Earthquakes Design Provision

8/8/2019 Earthquakes Design Provision

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OVERVIEWOVERVIEW

IntroductionIntroduction

Design forces calculations under UBC 97Design forces calculations under UBC 97

Seismic provisions of RC structures under ACISeismic provisions of RC structures under ACI


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EarthquakeEarthquake: a sudden shaking and breaking of the: a sudden shaking and breaking of the

earth and defined by:earth and defined by:

Focus ( Hypocenter)Focus ( Hypocenter): the point within the earth: the point within the earth

where the earthquake starts.where the earthquake starts.

EpicenterEpicenter: the location on the earth surface directly: the location on the earth surface directlyabove the focus.above the focus.

Measured by intensity and magnitude (Richter scale)Measured by intensity and magnitude (Richter scale)


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Global

Distribution of

Earthquakes


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What Controls the LevelofShaking?What Controls the LevelofShaking?

MagnitudeMagnitude

More energy releasedMore energy releasedDistanceDistance

Shaking decays with distanceShaking decays with distance

Local soilsLocal soils

amplify the shakingamplify the shaking


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Earthquakes included forces or displacements have exceededEarthquakes included forces or displacements have exceeded

the ultimate capacity of the structures.the ultimate capacity of the structures.

Earthquakes are extremely random and oscillatory in natureEarthquakes are extremely random and oscillatory in naturewhich causes the structures to largely deform in oppositewhich causes the structures to largely deform in opposite

directions which requires an understanding of the structuredirections which requires an understanding of the structure

behavior under cyclic loading.behavior under cyclic loading.

In order to do that we will employ the UBC 97 for the analysisIn order to do that we will employ the UBC 97 for the analysis

and the ACI for design.and the ACI for design.


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UBC 97UBC 97


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UBC 97UBC 97

In this section we will utilize the UBC (In this section we will utilize the UBC ( UUniformniform

BBuildinguilding CCode) to compute the design forces resultedode) to compute the design forces resulted

from seismic excitation.from seismic excitation.

The latest code for seismic provisions is IBCThe latest code for seismic provisions is IBC

((IInternationalnternational BBuildinguilding CCode), but we will use theode), but we will use the

UBC instead because it has been adapted by manyUBC instead because it has been adapted by manycountries around the globe as the roots of their localcountries around the globe as the roots of their local

codescodes


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UBC 97UBC 97

Definition of Structural ComponentsDefinition of Structural Components

Before we start we have to define our structuralBefore we start we have to define our structural

components that represent the earthquakescomponents that represent the earthquakesresistant system.resistant system.

For example the structural system may consist ofFor example the structural system may consist of

frames, shear walls , or combination of both ( dualframes, shear walls , or combination of both ( dualsystem) all connected by diaphragms.system) all connected by diaphragms.


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UBC 97UBC 97

Zoning and Response SpectraZoning and Response Spectra

The UBCThe UBC--code has four zoning regions according tocode has four zoning regions according totheir expected seismic hazard and intensity. Thetheir expected seismic hazard and intensity. The

zoning is based on 90% probability that the assignedzoning is based on 90% probability that the assigned

ground acceleration of each zone will not beground acceleration of each zone will not be

exceeded in 50 years. This probability implies aexceeded in 50 years. This probability implies a

return period of 475 years. The UBC four zones arereturn period of 475 years. The UBC four zones aredefined with a Zdefined with a Z--factor as given in UBC Tablefactor as given in UBC Table


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UBC 97UBC 97


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UBC 97UBC 97

For each zone, USCFor each zone, USC--code assigns two coefficients;code assigns two coefficients;one coefficient for acceleration (Cone coefficient for acceleration (Caa) and another for) and another forvelocity (Cvelocity (Cvv). C). Caa and Cand Cvv are given according to theare given according to thesoil profile at the site under considerationsoil profile at the site under consideration

WHERE:WHERE:

1. Soil Type SA:1. Soil Type SA: Hard rock.Hard rock.

2. Soil Type SB:2. Soil Type SB: Rock.Rock.

3. Soil Type Sc:3. Soil Type Sc: Very dense soil and soft rock.Very dense soil and soft rock.

4. Soil Type SD:4. Soil Type SD: Stiff soil profile.Stiff soil profile.5. Soil Type SE:5. Soil Type SE: Soft soil profile.Soft soil profile.

