Shear Walls. Shear Walls: Stiffness

Transcription

1 Shear Walls Load Distribtion to Shear Walls Shear wall stiffness Shear walls with openings Diaphragm tpes Tpes of Masonr Shear Walls Maxim Reinforcement Reqirements Shear Strength Example: simple bilding Shear Walls 1 Shear Walls: Stiffness Lateral Force Resisting Sstem d h h/d < < h/d < 4.0 h/d >4.0 stiffness predominates Both shear and bending stiffness are important stiffness predominates Shear Walls 2

2 Shear Wall Stiffness Deflection calclations shall be based on cracked section properties. Assmed properties shall not exceed half of gross section properties, nless a cracked-section analsis is performed. Cantileer wall cant k cant h L Emt h 4 L Fixed wall (fixed against rotation at top) fixed k fixed h L 2 Emt h L h = height of wall A = shear area; (5/6)A for a rectangle E = G = shear modls (modls of rigidit); gien as 0.4E m ( ) E =, where ν is Poisson s ratio t = thickness of wall L = length of wall Real wall is probabl between two cases; diaphragm proides some rotational restraint, bt not fll fixit. Shear Walls 3 T- or L- Shaped Shear Walls Section Wall intersections designed either to: a) : b) Connection that transfers shear: (st be in rnning bond) a) Fift percent of masonr nits interlock b) Steel connectors at max 4ft. c) Intersecting bond beams at max 4 ft. Reinforcing of at least 0.1in 2 per foot of wall Metal lath or wire screen to spport grot 2-#4 s 1 / 4 in. x 1 1 / 2 in. x 28in. with 2in. long 90 deg bends at each end to form U or Z shape Shear Walls 5

3 Effectie Flange Width ( ) Effectie flange width on either side of web shall be smaller of actal flange width, distance to a moement joint, or: Flange in compression: 6t Flange in tension: Unreinforced masonr: 6t Reinforced masonr: 0.75 times floor-to-floor wall height Shear Walls 6 Example: Flanged Shear Wall Gien: Fll groted shear wall Reqired: Stiffness of wall Soltion: Determine stiffness from basic principles. A 112in. Find centroid from oter flange srface 7.62in48in3.81in 7.62in40in20in in 2 671in I A 56in. 6t=48in. 40in. Eleation Plan Shear Walls 7

4 Example: Flanged Shear Wall P P 1 k in 3 3 Ph Ph 112in 112in in 305in 0.4E 2 3EmI n A E 3Em m E m Shear Walls 9 Shear Walls with Openings 1. Diide wall into piers. 2. Find flexibilit of each pier 3. Stiffness is reciprocal of flexibilit 4. Distribte load according to stiffness 3 h 2k t ktkb 2kb 3 12h f 1 6EI kt 2ktkb kb 5EA M M t b Vh k t Vh k t kt 1 kb 2k k k t b b kb 1 kt 2k k k t As the top spandrel decreases in height, the top approaches a fixed condition against rotation. b b k t k b h h t h h b h b h h t Top spandrel Bottom spandrel If h b = 0 3 h 2 k t 12h f 1 6EI 2kt 1 5EA M t Vh k k 1 2 t t M b 1 k t Vh 2kt 1 Qamarddin, M., Al-Oraimi, S., and Hago, A.W. (1996). Mathematical model for lateral stiffness of shear walls with openings. Proceedings, Seenth North American Masonr Conference, Shear Walls 10

5 Example - Shear Walls with Openings Gien: 4ft 4ft Reqired: A. Stiffness of wall B. Forces in each pier nder 10 kip horizontal load A B C 8ft 2ft 4ft 2ft 6ft 2ft Shear Walls 11 Example - Shear Walls with Openings Sample calclations: Pier B h = 4ft h t = 4ft h b = 8ft k t = h/h t = 4ft/4ft =.0 k b = h/h b = 4ft/8ft = tl I A tl 12 2 ftt f 3 h 2k t ktkb 2kb 3 12h 1 6EI kt 2ktkb kb 5EA k Et f 47.6 / Et 0210 Shear Walls 12

