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1
UNREINFORCEDMASONRYEXAMPLES
2013byInternationalMasonryInstitute
Allrightsreserved.
Thisprogramisintendedasapreliminarydesigntoolfordesignprofessionalswhoareexperiencedandcompetentinmasonrydesign.Thisprogramisnotintendedtoreplacesoundengineeringknowledge,experience,andjudgment.Usersofthisprogrammustdeterminethevalidityoftheresults.TheInternationalMasonryInstituteassumesnoresponsibilityfortheuseorapplicationofthisprogram.
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2
EXAMPLEA:
Given: 8inchmediumweighthollowCMU TypeNmasonrycement 12ftverticalspan(simplesupport) Designloadis5psfoutofplane SeismicDesignCategoryA Usethe2012IBC
Required:Determineifthewallisadequate.Solution:LoadCombinationA(0.6D+wL)willcontrol.Themaximummomentwilloccuratmidheightofthewall.
ftlbftftpsfhwM L /908125
8
22
Theaxialforce(wallweightis36psf)is:
ftlbftpsfhwP wall /6.129212366.0
26.0
Calculatetheflexuraltensionstress.
psi
ftinftlbft
ftinftlb
SM
AP
nn
01.9/0.81
/9012/0.30/6.12912
32
Theallowablestress,Ft,is12psi,sothewallisadequate.Ifthe2009IBChadbeenused,theallowableflexuraltensionstressis9psi,sothewallwouldnotbeadequate.Manydesignerswouldsaythat9.01psiiscloseenoughto9psithatthewallwouldbeOK.Theprogramroundstothenearest0.1psi,so9.01psiwouldroundto9.0psi,andtheprogramwouldindicatethewallisadequate.IfTypeSmasonrycementwereused,theallowableflexuraltensionstresswouldbe15psi,andthewallwouldbeadequate.Thereactionatthetopofthewall,Rtop,is:
ftlbftpsfhwR Ltop /302125
2
Asufficientanchoragewouldneedtobeprovidedatthetopofthewalltocarrythisoutofplanereactionforce.
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3
EXAMPLEB:
Given: 8inchmediumweighthollowCMU TypeNmasonrycement 12ftverticalspan(simplesupport) Designloadisahorizontalloadof50lb/ftataheightof36fromthefloor SeismicDesignCategoryA Usethe2012IBC
Required:Determineifthewallisadequate.Solution:LoadCombinationD(0.6D+HL)willcontrol.Themaximummomentwilloccuratthelocationofthehorizontalload.
ftlbftftftftftftlbhh
hhHM LLL /0.1245.31212
5.3/50
Theaxialforce(wallweightis36psf)is: ftlbftftpsfhhwP Lwall /6.1835.312366.06.0 Calculatetheflexuraltensionstress.
psi
ftinftlbft
ftinftlb
SM
AP
nn
2.12/0.81
/12412/0.30/6.18312
32
Theallowablestress,Ft,is12psi,sothewallisnotadequate.UsingaTypeSmasonrycementwouldincreasetheallowablestressto20psi,andthewallwouldbeOK.The50lb/ftat36abovethefloorisatypicalhandrailload.TheIBCandASCE7areunclearastowhetherahandrailloadshouldbecombinedwitha5psfoutofplaneload.Ourinterpretationisthatthe5psfoutofplaneloadisaminimumload,andthatitdoesnotneedtobecombinedwiththehandrailload.
Thereactionatthetopofthewall,Rtop,is: ftlbft
ftftlbhhHR LLtop /6.1412
5.3/50
Thisforceissmallerthanthe30lb/ftanchorageforcefroma5psfuniformlateralload(seeExampleA).Asufficientanchoragewouldneedtobeprovidedatthetopofthewalltocarry30lb/ft.
