Problems Problems on Bearing Capacity
Example # 01 •
A square footing 2.5 m by 2.5 m is built in a homogenous bed of sand density 2.0 t/m3 and having an angle of shearing 38o . The depth of a base of the footing is 1.5 m below the ground surface. Calculate the safe load that can be carried by a footing with a factor of safety 3 against complete shear failure. Use Terzaghi’s analysis.
Solution: B = 2.5 m D = 1.5 m = 2 t/m3 = 38o which is greater than 36o so general shear failure will occur. Values of bearing capacity factors from Fig. Nq = 47 N = 64 Since c = 0 qf = D Nq + 0.4 B N qnf = D (Nq -1) + 0.4 B N = 2 x 1.5 (47-1) + 0.4 x 2 x 2.5 x 64 = 138 + 128 = 366 t/m2 qs = qnf /F + D = 366/3 + 2 x 1.5 = 122 + 3 = 125 t/m2 Maximum safe load = B2 x qs = (2.5)2 x 125 =
Given:
780 t.
Example # 02 •
A strip footing 1 m wide at its base is located at a depth of o.8 m below the ground surface. The properties of the foundation soil are : = 1.8 t/m3 and = 20o and c = 3 t/m2 . Determine the safe bearing capacity, using a factor of safety 3 . Use Terzaghi’sanalysis. Assume soil’s local shear failure.
Solution:
qf = 2/3 c Nc + D Nq + 0.5 B N
For = 20o From table: Nc= 9.5, Nq= 3.5 N = 1.7 qf = (2/3 x 3 x 9.5) + (1.8 x 0.8 x 3.5) + 0.5 (1.8 x 1 x 1.7) = 19 + 5.04 + 0.9 = 24.94 t/m2 qnf = qf - D = 24.9- 1.8 x 0.8 = 23.5 t/m2 qs = qnf /F + D = 23.5/3 + (1.8 x
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Solve above example if the water table is located at a depth of 1.5 m below below the ground level. Rw1= 1 Rw2 =?
Zw2= depth of water table below the base of footing. = 1.5 – 0.8 = 0.7 ∴ Rw2 = 0.5 (1+Zw2 /B ) = 0.5 (1+ 0.7/1) = 0.85 q = 2/3 c N + D N f c q Rw1 + 0.5 B NRw2 •
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= 2/3 x 3 x 9.5 + (1.8 x 0.8 x 3.5 x 1)
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qnf = qf - D = 24.8 – 1.44 = 23.36
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qs = qnf /F + D = 23.36/3 + 1.44
= 7.76 + 1.44 = 9.2 t/m2
Example # 03 •
An R.C R.C column having square in shape is to rest 1.5 m below ground level. The total load to be transmitted including the weight of the column is 200 tons. As the area is subjected to frequent flooding, the friction of the footing along the sides is to be neglected and a factor of safety 2.5 is to be allowed. If the saturated density of the sand be 2.4 g/cc, angle of internal friction 33o and value of N = 33, Nq = 32, find the suitable size of the footing for the above condition.
Solution: Assume the size of footing as 2m x 2m. B = 2 m and D = 1.5 m B>D, foundation is shallow. qf =1.3 cNc + D Nq + 0.4 B N 3) Now c = 0 , N = 32, N = 33 = 2.4 g/cc (2.4 t/m q B = 2. qf = 0 + 2.4 x 1.5 x 32 + 0.4 x 2.4 x 2 x 33.
= 115.2 + 63.4 = 178.6 qs= 178.6/2.5 178.6/2.5 = 71.4 (F.S (F.S = 2.5)
Now area of footing 2x 2. ∴ Total load = 4 x 71.4 = 285.6 tones It is higher than the required 200 tones. it is safe. We can make it more economical by reducing the section and retrial may be made.
Example # 04 •
Size of an isolated footing is to be limited to 1.5 m square. Calculate the depth at which the footing should be placed to take a load of 200 tonnes with a factor of safety 3.The soil is having angle of internal friction = 30o , = 2.1 g/cc, weight of footing 5% of the external load. Nq= 22, N = 20.
