CHAPTER 3 Prepared by: Dr. Farouk Majeed Muhauwiss Civil Engineering Department College of Engineering Tikrit University EARING CAPACITY OF SHALLOW FOUNDATIONS 3. MODES OF FAILURE Failure is defined as mobilizing the full value of soil shear strength accompanied with excessive settlements. For shallow foundations it depends on soil type, particularly its compressibility, and type of loading. Modes of failure in soil at ultimate load are of three types; these are (see Fig..5): Mode of Failure Characteristics Typical Soils. General Shear failure Well defined continuous slip surface up to ground level, Heaving occurs on both sides with final collapse and tilting on one side, Failure is sudden and catastrophic, Ultimate value is peak value. Low compressibility soils Very dense sands, Saturated clays (NC and OC), Undrained shear (fast loading).. local Shear failure (Transition) Well defined slip surfaces only below the foundation, discontinuous either side, Large vertical displacements reuired before slip surfaces appear at ground level, Some heaving occurs on both sides with no tilting and no catastrophic failure, No peak value, ultimate value not defined. Moderate compressibility soils Medium dense sands, 3. Punching Shear failure Well defined slip surfaces only below the foundation, non either side, Large vertical displacements produced by soil compressibility, No heaving, no tilting or catastrophic failure, no ultimate value. High compressibility soils Very loose sands, Partially saturated clays, NC clay in drained shear (very slow loading), Peats. Fig. (3.): Modes of failure.
3. EARING CAPACITY CLASSIFICATION (According to column loads) Gross earing Capacity ( gross ): It is the total unit pressure at the base of footing which the soil can take up. P G.S. D f. D f D o t gross total pressure at the base of footing P footing / area.of. footing. where P footing p.( column.load ) + own wt. of footing + own wt. of earth fill over the footing. gross (P + s.d o..l + c.t..l)/. L gross P + s.d o + c.t....(3.).l Ultimate earing Capacity ( ult. ): It is the maximum unit pressure or the maximum gross pressure that a soil can stand without shear failure. Allowable earing Capacity ( all. ): It is the ultimate bearing capacity divided by a reasonable factor of safety. all. ult. F.S.........(3.) Net Ultimate earing Capacity: It is the ultimate bearing capacity minus the vertical pressure that is produced on horizontal plain at level of the base of the foundation by an adjacent surcharge. ult. net ult. Df.........(3.3) 57
Net Allowable earing Capacity ( ): It is the net safe bearing capacity or the ultimate bearing capacity divided by a reasonable factor of safety. Approximate: Exact: all. net all. ult. net F.S ult. D F.S f all. net..........(3.4) ult. net D f..........(3.5) F.S 3.3 FACTOR OF SAFETY IN DESIGN OF FOUNDATION The general values of safety factor used in design of footings are.5 to 3.0, however, the choice of factor of safety (F.S.) depends on many factors such as:. the variation of shear strength of soil,. magnitude of damages, 3. reliability of soil data such as uncertainties in predicting the ult., 4. changes in soil properties due to construction operations, 5. relative cost of increasing or decreasing F.S., and 6. the importance of the structure, differential settlements and soil strata underneath the structure. 3.4 EARING CAPACITY REQUIREMENTS Three reuirements must be satisfied in determining bearing capacity of soil. These are: () Adeuate depth; the foundation must be deep enough with respect to environmental effects; such as: frost penetration, seasonal volume changes in the soil, to exclude the possibility of erosion and undermining of the supporting soil by water and wind currents, and to minimize the possibility of damage by construction operations, 58
() Tolerable settlements, the bearing capacity must be low enough to ensure that both total and differential settlements of all foundations under the planned structure are within the allowable values, (3) Safety against failure, this failure is of two kinds: the structural failure of the foundation; which may be occur if the foundation itself is not properly designed to sustain the imposed stresses, and the bearing capacity failure of the supporting soils. 3.5 FACTORS AFFECTING EARING CAPACITY type of soil (cohesive or cohesionless). physical features of the foundation; such as size, depth, shape, type, and rigidity. amount of total and differential settlement that the structure can stand. physical properties of soil; such as density and shear strength parameters. water table condition. original stresses. 3.6 METHODS OF DETERMINING EARING CAPACITY (a) earing Capacity Tables The bearing capacity values can be found from certain tables presented in building codes, soil mechanics and foundation books; such as that shown in Table (3.). They are based on experience and can be only used for preliminary design of light and small buildings as a helpful indication; however, they should be followed by the essential laboratory and field soil tests. Table (3.) neglects the effect of: (i) underlying strata, (ii) size, shape and depth of footings, (iii) type of the structures supported by the footings, (iv) there is no specification of the physical properties of the soil in uestion, and (v) assumes that the ground water table level is at foundation level or with depth less than width of footing. Therefore, if water table rises above the foundation level, the hydrostatic water pressure force which affects the base of foundation should be taken into consideration. 59
Table (3.): earing capacity values according to building codes. Soil type Description earing pressure (kg/cm ) Notes Rocks. bed rocks.. sedimentary layer rock (hard shale, sand stone, siltstone). 3. shest or erdwas. 4. soft rocks. 70 30 0 3 Unless they are affected by water. Cohesionless soil. well compacted sand or sand mixed with gravel.. sand, loose and well graded or loose mixed sand and gravel. 3. compacted sand, well graded. 4. well graded loose sand. Dry 3.5-5.0.5-3.0.5-.0 0.5-.5 submerged.75-.5 0.5-.5 0.5-.5 0.5-0.5 Footing width.0 ms. Cohesive soil. very stiff clay. stiff clay 3. medium-stiff clay 4. low stiff clay 5. soft clay 6. very soft clay 7. silt soil -4-0.5-0.5-0.5 up to 0. 0.-0..0-.5 It is subjected to settlement due to consolidation (b) Field Load Test This test is fully explained in (chapter ). (c) earing Capacity Euations Several bearing capacity euations were developed for the case of general shear failure by many researchers as presented in Table (3.); see Tables (3.3, 3.4 and 3.5) for related factors. 60
