# lecture-09-design-of-retaining-wall

Arm or stem Surcharge Figure 1: Retaining Wall. key Page 1 of 28 Backfill Terms related to Retaining Walls: Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Continuous back drain, crushed stone Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Arm Weep Holes GL GL GL Section A-A Crushed stone Base Slab Counterfort key Backfill Tile drain Tile drain δ

Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Retaining Walls: Retaining walls are used to hold back masses of earth or other loose material where conditions make it impossible to let those masses assume their natural slopes. They are extensively used in the construction of railways, highways, bridges, canals, basement walls in buildings, walls of underground reservoirs, swimming pools etc. Terms related to Retaining Walls: Surcharge Backfill Arm or stem GL Heel Toe key Figure 1: Retaining Wall. Types of Retaining Wall: 1. Gravity Wall, 2. Cantilever Wall, 3. Counterfort Wall. â€˘ The gravity wall (Fig. 2a) retains the earth entirely by its own weight and generally contains no reinforcement. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 1 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar • Design of Retaining Walls The reinforced concrete cantilever wall (Fig. 2b) consists of the vertical arm that retains the earth and is held in position by a footing or base slab. In this case, the weight of the fill on top of the heel, in addition to the weight of the wall, contributes to the stability of the structure. Since the arm represents a vertical cantilever, its required thickness increases rapidly with increasing height. • To reduce the bending moments in vertical walls of great height, counterforts are used, spaced at distances from each other equal to or slightly larger than one-half of the height (Fig. 2c). Gravity walls are economical only for relatively low walls; possibly up to about 10 ft. Cantilever walls are economical for heights from 10 to 20 ft, while counterforts are used for greater heights. δ Backfill Arm Continuous back drain, crushed stone Crushed stone Tile drain GL GL Tile drain Toe (a) Heel Base Slab (b) Counterfort Weep Holes A A GL Section A-A key (c) Figure 2: Types of Retaining Walls, (a) gravity wall, (b) Cantilever wall, (c) Counterfort wall. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 2 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Soil parameters used in the analysis of retaining walls: Table 1 gives representative values for γs (unit weight of soil) and φ (angle of internal friction of soil) often used in engineering practice. (Note that the φ values do not account for probable additional pressures due to pore water, seepage, frost, etc). The table also contains values for the coefficient of friction “f” between concrete and various soils. The values of φ for soils 3 through 5 maybe quite unconservative; under saturated conditions, clays and silts may become entirely liquid (that is, φ = 0). Soils of type 1 or 2 should be used as backfill for retaining walls wherever possible. Table 1: Unit weight (γ), effective angles of internal friction (φ), and the coefficient of friction with concrete (f) Soil 1. Sand or gravel without fine particles, highly permeable 2. Sand or gravel with silt mixture, low permeability 3. Silty sand, sand and gravel with high clay content Unit Weight (γs), pcf φ (degree) f 110 to 120 33 to 40 0.5 to 0.6 120 to 130 25 to 35 0.4 to 0.5 110 to 120 25 to 30 0.3 to 0.4 4. Medium or stiff clay 100 to 120 25 to 35 0.2 to 0.4 5. Soft clay, silt 90 to 110 20 to 35 0.2 to 0.3 Earth pressure for normal conditions of loading: In computing earth pressures on walls, three common conditions of loading are most often met: (1) Horizontal surface of fill at the top of the wall (figure 3a), (2) Inclined surface of fill sloping up and back from the top of the wall (figure 3b), (3) Horizontal surface of fill carrying a uniformly distributed additional load (surcharge), such as from goods in a storage yard or traffic on a road (figure 3c). The increase in pressure caused by uniform surcharge s (case 3) is computed by converting its load into an equivalent imaginary height of earth (h') above the top of the wall such that, h′ = s / γs and measuring the depth to a given point on the wall from this imaginary surface. