2) Mechanical stabilization using reinforcing geotextile fabric. The required wall dimensions are 20 ft high by 100 ft long and the wall must support a uniform surcharge load of 600 psf on a level backfill. The wall facing material will be precast concrete modular blocks that are 1 ft high, and reinforcing is to be placed between every other course of facing block. Therefore the vertical spacing of reinforcement must be 2 ft. Layers of reinforcing material (either steel strips or geotextile fabric) will be placed at grade level (zero height), two-ft height, four-ft height, etc. continuing to 18-ft height, for a total of ten layers of reinforcement (no reinforcing at the 20-ft height). For ease of construction, you must limit variation in the design reinforcement embedment lengths; rather than varying the embedment lengths continually with depth, specify one length for the bottom layer of reinforcement, one length for the lower half of the wall, and one (longer) length for the upper half of the wall. Rankine active earth pressure coefficients are to be used to calculate horizontal earth pressures. Assume that ground water will not be present behind the wall. Calculations must demonstrate that the reinforcing steel strips and/or the geotextile fabric have sufficient strength to withstand maximum wall pressures. The calculations must also demonstrate an adequate factor of safety against translational movement and provide the necessary embedment length at the base of the wall. Finally, you must determine which wall reinforcing material (steel strips or geotextile fabric) is the least expensive given unit costs shown below. (Assume no wastage in either material and assume that the cost of soil backfill is identical for both alternatives.) The project parameters are as follows: Underlying Residual Clayey Soil: unit weight = y = 102 pcf angle of internal friction = Q = 15° cohesion = c = 3,800 psf, adhesion = a = 1,900 psf Backfilled Granular Soil: unit weight = y = 118 pcf angle of internal friction = Q = 32° cohesion c=0
Steel Reinforcing Strips: width = w = 4 inches thickness = t = 0.03 inches vertical spacing = height = h = 2 ft allowable stress = fs = 20,000 lbs/in? factor of safety against pullout = FSpullout = 1.5 factor of safety against base translation = (FS)translation = 1.8 angle of friction between soil and steel strips = 8 = 16° steel strip cost per lineal ft = $3.10 • Geotextile Fabric Available for this Project: tensile strength = Tultimate = 10,000 pounds per ft angle of friction between soil and geotextile = 8 = 16° vertical spacing = height = 2 ft (FS)geotextile installation damage = (FS)id = 1.3 (FS)geotextile creep = (FS)er = 1.4 (FS)geotextile chemical degradation = (FS)cd = 1.2 (FS)gcotextile biological degradation = (FS)bd = 1.6 (FS)geotextile spacing = (FS)spacing = 1.5 (FS)base translation = (FS)translation = 1.8 (FS)tensile force = (FS)tensile = 2.0 geotextile fabric cost per sq. ft. = $2.80 Required Task 1: Specify design lengths and horizontal spacings of steel strips (constant spacings but three different lengths; base, lower half, and upper half). Task 2: Specify required lengths of geotextile fabric (three different lengths; base, lower half, and upper half). Task 3: Determine the least expensive alternative reinforcing material (steel strips or geotextile fabric).
Find the available tensile strength of the provided geotextile, Tavailable Tavailable = Tultimate/[ (FS)ia * (FS). * (FS). * (FS)od ] = lbs/ft The required vertical spacing of geotextile layers, h, is predetermined to be 2 feet. This spacing remains constant for the entire wall height. Vertical spacing between layers = h = Tactual/ ( pressure * (FS)spacing ) = 2 ft Pressure varies with depth, maximum pressure occurs at maximum depth of 20 ft. Check to be sure that available geotextile strength is sufficient at max. pressure. as before: max. Psurcharge = K,* and max. Psoil = K, YH Rearranging the equation yields Tactual = [h * (Psurcharge + soil) * (FS)spacing ] = Ibs/ft Does Tavailable exceed Tactual? If so, the specified geotextile has sufficient tensile strength. If not, specify a different fabric or adjust safety factors. Next, find the length of embedment at the base ( Lbase ) to protect against translational movement of the wall. Must determine the total forces acting on the wall; soil plus surcharge. Calculate the areas of the stress diagrams, rectangular for surcharge, triangular for soil loading. As calculated in Task 1: max. Psoil = 0.5 y H’K max. Psurcharge = K, H Lbase = [(max. Psoil + surcharge) * (FS)translation ] / adhesion
Also unchanged from Task 1, document the length of strip (and/or geotextile fabric) embedment necessary to extend through the failure zone: LRankine = (height above base) * tan(45 – 0/2) Next, find the length of geotextile embedment necessary to extend past the failure zone: Lmin = (FS)tensile * (K) * h/[2 tan 8 ] For all distances above base of wall, vertical spacing h = 2 ft between layers, therefore Lmin remains constant throughout wall. Final summary: Lotal = LRankine +Lmin • Geotextile must extend some length over residual clay soil at base of wall for adequate resistance to translation. • From base of wall to mid-height, extend geotextile some length into soil. • From mid-height to top of wall, extend geotextile some further length into soil. Note that for overall stability, the geotextile embedment should never be less than 80% of the wall height. Finally calculate the cost of the geotextile reinforcing material. Determine the total square footage of geotextile material required and multiply by the given cost per sq. ft. then compare to steel strips cost. Choose must cost-effective option.
2) Mechanical stabilization using reinforcing geotextile fabric. The required wall dimensions are 20 ft high by 100 ft l
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2) Mechanical stabilization using reinforcing geotextile fabric. The required wall dimensions are 20 ft high by 100 ft l
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