Bearing capcity problems SOlved Provlems 1

Geotechnical

1. Design of Shallow Foundation Dr. Muhammad Rehan Hakro

2. 1. In recent studies, investigators have suggested that, foundations are considered to be shallow if [ Df ≤ (3→4)B], otherwise, the foundation is deep.

3. The contact pressure at which shear failure takes place within the surrounding soil is known as the ultimate bearing capacity ( qu )

4. For actual failures on the field, the soil is often pushed up on only one side of the footing with subsequent tilting of the structure as shown in figure below: GENERAL SHEAR FAILURE

5. PUN C HING SHEAR FAILURE

6. Foundation design is based on the properties of subsurface soil and loading on a foundation. Soil properties are highly related to the soil under drained or undrained conditions. Soil parameters, such as the cohesion and friction angle, Young’s modulus, Poisson’s ratio, compression index, unloading/reloading index, and coefficient of earth pressure at rest, are necessary for foundation design. In principle, these soil parameters should be obtained from related soil tests.

7. However, some parameters can also be estimated by correlation with other soil properties, such as unit weight, density, N value from the standard penetration test (SPT), cone tip resistance from the cone penetration test (CPT), etc. In fact, a foundation should be designed to avoid not only failure, that is, the ultimate limit state, but also excessive settlement, that is, the serviceability limit state. The design method that addresses both limits is often called limit state design.

8. Geotechnical analysis method Drained behavior and undrained behavior S aturated coarse-grained soil under normal. The spring represents the coarse-grained soil, which will deform as it is stressed. The excess pore water pressure is nil and the pore water pressure remains the same as the hydrostatic pressure. The applied pressure directly transmits to the soil, causing the effective stress in the soil to be increased, Such deformation behavior of the soil is called drained behavior.  

9. Drained behavior and undrained behavior Saturated fine-grained soil, that is, clay, under normal pressure. Under these conditions, when a pressure acts on a piston, the pore water flows out very slowly. Therefore, at the initial stage of loading, the pressure is borne entirely by the pore water, and the spring is unstressed. This condition is called undrained behavior or short-term behavior. Pore water pressure is thus the sum of hydrostatic water pressure and excess pore water pressure, while excess pore water pressure is . If there is a shear stress-induced pore water pressure, then the excess pore water pressure will be generated by normal pressure as well as the shear stress.  

10. Materials with drained behavior are called drained material. Conversely, materials with undrained behavior are called undrained material. For geotechnical analysis, the target soil belongs to drained materials or undrained materials based on the soil classification test results of the target soil and the description of the site investigation and perform drained analysis or undrained analysis accordingly.

11. Effective stress analysis and total stress analysis The effective stress analysis method and total stress analysis method are two types of analysis methods in geotechnical engineering. The effective stress analysis method treats the soil as a two-phase material, that is, soil particles and pore water. The concept of effective stress represents the contact stress in soil mechanics. Due to the large void ratio of the soil, when the volume of the soil is changed by force, the pore water will flow out of (or in) the soil unit, so there is no excess pore water pressure. Basically, when soil particles are immersed in pore water, the shear strength or deformation of the soil is related only to the effective stress, not to the total stress.

12. Therefore, the effective parameters, such as effective strength parameters , effective unit weight , effective Poisson's ratio , and effective Young's modulus ( , When fine-grained soil, such as clay, is stressed, due to very small clay voids, the pore water cannot be drained in a short period of time, and excessive pore water pressure is generated. Under the undrained condition, the amount of excess pore water pressure affects the effective stress of the soil, which, in turn, affects the stress‒strain behavior of the soil. Therefore, total stress analysis is also called undrained analysis.  

13. Generally, a foundation is designed for centric loading. However, in many cases, the foundation may also be subjected to eccentric loading, such as slabs of retaining walls and footings under moment loading. The eccentric loading will cause nonuniform contact pressure and reduce the bearing capacity. The third item concerns the foundation settlement, which is essential for the functionality of structures.

14. B earing capacity of a foundation is based on two criteria—the pressure that might cause shear failure of the foundation soil and the maximum allowable pressure such that the settlements produced are not more than the tolerable values.

15. Bearing pressure’, which is used for design of the foundation. The bearing capacity based on the settlement criterion may be determined from the field load tests or plate load tests (dealt with in the next section), standard penetration tests or from the charts prepared by authorities like Terzaghi and Peck, based on extensive investigations.