6. Soil Type SF:6. Soil Type SF: Needs special evaluation.Needs special evaluation.


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UBC 97UBC 97

It should be pointed out that if the soil profileIt should be pointed out that if the soil profile

is unknown, the code allows soil type So tois unknown, the code allows soil type So to

be used for seismic zones 3 and 4, whereas,be used for seismic zones 3 and 4, whereas,

soil type SE may be used for zones 1, 2A,soil type SE may be used for zones 1, 2A,

and 2S. It should also be noted that the codeand 2S. It should also be noted that the code

assigns a nearassigns a near--source factors Na and Nv forsource factors Na and Nv for

zone 4 only as defined in UBCzone 4 only as defined in UBC


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UBC 97UBC 97


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UBC 97UBC 97

EarthquakeEarthquake--induced Forcesinduced ForcesThe UBCThe UBC--code allows four methods of analysis tocode allows four methods of analysis to

evaluate earthquake-induced forces in structures,evaluate earthquake-induced forces in structures,

these are:these are:11-- Simplified Design Base Shear.Simplified Design Base Shear.

22-- Static force Procedure.Static force Procedure.

33-- Response Spectrum Analysis.Response Spectrum Analysis.

44-- TimeTime--history Analysis.history Analysis.


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UBC 97UBC 97

The UBCThe UBC--code also uses an importancecode also uses an importance

factor, I, to reflect the importance of thefactor, I, to reflect the importance of thestructure as given in UBC Table 16structure as given in UBC Table 16--K.K.


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UBC 97UBC 97

The code assign RThe code assign R--values for each group of thevalues for each group of the

types above as given in UBC Tablestypes above as given in UBC Tables 1616--N andN and 1616--PP

for buildings and nonfor buildings and non--buildings respectively.buildings respectively.

Structural systems are not listed can be used only ifStructural systems are not listed can be used only if

its seismic structural properties can be verified byits seismic structural properties can be verified by

approved cyclic test and adequate analysis to proveapproved cyclic test and adequate analysis to prove

its integrity.its integrity.


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UBC 97UBC 97

Regularity of StructuresRegularity of Structures

At this stage we will define if we have anyAt this stage we will define if we have anyirregularity in our structure using UBC Tableirregularity in our structure using UBC Table

1616--L for vertical irregularity, and in Table 16L for vertical irregularity, and in Table 16--

M for horizontal irregularityM for horizontal irregularity


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UBC 97UBC 97

Directional EffectDirectional Effect

Eh = Ex + 0.3 Ey .. case (1)Eh = Ex + 0.3 Ey .. case (1)

Eh = 0.3 Ex + Ey . case (2)Eh = 0.3 Ex + Ey . case (2)

Ev = 0.5 Ca I DEv = 0.5 Ca I D

E = Eh + EvE = Eh + Ev

Em=0 EhEm=0 Eh


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ACIACI

Both UBC and IBC codes adopt ACIBoth UBC and IBC codes adopt ACI--code as theircode as their

official provisions for reinforced concrete design.official provisions for reinforced concrete design.

The ACIThe ACI--code offers detailed seismic provisions incode offers detailed seismic provisions in

Chapter 21.Chapter 21.

The ACIThe ACI--code does not provide any seismic forcecode does not provide any seismic forcecalculations, however, it provides the detailingcalculations, however, it provides the detailing

required to achieve the properties of the differentrequired to achieve the properties of the different

reinforced concrete systems that are used andreinforced concrete systems that are used and

endorsed by UBC and IBC codes for seismic design.endorsed by UBC and IBC codes for seismic design.

