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# Lecture 6 7 Rm Shear Walls

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Dan Abrams + Magenes Course on Masonry

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### Lecture 6 7 Rm Shear Walls

1. 1. Reinforced Masonry Working Stress Design of flexural members b strain stress εm fm Cm = fmb kd/2 kd d n.a. M Ts = Asfs t εs fs/n grout As As ρ= unit bd 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 Masonry Structures, slide 1 Reinforced Masonry Working Stress Design of flexural members 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 from Assumption #5, at any particular fiber : ε si = ε mi f si f mi Es from Assumption s #3 and #4 : = f si = f mi = nf mi E s Em Em fm f n 1− k from geometry of stress distribution : = s f s = nf m kd d − kd k Masonry Structures, slide 2
2. 2. Reinforced Masonry Working Stress Design of flexural members from equilibrium, C = T : f m bkd 1−k = As f s = ρ bd f s = ρ bd n f m 2 k 1− k k / 2= ρ n k 2 + 2 ρnk − 2 ρn = 0 k from equilibrium: ∑ M about C m = 0 M s = As f s jd = ρbd 2 jf s where jd=d-kd/3 or j=1-k/3 If fs=Fs then moment capacity will be limited by reinforcement. Allowable reinforcement tensile stress per MSJC Sec.2.3.2: Fs=20 ksi for Grades 40 or 50; Fs=24 ksi for Grade 60 Fs=30 ksi for wire joint reinforcement Allowable reinforcement tensile stress per UBC Sec.2107.2.11 : Fs= 0.5fy < 24 ksi for deformed bars; Fs= 0.5fy < 30 ksi for wire reinforcement Fs= 0.4fy < 20 ksi for ties, anchors, and smooth bars Masonry Structures, slide 3 Reinforced Masonry Working Stress Design of flexural members from equilibrium: ∑ M about T s =0 M m = 0.5 f m bkdjd = 0.5 f m jkbd 2 If fm= Fb then moment capacity will be limited by masonry. UBC 2107.2.6 & MSJC Sec.2.3.3.2: Fb=0.33f’m Masonry Structures, slide 4
3. 3. Reinforced Masonry WSD: Balanced Condition fm = F b Definition: The balanced condition occurs kbd C = fmb kd/2 when the extreme fiber stress in the masonry is equal to the allowable compressive stress, Fb, d and the tensile stress in the reinforcement is equal to the allowable tensile stress, Fs. T = As fs fs/n = Fs /n • 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. Masonry Structures, slide 5 Reinforced Masonry WSD: Balanced Condition from geometry: from equilibrium: C=T Fs fm = Fb Fb + Fb n d = Fb b kb = ρb b d Fs kb d d 2 kbd Fk F n d Fb ρb = b b = b kb = 2 Fs 2 Fs ( n + Fs ) Fs Fb + Fb n fs/n = Fs /n n ⎡ n ⎤⎡ 1 ⎤ kb = ρb = ⎢ ⎥⎢ ⎥ n + Fs /Fb ⎣ n + Fs / Fb ⎦ ⎣ 2Fs / Fb ⎦ Masonry Structures, slide 6
4. 4. 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 Fb = 0.33 f'm = 667 psi Fs = 24 ksi for Grade 60 reinforcement E m = 750 f'm = 1500 ksi E s = 29,000 ksi E s 29,000 n = = = 19.3 Em 1500 ⎡ 19.3 ⎤⎡ 1 ⎤ ρb = ⎢ = 0.48% ⎣ 19.3 + 24/0.667 ⎥ ⎢ 2 x24/0.667 ⎥ ⎦⎣ ⎦ Masonry Structures, slide 7 Design Strategy for RM Flexural Design Procedure for sizing section and reinforcement for given moment. Calculate ρb knowing f’m and fy determine Fb from f’m determine Fs from fy determine Emfrom f’m determine n = Es/Em Size section for some ρ < ρb Note: Section must also be determine k and j sized for shear. bd2 = M/ρjFs select b and d using common units Size reinforcement Check design As = M/Fsjd Ms = AsFsjd > M select number and size of rebars fb = M/0.5jkbd2 < Fb Masonry Structures, slide 8
5. 5. 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. 1. From previous example, ρb = 0.48% 2. Estimate ρ to be slightly lower than ρ so steel will govern. b A good estimate is ρ = 0.4% 3. Solve for k : k 2 + 2 ρnk − 2 ρn = 0 2 ρn = 2( 0.004 )( 19.3 )= 0.154 k 2 + 0.154 k − 0.154 = 0 k = 0.323 j = 1 − k / 3 = 0.892 4. Solve for bd 2 : bd 2 = M / ρ j F s bd 2 = ( 370 kip − in ) /( 0.004 )( 0.892 )24 ksi ) = 4321 in 3 Masonry Structures, slide 9 Example: Reinforced Masonry 5. Select dimensions of beam using 8” CMU’s: b = 7.63” 4 - 8” CMU’s dreq’d = [4321 / 7.63]0.5 = 23.8” d=27.8” use four units and center bars in bottom unit, d = 27.8” 6. Estimate amount of reinforcem ent : As req' d = M / Fs j d 7.63” As req' d = (370 kip - in) / (24 ksi) (0.892) (27.8) = 0.62 in 2 use 2 #5' s (0.62 in 2 ) 7 . Check design: ρ = Αs / bd = 0.62 in 2 / (7.63quot; ) ( 27.8quot; ) = 0.00292 k 2 + 2 ρnk − 2 ρn = 0 k = 0.284 j = 1 − 0.284 / 3 = 0.905 Μ s = Αs Fs j d = ( 0.62 in 2 )( 24 ksi )( 0.905 )( 27.8quot; ) = 374 kip − in. > 370 kip − in. OK ( 370 kip − in x 1000 ) f m = Μ / 0. 5 j k d 2 = = 488 psi < 667 psi ok ( 0.5 )( 0.905 )( 0.284 )(7.63quot; )( 27.8quot; ) 2 Masonry Structures, slide 10
