Taylor (1984)

Soal :

Dik : R =9m(dari gambar)L A-B = 4.52m(dari gambar)L B-C = 9.56m(dari gambar)y1=3.9my2=7.6ml = 3.1mq =10kN/m2Luas Area =41.7m2(Bidang Longsor lapisan 1 dan lapisan 2)1 = 2 =20kN/m3Cu1 =20kN/m2Cu2 =40kN/m2

Dit : Safety Factor ?(Penyelesaian :

SF =[ R x (L A-B x Cu2) + (L B-C x Cu1)] / [(W x y1) + (Q x y2)]Q =0.5 x q x l x 1mQ =15.5kNW = Luas Area x x 1mW = 834kNJadi,SF =0.5395205317

BISHOPSoal : Dik :c =20kN/m2 =20 =16kN/m3

tan =0.3639702343

Dit :Safety Factor ?

Jawab :

IrisanbhWnnSin nCos nLnWn Sin nW Cos nNo.(m)(m)(KN/m)(deg)(m)(KN/m)(KN/m)1234567891011.4244.8690.9340.358441.82416.055245.9377.6530.7990.6026.8301.565227.245347.5480370.6020.7995288.871383.345447.6486.4240.4070.9144.4197.837444.349546.5416120.2080.9784.186.491406.909644.6294.400.0001.00040.000294.40074.21.7114.24-12-0.2080.9784.3-23.752111.744 kol. 8 = kol. 9 = kol. 10 =32.6892.8371884.047SF=( kol. 8 )( c ) + ( kol. 10) tan kol. 9

SF=(32.6 X 20) + (1884.047 X 0.364)892.837SF=1.50

FELLENIUSSoal : Dik :c =20kN/m2 =20 =16kN/m3

tan =0.3639702343

Dit :Safety Factor ?

Jawab :

IrisanbhWnnSin nCos nLnWn Sin nW Cos nNo.(m)(m)(KN/m)(deg)(m)(KN/m)(KN/m)1234567891011.4244.8690.9340.358441.82416.055245.9377.6530.7990.6026.8301.565227.245347.5480370.6020.7995288.871383.345447.6486.4240.4070.9144.4197.837444.349546.5416120.2080.9784.186.491406.909644.6294.400.0001.00040.000294.40074.21.7114.24-12-0.2080.9784.3-23.752111.744 kol. 8 = kol. 9 = kol. 10 =32.6892.8371884.047SF=( kol. 8 )( c ) + ( kol. 10) tan kol. 9

SF=(32.6 X 20) + (1884.047 X 0.364)892.837SF=1.50