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

2. Reinforcement Information - Code

3. 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

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

5. 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

6. 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

7. 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

8. 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%

9. 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”.

10. Background
Since the mid-1990s, dozens of research investigations tested nearly 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.

11. 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

12. 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

13. 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)

14. 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

15. 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…

16. 2009/2012 IBC Requirements
For allowable stress design: ld = 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%.

17. 2009/2012 IBC Requirements
For allowable stress design: ld = db fs For 2008 MSJC: Fs = 24,000 psi (Grade 60) with 50% length increase, then the above reduces to: ld = 72 db

18. 2009/2012 IBC Requirements
For allowable stress design: ld = 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.

19. 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.

20. 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.

21. 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…

22. 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

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

24. Typical Test Specimen – NCMA - WSU

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

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

27. 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

28. 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*

29. 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

30. 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 IBC (by reference).

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

32. Maximum reinforcement: TMS 402 Sec. 9. 3. 3
Maximum reinforcement: TMS 402 Sec 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 L QE
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.

33. Maximum reinforcement: TMS 402 Sec. 9.3.3.5
mu = clay
or 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

34. Clay masonry, f ʹm = 2000 psi, Grade 60 steel
Max Reinforcement for Beams and walls under flexure only: TMS 402 Section
Pu  0.05 An fm
εs  1.5 εy
Clay masonry, f ʹm = 2000 psi, Grade 60 steel
ρmax =
Concrete masonry, f ʹm = 2000 psi, Grade 60 steel
ρmax =
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.