Lesson10

Masonry Structures, lesson 10 slide 1

Seismic design and assessment ofMasonry Structures

Seismic design and assessment ofMasonry Structures

Lesson 10October 2004


Out-of-plane seismic response of urm walls

Parapet failure

(Newcastle Earthquake Study, The Institution of

Engineers, Australia, 1990) Out-of-plane damage(from the 1997 Umbria-Marche

earthquake, Italy, Blasi et al., 1999).



Global and partial overturning mechanisms of

façade walls


Tests performed at Ismes, Bergamo, (Benedetti et al. 1996)

Out-of-plane collapse followed by global collapse



Earthquake Excitation atfootings

In-plane shear walls response filters theground motion andtransmits to floor diaphragms

Floor diaphragm response amplifies accelerations andtransmits load to out-of-plane walls

Out-of-plane wall

Shakingmotion

Parapet wall

Seismic Load Path for Unreinforced Masonry Buildings

(Doherty 2000,adapted after Priestley)



Important issues:-Strength of wall against out-of-plane forces and relevant mechanisms of resistance

-Out-of-plane displacement capacity of walls

-Evaluation of out-of-plane dynamic response

-Definition of seismic demand on walls considering filtering effect of building and diaphragms


Out-of-plane strength of urm walls

∆c

Self weight above crack = W/2

Base reaction = O+W

(A) (B)OverburdenForce = O

One-way bending, before cracking

One-way bending, after cracking

Wall subjected to horizontal distributed loading (wind or inertia forces)


Out-of-plane strength of urm walls, low vertical stress

Mid-height Displacement (∆)

Semi-rigid Non-linear F-∆Relationship

∆instability

Un-cracked Elastic Capacity

Displacement Control Loading

Force Control/Dynamic Loading (Explosive)

App

lied

late

ralF

orce

, w (k

N/m

)

Un-cracked Linear Elastic Behaviour

Res(1)

Relastic

(Doherty 2000)


Out-of-plane strength of urm walls, high vertical stress

Mid-height Displacement (∆)

Semi-rigid nonlinear F-∆relationship

∆instability

Force/DynamicControl Loading

Un-cracked Linear Elastic Behaviour

Res(1)

Relastic

DisplacementControl Loading

App

lied

late

ralF

orce

, w (k

N/m

)

(Doherty 2000)


Out-of-plane flexural strength of nonloadbering walls

Two fundamental resistance mechanisms:

tensile strength of masonry to resist one-way and two-way

bending arching action


Flexural strength of nonloadbering walls

Vertical one-way flexure: already treatedClearly, when zero or very low vertical compression is present, tensile strength of bedjoints assume an importan role and should not be neglected.

Horizontal flexure:


Horizontal flexure

Empirical relations for brick masonry (single-wythe):

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

jt

njttp f

fCf σ1'

for toothed failure, where fjt = tensile stregth normal to bedjoints, C=2 for stress in MPa;

jttbtp fff 55.045.0' +=

for failure in units, where ftb = tensile strength of unit (can be assumed as 0.1 times the compressive strength)


Two-way flexure (single-wythe)

Most walls are supported on three or four sides. Evaluation is difficult because the walls are statically indeterminate and the material is anisotropic.Wall tend to show fracture lines whose pattern depends also on geometry of wall, degree of rotational restraint, presence of vertical compression, besides properties of materials.

Common design/assessment methods make use of •simplistic conservative approaches such as the crossed strip method or•yield line and fracture line approaches


Arching action


Arching action

tfC c )1( γ−=

)( 0∆−= tCM γResisting moment:

)(802 ∆−= t

hCp γ

According to EC6 and BS: (1-γ)=0.1

From which, neglecting deflection (ok for h/t < 25):2

72.0 ⎟⎠⎞

⎜⎝⎛=htfp c


Arching action with gap

Ltg

2)(4 γ≤

gLtg g

−

∆=

22γ

From geometry of similar triangles:

This can be inserted in the equation of the previous slide:

tgL

tgLg

g γγ 44)(≅

−=∆

Arch action can be activated only if

)(82 gt

hCp ∆−= γ


Influence of movement of supports on arching action

The thrust force C can produce some displacement of the supports.

The displacement can be treated as an equivalent gap. If the movement is a function of the thrust force, a number of trials or iterations may be required to determine the optimum stress level.

Axial shortening due to compression of the wall and corresponding deflection can be treated similarly as an equivalent gap.


