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Questions on Bearing Capacity (Terzaghi’s Theory) http://usefulsearch.org User friendly site(usefulsearch.org) (useful search)

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- 1. QUESTIONS ON BEARING CAPACITY usefulsearch.org (user friendly site for new internet user)
- 2. T E R Z A G H I ’ S E Q U A T I O N F O R G E N E R A L S H E A R F A I L U R E Q1- A strip footing, 1.5 m wide at its base is located at a depth of 1.30 m below the G.L. The properties of the foundation soil are Y = 21.5 kN/m3, φ = 34o. Determine the safe bearing capacity, using a factor of safety of 3. Use Terzaghi’s analysis. Solution- Given- A strip footing, B = 1.5 m, D = 1.30 m, Y = 21.5 kN/m3, φ = 34o, F.O.S. = 3 As we know when φ > 28o, soil generally fails due to General shear failure and therefore, Qu = cNc + YDNq + 0.5YBNY And in sandy soil c = 0, For φ = 34o, Nq = 36.5 & NY = 35 Qu = 0 + 21.5x1.3x36.5 + 0.5x21.5x1.5x35 => Qu = 0 + 1020.175 + 564.376 = 1584.55 kN/m2 Then, Qnu = Qu – YD => Qnu = 1584.55 - (21.5x1.3) = 1556.60 kN/m2 Qsafe = (Qnu/F.O.S.) + YD => Qsafe = 1556.60/3 + (21.5x1.3) = 546.81 KN/m2
- 3. Q2- A strip footing of width 1.2 m and depth below the ground surface is 1 m. The properties of the foundation soil are Y = 19 kN/m3, c = 30 kN/m2 and φ = 20o. Determine the safe bearing capacity, using suitable factor of safety. Solution- Given- A strip footing, B = 1.2 m, D = 1 m, Y = 19 kN/m3, φ = 200, c = 30 kN/m2 As we know when φ < 280, soil generally fails due to Local shear failure and therefore, Qu = (2/3)cNc + YDNq + 0.5YBNY For φ = 200, Nc = 11.8, Nq = 3.9 & NY = 1.70 Qu = (2/3)x30x11.8 +19x1x3.9 + 0.5x19x1.2x1.7 => Qu = 236 + 74.1 + 19.38 = 329.48 kN/m2 Then, Qnu = Qu - YD => Qnu = 329.48 - (19x1) = 310.48 kN/m2 Qsafe = Qnu/F.O.S. + YD => Qsafe = 310.48/3 + (19x1) = 122.49 kN/m2 T E R Z A G H I ’ S E Q U A T I O N F O R L O C A L S H E A R F A I L U R E
- 4. Q3- Determine the allowable gross load and the net allowable load for a square footing of 2 m side and with a depth of foundation of 1 m. Use Terzaghi’s theory and assume local shear failure. Take a factor of safety of 3. The soil at the site has Y = 18.0 kN/m3, c = 18 kN/m2 and φ = 250. Solution- Given- A strip footing, Size = 2x2 m, D = 1 m, Y = 18 kN/m3, φ = 250, c = 18 kN/m2, F.O.S. = 3 Here, cm = (2/3)c Qu = 1.3cmNc + YDNq + 0.4YBNY For φ = 250, Nc = 14.8, Nq = 5.6 & NY = 3.2 Qu = 1.3x12x14.8 + 18x1x5.6 + 0.4x18x2x3.2 => Qu = 230.88 + 100.8 + 46.08 = 377.76 kN/m2 Then, Qnu = Qu – YD => Qnu = 377.76 - (18x1) = 359.76 kN/m2 Qsafe = Qnu/FOS + YD => Qsafe = 359.76/3 + (18x1) = 137.92 kN/m2 Gross allowable load = QsafexArea of footing = 137.92x(2x2) = 551.68 kN T E R Z A G H I ’ S E Q U A T I O N F O R L O C A L S H E A R F A I L U R E I N S Q U A R E F O O T I N G
