Lesson8

Seismic design and assessment of
Seismic design and assessment of
Masonry Structures
Masonry Structures

Lesson 8
October 2004

Masonry Structures, lesson 8 slide 1

Strength design/assessment of urm walls

Recent masonry codes are increasingly adopting strength
design/assessment as a fundamental element within a
Limit State or a Performance Based approach, for both
unreinforced and reinforced masonry.
Eurocodes, as an example, are entirely based on a Limit
State approach, both for unreinforced and reinforced
masonry and for non-seismic and seismic
design/assessment.
In the following, attention is focused on strength
evaluation of urm walls subjected to in-plane forces (e.g.
shear walls).


Definitions

• Consider a wall with free lateral edges,
subjected to self weight and to forces
applied at top and bottom sections.
• For each section, by integrating the stresses
it is possible to define a resultant normal
(axial) force N, a resultant shear force V, a
moment M (which can be defined as a
product of N and its eccentricity e w.
respect to the vertical axis of the wall). The
resultant forces are contained within the
middle plane of the wall.
• The following equilibrium equations hold:

Ninf = Nsup + P

V ⋅ h = M sup + M inf = N sup esup + N inf einf


Flexural (tensile) cracking

• Section of a wall with thickness t , length l
subjected to axial force N and moment M M
• Tensile strength of bedjoint: fjt

N
2
⎛N ⎞ l t
M crack = N ⋅ e = ⎜ + f jt ⎟ ⋅
⎝ lt ⎠ 6
N
• If fjt = 0 the well known condition on fjt
maximum eccentricity to avoid cracking for σ
a no-tension material is found:
e
e =l/6

• Flexural cracking can be considered a damage or
serviceability limit state, not an ultimate limit state.


Flexural strength (ultimate)

• Section of a wall with thickness t , length l
subjected to axial force N and moment M
• Compressive strength of masonry : fu
• Average compression stress on the section: σ0
• Assume equivalent rectangular stress block

N
σ0 =
l ⋅t
κ (=k3 )= 0.75-1 a / x (= k1)= 0.67-0.85

N
a=
κf u t

⎛ l − a ⎞ Nl ⎛ N ⎞ σ 0 l 2t ⎛ σ 0 ⎞
Mu = N⎜ ⎟= ⎜1 − ⎟= ⋅ ⎜1 − ⎟
⎝ 2 ⎠ 2 ⎜ κf u lt ⎟
⎝ ⎠ 2 ⎜ κf u⎝
⎟
⎠



• Note: it is easy to verify that if σ0 is very low
compared to fu the equation reduces to the
overturning resistance of a rigid block.
• Example:
Equilibrium of rigid block around O:
N
l l
V Vu ⋅ h = ( N + P) ⋅ ⇒ Vu = ( N + P )
2 2h

P h
Using flexural strength equation :

⎛ l − a ⎞ σ 0l t ⎛ σ 0 ⎞ σ 0l 2t
2
l
Mu = N⎜ ⎟= ⋅ ⎜1 − ⎟≅ = N inf
⎝ 2 ⎠ 2 ⎜ κf u
⎝
⎟
⎠ 2 2
O
l
l l
M u = Vu ⋅ h ⇒ Vu = N inf = ( N + P)
2h 2h



• In-plane flexural/rocking cyclic response of a brick masonry wall
80

60

Note: for low 40

values of 20
force (kN)

mean 0
compression,
-20
flexural failure
mode is -40

somewhat -60

“ductile” -80
-15 -10 -5 0 5 10 15

displacement (mm)
(Anthoine, Magenes and Magonette, 1994)


Shear strength (ultimate)

• The so-called shear failure can be
associated to different cracking
patterns, but is essentially
generated by the effect of the
combination of shear stresses,
mostly due to horizontal loading,
with the normal stresses.
• Two main shear failure modes can
be defined: diagonal cracking (a)
or shear-sliding (b).
• Mixed modes are also possible

stepped crack w. w. shear/tensile
joint failure cracking of units


Definition of shear strength criteria
Problems:
- experimental data are highly scattered (typical of brittle failure modes)
- in a real wall the distribution of stresses is highly non uniform, and its evaluation
is difficult (squat elements where beam theory is questionable, subjected to tensile
cracking).