Table (3.): earing capacity euations by the several authors indicated. Terzaghi (see Table 3.3 for typical values for K P values) φ π [ 0. 75π. ( )].tan φ 80 ult. cnc.sc + N + 0.5...N. S e tan φ k P N ; N c (N ). cot φ ; N ( ) cos ( 45 + φ/ ) cos φ ( φ + 33) where a close approximation of k P 3.tan 45 +. Strip circular suare rectangular S c.0.3.3 (+ 0.3 / L) S.0 0.6 0.8 (- 0. / L) Meyerhof (see Table 3.4 for shape, depth, and inclination factors) Vertical load: Inclined load: ult. c.nc.sc.dc +.N.S.d + 0.5...N.S. d ult. c.nc.d c.ic +.N.d.i + 0.5...N.d. i N π.tan φ e tan ( 45 + φ / ) ; Nc (N ). cot φ ; N (N ).tan(. 4φ) Hansen (see Table 3.5 for shape, depth, and inclination factors) N For.. φ > 0 : ult. cncscdcicgcbc + NSdigb + 0.5...NSdigb For.. φ 0 : 5. 4S ( + S + d i b g ) π.tan φ e tan ( 45 + φ / ult. u c c c c c + ); Nc (N ). cot φ ; N. 5(N ). tan φ Vesic (see Table 3.5 for shape, depth, and inclination factors) Use Hansen's euations above N π.tan φ e tan ( 45 + φ / ) ; Nc (N ). cot φ ; N (N + ). tan φ All the bearing capacity euations above are based on general shear failure in soil. 6
Note: Due to scale effects, N and then the ultimate bearing capacity decreases with increase in size of foundation. Therefore, owle's (996) suggested that for ( > m), with any bearing capacity euation of Table (3.), the term ( 0.5.N Sd ) must be multiplied by a reduction factor: r 0. 5log ; i.e., 0.5.N Sd r (m).5 3 3.5 4 5 0 0 00 r 0.97 0.95 0.93 0.9 0.90 0.8 0.75 0.57 +.5 π + Table (3.3): earing capacity factors for Terzaghi's euation. φ,..deg N c N N K P 0 5.7 +.0 0.0 0.8 5 7.3.6 0.5. 0 9.6.7. 4.7 5.9 4.4.5 8.6 0 7.7 7.4 5.0 5.0 5 5..7 9.7 35.0 30 37..5 9.7 5.0 34 5.6 36.5 36.0 35 57.8 4.4 4.4 8.0 40 95.7 8.3 00.4 4.0 45 7.3 73.3 97.5 98.0 48 58.3 87.9 780. 50 347.5 45. 53. 800.0 Table (3.4): Shape, depth and inclination factors for Meyerhof's euation. For Shape Factors Depth Factors Inclination Factors D Any φ Sc + 0..K P d. K f c + 0 α P i L c i 90 D φ 0 S S + 0..KP d d. K f α + 0 P L i φ φ 0 S S. 0 d d. 0 i 0 Where: K P tan ( 45 + φ/ ) R α angle of resultant measured from vertical without a sign. α, L, D f width, length, and depth of footing. Note:- When φ triaxial is used for plan strain, adjust φ as: φps (. 0. ) φtriaxial L 6
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3.7 WHICH EQUATIONS TO USE? Of the bearing capacity euations previously discussed, the most widely used euations are Meyerhof's and Hansen's. While Vesic's euation has not been much used (but is the suggested method in the American Petroleum Institute, RPA Manual, 984). Use Terzaghi Meyerhof, Hansen, Vesic Table (3.6) : Which euations to use. est for Very cohesive soils where D/ or for a uick estimate of ult. to compare with other methods, Somewhat simpler than Meyerhof's, Hansen's or Vesic's euations; which need to compute the shape, depth, inclination, base and ground factors, Suitable for a concentrically loaded horizontal footing, Not applicable for columns with moment or tilted forces, More conservative than other methods. Any situation which applies depending on user preference with a particular method. Hansen, Vesic When base is tilted; when footing is on a slope or when D/ >. 3.8 EFFECT OF SOIL COMPRESSIILITY (local shear failure). For clays sheared in drained conditions, Terzaghi (943) suggested that the shear strength parameters c and φ should be reduced as: * φ c * 0.67c and φ tan ( 0.67 tan )......(3.6). For loose and medium dense sands (when D r 0. 67 ), Vesic (975) proposed: where * φ tan ( 0.67 + Dr 0.75Dr )tanφ.......(3.7) D is the relative density of the sand, recorded as a fraction. r Note: For dense sands ( D r > 0. 67 ) the strength parameters need not be reduced, since the general shear mode of failure is likely to apply. 64
EARING CAPACITY EXAMPLES () Example (): Determine the allowable bearing capacity of a strip footing shown below using Terzaghi and Hansen Euations if c 0, φ 30, D f.0m,.0m, soil 9 kn/m 3, the water table is at ground surface, and SF3. Solution: (a) y Terzaghi's euation: ult. cn c.s c + N +...N. S Shape factors: from table (3.), for strip footing S c S. 0 earing capacity factors: from table (3.3), for φ 30, N.5,..N 9. 7 0 +.0 (9-9.8).5 + 0.5x(9-9.8)9.7x.0 97 kn/m ult. all. 97/3 99 kn/m (b) y Hansen's euation: for.. φ > 0 : ult. cn cscdcic gcbc + N Sd i g b + 0.5..N Since c 0, any factors with subscript c do not need computing. Also, all S d i g b g i..and.. b factors are.0; with these factors identified the Hansen's euation simplifies to: ult. NSd + 0.5..N S d From table (3.5): for... φ 34..use.. φ ps φtr,. use. φ.5 7 for L/ >..use.. φ ps.5φtr 7 ps φtr use. φ.5φ 7,.5 x 30-7 8,. ps tr i 65
earing capacity factors: from table (3.4), for φ 8, N 4.7,..N 0. 9 Shape factors: from table (3.5), S S.0, Depth factors: from table (3.5), d + tanφ( sinφ ) D d +.tan 8( sin 8 ).9,, and d. 0.0 (9-9.8)4.7x.9 + 0.5x(9-9.8)0.9x.0 4.355 kn/m ult. all. 4.355/3 74.785 kn/m Example (): A footing load test produced the following data: D f 0.5m, 0.5m, L.0m, soil 9. 3 kn/m3, φ tr 4. 5, c 0, P ult. ( measured ) 863.kN, ult. ( measured ) 863 / 0.5x 863 kn/m. Reuired: compute ult. by Hansen's and Meyerhof's euations and compare computed with measured values. Solution: (a) y Hansen's euation: Since c 0, and all g i..and.. bi factors are.0; the Hansen's euation simplifies to: ult. NSd + 0.5..N S d From table (3.5): L/ /0.5 4 >..use.. φ ps.5φtr 7,.5 x 4.5-7 46.75 take... φ 47 earing capacity factors: from table (3.) N π.tanφ e..tan ( 45 + φ / ), N.5( N ) tanφ for φ 47 : N 87., N 99. 5 66
