# Lecture 6 7 Rm Shear Walls

date post

07-May-2015Category

## Business

view

2.778download

1

Embed Size (px)

description

### Transcript of Lecture 6 7 Rm Shear Walls

- 1.Reinforced Masonry Working Stress Design of flexural members b strainstress mfm Cm = fmb kd/2 kdd n.a.M Ts = Asfs t sfs/ngrout AsAs = 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 Assumptions1. plane sections remain plane after bending (shear deformations are neglected, strain distribution is linear with depth)2. neglect all masonry in tension3. stress-strain relation for masonry is linear in compression4. stress-strain relation for steel is linear5. 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 : si = mi f si f miEsfrom Assumption s #3 and #4 :=f si = f mi = nf mi E s Em Emfm f n 1 kfrom geometry of stress distribution : = s f s = nf mkd d kd kMasonry Structures, slide 2

2. Reinforced Masonry Working Stress Design of flexural membersfrom equilibrium, C = T :f m bkd1k= As f s = bd f s = bd n f m2 k1 k k / 2= nk 2 + 2 nk 2 n = 0 kfrom equilibrium: M about C m = 0 M s = As f s jd = bd 2 jf swhere jd=d-kd/3 or j=1-k/3If 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 60Fs=30 ksi for wire joint reinforcementAllowable reinforcement tensile stress per UBC Sec.2107.2.11 :Fs= 0.5fy < 24 ksi for deformed bars; Fs= 0.5fy < 30 ksi for wire reinforcementFs= 0.4fy < 20 ksi for ties, anchors, and smooth barsMasonry Structures, slide 3 Reinforced Masonry Working Stress Design of flexural members from equilibrium: M about Ts =0 M m = 0.5 f m bkdjd = 0.5 f m jkbd 2If fm= Fb then moment capacity will be limited by masonry.UBC 2107.2.6 & MSJC Sec.2.3.3.2: Fb=0.33fm Masonry Structures, slide 4 3. Reinforced Masonry WSD: Balanced Conditionfm = F bDefinition: The balanced condition occurskbdC = 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 fsfs/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 Conditionfrom geometry: from equilibrium: C=TFs fm = Fb Fb +Fbn d =Fb b kb = b b d Fskb dd 2 kbdFkF n dFbb = b b = bkb =2 Fs 2 Fs ( n + Fs ) Fs Fb +Fb n fs/n = Fs /n nn 1 kb =b = n + Fs /Fb n + Fs / Fb 2Fs / Fb Masonry Structures, slide 6 4. Example: Balanced Condition Determine the ratio of reinforcement that will result in a balanced condition per UBC. Given: fm = 2000 psi and Grade 60 reinforcement Fb = 0.33 f'm = 667 psiFs = 24 ksi for Grade 60 reinforcement E m = 750 f'm = 1500 ksi E s = 29,000 ksi E s 29,000 n = = = 19.3Em 1500 19.3 1 b = = 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 b knowing fm and fydetermine Fb from fmdetermine Fs from fydetermine Emfrom fmdetermine n = Es/Em Size section for some < b Note: Section must also be determine k and jsized for shear. bd2 = M/jFs select b and d using common unitsSize reinforcement Check designAs = M/Fsjd Ms = AsFsjd > Mselect number and size of rebarsfb = M/0.5jkbd2 < Fb Masonry Structures, slide 8 5. Example: Reinforced MasonryDesign a beam section for a moment equal to 370 kip-in. Prisms have beentested and fm is specified at 2000 psi. Use Grade 60 reinforcement and 8CMUs. 1. From previous example, b = 0.48% 2. Estimate to be slightly lower than so steel will govern.bA good estimate is = 0.4% 3. Solve for k : k 2 + 2 nk 2 n = 0 2 n = 2( 0.004 )( 19.3 )= 0.154 k 2 + 0.154 k 0.154 = 0 k = 0.323 j = 1 k / 3 = 0.8924. Solve for bd 2 :bd 2 = M / j F sbd 2 = ( 370 kip in ) /( 0.004 )( 0.892 )24 ksi ) = 4321 in 3 Masonry Structures, slide 9Example: Reinforced Masonry5. Select dimensions of beam using 8 CMUs: b = 7.63 4 - 8 CMUs dreqd = [4321 / 7.63]0.5 = 23.8 d=27.8 use four units and center bars in bottom unit, d = 27.86. Estimate amount of reinforcem ent : As req' d = M / Fs j d 7.63As req' d = (370 kip - in) / (24 ksi) (0.892) (27.8) = 0.62 in 2use 2 #5' s (0.62 in 2 )7 . Check design: = s / bd = 0.62 in 2 / (7.63quot; ) ( 27.8quot; ) = 0.00292k 2 + 2 nk 2 n = 0k = 0.284 j = 1 0.284 / 3 = 0.905 s = s 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 = / 0. 