In practical design/assessment, simplifications must be introduced, sometimes to
the detriment of accuracy.
Some common simplified criteria used in design/assessment:
- Maximum principal tensile stress criterion
- Coulomb-like criterion


Maximum tensile stress criterion
From experimental results of shear-compression tests on urm panels,
where it was observed that the attainment of shear strength
corresponds to the onset of diagonal cracks at the centre of the panel,
it was postulated that shear failure takes place when the principal
tensile stress attains a limit value ftu , which is assumed as the
referential or conventionall tensile strength of masonry.

2
principal tensile stress: ⎛σz ⎞ σ
σt = ⎜ ⎟ + τ zx − z = f tu
2

⎝ 2 ⎠ 2

at the centre of the panel :
2
N V ⎛σ0 ⎞ σ
σ =σ0 = ; τ = bτ 0 = b σt = ⎜ ⎟ + (bτ ) − 0 = f tu
2

lt lt ⎝ 2 ⎠ 2

f tu lt σ b varies with the aspect ratio h/l of the wall. A simple criterion to
Vu = 1+ 0 evaluate b is (Benedetti & Tomaževič, 1984): b= 1.5 when h/l ≥ 1.5
b f tu (slender walls), b = 1 when h/l≤1.0, and b=h/l per 1 < b < 1.5.



• In-plane cyclic response of a brick masonry wall failing in shear

100

80

60

40
force (kN)

20

0

-20

-40

-60

-80

-100
-8 -6 -4 -2 0 2 4 6 8

displacement (mm)

(Anthoine, Magenes and Magonette, 1994)


Coulomb-like criterion: τ = c + µσ
where the shear stress τ and the normal stress σ can have different meaning, depending on the
criterion. The approach followed by e.g. Eurocode 6 and the Italian code is that both stresses
must be considered average stresses on the compressed part of the wall, ignoring any part that is
in tension.

According to Eurocode 6, the characteristic shear strength of an urm wall is expressed via
the shear strength per unit area fvk times the compressed area of the wall:

V Rk = f vk ⋅ t ⋅ lc
where lc is the length of the compressed zone, t is the thickness of the wall and the
characteristic shear strength fvk is defined as :

fvk = fvk0 + 0.4 σd with fvk ≤ fvk,lim
σd : average normal stress on the compressed area
fvk0 : characteristic shear strength under zero compressive stress
fvk,lim : limit value of fvk depending on the type of units (e.g. 0.065fb where fb is the normalized
compressive strength of the units).


According to Eurocode 6, the length of the compression zone and the corresponding average
stresses can be evaluated as follows:
M= N . e lc /3 l lc l
e if e> = −e
6 3 2
N
therefore:

⎛1 e⎞
lc = β ⋅ l = 3 ⋅ ⎜ − ⎟ ⋅ l =
⎝2 l ⎠
V ⎛1 M ⎞ ⎛1 V ⎞
= 3 ⋅ ⎜ − ⎟ ⋅ l = 3 ⋅ ⎜ − αV ⎟ ⋅ l
lc= x ⎝ 2 Nl ⎠ ⎝2 N ⎠

l/2 l/2 M H0
where: αV = =
Vl l

is the shear ratio of the section and H0 is the shear span of the section


In this latter case, a closed form expression can be written for the shear strength
based on a Coulomb-like criterion as:
⎛ ⎞
⎜ ⎟
⎛ N ⎞ 1.5c + µσ 0 ⎟
Vu = βlt ⋅ ⎜ c + µ
⎜ ⎟ = lt ⋅ ⎜ = lt ⋅τ u
⎝ βlt ⎟
⎠ ⎜ cαV ⎟
⎜ 1+ 3 σ ⎟
⎝ 0 ⎠

Note: when this sort of criterion is applied to a wall, failure is always predicted at the
section with the highest moment, i.e. the section with the smallest compressed area
(scketch on blackboard).
This criterion seems more suitable to describe a sliding shear failure, rather than a
diagonal cracking failure.


