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Background Page 1 of 10

C OMPUTERS AND S TRUCTURES , INC ., BERKELEY , C ALIFORNIA S EPTEMBER 2010

CONCRETESHELLREINFORCEMENTDESIGN Technical Note

Design Information

BackgroundThe design of reinforcement for concrete shells in accordance with a prede-termined field of moments, as implemented in the software, is based on theprovisions in DD ENV 1992-1-1 1992 Eurocode 2: Design of Concrete Struc-tures.

Generally, slab elements are subjected to eight stress resultants. Using thesoftware terminology, those resultants are the three membrane force compo-nents f 11 , f 22 and f 12 ; the two flexural moment components m11 and m22 and thetwisting moment m12; and the two transverse shear force components V 13 andV 23 . For the purpose of design, the slab is conceived as comprising two outerlayers centered on the mid-planes of the outer reinforcement layers and anuncracked core this is sometimes called a "sandwich model." The covers of the sandwich model (i.e., the outer layers) are assumed to carry momentsand membrane forces, while the transverse shear forces are assigned to thecore, as shown in Figure 1. The design implementation in the software as-

sumes there are no diagonal cracks in the core. In such a case, a state of pure shear develops within the core, and hence the transverse shear force ata section has no effect on the in-plane forces in the sandwich covers. Thus,no transverse reinforcement needs to be provided, and the in-plane rein-forcement is not enhanced to account for transverse shear.

The following items summarize the procedure for concrete shell design, asimplemented in the software:

1. As shown in Figure 1, the slab is conceived as comprising two outer layerscentered on the mid-planes of the outer reinforcement layers.

2. The thickness of each layer is taken as equal to the lesser of the following:

Twice the cover measured to the center of the outer reinforcement.

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Concrete Shell Reinforcement Design Design Information

Background Page 2 of 10

Twice the distance from the center of the slab to the center of

outer reinforcement.

Figure 1: Statics of a Slab Element Sandwich Model

1

11111

d db f m +

min

max1212

d db f m +

TOP COVER

CORE

BOTTOM COVER

1

11111d

dt f m +

min

max1212

d dt f m +

2

22222

d dt f m +

2

22222

d db f m +

1

2

21 Ct Ct

21 d d

21 CbCb

1

11111

d db f m +

min

max1212

d db f m +

TOP COVER

CORE

BOTTOM COVER

1

11111d

dt f m +

min

max1212

d dt f m +

2

22222

d dt f m +

2

22222

d db f m +

1

2

21 Ct Ct

21 d d

21 CbCb

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Page 3 of 10

3. The six resultants, f 11 , f 22 , f 12 , m11 , m22 , and m12 , are resolved into pure mem-

brane forces N 11 , N 22 and N 12, calculated as acting respectively within thecentral plane of the top and bottom reinforcement layers. In transformingthe moments into forces, the lever arm is taken as the distance betweenthe outer reinforcement layers.

4. For each layer, the reinforcement forces NDes 1, NDes 2, concrete principalcompressive forces Fc 1, Fc 2, and concrete principal compressive stressesSc1 and Sc2, are calculated in accordance with the rules set forth in Euro-code 21992.

5. Reinforcement forces are converted to reinforcement areas per unit width Ast 1 and Ast 2 (i.e., reinforcement intensities) using appropriate steel stressand stress reduction factors.

Basic Equations for Transforming Stress Resultantsinto Equivalent Membrane ForcesFor a given concrete shell element, the variables h, Ct 1, Ct 2, Cb 1, and Cb 2, areconstant and are expected to be defined by the user in the area section prop-erties. If those parameters are found to be zero, a default value equal to 10percent of the thickness, h, of the concrete shell is used for each of the vari-ables. The following computations apply:

11 2Ct

hdt = ; 22 2

Ct h

dt = ; 11 2Cb

hdb = ; 22 2

Cbh

db =

111 CbCt hd = ; 222 CbCt hd = ;

d min = Minimum of d 1 and d 2

db max = Maximum of db1 and db2

dt max = Maximum of dt 1 and dt 2

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Equations for Design Forces and Corresponding Reinforcement Intensities Page 4 of 10

The six stress resultants obtained from the analysis are transformed into

equivalent membrane forces using the following transformation equations:

( ) 11 11 1111

topm f db

N d

+ = ; ( ) 11 11 111

1

botm f dt

N d

+ =

( ) 22 22 2222

topm f db

N d

+ = ; ( ) 22 22 222

2

botm f dt

N d

+ =

( ) 12 12 max12min

topm f db

N d

+ = ; ( ) 12 12 max12

min

botm f dt

N d

+ =

Equations for Design Forces and CorrespondingReinforcement IntensitiesFor each layer, the design forces in the two directions are obtained from theequivalent membrane forces using the following equations, in accordance withEurocode 2-1992. In the equations below, F 11 , F 22 , F 12 , NDes 1 , and NDes 2 aretemporary variables.

For reinforcement in top layer -

If ( )11 22top (top), N N then 11 11 (top);F N = 22 22 (top);F N = 12 12 (top)F N =

If ( )11 top N > 22 (top), N then 11 22 (top);F N = 22 11 (top);F N = 12 12 (top)F N =

If 11 12 ,F F then

12111 F F NDes +=

12222 F F NDes +=

( ) 12top 2Fc F =

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If 11F < 12 ,F then

01 = NDes

11

212

222 F F

F NDes +=

( ) [ ]2

1211

11

top 1F

Fc F F

= +

If ( )11 22top (top), N N then ( )1 1top NDes NDes= ; ( )2 2top NDes NDes=

If ( )11 22top (top), N N > then ( )1 2top NDes NDes= ; ( )2 1top NDes NDes=

For reinforcement in bottom layer -

If ( )11 22bot (bot), N N then 11 11 (bot);F N = 22 22 (top);F N = 12 12 (bot)F N =

If ( )11 bot N > 22 (bot) N then 11 22 (bot);F N = 22 11 (bot);F N = 12 12 (bot)F N =

If 11 12 ,F F then

12111 F F NDes +=

12222 F F NDes +=

( ) 122 F bot Fc =

If 1F < 12 ,F then

01 = NDes

11

212

222 F F

F NDes +=

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( ) [ ]

212

11 11

bot 1F

Fc F F

= +

If ( )11 22bot (bot), N N then ( )1 1bot NDes NDes= ; ( )2 2bot NDes NDes=

If ( )11 22bot (bot), N N > then ( )1 2bot NDes NDes= ; ( )2 1bot NDes NDes=

Following restrictions apply if NDes 1 or NDes 2 is less than zero:

If ( )1 top 0, NDes < then ( )1 top 0 NDes =

If ( )2 top 0, NDes < then ( )2 top 0 NDes =

If ( )1 bot 0, NDes < then ( )1 bot 0 NDes =

If ( )2 bot 0, NDes < then ( )2 bot 0 NDes =

The design forces calculated using the preceding equations are converted intoreinforcement intensities (i.e., rebar area per unit width) using appropriatesteel stress from the concrete material property assigned to the shell elementand the stress reduction factor, s . The stress reduction factor is assumed toalways be equal to 0.9. The following equations are used:

( )( )1

1top

top0.9( ) y

NDes Ast

f = ; ( )

( )11

botbot

0.9( ) y

NDes Ast

f =

( )( )2

2top

top0.9( ) y

NDes Ast

f = ; ( )

( )22

botbot

0.9( ) y

NDes Ast

f =

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Principal Compressive Forces and Stresses in Shell Elements Page 7 of 10

Principal Compressive Forces and Stresses in ShellElementsThe principal concrete compressive forces obtained earlier are used to com-pute the principal compressive stresses to top and bottom layers as follows:

( )( )

1 2

toptop

2 Minimum( , )

FcSc

Ct Ct = ;

( )( )

1 2

botbot

2 Minimum( , )

FcSc

Cb Cb=

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Notations Page 8 of 10

NotationsThe algorithms used in the design of reinforcement for concrete shells are ex-pressed using the following variables:

Ast 1(bot) Reinforcement intensity required in the bottom layer in localdirection 1

Ast 1(top) Reinforcement intensity required in the top layer in local direc-tion 1

Ast 2(bot) Reinforcement intensity required in the bottom layer in localdirection 2

Ast 2(top) Reinforcement intensity required in the top layer in local direc-tion 2

Cb 1 Distance from the bottom of section to the centroid of the bot-tom steel parallel to direction 1

Cb 2 Distance from the bottom of the section to the centroid of thebottom steel parallel to direction 2

Ct 1 Distance from the top of the section to the centroid of the topsteel parallel to direction 1

Ct 2 Distance from the top of the section to the centroid of the topsteel parallel to direction 2

d 1 Lever arm for forces in direction 1

d 2 Lever arm for forces in direction 2

db 1 Distance from the centroid of the bottom steel parallel to direc-tion 1 to the middle surface of the section

db 2 Distance from the centroid of the bottom steel parallel to direc-tion 2 to the middle surface of the section

dbmax Maximum of db 1 and db 2

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Notations Page 9 of 10

d min Minimum of d 1 and d 2

dt 1 Distance from the centroid of the top steel parallel to direction1 to the middle surface of the section

dt 2 Distance from the centroid of the top steel parallel to direction2 to the middle surface of the section

dt max Maximum of dt 1 and dt 2

f 11 Membrane direct force in local direction 1

f 12 Membrane in-plane shear forces

f 22 Membrane direct force in local direction 2

Fc( bot ) Principal compressive force in the bottom layer

Fc( top ) Principal compressive force in the top layer

f y Yield stress for the reinforcement

h Thickness of the concrete shell element

m11 Plate bending moment in local direction 1

m12 Plate twisting moment

m22 Plate bending moment in local direction 2

N 11(bot) Equivalent membrane force in the bottom layer in local direc-tion 1

N 11(top) Equivalent membrane force in the top layer in local direction 1

N 12(bot) Equivalent in-plane shear in the bottom layer

N 12(top) Equivalent in-plane shear in the top layer

N 22(bot) Equivalent membrane force in the bottom layer in local direc-tion 2

N 22(top) Equivalent membrane force in the top layer in local direction 2

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References Page 10 of 10

NDes 1(top) Design force in the top layer in local direction 1

NDes 2(top) Design force in the top layer in local direction 2

NDes 1(bot) Design force in the bottom layer in local direction 1

NDes 2(bot) Design force in the bottom layer in local direction 2

Sc(bot) Principal compressive stress in the bottom layer

Sc(top) Principal compressive stress in the top layer

s Stress reduction factor

ReferencesDD ENV 1992-1-1: 1992 Eurocode 2: Design of Concrete Structures, Part 1.

General rules and rules for buildings

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