3. A square foundation is constructed in a soil deposit as shown in the figure below. Assumethat the groundwater table is 6 meters below the foundation. The applied load on the shallowallowable load. (Use general bearing capacity equation.)square foundation makes an angle of 10° with the vertical. Use FS3 and determine the2 m6 m4 my = 17.5 kN/m³4' = 33°c' = 20 kN/m²Ysat =20k N/m³Groundwater table

A square footing of dimension B = 4 m Depth of footing, D = 2 m Factor of safety, FOS = 3 Load is…

3. A square foundation is constructed in a soil deposit as shown in the figure below. Assume that the groundwater table is 6 meters below the foundation. The applied load on the shallow square foundation makes an angle of 10° with the vertical. Use FS = 3 and determine the allowable load. (Use general bearing capacity equation.) 2 m 6 m 4 m y = 17.5 kN/m³ p' = 33° c' = 20 kN/m² Ysat = 20k N/m³ Groundwater table

3. A square foundation is constructed in a soil deposit as shown in the figure below. Assume that the groundwater table is 6 meters below the foundation. The applied load on the shallow square foundation makes an angle of 10° with the vertical. Use FS = 3 and determine the allowable load. (Use general bearing capacity equation.) 2 m 6 m 4 m y = 17.5 kN/m³ p' = 33° c' = 20 kN/m² Ysat = 20k N/m³ Groundwater table

Principles of Foundation Engineering (MindTap Course List)

8th Edition

ISBN:9781305081550

Author:Braja M. Das

Publisher:Braja M. Das

Chapter7: Settlement Of Shallow Foundations

Section: Chapter Questions

Problem 7.9P: Solve Problem 7.8 using Eq. (7.29). Ignore the post-construction settlement. 7.8 Solve Problem 7.4...

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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)]
3. A square foundation is constructed in a soil deposit as shown in the figure below. Assume that the groundwater table is 6 meters below the foundation. The applied load on the shallow square foundation makes an angle of 10° with the vertical. Use FS = 3 and determine the allowable load. (Use general bearing capacity equation.) 2 m 6 m 4 m y = 17.5 kN/m³ p' = 33° c' = 20 kN/m² Ysat = 20k N/m³ Groundwater table
Question 2) For a shallow foundation shown below: A. Estimate the ultimate bearing capacity when the water table located at a depth of 2 m below the ground surface. B. Estimate the moments about the x- and y-axis; assume that the foundation is subjected to a vertical load and a moment. If eg and eL is 0.33 m and 0.12 m, respectively. G.S Iz 2 m (2 m x 3 m) Silty clay Yo=17 kN/m³ , Ysat = 20 kN/m3 %3D 6 m c'=78 kN/m? 0'=35° Shear modulus=250 kN/m? CS Scanned with CamScanner
2 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 H
A square shallow foundation (B × B) is planned to be constructed on a normality consolidated (NC) clay soil as shown in the below figure. The maximum acceptable settlement for the foundation is equal to 2.0 inches (5 cm), and the safety factor against bearing capacity is FS = 4. Determine the size of foundation. (Note: To simplify the calculations, ignore both the elastic settlement and secondary compression settlement. Also consider 4o'ave = 40'm) Q = 500 kN Ysat = 19.24 kN/m³ en = 0.8 C. = 0.25 p'= 0 c'= 25 kPa 2 m B ×B FS again Bearing Capacity = 4 Acceptable settlement = 2.0 inches 10 m
A square shallow foundation (B × B) is planned to be constructed on a normality consolidated (NC) clay soil as shown in the below figure. The maximum acceptable settlement for the foundation is equal to 2.0 inches (5 cm), and the safety factor against bearing capacity is FS = 4. Determine the size of foundation. (Note: To simplify the calculations, ignore both the elastic settlement and secondary compression settlement. Also consider Ao'ave = 4o'm) Q = 500 kN Ysat = 19.24 kN/m³ eo = 0.8 C. = 0.25 p'= 0 c' = 25 kPa FS again Bearing Capacity = 4 Acceptable settlement = 2.0 inches 2 m В ХВ 10 m
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
Solve 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).
A 4-m square footing is shown in figure B which supports a load Q = 5000 kN. In this problem H1 = 8 m and H2 = 12 m. Calculate: a. The pressure at the base of the footing b. The pressure at the mid height of the clay layer assuming that the stress beneath the footing is spread at a slope of 1H to 2V c. The pressure at the mid height of the clay layer [below the center of the footing using Bousinessq equation assuming point load
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…
o 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 < II
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