Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN: 9781305081550
Author: Braja M. Das
Publisher: Cengage Learning
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Chapter 5, Problem 5.5P
To determine
Find the gross allowable bearing capacity of the clay.
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Refer to Figure 5.9. For a continuous foundation in two-layered clay, given:●● γ1 = 121 lb/ft3, cu(1) = 1000 lb/ft2, Φ1 = 0●● γ2 = 115 lb/ft3, cu(2) = 585 lb/ft2, Φ2 = 0●● B = 3 ft, Df = 1.65 ft, H = 1.65 ftFind the gross allowable bearing capacity. Use a factor of safety of 3.
9.3 Given: T- shape foundation as shown in figure is loaded with a uniform load of 120 kN/m².
Required: the increment in vertical stress at point (P) at a depth of 5m.
0.6
0.6
0.6
3m
m
n
I₁
1.5
0.3
0.0629
0.1431
9m
1.2
3m
0.6
0.1069
Refer to Figure 5.12. For a rectangular foundation on layered sand, given:●● B = 4 ft, L = 6 ft, H = 2 ft, Df = 3 ft●● γ1 = 98 lb/ft3, Φ'1 = 30º, c'1 = 0●● γ2 = 108 lb/ft3, Φ'2 = 38º, c'2 = 0Using a factor of safety of 4, determine the gross allowable load the foundation can carry.
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Principles of Foundation Engineering (MindTap Course List)
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- Consider a continuous foundation of width B = 1.4 m on a sand deposit with c = 0, = 38, and = 17.5 kN/m3. The foundation is subjected to an eccentrically inclined load (see Figure 6.33). Given: load eccentricity e = 0.15 m, Df = 1 m, and load inclination = 18. Estimate the failure load Qu(ei) per unit length of the foundation a. for a partially compensated type of loading [Eq. (6.89)] b. for a reinforced type of loading [Eq. (6.90)]arrow_forward2 ft 2 ft 24 ft 24 ft 24 ft Problem 4 B D E F G | 3 ft DL=100 kip DL=180 kip LL = 60 kịp LL = 120 kip DL=190 kip DL=110 ki • The plan of a mat foundation with column loads is shown in Figure 2. Use the rigid method to calculate the soil pressures at point A, B, C, D, E, F, G, H, , J, K, L, M and N. The size of the mat is 76 ft x 96 ft, all columns are 24 in x 24 in in section, and qlnet = 1.5 kip/ft². Verify that the soil pressures are less than the net allowable bearing capacity. LL = 120 kip LL = 70 ki 30 ft DL=180 kip DL=400 kip DL=200 kip LL = 250 kip LL = 120 kip DL=360 kip LL = 120 kip LL = 200 kip ex 30 ft DL-190 kip DL=500 kip LL = 130'kip LL = 240 kip DL=T10 kip DL=200 kip LL =300 kip LL =120 kip 30 ft DL=180 kip DL=120 kip LL =120 kip L =70 kip x' 3 ft IDL=120 kip DL=180 kip ILL =70 kip LL =120 kip J Figure 2: Plan of a Mat Foundation M L K Harrow_forward7.7 Refer to Figure P7.7. Using the procedure outlined in Section 7.10, determine the average stress increase in the clay layer below the center of the foundation due to the net foundation load of 50 tons. [Use Eq. (7.26).] 450 10 A 50 kin tet load Figure P7.7 5nxsn Sand 7-122 1² Sand y 100 lb/m Yu120 vn =0.7 C=0.25 C-0.06 Water table Clay Preconsolidation pressure-2000 lb/n² Carrow_forward
- 10. A flexible foundation is subjected to a uniformly distributed load of q-500 kN/m². Table 3 could be useful. Determine the increase in vertical stress, in kPa, Aoz at a depth of z=3m under point F. B 4m 3m 6m E 10m Table 10.3 Variation of I, with m and n m 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.1 0.0047 0.0092 0.0270 0.0279 0.2 0.0132 0.0092 0.0179 0.0259 0.0132 0.0259 0.0374 0.0222 0.0242 0.0435 0.0474 0.0629 0.0686 0.0258 0.0504 0.0528 0.0547 0.3 0.0731 0.0766 0.0794 0.4 0.1013 0.5 0.0198 0.0387 0.1202 0.6 0.0222 0.0435 0.7 0.0242 0.0474 0.0947 0.1069 0.1168 0.1247 0.1311 0.1361 0.1365 0.1436 0.1491 0.1537 0.1598 0.0168 0.0198 0.0328 0.0387 0.0474 0.0559 