Solution:
A soil is sandy so no cohesion Allowable bearing capacity for square square footing: q = 1/F { 1.3 cN + D (N s c q – 1) + 0.4 B N } + D Now total load = 200 + 5/100 x 200 = 210 tonnes q = 210/1.5 x 1.5. s
Put values in the above eq. 210/1.5 x 1.5 = 1/3 { 2.1 x D (22-1) + 0.4 x 2.1 x 1.5 x 20} + 2.1 x D = 1/3 {2.1 x 21 x D + 25.2} 2.1 D = 14.7D + 8.4 + 2.1D 16.8D = 93.4 – 8.4 = 85 ∴ D = 85/16.8 = 5.6 meters ∴ The footing will be placed at 5.6 m below the
Example # 05
A 3.0 m strip footing rests 2.5 m below ground level over sandy clay having unit weight of 2250 kg/m 3 . Quick tests in a shear box apparatus gave shear strength of 0.37 and 0.50 kg/cm2 for normal stress of 0.75 and 1.5 kg/cm2 respecti respe ctivel vely y. Find out cohesion and angle of internal friction. From Terzaghi’s formula, calculate the ultimate load per meter run of the foundation.
Solution: •
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As per Coulomb’s eq. = c + tan -------------(A) (A) c = ? And = ? Two Two equations equatio ns can determine c and . Given: 1 = 0.37 kg/cm2 2 = 0.5 kg/cm2 1= 0.75 kg/cm2 2= 1.5 kg/cm2
Put values in eq. (A) 0.37 0. 37 = c + 0. 0.75 75 tan tan. ----------------(i (i)) 0.5 0. 5 = c + 1. 1.5 5 tan tan.--.----------------( -(ii ii)) Solve simultaneously simultane ously.. c = 0.25 kg/cm2 tan = 0.1733 = o o 9.83 10 . Assume local shear failure. Nq= 1.9 N = 0.5 ∴ Nc = 8 ,
qf = 2/3 c Nc + D Nq + 0.5 B N
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qf = 2/3 x 0.25 x 8 + 2250/10 9 x 2.5 x 102 x1.9 + 0.5 x 2250/10 9 x 3.0x102 x0.5 = 1.333 + 0.01069 + 0.001688 = 1.345 kg/cm2 = 13.45 t/m2 Ultimate load per meter run of foundation = 13.45 x 3 = 40.37 Tonnes Tonnes/met /meter er length. length.
Example # 06 •
Compute the allowable bearing pressure using the Terzaghi’s erzaghi’s equation for the footing and soil parameters shown in Fig. Fig. Use factor of safety 3 to obtain qa . The soil data are obtained from a series of γ = 110 D 4 = U-triaxial tests.ft Is the soil saturated pcf = 20 B ? C = 300 o
psf
Solution: •
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Since a U –test gives a angle. The soil is not saturated. For = 20o , Assuming square foundation general shear failure. qult =1.3 cNc + q Nq + 0.4 B N Now for = 20o , Nc= 17.7, Nq= 7.4 N = 5.0 Put values. qult =1.3 x 300/1000 x 17.7 + 4 x 110/1000 x 7.4 + 0.4 x 110/1000 x B x 5.0 = (10.2 + 0.22B) ksf Qa = qult /F.S = 10.2 0.22B /3 = (3.4 + 0.07B) ksf Normally B varies between 5 to 10 ft. For B = 5ft , 3.4 + 0.07 x 5 = 3.75 ksf = 10 ft, 3.4 +0.07 x 10 = 4.10 ksf
Recommended is
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3.5 ksf
Example # 07 •
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The results of full scale tests conducted by H.Muhs in Berlin reported by J.B.Hansen. The pertinent data associated with this test as reported by Hansen are footing dimension L = 2 m, B 0.5 m , D = 0.5 m dense sand = 0.95 t/m3 , c = 0 . Failure load Q 190 o . Ton Ton. . = 47 Compare Compare the results with Terzaghi erzaghi Analysis and Hansen.
