3-BEARING CAPACITY OF SOILS INTRODUCTION ultimate bearing capacity, which considers soil strength, where qa=qult/F.Where F is the factor of safety, which typically is taken as 3 for

MAYSAN UNIVERSITY FOURTH YEARCOLLEGE of ENGINEERING FOUNDATION DESIGN CE402CIVIL ENGINEERING DEPARTMENT Dr. RAID AI SHADIDI

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3-BEARING CAPACITY OF SOILS

INTRODUCTIONThe soil must be capable of carrying the loads from any engineered structure placed upon itwithout a shear failure and with the resulting settlements being tolerable for that structure.The evaluation, of the limiting shear resistance or ultimate bearing capacity qult of the soilunder a foundation load, depends on the soil strength perimeters.The recommendation for the allowable bearing capacity qa to be used for design is based ontheminimum of either1. Limiting the settlement to a tolerable amount.2. The ultimate bearing capacity, which considers soil strength, where qa= qult / F. Where Fis the factor of safety, which typically is taken as 3 for the bearing capacity of shallowfoundations.

Probably, one of two potential mode of failure may occur when a footing loaded to producethe ultimate bearing capacity of the soil.

a. Rotate about some center of rotation (probably along the vertical line oa) with shearresistance developed along the perimeter of the slip zone shown as a circleb. Punch into the ground as the wedge agb.

a. Footing on = 0 Soil

Upper Bound Solution∑ = 0 ↷× 2 + × − × 2 = 0 ⎯⎯⎯ = +Lower Bound SolutionAt point o, left side element 1 and adjacent right side element 2, the horizontal stressesshould be equal 3, 1= 1, 2

45 − ∅2 − 2 45 − ∅2 = 45 + ∅2 + 2 45 + ∅2For =0, tan 45°=1 then = +For the case of =0, i.e. foundation on ground surface, the average value is = .


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b. Footing on φ-c Soil


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BEARING CAPACITY EQUATIONS

Terzaghi Bearing-Capacity EquationOne of the early sets of bearing-capacity equations was proposed by Terzaghi (1943) forShallow (D ≤ ) strip footing. Terzaghi's equations were produced from a slightly modifiedbearing-capacity theory developed by Prandtl (1920) but Terzaghi used shape factors.= + + .WhereC= cohesion of soil

= angle of internal friction of soilB= least lateral dimension of footing (diameter for round footing).

= unit weight of soil (use buoyant (submerged) unit weight for soil below water table.= D= acos (45 + ∅/2) = .= − 1 ∅= ∅2 ( ∅ − 1)

Conclusions

1. The ultimate bearing capacity increases with the depth of footing.2. The ultimate baring capacity of cohesive soil ( =0)is independent of footing size; i.e.,

at ground level qult = 5.7 c3. The ultimate bearing capacity of a cohesionless soil(c=0) is directly dependent on the

footing size, but the depth of footing is more important than size.

Meyerhof’s Bearing-Capacity EquationMeyerhof (1951, 1963) proposed a bearing-capacity equation similar to that of Terzaghi butincluded a shape factor Sq with the depth term Nq. He also included depth factors di, andinclination factors ii, for cases where the footing load is inclined from the vertical.Up to a depth of D « B ≈ the Meyerhof qult is not greatly different from the Terzaghi value.The difference becomes more pronounced at larger D/B ratios.= + + . Vertical load= + + . Inclined load

°0 5.7 1.0 0.05 7.3 1.6 0.510 9.6 2.7 1.215 12.9 4.4 2.520 17.7 7.4 5.025 25.1 12.7 9.730 37.2 22.5 19.734 52.6 36.5 36.035 57.8 41.4 42.440 95.7 81.3 100.445 172.3 173.3 297.448 258.3 287.9 780.150 347.5 415.1 1153.2

