CIVI_453 Tutorial 12 -Ductile shear wall OK

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CIVI 453 - Design of Ductile Shear Walls Romaric Leo Esteban Desbrousses Fall 2019
Design of Ductile Shear Walls CIVI 453 Contents 1 Theoretical Background 3 1.1 Plastic Hinge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Design for Flexure and Axial Load . . . . . . . . . . . . . . . . . 3 1.2.1 Plastic Hinge Region . . . . . . . . . . . . . . . . . . . . . 3 1.2.2 Other Regions . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2.3 Tie Requirements . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 Preliminary Selection of Vertical Reinforcement . . . . . . . . . . 5 1.4 Moment Resistance at the Base of the Wall . . . . . . . . . . . . 6 1.5 Ductility of Ductile Shear Walls . . . . . . . . . . . . . . . . . . . 6 1.6 Check Wall Thickness for Stability . . . . . . . . . . . . . . . . . 8 1.7 Buckling Prevention Ties for Concentrated Reinforcement . . . . 8 1.8 Shear Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.9 Sliding Shear Resistance . . . . . . . . . . . . . . . . . . . . . . . 9 2 Problem 10 3 Solution 10 3.1 Earthquake Loading . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.2 Select the Wall Thickness . . . . . . . . . . . . . . . . . . . . . . 10 3.3 Find the Concentrated and Distributed Reinforcement . . . . . . 11 3.3.1 Concentrated Reinforcement . . . . . . . . . . . . . . . . 11 3.3.2 Distributed Reinforcement . . . . . . . . . . . . . . . . . . 12 3.4 Moment Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.5 Wall Ductility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.6 Buckling Prevention for Concentrated Reinforcement . . . . . . . 14 3.7 Shear Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3.8 Sliding Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3.9 Axial Load Resistance . . . . . . . . . . . . . . . . . . . . . . . . 15 Romaric Leo Esteban Desbrousses Page 2
Design of Ductile Shear Walls CIVI 453 1 Theoretical Background Notes adapted from [1] and [2]. 1.1 Plastic Hinge Provided that the building is substantially uniform, it is possible to design a wall so that there is one well-defined plastic hinge region in the building (see Cl.21.6.2.2 ). The length of the plastic hinge in a wall is expected to be about equal to the length of the wall. The length over which special detailing must be provided is 1.5 times the length of the wall l w . Figure 1: Plastic Hinge Region 1.2 Design for Flexure and Axial Load 1.2.1 Plastic Hinge Region Distributed reinforcement: Romaric Leo Esteban Desbrousses Page 3
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Design of Ductile Shear Walls CIVI 453 Cl.21.6.5.1 ρ 0 . 0025 for both vertical and horizontal distributed rein- forcement Cl.21.6.5.2 The spacing of ditributed reinforcement in the plastic hinge region must not exceed: s 300 mm If A s,vert > 0 . 005 A g or if the maximum bar size is greater than 15M, the vertical distributed reinforcement must be tied as per Cl.21.6.6.9 Cl.21.6.5.5 Horizontal reinforcement shall be anchored within the region of concentrated reinforcement to develop 1 . 25 f y Concentrated reinforcement: Cl.21.6.6.1 Concentrated reinforcements are needed at then ends of walls Cl.21.6.6.2 Concentrated reinforcement must resist the portion of fac- tored load effects including earthquake not resisted by distributed vertical reinforcement Cl.21.6.6.4 The minimum area of concentrated reinforcement in plastic hinge regions must be (at each end of the wall): A s,min 0 . 