6 Example - Shear Walls with Openings Pier h (ft) h t (ft) h b (ft) k t k b I (ft 3 ) A (ft) f s f k A inf 0.667t 2t /Et Et B t 2t /Et Et C t 2t /Et Et f is flexral deformation; s is shear deformation Total stiffness is Et Aerage stiffness from finite element analsis Et Solid wall stiffness Et (free at top); 0.25Et (fixed at top) Shear Walls 14 Example - Shear Walls with Openings Sample calclations: Pier B M M b t Vh k Vh k t t kt 1 kb 2k k k t b kb 1 kt 2k k k t b b b kb Et Vb V 10k 4. 66kips k Et Pier V (kips) M t (k-ft) M b (k-ft) A B C Shear Walls 15

7 Example - Shear Walls with Openings To find axial force in piers, look at FBD of top 4ft of wall. 10k Moment direction on piers 4ft 0.68k 4.66k 4.66k 3.50k-ft 11.19k-ft 11.19k-ft 2ft 4ft 2ft 6ft 2ft This 65.9k-ft moment is carried b axial forces in the piers. Axial forces are determined from Mc/I, where each pier is considered a concentrated area. Find centroid from left end. A i A i i Shear Walls 17 Example - Shear Walls with Openings The axial force in pier A is: FBD s of piers: 4.45k 3.50k-ft 0.68k 0.45k 11.19k-ft 4.66k 4.90k 11.19k-ft 4.66k B C A 0.45k 4.66k 7.46k-ft 4.90k 4.66k 7.46k-ft 4.45k 0.68k 4.66k-ft Shear Walls 19

8 Example - Shear Walls with Openings FBD s of entire wall to obtain forces in bottom right spandrel: 10k F x 10k 0.68k Fx 0 Fx 9. 32k F 4.45k F 0 F 4. 45k x A B C M z 10k(16 ft) (4.66k ft) 4.45k(1 ft) 4.45k(11ft) Mz 0 Mz 110.8k ft Z 4.45k 0.68k 4.66k-ft F=4.45k Fx=9.32k Mz=110.8k-ft Shear Walls 21 Shear Walls: Bilding Laot 1. All elements either need to be isolated, or will participate in carring the load 2. Elements that participate in carring the load need to be properl detailed for seismic reqirements 3. Most shear walls will hae openings 4. Can design onl a portion to carr shear load, bt need to detail rest of strctre Shear Walls 23

9 Copled Shear Walls Shear Wall Shear Wall Shear Walls 24 Diaphragms Diaphragm: sstem that transmits forces to the ertical elements of the lateral load resisting sstem. Diaphragm classification: : distribtion of shear force is based on tribtar (wind) or tribtar (earthqake) : distribtion of shear force is based on relatie. Lateral Force Resisting Sstem Tpical classifications: : Precast planks withot topping, metal deck withot concrete, plwood sheathing : Cast-in-place concrete, precast concrete with concrete topping, metal deck with concrete Shear Walls 25

10 Rigid Diaphragms Direct Shear: F V RRi RR i RRid i Torsional Shear: Ftt Ve 2 RR d i i V = total shear force RR i = relatie rigidit of lateral force resisting element i d i = distance from center of stiffness e = eccentricit of load from center of stiffness Shear Walls 26 Diaphragms: Example Gien: The strctre shown is sbjected to a 0.2 kip/ft horizontal force. Relatie rigidities are gien, where the relatie rigidit is a normalized stiffness. Reqired: Distribtion of force assming: flexible diaphragm rigid diaphragm. RR = 4 50 ft 50 ft RR = kip/ft RR = 1 PLAN VIEW Shear Walls 27

11 Flexible Diaphragms: Example Soltion: Flexible diaphragm wind Distribte based on tribtar area 50 ft 50 ft For seismic, the diaphragm load wold be distribted the same (assming a niform mass distribtion), bt when wall weights were added in, the forces cold be different. RR = 4 RR = 5 RR = 1 PLAN VIEW Shear Walls 28 Rigid Diaphragms: Example Soltion: Rigid diaphragm 50 ft 50 ft Wall x RR x(rr) RR = 4 Wall 1 RR = 5 Wall 2 RR = 1 Wall 3 Total Center of stiffness = 350/10 = 35 ft x Wall RR d (ft) RR(d) RR(d 2 ) F F t F total Total Shear Walls 29