Shape Factor Strip Round Square Rectangle1 1.3 1.3 1+ 0.3B/L1 0.6 0.8 1- 0.2B/L


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= ( ) 45 + 2= − 1 ∅= − 1 tan(1.4∅)Shape, depth and inclination factors

Hansen's Bearing-Capacity MethodHansen (1970) proposed the general bearing-capacity case and N factor equations. Thisequation is readily seen to be a further extension of the earlier Meyerhof work. Theextensions include base factors for situations in which the footing is tilted from thehorizontal bi and for the possibility of a slope of the ground supporting the footing to giveground factors gi. The Hansen equation implicitly allows any D/B and thus can be used forboth shallow and deep foundation.= + + + .When =0 = . ( + ′ + ′ − ′ − − ) += ( ) 45 + 2= − 1 ∅= 1.5 − 1 tan(∅)su =undrained =0 ultimate shear strengthVesic's Bearing-Capacity EquationsThe Vesic (1973, 1975) procedure is essentially the same as the method of Hansen (1961)with select changes. The Nc and Nq terms are those of Hansen but N is slightly different.There are also differences in the ii, bi, gi terms. The Vesic equation is somewhat easier to usethan Hansen's because Hansen uses the / terms in computing shape factors si whereas Vesicdoes not. = + + + .

= ( ) 45 + 2= − 1 ∅= 2 + 1 tan(∅)


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Which Equations to UseThe bearing capacity can be obtained—often using empirical SPT or CPT data directly—to asufficient precision for most projects.The Terzaghi equations, being the first proposed, have been very widely used because oftheir greater ease of use. They are only suitable for a concentrically loaded footing onhorizontal ground. Both the Meyerhof and Hansen methods are widely used. The Vesicmethod has not been much used

Use Best for

Terzaghi

Very cohesive soils where D/B < 1 or for a quick estimate ofquit to compare with other methods. Don’t use for footingswith moments and/or horizontal forces or for tilted basesand/or sloping ground.

Hansen, Meyerhof, Vesic Any situation that applies, depending on preference orfamiliarity with a particular method.

Hansen, Vesic When base is tilted; when footing is on a slope or when D/B> 1.


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ADDITIONAL CONSIDERTIONS1. A reduction factor is applicable for the third term of B effect as follows;

r = 1 - 0.25 ( ) for B≥2. Which φThe soil element beneath the centerline of a strip footing is subjected to plane strainloading, and therefore, the plain strain friction angle must be used in calculating its bearingcapacity, the plane strain friction angle can be obtained from a plane strain compressiontest. The loading condition of a soil element along the vertical centerline of a square orcircular footing more closely resembles axisymmetric loading than plane strain loading, thusrequiring a triaxial friction loading, which can be determined from a consolidated drained orundrained triaxial compression test.Meyerhof proposed the corrected friction angle for use with rectangular footing as:

∅ = (1.1 − 0.1 )∅Example: Compute the allowable bearing

pressure using the Terzaghi equation forthe footing and soil parameters shown.

Use a safety factor of 3 to obtain qa?

Critical Thinking:F=3 is recommended for cohesive soil

Solution:

Find the bearing capacity. (B not known but D is)From equations and tablesNc =17.7 Nq=7.4 Nϒ=5.0Sc =1.3 Sϒ = 0.8= + + .

= 20 x 17.7 x 1.3 + 17.30 x 1.2 x 7.4 + 0.5 x B x 17.3 x 5.0 x 0.8 = 613.8 +34.6 B kPaqa= qult / F.qa= 613.8 +34.6B)/3 = 205 +11.5BLikely B ranges from 1.5 to 3m, with B=3m r = 1 - 0.25 log( ) = 0.95qa = 205 +11.5x 1.5 = 220 kPaqa = 205 +11.5x 3 x 0.95 = 240 kPa consider qa =220 kPa