Solution: •
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Terzaghi Terzaghi’seq. ’seq. As 4P = 2/0.5 =4, use strip footing formula. qu = c Nc + D Nq + 1/2 B N. For = 47o , Nq = 246 , N = 585 = 0 + 0.95 x 0.5 x 246 + ½ x 0.95 x 0.5 x 585 =116.85 + 138.9 = 255.78 t/m 2. More than actual. Test result = Q/A =190/2 x 0.5 = 190 T/m 2 same if checked by Hansen eq. = 180.6 ton/m2
Example # 08 •
A series of large scale footing bearing capacity tests were performed on soft Bangkok Bangkok clay (Brand et al 72). One of the tests consisted of a 1.05 m square sq uare footing at a depth of 1.5 m. at a 1 inch settlement the load was approximately 14.1 Tons. From the interpretation of the given load – settlement curve. Uncofined compression and vane shear tests on the same soil gave a strength values and other properties as follows:
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qu = 3.0 t/m2 Su van = 2.4 t/m2 Plasticity data: wL= 80% and wp = 35% Compute the ultimate bearing capacity by Terzaghi’s eq. and compare with load test values. For field van , su desi = design gn = suvan reduction reductio n factor. factor. , from graph I p vs (farmula) For Ip= 80 – 35 = 45, = 0.8 2. Su desi = 0.8 x 2.4 = 1.92 t/m design gn (c) For = 0 , Nc=5.7, Nq=1 N = 0 qu(net) = 1.3 c Nc = 1.3 x 1.92 x 5.7 = 10.94
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Correction qu= 1.3 c Nc = 2.4 x 5.7 x 1.3 = 13.68 x 1.3 = 17.78 From unconfined unconfined compression compression test , qu = 1.3 c Nc c = ½ qu = 3.0/2 = 1.5 t/m2 qu = 1.5 x 5.7 x 1.3 = 11.115 t/m2 From load test Load/Area = 14.1/ (1.05)2 =12.8 t/m2 Load Test Unconfi Unconfined ned comp. comp . Vane shear t/m2 12.8 11.115
Example # 09 •
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A given series of values for q u in the of interest from SPT (unconfined) samples from a boring log gives average 200 kpa. Estimate the allowable bearing capacity for square footing by Ter Terzag zaghi’ hi’sform sformula ula.. Use F.S =3. (unconfined only for clay)
Solution: •
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Now
c = qu /2 = 200/2 = 100 kPa.
Terzaghi’seq. for square footing- =0 qu = 1.3 c Nc = 1.3 x 100 x 5.7 = 741 kPa qa= 741/3 = 247 kPa Normally qu from SPT is not reliable. UDS are taken. Or in case of SPT q result F.S = 4 to
Example # 10 •
A 30 cm square bearing plate settles by 1.5 cm in a plate loading test on a cohesion less soil when the intensity of loading is 2 kg/cm 2. What will be the settlement of a prototype shallow footing 1.0 m square under the same intensity of loading.
Solution: •
Given Bp= 30 cm
B = 100 cm
p= 1.5 cm.(settlement of
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plate) •
= 1.5 [ 100 (30 + 30.5)/30
(100 +
30.5)]2 = 3.6 cm
Example # 11 •
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The results of a plate load test in a sandy soil are given below. The size of the plate is 0.305 x .305 m. determine the size of square footing of column foundation that should carry a load of 2500 kN with Load/unit area settlement 200 400 of 6002570mm. 0 a maximum kN/m2 Settlement (m (mm) 5
12.5
28
60
Solution: Given:
Size of the plate 0.305 x 0.305 m. Load v/s settlement , draw a curve or plot the data. Load on footing------2500 kN 2 200 400 600 800 kN/m Max. settlement = 25mm 10
30
50 70
Now settlement of footing: Now by trial and error Procedure assume the size of footing and find the (Size in m) amount of settlement 2 2 B ( B p + 30.48) where the trial agrees S F = S p F B p ( B F + 30.48) with the conditions Or given will be the size 2 2 of footing. B ( 3.2 B p + 1)
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S F
= S p F ( 3.2 B + 1) B p F
Qo , kN 2500 2500 2500 2500
Assumed width, BF 400 (cm) 300 320 310
qo Qo /BF 2 , 156.25 2 kN/m 277.8 244.10 260.10
S p correspondi 4 to Q ng o 8 6 .8 (mm) 7 .2
SF from above Eq. 13.8 (mm) 26.35 22.70 23.86
Qo , kN 2500 2500 2500 2500
Assumed width, BF 400 (cm) 300 320 310
qo Qo /BF 2 , 156.25 2 kN/m 277.8 244.10 260.10
S p correspondi 4 to Q ng o 8 6 .8 (mm) 7 .2
SF from above Eq. 13.8 (mm) 26.35 22.70 23.86
Example # 12 •
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The results of two plate load tests are given in the following table: Plate dia. (B) (m) 0.305 0.610
Total Load, Q (kN) 32.2 71.8
Settlement (mm) 20 20
Determine the size of square footing to carry a load of 715 kN with tolerable settlement as 20mm. Solve for foundation.
