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1 Rbq^fkfkd=t^ii=abpfdk MORGAN STATE UNIVERSITY SCHOOL OF ARCHITECTURE AND PLANNING LECTURE VIII Dr. Jason E. Charalambides = = elt=albp=^=`^kqfibsbr= Rbq^fkfkd=t^ii=tlRh\ The main function of a cantilever retaining wall is to constrain soil grade changes. It is a very practical design particularly for heights between 6`-18`. The stem of the wall is essentially a slab acting as a cantilever to the lateral load applied by the soil on it s side, and it is anchored to the base. The base is a one-way footing that counteracts direct loads and moments generated from the stem and from the soil that is retained. A drainage system should always be accommodated within the design in order to minimize the hydrostatic lateral pressure against the stem.

2 mrlmlrqflkp=^ka= il^afkd The maximum vertical reaction pressure against the toe should remain within the limits of the Allowable Soil Pressure, and all surface of the base is under positive pressure. The force displacement graph for soils is a non linear relationship where the slope of the graph decreases as the applied force increases. When the soil fails, the value of stiffness is zero but the displacement is too large to be tolerated if the wall is to function under anticipated loads. The initial stiffness of the soil (0%- 5%) should be applied as basis for allowable stress when the wall is loaded. pqrbppbp=^mmifba The soil pressure against the stem is regarded as a hydrostatic pressure applied by a fluid with density identical to the density of the soil. Totally saturated soil (no air gaps) have a lateral pressure coefficient as high as 1.0. Soils with a relative cohesion carry coefficients at a scale between 35% and 40%, while sand may carry a coefficient at approximately 30%. All these factors need to be obtained by specializing geotechnical engineer reports.

3 bu^jmib Assuming a standard surcharge height of, calculate the forces, moments, eccentricity, safety factor, and bearing stress of the retaining wall indicatedk h base =10 inch Length 1 =6 inch Length =60 inch γ soil =15 pcf γ conc =150 pcf k.s=0.4 L=144 inch B=96 inch w stem =B (Length 1 +Length )=10 inch Z =L h base =134 inch bu^jmib F stemv =Z w stem 1 γ.conc=1.4 kip F basev =h base B 1 γ.conc=1 kip F surchv =s Length 1 γ.soil=1.5 kip F earthv =Z Length 1 γ.soil=6.98 kip F stemv =Z w stem 1 γ.conc=1,4 kip

4 bu^jmib bu^jmib The Safety Indicator, a factor that determines whether the design is approved or if it should be adjusted based on geometric forms and loading generated, is given by the following formula: In this case, the result is.445, thus we are ok to proceed.

5 bu^jmib The distance from the heel to the resultant vertical force equals the moment of all forces divided by the vertical force and has a value of 53.73k`/10.6k = The center of the base is 4 from the heel and the eccentricity of the resultant force is = 1.06 from the center of the base. Since the eccentricity of force is located within the kern (middle third of the base), it is rational to assume that the base is bearing against soil. bu^jmib The bearing stresses are given by the following: If the resultant vertical force was outside the kern, the pressure distribution would be treated as a triangular form. Naturally, that triangular diagram would have a length 3 times the distance from the toe to the resultant force.

6 Analysis of Cantilever Retaining Wall Analyze and verify the capacity and the loads applied on the indicated retaining wall of the figure below given that it retains granular soil (Use the Rankine Formula for Active soil pressure). For the soil, a triaxial test has been conducted and the results of the axial and radial stress are provided. Loads, dimensions and all necessary data are given. σ A := 3.5 tonf σ R := 1.75 tonf γ conc := 150 lbf 3 C A := 950 lbf Cl cover := in CA is coeff of adhesion f y := 60ksi ASP := tonf = 4000 psf f' c := 3ksi γ soil := 15 lbf d 3 b_ini := 1in δ := 17deg b := 1 we consider strips of 1 λ := 90deg s:= 0in O:= 3 E := 5.5 V := 1.5 U:= β := 18.4deg L:= 15 T:= 1 F := 1.5 Base := O + E + F F' := F T 1) Determining angle of Internal friction, beta angle, angle of failure plain, the Rankine Active Pressure and the Total Active Pressure Resultant: Φ := asin σ A σ A + σ R H:= L + V + E tan( β) β := atan H L V E H = β = 18.4 deg These are Rankine's formulas. Another one is by Coulomb that is applied below sin( λ + Φ) bk a γ soil H K a := K sin( λ) a = R a := sin( Φ + δ) sin( Φ β) sin( λ δ) 1 + R sin( λ δ) sin( λ + β) a = lbf