0015 b w l w A s 0 . 06 × area of concentrated region Cl.21.6.6.8 The concentrated reinforcement must be tied with buckling prevention ties as per Cl.21.6.6.9 and must also satisfy Cl.7.6 Cl.21.6.6.7 Not more than 50% of the reinforcement at each end of the walls shall be spliced at the same location 1.2.2 Other Regions Distributed reinforcement: Cl.21.6.5.1 ρ 0 . 0025 for both vertical and horizontal distributed rein- forcement. The spacing of distributed reinforcement must not exceed: s 450 mm Vertical reinforcement must be tied as per Cl.7.6.5 unless A s,vert < 0 . 005 A g and the maximum bar size is 20M or smaller Cl.21.6.5.4 Horizontal reinforcement must extend to the ends of the wall and contained at each end of the wall within a region of concentrated reinforcement as specified in Cl.21.6.6 Concentrated reinforcement: Cl.21.6.6.1 Concentrated reinforcements are needed at then ends of walls Romaric Leo Esteban Desbrousses Page 4
Design of Ductile Shear Walls CIVI 453 Cl.21.6.6.3 The minimum area of concentrated reinforcement must not be less than (at each end of the wall): A s = 0 . 001 b w l w Cl.21.6.6.8 Concentrated reinforcement must be tied as specified in Cl.7.6 Splice: 1 . 5 l d and 100% at the same location 1.2.3 Tie Requirements Cl.21.6.6.9: If ρ > 0 . 005 and in the plastic hinge zone, the tie spacing must not exceed: s min of 6 d b,l 24 d tie 1 / 2 of least dimension of the member s from Cl. 21 . 6 . 7 Cl.7.6.5.2: If ρ > 0 . 005 outside the plastic hinge region, the tie spacing must not exceed: s min of 16 d b,l 48 d tie Least dimension of member 300 mm for bundled bar 1.3 Preliminary Selection of Vertical Reinforcement In regions of plastic hinging, the minimum area of concentrated reinforcement is (at least four bars): A s,min = 0 . 0015 b w l w Outside regions of plastic hinging, the minimum area of concentrated reinforce- ment becomes: A s,min = 0 . 001 b w l w Cl.21.6.4.3 states that the reinforcement ratio in any region of concentrated reinforcement must not exceed ρ = 0 . 06 (even in regions containing lap splices). Romaric Leo Esteban Desbrousses Page 5
Design of Ductile Shear Walls CIVI 453 Cl.21.6.4.4 states the diameter of the bars used in a wall cannot exceed ( b w is the thickness): d b = b w 10 Cl.21.6.5.1 and Cl.21.6.5.2 state the reinforcement ratio for distributed re- inforcement cannot be less than ρ = 0 . 0025 and that the maximum spacings are: s = 300 mm in the plastic hinge 450 mm in other regions Note that the wall moments are mainly resisted by the concentrated reinforce- ment. Walls designed with distributed steel only fail by rupture of the edge tension reinforcement prior to developing significant ductility. 1.4 Moment Resistance at the Base of the Wall Once the factored moment applied on the wall is know, it must be compared to the wall’s moment resistance ( M r ), given by the following equation: M r = 0 . 5 φ s A s f y l w (1 + P f φ s A s f y )(1 - c l w ) + φ s A s,conc f y l Where c/l w is found by using Cardenas and Magura’s simplified formulae: c l w = ω + α 2 ω + α 1 β 1 ω = φ s A s f y φ c f 0 c l w t α = P f φ c f 0 c l w t 1.5 Ductility of Ductile Shear Walls Find the elastic deformation Δ f from the dynamic analysis on ETABS or SAP2000 and use it to calculate the inelastic rotational demand θ id as spec- ified in Cl.21.6.7.2 . Romaric Leo Esteban Desbrousses Page 6