When walls are subjected to low vertical forces and high horizontal cyclic forces,
horizontal cracks may develop all along the whole section of the wall, typically in
bedjoints, and the resistance to sliding would then be expressed as:

⎛ N ⎞
Vu = β lt ⋅ ⎜ µ sl
⎜ ⎟ = µ sl ⋅ N
⎝ βlt ⎟
⎠

Note: the application of such strength formula has proven to give too conservative
and unrealistically low strength predictions. This is especially the case for the upper
stories of a building. In reality:
- the low tensile strength of bedjoints, usually neglected in computations, can
prevent the formation of fully cracked sections at upper stories;
- when a flexural cracks closes in compression it is arguable to assume that the
residual strength is based on friction only; a more realistic approach should assume a
residual cohesion.


Flanged sections (urm walls)
See Tomaževič, pages 151-154.


Response of Building Systems


Response of building systems
Response of an elementary cell to horizontal loading:
role of diaphragms and ring beams

Out-of-plane collapse Out-of-plane collapse Out-of-plane collapse
mechanism prevented, presence of ring prevented, rigid
beam (tie-beam), flexible diaphragm
diaphragm



Response to horizontal loading : role of diaphragms and ring beams
• Ring beams and diaphragms contribute to restrain the out-of-
plane deflections of walls and to avoid out-of-plane collapse (i.e
they “hold the box together”).
• To exert an effective restraint ring beams and diaphragm must be
able to transmit tensile forces and tensile stresses and must be
effectively connected to the walls.
• A ring beam connecting walls in the same plane confers
robustness and redundancy to the system, allowing force
redistribution among walls. (Note: a similar role, in part, is played by
tie-rods, used also in ancient buildings)
• Rigidity of diaphragms affects horizontal load distribution among
shear walls


Response to horizontal loading: role of tie-rods to prevent out-of-plane
collapse


Response to horizontal loading: role of tie-rods to improve in-plane response

urm with tie-rods

tie-rods to
restrain
overturning and
mobilitate in-
plane resistance
of walls.

urm without tie-rods (Giuffré, 1993)


Example of reinforced concrete ring beam: PLAN VIEW
at corners

longitudinal
reinforcement,
min. 6-8 cm2

stirrups 6-8 mm dia. min., max.
spacing 25-30 cm

⎧ 2 Prescriptions suitable for general (non-
⎪ t ⎧h
b0 ≥ ⎨ 3
seismic) design, low-rise buildings.
h0 ≥ ⎨ More stringent requirements are
⎪12 cm
⎩ ⎩t/2 suggested for seismic design.


Flexible Diaphragms
Distribution of Lateral Force to Shear Walls
L1 L2
F3

wall “1”
F1

wall “3”
F2 flexible

wall “2”
diaphragm

wind or earthquake forces = w

For fully flexible diaphragms, walls attract lateral forces based on tributary areas

F1 = wL1/2
F2 = wL1/2 + wL2/2
F3 = wL2/2

I.e. horizontal forces are resisted by the shear walls are taken from the floors to
which they are directly connected, unless a “semi-rigid” analysis is carried out.