Solution:
Qo = Am + Pn Put the values in the above Eq. to find m and n 32.2 = /4 (0.305)2 m + (0.305)n.---------(i) (0.305)n.-- -------(i) 71.8 = /4 (0.610)2 m + (0.610)n.--------(0.610)n.- -------(ii) By solving simultaneously we can find: m = 50.68 kN/m 2 n = 29.75 kN/m2. Now again use the above Eq.with these values to find the size of square footing. Load on foundation = 715 kN A = Bf 2 , P = 4 Bf ; Put values 715 = B 2 x 50.68 + 4 B f f x 29.75 •
Example # 13 •
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Plate load test performed on a uniform deposit of sand and the following observations were recorded. Load t/m2 Settlement (mm)
5 10 4.5 8.5
20 16
30 40 50 60 31.3 50 74 104
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The size of the plate was 30 x30cm. x 30cm. Plot the load-settlement curve and determine the load on a footing 1.5m x 1.5m. Can it carry safely if the settlement is not to exceed 50 mm.
Solution: •
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Given: Size of plate 30 x 30 cm Footing size 1.5 x 1.5 m Settlement of footing 50 mm Plot Load vs settlement
10 10
30 50 70 90
20
30
40
50
60
B f B p + 30.48 S F = S p B f + 30.48 B p
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or put the values to calculate S p .
150 30 + 30.48 50 = S p 30 150 + 30.48 S p = 18 mm •
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Now from load – settlement curve the load corresponding to this settlement = 20 t/m 2. Safe on this footing for 50mm settlement = 1.5 x 1.5 x 20 = 45 tons.
Example # 14 •
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A load test was made with a 35 cm square plate at a depth of one meter below the ground level in soil with = 0. The water table was located at a depth of 5m below b elow the ground level. Failure occurred at a load of 5200 kg. What would be the ultimate bearing capacity per unit area for a 1.6m wide continuous footing with its base loaded at the same depth in the same soil. Unit weight of soil was 1.9 g/cc above water table. For = 0, Nc= 5.7, Nq= 1 , N = 0. Assume general shear failure.
Solution: •
For square footing, qf =1.3 c Nc + D Nq + 0.4 B N Taking Taking the case ca se of load test, B = 0.35cm , D = 1. Since = 0 , N = 1 and N q = 0 q =1.3 c N + D N u c q 5200/.35 x .35 x1000 = 1.3 x c x 5.7 + 100 x 1.9 x1.0. 7.42 c = 42.5 – 1.9 = 40.6 40.6 c = 40.6/7.42 = 5.47 tonnes/m2 . Now for contineous footing strip of width 1.0m q = c N + D N f c q + 0.5 B N = 5.47 x 5.7 + 1.9 x 1 x 1 + 0
33.1 ton/m2
Example # 15 (with inclination and eccentricity)
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A 24 m long retaining wall has a 6m wide foundation at a depth of 1.5m in a silty sand having c = 1.5 t/m 2 , = 25o , = 2 t/m3 . The wall carries a horizontal load of 25 ton/m run at a point 2m above the base and a centrally applied load of 100 t/m run. Determine the safety factor against general shear failure of the wall, (tak ( take e N c = 20.7 , Nq =
Solution: •
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e/Z e/Z = tan tan = H/V H/V = 2 25/ 5/10 100 0 = 0. 0.25 25 , Z =2 m. e/Z = 0.25 , e = 0.25 x 2 = 0.5 , = 14 o B = B- 2e = 5 , L = 24m A = 5 x24 2 =120 m2
H I c = I q = 1 − ′ cot V A c φ +
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H = 25 x 24 V =100 x 24 Ic = Iq = 0.616 I = Iq2 = 0.379 qf = c Nc Ic+ D Nq Iq+ 0.5 B N I
Example # 16 •
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At a certain site subsoil soil consists of a thick layer of soft clay (cu = 20 kN/m2, u = 0) which is overlain by a stiffer clay (cu = 87.5 kN/m2, u = 0) of a variable thickness. The ground g round surface is horizontal and the water table is at a considerable depth. It is proposed to install widely spaced footings 1.5 m square at a depth of 1 m below ground level in the upper up per clay material. Estimate the net working load on the footings, using a load factor of 3.