Such systems include ordinary, intermediate, andSuch systems include ordinary, intermediate, and

special moment frames, in addition, they includespecial moment frames, in addition, they include

ordinary and special shear walls.ordinary and special shear walls.


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OUR

PROBLEM

The residentialbuilding shown is aThe residentialbuilding shown is a

bearingbearing--wall type building subjectedwall type building subjected

to a horizontal earthquake excitationto a horizontal earthquake excitation

in direction y. The building is sixin direction y. The building is six--

story with 200 mm solid onestory with 200 mm solid one--waywayslabs spanning in direction x. Theslabs spanning in direction x. The

slabs are supported by ten shearslabs are supported by ten shear

walls as shown in the same figurewalls as shown in the same figure..

Each wall has 300 mm width with theEach wall has 300 mm width with the

following material properties:following material properties:

fc' = 30 MPa,fc' = 30 MPa,

fy = 420MPa,fy = 420MPa,


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The following loading has tobe considered in the design oftheThe following loading has tobe considered in the design ofthebuilding:building:

1. Own weight (slabs and shear walls).1. Own weight (slabs and shear walls).

2. Superimposed dead load,2. Superimposed dead load,

qSDL = 2 kN/m2qSDL = 2 kN/m2

3. Live load,3. Live load, qL = 3 kN/m2qL = 3 kN/m2

4.Wind load,4.Wind load, qw = 1 kN/m2qw = 1 kN/m2

5. Seismicloading according to UBC code with the following5. Seismicloading according to UBC code with the followingparameters:parameters:

5.1. Seismic zone factor: Z = 0.35.1. Seismic zone factor: Z = 0.3

5.2. Site soil profile:5.2. Site soil profile: SDSD

Ifthe gravityloads are assumed tobe tota

lly carried by theIfthe gravity

loads are assumed tobe tota

lly carried by thewalls in direction y; and their reactions on the walls arewalls in direction y; and their reactions on the walls are

uniformly distributed over the wallcross section, Design oneuniformly distributed over the wallcross section, Design oneinterior shear wall (#3 on the plan) using ACIinterior shear wall (#3 on the plan) using ACI--code provisionscode provisions

for ordinary shear walls (OSW) and special shear walls.for ordinary shear walls (OSW) and special shear walls.


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Solution willbe carried out by design for theSolution willbe carried out by design for theseismicload case (1.2D + 1.0L + 1.0E), and thenseismicload case (1.2D + 1.0L + 1.0E), and then

checkfor the basicload (1.2D + 1.6L)checkfor the basicload (1.2D + 1.6L)

(1) Gravity loads:(1) Gravity loads:

Slabs:Slabs:

qqDD = h = 25 (0.2) = 5 KN/m2= h = 25 (0.2) = 5 KN/m2

qqSDLSDL = 2 KN/m2= 2 KN/m2

qqLL = 3 KN/m2= 3 KN/m2

Walls:Walls:

qqwallwall = b = 25 (0.3) = 7.5 KN/m2= b = 25 (0.3) = 7.5 KN/m2

Solution of our prob.Solution of our prob.


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(2) Mass weight:(2) Mass weight:

W = D + 0.25 warehouse + partitions + equipmentW = D + 0.25 warehouse + partitions + equipment

WhereWhere

D = slabs + wa

llsD = s

labs + wa

lls

D= 7 (10)(15) (6) + 7.5 (26)(20)D= 7 (10)(15) (6) + 7.5 (26)(20)

= 6'300 + 3'900 = 10'200 KN= 6'300 + 3'900 = 10'200 KN

Since there are nowarehouse loads, partitions, orSince there are nowarehouse loads, partitions, or

equipment given, thenequipment given, then

W = 10'200 KNW = 10'200 KN


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Since the building is regular, the seismicforce may beSince the building is regular, the seismicforce may be

calculated using the staticforce procedures. For thiscalculated using the staticforce procedures. For this

system and soil profile, the following parameters may besystem and soil profile, the following parameters may beobtained from UBC tables listed before, henceobtained from UBC tables listed before, hence

