Subsoil Exploration. Foundation Engineering. Solution Givens:

Size: px

Start display at page:

Download "Subsoil Exploration. Foundation Engineering. Solution Givens:"

Gyles Blake
5 years ago
Views:

1 Problems: 1. Site investigation is to be made for a structure of 100m length and 70m width. The soil profile is shown below, if the structure is subjected to 200 KN/m 2 what is the approximate depth of borehole (Assume γ w =10KN/m 3 ). γ = 18KN/m Solution Givens: q = 200KN/m, structure dimensions = (70 100)m P = 200 (100 70) = KN. D = 0.0 (Structure exist on the ground surface), γ = 18KN/m. D = 130m (distance from the lower face of structure to the bedrock). 1. Calculating the depth (D 1 ) at which σ D1 q = 200 = 20KN/m. = 1 q : 10 The following figure showing the distribution of stress under the structure at depth (D 1 ): Page (12)

2 The increase in vertical stress ( σ ) at depth (D ) is calculated as follows: σ = P A = (100 + D ) (70 + D D σ = q. ( ) ( ) = 20 D = 180 m. 2. Calculating the depth (D 2 ) at which σ σ o = The effective stress(σ ) at depth D 2 is calculated as following: σ, = (γ γ ) D σ, = (18 10) D σ, = 8D. The increase in vertical stress ( σ ) at depth (D ) is calculated as follows: σ = P A = (100 + D ) (70 + D D. = 0.05 = 0.05 (8D ( ) ( ) ) D = m So, the value of (D) is the smallest value of D 1, D 2, and D 3 D = D = m. D = D + D D = = m. Page (13)

3 2. (Mid 2005) Site investigation is to be made for a structure of 100m length and 70m width. The soil profile is shown below. Knowing that the structure exerts a uniform pressure of 200 KN/m 2 on the surface of the soil, and the load transports in the soil by 2V:1H slope. What is the approximate depth of borehole? (Assume γ w =10KN/m 3 ). γ = 17KN/m γ = 19KN/m Solution Givens: q = 200KN/m, structure dimensions = (70 100)m P = 200 (100 70) = KN. D = 0.0 (Structure exist on the ground surface). D = 130m (distance from the lower face of structure to the bedrock). 1. Check if (D 1 <30m or D 1 depth D=30 m if σ < q D < 30m, elsed > 30m Because the magnitude of ( σ ) decreased with depth. Page (14)

4 1 10 q = = 20KN/m. The following figure showing the distribution of stress under the structure at depth (30m): The increase in vertical stress ( σ ) at depth (30m) is calculated as follows: σ = P A = ( ) ( ) = KN/m. σ > 1 10 q D > 30m. 2. Calculating the depth (D 1 ) at which σ D q = = 20KN/m. = 1 q : 10 The increase in vertical stress ( σ ) at depth (D ) is calculated as follows: σ = P A = (100 + D ) (70 + D D σ = q. ( ) ( ) = 20 D = 180 m. Page (15)

5 3. Check if (D 2 <30m or D 2 depth D=30 m if σ < 0.05 D < 30m, elsed > 30m o Because the magnitude of decreased with depth. σ = KN/m (as calculated above). σ o The effective stress at depth (30m) is calculated as follows: = (γ γ ) 30 σ, σ, = (17 10) 30 σ, = 210KN/m σ σ o = = 0.51 > 0.05 D > 30m. 4. Calculating the depth (D 2 ) at which σ σ = o Let D = 30 + X (X: distance from layer (2)to reach(d ). The effective stress(σ ) at depth D 2 is calculated as following: = (17 10) 30 + (19 10) X σ, σ, = X. The increase in vertical stress ( σ ) at depth (D ) is calculated as follows: σ = P A = (100 + D ) (70 + D ), but D = 30 + X σ = (130 + X) (100 + D = 0.05 X = 69 m D = = 99 m () () = 0.05 ( X) So, the value of (D) is the smallest value of D 1, D 2, and D 3 D = D = 99 m. D = D + D D = = 99 m. Page (16)

6 3. (Mid 2013) For the soil profile shown below, if D 1 =10m and D 2 =2D 1. A- Determine the dimensions of the foundation to achieve the required depth of borehole. B- Calculate the load of column which should be applied on the foundation to meet the required depth of boring. γ = 18 KN/m γ = 22 KN/m Solution Givens: D = 10m, D = 2D D = 2 10 = 20m, D = 2m D = 40m (distance from the lower face of foundation to the bedrock) A. σ = q Page (17)

7 The following figure showing the distribution of stress under the structure at depth (D 1 =10m): The increase in vertical stress ( σ ) at depth (D = 10m) is calculated as follows: σ = = σ () () = () Eq.1 q = = ( ) q = ( ) Eq.2 By equal 1&2 () = B =4.62m. B. (P=??) D = 2D D = 2 10 = 20m, B = σ = 0.05 σ, σ = = σ () () = (.) Eq.1 Page (18)