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If the footings are fully loaded at this pressure, pressure, estimate the thickness of stiff clay required below them in order to ensure that the load factor against shear failure in the soft clay is at least 3. (Ignore settlement effects and assume a load dispersion at 30oto the vertical. Take Nc = 7.5 on the lower clay and assume Terzaghi’s bearing capacity equation to apply)
Solution: •
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Upper layer cu u
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G.L
= 87.5 kN/ m2 = 0
Lower layer:
cu = 20 kN / u = 0
Nc = 7.5
cu
m2
u
1 m
= 87. 5 kN /m2 = 0
Z
30o
Z/ 3
B = 1.5m square footing. F.S = 3
1.5m cu = 20 kN /m2 u = 0
Z/3
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Terzaghi’sEq. Nc = 5.7
nf q=1.3
c Nc where
For upper layer qnf =1.3 x 87.5 x 5.7 = 648.375 kN/m2 648 kN/m2 Net unit working load = 648/3 = 216 kN/m2 Net working load on footing (1.5m)2 = 216 x 1.5 x 1.5 = 486 kN. This load is dispersed at 30o angle,
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Pressure on top of lower clay stratum Pressure Load /Area = 486/
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(1.5 + 2Z/ 3)2 Now allowable pressur p ressure e on lower stratum: Qf /F = cNc /F = 20 x 7.5/3 =50 Equating these two pressures: pressures: 50 = 486/ (1.5 + 2 Z/3)2 112.5 + 173.2 Z + 66.7 Z 2= 486 Z2 + 2.6 Z – 5.6 = 0
Z = 1.4 m
Example # 17 •
Determine the net bearing pressure for a 3m x 3m footing at a depth of 2 m in a medium dense sand so that the total settlement does not exceed exceed 25mm. The average SPT blows below the footing (upto B ) 3m are 28/30 cm. the average moist density is 1.75 t/m 3. The water table is more than 3 m below the footing.
Solution: •
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o at B/2 below footing
1.75(2+1.5) = 6.125 t/m 2 Cn = 0.77 log 200/ o = 0.77 log 200/ 6.125 = 1.165 Nn = Cn N = 1.165 x 28 =33 q25 = 0.041 Nn S = 0.041 x 33 x 25 =
33.8 t/m2
Example # 18 •
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A water tank has a concrete foundation slab, 5m wide by 20m long, constructed at a depth of 1m in medium dense sand where water table is 2m below the ground surface. The depth of the water in the tank is 8m. The density of sand above the water table is 1.7 t/m 3 and the submerged density is 0.9 ton/m3 The SPT blows in a bore hole are given in table below. Examine whether the total settlement will not exceed
Depth N below G.S 6 1 (m) 2(W.T) 8 3 10 4 12 5 12 6 14
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1.7 3.4 4.3 5.2 6.1 7.0
t/m2 Cn 1.59 1.36 1.28 1.22 1.17 1.12
Nn 10 11 13 15 14 16
Average Nn
13
Solution: •
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Cw = 0.5 + 0.5 D w/D + B = 0.67 { Dw =2 , D = 1, B = 5} q25 = 0.041 Nn Cw S = 0.041 x 13 x 0.67 x 25 = 8.9 ton/m 2 Actual net pressure = q- o = 8 x 1 – 1x 1.7 = 6.3 t/m 2. Net pressure pressure applied is less than the allowable net pressure for 25mm settlement. Hence the settlement will remain less than 25mm.
