Chapter (4) Ultimate Bearing Capacity of Shallow Foundations (Special Cases)

Transcription

1 Chapter (4) Ultimate earing Capacity of Shallow Foundations (Special Cases)

2 Ultimate.C. of Shallow Foundations (Special Cases) Introduction The ultimate bearing capacity theories discussed in Chapter 3 assumed that the soil supporting the foundation is homogeneous (i.e. one layer) and extends to a great depth below the bottom of the foundation. They also assume that the ground surface is horizontal. However, that is not true in all cases: It is possible to encounter a soil may be layered and have different shear strength parameters, and in some cases it may be necessary to construct foundations on or near a slope. All of above cases are special cases from Chapter 3, and will be discussed in this Chapter. earing Capacity of Layered Soils: Stronger soil Underlain by Weaker Soil The bearing capacity equations presented in Chapter 3 involved cases in which the soil supporting the foundation is homogeneous and extend to a great depth (i.e. the cohesion, angle of friction, and unit weight of soil were assumed to remain constant for the bearing capacity analysis). However, in practice, layered soil profiles are often encountered (more than one layer). In such instances, the failure surface at ultimate load may extend in two or more soil layers. This section features the procedures for estimating the bearing capacity for layered soils (stronger soil layer, underlain by a weaker soil layer that extends to a great depth). Notes: 1. Always the factors of top soil are termed by (1) and factors of bottom soil are termed by (2) as shown in the following table: Soil Properties Layer Unit Friction weight angle Cohesion Top γ ϕ c ottom γ ϕ c 2. The equation will be derived for continuous or strip footing and then will be modified to be valid for rectangular, square, and circular footings. Page (67)

3 Ultimate.C. of Shallow Foundations (Special Cases) Let the depth,h, is the distance from the bottom of the foundation to the top of weaker soil (bottom soil layer) and,, is the width of continuous or strip footing (i.e. equation will be derived for continuous footing), the failure surface in layered soil below the foundation may have two cases: Case I: If the depth H is relatively small compared with the foundation width (upper layer can t resist overall failure due to its small thickness), a punching shear failure will occur in the top soil layer, followed by a general shear failure in the bottom soil layer (due to its large extend downward), so the ultimate bearing capacity in this case will equal the ultimate bearing capacity of bottom layer (because general shear failure occur on it) in addition to punching shear resistance from top layer. q = q + Punching shear resistance from top layer (q ) q can be calculated as following (see the above figure): q = (2C + 2P sin δ) 1 H C = adhesive force (between concrete and soil) C = c H c = adhesion between concrete and soil along the thickness H Page (68)

4 Ultimate.C. of Shallow Foundations (Special Cases) δ = inclination of the passive, P, force with the horizontal P = passive force per unit length along the thickness H applied from soil to the foundation and can be calculated as following: P = 1 H vertical effective stress K 2 1 H vertical effective stress = area of the vertical pressure diagram 2 vertical effective stress = γ H K = Kcosδ K = K cosδ K = Coefficient used to transform vertical pressure to the direction of passive force K = Horizontal component of passive earth pressure coefficient(k) Now the equation of P will be: P = 1 2 H (γ H) K = 1 2 γ H K cosδ Now substitute in equation ofq : q = 2c H q = 2c H γ H K sin δ cosδ H + γ H K tanδ H Now correction for depth factors (according Terzaghi assumption) should be established. This modification will be in punching shear term as following: q = 2c H + γ H 1 + 2D H K tanδ H From several experiments, investigators found that K tanδ = K tanϕ K = Punching shear coefficient q = 2c H + γ H 1 + 2D H K tanϕ H Now substitute in equation of (q ): q = q + 2c H + γ H 1 + 2D H K tanϕ H Page (69)

5 Ultimate.C. of Shallow Foundations (Special Cases) Case II: If the depth,h, is relatively large (thickness off top layer is large), then the failure surface will be completely located in the top soil layer and the ultimate bearing capacity for this case will be the ultimate bearing capacity for top layer alone (q ). q = q = c N () + q N () + 0.5γ N () N (), N (), N () = Meyerhof bearing capacity factors (for ϕ )(Table3. 3) All depth factors will equal (1) because their considered in punching term. All shape factors will equal (1) because strip or continuous footing. Assume no inclination so, all inclination factors equal (1). Combination of two cases: As mentioned above, the value of q is the maximum value of q can be reached, so it should be an upper limit for equation of q : q = q + 2c H + γ H 1 + 2D H K tanϕ H q The above equation is the derived equation for strip or continuous footing, but if the foundation is square, circular and rectangular the equation will be modified to be general equation for all shapes of footings: Page (70)