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 3 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls The distributions of pressure for cases 1 to 3 are shown in Figure 3. The total earth thrust (P) per linear foot of wall is equal to the area under the pressure distribution diagram, and its line of action passes through the centroid of the pressure distribution diagram. Figure 3 gives information on magnitude, point of action, and direction of (P) for these three cases. s δ P h P P= h 3 1 2 K γh 2 ah s h P y Kah γs h K ah γs h y= h δ y s h' = w h y= 3 1 2 K γh P= 2 ah s For δ = φ, Ka = cosφ (a) (b) y Kah γs (h+h') 2 y= P= h + 3hh' 3(h+2h') 1 K γ h(h+2h') 2 ah s (c) Figure 3: Earth pressures for (a) horizontal surface; (b) sloping surface; (c) horizontal surface with surcharge s. Failure of Retaining Walls: A wall may fail in three different ways: 1. Its individual parts may not be strong enough to resist the acting forces, such as when a vertical cantilever wall is cracked by the earth pressure acting on it, 2. The wall as a whole may be bodily displaced by the earth pressure, without breaking up internally. 3. The soil beneath the wall may fail. To design against the first possibility requires the determination of the necessary dimensions, thicknesses, and reinforcement to resist the moments and shears; this procedure, then, is in no way different from that of determining required dimensions and reinforcement of other types of concrete structures. The usual load factors and strength reduction factors of the ACI Code may be applied. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 4 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls To safeguard the wall against bodily displacements, i.e., to ensure its external stability, requires special consideration. Consistent with current practice in geotechnical engineering, the stability investigation is based on actual earth pressures (as nearly as they may be determined) and on computed or estimated service dead and live loads, all without load factors. Computed bearing pressures are compared with allowable values, and overall factors of safety evaluated by comparing resisting forces to maximum loads acting under service conditions. Factor of safety against overturning: (FOS)OT = Stabilizing moment / overturning moment = Rva / Py ≥1.5 Factor of safety against sliding: (FOS)S = µRv / P ≥1.5 P GL R v = ΣW a R y µRv = µ(ΣW) Figure 4: Rv = Summation of vertical forces acting on wall, P = Earth pressure, R = Resultant of P and Rv. Next, it is necessary to ensure that the pressure under the footing does not exceed the permissible bearing pressure for the particular soil. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 5 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Bearing pressure of soil under various conditions of horizontal and vertical loading: It will be found that if the resultant is located within the middle third (a > l/3, where a = horizontal distance of resultant from edge of toe slab), compression will act throughout the section, and the maximum and minimum pressures can be computed from the equations shown in figure 5a. If the resultant is located just at the edge of the middle third (a = l/3), the pressure distribution is as shown in figure 5b. Rv b R l/3 q1 = (4l - 6a) R2v l a a q2 q1 q 2 = (6a - 2l) R2v l When a = l/2; q1 = q 2 = l Rv l2 (a) Rv b R l/3 a a q2 q1 q1 = 2Rv l q2= 0 (b) Rv b R l/3 a a 3a q1 = 2Rv 3a q1 (c) Figure 5: Bearing pressure for different locations of resultant. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 6 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls If the resultant were located outside the middle third (a < l/3), equations (qmax, min = N/A ± Mc/I) would indicate tension at and near point “a”. Obviously, tension cannot be developed between soil and a concrete footing that merely rests on it. Hence, in this case the pressure distribution of figure 5c will develop, which implies a slight lifting off the soil of the rear part of the footing. Equilibrium requires that Rv pass through the centroid of the pressure distribution triangle, from which the formula for q1 for this case can easily be derived. See detailed proof of bearing pressures are available in appendix A, at the end of this document. It is good practice, in general, to have the resultant located within the middle third. This will not only reduce the magnitude of the maximum bearing pressure but will also prevent too large a nonuniformity of pressure. If, as is mostly the case, the resultant strikes within the middle third, adequate safety against overturning exists and no special check need be made. If the resultant is located outside the middle third, a factor of safety of at least 1.5 should be maintained against overturning; i.e., the restoring moment should be at least 1.5 times the overturning moment. Basis of structural design: • The structural design of retaining wall should be consistent with methods used for all other types of members, and thus should be based on factored loads in recognition of the possibility of an increase above service loading. ACI load factors relating to structural design of retaining walls are summarized below (ACI 9.2): i. If resistance to earth pressure H is included in the design, together with dead loads D and live loads L, the required strength U shall be at least equal to: U = 1.2D + 1.6L + 1.6H ii. Where D or L reduces the effect of H. the required strength U shall be at least equal to: U = 0.9D + 1.6H iii. For any combination of D, L, and H, the required strength shall not be less than: U = 1.2D + 1.6L • In the investigation of a retaining wall for external stability, it is the current practice to base the calculations on actual earth pressures, and on computed or estimated Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 7 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls service dead and live loads, all with load factors of 1.0 (i.e., without load increase to account for a hypothetical overload condition). • Computed soil bearing pressures, for service load conditions, are compared with allowable values set suitably lower than ultimate bearing values. Factors of safety against overturning and sliding are established, based on service load conditions. • Following the ACI Code, lateral earth pressures are multiplied by a load factor of 1.6. • In general, the reactive pressure of the soil under the structure at the factored load stage is taken equal to 1.6 times the soil pressure found for service load conditions in the stability analysis. • For cantilever retaining walls, the calculated dead load of the toe slab, which causes moments acting in the opposite sense to those produced by the upward soil reaction, is multiplied by a factor of 0.9, ACI R9.2. (see fig. 10 below) • For the heel slab, as the dead load of heel tends to increase the design moment, the required moment capacity is based on the dead load of the heel slab multiplied by 1.2, while the downward load of the earth is multiplied by 1.6. • The