Modeling of Reinforced Concrete Beam

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Practical Application of Finite Element Analysis

Modeling of Reinforced Concrete Beam Using ANSYS software

BSc. Tu Trung Nguyen

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Table of Content

I. Introduction................................................................................................................................ 3 II. Specification ............................................................................................................................. 3 III. Data......................................................................................................................................... 3 IV. Determination of maximum load and deflection...................................................................... 6

IV.1. Shear capacity.................................................................................................................. 6 V.2. Deflection .......................................................................................................................... 8

V. Using element types ................................................................................................................ 9 VI. Assumptions of element type and disadvantages................................................................. 10

VI.1. SOLID65......................................................................................................................... 10 VI.2. BEAM23 ......................................................................................................................... 10

VII. Simplified ANSYS model ..................................................................................................... 11 VII.1. Element types................................................................................................................ 11 VII.2. Real Constants.............................................................................................................. 12 VII.3. Material properties......................................................................................................... 12 VII.4. Modelling ....................................................................................................................... 13 VII.5. Creating element ........................................................................................................... 14 VII.6. Applying boundary condition ......................................................................................... 16 VII.7 Solution (Solve current LS) ............................................................................................ 17 VII.8. Results .......................................................................................................................... 17

VIII. Discussion........................................................................................................................... 21 VIII.1 Deflection ...................................................................................................................... 21 VIII.2 Result ............................................................................................................................ 22 VIII.3 Bending moment and Shear force................................................................................. 22 VIII.4 Structural model ............................................................................................................ 22 VIII.5 Convergence ................................................................................................................. 24

IX. Conclusion ............................................................................................................................ 25 Reference:.................................................................................................................................. 26 Appendix A – Log file without bearing plate ............................................................................... 27 Appendix B – Log file with bearing plate .................................................................................... 28 Appendix C – LINK8 (3D Spar) .................................................................................................. 28

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I. Introduction

This purpose is to investigate the critical of shear strength of reinforced concrete (RC) beams is

very significant particularly when this value is used in the practical design. A value of the shear

capacity may lead to an unpredicted and at stage brittle collapse of the structural RC beams.

Therefore, this modelling of shear crack is predicted by using Finite-Element analysis in the

purpose. In this aim, ANSYS, which is finite-element software, is applied to confirm and verify

that this RC simply supported RC beam can be achieved.

II. Specification

- Using eight-node SOLID65 and 2 or 3D Spar or BEAM element in ANSYS is as reinforced

concrete element to confirm and verify the shear capacity led to an unexpected and crack.

- Material properties:

o Linear Elastics

o Non-linear (Stress-Strain curves for concrete and steel, cracking for concrete

element)

- Real Constants: Any general geometric properties which are applicable to any element.

- Model details:

o Yield stress of concrete: fcu = 30 N/mm2

o Yield stress of Steel fy = 460 N/mm2

- Geometry of simply supported beam shown below:

Figure 1

Where: B: Breadth H: Height L: Length

III. Data

- Breadth of the beam (B) 250mm

- Height of the beam (H) 400mm

- Length of the beam (L) 5500mm

- Bar diameter φ 12mm

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- Number of bars 4

Figure 2

- Compressive strength of concrete: fcu = 30 N/mm2

- Yield strength of structural steel fy = 460 N/mm2

- Young’s modulus of steel: E = 200000 N/mm2

- Cross-sectional area of one bar: A = 4

012.0 2

×π = 1.13x10-4 m2

- Moment of inertia of area of bar: I = 64012.0 4

×π = 1x10-9 m4

- Determine stress-strain curve and Young’s modulus of concrete:

In this situation, the ratio between stress and strain must be equal to Young’s module at the first

point of stress-strain curve, and then this ratio is decreased to the last data when the

compressive strength increases. As the figure below shown, the cross-area is safe-area, where

the reinforced concrete does not crack or crush.

Figure 8

A: Safe area, B: Start cracking, C: Totally collapsed

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))()(2(' 2

00 εε

εεσσ −= cucu (Design stress-strains for concrete Clause 3.1.7)

cuσ = fcu when 20 cuc εεε ≤≤

Where: cuσ : stress, cu'σ : maximum stress, ε : strains, 0ε : maximum strain.