6. 6. Flexural Capacity of Partially Grouted Masonry Case A: neutral axis in flange * per MSJC Sec. 2.3.3.3 flange b = 6t or 72” or s* b tf kd neutral d t axis As As per width b 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: As k 2 + 2 ρnk − 2 ρn = 0 ρ= bd If kd < tf assumption is valid, determine moment capacity as for rectangular section. If kd > tf assumption is not valid, need to consider web portion. Masonry Structures, slide 11 Shear Design of Reinforced Masonry s Cm Vm d Vs Vext. Viy Asfs Vd R Basic shear mechanisms: before cracking: Vext = Vint = Vm + Vd + Viy + Vs Once diagonal crack forms: • Vm reduces • flexural stresses increase • dowel action invoked • fsa is related to Mb Presence of shear reinforcement will: • restrict crack growth • resist tensile stress • help dowel action Masonry Structures, slide 12
7. 7. Shear Design of Reinforced Masonry after cracking: Vext = Vint = Vs = nAvfs where n is the number of transverse bars across the diagonal crack. Assuming a 45 degree slope, n=d/s Vs = (d/s)Avfs Av Vs Vs UBC Sec. 2107.2.17 (Eq. 7.38) = = MSJC Sec. 2.3.5.3 (Eq. 2-26) s df s dFs Masonry Structures, slide 13 Shear Design of Reinforced Masomry Flexural shear stress dx C M C + dC M + dM fvbdx na jd T T + dT T T + dT fvb dx = dT = dM/jd fv = (dM/dx)/bjd V fv = UBC Sec. 2107.2.17 (Eq. 7-38) bjd V fv = bd MSJC Sec. 2.3.5.2.1 (Eq. 2-19) Masonry Structures, slide 14
8. 8. 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 Fv = 1.0 f'm < 50 psi 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 Fv = 3.0 f'm < 150 psi UBC Eq. 7-18; MSJC Eq. 2-23 Masonry Structures, slide 15 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 M 1 M M for <1 F v = ( 4 − ) f 'm <( 80 − 45 ) psi UBC Eq. 7-19; MSJC Eq. 2-21 Vd 3 Vd Vd M for ≥1 F v = 1.0 f 'm < 35 psi UBC Eq. 7-20; MSJC Eq. 2-22 Vd 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 M 1 M M for <1 Fv = ( 4 − ) f'm < ( 120 − 45 )psi UBC Eq. 7-21; MSJC Eq. 2-24 Vd 2 Vd Vd M for ≥1 F v = 1.5 f 'm < 75 psi h UBC Eq. 7-22; MSJC Eq. 2-25 Vd Masonry Structures, slide 16
9. 9. Shear Design of Reinforced Masonry Moment-to-Shear Ratios For a single-story For piers between openings cantilevered shear walls M d V V d h h M M M Vh h M Vh / 2 h = = = = Vd Vd d Vd Vd 2d Masonry Structures, slide 17 Shear Design of Reinforced Masonry Additional MSJC Requirements MSJC Sec. 2.3.5.3.1 smax = d/2 or 48” MSJC Sec. 2.3.3.4.2 minimum reinforcement perpendicular to shear reinforcement = Av/3 smax = 8 ft MSJC Sec. 2.3.5.5 design for shear force at distance “d/2” out from support Vdesign Vdesign d/2 Masonry Structures, slide 18
10. 10. Shear Design of Reinforced Masonry Shear Design Strategy for Reinforced Sections Start Determine Flexural Tension Stress Determine Fv Assuming Shear ft= -P/A+Mc/I Reinforcement to take 100% of Shear consider as no no is ft>Ft? Resize unreinforced is fv<Fv? Section yes yes Determine Maximum Design Shear Provide Reinforcement to Take 100% of Shear Determine Shear Stress Av V V V = s fv = or s dFs bjd bd no yes Determine Fv Assuming No Shear is fv<Fv? End Reinforcement Masonry Structures, slide 19 Example: Design of RM Shear Wall Determine the maximum lateral force, Hwind per UBC and MSJC 6’-8” 8” CMU wall 120 psi Type S - PCL mortar solidly grouted f’m=3000 psi #4 @ 32” 2 - #8’s each end of wall 8’-0” Case A: neglect all reinforcement Case B: consider vertical reinf., neglect horizontal reinf. Case C: consider vertical and horizontal reinf. 6’-4” Case D: design horizontal reinforcement for max. shear Masonry Structures, slide 20
11. 11. Example: Design of RM Shear Wall Case A: neglect all reinforcement per UBC: 7.63 × 80 2 Sg = = 8139 in3 flexure 6 M 96 × H - fa + = Ft − 120 + = 40 x 1.33 H = 14 ,684 lbs. = 14.7 kips S 8139 shear Fv = [ 34 psi + 0.2 fa dead ] x 1.33 = [ 34 + 0.2 ( 120 )] x 1.33 = 77.1 psi 77.1 psi Vmax = Ae Fv = ( 7.63 x 80.0 )( ) = 47.1 kips 1000 per MSJC: flexure 96 × H - fa + M / S = Ft − 120 + =0 H = 10 ,174 lbs . = 10.2 kips shear 8139 Fv = 60 + 0.45 ( 120 ) = 114 psi > 1.5 f'm = 82.2 psi Fv = 1.33 × 82.2 psi = 109 psi 2 2 Vmax = Fv bt = ( 109 psi )( 7.63 x 80 ) = 44.3 kips 3 3 Masonry Structures, slide 21 Example: Design of RM Shear Wall Case B: consider only vertical reinforcement Flexure by UBC or MSJC: neglecting fa Ms = AsFsjd = 2 x 0.79 in2 (1.33 x 24 ksi) (0.9 x 72.0”) = 3268 k-in Hwind = 34.0 kips lumping 2 - #8’s ave. d for 2 bars Shear per UBC Sec.2107.2.9 or MSJC Sec.2.3.5.2 8.0' M/Vd = = 1.33 > 1 6.0' M for > 1 Fv = 1.0 f 'm < 35 psi Fv = 1.33 x 35 psi = 46.6 psi Vd for UBC Vmax = bjdFv = ( 7.63quot; )( 0.9 )( 72quot; )( 46.6 psi ) / 1000 = 23.0 kips governs for MSJC Vmax = bdFv = ( 7.63quot; )( 72quot; )( 46.6 psi ) / 1000 = 25.6 kips governs Masonry Structures, slide 22
