wall design example 2009

Wall boundary element design exampleLuis E. Garcia

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DESIGN EXAMPLE OFDESIGN EXAMPLE OFDESIGN EXAMPLE OF STRUCTURAL WALL

BOUNDARY ELEMENTS

DESIGN EXAMPLE OF STRUCTURAL WALL

BOUNDARY ELEMENTSby:by:by:

Luis Enrique GarcíaPresident American Concrete Institute – ACI – 2008-2009Partner Proyectos y Diseños Ltda., Consulting Engineers Professor of Civil Engineering, Universidad de los Andes,

Bogotá, Colombia

by:Luis Enrique García

President American Concrete Institute – ACI – 2008-2009Partner Proyectos y Diseños Ltda., Consulting Engineers Professor of Civil Engineering, Universidad de los Andes,

Bogotá, Colombia

The ExampleThe ExampleWall section 3 m by 0.30 m subjected to factored Pu = 1750 kN, Mu = 2000 kN-m, and Wall section 3 m by 0.30 m subjected to factored Pu = 1750 kN, Mu = 2000 kN-m, and acto ed u 50 , u 000 , a dVu = 600 kN, which include gravity and seismic forces.The wall has 10 stories with each floor having a 3 m height from finish to finish for a 30 m total.

acto ed u 50 , u 000 , a dVu = 600 kN, which include gravity and seismic forces.The wall has 10 stories with each floor having a 3 m height from finish to finish for a 30 m total. Expected lateral displacement under seismic design forces at the roof is δu = 0.20 m.Structure is being designed as a Special Wall under ACI 318-08.

Expected lateral displacement under seismic design forces at the roof is δu = 0.20 m.Structure is being designed as a Special Wall under ACI 318-08.

The ExampleThe Example

hw = 30 m(10 stories with 3 m finish-to-finish each)

hw = 30 m(10 stories with 3 m finish-to-finish each)

w = 3 mw = 3 m

ExampleExampleThe appropriate wall reinforcement to resist

these forces must be determined. OnceThe appropriate wall reinforcement to resist

these forces must be determined. Oncethese forces must be determined. Once reinforcement is found the need for boundary elements must be investigated, and case they are needed, their design must be done in accordance with:

(a) Old pre-ACI 318-99 procedure.

these forces must be determined. Once reinforcement is found the need for boundary elements must be investigated, and case they are needed, their design must be done in accordance with:

(a) Old pre-ACI 318-99 procedure. (b) Using 21.9.6.2 (displacement-controlled) of

ACI 318S-08. For this case δu = 0.20 m(c) Using 21.9.6.3 (stress-controlled) of ACI

318S-08.

(b) Using 21.9.6.2 (displacement-controlled) of ACI 318S-08. For this case δu = 0.20 m

(c) Using 21.9.6.3 (stress-controlled) of ACI 318S-08.


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SolutionSolution

Flexural design of wallFlexural design of wallMinimum steel ratio is 0.0025 for longitudinal and transverse directions.Minimum steel ratio is 0.0025 for longitudinal and transverse directions.

Smaller (Section 14.3) steel ratio is permitted is shear force does not exceed .

For this case: and

Smaller (Section 14.3) steel ratio is permitted is shear force does not exceed .

For this case: and

Then

therefore the minimum steel ratio is 0.0025 and bar spacing cannot be larger than 450 mm.

Then

therefore the minimum steel ratio is 0.0025 and bar spacing cannot be larger than 450 mm.

Flexural design of wallFlexural design of wall14.3.4 requires that wall with thickness grater than 250 mm must have their reinforcement in 14.3.4 requires that wall with thickness grater than 250 mm must have their reinforcement in both directions placed in two layers.

Reinforcement area is then:

7 7 1( ) i l

both directions placed in two layers.

Reinforcement area is then:

7 7 1( ) i l7.7.1(c) requires a concrete cover greater or equal to 20 mm. Bar spacing is computed as follows for different bar diameter alternatives:

7.7.1(c) requires a concrete cover greater or equal to 20 mm. Bar spacing is computed as follows for different bar diameter alternatives:

Flexural design of wallFlexural design of wallφ 9.5 mm (3/8”) diameter bars (db = 9.5 mm, Ab = 71 mm2)

Number of bars = 2 250/71 = 32 in two vertical layers with 16 bars each Vertical bar spacing = (3 000 – 2 · 20 – 9.5)/15 = 197 mm < 500 mm. ok!