16. Plate Load Test Perhaps the most direct approach to obtain information on the bearing capacity and the settlement characteristics at a site is to conduct a load test. As tests on prototype foundation are not practicable in view of the large loading required, the time factor involved and the high cost of a full-scale test, a short-term model loading test, called the ‘plate load test’ or ‘plate bearing test’, is usually conducted. This is a semi-direct method since the differences in size between the test and the structure are to be properly accounted for in arriving at meaningful interpretation of the test results.

17. The test essentially consists in loading a rigid plate at the foundation level, increasing the load in arbitrary increments, and determining the settlements corresponding to each load after the settlement has nearly ceased each time a load increment is applied. The nature of the load applied may be gravity loading or dead weights on an improvised platform or reaction loading by using a hydraulic jack. The reaction of the jack load is taken by a cross beam or a steel truss anchored suitably at both ends.

18. From the results of the test, a plot should be made between pressure and settlement, which is usually referred to as the ‘‘load-settlement curve

19. AN OVERVIEW OF BEARING CAPACITY THEORIES The ultimate bearing capacity, or the allowable soil pressure, can be calculated from Bearing capacity theories (Each theory has its own good and bad points) Field Tests. No exact analytical solution for computing bearing capacity of footings is available at present. Prandtl developed an equation based on his study of the penetration of a long hard metal punch into softer materials for computing the ultimate bearing capacity.

20. TERZAGHI’S BEARING CAPACITY THEORY Terzaghi (1943) used the same form of equation(semiempirical) as proposed by Prandtl (1921) for computing the ultimate bearing capacity of strip footings by taking into account The weight of soil and the effect of soil above the base of the foundation on the bearing capacity of soil.

21. Assumptions Terzaghi made the following assumptions for developing an equation for determining qu for C- Ø soil. The soil is semi infinite, homogeneous and isotropic The foundation is shallow(strip) and base of the footing is rough The failure of the soil is similar to general shear failure (i.e. equation is derived for general shear failure) . The load is vertical and symmetrical The ground surface is horizontal The overburden pressure is replaced by an equivalent surcharge load q' = γ D f Coulomb’s law is strictly valid, that is, τ = c + σ tan Ø.

22. Terzaghi’s Bearing Capacity Equations As mentioned previously, the equation was derived for a strip footing and general shear failure, this equation is: q u = cNc + qNq + 0.5B γ N γ ( for continuous or strip footing) Or q u = = CNc + γ D f Nq+ B γ N γ Where q u = Ultimate bearing capacity of the 𝐮𝐧𝐝𝐞𝐫𝐥𝐲𝐢𝐧𝐠 soil (KN/m 2 ) C = Cohesion of 𝐮𝐧𝐝𝐞𝐥𝐲𝐢𝐧𝐠 soil (KN/m 2 ) q = 𝐄𝐟𝐞𝐞𝐜𝐭𝐢𝐯𝐞 stress at the bottom of the foundation (KN/m 2 ) Nc, Nq, N 𝜸 = Bearing capacity factors (nondimensional) and are functions 𝐨𝐧𝐥𝐲 of the 𝐮𝐧𝐝𝐞𝐫𝐥𝐲𝐢𝐧𝐠 soil friction angle, Ø ..  

23. Local

24. Terzaghi’s Bearing Capacity Equations The variations of bearing capacity factors and underlying soil friction angle are given in (Table) for general shear failure

25. Terzaghi’s Bearing Capacity Equations The bearing capacity factors are expressed by the following equations

26. Terzaghi’s Bearing Capacity Equations There are three terms that are added together to obtain the ultimate bearing capacity of the strip footing. q u = CNc + qNq + 0.5B γ N γ qu = CNc + γ D f Nq+ B γ N γ CNc The first term accounts for the cohesive shear strength of the soil located below the strip footing. If the soil below the footing is cohesionless( i.e , C=0), then this term is zero. γ D f Nq This Second term accounts for soil located above the bottom of footing. The value of 𝜸 times D f represent a surcharge pressure that help to increasing bearing capacity of the footing. If the footing was constructed at ground (i.e. D f =0), then this term would equal zero. This second term indicates that the deeper the footing, the greater the ultimate bearing capacity of the footing. In this term represents the total unit weight of the soil located above the bottom of footing. The total unit weight above and below the footing bottom may be different, in which case different values are use in the second third term of equation.  