upward pressure of the soil under the heel slab is taken equal to zero, recognizing that for the severe overload stage a nonlinear pressure distribution will probably be obtained, with most of the reaction concentrated near the toe. • Surcharge, if present, is treated as live load with a load factor of 1.6. • Similar assumptions appear to be reasonable in designing counterfort walls. • In accordance with ACI 14.1.2, cantilever retaining walls are designed following the flexural design provisions, with minimum horizontal reinforcement provided in accordance with ACI 14.3.3, which stipulates a minimum ratio of horizontal reinforcement area to gross concrete as: ¾ 0.0020 for deformed bars not larger than No.5 (No. 16) with a specified yield strength not less than 60,000 psi, ¾ 0.0025 for other deformed bars, ¾ 0.0020 for welded wire reinforcement not larger than W31or D31”. • Minimum main reinforcement is according to ACI 10.5. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 8 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Drainage and other details: Failures or damage to retaining walls, in most cases, occur due to one of two causes: i. Overloading of the soil under the wall with consequent forward tipping, and ii. Insufficient drainage of the backfill. The two causes are often interconnected, since large thrusts correspondingly increase the bearing pressure under the footing. Therefore, allowable bearing pressures should be selected with great care. It is necessary, for this purpose, to investigate not only the type of soil immediately underlying the footing, but also the deeper layers. Drainage can be provided in various ways. Weep holes consisting of 6 or 8 in. pipe embedded in the wall, as shown in figure 2c, are usual1y spaced horizontally at 5 to 10 ft. Care must be taken that the outflow from the weep holes is carried off safely so as not to seep into and soften the soil underneath the wall. To prevent outflow to seep into the soil underneath the wall, instead of weepers, longitudinal drains embedded in crushed stone or gravel can be provided along the rear face of the wall (figure 2b) at one or more levels; the drains discharge at the ends of the wall or at a few intermediate points. The most efficient drainage is provided by a continuous back drain consisting of a layer of gravel or crushed stone covering the entire rear face of the wall (figure 2a), with discharge at the ends. Such drainage is expensive, however, unless appropriate material is cheaply available at site. In long walls, provision must be made against damage caused by expansion or contraction from temperate changes and shrinkage. The AASHTO Standard Specifications for Highway Bridges require that for gravity walls, as well as reinforced concrete walls, expansion joints be placed at intervals of 90 ft or less, and contraction joints at not more than 30 ft. Allowable bearing pressures should be selected with great care. It is necessary, for this purpose, to investigate not only the type of soil immediately underlying the footing, but also the deeper layers. Unless reliable information is available at the site, subsurface borings should be made to a depth at least equal to the height of the wall. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 9 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Design of Cantilever Retaining Wall: A cantilever retaining wall is to retain a bank 11 ft 6 in. high whose horizontal surface is subject to a live load surcharge of 400 psf. The soil is a sand and gravel mixture with a rather moderate amount of fine, silty particles. It can, therefore, be assumed to be in class 2 of Table 1 given above, with the following characteristics: Unit weight γs = 120 pcf, φ = 30° (with adequate drainage to be provided), and base