Module 7 (Lecture 25) RETAINING WALLS Topics Check for Bearing Capacity Failure Example Factor of Safety Against Overturning Factor of Safety Against Sliding Factor of Safety Against Bearing Capacity Failure 1.6 OTHER TYPES OF POSSIBLE RETAINING WALL FAILURE Check for Bearing Capacity Failure The vertical pressure as transmitted to the soil by the base slab of the retaining wall should be checked against the ultimate bearing capacity of the soil. The nature of variation of the vertical pressure transmitted by the base slab into the soil is shown in figure 7.10. Note that q toe and q heel are the maximum and the minimum pressures occurring at the ends of the toe and heel sections, respectively. The magnitudes of q toe and q heel can be determined in the following manner.
Figure 7.10 Check for bearing capacity failure The sum of the vertical forces acting on the base slab is Σ VV (see column, table 2), and the horizontal force is PP aa cos αα. Let R be the resultant force, or RR = Σ VV + (PP aa cos αα) [7.15] The net moment of these forces about point C (figure 7.10) is MM net = Σ MM RR Σ MM OO [7.16] Note that the values of Σ MM RR and Σ MM OO have been previously determined (see column 5, table 2 and equation ()]. Let the line of action of the resultant, R, interest the base slab at E, as shown in figure 7.10. The distance CCCC then is CCCC = XX = MM net Σ VV [7.17] Hence the eccentricity of the resultant, R, may be expressed as ee = CCCC [7.18] 2 The pressure distribution under the base slab may be determined by using the simple principles of mechanics of materials:
qq = Σ VV AA ± MM net yy II [7.19] Where MM net = moment = (Σ VV)ee II = moment of inertia per unit length of the base section = 1 12 (1)(2 ) For maximum and minimum pressures, the value of y in equation (19) equals /2. Substituting the preceding values into equation (19) gives qq max = qq toe = Similarly, Σ VV ()(1) ee(σ VV) 2 + = Σ VV 1 12 (2 ) 6ee 1 + [7.20] qq min = qq heel = Σ VV 6ee 1 [7.21] Note that Σ VV includes the soil weight, as shown in table 2, and that, when the value of the eccentricity, e, becomes greater than /6, qq min becomes negative [equation (21)]. Thus, there will be some tensile stress at the end of the heel section. This stress is not desirable because the tensile strength of soil is very small. If the analysis of a design shows that ee > /6, the design should be reproportioned and calculations redone. The relationships for the ultimate bearing capacity of a shallow foundation were discussed in chapter. Recall that qq uu = cc 2 NN cc FF cccc FF cccc + qqqq qq FF qqqq FF qqqq + 1 2 γγ 2 NN γγ FF γγγγ FF γγγγ [7.22] Where qq = γγ 2 DD = 2ee FF cccc = 1 + 0.4 DD FF qqqq = 1 + 2 tan φφ 2 (1 sin φφ 2 ) 2 DD FF γγγγ = 1 FF cccc = FF qqqq = 1 ψψ 90 2 FF γγγγ = 1 ψψ 2 φφ 2
ψψ = tan 1 PP aa cos αα Σ VV Note that the shape factors FF cccc, FF qqqq, and FF γγγγ given in chapter are equal to 1 because they can be treated as a continuous foundation. For this reason, the shape factors are not shown in equation (22). Once the ultimate bearing capacity of the soil has been calculated by using equation (22), the factor of safety against bearing capacity failure can be determined. FFFF (bearing capacity ) = qq uu qq max [7.2] Generally, a factor of safety of is required. In chapter we noted that the ultimate bearing capacity of shallow foundations occurs at a settlement of about 10% of the foundation width. In the case of retaining walls, the width B is large. Hence the ultimate load qq uu will occur at a fairly large foundation settlement. A factor of safety of against bearing capacity failure may not ensure, in all cases, that settlement of the structure will be within the tolerable limit. Thus this situation needs further investigation. Example 1 The cross section of a cantilever retaining wall is shown in figure 7.11. Calculate the factors of safety with respect ot overturning and sliding and bearing capacity. Figure 7.11
