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Unreinforced Masonry Under Lateral Loads Durgesh C. Rai Assistant Professor Department of Civil Engineering Indian Institute of Technology Kanpur Kanpur - 208 016

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Unreinforced Masonry Under Lateral Loads

Durgesh C. RaiAssistant Professor

Department of Civil EngineeringIndian Institute of Technology Kanpur

Kanpur - 208 016

Axial Compression

Unity Formula for Interaction

Out-of-Plane Lateral LoadsInteraction of Axial & Flexural Stresses

In-Plane Lateral LoadsPier Analysis of Shear Walls

DiaphragmFlexible and Rigid

Distribution of Forces

Walls have 3 functions• Resist vertical compression

• Resist bending stresses from

• Eccentric vertical loads

• Transverse loads due to wind, earthquake, or blast loads.

• Referred as out-of-plane loads

• Resist in-plane shear and bendingfrom lateral loads applied to building system in direction parallel to plane of wall.

Unreinforced Bearing & Shear Walls

Floor or Roof loads

In-plane Shear & Moment

L

t

Historically, URM walls have been sized in terms of h/t ratio and is limited to 25.

Limiting Slenderness of URM Walls

3.25

72.035.0

68

68

68

2

2

22

×==

===

th

MPaMPa

wF

th

tFSF

whM

b

bb

1327< 2 Storeys

2027> 2 Storeys

LMCM

Max. slenderness ratio for load bearing wall

h

t

Allowable Compressive Stress

AI

klE

klIE

Pcr mcr

m2

2

2

2

)(,

)(πσπ

==

h’=kl

tP

P

SRth

alsothf

Then

klhandfEIftkl

E

tkl

Et

klE

tt

btbt

AI

r

btAklbt

I

mcr

mm

mmmcr

==

==

=

==∴

====

==

')/'('615

,

''750

82.012

)289.0()(

289.012

1.

12

,)(

2

22

22

2

2

23

2

3

σ

ππσ

For rectangular cross-section

Allowable Compressive Stress …

2

2

3

70'25.0,99

'

140'

1'25.0,99'

'00682.029.0'

]027.017.1['25.0

42'

1'20.0

=>

−=≤

−=

⋅−=

−=

hr

mfFrh

rh

fFrh

ACIth

f

SRfF

IS

th

fF

UBC

a

ma

m

ma

ma

0 25 50 75 100h’/t

σcr

0.2

0.4

0.6

0.8

f’m h’/t = 24.8

2)/'('615

thf

Euler m

ACIUBC

IS:1905

(IS : 1905, Table ; §4.3.1)

Effective Height of Wall

Lateral+Rotation

Lateral+Rotation Full Restrainedk = 0.75

0.75H

H

0.85H

Lateral

Lateral+Rotationk=0.85 (0.70 UBC)

Lateral only

Lateral onlyk=1

Lateral+Rotationk=1.5 (2.0 UBC)

Free

1.5H

Singlewythe thickness = t = specified• Multiwythe.

• Cavity Walls Both wythes loaded

Mortar grout filled

tt=te

Each considered separately

t1 t2

t1 t2

)(32

21

222

1

tt

ttteff

+=

+=

Wythes connected w/wire joint renf

Effective Wall Thickness

Effective Wall Area

Minimum area of mortal bed joints plus any grouted area.

Allowable Compressive Stress …

Example:Determine allowable vertical load capacity of UR cavity wall as per IS: 1905.

Design for Compression Load

mMNmMPaP

MPaMPafkF

Tablek

okth

k

mmt

bricksforMPafblocksforMPaf

a

msa

s

mm

/26.1)0.130.02)(1.2(

1.2)1525.0(56.0)'25.0(

)9(56.0

.2713.223.203

4500',1

3.203)115190(32

.20'15'

2 =××=

=×==≅

<===

=+=

==

Case A: Prisms have been tested

4.5m

ungrouted

30mm

air

190 50 115

mMNP

MPaMPaF

TableMPaF

H

a

a

c

/40.0

67.019.156.0

)8(19.1

)214:2

1:1(2

==×=

=

Case B: Prisms have not been tested

Example contd… Determine allowable vertical load capacity of UR cavity wall as per IS: 1905.

Design for Compression Load

4.5m

ungrouted

30mm

air

190 50 115

Depends on end conditions:• Free end

• Continuity of the wall

• Support of room cross walls/piers/buttresses

• Openings.