Shape factors: from table (3.5), 0.5 0.5 S + tanφ + tan 47.7, S 0.4 0.4 0. 9 L.0 L.0 Depth factors: from table (3.5), d D 0.5 + tanφ( sinφ ), d + tan 47( sin 47 ). 55, d. 0 0.5 0.5 (9.3)87.x.7x.55 + 0.5x0.5(9.3)99.5x0.9x.0 905.6 kn/m ult. versus 863 kn/m measured. (b) y Meyerhof's euation: From table (3.) for vertical load with c 0: From table (3.4): φ (. 0. φ L earing capacity factors: from table (3.) N ult. NSd + 0.5..N S d ps ) tr, (. - 0.. 0 0. 5 )4.5 45.7, take... φ 46 π.tanφ e..tan ( 45 + φ / ), N ( N )tan(.4φ ) for φ 46 : N 58. 5, N 38. 7 Shape factors: from table (3.4) 0.5 K p tan ( 45 + φ / ) 6.3, S S + 0..K p + 0.( 6.3 ). 5 L.0 Depth factors: from table (3.4) D 0.5 K p.47, d d + 0.. K p + 0.(.47 ). 5 0.5 ult. 0.5(9.3)58.5x.5x.5 + 0.5x0.5(9.3)38.7x.5x.5 60.4 kn/m versus 863 kn/m measured oth Hansen's and Meyerhof's es. give over-estimated ult. compared with measured. 67
Example (3): A.0x.0m footing has the geometry and load as shown below. Is the footing Solution: adeuate with a SF3.0?. We can use either Hansen's, or Meyerhof's or Vesic's euations. An arbitrary choice is Hansen's method. P 600 kn H 00 kn m Check sliding stability: H P η 0 D 0.3m 7.5 kn/m 3 φ 5 ; c 5 kn/m use δ φ; C a c and A x f 4m H. A C + V tanδ 4x5 + 600 tan 5 80 > 00 kn (O.K. for sliding) max f a earing capacity y Hansen's euation: with..inclination.. factors..all..s i.0 ult. cn c.d c.i c.b c + N.d.i.b + earing capacity factors from table (3.): 0.5..N.d.i. b N c π.tanφ ( N ).cot φ, N e..tan ( 45 + φ / ), N.5( N ) tanφ for φ 5 : N c 0. 7, N 0. 7, N 6. 8 Depth factors from table (3.5): for D 0.3m, and m, D/ 0.3/0.5 <.0 (shallow footing) D D d c + 0.4 + 0.4(0.5 ).06, d + tanφ ( sinφ ) + 0.3( 0.5). 05, d.0 68
Inclination factors from table (3.5): 0.5H 5 0.5x00 5 i ( ) ( ) V + A.c.cot φ 600 + 4x5x cot 5 i f ( i ) 0.5 i 0.5 0.47, ( N ) 0.7 c 0.5, (0.7 η / 450 )H 5 (0.7 0 / 450 )00 5 for.. η > 0 : i ( ) ( ) 0. 40 V + A.c.cot φ 600 + 4x5x cot 5 f The base factors for.. η 0 (0.75..radians ) from table (3.5): η 0 b c 0.93, 47 47 ( η tanφ ) ( ( 0.75 )tan 5 ) b e e (.7η tanφ ) (.7( 0.75 )tan 5 ) 0.85, b e e 0. 80 ult. 5(0.7)(.06)(0.47)(0.93) + 0.3(7.5)(0.7)(.05)(0.5)(0.85) all. all. 304 / 3 0.3 kn/m all. f + 0.5(7.5)(.0)(6.8)()(0.40)(0.80) 304 kn/m P. A 0.3(4) 405. kn < 600 kn (the given load), m is not adeuate and, therefore it must be increased and P all. recomputed and checked. 3.9 FOOTINGS WITH INCLINED OR ECCENTRIC LOADS INCLINED LOAD: If a footing is subjected to an inclined load (see Fig.3.7), the inclined load Q can be resolved into vertical and horizontal components. The vertical component Qv can then be used for bearing capacity analysis in the same manner as described previously (Table 3.). After the bearing capacity has been computed by the normal procedure, it must be corrected by an R i factor using Fig.(3.7) as:.x. R.......(3.8) ult.( inclined..load ) ult.( vertical..load ) i 69
(b) b) Inclinedd foundatioon (a) horiizontal fou undation Figure (3.7): Incllined load reduction factors. I Importan nt Notes: Reemember that t in this case, Meyyerhof's beearing capaacity euattion for incclined loadd (fr from Table 3.) can be b used dirrectly: ult.( inclined...(3.9)) i..load ) cn c d c i c + N d i + 0.5..N d i. Thhe footings stability with regarrd to the innclined loaad's horizoontal compponent alsoo m be checked by caalculating the must t factor of safety against a slidding as folllows: Fs( slididding ) H max m.........(3.0)) H w where: H the inclineed load's horizontal h c component t, H maxx. the. ma ax imum.ressisting. force A f.c a + σ tan δ. for ( c φ ) soills; or H maxx. A f.c a. for thhe undrainned case inn clay ( φ u 0 ); or 70
H max. σ tanδ. for a sand and the drained case in clay ( c 0 ). A f effective..area.l C adhesion α. a C u where... α.0... for.soft.to.medium.clays.; and. α 0.5...for.stiff.clays. σ the net vertical effective load v f f Q v D f. ; or σ ( Q D. ) u. A (if the water table lies above foundation level) δ the skin friction angle, which can be taken as eual to (φ ),and u the pore water pressure at foundation level. ECCENTRIC LOAD: Eccentric load result from loads applied somewhere other than the footing's centroid or from applied moments, such as those resulting at the base of a tall column from wind loads or earthuakes on the structure. To provide adeuate SF ( against.lifting ) of the footing edge, it is recommended that the eccentricity ( e / 6 ). Footings with eccentric loads may be analyzed for bearing capacity by two methods: () the concept of useful width and () application of reduction factors. () Concept of Useful Width: In this method, only that part of the footing that is symmetrical with regard to the load is used to determine bearing capacity by the usual method, with the remainder of the footing being ignored. First, computes eccentricity and adjusted dimensions: M y M x ex ; L L ex ; e y V V ; e y ; A f A. L 7
Second, calculates ult. from Meyerhof's, or Hansen's, or Vesic's euations (Table 3.) using in the (.. N ) term and or/and L in computing the shape factors and not in computing depth factors. () Application of Reduction Factors: First, computes bearing capacity by the normal procedure (using euations of Table 3.), assuming that the load is applied at the centroid of the footing. Then, the computed value is corrected for eccentricity by a reduction factor ( R e ) obtained from Figure (3.8) or from Meyerhof's reduction euations as: Re Re - (e/).....for..cohesive..soil /..(3.) - (e/)...for.cohesionless.soil ult.( eccentric ) ult.( concentric ). x.r e.......(3.) Figure( (3.8): Eccentric load reduction factors. 7
EARING CAPACITY EXAMPLES () Footings with inclined or eccentric loads Example (4): A suare footing of.5x.5m is subjected to an inclined load as shown in figure below. What is the factor of safety against bearing capacity (use Terzaghi's euation). G.S. α 30 80 kn Solution: y Terzaghi's euation: ult. cn c.s c + N +...N. S Shape factors: from table (3.) for suare footing S c.3; S 0. 8, c u / 80 kpa earing capacity factors: from table (3.3) for φ u 0 : N c 5.7,..N.0,..N 0 ult.( vertical.load ) 80(5.7)(.3)+0(.5)(.0) + 0.5(.5)(0)(0)(0.8) 6.8 kn/m From Fig.(3.7) with α 30 and cohesive soil, the reduction factor for inclined load is 0.4. 6.8(0.4) 6.576 kn/m ult.( inclined.load ) Q v Q.cos 30 80 (0.866) 55.88 kn Qult. 6.576(.5 )(.5 ) Factor of safety (against bearing capacity failure) 3. 77 Qv 55.88 Check for sliding: Q h Q.sin 30 80 (0.5) 90 kn H A.C + σ tanδ (.5)(.5)(80) + (80)(cos30)(tan0)80 kn max. f a D f.5m 0 kn/m 3 4 m.5m W.T. H max. 80 Factor of safety (against sliding). 0 (O.K.) Qh 90 u 60 kpa 73
