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# Question and Answers on Terzaghi’s Bearing Capacity Theory (usefulsearch.org) (useful search)

Question and Answers on Terzaghi’s Bearing Capacity Theory (usefulsearch.org) (useful search)

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### Question and Answers on Terzaghi’s Bearing Capacity Theory (usefulsearch.org) (useful search)

1. 1. A POWERPOINT PRESENTATION ON PREVIOUS YEAR SOLVED QUISTION PAPER Question on Bearing Capacity Usefulsearch.org
2. 2. Q.1 A strip footing 1m wide is laid at a depth of 2m in c-ϕ soil having the following characteristics. calculate the ultimate bearing capacity qf by terzaghi analysis, take properties of soil c=20kN/m²,ϓ=18kN/m³and ϕ=28º give terzaghi bearing capacity factors (ϕ=28º) as Nc=32,Nq=18,Ny=16. (june 2010) sol: given: b=1m d=2m ϓ=18kn/m² ϕ=28º c=20kn/m² Nc=32,Nq=18,Ny=16 Since ϕ=28º, footing is likely to fall by general shear failure. Hence,ultimate bearing capacity
3. 3. qf=cNc +ϓDNq+0.5ϓBNy =20X32+18X2X18+0.5X18X1X16 =1432kN/m² qnf=qf-ϓD = 1432-18*2 =1396kN/m² if factor of safety F=3 qs=qnf/F+ ϓD =1396/3+18X2 qs=501.33kN/m²
4. 4. Q2. A strip footing is to be designed to carry a total load of 1000 kN/m at a depth of 1 m in sandy soil. The soil has C=0, ϕ=36º . Unit weight of soil below water table = 20kN/m³ Unit weight of soil above water table = 18kN/m³ Groundwater table exist at foundation level. Given for ϕ=36º Nq=47, Ny=43. Assume unit weight weight of water = 10 kN/m³ factor of safety = 3. ( june 2008 ) Solⁿ: Given : load P = 1000kN/m Depth D =1m Cohesion C = 0 ( for sandy soil) Φ = 36° Net ultimate stress qnf=cNc+Rw1 (Nq-1)σ +0.5 Rw2 Ny bΎ Rw1= 0.5(1+Zw1/D) Rw2= 0.5(1+Zw2/B)
5. 5. here Ground water table is at foundation level Hence Zw1=D=1m Zw2=0 B Then Rw1=1 D Zw1 Rw2=0.5 B Zw2 qnf= 0+(47-1)18+0.5*0.5*43*20* B = 828+215B qs = qnf/3+ΎD = (828+215B)/3 +18 = 294+71.6B Safe stress= load/area = 1000/B Hence 1000/B = 294+71.6B B = 2.2m
6. 6. Q3. Calculate the depth at which the footing (2mX2m) should be placed to transfer total load of 200 tons with a factor of safety3. The soil is sandy having Φ = 30° and unit weight 2 gms/cm³. Ground Water level is too deep. Given Nq=22 and Ny=20 for Φ = 30° . ( Dec 2008 , June 2011 , June 2012 ) Solⁿ: Given : load P = 200 tons = 200 x 1000 kg Depth D =? Cohesion C = 0 ( for sandy soil) Φ = 30° Sy =0.8 (for square footing) Ύ = 2 gms/cm³ = 2000 kg/m³ Net ultimate stress qnf=cNc+ (Nq-1)σ +0.5 Sy Ny bΎ = 0+ (22 -1)X 2000XD + 0.5X0.8X20X2X2000 =42000D + 32000
7. 7. safe bearing capacity(kg/m²) = qns = qnf/3 +ΎD = (42000D +32000)/3 + 2000D = 16000D +10666.6 Safe stress( kg/m²) = load / area = 200X 1000/ 2X2 = 50000 Hence 50000 = 16000D+10666.6 D = 2.458 m
