4- Given the following findA- Ultimate point load using MeyerhofB- Ultimate point load using Vesic with I, = 50C-Qsusing Eq 12.42D- The allowable load with a FOS of 4YE15.7 KN/m3,$=32', c=oYSAT = 18,2 Klm3, $=32", c =03m0.381meClosed ended pileYSAT = 19.2 KNlm3, $=40°, C=o

Answered: Image /qna-images/answer/9554bd21-e445-4084-94f1-7a0c095e6b18.jpg

4-Given the following find A- Ultimate point load using Meyerhof B-Ultimate point load using Vesic with - 50 C-Qs using Eq 12.42 D-The allowable load with a FOS of 4 Y=15.7 KN/M3, $= 32 , C=o YSAT = 18.2 KAlum3, o=32°, Cao 3m 0.381me Closed ended eile YSAT= 19.2 KNlm3, =40", C=0

4-Given the following find A- Ultimate point load using Meyerhof B-Ultimate point load using Vesic with - 50 C-Qs using Eq 12.42 D-The allowable load with a FOS of 4 Y=15.7 KN/M3, $= 32 , C=o YSAT = 18.2 KAlum3, o=32°, Cao 3m 0.381me Closed ended eile YSAT= 19.2 KNlm3, =40", C=0

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

4- Given the following find
A- Ultimate point load using Meyerhof
B- Ultimate point load using Vesic with I, = 50
C-Qsusing Eq 12.42
D- The allowable load with a FOS of 4
YE15.7 KN/m3,$=32', c=o
YSAT = 18,2 Klm3, $=32", c =0
3m
0.381me
Closed ended pile
YSAT = 19.2 KNlm3, $=40°, C=o

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