5.13 The shallow foundation shown in Figure 5.29 measures 1.5 mx 2.25 m and is subjected to a centric load and a moment. If e = 0.12 m, e, = 0.36, and the depth of the foundation is 0.8 m, determine the allowable load the foun- dation can carry. Use a factor of safety of 4. For the soil, we are told that the unit weight y 17 kN/m³, friction angle d' = 35°, and cohesion c' = 0. =

5.13 The shallow foundation shown in Figure 5.29 measures 1.5 mx 2.25 m and is subjected to a centric load and a moment. If e = 0.12 m, e, = 0.36, and the depth of the foundation is 0.8 m, determine the allowable load the foun- dation can carry. Use a factor of safety of 4. For the soil, we are told that the unit weight y 17 kN/m³, friction angle d' = 35°, and cohesion 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

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

5.13 The shallow foundation shown in Figure 5.29 measures
1.5 mx 2.25 m and is subjected to a centric load and a
moment. If e, = 0.12 m, e, = 0.36, and the depth of the
foundation is 0.8 m, determine the allowable load the foun-
dation can carry. Use a factor of safety of 4. For the soil, we
are told that the unit weight y = 17 kN/m³, friction angle
o'= 35°, and cohesion c' = 0.

(a)
L'-
B
y
M
W<
B-
(b)
BXL
M₂
"0
"W
(c)
(d)
eL

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Step 2: Determine eccentricity ratio

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Step 3: Determine The value of L1 and L2

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Step 4: Determine effective area and effective width

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Example 5.7 Consider a rectangular foundation 2 mx 4 m in plan at a depth of 1.2 m in a sand deposit, as shown in Figure 5.23a. Given: y = 17.5 kN/m³; ā = 145 kN/m², and the following approximated variation of qc with z: 1.2 m q=145 kN/m² ++++y=17.5 kN/m³ z (m) 9c (kN/m²) B=2m- 0-0.5 2250 L=4 m 0.5-2.5 3430 2.5-5.0 2950 Estimate the elastic settlement of the foundation using the strain influence factor method.
An eccentrically loaded continuous foundation is shown in Figure P6.18. Determine the ultimate load Qu per unit length that the foundation can carry. Use the reduction factor method [Eq. (6.67)]. 4 ft 2 ft Figure P6.18 Qu 2 ft → -5 ft Y = 105 lb/ft³ Groundwater table Ysat 118 lb/ft³ c' = 0 $' = 35° =
9.6 A mat foundation is shown in Figure P9.6. The design considerations are L = 12 m, B = 10 m, D = 2.2 m, Q = 30 MN, x₁ = 2 m, x₂ = 2 m, x3 = 5.2 m, and precon solidation pressure o 105 kN/m². Calculate the conso dation settlement under the center of the mat. D₁ X3 Figure P9.6 Size of mat = B x L Vo Clay Ysat = 17.5 kN/m³ €= 0.88 C=0.38 C₁ = 0.1 Sand y= 16.0 kN/m³ Water table Sand Ysat = 18.0 kN/m³
Figure 5.16 shows a foundation of 6.25 ft x 10ft resting on a sand deposit. The net load per unit area at the level of the foundation, qo, is 3000 lb/ft. For the sand, u, =0.3, E, = 3200 lb/in', Dr= 2.5 ft and H= 32 ft. Assume that the foundation is rigid, and determine the elastic settlement the foundation would undergo. Use Eqs. (5.33) and (5.41); note that these equations are in the class note posted on the Blackboard. Eq (5.33) = Se=20(aB').-us Foundation B XL Es 2o= net applied pressuve on fandadien Ms = Poisson's ratio of soil Es = average modulus of elaslisity. B= B½ fer centr of feundatien Rigid foundation settlement Plexible foundation settlement Is = shape facker A Poisson's ratio E, - Modulus of elasticity Eq (5.41) = Secrigid)0.93 Secflekible, center) Soil Rock Figure 5.16
For the embedded strip footing (infinitely long in the out-of-plane direction) shown below, the maximum vertical pressure that the soil can bear before failure is 100 kPa (i.e., qmax should not exceed 100 kPa). What is the maximum overall eccentricity of the foundation in mm before failure? (answer tolerance = 2%). Consider γconcrete = 25 kN/m3, γsoil = 18 kN/m3 and assume that the width of the embedded column is negligible, and the entire top of the foundation is covered with soil. Hint: for a strip footing, the calculations should be conducted assuming a 1-m long footing in the out-of-plane direction.
4. A flexible foundation is shown below, determine the immediate settlement below the center of the foundation. Assume the thickness of the soil below the foundation is 20 meters. Following is the variation of the modulus of the soil below the foundation. Es (kN/m²) Depth below the foundation(m) 0-4 4-8 8-20 >20 1.2m 10000 8000 12000 ∞ 90 = BXL = 2m x 2m Us = 0.3 150kPa
i5:) For the continuous foundation show below, determine the amount of ultimate load for each meter length of the foundation. Consider the one-dimension eccentricity. : -) = 105 Ib/ft: Groundwater table 2 ft 4 ft Ysat = I18 lb/ft c' = 400 lb/ft? +2 ft - 25° 5 ft
THE ANSWER MUST BE EXACT: Srigid = 4.93 mm KINDLY HAVE A COMPLETE SOLUTION WITH EXPLANATION WILL DO