Lecture 6 7 Rm Shear Walls
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Lecture 6 7 Rm Shear Walls

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Dan Abrams + Magenes Course on Masonry

Dan Abrams + Magenes Course on Masonry

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Lecture 6 7 Rm Shear Walls Lecture 6 7 Rm Shear Walls Presentation Transcript

  • Reinforced Masonry Working Stress Design of flexural members strain Ref: NCMA TEK 14-2 Reinforced Concrete Masonry BIA Tech. Note 17 Reinforced Brick Masonry - Part I BIA Tech. Note 17A Reinforced Brick Masonry - Materials and Construction T s = A s f s C m = f m b kd/2 M b d t A s grout unit n.a. kd f m f s /n stress
  • Assumptions 1. plane sections remain plane after bending (shear deformations are neglected, strain distribution is linear with depth) 2. neglect all masonry in tension 3. stress-strain relation for masonry is linear in compression 4. stress-strain relation for steel is linear 5. perfect bond between reinforcement and grout (strain in grout is equal to strain in adjacent reinforcement) 6. masonry units and grout have same properties Reinforced Masonry Working Stress Design of flexural members
  • Reinforced Masonry Working Stress Design of flexural members If f s =F s then moment capacity will be limited by reinforcement. Allowable reinforcement tensile stress per MSJC Sec.2.3.2: F s =20 ksi for Grades 40 or 50; F s =24 ksi for Grade 60 F s =30 ksi for wire joint reinforcement Allowable reinforcement tensile stress per UBC Sec.2107.2.11 : F s = 0.5 f y < 24 ksi for deformed bars; F s = 0.5 f y < 30 ksi for wire reinforcement F s = 0.4 f y < 20 ksi for ties, anchors, and smooth bars from equilibrium: where jd=d-kd/3 or j=1-k/3
  • Reinforced Masonry Working Stress Design of flexural members If f m = F b then moment capacity will be limited by masonry. UBC 2107.2.6 & MSJC Sec.2.3.3.2: F b =0.33f’ m from equilibrium:
  • WSD: Balanced Condition
    • For any section and materials, only one unique amount of balanced reinforcement exists.
    • Although balanced condition is purely hypothetical case, it is useful because it alerts the engineer to whether the reinforcement or the masonry stress will govern the design. Balanced stresses are not a design objective.
    Reinforced Masonry Definition: The balanced condition occurs when the extreme fiber stress in the masonry is equal to the allowable compressive stress, F b , and the tensile stress in the reinforcement is equal to the allowable tensile stress, F s . C = f m b kd/2 f s /n = F s /n d k b d f m = F b T = A s f s
  • WSD: Balanced Condition Reinforced Masonry from geometry: f s /n = F s /n d k b d f m = F b from equilibrium: C=T
  • Example: Balanced Condition Determine the ratio of reinforcement that will result in a balanced condition per UBC. Given: f’ m = 2000 psi and Grade 60 reinforcement
  • Design Strategy for RM Flexural Design Procedure for sizing section and reinforcement for given moment. Calculate  b knowing f’ m and f y determine F b from f’ m determine F s from f y determine E m from f’ m determine n = E s /E m Size section for some  b determine k and j bd 2 = M/  jF s select b and d using common units Size reinforcement A s = M/F s jd select number and size of rebars Check design M s = A s F s jd > M f b = M/0.5jkbd 2 < F b Note: Section must also be sized for shear.
  • Example: Reinforced Masonry Design a beam section for a moment equal to 370 kip-in. Prisms have been tested and f’ m is specified at 2000 psi. Use Grade 60 reinforcement and 8” CMU’s.
