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DCP NRC 002728December 30, 2009

ENCLOSURE 1

"In-Plane Behavior of Concrete Filled Steel (CFS) Elements" Presentation

0381 Ijb.doc 12/30/2009 12:15 I'M

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IN-PLANE BEHAVIOR OF

CONCRETE FILLED STEEL (CFS) F

ELEMENTS(by Amit H. Varma and Sanjeev R. Malushte !4

d, Purdue University and Bechtel Power

I r IIIj

COPYRIGHT

o This source material is copyrighted by Prof. Amit H.Varma, School of Civil Engineering, PurdueUniversity, West Lafayette, IN 47906

o It is excerpted from the report:

o Varma, A.H., and Malushte, S. (2009). "In PlaneBehavior, Analysis, and Design of Concrete FilledSteel Wall Panels." Bowen Laboratory ResearchReport No. 2009- 03, School of Civil Engineering,Purdue University, West Lafayette, IN 27906, under Kreview by sponsor.

© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906iiid

RESULTS FROM LEVEL 2 ANALYSIS

" The level 2 analysis of the structure will result inforce and deformation demands on the components. o

" In the spirit of LRFD, the level 2 analysis will beelastic that accounts for the effects of: (a) concretecracking, (b) slip between the steel plates andconcrete infill, and (c) connection flexibility.

" The results from the analysis will include for thecylindrical shield building the following demands:o In plane Force demands (Sx, Sy, and Sxy)

o Moment (Mx, My, and Mxy)

o Out of Plane shear (Vxz and Vyz)

© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906- - -iLE J

IN PLANE BEHAVIOR" The effects of these demands on the behavior of the

steel-concrete composite (SC) structure wasinvestigated

" To begin, lets focus on the behavior for in-plane forcesSx, Sy, and Sxy.

o These are forces per unit length typically in kips/ft

" Just to give a sense of quantityo The pure tension strength of the AP1000 design = 0.75 x 2 x

12 x 50 ksi = 900 kips/fto The pure compression strength (squash load) - 0.75 x 2 x 12

x 50 ksi + 34.5 x 12 x 6 ksi 3384 kips/ft

© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906

IN PLANE BEHAVIOR

o Consider a concrete filled steel (CFS) element takenfrom the SC structure

o It is subjected to in-planeunit length (in.).

(a) isometric view

forces Sx, Sy, and Sxy per

(b) plan view

SvSysteel plate

-I ..Sx

Sxy

,Sxy

Sx

Sxy

SxSx

SxySy

Nssteel plate 05

--------- middle plane

(0C Amit H Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906

IN PLANE BEHAVIOR

o These in-plane forces will cause some deformations inthe element. These deformations can be denoted by

t'z

the strain terms: EX, Eyý xy

(a) isometric view (b) x-y strains

Systeel plate•A

xv

Sx

Sxy

1SxY

Sx

y.fl

Sys l

steel plate

--------- middle plane

©c Amit H. Vnrma , School of Civil Engineering, Purdue lUnivprsityv, West Lafayette, IN 47906

ONE ASSUMPTION

o Let us assume that the strains E x, E y, Yxy are compatiblebetween the steel plate and the concrete infill.

o We will revisit this assumption later.

o The AP 1000 design will have to demonstrate that theshear stud spacing (8.5in.) is adequate to achieve this asan engineering approximation for the load cases. L

41!1JIt © Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906

I

Two FACTSo If 6 is the principal direction

o Stress Transformation relates the principal stresses towthe x-y stresses,

[T]a -211 + cos(20p)

1- cos(20p)

--sin(2Op)

1 - cos(20p)

1 + cos(20p)

sin(2Op)

2 sin(2Op)

-2 sin(2Op)

2cos(2Op)

o Strain Transformation relates the principal strains tothe x-y strains

[T]~12•

1 + cos(2&p)

1- cos(206)

C-'2 sin(260

1- cos(20p) sin(20-)

1 + cos(2Op) -in(20

2sin(26 2cos(2Op)0

© Amit H. Varma, School of Civil Engineering Purdue University, West Lafayette, IN 47906

PRINCIPAL IN-PLANE FORCES SPi AND SP2

o The difference between the two transformationsis due to the definition of engineering shearstrain r xy which is twice the shear strain E xy

0 F V

ii

II

Ii[1,


PRINCIPAL IN-PLANE FORCES (SPI, SP2)

o The in-plane forces (Sx, Sy, and Sxy) can be used tocompute the principal direction (0 p) and principalCmembrane forces Spl and Sp2 per unit length usingEquations belowy:

Sp2

Sxy P

. I OP 2S,."sx Sx i tan(20,,)- . ,

Sp2 X - 1

Sy

S(1 + cos(20p) (I - cos(2O,) SY +sin(20 SSP -2 2 2

(A - cos(201 ,) (1 + cos(20,)S =. . . SX + ...g S. - sin(20 tSN4o6©c Amit H./Varma , School'f Civil Engine~erin'g, Purd~t. University, Y-West LafayettiW, _ 0 0

V

CONCRETE CRACKING

" Before concrete cracking, the principal forces areresisted by the composite section (steel and concrete)

" Concrete cracking occurs after the principal membrane§force (S 1 or S 2) exceeds S,

4 (f shEc) Tc + E 2t• kips/in.Sc -(1000 Ecs

" Set can be reduced to account for the effects of concreteshrinkage tensile strain (E sh)

" This point is variable because the tensile strength ofconcrete can be highly variable and shrinkage effectsare not easy to characterize.

" Cracking can occur due to one or both principal force,The cracks will be oriented in the plane perpendicular-t~~"n forcegr eet

© _, versity, ' est, IN 47906

POST-CRACKING BEHAVIOR

" Post-cracking the concrete will offer very littlestiffness and stress capacity in the principal directionperpendicular to the plane of cracking.

" However, it will have stiffness and strength in theprincipal direction parallel to the plane of cracking.This can be referred as cracked orthotropic behavior

" Japanese researchers recommend that the stiffness inthe direction parallel to the plane of cracking can beassumed to be 70% of the uncracked stiffness (Ec).

" Lets assume the stiffness and strength perpendicularto the direction of cracking are assumed to be zero,which is a conservative assumption.

© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, !N 47906ii

PRINCIPAL STRAINS

o Assume that the principal strains (E.pl and Fp2) are inthe same direction as the principal forces (Spz and Sp2).

o They are related to the x-y strain ( E, y-, and rx) bythe strain transformation rules. In these equations 0p isthe principal direction computed earlier.

11

I•

Ep2 LJ]E

0

yxYX 1

1 + cos(20p)

1- cos(2Op)

-2 sin(2OP )

1- cos(2Op)

1 + cos(2Op)

2 sin(2Op)

sin(2Op)

-sin(2Op)

2cos(2Op)


CRACKED CONCRETE STRESS-STRAINRELATIONSHIP

o The concrete infill will have principal stresses (, c-p1 and c, p)

corresponding to the principal strains (,Fpl and 6p). Theprincipal stresses will be related to the strains via the reducedelastic modulus (Ec)

(a) principal strains (b) principal stresses

E2 , (T2 CU' 0~p or10"

E, UpI. ,,Cp2 = ,EL 0 Oorl 0 Ep2

ep Ip EP J{2) },lgi o o o0

e,,/ 0

© Amit H. Varma. School of Civil Engineering, Purdue University, West Lafayette, IN 47906

CRACKED CONCRETE STRESS-STRAINRELATIONSHIP

o The concrete principal stresses (, O-pl and c cp 2 ) can betransformed back to stresses in the x-y directions

.0y, and C-;as shownUC

) C

(Y' V

a.C 49pl

\C Up 2

1 + cos(20,) 1 - cos(26p)C ax" CrYP 1 [T]cos(20p) I + cos(20p)

20 - sin(20p) sin(20p)

© Amit H. Varma, School of Civil Engineering, Purdue 1Iniver.ity, West Lafayette, IN 47906

2 sin(2Op)

-2 sin(20p)

2 cos(2Op)

CRACKED CONCRETE STRESS-STRAINRELATIONSHIP.

o Collecting all the terms that we have developed so farwe get the stress-strain relationship for crackedconcrete: I r

L'w9

Cz

C a x x

C ('Y = [K]c EYCT YX),

[K] = '

0 or Ec

0

0

0 0

0 or E,

0

[T']E0

0

sin(20p) 1-sin(20p)

2 cos(20P )

L! iL

[K]. -

I + cos(20p)

1 - cos(20,,)

--sin(20p)

I - cos(2Op)

I + cos(20p)

sin(20p)

2 sin(2Op)

-2 sin(20p)

2cos(20,) I.[O or 1

0

0

0

0 or 1

0

0

0

0

I + cos(2Op)

1 - cos(2Op)

-2 sin(2Op)

1- cos(2Op)

1. + cos(2Op)

2 sin(20p)

I

© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafavette, IN 47906

STEEL PLATE STRESS-STRAINRELATIONSHIP

o The steel plates will have plane stress behavior, and thecorresponding elastic isotropic behavior can be used torelate the steel plate membrane stresses (, ox, s uy, and

sz7) to the composite section strains ( E x E y, and Yxy)

S X

STEXy

I VV I

00

[ K]s -v

Eu

Eyx0 0 1-v2


FORCE EQUILIBRIUM OF COMPOSITESECTIONo Static force equilibrium diagram for the CFS composite panel /

element subjected to membrane forces (Sx, Sy, and Sxy) per unitIp n xfth. (in -)

cyt~T 4 ý

Sx

Sxy

0F 00poit

~sxy

S x

,. o y'V V

c'17

A07 )r

C- V t,.

VT.

S T y '

rF- ý' ( QtS T t S IS, t

Wc'4 afiyte IN 47906© Amit H. Varnma, School of Civil Engineering, Purh{* University,

FORCE EQUILIBRIUM OF COMPOSITE SECTION

saox C 0So Say x2t + c xy T

StY S y XY

SS c co~b

Sy - 2ts [K]s •y+ [K]c cy

Yxy Sxry

o For the applied Sx, Sy, and.Sxy, solve the equation© AabOmaetmcgetf c& iIngpen•[•Mn duef*,rsity, West Lafayette, IN 47906 i. i.

d LL

FORCE EQUILIBRIUM OF COMPOSITESECTION

o After obtaining the strains, calculate the steel platesstresses and concrete stresses using the followingequations:

0•

scx

sGy -[K]sS•X yxy

Oorl 0 0

E 0Oorl 0

0 0 0

sap 2 -[T],.

0

S x~

s'VS

S

d

7

C UPI

c ap2

0

[T]eyX

E'y


DESIGN CHECKS AND LOAD RATIOS

o Establish yielding of the steel plates. Calculate the VonMises stress caused by the loading

UVM SOUp 1 sOp2 s Tpl * s ap2

o The load ratio to cause yielding = Fy /OuvM

o Since the concrete was assumed to remain elastic, checkthat the minimum principal stress (compressive) is still inthe elastic range, i.e.,

cyp2,p 1 0.70fc:

o This entire process can be automated easily to determinethe behavior of CFS panels subjected to a variety ofin-plane membrane force combinations

I,9

0,


final-interaction.html IF -- '6/28/0

ts := 0.3875

T..:= 30

Es := 29000

v:= 0.3

fc := 6000

Material Properties Section

Fy:= 50

Ec := 57 . 60000.5 --- 4415.2010146764552491 CDcc

Unitload:= ts

Sxy:= Unitload

Sx:= 0.0000000001

Sy:= 0

Applied Loads

:=2 -Sxy - 7.75e9 Fixed crack orientationSx-ly)

0:= - atan() - 0.785398163332932180582

SXpl :-(Sx Sy) + ' + Sxy2 -- 0.387500000052 [2

Sxp2 (Sx: + Sy) -lY) 2 +S 2 -0.387499999952 J- 2 j

La:= 10 if Sxpl>0

1 otherwise

b:= 10 if Sxp2>0

1 otherwise

© Amit HfYTr a, 01-W of Civil Engineering, Purdue University, West Lafayette, IN 47906

Craciconcrete 0.[ b0 0I _3> f 1/ Concrete contribution in principal directions0ou 0) o oo)

final-interaction.html

U L -J

E

t s x1Strains := Totstiffess- I . Sy

Sxy

(0.00001341364243137941753194

0.00001341364242623966630965

0.00003983307197549749994407)

Strains from Force Equilibrium

0.555708043536580445407717718

Steelstress := - . Strains -- 0.555708043421924456602857018ts 0.444291956649779807062344912

(1.00000000012903225307151014Steeltxans Stresstrans Steelstress -+ 0.1114160363294726439391346661 1

K6.98147263681860780342587348e-21

C-0.0143557911130233281729035635Ccncretest~iffe ssK

Concretestress Strains -- -0.0143557911217330484621370571TT0.01435579111988063331736342772

-3.2717216691711313940852610208e-23

Concprinstress Stresstrans Concretestress --- j -0.023711582239761376635007908383

\-1.3311083432742040945524135676e-23

Steel Stresses

Steel Principal Stresses

Concrete Stresses

Concrete Principal Stresses

N.

Co

Steeleffstress j(Steeltans0 )2 + (Steeltrans1 )2 - Steeltranso0 Steeltrans -- 0.94920••5•90230090630320192014Calculating the von-mises stress

Load ratio to failure hy yieldingLoadrratio := - 52.6754460107144273242333455973

Steeleffstress

JLI© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906

PURE SHEAR BEHAVIOR

o For pure in-plane shear loading, the appliedmembrane forces Sx and S will be equal to zero, and Q0

the in-plane shear loading will be equal to Sxy.a The principal direction ( 80) =tan-1 (inf.) = 450

0 The principal forces will be equal to Sxy inmagnitude; SP, will be in tensile and S will becompressive.

o Concrete cracking will occur when the applied in-plane shear (Sxy) exceeds the tensile crackingstrength (Set) of the composite CFS section

4 f'Sc, =(41 hE,) T, + E 2ts X 12 kipsift

0 Amit H. Varma, School of Civl IEngineering, Purdue University, West Lafayette, !N 47906

PURE SHEAR BEHAVIOR

" The in-plane shear stiffness before the concrete.Ncracking limit state can be calculated as:

Kcr = GsAs + GcAc = E t ECC kips/in2(1 + vs) 2(1 + vc)

" The concrete cracking planes will be perpendicular todirection of the principal stress Sf1, i.e., at 1350.

" The principal direction Op will be equal to 450.Substituting this value in the equation for [K],

© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, !N 47906

PURE SHEAR BEHAVIOR

F 7-

sin(% )

- sin(20p)E"4[I + cos(2Op)

- cos(2Op)- sin(2O0,)

1 - cos(20p)

I + cos(20p)

2 sin(20p)

-2sin(20p) 1-Oorl 0 0

0 Oorl 0

0 .0 0

1 + cos(20p)

1 - cos(20p)

-2 sin(20)

1 - cos(2Op)

1 + cos(2Op)

2sin(20,,)sin(20p) 2 cos(20p)

o Substituting 0 p=450 The concrete [K], becomes

2cos(20p)

F]

[K]cc'

4

11

11

-1-11

cory =[K]cLx

EY

Try-1 -1

© Amit H, Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906

PURE SHEAR BEHAVIOR

o Force Equilibrium for Pure Shear 0.

00

S -9

2Est

1- v 2

I VV I

00

+ T 1-•-•_11

11 11j

Lx

Ey

Y-r0 0 1-v2

o Solve simultaneous equations to get the strainsand calculate stresses etc. for pure shear case

• Sxy Ka2ts Ka+ K0, Ox~ •,, -c• K, K7

1Ka = Gs2ts ,P 4 +2(1-v)

ivil Engineering, Purdue UniverquR W.est L,•2qtstte, 'N 47906ii [Sii© Amit H. Varma, School of

PURE SHEAR BEHAVIOR

" Use the stresses to calculate von-mises stress andestablish the yielding of the steel plate.

" The yield load corresponds to:

v Ka+K (2tF)

x

CU" The shear stiffness post-CCD

cracking corresponds to; K) +

SXY = (Ka + Kp) yx.y-

© Amit H. Varma, School of Civil Engineering, Purdue University, West LafaytaTAT AeQan6 Deformation Y.,*

PURE SHEAR BEHAVIOR - PARAMETRIC STUDY

Effect of Concrete Thickness on In-plane Shear Strength and Stresses

Assumed 0.5 in. steel plate thickness with Fy = 50 ksi

____ "9y __'._ -lay- __T C a ,Concrete Ka KP AsF,

Tr (in.) (kips-ft/ft) (kips-ft/ft) s y y Ff f

18 133846 99852 0.93 0.40 0.53 0.24 0.48

20 133846 106206 0.94 0.42 0.52 0.23 0.45

22 133846 112040 0.95 0.44 0.52 0.21 0.43

24 133846 117414 0.97 0.45 0.52 0.20 0.41

26 133846 122381 0.98 0.47 0.51 0.19 0.39

28 133846 126986 0.99 0.48 0.51 0.19 0.37

30 133846 131266 1.00 0.49 0.50 0.18 0.36

32 133846 135256 1.00 0.50 0.50 0.17 0.34

o0,

34 133846 138983 1.01 0.51 0.50 0.16 0.334 4 I 4 4 4 4

36 133846 142472 1.02 0.52 0.49 0.16 0.32


COMPARISON WITH EXPERIMENTALRESULTS

o Japanese Tests by Ozaki et al. (2004)Aval•oble onlonem o wwwscencedh'ecr,cor Nuclear

SC19eNCE CkltCTO

IR'° Engineenng7 . .. . .and DesignELSEVI ER Nuclear Ewtgimecting* ad Design 22A k0l(-) 225-244 Design

Study on steel plate reinforced concrete panelssubjected to cyclic in-plane shear

Masahiko Ozaki a, Shodo Akita b, Hiroshi Osuga *,

Tatsuo Nakayamad, Naoyuki AdachicOffiCe #fN Uin•rt erq derwd Arhiee.•re HeMd*ani;m Te, Katai Eteerfiter Cr. ieC.. O.Meaka, Japant

-,iture.a Facilhies Di••siont, Otxvas, CS fA • ,i ta-K, Taklo, J.h,.r.

Received 19 Febwarv 2002; received in revised form 8 June 2003- accepted 10 June 2003

COMPARISON WITH EXPERIMENTAL RESULTS

o Japanese tests of Ozaki et al. (2004)'Table 2Test specimen (research program 1)

-7

Specimens Surface steelplate (1) (mm)

Headed stud bolt Nodal force(MPa)

ND

Pitch in welding (il) (mm) Diameters (mmm) H11

S2-ONN' 763 430 -60.0 I~S2 KSN -- L47$2-30NN• 2.94

S3.3ONN 2.94 -

S3-00PS 01.0 Studs were weldedS3-00PN Without studs

. .. . . 3 . . . . . ..13. . . . . 7

Table 4-...."'.-- .- ... ..-.. ............................... ... -..-- .- ,- . ..-- - - - -.... ....... .............--.......- ' ......--- -- ................................ ............... .. .................. -. -

Experimental results

Specimen Steel Concrete Elastic shear ost-or,-c king Cracking Yield strength Mmximurt stnngthmWdulus sgemny modulus streng__ __

Yield stress, A.,. x A,, Compressive stringth. G, QG. y, ., (ys) y, (x10-3) Q. (kN) y", (Yio",)Yuung's (cm;) tangential stiffness (x I W MPa) (x 103 MPa) (kN) (x l10')modulus (MP;" (MI'

*FS2-OONN 340 (1.97xl06); 53.5 (17.1) 42.2 (2.72 x 10ý') 12.4 4.16 293 0.115 2290 (-2110) 2.50 (-1.99) 29&60 (-2780) 9.41 (-6.12)"S2-5NN... . .. .41-6 (277 k10-) 13.2. ....... 44--:--.-- -- 433.... 0133- .... 2330 (-2290 2.71'2.21)• 10 -- 2931) -1)0'(-6.2)S2-30NN 42.0 (2.79 x 101) 16.4 3.69 542 0.168 2491) (-25710) 3.01 (-2.41) 3110 (-3200)) 10.48 (-6.03)

S3ON S(L"9X 10') 75.4(16.9) 41.9 (2.71 x 10") 12.9 4A8 311 0.134 -351)0 (-3070) 3.01 (-2.W1) 3610 (.-34-30) 6.05(-.3SS3.1SNN~ .................................. . • E-4r6-2.67-×-0'• ..... 13:1 .......... 429 " ............. 384 .... 0.141 ....... 3E13(-(=3t20 .. 2L99 (•.3.01V[) 3760 ('=3330) ~"4•"¢9 f<.6.01IY

S3-30NN 40.1 (2.70x 10') 11.9 4.67 385 0.186 3171) (-3080) 2.80 (-2.96) 3730 (-3551)) 5.57 (-5.635S3-0OPS 75.4 (25.4) 41.9 (2.71 K 104) 13.1 5.81 350 0.141 2680 (-2640) 1.93 (-1.97) 3580 (-3220) 10.87 (f-5.98)S3-OOPN 39.9 (2.72 x 104) 16.4 4.92 271 0.113 2350 (-2390) 2.01 (-2.03) 3510 (-30605) 17T0 (-6.02)


SZ-OONp

SZ-I NN

SZ3m

S3-15NN

S3-3ONN

S4-OOdN

-4

4000

3,00

SZ-OOff#4

40 4St

5s 55 so 65Crackngmii( M&

V: -O wbor Sr-otO•Y" WIt j

o) Maamn0a Ii

Fig. 7. Cracking angles.

S00S3 OONN $4 000

I xrk r mijilat4000 a..

34o10 .1 ,I V

I0

//I100

LWxff7mA-7

to

I

'2Of 0'

0 IsD5 10 is

I


Yield Force,_II111HII . ,. , . , ,, , ,.> , ,.

3000

0(

-77S.......... .. .........

0LO

II

0

7nn

0OD 400 600 800 1I000CacIM1i419 QcekN)

1'0Q00 2000 1000 40,00 R(Calculate~d "ýk]LN

00

tuo .......... .(3.'•............ I ........... •• ....... 7+b150..443

0 0

0 510 10t 150 200Crlaclkted G StX iffne

Pre-Cracking Stiffness

d

x

1001 . _ '- • '-- -• + "•

40 7

r i + ,,,

0 ,____ _____ i i0 20 40 60 go 100

Post-Cracking Stiffness

Ii


o Tests conducted by Lee, Choi, Hong, and Lee in KoreaThe 5' International Symposium on Steel Structures

March 12-14, 2009, Seoul, Korea0

In-Plane Shear Behavior of Composite Steel Concrete Walls

Myung-Jae [email protected]

Department ofArchitecture, ChungAng University, Seoul, Korea

Byung-Jung Choi

Department of Architecture, Gyungi University, Seoul, Korea

Sung-Gul Hong and Eo-Jin Lee§glhon.•[email protected]

Department of Architecture, Seoul National University, Seoul, Korea

~~~~1

Designation Thickness Diameter of Spacing Amal Force Axial

(mm) Stud (amm) Ratio ForceI (,p-rm•) N% (kN)S30/400 F3.2

"lTAA'l 10 0

S30/400 F3.2 0 0NO0-2 3.2 6-57 100N8U14UU kJ.2N08

6

H~


o Setup used by these researchers tries to simulate a large panel zonein beam-to-column cru

o The setup worked ok.

iciform

First specimen failed due to welding failurewCDQ0o

. .. . - - - --

-... -: ,,, --- --- ---I:J)

II)-c(I)

i* 'p S I 1

...... i .. .. .. . . . . . . . .. . . . . . .

* I I* I I I

* I I I

* I

-- M./40 FS .2 MOS

...... 3ffi1400 F-5.2 Nac0-2

3•J400 F- .2 NO0-1

Shear angle( -x 10- Rad )

Figure 3. Shear Behavior of SC walls without stiffenerFigure 1. Loading Set-up for Pure Shear

P•


iotograph showing 45 deg. cracks in concrete for the pure shear specimen

bF

Figure 5. Diagonal Cracks in Con~te

COMPARISON WITH EXPERIMENTAL

RESULTS

Table 3. Strength of Non-Stiffened Steel Concrete panel

Max MaterialUltimate Shear Yield Stregth

Specimen Shear Angle Strength Specification s(Q kN) (rad (kN) (kN) steel conc.

x 10-3) ...... (MPa) (MPa)S30/400

F3.2NOO-1

2019 4.39 1601 2233 297 36

S30/400F3.2 257 6.46 1601 2233 297 36Noo-2

F3.2 N082900 13.6 1601 2262 297 36

SAs Fy- 1200 mm x 3.2 mm x 2 x 297 N/mm 2 2 N

IN-PLANE BEHAVIOR

" Used the Mathcad program to analyze the CFS NDelement for different combinations of membraneforce (tension or compression) and shear force.

" The limit state was always yielding of the steel plate,and the concrete stresses were limited to the elasticrange (less than 0.70 e)

" The last point on the compression + shear portion ofthe curve had concrete stress equal to 0.70 fc

© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906di

Combined Membrane Forces (Sy and Sxy)

1.6Compression +'Shear

0E-

I-

'.

w

a_C,'U

Corea

ncrete stress 1.2

Lched 0.7f,

Tension.... ... ~ 0.6 ...

eel yielding governs iailurepoai~rnglfr~Liniits !

csthak 0 7f 0.2nerete stress- less ta0 f

0O

+ Shear

St

Cc

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5

Axial Force / As Fy (Positive Tension, Negative Compression)

© Amit H. Varma , School of Civil Enoxinpering, Pirdue University, West. Lafavptte, IN 47906

IN-PLANE BEHAVIOR WITH SY AND SxY

o Combined effects of axial force (Sy) and in-plane forceSxy.

o Results obtained using Mathcad sheet.o Assumptions - Concrete remains elastic (checked by

making sure that concrete principal stress is less than0.7fc).Concrete has no tension strength

o Note the change in shear strength with axial tensionand axial compression.

o Axial compression increases the in-plane shearstrength.

o Need more sophisticated analysis to investigate thebehavior of CFS panels elements to membrane forces".

© Amit H. Varmna , School of Civil Engineering, Purdue University, West Lafayette, IN 47906--

LIMITATIONS OF MATHCAD APPROACH

" The concrete material model was elastic with nocompression yielding possible. This limitationbecomes significant for cases with largercompression + shear.

" The mathcad program assumes zero tensionstiffness and strength. This is conservativeassumption but needs refinement

a Theory is not verified using independent finiteelement analyses

© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906I

NONLINEAR FINITE ELEMENT ANALYSISOF CFS PANELS TO IN-PLANE FORCES

" Three different types of modelso (1) Shell Solid (SSo) Model

o (2) Shell Shell (SSh) Modelo (3) Layered Composite (LCS) Shell Model

" Three different concrete material modelsS(1) Smeared cracking - linear Drucker-Prager -

associated flow rule concrete modelo (2) Concrete damaged plasticity model - smeared

cracking - parabolic Drucker - Prager - non-associated flow rule - and damaged elasticity

o (3) Linear elastic orthotropic model for crackedconcrete


Fr

C

p1!'IU

INPUT PARAMETERS

o Steel stress-strain curveN.

ND

C4-

.C

ho

C

w

100

90

80

70

60

50

40

30

20

10

0

U,

InU,a,1..4.

a,

1~I-

100

90

80

70

60

50

40

30

20

10

0

r-- - --

m,---

+ E) "eI r = •(1 + e) -

4

i

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

Engineering Strain (in./in.)

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

True Plastic Strain (in./in.)

0d© Amit H. Varma , School of Civil Engineering, Purdue University, West Lafayette, !N 47906

INPUT PARAMETERS

o Concrete compression and Tension behavior

U)

0

CL

E0a)

6.0

5.0

4.0

3.0

2.0

1.0

0.0

350

300

" 250

200I_

150

100a)

Tension stiffening stress-strainusing Chang and Mander (1987)

Tension stress-strai,,nfrom fracture energy0u

0 'U

0 0.001 0.002 0.003 0.004 0.005

Compressive Strain (in/in)

G

Tensile Strengtho

0.0000 0.0002 0.0004 0.0006 0.0008 0.0010

Tensile Strain (in./in.)

u

Frastuu.= 2G 1/cYio

IlL

n) Amit H. Vqrmrq School of Civil Engineerinpg Piirdiipe University ing displacementu,

R FF-

INPUT PARAMETERS

o Concrete Smeared Cracking Modelc e s

'crack detection' surface Q

0

4a)

CO)

Failure Surfaces in p-q space

'crack detection' surface biaxialtension

1*1-compression' surface

I p

hydrostatic pressure axis0

ii© Amrit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906

INPUT PARAMETERS

o Concrete Damaged Plasticity

-S 2 -S" I

,Kc= 1biaxialtension

OL(&- 3a • )= o

Figure 4.5.2-5 Yield surface in plane stress.

© Anit H. Varma, School of Civl! Engineering, Purdue University, West Lafayette, IN 47906Ii7 Jid I . .. I

NONLINEAR FINITE ELEMENT ANALYSIS(a) (b) (c) I--

SteP: Step-1 Step: Step-1 SteP: St p-iIncrement 100: 1AAc 10.00 !Increment 100. Arc M 000 I1ncremen.t 100. Arc =10,00

I

I

(d) (e) (f)

Figure 5-7. Details of SSo model:(a) Model with mesh and boundary conditions.(b) Steel plates with shell (S4R) elements, fully tied to concrete elements(c) Concrete infill modeled using solid (C3D8R) elements(d) Loading applied as uniformly distributed shear traction on the concrete element surfaces(f.•teel retprf tied toP Uioncd tU st.. .

I

NONLINEAR FINITE ELEMENT ANALYSIS(a) (b) (c)

V1VVJlJO0: 1 ODH: C: l(Itill -,;0 SSO StIlVdi f2d S44111.0(th Viewliml: 2 ODD: C: Teitip A) SSo smemed stiim.odli ViiAupoit: 3 ODH: C: I emp 3D SSo striedi ed sliain.o(lb

8, Mises

SNEG, (fraction 0 -10)(Avg: 75%)

+6 521e+01+6 520e+01+6.517e+01+6 516e+01S+6 515e+01

S, Min. Principal(Avgt 75%)

-2.346e+00-2.354e+00-2.362*+(00-2.371e+00-2.379e+00-2.388e+00

C.Q0

Step, Step-1Increment

SII,ý Arc 48- Ar, 48A A

CI

1.00

0.80

0.60

0.40

0.20

1.00

I 0.80

0.60

0.40

0.20

n nn

1.00 0

0.80

0.60

0,40

0.20

0,00. I*-

-3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0. 0Concrete Stress (Min and Max Principa 0

smaxn-lpf-nsmeared- strain

smin-lpf-smeared-strai7

I . .

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 [x. E-3J

Shear Strain (e12)

1 e12-lpf-sr..e. red-strain

0. 10, 20. 30. 40, 50. 60, FyVon MWses Stress (Steel Plate)

mises-.pf- smeared-,st-a.

(d) (e)

Figure 5-8. Results from SSo model with smeared concrete cracking model (tension stiffening stress strain curve):(a) Model with boundary conditions,(b) Von Mises stresses in steel plate when shear force ratio (Sxy/AsFy)=1.0,(c) Minimum principal stress in the concrete when shear force ratio=1.0,(d) Shear force ratio (Sxy/AsFy) vs. shear strain response,(e) Shear force ratio (Sxy/AsFy) vs. Von Mises stress in steel plate, and

(f) Shear force ratio vs. minimum and maximum principal stress in concrete.© Amit H. Varma, School of Civil Engineering, Purdue University, West Lafayette, IN 47906

I

NONLINEAR ANALYSIS - SSH MODEL

djewpoll: I 0DR Temp-11) ssh Model (111m, sinealml sti'llmodh vivv4poll: / 00t]: C: lemj)'31) syl Model Colic Y'llvalod %h'timodh vlowport: R 01)[1: (!:ý Ivillp ýfl) ssh Model Conc sIrIPM (41 sl[AM

I

Lsz~L\Z~

0

Step; Step-i am I

I

I I

(d) (e) (f)Figure 5-11. Details of SSh model:(a) Model with mesh and boundary conditions.(b) Steel plates with shell (S4R) elements, fully tied to concrete elements(c) Concrete infill modeled using shell (S4R) elements, boundary conditions included(d) Loading applied as uniformly distributed shear traction on the concrete shell element edges(e) Steel plates tied to concrete(f) Concrete surfaces tied to steel plates

C Amit. H. Varm School of Civil Enginemring, Purdue University, West Lafayette, IN 47906

p

NONLINEAR ANALYSIS - SSH MODEL

-V"

0.8

0.6

0.4

0.2

0.0 L L L0.00 0.05 0.10 0.15 0,20 0.

a1.0

0.8>

0.6

0.4

0. 10. 20. 30., 40. 50. 60.

Stress Fy

1

0

0

0

0

0,

.20

.4

.2 0 - 20 -1 1 5 Q

IC'-4C12

a)25 Stress -V¸Displacement

I-umag-Ipf-shell- smeared -strainSsmn-Ipf- shell- smeared- strainsomn-lpf-shell-conc-smeared-sttrn

I I

U

Figure 5-13. Results from SSh model with concrete smeared cracking model (tension stiffening stress-strain curve):(a) Von Mises stresses in steel plate when shear force ratio (Sxy/AFy)=1.0,(b) Minimum principal stress in the concrete when shear force ratio=1.0,(c) Maximum principal stress in the concrete when shear force ratio=1.0(d) Shear force ratio (Sxy/AsFy) vs. shear strain response, 0(e) Shear force ratio (Sxy/A6 Fy) vs. Von Mises stress in steel plate, and(f) Shear force ratio vs. minimum and maximum principal stress in concrete.

© Amit H. Varms , School of Civil Engineering, Purdue University, West Lafayette, IN 47906

NONLINEAR ANALYSIS - LCS MODEL

Ede todel Vieot y•ew Matria ••tion Prof•e Composite Assign Special Featwe Tools Plug-ins Help IN?

-1 w~ + " ~ I :xh4 ý 11 .W A

1.1i , 2 3 4 Propertydefiaut• F'• fl •Name: CompositeLayup- !C.

Module Property " Model: Model-I v Part: CFS-composite v Element type: Conventional Shel Description:.

•" •Layup Or rntaton

:k*•Definition: Par t glba

•'4 Section wiegratr,: m During analysis :'2 Before analysis

teRot integratra rule: Co) Simpso r

Ply Nai Region Maer-ial Thickness CSYS Roating oInter

"I V Ply-I (Picked) steel 0.1875 <Layujp> 0 3

2 V Ply-2 (Posked) doneraede-Smeared-Siraahn 13.125 <tayted> hs9

+3 V" Ply-3 (Pick~ed) Steel 0.1875 <Layup> 0 3

Ft i he copspteifi upedvitwort wedpinealoiegV\ektp3S~-oltfThe model database "C \ýTemp\3D-concrete-smeared-092109.cae" has been opened

Th peci f ead viewports were pr inted to f ile "Y Desktop\3DSSo-de t if "

The model database "C \Teap\3D-SSh-xodel cae" has been openedThe specified vievports were printed to file "Y \Desktop\3DSSh-model tif"The model database "C \Teop\3D-Layered-Shell cae" has been opened

Figure 5-15. Layered Composite Shell (LCS) model with ply details, boundary conditions, and shell edge loading.

) Amit H. Varma , S'hool of Civil Engineering, Purdue University, West Lafayette, IN 47906

NONLINEAR ANALYSIS - LCS MODELSTENSION STIFFENING AND FRACTURE ENERGY

oL4 C,

I-

-3.0 -2.5 -2.0 -1.5 -1.0 -0.5 "0J

Concrete Stress (Min and Max. Princip

srax-lpf-LCS-smeared- strainsrnx -Ipf-LSM-smeared-frauctu

srnin-lpf-LCS-srnear-strain

srin-lpf-LSM-smeared-fractur.

Figure 5-16. Results from LCS model with concrete smeared cracking model (tension stiffening and fracture energy):(a) Model with boundary conditions(b) Deformed shape and Von Mises stresses in steel plate when shear force ratio (Sxy/AsFy)=1.0(c) Minimum principal stress in the concrete when shear force ratio=1.0(d) Shear force ratio (Sxy/AsFy) vs. shear strain response(e) Shear force ratio (Sxy/AsFy) vs. Von Mises stress in steel plate

(f) Shear force ratio vs. minimum and maximum principal stress in concrete.

© Amit H4 Varm, School of Civil Engineering, Purdue University, West Lafayette, IN 47906

I

NONLINEAR ANALYSIS - LCS MODEL WITH

ORTHOTROPIC ELASTIC MODEL

1.2

0.8

0.4

T I I I I

0.4

1. 2

0.4

C-o3ne0 -2.5 -2.0 -1.5 -1.0 -0.5 .0 CConcrete Stress (Mmn and Max. PrncipI .. .. ,-.. -... . .. . . . . "0.0 1.0 2,0 3.0 4.0 5.0 (x1.E-3J

Shear Strain (e12)

e2-1pf-LCS-smeared-strai.e12-1 pl _rtho

0.10. 20o 30. 40. 50, o60 y

Von Mises Stress (Steel Plate)- mises-Ipf-LCS-smeared-strain I

M rises-lpf-ortho I-ma ×-prC-L.-sme are - stra

$na .-Ipf-ortho

- -sm~n-Ipf-LCS-smear-strains-nin-lpf-ortho

I

Figure 5-18. Results from LCS model with orthotropic elastic concrete model:(a) Model with boundary conditions(b) Deformed shape and Von Mises stresses in steel plate when shear force ratio (Sxy/AsFy)=1.0(c) Minimum principal stress in the concrete when shear force ratio=1.0(d) Shear force ratio (Sxy/AsFy) vs. shear strain response(e) Shear force ratio (Sxy/A 0Fy) vs. Von Mises stress in steel plate

(f) Shear force ratio vs. minimum and maximum principal stress in concrete.


NONLINEAR ANALYSIS USING ANALYTICAL

MODEL WITH ORTHOTROPIC CRACKED CONCRETt9


1.60

or

0LL

4-

M

0-

a.

1.4

1.2

0

-0.5

1 -1

LL

'C- -2C.

-2.5

In-Plane Interaction in Spl - Sp2 SpaU!

0.5 1 1.5 2 0 2.5

0.8

0.6

0.4

0.2

-30

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5

Axial Force / As Fy (Positive Tension, Negative Compression)

Principle Force 1

28S3tan(2Op) = -

St-S'

(l + cos(261 ,)2

(1 -cos(20 )+ 2 SY. + sin(20p ) S,.V2-"

S•2 (1-cos(2OP)2 (l±cos(2Op),6'2= S+(I S. - sin(20p ) S-,,f Amit H. Varma ,School of Civil Enr.gneer.ng, Purdue University, West Lafayette, IN .7906

FEM ANALYSIS FOR TENSION + SHEAR

Applied Tension - Shear - As Fy: Analysis using modified RikE

= U R I M M U M M E e a r k~ r0.70 S, Mises LE, LE12fraction = -0.979167, PLY-i (middle)

0.60 States shown (Avg: 75%) (Avg: 75%)050 +5 001e+01 +1 ,981e-03

0.5 •+5 .001e+ 01

+1:974e-030. 40•

+ 5 00 1e + 0 1 + 1 9 6 7e -0

0.0•+5.0O01e+ Ol

+1 .959e-03+5.001e+01

+1.952e-03

0 .3 0 + -+5 .0 0 0 e + 0 1+.O l 9 4 5 e -0 3

0.10 Principal stref

• 0.20

0.00 d irection in0.0o00 0.002 0.004 0.006 0.008 1 "-1- o

method

MVr

Mosom

1100mr"RENk

IML

MAIft,IMMýjMMMIL

A Off ii FRO MR

© Amit H. Vqrmq , Schnnl of Civil Engineering, Purdue University, West Lafayette, IN 47906

FEM ANALYSIS FOR TENSION + SHEAR

CRACK ORIENTATION FOR CASES (PROPORTIONAL LOADINI3)TIV=1/8 TIV=0.75 T/V=0.5 T/V=0.75

I I Ifraction = -0.979167,

S, Ma., In-PMin. In AI

S, Out-of-PM

fraction = -0.97916,

• S Max. In-S, Min. I160

S, Out-o9

fraction = -0.979ý

S , Max. In;

S, Min. Ir

s, Out-o•I

fraction = -0.979:

SS MaxnnS Min. Ir

S, Out o~I

ýx Step: StepIncrement 100: Arc Length = 49.15S Stepr Stepent i P00rLt I • ntep Ste P

Increment 100: Arc Length 49.15 Inecrement 100: Arc Length = 49.15 I ýx Step: Step 0 IIncreent 100: Arc Length = 45.83

viwoi 0)1 :le p11VlmhVe pit W : e pIIV i~I iwpi:8 I : I IIIVl lod iwol:9 OD : iiil tdfraction = -0.9791

SS, Max. In-nS, Min. IrS, Out-off

L x Step: Step-OIncrement 100: Arc Length = 41.47

fraction = -0.9791

S Max, In

S: Min. Inj

S Outof- f

fraction = -0.979

L S, Max. In-1S, Min. In-IS, Out-of-l

ýX Step: Step-O

Increment 100: Arc Length

fraction = -0.979:

S,0 ut-of-P:

Lx Step: Step-C49.15 I Increment 100: Arc Length = 49.15Lx SteIp: Step-m Arc Lngth= 41• Increment 100: Arc Length = 49.15

T=V T=2V T=4VAs shown, the crack orientation changes slowly with increasing T.As T/V changes from 0 to 1, there is little change in crack orientation.C~Rto-F*i t,.AqbaWX9 ._gr TOeA, T.,tWAt, _ s from 1 to inf.

RESULTS FROM FINITE ELEMENT ANALYSIS

TENSION + SHEAR (PROPORTIONAL LOADING)

1.1 Points for T-V1.0 Interactio.

Lz0

.. ,

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

T= 0. 125 V

T=0.25 V

T=0.5 V

-.- T=0.75 V

-T=1.33 V

-"-T=2 V 00 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 0.005 0.0055 0.006

OC Amit H. Varma School of Civil Engine rr!• • iirlilkt L[afayette, IN 47906

RESULTS FROM TENSION + SHEAR ANALYSIS

Von Mises stress in steel plates corresponding to the points of yieldingi.e,. turning points in the shear force-strain responses on previous slide

|

iii

T/V=0.77T/V=1/8 T/V=0.75 T/V=0.5viewpoll:~~~~I 1 c" :fli' ioV1.1 Vepr:3OD :Ttp n 1mjiEeplI40)3 :Iefl)1112V10hvepr:5 D:C fl) 10V10l

S, MSs$PLY-1 (middle)(Avg: 75%)i +5.001e+0l

+5,001e+01+5.001e+01+5,001e+01+5.001e+01

yi-x Step: Step-0Increment

I

S, MisesPLY-i (middle)(Avg: 75%)

+5.001e+(+5.001e+(+5.001e+C+5.001e+C+5.001e+(+5.001e+I

Lx Step: Step-0Increment

S, MisesPLY-1 (middle)(Avg: 75%)

+5.001e++5.001e++5.001e++5.001e++5,001e++5,001e+;

Lx Step; Step-10Increment 18: Arc Lengith 8.150 I

S, Misesfraction = -0.977(Avg: 75%)

+5.,000e+S+5,000e++5,000e++5.000e++5.000e++5.000e+

L x Step: Step-0Increment 16: Arc Length = 7.15018: Arc Length = 8.150 18: Arc Length = 8.150

Vie poil: O R C fenpi1 11)(1) Vew ort I DH:C:eru 1 V inmdbVie pot: OD C:I e p 1 li 1. 1)Vie por: 9 0D: CJeip II fit~mllS, Misesfraction = -0.9(Avg: 75%)

S+5.001e++5.001e++5.001e++5,001e++5.001e++5.000e+

Lx Ste P: Step-i

Inre en 111001171 ""41140m,

I lmmbb..

I-logE

I


+5.000e++5.000e++5.000e++5.000e++5.000e++5.000e+

LxStep; Step-In crement

7

*1

I


+5.000e+S+5,000e++5.000e++5.000e++5.000e++5,000e Fl

S, Misesfraction = -0.9(Avg: 75%')

+5 001e++5.001e++5.001e++5.001e++5.001e++5.001e+

rimilm

EMPOP"nP

0 0v V

LX Step: Step-0Increment 26: Are Length = 12.15 Lx Step: Step-0Increment 28: Arc Length = 13.1516: Arc Length = 7.150 22: Arc Length = 10.15

.. - h .. n l l .

T=V© Amit H. Vqrmq - School

T=2Vof Civil Enginpering, Pilirdiwt

T=4VUniversity, Wt-st Lafayette, IN 47906

T=8V


Minimum principal stress in concrete at the onset of yielding

TIV=1/8 TV=0.75 TIV=0.5 T/V=0.75

S, Min. In-Plane PrincipalPLY-2 (middle)(Avg: 75%)• -2,032e+00

.2, 034e+00-2.036e+00-2 038e+00-2 040e+00-2. 042e+00

L x Step: Step-0Increment 18: Arc Lenath = 8.150

S, Min. In-Plane PrincipalPLY 2 (middle)(Avg: 75%C-1 880e+0

1.884e+01 888e+0

-1.892e+0-1.896e+0-1.900e+0

S, Min. In-Plane PrincipalPLY-2 (middle)(Avg: 75%)

-1.632e+01.639e+0

-1.647e+O-1.654e+0-1.661e+0-1.668e+01


-1.344e+011.352e+01

-1.359e+01-1.367e+01-1.375e+0

-1.383e+01

Step: Step-0Increment 18: Arc Lenoth = 8.150

Lx Step: Step-0Increment 18: Arc Lenath = 8.150

LX Step: Step-0Increment 16: Arc Lenath = 7.150


-1.2lie+O!-1 ,211e+0-1.220e+01

-1.229e+0-1.238e+01-1.247e+0

-1.256e+01

S, Min. In-PlanePLY-2 (middle)(Avg: 75%)

-8.375e-0-8.441e-0-8.507e-0-8.573e-0-8.639e-0-8,705e-0

e Principal S, Min. In-Plane PPLY- 2(middle)f(Avg: 75%)

-6.379e-0-6.459e-0-6.540e-0-6.620e-0-6.701e-0-6.781e-01

S, Min. In-PlatPLY-2 (middle(Avg: 75%)

-5.901e--5.963e--6.025e--6.087e--6.150e--6.212e-

--X Step: Step-0Increment 16: Arc Length = 7.150

Lx Step: Step-0Increment

L x Step: Step-0Increment 26: Arc Length =

L Step: Step-0Increment 28: Arc22: Arc Length = 10.15 12.15

k"w111111111111111

T=V T=2V T=4V

Concrete contribution (min. princ. stress) decreases as the TN increases.. Concrte v- t-ribution ishl hish for te?,iih 7se.©c Amit H. Varmn W,." I'cha oF .wl•ntgneering, Yurdjt-e Uiv-rs.-yr, W.. .ay•e yh -T790W"


Shear strain at onset of yielding decreases as the T/V increases.Largest shear strain at yielding occurs for the pure shear case.

T/V=1/8 T/V=0.75 TIV=0.5 TIV=O. 7I iw ol DI :TmpIfnlVt di e pI 3IM :11111111 .11 i iI:I 0D :C erp1112VI~d iw ol , 0M : m 10V101) .LE. LE12PLY-2 (mnddle)(Ang: 75%)

+2,.351e-03+2 350e 03+2 349e- 03+2Y348e-03+2 347e-03+2 346e- 03

LE, LE12PLY-2 (middle)(Avg: 75%)

+2.326e-:+2.324e+2.322e-C+2.320e+2.318e-+2.316e(


+2.280e-02.284e+2.276e -0

+2.2-72e -0+2.268e-0O+2.264e-0


+1.997e-0:+1.991e-0+1.986e-0+ 1.980e-0+ 1.974e-0+1.969e-0

Step: Step-0Increment 18: Arc Length = 8.150

Step: Step-0Increment 18: Arc Length = 8.150- Ste p: Step-O

IncrementStep: Step-0Increment

18: Arc Length = 8.150 16: Arc Length = 7.150

LE, LE12 LE, LE12 LE, LE12 LE, LE12PLY-2 (middle) PLY-2 (middle) PLY-2 (middle) PLY-2 (middle)(Avg; 75%) (Avg: 75%) (Avg: 75%) (Avg: 75%)

+ 1.974e-0 +1.440e-0 +9.037e- +5.025e-+1.974e-0 +1.450e -0 +9.137e- +5 082e-

+ 1.967e-0 +1.43le-0 +8.961e- +4.967e-+ 1.959e-0 +1.42le-0 +8.874e- +4.909e-+ 1.952e-0 +1.412e-0 +8.786e-I +4.85le-+ 1.945e-0 +1.402e-0 +8.698e- +4.793e- 4

Step: Step-0 L Step: St.p-oS L P step: Step-o L5 Step: Step-0Increment 16; Arc Length = 7,150 Increment 22: Arc Length = 10.15 increment 26: Arc Length 12.15 Increment 28: Arc Length = 13.15

T=V T=2V T=4V ToIAmit H Varma, School of Civil Engineering, Puirdue University, West Lafayette, IN 47906

SUMMARY FROM FEM ANALYSIS RESULTS

" The steel plates govern the behavior even more fortension + shear loading cases.

" The cracking direction does not change until there issignificant tension (T/V greater than or equal to 1)

" The concrete contribution to the shear behaviordecreases with increasing tension. It is maximum forthe pure shear case (as compared to tension + shear).

o Compression + shear results in significant increase inthe in-plane shear strength of the composite design.

0 Amit H. Varma , School of Civil Engineering, Purdue University, West Lafayette, IN 47906

ANALYSIS FOR IN-PLANE FORCE


LL

0EL,..

M

M

-*-Compression + Sheart

1.4 a-*Tension + Shear

-+-Results from FEM1.2

I FEM

0.8

Theory0.6

0.4

0.2

0

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5

am~t H v•m• •x ,l5e,,4 As FyýPostv Tpni pegtv omrsin...... -n•,in-lF~n moor ,n , lr!eT _•i~vj ers!i~y n esv N t~offo•I qq

RESULTS IN PRINCIPAL FORCE SPACE

o Take the LCS model with smeared cracking concretemodel and fracture energy.

o Analyze for combinations of axial and shear forceo Take the results for Sy and Sxy, and convert into or

calculate Spl and Sp2.

SPl 2p2 + L::j' )j+(Sxy

o The principal forces were plotted from the results.


FAILURE SURFACE IN SP1-SP2 SPACEFailure Surface for In-Plane Forces in

Principal ForcesI.a

['3['

0Principal Force 1 (Spl kips/ft)


CAPACITY SURFACE IN PRINCIPAL FORCE SPACE

Failure Surface for In-Plane Forces in Principal Forces

xial Tens

.)4

.. 44

Pure C

Bia:

Tensit-_ -P "E

Y

Sheai

I a I,,B.. % J Ja 'I

CD Amit H Varma, Schono of Civil Envineering, Plirtim- LTnivir-itv Wpqt Lfnyetti, TN 47q06

DESIGN CAPACITIES

o The phi-factors should be representative of thefailure mode (brittle and ductile)

o Creep and shrinkage effects can be included byusing FR (residual stress).

o The compression capacity can be reduced toinclude the effects of creep, shrinkage, localbuckling (if any), and locked in stresses

o For example, As (Fy-Fr) + /3 fc Ac


enclosure 1 to dcp nrc 002728, 'in-plane behavior of ... · dcp nrc 002728 december 30, 2009...

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