Chapter 5: Settlement of BuildingsEXE-3)Determine the total immediate settlement of the rectangular footing shown in figurebelow after 2 months.1200 kNG.S.1.0m3m x 4m8000.kN/m²,..Y = 20.kN/m0.5B =1.5m0.62.0mClay0.533Iz(avg.)Sand3.0mE = 20000.kN/m?4.5m0.133Rock2B = 6m2B – 0.6 Izredo problem (5.3) but with sand instead of clay asshown in the figure below (Ans:Si(total)=5.75mm

The given data is from the figure: Df/B=1/3=0.33L/B=4/3=1.33μo=0.93AndFor…

Chapter 5: Settlement of Buildings X6-3) Determine the total immediate settlement of the rectangular footing shown in figure below after 2 months. 1200 kN G.S. 1.0m 3m x 4m 2.0m 9c = 8000.kN/m",..y = 20.kN/m 0.5B =1.5m 0.6 Clay 0.533

Chapter 5: Settlement of Buildings X6-3) Determine the total immediate settlement of the rectangular footing shown in figure below after 2 months. 1200 kN G.S. 1.0m 3m x 4m 2.0m 9c = 8000.kN/m",..y = 20.kN/m 0.5B =1.5m 0.6 Clay 0.533

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

Load and resistance factor design

Load and resistance factors design (LRFD), also known as Limit state design (LSD), refers to the design method used in building engineering. A boundary condition is a condition of a structure where it can no longer meet the relevant design requirements. The condition may refer to the load level or other actions on the structure, while the terms refer to the integrity of the structure, suitability for use, durability, or other design requirements. A structure designed for LRFD is limited to keeping track of all possible actions during its design time and is always ready for use, with the appropriate level of reliability in each boundary condition. LRFD-based construction codes clearly define the appropriate levels of reliability in their instructions.

Question

Chapter 5: Settlement of Buildings
EXE-3)
Determine the total immediate settlement of the rectangular footing shown in figure
below after 2 months.
1200 kN
G.S.
1.0m
3m x 4m
8000.kN/m²,
..Y = 20.kN/m
0.5B =1.5m
0.6
2.0m
Clay
0.533
Iz(avg.)
Sand
3.0m
E = 20000.kN/m?
4.5m
0.133
Rock
2B = 6m
2B – 0.6 Iz
redo problem (5.3) but with sand instead of clay as
shown in the figure below (Ans:Si(total)=5.75mm

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