27. Terzaghi’s Bearing Capacity Equations qu = CNc + γ D f Nq+ B γ N γ 1/2 𝜸B N γ , This third term account for the frictional shear strength of the soil located below the strip footing. the frictional angle is not included in this term, but is accounted for by the bearing capacity factor N γ . Note that 𝜸 represent the total unit weight of the soil located below the footing In order to calculate allowable bearing pressure qa which is use to determine the size of footing, the following equation used: q a = F=factor of safety. For bearing capacity analysis, the commonly use factor of safety is equal to 3.  

28. Effect of Shape of Foundations The Shape of the footing influences the bearing capacity of soil

29. Equations for Square, Circular, and Rectangular Foundations The terzahi and other contributors have suested the correction to the bearin capacity equations for shapes. q u = CNc + qNq + 0.5B γ N γ ( for strip footing) The above equation (for strip footing) was modified to by using the shape factors : Strip foundation q u = CNc + qNq + 0.5B γ N γ Square foundations q u = 1.3CNc + γ D f Nq + 0.4 γ B N γ Circular foundations q u = 1.3CNc + γ D f Nq + 0.3 γ B N γ Rectanular foundations q u = CNc + γ D f Nq + BN γ Where B= width and L= length Note: These two equations are also for general shear failure , and all factors in the two equations (except, B,) are the same as explained for strip footing.  

30. Ultimate bearing capacity for local shear failure .

31. Ultimate bearing capacity for local shear failure . When a soil fails by local shear, the actual shear parameters C and Ø are to be reduced as per terzaghi . Now for local shear failure the above three equations were modified to be useful for local shear failure as following: q u =2/3cNc′ + qNq ′ + 0.5B 𝜸 N 𝜸 ′ (for continuous or strip footing) q u = 0.867CNc′ + qNq ′ + 0.4B γ N γ ′ (for square footing) q u = 0.867CNc′ + qNq ′ + 0.3B γ N γ ′ (for circular footing) Nc′, Nq′ , N γ ′ = Modified bearing capacity factor Where Nc ′,Nq ′ ,N 𝜸 ′ Modified or reduced bearing capacity factors for local shear failure. These factors may be obtained from table. or fig by making use of the frictional angle Ø.

32. General Shear Failure and Local Shear failure (C-Ø Soil) General Shear Failure Strip foundation q u = CNc + qNq + 0.5B γ N γ Square foundation q u = 1.3CNc + γ D f Nq + 0.4 γ B N γ Circular foundation q u = 1.3CNc + γ D f Nq + 0.3 γ B N γ Nc, Nq, N 𝜸 = Bearing capacity factors Local Shear failure Strip foundation q u = 2/3cNc′ + qNq ′ + 0.5B γ N γ ′ Square foundations q u = 0.867CNc′ + qNq ′ + 0.4B γ N γ ′ Circular foundation q u = 0.867CNc′ + qNq ′ + 0.3B γ N γ ′ Nc′,Nq ′ , N γ ′ = Modified bearing capacity factor

33. Ultimate Bearing Capacity qu in Purely Cohesionless and Cohesive Soils Under General Shear Failure Equations for the various types of footings for (C - Ø) soil under general shear failure have been given earlier are . Strip foundation q u = CNc + qNq + 0.5B γ N γ Square foundations q u = 1.3CNc + γ D f Nq + 0.4 γ B N γ Circular foundations q u = 1.3CNc + γ D f Nq + 0.3 γ B N γ Rectanular foundations q u = CNc + γ D f Nq + B N γ Where B= width and L= length The same equations can be modified to give equations for cohesionless soil (for c = 0) and cohesive soils (for Ø = 0) as follows. It may be noted here that for C= 0, the value of Nc =0 for Ø = 0 the value of Nc = 5.7, N γ =0 for a strip footing and Nq=1.   Strip footing For C=0 qu = γ D f Nq + B γ N γ For Ø=0 qu = 5.7c + γ D f Square footing For C=0 qu = γ D f Nq + 0.4 B γ N γ For Ø=0 qu = 7.4c + γ D f For circular footing For C=0 qu = γ D f Nq + 0.3B γ N γ For Ø=0 qu = 7.4c + γ D f For rectangular footing For C=0, qu= γ Df Nq + B γ N γ For Ø=0, qu=5.7c + γ Df Similar types of equations as presented for general shear failure can be developed for local shear failure also.  