friction coefficient f = 0.5. With sin 30o = 0.5, the soil pressure coefficients are Kah= 0.333 and Kph= 3.0. The allowable bearing pressure is assumed to be 8000 psf. This coarse-grained soil has little compressibility, so that the resultant can be allowed to strike near the outer-third point. The depth of footing is 3′-6″. The weight of the concrete is γc =150 pcf. Concrete compressive strength = 4500 psi and steel yield strength = 60,000 psi will be used. Given Data: Height of retaining wall above ground level = 11′-6″ Depth of footing = 3′-6″ Live load surcharge over backfill = 400 lb/ft2 Unit weight of soil (γs) = 120 pcf Angle of internal friction of soil (φ) = 30° (with adequate drainage to be provided) Base friction coefficient (f) = 0.5 Kah = 0.333 Kph = 3.0 Allowable bearing pressure = 8000 psf Weight of concrete (γc) = 150 pcf Concrete compressive strength (fc′) = 4500 psi Steel yield strength (fy) = 60,000 psi Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 10 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Solution: Step No 1: Sizes. 8" to 12" 11'-6" h = 15'-0" B/3 GL 3'-6" h/15 to h/10 D = h/15 to h/10 B = 0.4h to 0.65h Figure 6: Trial dimensions for cantilever retaining wall (reference: topic 24.7, pg 705, Treasures of R.C.C Design by Sushil Kumar). Let, B = 0.65h = 0.65 x 15 = 9.75′ D = h/10 = 15/10 = 1.5′ Top width of arm of retaining wall = 8″ Width of arm at bottom = h/12 = 15/12 = 1.25′ = 15″ Length of toe = B/3 = 3.25′ Equivalent depth of surcharge (h′) = s / γs = 0.4 / 0.120 = 3.33′ Stability checks: 1. Check for overturning: Active earth pressure at base (see fig. 3c) of retaining wall is given as: (Pa) = (1/2){Kahγsh(h + 2 h′)} Here h = total height of retaining wall = 15′ Pa = (1/2)(0.333 x 0.120 x 15)(15 + 2 x 3.33) = 6.43 k Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 11 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls 8" 3'-4" a b W5 11'-6" W6 W1 3.25' 3'-6" W2 W4 1.5' W3 15" 9.75' Figure 7: Cross section of retaining wall. Location of resultant (see fig. 3c) from base of retaining wall is: y = (h2 +3hh′)/3(h+2h′) = (152 + 3 x 15 x 3.33)/3 x (15 + 2 x 3.33) = 5.77′ Overturning moment (OTM) = Pay = 6.43 x 5.77 = 37.10 ft-k Calculate the weights shown in figure 7 and take their moment about toe edge. S.No. 1 2 3 4 5 6 Table 2: Weights and moments about front edge (toe) of retaining wall. γ Area (A) W = γA x (from toe) 0.15 (8/12) x 13.5 = 9 1.35 3.58 0.15 (1/2)(7/12)(13.5) = 3.94 0.591 4.11 0.15 9.75 x 1.5 = 18 2.2 4.875 0.12 3.25 x 2 = 6.5 0.78 1.625 0.12 (1/2)(7/12)(13.5) = 3.94 0.4728 4.31 0.12 5.25 x 16.83 = 88.35 10.602 7.125 ∑W=Rv =16.00 k Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Wx 4.83 2.43 10.73 1.27 2.04 75.54 ∑(Wx)= 96.84 k Page 12 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls F.O.S = ∑(Wx) / OTM = 96.84 / 37.10 = 2.61 > 1.5 O.K. 2. Check for sliding: Total horizontal force sliding the wall (Pa) = 6.43 k Now, µ = tanφ ≈ 0.577 Resistance to sliding = µRv = 0.577 x 16.00 = 9.23 k Factor of safety against sliding = µRv / Pa = 9.23 / 6.43 = 1.44. This is slightly less than the recommended value of 1.5 and can be regarded as adequate. However, if key is required, it shall be designed as follows. Design of key: Force resisted by the key = 1.6Pa – µRv = 1.6 x 6.43 – 9.23 = 1.058 k If “z” is the height of key, then the passive earth pressure (Pp) on key is: Pp = (1/2)γsKphz2 = (1/2) x 0.120 x 3 x z2 Now, Pp = Force resisted by key = 0.768 k 1.058 = (1/2) x 0.120 x 3 x z2 z = 2.42 ≈ 2.5 ′ To find width of key, equating the force resisted by key to shear capacity of concrete (ΦVc). ΦVc = 1.058 k 1.058 = Φ2√(4500) x 12 x d/1000 d = 0.88 ′ Therefore b = d + 2.5″ = 0.88 x 12 + 2.5 = 13.06″ ≈ 14″ Therefore, provide a key of 14″ width and 2.5′ depth. 