COS12)

MH (PaH*LENGAN)

0,5 1 ) * H3^2 * Kp0,5 * air * H3^2

H = MH =

PERHITUNGAN GAYA VERTIKAL

No RUMUS Pa LENGAN

1 q * Ka1 * ( H1 + H5 ) 6.630 1.378 37 51.0032 q * Ka2 * H2 5.085 1.057 37 39.1183 q * Ka3 * H3 3.490 0.726 37 26.8484 9.083 1.889 37 69.8755 27.866 5.794 37 214.3686 28.688 5.965 37 220.6917 5.492 1.142 37 42.2478 11.308 2.351 37 86.9869 0.838 0.174 37 6.443

10 4.500 0.936 37 34.61721.411 792.196

Jumlah MV = 6052.983Jumlah MH = 294.795Jumlah V = 462.397Jumlah H = 107.577

Faktor keamanan trhahadap geser

SFs = = 2.003 Syarat aman SFs > 2 ..OK!

Faktor keamanan terhadap guling

SFo =MV

= 20.533 Syarat aman SFs > 2 ..OK!MH

Faktor keamanan terhadap daya dukung

SF = = 18.9813324

1 +6e

= 24.752B B

1 -6e

= 0.056B B

= 14.959= 469.828

e = ((a+b+c+d)/2) - ((MV-MH)/V) = 6.047

Diamana : 25 e = Tekanan tanah yang terjadiMaka : Nc = 25.1 Daya dukung

Nq = 12.7 B = Lebar dasar dinding penahan9.7

PaV (Pa*Sin12)

MV (PaV*LENGAN)


V = MV =

V * tg25H

qultqmax

qmax =V

qmin =V

qrata-rata = (1 * (H1+H5)) + (2 * H2) + ((3-1) * H3)q ult = (c4 * Nc) + (qrata-rata * Nq) +(0,5 * (4 - 1) * (a+b+c+d) * N))

=q ult =

N =

TOE

a = 10.5 1.7 H1 = 3 c1 = 3b = 7.5 1.8 H2 = 3 c2 = 2.5c = 10 1.6 H3-TEBAL 1.3 c3 = 3.5d = 9 1.6 TEBAL = 0.7 c4 = 4

H5 = tan 12 * d 1.917 vi1 = 25 Ka1 = 0.442q 2.5 vi2 = 15 Ka2 = 0.678

vi3 = 30 Ka3 = 0.698vi4 = 13 Kp = 1.433

No Rumus Lengan Mv

1 a * tebal * 2.4 4.410 5.250 23.1534 8.190 5.250 42.998q1 a * 18,370 192.885 5.250 1012.646q2 0,5 * a * 2,497 13.109 3.500 45.882

-193.394 -992.379

PERHITUNGAN GAYA HORISONTAL

No Rumus Pp Lengan MH

Pp1 -0.147 0.667 -0.098Pp2 -0.245 0.667 -0.163

-0.392 -0.261

Jumlah MV = -992.379Jumlah MH = -0.261Jumlah V = -193.394Jumlah H = -0.392

SFs = = 229.902

SFo =MV

= 3797.367MH

SF = = -7.428

1 +6e

= -19.681a a

1 -6e

= -17.156a a= 1.200= 146.195

e = (a/2) - ((MV-MH)/V) = 0.120

Diamana : 25Maka : Nc = 25.1

Nq = 12.79.7

1 =2 =3 =4 =

Gaya berat (W)

a * (h3-tebal) * (3 - 1)

V = MV =

0,5 * (3-1) * TEBAL^2 * 10,5 * air * TEBAL^2

H = MH =

V * tg25H

qultqmax

qmax =V

qmin =V

qrata-rata = (3-1) * H3q ult = (c4 * Nc) + (qrata-rata * Nq) +(0,5 * (4 - 1) * (a* N))

=

N =

STEEM


H5 = tan 12 * d 1.917 vi1 = 25 Ka1 = 0.442q 2.5 vi2 = 15 Ka2 = 0.678

vi3 = 30 Ka3 = 0.698vi4 = 13 Kp = 1.433

No Rumus Lengan Mv

2 0,5 b * (H1+H2+(H3-TEBAL)) 27.375 9.000 246.3753 c * ( H1 + H2 + ( H3-tebal ) 73.000 0.900 65.7005 0,5 * 2,1 * (H3-TEBAL) 1.365 1.250 1.706