34. Ultimate Bearing Capacity qu in Purely Cohesionless and Cohesive Soils General Shear Failure Strip footing For C=0 qu = γ D f Nq + B γ N γ For Ø=0 qu = 5.7c + γ D f Square footing For C=0 qu = γ D f Nq + 0.4 B γ N γ For Ø=0 qu = 7.4c + γ D f Circular footing For C=0 qu = γ D f Nq + 0.3B γ N γ For Ø=0 qu = 7.4c + γ D f Rectangular footing For C=0, qu= γ Df Nq + B γ N γ For Ø=0, qu=5.7c + γ Df Similar types of equations as presented for general shear failure can be developed for local shear failure also.   Local Shear Failure Strip footing For C=0 q u = qNq ′ + 0.5B γ N γ ′ For Ø=0 q u = 3.8cNc′ + qNq ′ Square footing For C=0 q u = qNq ′ + 0.4B γ N γ ′ For Ø=0 q u = 4.9CNc′ + qNq ′ Circular footing For C=0 q u = qNq ′ + 0.3B γ N γ ′ For Ø=0 q u = 4.9CNc′ + qNq ′ Rectangular footing For C=0, qu= γ Df Nq + B γ N γ For Ø=0, qu=5.7c + γ Df Similar types of equations as presented for general shear failure can be developed for local shear failure also.  

35. Bearing Capacity Factors. Table presents bearing capacity factors Nc , Nq and N γ (based on Terzaghi,1943). There are many other charts, graphs, and figures that present bearing capacity factors developed by other engineers and researchers based on varying assumptions(Vesic,1975; Myslivec and kysela,1978). As indicated in table the bearing capacity factors are directly related to the friction angle of the soil . A dense cohesionless soil would tend to have a high friction angle and high bearing capacity factors, resulting in a large ultimate bearing capacity, on the other hand, a loose cohesionless soil would tend to have a lower friction angle and lower ultimate bearing capacity. Thus a major disadvantage of building code values (such as those in table is that they consider only the material type, and not the density condition of the soil which influence the friction angle and bearing capacity factors.

36. Modification of Bearing Capacity Equations for Water Table The theoretical equations developed for computing the ultimate bearing capacity qu of soil are based on the assumption that the water table lies at a depth below the base of the foundation equal to or greater than the width B of the foundation. otherwise the depth of the water table from ground surface is equal to or greater than ( D f +B ). In case the water table lies at any intermediate depth less than the depth ( D f +B ), the bearing capacity equations are affected due to the presence of the water table.

37. Modification of Bearing Capacity Equations for Water Table Terzaghi equations give the ultimate bearing capacity based on the assumption that the water table is located well below the foundation. However, if the water table is close to the foundation, the bearing capacity will decreases due to the effect of water table, The values which will be modified are: 1. (q for soil above the foundation) in the second term of equations. 2. ( γ for the underlying soil) in the third (last) term of equations . There are three cases according to location of water table: Case I. The water table is located so that 0 ≤ D1 ≤ Df as shown in the following figure: The factor ,q, (second term) in the bearing capacity equations will takes the following form: (For the soil above the foundation) q = effective stress at the level of the bottom of the foundation q = D 1 × γ + D 2 × ( γ sat − γ w ) The factor , γ , (third term) in the bearing capacity equations will takes the following form: (For the soil under the foundation) γ = effective unit weight for soil below the foundation γ ′ = γ sat − γ w

38. Case II. The water table is located as shown in the figure: The factor ,q, (second term) will take the following form: (For the soil above the foundation) q = effective stress at the level of the bottom of the foundation q = Df × γ The factor , γ , (third term) in the bearing capacity equations will takes the following form: (For the soil under the foundation) ’ =dx γ +(B-d)x( γ sat - γ w )  

39. EFFECT OF WATER TABLE ON BEARING CAPACITY Case III. The water table is located so that d ≥ B, in this case the water table is assumed have no effect on the ultimate bearing capacity.

40. EFFECT OF WATER TABLE ON BEARING CAPACITY Two cases may be considered here Case 1: When the water table lies above the base of the foundation. Case 2: When the water table lies within ( D f +B ) depth below the base of the foundation. We will consider the two methods for determining the effect of the water table on bearing capacity as given below. Method 1 For any position of the water table within the depth ( Df + B) we may write Eq. (5.6) as qu = CN c + 𝜸 D f NqRw1 + B 𝜸 N γ Rw2 Rw1 = reduction factor for water table above the base level of the foundation Rw2 = reduction factor for water table below the base level of the foundation γ = γ sat for all practical purposes in both the second and third terms of Eq.  

