EBK PRINCIPLES OF FOUNDATION ENGINEERIN
8th Edition
ISBN: 8220100547058
Author: Das
Publisher: CENGAGE L
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Chapter 4, Problem 4.5P
A column foundation (Figure P6.9) is 3 m × 2 m in plan. Given: Df = 1.5 m, ф′ = 25°, c′ = 70 kN/m2. Using Eq. (6.28) and FS = 3, determine the net allowable load [see Eq. (6.24)] the foundation could carry.
Figure P6.9
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A column foundation (Figure P4.5) is 3 m x 2 m in plan. Given: Df = 1.5 m, Φ' = 25°, c' = 70 kN/m2. Using Eq. (4.26) and FS = 3, determine the net allowable load [see Eq. (4.22)] the foundation could carry.
A column foundation (Figure P3.5) is 3 m x 2 m in plan. Given: D; = 2 m,
o' = 25°, c' = 50 kN/m². Using Eq. (3.23) and FS = 4, determine the net
allowable load [see Eq. (3.15)] the foundation could carry. Use bearing capac-
ity, shape, and depth factors given in Şection 3.6.
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°
=
Chapter 4 Solutions
EBK PRINCIPLES OF FOUNDATION ENGINEERIN
Ch. 4 - For the following cases, determine the allowable...Ch. 4 - A square column foundation has to carry a gross...Ch. 4 - Prob. 4.3PCh. 4 - The applied load on a shallow square foundation...Ch. 4 - A column foundation (Figure P6.9) is 3 m × 2 m in...Ch. 4 - Prob. 4.6PCh. 4 - For the design of a shallow foundation, given the...Ch. 4 - An eccentrically loaded foundation is shown in...Ch. 4 - Prob. 4.9PCh. 4 - For an eccentrically loaded continuous foundation...
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- A continuous foundation with a width of 1 m is located on a slope made of clay soil. Refer to Figure 5.19 and let Df = 1 m, H = 4 m, b = 2 m, γ = 16.8 kN/m3, c = cu = 68 kN/m2, Φ= 0, and β = 60°.a. Determine the allowable bearing capacity of the foundation. Let FS = 3.b. Plot a graph of the ultimate bearing capacity qu if b is changed from 0 to 6 m.arrow_forwardProb. 3): A square shallow foundation is shown below. If the load eccentricity is 0.3 m, determine the maximúm allowable load that the foundation can carry. Use Mayerhof's method, and FS as 4. (Eccentricity in one direction only) e = 0.3 m Qal Y = 16.3 kN/m3 c' = 20 kN/m? p'=28° 1.0 m 1.5 m X 1.5 m Centerlinearrow_forwardA square column foundation has to carry a gross allowable load of 1805 kN (FS = 3). Given: Df = 1.5 m, γ = 15.9 kN/m3 ϕ = 34 and c′ = 0. Use Terzaghi’s equation to determine the size of the foundation (B). The applied load on a shallow square foundation makes an angle of 15° with the vertical. Given: B = 1.83 m, Df = 0.91 m, γ = 18.08 kN/m3 ,ϕ=25° , and c′ = 23.96 kN/m2 . Use FS = 4 and determine the gross allowable (vertical component) load. Use Eq. (16.9).arrow_forward
- An eccentrically loaded foundation is shown in Figure P3.9. Use FS of 4 and determine the maximum allowable load that the foundation can carry. Use Meyerhof's effective area method and the bearing capacity, shape, and depth factors given in Section 3.6.arrow_forward" (a) Qu 1 Qu 3 B (b) M M → X M Qu M (c) Figure 4.24 Analysis of foundation with two-way eccentricity 7 = 17kN/m³ friction angle = : 35⁰° , and cohesion c = 0 Qu (d) The shallow foundation is shown in Figure 4.24 measures 1.5 m X 2.25 m and is subjected to a centric load and a moment. If ев = = 0.12m e₁ eL = 0.36m and the depth of the foundation is 0.8 m, determine the allowable load the foundation can carry. Use a factor of safety of 4. For the soil, we are told that unit weightarrow_forwardQ6. A column foundation (Figure below) is 3 m X 2 m in plan. Given: De = 1.5 m, o = 25°, c= 70 kN/m . Terzaghi's equation and assume general shear failure in soil and FS = 3, determine the net alowable koad. y = 17 kN/m 1.5 m 3 m x 2 m Yat = 19.5 kN/m Groundwater levelarrow_forward
- A square foundation is 1.5 m x 1.5 m in plan. The soil supporting the foundation has a friction angle = 32 deg. and c = 21 kPa. The unit weight of soil is 17.5 kN/m3. Assume that the depth of foundation Df is 1 meter, Use Nc = 44.14 ; Nq = 28.52 and N? = 26.87. a).Determine the allowable bearing capacity on the foundation with a factor of safety of 3.0 b.)Determine the allowable gross load on the foundation with a factor of safety of 4.0 .arrow_forwardConsider a continuous foundation of width B = 1.4 m on a sand deposit with c' = 0, Φ' = 38° and γ = 17.5 kN/m3. The foundation is subjected to an eccentrically inclined load (see Figure 4.31). Given: load eccentricity e = 0.15 m, Df = 1 m, and load inclination β = 18°. Estimate the failure load Qu(ei) per unit length of the foundation a. for a partially compensated type of loading [Eq. (4.85)] b. for a reinforced type of loading [Eq. (4.86)]arrow_forwardA square column foundation has to carry a gross allowable load of 1805 kN ( FS = 3). Given: D f = 1.4 m, γ = 15.4 kN/m 3 , ϕ ′ = 34 ° , and c ′ = 0. Use Terzaghi's equation to determine the size of the foundation ( B ). Assume general shear failure.arrow_forward
- 1. For the following cases, determine the allowable gross vertical load bearing capacity of the foundation. Use Terzaghi's equation. Part В D; Foundation Type 3 ft 3 ft 28° 400 psf 110 pcf Continuous a b 1.5 m 1.2 m 35° 17.8 kN/m³ Continuous 3 m 30° 30° 16.5 kN/m³ Square 2. A square foundation has to carry gross allowable load of 1805 kN (FS=3). Given: D; = 1.5 m, y=15.9 kN/m³, 0=34°, and c = 0. Use Terzaghi's equation to determine the size of the foundation (B).arrow_forward7.5 A 2.0 m wide square foundation is placed at 0.5 m depth in a saturated clay where c, = 40 kN/m² and y = 19.0 kN/m³. There is a very stiff stratum present at 1.0 m below the foundation. Determine the ultimate bearing capacity using Buisman's (1940) equation.arrow_forward10. 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_forward
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