CEE 626 MASONRY DESIGN SLIDES Slide Set 4b Reinforced Masonry Spring 2015

Reinforcement Information - Code

SD Limits on Strength of Reinforcement: TMS 402 Sec. 9.1.9.3 Reinforcement for In-Plane Flexural Tension and Flexural Tension Perpendicular to Bed Joints fy  60 ksi Actual yield strength shall not exceed 1.3 times the specified value Reinforcement for In-Plane Shear and Flexural Tension Parallel to Bed Joints fy  60 ksi for reinforcing bars fy  85 ksi for reinforcing wire

More on Min and Max reinforcing (except Special Shear Walls) Seismic Provisions may require min, As. ASD and SD SD ONLY - For beams 9.3.4.2.3 requires Mn > 1.3 Mcr unless Mn> 1.33 x Mu. Beam example on Board

Reinforcement requirements and details: TMS 402 Section 6 Reinforcement requirements and details: TMS 402 Section 6.1 (ASD and SD) Bar diameter  # 11,1/8 x nominal wall thickness, 1/2 x cell dimension (smallest). Bar area ≤0.4% cell area Wire diameter for joint reinf. ≤ ½ joint thickness. Clear distance – not less than db or 1” in columns or pilasters 1.5 db or 1.5” in Clear space between bars and units ¼” between bar and cell edge – fine grout 1/2” between bar and cell edge – coarse grout

Reinforcement requirements and details: TMS 402 Section 6 Reinforcement requirements and details: TMS 402 Section 6.1 (ASD and SD) Cover – Protection 2” exposed for larger than #5 1.5” exposed for #5 and smaller 1.5” not exposed Includes masonry unit elements

Reinforcement requirements and details: TMS 402 Section 9.3.3 SD Addl. Bar diameter  1/8 nominal wall thickness, ¼ cell dimension or #9. (smallest) Standard hooks and development length development length based on pullout and splitting (Same ASD and SD) – More later In walls, shear reinforcement must be bent around extreme longitudinal bars – Splices lap splices based on required development length welded and lap splices must develop 1.25 fy

Development Length ASD 8.1, SD 9.3 Required embedment length in tension addresses splitting from bar to surface and bar to bar For epoxy – coated bars or wires, increase the above values by 50%

Development of reinforcement embedded in grout: TMS 402 Sec. 8.1.6 Other requirements for flexural reinforcement Equation (8-12) used for lap splices Can decrease lap splice using confining steel. SD requirements the same. Must space laps within 8”.

Background Since the mid-1990s, dozens of research investigations tested nearly 1000 specimens to better understand the performance and strength of lap splices in masonry construction. The resulting code design equations have fluctuated considerably as a result.

How Did We Get Here? Early phases of research answered many questions, and raised others. As bars approached yield (no pullout), longitudinal splitting of the masonry was dominate failure mode. Short lap: fs < fy Long lap: fs > fy Wow lap: fs ~ fu Target: fs > 1.25fy

ACI 318 Equation (general) Cb is the smallest of side cover to bar center or ½ c-c spacing (Cb +Ktr)/db= a confinement term ≤ 2.5

ACI 318 Equation (general) t is location factor = 1 for not top e is coating factor = 1 for uncoated s is size factor = .8 for #6 or smaller  Concrete type factor = 1.0 for normal (Cb +Ktr)/db at least 1.0 (min cover, etc)

ACI 318 Equation (general) If clear cover is db and Spacing is 2 db & fy = 60,000 psi and f’c = 4000 psi ld = 47 db or for lower cover or spacing ld = 71 db can also reduce ld by excess As except in SDC D and above

2008/2011 MSJC Equation Key Features: Smaller bars sizes and larger cover distances result in smaller lap splice lengths. However, the MSJC lap splicing requirements are ‘overridden’ by the IBC…

2009/2012 IBC Requirements For allowable stress design: ld = 0.002 db fs Where: fs = calculated stress in reinforcement In cases where the stress in the reinforcement is greater than 80% of the allowable steel stress (Fs), the lap length is increased 50%.

2009/2012 IBC Requirements For allowable stress design: ld = 0.002 db fs For 2008 MSJC: Fs = 24,000 psi (Grade 60) with 50% length increase, then the above reduces to: ld = 72 db

2009/2012 IBC Requirements For allowable stress design: ld = 0.002 db fs For 2011 MSJC: Fs = 32,000 psi* (Grade 60) with 50% length increase, then the above reduces to: ld = 96 db *The 2011 MSJC as recalibrated the allowable stresses.

2009/2012 IBC Requirements Because the stress in a bar varies depending upon its location within an element, the common default is to detail laps for the largest stress possible. Bar Size 72 db 96 db 4 36 in. 48 in. 5 45 in. 60 in. 6 54 in. 72 in. 7 63 in. 84 in. 8 96 in. 9 81 in. 108 in. 10 90 in. 120 in. 11 99 in. 132 in.

