m v = 1.04 x 10-4 m 2 /kn, C v = 1.29 x 10-2 cm 2 /min

Transcription

1 2.10 Problems Example 2.1: Design of Shallow Foundation in Saturated Clay Design a square footing to support a column load of 667 kn. The base of the footing will be located 1 m below the ground level and the soil is saturated clay having the following properties: C u = 53.0 kpa, = 0, γ = 17.3 kn/m 3 m v = 1.04 x 10-4 m 2 /kn, C v = 1.29 x 10-2 cm 2 /min i. Determine the safe bearing capacity of the clay and hence obtain a suitable sized footing. ii. If the clay extends 5 m below the base of the footing, calculate the oedometer settlement assuming that the load is distributed at 2 to 1 and that m v is constant over the stratum. (see note below.) iii. When is the oedometer settlement 75% complete? Assume that the initial pore pressure is uniform over the stratum and that the clay is underlain by an incompressible pervious layer. 1

2 2.1a Fig Ex a) Skempton s N c values depend upon. Hence the footing size will have to be determined by trial and error. Try B = 2.4 m = 0.42 N c = 7.0 (from Graph Fig Ex 2.1a) Net Net = 123 kpa The area of footing required = = 5.42m 2 A 2.4m square footing is satisfactory in bearing. 2

3 b) The solution given below is intended to illustrate the basic approach to settlement calculation. There are however several alternative methods in common use. Divide layer into 5-1 m strips. The corresponding oedometer settlement calculations are given below. 2.4 m 667 kn 1 m 6 m Fig E 2.1b Layer number Mean Depth m Area Mid Section m 2 due to Load kpa

4 Settlement = 1.93 cm (0.76 ins.) The oedometer settlement is 75% complete when the time factor T The drainage path, H, cannot be defined precisely in the above problem. The minimum time for 75% consolidation would be given by assuming H = 2.5 m. T = 0.42 = 0.48 = t= Example 2.2: Check of Transcona Grain Elevator Failure (field data approx. only) Problem:For the soil and building details given in Fig Ex. 2.2 determine Factor of Safety against bearing capacity failure. i. Using Skempton s formula. ii. Using circular arc method. Dead load = 20,000 tons Live load = 26,000 tons Total = 46,000 tons Q gross applied = = 3.06 tsf 4

5 77' x 195' O q gross R = 95' approx. 12' 7' E 30' A B = 103 pcf Cu= 0.5tsf 20' C Cu= 0.3tsf D Fig Ex. 2.2 Method (i) (Using average value of C u = 0.4 tsf) (q gross ult = = 2.82 tsf) Net applied building load = 3.6 soil removed. F.S. 5

6 and failure would (did) occur. [i.e., Maximum allowable building load should have been limited to (for F.S. = 3)] Method (ii) Assuming the centre of rotation of the probable failure arc to be located at 0 with R = 95 ft. approx., Then AC = 25 ; CD = 127 ; DE = 40 (for strip footing loading) (216 kpa) For the grain elevator, , use shape factor Therefore This value is to be compared with using Skempton s formula. The difference is mainly due to the greater distance of the failure arc located in the soft stratum. Example 2.3: Given: A square footing as shown in the Fig Ex. 2.3a is to carry the loading indicated. The water table will be temporarily lowered during construction. The average S.P.T. N value is 25 as measured in the field. of the sand is 18.0 kn/m 3 and is 11.0 kn/m 3. Determine: The size of a square footing to ensure that the settlement does not exceed 25 mm. 6

7 D.L. = 356 kn L.L. = 712 kn 1.80 m 1.52 m B Fig Ex. 2.3a Fig Ex. 2.3b: Chart for correction of N-values in sand for influence of overburden pressure Solution: Neglect the difference between the unit weights of soil and concrete and compute the original value of effective vertical stresses at a depth of 0.5B below the base of the footing. Assume B = 2m. Then, 7

8 Also, 25 mm = = Area of footing required = = or B = 2.16m (If the original assumption regarding the value of B was not close another iteration would be required) Example 2.4: Immediate Settlement Calculation of Foundation on Saturated Clay Given: The continuous wall footing, shown in the Fig Ex.2.4a, is 17m long and is founded on a deposit of stiff saturated clay as shown. Determine: The immediate settlement of the footing. 8

9 255 kn/m 1 m 0.3 m 0.3 m B= 1.7 m 17 m Clay: s = 20 kn/m3 Eu = kn/m2 concrete = 23.6 kn/m 3 = 0.5 Fig Ex.2.4a L = length D q = 0.5 B 1.0 H = Average settlement 0 1 (qb D/B 9

10 Fig Ex.2.4b: Values of and for settlement calculations Solution: The net pressure applied by the footing is the pressure applied by the footing in excess of the pressure that existed at the depth of the footing before the footing was constructed. i.e. kn/m (Usually the difference between the unit weights of soils and concrete is neglected) Now = 1.3 and = 1.0 (from Fig Ex.2.4b) Then, 10

11 Example 2.5: Given: (i) the continuous wall footing, shown in the Fig Ex.2.5, is 17m long and is founded on a deposit of stiff saturated clay as shown. (ii) the soil is sand with a value of = 36 (iii) the depth to the water table = 2m (iii) kn/m 3 ; kn/m 3 Determine: Factor of Safety against bearing capacity failure. 255 kn/m 1 m 2 m 0.3 m 0.3 m B= 1.7 m 17 m N.T.S Fig Ex.2.5a 11

12 Fig Ex.2.5b Solution: Using the following equation, And from the Fig Ex.2.5a, ; Where, = = 14.7 kn/m 3 12

13 Also, F.S. = 13