φ 9.5 mm (3/8”) diameter bars (db = 9.5 mm, Ab = 71 mm2) Number of bars = 2 250/71 = 32 in two vertical layers with 16 bars each Vertical bar spacing = (3 000 – 2 · 20 – 9.5)/15 = 197 mm < 500 mm. ok!Horizontal bar spacing = (71 · 2)/(300 · 0.0025) = 189 mm. ok!

φ 12.5 mm (1/2”) diameter bars (db = 12.5 mm, Ab = 129 mm2) Number of bars = 2 250/129 = 18 in two layers with 9 bars each. Vertical bar spacing = (3 000 – 2 · 20 – 12.5)/8 = 368 mm < 500 mm. ok!Horizontal bar spacing = (129 · 2)/(300 · 0.0025) = 344 mm. ok!

φ 16 mm (5/8”) diameter bars (db = 16 mm, Ab = 199 mm2) Number of bars = 2 250/199 = 12 in two vertical layers with 6 bars each

Horizontal bar spacing = (71 · 2)/(300 · 0.0025) = 189 mm. ok!φ 12.5 mm (1/2”) diameter bars (db = 12.5 mm, Ab = 129 mm2)

Number of bars = 2 250/129 = 18 in two layers with 9 bars each. Vertical bar spacing = (3 000 – 2 · 20 – 12.5)/8 = 368 mm < 500 mm. ok!Horizontal bar spacing = (129 · 2)/(300 · 0.0025) = 344 mm. ok!

φ 16 mm (5/8”) diameter bars (db = 16 mm, Ab = 199 mm2) Number of bars = 2 250/199 = 12 in two vertical layers with 6 bars eachbars each. Vertical bar spacing = (3 000 – 2 · 20 – 16)/5 = 589 mm > 500 mm. Doesn’t work!Horizontal bar spacing = (199 · 2)/(300 · 0.0025) = 530 mm. Doesn’t work!

The φ 12.5 mm (1/2”) bar diameter option is selected.

bars each. Vertical bar spacing = (3 000 – 2 · 20 – 16)/5 = 589 mm > 500 mm. Doesn’t work!Horizontal bar spacing = (199 · 2)/(300 · 0.0025) = 530 mm. Doesn’t work!

The φ 12.5 mm (1/2”) bar diameter option is selected.


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Diagrama de Interacción - Muro Tarea 5 - 2006

15000

20000

Pn-Mn

Interaction diagram – 18 φ 12.5 (1/2”) diameter bars

2000, 1750 3130.3, 17505000

10000

Axi

al (k

N)

φPn-φMn

3638.1, 1750

-5000

0

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000Momento (kN x m)

Flexural design of wallFlexural design of wallFor axial force of 1 750 kN, moment φMn = 3 130 kN > 2 000 kN, which meet the design objective. For axial force of 1 750 kN, moment φMn = 3 130 kN > 2 000 kN, which meet the design objective.

Moment strength is Mn = 3 638 kN and neutral axis depth when reaching the moment strength is k = 0.132 for Pu = 1 750 kN

Therefore, c = 0.132 x 3 000 = 396 mm (this information is needed for the design following

Moment strength is Mn = 3 638 kN and neutral axis depth when reaching the moment strength is k = 0.132 for Pu = 1 750 kN

Therefore, c = 0.132 x 3 000 = 396 mm (this information is needed for the design followinginformation is needed for the design following ACI 318-08).

Note: the load factors and φ factors of Appendix C of ACI 318-08 were used.

information is needed for the design following ACI 318-08).

Note: the load factors and φ factors of Appendix C of ACI 318-08 were used.

Shear design of wallShear design of wallNow we check the shear strength for the minimum reinforcement used for flexural design Now we check the shear strength for the minimum reinforcement used for flexural design gaccording to Section C.21.9.4 of ACI 318-08:

gaccording to Section C.21.9.4 of ACI 318-08:

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99Now we investigate if the wall needs boundary elements following the pre-ACI 318-99 procedure (the checking is the same prescribed by ACI 318-

S

Now we investigate if the wall needs boundary elements following the pre-ACI 318-99 procedure (the checking is the same prescribed by ACI 318-

S08 Section 21.6.3.This section requires the use of boundary elements if the extreme fiber in compression stress exceedsThe stress must be computed using factored forces that include seismic effects (Pu y Mu), a linear elastic model and gross section

08 Section 21.6.3.This section requires the use of boundary elements if the extreme fiber in compression stress exceedsThe stress must be computed using factored forces that include seismic effects (Pu y Mu), a linear elastic model and gross sectionlinear elastic model and gross section dimensions.

Stress is obtained from:

linear elastic model and gross section dimensions.

