Question

A circular footing carries an ultimate load of 2125 kN in cohesionless soil. The soil
has an angle of shearing resistance o = 36° and bulk unit weight 18 kN/m, and
saturated unit weight 20 kN/m². The depth of footing is 1.5 m. Determine the
diameter of footing. Water table is at a depth of 1 m below the ground surface. No =
49.38, Ny = 54. Use Terzaghi's theory. Assume yw = 10 kN/m
a) 1.41 m
b) 1.67 m
c) 1.24 m
d) 2.14 m
A
B​

Answer 1

Answer:

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.  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. 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. 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. Now actual footing load = qa = 400 kN/m²  F.S = qf/qc = 1546.9/400 3.87

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. 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. 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. 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