12 Soltion: Forces are shown for a rigid diaphragm. The moment is generall taken throgh chord forces, which are simpl the moment diided b the width of the diaphragm. In masonr strctres, the chord forces are often take b bond beams. Diaphragms Design Forces V (k) kip/ft 50 ft 50 ft 3.9 k 12.2 k 3.9 k 19.5 ft M (k-ft) -55 Shear Walls 30 Drag Strts and Collectors Shear forces: generall considered to be niforml distribted across the width of the diaphragm. Drag strts and collectors: transfer load from the diaphragm to the lateral force resisting sstem. PLAN VIEW w L/3 L/3 L/2 L/2 L/2 L WEST WALL ELEVATION EAST WALL ELEVATION Shear Walls 31

13 Diaphragm Behaior Three lateral force resisting sstems: Length=20 ft; Height=14 ft W24x68 W16x40 W14x68 Moment Resisting Frame k = 83.2 kip/in W14x68 L 4x4x5/16 Braced Frame k = 296 kip/in E = 1800 ksi W16x40 Face shell bedding End cells fll groted L 4x4x5/16 Masonr Shear Wall k = 1470 kip/in Lateral force resisting sstems at 24 ft o.c. Diaphragm assmed to be concrete slab, E=3120ksi, ν=0.17, ariable thickness, load of 1 kip/ft. Diaphragm Lateral Force Resisting Sstem 24 ft 24 ft Shear Walls 32 Diaphragm Behaior Shear Walls 33

14 Diaphragm Behaior Shear Walls 34 Diaphragm Behaior Shear Walls 35

15 Shear Walls (nreinforced) shear wall ( ): Unreinforced wall (nreinforced) shear wall ( ): Unreinforced wall with prescriptie reinforcement. Within 16 in. of top of wall Strctrall connected floor and roof leels 40d b or 24 in. 8 in. Corners and end of walls 8 in. Control joint 16 in. 10 ft. Joint reinforcement at 16 in. o.c. or bond beams at 10 ft. Reinforcement not reqired at openings smaller than 16 in. in either ertical or horizontal direction Reinforcement of at least 0.2 in 2 Shear Walls 36 Shear Walls reinforced shear wall ( ): Reinforced wall with prescriptie reinforcement of detailed plain shear wall. reinforced shear wall ( ): Reinforced wall with prescriptie reinforcement of detailed plain shear wall. Spacing of ertical reinforcement redced to 48 inches. reinforced shear wall ( ): 1. Maxim spacing of ertical and horizontal reinforcement is min{1/3 length of wall, 1/3 height of wall, 48 in. [24 in. for masonr in other than rnning bond]}. 2. Minim area of ertical reinforcement is 1/3 area of shear reinforcement 3. Shear reinforcement anchored arond ertical reinforcing with standard hook 4. Sm of area of ertical and horizontal reinforcement shall be times gross cross-sectional area of wall 5. Minim area of reinforcement in either direction shall be times gross cross-sectional area of wall [ for horizontal reinforcement for masonr in other than rnning bond]. Shear Walls 37

16 Minim Reinforcement of Special Shear Walls Reinforcement Ratio 8 in. CMU wall 12 in. CMU wall A s (in 2 /ft) Possibilities A s (in 2 /ft) Possibilities #4@32 #5@ #4@24 #5@32 #6@48 #4@24 #4@ #5@ #5@24 #6@40 #6@ #4@16 #5@32 #6@ #5@16 #6@24 Use specified dimensions, e.g in. for 8 in. CMU walls. Shear Walls 38 Special Shear Walls: Shear Capacit Minim shear strength ( ): Design shear strength, V n, greater than shear corresponding to 1.25 times nominal flexral strength, M n Except V n need not be greater than 2.5V. Normal design: φv n has to be greater than V. Ths, V n has to be greater than V /φ = V /0.8 = 1.25V. Ths, the reqirement approximatel dobles the shear. Shear Walls 39

