Date July, 2015 PCA Work shop W. Mark McGinley Ph.D, PE ASD Wall Design for a Single Story Masonry Building

Look at a Single Wythe single story CMU Building Example Slide 2

3 Plan of Typical Big Box - single story Flexible diaphragm reinforced 8” CMU, Load bearing and non loadbearing walls. Flex Diaphragm- High Seismic Single Story Building - Ex1 ItemValue Roof Live Load20 psf Roof Dead Load (including joist weight) 20 psf 8” CMU wall weight (grouted at 2’ OC) 60 psf 8” CMU wall weight (fully grouted) 80 psf Weight of Glass doors and store front 10 psf 4” brick wall weight40 psf Door dead load5 psf Wind Zone 150 mph, Risk Category II Wind ExposureC Soil Site ClassD Seismic S Seismic S s 1.50

4 Single Story Building Ex1 Elevations

5 Ex1 Loads Roof Load:Ρ DL = 11,900 lb Ρ LL = 11,500 lb Ρ uplift = 31,700 lb

6 Ex1 Loads Using loads determine Critical P, M and V at support and at mid-height of walls for all load cases

7 Capacity of Wall – Single Wythe Reinforced CMU’s

TMS 402 Design Provisions Two Rational Design Methods Allowable Stress Design (ASD) – stresses under service level loads must be less than allowable values. Strength Design (SD) – Capacities must be greater than load effects under strength level loads (ultimate levels). Both methods can be applied to unreinforced or reinforced masonry. Today Address Reinforced ASD. Slide 8

9 ASD Load Combinations – IBC 2012 /ASCE 7-10 D +F D + H+ F+L D + H +F+(L r or S or R) D +H+F+0.75(L) (L r or S or R) D +H+F+(0.6W or 0.7E) D +H+F+0.75(0.6W) L+0.75(L r or S or R) D +H+F+0.75(0.7E) L+0.75(S) 0.6D +0.6W + H 0.6(D+F) +H + 0.7E No increase for E or W any more with Stress Recalibration – Even with alternative load cases

ASD - Reinforced Masonry: TMS 402 Section 8.3 Masonry in flexural tension is cracked Reinforcing steel is needed to resist tension Linear elastic theory No minimum required steel area and A s limited only on special shear walls. Wire joint reinforcement can be used as flexural reinforcement No unity interaction equation – Combined loads must used interaction diagram Slide 10

Allowable Stresses for Steel Reinforcement: TMS 402 Sec Tension Grade 40 or 5020,000 psi Grade 6032,000 psi Wire joint reinforcement30,000 psi Compression Only reinforcement that is laterally tied (Section ) can be used to resist compression Allowable compressive stress = tensile values above Slide 11

Allowable Axial Compressive Capacity For reinforced masonry, allowable compressive capacity is expressed in terms of force rather than stress Maximum compressive stress in masonry from axial load plus bending must not exceed 0.45f’ m Axial compressive stress alone must not exceed allowable axial stress from TMS 402 Section (unreinforced F a ). Slide 12

Allowable Axial Compressive Capacity: TMS 402 Section TMS 402 Equations (8- 21) and (8 - 22) define axial force limits (ASD). Slenderness reduction coefficients are identical to those used for unreinforced masonry. Allowable capacity is sum of capacity of masonry plus compressive reinforcement Slide 13

TMS 402 Section – Shear in Reinforced Masonry... In design we generally: Check if shear can be resisted entirely by masonry. If not, increase cross – section or, Add shear reinforcement. Check shear stress. If still no good, increase cross – section. Next slide shows changes from past code edition! Slide 14

Masonry shear stress: TMS 402 Section Shear stress is computed as: Allowable shear stresses for partially grouted shear walls, 1.0 otherwise. Slide 15

Masonry Shear Stress: TMS 402 Section Allowable shear stresses limits: M / Vd v ≤ 0.25 M / Vd v  1 Can linear interpolate between limits Slide 16

Masonry Shear Walls: TMS 402 Section Allowable Shear Stress Resisted by the Masonry Special Reinforced Masonry Shear Walls All other masonry M/Vd v is positive and need not exceed 1.0. Slide 17