Example # 19 •
Determine the net and gross g ross bearing pressure pressure for a raft, 8m x 15m placed at a depth of 3 m in coarse sand underlain by fine silty sand below a depth of 7.5 m. The water table is 3m below ground surface. SPT was performed in 5 bore holes. The SPT blows for one of the bore holes representing the wekest spot are given given below. below. above water table = 1.9 t/m3 , sub = 1.12 t/m3 (coarse sand below water table) sub = 1.05 t/m3 for silty sand. The maximum settlement is limited to 50 mm. Use
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N = cor corrrecti ection on appl applie ied d for for su subm bmer erge genc nce e (corrected (corrected value) v alue) N = 15 + ½ (N-15) This correction is applied on very fine or silty sand below water table when the measured N value is greater than 15 (overburden correction correction is applied first if both corrections are applied). Now Total N = 177 Average blows = 177/12 =15 Cw = 0.5 + 0.5 (3)/3+8 = 0.64 q50 = 2.05 x N x Cw = 2.05 x 15 x 0.64 = 19.68 t/m2 q
(Gross)
= 19.68 + D =
2538 t/m2
Example # 20 •
A proposed strip footing that is 4 ft(1.2m) wide will be located 2 ft(0.6m) below ground surface. The soil type is uniform dense sand that has a friction angle = 35o. The total unit weight of the soil is equal to 125 pcf (19.7 kN/m3). The ground is water table is well below the bottom of the footing and will not be a factor in the bearing capacity analysis. Using a factor of safety of 3, calculate the allowable bearing pressure using Figs. 6.5 and 6.6.
Solution: •
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From Fig. 6.5, for = 35o, N=37 and Nq = 33. Using Eq. 6.1 with c = 0: qult = ½ t B N + t Df Nq = ½ (125)(4) (37) + (125)(2)(33) = 17500 psf (840 kPa) Using F = 3, qall = qult /3 = 17500/3 = 5800 psf (280 kPa) From Fig.6.6, ig.6.6, for = 35o , N= 40 and Nq = 36 Using Eq. 6.1 with c = 0:
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qult = ½ t B N + t Df Nq = ½ (125) (4)(40) + (125)(2)(36) = 19000 psf (910 kPa)
Example # 21 •
Use the same data as the above example problem, but assume silty sand with a friction angle of 30o. Also assume the footing will not punch into the soil.
Solution: •
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From Fig. 6.5, for = 30o, N=15 and Nq = 19. Using Eq. 6.1 with c = 0: qult = ½ t B N + t Df Nq = ½ (125)(4)(15) + (125)(2)(19 = 8500 psf (400 kPa) Using F = 3, qall = qult /3 = 8500/3 = 2800 psf (130 kPa) From Fig.6.6, ig.6.6, for = 30o , N = 17 and Nq = 20. Using Eq. 6.1 with c = 0:
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qult = ½ t B N + t Df Nq = ½ (125)(4)(17 + (125)(2)(20) = 9250 psf (440 kPa) Using F = 3, 3, qall = qult /3 = 9250/3 = 3100 psf (150 kPa)
Example # 22 •
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Refer to Fig. 3.21. a foundation 1.5m x 1m is located at a depth, Df , of clay. A soft clay layer is located at a depth, H, of 1 m measured from the bottom foundation. For the top clay layer Undrained shear strength = 120 kN/m2 Unit weight = 16.8 kN/m 3 = 16.kN/m3 and for the bottom clay layer Undrained shear strength = 148 kN/m 2 Unit weight = 16.2 kN/m 3 Determine the gross allowable load for the foundation with an FS of 4.