Where: 0ε =0.002, 2cuε = 0.0035

(Table 3.1 Strength classes for concrete – BS EN 1992-1-1:2004 – EC2-1-1)

Figure 9 – Stress-strains curve of concrete

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Table of stress-strain curve of concrete

No. Strains Stress (N/mm2) Young’s Modulus (N/mm2)

1 0.0001 2.925 29250 2 0.0002 5.7 28500 3 0.0003 8.325 27750 4 0.0004 10.8 27000 5 0.0005 13.125 26250 6 0.0006 15.3 25500 7 0.0007 17.325 24750 8 0.0008 19.2 24000 9 0.0009 20.925 23250

10 0.001 22.5 22500 11 0.0011 23.925 21750 12 0.0012 25.2 21000 13 0.0013 26.325 20250 14 0.0014 27.3 19500 15 0.0015 28.125 18750 16 0.0016 28.8 18000 17 0.0017 29.325 17250 18 0.0018 29.7 16500 19 0.0019 29.925 15750 20 0.002 30 15000 21 0.0021 30 15000 22 0.0022 30 15000 23 0.0023 30 15000 24 0.0024 30 15000 25 0.0025 30 15000 26 0.0026 30 15000 27 0.0027 30 15000 28 0.0028 30 15000 29 0.0029 30 15000 30 0.003 30 15000 31 0.0031 30 15000 32 0.0032 30 15000 33 0.0033 30 15000 34 0.0034 30 15000 35 0.0035 30 15000

IV. Determination of maximum load and deflection

IV.1. Shear capacity - Assume that effective length is 5.5m long.

When the concrete includes rebars to prevent tension at the bottom of concrete block, the

neutral axis now located on a different place from the centroid of concrete section. The reason

for this is reinforced concrete assumes that the concrete is cracked in location of tensile strains.

Therefore, after cracking, all of the tension is carried by the reinforcement. In addition,

according to the reinforcement is good at tension and compression. The tensile capacity of

concrete redistributes. Consequently, there must be linear distribution of strains to make the

compressive and tensile forces in equilibrium state on the section.

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Figure 3

+ For equilibrium of the compressive and tensile forces on the section due to rectangular parabolic stress block:

Fcc = Fst

Where: Fcc compressive force of concrete, Fst: tensile force of reinforcement. Therefore:

0.459fcub x = sγ

1fyAs ∴ 0.459x30x250x x = 0.87x460x452

The area of tension reinforcement: As = 4x 2r×π = 4x4

122

×π = 452 mm2

∴ x = 53 mm ≤0.617h = 0.617x400 = 250mm This neutral axis is at ultimate moment of resistance of the cross-section. Where: cγ is the usual partial safety factor for the strength of concrete.

α is the factor allowing the different between bending strength and the cylinder crushing strength of the concrete.

sγ is the usual partial safety factor for the strength of steel. Therefore: s = x x0.8 = 42.4mm Ultimate moment of resistance of the section is at final collapse:

M = Fst x z = 0.87xfyx As(h-2s

) = 0.87x460x452x(400-2

4.42) x10-6= 68500 Nm

(5.2.4.1 ( Bending) in IStructE manual for the design of concrete building structures to Eurocode 2) + Determine maximum concentrated load: Assume that there are 2 concentrated loads on the beam as shown in figure below:

Figure 4

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Maximum moment: M = 3

PL = 68500 Nm ∴ P =

5.5685003×

= 37400N [1]

Therefore, maximum concentrated load making the RC crack is 37400N Say 40000N to confirm that the RC is completely collapsed.

V.2. Deflection

To determine deflection of reinforced concrete, the section assumes that this loading is long-

term duration and reinforcement and concrete work together. Then calculation of deflection

needs to be determined the curvature of section. (Bill Mosley and John Bungey and Ray Hulse,

2007, p136). Therefore, the deflection is affected by the type and size of reinforcement and

stiffness of concrete is also assumed that it has small effects (Ec,eff) to make reinforcement and

concrete co-operate.

