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METODO DE FELLENIUS.EJEMPLO =20KN/m3 =32c =20KN/m2ru =0.4b=2Rebanadab (m)h (m)()l (m)W (KN)Wsen(a)Wcos(a)u (KN/m2)Wcos(a)-u*l121.3-142.0652-12.650.510.429.0222.5-42.00100-7.099.82059.7323.682.0214420.0142.628.884.4424.6182.1018456.9175.036.897.6524.4362.47176103.5142.435.255.4622.2583.778874.646.617.6-19.814.43235.4306.3
FS =2.039
EJERCICIO PROPUESTO =16KN/m3 =35c =10KN/m2ru =0.4b=3.15REBANADAb (m)h (m)()l (m)W (KN)Wsen(a)Wcos(a)u (KN/m2)Wcos(a)-u*l13.1516.53.1750.45.750.16.429.823.153.0819.53.34155.23251.8146.319.71280.533.154343.80201.6112.7167.125.669.943.152.751.55.06136.08106.584.717.28-2.715.37276.8177.4FS =1.004
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