17 Seismic Design Categor Seismic Design Categor A or B C D and higher Response modification factor: Seismic design force diided b response modification factor, which acconts for dctilit and energ absorption. Knoxille, Tennessee Categor C for Use Grop I and II Categor D for Use Grop III (essential facilities Allowed Shear Walls Shear Wall R Ordinar plain 1.5 Detailed plain 2 Ordinar reinforced 2 Intermediate reinforced 3.5 Special reinforced 5 Shear Walls 40 Maxim reinforcing ( ) No limits on maxim reinforcing for following case ( ): M 1 and R 1.5 Sqat walls, not designed for dctilit V d In other cases, can design b either proiding bondar elements or limiting reinforcement. Bondar element design ( ): More difficlt with masonr than concrete Bondar elements not reqired if: P P f 0.1 ma g 0.05 fma g geometricall smmetrical sections geometricall nsmmetrical sections AND M V l w 1 M OR V 3 3An f m AND V l w Shear Walls 41

18 Maxim reinforcing ( ) Reinforcement limits: Calclated sing Maxim stress in steel of f Axial forces taken from load combination D+0.75L+0.525Q E Compression reinforcement, with or withot lateral ties, permitted to be inclded for calclation of maxim flexral tensile reinforcement Uniforml distribted reinforcement = 1.5 ordinar walls = 3 intermediate walls = 4 special walls Compression steel with area eqal to tension steel As d As bd 0.64 f mb f f P d fmb 0.64 d min d P bd, Es Shear Walls 42 Maxim reinforcing Consider a wall with niforml distribted steel: f Strain Steel in tension Stress 0.8f m Steel in compression A s taken as total steel d is actal depth of masonr C m C s T s P Cm fm d b T s f As Portion of steel in tension C s Yielded steel 0.5 f As Shear Walls Elastic steel 1 f As f As

19 Shear Walls 44 Maxim reinforcing P C C T m s s s s s s s A f A f A f C T P C P b d f A f C T m m s s s m s f d P bf d A 0.64 Shear Walls 45 Maxim reinforcing, ε s = 4ε

20 Maxim reinforcing, ε s = 3ε Shear Walls 46 Maxim reinforcing, ε s = 1.5ε Shear Walls 47

21 V n nm ns g Shear Strength ( ) V V 0. 8 γ g = 0.75 for partiall groted shear walls and 1.0 otherwise M Vm An fm Vd P M /V d need not be taken > 1.0 P = axial load V s A 0. 5 f d s Vertical reinforcement shall not be less than onethird horizontal reinforcement; reinforcement shall be niforml distribted, max spacing of 8 ft ( ) Maxim V n is: V n A n f m g V A f Interpolate for ales of M /V d between 0.25 and 1.0 n 6 ( M / V d ) n m g 4 ( M / V d ) 1. 0 V n 4 M Vd A n f m g Shear Walls 48 Partiall Groted Walls Method Mean V exp / V n St. De. Partiall Groted Walls (Minaie et al, 2010; 60 tests) 2008 Proisions Mltipl shear strengths b A g /A n Using jst face shells Fll Groted Walls (Dais et al, 2010; 56 tests) 2008 Proisions /1.16 = 0.776; ronded to 0.75 Shear Walls 49

22 Design of Shear Walls with Single Laer of Reinforcement 1. Determine a, depth of compressie stress block a d d 2 2 P d lw / f t m sp M 2. Sole for A s A s 0.8 fm tspa P / f 3. Check axial capacit 4. Check maxim reinforcement 5. Check shear Shear Walls 50 Example Gien: 2 ft long, 8 ft high CMU pier; Tpe S masonr cement mortar; Grade 60 steel; fll groted. P = 11 kips, V = 7 kips, M = 28 k-ft Reqired: Reqired amont of steel Soltion: Choose/determine material properties. fʹm = 2000 psi; f = 60,000 psi 20 in. 24 in. a d d 2 2 P d lw / 2 M 0.8 f t m sp A s 0.8 fm tspa P f / Shear Walls 51