8. 8. Q.4 A r.c.c. column has a square footing founded at a 3.0 m depth below the surface of clay soil of average density of18 kN/m³ and a shear strength of 80 kN/m² . The wall cohesion on the vertical sides of the footing which is acting over the lower half portion only may be taken as one half of the shear strenth . The total load applied to the soil is 2000 kN . Calculate the size of footing for a safety factor of 3. (june 2007) sol: given: b=? d=3m ϓ=18 kN/m³ ϕ=0 c=(80/2)= 40kN/m² Nc=5.7, Nq=1, Ny=0 ( for ϕ=0 ) qnf=cNc+ (Nq-1)σ +0.5 Sy Ny bΎ = 40X5.7+0+0 =228 kN/m²
9. 9. safe bearing capacity(kN/m²) = qns = qnf/3 +ΎD = 228/3 + 18X 3 = 120(kN/m²) Safe stress( kg/m²) = load / area = 2000/B² hence 2000/B² = 120 B = 4.08 m
10. 10. Q.5 A strip footing 1.5 m wide at its base is located at a depth of 1 m below the ground surface. The foundation level are Ύ = 18 kN/m³, C = 30 kN/m² , Φ =20°. Determine the safe bearing capacity using a factor of safety of 2.5. Use Terzaghi analysis and assume soil fails by local shear.[ Given for Φ =20°. bearig capacity factors Nc’ = 11.8, Nq’ = 3.9, Ny’ = 1.7 ]. (June 2009 ) sol:given: b=1.5m d=1m ϓ=18kn/m² ϕ=20º c=30kn/m² Nc’=11.8, Nq’=3.9, Ny’=1.7 Since ϕ=28º, footing is likely to fail by local shear failure. Hence,ultimate bearing capacity
11. 11. qnf=(2/3)cNc’+ (Nq’-1)σ +0.5 Ny’ bΎ = (2/3) 30X11.8 + (3.9-1) X 18 X1 + 0.5X 1.7 X 1.5 X18 = 311.15 kn/m² qs = qnf /2.5 + ΎD = 311.15/2.5 +18X1 qs = 142.46 kn/m²
12. 12. Q6. Calculate the net ultimate bearing capacity of a rectangular footing 2 m X 4m in plan founded at a depth of 1.5m below the ground surface. The load on the footing acts at an angle of 15º to the vertical and is eccentric in the direction of width by 15 cm. The saturated unit weight of soil is18 kN/m³ . The rate of loading is slow and hence the effective stress shear strength parameters can be used in the analysis . C’ = 15 kN/m² and ϕ’=25º . Natural water table is at a depth of 2 m below the ground surface. ( for ϕ’=25º ,Nc’=20.7, Nq’=10.7, Ny’=10.9) Use IS : 6403 – 1981 recommendations. sol:given: b X L=2m x 4m d=1.5m ϓ= ϓsat= 18 kN/m³ ϕ =ϕ’=25º c=c’=15 kN/m²
13. 13. Nc’=20.7, Nq’=10.7, Ny’=10.9 Since ϕ=28º, footing is likely to fail by local shear failure. Hence,ultimate bearing capacity qnf = c Nc Sc dc ic + σ ( Nq -1) Sq dq iq + 0.5 Ύ B Ny Sy dy iy W’ for D’w/B = 0.25 (D’w=2 – 1.5 = 0.5m), W’ = 0.625 eₓ = 0.15m ; effective width B’ = B -2 eₓ = 2-0.3 = 1.7m Sc = Sq = 1 + 0.2B’/L = 1+ 0.2(1.7/4) = 1.025 Sy = 1 - 0.4B’/L = 1 - 0.4(1.7/4) = 0.83 dc = 1 + 0.2 (Df/B’)tan(45º+ ϕ/2) = 1.28 dq = dy = 1 + 0.1 (Df/B’)tan(45º+ ϕ/2) = 1.14
14. 14. ic = iq = ( 1 – α/90 )² = ( 1- 15/90 )² = 0.69 iy = ( 1 – α/ ϕ )² = ( 1- 15/25 )² = 0.16 substituting these values qnf = 15 X 20.7 X 1.085 X 1.28 X0.69 + 27 X 9.7 X 1.085 X 1.14X 0.69 + 0.5 X 18 X 1.7X 10.9 X 0.83 X 1.14 X0.16 X0.625 qnf = 536.8 kN/m²