Flexible Diaphragms
Diaphragm Deflections
A
A
w
Em
∆
L1 floor or roof
d 6t
Ei

L2
d brick or
A block
5 wL4 wall
∆= 2
t
384 E m I

bL3 Ei L1 Section A-A
I =n 1
+ ∑ Ad 2 n= d=
12 Em 2


Flexible Diaphragms
Imposed Deflections on Out-of-Plane Walls

∆ = imposed from flexible diaphragm
H
Hh 3
∆=
3 EI
h 3 E m I∆
M = Hh =
h2

3 E m I∆ t
3 Em ∆
fb =
M
= h2 = 2 =F + f
2 t a
S S h
t P
M ⎛ h⎞
2
A
( Ft + f a )⎜ ⎟
∆ ⎝t⎠ allowable
=
t 1.5 E m tensile stress


Flexible Diaphragms
Imposed Deflections on Out-of-Plane Walls

Values of allowable diaphragm deflection, ∆, divided by wall thickness, t*

h/t
Ft + fa
(psi) 10 15 20 25 30
20 0.00133 0.00300 0.00533 0.00833 0.01200
40 0.00267 0.00600 0.01067 0.01666 0.02400
60 0.00400 0.00900 0.0160 0.02500 0.03600
80 0.00533 0.01200 0.02133 0.03333 0.04800
100 0.00667 0.01500 0.02667 0.04166 0.06000
120 0.00800 0.01800 0.03200 0.05000 0.07200
140 0.00933 0.02100 0.03730 0.05833 0.08400

*based on Em = 1,000 ksi


Example: Flexible Diaphragms
Determine deflections of the diaphragm in the north-south direction
north 80’-0”

two-wythe URM brick walls
10’-0” story height

30’-0”
timber floor f’m = 2000 psi
7.63” Type N mortar with masonry cement
fa dead = 47 psi

3 w = 400 lb/ft
nbl 1 5 wl24
I= + ∑ Ad 2 ∆=
12 384 EI
6 ( 7.63quot; )2
A = 6t 2 = = 2.43 ft 2 5 0.400 kip / ft ( 80' )4 x 12
144 ∆= = 0.0108quot;
384 1500 ksi x 144 ( 1094 ft 4 )
E m = 750 f 'm = 750 × 2.0 ksi = 1500 ksi
30'
d= = 15' ; ∑ Ad 2 = 2( 2.43 x 15 2 ) = 1094 ft 4
2
bl13
neglecting for simplicity I = 1094 ft 4
12


Example: Flexible Diaphragms
Check Cracking of the Out-of-Plane Walls as per UBC

h 120quot;
allowable deflection per Table for = = 15.7
t 7.63quot;
Ft = 15 psi, f a + Ft = 62 psi

∆
= 0.00900 for E m = 1000 ksi
t

for E m = 1500 ksi
∆ 1000 ksi
= 0.00900 = 0.0060 ; ∆ = 0.006 x 7.63 = 0.046quot; > 0.0108quot; ok
t 1500 ksi


Flexible Diaphragms
Reducing Flexibility with Drag Struts

without strut with strut

B
B
A A drag strut

C C

w w


Flexible Diaphragms
Determine the lateral force attracted to each wall (case 1).

A w = 1.0 kips/ft

lateral force, kips
24’-0”

wall w/o strut w/strut
drag strut A 33.0 12.0
B B 0.0 33.0
C 33.0 21.0
42’-0”

total 66.0 66.0
C

30’-0” 32’-0”


Flexible Diaphragms
Determine the lateral force attracted to each wall (case 2).

w = 1.0 kips/ft
A

lateral force, kips
24’-0”

wall w/o strut w/strut
drag strut A 25.4 12.0
B 12.0 33.0
B
C 28.6 21.0
42’-0”

total 66.0 66.0
C

30’-0” 32’-0”


Rigid Diaphragms
Translation Without Rotation
symmetrical
about centerline analogy: springs in series
F1 F2 F1
∆
wall “2”
wall “l”

wall “l”

k1 k2 k1
∆

H
wind or earthquake forces
equilibrium
k i = lateral stiffness of wall
H = F1 + F2 + F1 = ∑ Fi
bE m
ki = for solid, cantilever ed wall = k1∆ + k2 ∆ + k1∆ = ∑ ki ∆
⎛ h ⎞⎡ ⎛ h ⎞ ⎤
2
⎜ ⎟ ⎢ 4⎜ ⎟ + 3⎥ Η ki
⎝ L ⎠⎢ ⎝ L ⎠ ∆= Fi = ki ∆ = H
⎣ ⎥
⎦ ∑ ki ∑ ki
or, summation of pier stiffnesses for perforated shear wall