Example (5): A.5x.5m suare footing is subjected to eccentric load as shown below. What is the safety factor against bearing capacity failure (use Terzaghi's euation): (a) y the concept of useful width, and (b) Using Meyerhof's reduction factors. P 330 kn G.s..m Centerline of footing 0 kn/m 3 u 90 kn/m e x 0.8.5m.5m e x 0.8.5m Solution: () Using concept of useful width: from Terzaghi's euation: ult. cnc.s c + N +...N. S Shape factors: from table (3.) for suare footing S c.3; S 0. 8, c / 95 kpa earing capacity factors: from table (3.3) for φ u 0 : N c 5.7,N.0,N 0 The useful width is: ex.5 (0.8 ).4m.5-(0.8).4m.5m ult. 95(5.7)(.3)+0(.)(.0) + 0.5(.4)(0)(0)(0.8) 77.95 kn/m Qult. 77.95(.4 )(.5 ) Factor of safety (against bearing capacity failure) 3. 77 Qv 330 u 74
() Using Meyerhof's reduction factors: In this case, ult. is computed based on the actual width:.5m from Terzaghi's euation: ult..3cnc + N + 0.4.. N ult.( concentric.load ).3(95)(5.7) +0(.)(.0) + 0.4(.5)(0)(0) 77.95 kn/m For eccentric load from figure (3.8): e 0.8 with Eccentricity ratio x 0. ; and cohesive soil R e 0.76.5 ult.( eccentric.load ) 77.95 (0.76) 553.4 kn/m Qult. 553.4(.5 )(.5 ) Factor of safety (against bearing capacity failure) 3. 77 Qv 330 Example (6): A suare footing of.8x.8m is loaded with axial load of 780 kn and subjected to M x 67 kn-m and M y 60. kn-m moments. Undrained triaxial tests of unsaturated soil samples give φ 36 and c 9. 4 kn/m. If D f.8m, the water table is at 6m below the G.S. and 8. kn/m 3, what is the allowable soil pressure if SF3.0 using (a) Hansen bearing capacity and (b) Meyerhof's reduction factors. Solution: 67 60. e y 0.5m ; e x 0.09m 780 780 e y.8 (0.5 ).5m ; L L ex.8 (0.09 ).6m (a) Using Hansen's euation: ( with...all...i i,gi..and...b i..factors...are...0 ) ult. cn c.s c.d c + N.S.d + 0.5..N earing capacity factors from table (3.):.S. d N c π.tanφ ( N ).cot φ, N e..tan ( 45 + φ / ), N.5( N ) tanφ 75
for φ 36 : N c 50. 6, N 37. 8, N 40 Shape factors from table (3.5): S N 37.8.5.5 + +.69, S tan tan 36. 673 N L + φ + 50.6.6 L.6 c c.5 S 0.4 0.4 0.69 L.6 Depth factors from table (3.5): for D.8m, and.8m, D/.0 (shallow footing) D d c + 0.4 + 0.4(.0 ).4, D d + tanφ( sinφ ) + tan 36( sin 36 ) (.0 ).46 ult. 9.4(50.6)(.69)(.4) +.8(8.)(37.7)(.673)(.46) all. 408.635 / 3 34.878 kn/m + 0.5(8.)(.5)(40)(0.69)() 408.635 kn/m, d. 0 Actual soil pressure ( act. ) 780/(.5)(.6) 73.50 < 34.878 (O.K.) (b) Using Meyerhof's reduction: e 0.09 e x / 0.5 y / 0.5 0.5 Rex ( ) ( ) 0.78 ; Rey ( ) ( ) 0. 7 L.8.8 Recompute ult. as for a centrally loaded footing, since the depth factors are unchanged. The revised Shape factors from table (3.5) are: S N 37.8.8.8 + +.75 ; S + tanφ + tan 36. 73 N L 50.6.8 L.8 c c.8 S 0.4 0.4 0.60 L.8 ult. cn c.s c.d c + N.S.d + 0.5..N.S. d 76
ult. 9.4(50.6)(.75)(.4) +.8(8.)(37.7)(.73)(.46) + 0.5(8.)(.8)(40)(0.60)() 4.403 kn/m 4.403 / 3 404.34 kn/m all.centrally.loaded.footing all.eccentric.loaded..footing all.centrally.loaded.footing ( R ex )( R ey ) 404.34(0.78)(0.7) 788.35 kn/m (very high) Actual soil pressure ( act. ) 780/(.8)(.8) 549.383 < 788.35 (O.K.) 3.0 EFFECT OF WATER TALE ON EARING CAPACITY Generally the submergence of soils will cause loss of all apparent cohesion, coming from capillary stresses or from weak cementation bonds. At the same time, the effective unit weight of submerged soils will be reduced to about one-half the weight of the same soils above the water table. Thus, through submergence, all the three terms of the bearing capacity (.C.) euations may be considerably reduced. Therefore, it is essential that the.c. analysis be made assuming the highest possible groundwater level at the particular location for the expected life time of the structure. Case (): m G.S. W.T. Case (5) D f W.T. W.T. d w W.T. W.T. Case (4) Case (3) Case () Case () If the water table (W.T.) lies at or more below the foundation base; no W.T. effect. D D 77
Case (): (from Ref.;Foundation Engg. Hanbook): if the water table (W.T.) lies within the depth (d w <) ; (i.e., between the base and the depth ), use av. in the term.. N as: av. + ( d w / )( m ).....(from Meyerhof) (from Ref.;Foundation Analysis and Design): if the water table (W.T.) lies within the wedge zone { H 0.5.tan( 45 + φ / ) }; use av. in the term.. N as: d w av. ( H d w ). wet + ( H d w ).(from,owles) H H where: H 0.5.tan( 45 + φ / ). submerged unit weight ( sat. w ), d w depth to W.T. below the base of footing, m wet moist or wet unit weight of soil in depth ( d w), and Snice in many cases of practical purposes, the term.. N can be ignored for conservative results, it is recommended for this case to use.. N instead of av. ( < av.( from..meyerhof ) < av. ( from..owles )) in the term Case (3): if d w 0 ; the water table (W.T.) lies at the base of the foundation; use Case (4): if the water table (W.T.) lies above the base of the foundation; use: t.d( above.. W.T.) +. D( below.. W.T.) and in.. N term. 78
Case (5): if the water table (W.T.) lies at ground surface (G.S.); use:.d f and in.. N term. Note: All the preceding considerations are based on the assumption that the seepage forces acting on soil skeleton are negligible. The seepage force adds a component to the body forces caused by gravity. This component acting in the direction of stream lines is eual to ( i. w ), where i is the hydraulic gradient causing seepage. Example (7): A (.x4.)m rectangular footing is placed at a depth of ( D f m) below the G.S. in Solution: clay soil with φ u 0, 8 kn/m 3, which can be applied under the following conditions: (a) W.T. at base of footing with sat 0 kn/m3, (b) W.T. at 0.5m below the surface and sat 0 kn/m3, C u kn/m. Find the allowable maximum load (c) If the applied load is 400kN and the W.T. at the surface what will be the factor of safety of the footing against.c. failure. Q all.? L/ 4./. 3.5 < 5 rectangular footing, D/ /. 0.833 <.0 shallow footing; therefore Terzaghi's euation is suitable. y Terzaghi's euation: G.S. D f.0m ult. cn c.s c + N +.m...n. S Shape factors: from table (3.), for rectangular footing S c (+ 0.3 ); S ( 0. ) L L earing capacity factors: from table (3.3), for φ 0, N c 5. 7, N.0,..N 0 8 kn/m 3 u 0 c kn/m φ 79