12. 12. Example: Design of RM Shear Wall Case C: consider all reinforcement Flexure by UBC or MSJC: same as case B Shear per UBC Sec. 2107.2.17 or MSJC Sec.2.3.5.3 Vmax= Vs=(Av/s)Fsd = (0.20 in2/32”)(24 ksi x 1.33)(72”) = 14.4 kips governs Overall shear per UBC Sec. 2107.2.9.C or MSJC Sec. 2.3.5.2.3 (b) M for > 1 F = 1.5 f'm ≤75 psi F = 1.5 3000 = 82.2psi>75 psi v v Vd Fv = 1.33x 75 psi= 100 psi V (14.4 kips x 1000) UBC f v = = = 30.7 psi < 100 psi okay bjd (7.63)( 0 .9 )(72) V (14.4 kips x 1000) MSJC f v = = = 27.7 psi < 100 psi okay bd (7.63)(72) Masonry Structures, slide 23 Example: Design of RM Shear Wall Case D: design horizontal reinforcement for maximum shear strength Vmax = Fvbjd = ( 100 psi)( .63)(0.9 )(72)/ 1000 = 49.4 kips > 34 kips oka govern 7 y s Av /S = Vmax/Fsd = 49.4 kips/( .33 x 24 ksi)( ) = 0.0215in2 per in. 1 72 u sing #4 rebars(Av = 0.20 in2 ) s = 0.20 / 0.0215 = 9.3quot; use # @ 8 in. horiz 4 ontal Summary: Hmax, kips Case Consideration UBC MSJC A No steel No steel 14.7* 14.7* 10.2* 10.2* vertical steel 24.3 27.0 B vertical steel 23.0 25.6 no horizontal steel no horizontal steel vertical steel and 15.2 15.2 C vertical steel and 14.4 14.4 #4 @ 32” horizontal steel #4 @ 32” horizontal steel D #4#4 @ 8” horizontal @ 8” horizontal 34.0* 34.0* 34.0* *flexure governs Masonry Structures, slide 24
13. 13. Flexural Bond Stress M = Tjd M + dM = (T + dT)jd dM = dT jd dx dT = dM/jd C + dC U = bond force per unit length for group of bars C U dx = dT = dM/jd U = (dM/dx)/jd = V/jd M U dx jd u = flexural bond stress = M + dM ∑o where ∑ o = sum of perimeters of all bars in group T T + dT V u= UBC Sec. 2107.2.16 Eq. 7-36 dx Σ ο jd U T + dT allowable bond stress per UBC Sec.2107.2.2.4: T 60 psi for plain bars 200 psi for deformed bars dx 100 psi for deformed bars w/o inspection Masonry Structures, slide 25 Development Length uπ d b db As fs ld As fs = uπdb ld πd b 2 fs = uπdb ld 4 f s db ld = = 0.002 db fs for u = 125 psi UBC Sec. 2107.2.2.3 Eq.7 - 9 4u ld = 0.0015 db Fs for u = 167 psi MSJC Sec . 2.1.8.2 Eq . 2 − 8 Masonry Structures, slide 26
14. 14. 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 “12db” past the theoretical cutoff point Rule #2: extend bars a distance of “ld” past the point of maximum stress Example for shear wall: bars “a” Moment Diagram (#2) > ld (#1) d or 12db theoretical cutoff point capacity with bars “a” bars “b” (#2) > ld moment capacity with bars “a” and “b” Masonry Structures, slide 27 Combined Bending and Axial Loads Code Requirements UBC Sec. 2107.1.6.3 fa fb use unity formula to check compressive stress: + < 1.0 Fa Fb UBC Sec. 2107.1.6.1 UBC Sec. 2107.2.15 P M Note: unity formula is conservative - better approach is to use P-M interaction diagram. M As fs fb = ( Eq .7 − 31 ) 2 jkbd 2 fa = P/Ae kd jd 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 fa + fb < 1/3 f’m provided that fa < Fa In lieu of approximate method, use an axial-force moment interaction diagram. Masonry Structures, slide 28
15. 15. Axial Force-Moment Interaction Diagrams General Assumptions 1. plane sections remain plane after bending • shear deformations neglected • strain distribution linear with depth 2. neglect all masonry in tension 3. neglect steel in compression unless tied Strain Stress 4. stress-strain relation for masonry is linear in compression εm fm 5. stress-strain relation for steel is linear 6. perfect bond between reinforcement and grout Cs • strain in grout is equal to strain in adjacent reinforcement 7. grout properties same as masonry unit properties P M εs Ts=Asfs Masonry Structures, slide 29 Axial Force-Moment Interaction Diagram Out-of-Plane Bending of Reinforced Wall Pa Mb d = t/2 Range “a”: large P, small M, e=M/P < t/6 unit width = b Pa = 0.5(fm1 + fm2)A Ma= 0.5(fm1 - fm2)S where S = bt2/6 fm2 fm1 em Cm Masonry Structures, slide 30