7 ) Solving for loads and moments generated: Vertical components: 1 ( H V L) E γ soil P E sl_v := P sl_v = 0.63 kip M sl_v := P sl_v 3 M sl_v = 1.15 k' P sl_v1 := 1 ( L) ( E) γ soil P sl_v1 = kip E M sl_v1 := P sl_v1 M sl_v1 = 8.36 k' P Base := 1 V Base γ conc P Base =.5 kip Base M Base := P Base M Base = 11.5 k' P Stem_ 1 F' F' := L γ conc P Stem_ = 0.56 kip M Stem_ := P Stem_ E + T + 3 M Stem_ = 3.75 k' P Stem_1 := 1 T L γ conc T P Stem_1 =.5 kip M Stem_1 := P Stem_1 E + M Stem_1 = 13.5 k' P sc_toe := 1 O U γ soil P sc_toe = 0.75 kip O M sc_toe := P sc_toe Base M sc_toe = 6.38 k' P sc_v := 1 s E γ soil P sc_v = 0 kip E M sc_v := P sc_v M sc_v = 0k' P vert := P sl_v1 + P sl_v + P sc_v + P Base + P Stem_1 + P Stem_ + P sc_toe P vert = kip M Vtot := M Stem_1 + M Stem_ + M Base + M sc_v + M sl_v1 + M sl_v + M sc_toe M Vtot = k' Horizontal components: P sc_h := 1γ soil K a s L P sc_h = 0 kip L M sc_h := P sc_h M sc_h = 0k' L L P soil_h := 1γ soil K a P soil_h = kip M soil_h := P soil_h 3 M soil_h = 8.03 k U U P soil_t := 1γ soil K a P soil_t = kip M soil_t := P soil_t 3 M soil_t = k P horiz := 1P soil_h + P sc_h + P soil_t M Htot := M sc_h + M soil_h + M soil_t Summing up moments for overturning: P horiz = 5.51 kip M Htot = 7.97 k' M overt := M Vtot + M Htot M overt = 9.35 k' 3) We continue with calculations for overturning and sliding at this stage:

8 Estimating safety indicator against overturning and verifying that that the resultant is within kern (middle third of footings length): ( ) P vert M Vtot SI ot = B > M Htot ( P vert ) ( ) P vert M Vtot Base M = Htot ( P vert ) 3.69 SI ot = "OK" M overt P vert = 5.51 Or Percentagewise... M overt P vert Base = Res 3rd = "Kern" R av := R a sin( δ) R av = lbf R ah := R a cos( δ) R ah = lbf Estimating safety against sliding (Use delta in tan for keyless, phi for keyed): ( ) tan δ P vert + R av ( ) + C A Base SI slide = > 1.5 R ah ( ) tan δ P vert + R av ( ) + bc A Base = 1.9 R ah SI slide = "OK" 4) Calculating eccentricity and bearing stress: First, the distance from the heel to the resultant of vertical forces, then the eccentricity, and then the FP bearing. V u := P Stem + P Base + P sc_v + P sl_v1 + P sl_v + P sc_toe V u = kip M overt d heel_res := d V heel_res = 5.51 u A F := Base 1 A F = 10 Base ecc := d heel_res ecc = 0.51 V u 6V u ecc F p_toe := + A F 1 Base F p_toe =.19 ksf V u 6V u ecc F p_heel := A F 1 Base F p_heel = 1.16 ksf 5) Sizing the depth of the stem: Note: For the size of the stem we do not apply safety factors as we would do for "strength design", that is for rebars in the concrete. Instead, we use the loads as estimated because that will be sufficient for deformation and displacement. In the following formula we see the 1 multiplied by the specific weight of soil and then multiplied by the K factor and in the larger parenthesis we see the following: Surcharge height multiplied by wall height and then by midheight where the resultant of that rectangular form would be located Wall height multiplied by wall height (as rotated horizontally) divided by for rectangular form, multiplied by 1/3 of wall height which will be the location of the resultant. Triangular formation of Soil above the wall is treated in a fashion similar to the surcharge.