(a) for W.T. at base of footing: ult. ()(5.7) (+0.30 all. 54.48 /3 5.388 kn/m. ) +.0(8)() 4. Q all. 5.388(.x4.) 58.970 kn (b) for W.T. at 0.5m below the surface:. + 0.5(.)(0-0)(0)(-0.0 ) 54.48 kn/m 4. t.d( above.. W.T.) +. D( below.. W.T.) D 0.5 and D 0. 5 ; 8(0.5 ) + ( 0 0 )(0.5 ) 4 kn/m ult. ()(5.7) (+0.30 all. 50.48 /3 50.049 kn/m. ) +.0(4)() 4. Q all. 50.049(.x4.) 5.49 kn. + 0.5(.)(0-0)(0)(-0.0 ) 50.48 kn/m 4. (c) If the applied load is 400kN and the W.T. at the surface what will be the factor of safety of the footing against.c. failure?. Q all. 400 kn; all. 400/(.(4.) 79.36 kn/m ; D f. ()(0-0)0 kn/m ult. ()(5.7) (+0.30.. ) + 0() + 0.5(.)(0-0)(0)(-0.0 ) 46.4 kn/m 4. 4. SF ult. all. 46.4 79.36.8 80
3. earing Capacity For Footings On Layered Soils Stratified soil deposits are of common occurrence. It was found that when a footing is placed on stratified soils and the thickness of the top stratum form the base of the footing ( d or H ) is less than the depth of penetration [ H crit. 0.5 tan( 45 + φ / )]; in this case the rupture zone will extend into the lower layer (s) depending on their thickness and therefore reuire some modification of ultimate bearing capacity ( ult.). Several solutions have been proposed to estimate the bearing capacity of footings on layered soils, however, they are limited for the following three general cases: Case (): Footing on layered clays (all φ 0): (a) Top layer stronger than lower layer (C C / C ). (b) Top layer weaker than lower layer (C C / C > ) ). For clays in undrained condition ( φ u 0), the undrainedd shear strength ( S u or c u ) can be determined from unconfined compressive ( u ) tests. So that assuming a circular slip surface of the soil shear failure pattern, may give reasonably reliable results (see figure (3.9)). Figure (3.9): Footings on layered clays. 8
The first situation occurs when the footing is placed on a stiff clay or dense sand stratum followed by a relatively soft normally consolidated clay. The failure in this case is basically a punching failure. While, the second situation is often found when the footing is placed on a relatively thin layer of soft clay overlying stiff clay or rock. The failure in this case occurs, at least in part by lateral plastic flow (see Fig.(3.0)). G.S. G.S. φ Soft layer c, H φ Stiff layer c, H φ Stiff layer c, φ Soft layer c, Hansen Euation (Ref., owles's ook, 996) For both cases (a and b), the ultimate bearing capacity is calculated from Table (3.) for (φ 0) as: ult. Su Nc( + Sc + dc ic bc gc ) +.......(3.5) If the inclination, base and ground effects are neglected, then euation (3.5) will be:- where: 996): S u and (a) Figure (3.0): Typical two-layer soil profiles. ult. Su Nc( + Sc + dc ) +.......(3.6) Nc can be calculated by the following method (From owles's ook, In this method, S u is calculated as an average value C avg. depending on the depth of penetration ( H crit. 0.5tan( 45 + φ / ), while N c 5.4. So that, euation (3.6) is written as: ult. 5.4Cavg. ( + Sc + dc ) +......(3.6b) (b) 8
where: S u C avg. S c 0. ; L C H + C [Hcrit - H] ; Hcrit Df d c 0.4 for Df ; and D dc 0.4 tan for (D >) Case (): Footing on layered c φ soils as in Fig.(3.): (a) Top layer stronger than lower layer (C /C ). (b) Top layer weaker than lower layer (C /C > ). Figure (3.8) shows a foundation of any shape resting on an upper layer having strength parameters c, φ and underlain by a lower layer with c, φ. G.S. D f H or d d, φ c,, c, φ Layer () Layer () Figure ( 3.): Footing on layered c φ soils. Hansen Euation (Ref., owles's ook, 996) () Compute Hcrit. 0.5tan( 45 + φ / ) using φ for the top layer. () If H crit. > H compute the modified values of c andφ as: Hc + ( H crit. H )c c* ; H crit. Note: A possible alternative for φ* Hφ + ( H H crit. crit. H ) φ c φ soils with a number of thin layers is to use average values of c andφ in bearing capacity euations of Table (3.) as: 83
c + + + H ch... cnh c n av. ; H i φ av. tan H tanφ + H tanφ +... + Hi (3) Use Hansen's euation from Table (3.) for ult. with c * and φ * as: ult. c* NcScdcic gcbc + NSdi gb + 0.5 H n tanφ N S d i g b...(3.7) If the effects of inclination, ground and base factors are neglected, then euation (3.7) will takes the form: where: ult. c* NcScdc + NSd + 0.5 N S d.........(3.8) earing capacity factors: from table (3.) N π.tanφ* e tan ( 45 + φ * / ), Nc ( N )cotφ *, N.5( N )tanφ * N Shape factors from table (3.6): Sc +, S + tanφ *, Nc L L Depth factors: from table (3.6) d c + 0.4k, d + tan φ * ( sin φ*) k, d. 0 S 0.4 n L where: D k for Df D or k tan ( radian ) for Df > Case (3): Footing in layered sand and clay soils: (a) Sand overlying clay. Hansen Euation (Ref., owles's ook, 996) (b) Clay overlying sand. () Compute Hcrit. 0.5tan( 45 + φ / ) using φ for the top layer. () If H crit. > H, for both cases; sand overlying clay or clay overlying sand, estimate ult. as follows: ult. b p.pv.k.tanφ + + A s f p.d A f c t...(3.9) where: t, b ultimate bearing capacities of footing with respect to top and bottom soils, 84
for φ > 0 (sand or clay) t c N c S c d c + D N S d + 0.5 N f S d........(3.9a) b c N c S c d c + ( D + H )N S d + 0.5 N f S d.....(3.9b) for φ u 0 (clay in undrained condition) t 5.4Su(+ Sc + dc ) + D f.............(3.9c) 5.4S ( + S + d ) + ( D H ).............(3.9d) b u c c f + Hansen's bearing capacity factors from table (3.) with ( φ φi ) : N π.tanφ e tan ( 45 + φ / ), Nc ( N )cotφ, N.5( N ) tanφ Shape factors from table (3.5): Depth factors from table (3.5): N Sc +, S + tanφ, Nc L L S 0.4 d c + 0.4k, d + tanφ( sinφ ) k, d. 0 L where: D k for Df D or k tan ( radian ) for p total perimeter for punching (+L) or π. D (diameter), Df > P v total vertical pressure from footing base to lower soil computed as: d 0 h.dh + d d + D f. d K s lateral earth pressure coefficient, which may range from tan ( 45 ± φ / ) or use K o sinφ, tan φ coefficient of friction between Pv Ks and perimeter shear zone wall, pd c cohesion on perimeter as a force, A f area of footing. (3) Otherwise, if ( H / ) crit. ( H / ),then ult. is estimated as the bearing capacity of the first soil layer whether it is sand or clay. 85
EARING CAPACITY EXAMPLES (3) Example (8): (footing on layered clay) Footings on layered soils Prepared by: Dr. Farouk Majeed Muhauwiss Civil Engineering Department College of Engineering Tikrit University A rectangular footing of 3.0x6.0m is to be placed on a two-layer clay deposit as shown in figure below. Estimate the ultimate bearing capacity. G.S. P Clay () Clay () H.5m 3m c S u.83m.m 5 kpa c S u 77 kpa φ 0 Solution: H crit. 0.5 tan( 45 + φ / ) 0.5(3) tan45.5m >.m the critical depth penetrated into the nd. layer of soil. For case(); clay on clay layers using Hansen's euation: From owles's ook, 996: where: S u C avg. ult. 5.4.C avg. ( + Sc + dc ) + C H + C [Hcrit - H] Hcrit 77(.) + 5 (.5 -.) 84.093.5 S c 0. / L 0.( 3 / 6 ) 0.; for Df / : d c 0.4D / 0.4(.83 / 3 ) 0. 4 ult. 5.4(84.093)(+0.+0.4)+.83(7.6 ) 60.784 kpa 86
