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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online