Calculating the curvature of the cracked section the moment of inertia value of the transformed

concrete section must be recalculated. Taking area moments about the neutral axis

∴ )(2

xdAaxxb se −=×

(Bill Mosley and John Bungey and Ray Hulse, 2007, p136~138)

Where: ae is the modular ratio equal to the ratio of the elastic modulus of the reinforcement to

that of the concrete.

ae = effC

S

EE

,=

3.29200

= 6.82

Where: effective modulus: Ec,eff = 29.3/(1+0) = 29.3 kN/mm2 (Assume that creep ignored therefore ),( 0t∞φ = 0, Ec,eff = Ecm/(1+ ),( 0t∞φ )

Therefore: 250x2

2x= 6.82x452(350-x) Because of 50mm cover, d = h-50 = 350

Neutral axis: x = 81.4 mm (Due to the reinforced concrete beam is cracked in serviceability limit state). This neutral axis can compare with neutral axis of uncracked section in serviceability limit state calculated below:

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xSLS = 45282.6400250

35045282.62

4004002502

×+×

××+××=

+

+×

se

se

Aabh

dAahbh = 204mm

In comparison of neutral axis, the neutral axis of cracked section moved up to the top surface of

reinforced concrete. Concrete does not co-operate now with the reinforcement, when the

cracked concrete occurs at the bottom. The area of concrete will be now smaller than before in

period of crack and crush occurring. Therefore, neutral axis is move to compressive region of

reinforced concrete. This reinforcement will displace, the deflection is now calculated due to

curvature of cracked section.

Moment of inertia: ICr = 23

)(3

xdAaxb se −+×

= 23

)82350(45282.63

82250 −×+× = 2.9x109 mm4

Calculate the curvature of the cracked section:

CreffC IE

Mr ,

1= = 93

6

109.2103.29105.68

××××

= 8.06x10-6 (mm)x

Calculate deflection of serviceability limit state due to cracked section:

)/1(21 rLf=Δ = 62 1006.855001065.0 −××× = 26mm

Where: f1 = )6

125.0(2α

− is deflection coefficient for various loading and restraint condition.

(Concrete Building Design, Table D2.4 Deflection coefficient f1 various loading and restraint condition)

V. Using element types

There are 2 options of element types to be used in this solution.

- SOLID65 is used for the three-dimensional modelling of solids with or without reinforcing bars

(rebars). The solid is capable of cracking in tension and crushing in compression. In concrete

applications, for example, the solid capability of the element may be used to model the concrete

while the rebar capability is available for modelling reinforcement behaviours. Other cases for

which the element is also applicable would be reinforced composites (such as fiberglass), and

geological materials (such as rock). The element is defined by eight nodes having three

degrees of freedom at each node: translations in the nodal x, y, and z directions. Up to three -

different rebar specifications may be defined. (ANSYS help, version 10 Ed)

In addition, SOLID65 element has a special cracking and crushing capabilities. However, the

most important aspect of this element is the treatment of nonlinear material properties. The

concrete is capable of cracking in three orthogonal directions, crushing, plastic deformation, and

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creep. The rebars are capable of tension and compression, but not shear. They are also

capable of plastic deformation and creep. - BEAM23 is a uniaxial element with tension-compression and bending capabilities. The

element has three degrees of freedom at each node: translations in the nodal x and y direction

and rotation about the nodal z-axis. Moreover, the element has plastic, creep, and swelling

capabilities. Therefore, the rebars are modelled by BEAM23 in this problem. (ANSYS help,

version 10 Ed). This element can compare to 3D spar (LINK8) for modelling reinforcements in

concrete.

VI. Assumptions of element type and disadvantages

VI.1. SOLID65 - Volume elements are not allowed to be equal to zero.

- Elements may be numbered planes (ANSYS help – Solid65). Also, the element may not be

warped such that the element has two separate volumes. This occurs most frequently when the

elements are not numbered properly.

- All elements must have eight nodes.

- A tetrahedron shape is also available.

- The extra shapes are automatically deleted for tetrahedron elements.

- The rebar capability of the element is used, the rebars are assumed to be smeared throughout

the element. The sum of the volume ratios for all rebars must not be greater than 1.0.

