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Numerical problem on bearing capacity is code terzaghi water table (usefulsearch.org) (useful search)

Numerical problem on bearing capacity is code terzaghi water table (usefulsearch.org) (useful search)

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Numerical problem on bearing capacity is code terzaghi water table (usefulsearch.org) (useful search)

  1. 1. Q.No.1- A square footing of 2.5m by 2.5m is built in a homogeneous bed of sand of unit weight of 20kN/m3 aand an angle of shearing resistance of 36°. The depth of the base of the footing is 1.5m below the ground surface. Calculate the safe load that can be carried by a footing with a f.o.s of 3 against complete shear failure. Use Terzaghi’s analysis. Solution- Given that, B=2.5m D= 1.5m y= 20kN/m3 Ø= 36°, Since the soil is dense, the footing is likely to fail by general shear failure. From table, the values of bearing capacty factors are: Nc= 65.4, Nq= 49.4, N= 54 Also,  = D Since, c= 0, qf =  D Nq + .04 B N Or, qnf =  D (Nq – 1) + 0.4  B N  = 20 * 1.5(49.4 - 1) + 0.4 * 20 * 2.5 * 54 = 2532 kN/m3
  2. 2.  Qs = qnf/F +  D = 2532/3 + (20 * 1.5) = 874 kN/m2 Maximum safe load = B² * qs = (2.5)² * 874 = 5462.5 kN
  3. 3. Q.No 2- A strip footing, 1m wide at its base is located at a depth of 0.8m below the ground surface. The properties of the foundation soil are:  = 18kN/m³, c = 30kN/m² and ø = 20°. Determine the safe bearing capacity, using a factor of safety of 3. Solution- The bearing capacity is given by- qf = 2/3 cNc’ +  D Nq’ + 0.5  B N’ Taking  =  D, For ø = 20°, the bearing capacity factor taken from table are- Nc’ = 11.8 ; Nq’ = 3.9 ; N’ = 1.7 qf = (2/3 * 30 * 11.8) + (18 * 0.8 * 3.9) + (0.5 * 18 * 1 * 1.7) =307.5kN/m² qnf = qf -  d = 307.5 – 18 * 0.8 = 293.1 kN/m² qs = qnf/F +  D = 293.1/3 + (18 * 0.8) = 112.1 kN/m²
  4. 4. Q.No.3- A strip footing 2m wide carries a load intensty of 400 kN/m² at adepth of 1.2m in sand. The saturated unit weight of the sand is 19.5 kN/m³ and unit weight above the water table is 16.8 kN/m³. the shear strength parameters are c = 0 and ø = 35°. Determine the factor of safety with respect to shear failure for the water table 4m below G.L. For a strip footing the bearing capacity is given by- qf = cNc +  Nq + ½ B  N Taking into account the water reduction factor, we have qf = cNc +  Nq + ½ B  N. Rw2 For the present case, c = 0  qf = 41.4 * 1.2 * .Rw1 + ½ * 2 * 42.2  Rw2 qf = 49.68 . Rw1 + 42.3 . Rw2 Zw2 = 4- 1.2 = 2.8m Rw1 = 1 since Zw2 > B, Rw2=1 Hence there will be no effect of water table qf = 49.68 * 16.8 * 1 + 42.4 * 16.8 * 1 =1546.9 kN/m²
  5. 5. Now actual footing load = qa = 400 kN/m²  F.S = qf/qc = 1546.9/400 3.87
  6. 6. Q.No.4- Design a strip footing to carry a load of 750 kN/m at a depth fo 1.6m in a c-ø soil having a unit weight of 18 kN/m³ and shear strength parameters as c = 20 kN/m³ and ø = 25°. Determine the width of footing, using a factor of safety of 3 against shear failure. solution- assume general shear failure, qf = cNc + Nq + 0.5  B N for ø = 25°, we have, Nc = 25.1 ; Nq = 12.7 and N = 9.7 also, we have, = 18 * 16 substituting the values, we get qf = (20 * 25.1) + (18 * 1.6) * 12.7 + 0.5 * 18 * B * 9.7 qf = 867.8 + 87.3 B  intensity of pressure, at F.S. of 3 at footing level = qf/3 = (867.8 + 87.3 B)/3 kN/m²  =  D,
