Question

The following figure shows a section of a long reinforced concrete cantilever wall with unit weight of 23.5 kN/m”. The distributed surcharge on the back of the wall is a live load. The following properties are known for the backfill: unit weight saturated unit weight shear strength parameters 7 = 17 kN/m3 7sat = 20 kN/m3 d =0 Ó' = 25° d = 20° friction angle between wall and soil a. Determine the factor of safety against overturning (about the toe) and sliding, assuming x = 0 m (simple gravity wall). b. Determine x to satisfy a minimum required factor of safety of 3 against both overturning and sliding. 20 kN/m2 7.0 m 0.4 m A r- 5.0 m 0.4 m 0.6 m

Answer 1

ANSWER:-

For the reinforced concrete retaining wall shown above, calculation of the factor of safety against overturning and sliding given the following parameters:

Cohesionless soil : c=0

Unit Weight of Backfill: 17 kN/m3

Angle of Internal Friction for Soil: 25 degrees

Friction angle Between Concrete and Soil: 20 degrees

Weight of Plain Concrete: 25 kN/m3

Saturated unit weight: 20 kN/m3

Step 1:-

The first step is to calculate the coefficient of active earth pressure, k_a, using the Rankine theory. Since the backfill profile is horizontal behind the cantilever retaining wall, the Rankine equation for calculating the coefficient of active earth pressure is greatly simplified. Therefore, The coefficient of active earth pressure is equal to one minus the sine of the angle of internal friction divided by one plus the sine of the angle of internal friction.

Step 2:-

The second step is to compute the lateral force resultant, R_a, due to the active earth pressure. In order to find this, the active earth pressure along the depth of the cantilever retaining wall must be know. The active earth pressure is equal to vertical soil pressure times the coefficient of active earth pressure minus two times the soil cohesion times the square roof of the coefficient of active earth pressure. The problem statement mentioned that the backfill soil is cohesionless, so the active earth pressure along the depth of the cantilever retaining wall is reduced to the vertical soil pressure multiplied by the coefficient of active earth pressure.The vertical soil pressure at any depth along the cantilever retaining wall is equal to the unit weight of the backfill soil multiplied by the vertical distance from the top of the backfill to the location of interest resulting in a triangular pressure distribution. To calculate the lateral force resultant, the area of the triangle is found and multiplied by the coefficient of active earth pressure. Therefore, the lateral force resultant is equal to total cantilever retaining wall height squared times times one-half times the unit weight of the backfill soil times the coefficient of active earth pressure.

Step 3:-

The third step is to find the overturning moment, M_o, caused by the lateral force resultant due to the active earth pressure. It is equal to the lateral force resultant multiplied by the moment arm of the resultant force. The moment arm is equal to the perpendicular distance from the toe of the cantilever retaining wall to the centroid of the lateral pressure distribution due to the active earth pressure. Because the lateral pressure distribution is triangular, the distance from the toe of the cantilever retaining wall to the centroid is simply the total height of the cantilever retaining wall divided by three.

Step 4:-

The fourth step is determine the balancing moment that is responsible for resisting the overturning moment on the cantilever retaining wall. To achieve this, the weights and centroids of both the cantilever retaining wall and backfill soil must be known; however, in most cases, the cantilever retaining wall and soil are broken up into simpler geometric components, and the weights and centroids of these components are found individually. After finding the self-weights and centroids of the cantilever retaining wall and soil, the moments their self-weights exert about the toe of the cantilever retaining wall are tabulated and added together to arrive at the final balancing moment quantity.

Step 5:-

The fifth step is calculate the factor of safety against overturning. It is equal to the balancing moment divided by the overturning moment and should be greater than or equal to two.

Step 6:-

The sixth step is to find the factor of safety against sliding. It is equal to the coefficient of friction between the concrete and soil times the total weight of the cantilever retaining wall and soil divided by the lateral force resultant due to the active earth pressure. Since we decided to ignore the effects of the passive earth pressure, a minimum required factor of safety against sliding of 1.5 is acceptable; otherwise, a factor of 2.0 is typically used.


Determination of the factor of saftey agaist Ovesturing la toe) and sliding, assuming x=0 m į gravity wall: Ovestering about the * frist we have to calculate magnitude and point of application of the resultant active thrust on the wall - 20 kN/m2 LULUWU----r- 2m Y = 17kN/m2 $=250 I - ---The I sat = 20KN/m3 keket 5m Coefficient of active earth pressure ka= (1-Singº ) 0.405 11+ Sin25 submerged density of the soil below W.T You - Ysat - Yw = 20-10 = 101W/m 3 Active pressure due to moist soil above WoT - = Kayz = 4 0.405 x 17x2 = 13.8 kN/m2

Active pressure due to suschera 0.405 x 20 It is the = 8011 kN/m2 Active pressure below WoT H due to submerged soil Ka Yo z = 0.405x10x5 - 20:29 kN/m2 Lateral pressure exerted by water : DY, Z = 10x5 = 50 kN/m2 Now, 2- 7 2 = 3-5 m + X=5+ = 5.67 m P = 8.11x7. = $6.77 KN P = $x 13.8 x 2 = 1378 KN P = 13.865 1. = 69 KN 14 P4 - śl so +20129) 30720 23 - ? = 2.5 m 43 = 1,67 m } - 35:145 kn Ral+ P + P 27 P4 = 56.77 +13-8769+35.145 Resultant thrust. = 174.75 EN

Val The point of application of this thrust from bottom of was x= Pix, + P2 X₂ + P₂ X ₂ + Poleg - Piept Pzt Pq x = 56.77X3.5 + 13.8 x 5.67 +69x2.5 +35145716 174.715 T2 = 8.90 min Pa = 174.715 wall ă= 2.90 Calculation of Overturing and resistfug moments : (x=0) Ž → W, OS

Mo (overturing Moment) = PA x ł = 174.715x2.9 = 506.67 kNm Now to calculate Me we divided the soil and the concrete into rectangles and triangles to find the area epily and to find the arm from the center of each area to point O. (Assoning Y= 2560). Section | Area I weight/ unit lengt/ Moment carm Moment of the wall measured from of about 1 10:4x70.4 x7x25 06+004 56 370 -0.8 2 x0647 5x0.6x7x25 x06 - 52.5 .0.4 Ord | 2 weight of sail Not coneidered as azo SV -122.5 KN EMR = 77 kNm

Factor of Por of safty agaist Overturing sat - Mollesisting moment) 1. Mo (Overturing moment = 770 506.67 Fos. = 1.52 Factor of safty agaist sliding: Fos = 4 EV = (tangoº ) x 122-5x10 17:45 = 2.56 I FOS, 32
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