Lesson8

Masonry Structures, lesson 8 slide 1

Seismic design and assessment ofMasonry Structures

Seismic design and assessment ofMasonry Structures

Lesson 8October 2004


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

infinfsupsupinfsup eNeNMMhV +=+=⋅


Flexural (tensile) cracking

• Section of a wall with thickness t , length lsubjected to axial force N and moment M

• Tensile strength of bedjoint: fjt

M

N

fjt

σ

N

e

6

2tlfltNeNM jtcrack ⋅⎟

⎠⎞

⎜⎝⎛ +=⋅=

• If fjt = 0 the well known condition on maximum eccentricity to avoid cracking for a no-tension material is found:

6/le =

• 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 lsubjected to axial force N and moment M

• Compressive strength of masonry : fu• Average compression stress on the section: σ0• Assume equivalent rectangular stress block

tfNauκ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=⎟

⎠⎞

⎜⎝⎛ −

=uu

u ftl

ltfNNlalNM

κσσ

κ0

20 12

122

tlN⋅

=0σ

κ (=k3 )= 0.75-1 a / x (= k1)= 0.67-0.85



• 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:

NV

l

h

O

P

221

22 inf

200

20 lNtl

ftlalNM

uu =≅⎟⎟

⎠

⎞⎜⎜⎝

⎛−⋅=⎟

⎠⎞

⎜⎝⎛ −

=σ

κσσ

hlPN

hlNVhVM uuu 2

)(2

inf +==⇒⋅=

Equilibrium of rigid block around O:

hlPNVlPNhV uu 2

)( 2

)( +=⇒⋅+=⋅

Using flexural strength equation :



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

-80

-60

-40

-20

0

20

40

60

80

-15 -10 -5 0 5 10 15

displacement (mm)

forc

e (k

N)

(Anthoine, Magenes and Magonette, 1994)

Note: for low values of

mean compression,

flexural failure mode is

somewhat “ductile”


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. joint failure

w. shear/tensile cracking of units


Shear strength (ultimate)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


Shear strength (ultimate)Maximum tensile stress criterionFrom 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.

1 0

tu

tuu fb

ltfV σ+=

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

tuz

zxz

t f=−+⎟⎠⎞

⎜⎝⎛=

222

2 στσσ

tut fb =−+⎟⎠⎞

⎜⎝⎛=

2)(

202

20 στσσ

ltVbb

ltN

==== 00 ; ττσσ

principal tensile stress:

at the centre of the panel :


Shear strength (ultimate)

-100

-80

-60

-40

-20

0

20

40

60

80

100

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

displacement (mm)

forc

e (k

N)

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

(Anthoine, Magenes and Magonette, 1994)


Shear strength (ultimate)Coulomb-like criterion:

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.

µστ += c

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:

cvkRk ltfV ⋅⋅=

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,limσd : average normal stress on the compressed areafvk0 : characteristic shear strength under zero compressive stressfvk,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).


Shear strength (ultimate)According to Eurocode 6, the length of the compression zone and the corresponding average stresses can be evaluated as follows:

e

N

l/2l/2

lc= x

lc /3M= N . e

V

ellc −=23

if 6le >

lNVl

NlM

llell

V

c

⋅⎟⎠⎞

⎜⎝⎛ −⋅=⋅⎟

⎠⎞

⎜⎝⎛ −⋅=

=⋅⎟⎠⎞

⎜⎝⎛ −⋅=⋅=

α

β

213

213

213

lH

VlM

V0==α

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

where:

therefore:


Shear strength (ultimate)In this latter case, a closed form expression can be written for the shear strength based on a Coulomb-like criterion as:

uV

u ltcclt

ltNcltV τ

σαµσ

βµβ ⋅=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+

+⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛+⋅=

0

0

31

5.1

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.


Shear strength (ultimate)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:

Nlt

NltV slslu ⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅= µ

βµβ

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 systemsResponse of an elementary cell to horizontal loading: role of diaphragms and ring beams

Out-of-plane collapse mechanism

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

diaphragm

Out-of-plane collapse prevented, rigid

diaphragm


Response of building systems

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 beeffectively 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 of building systemsResponse to horizontal loading: role of tie-rods to prevent out-of-plane collapse


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

urm without tie-rods

urm with tie-rods

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

(Giuffré, 1993)


Response of building systemsExample of reinforced concrete ring beam:

PLAN VIEW at corners

⎪⎩

⎪⎨⎧

≥cm 12

t32

b0

⎩⎨⎧

≥t/2h

h0

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

Prescriptions suitable for general (non-seismic) design, low-rise buildings. More stringent requirements are suggested for seismic design.

