Principles of Foundation Engineering (MindTap Course List)
9th Edition
ISBN: 9781337705028
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 9, Problem 9.12P
To determine
Calculate the settlement of a 2 m wide square foundation.
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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.
Problem 1. A column foundation (Figure below) is 3 m × 2 m in plan. The load on the column,
including the weight of the foundation is 4500 kN. Determin the average vertical stress increase 4
m beneath the corner of the foundation in the soil layer due to the foundation loading by:
a) Boussinesq equations
b) 2:1 method
Given: Df = 1.5 m, Ø'= 25°, c'= 70 kN/m².
1.5 m 1 m
3m x 2m
y = 17 kN/m³
Water level
Ysat 19.5 kN/m³
A shallow foundation measuring 1.75 m x 1.75 m is to be constructed over a layer of
sand to support a net vertical load 400 kN. Given: D= 1 m; average No =12. The sand is
normally consolidated. Estimate the elastic settlement of the foundation. Using Modified
Meyerhof's method.
Chapter 9 Solutions
Principles of Foundation Engineering (MindTap Course List)
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- 7.14 Refer to Figure 7.15. For a foundation on a layer of sand, given: B = 5 ft, L = 10 ft, d = 5 ft, B = 26.6°, e = 0.5 ft, and & = 10°. The Pressuremeter testing at the site pro- duced a mean Pressuremeter curve for which the pam) versus AR/R, points are as follow. AR/R. (1) P,(m) (lb/in.?) (2) 0.002 7.2 0.004 24.2 0.008 32.6 0.012 42.4 0.024 68.9 0.05 126.1 0.08 177.65 0.1 210.5 0.2 369.6 What should be the magnitude of Q, for a settlement (center) of 1 in.? Foundation BxL В Figure 7.15 Definition of parameters-B,arrow_forward8.4 A rectangular foundation is shown in Figure P8.2, given B=2m, L=4m q = 240 kN/m², H = 6m, and D; = 2 m. (a) Assuming E = 3800KN/m², calculate the average elastic settlement. Use Eq. (8.24). (b) If the clay is normally consolidated, calculate the consolidation settlement. Use Eq. (8.35) and y,t = 17.5 kN/m’, C, = 0.12, and e, = 1.1. %3D G.W.T. D,=2 m = 240 kN/m² Clay e. = .IO H= 6 m 1. Rock Figure P8.2 S,(average) = µ,M0 qB (v = 0.5) E (8.24) (8.35)arrow_forward|A rigid shallow foundation 1m x 1m in plan is shown in the following figure. Calculate the elastic settlement at the center and corner of the foundation. Ao = 200 kN/m2 1 m 1 m X 1 m E, (kN/m?) + 8000 Hy = 0.3 + 6000 3 10,000 Rock (m).arrow_forward
- A continuous foundation is shown in Figure is to be constructed on a sand deposit. If the load eccentricity is 0.2 m, determine the ultimate load, Qu, per unit length of the foundation. Use Meyerhof's effective area method. 1.5.m -2 m Sand $' = 40° -0 y=16.5 kN/m³arrow_forward8.4 A rectangular foundation is shown in Figure P8.2, given B= 2 m, L=4m q=240 kN/m², H=6m, and D; =2 m. (a) Assuming E = 3800KN/m², calculate the average elastic settlement. Use Eq. (8.24). (b) If the clay is normally consolidated, calculate the consolidation settlement. Use Eq. (8.35) and yat = 17.5 kN/m², C¸ = 0.12, and e, = 1.1.arrow_forwardProblem 2: A square foundation of B = 4 m applies a uniform pressure of 17.5 kN/m² to the underlaying ground. Determine the vertical stress increase using at a depth of 1m below the center using: a) 2:1 method b) m and n method. c) Stress isobars d) Newmark Methodarrow_forward
- 4. For the stressed soil element shown. use the pole method and determine: 1) the major and minor principal stresses, 2) the normal and shear stresses on the plane oriented 0 to the horizontal. O is 12⁰. Gy = 50 kN/m² Tyx = 50 kN/m² 0 Txy = 50 kN/m² Ox= 150 kN/m²arrow_forwardA 2-m diameter flexible foundation applies a uniform pres- sure to the underlying soil of 200 kN/m. Plot the varia- tion of the vertical stress increase below the center of the foundation as determined using both the Boussinesq (Aop) and Westergaard (Aow) solutions up to a depth of z = 5 m. (Note: p, 0.2.) 8.14arrow_forwardo T ☺ ť (Q4): For the soil profile shown in figure (2) find the difference in stress in (kPa) at point located at a depth (10m) under the center of the circular foundation when using the below formulas and when using the approximation method 1 Ao, = q [(R/2)* + 1] Dia4m q0.21 N/mm N.G.S Sand e0.8 G.-2.6 3m W.L e0.7 G-2.7 5m Clayey - Sand e045 G.-2.68 مر ر ل لأعلى لاستعراض الفلاتر إضافة شرح. .. mum < IIarrow_forward
- A footing 2.25 m square is located at a depth of 1.5 m in a sand of unit weight 18 kN/m³. The shear strength parameters are c' = 0 and 6 = 36°. Calculate the safe load carried by the footing against complete shear failure. Factor of safety against shear failure is 3. Use Terzaghi's analysis. (N. = 65.4, N, = 54.0) = 49.4, b. %3D N.arrow_forwardA rigid foundation is subjected to a vertical column load, P = 355 kN, as shown in Figure 1. Estimate the elastic settlement due to the net applied pressure, Ao, on the foundation. Given: B = 2m; L= 3m; Df=1.5m; H = 4m; Es = 13,500 kN/m²; and µs = 0.4. P Foundation Ao. B× L Soil µ = Poisson's ratio E, modulus of elasticity: H Rockarrow_forward2. As shown in the figure two adjacent foundations A and B, its size, relative position and the distribution of additional pressure on the substrate as shown in the figure, the role of the uniform load on the foundation A and B is POA POB = 100 kPa, find Ов (1) Without considering the effect of foundation B, the additional stress point and center point of foundation A at a depth of 2m. (2) Considering the influence of B foundation, the additional stress at the depth of 2m under the center point of A foundation. Value of vertical additional stress coefficient under corner points when rectangular area is subjected to uniform load 1 2 0.249 0.249 0.240 0.244 0.223 0.233 0.175 0.200 0.084 0.120 zb 0.2 0.4 0.6 1.0 2.0 L/b A 0 | 2m |10| = B 2 m 3 m 3 0.249 0.244 0.234 0.203 0.131 under the corner 4 0.249 0.244 0.234 0.204 0.135arrow_forward
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