2009/2012 IBC Requirements Alternatively, the stress in the steel can be ‘capped’ at 80% of the allowable stress, which reduces maximum laps by 50%...but results in design conservatism for assembly strength. Bar Size 48 db 64 db 4 24 in. 32 in. 5 30 in. 40 in. 6 36 in. 48 in. 7 42 in. 56 in. 8 64 in. 9 54 in. 72 in. 10 60 in. 80 in. 11 66 in. 88 in.

2009/2012 IBC Requirements The strength design provisions in the IBC are a little easier to implement. They simply require the use of the MSJC lap splice equation, but cap it at 72 db. Compared to the MSJC directly…

2009/2012 IBC Requirements Typical 8 inch Concrete Masonry Unit Lap Lengths – Strength Design Bar Size Lap Length, MSJC (in.) Lap Length, IBC – 72db Cap (in.) No. 3 12.0 No. 4 14.1 No. 5 22.5 No. 6 42.8 No. 7 59.4 No. 8 91.2 72.0 No. 9 117.6 81.0

New Design Provisions The 2011 MSJC (and by reference the 2012 IBC) introduced a new coefficient that permits lap splices to be reduced when ‘confined’ by transverse reinforcement.

Typical Test Specimen – NCMA - WSU

Observations Significant difference in the performance and strength of lap splices confined with transverse reinforcement. With Bond Beams Without Bond Beams

Modified Code Equation Where the ‘confinement’ factor is: and

Limits on Confinement Asc < 0.35 in.2 but greater than 0.11(#3) Transverse Offset < 1.5 in. Longitudinal Offset < 8 in. Lap Splice > 36db

Modified Code Equations Typical 8 inch Concrete Masonry Unit Lap Lengths Centered 8 in units, f’m = 1500 psi Bar Size MSJC Lap Length, No Confinement (in.) MSJC Lap Length, With No. 5 Confinement (in.) No. 3 12.0 13.51 No. 4 14.1 18.01 No. 5 22.5 22.51 No. 6 42.8 27.01 No. 7 59.4 31.51 No. 8 91.2 36.01 No. 9 117.6* 55.2* No. 10 148.1* 87.6* No. 11 182.8* 124.0*

Modified Code Equations Typical 8 inch Concrete Masonry Unit Lap Lengths Centered 8 in units, f’m = 1500 psi Bar Size MSJC Lap Length, No Confinement (in.) f’m 1500 psi MSJC Lap Length, No Confinement (in.) f’m 2500 psi No. 3 12.0 12 No. 4 14.1 No. 5 22.5 17.4 No. 6 42.8 33.2 No. 7 59.4 46.0 No. 8 91.2 70.6 No. 9 117.6 91.1 No. 10 148.1 114.7 No. 11 182.8 141.6

Application of Confinement The use of confinement reinforcement is an option – consider using with large bars (ignore for small bars). Most beneficial when bond beam reinforcement is already a part of the design. Incorporated into 2011 MSJC and 2012 IBC (by reference).

Development of reinforcement embedded in grout: Standard Hooks le= 13 db

Maximum reinforcement: TMS 402 Sec. 9. 3. 3 Maximum reinforcement: TMS 402 Sec. 9.3.3.5 STRENGTH (SD) ONLY Max Reinf. No upper limit when Mu/(Vudv) ≤ 1 and R ≤ 1.5 Other members, maximum area of flexural tensile reinforcement determined based on: Strain in extreme tensile reinforcement = 1.5 εy Axial forces determined from D + 0.75L + 0.525QE Compression reinforcement, with or without lateral restraining reinforcement, permitted to be included. Intermediate shear walls with Mu/(Vudv) ≥ 1, strain in extreme tensile reinforcement = 3εy Special shear walls with Mu/(Vudv) ≥ 1, strain in extreme tensile reinforcement = 4εy The requirements for shear walls only apply to in-plane loads.

Maximum reinforcement: TMS 402 Sec. 9.3.3.5 mu = 0.0035 clay or 0.0025 concrete s =  y  fy Reinforcement Axial Load 0.80 fm 1 = 0.80 Locate neutral axis based on extreme - fiber strains Calculate compressive force, C Reinforcement + Axial Load = C

Clay masonry, f ʹm = 2000 psi, Grade 60 steel Max Reinforcement for Beams and walls under flexure only: TMS 402 Section 9.3.3.5.1 Pu  0.05 An fm εs  1.5 εy Clay masonry, f ʹm = 2000 psi, Grade 60 steel ρmax = 0.01131 Concrete masonry, f ʹm = 2000 psi, Grade 60 steel ρmax = 0.00952 Maximum reinforcement ratios are given for clay and concrete masonry for a common f’m. The maximum reinforcement ratio corresponds to 84% of rho-balanced for clay masonry, and 82% of rho-balanced for CMU.