Example (9): (footing on sand overlying clay) A.0x.0m suare footing is to be placed on sand overlying clay as shown in figure below. Estimate the allowable bearing capacity of soil?. P G.S. W.T. Sand Clay m x m H.88m S u.50m c 0 kpa φ 34 7.5 0.60m / 75 kpa u Solution: Hcrit. 0.5tan( 45 + φ / ) 0.5( )tan( 45 + 34 / ). 88m > 0.6m the critical depth H crit. > H penetrated into the nd. layer of soil. For case (3); sand overlying clay using Hansen's euation: ult. b + p.pv.k A s f.tanφ + p.d A f c t where: for sand layer: t D N S d + 0.5 N f S d Hansen's bearing capacity factors from Table (3.) with ( φ 34 ) : tan34 N e π tan ( 45 + 34 / ) 9.4, N.5( 9.4 )tan 34 8. 7 Shape factors from Table (3.5): S + tanφ. 67, S 0.4 0. 6 L L 87
Depth factors from Table (3.5): D f.5 d + tanφ ( sinφ ) + tan34( sin34)., d.0 t 7.5(.5)(9.4)(.67)(.)+ 0.5()7.5)(8.7)(0.6)(.0) 8.5 kpa for clay layer: S c 0. 0. 0. ; L b 5.4Su ( + Sc + dc ) + for Df > : D f.5 + 0.6 dc 0.4 tan 0.4 tan ( ) 0. 3 S d ; b 5.4(75)(+0.+0.3)+(.5+0.6)(7.5) 6 kpa Now, obtain the punching contribution: P v d 0 0.6 d 0.6 + D f d 7.5 + 7.5(.5 )(0.6 ) 8. 6 kn/m 0 h.dh + d K o sinφ sin 34 0.44, ( + )(8.6 )(0.44 )tan34 ( + )(0.6 )(0 ) ult. 6 + + 633 kpa < x x ult. 8.5 kpa all. 633 / 3 kpa 88
Example (0): (footing on c φ soils) Check the adeuacy of the rectangular footing.5x.0m shown in figure below against shear failure (use F.S. 3.0), w 0 kn/m 3. P 300 kn parameter Soil () Soil () Soil (3) Gs.70.65.75 e 0.8 0.9 0.85 c (kpa) 0 60 80 φ 35 0 0 Soil () Soil () Soil (3) G.S. 0.8m W.T..5 x m 0.4m 0.5m Solution: G s..70(0 ) w d 5 kn/m 3 + e + 0.8 ( G s + e ) w (.70 + 0.8 )0 sat + e + 0.8 9.4 kn/m 3 G s..65(0 ) w d 8.7 kn/m 3 + e + 0.9 (.75 + 0.85 )0 sat 9.45 kn/m 3 + 0.85 H crit. 0.5 tan( 45 + φ / ) 0.5(.5) tan45 0.75m > 0.50m the critical depth penetrated into the soil layer (3). Since soils () and (3) are of clay layers, therefore; by using Hansen's euation: From owles's ook, 996: ult. 5.4Cavg. ( + Sc + dc ) + 89
where: C avg. C H + C [Hcrit - H] Hcrit S c 0. / L 0.(.5 / ) 0.5; 60(0.5) + 80 (0.75-0.50) 66.67 0.75 for Df / d c 0.4D / 0.4(. /.5 ) 0. 3 ult. 5.4(66.67)(+0.5+0.3)+0.8(5)+0.4(9.45-0) 59.5 kpa 59.5 all ( net ) 5.78 57.4 kpa 3 300 applied 00 kpa < all ( net ) 57. 4 kpa (O.K.).5x Check for sueezing: For no sueezing of soil beneath the footing: ( ult. > 4c + ) 4c + 4(60)+ 0.8(5)+0.4(9.45-0) 55.78 kpa < 59.5 kpa (O.K.) 3. Skempton's earing Capacity Euation Footings on Clay and Plastic Silts: From Terzaghi's euation, the ultimate bearing capacity is: ult. cn c.s c + N +...N. S......(3.) For saturated clay and plastic silts: ( φ u 0 and N c 5.7,N.0,.and.N 0 ), For strip footing: S c S. 0 ult. cnc +.........(3.30) ult. all. and all all. 3.( net ) all.( net ) ult. cn cn c + c + ( ).....(3.30a) 3 3 3 3 90
where: N c bearing capacity factor obtained from figure D f footing and. ( ) is a small value can be neglected. 3 (3.) depending on shape of (τ) C φ Soil S Cu + σ. tan φ for c φ soil: σ for UCT: or φ u 0 ; all.( net ) σ 3 σ u andσ 3 0; tan ( 45 + φ / ) + c tan( 45 + φ / then u c tan( 45 c u and euation (3.30a) will be: + φ / N c u.......(3.30b) 6 ) ) C u C u σ 3 0 (τ) σ 3 0 θ σ u Pure Cohesive Soil u S Cu, φu 0 σ u (σ) (σ) D f From figure (3.) for 0: N c 6. for suare or circular footings; 5.4 for strip or continuous footings If N c 6. 0, then: al l.( net ) u......(3.3) See figure (3.3) for net allowable soil pressure for footings on clay and plastic silt. N c Figure (3.): N c bearing capacity factor for Footings on clay under φ 0 conditions (After Skempton, 95). N c (net) Suare and circular / L Continuous / L 0 D f / N c (strip) ( + 0. ) or L Net allowable Soil pressure (kg/ cm ) 0 8 6 4 0.0.8.6.4..0.8.6.4 0 D f / D f / 4 D f / 4 6 D f / 0.5 D f / 0...4.6.8.0..4.6.8.0 Unconfined compressive strength (kg/ cm ) Figure (3.3): Net allowable soil pressure for footings on clay and plastic silt, determined for a factor of safety of 3 against bearing capacity failure ( φ 0 conditions). Chart values are for strip footings (/L 0); and for other types of footings multiply values by (+ 0./L). N c (net) N c(suare) ( 0.84 + 0.6 ) L 9 8 0
Example (): (footing on clay) Determine the size of the suare footing shown in figure below. If Q 000 kn G.S. u 00 kpa and F.S. 3.0? m? 0.4m kn/m3 kn/m3 soil 0 conc. 4 Solution: Assume 3.5m, ult. cnc + D / /3.5 0.57 then from figure (3.): N c 7. 3 50(7.3) + (0) 405 kpa 405 ult. 0(.6 ) 4(0.4 ) 93.4 kpa 3 3 all.( net ) Area000/93.4 0.7 m ; for suare footing: take 3.5m, and 0.7 3.7 < 3.5m D / /3.5 0.6 then from figure (3.5): N c 7. 5 ult. cnc + 50(7.5) + (0) 45 kpa 45 ult. 0(.6 ) 4(0.4 ) 96.73 kpa 3 3 all.( net ) Area000/96.73 0.34 m ; use x (3.5 x 3.5)m 0.34 3. 3.5m (O.K.) Example (): (footing on clay) For the suare footing shown in figure below. If u 380 kpa and F.S. 3.0, determine all. and D f (min.) which gives the maximum effect on all.?. Q G.S. D f? 0.9x0.9m u 380 kn/m 9
Solution: From Skempton's euation: For strip footing: all.( net ) cn 3 c cn For suare footing: c all.( net ) x. 3 From Skempton's figure (3.) at D f / 4 and /L (suare footing): N c 9 380 (9) all.(net) 570 kpa and D 3 f 4(0.9) 3.6m Rafts on Clay: Q Total.load( D.L. + L.L.) If b > all. use pile or floating foundations. A area From Skempton's euation, the ultimate bearing capacity (for strip footing) is: ult. cnc +............(3.30) ult.( net ) cnc, cn c all.( net ) or F.S. F.S. cnc all.( net ) Net soil pressure b D f. Notes: cn F.S. c.........(3.3) Df. b () If b D f. (i.e., F.S. ) the raft is said to be fully compensated foundation (in this case, the weight of foundation (D.L.+ L.L.) the weight of excavated soil). () If b > D f. (i.e., F.S. certain. value ) the raft is said to be partially compensated foundation such as the case of storage tanks. 93