Solution Referring to figure 7.11, HH = HH 1 + HH 2 + HH = 2.6 tan 10 + 6 + 0.7 = 0.458 + 6 + 0.7 = 7.158 m The Rankine active force per unit length of wall= PP pp = 1 2 γγ 1HH 2 KK aa. For φφ 1 = 0, αα = 10, KK aa is equal to 0.50 (table 2 from chapter 6). Thus, PP aa = 1 2 (18)(7.158)2 (0.5) = 161.4 kn/m PP vv = PP aa sin 10 = 161.4(sin 10 ) = 28.0 kn/m PP h = PP aa cos 10 = 161.4(cos 10 ) = 158.95 kn/m Factor of Safety Against Overturning The following table can now be prepared for determination of the resisting moment: Section no. Area (mm 2 ) Weight/unit length (kn/m) Moment from C (kn/m) Moment (kn m) 1 6 0.5 = 70.74 1.15 81.5 2 1 2 (0.2)6 = 0.6 14.15 0.8 11.79 4 0.7 = 2.8 66.02 2.0 12.04 4 6 2.6 = 15.6 280.80 2.7 758.16 5 1 2 (2.6)(0.458) = 0.595 10.71.1.52 PP vv = 28.0 4.0 112.12 Σ 1128.98 ΣVV = 470.45 = Σ MM RR For section numbers, refer to figure 7.11, γγ concrete = 258 kn/m
The overturning moment, MM OO MM OO = PP h HH = 158.95 7.158 = 79.25 kn m FFFF (overturning ) = ΣMM RR 1128.98 = = 2.98 > 2 OK MM OO 79.25 Factor of Safety Against Sliding From equation (11) FFFF (sliding ) = (Σ VV) tan (kk 1φφ 2 )+ 2 cc 2 +PP pp PP aa cos αα Let kk 1 = kk 2 = 2 Also PP pp = 1 2 KK ppγγ 2 DD 2 + 2cc 2 KK pp DD KK pp = tan 2 45 + φφ 2 2 = tan2 (45 + 10) = 2.04 DD = 1.5 m So PP pp = 1 2 (2.04)(19)(1.5)2 + 2(40)( 2/04) (1.5) = 4.61 + 171.9 = 215 kn/m Hence FFFF (sliding ) = = 111.5+106.67+215 158.95 (470.45)tan 2 20 +(4) 2 (40)+215 158.95 = 2.7 > 1.5 OK Note: For some designs, the depth D for passive pressure calculation may be taken to be equal to the thickness of the base slab. Factor of Safety Against Bearing Capacity Failure Combining equations (16, 17 and 18), ee = 2 Σ MM RR Σ MM OO Σ VV = 4 2 1128.98 79.25 470.45 = 0.406m < 6 = 4 6 = 0.666 m
Again, from equations (20 and 21) toe qq heel = Σ VV 1 ± 6ee = 470.45 4 1 ± 6 0.406 = 189.2 kn/m2 (toe) 4 45.9 kn/m 2 (heel) The ultimate bearing capacity of the soil can be determined from equation (22): qq uu = cc 2 NN cc FF cccc FF cccc + qqqq qq FF qqqq FF qqqq + 1 2 γγ 2 NN γγ FF γγγγ FF γγγγ For φφ 2 = 20 (table 4 from chapter ), NN cc = 14.8, NN qq = 6.4 and NN γγ = 5.9. Also qq = γγ 2 DD = (19)(1.5) = 28.5 kn/m 2 = 2ee = 4 2(0.406) =.188 m FF cccc = 1 + 0.4 DD 1.5 = 1 + 0.4 = 1.188.188 FF qqqq = 1 + 2 tan φφ 2 (1 sin φφ 2 ) 2 DD 1.5 = 1 + 0.15 = 1.148.188 FF γγγγ = 1 FF cccc = FF qqqq = 1 ψψ 90 2 ψψ = tan 1 PP aa cos αα = Σ VV tan 1 So 158.95 470.45 FF cccc = FF qqqq = 1 18.67 = 0.628 90 FF γγγγ = 1 ψψ φφ 2 = 1 18.67 20 0 Hence = 18.67 qq uu = (40)(14.8)(1.188)(0.628) + (28.5)(6.4)(1.148)(0.628) + 1 2 (19)(5.9)(.188)(1)(0) = 442.57 + 11.50 + 0 = 574.07 kn/m 2 FFFF (bearing capacity ) = qq uu = 574.07 =.0 > OK qq toe 189.2 Example 2 A concrete gravity retaining wall is shown in figure 7.12. Determine a. The factor of safety against overturning
b. The factor of safety against sliding c. The pressure on the soil at the toe and heel (Note: Unit weight of concrete = γγ cc = 150 lb/ft ). Solution HH = 15 + 2.5 = 17.5 ft KK aa = tan 2 45 φφ 1 2 = tan2 45 0 2 = 1 PP aa = 1 2 γγ(hh ) 2 KK aa = 1 2 (121)(17.5)2 1 = 6176 lb/ft = 6.716 kip/ft Since αα = 0 PP h = PP aa = 6.176 kip/ft PP vv = 0 Part a: Factor of Safety Against Overturning Figure 7.12