16. 16. Axial Force-Moment Interaction Diagram Out-of-Plane Bending of Reinforced Wall Pb Mb d = t/2 Range “b” medium P, medium M, e > t/6, As in compression unit width = b 0.5 < α < 1.0 for section with reinforcement at center t αt em = − 2 3 fm 1 fm1 Pb = C m = αtb em 2 Cm M b = C m em αt Masonry Structures, slide 31 Axial Force-Moment Interaction Diagram Out-of-Plane Bending of Reinforced Wall Pc Mc Range “c” small P, large M, e > t/6, As in tension d = t/2 α < 0.5 for section with reinforcement at center unit width = b t αt em = − 2 3 fm 1 Pc = C m − Ts Cm = αtb Ts = As fs 2 f s ⎡ d − αt ⎤ ⎡ 0. 5 − α ⎤ t = fm 1 = ⎢ f for d = fm1 n ⎢ αt ⎥ ⎣ ⎦ ⎣ α ⎥ m1 ⎦ 2 em Ts Cm t M c = C m em + Ts ( d − ) αt 2 Masonry Structures, slide 32
17. 17. Axial Force-Moment Interaction Diagram Out-of-Plane Bending of Reinforced Wall Range “a” Range “b” Range “c” e=0; M=0 fm1= fm2=Fa fm1= Fb= f’m/3 P=Fa A Reduce fm2 from Reduce α from Reduce α from 2Fa-Fb by 1.0 by 0.5 by Start increment increment increment compression controlling Determine P & M Determine P and Determine P and tension controlling per Range “a” M per Range “b” M per Range “c” no yes no is As in yes yes fm2 = 0? tension? M = 0? no no yes fs = Fs fs < Fs? fs < Fs fm1 < Fb fm1 = Fb Stop Masonry Structures, slide 33 Axial Force-Moment Interaction Diagram Out-of-Plane Bending of Reinforced Wall fm1 = Fb Fb fm1 = Fa fm2 = 2Fa - Fb Fa Fb Range “a” lim it b yu Axial Force Fb Range “b” nit 1 y for tension compression e mu fs/n controls controls Fs/n la Fb Fb balanced point Range “c” Fs Fs/n Moment fs = fm Masonry Structures, slide 34
18. 18. 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. Fs = 24 ksi for Grade 60 Fa = 0.25 f´ m = 625 psi without reduction factor Fb = 0.33f´ m = 833 psi E m = 750 f´ m = 1875 ksi per UBC E s = 29,000 ksi n = E s /E m = 15.5 per foot of wall : Ag = 7.63quot; x 12quot; = 91.6 in 2 ; S g = 12quot; x 7.63 2 / 6 = 116 in 3 As / ft = 0.20 x 12 / 16 = 0.150 in 2 ρ = 0.20 in 2 /( 16 x 3.81quot; ) = 0.0033 ρn = 0.0509 k = 0.272 j = 0.909 Masonry Structures, slide 35 Example: Interaction Diagram Fs ⎛ α ⎞ *masonry stress inferred from Fs and α: f m 1 = ⎜ ⎟ n ⎝ 0.5 − α ⎠ Range Case fm1 fm2 α Cm em Ts P=Cm- Ts M=Cm em (psi) (psi) (kips) (in.) (kips) (kips) (kip-in) 1 625 625 - 57.2 0 - 57.2 0 Compression Controls a 2 833 417 - 57.2 - - 57.2 24.1 3 833 0 - 38.1 1.27 - 38.1 48.4 4 833 - 0.75 28.6 1.91 - 28.6 54.5 b 5 833 - 0.50 19.1 2.54 - 19.1 48.5 6 833 - 0.33 12.6 2.97 0.9 11.7 37.4 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 9 for P = 0: Mm= 0.5Fbjkbd2 = 0.5(833 psi)(0.909)(0.272)(12)(3.81)2 = 17.9 c Controls 10 833 bal. - 0.175 6.7 3.37 3.6 3.1 22.5 Tension 11 664* - 0.150 4.6 3.43 3.6 1.0 15.7 12 check for P = 0: Ms = AsFsjd = (0.15 in2)(24 ksi)(0.909)(3.81”) = 12.5 Masonry Structures, slide 36
19. 19. Example: Interaction Diagram 50 1 2 6 833 0.9k 1 625 0.33t 40 2.0 k 2 833 7 833 Axial Force 3 417 0.25t kips 30 4 0 3 833 8 833 3.9 k > AsFs 20 5 0.167t 4 833 10 833 6 .75t 3.6 k = AsFs 10 10 0.175t 11 7 12 5 833 11 664 3.6 k = AsFs 10 20 30 40 50 0.15t 8 0.50t 9 Moment, kip-in Masonry Structures, slide 37 Flexural Capacity with Axial Compression Short Cut Method Out-of-Plane Bending, Reinforcement at Center fm d kd Cm d jd M Ts P fs/n f s /n f k f E stress compatibility: = m ; fm = ( s ) where n = s [1] d - kd kd 1-k n Em C m = 0.5 f m bkd [2] Ts = As f s = ρbdf s [3] Masonry Structures, slide 38
20. 20. Flexural Capacity with Axial Compression Short Cut Method P = C m - Ts [4] equilibrium: P = 0.5 f m bkd - ρbdf s [5] f ⎛ k ⎞ P = 0.5 ( s )⎜ ⎟ bkd - ρbdf s [6] n ⎝1-k ⎠ P k2 ⎛ 1 ⎞ = 0 .5 ⎜ ⎟− ρ [7] bdf s 1− k ⎝ n⎠ P if tension controls , f s = Fs set α = [8] bdFs 2 k ⎛ 1⎞ a = 0.5 ⎜ ⎟− ρ [9] (1 − k) ⎝ n ⎠ 1 k2 ρ +α = [10] 2n 1 − k k 2 + 2 n( ρ + α )k − 2 n( ρ + α ) = 0 [11] M = C m jd = 0.5 f m bkjd 2 where j = 1 − k / 3 [12] Masonry Structures, slide 39 Strength Design of Reinforced Masonry Ultimate Flexural Strength strains stresses As ε mu Cm c d n.a. t d Mn b εs > εy Ts = Asfy k3f’m k3f’m fm f’m k2c k2c c klc Cm = c Cm = k1k3f’mbc εm εmu Note: rectangular stress block can represent compressive stress distribution if k2/k1 = 0.5 Masonry Structures, slide 40
21. 21. Strength Design of Reinforced Masonry Measuring k1k3 and k2 Po P1 Po in displacement control P1 in force control summing moments about centroid: a increase P1 P1a = (Po + P1)g so that ∆ = 0 = (Po + P1)(c/2 - k2c) ∆ P1 a k 2 = 0.5 - c Po + P1 c strain g total compressive force: stress Po + P1 = k3f’m k1cb P0+P1 Po + P1 k2c k1 k 3 = f 'm bc k3f’m k1c Masonry Structures, slide 41 Strength Design of Reinforced Masonry Measured k1k3 and k2 values Sample experimentally determined constants k1k3, and k2 1 0.8 K1K3 K1K3 & K2 0.6 0.4 K2 0.2 0 0 0.001 0.002 0.003 0.004 0.005 0.006 Extreme Fiber Strain (in/in) Masonry Structures, slide 42