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EFFECT OF WATER TABLE ON BEARING CAPACITYThe effective unit weight of the soil is used in the bearing-capacity equations for computingthe ultimate capacity, in calculation for q in the qNq term and in the term 0.5yBNy.1. When the water table is below the wedge zone [depth approximately 0.5B tan (45 + /2)],the water table effects can be ignored for computing the bearing capacity.2. When the water table lies within the wedge zone, if B is known, computing the averageeffective unit weight to be used in the 0.5 BN term will be as follows= (2 − ) + ′ ( − )WhereH = 0.5 B tan (45 + /2)dw = depth of water table below base of footing= wet unit weight of soil in depth dw’ = submerged unit weight of soil below water table = −Example: for the square footing shown,what is the allowable bearing capacity?use F=2, Assume the soil above water tableis wet with normal water content wN =10%

and Gs=2.68

Critical Thinking:F=2 is recommended for cohesionless soilEffective unit weight should be calculated

Solution:

= (1.1 -0.1B/L) tri =35°H = 0.5 B tan (45 + /2) = 0.5x2.5xtan (45+35/2) = 2.4 mdw = 1.95-1.1= 0.85∴ Water table lies within the wedge zone= 1 + = 18.101 + 0.1 = 16.45 /= (9.81) = 16.452.68(9.81) = 0.626= 1 − = 1 − 0.626 = 0.374The saturated unit weight is the dry weight +weight of water in voids= 16.45 + 0.374(9.81) = 20.12 /= (2 − ) + ′ ( − )= (2 2.4 − 0.85) 0.852.4 18.1 + 20.12 − 9.812.4 (2.4 − 0.85) = 14.85 /


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To calculate q in the qNq term useq = 18.1 1.1 = 19.91 /In the 0.5 BN term use = 14.85 /1. Bearing Capacity by Terzaghi’s equations = 35 Nc = 57.8 Nq = 41.4 Nϒ = 42.4= + + . r = 1 - 0.25 log( . ) = 0.975

= 0 + 19.91 x 41.4 + 0.5x2.5x14.85x42.4x0.8x0.975 = 1438 kN/m2

qa=qult /2 =1438/2=719 kN/m2

2. Bearing Capacity by Meyerhof’s equations = 35 Nc = 46.1 Nq = 33.3 Nϒ = 37.1= + + .Sq = s =1+0.1Kp B/L =1+0.1x3.7x1=1.37 dq= d =1+0.1√3.7 x 1=1.08qult = 19.91x33.3x1.37x1.08 + 0.5x2.5x14.85x37.1x1.37x1.08x0.975=2000 kN/m2

qa=qult /2 =2000/2=1000 kN/m2

3. Bearing Capacity by Hansen’s equations

qa=qult /2 =1524/2=762 kN/m2

BEARING CAPACITY FOR FOOTINGS ON LAYERED SOILS

There are three general cases of the footing on a layered soil as follows:Case 1. Footing on layered clays (all = 0).a. Top layer weaker than lower layer (c1 < c2)b. Top layer stronger than lower layer (c1> c2)CR= strength ratio

If =CR ≤ 0.7 = 1.5 + 5.14 ≤ 5.14= 3 + 6.05 ≤ 6.05 &When 0.7≤CR ≤ 1 Reduce the aforesaid by 10%

IF =CR ≥ 1= 4.14 + 0.5 = 4.14 + 1.1


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= 5.05 + 0.33 = 5.05 + 0.66 &= ×+ × 2

Notice: When the top layer is very soft with a small d1/B ratio, one should give considerationeither to placing the footing deeper onto the stiff clay or to using some kind of soilimprovement method

Case 2. Footing on layered -c soils with a, b same as case 1.A modified angle of friction and soil cohesion may be used to determine the bearingcapacity when the depth of the top layer( ) under the base of foundation is less than(H).Calculate H ; H = 0.5 B tan (45 + /2)IF H > d1 then compute∅ = ∅ + ( − )∅ = + ( − )For case 1 and 2 ,if the top layer is soft the calculated bearing capacity should not be morethan qult= 4c1 +qCase3. Footing on layered sand and clay soils.a. Sand overlying clayb. Clay overlying sand

IF H > d1 then1. Find qult based on top-stratum soil parameters using one of bearing capacity equations.2. Assume a punching failure bounded by the base perimeter of dimensions B x L. Hereinclude the q contribution from d1, and compute q’ult of the lower stratum using the basedimension B.3. Compare qult to q'ult and use the smaller value.

