Draft

14.2BuildingsStructures 233

14.2 BuildingsStructures

11 Therearethreeprimarytypes of buildingsystems:

WallSubsytem: in which veryrigidwallsmadeupof solidmasonry, paneledor bracedtimber,

or steeltrussesconstitutea rigidsubsystem.Thisis onlyadequateforsmallrisebuildings.

VerticalShafts: madeupof foursolidor trussedwallsforminga tubularspacestructure.The

tubularstructuremay be interior(housingelevators,staircases)and/orexterior. Most

ecient forveryhighrisebuildings.

RigidFrame:which consistsof linearverticalcomponents (columns)rigidlyconnectedto sti

horizontalones(beamsandgirders).Thisis nota veryecient structuralformto resist

lateral(wind/earthquake) loads.

14.2.1 WallSubsystems

12 Whereasexteriorwallprovideenclosureandinterioronesseparation,bothof themcanalso

have a structuralrolein trnsferingverticalandhorizontalloads.

13 Wallsareconstructedoutof masonry, timber concreteor steel.

14 If thewallis bracedby
oors,thenit canprovideanexcellent resitanceto horizontalload

in theplaneof thewall(butnotorthogonalto it).

15 Whenshear-wallssubsytemsareused,it is bestif thecenterof orthogonalshearresistance

is closeto thecentroidof lateralloadsas applied. If thisis notthecase,thentherewillbe

torsionaldesignproblems.

14.2.1.1 Example: ConcreteShearWall

From (Linand Stotesbury 1981)

16 We considera reinforcedconcretewall20 ft wide,1 ft thick, and 120 ft highwitha vertical

loadof 400k actingonit at thebase. Asa resultof wind,we assumea uniformhorizontal

forceof 0.8kip/ftof verticalheight actingonthewall. It is requiredto computethe
exural

stressesandtheshearingstressesin thewallto resistthewindload,Fig.14.5.

1. Maximumshearforceandbendingmoment at thebase

Vmax = wL= (0:8)k.ft(120)ft= 96k (14.8-a)

Mmax =

wL

2

2

=

(0:8)k.ft(120)

2

ft

2

2

= 5; 760 k.ft (14.8-b)

2. Theresultingeccentricity is

eActual=

M

P

=

(5;760)k.ft

(400)k

= 14: 4 ft (14.9)

3. Thecriticaleccentricity is

ecr=

L

6

=

(20)ft

6

= 3: 3 ft< eActualN:G: (14.10)

thus therewillbe tensionat thebase.