9 ( L + V) ( L + V) M u b γ soil K a s ( L + V) ( H L V) L + V ( ) := + + M 3 u = 43.5 k' Determine the β1 value and the coefficient k10 for the given concrete and steel grades: β 1 = 0.85 k 10 :=.177 f' c β β 1 k 10 = 0.41 ksi ( ) Calculate the distance necessary to locate the rebars within the stem: (Note that for "b" we take 1 strips) bd M u M u > d := d = in k 10 bk 10 Given the above mentioned value for stem depth, we should also consider " for clear cover (The minimum cover for rebar #6 or above is in (ACI 7.7.1)) + the diameter of horizontal rebars + the radius of vertical rebars. We give approximate estimates for initial values of bar diameter to be used: d tot = 14 in 6) Designing and determining location for reinforcement. Here we consider safety factor of 1.6 for soil and overburden pressures that will be applied for the design for strength, i.e. the rebars that we will include in our final product. Φ := 1.6 f h := Φγ soil K a f h = 0.08 ksf Take y at top surface of base:.5 L We give an array of values of depth.5 L P u = f h ( s + y) y := based on the total depth divided in.75 L quarters: 1.0L P u := f h ( s + y) P u = ksf M y b f h s y y := M y = k' M y As_est := As_est = 1d f y b The above mentioned values for area of steel correspond to the depths given in the array of values "y". It is however necessary that we apply no value less than the minimum at any depth even though the calculations may yield cross sectional areas that are less than the As_min defined by the ACI codes. As_min :=.0033 d As_min = 0.41 in in

10 In this case, we can pick the maximum of the array of values to start with at the very bottom of the stem. A s_sel = in At various heignt (or depth) values of the stem we can we can specify different "ρ" distribution according to the cross sectional area of steel needed. Bar Designation Number Weight per foot (lbf) Diameter db Area As Perimeter A b6 :=.44in d b6 :=.76in A b7 :=.6in d b7 :=.875in A b6 A s_sel A b7 A s_sel = = Take 6'' c/c at full depth. At three quarters depth we see that the spacing can be modified to 15" c/c but that will not be easy to adjust with the 6" increments. Therefore, every other reba (i.e. at 1" c/c) will extend all the way to the top. Note: Larger spans with #8 were possible but it is prefferable to have smaller rebars at shorter distances that will better integrate the steel with the concrete. in in Selecting the remaining reinforcement (rebars and placement): We need to place vertical reinforcement to support the horizontal bars on the exterior face of the retaining wall. If Z>14 we use #5@18in c/c, otherwise we use #4@18in c/c. The horizontal shrinkage and temperature reinforcement required for the stem and the footing is given by the following: ( ).00 b d tot A sh := A sh = 0.34 in A b4 :=.in Walls thicker than 10 in require two layers of reinforcement. Therefore: Take 1 #4@10in c/c. d b4 :=.4in Given the above bar diameters we can establish the precise dimension of the cantileven wall's effective depth: d b6 d := d tot Cl cover d b4 d = 11. in The shear experienced at the base of the wall is: A b4 A sh = 7.14 in

11 L V u := f h sl + b V u = 8.97 kip Therefore, the shear capacity should exceed the experienced shear: Φ :=.75 ΦV n = Φ ( bd ) f' c ΦV n = kip V cond = "OK" Note: The initial wall thickness (initial guess) was 1 on top and 1.5 on the bottom. The 14" overall thickness will not produce any further stress than the initial estimation. F' P Stem := L γ conc P Stem =.81 kip P Stem_fin := 1 d tot L γ conc P Stem_fin =.6 kip