Problem 3 An eccentrically loaded foundation is shown in Figure 3. Use factor of safety (FOS) of 4.0, determine the maximum allowable load that the foundation can carry. Use Meyerhof's effective area method. For o' = 32°, Ne = 35.49, Ng 23.18, N,- 30.22. %3D %3D (Eccentricity in one direction only) 0.1 m Qall y = 17 kN/m c' = 0 d' = 32° 0,8 m 1.5 m X 1.5 m Centerline Figure 3

Problem 3 An eccentrically loaded foundation is shown in Figure 3. Use factor of safety (FOS) of 4.0, determine the maximum allowable load that the foundation can carry. Use Meyerhof's effective area method. For o' = 32°, Ne = 35.49, Ng 23.18, N,- 30.22. %3D %3D (Eccentricity in one direction only) 0.1 m Qall y = 17 kN/m c' = 0 d' = 32° 0,8 m 1.5 m X 1.5 m Centerline Figure 3

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

Problem 3
An eccentrically loaded foundation is shown in Figure 3. Use factor of safety (FOS) of
4.0, determine the maximum allowable load that the foundation can carry. Use
Meyerhof's effective area method. For o' = 32°, Ne = 35.49, Nq 23.18, N, 30.22.
!!
%3D
(Eccentricity
in one direction
only) 0.1 m
Qall
y = 17 kN/m
e' = 0
d' = 32°
0.8 m
1.5 m X 1.5 m
Centerline
Figure 3

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