longitudinal reinforcement, min. 6-8 cm2


Flexible Diaphragms

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

Distribution of Lateral Force to Shear WallsL1 L2

wal

l “1”

wal

l “2”

wal

l “3”

wind or earthquake forces = w

flexiblediaphragm

F1 F2

F3

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

w

L1d

L2

IEwL

3845

m

42=∆

∆

A

A

floor or roof

Ei

d brick or blockwall

6t

Em

A

t

Section A-A2Ld

EEnAd

12bLn I 1

m

i231 ==+= ∑


Flexible DiaphragmsImposed Deflections on Out-of-Plane Walls

h

∆ = imposed from flexible diaphragmH

t

EI3Hh3

=∆

M

2m

hIE3HhM ∆

==

at2

m2m

b fFh

2tE3

Sh

IE3

SMf +====

∆∆

m

2

at

E5.1th)fF(

t

⎟⎠⎞

⎜⎝⎛+

=∆

AP

allowable tensile stress


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

Ft + fa(psi)

h/t

10 15 20 25 30

20 0.00133 0.00300 0.012000.008330.00533

*based on Em = 1,000 ksi

Flexible DiaphragmsImposed Deflections on Out-of-Plane Walls

40 0.00267 0.00600 0.01067 0.01666 0.0240060 0.00400 0.00900 0.0160 0.02500 0.03600

80 0.00533 0.01200 0.02133 0.03333 0.04800100 0.00667 0.01500 0.02667 0.04166 0.06000120 0.00800 0.01800 0.03200 0.05000 0.07200

140 0.00933 0.02100 0.03730 0.05833 0.08400


Example: Flexible DiaphragmsDetermine deflections of the diaphragm in the north-south direction

two-wythe URM brick walls10’-0” story heightf’m = 2000 psiType N mortar with masonry cementfa dead = 47 psi

7.63” 30’-

0”

80’-0”

w = 400 lb/ft

timber floor

north

431 ft1094I simplicityfor

12blneglecting =

422mm

22

2

ft1094)15x43.2(2Ad'152

'30d

15000.2750'f750E

ft43.2144

)"63.7(6t6A

==∑==

=×==

===

;

ksi ksi

231 Ad

12nbl

I ∑+=EIwl

3845 4

2=∆

"0108.0)ft1094(144xksi1500

12x)'80(ft/kip400.0384

54

4

==∆


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

15.7 7.63"120"

th forTable per deflection allowable ==

psi 62Ff psi,15 F tat =+=

ksi1000E for0.00900 t m ==∆

ksi 1500E for m =

ok"0108.0"046.063.7x006.0 ;0060.0ksi1500ksi 100000900.0

t>==== ∆∆


Flexible DiaphragmsReducing Flexibility with Drag Struts

without strut

A

B

C

w

with strut

AB

C

w

drag strut


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

30’-0” 32’-0”

A

B

drag strut

C

42’-

0”24

’-0”

w = 1.0 kips/ft

lateral force, kips

wall w/o strut w/strut

A 33.0 12.0B 0.0 33.0C 33.0 21.0

total 66.0 66.0


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

30’-0” 32’-0”

A

B

drag strut

C

42’-

0”24

’-0”

w = 1.0 kips/ft

lateral force, kips

wall w/o strut w/strut

A 25.4 12.0B 12.0 33.0C 28.6 21.0

total 66.0 66.0


Rigid DiaphragmsTranslation Without Rotation

wallshear perforatedfor sstiffnessepier of summation or,

walledcantileversolid,for

wall of stiffness lateral ki

3Lh4

Lh

bE k2

mi

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

=

symmetricalabout centerline

wind or earthquake forces

wal

l “l”

wal

l “2”

wal

l “l”

∆F1 F2 F1

k1 k1k2

∆

H

analogy: springs in series

equilibrium

Hk

kkF k

kkkk FFFFH

i

iii

i

i121

i121

∑==

∑=

∑=++=∑=++=

∆Η∆

∆∆∆∆


Rigid DiaphragmsTranslation with Rotation

center of stiffnesswall “1”

y

ye

xP

yP xe x

wall “2”

yi

iyir

xiixi

r kxk

xk

yky ∑

∑=

∑∑

=

:stiffnessofcenter find

Plan View of Roof or Floor System

xr

xr1

xr2

yr

yr1

yr2

datumsometo"" wall of centroid from distance

direction y to parallel

"" wall of stiffness lateraldirectionx to parallel

"" wall of stiffness lateral:where

i y,x

ik

ik

ii

yi

xi

=

=

=



Θo

intyxxyext

oMePePM

0M=+=

=∑

ye

xP

yP

xeext

r

rixixi M

JykF = ext

r

riyiyi M

Jxk

F =

Θ2rx

Θ1ry

]xkyk[J where

JM

]x)xk(y)yk[(M

]xFyF[M

22r2y

2rl1xr

rint

2r2r2yrlrl1xint

2r2yrl1xint

+=

=

+=

+=

Θ

ΘΘ

:only stiffnessplane-ingconsiderin

rJM

=Θ

Θ= 2r2y2y xkF

Θ1r1x1x ykF =

)JM(xkxkF

)JM(ykykF

r2r2y2r2y2y

rrl1xrl1x1x

==

==

Θ

Θ



2riyi

2

rixir

yxxyr

riyiy

yi

yiyi

yxxyr

rixix

xi

xixi

xkykJ

)ePeP()J

xk(P)

kk

(F

)ePeP()J

yk(P)k

k(F

rotation ntranslatio

∑+∑=

++∑

=

++∑

=

Forces attracted to shear walls:


Example: Rigid DiaphragmsDetermine which wall is the most vulnerable to a NS wind load of 20 psf

A B

30’-0”

70’-0”

double wythebrick wall 8”

CMU

30’-

0”

10’

50’-

0”

floor plan

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.