3. Check for Allowable pressure: Rv = 16.50 k (from table 2) To find the point of action (a) of Rv, take moment about the toe of wall: aRv = ∑(Wx) – OTM a = {∑(Wx) – OTM}/Rv Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 13 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls a = (96.84 – 37.10)/ 16.00 a = 3.73′ B/3 = 9.75/3 = 3.25′ < 3.73′ Therefore, resultant lies within the middle third. Hence, equations of case a of figure 5 in this document are applicable. qmax = (4l – 6a) Rv/l2 qmin = (6a – 2l) Rv/l2 l = 9.75′; a = 3.73′; and Rv = 16.00 k qmax = (4 x 9.75 – 6 x 3.73) x 16.00/9.752 = 2.79 ksf qmin = (6 x 3.73 – 2 x 9.75) x 16.00/9.752 = 0.48 ksf Allowable pressure = 8000 psf or 8 ksf > qmax, O.K. If not, increase the width (B) of the retaining wall. 8" 11'-6" 15'-0" 3.25' 3'-6" 15" 1.5' 9.75' Figure 8: Selected dimensions of cantilever retaining wall. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 14 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Step No 2: Loads. i. Toe: Factored Self weight of toe = 0.9 x 0.15 x 1.5 x 1 = 0.203 k/ft Factored earth fill load = 1.6 x 0.12 x 2 x 1 = 0.384 k/ft Total load = 0.203 + 0.384 = 0.587 k/ft Factored soil pressure at exterior end of toe slab = 1.6 x 2.82 x 1= 4.512 k/ft Factored soil pressure at interior end of toe slab = 3.312 k/ft (see fig 10) ii. Heel: Factored Self weight of heel = 1.2 x 0.15 x 1.5 x 1= 0.27 k/ft Factored earth fill load = 1.6 x 0.12 x 13.5 x 1= 2.592 k/ft Factored surcharge load = 1.6 x 0.12 x 3.33 x 1 = 0.639 k/ft Total load = 0.27 + 2.592 + 0.639 = 3.50 k/ft Step No 3: Analysis. i. Arm: Analysis for flexure: Moment variation with height 0 0 274.418796 3 1241.311491 Height (ft) 3116.080343 6 6114.090995 10450.69187 9 16341.22241 24001.01702 12 33645.40707 45489.72204 0 10000 20000 30000 40000 50000 Moment (Mu), lb-ft Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 15 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls General equation of factored active earth pressure w.r.t bottom of arm is: Pa = 1.6 x (1/2)(0.333 x 0.120h)(h + 2 x 3.33) Taking moment at the base of the arm. Moment arm will be equal to: y = (h2 +3h x 3.33)/3(h+2 x 3.33) Mu = Pay (neglecting passive earth pressure for safety) = 1.6(1/2) (0.333 x 0.120h) (h + 2 x 3.33) (h2 +3h x 3.33)/3(h+2 x 3.33) = (0.0319h2 + 0.212h) (h2 +9.99h)/3(h + 6.66)……………..………..(i) Analysis for shear: Active earth pressure coefficient (Kah) = (1 – sinφ)/ (1 + sinφ) = 0.333 The arm is now checked for shear at a distance d above the base of arm: h = Height of section at distance d from base of arm For a cover of 2″ and 1″ dia bar, the effective width “d” at bottom of arm is given as: d = 15 – 2 – 0.5 = 12.5″. Therefore, h = 13.5 – 12.5/12 = 12.46′ Total shear at depth “h” is: V = (1/2)(Kah γs(h + h′ )+ Kah γs h′) (h) (Area of trapezoid) At h = 12.46′ V = (1/2) {0.333x0.120 x (12.46 + 3.33)+ 0.333 x 0.120 x 3.33) 12.46= 4.75 k Factored shear force: Vu = 1.6V = 1.6 x 4.75 = 7.6 k Equivalent height of surcharge (h') = 3.33' h = 12.46' Figure 9: Section in 13.5' Section at a distance "d" from base of arm arm where shear is checked. d = 12.5" Vu Kaγs (h+h') Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 16 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls ii. Toe: Analysis for flexure: Refer figure 10. Mu=[0.587x3.25x(3.25/2)–3.312x3.25x3.25/2–(1/2)1.2x3.25x(2/3)x3.25]x12 = – 223.90 in-k/ft Analysis for shear: Refer figure 10 Effective depth (d) = 18 – 3 – 0.5 = 14.5″ (for 1″ dia bar) Critical shear at distance “d” from section a-a shown in figure 10 is: Vu = 0.587x(3.25–14.5/12)–3.47x(3.25–14.5/12) – (1.228+ 0.68)x (3 – 14.5/12)/ 2 = (3.25 – 14.5/12) {0.587 – 3.47 – (1.228 + 0.68) / 2} = 6.51 k d = 14.5" 0.587 k/ft a a M u= 223 in-k/ft 4.512 k/ft 1.2 k/ft 3.312 k/ft 3'-3"= 3.25' Figure 10: Factored loads on Toe