Principles Of Foundation Engineering 9e
9th Edition
ISBN: 9781337705035
Author: Das, Braja M.
Publisher: Cengage,
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Chapter 16, Problem 16.10P
To determine
Find the Rankine active earth pressure coefficient
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Students have asked these similar questions
Q4: For the soil element shown, compute the stresses acting on the plane inclined by 40° with the horizontal
plane then draw Mohr circle and place the stresses with respect to O.P.
20kPa
35kPa
100kPa
300
40°
120 KN
50 KN/m
2.0 m
50 KN/m
2.0 m
B
A
4.0 m
4.0 m
a. Determine the location of the maximum deflection using double integration.
b. Determine the magnitude of the maximum deflection using double integration method using E = 200
x10^6 KPa and I = 1.440 x10^-5 m4
The location of trial failure surface on a slope is shown in Figure and the stress components for each slice are
listed in Table.
Slice
Shearing
component
_(kN.m·')
-0.280
-0.227
0.383
3.214
6.543
8.368
9.792
Normal
component
(kŇ.m*)
1.911
7.745
13.139
16.344
17.625
16.718
12.125
0.486
Length of trial failure surface=11m
Soil friction angle 6°
Cohesion, c=28kPa
No.
1
3
4
6.
4.228
Trial Fallure
Surface
(a) Compute the driving stress.
(b) Compute the stabilizing stress.
(c) Analyze the safety of the slope.
Chapter 16 Solutions
Principles Of Foundation Engineering 9e
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- a) As a geotechnical engineer, explain why construction of foundations and other loads (structures) leads to a compression of soil layers. b) During the construction of a retaining wall, it was noticed that the wall does not yield toward the backfill or away from the backfill. Explain this phenomenon. Sketch a diagram to explain your answer. c) Let us consider the retaining wall shown in Figure below. The height of the wall is 9.75 m, and the unit weight of the sand backfill is 18.7 kN/m³. Using Coulomb's equation, calculate the active force, Pa, on the wall for angle of wall friction 8'-14°. Also, comment on the direction and location of the resultant. Sand y = 18.7 kN/m³ 0 = 10° c' = 0 H= 9.75 m 6' = 34° 0 = 12° 8' (wall friction)arrow_forward'2\ Answer T for true statements or F for false ones: 1. Intact rock: solid rock sample containing cracks on eye naked. 2. Rock Mass: contains rock samples and crack surfaces on eye naked. 3. The rock sample will break if the applied force is vertical and associated with shear. 4. Unit weight (y) = gm./cm³. 5. Void Ratio (e) = Vs/Vv. 6. In point load test if the rock is fresh, ot= 5 * I, (50) 7. Brazilian Test provides tensile strength values closer to reality than Point Load Test. 8. Schmidt Hammer Test gives direct values for strength. 9. As more the surface is rough its strength for slipping decreases. 10.Fracture Intercept is the measured distance between irregular fractures !!arrow_forward3. The vertical stress o̟ under the corner of a rectangular area subjected to a uniform load of intensity q is given by the solution of Boussinesq's equation: 2mn /m² +n² +1 m² + n² +2 + sin 47 m² +n² +1+m²n² m² +n² +1 2mn/m² +n² +1 m² +n² +1+m²n² O = because this equation is inconvenient to solve manually, it has been reformulated as 0̟ = qf (m,n) where f.(m,n) is called the influence value and m and n are dimensionless ratios, with m=a/z and n=b/z and a and b are defined in the following figure. 1 The influence value is then tabulated, a portion of which is given in the following table n = 1.2 n = 1.4 0.03007 0.05894 0.08561 n = 1.6 0.1 0.02926 0.03058 0.2 0.05733 0.05994 0.3 0.08323 0.08709 0.4 0.10631 0.10941 0.11135 0.13003 0.14749 0.16199 0.5 0.12626 0.13241 0.15027 0.16515 0.17739 0.6 0.14309 0.7 0.15703 0.8 0.16843 0.17389 If a=4.6 and b=14,compute o̟ at a depth 10 m below the corner of a rectangular footing that is subject to a total load of 100 t (metric tons). Note that q is…arrow_forward
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