22. 22. Strength Design of Reinforced Masonry Ultimate Flexural Strength equilibrium C m + Ts = 0 k1 k 3 f 'm bc = As f y = ρbd f y fs ρdf y fy c = k1 k 3 f 'm summing moments about Cm εs M n = As f y ( d − k 2 c ) ρ df y M n = As f y ( d − k 2 ) k 1 k 3 f 'm ρf y M n = As f y d ( 1 − k 2 ) k 1 k 3 f 'm k2 fy if = 0.5 and k 3 = 0.85 then : M n = As f y d ( 1 − 0.59 ρ ) k1 f 'm Masonry Structures, slide 43 Strength Design of Reinforced Masonry Balanced condition with single layer of reinforcement strains stresses strain compatibility ε mu c ε mu ε mu = or c = d ε mu + ε y d ε mu + ε y k1c Cm c equilibrium n.a. C m + Ts = 0 d Mn ε mu k1 k 3 f 'm b d = ρ b bdf y ε mu + ε y k1 k 3 f 'm ε mu ρb = εs = ε y Ts= Asfy fy ε mu + ε y fy if k 3 = 0.85 ε mu = 0.003 ε y = E s = 29 ,000 ,000 psi : Es k1 ( 0.85 ) f 'm 0.003 0.85 k1 f 'm 87 ,000 ρb = = fy 0.003 + f y / E s fy 87 ,000 + f y Masonry Structures, slide 44
23. 23. Strength Design of Reinforced Masonry Balanced condition with single layer of reinforcement 0.85 k1 f 'm 87 ,000 ρb = ρ tb = As / bt for one layer of steel t = 2d k 1 = 0.85 fy 87 ,000 + f y Grade 40 Grade 60 f’m ρb ρ tb ρb ρ tb 1000 0.0124 0.0062 0.0071 0.0036 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 Masonry Structures, slide 45 Strength Design of Reinforced Masonry Balanced condition with multiple layers of reinforcement strains stresses b ε mu 0.85f’m d1 ε s1 Cs1 d2 0.85c ε s2 Cm=0.85f’mb(0.85c) Cs2 c d3 d4 ε s3 Ts3 Ts4 = Asbal fy ε s4 = ε y Asbal strain compatibility equilibrium d ε mu f si = E sε si < f y ε si < ε mu − ( i )( ε mu + ε sy ) c= d d4 ε mu + ε y C si or Tsi = Asbal i f si 60 ksi C m + ∑( C si + Tsi ) = 0 εy= = 0 .00207 (Grade 60) 29,000 ksi − 0.428 f 'm bd + Asbal ∑ f si = 0 if ε mu = 0.003 , then c = 0 .592 d solve for Asbal Masonry Structures, slide 46
24. 24. Example: Flexural Strength of In-Plane Wall Determine the maximum bar size that can be placed as shown. Maximum steel is equal to one-half of that resulting in balanced conditions. f’m= 1500 psi Grade 60 reinforcement special inspection 7.63” 0.003 0.85f’m ε s1 4.0” Cs1 20.0” Cm = 0.85f’mb(0.85c) εs 2 44.0” c 60.0” Cs2 5’-4” Pn = 0 n.a. εs 3 Ts3 Asbal ? Ts4 = Asbal fy ε s = ε y = 0 .00207 Masonry Structures, slide 47 Example: Flexural Strength of In-Plane Wall Determine the maximum bar size. c = 0.003/0.00507 (60.0”) = 35.5” Cm = 0.85f’mb(0.85c) = -0.85(1500)(7.63”)(0.85 x 35.5) = -294 k layer di εsi fsi ⎛ c − d1 ⎞ εi =⎜ ⎟ ( −0.003 ) ⎝ c ⎠ 1 4.0” -0.00261 (C) -60.0 2 20.0” -0.00131 (C) -38.0 f si = E sε si ≤ f y 3 44.0” 0.00072 (T) 20.8 4 60.0” 0.00207 (T) 60.0 without compression steel (neglect Cs1 andCs2 forces) Cm + Σ(Csi + Tsi) = -294 + Asbal (20.8+ 60.0) = 0 Asbal = 3.64 in2 Asmax = 1.82 in2 max. bar size is #ll (1.56 in2)* *bars larger than #9 are not recommended because of anchorage and detailing problems with compression steel (include Cs1 and Cs2 forces) Cm + Σ(Csi + Tsi) = -294 + Asbal (-60.0 - 38.0 + 20.8 + 60.0) = 0 Asbal = -17.1 in2 note: negative Asbal means that ΣC > ΣT , in such case no limit on tensile reinforcement Masonry Structures, slide 48
25. 25. Example: Flexural Strength of In-Plane Wall Determine flexural strength of wall. f’m= 1500 psi Grade 60 reinforcement special inspection 0.003 0.85f’m ε s1 Cs1 4.0” 20.0” c Cm = 0.85f’mb(0.85c) 44.0” n.a. Ts2 60.0” εs 2 5’-4” εs 3 Ts3 εs > ε y Ts4 = As fy #8 (typ) = 0.79 in2 x 60 ksi = 47.4 k 7.63” Masonry Structures, slide 49 Example: Flexural Strength of In-Plane Wall Determine flexural strength of wall. ⎛ c − di ⎞ εi =⎜ ⎟( −0.003) compressive strains = (-) fsi = Esεsi < f y Cm =8.27c ⎝ c ⎠ d1 = 4.0” d2 = 20.0” d3 = 44.0” c ε1 f1 Csl ε2 f2 Ts2 ε3 f3 Ts3 Cm ∑(C + T ) 0.00360 60.0 20.0 -0.00240 -60 -47.4 0 0 0 -165 -117.6 47.4 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 11.5 -0.00196 -56 -44.8 0.00222 60.0 47.4 0.00848 60.0 47.4 -95 +2.3 0.85 c close to zero, take c = 11.5” M n = ∑ { Asi f si ( d i − )} 2 = ( −44.8 )( 4.00 − 4.89 ) + ( 47.4 )( 20.0 − 4.89 ) + ( 47.4 )( 44.0 − 4.89 ) + ( 47.4 )( 60.0 − 4.89 ) = 5 ,222 kip − in Masonry Structures, slide 50