Solution:
B B 2ca H qu = 1 + ( 0.2) cu ( 2) N c + 1 + + γ 1 D f L L B Where B = 1 m H = 1 m Df L =1.5 m N c = 5.14 3 D f = 1 m γ 1 =16.8 kN / m From Figure 3.23,
= 48 / 120 = 0.4, the value of ca ( 2) / cu (1) ≈ 0.9. So
H=1 m
= (0.9)(120) = 108 kN / m2
B
=1 0 Su
1
2= 0 Su
2
1
for cu ( 2) / cu (1)
ca
B x L
Layer No: 1 Layer No: 2
2
Fig. 3.21
and an d
B B 2ca H ( ) 1 0 . 2 1 qu = + + γ 1 D f cu( 2) N c + + L L B 1 1 (2)(108 10 8)(1) qu = 1 + ( 0.2) (48)(5.14) + 1 + + (16.8)(1) 1.5 1.5 1 2 = 279.6 + 360 + 16.8 = 656 65 6.4 kN / m Check
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qu
B = 1 +( 0.2 ) cu (1) N c ++γ 1 D f L 1 = 1 +( 0.2 ) (120 )(5.14 ) +(16 .8 )(1) 1.5 =699 +16 .8 =715 .8 kN / m 2
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Thus qu = 656.4 kN/m2 (that is smaller of the two values calculated above) and qall = qu/FS = 656.4/4 = 164.1 kN/m2 The total allowable load is (qall )(1 x 1.5) = 246.15
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Example # 23 •
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Refer to Figure Figure 3.24. For sand = 117 lb/ft3 = 40o and for clay cu = 400 lb/ft2 For the foundation B = 3 ft L = 4.5 ft Df = 3 ft H = 4 ft Determine the gross ultimate bearing capacity of the foundation.
Qu
Sand
= 40o c = 0
B Sand = 40o c = 0
Clay
= 0 cu= 400 lb/ft2
Clay = 0 cu= 400 lb/ft2
FIGURE FIGURE 3. 24 24
Solution: •
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The foundation is rectangular the following equations will apply: ta nφ B B 2 2 D f tan K s + γ D f qu = 1 + 0.2 cu N c + 1 + γ H 1 + L L H B 1 B qu = 1 − ( 0.4) γ BN γ + γ D f N q 2 L
For = 40o , from table 3.4, N = 109.41 and ( 400 ) ( 5.14 ) cu N c 0.5γ BN γ
=
(0.5)(117)(3)(109.41)
= 0.107
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•
From Fig. 3.25, for cuNc/0.5BN = 0.107 and = 40o , the value of K s 2.5 Above Equation gives: 2 D
B B 2 tan φ f qu = 1 + ( 0.2 ) cu N c + 1 + γ H 1 + K s + γ D f L L H B 3 3 = 1 + ( 0.2) (400)(5.14) + 1 + (117)(4) 2 4.5 4.5 (2)(3) tan 40 o × 1 + + (117)(3) (2.5) 4 3 = 2330 + 5454 + 351 = 8135 lb / ft 2
•
Again, from Eq.2
•
•
qu
=
1 B 1 − ( 0.4 ) γ BN γ + γ D f N q 2 L
For φ = 40o N q
= 64.20 (table) and 1 3 qu = 1 − ( 0.4) (117)(3)(109.41) 2 4.5 + (117)(3)(64.20) = 14081 + 22534 = 36615 lb / ft 2
Hence, qu
= 8135 lb / ft 2
Example # 24 •
A rectangular footing, with a plan area of 1.4 m x 2 m is to be placed at a depth of 2 m below the ground surface. The footing would be subjected to a load inclined at 10o to the vertical. The subsoil is clayey, sandy silt with saturated unit weight of 18 kN/m3, and c= 10 kN/m2 and = 30o. Assuming the rate of loading is such that drained condition prevails, compute the magnitude of load the footing can carry if the water table is at the base of the footing. Take FS = 3.
Solution:
qnd = cNc sc d c ic + q(Nq -1)sq d q iq + 0.5 BN s d i W Here c = c= φ 10 kN/m2, = = N ) tan = (e 45 + =18 .38 o 30 2 π tan tan φ
2
o
q
= ( N q −1) cot φ =17 .38 cot 30 o = 30 .10 Nγ = 2( N q +1) tan φ = 22 .37 N c
sc
=1 + 0.2
sq
=1 + 0.2
B L B
sγ =1 − 0.4 d c
L B
=1 + 0.2
=1.14 =1.14 = 0.72
L D f B
tan( 45 o
+φ / 2) =1 + 0.2
2 1.4
×1.732 =1.5
= d γ =1 +0.1 D f tan(
d q ic
=iq = (1 −
α 90
)2
45
+φ / 2) =1 +
0.35 1.4
=1.25
= 0.79
iγ = 0.44 W ′ = 0.5
(10 × 30 .1 ×1.14 ×1.5 × 0.79 ) + (18 × 2 × (18 .32 −1) ×1.14 ×1.25 × 0.79 ) + ( 0.5 ×18 ×1.4 × 22 .37 × 0.72 ×1.25 × 0.44 × 0.5 ) q nd
=
= ∴
406 .6 + 704 .4 + 55 .8 = 1166 .8 kN .