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4
EXAMPLEC:
Given: 8inchmediumweighthollowCMU TypeNmasonrycement 12ftverticalspan(simplesupport) Designloadisaverticalloadof40lb/ftataneccentricityof4inchesoutsidethewall SeismicDesignCategoryA Usethe2012IBC
Required:Determineifthewallisadequate.Solution:LoadCombinationB(0.6D+PL)andC(D+PL)willbechecked.ThemaximummomentwillbePe.
ftlbftinftininftlbPeM /0.26
1214
2625.7/40
Theeccentricloadisassumedtooccuratthetopofthewall;hencethewallweightis0.Theaxialforcewilljustbetheappliedverticalload.Calculatetheflexuraltensionstress.
psi
ftinftlbft
ftinftlb
SM
AP
nn
5.2/0.81
/2612/0.30
/401232
Theallowablestress,Ft,is12psi,sothewallisnotadequate.Althoughtensionalmostalwayscontrolswithunreinforcedmasonry,theunityequationwillbecheckedforcompleteness.Thevalueofris3.21in.,makingh/r=144in/3.21in.=44.8
psiftin
ftlbAPf
na 3.10.30
/402
psiininpsi
rhfF ma 8.302.21.3
.14414014
135014014
22
psiftin
ftlbftSMfn
b 8.30.81/261212
3
psipsif
F mb 45031350
3
0.1013.0450
8.38.302
3.1
psipsi
psipsi
Ff
Ff
b
b
a
a OK
TheTMS402Codealsorequiresthatwhenusingallowablestressdesignthefollowingequationsbechecked: ePP 41
3
2
2
577.01
re
hIEP me
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5
Forthiswall, / =3.21in.Withe=7.81in.,10.577e/r=0.405.Thus,thewalldoesnotmeetthiscoderequirement.However,giventherelativelylightload,manydesignerswouldacceptthewallasadequate.Theeccentricverticalloadistypicalofalargemonitorthatisattachedtothewall.TheIBCandASCE7areunclearastowhetherthisloadshouldbecombinedwitha5psfoutofplaneload.Ourinterpretationisthatthe5psfoutofplaneloadisaminimumload,andthatitdoesnotneedtobecombinedwiththehandrailload.The5psfoutofplaneloadshouldalsobechecked(seeExampleA),andwouldcontrolinthiscase.
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6
EXAMPLED:
Given: 8inchmediumweighthollowCMU TypeNmasonrycement 12fthorizontalspan(simplesupport) Designloadis5psfoutofplane SeismicDesignCategoryA Usethe2012IBC
Required:Determineifthewallisadequate.Solution:LoadCombinationA(0.6D+wL)willcontrol.Themaximummomentwilloccuratmidlengthofthewall(hisusedforlength).
ftlbftftpsfhwM L /908125
8
22
Thereisnoaxialforceonthewall.Calculatetheflexuraltensionstress.
psi
ftinftlbft
SM
n
3.13/0.81
/9012123
Theallowablestress,Ft,is25psi,sothewallisadequate.Bytrialanderrorchangingofthewalllength,itcanbedeterminedthatthemaximumhorizontalspanforthiswallis16ftfora5psfoutofplaneload.
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7
EXAMPLEE:
Given: 8inchmediumweighthollowCMU TypeNmasonrycement 5fthighcantilever Designloadis5psfoutofplane SeismicDesignCategoryA Usethe2012IBC
Required:Determineifthewallisadequate.Solution:LoadCombinationA(0.6D+wL)willcontrol.Themaximummomentwilloccuratthebaseofthewall.
ftlbftftpsfhwM L /5.62255
2
22
Theaxialforce(wallweightis36psf)is: ftlbftpsfhwP wall /1085366.06.0 Calculatetheflexuraltensionstress.
psi
ftinftlbft
ftinftlb
SM
AP
nn
66.5/0.81
/5.6212/0.30
/1081232
Theallowablestress,Ft,is12psi,sothewallisadequate.
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8
EXAMPLEF:
Given: 8inchmediumweighthollowCMU TypeNPortlandcementlimemortar 12ftverticalspan(simplesupport) Minimumdesignloadis5psfoutofplane SDS=0.5;SeismicDesignCategoryC Usethe2012IBC
Required:Determineifthewallisadequate.Solution:CheckLoadCombinationG(0.6D+0.7E).Wallweightis36psf,andR=1.5forunreinforcedmasonry.
