CEE 626 MASONRY
DESIGN
SLIDES
Slide Set 4
Reinforced Masonry
Spring 2015
1
Reinforced Masonry
Allowable (ASD) & Strength (SD)
PLAN OVER THE NEXT FEW LECTURES  LOOK AT ASD AND STRENGTH DESIGN (SD) FOR
FLEXURE ONLY – BEAMS AND WALLS
 LOOK AT ASD & SD DESIGN FOR FLEXURE AND AXIAL
LOAD – WALLS AND COLUMNS
 LOOK AT SOME DETAILS – Ld and limits
 SHEAR – SHEAR IN BEAMS AND SHEAR WALLS FOR
ASD AND STRENGTH DESIGN
 DO EXAMPLES OF EACH IN PARALLEL As we go.
Design of Reinforced
Masonry – FYI

unreinforced versus reinforced masonry
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unreinforced masonry : masonry resists flexural tension ;
reinforcement is neglected
reinforced masonry : masonry does not resist flexural
tension ; reinforcement resists flexural tension
In Both Cases Direct tension must be resisted by the
reinforcement
allowable – stress versus strength design

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allowable – stress design : stress is proportional to strain
; stresses from (service) loads are compared against
allowable stresses
strength design : factored actions are compared against
design capacities
Design of Reinforced walls for Out of Plane Loads
Wall under out-ofplane Load
Walls Bending out of plane [Drysdale et-al
Masonry Design]
Section
Note bars usually
In center of wall
Allowable Stress Design Flexure
Ch 9 MDG 7 OOP (IP similar)
Flexural Tensile Reinforcement
Only single layer of steel - Strain
Distribution
Assumptions

Masonry in (net) flexural tension is
cracked

Reinforcing steel is needed to resist
tension

Linear elastic theory applies

No minimum As (except seismic- later)
required steel area and only one instance
max steel area – (equations do provide
cap)
M
d
kd
m
s
As Usually
in middle
Definition of Terms – In Code
C7—C24
t for OOP
ASD- Flexure Only OOP (IP similar)
Look at internal Equil.
M
SFvert = 0 C = T
d
C = ½ x b x fm x kd
T = Asx f s = C
kd
T
mx Em = fm
SM about C = T x d-1/3kd
s x Es = fs
C
Let d-1/3kd = d j where j = 1-k/3
The Ms = As x fs x j d
The Mm = ½ fm x b x k d x j x d
As
The Mm = ½ fm x b x k x j x d2
At the limit fs = Fs or fm = Fb
fb = is the fm due to just flexure
ASD - Flexure Only OOP
(single layer Reinf)


reinforced masonry Limits stresses in masonry
And Limits Stresses in the Steel
2M
f b  2  Fb
bd kj
k 2  2 nk  2 n  0
M
fs 
 Fs
As jd
Es
n
2
Em
k  (n )  2n  n
As

bd
ASD - Reinforced Walls
for Flexure

Allowable tensile stresses in reinforcement
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Allowable compressive stresses in masonry
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Grades 40 and 50 : Fs = 20,000 psi
Grade 60 :
Fs = 32,000 psi (was 24ksi)
welded – wire fabric :
Fs = 30,000 psi
Fb = 0.45fm’ (used to be 1/3f’m)
Allowable compressive stresses in Steel - Ignore
unless tied 5.3.1.4 – otherwise same as tension.
ASD DESIGN

Axial Compression in Bars Neglect unless tied
– Not usual except columns - Later
5.3.1.4 Lateral ties Lateral ties — Lateral ties shall conform to the following:

(a) Longitudinal reinforcement shall be enclosed by lateral ties at least 1/4
in. (6.4 mm) in diameter.

(b) Vertical spacing of lateral ties shall not exceed 16 longitudinal bar
diameters, 48 lateral tie bar or wire diameters, or least cross-sectional
dimension of the member.

(c) Lateral ties shall be arranged such that every corner and alternate
longitudinal bar shall have lateral support provided by the corner of a lateral
tie with an included angle of not more than135 degrees. No bar shall be
farther than 6 in. (152 mm) clear on each side along the lateral tie from
such a laterally supported bar. Lateral ties shall be placed in either a mortar
joint or in grout. Where longitudinal bars are located around the perimeter
of a circle, a complete circular lateral tie is permitted. Lap length for circular
ties shall be 48 tie diameters.

d) Lateral ties shall be located vertically not more than one-half lateral tie
spacing above the top of footing or slab in any story, and shall be spaced as
provided herein to not more than one-half a lateral tie spacing below the
lowest horizontal reinforcement in beam, girder, slab, or drop panel above.