Shear Design of Reinforced Masonry: TMS 402 Section (continued) If allowable shear stress in the masonry is exceeded then: design shear reinforcement using Equation 8-30 and add F vs to F vm Shear reinforcement is placed parallel the direction of the applied force at a maximum spacing of d/2 or 48 in. One - third of A v is required perpendicular to the applied force at a spacing of no more than 8 ft. Slide 18

19 C M V T jd ASD Reinforced Masonry- Singly Reinforced - FLEXURE n = E s /E m and from equil. Ms= A s f s jd (at the limit) = A s F s jd Mm = ½ bjkd 2 f m (at limit)= ½ bjkd 2 F b f m ≤ Fb f s /n ≤ Fs/n kd

20 ASD Interaction Diagrams- Flexural Compression Members To design reinforced walls under combined loading must construct interaction diagram stress is proportional to strain ; assume plane sections remain plane ; vary stress ( stress ) gradient to maximum limits and position of neutral axis and back calculate combinations of P and M that would generate this stress distribution

21 Allowable Stress Interaction Diagrams Assume single reinforced Out-of-plane flexure Grout and masonry the same Solid grouted Steel in center CLCL

22 ASD Interaction Diagrams Walls - Singly Reinforced allowable – stress interaction diagram Linear elastic theory – tension in masonry it is ignored- Plane sections remain plane Limit combined compression stress to F b = 0.45 F’ m P ≤ P a d usually = t/2 - ignore compression steel since not tied.

23 Allowable Stress Interaction Diagrams Walls - Singly Reinforced Assume Stress gradient- Range A – All the Section in compression Get equivalent force-couple about center line P a = 0.5(f m1 +f m2 )A n M a = (f m1 -f m2 )/2 (S) S = bd 2 /6 Note at limit – f m1 and f m2 ≤F b (set f m1 =F b ) Note much of this is from Masonry Course Notes By Dan Abrams Also P a must less than Eq 8-21or 8-22 b eff

Distribution of concentrated loads, running bond: TMS 402 Section Slide 24

Distribution of concentrated loads, running bond: TMS 402 Section Load 1 2 Effective Length Load 1 2 Effective Length Slide 25

26 Allowable Stress Interaction Diagrams Walls - Singly Reinforced Assume Stress gradient- Range B – Not All Section in compression- but no tension in steel Get equivalent force-couple about center line P b = Cm = 0.5(fm1)  tb M b = e m x C m e m =d-  t/3 = t/2-  t/3 =t(1/2-  /3) Note that  t = kd This is valid until steel goes into tension Set f m1 = F b at limit b eff

27 Allowable Stress Interaction Diagrams Walls - Singly Reinforced Assume Stress gradient- Range C – Section in compression- tension in steel Get equivalent force-couple about center line e m =d-  t/3 = t/2-  t/3 =t(1/2-  /3) C m = 0.5(f m1 )  tb P c = C m –Ts & T= A s x f s From similar triangles of the stress diagram f s /n= ([d-  t]/  t)f m1 M b = e m x C m –T s (d-t/2) note that d=t/2 usually so second term goes to zero At limit f s = F s and f m1 = F b one or the other governs Note that at = kd b eff tt fs/n

Effective compression width per bar: TMS 402 Section For running - bond masonry, or masonry with bond beams spaced no more than 48 in. center – to – center, the width of compression area per bar for stress calculations shall not exceed the least of: Center - to - center bar spacing Six times the wall thickness (nominal) 72 in. Slide 28

29 Axial Load P Moment M Capacity envelop letting f m1 = F b Range B Range A Capacity envelop letting fm1 = Fb Range C Capacity envelop letting fs = Fs P cut off Eq 8-21 or 8-22 Ms Mm Can get a three point interaction diagram easily Most walls have low axial loads Allowable Stress Interaction Diagrams Walls - Singly Reinforced NO GOOD ABOVE P CUT OFF

30 West Wall Design for Out-of- Plane Loads Guess at wall unit size –Usually 8” CMU. Guess at bar size location and spacing – can use max. Moment and assume steel stress governs Create interaction diagram for wall. Plot diagram and critical P &M values – If all load effects within capacity envelope wall is OK. No P-delta, or min. As, or max. As.