Stress is obtained from:


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(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99Pu = 1 750 kNMu = 2 000 kN·mPu = 1 750 kNMu = 2 000 kN·mMu 2 000 kN mAg = 3 000 x 300 = 900 000 mm2

w = 3 000 mmIg = (1/12) x bw x w

3 = 675 x 109 mm4

Mu 2 000 kN mAg = 3 000 x 300 = 900 000 mm2

w = 3 000 mmIg = (1/12) x bw x w

3 = 675 x 109 mm4

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99

Therefore, the wall requires boundary l t !

Therefore, the wall requires boundary l t !elements!

The former pre-ACI 318-99 procedure required the boundary elements alone to be able to resist all forces derived from flexural response including factored gravity loads and factored seismic forces

elements!The former pre-ACI 318-99 procedure required the boundary elements alone to be able to resist all forces derived from flexural response including factored gravity loads and factored seismic forcesgravity loads and factored seismic forces.The factored axial force Pu includes gravity loads and factored wall moment Muincludes seismic forces.

gravity loads and factored seismic forces.The factored axial force Pu includes gravity loads and factored wall moment Muincludes seismic forces.

((a) Design using pre-ACI 318-99((a) Design using pre-ACI 318-99

w

h

w

h

Pu

Mu

hbe

Pu

Mu

hbe

( )u u

cuw be

P MPh2

= +−

( )u u

tuw be

P MPh

02

= − ≤−

u

n c g st st yP f A A A f0 [0.85 ( ) ]φ φ ′⋅ = ⋅ ⋅ ⋅ − + ⋅

cu n nP P P(max) 00.80φ φ≤ ⋅ ≤ ⋅ ⋅tu tn st yP P A fφ φ≤ ⋅ = ⋅ ⋅

( )u u

cuw be

P MPh2

= +−

( )u u

tuw be

P MPh

02

= − ≤−

u

n c g st st yP f A A A f0 [0.85 ( ) ]φ φ ′⋅ = ⋅ ⋅ ⋅ − + ⋅

cu n nP P P(max) 00.80φ φ≤ ⋅ ≤ ⋅ ⋅tu tn st yP P A fφ φ≤ ⋅ = ⋅ ⋅

((a) Design using pre-ACI 318-99((a) Design using pre-ACI 318-99A initial trial size of 300 by 300 mm is given to the boundary elements. A initial trial size of 300 by 300 mm is given to the boundary elements. Therefore, hbe = 300 mm, and Ag = 300 x 300 = 90 000 mm2.Boundary element tension axial force is:

Therefore, hbe = 300 mm, and Ag = 300 x 300 = 90 000 mm2.Boundary element tension axial force is:

It is in compression!It is in compression!


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(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99Axial force in the boundary element in compression is: Axial force in the boundary element in compression is: p

Ast is then:

p

Ast is then:


This area is less that the required for tension and 1 877 mm2 must be used. Using φ 16 mm (5/8”) diameter bars having A = 199 mm2 the

This area is less that the required for tension and 1 877 mm2 must be used. Using φ 16 mm (5/8”) diameter bars having A = 199 mm2 the(5/8 ) diameter bars having Ab = 199 mm , the number of bars is 1 877/199 = 9.4, and 10 φ 16 mm (5/8”) bars are selected.

The vertical steel ratio for the boundary element is then: ρ = (10 x 199)/(300 x 300) =

(5/8 ) diameter bars having Ab = 199 mm , the number of bars is 1 877/199 = 9.4, and 10 φ 16 mm (5/8”) bars are selected.

The vertical steel ratio for the boundary element is then: ρ = (10 x 199)/(300 x 300) =element is then: ρ (10 x 199)/(300 x 300) 0.022, This is 2.2% of the boundary element area. This steel ratio indicates that a smaller section boundary element may not have worked.

element is then: ρ (10 x 199)/(300 x 300) 0.022, This is 2.2% of the boundary element area. This steel ratio indicates that a smaller section boundary element may not have worked.


Boundary elements, when required, must have hoop transverse reinforcement as forBoundary elements, when required, must have hoop transverse reinforcement as forhave hoop transverse reinforcement as for special moment frame columns.

Sections 21.6.4.2 through 21.6.4.4, except Eq. (21-4) need not be satisfied and the transverse reinforcement spacing limit of

have hoop transverse reinforcement as for special moment frame columns.

Sections 21.6.4.2 through 21.6.4.4, except Eq. (21-4) need not be satisfied and the transverse reinforcement spacing limit oftransverse reinforcement spacing limit of 21.6.4.3(a) shall be one-third of the least dimension of the boundary element.

transverse reinforcement spacing limit of 21.6.4.3(a) shall be one-third of the least dimension of the boundary element.