(e) Where beams or brackets frame into a column from four directions,
lateral ties may be terminated not more than 3 in. (76.2 mm) below the
lowest reinforcement in the shallowest of such beams or brackets.
Basic Steps of ASD Design for
Flexure (ONLY) OOP – Single layer
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Guess unit size and f’m – (8” cmu , 1500 psi)
Assume that stress in reinforcement
governs . Approximate j = .9 , compute
required area of reinforcement .
Check stress in masonry. If it is close to
the allowable value , compute values of k
and j , re – check stresses in reinforcement
and masonry or get Ms &Mm.
Reinforced Masonry ASD
OOP – Single layer



Limit tension or Compression (only if tied)
stress in steel to allowable value
Limit compression stresses due to bending to 
0.45 f’m
In ASD generally no deflection check unless
supporting unreinforced Masonry on reinforced
masonry beams
Design Masonry Flexure
13.5 ft
W = 33.33 psf
Try a reinforced 8” CMU – on board problem
Try 1 ft design width assume 0.6 D + 0.6 W governs
Get max moment Mmax = (13.5ft)2(33.33psf) (0.6) (1ft)/8=455.6
lb.ft
5.2 –Masonry Beams (lintels)
Some details
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span length equals clear span plus depth , but not
more than distance between support centers
minimum bearing distance = 4 in .
lateral support on compression face required at a
maximum spacing of 32 times the beam thickness (
nominal ) or 120b2/d
must meet deflection limits of Code 5.2.1.4


If supporting unreinforced masonry D+L delf<L/600
Look at code Provisions – Talk about Deep beams
Design Masonry Flexure (ASD)
Given A lintel – Over a door in a 8 CMU
wall – Max moment = 370 kip.in V = 8
kips Assume by tests f’m = 2000 psi
D + L governs


As req = M/Fsjd = 370000/
(32000x0.9x27.8) = 0.462 in2
Try 2 - #5 rebar As =0.62 in2
Guess about 0.9
Design Masonry Flexure (ASD)
Given A lintel – Over a door in a 8 CMU
wall – Max moment = 370 kip.in V = 8
kips Assume by tests f’m = 2000 psi
32 ksi
499.0 kip.in
.45 (2000) = 900 psi
OR
The Mm = ½ Fb x b x k x j x d2
Mm = ½ (0.45)2000 x 7.625 x 0.284 x 0.905 x (27.8)2
Mm = 681.6 kips.in - >>> 499 - Steel stress governs
More on ASD Walls Loaded Out-of-Plane

From Dan Abrams- Masonry Structures Class Notes
More on ASD Walls Loaded Out-of-Plane

From Dan Abrams- Masonry Structures Class Notes
More on ASD Walls Loaded Out-of-Plane

From Dan Abrams- Masonry Structures Class Notes
More on ASD Walls Loaded Out-of-Plane
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From Dan Abrams- Masonry Structures Class Notes – next 10 slides
More on ASD Walls Loaded Out-of-Plane
.45 (f’m)
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More on ASD Walls Loaded Out-of-Plane
More on ASD Walls Loaded Out-of-Plane
Strength Design of Reinforced Masonry
for Flexure Ch 10 MDG – OOP Single layer
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tensile strength of masonry is neglected
continuity between reinforcement and grout ,
equilibrium , plane sections remain plane
elasto – plastic stress – strain curve for
reinforcement
equivalent rectangular compressive stress block
in masonry , with a height of 0.80 fm’ and a
depth of 0.80 c
mu = 0.0035 for clay masonry , 0.0025 for
concrete masonry
Single Reinforcing –
Strength design
No axial load
M
s  y
c
mu = 0.0035 clay, 0.0025 concrete
Reinforcement
fy
0.80 f’m
1 = 0.80 where a =1 c
a

C
Ti
Locate neutral axis based on extreme fiber strains

Calculate compressive force , C

0 = C - T = .8 f’m a b - As fy
a = As fy/ (.8 f’m b)
d

M = As fy (d-a/2)
Basic Steps of Strength Design
for Flexure only
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Try 8” cmu or size section for no shear
reinforcing (if possible)
Assume that stress in reinforcement
governs . Approximate internal lever arm
as 0.9 d , compute required area of
reinforcement .
Re – check with exact location of neutral
axis once As chosen
Check Maximum Steel Area
ASCE 7-10 ASD Load
Combinations Note
code references ASCE
7-10 or Bld. Code.
IBC 2012
30
Design Masonry Flexure
13.5 ft
W = 33.3 psf
Try a reinforced 8” CMU – on board problem
Try 1 ft design width - Assume it is all wind and .9D + W governs
Get max moment Mmax = (13.5ft)2(33.3psf x)(1ft)/8=759. lb.ft