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Page 2

B = 453 mm

1. One way action

diambil lebar pondasi (B)= 500 mm

qu= Pu/B^2 = 0.22826 N/mm
2

Vu < Vc

qu*B(0.5*B-0.5*b-d)0.6*1/6(sqrt(f'c))*B*d=

Tebal pondasi (d)  148.862 mm

diambil tebal pondasi (d)= 150 mm

qu= Pu/B^2<s'lentur = 0.228 N/mm
2

< 0.278 N/mm
2

Ok!

2. Two way action

bc =B/b = 2.5

bo =2*((B+d)+(B+d)) = 2600 mm

Vc =(1+2/bc)*2*sqrtf'c*bo*d = ####### N

Vcmax=4*sqrt*f'c*bo*d = 7148818 N

Vc < Vcmax……..ambil Vc = ####### N

fVc=0.6*Vc = 3860362 N

Vu =qu*((B^2-(B+d)*(b+d)) = 5135.76 N < 3860362 N Ok!

3. Tinjauan terhadap momen lentur

M1 (momen akibat reaksi kolom) = 70000 kgmm

M2 (momen akibat gaya horisontal) = 0 kgmm

AS = Mu/f0,9*d*fy = 27.0062 mm
2

dicoba tulangan f = 8 mm

Jarak = 200 mm

As perlu = 251 mm
2

r =As perlu/B*d = 0.00335 > r min = 0.0006
r min =1.4/FY = 0.00058

rb =b1*((0.85*f'c)/fy)+(600/(600+fy)) = 0.26322 > r = 0.0034

r max =0.75*rb = 0.19741

Lengan momen (a)=(Aperlu*fy)/(0.85*f'c*b)= 6.75838 mm

fMn =fAs*fy(d-0.5a) = 7075167 Nmm

= 707.517 kgm > Mu 70 kgm Ok!

Mydoc/titip/jal/design/pondasi.xls

Page 9

Pu/B^2 <s'lentur jumlah gelam total 4

B^2  Pu/s'lentur, B^2 = 10375.9 mm
2

B = 102 mm

1. One way action

qu= Pu/B^2 = 0.27785 N/mm
2

Vu < Vc

qu*B(0.5*B-0.5*b-d)0.6*1/6(sqrt(f'c))*B*d=

Tebal pondasi (d)  -97.051

diambil lebar pondasi (B)= 1000 mm

qu= Pu/B^2<s'lentur = 0.003 N/mm
2

< 0.278 N/mm
2

Ok!

Vu < Vc
qu*B(0.5*B-0.5*b-d)0.6*1/6(sqrt(f'c))*B*d=

Tebal pondasi (d)  2.77576 mm

diambil tebal pondasi (d)= 200 mm

2. Two way action

bc =B/b = 5

bo =2*((B+d)+(B+d)) = 4800 mm

Vc =(1+2/bc)*2*sqrtf'c*bo*d = ####### N

Vcmax=4*sqrt*f'c*bo*d = 1.6E+07 N

Vc < Vcmax……..ambil Vc = ####### N

fVc=0.6*Vc = 6746826 N

Vu =qu*((B^2-(B+d)*(b+d)) = 1499.13 N < 6746826 N Ok!

3. Tinjauan terhadap momen lentur

AS = Mu/f0,9*d*fy = 0 mm
2

dicoba tulangan f10-150

As perlu = 524 mm
2

r =As perlu/B*d = 0.00262 > r min = 0.0006
r min =1.4/FY = 0.00058

rb =b1*((0.85*f'c)/fy)+(600/(600+fy)) = 0.25268 > r = 0.0026

r max =0.75*rb = 0.18951

Lengan momen (a)=(Aperlu*fy)/(0.85*f'c*b)= 8.45445

fMn =fAs*fy(d-0.5a) = 2E+07 Nmm

= 1969.63 kgm > Mu 0 kgm Ok!

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