8 The effective stress(σ ) at depth (D 2 =20m) is calculated as following: σ is calculated from the ground surface = (22 10) 10 = 336KN/m σ = 0.05 σ, P = 10,183.2 KN. P = ( ) 4. Site investigation is to be made for 2500 KN load carried on (3.0 m x 2.0 m) footing. The foundation will be built on layered soil as shown in the figure below, estimate the depth of bore hole. (Assume γ w = 10KN/m 3 ). Sand γ = 17 KN/m Sand γ = 18.5 KN/m Clay γ = 16.9 KN/m Solution Givens: P = 2500 KN, foundation dimensions = (3 2)m q = P A = = KN/m, D = 1.5m D = = 98.5m Page (19)

9 Without check, it s certainly the values of D 1 & D 2 > 3.5m, but if you don t sure you should do the check at every change in soil profile (like problem 2). 1. Calculating the depth (D 1 ) at which σ D1 q = = 41.67KN/m. = 1 q : 10 The following figure showing the distribution of stress under the foundation at depth (D 1 ): The increase in vertical stress ( σ ) at depth (D ) is calculated as follows: σ = P A = 2500 (3 + D ) (2 + D D σ = q ( ) ( ) = D = 5.26 m. 3. Calculating the depth (D 2 ) at which σ σ o = Let D = X (X: distance from the clay layer to reach(d ). The effective stress(σ ) at depth D 2 is calculated as following: = ( ) ( ) X σ, Page (20)

10 σ, σ, = X σ, = (D 3.5) = D The increase in vertical stress ( σ ) at depth (D ) is calculated as follows: σ = P A = 2500 (3 + D ) (2 + D D 2500 = 0.05 = 0.05 ( D (3+D 2 ) (2+D 2 ) 2) D = m So, the value of (D) is the smallest value of D 1, D 2, and D 3 D = D = 5.26 m. D = D + D D = = 6.76 m. Page (21)

11 Ultimate Bearing Capacity of Shallow Foundations Problems The square footing shown below must be designed to carry a 2400 KN load. Use Terzaghi s bearing capacity formula and factor of safety = 3. Determine the foundation dimension B in the following two cases: 1. The water table is at 1m below the foundation (as shown). 2. The water table rises to the ground surface kn ϕ = 32 C = 50 kn/m 2 γ d = kn/m 3 γ s = 19.5 kn/m 3 1. q u = 1.3cN c + qn q + 0.4BγN γ Solution q u = q all FS (q all = Q all, FS = 3) Area Applied load Q all Q all = 2400kN q all = Q all Area = 2400 B 2, FS = 3 q u = B 2 c = 50 kn/m 2 q(effective stress) = γ D f = = 34.5 kn/m 2 Since the width of the foundation is not known, assume d B Page (45)

12 Ultimate Bearing Capacity of Shallow Foundations γ = γ = γ + d (γ γ ) B γ = γ sat γ w = = 9.5kN/m 3, d = 3 2 = 1m 1 ( ) γ = γ = B B Assume general shear failure Note: Always we design for general shear failure (soil have a high compaction ratio) except if we can t reach high compaction, we design for local shear (medium compaction). For ϕ = 32 N c = 44.04, N q = 28.52, N γ = (Table 3. 1) Now substitute from all above factors on terzaghi equation: 7200 B 2 = B ( B ) = B B 2 Multiply both sides by (B 2 ) B B = 0.0 B = 1.33m. 2. All factors remain unchanged except q and γ: q(effective stress) = ( ) 2 = 19 kn/m 2 γ = γ = = 9.5 kn/m 3 Substitute in terzaghi equation: 7200 B 2 = B B 2 = B Multiply both sides by (B 2 ) B B = 0.0 B = 1.42m. Note that as the water table elevation increase the required width (B) will also increase to maintain the factor of safety (3). Page (46)

13 Ultimate Bearing Capacity of Shallow Foundations Determine the size of square footing to carry net allowable load of 295 KN. FS=3. Use Terzaghi equation assuming general shear failure. ϕ = 35 C = 0.0 γ d = kn/m 3 Solution Q all,net = 295 kn and we know q all,net = Q all,net Area q all,net = 295 B 2 Also, q all,net = q u q FS q(effective stress) = γ D f = = kn/m 2, FS = B 2 = q u ϕ = 25 C = 50 kn/m 2 γ d = 20 kn/m 3 q u = (1) B2 q u = 1.3cN c + qn q + 0.4BγN γ c = 50 kn/m 2 q(effective stress) = kn/m 2 γ = 20 kn/m 3 (for underlying soil) For ϕ = 25 N c = 25.13, N q = 12.72, N γ = 8.34 (Table 3. 1) Substitute from all above factor in Terzaghi equation: Page (47)

14 Ultimate Bearing Capacity of Shallow Foundations q u = B q u = B Substitute from Eq. (1): = B B2 Multiply both side by B 2 : B B = 0.0 B = 0.68 m. Page (48)

Chapter (9) Sheet Pile Walls

Chapter (9) Sheet Pile Walls Chapter (9) Introduction Sheet piles are a temporary structures used to retain a soil or water for a specific period of time, to build a structure in the other side of this wall. For example; if we want