Example 2.A Width is 1 m. Find: F and the position that the fluid pressure exerted. ΣF=0=Fy-Fy - W FH -2 m Fv tw Fx F 4m y Fy=pA = 9810*4*2*1 = 78.5 kN π 2 W = WV = y = y₁² 4 = 9810*0.25 **4*1 = 30.8 kN F₁ = 30.8+78.5 y F₁ = 109.3 KN *1 ΣF=0=FH-Fx Fx = PA X = 5*9810*2*1 Fx = 98.1kN
Example 2.A Width is 1 m. Find: F and the position that the fluid pressure exerted. ΣF=0=Fy-Fy - W FH -2 m Fv tw Fx F 4m y Fy=pA = 9810*4*2*1 = 78.5 kN π 2 W = WV = y = y₁² 4 = 9810*0.25 **4*1 = 30.8 kN F₁ = 30.8+78.5 y F₁ = 109.3 KN *1 ΣF=0=FH-Fx Fx = PA X = 5*9810*2*1 Fx = 98.1kN
Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN:9781305081550
Author:Braja M. Das
Publisher:Braja M. Das
Chapter14: Sheet-pile Walls
Section: Chapter Questions
Problem 14.5P
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