Example (3): (raft on clay) Determine the F.S. for the raft shown in figure for the following depths: D f m,m, and 3m?. Solution: G.S. Q 0 000 kn F.S. b cnc Df. For D f m: D f 0 x 0 m soil 8 kn/m3 u 00 kn/m From figure (3.) D f / /0 0. and / L 0: 0 Nc strip 5.4 and N crec tan gular Nc strip ( + 0. / L ) 5.4 (+ 0. ) 5.94 0 cn (00 / )5.94 50( 5.94 ) F.S. c 3. 6 b Df. 0000 00 8 (8 ) 0x0 For D f m: From figure (3.) D f / /0 0. and / L 0 : 0 Nc strip 5.5 and N crec tan gular 5.5 (+ 0. ) 6.05 0 cn (00 / )6.05 50(6.05 ) F.S. c 4. 7 b Df. 0000 00 36 (8 ) 0x0 For D f 3m: From figure (3.) D f / 3/0 0.3 and / L 0: 0 Nc strip 5.7 and N crec tan gular 5.7 (+ 0. ) 6.7 0 cn (00 / )6.7 50(6.7 ) F.S. c 6. 8 b Df. 0000 00 54 3(8 ) 0x0 94
3.3 Design Charts for Footings on Sand and Nonplastic Silt Notes: From Terzaghi's euation, the ultimate bearing capacity is: ult. cnc.s c + N +...N. S..........(3.) For sand ( c 0 ) and for strip footing ( S c S. 0 ), then, E.(3.) will be: ult. N +.. N.........(3.33) ult.( net ) ult.( net ) ult.( net ) N +..N D f..n +..N D f. D f. D f. ( N ) +..N ( N ) +. N D f. all.( net ) ( N ) +.N.........(3.34) F.S. () the allowable bearing capacity shown by (E.3.34) is derived from the frictional resistance due to: (i) the weight of the sand below the footing level; and (ii) the weight of the surrounding surcharge or backfill. () the ult. of a footing on sand depends on: (a) width of the footing, (b) depth of the surcharge surrounding the footing, (c) angle of internal friction, φ (d) relative density of the sand, D r (e) standard penetration resistance, N-value and (f) water table position. D f 95
(3) the wider the footing, the greater ult. /unit area. However, for a given settlement S i such as ( inch or 5mm), the soil pressure is greater for a footing of intermediate width b than for a large footing with a width footing with width a (see figure 3.4a). c or for a narrow D f (4) for constant and a given settlement on sand, there is an actual relationship between all. and represented by (solid line) (see figure 3.4b). However, as basis for design a substitute relation (dashed lines) can be used as shown in (figure 3.4c). The error for footings of usual dimensions is less than ± 0%. The position of the broken line efg is differs for different sands. Q Q Q3 a b (a) Footings of different widths. Soil Pressure, c Settlement, Si d Given c a b Settlement Narrow footing Wide footing Intermediate footing (b) Load-settlement curves for footings of increasing widths. Soil pressure, e b a c g f d Width of footing, (c) Variation of soil pressure with for given settlement, Si. Figure (3.4): Footings on sand. 96
(5) the design charts for proportioning shallow footings on sand and nonplastic silts are shown in Figures (3.5, 3.6 and 3.7). D f / D f / D f / Net soil pressure (kg/ cm ) 6 5 4 3 0 N 50 N 40 N 30 N 0 N 5 N 0 N 5 6 5 4 3 0 6 N 50 N 50 5 N 40 N 40 4 N 30 N 30 3 N 0 N 0 N 5 N 5 N 0 N 0 N 5 N 5 0 0.0 0.3 0.6 0.9. 0.0 0.3 0.6 0.9. 0.0 0.3 0.6 0.9..5.8 Width of footing,, (m) Fig.(3.5): Design charts for proportioning shallow footings on sand. Correction factor C N Effective vertical overburden ressure (kn/ m ) 0.4 0.6 0.8.0..4.6.8.0 0 50 00 50 00 50 300 350 400 450 500 Fig.(3.6): Relationship between bearing capacity factors and φ. Fig.(3.7): Chart for correction of N-values in sand for overburden pressure. 97
Limitations of using charts (3.5, 3.6 and 3.7): These charts are for strip footing, while for other types of footings multiply all. by (+ 0. /L). The charts are derived for shallow footings ( D f / ); 00 Ib/ft 3 ; settlement.0 (inch); F.S..0; no water table (far below the footing); and corrected N-values. N-values must be corrected for: (i) overburden pressure effect using figure (3.7) or the following formulas: 0 CN 0.77 log or Po (Tsf ) 000 CN 0.77 log Po ( kpa ) If p o < 0.5(Tsf ) or < 5( kpa ), (no need for overburden pressure correction). (ii) and water table effect: C w D 0.5 + 0.5 w + D f D f W.T. D w G.S. N Example (4): (footing on sand) Determine the gross bearing capacity and the expected settlement of the rectangular footing shown in figure below. If N avg. (not corrected) and the depth for correction 6m?. Q G.S. Solution: P o 0.75(6) + 5.5(6-9.8) 44.5 kpa >5 kpa 0.75m 0.75x.5m W.T. 6 kn/m 3 98
000 000 CN 0.77 log 0.77 log.66 Po ( kpa ) 44.5 C D 0.5 + 0.5 w + D 0.75 0.5 + 0.5 0.75 + 0.75 w f 0.75 N corr. (.66)(0.75) 0.8 (use N 0) From figure (3.5) for footings on sand: at D f / and 0.75m (.5ft) and N 0 for strip footing:.( Tsf )x05.594 3. 307 kpa all.( net ) for rectangular footing: all 3. 307 x (+0./L) 55.538 kpa.( net ) D. 55.538 + 0.75(6) 67.538 kpa gross all.( net ) + f And the maximum settlement is not more than ( inch or 5mm). Example (5): (bearing capacity from field tests) SPT results from a soil boring located adjacent to a planned foundation for a proposed warehouse are shown below. If spread footings for the project are to be found (.m) below surface grade, what foundation size should be provided to support (800 kn) column load? Assume that 5mm settlement is tolerable, W.T. encountered at (7.5m). P800 kn SPT sample depth N (m) 0.3 9. 0.4 5 3.6 4.8 9 6 9 7.5 33 0 7 Solution: field Find σ o at each depth and correct 7.5m G.S. D f.m W.T.? N field values. Assume.4 m 7 kn/m 3 0 kn/m 3 99
At depth below the base of footing (.+.4) 3.6m; ( 5 + 9 + 5 ) / 3 0 N avg. For 0, and D f / 0.5; all.. T/ft 3.3 kpa from Fig.(3.5). N avg. SPT sample depth (m) N field σ o (kn/m ) σ o (T/ft ) C N (Fig.3.7) N C N. N field 0.3 9. 0 0.4 0..55 5.4 5 40.8 0.43.8 9 3.6 6. 0.64.5 5 4.8 9 8.6 0.85.05 0 6 9 0.07 0.95 7 7.5 33 7.5.33 0.90 30 0 7 5.5.59 0.85 3 P 800 Say.5 m, all., L 3.0m.x.L 3.3X.5 Rafts on Sand:, use (.5 x 3.5)m footing. For allowable settlement (inch) and differential settlement >3/4 (inch) provided that D f ( 8 ft ).or.(.4m )min. the allowable net soil pressure is given by: G.S. Q D f Raft foundation D w W.T. D f D w N Sand If S ( N ) C all. all.( net ) w.. for5 N 50......(3.35) 9.0( N ) C w and S all. ; then all.( net ).0 0.N( Tsf ) 3.3N( kpa ) 9 00