0.0168 0.0328 0.0474 0.0602 0.0711 0.0801 0.0873 0.0931 0.0977 0.0559 0.0711 0.0840 0.0947 0.1034 0.1104 0.1158 0.0629 0.0801 0.0686 0.0873 0.1034 0.8 0.0258 0.0504 0.0731 0.0931 0.1104 0.9 0.0270 0.0528 0.0766 0.0977 0.1158 0.0794 0.1013 0.1202 0.0832 0.1263 1.4 0.1300 1.6 0.0306 0.0599 0.0871 0.1114 0.1324 1.8 0.0309 0.0606…arrow_forwardWith the same load acting on the same size of foundation, calculate the settlement (in.) that will occur in the clay layer as shown below, using Factor of Safety of 1.5 Please use 2:1 method to find stress increase under a foundation Assume: OCR = 1 EL.+450 FT. Sand Y = 110 pcf EL. +445 FT. GWT: +442 FT. Y = 120 pcf %3D Clay eo = 0.3 Cc = 0.250 Cr = 0.025 EL. +435 FT. ROCK (Picture is not in scale)arrow_forwardQ.8 A long foundation 0.6 m wide carries a line load of 100 kN/m. Calculate the vertical stress, at a point P, the coordinates of which are x = 2.75 m, and z = 1.5 m, where the x coordinate is normal to the line load from the central line of the footingarrow_forward
- Q1.1// The figure below shows the plan of a rectangular raft foundation which supports a total load of 40500 kN. The foundation is located at 3 m from the surface. Evaluate the increase in the effective stress at a point 8 m below the surface and having x, y coordinates as (-9, -6.5) with respect to the central axis of the footing. X 15 m w 6arrow_forwardProblem II. The initial principal stresses at a certain depth in a clay soil are 100 kPa on the horizontal plane and 50 kPa on the vertical plane. Construction of a surface foundation induces additional stresses consisting of a vertical stress of 45 kPa, a lateral stress of 20 kPa, and a counterclockwise (with respect to the horizontal plane) shear stress of 40 kPa. a. Plot Mohr's circle (1) for the initial state of the soil and (2) after construction of the foundation. b. Determine the change in magnitude of the principal stresses. C. the change in maximum shear stress d. the change in orientation of the principal stress plane resulting from the construction of the foundation.arrow_forwardPROBLEMS 8.1 Refer to Figure 8.3. For a flexible load area, given: B= 3 m, L=4.6m, q= 180KN/m², D; =2m, H = 00, v= 0.3, and E = 8500KN/m³. Estimate the elastic settlement at the center of the loaded area. Use Eq. (8.14). %3D Foundation B×L Rigid :foundation Flexible foundation H settlement settlement v = Poisson's ratio E = Modulus of elasticity Soil Rock Figure 8.3 Elastic settlement of flexible and rigid foundations. (8.14)arrow_forward
- Question 1) For a shallow foundation measuring (1.7 m x 2.2 m) as shown below: , A. Estimate the elastic settlement proposed by Mayerhof. Then, B. Estimate the elastic settlement proposed by Bowles, if the water table rises 1.5 m. Then, Use yw=10 kN/m³ qnet= 1.2 MN/m2 G.S 1.5 m Sand Yd=16 kN/m³ Ysat= 17 kN/m3 %3D 2.5 m N60=52 V W.T. Silty Sand Ya=18 kN/m³ Ysat = 18.5 kN/m? N60=52 3.5 m Sand Ya=19 kN/m3 Ysat = 22 kN/m³ e, = 0.4, Ae=0.04 , o'= 194 kN/m2 5 m Cc= 0.3, Cs= 0.2 , Ca= 0.05 N60=60 CS Scanned with CamScannerarrow_forwardSolve Problem 7.8 using Eq. (7.29). Ignore the post-construction settlement. 7.8 Solve Problem 7.4 with Eq. (7.20). Ignore the correction factor for creep. For the unit weight of soil, use γ = 115 lb/ft3. 7.4 Figure 7.3 shows a foundation of 10 ft × 6.25 ft resting on a sand deposit. The net load per unit area at the level of the foundation, qo, is 3000 lb/ft2. For the sand, μs = 0.3, Es = 3200 lb/in.2, Df = 2.5 ft, and H = 32 ft. Assume that the foundation is rigid and determine the elastic settlement the foundation would undergo. Use Eqs. (7.4) and (7.12).arrow_forward
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