- The element is nonlinear and requires a repeating solution. When both cracking and crushing

are used together, concern must be taken to apply the load slowly to prevent possible

pretended crushing of the concrete before proper load transfer can occur through a closed

crack. This usually happens when excessive cracking strains are coupled to the orthogonal

uncracked directions through Poisson's effect. Also, at those integration points where crushing

has occurred, the output plastic and creep strains are from the previous converged sub-step.

Furthermore, when cracking has occurred, the elastic strain output includes the cracking strain.

The lost shear resistance of cracked and/or crushed elements cannot be transferred to the

rebars, which have no shear stiffness.

There are some disadvantages following the options above are presented in case of cracking or

crushing nonlinearities:

- Stress-stiffening effects.

- Large strain and large deflection. The results may be incorrect, especially if significantly large

rotation is involved.

VI.2. BEAM23 - The beam element must lie in an X-Y plane and must not have a zero length or area.

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- The height is used in calculating the bending and thermal stresses and for locating the

integration points.

- For the rectangular section, the input area, moment of inertia, and height should be consistent

with each other.

- The effect of implied offsets on the mass matrix is ignored.

VII. Simplified ANSYS model

- ANSYS 10 or 11 ED (Education version or Academic version) will be used for modelling the

structure. A disadvantage of this software is the limitation of nodes (10000 nodes) and the

maximum amount of elements (1000 elements). Therefore, the reinforced concrete is restricted

to model in the range of element given. The results may be acceptable in this situation.

Figure 5

- The model assumes that there is no cover at the head of beam, it means the length of

reinforcement is same the length of the beam (L= 5.5m).

- The beam will be modelled with one-half of the beam.

VII.1. Element types Preprocessor -> Element type -> Add/Edit/Delete -> Add

Choose Concrete 65 (SOLID65)

Figure 6

Similarly to choose: BEAM -> PLASTIC 23 (BEAM23)

In the OPTION of BEAM23, choose ROUND SOLID BAR at Cross-section K6

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VII.2. Real Constants

Preprocessor -> Real Constants -> Add/Edit/Delete -> Add

- Choosing SOLID65 as SET 1 and no input data at here because the rebar will be modelled as

BEAM23. In addition, SOLID65 element only supports 3 rebars however there are 4 rebars in

this problems.

- Similarly to choose BEAM23 as SET 2: OUTER DIAMETER OD: 0.012

VII.3. Material properties - Structural -> Nonlinear -> Inelastic -> Rate Independent -> Isotropic Hardening Plasticity ->

Mises Plasticity -> Multilinear: Please see table of stress-strain curve of concrete (Previous

calculation)

- Structural -> Nonlinear -> Inelastic -> Non-linear Metal Plasticity -> Concrete

o Shear transfer coefficients for an open crack (ShrCf-Op): 1

o Shear transfer coefficients for a closed crack (ShrCf-Cl): 1

o Uniaxial tensile cracking stress (UnTensSf): 1E6

o Uniaxial crushing stress (positive) (UnComSt): -1

There are 2 material properties needing to be input. One is concrete, one is rebar.

Preprocessor -> Material Props -> Material Models

Figure 7

+ Concrete (Material Model Number 1):

- Structural -> Linear -> Elastics -> Isotropic:

o EX (Young’s modulus): 2.9250E10 N/m2 (Please see previous calculation)

o PRXY (Poisson’s ratio): 0.2

+ Rebar (Material Properties 2):

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- Structural -> Linear -> Elastics -> Isotropic:

o EX (Young’s modulus): 2E11

o PRXY (Poisson’s ratio): 0.3

- Structural -> Nonlinear -> Inelastic -> Rate Independent -> Isotropic Hardening Plasticity ->

Mises Plasticity -> Bilinear

o Yield Stress: 460 N/mm2

o Tang mod: 0

Figure 9

VII.4. Modelling The beam given is symmetrical geography and concentrated load, therefore, one half of the beam will be taken for simplification of computer model. L = 5.5/2 = 2.75mm D = 0.4m

B = 0.25m There are 4 rebars, the cover is 0.05m Therefore, the model will have 936 nodes (6 nodes in Z direction, 6 nodes in Y direction, 26

nodes in X direction) then the number of element is 5x5x25 = 625 elements < 1000 elements.

Structure is modelled with first-six-nodes in Z direction, after that using COPY function to finish

the model.