The following table can now be prepared to obtain Σ MM RR : Area (from figure 7. 12) Weight (kip) Moment arm from C(ft) Moment about C (kip/ft) 1 1 2 (0.8)(15)(γγ cc) = 0.9 1.25 + 2 (0.8) = 1.78 1.605 2 (1.5)(15)(γγ cc ) =.75 1.25 + 0.8 + 0.75 = 2.8 9.45 1 2 (5.25)(15)(γγ cc) = 5.906 1.25 + 0.8 + 1.5 + 5.25 = 5. 1.0 4 (10.)(2.5)(γγ cc ) =.86 10. 2 = 5.15 19.89 5 1 2 (5.25)(15)(0.121) = 4.764 1.25 + 0.8 + 1.5 + 2 (5.25) = 7.05.59 6 (1.5)(15)(0.121)= 2.72 The overturning moment 21.51 MM OO = HH PP aa = 17.5 (6.176) = 6.0 kip/ft FFFF (overturning ) = 121.84 6.0 =.8 Part b: Factor of Safety Against Sliding 1.25 + 0.8 + 1.5 + 5.25 + 0.75 = 9.55 26.0 121.84 = MM RR From equation (11), with kk 1 = kk 2 = 2 and assuming that PP pp = 0, FFFF (sliding ) = Σ VV tan 2 φφ 2+ 2 cc 2 PP aa
= 21.51 tan 2 20 +10. 2 (1.0) = 5.1+6.87 6.176 6.176 = 1.94 Part c: Pressure on the Soil at the Toe and Heel From equations (16, 17 and 18), ee = 2 Σ MM RR Σ MM OO Σ VV = 10. 121.84 6.0 = 5.15.99 = 1.16 ft 2 21.51 qq toe = Σ V B 1 + 6e = 21.51 10. (6)(1.16) 1 + =.5 kip/ft 2 10. qq heel = Σ V B 1 6e = 21.51 10. (6)(1.16) 1 = 0.678 kip/ft 2 10. Example Repeat example 2 and use Coulomb s active pressure for calculation and δδ = 2φφ/. Solution Refer to figure 7.1 for the pressure calculation: δδ = 2 φφ = 2 (0) = 20 Figure 7.1 From table 5 (chapter 6), KK aa = 0.4794(αα = 0, ββ = 70 ), so
PP aa = 1 2 (0.121)(17.5)2 (0.4794) = 8.882 kip/ft PP h = PP aa cos 40 = (8.882)(cos 40) = 6.8 kip/ft PP vv = PP aa sin 40 = 5.71 kip/ft Part a: Factor of Safety Against Overturning Refer to figure 7. 14 and 12. Area (from figure 7. 12 and 14) Weight (kip) Moment arm from C(ft) Moment about C (kip/ft) 1 0.9 aa 1.78 aa 1.605 2.75 aa 2.8 aa 9.46 5.906 aa 5. aa 1.0 4.86 aa 5.15 aa 19.89 PP vv = 5.71 19.75 1.25 + 0.8 + 1.5 + 5.25 121 = 7.59 4.4 105.6 Same as in example 2 The overturning moment is MM OO = PP h HH Hence = (6.8) 17.5 = 9.67 kip/ft FFFF (overturning ) = 105.6 9.67 = 2.66
Part b: Factor of Safety Against Sliding FFFF (sliding ) = Σ VV tan 2 φφ 2+ 2 cc 2 PP h Figure 7.14 = 19.75 tan 2 (20)+10. 2 (1.0) 6.8 = 1.7 Part c: Pressure on the Soil at the Toe and Heel ee = 2 Σ MM RR Σ MM OO Σ VV = 10. (105.6 9.67) = 1.8 ft 2 19.67 qq toe = 19.75 10. (6)(1.8) 1 + =.9 kip/ft2 10. qq heel = 19.75 (6)(1.8) 1 = 0.09 10. 10. kip/ft2 0 OTHER TYPES OF POSSIBLE RETAINING WALL FAILURE In addition to the three types of possible failure for retaining walls discussed in section 4, two other types of failure could occur: shallow shear failure and deep shear failure. Shallow shear failure in soil below the base of a retaining wall takes place along a cylindrical surface aaaaaa passing through the heel, as shown in figure 7.15a. The center of the arc of the circle aaaaaa is located at OO, which is found by trial and error (corresponds to the minimum factor of safety). This type of failure can occur as the result of excessive induced shear stress along the
cylindrical surface in soil. In general, the factor of safety against horizontal sliding is lower than the factor of safety obtained by shallow shear failure, if FFFF (sliding ) is greater than about 1.5, shallow shear failure under the base may not occur. Figure 7.15 (a) Shallow shear failure; (b) deep shear failure Deep shear failure can occur along a cylindrical surface aaaaaa as shown in figure 7.15b, as the result of the existence of a weak layer of soil underneath the wall at a depth of about 1.5 times the width of the retaining wall. In such