( q net net ) safe
=
1166 .8 / 3 = 388 388 .9 ≅ 380 kN/m 2
Hence , safe load
=
380 380 ×1.4 × 2 = 1064
≅
1060 kN
Example # 25 •
•
What will be the safe load in Example # 24 if undrained condition prevails? Take cu =30 kN/m2, u = 0, Nc = 5.14, Nq = 1, and N = 0.
Solution: qnd = cu Nc sc d s is
•
= 30 x 5.14 x 1.14 x 1.5 x
•
0.79 •
•
•
(qnet )safe
= 208.3 kN/m2 = 208.3/3 = 69.4 70 kN/m2
Safe load = 70 x 1.4 x 2 = 196 kN
TEST Bearing capacity Method 1.5 2 0.5 D = 0.0 m 0 03.5 B =60.5m 07.5 08.5 1.0 L = 2.0 m 2.0 2.0 1.0 = 15.69 16.38 17.06 17.06 kN/m = 3 35.5(36.2 38.5(40.75 38.5 )7.8 37 c =o(38.5o) 5) 3.92 7.8 6.37kPa
0 .4 0.71 0.71 17.65 22 12.75
4.5 0 0.71 0.71 17.65 25 14.7
0.0 5 0.3 0.71 0.71 0.71 0.71 17.06 17.06 20 20 9.8 9.8
Contd…
Example # 26 •
A square column foundation to be constructed on a sandy soil to carry a gross allowable total load of 150 kN. The depth of the foundation will be 0.7 m. the load will be inclined at an angle of 20 o to the vertical (Figure). The standard penetration resistances NF , obtained from field Depth(m) 1.5 3 4.5 6 7.5 9 exploration are N 3 6 9 10 10 8 F
•
•
Assume that the unit weight of the soil is 18kN/m3. Determine the width of the foundation, B . Use Eq. (3.17) and a factor of safety of 3.
20o 0. 7 m
c = 0 = 30o = 18 kN/m3 B
Example # 27 •
The subsoil at a building site consists of medium sand with = 18 kN/m3 , c = 0, = 32o and water table at the ground surface. A 2.5 m square footing is to be placed at 1.5 m below ground surface. Compute the safe bearing capacity of the footing. What would be safe bearing pressure if the water table goes down to 3 m below G.L?
Solution: •
•
Since lies between local and general shear failure conditions. Referring to Fig. 6.7, for =32 o,Nq = 28, Nq = 10, N = 30, N = 6 18( 32 − 28) = 20.3 N q = 10 + 35 − 28 24( 32 − 28) =19.7 N γ = 6 + 35 − 28
•
•
•
•
•
•
•
•
•
Case I (water table at ground level) Using Eq. (6.27) modified for square footing, ultimate gross bearing capacity (qult )gross = Df Nq + 0.4 B N = 8.0 x 1.5 x 20.3 + 0.4 x 8.0 x2.5 x 19.7 = 243.6 + 157.6 = 401.2 kN/m2. and ultimate net bearing capacity, (qult )net = (qult )gross - Df = 401.2- 18 x 1.5 = 384.2 kN/m2 Safe net bearing capacity, (qsafe )net = 384/2.5 = 152 kN/m2 (Fs = 2.5) and gross bearing capacity, (q ) = 152 + 18 x 1.5
Case II (Water table at 3 m below G.L., that is at a depth greater than width of footing)
(qult )gross = 18 x 1.5 x 20.3 + 0.4(14 x 2.5)19.7
= 548.1 + 275.8 = 823.9 kN/m 2
(qult )net = 823.9 – 18 x 1.5 = 796 .9
kN/m2
Safe net bearing capacity, (qsafe )net = 796.9/2.5 = 319
320
kN/m2
and safe gross bearing capacity, (qsafe )gross = 320 + 18 x 1.5 = 347