  • Example: Reinforced Masonry 5. Select dimensions of beam using 8” CMU’s: b = 7.63” d req’d = [4321 / 7.63] 0.5 = 23.8” use four units and center bars in bottom unit, d = 27.8” 4 - 8” CMU’s d=27.8” 7.63”
  • Flexural Capacity of Partially Grouted Masonry Case A: neutral axis in flange * per MSJC Sec. 2.3.3.3 If kd < t f assumption is valid, determine moment capacity as for rectangular section. If kd > t f assumption is not valid , need to consider web portion. t t f A s per width b b = 6t or 72” or s * A s flange b d kd neutral axis If neutral axis is in flange, cracked section is the same as a solid rectangular section with width “b.” Therefore, depth to neutral axis from extreme compression fiber may be calculated using:
  • Shear Design of Reinforced Masonry d s
    • V m reduces
    • dowel action invoked
    Once diagonal crack forms:
    • flexural stresses increase
    • f sa is related to M b
    A s f s C m R Basic shear mechanisms: before cracking: V ext = V int = V m + V d + V iy + V s V d V m V ext. V iy V s Presence of shear reinforcement will:
    • restrict crack growth
    • help dowel action
    • resist tensile stress
  • Shear Design of Reinforced Masonry after cracking: V ext = V int = V s = nA v f s where n is the number of transverse bars across the diagonal crack. Assuming a 45 degree slope, n=d/s V s = (d/s)A v f s UBC Sec. 2107.2.17 (Eq. 7.38) MSJC Sec. 2.3.5.3 (Eq. 2-26)
  • Shear Design of Reinforced Masomry Flexural shear stress f v b dx = dT = dM/jd f v = (dM/dx)/bjd dx UBC Sec. 2107.2.17 (Eq. 7-38) MSJC Sec. 2.3.5.2.1 (Eq. 2-19) M M + dM C jd T C + dC T + dT na T T + dT f v bdx
  • Shear Design of Reinforced Masonry Allowable shear stresses for flexural members per UBC and MSJC UBC Sec. 2107.2.8.A and MSJC Sec. 2.3.5.2.2(a): members with no shear reinforcement UBC Eq. 7-17; MSJC Eq. 2-20 UBC Sec. 2107.2.8.B and MSJC Sec. 2.3.5.2.3(a): members with shear reinforcement designed to take the entire shear UBC Eq. 7-18; MSJC Eq. 2-23
  • Shear Design of Reinforced Masonry Allowable shear stresses for shear walls per UBC and MSJC UBC Sec. 2107.2.9.i and MSJC Sec. 2.3.5.2.2(b): walls with in-plane flexural reinforcement and no shear reinforcement UBC Sec. 2107.2.9.ii and MSJC Sec. 2.3.5.2.3(b): walls with in-plane flexural reinforcement and shear reinforcement designed to take 100% of shear UBC Eq. 7-19; MSJC Eq. 2-21 UBC Eq. 7-21; MSJC Eq. 2-24 UBC Eq. 7-20; MSJC Eq. 2-22 UBC Eq. 7-22; MSJC Eq. 2-25
  • Shear Design of Reinforced Masonry Moment-to-Shear Ratios d V M For a single-story cantilevered shear walls h V M M For piers between openings h d
  • Shear Design of Reinforced Masonry MSJC Sec. 2.3.5.3.1 s max = d/2 or 48” d/2 V design Additional MSJC Requirements V design MSJC Sec. 2.3.3.4.2 minimum reinforcement perpendicular to shear reinforcement = A v /3 s max = 8 ft MSJC Sec. 2.3.5.5 design for shear force at distance “d/2” out from support
