Prob 25.6 - CEE 452 FOUNDATIONS Problem 25.6 Fall 2009 Page...

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CEE 452 FOUNDATIONS Fall 2009 Problem 25.6 Page 1 of 3 Problem 25.6 Refer to Cantilever Wall Design summary sheet (attached at end of file): Given: 18 ft tall wall, dry soil, ’ = 37 o , = 126 lb/ft3: For ’ = 37 o : K A = 0.25 , K P = 4.02 Using FS = 2 (using K P = 4.02/2 = 2.01): A = K A H = 0.25 (18) 126 = 567 L 1 = A / [(K P -K A ) ] = 567 / [(2.01-0.25) 126] = 567 / 221.76 = 2.56 ft P I = A (H) / 2 = 567 (18) / 2 = 5103 lb / ft P II = A (L 1 ) / 2 = 567 (2.56) / 2 = 725.8 lb / ft Force Equilibrium gives: P I + P II + P III + P IV = 0 P III = - B (L 2 ) / 2 and B = [(K P -K A ) ]L 2 /2 , So P III = - (K P -K A ) (L 2 ) 2 / 4 = -55.4 (L 2 ) 2 P IV = C (L 3 ) / 2 and C = [(K P -K A ) ]L 3, So P IV = [(K P -K A ) ](L 3 ) 2 / 2 = 110.9 (L 3 ) 2 Combining terms gives: 5103 + 725.8 + 110.9 (L 3 ) 2 = 55.4 (L 2 ) 2 which reduces to 105.2 + 2(L 3 ) 2 = (L 2 ) 2 Moment Equilibrium (Summing Moment about the point of application of P III ) gives: (P I ) x (L 2 /2 + 2.56 + 6) + P II x [L 2 /2 +2(2.56)/3] = P IV x (2L 3 /3)= 0, which reduces to: 2551.5 L 2 + 43681.7 + 362.9 L 2 + 1238.7 = 73.93 (L 3 ) 3 2914.4 L 2 + 44920.4 = 73.93 (L 3 ) 3 or L 2 = 0.02537 (L 3 ) 3 - 15.413
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This note was uploaded on 11/29/2010 for the course CEE 452 taught by Professor Zapata during the Spring '10 term at University of Arizona- Tucson.

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Prob 25.6 - CEE 452 FOUNDATIONS Problem 25.6 Fall 2009 Page...

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