23 Example Tr #4 bars Consider second laer of steel: P C n m T 1 T 2 20 in. 24 in. Sole for c sch that P n = 11 kips/0.9 = 12.2 kips c = inches; C m = 29.14kips; T = 12kips; T 1 = 4.92kips M n 334k in 27.9k ft Shear Walls 53 Example 3. Check axial load 1 3 I 12 bt 1 2 r 12 t 0.289t A bt 7.625in. 2. in h r in. 8 ft 12 ft 2.20in P n f m A n A st 2.0ksi7.62in. 24in. 0 60ksi0 k k f A st 2 h 1 140r k 190. k P n k < 190k OK Applied load is 6% of axial capacit Shear Walls 55

24 Example 4. Check maxim reinforcement: Fll groted with eqal tension and As bd P f 0.64 m bd d d compression reinforcement f min, Es Assme axial force is from 0.9D, so P for maxim reinforcement is 11k/0.9 = 12.2 kips. For an ordinar wall: ksi in. 60ksi min in in. 20in. 12.2k, ksi in. 20in. 3. A s, max bd in Shear Walls 56 Example Maxim Reinforcement Axial Force, P A s,reqd Ordinar Intermed. Special 0 kips 0.32 in in in in 2 11 kips (6% of axial capacit 0.21 in in in in 2 19 kips (10% of axial capacit) 0.13 in in in in 2 38 kips (20% of axial capacit) in in in 2 Shear Walls 57

25 Example 5. Check Shear: M V d 28k ft 7k 2 ft 2 Use M /(V d ) = 1.0 V nm M Vd A in. 24in psi 1ksi k 18.4k 2.8k 21.2k n f m 0.25P 1000 psi V 21.2k 16.9k V k nm OK Shear Walls 58 Example Gien: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, f m =2000psi; 2-#5 each end; #5 at 48in. (one space will actall be 40 in.) Reqired: Interaction diagram Soltion: Illstrate with one point, c = 54 inches, a = 0.8(54) = 43.2 in. Shear Walls 59

26 Example T 1 T 2 T 3 N.A. T 4 C 2 C in. 8 in 188 in 140 in 92 in 54 in Strain Shear Walls 60 Example P n Sm moments abot middle of wall (8 ft from end) to find M n M n Shear Walls 62

27 Mltiple Laers of Reinforcement: Design eqations for single laer of reinforcement can be sed for preliminar steel estimates Spacing of intermediate reinforcing bars often controlled b ot-of-plane loading Shear area for a partiall groted shear wall is area of face shells and groted cells Shear Walls 64 Example Gien: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Tpe S mortar; f m =2000psi; sperimposed dead load of 1 kip/ft. In-plane seismic load (from ASCE 7-10) of 90 kips. S DS = 0.4 Reqired: Design the shear wall; ordinar reinforced shear wall Soltion: Check sing 0.9D+1.0E load combination. Need to know weight of wall to determine P. Need to know reinforcement spacing to determine wall weight Estimate P b ignoring wall weight and seismic ertical force P = 0.9(1k/ft)(16ft) = 14.4k M = (90k)(10ft) = 900k-ft Shear Walls 65

28 Estimate reqired reinforcement: Example 1. Use eqations for single laer of reinforcement (d ~ d - 4 =188 in.): a = 6.0 in., A s = 0.95 in 2 This is abot 3.1 #5 bars; tr 2-#5 at end and 64 in. 2. Use spreadsheet/compter program and interaction diagram: oerstressed b 3.3%; ma work Shear Walls 66 Example Weight of wall: 36psf(10ft)(16ft) = 5760 lb Lightweight nits, grot at 64 in. o.c. 36 psf (estimated) P = [ (S DS )]D = [ (0.4)](1k/ft(16ft)+5.8k) = 17.8 kips With slightl higher axial force M n = 995 k-ft M n has increased to 896 k-ft Wall weight ma be slightl greater de to bond beams Shear Walls 67

29 Example Calclate net area, A n, inclding groted cells. A n M / V d Maxim V n V n Shear Walls 68 Example Top of wall (critical location): P = ( S DS )D = 0.82(1k/ft)(16ft) = 13.1 kips M Vnm An fm 0. 25P Vd V n V V V V nm ns g ns Use #5 bars in bond beams. Determine spacing. V ns A 0.5 s f d s ASD: s min{d/2, 48 in.} = min{94 in., 48 in.} = 48 in. Code In strength design, this proision onl applies to beams ( (e)) Sggest that minim spacing also be applied to shear walls. Use #5 bars at 16 inches Shear Walls 70