41. EFFECT OF WATER TABLE ON BEARING CAPACITY

42. EFFECT OF WATER TABLE ON BEARING CAPACITY Case 1: When the water table lies above the base level of the foundation or when 1the equation for R w1 may be written as R w1 = (1+ For =0 we have R w1 =0.5 and for =1.0, we have R w1 =1.0   qu = CN c + 𝜸 D f NqRw1 + B 𝜸 N γ Rw2  

43. EFFECT OF WATER TABLE ON BEARING CAPACITY Case 2: When the water table lies below the base level 0r when <=1 the equation for R w2 is R w2 = (1+ For =0, we have R w2 =0.5and for =1.0, we have R w2 =1.0   qu = CN c + 𝜸 D f NqRw1 + B 𝜸 N γ Rw2  

44. EFFECT OF WATER TABLE ON BEARING CAPACITY qu = CN c + 𝜸 D f NqRw1 + B 𝜸 N γ Rw2  

45. Problem: A strip footing of width 3m is founded at a depth of 2m below the ground surface in a (C- Ø) soil having a cohesion c = 30 KN/m 2 and angle of shearing resistance Ø = 35°. The water table is at a depth of 5 m below ground level. The moist weight of soil above the water table is 17.25 KN/m 3 . Determine The ultimate bearing capacity of the soil The net bearing capacity The net allowable bearing pressure load/m for a factor of safety of 3. Use the general shear failure theory of Terzaghi. Solution q u = cNc + γ D f Nq+ B γ N γ For Ø= 35° Nc= 57.75 Nq= 41.44 N γ = 45.41 = 30x57.75+ 17.25x2x41.44+ q nu = q u - γ D f =4259-17.25x2=4225kN/m 2 q na = = = 1408kN/m 2 Qa = q nu B =1408x3=4225kN/m ∵ q u = = cNc + γ D f Nq+ B γ N γ  

46. If the soil in Ex. 5.1 fails by local shear failure, determine the net safe bearing pressure. All the other data given in Ex. 5.1 remain the same. Solution For local shear failure: tan Ø’= 0.67tan Ø’=tan -1 ( 0.67tan35 =25 C’=0.67C=0.67x30=20kN/m 2 From Table (general shear failure) for Ø’= 25 N’ c =25.13 N’ q =12.72 N γ ’ =8.34 Now form Eq. q u = CNc + γ Nq+ B γ N γ q u =20x25.13+17.25x2x12.72+ 17.25x3x8.34=1191KN/m 2 q nu =1191-17.25x2=1156.5KN/m 2 q na = =385.5KN/m 2 Qa =385.5x3=1156.5KN/m  

47. If the water table in previous Example rises to the ground level, determine the net safe bearing pressure of the footing. All the other data given in Ex. 5.1 remain the same. Assume the saturated unit weight of the soil γ sat = 18.5 kN /m 3 . Solution When the WT is at ground level we have to use the submerged unit weight of the soil There fore γ b = 𝜸 sat - γ w = 18.5-9.81=8.69kN/m 3 The net ultimate bearing capacity is q nu = ( qu - γ D f ) = cNc + γ D f (Nq-1)+ B γ N γ q nu =30x57.75+8.69x2(41.44-1)+ x8.69x3x45.41=2992KN/m 2 q na = =997.33 KN/m 2 Qa =997.33kN/m  

48. If the water table in Ex, 5.1 occupies any of the positions: (a) 1.25 m below ground level (b) 1.25 m below the base level , what will be the net safe bearing pressure? Assume 𝜸 sat =18.5 KN/m 3 , 𝜸 (above WT) = 17.5 KN/m 3 . All the other data remain the same as given in Ex. 5.1. Solution (Method 1) By making use of reduction factor Rw1=Rw2 and using Equ q nu = CNc + γ Df (Nq-1)Rw1 + 1/2B γ N γ Rw2 given: for Ø= 35° Nq=41.44, N γ =45.41 and Nc=57.75 Case 1: When the WT is 1.25 m below the gL From Eq.(5.24), we et R w1 =0.813 for =0.625, R w2 =0.5 for =0 R w1 = (1+ R w2 = (1+ By Substituting the known value in the equation for q nu , we have qu = 30x57.75+18.5x2x40.4x0.813+ x18.5x3x45.41x0.5=3538KN/m 2 q na = = 1179KN/m 2 Case 2: When the WT is 1.25 m below the base R w1 =1.0 for =1, R w2 =0.71 for =0.42 R w1 = (1+ R w2 = (1+ Now the net bearing capacity is q u = 30x57.75+18.5x2x41.44x1+ 18.5x3x45.41x0.71=4064KN/m 2 q na = = 1355KN/m 2  

