CEE 626 MASONRY
DESIGN
SLIDES
Slide Set 4b
Reinforced Masonry
Spring 2015
1
Reinforcement Information - Code
Slide 2
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
Slide 3
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
Slide 4
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
Slide 5
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

Slide 6
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
Slide 7
Development Length ASD 8.1, SD 9.3

Required embedment length in tension
addresses splitting from bar to surface and bar
to bar
ld 

0.13 d b2 f y 
K
f m'
 12 in. (bars) , 6 in. (wires) (8 - 12)
For epoxy – coated bars or wires, increase the
above values by 50%
Slide 8
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.
  1.0 
2
2.3 Asc
fy 
db
2.5

SD requirements the same.

Must space laps within 8”.
(8 - 13)
Slide 9
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.

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
3
ld 
40
f y t e s
 cb  K tr
 f 'c 
 db



db
Cb is the smallest of side cover to bar
center or ½ c-c spacing
(Cb +Ktr)/db= a confinement term ≤ 2.5
40 Atr
K tr 
sn
3
ld 
40
f y t e s
 cb  K tr
 f 'c 
 db



db
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)
3
ld 
40
f y t e s
 cb  K tr
 f 'c 
 db



db
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
0.13d b f y
2
ld 
K f 'm
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…
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%.
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
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.
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
72 in.
96 in.
9
81 in.
108 in.
10
90 in.
120 in.
11
99 in.
132 in.
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
48 in.
64 in.
9
54 in.
72 in.
10
60 in.
80 in.
11
66 in.
88 in.
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…
Typical 8 inch Concrete Masonry Unit Lap
Lengths – Strength Design
Bar
Lap Length,
Lap Length, IBC –
Size
MSJC (in.)
72db Cap (in.)
No. 3
12.0
12.0
No. 4
No. 5
No. 6
No. 7
14.1
22.5
42.8
59.4
14.1
22.5
42.8
59.4
No. 8
No. 9
91.2
117.6
72.0
81.0
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.
Longitudinal
Split
Lap Spliced Reinforcement
Horizontal
Reinforcement
Significant
difference in the
performance and
strength of lap
splices confined
with transverse
reinforcement.
With Bond Beams
Without Bond Beams
 0.13d b f y
ld  
 K f'
m

2
Where the
‘confinement’
2.3 Asc
  1is:

factor
2.5
db
and




2.3 Asc

1
.
0
2 .5
db

Asc < 0.35 in.2 but greater than 0.11(#3)
Transverse Offset < 1.5 in.
 Longitudinal Offset < 8 in.
 Lap Splice > 36db

Typical 8 inch Concrete Masonry Unit Lap Lengths
Centered 8 in units, f’m = 1500 psi
MSJC Lap Length,
MSJC Lap Length,
Bar Size
No Confinement
With No. 5
(in.)
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*
Typical 8 inch Concrete Masonry Unit Lap Lengths
Centered 8 in units, f’m = 1500 psi
MSJC Lap Length, MSJC Lap Length, No
Bar Size
No Confinement
Confinement (in.)
(in.) f’m 1500 psi
f’m 2500 psi
No. 3
12.0
12
No. 4
14.1
12
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
 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
Slide 31
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
Slide 32
Maximum reinforcement:
TMS 402 Sec. 9.3.3.5
Axial

Locate neutral axis
based on extreme fiber strains

Calculate
compressive force, C

Reinforcement +
Axial Load = C
Load
s =  y
Reinforcement
 fy
mu = 0.0035 clay
or
0.0025 concrete
0.80 fm
1 = 0.80
Slide 33
Max Reinforcement for Beams and walls under
flexure only: TMS 402 Section 9.3.3.5.1
Pu  0.05 An fm
 max
0.80.8 f m

fy

m

   1.5
y
 m





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
Slide 34