  7. 7. Applied load intensity = 750/(B * 1) = 750/B kN/m² Equating the two, we get 750/B = (867. + 87.3 B)/3 87.3B² + 867.8 B – 2250 = 0 B² + 9.94 B – 25.77 = 0 From which we get, B = 2.134 m
  8. 8. Q5-Determine the ultimate bearing capacity of the footing in last question if the ground water table is located a)at a depth of 0.5 m below the ground surface , b)at a depth of 0.5 m below the base of the footing ᵞsat=20KN/mᶟ . Use Terzaghi’s theory. Sol- qᵤ=q Nq+0.5 ᵞ B Ny (where q is the effective surcharge and ᵞ is the effective unit weight of the soil beneath the footing. a) Q is calculated by assuming the unit wt. to the top 0.5m soil to be unchanged , i.e.=17kN/mᶟ and submerged unit wt. of rest of the soil=10 (20-10) q=0.5x17+.5x10 = 13.5kN/m² y=y’ so qu=13.5x60+.5x10x1.5x75 = 1372.5kN/m²
  9. 9. b)For this case , q=17kN/m² In the term 0.5yBNy , y is given by y = y’+(Dw’) (Yt-Y’) B Dw’=0.5m , Yt=17kN/mᶟ and Y’=10kN/mᶟ Y=10+0.5(17-10)=12.3kN/mᶟ qᵤ=17x60+0.5x12.3x1.5x75 =1711.9kN/m²
  10. 10. Q6-Determine the ultimate bearing capacity of a strip footing 2m in width , with its base at a depth of 1.5m below ground surface and resting on a saturated clay soil with the following properties. ᵞsat=20kN/mᶟ , cᵤ=40kN/m²,φᵤ=0, C’=10kN/m²,φ’=20⁰.The natural water table is at a 1m depth below ground level . Ignore depth factor. Sol- a) undrained condition φᵤ=0 Hence qᵤ=Cu Nc + q since Nc=1 and Ny=) in Terzaghi’s ,Hansen’s etc recommendations. q = 20x1+10x0.5=25kN/m² , Cu=40kN/m² qᵤ=40x5.7+25=253kN/m² (Terzaghi) qu=40x5.14+25=231kN/m² (Meyeroff and Hansen’s)
  11. 11. Drained conditions- qu=c’Nc+qNq+0.5yBNy q=25kN/m²,y=y’=10kN/m² Appropriate Nc , Nq , Ny values are substituted for the different cases for φ’=20⁰ qu=10x17.7+25x7.4+0.5x10x2x5 = 412kN/m² (Terzaghi) qu=10x14.8+25x6.4+0.5x10X2x2.9 = 337kN/m² ( Mayerhof and Hansen) qu =10x14.8+25x6.4+0.5x10x2x5.4 = 362kN/m² (IS Code) Drained condition > undrained condition
  12. 12. Q7- Calculate the net ultimate bearing capacity of a rectangular footing 2mX4m in plan , founded at a depth of 1.5 m below the ground surface. The load on the footing acts at an angle of 15⁰ to the vertical and is eccentric in the direction of width by 15cm. The saturated unit weight of the soil is 18kN/mᶟ . The rate of loading is slow and hence the effective stress shear strength parameters can be used in the analysis . c’=15kN/m² and φ’=25⁰.Natural water table is at a depth of 2m below the ground surface. Use IS-Code recommendations. Sol- qv =c Nc Sc dc ic+q(Nq-1)sq dq iq+0.5 Y B Ny Sy dy iy W’ C = c’= 15kN/m² φ = φ’ = 25⁰, Nc=20.7,Nq=10.7 and Ny=10.9 Y = Ysat = 18kN/mᶟ q=18x1.5 = 27 kN/m² For Dw’ = 0.25 (Dw’=2-1.5=0.5m), W’=0.625 B
  13. 13. Sc = Sq = 1+0.2 B’ =1+0.2x (1.7/4) = 1.025 L Sy = 1- 0.4 B’ = 1- 0.4 (1.7/4) = 0.83 L dc = 1+ 0.2 ( Df/B’) tan (45⁰+ φ/ 2) = 1+0.2(1.5/1.7)tan 57.5⁰ = 1.28 dq = dy = 1 +0.1 (Df/B’) tan (45⁰ + φ /2) = 1+0.1(1.5/1.7) tan 57.5⁰ = 1.14 ic = iq = (1-ᾳ/φ)² = (1-15/90)² = 0.69 iy = (1-ᾳ/φ)² = (1-15/25)² = 0.16 Substituting these values , qnu = 15 X 20.7 X1.085 X 1.28 X 0.69 + 27 X9.7 X 1.085 X 1.14 X 0.69 + 0.5 X 18 X 1.7 X 10.9 X 0.83 X 1.14 X 0.16 X 0.625 = 297.5 + 223.5 +15.8 = 536.8 kN /m²

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