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Design of Ductile Shear Walls CIVI 453 θ id = Δ f R o R f - Δ f γ w h w - l w 2 0 . 004 The value of 0.004 is a minimum rotational demand. Where: Δ f R o R d is the design displacement Δ f γ w is the elastic portion of the displacement l w is the length of the longest wall in the direction being considered γ w is the wall overstrength factor and is defined as the ratio between the nominal moment resistance of the wall to the factored moment on the wall. It is typically taken as 1.3 Figure 2: Inelastic Rotational Demand The inelastic rotational capacity ( θ ic ) is then calculated and compared to the inelastic rotational demand to make sure that θ ic θ id . Cl.21.6.7.3: θ ic = cu l w 2 c - 0 . 002 0 . 025 c is the distance to the neutral axis. Romaric Leo Esteban Desbrousses Page 7
Design of Ductile Shear Walls CIVI 453 1.6 Check Wall Thickness for Stability Cl.21.6.3.2 indicates that the wall thickness within a region of plastic hinging is: t w l u 14 The wall thickness is generally: t l u 10 Cl.21.6.3.4 states that if c < 4 b w or c < 0 . 3 l w , t l u 10 does not need to be applied. Note that l u is the unsupported length of the wall. 1.7 Buckling Prevention Ties for Concentrated Reinforce- ment Cl.21.6.6.9 states that the tie spacing must not exceed the smallest of: s 6 d b,l,concentrated 24 d b,hoop 0 . 5 t w 1.8 Shear Design Walls must resist the shear force corresponding to the formation of a plastic hinge at the base. The design base shear is given by: V = V f × M pw M f Where M pw is the wall’s probable moment resistance. The probable moment resistance is calculated in a fashion similar to that of the regular moment resis- tance. The distance arises from the value of φ s , φ c and f y used to calculate the moment resistance. To get M pw , φ c and φ s are taken as unity while f y is taken as 1 . 25 f y . Alternatively, the probable moment resistance can be estimated as M pw 1 . 47 M r . Once the design base shear is known, V max must be calculated to ensure that V < V max . V max = 0 . 15 φ c f 0 c b w d v Romaric Leo Esteban Desbrousses Page 8
Design of Ductile Shear Walls CIVI 453 Calculate the wall’s shear resistance given by concrete and steel ( Cl.11.3.4 and Cl.11.3.5.1 ): V r = V c + V s = φ c β q f 0 c b w d v + φ s A v f y d v cot ( θ ) s Where β = 0 . 18 and θ = 45 (given that θ id 0 . 005, see Cl.21.6.9.6 ). Check that V r V f . 1.9 Sliding Shear Resistance The factored shear stress resistance is given by: v r = φ c ( c + μ ( ρ v f y + N A g )) Where N is the factored axial force and ρ v is the percentage of shear friction reinforcement. Check that: V r = v r t w l w V Romaric Leo Esteban Desbrousses Page 9
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Design of Ductile Shear Walls CIVI 453 2 Problem This problem is taken from [3] . Design a ductile shear wall of length l w = 6 , 950 mm located in a 9-storey building with a floor height of 3.5m. The floors are 220mm thick flat plates. The wall is subjected to the following loads at its base: Axial load only: P f = 10 , 683 kN Axial load and moment: P f = 5 , 466 kN , M f = 16 , 523 kN.m and V f = 958 kN 3 Solution 3.1 Earthquake Loading For a ductile shear wall: R d = 3 . 