Example: A footing of B = 3, L = 6 m is to beplaced on a two-layer clay deposit as shown.

Estimate the ultimate bearingCapacity using Hansen’s equation?

Critical Thinking:Footing on layered caly soil =0


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Solution:Check H ; H = 0.5 B tan (45 + /2)

H=0.5 x 3 tan 45°= 1.5 m > (d1=1.22 m)

= = 11577 = 1.5 > 1Nc1= 4.14 +0.5x3/1.22= 5.37 Nc2= 4.14 + 1.1x3/1.22= 6.84

= ×+ × 2 = 5.37 × 6.845.37 + 6.84 × 2 = 6= . ( + ′ + ′ − ′ − − ) +

qult= 6 x 77 (1 + (0.2x3/6) + (0.4x1.83/3) ) + 17.26 x1.83 x1x1 = 652 kPa

BEARING CAPACITY FOR FOOTINGS WITH ECCENRIC LOADINGTwo methods can be followed to solve this caseMethod 1. Use either the Hansen or Vesic bearing-capacity equation with the followingadjustments:a. Use B' in the yBNy term.b. Use B' and L' in computing the shape factors.c. Use actual B and L for all depth factors.qa = qult/F and Pa = qaB'L'

Method 2. Use the Meyerhof general bearing-capacity equation and a reduction factor Reused asquIt, des = quIt, comp x ReRe = 1 - 2e/B (cohesive soil)Re = 1 - √e/B (cohesionless soil and for 0 < e/B < 0.3)

Alternatively, one may directly use the Meyerhof equation with B' and L’ used to computethe shape and depth factors and B’ used in the 0.5 B'N term.


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Example: A square footing is 1.8 X 1.8 mwith a 0.4 X 0.4 m square column. It isloaded with an axial load of 1800 kNand Mx = 450 kN m; My = 360 kN m.Undrained triaxial tests (soil notsaturated) give = 36° and c = 20 kPa.The footing depth D = 1.8 m; the soilunit weight y = 18.00 kN/m3; thewater table is at a depth of 6.1 mfrom the ground surface. What isthe allowable soil pressure, if F = 3.0,using the Hansen bearing-capacityequation with B', L', Meyerhof'sequation; and the reduction factor Re?

Solution:ex = My/V= 360/1800 = 0.2 < B/6ey = Mx/V = 450/1800 = 0.25 < L/6Bmin = 4ex +wx = 4x0.2+0.4 = 1.2 <1.8Lmin = 4ey +wy =4x0.25+0.4= 1.4 <1.8B’ = B - 2ex = 1.8-2x0.2 = 1.4mL’ = L - 2ey = 1.8-2x0.25 = 1.3m

Hansen’s bearing-capacity equation

Nc = 51 Nq= 38 N = 40Compute shape factors using B’, L’sc= 1+ (Nq/Nc) (B’/L’)= 1.69sq = 1+ ( B’/L’) sin =1.55s = 1-0.4B’/L’=0.6Compute depth factors using B , Ldc= 1+0.4D/B = 1.4dq = 1+2tan (1-sin )2D/B=1.25d = 1 = + + + .