Now the hoop transverse confinement Now the hoop transverse confinement reinforcement for the boundary element is defined.

Distributed horizontal reinforcement for the wall was determined to be φ 12.5 mm

reinforcement for the boundary element is defined.

Distributed horizontal reinforcement for the wall was determined to be φ 12.5 mm φ(1/2”) bars. The same bar diameter is used for the hoop transverse confinement reinforcement of the boundary element.

φ(1/2”) bars. The same bar diameter is used for the hoop transverse confinement reinforcement of the boundary element.


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The layout of the hoop follows the same rules that for column ties in Section 7 10 5The layout of the hoop follows the same rules that for column ties in Section 7 10 5rules that for column ties in Section 7.10.5 of ACI 318-08: all vertical bars must be at the corner of a tie or crosstie, unless the clear distance to next bar which is at a tie or crosstie corner is less than 150 mm. On the other hand, Section 21.6.4.2 requires

rules that for column ties in Section 7.10.5 of ACI 318-08: all vertical bars must be at the corner of a tie or crosstie, unless the clear distance to next bar which is at a tie or crosstie corner is less than 150 mm. On the other hand, Section 21.6.4.2 requires that the horizontal spacing of crossties or legs of rectilinear hoops, hx, within a cross section of the member shall not exceed 350 mm on center.

that the horizontal spacing of crossties or legs of rectilinear hoops, hx, within a cross section of the member shall not exceed 350 mm on center.

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99φ 12 mm (1/2”) spaced 340 mm vertically..This reinforcement must be anchored in the boundary element10 φ 16 mm (5/8”)

φ12 mm (1/2”) spaced 340 mm horizontally

the boundary element.10 φ 16 mm (5/8 )

Confinement hoops φ 12 mm (1/2”)

300 mm

300 mm.φ 12 mm (1/2”) spaced 340 mm vertically.

This reinforcement must be anchored in the boundary element.


Now in order to obtain the vertical spacing Now in order to obtain the vertical spacing of the confining hoops we must use Eq. (21-5). We solve it for the vertical spacing because we know all the other variables.

of the confining hoops we must use Eq. (21-5). We solve it for the vertical spacing because we know all the other variables.


In order to apply this equation it must be t k i t t th t A i th fIn order to apply this equation it must be t k i t t th t A i th ftaken into account that Ash is the area of all hoop legs including crossties that are perpendicular to the direction under study.

bc is the dimension of the confined core

taken into account that Ash is the area of all hoop legs including crossties that are perpendicular to the direction under study.

bc is the dimension of the confined corebc is the dimension of the confined core perpendicular to the hoop and crosstie legs and measured center to center of the outer hoop leg.

bc is the dimension of the confined core perpendicular to the hoop and crosstie legs and measured center to center of the outer hoop leg.


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(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99The concrete cover to the hoop is 40 mm (Section 7 7 1(c) of ACI 318-08)The concrete cover to the hoop is 40 mm (Section 7 7 1(c) of ACI 318-08)(Section 7.7.1(c) of ACI 318 08). Using this value, the distance from the wall edge to the center of the outer hoop bar is: 40 + 12.5/2 = 46 mm. Therefore, in this case hc for both directions is: 300 – 46 x 2 = 208 mm

(Section 7.7.1(c) of ACI 318 08). Using this value, the distance from the wall edge to the center of the outer hoop bar is: 40 + 12.5/2 = 46 mm. Therefore, in this case hc for both directions is: 300 – 46 x 2 = 208 mmdirections is: 300 46 x 2 = 208 mm. Ab for φ 12.5 mm (1/2”) diameter bars is 129 mm2.

directions is: 300 46 x 2 = 208 mm. Ab for φ 12.5 mm (1/2”) diameter bars is 129 mm2.

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99First we check in the direction parallel to wall length w . First we check in the direction parallel to wall length w . g w

Variable values are as follows:Ash = 3 x 129 = 387 mm2

fyt = 420 MPahc = 208 mm

28 MP

g w

Variable values are as follows:Ash = 3 x 129 = 387 mm2

fyt = 420 MPahc = 208 mm

28 MP= 28 MPa= 28 MPa

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99Now we apply the other rules for spacing of hoops:Now we apply the other rules for spacing of hoops:p

From Eq. (21-5) 310 mm1/3 of the minimum section dimension 300/3 -> 100 mmand 100 mm.

Therefore, for this direction the maximum hoops vertical spacing corresponds to the lesser of these values = 100 mm

pFrom Eq. (21-5) 310 mm1/3 of the minimum section dimension 300/3 -> 100 mmand 100 mm.

Therefore, for this direction the maximum hoops vertical spacing corresponds to the lesser of these values = 100 mm

For the other direction the variables have the same values thus the vertical spacing is the same.

For the other direction the variables have the same values thus the vertical spacing is the same.


Now we need to check shear for the situation where the wall is at flexural strength. We have Now we need to check shear for the situation where the wall is at flexural strength. We have gtwo alternatives, either to design for the shear force that corresponds to flexural strength or to use Section 9.3.4(a) where a reduced φ factor of 0.6 is required when not designing for the shear corresponding to flexural strength.

gtwo alternatives, either to design for the shear force that corresponds to flexural strength or to use Section 9.3.4(a) where a reduced φ factor of 0.6 is required when not designing for the shear corresponding to flexural strength.

The intent is to prevent a premature shear failure when the wall is responding in the nonlinear range.

The intent is to prevent a premature shear failure when the wall is responding in the nonlinear range.


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(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99The flexural strength already computed didn’t take into account the vertical reinforcement of the b d l t It t b t d f

The flexural strength already computed didn’t take into account the vertical reinforcement of the b d l t It t b t d fboundary elements. It must be computed for a value of φ = 1.0. ACI 318-08 does not require the use of the strain-hardening stress for walls now but used to require it in the pre-ACI 318-99 versions.The moment-curvature diagram for an axial load of 1 750 kN, including the boundary element

boundary elements. It must be computed for a value of φ = 1.0. ACI 318-08 does not require the use of the strain-hardening stress for walls now but used to require it in the pre-ACI 318-99 versions.The moment-curvature diagram for an axial load of 1 750 kN, including the boundary element , g yvertical reinforcement and allowing strain-hardening with a yield strength for the reinforcement of 1.25 fy was computed. The probable moment strength (1.25fy y φ = 1) is of the order of 5 550 kN·m.

, g yvertical reinforcement and allowing strain-hardening with a yield strength for the reinforcement of 1.25 fy was computed. The probable moment strength (1.25fy y φ = 1) is of the order of 5 550 kN·m.

Diagrama Momento Curvatura - Para Carga Axial de 1750 kN

5000.0

6000.0Mp = 5550 kNxm

M-φ diagram including boundary elements

2000.0

3000.0

4000.0

Mom

ento

(kN

x m

)

0.0

1000.0

0.000000 0.000002 0.000004 0.000006 0.000008 0.000010

Curvatura (1/mm)

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99The value of the shear reached when responding in the nonlinear range at The value of the shear reached when responding in the nonlinear range at p g gmoment strength is obtained proportional to the one from analysis as given in the data for the example:

p g gmoment strength is obtained proportional to the one from analysis as given in the data for the example:

Shear strength computed previously was: φVn = 1478 kN.Shear strength computed previously was: φVn = 1478 kN.

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99The horizontal reinforcement steel ratio is obtained by solving the equation for shearThe horizontal reinforcement steel ratio is obtained by solving the equation for shearobtained by solving the equation for shear strength for variable ρt:obtained by solving the equation for shear strength for variable ρt:


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(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99The steel ratio required for shear is:The steel ratio required for shear is:

Now we solve for the vertical spacing for 2 Now we solve for the vertical spacing for 2 φ 12.5 mm (1/2”) diameter bars :φ 12.5 mm (1/2”) diameter bars :

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99From the construction point of view this to close spacing Using φ 16 mm (5/8”)From the construction point of view this to close spacing Using φ 16 mm (5/8”)close spacing. Using φ 16 mm (5/8 ) diameter bars the following vertical spacing is obtained (this reinforcement can be used in the zone critical for shear and change to the other reinforcement in the upper floors):

close spacing. Using φ 16 mm (5/8 ) diameter bars the following vertical spacing is obtained (this reinforcement can be used in the zone critical for shear and change to the other reinforcement in the upper floors):


A horizontal reinforcement verticalA horizontal reinforcement verticalA horizontal reinforcement vertical spacing of 85 mm will be used.

This type of reinforcement must be used until the compressive stress at the wall

A horizontal reinforcement vertical spacing of 85 mm will be used.

This type of reinforcement must be used until the compressive stress at the wall pedge reaches a value of

pedge reaches a value of

(a) Design using pre-ACI 318-99(a) Design using pre-ACI 318-99φ 16 mm (5/8”) spaced 85 mm vertically..This reinforcement must be anchored in the boundary element10 φ 16 mm (5/8”)

φ12 mm (1/2”) spaced 340 mm horizontally

the boundary element.10 φ 16 mm (5/8 )

Confinement hoops φ 12 mm (1/2”) spaced at 100 mm vertically

300 mm




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(b) Using 21.9.6.2 of ACI 318-08(b) Using 21.9.6.2 of ACI 318-08According to this Section of ACI 318-08 zones in compression must be reinforced with special b d l t h

According to this Section of ACI 318-08 zones in compression must be reinforced with special b d l t hboundary elements when:

( 21-8)

where c in Eq. (21-8) corresponds to the largest

boundary elements when:

( 21-8)

where c in Eq. (21-8) corresponds to the largest neutral axis depth for the range of factored axial loads at moment strength consistent with the design displacement. Quotient in Eq. (21-8) must not be taken less than 0.007.

neutral axis depth for the range of factored axial loads at moment strength consistent with the design displacement. Quotient in Eq. (21-8) must not be taken less than 0.007.

(b) Using 21.9.6.2 of ACI 318-08(b) Using 21.9.6.2 of ACI 318-08First we compute the quotient:First we compute the quotient:

Since it is less than 0.007, we take this last value as the value for the quotient.Since it is less than 0.007, we take this last value as the value for the quotient.

Dimension w = 3 000 mm and c was obtained before as c = 396 mm.Dimension w = 3 000 mm and c was obtained before as c = 396 mm.

(b) Using 21.9.6.2 of ACI 318-08(b) Using 21.9.6.2 of ACI 318-08Applying Eq. (21-8) :Applying Eq. (21-8) :

Therefore, c = 396 mm < 714 mm, then no Therefore, c = 396 mm < 714 mm, then no boundary elements are required!boundary elements are required!

(b) Using 21.9.6.2 of ACI 318-08(b) Using 21.9.6.2 of ACI 318-08The other condition required is that when no boundary elements are needed, in Section 21 9 6 5 th t l ti f ti l i f t t

The other condition required is that when no boundary elements are needed, in Section 21 9 6 5 th t l ti f ti l i f t t21.9.6.5 the steel ratio of vertical reinforcement at the edge of wall is greater than confinement reinforcement with a maximum spacing of 200 mm must be provided.

Also requires that if no boundary elements are required and shear is greater than the

21.9.6.5 the steel ratio of vertical reinforcement at the edge of wall is greater than confinement reinforcement with a maximum spacing of 200 mm must be provided.

Also requires that if no boundary elements are required and shear is greater than therequired and shear is greater than the horizontal wall reinforcement must be anchored at the edge by hook or additional U shaped reinforcement.

required and shear is greater than the horizontal wall reinforcement must be anchored at the edge by hook or additional U shaped reinforcement.


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(b) Using 21.9.6.2 of ACI 318-08(b) Using 21.9.6.2 of ACI 318-08The vertical distributed steel ratio at the edge is 0.0025, and this steel ratio is less than:The vertical distributed steel ratio at the edge is 0.0025, and this steel ratio is less than:

No need for the reinforcement in 21.9.6.5(a).

Vu = 600 kN >

No need for the reinforcement in 21.9.6.5(a).

Vu = 600 kN >

therefore it is required for the distributed horizontal reinforcement ending at the edge to have a standard hook engaging the vertical reinforcement as required by 21.9.6.5(b).

therefore it is required for the distributed horizontal reinforcement ending at the edge to have a standard hook engaging the vertical reinforcement as required by 21.9.6.5(b).

(b) Using 21.9.6.2 of ACI 318-08(b) Using 21.9.6.2 of ACI 318-08φ 12.5 mm (1/2”) spaced at 340 mm vertically.This reinforcement must end in a standard hook at the wall edge

.

φ 12.5 mm (1/2”) spaced at350 mm horizontally

hook at the wall edge.

300 mm

.φ 12.5 mm (1/2”) spaced at 340 mm vertically.This reinforcement must end in a standard hook at the wall edge.

(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08Section 21.9.6.3 del ACI 318-08 prescribes the need to use boundary elements in anSection 21.9.6.3 del ACI 318-08 prescribes the need to use boundary elements in anthe need to use boundary elements in an similar way than the old pre-ACI 318-99 procedure.

The main difference is that under ACI 318-08 there is no need to resist all flexural

the need to use boundary elements in an similar way than the old pre-ACI 318-99 procedure.

The main difference is that under ACI 318-08 there is no need to resist all flexural08 there is no need to resist all flexural effect with just the boundary elements and the rest of the wall can contribute to strength for flexure.

08 there is no need to resist all flexural effect with just the boundary elements and the rest of the wall can contribute to strength for flexure.

(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08The size of the boundary element section is defined by geometric rather than t th i t

The size of the boundary element section is defined by geometric rather than t th i tstrength requirements.

The depth c of the neutral axis is needed here also. It was established before to be 396 mm.The horizontal extension of the boundary l t i th l f

strength requirements. The depth c of the neutral axis is needed here also. It was established before to be 396 mm.The horizontal extension of the boundary l t i th l felement is the larger of:element is the larger of:


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(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08A dimension of 200 mm complies with minimum 198 mm. There is no special requirement for the vertical reinforcement

A dimension of 200 mm complies with minimum 198 mm. There is no special requirement for the vertical reinforcementrequirement for the vertical reinforcement, so the same bars φ 12.5 mm (1/2”) diameter may be used.

The boundary element must comply with the requirement for transverse

requirement for the vertical reinforcement, so the same bars φ 12.5 mm (1/2”) diameter may be used.

The boundary element must comply with the requirement for transverse qreinforcement in special frame columns.

6 φ 12.5 mm (1/2”) diameter bars will be used with the hoop layout shown.

qreinforcement in special frame columns.

6 φ 12.5 mm (1/2”) diameter bars will be used with the hoop layout shown.

(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08

φ 12.5 mm (1/2”) spaced .

φ 12.5 mm (1/2”) spaced 350 mm horizontally

340 mm vertically6 φ 12.5 mm (1/2”)

φ 12.5 mm (1/2”) hoops 300 mm

200 mm.φ 12.5 mm (1/2”) spaced

340 mm vertically


Now we define the confinement hoops asNow we define the confinement hoops asNow we define the confinement hoops as required by ACI 318-08.Using a concrete cover of 40 mm the distance from wall edge to center of boundary element vertical bar is 40 + 12.5/2 = 46 mm. Then bc for analysis in the direction parallel to is 300 – 46 x 2 = 208

Now we define the confinement hoops as required by ACI 318-08.Using a concrete cover of 40 mm the distance from wall edge to center of boundary element vertical bar is 40 + 12.5/2 = 46 mm. Then bc for analysis in the direction parallel to is 300 – 46 x 2 = 208direction parallel to w is 300 – 46 x 2 = 208 mm and in the direction perpendicular to

w is 200 – 46 x 2 = 108 mm Ab for f 12.5 mm (1/2”) bars is 129 mm2.

direction parallel to w is 300 – 46 x 2 = 208 mm and in the direction perpendicular to

w is 200 – 46 x 2 = 108 mm Ab for f 12.5 mm (1/2”) bars is 129 mm2.

(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08Minimum hoop spacing in ACI 318-08:

21.6.4.3 — Spacing of transverse reinforcement along the

Minimum hoop spacing in ACI 318-08:

21.6.4.3 — Spacing of transverse reinforcement along the 6 3 Spac g o t a s e se e o ce e t a o g t elength lo of the member shall not exceed the smallest of (a), (b), and (c):

(a) One-quarter of the minimum member dimension;(b) Six times the diameter of the smallest longitudinalbar; and(c) so, as defined by Eq. (21-2)

6 3 Spac g o t a s e se e o ce e t a o g t elength lo of the member shall not exceed the smallest of (a), (b), and (c):

(a) One-quarter of the minimum member dimension;(b) Six times the diameter of the smallest longitudinalbar; and(c) so, as defined by Eq. (21-2)

(21-2)

The value of so shall not exceed 150 mm and need not be taken less than 100 mm.

(21-2)

The value of so shall not exceed 150 mm and need not be taken less than 100 mm.


Page13

(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08Now we obtain the vertical spacing of hoops in the boundary elements. First in th di ti f th h l ll l t

Now we obtain the vertical spacing of hoops in the boundary elements. First in th di ti f th h l ll l tthe direction of the hoop legs parallel to wall length w

Ash = 3 x 129 = 387 mm2

fyt = 420 MPa

the direction of the hoop legs parallel to wall length w

Ash = 3 x 129 = 387 mm2

fyt = 420 MPay

bc = 208 mm= 28 MPa

y

bc = 208 mm= 28 MPa


Now in the direction of the hoop legs perpendicular to the wall length:Ash = 2 x 129 = 258 mm2

fyt = 420 MPab = 108 mm

Now in the direction of the hoop legs perpendicular to the wall length:Ash = 2 x 129 = 258 mm2

fyt = 420 MPab = 108 mmbc = 108 mm

= 28 MPabc = 108 mm

= 28 MPa

(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08The smaller of the two is 310 mm. Now we check the limits of ACI 318-08:The smaller of the two is 310 mm. Now we check the limits of ACI 318-08:1/3 of the smallest dimention of the

element = 200/3 = 66 mm6db of longitudinal reinforcement = 6 x

12.5 = 75 mmso according to Eq. (21-2). This Eq.

1/3 of the smallest dimention of the element = 200/3 = 66 mm

6db of longitudinal reinforcement = 6 x 12.5 = 75 mm

so according to Eq. (21-2). This Eq. o g q ( ) qincreases over a minimum of 100 mm, it is not relevant in this case.

The smallest value is 66 mm.

o g q ( ) qincreases over a minimum of 100 mm, it is not relevant in this case.

The smallest value is 66 mm.

(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08This spacing is controlled by the minimum dimension of the boundary element (200

) A l di i h b

This spacing is controlled by the minimum dimension of the boundary element (200

) A l di i h bmm). A larger dimension may have been used, say 300 mm, and the spacing would be larger and equivalent to the one obtained for the pre-ACI 318-99 part of the example (100 mm).

mm). A larger dimension may have been used, say 300 mm, and the spacing would be larger and equivalent to the one obtained for the pre-ACI 318-99 part of the example (100 mm).

The vertical reinforcement may have been kept using φ 12.5 mm (1/2”) bars. The vertical reinforcement may have been kept using φ 12.5 mm (1/2”) bars.


Page14


Now we have to check that shear strength at moment strength is appropriateNow we have to check that shear strength at moment strength is appropriateat moment strength is appropriate.

The flexural strength of the wall is computed including the boundary elements vertical reinforcement. This is done for φ = 1 0 but without the strain

at moment strength is appropriate.

The flexural strength of the wall is computed including the boundary elements vertical reinforcement. This is done for φ = 1 0 but without the straindone for φ = 1.0 but without the strain-hardening requirement. done for φ = 1.0 but without the strain-hardening requirement.

Diagrama Momento Curvatura - Para Carga Axial e 1750 kN

3500

4000

4500M n = 4240 kN x m

M-φ diagram including boundary elements

1500

2000

2500

3000

3500

Mom

ento

(kN

x m

)

0

500

1000

0 0.000002 0.000004 0.000006 0.000008 0.00001 0.000012

Curvatura (1/mm)

(c) Using 21.9.6.3 of ACI 318-08(c) Using 21.9.6.3 of ACI 318-08The shear at flexural strength is obtained proportional to the one for the loads

i

The shear at flexural strength is obtained proportional to the one for the loads

igiven:

This value is less than the shear strength bt i d i iti ll f th i i

given:

This value is less than the shear strength bt i d i iti ll f th i iobtained initially for the minimum

reinforcement (1 272 kN < 1 478 kN) therefore no change in reinforcement is needed.

obtained initially for the minimum reinforcement (1 272 kN < 1 478 kN) therefore no change in reinforcement is needed.


N° 4 a 340 mm verticalmente.E f d b d l d l ú l

N° 4 a 350 mmhorizontalmente

Este refuerzo debe quedar anclado en el núcleo del elemento de borde 6 N° 4

Estribos de confinamiento de barra N° 4

espaciados a 50 mm

verticalmente

300 mm

200 mmN° 4 a 340 mm verticalmente.Este refuerzo debe quedar anclado en el núcleo del elemento de borde


Page15


φ 12.5 mm (1/2”) spaced .

φ 12.5 mm (1/2”) spaced 350 mm horizontally


φ 12.5 mm (1/2”) hoops

spaced vertically at 65 mm

300 mm

200 mm.φ 12.5 mm (1/2”) spaced

340 mm vertically

ComparisonComparison

Pre-ACI 318-99Pre-ACI 318-99φ12 mm (1/2”) spaced 340 mm horizontally

φ 16 mm (5/8”) spaced 85 mm vertically..This reinforcement must be anchored in the boundary element.10 φ 16 mm (5/8”)

Confinement hoops φ 12 mm (1/2”) spaced at 100 mm vertically

300 mm

ACI 318-08 - DisplacementACI 318-08 - Displacement

φ 12.5 mm (1/2”) spaced 340 mm vertically

.

vertically



φ 12.5 mm (1/2”) spaced at350 mm horizontally

φ 12.5 mm (1/2”) spaced at 340 mm vertically.This reinforcement must end in a standard hook at the wall edge.

.

300 mm

.φ 12.5 mm (1/2”) spaced at 340 mm vertically.This reinforcement must end in a standard

ACI 318-08 - StressACI 318-08 - Stressφ 12.5 mm (1/2”) spaced 350 mm horizontally


φ 12.5 mm (1/2”) hoops

spaced vertically at 65 mm

300 mm

200 mm.φ 12.5 mm (1/2”) spaced

340 mm vertically

This reinforcement must end in a standard hook at the wall edge.

The EndThe End