  • Shear Design of Reinforced Masonry consider as unreinforced End Resize Section Determine F v Assuming Shear Reinforcement to take 100% of Shear Determine F v Assuming No Shear Reinforcement Determine Maximum Design Shear Shear Design Strategy for Reinforced Sections Determine Flexural Tension Stress f t = -P/A+Mc/I Start Provide Reinforcement to Take 100% of Shear Determine Shear Stress is f v <F v ? is f t >F t ? yes no is f v <F v ? no yes no yes
  • Example: Design of RM Shear Wall Determine the maximum lateral force, H wind per UBC and MSJC 8” CMU wall Type S - PCL mortar solidly grouted f’ m =3000 psi #4 @ 32” 2 - #8’s each end of wall 6’-8” 6’-4” 8’-0” 120 psi Case A: neglect all reinforcement Case B: consider vertical reinf., neglect horizontal reinf. Case C: consider vertical and horizontal reinf. Case D: design horizontal reinforcement for max. shear
  • Example: Design of RM Shear Wall Case A: neglect all reinforcement shear per UBC: flexure per MSJC: flexure shear
  • Example: Design of RM Shear Wall Case B: consider only vertical reinforcement Shear per UBC Sec.2107.2.9 or MSJC Sec.2.3.5.2 Flexure by UBC or MSJC: neglecting f a lumping 2 - #8’s ave. d for 2 bars M s = A s F s jd = 2 x 0.79 in 2 (1.33 x 24 ksi) (0.9 x 72.0”) = 3268 k-in H wind = 34.0 kips
  • Example: Design of RM Shear Wall Flexure by UBC or MSJC: same as case B Case C: consider all reinforcement Overall shear per UBC Sec. 2107.2.9.C or MSJC Sec. 2.3.5.2.3 (b) UBC MSJC V max = V s =(A v /s)F s d = (0.20 in 2 /32”)(24 ksi x 1.33)(72”) = 14.4 kips governs Shear per UBC Sec. 2107.2.17 or MSJC Sec.2.3.5.3
  • Example: Design of RM Shear Wall Case D: design horizontal reinforcement for maximum shear strength Summary: H max, kips 10.2* 27.0 15.2 34.0* No steel vertical steel no horizontal steel vertical steel and #4 @ 32” horizontal steel #4 @ 8” horizontal 14.7* 24.3 15.2 34.0* Case A B C D Consideration UBC MSJC *flexure governs Case A B C D Consideration UBC MSJC No steel 14.7* 10.2* vertical steel no horizontal steel 23.0 25.6 vertical steel and #4 @ 32” horizontal steel 14.4 14.4 #4 @ 8” horizontal 34.0* 34.0*
  • Flexural Bond Stress M = Tjd M + dM = (T + dT)jd dM = dT jd dT = dM/jd allowable bond stress per UBC Sec.2107.2.2.4: 60 psi for plain bars 200 psi for deformed bars 100 psi for deformed bars w/o inspection dx U T T + dT dx M C C + dC T jd T + dT M + dM dx dx U = bond force per unit length for group of bars U dx = dT = dM/jd U = (dM/dx)/jd = V/jd u = flexural bond stress = where sum of perimeters of all bars in group UBC Sec. 2107.2.16 Eq. 7-36
  • Development Length d b l d
  • Embedment of Flexural Reinforcement UBC Sec. 2106.3.4 and MSJC Sec.2.1. 8.3 Rule #1: extend bars a distance of “d” or “12d b ” past the theoretical cutoff point Rule #2: extend bars a distance of “l d ” past the point of maximum stress theoretical cutoff point capacity with bars “a” Moment Diagram moment capacity with bars “a” and “b” (#1) d or 12d b (#2) > l d (#2) > l d bars “b” bars “a” Example for shear wall:
  • Combined Bending and Axial Loads Code Requirements UBC Sec. 2.14.2 if h’/t >30 then analysis should consider effects of deflections on moments MSJC Sec. 2.3.3.2.2 f a + f b < 1/3 f’ m provided that f a < F a In lieu of approximate method, use an axial-force moment interaction diagram. UBC Sec. 2107.1.6.3 use unity formula to check compressive stress: UBC Sec. 2107.1.6.1 P f a = P/A e Note: unity formula is conservative - better approach is to use P-M interaction diagram. M UBC Sec. 2107.2.15 A s f s jd kd
  • Axial Force-Moment Interaction Diagrams General Assumptions
    • 1. plane sections remain plane after bending
      • shear deformations neglected
      • strain distribution linear with depth
    Strain Stress  m  s f m T s =A s f s C s P M 2. neglect all masonry in tension 3. neglect steel in compression unless tied 4. stress-strain relation for masonry is linear in compression 5. stress-strain relation for steel is linear 6. perfect bond between reinforcement and grout • strain in grout is equal to strain in adjacent reinforcement 7. grout properties same as masonry unit properties
  • Axial Force-Moment Interaction Diagram Range “a”: large P, small M, e=M/P < t/6 C m f m2 f m1 Out-of-Plane Bending of Reinforced Wall unit width = b e m P a = 0.5(f m1 + f m2 )A M a = 0.5(f m1 - f m2 )S where S = bt 2 /6 M b P a d = t/2
  • Axial Force-Moment Interaction Diagram Out-of-Plane Bending of Reinforced Wall f m1 unit width = b Range “b” medium P, medium M, e > t/6 , A s in compression d = t/2 0.5 <  < 1.0 for section with reinforcement at center e m C m P b M b
  • Axial Force-Moment Interaction Diagram Out-of-Plane Bending of Reinforced Wall Range “c” small P, large M, e > t/6 , A s in tension  < 0.5 for section with reinforcement at center unit width = b d = t/2 f m1 e m C m T s M c P c
  • Axial Force-Moment Interaction Diagram f s < F s f m1 = F b f s = F s f m1 < F b Out-of-Plane Bending of Reinforced Wall e=0; M=0 f m1 = f m2 =F a P=F a A Start Stop Determine P and M per Range “b” Reduce f m2 from 2F a -F b by increment Determine P & M per Range “a” Determine P and M per Range “c” f s < F s ? M = 0? f m2 = 0? no yes is A s in tension? no yes Reduce  from 1.0 by increment Range “b” Range “a” f m1 = F b = f’ m /3 Reduce  from 0.5 by increment Range “c” no no yes tension controlling compression controlling yes
  • Axial Force-Moment Interaction Diagram Out-of-Plane Bending of Reinforced Wall Axial Force F b Range “b” f m2 = 2F a - F b f m1 = F a f m1 = F b Range “a” F b F b F a limit by unity formula e 1 f s = F s F s /n f m Moment tension controls Range “c” balanced point F s /n F b compression controls f s /n F b
  • Example: Interaction Diagram Determine an axial force-moment interaction diagram for a fully grouted 8” block wall reinforced with #4 @ 16”. Prism compressive strength has been determined by test to be equal to 2500 psi. Reinforcement is Grade 60. Height of wall is 11.5 feet.