26. 26. Example: Flexural Strength of In-Plane Wall Approximate flexural strength of wall. neglecting C sl and Ts 2 , and lumping As 3 and As 4 ( 60 + 44 ) 2 × 0.79 d= = 52.0quot; ρ= = 0.00398 2 7.63 × 52 fy 5’-4” M n = As f y d ( 1 − 0.59 ρ ) f 'm 60 = 2( 0.79 in 2 )( 60 )( 52.0 )( 1 − 0.59 x 0.00398 x ) 1.50 = 4 ,467 kip − in 86% of answer #8 (typ) 7.63” Masonry Structures, slide 51 Slender Wall Design Limitations of Method: UBC Sec. 2108.2.4 (a) for out-of-plane bending of solid, reinforced walls lightly stressed under gravity loads (b) limited to: Pw + Pf ≤ 0.04 f'm ( 8 − 19 ) Sec. 8.2.4.4 210 where f'm <6000 psi Ag Pw + Pf h' Note : when 0.04f'm < < 0.20f'm , method still can be used providing that < 30 Ag t (c) ρg= As/bt < 0.5 ρbal (d) special inspection must be provided during construction (e) t > 6” Sec. 2108.1.3: Load factors U = 1.4 D + 1.7 L U = 0.9 D ± 1.4 E U = 1.4( D + L + E ) U = 0.9 D ± 1.3W U = 0.75( 1.4 D + 1.7 L + 1.7W ) Ref: NCMA TEK 14-11A Strength Design of Tall Concrete Masonry Walls Masonry Structures, slide 52
27. 27. Slender Wall Design Required Flexural Strength: UBC Sec. 2108.2.4.4 e eccentric transverse P∆ load load Puf Pufe t h/2 Puw wu h wuh2/8 (Puw + Puf)∆ u Pufe/2 h/2 Pu = Puf + Puw wu h2 Puf e Mu = + + ( Puw + Puf )∆u ( 8 − 20 ) 8 2 Masonry Structures, slide 53 Slender Wall Design Design Considerations Design strength: Sec. 2108.2.4.4 Mu < φ Mn (8 - 22) Strength reduction factor: flexure φ = 0.8 Sec. 2108.1.4.2.1 Assumptions for ultimate flexural strength (Sec. 2108.2.1.2) 1. equilibrium 2. strain compatibility 3. εmu = 0.003 4. fs = Esεs < fy 5. neglect masonry strength in tension 6. rectangular stress block, k1 = 0.85, k3 = 0.85 Masonry Structures, slide 54
28. 28. Slender Wall Design Equivalent area of reinforcement, Ae for single wythe construction reinforced at center: d As a = 0.85c b 0.85f’m Ts = Asfy d c Pu Cm =0.85f’mb(0.85c) Ts = Asfy Pu = Cm - Asfy Cm Cm = Pu + Asfy= Asefy ( Pu + As f y ) Pu Ase = Eq. (8-24) fy flexural strength a ( Pu + As f y ) M n = Ase f y ( d − ) Eq. (8 - 23) where a = Eq. (8 - 25) 2 0.85 f 'm b Masonry Structures, slide 55 Slender Wall Design Lateral Deflections 5 wh 4 5 wh 2 h 2 5 Mh 2 ∆= = = M 384 EI 48 8 EI 48 EI 5 M s h2 for Ms < Mcr ∆s = (8-28) My 48 E m I g ∆y 5 M cr h 2 5 ( M s − M cr )h 2 Ms for Ms > Mcr ∆s = + (8-29) 48 E m I g 48 E m I cr ∆s Mcr b( kd )3 where I cr = nAse ( d − kd )2 + ∆cr 3 Mcr = fr S ∆ (note “kd” may be replaced by “c” for simplicity) Modulus of Rupture, fr Eqs. 8-31, 32, 33 fully grouted partially grouted hollow unit 4.0 f'm < 235 psi 2.5 f 'm < 125 psi 2-wythe brick 2.5 f 'm < 125 psi not allowed Masonry Structures, slide 56
29. 29. Slender Wall Design Design Considerations Serviceability Criteria ∆s ≤ 0.007 h ( 8 − 27 ) Strength Criteria w h 2 Puf e Mu = u + + ( Puw + Puf )∆u 8 2 5 M cr h2 5 ( M u − M cr )h2 ∆u = + 48 E m I g 48 E m I cr Masonry Structures, slide 57 Example: Slender Wall Design Determine the maximum wind load, w, per UBC and MSJC 3.5” 500 lbs/ft dead 200 lbs/ft live 3’-0” P w 8” CMU, partially grouted f’m = 2000 psi, Grade 60 7 .63quot; e = 3 .50quot; + = 7 .31quot; 2 12quot; in 2 As = 0 .20 in 2( ) = 0 .075 32quot; ft 20’-0” #4 @ 32” As 0 .075 1 1 ρ = = = 0 .000164 < ρbal = ( 0.0143)=0.0072 ok bd 12× 3.81 2 2 Masonry Structures, slide 58
30. 30. Example: Slender Wall Design Flexural Strength per UBC U = 0.75 (1.4D + 1.7L + 1.7W) w h 2 Puf e a Mu = u + + ( Puw + Puf )∆u < φ M n = φ Ase f y ( d − ) 8 2 2 Pu = Puf + Puw Puf = 0.75 ( 1.4 x 500 + 1.7 x 200 ) = 780 lbs . Puw = 0.75 ( 1.4 x 64 psf x 13' ) = 874 lbs . Pu = 1654 lbs . ( Pu + As f y ) ( 1.65 kips + 0.075 x 60 ksi ) Ase = = = 0.103 in 2 fy 60 ksi ( Pu + As f y ) 0.103 ( 60 ) a= = = 0.302quot; 0.85 f 'm b ( 0.85 x 2.0 ksi x 12quot; ) 0.302quot; c= = 0.355quot; < 1.2quot; neutral axis within face shell, treat as rectangular section 0.85 Mu=φMn = φAsefy(d - a/2)=0.8(0.103in2)(60ksi)(3.81in - 0.302in/2)=18.1 kip-in Masonry Structures, slide 59 Example: Slender Wall Design Flexural Strength per UBC Es 29,000 Em =750 f 'm = 1500ksi n= = = 19.3 Em 1500 Icr = nAse ( d − c )2 + bc3 3 ( ) = 19.3 0.103in2 ( 3.81− 0.355 )2 + 12( 0.355 )3 3 quot; = 23.9 in4 2 2 5 Mcrh 5 ( Mu − Mcr )h ∆u = + 48 Em I g 48 Em Icr 12quot; x7.632 Mcr = fr Sg fr = 2.5( f 'm )0.5 = 112 psi Mcr = 0.112ksi ( ) = 13.1 k − in. 6 for simplicity, use gross section even though partially grouted to avoid iteration, assume Mmax = Mu 5 ( 13.1 )( 20 x 12 )2 5 ( 18.1− 13.1 )( 20 x 12 )2 ∆u = + = 0.118 + 0.837quot; = 0.955 quot; quot; 48 ( 1500)( 444 ) 48 ( 1500)( 23.9 ) w ( 20 )2 0.780 7.31 / 2 ) ' ( quot; 0.955 Mu = u + +( 1.654)( )=181k − in. / 12 . quot; 8 12 12 wu ws = = 17.8psf wu=22.7 psf 0.75x1.7 Masonry Structures, slide 60