For soil SD: CFor soil SD: Caa = 0.36, C= 0.36, Cvv = 0.54, I= 1, R = 4.5= 0.54, I= 1, R = 4.5

Period, T: T = CPeriod, T: T = Ctt ( h( hnn)^)^3/43/4 = 0.048'8 ( 20)^= 0.048'8 ( 20)^3/43/4 = 0.46 sec= 0.46 sec

Therefore,Therefore,

V = 0.2 W = 0.2 (10'200) =V = 0.2 W = 0.2 (10'200) = 2'040 KN2'040 KN


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((44) Wind effect:)Wind effect:

Totalwind force:Totalwind force:

Fw =Fw = 11 ((1515) () (2020) =) = 300300 KN < V =KN < V = 22''040040 KNKN

Therefore, seismic force controls designTherefore, seismic force controls design

((55) Distribution ofbase shear tofloors:) Distribution ofbase shear tofloors:

Since T =Since T = 00..4646


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It can be noted that the distribution above is a triangleIt can be noted that the distribution above is a triangle

with zero value at the base, therefore, the force may bewith zero value at the base, therefore, the force may be

treated as a distributed triangular load as shown withtreated as a distributed triangular load as shown with

intensity at the top equals tointensity at the top equals to

Fx = 2'040 (2)/20 = 204 KN/mFx = 2'040 (2)/20 = 204 KN/m

(6) Torsional effect:(6) Torsional effect:

Since the building is symmetric, the onlySince the building is symmetric, the only

torsion that exists is the accidental torsion.torsion that exists is the accidental torsion.MT = V.MT = V.eeminmin

= 2'040 (0.05 x 15)= 2'040 (0.05 x 15)

= 1'530 kN.m= 1'530 kN.m


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Distribution oftorsional moment to shear walls is given asDistribution oftorsional moment to shear walls is given as

k kjj rrjj^22 = k (1.52 + 7.52) (2) = 117 k= k (1.52 + 7.52) (2) = 117 k

Where Therefore,Where Therefore,

QQ33 = (1530/117 k) (k) (1.5) = 20 KN= (1530/117 k) (k) (1.5) = 20 KN

(7) Vertical seismicforce:(7) Vertical seismicforce:

EEVV = 0.5 Ca I D= 0.5 Ca I D

= 0.5 (0.36) (1) D = 0.18 D= 0.5 (0.36) (1) D = 0.18 D

Total verticalload,Total verticalload,

Qv = 0.18 (10'200) =1'836 KNQv = 0.18 (10'200) =1'836 KN


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(8) Reliability / Redundancy factor:(8) Reliability / Redundancy factor:

Area ofground floor:Area ofground floor:

AB = 15 (10) = 150 m2AB = 15 (10) = 150 m2

rrmaxmax = V= V11 (3.05/ L w ) / V(3.05/ L w ) / Vtottot

= 1(3.05/3) /4 = 0.254= 1(3.05/3) /4 = 0.254

hencehence

Since < 1,Since < 1, = 1 = 1

04.0961.12

150254.0

1.62

!!!V


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(9)Wall share from seismicforce:(9)Wall share from seismicforce:

Horizontalforce acting at (2/3) from the base:Horizontalforce acting at (2/3) from the base:

VE

VE

= V /4 + Q3= V /4 + Q3= 2'040/4 + 20= 2'040/4 + 20

= 510 + 20 == 510 + 20 = 530530KNKN

Verticalforce acting at centerVerticalforce acting at center

PEPE== QvQv / 4 = 1 '836 / 4 =/ 4 = 1 '836 / 4 = 460460KNKN

The effect ofthese forces is shownThe effect ofthese forces is shown

(10) Design forces(10) Design forces

Axialloads:Axialloads:

PD = (qD + qSDL) (Tributary area) (No. offloors)PD = (qD + qSDL) (Tributary area) (No. offloors)

= ( 7 + 7.5) {(10)(1.5+3)} (6) = 2'340 KN= ( 7 + 7.5) {(10)(1.5+3)} (6) = 2'340 KN

PL = (qL) (Tributary area) (No. offloors)PL = (qL) (Tributary area) (No. offloors)