Post-cracking behaviour loadbearing walls in single bending

h

h/2

∆

∆/2

P

W/2

W/2

P+W

H=∆W/2h

H=∆W/2h

w h/2

P

H=∆W/2h

W/2

∆o

R

x

w = ma

Priestley, 1985

hWH⋅∆⋅

=2

From moment equilibrium about R: ∆⋅+

⋅=⋅

⋅∆⋅

+∆⋅+∆⋅+

⋅=⋅ Rhwh

hWPWhwxR

822228

22

From moment equilibrium about O:

( )∆−⋅⋅= xRh

w 2

8

Static instability occurs when ∆=x


Post-cracking behaviour in single bending: statics

Calculation of force-deflection curve: first, calculate curvature at midheight section and the associated moment, then evaluate displacement at midheight assuming a given curvature distribution.

R

fcrack

x

t

Mcrack=Rt/6

fcrack=2R/t

φcrack=fcrack/Et

x=t/6

2

2 88 h

MwhwM crackcrack

crackcrack

⋅=⇒

⋅= JE

hwcrackcrack ⋅⋅

⋅⋅=∆

3845 4



After cracking, calculation behaviour is non linear and the following relationship are used for a given value of deformation/stress at the right edge of the section, from which the corresponding value of x is found:

f

xx

t

R

M =Rx

f =2R/(3(t/2-x))

φ =f/(3E(t/2-x))

crackcrack

∆⋅=∆φφ

Conservatively, it can beassumed that displacement si proportional to curvature at midheight :

At ultimate, a rectangular stress block is assumed and the corresponding curvature is evaluated, from which the displacement at midheight.



Having determined the displacement ∆ at midheight corresponding to a givenvalue of x, from the equation:

( )∆−⋅⋅= xRh

w 2

8

the corresponding value of distributed horizontal force is determined, and the nonlinear force-displacement curve is determined point by point.

Note: static instability may occur before the attainment of the ultimate stress. The behaviour is elastic nonlinear, i.e. the wall will load and unload alongthe same curve.


Note: if the two halves of the cracked wall were considered as rigid blocks, the force-displacement curve would be given by the blue line, which can be obtained by simple equilibrium of rigid blocks.:

Post-cracking behaviour in single bending: rigid block behaviour

App

lied

Late

ral F

orce

F0 Bi-linear F- ∆Relationship

Real semi-rigid Non-linear F- ∆Relationship

∆u=∆instability

K0

'Semi-rigid thresholdresistance'


∆

∆

Post-cracking behaviour in single bending: rigid block behaviour


1. LVP(TWD)

2 LVP (MWD)

4. INSTRON

3. LVDT (BWD)

5. ACCEL(TA)

6. ACCEL(TTA)

7. ACCEL(TWA)

8. ACCEL(MWA)

9. ACCEL(BWA)

Stationary reference frame: Rigid connection to strong floor

Wall specimen

Timber wall catch

Tests performed at University of Adelaide, Australia (Doherty et al. 2000)

Out-of-plane dynamic experimental behaviour


Braced steel frame (Representing URM shear wall action)

Cornice support

Table bearing support rails

10mm stiff rubber spacer

Wall Specimen

(after Doherty, 2000)


300% NAHANNI DISPLACEMENTS

-10

-5

0

5

10

15

0.01

0.62

1.23

1.84

2.45

3.06

3.67

4.28

4.89 5.5 6.11

6.72

7.33

7.94

8.55

9.16

9.77

10.4

TIME (secs)

DISP

LACE

MENT

(mm)

300% NAHANNI ACCELERATIONS

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

00.5

61.1

1

1.67

2.22

2.78

3.33

3.89

4.44 5

5.55

6.11

6.66

7.22

7.77

8.33

8.88

9.44

9.99

10.5

TIME (secs)

ACCE

LERA

TION (

g)

300% NAHANNI VELOCITIES

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.01

0.57

1.13

1.69

2.25

2.81

3.37

3.93

4.49

5.05

5.61

6.17

6.73

7.29

7.85

8.41

8.97

9.53

10.1

10.6

TIME (secs)

VELO

CITY (

m/S)