101.740 313.781PERHITUNGAN GAYA VERTIKAL

No RUMUS Pa LENGAN

1 q * Ka1 * ( H1 + H5 ) 5.433 1.130 0.8 0.9042 q * Ka2 * H2 5.085 1.057 0.8 0.8463 q * Ka3 * (H3-TEBAL) 2.269 0.472 0.8 0.3774 9.083 1.889 0.8 1.5115 27.866 5.794 0.8 4.6356 28.688 5.965 0.8 4.7727 5.492 1.142 0.8 0.9138 11.308 2.351 0.8 1.8819 0.354 0.074 0.8 0.05910 1.369 0.285 0.8 0.228

20.156 16.125

PERHITUNGAN GAYA HORISONTAL

No RUMUS Pa LENGAN

1 q * Ka1 * ( H1 + H5 ) 5.433 5.315 4.959 26.3522 q * Ka2 * H2 5.085 4.974 6.639 33.0223 q * Ka3 * (H3-TEBAL) 2.269 2.219 2.300 5.1044 9.083 8.885 6.639 58.9865 27.866 27.257 3.500 95.4016 28.688 28.061 1.000 28.0617 5.492 5.372 3.000 16.1158 11.308 11.061 1.000 11.0619 0.354 0.346 0.433 0.15010 0.845 0.827 0.433 0.358

94.316 274.609

1 =2 =3 =4 =

Gaya berat (W)

= =

PaV (Pa*Sin12

)

MV (PaV*LEN

GAN)


= =

PaH (Pa*COS

12)

MH (PaH*LEN

GAN)

0,5 * 1 * ( H1 + H5)^2 * Ka11 * ( H1 + H5 )^2 * Ka21 * ( H1 + H5 )^2 * Ka30,5 * 2 * H2^2 * Ka22 * H2^2 * Ka30,5 1 ) * (H3-TEBAL)^2 * Ka30,5 * air * (H3-TEBAL)^2

= =


SFs = = 0.659

SFo =MV

= 1.201MH

SF = = 13.513

1 +6e

= 26.779b + c b + c

1 -6e

= -12.848b + c b + c

= 14.539= 361.856

e = ((b+c+)/2) - ((MV-MH)/V) = 8.296


Nq = 16.61913.699

V * tg27H

qultqmax

qmax =V

qmin =V

qrata-rata = (1 * (H1+H5)) + (2 * H2) + ((3-1) * H3)q ult = (c4 * Nc) + (qrata-rata * Nq) +(0,5 * (4 - 1) * (b+c+) * N))

=

N =

HEEL


H5 = tan 12 * d 1.917 vi1 = 25 Ka1 = 0.442q 2.5 vi2 = 15 Ka2 = 0.678

vi3 = 30 Ka3 = 0.698vi4 = 13 Kp = 1.433

No Rumus Lengan Mv

1 d * tebal * 2,4 15.120 18.500 279.7206 14.665 0.900 13.1997 2.550 1.250 3.1888 48.600 0.400 19.4409 7.020 0.086 0.604q3 d * 15,934 143.406 32.500 4660.695q4 0,5 * d * 10,398 46.791 28.750 1345.241

278.152 6322.086PERHITUNGAN GAYA VERTIKAL

No RUMUS Pa LENGAN

3 q * Ka3 * tebal 1.222 0.254 7.300 1.8549 0.103 0.021 7.300 0.15610 0.245 0.051 7.300 0.372


No RUMUS Pa LENGAN

3 q * Ka3 * tebal 1.222 1.195 2.800 3.3459 0.103 0.100 0.433 0.04310 0.245 0.240 0.433 0.104

1.535 3.493

1 =2 =3 =4 =

Gaya berat (W)

0,5 * H5 * d * 1d * H1 * 1d * H2 * 2d * ( H3-tebal ) * ( 3-1 )

= =

PaH (Pa*SIN12)

Mv (PaH*LENGAN)

0,5 1 ) * tebal^2 * Ka30,5 * air * tebal^2

= =

PaH (Pa*COS12)

MH (PaH*LENGAN)

0,5 1 ) * tebal^2 * Ka30,5 * air * tebal^2

= =


SFs = = 37.154

SFo =MV

= 1810.717MH

SF = = -6.012

1 +6e

= -60.187b + c b + c

1 -6e

= 92.013b + c b + c

= 14.539= 361.856

e = ((b+c+)/2) - ((MV-MH)/V) = -13.948


Nq = 12.79.7

V * tg27H

qultqmax

qmax =V

qmin =V

qrata-rata = (1 * (H1+H5)) + (2 * H2) + ((3-1) * H3)q ult = (c4 * Nc) + (qrata-rata * Nq) +(0,5 * (4 - 1) * (b+c+) * N))

