Please explain these slides step by step so I can understand Y cpFHWLIXFyFExample 2.A4 mx cpx cpYcp = y +xwIJAXcpFy = Fy *1+W* xwср у=4r3π5+0.067 5.067m== 0.957 m0.849 m78.5*1+30.8*0.849109.3 Example 2.AWidth is 1 m. Find: F and the position that the fluid pressure exerted.ΣF=0=Fy-Fy – WFHFxWF4myFy = PA= 9810*4*2*1= 78.5 kNTW = y = y ²¹ r² *14= 9810*0.25*π *4*1= 30.8 kNFy = 30.8+78.5F₁, = 109.3 kNΣF=0= FH - FxXFx = pA= 5*9810*2*1F₂ = 98.1kNX

Given: The width of the surface = 1 m Depth to the start of the curve = 4 m Curve radius = 2 m…

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 = 9810421 = 78.5 kN π 2 W = WV = y = y₁² 4 = 98100.25 **41 = 30.8 kN F₁ = 30.8+78.5 y F₁ = 109.3 KN 1 ΣF=0=FH-Fx Fx = PA X = 598102*1 Fx = 98.1kN

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Braced cuts

The braced cut is a method of providing lateral support to the soil in foundation trenches, to prevent the trenches from deformation by the caving in of soil and resisting it from moving laterally. During shallow excavation in firm soil, generally, slopes are provided to resist the caving in of soil. During shallow excavation in soft soil and deep excavation, the chances of caving in of soil increase. Hence, it becomes necessary to make a provision to resist the deformation of trenches due to caving in of soil. The braced cut is the method used for such conditions. The excavation using braced cuts is a bit uneconomical as compared to the traditional open-cut excavation method but provides better safety than the cut excavation method.

Question

Please explain these slides step by step so I can understand

Y cp
FH
W
LIX
Fy
F
Example 2.A
4 m
x cp
x cp
Ycp = y +
xw
I
JA
XcpFy = Fy *1+W* xw
ср у
=
4r
3π
5+0.067 5.067m
=
= 0.957 m
0.849 m
78.5*1+30.8*0.849
109.3

Example 2.A
Width is 1 m. Find: F and the position that the fluid pressure exerted.
ΣF=0=Fy-Fy – W
FH
Fx
W
F
4m
y
Fy = PA
= 9810*4*2*1
= 78.5 kN
T
W = y = y ²¹ r² *1
4
= 9810*0.25*π *4*1
= 30.8 kN
Fy = 30.8+78.5
F₁, = 109.3 kN
ΣF=0= FH - Fx
X
Fx = pA
= 5*9810*2*1
F₂ = 98.1kN
X

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