The braced cut of two layers of clay is shown in the figure.1.0000AH; = 3 m; Y1 = 16 kN/m² c; = 21 kN/m²H; = 3 m; Y2= 17 kN/m³ cz= 22 kN/m²H3 = 2 m; Y3 = 18 kN/m² c3 = 23 kN/m²3.00002.0000Clay 1The struts are located 4 m on center in the plan.2.0000Determine the strut load at C.3.0000- Clay 22.00001.0000Clay 3.A332.63 kNВ333.36 kN332.36 kND322.36 kN

According to given data he braced cut of two layers of clay is shown in the figure. H = 3 m: Y1 =…

The braced cut of two layers of clay is shown in the figure. 1 H = 3 m; y; = 16 kN'm² c; = 21 kN/m² H; = 3 m; y2= 17 kN/m² ¢z= 22 kN/m² H, = 2 m; Ys = 18 kN'm² c3= 23 kN/m² 20 The struts are located 4 m on center in the plan. 20 Determine the strut load at C.

The braced cut of two layers of clay is shown in the figure. 1 H = 3 m; y; = 16 kN'm² c; = 21 kN/m² H; = 3 m; y2= 17 kN/m² ¢z= 22 kN/m² H, = 2 m; Ys = 18 kN'm² c3= 23 kN/m² 20 The struts are located 4 m on center in the plan. 20 Determine the strut load at C.

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

The braced cut of two layers of clay is shown in the figure.
1.0000
A
H; = 3 m; Y1 = 16 kN/m² c; = 21 kN/m²
H; = 3 m; Y2= 17 kN/m³ cz= 22 kN/m²
H3 = 2 m; Y3 = 18 kN/m² c3 = 23 kN/m²
3.0000
2.0000
Clay 1
The struts are located 4 m on center in the plan.
2.0000
Determine the strut load at C.
3.0000
- Clay 2
2.0000
1.0000
Clay 3.
A
332.63 kN
В
333.36 kN
332.36 kN
D
322.36 kN

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