  • Example: Interaction Diagram Case f m1 (psi)  C m (kips) f m2 (psi) e m (in.) T s (kips) P=C m - T s (kips) M=C m e m (kip-in) 9 for P = 0: M m = 0.5F b jkbd 2 = 0.5(833 psi)(0.909)(0.272)(12)(3.81) 2 = 17.9 Range 2 833 417 - 57.2 - - 57.2 24.1 3 833 0 - 38.1 1.27 - 38.1 48.4 5 833 - 0.50 19.1 2.54 - 19.1 48.5 7 833 - 0.25 9.5 3.18 2.0 7.5 30.2 8 833 - 0.167 6.4 3.39 3.9 2.5 21.5 11 664* - 0.150 4.6 3.43 3.6 1.0 15.7 4 833 - 0.75 28.6 1.91 - 28.6 54.5 b 12 check for P = 0: M s = A s F s jd = (0.15 in 2 )(24 ksi)(0.909)(3.81”) = 12.5 1 625 625 - 57.2 0 - 57.2 0 a Compression Controls 6 833 - 0.33 12.6 2.97 0.9 11.7 37.4 10 833 bal. - 0.175 6.7 3.37 3.6 3.1 22.5 c Tension Controls *masonry stress inferred from F s and 
  • Example: Interaction Diagram Axial Force kips 50 40 30 20 10 10 20 30 40 50 Moment, kip-in 1 12 10 10 833 3.6 k = A s F s 0.175t 11 11 664 3.6 k = A s F s 0.15t 1 625 417 2 833 2 0 3 833 3 4 833 .75t 4 5 833 0.50t 5 6 833 0.9k 0.33t 6 7 833 2.0 k 0.25t 7 8 8 833 3.9 k > A s F s 0.167t 9
  • Flexural Capacity with Axial Compression Short Cut Method Out-of-Plane Bending, Reinforcement at Center stress compatibility: [1] d T s C m jd M P kd f s /n f m d [2] [3]
  • Flexural Capacity with Axial Compression equilibrium: Short Cut Method [4] [5] [6] [7] [8] [9] [10] [11] [12]
  • Strength Design of Reinforced Masonry Ultimate Flexural Strength A s t b Note: rectangular stress block can represent compressive stress distribution if k 2 /k 1 = 0.5 stresses C m M n T s = A s f y strains c n.a. d d f’ m f m c k 3 f’ m k 2 c C m c k 2 c C m = k 1 k 3 f’ m bc k l c k 3 f’ m =
  • Strength Design of Reinforced Masonry Measuring k 1 k 3 and k 2  increase P 1 so that  = 0 P 0 +P 1 stress k 2 c k 3 f’ m k 1 c summing moments about centroid: P 1 a = (P o + P 1 )g = (P o + P 1 )(c/2 - k 2 c) total compressive force: P o + P 1 = k 3 f’ m k 1 cb a P o in displacement control P 1 in force control P o P 1 strain c g
  • Strength Design of Reinforced Masonry Sample experimentally determined constants k 1 k 3 , and k 2 0 0.2 0.4 0.6 0.8 1 0 0.001 0.002 0.003 0.004 0.005 0.006 Extreme Fiber Strain (in/in) K1K3 & K2 K1K3 K2 Measured k 1 k 3 and k 2 values
  • Strength Design of Reinforced Masonry Ultimate Flexural Strength f s summing moments about C m f y equilibrium
  • Strength Design of Reinforced Masonry Balanced condition with single layer of reinforcement d strains c n.a. stresses k 1 c C m T s = A s f y M n strain compatibility equilibrium
  • Strength Design of Reinforced Masonry f’ m Grade 40 Grade 60 Balanced condition with single layer of reinforcement 1000 0.0124 0.0062 0.0036 0.0071 2000 0.0247 0.0124 0.0143 0.0072 3000 0.0371 0.0186 0.0214 0.0107 4000 0.0495 0.0247 0.0285 0.0142 5000 0.0619 0.0309 0.0356 0.0178 6000 0.0742 0.0371 0.0428 0.0214