Q and gross all.( net ) + D f. Area where: D f. Dw + ( D f Dw )( w ) + ( D f Dw ) w C w D 0.5 + 0.5 w (correction for water table) + D f N SPT number (corrected for both W.T. and overburden pressure). Hint: A raft-supported building with a basement extending below water table is acted on by hydroustatic uplift pressure or buoyancy eual to ( D f Dw ) w per unit area. Example (6): (raft on sand) Determine the maximum soil pressure that should be allowed at the base of the raft shown in figure below If N avg. (corrected) 9?. G.S. Q D f 3m 9 mx5m W.T. Solution: 9 m Very fine sand 5.7 kn/m 3 ; N avg. 9 blow/30cm Rock For raft on sand: all.( net ) 3.3N( kpa ) 3.3(9) 44.37 kpa Dw 3 Correction for water table: Cw 0.5 + 0.5 0.5 + 0.5 0. 65 + D f 9 + 3 44.37(0.65 ) 75. 856 kpa all.( net ) The surcharge D f. 3(5.7) 47. kpa and gross all.( net ) + D f. 75.856+ 47. 33 kpa 0
3.4 earing Capacity of Footings on Slopes If footings are on slopes, their bearing capacities are less than if the footings were on level ground. In fact, bearing capacity of a footing is inversely proportional to ground slope. Meyerhof's Method: In this method, the ultimate bearing capacity of footings on slopes is computed using the following euations: ( ult. ) continuous. footing.on.slope cnc +.. N ( where: Notes: ) ult. c.or.s. footing.on.slope ( ult.........(3.36) ) continuous. footing.on.slope ( ( ult. ) ) c.or.s. footing.on.level.ground ult. continuous. footing.on.level.ground..(3.37) N c and N are bearing capacity factors for footings on or adjacent to a slope; determined from figure (3.8), c or s footing denotes either circular or suare footing, and ( ult. ) of footing on level ground is calculated from Terzaghi's euation. () A φ triaxial should not be adjusted to φ ps, since the slope edge distorts the failure pattern such that plane-strain conditions may not develop except for large ratios. b / () For footings on or adjacent to a slope, the overall slope stability should be checked for the footing load using a slope-stability program or other methods such as method of slices by ishop's. 0
(a) on face of slope. earing capacity factor, earing capacity factor, Distance of foundation from edge of slope b/ (for Ns 0) or b/h (for Ns > 0). Distance of foundation from edge of slope, b/ (b) on top of slope. Figure (3.8): bearing capacity factors for continuous footing (after Meyerhof). 03
EARING CAPACITY EXAMPLES (4) Footings on slopes Prepared by: Dr. Farouk Majeed Muhauwiss Civil Engineering Department College of Engineering Tikrit University Example (7): (footing on top of a slope) A bearing wall for a building is to be located close to a slope as shown in figure. The ground water table is located at a great depth. Determine the allowable bearing capacity by Meyerhof's method using F.S. 3?. Q.5m G.S. 6.m.0m D f.0m Solution: ( ult. ) continuous. footing.on.slope cnc +.. N........(3.36) b.5 D f.0 From figure (3.8-b): with φ 30, β 30,. 5, and. 0 (use the.0.0 dashed line) N 40 ( ult. ) continuous. footing.on.slope (0 )Nc + (9.5)(.0)(40) 390 kn/m 390 / 3 30 kn/m. all. 30 Cohesionless Soil 9.5 kn/m 3, c 0, φ 30 Example (8): (footing on face of a slope) Same conditions as example (6), except that a.0m-by.0m suare footing is to be constructed on the slope (use Meyerhof's method). D f.0m 30.0mx.0m Cohesionless Soil 9.5 kn/m 3, c 0, φ 30 04
Solution: ( ) ult. c.or.s. footing.on.slope ( ult. ) continuous. footing.on.slope ( ( ult. ) ) c.or.s. footing.on.level.ground ult. continuous. footing.on.level.ground..(3.37) ( ult. ) continuous. footing.on.slope (0 )Nc + (9.5)(.0)(5) 43.75kN/m ( ult. ) of suare or strip footing on level ground is calculated from Terzaghi's euation: ult. cnc Sc + N +...N. S earing capacity factors from table (3.3): for φ 30 ; N 37.,..N.5,..N 9. 7 c Shape factors table (3.): for suare footing S c. 3, S 0. 8 ; strip footing S c S. 0 ( ult ) 0 +.0 (9.5)(.5) + 0.5(.0)(9.5)(9.7)(0.8) 59.4 kn/m. suare. footing.on.level.ground ( ult ) 0 +.0 (9.5)(.5) + 0.5(.0)(9.5)(9.7)(.0) 630.8 kn/m. continuous. footing.on.level.ground 59.4 ( ult.) suare.footing.on. slope 43.75 8. 9 kn/m 630.8 8.9 and ( all. ) suare.footing.on. slope 76 kn/m 3 Example (9): (footing on top of a slope) A shallow continuous footing in clay is to be located close to a slope as shown in figure. The ground water table is located at a great depth. Determine the gross allowable bearing capacity using F.S. 4 Q 0.8m G.S. 6.m.m D f.m 30 Clay Soil 7.5 kn/m 3, c50kn/m, φ 0 05
Solution: Since <H assume the stability number N s 0 and for purely cohesive soil, φ 0 ( ult. ) continuous. footing.on.slope cn c b 0. 8 From figure (3.8-b) for cohesive soil: with φ 30, N s 0, 0. 67, and. D f..0 (use the dashed line) N c 6.3. kn/m ( ult. ) continuous. footing.on. slope ( 50 )(6.3 ) 35 35 / 4 78.8 kn/m. all. 3.5 Foundation on Rock It is common to use the building code values for the allowable bearing capacity of rocks (see Table 3.8). However, there are several significant parameters which should be taken into consideration together with the recommended code value; such as site geology, rock type and uality (as RQD). Usually, the shear strength parameters c and φ of rocks are obtained from high Pressure Triaxial Tests. However, for most rocks φ 45 except for limestone or shale φ ( 38 45 ) can be used. Similarly in most cases we could estimate c 5 MPa with a conservative value. Table (3.8): Allowable contact pressure all. of jointed rock. RQD % all. (T/ft ) all. (kn/m ) Quality 00 300 3678 Excelent 90 00 9 Very good 75 0 67 Good 50 65 6864 Medium 5 30 368 Poor 0 0 056 Very poor.0 (T/ft ) 05.594 (kn/m ) 06
Notes: () If all.( tabulated ) > u ( unconfined..compressive..strength ) of intact rock sample, then take. all. u () The settlement of the foundation should not exceed (0.5 inch) or (.7mm) even for large loaded area. (3) If the upper part of rock within a depth of about /4 is of lower uality, then its RQD value should be used or that part of rock should be removed. Any of the bearing capacity euations from Table (3.) with specified shape factors can be used to obtain ult. of rocks, but with bearing capacity factors for sound rock proposed by ( Stagg and Zienkiewicz, 968) as: Nc 4 6 5 tan ( 45 + φ / ), N tan ( 45 + φ / ), N N + Then, ult. must be reduced on the basis of RQD as: ult. ult. ( RQD ) ult. ( RQD ) and all. F.S. where: F.S.safety factor dependent on RQD. It is common to use F.S. from (6-0) with the higher values for RQD less than about 0.75. Rock Quality Designation (RQD): It is an index used by engineers to measure the uality of a rock mass and computed from recovered core samples as: lengths..of..int act..pieces..of..core 00mm RQD > length..of..core..advance 07