31. 31. Example: Slender Wall Design Check Service Load Deflections per UBC ∆s < 0.007 h 5 M cr h 2 5 ( M s − M cr )h 2 ∆s = + 48 E m I g 48 E m I cr w s h2 Ms = + Po e / 2 + ( Pw + Po ) ∆s 8 ⎡ ( 17.8 psf )( 20 )2 ⎤ 700 ( 7.31quot; ) =⎢ ⎥ x 12 + + 1532 ∆s = 13 ,239 + 1532 ∆s ( lb . − in .) ⎢ ⎣ 8 ⎥ ⎦ 2 5 ( 13.239 + 1.532 ∆s − 13.1 )( 20 x 12 )2 ∆s = 0.118quot; + 48 ( 1 ,500 )( 23.9 ) ∆s = 0.19quot; < 0.007 h = 0.007 (20' x 12 ) = 1.68quot; ok ) Masonry Structures, slide 61 Example: Slender Wall Design Maximum Wind Load per MSJC d = 3.82” 1.25” w h2 P e M s = s + os + ( Pws + Pos )∆s 8 2 #4 @ 32” w s ( 20' )2 = x 12 + 700 lbs ( 3.66quot; ) + ( 1532 lbs ) ∆s 8 = 600 w s + 2562 lb − in . + 1532 ∆s Determine Icr considering axial compression k 2 + 2 n( ρ + α )k − 2 n( ρ + α ) = 0 0.075 in 2 P 1.532 k n = 19.3 ρ= = 0.00164 α= = = 0.00104 12quot;×3.82quot; bdFs 12quot; × 3 .82quot; × 32 ksi k 2 + 0.0104 k − 0.0104 = 0 k = 0.275 kd = 1.05quot; < face shell thickness, ok k j = 1 − = 0.908 jd = 3.47quot; 3 b( kd )3 12quot; ( 1.05quot; )3 I cr = + nAs ( d − kd )2 = + 19.3( 0.075 in 2 )( 3.82quot; −1.05quot; )2 = 15.7 in 4 3 3 Masonry Structures, slide 62
32. 32. Example: Slender Wall Design Maximum Wind Load per MSJC M s = As Fs + ( Pws + Pos ) jd = [( 0.075 × 32 ksi) + 1.532 ] 3.47quot; = 13.64 kip - in 0.600 w s + 2.56 kip - in . + 1.532 ∆s = 13.64 kip - in w s = 18.47 − 2.55 ∆s ( w s is in psf) 2 5 M cr h 5 ( M s − M cr )h 2 ∆s = + 48 E m I g 48 E m I cr 5 ( 0.600 w s + 2.56 kip - in . + 1.532 ∆s − 13.1 )( 20 × 12 )2 ∆s = 0.118quot; + 48 ( 1500 ksi)(15.7 in 4 ) ∆s = 0.118quot; +0.153 w s + 0.652 + 0.390 ∆s − 3.34 ∆s = 0.251 w s − 4.21 w s = 18.47 − 2.55 ∆s ( 0.251 w s − 4.21 ) w s = 17.8 psf Note: same wind load as by UBC slender wall design procedure. Should also check compressive stress with an axial force-moment interaction diagram Masonry Structures, slide 63 Strength Design of RM Shear Walls UBC Requirements 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 Sec. 2108.1.2: Special inspection must be provided during construction. Prisms should be tested or unit strength method should be used. UBC Sec.2108.1.3: Shear wall design procedure 1. Required strength A. earthquake loading: U = 1.4 (D+L+E) (12-1) U = 0.9D + - 1.4E (12-2) 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) 2. Design strength φ = 0.65 for φPn > 0.1 f 'm Ae or φPn > 0.25 Pb (see next slide) A. axial load and flexure φ = 0.85 for φPn = 0 B. shear φ = 0 .60 for shear limit state φ = 0.80 for flexure limit state Masonry Structures, slide 64
33. 33. Strength Design of RM Shear Walls Definition of Balanced Axial Load, Pb ε mu 0.85f’m Cs1 a b = 0.85 c Cm = 0.85f’mbab c Cs2 Lw d Pb n.a. Ts3 Ts4 = Asbalfy ε s = ε y = 0.00207 b assume ΣC si = ΣTsi so Pb = C m emu for solidly grouted walls : Pb = 0.85 f 'm bab where ab = 0.85 d fy ( emu + ) Es Masonry Structures, slide 65 Strength Design of RM Shear Walls UBC Requirements 3. Design assumptions (same as for Slender Wall Design Procedure, UBC Sec. 2108.2.1.2) 1. equilibrium 2. strain compatibility 3. εmu = 0.003 4. fs = Esεs < fy 5. neglect masonry tensile strength 6. use rectangular stress block, k1 = 0.85, k3 = 0.85 7. 1500 psi < f’m < 4000 psi Masonry Structures, slide 66
34. 34. Strength Design of RM Shear Walls UBC Requirements 4. Reinforcement per UBC Sec. 2108.2.5.2 Mn > Mcr 1. minimum reinforcement M ductile ρ v + ρ h ≥ 0.002 ρ v and ρ h ≥ 0.0007 Mcr spacing ≤ 4' −0quot; Mn < Mcr nonductile 2. for flexural failure mode Mn > = 1.8 Mcr for fully grouted wall ∆ Mn > = 3.0 Mcr for partially grouted wall 3. anchor all continuous reinforcement 4. As vertical > 1/2 As horizontal 5. maximum spacing of horizontal reinforcement within plastic hinge region = 3t or 24” 5. Axial strength (no flexure) Po = 0.85 f’m(Ac-As) + fyAs Pu < = φ (0.80)Po Masonry Structures, slide 67 Strength Design of RM Shear Walls UBC Requirements 6. Shear Strength UBC Sec. 2108.2.5.5 1. maximum nominal shear: M Vn = 6.0 Ae f 'm ≤ 380 Ae for ≤ 0.25 Vd M = 4.0 Ae f 'm ≤ 250 Ae for ≥ 1.0 (Table 21 - J ) Vd Amv t 2. for walls limited by shear strength: Lw Vn = Vm + Vs where Vm = C d Amv f 'm (8 - 37) Vu M and C d = 2.4 for ≤ 0.25 Vd Amv = net area of masonry M wall section bounded by = 1.2 for ≥ 1.0 (Table 21 - K) Vd wall thickness and length of section in direction of shear Masonry Structures, slide 68