30 Joint Reinforcement Maxim specified ield strength of 85,000 psi (b) Minim diameter of 3/16 in. diameter Anchor arond edge reinforcing bar in the edge cell b bar placement between adjacent crosswires, or 90 bend in longitdinal wires with 3-in. bend extensions in mortar or grot Seismic Reqirements Seismic Design Categories (SDC) A and B At least two 3/16 in. wires; Maxim spacing of 16 in. SDC C, D, E, and F; partiall groted walls At least two 3/16 in. wires; Maxim spacing of 8 in. SDC C, D, E, and F; fll groted walls At least for 3/16 in. wires; Maxim spacing of 8 in. Shear Walls 72 Eqialent Joint Reinforcement Options: Joint Reinforcement Joint Reinforcement Eqialent Bar Reinforcement Replaces this Reinforcement 2-3/16 in. wires at 16 in in 2 /ft 56 in.; 80 in. 2 3/16 in. wires at 8 in in 2 /ft 32 in.; 40 in. 4 3/16 in. wires at 8 in in 2 /ft 16 in.; 24 in. Bar reinforcement ield stress = 60 ksi Joint reinforcement ield stress = 70 ksi Instead of bond beams in example, cold se 4 3/16 in. wires at 8 in. bt no manfactrer makes this Smaller distribted reinforcement generall reslts in better behaior. Shear Walls 73

31 Detailing Sggestion Add 90 corner bar Tension Steel Tie Conentional Model Moment Diagram Strt and Tie Model Eleation of Shear Wall Shear Walls 74 Example: Special Wall Gien: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Tpe S mortar; f m =2000psi; sperimposed dead load of 1 kip/ft. In-plane seismic load (from ASCE 7-10) of 90 kips. S DS = 0.4 Reqired: Design the shear wall; special reinforced shear wall Soltion: Check sing 0.9D+1.0E load combination. Shear capacit design proisions (Section ) V n shear corresponding to 1.25M n. Minim increase is 1.25/0.9 = 1.39 M n = 995 k-ft; 1.25M n = 1243 k-ft; design for shear of 124 kips V n need not exceed 2.5V Normal design V n V / = V /0.8 = 1.25V Increases shear b a factor of 2 Fll grot wall (Max V n was 91.9 kips) A n 7.625in. 192in in Shear Walls 75

32 Example: Special Wall Design for shear of 124 kips Bt wait, spacing of #5 s was decreased to 48 inches M n = 1124 k-ft, 1.25M n = 1405 k-ft, Design for 140 kips Bt wait, wall is fll groted. Wall weight has increased to 75 psf For P = 23.0k, fll groted, M n = 1167 k-ft, 1.25M n = 1459 k-ft Design for 146 kips Bt wait, Section has maxim spacing reqirements: 1/3 length of wall = (192 in.)/3 = 64 in. 1/3 height of wall = (120 in.)/3 = 40 in. 48 in. for rnning bond; 24 in. for not laid in rnning bond Decrease spacing of ertical reinforcement from 48 in. to 40 in. For 2-#5 at end, and 40in., M n = 1269 k-ft (P = 23.0k) 1.25M n = 1586 k-ft; Design for 159 kips Bottom line: an change in wall will change M n, which will change design reqirement; often easier to jst se 2.5V. Shear Walls 76 Example: Special Wall V nm M Vd 0.8 A n f m 0.25P 2 1kip in 2000 psi k 154.8kips 1000lb V V V 158.6k 154.8k 0.8 nm ns 4. 7 k 0.31in60ksi192in A A f d Vns 0.5 f d s 383in s V 4.7k Use maxim spacing of 1/3(height) = 40 in. ns Use #5 bars in bond beams. Determine spacing. Section (d): Shear reinforcement shall be anchored arond ertical reinforcing bars with a standard hook. Shear Walls 77