- 5. Q4- A rectangular footing 2 m by 3.5 m rests on soil at 1.7 m below the ground surface. Calculate the ultimate bearing capacity. The soil parameters Y = 18 kN/m3, c = 10 kN/m2 and φ = 300. Solution- Given- A rectangular footing, Size = 2x3.5 m, D = 1.7 m, Y = 18 kN/m3, φ = 300, c = 10 kN/m2, For φ = 30o, Nc = 37.2 , Nq = 22.5 & NY= 19.7 Qu = cNc{1 + 0.3(B/L)} + YDNq + 0.5YBNγ{1 - 0.2(B/L)} => Qu = 10x37.2{1 + 0.3(2/3.5)} + YDNq + 0.5x18x2x19.7{1 - 0.2(2/3.5)} => Qu = 435.77 + 688.5 + 314.07 = 1438.34 kN/m2 T E R Z A G H I ’ S E Q U A T I O N F O R R E C T A N G U L A R F O O T I N G
- 6. Q5- A square footing 2x2 m is located at a depth of 1 m below the ground surface. The soil has the following properties Y = 18 kN/m3, c = 19 kN/m2 and φ = 200. Using Meherhof ‘s equation. Calculate the ultimate bearing capacity. Solution- Given- A square footing, Size = 2x2 m, D = 1 m, Y = 18 kN/m3, φ = 200, c = 19 kN/m2, For φ = 200, Nc = 14.83, Nq = 6.4 & NY = 2.9 Qu = cNcScDcIc + YDNqSqDqIq + 0.5YBNγSγDγIγ Sc = 1 + 0.2tan2(45 + φ/2)(B/L) => Sc = 1 + 0.2tan2(45 + 20/2)(2/2) = 1.4 Sq = SY = 1 + 0.1tan2(45 + φ/2)(B/L) => Sq = SY = 1 + 0.1tan2(45 + 20/2)(2/2) = 1.2 Dc = 1 + 0.2tan(45 + φ/2)(D/L) => Dc = 1 + 0.2tan(45 + 20/2)(1/2) = 1.14 Dq = DY = 1 + 0.1tan(45 + φ/2)(D/B) => Dq = DY = 1+0.1 x tan(45+20/2) (1/2)=1.07 Here, Ic = Iq = Iγ = 1 (since loading is vertical) Qu = 19x14.83x1.4x1.14x1 + 18x1x6.4x1.2x1.07x1 + 0.5x18x1x2.9x1.2x1.07x1 Qu = 449.7 + 147.91 + 33.51 = 631.12 kN/m2 M E Y E R H O F ’ S E Q U A T I O N F O R S Q U A R E F O O T I N G
- 7. T E R Z A G H I ’ S E Q U A T I O N F O R W A T E R T A B L E E F F E C T Q6- A square footing, 3x3 m, is made on cohesion less soil, 3 m below the ground level. Water table is situated at 1 m below the ground level. Find out the allowable bearing capacity of soil if F.O.S. = 3, Ydry = 9 kN/m3, Ysaturated = 16 kN/m3 & φ = 300. Solution- From Table, Nc = 37.2, Nq = 22.5, NY = 19.7 for φ = 300, c = 0 (since soil is cohesion less) Effect of Water Table:- Rw1 = 0.5(1 + Zw1/D) = 0.5(1 + 1/3) = 0.67 Rw2 = 0.5(1 + Zw2/B) = 0.5(1 + 3/3) = 1 Determine ultimate soil bearing capacity using Terzaghi’s bearing capacity equation for strip footing. Qu = 1.3cNc + (YdryD1 + YsaturatedD2)NqRw1 + 0.4YsaturatedBNγ Rw2 => Qu = 0x37.2 + (9x1 + 16x2)x22.5x0.67 + 0.4x16x3x19.7x1 => Qu = 996.315 kN/m2 Allowable soil bearing capacity, Qa = Qu/F.O.S. = 996.315/3 = 332.105 kN/m2

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