49. A square footing fails by general shear in a cohesionless soil under an ultimate load of Qult =1687.5 kips. The footing is placed at a depth of 6.5 ft below ground level. Given: Ø= 35°, and γ = 110 Ib /ft 3 , determine the size of the footing if the water table is at a great depth (Fig. Ex. 5.5). Solution For a square footing. Eq. for C=0, we have q u = γ Nq+0.4 γ BN γ For ø=35 q u = = By substituting known values, we have =110x6.5x41.4+0.4x110x42.4B = (29.601+1.866B)10 3 B3+ 15.863B2-904.34=0 Solving this equation yields B=6.4  

50. A rectangular footing of size 10 x 20 ft is founded at a depth of 6 ft below the ground surface in a homogeneous cohesionless soil having an angle of shearing resistance Ø = 35°. The water table is at a great depth.' The unit weight of soil γ = 114 lb/ ft 3 . Determine: (1) Net ultimate bearing capacity, (2) Net allowable bearing pressure for Fs = 3 (3) Allowable load Qa the footing can carry. Use Terzaghi’s theory (Refer to Fig. Ex. 5.6). q nu = γ (Nq-1)+ B γ N γ (1-0.2 ) For ø=35 By substituting the known values. q nu = 114x6(41.44-1)+ (1-0.2x )=49385lb/ft 2 q na = =16462lb/ft 2 Qa =(B x L) q na =10x20x16462=3292x10 3 lb  

51. If the soil in Ex. 5.6 is cohesionless (c = 0), and fails in local shear, determine: (I) The ultimate bearing capacity (II) the net bearing capacity (III) the net allowable bearing pressure. All the other data remain the same. Solution From Eqs , (5.15)and(5.20), the net bearing capacity for local shear failure for C=0 is q nu =( q u - γ )= γ (Nq’-1)+ B γ N γ ’(1-0.2 ) Where Ø’=tan-1 (0.67) tan35 =25 By Substituting the known value , we have q nu =114x6(12.72-1)+ x 10x8.34(1-0.2 )=12979lb/ft 2 q na =  

52. Transition from Local to General Shear Failure in Sand As already explained, local shear failure normally occurs in loose and general shear failure occurs in dense sand. There is a transition from local to general shear failure as the state of sand changes from loose to dense condition. There is no bearing capacity equation to account for this transition from loose to dense state. Peck et al, (1974) have given curves for N𝜸 and Nq which automatically incorporate allowance for the mixed state of local and general shear failures as shown in Fig. Terzaghi’s bearing capacity factors which take care of mixed state of local and general shear failures in sand (Peck etal , 1974).

53. The curves for Nq and N𝜸 are developed on the following assumptions. 1. Purely local shear failure occurs when Ø ≤ 28°. 2. Purely general shear failure occurs when Ø ≥ 38°. 3. Smooth transition curves for values of Ø between 28° and 38° represent the mixed state of local and general shear failures. Figure 5.7 also gives the relationship between SPT value Ncor and the angle of internal friction Ø by means of a curve. This curve is useful to obtain the value of Ø when the SPT value is known.

54. SKEMPTON’S BEARING CAPACITY FACTOR Nc

55. SKEMPTON’S BEARING CAPACITY FACTOR Nc Skempton (1951) has showed that the bearing capacity factors in Terzaghi's equation tends to increase with depth for a cohesive soil. For saturated clay soils, Skempton (1951), proposed the following equation for a strip foundation q u = C N c + γ D f q nu = qu - γ D f = cNc q an = = The Nc values for strip and square (or circular) foundations as a function of the Df /B ratio are given in Fig. The equation for rectangular foundation may be written as follows (N C ) R =(0.84+0.16x )(N C ) S Here (N C ) R = N C for rectangular foundation, (N C ) s= Nc for square foundation.  