=

N =

SELURUH KONSTRUKSI

a = 19 1.7 H1 = 3 c1 = 3b = 16 1.8 H2 = 3 c2 = 2.5c = 16 1.6 H3-TEBAL 1.3 c3 = 3.5d = 15 1.6 TEBAL = 0.7 c4 = 4

H5 = tan 12 * d 0.744 vi1 = 25 Ka1 = 0.442q 2.5 vi2 = 15 Ka2 = 0.678

vi3 = 30 Ka3 = 0.698vi4 = 13 Kp = 1.433

PERHITUNGAN AKIBAT MOMEN VERTIKAL ( BEBAN DINDING PENAHAN + TANAH )

No Rumus Gaya berat (W) Lengan Mv

1 ( a + b + c + d ) * tebal * 2,4 110.880 33.000 3659.0402 0,5 * b * ( H1 + H2 + ( H3-tebal ) * 2,4 140.160 0.900 126.1443 c * ( H1 + H2 + ( H3-tebal ) * 2,4 280.320 1.250 350.4004 14.820 0.400 5.9285 0.819 0.086 0.0706 9.486 57.000 540.7027 76.500 50.750 3882.3758 81.000 50.750 4110.7509 11.700 50.750 593.775

10 PaV1 1.378 37.000 51.00311 PaV2 1.057 37.000 39.11812 PaV3 0.726 37.000 26.84813 PaV4 1.889 37.000 69.87514 PaV5 5.794 37.000 214.36815 PaV6 5.965 37.000 220.69116 PaV7 1.142 37.000 42.24717 PaV8 2.351 37.000 86.98618 PaV9 0.174 37.000 6.443


No RUMUS Pa PaH (Pa*COS12) LENGAN

1 q * Ka1 * ( H1 + H5 ) 6.630 6.485 4.959 32.1562 q * Ka2 * H2 5.085 4.974 6.639 33.0223 q * Ka3 * H3 3.490 3.414 3.500 11.9484 9.083 8.885 6.639 58.9865 27.866 27.257 3.500 95.4016 28.688 28.061 1.000 28.0617 5.492 5.372 3.000 16.1158 11.308 11.061 1.000 11.0619 0.838 0.819 0.667 0.546

10 4.500 4.402 0.667 2.934100.729 290.231

PERHITUNGAN GAYA VERTIKAL

No RUMUS Pa PaV (Pa*Sin12) LENGAN

1 q * Ka1 * ( H1 + H5 ) 6.630 1.378 37 51.003

1 =2 =3 =4 =

a * ( H3-tebal ) * (3-1)0,5 * 2,1 * ( H3-tebal ) * ( 3-1 )0,5 * H5 * d * 1d * H1 * 1d * H2 * 2d * ( H3-tebal ) * ( 3-1 )

V = MV =

MH (PaH*LEN

GAN)


H = MH =

MV (PaV*LEN

GAN)

2 q * Ka2 * H2 5.085 1.057 37 39.1183 q * Ka3 * H3 3.490 0.726 37 26.8484 9.083 1.889 37 69.8755 27.866 5.794 37 214.3686 28.688 5.965 37 220.6917 5.492 1.142 37 42.2478 11.308 2.351 37 86.9869 0.838 0.174 37 6.443

10 4.500 0.936 37 34.617


Faktor keamanan trhahadap geser

SFs = = 1.578Syarat am SFs > 2 .not OK!

Faktor keamanan terhadap guling

SFo =MV

= 48.330Syarat am SFs > 2 ..OK!

MH

Faktor keamanan terhadap daya dukung

SF = = 5.085478

1 +6e

= 26.301B B

1 -6e

= -3.690B B

= 12.965= 133.753

e = ((a+b+c+d)/2) - ((MV-MH)/V) = 14.590


Nq = 3.721.98


V * tg12H

qultqmax

qmax =V

qmin =V

qrata-rata = (1 * (H1+H5)) + (2 * H2) + ((3-1) * H3)q ult = (c4 * Nc) + (qrata-rata * Nq) +(0,5 * (4 - 1) * (a+b+c+d) * N))

=

N =

25 9.7

26.25 12.2

26.875 13.45

26.95313 13.6062526.992 13.684

26.99707 13.69414

27.000 13.699

27.002 13.70427.012 13.723

27.03125 13.762527.1875 14.075

27.5 14.7

30 19.7

PONDASI TIANG PANCANG KELOMPOK

Data-data yang diketahui :1. Tanah keras = 7 m2. Muka air tanah = 3 m dari muka tanah3. Data Cone Penetration Test : PK = Pt = 140 kg/cm2