= ( 3) {(10)(1.5+3)} (6) = 810 KN= ( 3) {(10)(1.5+3)} (6) = 810 KN

PE = 460 KNPE = 460 KN


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Maximum shear force at the base:Maximum shear force at the base:

VVDD = 0, V= 0, VLL = 0,= 0, VVEE = 530 KN= 530 KN

Maximum bending moment at the base:Maximum bending moment at the base:MMDD = 0, M= 0, MLL = 0= 0

MMEE = 530 (2/3 )(20) = 7'067 KN.m= 530 (2/3 )(20) = 7'067 KN.m

(11) Flexure design:(11) Flexure design:

CheckbeamCheckbeam--column action:column action:

Limit = 0.1 fc' Ag = 0.1 (30) ( 300)(3000) = 2'700 KNLimit = 0.1 fc' Ag = 0.1 (30) ( 300)(3000) = 2'700 KN

PPUU = 1.2 P= 1.2 PDD + P+ PLL + P+ PEE = 1.2 (2340) + 810 + 460 = 4'078 KN= 1.2 (2340) + 810 + 460 = 4'078 KN

SinceSince PUPU is more than theis more than the limitlimit above, the wall shallbeabove, the wall shallbedesigned as beam - column, i.e. include the effect ofaxialdesigned as beam - column, i.e. include the effect ofaxialload.load.


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Pn = Pu / = 4'078/0.65 = 6'274 KNPn = Pu / = 4'078/0.65 = 6'274 KN

Mu = 1.2 MD + ML + MEMu = 1.2 MD + ML + ME

= 1.2 (0) + 0 + 7'067 = 7'067 kN.m= 1.2 (0) + 0 + 7'067 = 7'067 kN.m

Mn =Mu / = 7'067/0.9 = 7'852 kN.mMn =Mu / = 7'067/0.9 = 7'852 kN.m

Therefore:Therefore:230.0

)3000()300(30

106274 3!

x

096.0)3000()300(30

1078522

6

!

x


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Using interaction diagram for = 0.8, results inUsing interaction diagram for = 0.8, results in

= 0.05, = 0.05,

SINCE = fSINCE = fyy/ 0.85 f/ 0.85 fcc' = 420/ (0.85 x 30)' = 420/ (0.85 x 30)

= 16.47, then= 16.47, then

= 0.05/16.47 = 0.003 (< 0.01, therefore, use= 0.05/16.47 = 0.003 (< 0.01, therefore, use 0.010.01))

Steel :Steel :

AAstst = . b . h = 0.01 (300) (3'000) = 9'000 mm^= . b . h = 0.01 (300) (3'000) = 9'000 mm^22 (1025)(1025)

= 4'500 mm^= 4'500 mm^22 . each face (525). each face (525)


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(12) Shear design:(12) Shear design:

The shear strength ofwalls is usually controlled byThe shear strength ofwalls is usually controlled by

minimum shear reinforcement; therefore, it is goodminimum shear reinforcement; therefore, it is goodpractice to provide the wallwith minimum reinforcementpractice to provide the wallwith minimum reinforcement

first, and then, check its capacity accordingly.first, and then, check its capacity accordingly.

Reference:Reference:

VVcvcv = 0.75 (822) = 616 KN,= 0.75 (822) = 616 KN, VVcvcv = 308 KN= 308 KN

VVuu = 1.2 V= 1.2 VDD + V+ VLL + V+ VEE = 1.2 (0) + 0 + 530 = 530 KN.m= 1.2 (0) + 0 + 530 = 530 KN.m

VVnn = V= Vuu / = 530 /0.75 = 707 KN/ = 530 /0.75 = 707 KN

Since { (VSince { (Vuu = 530 KN) is (< V= 530 KN) is (< Vcvcv = 616, >= 616, > VVcvcv = 308) },= 308) },the minimum reinforcement is given as:the minimum reinforcement is given as:

N822000)3000()300(6

30!!