6 Ultimate.C. of Shallow Foundations (Special Cases) q = q γ H 1 + L 1 + 2D H K tanϕ H q q = c N () F () + q N () F () + 0.5γ N () F () q = effective stress at the top of layer(1) = γ D q = c N () F () + q N () F () + 0.5γ N () F () q = effective stress at the top of layer(2) = γ (D + H) All depth factors will equal (1) because their considered in punching term. Assume no inclination so, all inclination factors equal (1). Note: All factors and equations mentioned above are based on Meyerhof theory discussed in Chapter 3. All of above factors are known except K andc K = f q q, ϕ and c c = f q q to find K andc : q q must be known. Calculating of q andq is based on the following three main assumptions: 1. The foundation is always strip foundation even if it s not strip 2. The foundation exists on the ground surface (D = 0.0) and the second term on equation will be terminated. 3. In calculating q we assume the top layer only exists below the foundation to a great depth, and the same in calculating of q. q = c N () + 0.5γ N () q = c N () + 0.5γ N () = Calculating of K : K = f q q, ϕ Page (71)

7 Ultimate.C. of Shallow Foundations (Special Cases) K can be calculated easily from (Figure 4.9) according the values of andϕ Calculating of c : c = f q c q can be calculated easily from (Figure 4.10) according the value of c c = and c = c = Important Notes: 1. If there is a water table near the foundation (above or below foundation), the three cases discussed in Chapter 3 should be considered (i.e. the factor q for top and bottom layers may be modified and γ and γ for top and bottom layers may also be modified according to the existing case of water table. 2. If the strong layer and the weak layer are not clear (cohesion and friction angle for each layer are convergent), to know the strong and the weak layer do the following: Calculate q and q and then calculate If < 1 The top layer is the stronger layer and the bottom is the weaker layer If > 1 The top layer is the weaker layer and the bottom is the stronger layer. 3. Any special cases can be derived from the general equation above. Page (72)

8 Ultimate.C. of Shallow Foundations (Special Cases) earing Capacity of Layered Soils: Weaker soil Underlain by Stronger Soil D Case I Case II Let the depth,h, is the distance from the bottom of the foundation to the top of stronger soil (bottom soil layer) and,, is the width of the foundation and,d, is the depth of failure beneath the foundation. As shown on the above figure, there are two cases: Case I: For (H < D < 1) The failure surface in soil at ultimate load will pass through both soil layers (i.e. the ultimate bearing capacity of soil will be greater than the ultimate bearing capacity for bottom layer alone). Case II: (H > D > 1) The failure surface on soil will be fully located on top,weaker soil layer, (i.e. the ultimate bearing capacity in this case is equal the ultimate bearing capacity for top layer alone). Page (73)

9 Ultimate.C. of Shallow Foundations (Special Cases) For these two cases, the ultimate bearing capacity can be given as following: For (H D 1) q = q + (q q ) 1 H D Note that if = 1 the value of q will equal q, and this is logical, because in this special case the failure surface will be exist on whole depth of top (weaker layer). For (H > D > 1) q = q ecause failure surface is fully located on top (weaker soil). q, = c N () F () + q N () F () + 0.5γ N () F () q = effective stress at the top of layer(1) = γ D q, = c N () F () + q N () F () + 0.5γ N () F () q = effective stress at the top of layer(1) by assuming the foundation is located directly above stronger soil layer at depth of D q = γ D Important Note: D = (for loose sand and clay) D = 2 (for dense sand ) Page (74)

10 Ultimate.C. of Shallow Foundations (Special Cases) earing Capacity of Foundations on Top of a Slope In some instances, foundations need to be constructed on top of a slope, thus calculating of bearing capacity of soil under such conditions will differ from Chapter 3. This section explains how we can calculate the bearing capacity of soil under these conditions. H = height of slope, β = angle between the slope and horizontal b = distance from the edge of the foundation to the top of the slope The ultimate bearing capacity for continuous or strip footing can be calculated by the following theoretical relation: q = cn + 0.5γN For purely granular soil (c = 0.0): q = 0.5γN For purely cohesive soil (ϕ = 0.0): q = cn Calculating of N γq : The value of N can be calculated from (Figure 4.15 P.204) according the following steps: 1. Calculate the value of. 2. If = 0.0 use solid lines on the figure. Page (75)

