1

UNREINFORCEDMASONRYEXAMPLES

2013byInternationalMasonryInstitute

Allrightsreserved.

Thisprogramisintendedasapreliminarydesigntoolfordesignprofessionalswhoareexperiencedandcompetentinmasonrydesign.Thisprogramisnotintendedtoreplacesoundengineeringknowledge,experience,andjudgment.Usersofthisprogrammustdeterminethevalidityoftheresults.TheInternationalMasonryInstituteassumesnoresponsibilityfortheuseorapplicationofthisprogram.

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.

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.

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

5

Forthiswall, / =3.21in.Withe=7.81in.,10.577e/r=0.405.Thus,thewalldoesnotmeetthiscoderequirement.However,giventherelativelylightload,manydesignerswouldacceptthewallasadequate.Theeccentricverticalloadistypicalofalargemonitorthatisattachedtothewall.TheIBCandASCE7areunclearastowhetherthisloadshouldbecombinedwitha5psfoutofplaneload.Ourinterpretationisthatthe5psfoutofplaneloadisaminimumload,andthatitdoesnotneedtobecombinedwiththehandrailload.The5psfoutofplaneloadshouldalsobechecked(seeExampleA),andwouldcontrolinthiscase.

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.

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.

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:

9

ftlbftpsfhwR Et /5.602124.147.0

27.0

Sufficientanchorsneedtobeprovidedatthetopofwalltocarrythisforce.Ifthispartitionwallwerepartofanegressstairway,theimportancefactorwouldbe1.5,whichincreasestheseismicload,wE,to21.6psf,themomentto272ftlb/ft,andtheflexuraltensilestressto36.5psi.Unreinforced,ungroutedmasonrywillnolongerwork,evenwithTypeSPortlandcementlimemortar.

VERIFICATThefollowundervarThemaximAllofthetheresultCASEA:mh=12ftHL=40lb/hL=10ftPL=100lbe=6inwE=14.4

Rb

7.0

7.0

IsRb

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

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)

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