Preprocessor -> Modelling -> Create -> Nodes -> In Active CS

Node X Y Z

1 0.00 0.00 0.00

2 0.00 0.00 0.05

3 0.00 0.00 0.10

4 0.00 0.00 0.15

5 0.00 0.00 0.20

6 0.00 0.00 0.25

- These nodes need to copy to become the structural model.

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Due to the cover of concrete is 50mm, so that there are 2 differences of modelling concrete on

Y direction.

Co-ordinate Distance from NODE I to NODE J

Axis X 0.11

Axis Y 0.1 and 0,5

Axis Z 0.05

+ Generating node in Y direction for the reinforcement

Modelling -> Create -> Copy -> Nodes -> Copy

- ITEM NUMBER OF COPIES: 2

- DX (X-offset in active CS): 0

- DY (X-offset in active CS): 0.05

- DZ (X-offset in active CS): 0

+ Generating node in Y direction for the rest of concrete. Picking node from 7 to 12.






- INC (Node number increment): 6

+ Generating node in Y direction for the rest of concrete. Picking node from 31 to 36.






- INC (Node number increment): 6

+ Generating node in X direction


- DX (X-offset in active CS): 0.11

- DY (X-offset in active CS): 0

- DZ (X-offset in active CS): 0 - INC (Node number increment): 30

VII.5. Creating element SOLID65 will be created with all nodes. The node list should be opened to simply create each

element.

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Element Attributes of SOLID65:

- Element type of number : SOLID65

- Material Number: 1

- Real Constant set number: 1

Creating SOLID65 element, Command-line should be input E,1,2,38,37,7,8,44,43 because of a

simple creation in three-dimension (3D). Similar way to the other SOLID65 element.

Element Input Command-line

Concrete block 1 E,1,2,38,37,7,8,44,43

Concrete block 2 E,2,3,39,38,8,9,45,44

Concrete block 3 E,3,4,40,39,9,10,46,45

Concrete block 4 E,4,5,41,40,10,11,47,46

Concrete block 5 E,5,6,42,41,11,12,48,47

+ Generating elements in Y direction


- NINC (Node number increment): 6

+ Generating elements in X direction


- NINC (Node number increment): 36

Element Attributes of BEAM23:

- Element type of number : BEAM23

- Material Number: 2

- Real Constant set number: 2

Element Node I Node J Comment on creating

Rebar 1 8 44

Rebar 2 9 45

Rebar 3 10 46

Rebar 4 11 47

To simply create element in 3D, at command-line: e,8,44 for

Rebar 1.

Similarly to creating node, the rebar 1 should be copy to the

end of the beam: ITEM NUMBER OF COPIES: 25, and NODE

NUMBER INCREMENT: 36

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1

FEB 26 201012:01:25

ELEMENTS

Figure 11 – Structural Model finished

VII.6. Applying boundary condition - Solution Type

o Solution -> Analysis Type -> New Analysis -> Choose Structural

o Solution -> Sol’n Controls

Frequency: Write every substep (Investigation cracks start to take shape in

the reinforced concrete)

Automatic time stepping: ON

- Define loads:

o Solution -> Define Loads -> Apply -> Structural -> Displacement -> On Node

- UX is applied for nodes from 901 to 936 at the end of the structural model. In fact

that when half-beam is modelled, the middle of the beam can not move in horizontal

direction because of rigid connection however that point can displace in vertical

direction. Therefore, UX will restrain movement in horizontal direction (X axis).

- UY and UZ is applied for nodes 1, 2, 3, 4, 5 and 6. Indeed, when the one side of

the beam is to be put on the brick pad. Therefore, the beam can move in X direction,

however it can not move in Y direction and Z direction. That UX, UZ will restrain

movement in the two-direction.

Reinforcement

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1

X

Y

Z

FEB 26 201012:02:51

ELEMENTS

U

PRES-NORM.150E+07

Figure 12

o Solution -> Define Loads -> Apply -> Pressure -> On Elements (External load

applieds for investigating cracks and crush of concrete at L/3 = 1.8666(m))

The 40000N applies at the sixteenth element on top of the reinforced concrete.