Rigid Diaphragms
Translation with Rotation
y xr
xr1 center of stiffness
wall “1”
find center of stiffness :
yr1 ∑ k xi yi ∑ k yi xi
yr = xr =
∑ k xi ∑ k yi
ey
yr2 where :
yr Px k xi = lateral stiffness of wall quot; iquot;
parallel to x direction
wall “2”
k yi = lateral stiffness of wall quot; iquot;
xr2 x parallel to y direction
Py ex
xi , yi = distance from centroid of wall
quot; iquot; to some datum
Plan View of Roof or Floor System


Rigid Diaphragms
Translation with Rotation ∑ Mo = 0
M ext = Py e x + Px e y = M int
y r 1Θ
considering in - plane stiffness only :
M int = [ Fx 1 y rl + F y 2 x r 2 ]
Fx 1 = k x 1 y r 1Θ
M int = [( k x 1 y rl Θ ) y rl + ( k y 2 x r 2Θ ) x r 2 ]
M int = J rΘ
Θ where J r = [ k x 1 y rl 2 + k y 2 x r 2 2 ]
ey o M
Θ=
Jr
Px
M
Fx 1 = k x 1 y rlΘ = k x 1 y rl ( )
Jr
x r 2Θ
M
F y 2 = k y 2 x r 2Θ = k y 2 x r 2 ( )
Py Jr
ex k xi yri k yi xri
Fy 2 = k y 2 x r 2 Θ Fxi = M ext Fyi =
Jr
M ext
Jr


Rigid Diaphragms
Translation with Rotation

Forces attracted to shear walls:
translatio n rotation
k xi k y
Fxi = ( ) Px + ( xi ri ) ( Py e x + Px e y )
∑ k xi Jr
k yi k yi x ri
Fyi = ( ) Py + ( ) ( Py e x + Px e y )
∑ k yi Jr
2 2
J r = ∑ k xi y ri + ∑ k yi x ri


Example: Rigid Diaphragms
Determine which wall is the most vulnerable to a NS wind load of 20 psf
A B
north 4” brick
70’-0”
collar joint filled
a with mortar
30’-0”
1
9.25”
10’

a Section a-a
double wythe
brick wall 8”
50’-0”

CMU 8” concrete block
30’-0”

fully grouted
type N mortar

b b
7.63”

2 Section b-b

floor plan 10’

Walls of the one-story building are 15’ tall and the roof system has been strengthened to be rigid.
Assume f’m equal to 5333 psi for the brick, and 2667 psi for the block.


Determine center of stiffness.
wall Em b h/L kxi kyi yi xi kxyi kyxi
ksi inches kip/in. kip/in. feet feet kip-feet/in. kip-feet/in.

A 3000 9.25 1.5 - 1542 - 0 - 0

B 2000 7.63 0.5 - 7630 - 70 - 534,100
1 3000 9.25 0.5 13,875 - 50 - 693,750 -

2 2000 7.63 1.5 848 - 0 - 0 -

y ∑ kxi = 14,723 ∑ kyi = 9172 ∑ k xi yi = 693,750 ∑ k yi xi = 534,100
A B
70’
1 ∑ k yi xi 534,100
ki =
bEm xr = = = 58.2'
∑ k yi 9172
⎛ h ⎞⎡ ⎛ h ⎞ ⎤
2

⎜ ⎟ ⎢ 4 ⎜ ⎟ + 3⎥
50’

⎝ L ⎠⎢ ⎝ L ⎠
⎣ ⎥
⎦ ∑ k xi yi 693,750
yr yr = = = 47.1'
xr x Em = 750 f 'm < 3000 ksi ∑ k xi 14,723
2


Determine torsional constant.