  • Strength Design of Reinforced Masonry Balanced condition with multiple layers of reinforcement c A sbal b strain compatibility equilibrium strains d 4 d 3 d 2 d 1 0.85c stresses 0.85f’ m C s1 C m =0.85f’ m b(0.85c) C s2 T s3 T s4 = A sbal f y
  • Example: Flexural Strength of In-Plane Wall Maximum steel is equal to one-half of that resulting in balanced conditions. f ’m = 1500 psi Grade 60 reinforcement special inspection Determine the maximum bar size that can be placed as shown. A sbal ? 7.63” 5’-4” T s3 T s4 = A sbal f y P n = 0 C s2 C s1 C m = 0.85f’ m b(0.85c) 0.85f’ m 60.0” c n.a. 0.003 4.0” 44.0” 20.0”
  • Example: Flexural Strength of In-Plane Wall c = 0.003/0.00507 (60.0”) = 35.5” C m = 0.85f’ m b(0.85c) = -0.85(1500)(7.63”)(0.85 x 35.5) = -294 k Determine the maximum bar size. layer d i  si f si with compression steel (include C s1 and C s2 forces) C m +  (C si + T si ) = -294 + A sbal (-60.0 - 38.0 + 20.8 + 60.0) = 0 A sbal = -17.1 in 2 note: negative A sbal means that  C >  T , in such case no limit on tensile reinforcement 1 4.0” -0.00261 (C) -60.0 2 20.0” -0.00131 (C) -38.0 3 44.0” 0.00072 (T) 20.8 4 60.0” 0.00207 (T) 60.0 *bars larger than #9 are not recommended because of anchorage and detailing problems without compression steel (neglect C s1 andC s2 forces) C m +  (C si + T si ) = -294 + A sbal (20.8+ 60.0) = 0 A sbal = 3.64 in 2 A smax = 1.82 in 2 max. bar size is #ll (1.56 in 2 )*
  • Example: Flexural Strength of In-Plane Wall Determine flexural strength of wall. 7.63” 5’-4” f ’m = 1500 psi Grade 60 reinforcement special inspection #8 (typ) 44.0” 20.0” 4.0” 60.0” c n.a. 0.003 T s4 = A s f y = 0.79 in 2 x 60 ksi = 47.4 k 0.85f’ m T s3 T s2 C s1 C m = 0.85f’ m b(0.85c)
  • Example: Flexural Strength of In-Plane Wall c d 1 = 4.0” f 1 C sl d 2 = 20.0” f 2 T s2 d 3 = 44.0” f 3 T s3 C m Determine flexural strength of wall. -0.00240 -60 -47.4 0 0 0 0.00360 60.0 47.4 20.0 -165 -117.6 c 85 . 0 in kip 222 , 5 ) 89 . 4 0 . 60 )( 4 . 47 ( ) 89 . 4 0 . 44 )( 4 . 47 ( ) 89 . 4 0 . 20 )( 4 . 47 ( ) 89 . 4 00 . 4 )( 8 . 44 ( )} 2 d ( f A { M i si si n               15.0 -0.00220 -60 -47.4 0.00100 29.0 22.9 0.00580 60.0 47.4 -124 -54 11.0 -0.00191 -55 -43.7 0.00245 60.0 47.4 0.00900 60.0 47.4 -91 +7.5 close to zero, take c = 11.5” 11.5 -0.00196 -56 -44.8 0.00222 60.0 47.4 0.00848 60.0 47.4 -95 +2.3
  • Example: Flexural Strength of In-Plane Wall Approximate flexural strength of wall. 7.63” 5’-4” #8 (typ)
  • UBC Sec. 2108.2.4 Limitations of Method: Slender Wall Design (a) for out-of-plane bending of solid, reinforced walls lightly stressed under gravity loads (c)  g = A s /bt < 0.5  bal Ref: NCMA TEK 14-11A Strength Design of Tall Concrete Masonry Walls (d) special inspection must be provided during construction (e) t > 6” Sec. 2108.1.3: Load factors (b) limited to:
  • Required Flexural Strength: UBC Sec. 2108.2.4.4 Slender Wall Design e h P uf t P uw w u h 2 /8 transverse load w u (P uw + P uf ) u h/2 h/2 P  eccentric load P uf e/2 P uf e