slab. iii. Heel: Analysis for flexure: Refer figure 11. Mu = 3.50 x 5.25 x (5.25/2) x 12 = 578.8 in-k/ft Analysis for shear: Refer figure 11. Effective depth (d) = 18 – 3 – 0.5 = 14.5″ (for 1″ dia bar) Shear at critical section is given as: Vu = 3.50 x (5.25 – 14.5/12) = 14.14 k Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 17 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls d = 14.5" 3.50 k/ft M u= 578.8 in-k/ft 5'-3" = 5.25' Figure 11: Factored loads on Heel slab. Step No 4: Design. i. Arm: Design for flexure: For h = 13.5′ Mu = 45489.7 ft-lb/ft = 545.87 in-k/ft Asmin = 3{√(4500)}/60000)bd = 0.0034 x 12 x (15 – 2 – 0.5) = 0.51 in2/ft a = Asminfy/ (0.85fc′b) = 0.51 x 60/ (0.85 x 4.5 x 12) = 0.66″ Φ Mn = ΦAsminfy (d-a/2) = 0.9 x 0.51 x 60 x (12.5 - 0.66/2) = 469.7 in-k/ft (39142 ft-lb/ft) Φ Mn calculated from Asmin is less than maximum moment in the arm. Therefore using trial method. Trial i: As = Mu/ 0.9fy(d – a/2) Let a = 0.2d = 0.2 x 12.5 = 2.5″ As = 545.87/ {0.9 x 60 x (12.5 – 2.5/2} = 0.90 in2 Trial ii: a = Asfy/ (0.85fc′b) = 0.90 x 60/ (0.85 x 4.5 x 12) = 1.17 in As = 545.87/ {0.9 x 60 x (12.5 – 1.17/2)} = 0.848 in2 Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 18 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Trial iii: a = Asfy/ (0.85fc′b) = 0.848 x 60/ (0.85 x 4.5 x 12) = 1.11 in As = 545.87/ {0.9 x 60 x (12.5 – 1.11/2)} = 0.846 in2, O.K. Using 1″ Φ (#8), with bar area Ab = 0.79 in2 Spacing =Area of one bar (Ab)/As = (0.79 in2/0.848 in2/ft) x 12 = 11.17″ ≈ 11″ • Maximum spacing for main steel reinforcement is: (i) 3h = 3 x 15 = 45″ (ii) 18″ Finally use #8 @ 9″ c/c. Similarly for other depths, the design is given in tabular form as below: Depth (h), ft Thickness of arm (w), in 0 3 6 9 12 13.5 8.00 9.56 11.11 12.67 14.22 15.00 Table 3: Design of main bars in arm of retaining wall. Moment Governing Governing from Design Area of equation (i) Asmin ΦMn,min Moment spacing (M) (in-k/ft) steel (in-k/ft) 0 0.224 58.81 58.81 0.2244 42 14.89 0.287 98.98 98.98 0.28787 32 73.36 0.351 149.52 149.52 0.35133 26 196.09 0.414 210.45 210.45 0.39 24 403.74 0.478 281.75 403.74 0.66 14 545.87 0.51 321.3 545.87 0.846 11 Maximum spacing allowed by ACI Final Spacing for #8 Bars 18 18 18 18 18 18 18 18 18 9 9 9 Therefore, from a depth of 13.5 ft to 9 ft, provide #8 @ 9″. And from 9 ft to 0 ft, provide #8 @ 18″. Horizontal Bars: According to ACI 14.3.3, Ast = 0.0025bh (for deformed bars larger than #5) Ast = 0.0025 x 12 x 15 = 0.45 in2/ft Using 3/4″ dia bar with area Ab = 0.44 in2 Spacing =Area of one bar (Ab)/Ast = (0.44 in2/0.45 in2/ft) x 12 = 11.73″ ≈ 9″ Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 19 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Although not required by the Code for cantilever retaining walls, vertical steel equal to 0.0012 times the gross concrete area will also be provided on exposed face of wall. As = 0.0012 x 12 x 15 = 0.216 in2 Using 3/4″ dia bar with area Ab = 0.44 in2 Spacing =Area of one bar (Ab)/Ast = (0.44 in2/0.216 in2/ft) x 12 = 24.44″ ≈ 24″ Design for Shear: Vu = 7.6 k Shear Capacity is given as: ΦVc = Φ2√(4500) x 12 x 12.5/1000 = 15.09 k > 7.6 k O.K. ii. Toe: Design for flexure: Mu = 223.90 in-k/ft Asmin = 3{√(4500)/60000)bd = 0.0034 x 12 x (18 – 3 - 0.5) = 0.592 in2/ft a = Asminfy/ (0.85fc′b) = 0.592 x 60/ (0.85 x 4.5 x 12) = 0.774″ Φ Mn = ΦAsminfy (d-a/2) = 0.9 x 0.592 x 60 x (14.5- 0.774/2) = 451.16 in-k/ft Φ Mn calculated from Asmin is greater than maximum moment in the toe. Therefore As = Asmin = 0.592 in2/ft Using 1″ Φ (#8), with bar area Ab = 0.79 in2 Spacing =Area of one bar (Ab)/As = (0.79 in2/0.592 in2/ft) x 12 = 16.01″ ≈ 16″ • Maximum spacing for main steel reinforcement: (i) 3h = 3 x 18 = 54″ (ii) 18″ Finally use #8 @ 16″ c/c. Also provide #4 @ 18″ c/c as supporting bars for main bars. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 20 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Design for Shear: Vu = 6.794 k ΦVc =2 x 0.75 x √(4500) x 12 x 14.5/ 1000 = 17.51 k > 6.794 k, O.K. If not O.K, then increase thickness of toe. iii. Heel: Design for Flexure: Mu = 578.8 in-k/ft Asmin = 3{√(4500)/60000)bd = 0.0034 x 12 x (18 – 3 - 0.5) = 0.592 in2/ft a = Asminfy/ (0.85fc′b) = 0.592 x 60/ (0.85 x 4.5 x 12) = 0.774″ Φ Mn = ΦAsminfy (d-a/2) = 0.9 x 0.592 x 60 x (14.5- 0.774/2) = 451.16 in-k/ft Φ Mn calculated from Asmin is less than maximum moment in the Heel. Therefore using trial method. Trial i: As = Mu/ 0.9fy(d – a/2) Let a = 0.2d = 0.2 x 14.5 = 2.9″ As = 578.8/ {0.9 x 60 x (14.5 – 2.9/2} = 0.821 in2 Trial ii: a = Asfy/ (0.85fc′b) = 0.821 x 60/ (0.85 x 4.5 x 12) = 1.073 in As = 578.8/ {0.9 x 60 x (14.5 – 1.073/2)} = 0.767 in2 Trial iii: a = Asfy/ (0.85fc′b) = 0.767 x 60/ (0.85 x 4.5 x 12) = 1.003 in As = 578.8/ {0.9 x 60 x (14.5 – 1.003/2)} = 0.765 in2, O.K. Using 1″ Φ (#8), with bar area Ab = 0.79 in2 Spacing =Area of one bar (Ab)/As = (0.79 in2/0.765 in2/ft) x 12 = 12.39″ ≈ 12″ • Maximum spacing for main steel reinforcement: (i) 3h = 3 x 18 = 54″ (ii) 18″ Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 21 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Finally use #8 @ 12″ c/c. Also provide #4 @ 18″ c/c as supporting bars for main bars. Design for Shear: Vu = 14.14 k ΦVc =2 x 0.75 x √(4500) x 12 x 14.5/ 1000 = 17.51 k > 14.14 k, O.K. If not, increase thickness of heel. Development length check of Arm reinforcement in Heel slab Thickness if heel slab = 1′-6″ = 18″ ldh =(0.02βλfy/√fc′)db > 8db or 6 in (whichever is greater) • (ACI 12.5) β taken as 1.2 for epoxy-coated reinforcement, and λ taken as 1.3 for lightweight aggregate concrete. • For other cases, b and λ shall be taken as 1.0. ldh = (0.02×1×60000/ √(4500))×(8/8) = 17.88″ <hheel OK. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 22 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls 8" Step No 5: Drafting. #6 @ 24" c/c #8 @ 18" c/c 9'-0" to 0'-0" 11'-6" #6 @ 9" c/c 15'-0" #8 @ 9" c/c #4 @ 18" c/c 15" 3'-6" 13'-6" to 9'-0" #4 @ 18" c/c (supporting bars) 1'-6" 9'-9" Prof Dr. Qaisar Ali (http://www.eec.edu.pk) #8 @ 16" c/c Page 23 of 28 #8 @ 12" c/c Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Appendix A Bearing pressures for different locations of resultants: Refer figure 5a, qmax, min = N/A ± Mc/I = Rv/l ± Rv(l/2 – a) (l/2)/(l3/12) = Rv/l ± 3 Rv (l – 3a)/l2] = (Rv/l)[1 ± 3(l – 3a)/l] = (Rv/l)[l ± 3(l – 3a)]/l = (Rv/l2)[l ± 3(l – 3a)] qmax = (Rv/l2)(4l – 6a) qmin = (Rv/l2)(6a – 2l) Refer figure 5b, qmax, min = N/A ± Mc/I = Rv/l ± Rv(l/2 – l/3) (l/2)/(l3/12) = Rv/l ± Rv(l/6) (l/2)/(l3/12) = Rv/l ± Rv/l qmax = 2Rv/l qmin = 0 Refer figure 5c, Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 24 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls qmax, min = N/A ± Mc/I = Rv/3a ± Rv(3a/2 – a) (3a/2)/{(3a)3/12} = Rv/3a ± Rv(3a2/4)/{(3a)3/12} = Rv/3a ± Rv/3a qmax = 2Rv/3a qmin = 0 Related Images: Figure 12: Retaining wall stem under construction. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 25 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Figure 13: Gravity retaining wall. Figure 14: Gravity retaining wall. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 26 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls Figure 15: Cantilever retaining wall under construction. Figure 16: Gravity retaining wall. References Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 27 of 28 Department of Civil Engineering, N-W.F.P UET, Peshawar Design of Retaining Walls 他 Design of Concrete Structures by Nilson, Darwin and Dolan (13th ed.), 他 ACI 318-02/05. Prof Dr. Qaisar Ali (http://www.eec.edu.pk) Page 28 of 28