Ex 1 Out-of-Plane Wall Design Slide 31 Flexural Design of the Wall Trial reinforcing steel: D -.7E produces the maximum moment. Assume d =7.63/2 = 3.81 in. P = 3580 lb/ftM = 16,400 lb-in /ft Assume j = 0.9 A s = M / f s jd = 16,400 lb-in /(32,000 psi×0.9×3.81 in.) = 0.15 in. 2 /ft Try a No in. on center. This provides A s =0.232 in. 2 /ft

Ex 1 Out-of-Plane Wall Design Slide 32 Range C Range B

Ex 1 Out-of-Plane Wall Design Slide 33 Fs = 32,000 f s /n [  t/ ([d-  t]) =f b 32000/16.11 [  (3.81/ ([ (3.81  ]) C mas =0.5x0.1(3.81)(12)(221) P=221 – 0.23 x M b = e m x C m –T s (d-t/2) =(3.81-( 0.1 x 3.81/3)x32000x0.23

Ex 1 Out-of-Plane Wall Design Slide 34 fb=Fb = 900 f s = [d-  t/ ([  t]) nf b [(  (3.81/ (0.4(3.81  ])(900 x C mas =0.5x0.4(3.81)(12)(900) P=221 – 0.23 x M b = e m x C m –T s (d-t/2) =(3.81-( 0.1 x 3.81/3)x32000x0.23

Ex 1 Out-of-Plane Wall Design Slide 35

Ex 1 Out-of-Plane Wall Design Slide 36

37 A E V D1N2 V D2N1 V D1N1 V D1E1 V D1E2 V D2W V D2E1 North Wall 2 North Wall 1 South Wall East Wall 2 East Wall 1 West Wall Diaphragm 2 Diaphragm 1 V D2S Plan of Typical Big Box - single story Flexible diaphragm See MDG for Load determination and distribution to shear wall lines - Flex Diaphragm – SDC- D Diaphragm2 West Wall North Wall 1 East Wall 1 East Wall 2 West wall 2 Example Shear wall in a single story Building Shear Wall Ex2

38 To check wall segments under in -plane loads must first Distribute Load to Shear wall lines – Either by Trib. Width or Rigid Diaphragm analysis. Distribute Line load to each segment w.r.t. relative rigidity. Look at Shear Wall Design

39 Shear Wall Loads Distribution Segments get load w.r.t. relative k kips Diaphragm Shear due to seismic

40 Shear Wall Loads Distribution Segments get load w.r.t relative k For Cantilevered Shear wall segments For Fixed-Fixed Shear wall segments

41 Shear Wall Load Distribution

42 Shear Wall Load Distribution Segment 2 Designed in later Example

43 Design of Reinforced Masonry (ASD) In Plane Loading (shear Walls) h V L Axial Force

44 ASD Design of Reinforced Masonry -In Plane Loading (Shear Walls) Still use interaction diagrams Axial Load is still dealt with as out of plane (M=0) In plane load produces moment and thus moment capacity is dealt with slightly differently

Initially assume f m = F b and Neutral Axis then same as out-of-plane but area and S are based on Length = d and t = b use OOP equations in Range A and B. Adjust  L as before until rebars start to go into tension. Note that  L = kd Determine f si from similar triangles & get T i Check extreme f si ≤ F s & f m ≤ F b Cm =  L x b x ½ F b (or f m when f sn = Fs) M capacity (Σabout center)= Σ (T i x (d i -L/2) + C m x(L/2 –  L/3)) 45 P-M Diagrams ASD-In Plane

46 Reinforced Masonry Shear Walls - ASD h V L P V= base shear M = overturning moment Axial Force ~ 0

Flexure Only P = 0 on diagram h V L P- self weight only ignore V= base shear M = over turning moment Multiple rebar locations Reinforced Masonry Shear Walls - ASD 47

48 L V= base shear M = over turning moment fm k*d* F sc /N TiTiTiTi F s1 /N f si /N f sn /N<= Fs/N di – location to centroid of each bar Cm T sn = Fs As TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi T1T1T1T1 Tension Compression f si /N Reinforced Masonry Shear Walls – ASD (P =0) Can use the singly reinforced equations d*– location centroid of all bars in tension f* si /N

49 Moment only ASD In Plane To locate Neutral Axis – Guess how many bars on tension side – As* Find d* (centroid of tension bars) and  *=A s */bd* Get k* = ((  *n)  *n) 1/2 – n  * Unless tied ignore compression in steel.