  • Slender Wall Design Design Considerations 2. strain compatibility 3.  mu = 0.003 4. f s = E s  s < f y 5. neglect masonry strength in tension 6. rectangular stress block, k 1 = 0.85, k 3 = 0.85 Design strength: Sec. 2108.2.4.4 Assumptions for ultimate flexural strength (Sec. 2108.2.1.2) 1. equilibrium Strength reduction factor: flexure  = 0.8 Sec. 2108.1.4.2.1
  • Slender Wall Design Equivalent area of reinforcement, A e for single wythe construction reinforced at center: b A s d Eq. (8-24) flexural strength P u = C m - A s f y C m = P u + A s f y = A se f y T s = A s f y T s = A s f y P u P u c d a = 0.85c 0.85f’ m C m =0.85f’ m b(0.85c) C m
  • Slender Wall Design Lateral Deflections M M y M cr M s  s  y  cr Modulus of Rupture, f r Eqs. 8-31, 32, 33 fully grouted partially grouted hollow unit 2-wythe brick for M s < M cr (8-28) for M s > M cr (note “kd” may be replaced by “c” for simplicity) M cr = f r S (8-29)
  • Slender Wall Design Design Considerations Strength Criteria Serviceability Criteria
  • Example: Slender Wall Design Determine the maximum wind load, w, per UBC and MSJC 8” CMU, partially grouted f’ m = 2000 psi, Grade 60 500 lbs/ft dead 200 lbs/ft live #4 @ 32” 20’-0” 3’-0” 3.5” P w
  • Example: Slender Wall Design U = 0.75 (1.4D + 1.7L + 1.7W) Flexural Strength per UBC M u =  M n =  A se f y (d - a/2)=0.8(0.103in 2 )(60ksi)(3.81in - 0.302in/2)=18.1 kip-in
  • Example: Slender Wall Design to avoid iteration, assume M max = M u Flexural Strength per UBC w u =22.7 psf for simplicity, use gross section even though partially grouted
  • Example: Slender Wall Design Check Service Load Deflections per UBC
  • Example: Slender Wall Design Maximum Wind Load per MSJC Determine I cr considering axial compression 1.25” #4 @ 32” d = 3.82”
  • Example: Slender Wall Design Note: same wind load as by UBC slender wall design procedure. Should also check compressive stress with an axial force-moment interaction diagram Maximum Wind Load per MSJC
  • Strength Design of RM Shear Walls UBC Sec. 2108.1.1: Strength procedure may be used as an alternative to Sec. 2107 for design of reinforced hollow-unit masonry walls. UBC Requirements UBC Sec. 2108.1.2: Special inspection must be provided during construction. Prisms should be tested or unit strength method should be used. B. gravity loading: U = 1.4D + 1.7E (12-3) C. wind loading: U = 0.75(1.4D + 1.7 L + 1.7W) (12-4) U = 0.9D + - 1.3W (12-5) D. earth pressure: U = 1.4D + 1.7L + 1.7H (12-6) B. shear 2. Design strength A. axial load and flexure (see next slide) 1. Required strength A. earthquake loading: U = 1.4 (D+L+E) (12-1) U = 0.9D + - 1.4E (12-2) UBC Sec.2108.1.3: Shear wall design procedure
  • Strength Design of RM Shear Walls Definition of Balanced Axial Load, P b b L w C m = 0.85f’ m ba b d 0.85f’ m C s1 C s2 P b T s3 T s4 = A sbal f y c n.a.