5 j k d 2 == 488 psi < 667 psi ok ( 0.5 )( 0.905 )( 0.284 )(7.63quot; )( 27.8quot; ) 2Masonry Structures, slide 10 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*btf kd neutral dtaxis As As per width bIf 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:Ask 2 + 2 nk 2 n = 0 =bdIf 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 MasonrysCm Vm dVs Vext.ViyAsfs VdR Basic shear mechanisms:before cracking: Vext = Vint = Vm + Vd + Viy + Vs Once diagonal crack forms: Vm reducesflexural stresses increase dowel action invokedfsa is related to MbPresence of shear reinforcement will: restrict crack growth resist tensile stress help dowel actionMasonry Structures, slide 12 7. Shear Design of Reinforced Masonryafter cracking: Vext = Vint = Vs = nAvfs where n is the number of transversebars across the diagonal crack.Assuming a 45 degree slope, n=d/sVs = (d/s)Avfs Av Vs VsUBC Sec. 2107.2.17 (Eq. 7.38) = = MSJC Sec. 2.3.5.3 (Eq. 2-26) s df s dFsMasonry Structures, slide 13 Shear Design of Reinforced MasomryFlexural shear stress dx C M C + dC M + dMfvbdx najd TT + dT T T + dT fvb dx = dT = dM/jdfv = (dM/dx)/bjd Vfv =UBC Sec. 2107.2.17 (Eq. 7-38)bjdV fv =bd MSJC Sec. 2.3.5.2.1 (Eq. 2-19) Masonry Structures, slide 14 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 reinforcementFv = 1.0 f'm < 50 psiUBC 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 shearFv = 3.0 f'm < 150 psiUBC 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 MSJCUBC Sec. 2107.2.9.i and MSJC Sec. 2.3.5.2.2(b): walls with in-plane flexural reinforcement and no shear reinforcement M1 M M for 1 6.0'Mfor> 1 Fv = 1.0 f 'm < 35 psi Fv = 1.33 x 35 psi = 46.6 psiVd 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. 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.2.3.5.3Vmax= 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)Mfor> 1 F = 1.5 f'm 75 psi F = 1.5 3000 = 82.2psi>75 psivvVd Fv = 1.33x 75 psi= 100 psiV (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 strengthVmax = Fvbjd = ( 100 psi)( .63)(0.9 )(72)/ 1000 = 49.4 kips > 34 kips oka govern 7ysAv /S = Vmax/Fsd = 49.4 kips/( .33 x 24 ksi)( ) = 0.0215in2 per in. 172u sing #4 rebars(Av = 0.20 in2 ) s = 0.20 / 0.0215 = 9.3quot; use # @ 8 in. horiz4 ontal Summary: Hmax, kipsCaseConsideration UBC MSJCANo steelNo steel14.7*14.7* 10.2*10.2*vertical steel24.327.0Bvertical steel 23.025.6no horizontal steel no horizontal steelvertical steel and 15.2 15.2Cvertical steel and14.4 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. Flexural Bond Stress M = Tjd M + dM = (T + dT)jd dM = dT jddx dT = dM/jdC + dCU = bond force per unit length for group of bars C U dx = dT = dM/jd U = (dM/dx)/jd = V/jd MU dxjdu = flexural bond stress =M + dMo where o = sum of perimeters of all bars in groupTT + dTVu= UBC Sec. 2107.2.16 Eq. 7-36dx jd UT + dT allowable bond stress per UBC Sec.2107.2.2.4: T 60 psi for plain bars 200 psi for deformed barsdx 100 psi for deformed bars w/o inspection Masonry Structures, slide 25 Development Lengthu d b dbAs fs ldAs fs = udb ld d b2 fs = udb ld4 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 8Masonry Structures, slide 26 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 aMoment Diagram(#2)> ld (#1)d or 12dbtheoretical cutoff pointcapacity with bars a bars b (#2) > ld moment capacity with bars a and bMasonry Structures, slide 27 Co

*View more*