PERHITUNGAN PONDASI TELAPAK.xls

32750 0 5230 30 240 300 1200

kg kg kgm Mpa Mpa mm mm

Pu

Ph

1200 mm

= = = = = = =

Mu

= 1100 mm = 0.3 mm = dia16 - 250 = 5 bh

300 mm

d=0.3

Design Pondasi Gudang Oli PH 4,5 A. Data teknis Gaya aksial Perlu/vertikal (Pu) Gaya aksial Perlu/horizontal (Ph) Momen perlu (M1) f'c fy Lebar kolom pondasi(b) Tinggi kolom pondasi (h) B. Hasil Analisa Lebar pondasi (B) Tebal pondasi (d) Tulangan Jumlah tul.per lebar pondasi C. Analisa Data penyelidikan tanah g (Wet density) f (angle of internal friction)

=

c (cohesion)

=

2 0 kg/cm

t

=

0 cm

(timbunan tanah merah)

gelam f10-6m

3 1700 t/m

=

45

o

B=1100

Analisa Terzaghi qult = 1,3.c.Nc.+q.Nq+0,4.g.B.Ng gelam f10-6m

=

0

5

Nc

=

5.7

7.3

Nq Ng

= =

1 0

1.6 0.5

Untuk sudut geser (f)

=

45

Nc

=

20.1

Nq Ng

=

6.4

=

4.5

o

2 336.600 kg/cm

qult = 1,3.c.Nc.+q.Nq+0,4.g.B.Ng

=

SF ( safety factor)

=

qa tanah = qult/SF Daya dukung gelamf10-600

=

Rencana jumlah gelam Daya dukung gelam totalf10-600

= =

0 bh 2 0 kg/cm

qa tanah + gelam

=

2 224.400 kg/cm

=

B^2 = B =

B=1100

1.5 2 224.400 kg/cm 2 0 kg/cm

Pu/B^2
b

2 14594.47415 mm

121 mm

Page 1 of 2

=

2 22.44 N/mm

B=1100

f

b

Tabel Tersaghi untuk nilai :f,Nc,Nq,Ng

1. One way action diambil lebar pondasi (B) =

1100 mm 2 0.270661157 N/mm

qu= Pu/B^2 = Vu < Vc qu*B(0.5*B-0.5*b-d)0.6*1/6(sqrt(f'c))*B*d = Tebal pondasi (d)  390.7598194 mm diambil tebal pondasi (d) = 400 mm qu= Pu/B^2
=

2 0.271 N/mm

<

2 22.440 N/mm Ok!

<

2.4E+07 N

2. Two way action bc =B/b bo =2*((B+d)+(B+d)) Vc =(1+2/bc)*2*sqrtf'c*bo*d Vcmax=4*sqrt*f'c*bo*d Vc < Vcmax……..ambil Vc fVc=0.6*Vc

= = = = = =

3.666666667 6000 mm 40631055.2 N 52581365.52 N 40631055.2 N 24378633.1 N

Vu =qu*((B^2-(B+d)*(b+d))

=

43305.78512 N

Ok!

3. Tinjauan terhadap momen lentur M1 (momen akibat reaksi kolom)

=

5230000 kgmm

M2 (momen akibat gaya horisontal) = AS = Mu/f0,9*d*fy dicoba tulangan f Jarak

= = =

As perlu

=

0 kgmm 2 756.6550926 mm 16 mm 250 mm 2 804 mm

r =As perlu/B*d r min =1.4/FY

= =

0.001827836 0.000583333

>

r min = 0.0006

rb =b1*((0.85*f'c)/fy)+(600/(600+fy)) = 0.2903125 > r = r max =0.75*rb = 0.217734375 Lengan momen (a)=(Aperlu*fy)/(0.85*f'c*b) = 6.881263909 mm fMn =fAs*fy(d-0.5a) = 61234937.73 Nmm = 6123.493773 kgm > Mu

Page 2 of 2

0.0018

5230 kgm

Ok!