However, the top surface is a rectangular with 0.11x0.25, therefore the pressure

applies for the area is:

Pressure = 25.011.0

40000×

= 1.5x106 N/m2 = 0.15x107N/m2

VII.7 Solution (Solve current LS) To analyse crack and crush in concrete: Solution -> Analysis Type -> New Analysis: Static has to be chosen. Sol’n Controls: - Time at end of loadstep: 1 - Automatic time stepping: ON - Number of substeps is to be chosen - Number of substeps: 1 - Frequency: Write every substep. The solution will be terminated at 99% as shown in time-line history (figure 13) below

VII.8. Results The result will show 38 sub-steps from 45% up to 99%. to view the region of crack and crush: - General PostProc -> Read Results -> By Pick,

Beam support

Pressure

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- General PostProc -> Plot Results -> Concrete Plot -> Crack/Crush

o Plot symbols are located at: Integration pts o Plot crack faces for: any cracks

Time history:

Figure 13 – Time-line history

Investigation of a cracked line is at 0.45 of time-line:

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1

X

Y

Z

FEB 17 201022:51:17

CRACKS AND CRUSHING

STEP=1SUB =1TIME=.45

Figure 14 – Crack and Crush in half-beam

However, crack and crush start occurring in the head of concrete block at the last step:

X

Y

Z

FEB 17 201022:47:14

CRACKS AND CRUSHING


Figure 15 – More crack and crush at the head of block

Crack and crush appear more the last step as shown in figure below:

Region of crack and crush

Region of crack and crush

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1

X

Y

Z

FEB 26 201012:24:44

CRACKS AND CRUSHING


Figure 16 –The region of crack and crush

Deformed Shape:

The displacement of the beam is increasing when the time is going up.

At 0.45: the deflection is 0.0136 m = 13.6 mm:

1

X

Y

Z

FEB 17 201022:53:17

DISPLACEMENT

STEP=1SUB =1TIME=.45RSYS=0DMX =.013588

Figure 17: Deformed shape at 0.45

At the last step, the deflection is now 0.043747 (m) = 44 mm

More cracks and crushes

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1

X

Y

Z

FEB 26 201012:26:08

DISPLACEMENT

STEP=1SUB =37TIME=.980747RSYS=0DMX =.043747

Figure 18 – Last step of deflection

VIII. Discussion

VIII.1 Deflection The deformed shaped of reinforced concrete calculated by hand has a small difference from

deflection ANSYS. Indeed, the result of hand-calculation is significantly smaller than the

deflection at time at 0.45, 26mm and 13.6mm respectively. This can explain that 26 mm is a

deflection of reinforced concrete in serviceability limit state. However, the time at 0.45 given by

computer analysed is the first step that computer can shows. The deflection of this step shows a

wide region of crack and crush, so that it is not a first step when the reinforced concrete starts

showing cracks and crushes. In addition, when the lateral load impacts on the reinforced

concrete up to reinforced concrete appearing cracks, the deformation of reinforced concrete is

given Δmm. Therefore, when the reinforced concrete is completely collapsed, the deflection is

more significantly than hand-calculation.

There is one disadvantage which is number of stepping. This step started at 0. 3 and then it

down to 0.05, however it increases slowly to 0.99. However, the result will show at 0.45. In this

situation, the sub-step could not be in control of result below 0.45. The programme will be

terminated at 0.99 in comparison with 99%.

ANSYS Hand-calculation

13.6mm 26 mm

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VIII.2 Result The solution was solved in different cases:

1. Tang Mod of steel in stress-strain diagram is 2E9. In this case, the result will be terminated at

43%, sometimes it could not reach 30%.

2. The results will be converged at 83% when the support in Z direction (UZ) is to be removed.

3. Model with whole of beam, this situation is very good at result with 100%. However, element

of model is out of limitation and can not take full benefit of maximum element in the software.

The reason for this is the number of elements of the beam is more than half-beam (1250 and

625 respectively). The element of the software is given 1000 elements.

4. Element of reinforcement is used in 2 different ways. One model is with BEAM23, the other

uses LINK8 (3D Spar) for modelling. However, there is no difference between both models.

They show the same results in comparison (please see appendix for information of LINK8).