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Horizontal reinforcement:Horizontal reinforcement:

hh = 0.002'5 spaced at S= 0.002'5 spaced at S22 such that:such that:

SS22 LLww /5 = 3'000/5 = 600 mm/5 = 3'000/5 = 600 mm 3 h = 3 (300) = 900 mm3 h = 3 (300) = 900 mm

500 mm500 mm

As = 0.002'5 (300) (1 '000) = 750 mm2 /mAs = 0.002'5 (300) (1 '000) = 750 mm2 /m

UseUse 2 Layers .. 510 /m each2 Layers .. 510 /m each

Vertical reinforcement:Vertical reinforcement:

vv = 0.002'5 + 0.5 (2.5= 0.002'5 + 0.5 (2.5 -- [h[hWW/L/LWW])( ])( hh -- 0.002'5)0.002'5) hh0.002'50.002'5 spaced at Sspaced at S11 such that:such that:

SS11 LLWW/3 = 3000/3 = 1000 mm/3 = 3000/3 = 1000 mm

3h = 3(300) = 900 mm3h = 3(300) = 900 mm

500 mm500 mm

As = 0.002'5 (300) (1'000) = 750 mm2/mAs = 0.002'5 (300) (1'000) = 750 mm2/m

UseUse 2 Layers.. 510 /m each2 Layers.. 510 /m each


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Final details are shownFinal details are shown


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At this point we have designed the shearAt this point we have designed the shear

wallwith OSW detailing.wallwith OSW detailing.

Nowwe will design using special shear wallNowwe will design using special shear wall

provisions as follows:provisions as follows:


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(13) Length ofboundary elements:(13) Length ofboundary elements:

The length ofthe boundaryThe length ofthe boundary

element is function ofthe neutralelement is function ofthe neutral

axis depth, C, as shown.axis depth, C, as shown.

Therefore, depth (C) is needed which is calculated using PuTherefore, depth (C) is needed which is calculated using Pu

as required by ACI,as required by ACI,

Tocalculate (C), equilibrium offorces shown requires:Tocalculate (C), equilibrium offorces shown requires:

it willbe easier and more conservative ifthe compressionit willbe easier and more conservative ifthe compression

steel is neglected since this neglect produces higher valuessteel is neglected since this neglect produces higher values

ofcofc..


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Therefore, the length ofthe boundary element,Therefore, the length ofthe boundary element,

Therefore, use LT

herefore, use Lbb

= 620 mm= 620 mm


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((1414) Check the need for specialboundary elements (SBE):) Check the need for specialboundary elements (SBE):

Using stress method,Using stress method,


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The SBE must extend over the area where the stress is inThe SBE must extend over the area where the stress is in

excess ofexcess of0.15 f0.15 fcc''== 0.15 (30)0.15 (30)= 4.5= 4.5MPa.MPa.

By inspection offBy inspection offcmcm equation from previous, it can be seenequation from previous, it can be seen

that the axialload alone produces 4.53that the axialload alone produces 4.53MPaMPa, and therefore,, and therefore,

the SBE must extend over the entire wall, i.e. hthe SBE must extend over the entire wall, i.e. hbb = 20 m= 20 m

(16) Detailing ofSBE:(16) Detailing ofSBE:

Minimum hoops:Minimum hoops:

The maximum spacing ofthe transverse reinforcement isThe maximum spacing ofthe transverse reinforcement is

given as:given as:


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Where:Where:

Since Sx must be larger than 100, the 75 mm controls andSince Sx must be larger than 100, the 75 mm controls and

there is no need tocalculatethere is no need tocalculate ssxx, hence,, hence,

Minimum amount ofhoops:Minimum amount ofhoops:

-- In the transverse direction:In the transverse direction:


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-- In the longitudinal direction:In the longitudinal direction:

use the same equation as previous, or by proportionality touse the same equation as previous, or by proportionality to

dimensions, i.e.dimensions, i.e.

Details are shownDetails are shown

Documents

Earthquakes Design Provision