cases, the critical cylindrical failure surface aaaaaa has to be determined by trial and error with various centers, such as O (figure 7.15b). The failure surface along which the minimum factor of safety is obtained is the critical surface of sliding. For the backfill slope with αα less than about 10, the critical failure circle apparently passes through the edge of the heel slab (such as dddddd in figure 7.15b). In this situation, the minimum factor of safety also has to be determined by trial and error by changing the center of the trial circle. The following is an approximate procedure for determining the factor of safety against deepseated shear failure for a gently sloping backfill (αα < 10 ) developed by Teng (1962). Refer to 7.16.
Figure 7.16 Deep shear failure analysis 1. Draw the retaining wall and the underlying soil layer to a convenient scale. 2. For a trial center O, draw an arc of a circle aaaaaaaa. For all practical purposes, the weight of the soil in the area aaaaaaaaaa is symmetrical about the vertical line drawn through point O. let the radius of the trial circle be r.. To determine the driving force on the failure surface causing instability (figure 7.16a), divide the area in the zone eeeeeeh into several slices. These slices can be treated as rectangles or triangles, as the case may be. 4. Determine the area of each of these slices and then determine the weight W of the soil (and/or concrete) contained inside each slice (per unit length of the wall). 5. Draw a vertical line through the centroid of each slice, and locate the point of intersection of each vertical line with the trial failure circle. 6. Join point O (that is, the center of the trial circles) with the points of intersection as determined in step 5. 7. Determine the angle, ωω, that each vertical line makes with the radial line.
8. Calculate WW sin ωω for each slice. 9. Determine the active force PP aa on the face dddd, 1 2 γγ 1HH 2 KK aa. 10. Calculate the total driving force: Σ (WW sin ωω) + PP aa XX rr [7.24] Where XX = perpendicular distance between the line of action of PP aa and the center OO 11. To determine the resisting force on the failure surface (figure 7.16b), divide the area in the zones aaaaaa and iiiiiiiiii into several slices, and determine the weight of each slice, WW 1 (per unit length of the wall). Note that points bb and ii are on top of the soft clay layer; the weight of each slice shown in figure 7.16b is WW 1 is contrast to the weight of each slice W, as shown in figure 7.16a. 12. Draw a vertical line through the centroid of each slice and locate the point of intersection of each line with the trial failure circle. 1. Join point O with the points of intersection as determined in step 12. Determine the angles, ωω 1, that the vertical lines make with the radial lines. 14. For each slice, obtain WW 1 tan φφ 2 cos ωω 1 15. Calculate cc 2 ll 2 + cc ll 2 + cc 2 ll Where ll 1, ll 2, and ll are the lengths of the arcs aaaa, bbbb, and iiii 16. The maximum resisting force that can be derived along the failure surface is Σ (WW 1 tan φφ 2 cos ωω 1 ) + cc 2 ll 2 + cc ll 2 + cc 2 ll [7.25] 17. Determine the factor of safety against deep shear failure for this trial failure surface: FFFF (deep shear failure ) = Σ(WW 1 tan φφ 2 cos ωω 1 )+cc 2 ll 2 +cc ll 2 +cc 2 ll Σ (WW sin ωω)+ PP aa XX rr [7.26] Several other trial failure surfaces may be drawn, and the factor of safety can be determined in a similar manner. The lowest value of the factor for safety obtained from all trial surfaces is the desired factor of safety.