5 and R o = 1 . 6. As such: V = S ( T a ) M v I E W R d R o = 0 . 109 × 1 × 1 × W 3 . 5 × 1 . 6 = 0 . 0195 W V = 0 . 0195 × 58 , 619 = 1 , 143 kN Floor h (m) V (kN/Wall) M (kN.m/Wall) Roof 31.5 0 0 8 28 138 484 7 24.5 273 1441 6 21 391 2810 5 17.5 492 4511 4 14 577 6440 3 10.5 644 8548 2 7 695 10771 1 3.5 728 13046 0 0 745 15314 Recall that provided that the wall section remains uniform and that the rein- forcement is properly curtailed, a plastic hinge will form near the base of the wall with a height of 1 . 5 l w 3.2 Select the Wall Thickness Cl.21.6.3.2 : t l u 10 = 3 , 500 - 220 10 = 328 mm Use t = 350 mm . Romaric Leo Esteban Desbrousses Page 10
Design of Ductile Shear Walls CIVI 453 3.3 Find the Concentrated and Distributed Reinforcement 3.3.1 Concentrated Reinforcement In the plastic hinge, the minimum area of concentrated steel is: A s,min = 0 . 0015 b w l w = 0 . 0015 × 350 × 6 , 950 = 3 , 648 . 75 mm 2 Use 6-30M, giving an area of steel of 4 , 200 mm 2 at both ends of the wall. Figure 3: Concentrated Reinforcement Outside the plastic hinge region, the minimum area of concentrated reinforce- ment becomes: A s,min = 0 . 001 b w l w = 2 , 432 . 5 mm 2 Use 4-30M to get an area of steel of 2 , 800 mm 2 . Cl.21.6.4.3 states the ρ 6% in the concentrated reinforcement region even in the event of splicing. To size the length (m) of the concentrated region, start by considering the 6-30M bars (see Figure 4) used in the plastic hinge region: ρ = 4 , 200 350 × m 0 . 06 m 4 , 200 350 × 0 . 06 = 200 mm Now consider the case where 50% of the bars are spliced in the concentrated reinforcement region (see Figure 5). Since 3 bars are being spliced, there is an additional 3 bars to consider when looking at a plane orthogonal to the bars, giving: ρ = (3 + 6) × 700 350 × m 0 . 06 m (3 + 6) × 700 350 × 0 . 06 = 300 mm Use m=350mm. The maximum allowable bar size is given by Cl.21.6.4.4 as: d b = 350 10 = 35 mm Romaric Leo Esteban Desbrousses Page 11
Design of Ductile Shear Walls CIVI 453 Figure 4: Concentrated Reinforcement Figure 5: Splicing in the Concentrated Region 3.3.2 Distributed Reinforcement In the plastic hinge region, the maximum allowable spacing is 300mm. That spacing becomes 450mm in other regions. The reinforcement ratio ρ must be greater than 0.0025. Assuming 10M bars are used, the area of steel needed to get ρ = 0 . 0025 is: A s,min = 0 . 0025 × 350 × 1 , 000 = 875 mm 2 /m The thickness of the wall is greater than 250mm. As such, the horizontal and vertical reinforcement are placed in two layers. The number of bars per layer can be found using: # bars = 8755 mm 2 /m 2 × 100 mm 2 /bar = 4 . 375 bars/m ( per layer ) Romaric Leo Esteban Desbrousses Page 12
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Design of Ductile Shear Walls CIVI 453 The resulting spacing is: s = 1 , 000 4 , 375 = 228 . 57 mm c/c Use two layers of 10M@220mm. This spacing gives the following total number of bars per layer: (6 , 950 - 350) = 220 x x = 28 . 4 Two layers of 29-10M. The total area of steel is A s = 2 × 29 × 100 = 5 , 800 mm 2 . The vertical reinforcement are shown in Figure 6. Figure 6: Vertical Reinforcement (Red: Concentrated, Blue: Distributed) 3.4 Moment Resistance The moment resistance is calculated using Cardenas and Magura’s equations. The first step is to locate the depth of the neutral axis (c). ω = φ s A s f y φ c f 0 c l w t = 0 . 85 × 5 , 800 × 400 0 . 65 × 30 × 6 , 950 × 350 = 0 . 0415 α = P f φ c f 0 c l w t = 5 , 466 × 10 3 0 . 65 × 30 × 6 , 950 × 350 = 0 . 115 c l w = ω + α 2 ω + α 1 β 1 = 0 . 115 + 0 . 0415 2 × 0 . 0415 + 0 . 805 × 0 . 895 = 0 . 195 The moment resistance is then calculated using: Romaric Leo Esteban Desbrousses Page 13