JHL = 780 kg/cm24. Data laboratorium : C = 4 kg/cm2

30 1.75 kg/cm22.6 kg/cm2

5. Beban yang bekerja : Momen = 60 TmG. V = 75 TonG. H = 25 Ton

Penyelesaian :

Diasumsi :Diameter tiang = 0.3 mJumlah tiang = 9 buah

YMx dan My

25 T0,5 m

Y

X7 m

Y0,3 m

X X

X = 1 mY = 1 md = 0.3 m

=sub =sat =

V

BEBAN YANG BEKERJA PADA TIAP-TIAP TIANG

n = 9Pv = 75 TonMx = 60 Tm

6 * 1 ^2 = 66 *1 ^2 = 6

Pv=

75= 8.333 Tonn 9

No X (m) Y (m) X^2 Y^2 My*X Mx*YMy*X Mx*Y

P = (Pv/n)+8+9

1 2 3 4 5 6 7 8 9 10

P1 -1 1 1 1 0 60 0 12.346 20.679P2 0 1 0 1 0 60 0 12.346 20.679P3 1 1 1 1 0 60 0 12.346 20.679P4 -1 0 1 0 0 0 0 0 8.333P5 0 0 0 0 0 0 0 0 8.333P6 1 0 1 0 0 0 0 0 8.333P7 -1 -1 1 1 0 -60 0 -12.346 -4.012P8 0 -1 0 1 0 -60 0 -12.346 -4.012P9 1 -1 1 1 0 -60 0 -12.346 -4.012

Pmax 1 1 1 1 0 60 0 12.346 20.679

Tulangan : 25 = 2.5 cmTegangan yang dianjurkan pada :

50 kg/cm^21200 kg/cm^2

x^2 =y^2 =

x^2 y^2

8

Beton =Baja =

DAYA DUKUNG TIANG INDIVIDU

A. Beradasarkan kekuatan bahanFb + Fy

== (1/4 * 3,14 * 30^2) + {15 * ( 8 * 1/4 * 2,5^2 )}= 894 cm^2

= 50 * 894= 44700 kg= 44.7 Ton

B. Berdasarkan kekuatan tanah

+O * C

3 5Dimana :

1/4 * 3,14 * 30^2= 706.5

= 2 * 3,14 * 15= 94.2

706,5*140+

94,2 * 7803 5

= 47665.2 kg= 47.6652 Ton

Jadi daya dukung tiang individu ( Qsp ) = 44.7 Ton

Atiang =(1/4 * * d^2 ) + {15 * ( 8 * 1/4 * * dtul^2 )}

Ptiang = beton * Atiang

Qtiang =Atiang * Pt

Atiang =

Kelltiang (O)= 2 * * r

Qtiang =

DAYA DUKUNG TIANG KELOMPOK

=

= 0.667

Daya dukung tiap tiang = 0,667 * Qsp= 0,667 * 44,7= 29.815

Daya dukung kelompok = 9 * 29,815= 268.334

Kontrol : Pmax < Qsp'20.67901 < 29.815 ..OK!

2148' 5''

Eff = 1 90 {( n 1 ) m + ( m 1 ) nm n }

= arc tg DS= arc tg 40

100

21o 48 ' 5 } over { 90 } } ` left lbrace { { \( `3` - `1` \) `*`3`+` \( `3` - `1` \) `*`3} over {3``*`3} } right rbrace } {}

Eff = 1

KEMAMPUAN TIANG MENAHAN BEBAN HORISONTAL

Diketahui : e = 0.5 tinggi dari muka tanah sampai titik yang diberi beban horisontalL = Kedalaman pondasiB = Tebal pondasi

Ca = Kohesi kedalaman, misal = 3

22.56

Hu > 10 t ..OK!Jadi tiang aman menahan gaya horisontal sebesar = 10 ton

LB=

70 .4

= 17 .5

eB=

0 .50 .4

= 1.25

HuCuB2

= 47 (dari grafik halaman III 6 )

Hu3 0 .42

= 47

Hu= 22 . 56 ton

PENURUNAN YANG TERJADI

Tanah dibagi menjadi 6 bagian, dengan lapisan 1 - 5 masing-masing = 1 m dan lapisan 6 = 2 m

Diketahui :

11.75 kg/cm22.6 kg/cm2

20.75

37 m

4

5

6

Po' pada tiap-tiap lapisan :

Lapisan 1Po' = 1,75 * 50

= 87.5 kg/cm

Lapisan 2Po' = 87,5 + ( 1,75 * 100 )