35. 35. Strength Design of RM Shear Walls UBC Requirements t 6. Shear Strength UBC Sec. 2108.2.5.5.2 (continued) Vs = Amv ρ n f y (8 − 38) Avertical plane where ρ n = As horizontal / Avertical plane h As horizontal Vs = Amv ( )fy Avertical plane As horizontal As horizontal Lw Vs = Lw t ( ) fy = ( As horizontal ) f y ht h Lw Lw Lw As fy h As fy As fy Masonry Structures, slide 69 Strength Design of RM Shear Walls UBC Requirements 6. Shear Strength UBC Sec. 2108.2.5.5: continued 3. for walls limited by flexural strength: within hinge region, distance of Lw above base: V n = V s = Am ρ n f y (8 − 39) (Vu determined at Lw/2 from base) above hinge region: V n = Vm + V s Masonry Structures, slide 70
36. 36. Strength Design of RM Shear Walls UBC Requirements Boundary Members: Sec.2108.2.5.6 1. Provide boundary members when the extreme fiber strain exceeds 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. #3 @ 8” wall > 3t centroid min. t 0.0015 εmu > 0.0015 Section at Base of Wall Masonry Structures, slide 71 Example: Strength Design Determine the maximum wind force, H, and design horizontal reinforcement to develop the wall flexural strength. Consider: zero vertical load and 5’-4” H Pdead = 40 kips and Plive = 30 kips 8” concrete block, fully grouted Grade 60 reinforcement , f’m= 1500 psi from previous example: 10’-8” Mn = 5 ,222 kip − in. Mu = φ Mn = 0 .85( 5 ,222) = 4 ,439 kip − in. check cracking moment per Eq. 8 - 30 M cr = f r S g f r = 4.0 f'm < 235 psi f r = 4.0 1500 = 155 psi 64 2 M M cr = ( 0.155 ksi)(7.63 × ) = 807 kip - in < n ok 7.63” 6 1.8 Mu H 4 - #8’s if flexure limit state exists : H u = = 34.7 kips H = u = 26.7 kips 128quot; 1.3 Masonry Structures, slide 72
37. 37. Example: Strength Design Shear Reinforcement (neglecting vertical force) U = 1.3W shear design within Lw (5’-4”) of base: L Vn = Vs = Amv ρ n f y = ( w ) As horiz f y = ( 0.5 ) As horiz ( 60 ksi) h Vu = H u = 34.7 kips = φ Vn = 0.80[(0.5) (Ahoriz ) (60 ksi)] As horiz = 1.45 in 2 smax = 24quot; use #4' s @ 8quot; for bottom 8 courses As horiz provided = 8(0.20 in 2 ) = 1.60 in 2 > 1.45 in 2 shear design for top 5’-4” of wall: Vn = Vm + Vs = C d Amv f'm + Am ρ n f y L = 1.2(7.63 x 64.00) 1500 + ( w )(As horiz )f y = 22.7 kips + (0.5)(As horiz )(60 ksi) h Vu = H u = 34.7 kips = φ Vn = 0.80[22.7 + (0.5)(As horiz )(60 ksi)] As horiz = 0.69 in 2 smax = 48quot; use #4' s @ 16quot; for top 8 courses As horiz provided = 4(0.20 in ) = 0.80 in 2 > 0.69 in 2 2 Masonry Structures, slide 73 Example:Shear Wall Strength Design Confinement Reinforcement (neglecting vertical force) Confinement requirements for vertical reinforcement per Sec. 2108.2.5.6 Mu = 4,439 kip-in. 5’-4” 7.63” 3t > 5.7” #8 #3 @ 8” bottom 8 courses 11.5” ε = 0.003 ε = 0.0015 5.75” Strain Diagram per Previous Example Masonry Structures, slide 74
38. 38. Example: Strength Design Flexural Strength considering Vertical Loads Case 1: Pu = 0.75(1.4 x 40 + 1.7 x 30) = 80.3 kips perhaps maximum flexural capacity and critical for shear design Case 2: Pu = 0.9(40) = 36.0 kips perhaps minimum flexural capacity and lowest Hu capacity reduction factors Pb = 0.85f'm bab ε mu 0.003 ab = 0.85 d = 0.85 60quot; = 30.2quot; fy 0.003 + 0.00207 ε mu + Es Pb = 0 .85( 1 .5 ksi)( 7 .63quot;)( 30 .2quot;) = 294 kips 0 .25 Pb = 73 .5 kips considering reinforcement : Pb = −294 + 0 .79 in 2 ( − 60 .0 - 36 .0 + 20 .2 + 60 .0 ) = −306 kips Case 1: Pu = 80.3 kips > 0.25 Pb = 73.5 kips φ = 0.65 36.0 Case 2: Pu = 36.0 kips < 0.25 Pb = 73.5 kips φ = 0.65 + ( )0.20 = 0.75 73.5 Masonry Structures, slide 75 Example: Strength Design Flexural Strength considering Vertical Loads ⎛ c − di ⎞ εi = ⎜ ⎟ ( − 0.003 ) f si = E s ε si < f y C m = 8.27 c ⎝ c ⎠ d1 = 4.0” d2 = 20.0” d3 = 44.0” ε1 f1 Cs1 ε2 f2 Ts2 ε3 f3 Cs3 Cm Pn Mn (ksi) (kips) (ksi) (kips) (ksi) (kips) (kips) (kips) (kip-in) 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 0.85c ∑ M cl = C m (32.0quot; - ) + C sl (28.0quot; ) - Ts2 ( 12.0quot; ) + Ts3 ( 12.0quot; ) + Ts4 ( 28.0quot; ) 2 Masonry Structures, slide 76
39. 39. Example: Strength Design Flexural Strength considering Vertical Loads Case 1: M u = φ M n = 0.65(6450) = 4,192 kip - in. Axial Compressive Force, kips 140 6450 kip-in. 4,192 120 Hu = = 32.7 kips (10.67 x 12) 100 Hu H= = 25.7 kips governs ( 0.75 x 1.7) 5820 kip-in. 80 80.3 kips Case 1 60 Case 2: M u = φ M n = 0.75(5820) = 4,365 kip - in. 40 4,365 Case 2 Hu = = 34.1 kips 36.0 kips (10.67 x 12) 20 Hu H= = 26.2 kips 1.3 5500 6000 6500 7000 Moment, Mn kip-in. Hu = 34.1 kips (Case 2) ~ 34.7 kips (w/o vertical force). Use same shear design as for first part of problem. Mu = 4,365 kip-in. (Case 2) ~ 4,439 kip-in. (w/o vertical force). Use same boundary members as for first part of problem. Masonry Structures, slide 77