Effective Length of Wall

L

x xy y

Ll

Hy

Hx

8.06

,8

=

≥≥

L

x x y

Ll

Hy

Hx

9.06

,8

=

≥≥

xy x

Ll

Hy

Hx

=

≥≥6

,8

Effective Length of Wall…

Ll

Hy

Hx

26

,8

=

≥<

L

x y

L1

L2 xy

25.16

,8

Ll

Hy

Hx

=

≥<

x L

8H

x <

L

x y

Ll

Hy

Hx

3.16

,8

=

≥≥

Slenderness determined by height

Case A Full Restrained at Top & Bottom

Effective Height of Wall with Openings

H1

H

Effective height :hxx=0.75H+0.25H1hyy=H

xxy

y

Column formed by opening.

H1

H

xxy

y

For H1≤0.5Hhxx=Hhyy=2H

For H1>0.5Hhxx=2Hhyy=2H

and

Case B Partial Restrained at Top & Bottom

Stiffening Effect of Cross-Walls

• Effective thickness of walls stiffened by piers/cross walls/buttresses is obtained by multiplying the actual thickness by a stiffening coefficient.

• Cross wall is replaced by an equivalent pier equal to thickness of the cross wall and thickness equal to three times the thickness of the wall.

Cross Walls

Sp

wp

tp

1.01.01.0>20

1.21.11.015

1.41.21.010

1.71.31.08

2.01.41.06

Stiffening coefficient

p

p

wS

1/ =wp tt 2/ =wp tt 3/ ≥wp tt

Table 6 : 1905-1987

tw

URM Bearing Wall Subjected to Eccentric Axial Compression.

Eccentric Axial CompressionM

omen

t dia

gram

(on

tens

ion

side

)

P.ee

h

t

Bending stress

TensionM = P.e

fb = Mc/I = M/S f = P/A (+/-) M/S

f = fa+fb

f = fa - fb

Combined stressAxial stress

fa = P/A

P

Comp.

UBC & ACI Approach :

Eccentric Axial Compression ….

].//&[

,.)2

).' (..lim

)'33.0( ,

1.)1

wallofplanetooristensiontypeMortar

stresstensileallowableFffstresstensileLimiting

sesmallforcontrolsstresscompitingforformulaunity

fstressecompressivbendingallowableF

stressecompressivaxialallowableFwhereFf

Ff

tba

mb

ab

b

a

a

⊥=<+−

==

=<+

Eccentric Axial Compression ….

ignored.istensionunderarea(butstresscomp.basic1.25stressecompressivAllow.

1/6)(etyeccentricilargeFor2.)

FAP

alsoF1.25SM

AP

stress.ecompressivbasic1.25stress ecompressiv Allow.1/6)ebut1/24(etyeccentricilowFor1.)

aa

×=>

<≤+∴

×=<>

IS: 1905 – 1987 Approach

Eccentricity Appendix A, IS : 1905

Eccentricity depends on

• extent of bearing

• magnitude of loads

• relative stiffness of slab or beam & wall

• degree of fixity at the support.

• requires judgment

Code provides guidelines :

Determining Eccentricity

Exterior Wall

Determining Eccentricity …

For flexible diaphragms when span is less than 30 times thickness

Eccentricity = (Bearing width /6) due to sagging moment even with full bearing

For rigid diaphragms when span is less than 30 times thickness

Eccentricity = (Bearing width /12)= 0 for full bearing

For span is greater than 30 times thickness, eccentricity must be calculated.

Interior Wall : (Unequal Spans)

Determining Eccentricity …

Long Span

2R1

a a

2/3 a

Short Span

If R1/R2 lies 0.85-1.15, consider load to be axial else floor load is assumed to act at a distance equal to t/6and then overall eccentricity is determinedR2

Kern Distance For URM Wall

Eccentric Axial Compression

60

)6/(

6&

12,

,0

0.tan

2

23

te

btPe

AP

btS

btIbtA

SM

AP

ff

strengthtensilezeroforcedisKern

ba

=⇒=+−∴

===

=+−

=+−

eP

fa

+

fb

fb

fa+fbfa+fb=

0

Kern Distance For URM Wall …

Eccentric Axial Compression

Quick Check:If eccentricity is less than t/6then there will be no tensile stress.