= 262.5 kg/cm

Lapisan 3Po' = 262,5 + ( 1,75 * 100 )

= 437.5 kg/cm

Lapisan 4Po' = 437,5 + ( 1,75 * 50 ) + (( 2,6 - 1 ) * 50 )

= 605 kg/cm

Lapisan 5Po' = 605 + (( 2,6 - 1 ) * 100 )

= 765

Lapisan 6Po' = 852,5 + (( 2,6 - 1 ) * 150 )

= 1005 kg/cm

sub =sat =

eo =

Akibat beban vertika = 75 ton maka dibuat beban merata

q =75

=75000

= 1.875 kg/cm22 * 2 200 * 200

Lapisan 1B = 200 cm

m =200

= 4L = 200 cm 50Z = 50 cm

n =200

= 450Maka didapat nilai :

I = 0.250,25 * 3,75

= 0.9375 kg/cm2

Lapisan 2B = 200 cm

m =200

= 1.333L = 200 cm 150Z = 150 cm

n =200

= 1.333150Maka didapat nilai :

I = 0.220,22 * 3,75

= 0.825 kg/cm2

Lapisan 3B = 200 cm

m =200

= 0.800L = 200 cm 250Z = 250 cm

n =200


I = 0.180,18 * 3,75

= 0.675 kg/cm2

Lapisan 4B = 200 cm

m =200

= 0.571L = 200 cm 350Z = 350 cm

n =200


I = 0.140,14 * 3,75

= 0.525 kg/cm2

Lapisan 5B = 200 cm

m =200

= 0.444L = 200 cm 450Z = 450 cm

n =200


I = 0.120,12 * 3,75

= 0.45 kg/cm2

Mencari P untuk masing-masing lapisan, halaman 14 buku mektan II

P =

P =

P =

P =

P =

Lapisan 6B = 200 cm

m =200

= 0.364L = 200 cm 550Z = 550 cm

n =200


I = 0.090,09 * 3,75

= 0.3375 kg/cm2Penurunan yang terjadi :

Si = qn * ( 2B )* * Ip

qn = Berat total = Gaya vertikal + ( BS tiang * jumlah tiang ) + berat poer

BS tiang :Luas Penampang tiang = 1/4 * 3,14 * 40^2

= 1256 cm2= 0.1256 m2

BS tiang = 6*0,1256*2,4= 1.80864 ton

Berat poer = 0,5 * 2,8 * 2,8 * 2,4= 9.408

qn = 150 + ( 1,80864 * 9 ) + 9,408= 175.68576

I = Faktor pengaruh

Modulus deformasi menurut Butter400 * C C = 4

= 400 * 4 = 1600

Jadi penurunan yang terjadi

C1 = 0,5 = tanah tliatIp = Bujur sangkar = 0,82

0.5

P =

I - ^2 E

E =E =

Tanah liat =

I = (1 L8 Bq ) 0,5 Bq= Lebar poer = 2 .8I = (1 782 . 8 ) 0,5 L = Pangjang tiang pancangI = 0,688 0,5 . . .. . . .. . . . OK !

Si = qn 2 B I 2

E Ip

Si = 175 ,737( 22,8 ) 0 . 688 0 .52

1600 0. 82

Si = 0 . 221 cm 10 cm . .. . . .. . . .. . OK !

Sc = Cc H1 + eo

Log Po + PPo

Pada halaman VII 12 buku pondasi , harg aCc :Cc = 0 . 208 eo + 0 . 0083Cc = 0 . 208 0 .75 + 0 . 0083Cc = 0 . 164

Diketahui :0.75

Cc = 0.164

Lapisan Sc

1 100 87.5 0.9375 0.043372 100 262.5 0.825 0.012773 100 437.5 0.675 0.006274 100 605 0.525 0.003535 100 765 0.45 0.002396 100 1005 0.3375 0.00137

Jumlah penurunan = 0.06971

Jadi Stotal = Si + Sc= 0,221 + 0,6971= 0.29071 cm

eo =

H ( cm )

Po' ( kg/cm )

P ( kg/cm )

Sc = Cc H1 + eo

Log Po + PPo

Pada halaman VII 12 buku pondasi , harg aCc :Cc = 0 . 208 eo + 0 . 0083Cc = 0 . 208 0 .75 + 0 . 0083Cc = 0 . 164

POND. TELAPAKPOND. GABUNGANDD. PENAHANSELURUH KONStoeSTEEMHEELGAMBARSheet1TIANG PANCANGDAYA DUKUNG TIANG INDIVIDUDAYA DUKUNG TIANG KELOMPOKBEBAN HORISONTALPENURUNAN

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