note: please expert answer only the letter d in the problem 3. A dnven closed-ended pile, circular in cross section, is shown.Calculate the following.a. The ultimate point load using Meyerhof's procedure.b. The ultimate point load using Vesic's procedure. Take 4= 50.c. An approximate ultımate point load on the basıs of parts (a) and (b).d. The ultimate frictional resistance Qs, K= 1.4; 8'-0.8Ø"e. The alowable load of the pile (use FS= 4).y 15.7 kN/m= 32°3 mGroundwatertableYut = 18.2 kN/m' = 32°c' = 03 mYut = 19.2 kN/m= 40"c' = 015 m381 mm

The values of k and properties of soil are given

d. The ultimate fnctional resistance Qs, K= 1.4; 8'-0.8Ø e. The allowable load of the pile (use FS= 4). 15.7 kN/m - 32 3 m Groundwater table Yat - 18.2 kN/m '= 32° c' -0 3 m Yut - 19.2 kN/m - 40" e' -0 15 m 381 mm

d. The ultimate fnctional resistance Qs, K= 1.4; 8'-0.8Ø e. The allowable load of the pile (use FS= 4). 15.7 kN/m - 32 3 m Groundwater table Yat - 18.2 kN/m '= 32° c' -0 3 m Yut - 19.2 kN/m - 40" e' -0 15 m 381 mm

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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Alloys

An alloy is an admixture of two or more metals or a metal and a non-metal. Since ancient times, man has been using alloys. Alloying helps to retain the properties of the parent matter, such as hardness, strength, toughness, ductility and is corrosion-resistant. Copper alloys such as brass have antimicrobial properties that help to fight against (Coronavirus disease 2019) COVID-19 and other hospital-acquired infections. Some common alloys are brass, bronze, solder steel, etc. Most alloys have a lower melting point than their constituents. This is due to the presence of a nonmetal as one of the constituents that behaves as an impurity and restricts the raise of melting point.

Question

note: please expert answer only the letter d in the problem

3. A dnven closed-ended pile, circular in cross section, is shown.
Calculate the following.
a. The ultimate point load using Meyerhof's procedure.
b. The ultimate point load using Vesic's procedure. Take 4= 50.
c. An approximate ultımate point load on the basıs of parts (a) and (b).
d. The ultimate frictional resistance Qs, K= 1.4; 8'-0.8Ø"
e. The alowable load of the pile (use FS= 4).
y 15.7 kN/m
= 32°
3 m
Groundwater
table
Yut = 18.2 kN/m
' = 32°
c' = 0
3 m
Yut = 19.2 kN/m
= 40"
c' = 0
15 m
381 mm

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