= 20x51x1.69x1.4 + 18x1.8x38x1.55x1.25 + 0.5x1.4x18x40x0.6x1= 5100 kPaqa = 5100/3 =1700kpaThe actual soil pressure is 1800/ (1.4x1.3) = 989 kPa


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Meyerhof’s equation

Kp =tan2 (45+ /2) =3.85Nc =51 Nq=38 N =44sc = 1+ (0.2Kp B/L) = 1.77sq = s = 1+ (0.1Kp B/L) =1.39dc = 1+ (0.2√Kp D/B) = 1.39dq = d = 1+ (0.1√Kp D/B) =1.20

qult = 20x51x1.77x1.39 + 18x1.8x38x1.39x1.2 + 0.5x18x1.8x44x1.39x1.2 =5752 kPaRe=1-(2e/B) cohesive soilRex=1-(2x0.2/1.8) = 0.78 Rey= 1-(2x0.25/1.8) = 0.72qult= 5752x0.78x0.72 = 3230 kPaqa= 3230/3 =1073 kPaThe actual soil pressure is 1800/ (1.8x1.8) = 555 kPaRe= 1- √(e/B) cohesionless soil (This can be used since the cohesion of the soil is low (20)Rex = 1- √(e/B) = 1 - √0.2/1.8 = 0.67Rey= 1- √(e/B)= 1 - √0.25/1.8 = 0.63qult= 5752x0.67x0.63=2428 kPaqa=2428/3 =809 kPa

BEARING CAPACITY FOR FOOTINGS WITH INCLINED LOADInclined loads are produced when the footing is loaded with both a vertical V and ahorizontal component(s) Hi of loading (refer to Table 4-5 and its figure). This loading iscommon for many industrial process footings where horizontal wind loads are incombination with the gravity loads. In any case the load inclination results in a bearingcapacity reduction over that of a foundation subjected to a vertical load only.

Safety Factor Against SlidingThe footing must be stable against both sliding and bearing in case it subjected to ahorizontal load component, for sliding the general equation for the factor of safety is;= + ∁ +Example: The data obtained by theLoad test is Vult= 1060 kN

HL ult= 382 kNFind the ultimate bearing capacity byHansen and Vesic’s equations?

Critical Thinking:Change triaxial to plane strain


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Solution:plane strain = (1.1 – 0.1B/L) triaxial =(1.1– 0.1x 0.5/2)43° = 47°

Hansen’s Bearing Capacity Equation= + + + .From Equations ;Nq = 187 N = 300dqB =1+2 tan (1-sin )2 D/B = 1.155dqL =1+2 tan (1-sin )2 D/L = 1.04d B = d L =1

iqB = (1-(0.5HB / (V+Af Ca Cot )))2.5 = 1 Horizontal force parallel to B =0i B = (1-(0.7HB / (V+ Af Ca Cot )))3.5 = 1

iqL = (1-(0.5HL / (V+Af Ca Cot )))2.5 = (1 – 0.5x 382/1060)2.5 = 0.608i L = (1-(0.7HL / (V+Af Ca Cot )))3.5 = = 0.361

SqB =1+ sin B’ iqB / L’ = 1.18SqL =1+ sin L’ iqL /B’ = 2.78

S B =1 -0.4 B’ i B /( L’ i L) = 0.723S L =1- 0.4 L’ i L /( B’ i B) = 0.422 Use 0.6 both should be equal or more than 0.6

Substitute the factors obtained for each direction in Hansen’s equation and take thesmaller ultimate bearing capacity

qult = 9.43 x 0.5x187x1.18x1.16x1 +0.5x0.5x9.43x300x0.732x1x1 =1700 kPaqult = 9.43 x 0.5x187x2.78x1.04x0.608 +0.5x2.0x9.43x300x0.6x1x0.361 =2150 kPa