11 Ultimate.C. of Shallow Foundations (Special Cases) 3. If = 1 use dashed lines on the figure. 4. Calculate the value of which the horizontal axis aof the figure. 5. According the values of (ϕ, β and factors mentioned above ) we can calculate the value of N on vertical axis of the figure. Note: If the value of Calculate N at = 1. is in the following range: 0 < < 1 do the following: Calculate N at = 0. Do interpolation between the above two values of N to get the required value of N. Calculating of N cq : The value of N can be calculated from (Figure 4.16 P.205) according the following steps: 1. Calculate the value of. 2. If = 0.0 use solid lines on the figure. 3. If = 1 use dashed lines on the figure. 4. Determining the horizontal axis of the figure: If < H the horizontal axis of the figure is If H the horizontal axis of the figure is 5. Calculating the value of stability number for clay (N ): If < H use N = 0.0 in the figure If H calculate N from this relation N = to be used in the figure. 6. According the values of (ϕ, β and factors mentioned above) we can calculate the value of N on vertical axis of the figure. Note: If the value of is in the following range: 0 < < 1 Do interpolation as mentiond above. Page (76)

12 Ultimate.C. of Shallow Foundations (Special Cases) Problems 1. The figure below shows a continuous foundation. a) If H=1.5 m, determine the ultimate bearing capacity,q b) At what minimum depth,h, will the clay layer not have any effect on the ultimate bearing capacity of the foundation? Sand γ = 17.5 kn/m ϕ = 40 C = 0 C = 30 kn/m Solution The first step in all problems like this one is determining whether the two soils are stronger soil and weaker soil as following: q = c N () + 0.5γ N () (c = 0.0) q = 0.5γ N () = 2m, γ = 17.5 kn/m For ϕ = 40 N () = (Table3. 3) q = = kn/m q = c N () + 0.5γ N () (ϕ = 0.0) q = c N () c = 30 kn/m, For ϕ = 0 N () = 5.14 (Table3. 3) q = = kn/m Clay γ = 16.5 kn/m ϕ = 0.0 Page (77)

13 Ultimate.C. of Shallow Foundations (Special Cases) q = = 0.08 < 1 The top layer is stronger soil and bottom q layer is weaker soil. 1. For strip footing: q = q + 2c H + γ H 1 + 2D H K tanϕ H q q = c N () + q N () + 0.5γ N () c = 0.0, q = γ D = = 21kN/m, = 2m For ϕ = 40 N () = 75.31, N () = 64.2, N () = (Table3. 3) q = = kn/m q = c N () + q N () + 0.5γ N () c = 30, q = γ (D + H) = 17.5 ( ) = kn/m For ϕ = 0 N () = 5.14, N () = 1, N () = 0 (Table3. 3) q = = kn/m Calculating of c a : q q = 0.08 From figure (4.10) = 0.7 c = = 0 Calculating of K s : q = 0.08 q From figure (4.9) K = 2.4 q = tan q = 278 kn/m The minimum depth that make the clay layer have no effect on q is occur when q = q and q = 0.0 q = q = 0 + 2c H = H H = m. + γ H 1 + 2D H K tanϕ H 2.4 tan40 H H Page (78)

14 Ultimate.C. of Shallow Foundations (Special Cases) 2. Solve examples 4.4 and 4.5 in your text book. 3. Solve example 4.6 in your text book, but use this equation for calculating(q ): q = q + (q q ) 1 H D ecause the equation in text book for this case doesn t true. 4. For the soil profile shown below, determine the ultimate bearing capacity of the continuous footing. Solution From the figure: = 2.5m, b = 1.25 m, H = 5m, D = 2.5m, β = 45 q = cn + 0.5γN butϕ = 0.0 q = cn c = 40 kn/m γ = 17.5 kn/m C = 40 kn/m ϕ = 0.0 Calculating of N cq (Figure 4.16): D = 2.5 = 1 use dashed lines on figure 2.5 = 2.5 < H = 5 the horizontal axis of the figure is b 1.25 = 2.5 = 0.5 = 2.5 < H = 5 use N = 0.0 in the figure From the figure, the value of N 5.7 q = = 228 kn/m. Page (79)

15 Ultimate.C. of Shallow Foundations (Special Cases) 5. For the soil profile shown below, determine the ultimate bearing capacity of the continuous footing. γ = 20 kn/m C = 25 kn/m ϕ = 0.0 Solution From the figure: = 3m, b = 1.25 m, H = 2.5m, D = 0.0m, β = 45 q = cn + 0.5γN butϕ = 0.0 q = cn c = 25 kn/m Calculating of N cq (Figure 4.16): D = 0 = 0 use solid lines on figure 2.5 = 3 > H = 2.5 the horizontal axis of the figure is b 1.25 = H 2.5 = 0.5 = 3 > H = 2.5 use N = γ H = = 2 in the figure c 25 From the figure, the value of N 3 q = 25 3 = 75 kn/m. Page (80)