10’

north

1

2

a

a Section a-a

4” brickcollar joint filledwith mortar

9.25”

b b

Section b-b

8” concrete blockfully groutedtype N mortar

7.63”


Determine center of stiffness.Example: Rigid Diaphragms

y

xry

rx

70’

50’

wall Em b h/L kxi kyi yi xi kxyi kyxiksi inches kip/in. kip/in. feet feet kip-feet/in. kip-feet/in.

917214,723 =∑=∑ yixi kk 534,100693,750 =∑=∑ iyiixi xkykB

B 2000 7.63 0.5 - 7630 - 70 - 534,100

1

1 3000 9.25 0.5 13,875 - 50 - 693,750 -

2

2 2000 7.63 1.5 848 - 0 - 0 -

58.2'9172

534,100==

∑∑

=yi

iyir

kxk

x

47.1'14,723

693,750==

∑∑

=xi

ixir k

ykyksi3000'f750E

3Lh4

Lh

bEk

mm

2m

i

<=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

A

A 3000 9.25 1.5 - 1542 - 0 - 0


2riyi

2rixir xkyk J stiffness torsional ∑+∑==

Example: Rigid DiaphragmsDetermine torsional constant.

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

Jr = 8,283,426

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


)ePeP)(J

xk(P)

kk(F

)ePeP)(J

yk(P)k

k(F

yxxyr

riyiy

i

iyi

yxxyr

rixix

i

ixi

++∑

=

++∑

=

Example: Rigid DiaphragmsDetermine wall shear forces.

'2.23'0.35'2.58ekips5.10P

psf20psfinpressurewindwwherew525w)'70)(2

'15(P

0P

xy

y

x

=−==

====

=

Wind in north-south direction will result in larger wall shears than for wind in east-west direction because of larger wind surface area.

in/ft 004,341

.)in/ftkip426,283,8()'2.23)(kips5.10(

JeP

2r

xy =−

=


Example: Rigid DiaphragmsDetermine wall shear forces.

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

kips kips kips inches psi psi

A 4.41 1152 3.8 29.1 7.7governs

76.1)5.10(91721542

= 64.2004,34

)2.58(1542=

73.8)5.10(91727630

= 64.2004,34

)9.11(7630−=

−B 6.09 2776 2.2 30.6 13.6

18.1004,34

)9.2(875,13=1 0 1.18 3372 0.35 29.1 83.1

18.1004,34

)1.47(848−=

−2 0 1.18 945 1.25 30.6 24.5

(1)Fv = 1.33 x 0.3 f’m0.5 = 29.1 psi < 80 psi for brick wall

Fv = 1.33 x 23 psi = 30.6 psi for concrete block with Type N mortar


Example: Rigid DiaphragmsDetermine wall flexural tensile stresses.

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

wall Sg shear fbi = Hih/Sg fa dead -fa + fbinches3 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


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

tottotp

ii

tottotp

Riyiytot

toty

yiiy

tottotp

Rixixtot

totx

xiix

MJK

T

MJ

xxKV

KK

V

MJ

yyKV

KK

V

⋅=

⋅−⋅

+⋅=

⋅−⋅

−⋅=

,

,,

,

,,

,

;)(

;)(

θ

Shear force in wall i :

∑∑∑

∑∑

+−⋅+−⋅=

==

ii

iRiyi

iRixitotp

iyitoty

ixitotx

KxxKyyKJ

KKKK

θ22

,

,,

)()(

; ;

where:are respectively:

total translational stiffnesses

total torsional stiffness

yVxtotxVytotRCxtotRCytottot eVeVyyVxxVM ,,,,,, )()( ⋅−⋅=−⋅−−⋅=


Vertical structures: degree of coupling

(a) (b)

(c)

Role of coupling provided by floors and/or spandrel beams

deflected shape and crack pattern

shears and momentsdeflected shape and

crack patternshears and moments

deflected shape and crack pattern

shears and moments


Vertical structures: modelingSome possible modelling approaches for multistorey masonry walls

a) cantilever model b) equivalent frame c) equivalent frame with rigid offsets

d) 2-D or 3-D finite element modelling

Lesson8

Economy & Finance

Transcript of Lesson8