Vesic’s Bearing Capacity Equation= + + + .From EquationsNq = 187 N = 404Sq= 1 + B/L tan = 1 + (0.5/2) tan 47 = 1.27S = 1 – 0.4 B/L = 1 – 0.4 (0.5/2) = 0.9 ≥ 0.6 OKdq = 1 + 2 tan (1 – sin )2 k = 1.16d = 1m=(2 +(L/B)) / (1+(L/B))iqL = (1-(HL / (V+Af Ca Cot )))m = (1 – 382/1060)1.2 = 0.585i L = (1-(HL / (V+Af Ca Cot )))m+1 = = 0.374

qult = 9.43 x 0.5x187x1.27x1.16x0.585 +0.5x0.5x9.43x404x0.9x1x0.374 =1080 kPa


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BEARING CAPACITY OF FOOTINGS ON SLOPES

Example: A 2 x 2 m square footing has a groundslope of = 0° and base slope of =10°. Are thefooting dimensions adequate for the given loadwith a factor of safety 3?Use Hansen’s and Vesic’s equationsCritical Thinking:Assume angle of friction between concrete and

soil ( )= =25°Base adhesion Ca =0.6 to 1 x C = 25 kPaNeglect passive earth pressure for D = 0.3 m

Solution:Factor of safety against sliding must be checked

= + ∁ + = 600 tan 25 + 25 × 2 × 2200 = 1.9Hansen’s Bearing Capacity Equation= + + + .From table Nc = 20.7 Nq= 10.7 N = 6.8dc =1+ 0.4 D/B = 1.06dq =1+2 tan (1-sin )2 D/B = 1.05d =1

iqB = (1-(0.5HB / (V+Af Ca Cot )))3 = 0.675i B = (1-((0.7- /450)HB / (V+Af Ca Cot )))4 = 0.481i L = 1ic = iq – (1-iq)/(Nq-1) = 0.641

ScB =1+ (Nq/Nc)( B’ iqB / L) = 1.329SqB =1+ sin B’ iqB / L = 1.285S B =1 -0.4 B’ i B / L i L = 0.808

= 10° = 0.175 radbcB = 1 – /147 = 0.93bqB = exp( -2 tan ) = 0.849b B = exp( -2.7 tan ) = 0.802

Ground factors gi =1 for =0qult = 25x20.7x1.329x1.06x0.641x0.93 + 17.5x0.3x10.7x1.285x1.05x0.675x0.848 +

0.5x17.5x2x6.8x0.808x1x0.481x0.802 = 515 kPa

Pa = A x qa = 2x2 x 515/3 = 686 kN > 600 OK


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BEARING CAPACITY FROM IN SITUE SPT

= (1 + 0.33 ) ≤ 1.2= ( + 0.3) (1 + 0.33 ) > 1.2= (1 + 0.33 )

qa = allowable bearing pressure for ∆Ho = 25-mm or 1-in. settlement, kPa ,for any othervalue of settlement ∆Hi , qai=∆Hi/∆Ho x qaKd = 1 + 0.33 D/B ≤ 1.33N= is the statistical average value of blows for the footing influence zone of about 0.5Babove footing base to at least 2B below. To convert N’70 x 70 = N55 x 55

Example: Find the allowable bearing pressure for a soil of N70 =24, if the foundation depth=1m and width=3m for a tolerable settlement of 15 mm?

Solution:

= 240.06 (3 + 0.33 ) (1 + 0.33 13) = 537qa for 15mm settlement= 537 x 15/25 = 322 kPa

BEARING CAPACITY FROM IN SITUE CPT

An approximation for the bearing capacity should be applicable for D/B < 1.5 from anaveraged value of qc (cone point resistance in kg/ cm2) over the depth interval from aboutB/2 above to 1.1B below the footing base. For cohesionless soils one may use= 28 − 0.0052(300 − ) . /= 48 − 0.009(300 − ) . /For cohesive soils one may use= 2 + 0.28 /= 5 + 0.34 /

N55 N’70

F1 0.05 0.04F2 0.08 0.06

3-BEARING CAPACITY OF SOILS INTRODUCTION ultimate bearing capacity, which considers soil strength, where qa=qult/F.Where F is the factor of safety, which typically is taken as 3 for

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