Theseismicloadis psfpsf
IR
WSw
p
p
pDSE 4.14
0.15.1365.02.12.1
Since0.7(14.4psf)=10.1psf,seismicwillcontrol.Themaximummomentwilloccuratmidheightofthewall.
ftlbftftpsfhwM E /181
8124.147.0
87.0 22
Theaxialforceis:
ftlbftpsfhwSP wallDS /5.114212365.02.07.06.0
22.07.06.0
Calculatetheflexuraltensionstress.
psi
ftinftlbft
ftinftlb
SM
AP
nn
1.23/0.81
/18112/0.30/5.11412
32
Theallowablestress,Ft,is25psi,sothewallisadequate.PartitionwallsinSDCarerequiredtohaveprescriptiveseismicreinforcementineitherthehorizontalORverticaldirectioninaccordancewiththefollowing:
(a)HorizontalreinforcementTwolongitudinalwiresofW1.7(9gage)bedjointreinforcementspacednotmorethan16in.oncenter,orNo.4barsspacednotmorethan48in.oncenter.Horizontalreinforcementneedstobeprovidedwithin16in.ofthetopandbottomofthewall.
(b)VerticalreinforcementNo.4barsspacednotmorethan120in.oncenter.Verticalreinforcementneedstobeprovidedwithin16in.oftheendsofthewall.
ItisrecommendedthatW1.7(9gage)bedjointreinforcementat16in.(everyothercourse)beprovidedtomeettheprescriptiveseismicrequirements.Duetothepermittedonethirdstressincreaseinthe2009IBC(2008MSJC),theallowableflexuraltensionstresswouldbethesame,andthiswallwouldbeadequateusingthe2009IBC.Thereactionatthetopofwall,Rt,is:
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9
ftlbftpsfhwR Et /5.602124.147.0
27.0
Sufficientanchorsneedtobeprovidedatthetopofwalltocarrythisforce.Ifthispartitionwallwerepartofanegressstairway,theimportancefactorwouldbe1.5,whichincreasestheseismicload,wE,to21.6psf,themomentto272ftlb/ft,andtheflexuraltensilestressto36.5psi.Unreinforced,ungroutedmasonrywillnolongerwork,evenwithTypeSPortlandcementlimemortar.
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VERIFICATThefollowundervarThemaximAllofthetheresultCASEA:mh=12ftHL=40lb/hL=10ftPL=100lbe=6inwE=14.4
Rb
7.0
7.0
IsRb
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CASEB:mh=12ftHL=40lb/hL=5ftPL=100lbe=6inwE=14.4
Rb
7.0
7.0
Is0.75*0. 7.075.0 w7.075.0
Yes,37.8lhhx
Maximum
RM
6
max
maximummom
/ft
b/ft
psf
p
hwE
24.147.05
27.05
7wEhLRb
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CASEC:mh=12ftHL=40lb/hL=3ftPL=100lbe=6inwE=14.4
Rb
7.0
7.0
Is0.75(0.7 w7.075.0 w7.075.0
Yes,52.7lRhx0
Maximum
ft
lb
RM
201
0.71
max
maximummom
/ft
b/ft
psf
p
hwE
24.147.05
27.05
7wEhL+HL)
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CASED:mh=12ftHL=20lb/hL=3ftPL=200lbe=48inwE=14.4
Rb
7.0
7.0
IsRb>0.7 w7.075.0
Yes,106.6MaximumM 0max
maximummo
/ft
b/ft
psf
p
hwE
24.147.05
27.05
75(0.7wEh+HL)Hhw LE
6lb/ft>105.7mmomentocc
Pe 7.075.0
mentoccurs
ftpsfh
hhH LL
2012
)? 147.075.0
7lb/ft.CaseDcursattopof
ftlb /20075
attop.
ftftftlb
hePL
1212/0
ftpsf 124.4Dcontrols.xfwall.
fft 6004
13
ft 2003
ftlbt /20x=0
ftlbft /
ft
ftftlb12
4/0
flb /7.105
lb /6.106
ft
ft