56. A rectangular footing of size 10 x 20 ft is founded at a depth of 6 ft below the ground level in a cohesive soil (Ø= 0) which fails by general shear. Given: γ sat =114 lb/ft 3 , c=945 lb/ft 2 . The water table is close to the ground surface. Determine qu , qnu and qna by (a) Terzaghi’s method (b) Skempton’s method (use F.s = 3). (a) Terzaghi’s method For ø =0 q u = CN c (1+0.3x )+ 𝜸 b q u =945x5.7 (1+0.3x )+(114-62.4)x6=6504lb/ft 2 q nu =(qn- γ b )=6504-(114-62.4)x6=6195lb/ft 2 q na = = =2065lb/ft 2 Skempton’s methad From Eqs (5.22a)and(5.22b), we may write q n = CN cr + γ Where Ncr = bearing capacity factor for a rectangular foundation. Ncr =(0.84+0.16x ) xNcs Where Ncs = bearing capacity factor for a square foundation. From fig,Ncs =7.2 for =0.60. Ncr =(0.84+0.16x )x7.2=6.62 Now q u = 945x6.62+114x6=6940lb/ft 2 q nu =(qn- γ )=6940-114x6=6256lb/ft 2 q na = = =2085lb/ft 2 Note : Tirzahg's and Skempton’s values are in closed agreement for cohesive soils  

57. EFFECT OF SOIL COMPRESSIBILITY The change of failure mode is due to soil compressibility. Vesic (1973) proposed the following modification to the general bearing capacity equation: q u = c’ N c F cs F cd F cc + q N q F qs F qd F qc + 0.5 B  N  F  s F  d F  c where F cc , F qc , and F  c are soil compressibility factors. Steps for calculating the soil compressibility factors. Page 232

58. EFFECT OF SOIL COMPRESSIBILITY Table 6.8 Figure 6.17

59. EXAMPLE 6.6 Example 6.6

60. Pressure Distribution at Foundation Base • The foundation design in practice requires that the foundation dimensions are calculated for a given design column load, hence the bearing capacity calculation is used to check that the applied pressure at the base of the foundation can be resisted by the ground, i.e.

61. Pressure Distribution at Foundation Base • As with the base of a retaining wall or structures under lateral loading, foundations are subjected to moments in addition to vertical loading. In this situation, the distribution of bearing pressure beneath the foundation is no longer uniform..

62. The pressure distribution at the base of an elastic foundation is assumed to be;– Uniform, if the applied load is vertical and applied at the center of the foundation. – Irregular, if the applied load is eccentric, inclined and/or a moment is applied. • The magnitude of the maximum and minimum pressure applied at the foundation base is dependent on the eccentricity caused by inclined loads and/or moment. Pressure Distribution at Foundation Base

63. Eccentrically loaded foundations The distribution bearing pressure along the footing base varies with different eccentricities e. When e < B/6, the footing is fully in contact with the soil, and the shape of the bearing pressure distribution is trapezoidal. The bearing pressure at one edge is larger than that at the other edge. However, when e = B/6, the shape of the bearing pressure distribution is triangular, with 0 bearing pressure at one edge. At this time, the footing begins to lift off. When e > B/6, the shape of the bearing pressure distribution is also triangular; however, because the soil cannot sustain tension, the footing lifts off without full contact with the soil below.

64. To avoid a large amount of eccentricity from affecting the serviceability and stability of the footing, a common design requires that the maximum eccentricity should not be larger than 1/6 the width of the foundation slab under long-term load conditions and should not be larger than 1/3 the width of the foundation slab under short-term load conditions.

65. Ultimate Bearing Capacity under Eccentric Loading Two-Way Eccentricity Figure 6.25 Figure 6.25

66. Ultimate Bearing Capacity under Eccentric Loading Two-Way Eccentricity Figure 6.26 Figure 6.26

67. Ultimate Bearing Capacity under Eccentric Loading Two-Way Eccentricity Figure 6.27 Figure 6.27

68. Ultimate Bearing Capacity under Eccentric Loading Two-Way Eccentricity Figure 6.28 Figure 6.28 Figure 6.28

69. Ultimate Bearing Capacity under Eccentric Loading Two-Way Eccentricity Figure 6.29 Figure 6.29

70. Ultimate Bearing Capacity under Eccentric Loading Two-Way Eccentricity Figure 6.30 Figure 6.30 6.10 Table 6.10