Design of Ductile Shear Walls CIVI 453 M r = 0 . 5 φ s A s f y l w (1 + P f φ s A s f y )(1 - c l w ) + φ s A s,conc f y l Note that l is the lever arm between the centroid of the two concentrated rein- forcement regions ( l w - m = 6 , 950 - 350 = 6 , 600 mm ). M r = 0 . 5 × 0 . 85 × 5 , 800 × 400 × 6 , 950(1 + 5 , 466 × 10 3 0 . 85 × 5 , 800 × 400 )(1 - 0 . 195) +0 . 85 × 4 , 200 × 400 × 6 , 600 M r = 30 , 230 kN.m M f = 16 , 523 kN.m 3.5 Wall Ductility The ductile shear wall has the following factors: R d = 3 . 5, R o = 1 . 6. The dynamic analysis on ETABS gave Δ f = 14 mm . The inelastic rotational demand is given by Cl.21.6.7.2 : θ id = Δ f R d R o - Δ f γ w h w - l w 2 = 14 × 3 . 5 × 1 . 6 - 14 × 1 . 3 31 , 500 - 6 , 950 / 2 = 0 . 002148 < 0 . 004 Use θ id = 0 . 004. The inelastic rotational capacity is given by Cl.21.6.7.3 : θ ic = cu l w 2 c - 0 . 002 = 0 . 0035 × 6 , 950 2 × 1 , 355 . 25 - 0 . 002 = 0 . 00697 0 . 025 θ ic θ id 3.6 Buckling Prevention for Concentrated Reinforcement Cl.21.6.6.9 : s 6 d b,l = 6 × 30 = 180 mm 24 d tie = 240 mm 0 . 5 t w = 175 mm Use 10M ties spaced at 170mm. Horizontal reinforcement must develop 1 . 25 f y in the plastic hinging concen- trated region. As such their development length is: l d = 0 . 45 k 1 k 2 k 3 k 4 1 . 25 f y p f 0 c d b l d = 0 . 45 × 1 . 25 × 400 30 × 10 = 329 mm 350 mm Romaric Leo Esteban Desbrousses Page 14
Design of Ductile Shear Walls CIVI 453 3.7 Shear Design The probable moment resistance of the wall can be approximated as: M pw 1 . 47 M r = 1 . 47 × 30 , 230 = 44 , 438 . 1 kN.m V = V f × M pw M f = 958 × 44 , 438 . 1 16 , 523 = 2 , 576 . 51 kN V max = 0 . 15 φ c f 0 c b w d v = 0 . 15 × 0 . 65 × 30 × 350 × 0 . 8 × 6 , 950 V max = 5 , 962 kN V The wall’s shear resistance is: V r = φ c β q f 0 c b w d v + φ s A v f y d v cot ( θ ) s V r = 0 . 65 × 0 . 18 × 30 × 350 × 0 . 8 × 6 , 950 + 0 . 85 × 200 × 400 × 5 , 560 × 1 220 V r = 2 , 965 . 61 kN V 3.8 Sliding Shear v r = φ c ( c + μ ( ρ v f y + N A g )) Cl.11.5.2 gives the values of c and μ . For concrete placed against hardened concrete with the surface clean but not intentionally roughened: c = 0 . 25 MPa and μ = 0 . 60. v r = 0 . 65(0 . 25 + 0 . 60(0 . 0025 × 400 + 5 , 466 × 10 3 350 × 6 , 950 )) = 1 . 43 MPa V r = v r × t w × l w = 1 . 43 × 350 × 6 , 950 = 3 , 478 . 47 kN V r 3.9 Axial Load Resistance The wall’s axial load carrying capacity is: P r = 2 3 α 1 φ c f 0 c A g (1 - ( kh u 32 t ) 2 ) P r = 2 3 × 0 . 805 × 0 . 65 × 30 × 350 × 6 , 950 1 - ( 1 × 3 , 280 32 × 350 ) 2 P r = 23 , 272 . 86 kN P f Romaric Leo Esteban Desbrousses Page 15
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Design of Ductile Shear Walls CIVI 453 References [1] Khaled Galal. Design of Ductile Walls . Concordia University. [2] Canadian Standards Association. CSA A23.3-04 - Design of Concrete Struc- tures . English. 2004. [3] Khaled Galal. CIVI 453 - Chapter 7 - Example on Design of Ductile Shear Walls . Concordia Univeristy. 2018. Romaric Leo Esteban Desbrousses Page 16

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