Question

Chapter 13, Problem 4P Bookmark Show all steps: O ON Repeat Problem 1 using Coulomb's active earth pressure in your calculation and letting 6 = 2/3 01. Problem 1 A gravity retaining wall is shown in Figure 1. Calculate the factor of safety with respect to overturning and sliding, given the following data: Wall dimensions: H= 6 m, x1 = 0.6 m, x2 = 2 m, x3 = 2 m, x4 = 0.5 m, x5 = 0.75 m, x6 = 0.8 m, D= 1.5 m Soil properties: y1 = 16.5 kN/m3, 61 = 32, y2 = 18 kN/m3, 62 = 22, c2 = 40 kN/m2 Use the Rankine active earth pressure in your calculation. Figure 1

Answer 1

The dimensions and properties of the retaining wall and supported backfill soil are shown in the diagram. It is required to check the safety of the retaining wall with respect to overturning and sliding. % = 16.5 kN/m c', = 0 6 m ';= 32 90° - B 1.5 m 10.75 m 0.8 m 0.8 0.5 m2 m !0.6 m 2 m Yo = 18 kN/m o'= 22 C', = 40 kN/m

The coefficient of active earth pressure for the backfill soil can be determined from Coulomb's theory: sinº ( B+") K =- sinº Bsin(6-8) 14. sin(+') sin(09-a) .... (1) V sin(B-8") sin(a+b)] Here, Angle is 8', Effective angle of friction between soil and wall s'. Calculate 8: -21.33 Calculate : 71.5° Comment Step 3 of 9 A Substitute 21.330for s', 71.5*for p in equation (1) sinº (B+ ") K =- sinº B sin(B-8") 1 +, sin (0+")sin(-a) V sin(B-8") sin(a+b) sin (71.6° +32) sin?71.5°x sin(71.50-21.33°)x ++ sin(71.5-21.33")* sin(O®+71.5")] sin(32° +21.33")sin(32° -0°) 0.945 0.439

6:55 l 490 Step 4 of 9 A The stabilizing forces and moments can be calculated as shown below: Area V = Areaxy Portion (m) (kN/m) | (kN/m) Xw (m) MV, XX (kNm/m) 0.8x5.85 23.58 = 4.68 110.35 0.5x5.85 = 2.925 322.77 0.5x2x612358 3.1+*2.0 532.96 141.48 =6.00 = 3.767 2.5 +0.5x0.6 237.69 0.6x6 = 3.60 23.58 84.89 = 2.80 0.5x2x6 23.58 141.48 0.5+x2.0 259.33 =6.00 = 1.833 Total 478.20 1,352.75 Comment Step 5 of 9 A Calculate the total active force behind the wall: P. ==xy(H + x,)*xK Here, H is the height of stem H, xis the protrusion of the base into the active side. Substitute 0.75 m for xs, 6 m for H 2. = xy(H +x,)ºxK = x16.5x(6.0+0.8) *0.439 = 167.6 kN/m

Step 6 of 9 A Calculate the horizontal and vertical components of this destabilizing force: P.-P. Cos(90°- B+8") - 167,6x cos(90°-71.5° +21.33%) -128.88 kN/m P.-P.xsin(90°- B+8) - 167.6x sin(90° - 71.50 +21.33º) -107.1 kN/m Comment Step 7 of 9 Calculate the total stabilizing moment: M, - M +P.** Substitute 107.1 kN/m for P-, M, - M+P.xx -- 1,352.75+107.1(49-69 -0.8) van 1961) = 1,352.75 +107.1x4.412 -1,825 kN.m/m The overturning will be: - 128.888 - 292.13 kN.m/m Calculate the total vertical forces of the system: Ev - V. P. 478.2+107.1 - 585.3 kN/m

Step 8 of 9 Calculate the factor of safety against overturning: Substitute 292.13 kN-m/m for Mo, 1,825,3 kN • m'm for Me MO 1825 292.13 6.2 Therefore, the factor of safety 6.2 with respect to overturning is very high, which is typical in the case of gravity retaining walls. Comment Step 9 of 9 ^ Calculate the factor of safety against sliding: ES. V x tan 8'+c',XB 585,3x0.26+(x40x5.85 128.88 309.2 128.88 2.3 Therefore, the factor of safety 23 with respect to sliding shows that the retaining wall is safe.