PGD 14mm

PGV 207mm/sec

PGA 0.65g

300%

Nahanni



80% PACOIMA DAM DISPLACEMENTS

-50

-40

-30

-20

-10

0

10

20

30

40

0.02 0.6 1.18

1.76

2.34

2.92 3.5 4.08

4.66

5.24

5.82 6.4 6.98

7.56

8.14

8.72 9.3 9.88

10.5 11 11.6

12.2

12.8

13.4

TIME (secs)

DISPL

ACEM

ENT (

mm)

80% PACOIMA DAM ACCELERATIONS

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0

0.58

1.16

1.74

2.32 2.9 3.48

4.06

4.64

5.22 5.8 6.38

6.96

7.54

8.12 8.7 9.28

9.86

10.4 11 11.6

12.2

12.8

13.3

TIME (secs)

ACCE

LERA

TION (

g)

80% PACOIMA DAM VELOCITIES

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.02

0.62

1.22

1.82

2.42

3.02

3.62

4.22

4.82

5.42

6.02

6.62

7.22

7.82

8.42

9.02

9.62

10.2

10.8

11.4 12 12.6

13.2

TIME (secs)

VELO

CITY (

m/S)

PGD 43.2mm

PGV 245mm/sec

PGA 0.35g

80%

Pacoima Dam



Safety wih respect to collapse: force or displacement?

Clearly, any methodology should take into consideration the nonlinear nature of the response.

Earlier proposal by Priestley (1985): force based assessment based on equal energy equivalence;

Equivalent “resistance” of a linear response


Design accelerations to be compared with

“resistance”


(from Tena-Colunga & Abrams)

A more rigorous evaluation of the flexibility of diaphragms would be made by an adequate dynamic global model.


Force-based assessment with q-factor

Eurocode 8 & Italian code approach

The effect of the seismic action may be evaluated considering a force system proportional to the masses (concentrated or distributed) of the element; the resultant force (Fa) on the element, is computed as:

Fa = Wa Sa / qa

Wa is the weight of the element.qa is the behaviour factor, to be considered equal to 1 for cantilever elements (for example chimneys and parapets fixed only at the base) equal to 2 for non-structural walls, equal to 3 (Italian code) for structural walls which do not exceed specified slenderness limits.

Sa is the seismic coefficient to apply at the structural or non-structural wall, that considers dynamic interaction with the building.


Force-based assessment with q-factor

gSa

0.5)/TT(11

Z/H)3(1gSa

S g2

1a

ga ≥⎥

⎦

⎤⎢⎣

⎡−

−++

=

agS is the design ground accelerationZ is the height from the foundation of the centre of the mass of the elementH is the height of the structureg is the gravity accelerationTa is the first period of vibration of the wall element in the considered direction (out-of-plane), estimated also in an approximate way.T1 is the first period of vibration of the structure in the considered direction.


Recent trends: displacement based assessment

Mid-height Displacement

App

lied

Lat

eral

For

ce

Rigid bodyCalculated curve

Fu

∆u∆2

F0

Griffith et al., 2000-today

F0

∆1

K0

FU

∆2 ∆U

K1

∆MAX

K2 KS

Dynamic response is estimated via an equivalent secant stiffness (K2) and an equivalent s.d.o.f. system with suitable effective damping (5% suitable for one-way vertical bending).


Recent trends: displacement based assessment

Displacement demand must be evaluated using an elastic spectrum which takes into account the filtering effect of the structure, e.g.(Italian code)

( )( )

2s

s 1 d s g 2 2s 1

1 s1 s D d s g 2

1 DD s d s g 2

3 1 Z HTT < 1.5T ∆ (T ) = a S 0 54 1 1 T T

1 5T T Z1.5T T < T ∆ (T ) =a S 1 9 2 4H4

1 5T T ZT T ∆ (T ) =a S 1 9 2 4H4

.

. . .

. . .

⎛ ⎞+⎜ ⎟−⎜ ⎟π + −⎝ ⎠

⎛ ⎞≤ +⎜ ⎟π ⎝ ⎠

⎛ ⎞≤ +⎜ ⎟π ⎝ ⎠

agS is the design ground accelerationZ is the height from the foundation of the centre of the mass of the elementH is the height of the structureg is the gravity accelerationTa is the first period of vibration of the element in the considered direction, estimated also in an approximate way.T1 is the first period of vibration of the structure in the considered direction.

Lesson10

Documents

priestley masonry structures

fc h masonry structures

t g h2 masonry structures

diaphragmsmasonry structures

wmasonry structures

previous slide

plane shear walls response

plane forces