Download - Masonry Design DCR-1
Unreinforced Masonry Under Lateral Loads
Durgesh C. RaiAssistant Professor
Department of Civil EngineeringIndian Institute of Technology Kanpur
Kanpur - 208 016
Axial Compression
Eccentric Axial LoadsKern
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
hLateral loads
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
Euler Buckling Load
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
Both wythes loaded
t1 t2
One wythes loaded
)(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
mortarMGradeforMPaF
betterorMGradeforMPaF
t
t
2 ,05.0
1 ,07.0
==
MPafunitsforMPaF
MPafunitsforMPaF
utt
utt
5.7'10.0
10'14.0
>=>=
URM Wall Subjected to Transverse Loading
Transverse Loading (Out-of-plane)
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
Transverse Loading …
w
e
PM
=
e0
P
w
e1
V V
PM
e =1
URM Wall Subjected to Transverse Loading…
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.
Transverse Loading …
).0.20.1..(
...
andbetweenliesSF
loadslateraltoduemomentgoverturninloadsverticaltoduemomentgstabilizin
SF =
Example :Determine maximum transverse load w, as per IS 1905, for unreinforced clay tile wall as shown below :
Transverse Loading
w
300mm4m
75mm Wall cross-section(Hollow Clay Tiles)
150300
50 kN
Transverse Loading (Out-of-plane)…
HCTType S (H2)1 : 1/2 : 4-1/2
430075
33.133004000
52025.0.
20'
te
SRsslendernes
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
×+×=
+××=+=
×=×==
×=×=
Transverse Loading (Out-of-plane)…
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.
Transverse Loading (Out-of-plane)…
./&
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.
Transverse Loading (Out-of-plane)…
.
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 …
Transverse Loading (Out-of-plane)…
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
Transverse Loading (Out-of-plane)…
.,
0.1
momentson
deflectionofeffectsconsidercolumnsslenderFor
Ff
Ff
b
b
a
a <+