  • Strength Design of RM Shear Walls 3. Design assumptions (same as for Slender Wall Design Procedure, UBC Sec. 2108.2.1.2) 1. equilibrium UBC Requirements 2. strain compatibility 3.  mu  = 0.003 4. f s = E s  s < f y 5. neglect masonry tensile strength 6. use rectangular stress block, k 1 = 0.85, k 3 = 0.85 7. 1500 psi < f’ m < 4000 psi
  • Strength Design of RM Shear Walls 2. for flexural failure mode M n > = 1.8 M cr for fully grouted wall M n > = 3.0 M cr for partially grouted wall 3. anchor all continuous reinforcement M  UBC Requirements 4. A s vertical > 1/2 A s horizontal 5. maximum spacing of horizontal reinforcement within plastic hinge region = 3t or 24” M cr M n > M cr ductile M n < M cr nonductile 4. Reinforcement per UBC Sec. 2108.2.5.2 1. minimum reinforcement 5. Axial strength (no flexure) P o = 0.85 f’ m (A c -A s ) + f y A s P u < =  (0.80)P o
  • Strength Design of RM Shear Walls UBC Requirements A mv = net area of masonry wall section bounded by wall thickness and length of section in direction of shear 2. for walls limited by shear strength: 6. Shear Strength UBC Sec. 2108.2.5.5 1. maximum nominal shear: A mv L w V u t
  • Strength Design of RM Shear Walls UBC Requirements h t 6. Shear Strength UBC Sec. 2108.2.5.5.2 (continued) A vertical plane A s horizontal A s f y h L w L w A s f y L w A s f y
  • Strength Design of RM Shear Walls 6. Shear Strength UBC Sec. 2108.2.5.5: continued 3. for walls limited by flexural strength: UBC Requirements within hinge region, distance of L w above base: ( V u determined at L w /2 from base) above hinge region:
  • Strength Design of RM Shear Walls 1. Provide boundary members when the extreme fiber strain exceeds 0.0015. Boundary Members: Sec.2108.2.5.6 UBC Requirements Section at Base of Wall > 3t wall centroid #3 @ 8” min. 0.0015 t  mu > 0.0015 2. The minimum length of boundary members shall be 3t. 3. Boundary members shall be confined with a minimum of #3 bars @ 8” spacing, or equivalent confinement to develop an ultimate compressive masonry strain equal to 0.006.
  • Example: Strength Design Determine the maximum wind force, H, and design horizontal reinforcement to develop the wall flexural strength. Consider: zero vertical load and P dead = 40 kips and P live = 30 kips 7.63” 4 - #8’s 5’-4” H 10’-8” 8” concrete block, fully grouted Grade 60 reinforcement , f’ m = 1500 psi from previous example:
  • Example: Strength Design U = 1.3W Shear Reinforcement (neglecting vertical force) shear design within L w (5’-4”) of base: shear design for top 5’-4” of wall:
  • Example:Shear Wall Strength Design Confinement requirements for vertical reinforcement per Sec. 2108.2.5.6 Confinement Reinforcement (neglecting vertical force) 7.63” 5’-4” #8 M u = 4,439 kip-in. 11.5” 5.75”  0.0015 3t > 5.7”  0.003 #3 @ 8” bottom 8 courses Strain Diagram per Previous Example
  • Example: Strength Design Case 1: P u = 0.75(1.4 x 40 + 1.7 x 30) = 80.3 kips perhaps maximum flexural capacity and critical for shear design Flexural Strength considering Vertical Loads Case 2: P u = 0.9(40) = 36.0 kips perhaps minimum flexural capacity and lowest H u capacity reduction factors Case 1: Case 2:
  • Example: Strength Design Flexural Strength considering Vertical Loads 20.0 -0.00240 -60 -47.4 0 0 0 0.00360 60.0 47.4 -165 -118 6,983 15.0 -0.00220 -60 -47.4 0.00100 29.0 22.9 0.00580 60.0 47.4 -124 -54 6,126 16.8 -0.00229 -60 -47.4 0.00057 16.6 13.1 0.00185 53.8 42.5 -139 -83 6,463 13.1 -0.00208 -60 -47.4 0.00104 30.0 23.7 0.00708 60.0 47.4 -108 -37 5,794 d 1 = 4.0” d 2 = 20.0” d 3 = 44.0”   f  C s    f  T s    f  C s  C m P n M n (kips) (kip-in) (kips) (kips) (kips) (kips) (ksi) (ksi) (ksi)
  • Example: Strength Design Flexural Strength considering Vertical Loads H u = 34.1 kips (Case 2) ~ 34.7 kips (w/o vertical force). Use same shear design as for first part of problem. M u = 4,365 kip-in. (Case 2) ~ 4,439 kip-in. (w/o vertical force). Use same boundary members as for first part of problem. Axial Compressive Force, kips 140 120 100 80 60 40 20 5500 6000 6500 7000 Moment, M n kip-in. Case 2 6450 kip-in. 80.3 kips 5820 kip-in. Case 1 36.0 kips Case 2: Case 1: