e10eng

CivilFEM Manual of Essential Examples. 300609 .Ingeciber, S.A.

Chapter 10 Concrete Structures - Eurocode 2


CivilFEM Manual of Essential Examples Chapter 10 Table of Contents

Example 10.1 Axial+Bending Reinforcement Check According to EC2 ................. 1

Example 10.2 Axial+Bending Reinforcement Design According to EC2 ................ 7

Example 10.3 Biaxial Bending + Axial Force Checking According to EC2 ........... 11

Example 10.4 Biaxial Bending + Axial Force Design According to EC2............... 17

Example 10.5 Shear and Torsion Reinforcement Check According to EC2 ........ 23

Example 10.6 Shear and Torsion Reinforcement Design According to EC2 ....... 31

Example 10.1 Axial+Bending Reinforcement Design


10-1

Example 10.1 Axial+Bending Reinforcement Check According to EC2

Example Description

A double T cross section beam is loaded as shown bellow. For this beam check the reinforcement under axial+bending combined action according to Eurocode No.2 provisions.

Geometric Properties

a = 10 m

DEPTH = 1.10 m

TW = 0.15 m

BFTOP = 0.90 m

TFTOP = 0.18 m

BFBOT = 0.50 m

TFBOT = 0.20 m

Geometrical cover = 0.03 m

Loading q = Self Weigh

Px = 380 KN

Py = 250 KN

Material

Concrete = C40/50

Steel reinforcement = S500

a

q

Py

a/2

Px



10-2

Z

Y

BFBOT

TF

BO

T

TW

DE

PT

H

TF

TO

P

BFTOP

Figure 10.1-1 Problem Sketch

X

Y

1 2 3 4 5 7 8 9 10 111 2 3 4 7 8 9 10

Figure 10.1-2 Representative Finite Element Model

Analysis Assumptions and Modeling Notes

Meters and Newton are used as length and force units. For this example BEAM3 elements have been used.

Input Data Listing

FINISH

~CFCLEAR,,1

! Title

/TITLE, EC2: Axial+Bending Checking

! CivilFEM Setup: Code & Units

~UNITS,SI ! m, sg, N

~CODESEL,EC3,EC2,,,EC8 ! Concrete code: EC2

/PREP7

! CivilFEM Preprocessor

! --------------------------------------------------

! Materials

~CFMP,1,LIB,CONCRETE,EC2,C40/50,0,0,0

~CFMP,2,LIB,REINF,EC2,S500,0,0,0

! Element Types

ET,1,BEAM3 ! Type 1: 2D Beam

! Cross Section

~CSECDMS,1,I,1,1.1,0.15,0.9,0.18,0.5,0.2,0,0,0,0

! Constant reinforcement (Group 1)

~RNFDEF,1,1,2,1,1,0.03, , ,16.00,6,,,4

! Scalable reinforcement (Group 2)

~RNFDEF,1,2,2,2,0,0.03, , ,20.00,4,,,4



10-3

! Beam & Shell Property

~BMSHPRO,1,BEAM,1,1,,,3,1,0, ! Beam Property 1

! Ansys Preprocessor

! --------------------------------------------------

! Nodes

N,1 $ N,11,10 $ FILL,1,11

! Elements

REAL,1 $ E,1,2 $ EGEN,10,1,1,1

/SOLU

! Ansys Solution

! --------------------------------------------------

! Constrains

D, 1,UX,0,,,,UY

D,11,UY,0

! Apply Load

ACEL,,9.81,

F,6,FY,-250E3

F,6,FX,-380E3

! Solve

SOLVE

/POST1

! CivilFEM Postprocessor

! --------------------------------------------------

! Read results

~CFSET,,1,1 !First load case

! Axial+bending checking

~CHKCON,2DB,,0

! Plot elements OK and No OK

~PLLSCON,ELM_OK

! Plot axial+bending criterion

~PLLSCON,CRT_TOT

! Plot interaction diagrams

~PL2DINT,BEAM,6,J,,,,,,MZ ! Element: 6, section: J

CivilFEM Results



10-4

Figure 10.1-3 Elements OK and No OK

ANSYS 5.6

DEC 5 2000

11:09:14

PLOT NO. 1

LINE STRESS

STEP=1

SUB =1

TIME=1

CFETAB_ICFETAB_J

MIN =.515E-15

ELEM=10

MAX =1.313

ELEM=6

1

X

Y

Z

.515E-15

.145925

.29185

.437774

.583699

.729624

.875549

1.021

1.167

1.313

CivilFEM Example 0801. EC2: Axial+Bending Checking

Figure 10.1-4 Axial + Bending Criterion



10-5

Figure 10.1-5 Interaction Diagram (Element 6, Section J)



10-7

Example 10.2 Axial+Bending Reinforcement Design According to EC2

Example Description

A double T cross section beam is loaded as shown bellow. For this beam design the necessary reinforcement under axial+bending combined action according to Eurocode No.2 provisions.


a = 10 m

DEPTH = 1.10 m

TW = 0.15 m

BFTOP = 0.90 m

TFTOP = 0.18 m

BFBOT = 0.50 m

TFBOT = 0.20 m

Geometrical cover = 0.03 m

Loading q = Self Weigh

Px = 380 KN

Py = 250 KN

Material

Concrete = C40/50


a

q

Py

a/2

Px



10-8

Z

Y

BFBOT

TF

BO

T

TW

DE

PT

H

TF

TO

P

BFTOP


X

Y

1 2 3 4 5 7 8 9 10 111 2 3 4 7 8 9 10




Input Data Listing

FINISH

~CFCLEAR,,1

! Title

/TITLE, EC2: Axial+Bending Design


~UNITS,SI

~CODESEL,EC3,EC2,,,EC8

/PREP7


! --------------------------------------------------

! Materials



! Element Types


! Cross Section

~CSECDMS,1,I,1,1.1,0.15,0.9,0.18,0.5,0.2,0,0,0,0

! Constant reinforcement (Group 1)

~RNFDEF,1,1,2,1,1,0.03, , ,16.00,6,,,4

! Scalable reinforcement (Group 2)



10-9

~RNFDEF,1,2,2,2,0,0.03, , ,20.00,4,,,4

! Beam & Shell Property

~BMSHPRO,1,BEAM,1,1,,,3,1,0, ! Beam Property 1


! --------------------------------------------------

! Nodes

N,1 $ N,11,10 $ FILL,1,11

! Elements

REAL,1 $ E,1,2 $ EGEN,10,1,1,1

/SOLU

! Ansys Solution

! --------------------------------------------------

! Constrains

D, 1,UX,0,,,,UY

D,11,UY,0

! Apply Load

ACEL,,9.81,

F,6,FY,-250E3

F,6,FX,-380E3

! Solve

SOLVE

/POST1


! --------------------------------------------------

! Read results

~CFSET,,1,1

! Axial+bending design

~DIMCON,2DB,,0,0.25,5 ! Wmin=0.25, Wmax=5.0

! Plot reinforcement factor

~PLLSCON,REINFACT

CivilFEM Results



10-10

ANSYS 5.6

DEC 5 2000

11:15:25

PLOT NO. 1

LINE STRESS

STEP=1

SUB =1

TIME=1

CFETAB_ICFETAB_J

MIN =.25

ELEM=1

MAX =1.316

ELEM=6

1

X

Y

Z

.25

.368498

.486996

.605494

.723991

.842489

.960987

1.079

1.198

1.316

CivilFEM Example 0802. EC2: Axial+Bending Design

Figure 10.2-3 Reinforcement Factor

Example 10.3 Biaxial Bending+Axial Force Checking


10-11

Example 10.3 Biaxial Bending + Axial Force Checking According to EC2

Example Description

A frame having two different cross sections, one for the piers and other for the beams, is subjected to the loads shown bellow. Using the data provided, check this structure under axial+ biaxial bending.


H = 3 m

B = 4 m

D = 5 m

Loading FX = 6 Ton

FY = 15 Ton

FZ = 6 Ton

Material

Concrete = C30/37


FY

FXFZY

ZX

H

BD

FY

FXFZY

Z

Y

ZX

H

BD




10-12

Section A-A (Piers)

d = 0.30 m

b = 0.30 m

Bending Reinforcement of Section A-A

= 16 mm

No. Bars = 3

Gc = 3 cm

Y

Zd

b

Y

Zd

b

Figure 10.3-2 Section A-A

Section B-B (Beams)

d = 0.30 m

b = 0.25 m

Bending Reinforcement of Section B-B

Total Area at Top = 6E-3 m2

Total Area at Bottom = 1E-3 m2

Y

Zd

b

Y

Zd

b

Figure 10.3-3 Section B-B



10-13


Meters and Tons are used as length and force units. For this example BEAM4 elements have been used.

Input Data Listing

FINISH

~cfclear,,1

/TITLE, Biaxial bending + axial force checking


~UNITS,,LENG,M

~UNITS,,TIME,S

~UNITS,,FORC,MP


/PREP7


! --------------------------------------------------

! Material Definition



! Element Type Definition

ET,1,Beam4 ! Element Type 1: 3D Beam

! Cross-section definition

~CSECDMS,1,REC,1,0.30,0.30,0,0,0,0,0,0,0,0 !Piers

~CSECDMS,2,REC,1,0.30,0.25,0,0,0,0,0,0,0,0 !Beams

! Reinforcement definition

! Cross Section 1 longitudinal reinforcement

~RNFDEF,1,1,2,1,0,0.03, , ,16.00,3, ,,4

~RNFDEF,1,2,2,2,0,0.03, , ,16.00,3, ,,4

~RNFDEF,1,3,2,3,0,0.03, , ,16.00,3, ,,4

~RNFDEF,1,4,2,4,0,0.03, , ,16.00,3, ,,4


~RNFDEF,2,1,2,2,0,0.030,0.600E-02 !Top Reinf.

~RNFDEF,2,2,2,4,0,0.030,0.100E-02 !Bottom Reinf.

! Beam & Shell property

~BMSHPRO,1,BEAM,1,,,,4,1,0, ! Beam property 1



! --------------------------------------------------

! Model Construction

H=3 ! 3m

B=4 ! 5m

D=5 ! 4m

! View direction

/VIEW, 1 ,1,1,1

K,1 ! Keypoints input

K,2,0,H

K,3,0,H,-B

K,4,D,H,-B

K,5,D,H

K,6,0,0,-B

K,7,D,0,-B

! Lines

LSTR, 1, 2



10-14

LSTR, 2, 3

LSTR, 3, 4

LSTR, 2, 5

LSTR, 5, 4

LSTR, 6, 3

LSTR, 7, 4

! Piers Meshing

TYPE, 1

REAL, 1

ESIZE,1

LSEL,S,LENGTH,,H

LMESH,ALL

LSEL,ALL

! Beams Meshing

TYPE, 1

REAL, 2

ESIZE,1

LSEL,U,REAL,,1

LMESH,ALL

LSEL,ALL

FINISH ! Preprocessor output

/SOLU

! Ansys Solution

! --------------------------------------------------

! Constrains

KSEL,S,LOC,Y,0

DK,ALL,ALL

KSEL,ALL

! Loads

FY=-15.000

FZ=6.000

FX=6.000

FK,5,FX,FX

FK,5,FY,FY

FK,5,FZ,FZ

SOLVE ! Calculation

FINISH ! Finish calculation module

/POST1


! --------------------------------------------------

! Read results

~CFSET,,1,1 !Read first load step

! Axial+Biaxial bending checking according to EC2

~CHKCON,3DB

! Plot axial+biaxial bending total criterion

~PLLSCON,CRT_TOT

! Plot elements OK & NO OK

~PLLSCON,ELM_OK

! Plot 3D interaction diagram

~PLCSCON,BEAM,1,J

CivilFEM Results



10-15

ANSYS 5.6

DEC 5 2000

11:19:10

PLOT NO. 1

LINE STRESS

STEP=1

SUB =1

TIME=1

CFETAB_ICFETAB_J

MIN =.094579

ELEM=6

MAX =1.742

ELEM=7

1

X

Y

Z

.094579

.277606

.460634

.643661

.826689

1.01

1.193

1.376

1.559

1.742

CivilFEM Example 0803, Biaxial bending + axial force checking

Figure 10.3-4 Total Criterion

Figure 10.3-5 Elements OK & NO OK



10-16

Figure 10.3-6 3D Interaction Diagram

Example 10.4 Biaxial Bending+Axial Force Design


10-17

Example 10.4 Biaxial Bending + Axial Force Design According to EC2

Example Description

A frame having two different cross sections, one for the piers and other for the beams, is subjected to the loads shown bellow. Using the data provided, design the necessary reinforcement to resist the axial+ biaxial bending combined actions according to Eurocode No.2.


H = 3 m

B = 4 m

D = 5 m

Loading FX = 6 Ton

FY = 15 Ton

FZ = 6 Ton

Material

Concrete = C30/37


FY

FXFZY

ZX

H

BD

FY

FXFZY

Z

Y

ZX

H

BD




10-18

Section A-A (Piers)

d = 0.30 m

b = 0.30 m

Bending Reinforcement of Section A-A

= 16 mm

No. Bars = 3

Gc = 3 cm

Y

Zd

b

Y

Zd

b


Section B-B (Beams)

d = 0.30 m

b = 0.25 m

Bending Reinforcement of Section B-B

Total Area at Top = 6E-3 m2

Total Area at Bottom = 1E-3 m2

Y

Zd

b

Y

Zd

b




10-19


Meters and Tons are used as length and force units. For this example BEAM4 elements have been used.

Input Data Listing

FINISH

~cfclear,,1

/TITLE, Biaxial bending + axial force design


~UNITS,,LENG,M

~UNITS,,TIME,S

~UNITS,,FORC,MP


/PREP7


! --------------------------------------------------

! Material Definition



! Element Type Definition

ET,1,Beam4 ! Element Type 1: 3D Beam

! Cross-section definition

~CSECDMS,1,REC,1,0.30,0.30,0,0,0,0,0,0,0,0 !Piers

~CSECDMS,2,REC,1,0.30,0.25,0,0,0,0,0,0,0,0 !Beams

! Reinforcement definition


~RNFDEF,1,1,2,1,0,0.03, , ,16.00,3, ,,4

~RNFDEF,1,2,2,2,0,0.03, , ,16.00,3, ,,4

~RNFDEF,1,3,2,3,0,0.03, , ,16.00,3, ,,4

~RNFDEF,1,4,2,4,0,0.03, , ,16.00,3, ,,4


~RNFDEF,2,1,2,2,0,0.030,0.600E-02 !Top Reinf.

~RNFDEF,2,2,2,4,0,0.030,0.100E-02 !Bottom Reinf.

! Beam & Shell property




! --------------------------------------------------

! Model Construction

H=3 ! 3m

B=4 ! 5m

D=5 ! 4m

! View direction

/VIEW, 1 ,1,1,1

K,1 ! Keypoints input

K,2,0,H

K,3,0,H,-B

K,4,D,H,-B

K,5,D,H

K,6,0,0,-B

K,7,D,0,-B

! Lines

LSTR, 1, 2



10-20

LSTR, 2, 3

LSTR, 3, 4

LSTR, 2, 5

LSTR, 5, 4

LSTR, 6, 3

LSTR, 7, 4

! Piers Meshing

TYPE, 1

REAL, 1

ESIZE,1

LSEL,S,LENGTH,,H

LMESH,ALL

LSEL,ALL

! Beams Meshing

TYPE, 1

REAL, 2

ESIZE,1

LSEL,U,REAL,,1

LMESH,ALL

LSEL,ALL

FINISH ! Preprocessor output

/SOLU

! Ansys Solution

! --------------------------------------------------

! Constrains

KSEL,S,LOC,Y,0

DK,ALL,ALL

KSEL,ALL

! Loads

FY=-15.000

FZ=6.000

FX=6.000

FK,5,FX,FX

FK,5,FY,FY

FK,5,FZ,FZ

SOLVE ! Calculation

FINISH ! Finish calculation module

/POST1


! --------------------------------------------------

! Read results

~CFSET,,1,1 !Read first load step

! Axial+Biaxial bending design according to EC2

~DIMCON,3DB,,,1,3

! Plot axial+biaxial bending total criterion

~PLLSCON,REINFACT

! Plot elements designed and not designed

~PLLSCON,DS_OK

CivilFEM Results



10-21

ANSYS 5.6

DEC 5 2000

11:43:23

PLOT NO. 1

LINE STRESS

STEP=1

SUB =1

TIME=1

CFETAB_ICFETAB_J

MIN =1

ELEM=4

MAX =2.03

ELEM=7

1

X

Y

Z

1

1.114

1.229

1.343

1.458

1.572

1.686

1.801

1.915

2.03

CivilFEM Example 0804, Biaxial bending + axial force design

Figure 10.4-4 Reinforcement Factor

Figure 10.4-5 Designed elements

Example 10.5 Shear and Torsion Reinforcement Check


10-23

Example 10.5 Shear and Torsion Reinforcement Check According to EC2

Example Description

A rectangular beam having two different cross sections is loaded as shown bellow. Using the data provided, check the beam under shear and torsion.


a = 10 m

b = 3 m

Loading q = 2500 Nw

p = 50000 Nw

Material

Concrete = C25/30


X

Y

Z

a/2a

b

q

AA

B

B

p


Section A-A

d = 0.40 m

b = 0.50 m

Shear Reinforcement of Section A-A



10-24

Total Area = 16E-4 m2

Distance = 0.20 m

Torsion Transversal Reinforcement of Section A-A

= 12 mm

Distance = 0.20 m

Torsion Longitudinal Reinforcement of Section A-A

= 20 mm

No. Bars = 4

Z

Y

b

d


Section B-B

d = 0.40 m

b = 0.25 m

Shear Reinforcement of Section B-B


Distance = 0.20 m

Torsion Transversal Reinforcement of Section B-B

= 12 mm

Distance = 0.20 m

Torsion Longitudinal Reinforcement of Section B-B

= 16 mm

No. Bars = 4



10-25

Z

Y

b

d


X

Y

1 2 3 4 5 6 7 8 9 10 11

12

13

14

15

16

17

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16




Input Data Listing

FINISH

~CFCLEAR,,1 ! Not needed

! Title

/TITLE, EC2: Shear & Torsion Checking


~UNITS,SI ! m, s, N


/PREP7


! --------------------------------------------------

! Materials



! Element Types


! Cross Sections

~CSECDMS,1,REC,1,0.5,0.4,0,0,0,0,0,0,0,0

~CSECDMS,2,REC,1,0.25,0.4,0,0,0,0,0,0,0,0

! Reinforcement of rectangular section 1

! Shear reinforcement

~RNFDEF,1,SHEAR,2,0.000,0.000, , , ,16E-4,0.20



10-26

! Torsion transversal and longitudinal reinforcement

~RNFDEF,1,TORSION,2, , ,0.20,12, ,20,4



~RNFDEF,2,SHEAR,2,0.000,0.000, , , ,8E-4,0.20


~RNFDEF,2,TORSION,2, , ,0.20,12, ,16,4

!Beam & Shell Properties

~BMSHPRO,1,BEAM,1,1,,,4,1,0 ! Beam Property 1



! --------------------------------------------------

! View direction

/VIEW,1,1,1,1

/VUP,1,Z

! Nodes

N,1 $ N,11,10 $ FILL,1,11

N,12,5,0.5 $ N,17,5,3 $ FILL,12,17

! Elements

REAL,1 $ E,1,2 $ EGEN,10,1,1

REAL,2 $ E,6,12 $ E,12,13 $ EGEN,5,1,12

/SOLU

! Ansys Solution

! --------------------------------------------------

! Constrains

D, 1,ALL

D,11,ALL

! Apply Load

Q=2500.0

F=-50E3

SFBEAM,ALL,1,PRES,Q,Q

F,17,FZ,F

! Solve

SOLVE

/POST1


! --------------------------------------------------

! Read results


! Shear and torsion checking

~CHKCON,SHT,BOTH,1 ! Z direction

! Elements OK & no OK

~PLLSCON,ELM_OK

! Plot combined shear and torsion total criterion

~PLLSCON,CRT_TOT

! List results (element 1)

~PRCON,DETAILED,BOTH,1

CivilFEM Results



10-27

Figure 10.5-5 Elements OK and No OK



10-28

Figure 10.5-6 Total Criterion of Combined Shear and Torsion



10-29

Beam Element 1 - Node I

Result Value Concepts Description

VRD1 84.370E+03 Design shear resistance without considering the reinforcement

CRVRD1 488.920E-03 Ratio of the design shear force (VSd) to the resistance VRd1

VRD2 753.131E+03 Max shear without crushing of the concrete compressive struts


VWD 1.154E+06 Contribution of shear reinforcement to the shear resistance

VRD3 993.886E+03 Design shear resistance


CRTSHR 54.771E-03 Shear Criterion

TENS 312.544E+03 Tensile forces in the longitudinal reinforcement

TRD1 81.245E+03 Max resisted without crushing of the concr. compressive struts

CRTRD1 992.367E-03 Ratio of the design torsional moment (TSd) to TRd1

TRD2 56.586E+03 Maximum torsional moment resisted by the torsion

reinforcement

CRTRD2 1.425E+00 Ratio of the design torsional moment (TSd) to TRd2

ALT 1.257E-03 Required torsion longitudinal reinforcement

CRTALT 1.000E+00 Ratio of required to the torsion longitudinal reinforc.

CRTTRS 1.425E+00 Torsion Criterion

CRTCST 987.792E-03 Criterion based on shear-torsion resisted by compressive struts

CRT_TOT 1.425E+00 Global Criterion



10-30

Beam Element 1 - Node J

Result Value Concepts Description

VRD1 84.370E+03 Design shear resistance without considering the reinforcement


VRD2 753.131E+03 Max shear without crushing of the concrete compressive struts


VWD 1.154E+06 Contribution of shear reinforcement to the shear resistance

VRD3 993.886E+03 Design shear resistance


CRTSHR 51.452E-03 Shear Criterion

TENS 187.487E+03 Tensile forces in the longitudinal reinforcement

TRD1 81.245E+03 Max resisted without crushing of the concr. compressive struts

CRTRD1 992.367E-03 Ratio of the design torsional moment (TSd) to TRd1

TRD2 56.586E+03 Maximum torsional moment resisted by the torsion

reinforcement

CRTRD2 1.425E+00 Ratio of the design torsional moment (TSd) to TRd2

ALT 1.257E-03 Required torsion longitudinal reinforcement

CRTALT 1.000E+00 Ratio of required to the torsion longitudinal reinforc.

CRTTRS 1.425E+00 Torsion Criterion

CRTCST 987.439E-03 Criterion based on shear-torsion resisted by compressive struts

CRT_TOT 1.425E+00 Global Criterion

Example 10.6 Shear and Torsion Reinforcement Design


10-31

Example 10.6 Shear and Torsion Reinforcement Design According to EC2

Example Description

A rectangular beam having two different cross sections is loaded as shown bellow. Using the data provided, check the beam under shear and torsion.


a = 10 m

b = 3 m

Loading q = 2500 Nw

p = 50000 Nw

Material

Concrete = C25/30


X

Y

Z

a/2a

b

q

AA

B

B

p


Section A-A

d = 0.40 m

b = 0.50 m

Shear Reinforcement of Section A-A



10-32


Distance = 0.20 m

Torsion Transversal Reinforcement of Section A-A

= 10 mm

Distance = 0.20 m

Torsion Longitudinal Reinforcement of Section A-A

= 16 mm

No. Bars = 4

Z

Y

b

d


Section B-B

d = 0.40 m

b = 0.25 m

Shear Reinforcement of Section B-B


Distance = 0.20 m

Torsion Transversal Reinforcement of Section B-B

= 10 mm

Distance = 0.20 m

Torsion Longitudinal Reinforcement of Section B-B

= 16 mm

No. Bars = 4



10-33

Z

Y

b

d


X

Y

1 2 3 4 5 6 7 8 9 10 11

12

13

14

15

16

17

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16




Input Data Listing

FINISH

~CFCLEAR,,1 ! Not needed

! Title

/TITLE, EC2: Shear & Torsion Design


~UNITS,SI ! m, sg, N


/PREP7


! --------------------------------------------------

! Materials



! Element Types


! Cross Sections

~CSECDMS,1,REC,1,0.5,0.4,0,0,0,0,0,0,0,0

~CSECDMS,2,REC,1,0.25,0.4,0,0,0,0,0,0,0,0





10-34

~RNFDEF,1,SHEAR,2,0.000,0.000, , , ,3E-4,0.20


~RNFDEF,1,TORSION,2, , ,0.20,10, ,16,4



~RNFDEF,2,SHEAR,2,0.000,0.000, , , ,2E-4,0.20


~RNFDEF,2,TORSION,2, , ,0.20,10, ,16,4

!Beam & Shell Properties




! --------------------------------------------------

! View direction

/VIEW,1,1,1,1

/VUP,1,Z

! Nodes

N,1 $ N,11,10 $ FILL,1,11

N,12,5,0.5 $ N,17,5,3 $ FILL,12,17

! Elements

REAL,1 $ E,1,2 $ EGEN,10,1,1

REAL,2 $ E,6,12 $ E,12,13 $ EGEN,5,1,12

/SOLU

! Ansys Solution

! --------------------------------------------------

! Constrains

D, 1,ALL

D,11,ALL

! Apply Load

Q=2500.0

F=-50E3

SFBEAM,ALL,1,PRES,Q,Q

F,17,FZ,F

! Solve

SOLVE

/POST1


! --------------------------------------------------

! Read results


! Shear and torsion design

~DIMCON,SHT,BOTH,1 ! Z direction

! Plot design shear reinforcement

~PLLSCON,ASSH

! Plot design transverse torsional reinforcement

~PLLSCON,ASTT

! Plot design longitudinal torsional reinforcement

~PLLSCON,ASLT

! List results (element 1)

~PRCON,DETAILED,BOTH,1

CivilFEM Results



10-35

ANSYS 5.6

DEC 5 2000

12:03:44

PLOT NO. 1

LINE STRESS

STEP=1

SUB =1

TIME=1

CFETAB_ICFETAB_J

MIN =.255E-03

ELEM=5

MAX =.510E-03

ELEM=11

1

X Y

Z

.255E-03

.283E-03

.312E-03

.340E-03

.368E-03

.397E-03

.425E-03

.454E-03

.482E-03

.510E-03

CivilFEM Example 0806. EC2: Shear & Torsion Design

Figure 10.6-5 Design Shear Reinforcement Area by Length Unit

ANSYS 5.6

DEC 5 2000

12:03:45

PLOT NO. 2

LINE STRESS

STEP=1

SUB =1

TIME=1

CFETAB_ICFETAB_J

MIN =0

ELEM=11

MAX =.001032

ELEM=4

1

X Y

Z

0

.115E-03

.229E-03

.344E-03

.458E-03

.573E-03

.688E-03

.802E-03

.917E-03

.001032


Figure 10.6-6 Design Area of Torsional Transversal Reinforc. by Unit Length



10-36

ANSYS 5.6

DEC 5 2000

12:03:46

PLOT NO. 3

LINE STRESS

STEP=1

SUB =1

TIME=1

CFETAB_ICFETAB_J

MIN =0

ELEM=11

MAX =.001398

ELEM=7

1

X Y

Z

0

.155E-03

.311E-03

.466E-03

.622E-03

.777E-03

.932E-03

.001088

.001243

.001398


Figure 10.6-7 Total Area Design for Torsional Longitudinal Reinforcement

EC2 SHEAR & TORSION DESIGN

ELEMENT 1

RESULT N Value Concepts Description

-------- - -------------- ----------- --------------------------------------------------

VRD1 I 84.36960E+03 Design shear resistance without considering the

reinforcement

CRVRD1 I 488.92016E-03 Ratio of the design shear force (VSd) to the

resistance VRd1

VRD2 I 776.24998E+03 Max shear force allowed without crushing the concrete

CRVRD2 I 53.14010E-03 Ratio of the design shear force (VSd) to the

resistance VRd2

VWD I 41.25000E+03 Contribution of the necessary shear reinforcement

TRD1 I 83.73914E+03 Max torsional moment that can be resisted by the

concrete

CRTRD1 I 962.81141E-03 Ratio of the torsional moment (TSd) to the

Resistance TRd1

CRTCST I 929.82969E-03 Criterion for the ultimate compression strength of

the concrete

ASSH I 366.03010E-06 Area per unit length of defined shear reinforcement

ASTT I 1.03162E-03 Area per unit length of torsional transverse reinf.

defined

ASLT I 1.39842E-03 Total area of torsional longitudinal reinforcement

defined

VRD1 J 84.36960E+03 Design shear resistance without considering the

reinforcement

CRVRD1 J 459.28864E-03 Ratio of the design shear force (VSd) to the

resistance VRd1

VRD2 J 776.24998E+03 Max shear force allowed without crushing the concrete

CRVRD2 J 49.91949E-03 Ratio of the design shear force (VSd) to the

resistance VRd2

VWD J 38.75000E+03 Contribution of the necessary shear reinforcement

TRD1 J 83.73914E+03 Max torsional moment that can be resisted by the

concrete



10-37

CRTRD1 J 962.81141E-03 Ratio of the torsional moment (TSd) to the

resistanceTRd1

CRTCST J 929.49777E-03 Criterion for the ultimate compression strength of

the concrete

ASSH J 343.84646E-06 Area per unit length of defined shear reinforcement

ASTT J 1.03162E-03 Area per unit length of torsional transverse reinf.

defined

ASLT J 1.39842E-03 Total area of torsional longitudinal reinforcement

defined

Chapter 10 Concrete Structures - Eurocode 2Axial+Bending Reinforcement Check According to EC2Axial+Bending Reinforcement Design According to EC2Biaxial Bending + Axial Force Checking According to EC2Biaxial Bending + Axial Force Design According to EC2Shear and Torsion Reinforcement Check According to EC2

Beam Element 1 - Node IBeam Element 1 - Node JShear and Torsion Reinforcement Design According to EC2

e10eng

Documents

axial bending checking

ec2 example description

axial bending criterion

ec2 prep7

torsion reinforcement

axial bending combined

torsion reinforcement

civilfem preprocessor