33 Example: Special Wall Prescriptie Reinforcement Reqirements ( ) in each direction total Vertical: 8(0.31in 2 )/1464in 2 = (spacing of 40 in.) Horizontal: 3(0.31in 2 )/[120in(7.625in)] = Total = = OK Shear Walls 78 Example: Special Wall Section Maxim Reinforcement Since M /(V d ) < 1, strain gradient is based on 1.5ε. Strain c/d, CMU c/d, Cla 1.5ε ε ε c = 0.446(188in.) = 83.8 in. Calclate axial force based on c = 83.8 in. Inclde compression reinforcement P n = 732 kips Assme a lie load of 1 k/ft D L Q E = (1k/ft (1k/ft))16ft = 28 kips OK Shear Walls 79

34 Example: Special Wall Section : Maxim reinforcement proisions of do not appl if designed b this section (bondar elements) Special bondar elements not reqired if: P P f 0.1 ma g f 0.05 ma g geometricall smmetrical sections geometricall nsmmetrical sections AND M V d 1 M OR V 3 3An f m AND V d For or wall, M /V d < 1 P < 0.1fʹmA g = 0.1(2.0ksi)(1464in 2 ) = 293 kips OK Shear Walls 80 Special Walls: Smmar Prescriptie Reinforcement Reqirements ( ) in each direction total Spacing Reqirements ( ) Shear Capacit Design (Section ) V n shear corresponding to 1.25M n. V n need not exceed 2.5V Maxim Reinforcement Reqirements ( ; ) Shear Walls 81

35 Example: T-Wall Gien: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Tpe S mortar; f m =2000psi; sperimposed dead load of 1 kip/ft. In-plane seismic load (from ASCE 7-10) of 90 kips. S DS = 0.4; intersecting wall on one side. Reqired: Design the shear wall; ordinar reinforced shear wall Soltion: Check sing 0.9D+1.0E load combination. 16ft 1 k/ft Effectie flange width: 6t = 6(8in.) = 48 in. compression 0.75h = 0.75(120in.) = 90 in. tension Shear Walls 82 Example: T-Wall Flange in tension: Approximate as a single laer of reinforcement. Use design procedre for single laer of reinforcement with P = 18.1k, M = 900k-ft., b=7.625in., l w =200in. (16 ft + 8 in. flange); d=196in. Reqired reinforcement: a = 5.9in., A s = 0.87in 2 or 3 - #5. Flange in compression: Reinforcement will be approximatel the same as for a non-flanged wall. Increase in compression area will onl slightl redce reqired steel. Axial load on jst the web creates a moment with a small tension in the flange and compression in the web. Shear Walls 83

36 Example: T-Wall Trial Design Tension: 3 bars total; assme spacing of 48 in. for OOP 90 in. flange Compression: 3 groted cells within flange All reinforcement #5 bars Flange compression area: ( )(2.5) + 3(8)( ) = 382in 2 Eqialent thickness = 382in 2 /7.625in. = 50.1in. Compression Width Tension Width Shear Walls 84 Example: T-Wall Check maxim reinforcement with flange in tension: c/d = (α = 1.5) c = 0.446(196in.) = 87.4 in. P n = kips P = D+0.75L = (6.1+16) +0.75(16) = 34.1 kips Perhaps shold inclde steel jst otside effectie tension flange: P n = kips Shear Walls 87

37 Example: T-Wall Shear strength is based on the flange, and similar to preios example. Use 40 in. Shear Walls 88 Example: T-Wall Check shear at interface. Check sing intersecting bond beams. Reinforcement from #5@40in in 2 /ft close to 0.1in 2 /ft in 40in. 12in in / ft 1ft Shear at interface: Set eqal to tension force in reinforcement in flange. V = T = f A sf = 60000psi(3)(0.31in 2 ) = lb A n V 2 120in 3(8in)(7.62in 2.5in) in in nm M Vd 0.8 A n f in 2000psi 34050lb m 0.25P ft high wall 3 bond beams A n f m If sed γ = 0.75, 25,500 lbs Shear Walls 89

38 Example: T-Wall Need to fll grot interface: A n = 915in 2 ; V n = lb OK Additional cells to be groted Sggest adding shear straps between each bond beam (16 in. o.c.) Sggest sing bond beam nits and ctting ot half of face-shell in flange blocks to get good grot at interface. Shear Walls 90