KernKern for tensile stress ft

b/3b

t/3t

Column or pier section

For Specific Tensile Strength, Ft

Kern Distance

PbtFt

e

PbF

te

t

Fbt

PebtP

FSM

AP

t

t

t

t

66

61)6/(

2

2

2

+=⇒

=+−

=+−

=+−

fa+fb-fa+fb=Ft

Permissible Tensile Stresses

Tension Normal to bed joints Tension Parallel to bed joints

Strong units Weak units

Effective area

hollow solid

No direct tensile strength assumed normal to bed joint just shear strength along bed joints

Effective

t

t

2 ,05.0

1 ,07.0

==

MPafunitsforMPaF

MPafunitsforMPaF

utt

utt

5.7'10.0

10'14.0

>=>=

w

P.e

2Pe 8

2wh2h

e

h

tP

82,sec:arg

,sec:arg2whPe

MnightmidattioncriticalwelandPsmallFor

PeMwalloftopistioncriticalwsmallandPelFor

+=−

=

• Design & Analysis same as eccentrically loaded walls.

• Using concept of equivalent eccentricity, e0 , after bending moment is calculated

w

e

PM

=

e0

P

w

e1

V V

PM

e =1

Four Types of Walls Subjected to Transverse Loads:

• Load bearing wall

• Free standing wall

• Infill walls

Free standing walls need to be checked under stability against overturning.

).0.20.1..(

...

andbetweenliesSF

SF =

Example :Determine maximum transverse load w, as per IS 1905, for unreinforced clay tile wall as shown below :

w

300mm4m

75mm Wall cross-section(Hollow Clay Tiles)

150300

50 kN

HCTType S (H2)1 : 1/2 : 4-1/2

430075

33.133004000

52025.0.

20'

te

MPastresscompBasic

MPaf m

==

==

=×==

For a 1m width of wall

.1021018758

)4000(2

)75()1050(82

:

101506

30010006

103)3001000(

63

232

3522

25

wNmm

mmwmmNwhpeM

wallofheightmidatMoment

mmbt

S

mmA

×+×=

+××=+=

×=×==

×=×=

Tension criterion :

1.013kPaw

0.133w0.1250.16710150

w102101875mm103

N1050

joint)bedtonormal(Tension0.0931N/mm1.33FSM

AP

5

63

25

3

2t

=⇒

++−⇒×

×+×+×

×−

=×=+−

41kPa.w

5.73MPa

MPa)50.69(F1.331.25

0.133w)(0.1250.167SM

AP

a

=⇒<

×=××<

++=+

Compression criterion :

Strength of Walls with No Tensile Stress

Case A : Limiting Compressive Stress, f’m when Resultant Load is within Kern.Since all of section is subjected to compression, all of it will be considered effective.

./&

25.1.25.1

).24/16/1(,19871905:

:

61

62

a

am

bam

m

m

m

fAP

ff

stressecompressivbasicstressecompressivAllowable

ebute

casetyeccentricilowisthisISFor

ForFftrequiremenDesignte

btP

f

btPe

btP

f

SM

AP

f

<≤

×=><

−<

+=

+=

+=

fm

P

eP Wall centroid

Strength of Walls with No Tensile StressCase B :Limiting Compressive Stress f’m, When

Resultant Load is Outside of Kern.

.

23

22,

23

23

)6/1,arg19871905:(

25.1

.2

..

==

−=⇒−=

>−<<

==

et

b

PPb

fcombining

et

et

ecasetyeccentricielIS

fForF

stressedgecompP

bf

ncompressiounderthatis

parteffectiveOnlytensioninmasonryallNeglect

m

aba

m

η

ηη

η

eP

ηη/3

P

e

t/2

ηη

Case B …

NOTE:

1. This approach is beyond ACI & UBC and may give higher values than those based on limited allowable tensile strength.

2. Because wall is partially cracked, it is not prismatic along its height. Stability of wall must be checked based on Euler criteria modified to account for “cracked” masonry. (Ref: Structural Masonry, S. Sahlin)

eP

ηη/3

P

e

t/2

ηη

Combined Bending and Axial Loads

Unity Formula

.,

0.1

momentson

deflectionofeffectsconsidercolumnsslenderFor

Ff

Ff

b

b

a

a <+