2 2
torsional stiffness = J r = ∑ k xi y ri + ∑ k yi x ri

wall kxi yri kyi xri Jr
kip/in feet kip/in feet kip.ft2/in

A - - 1542 58.2 5,223,124

B - - 7630 11.8 1,062,401

1 13,875 2.9 - - 116,689

2 848 47.1 - - 1,881,212

Jr = 8,283,426


Determine wall shear forces.
ki k y
Fxi = ( )Px + ( xi ri )( Py e x + Px e y )
∑ ki Jr
ki k x ri
Fyi = ( ) Py + ( yi )( Py e x + Px e y )
∑ ki Jr

Wind in north-south direction will result in larger wall shears than for
wind in east-west direction because of larger wind surface area.
Px = 0
15'
Py = ( )( 70' )w = 525 w where w = wind pressure in psf = 20 psf
2
Py = 10.5 kips e x = 58.2' − 35.0' = 23.2'

Py e x ( 10.5 kips )( 23.2' ) 1
= = in/ft
Jr ( 8 ,283 ,426 kip − ft 2 / in .) 34 ,004


Determine wall shear forces.

wall direct shear torsional shear total shear Ae fv=V/Ae Fv(1) Fv/fv
kips kips kips inches psi psi

1542 1542( 58.2)
A (10.5) = 1.76 = 2.64 4.41 1152 3.8 29.1 7.7
9172 34,004 governs

7630 − 7630(11.9)
B (10.5) = 8.73 = − 2.64 6.09 2776 2.2 30.6 13.6
9172 34,004

13,875( 2.9)
1 0 = 1.18 1.18 3372 0.35 29.1 83.1
34,004

− 848( 47.1)
2 0 = − 1.18 1.18 945 1.25 30.6 24.5
34,004
(1)F = 1.33 x 0.3 f’m0.5 = 29.1 psi < 80 psi for brick wall
v
Fv = 1.33 x 23 psi = 30.6 psi for concrete block with Type N mortar


Determine wall flexural tensile stresses.

wall Sg shear fbi = Hih/Sg fa dead -fa + fb
inches3 kips psi psi psi

A 23,944 4.41 33.2 30.0 3.2

B 168,320 6.09 6.5 30.0 -23.5

1 204,967 1.18 1.0 30.0 -29.0

2 19,495 1.18 10.9 30.0 -19.1

worst case is flexural tension for wall A:
-30.0 + (w/20)(33.2) = Ft = 1.33 x 15.0 psi
maximum wind load, w= 30 psf


Rigid diaphragms: summary (elastic analysis)
External torque due to eccentricity of shear force Vtot w. respect to center of stiffness:

M tot = Vtot , y ⋅ ( xC − xR ) − Vtot , x ⋅ ( yC − y R ) = Vtot , y ⋅ eV , x − Vtot , x ⋅ eV , y

Shear force in wall i :

K xi K ⋅ ( yi − y R )
Vix = ⋅ Vtot , x − xi ⋅ M tot ;
K x,tot J p,tot
K yi K yi ⋅ ( xi − x R )
Viy = ⋅ Vtot , y + ⋅ M tot ;
K y ,tot J p ,tot
Kθi
Ti = ⋅ M tot
J p ,tot
where:
K x,tot = ∑ K xi ; K y ,tot = ∑ K yi ; are respectively:
total translational stiffnesses
i i
J p,tot = ∑ K xi ⋅ ( yi − y R ) 2 + ∑ K yi ⋅ ( xi − x R ) 2 + ∑ Kθi total torsional stiffness
i i i


Vertical structures: degree of coupling
Role of coupling provided by floors and/or spandrel beams

deflected shape and shears and moments
deflected shape shears and moments crack pattern
and crack pattern
(b)
(a)

deflected shape and shears and moments
crack pattern

(c)


Vertical structures: modeling
Some possible modelling approaches for multistorey masonry walls

a) cantilever model b) equivalent frame c) equivalent frame d) 2-D or 3-D finite
with rigid offsets element modelling


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