50 Moment only ASD In Plane Check k* d* to ensure assume tension bars correct – iterate if not Determine f si from similar triangles and T i M capacity (Σabout C)= Σ (T i x (d i – k*d*/3))

51 Shear Wall Example 3 Geometry: typical wall element: 25 ft – 4 in. total height 3 ft – 4 in. parapet 24 ft length between control joints 8 in. CMU grouted solid: 80 psf dead f’ m = 1500 psi VDVD VPVP 22’-0” 25’-4” 12’-8” 24’-0”

52 Shear Wall Example 3 West Wall seismic load condition V diaphragm = 25,800lb acting 22ft above foundation V pier = 8,100lb acting 12.7ft above foundation 8in. CMUgrouted solid (maximum possible dead load) P base = 80lb/ ft 2 x 25.3 ft x 24ft = 48,600lb Vertical Seismic: V pier =0.2 S DS D = (-0.2 (1.11)(48,600)=10,800lb ASD Load Combination 0.6 D E P = 0.6 x 48, x-0.2 (1.11) (48,600) =21,600 lb M = 0.6 x x (25,800lb x 22ft +8,100 lb x 12.7ft) =469,000 lb-ft =5,630,000 lb in. V = 0.6 x x (25,800lb +8, 100 lb) =23,700 lb

53 Shear Wall Example 3 #5 bar (typ) 24” 4”4” 8”8” 8”8” Assume the rebar in the wall are as shown – Axial load is negligible – ignore - To simplify assume that only three end bars are effective (only lap these to foundation)

54 Shear Wall Example 3 We are using 3 #5 Bars but if needed an estimate of A s can be determined by assuming j =.9 and applied moment, M

55 Shear Wall Example 3 Calculate j and k: You need to get the stress at the centroid based on the extreme bar f s =F s Check Masonry Compression Stresses Should get third bar stress then  Moments but M ≈

56 Shear Wall Example 3 Check Shear Stress Assume no shear reinforcing and thus

57 Shear Wall Example 3 Check Shear Stress Conservatively assume just face shell bedded areas resist shear

58 So the Final Design – can use the # 5 at the ends of the wall – ignoring any bars that will likely be there for out-of-plane loading #5 bar (typ) 24” 4”4” 8”8” 8”8” Shear Wall Design Check Prescriptive Seismic Reinforcing

Requirements for Detailed Plain SWs and SDC C: TMS 402 Section #4 bar (min) within 8 in. of corners & ends of walls roof diaphragm roof 48 in. max oc #4 bar (min) within 16 in. of top of parapet Top of Parapet #4 bar diaphragms continuous through control joint #4 bar (min) within 8 in. of all control joints control joint #4 10 ft oc or W1.7 joint 16 in. oc #4 10 ft oc 24 in. or 40 db past opening #4 bars around openings Slide 59

Seismic Design: TMS 402 Chapter 7 Seismic Design Category D Masonry that is part of the lateral force – resisting system must be reinforced so that  v +  h  0.002, and  v and  h  Type N mortar and masonry cement mortars are prohibited in the seismic force – resisting system Shear walls must meet minimum prescriptive requirements for reinforcement and connections (special reinforced) Other walls must meet minimum prescriptive requirements for horizontal and vertical reinforcement Slide 60

Requirements for Special Reinforced Shear Walls: TMS 402 Section roof diaphragm roof 48 in. max oc #4 bar (min) within 16 in. of top of parapet Top of Parapet #4 bar diaphragms continuous through control joint #4 bar (min) within 8 in. of all control joints control joint #4 4 ft oc #4 bar (min) within 8 in. of corners & ends of walls 24 in. or 40 db past opening #4 bars around openings Slide 61

Seismic Design Categories E and F: TMS 402 Section Additional reinforcement requirements for masonry not laid in running bond and used in nonparticipating elements Horizontal Reinforcement of at least A g Horizontal Reinforcement must be no more than 24 in. oc. Must be fully grouted and constructed of hollow open-end units or two wythes of solid units Slide 62