Type of models Success

BEAM23 full-beam modelled 100%

BEAM23 without brick plate 99%

BEAM23 with brick plate 83%

LINK8 Terminated at 75% due to error in symeqn called by dstmask

VIII.3 Bending moment and Shear force The results are similar. However, the result of ANSYS with Brick plate is slight different from

hand-calculation approximately 14%. As a consequence of divergence is at 83%, therefore the

result does not show a similarity.

Hand-calculation ANSYS without Brick

plate ANSYS with Brick plate

Bending moment 68.5 kNm 69.6kNm 58.9kNm Shear force 40 kN 40.8kN 34.5kN

VIII.4 Structural model The structure above is modelled with out bearing. Therefore, there are a lot of cracks and

crushes at the support as the result in figure 16. It can explain that there is no support It means

reaction at support is now considered for concentrated load, the cracking line now occurs at the

concentrated load with around 450.

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Figure 19 – Cracking line

However, the beam is modelled with a bearing plate, 0.11m long, 0.25m wide and 0.05m high

as shown in figure 20 below, it gives different result at approximately 83% of time-line (the

model of bearing plate is not converged at 82%) in comparison with the previous model. (figure

21). The area of reinforced concrete at bearing pad is now suffered by pressure as a

consequence. Therefore, the plate area shows cracks and crushes less than the previous

model

Figure 20.a – Concept of structural model

1

X

Y

Z

Crack and Crush Concrete

MAR 1 201021:00:01

ELEMENTS

U

PRES-NORM.150E+07

Figure 20.b – Structural model in ANSYS

Bearing plate modeled with SOLID45 as a brick (Please see appendix: Clear log file)

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1

X

Y

Z

MAR 1 201020:48:15

CRACKS AND CRUSHING


Figure 21.a – Model of bearing plate at 83%

1

X

Y

Z

MAR 1 201020:56:25

CRACKS AND CRUSHING


Figure 21.b – Model of no bearing plate at 83%

VIII.5 Convergence The solution was diverged. The reason for this is elements was solved with nonlinearity,

therefore they have to depend on properties such as large displacement analysis, plastic

material, gap elements, hook element and surface contacts. Then the solution will be

considered longer than a linear analysis. On the other hand, the arc-length method causes the

strain-stress curved equilibrium iterations to converge along an arc, thereby often preventing

divergence, even when the slope of the load versus deflection curve becomes zero or negative.

This iteration method is represented schematically as shown in figure below (Peter Budgell, ud).

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Figure 22: Converged diagram

Another reason for converged problems is “shear transfer coefficients for” an open crack and a

close crack. A number of preliminary analyses were attempted in this study with various values

for the coefficients within this range to avoid these problems (R.Santhakumar and et. al, 2007).

IX. Conclusion

The general explanation is that using 3D ANSYS modelling which is properly suggested the

nonlinear behaviour of RC beams with shear reinforcement.

ANSYS 3D concrete element is very good concerning the flexural and shear crack development

but poor concerning the crushing state. However this deficiency could be easier removed by

employing a certain multi-linear plasticity options available in ANSYS.

By using ANSYS smeared approach for beams with moderate shear span we are not able to

replicate satisfactory the softening due to big sliding emerging at the critical shear crack. That is

likely to be more realistically achieved by 3D discrete crack approach.

The assumptions suggest that we need some correction factors to adjust the values of material

parameters available from the experiment and convert them to effective parameters related to

the particular modelling.

An important feature of the present 3D modelling is that the flexural, shear/flexural and the

critical shear crack are prescribed, which means that their position and length must be

determined in advance using a certain method. In this document, the linear fracture mechanics.

A 3D isoperimetric ANSYS element with eight-node solid is used to represent the concrete

continuum and a beam element with plasticity options to model the reinforcement bars.

TU NGUYEN 26

Reference:

1. ANSYS help – Concrete (SOLID65), Plastic beam (BEAM23), Brick (SOLID45)

2. ANSYS Theory Reference - Section 14.65

3. ANSYS help: Chapter 9: Hourglass stiffness factor

4. Bill Mosley and John Bungey and Ray Hulse (2007) Reinforced Concrete Design to

Eurocode 2, Palgrave Macmillan.

5. Chanakya Arya (2004) Design of Structural Elements – Concretes, steelworks, masonry

and timber design to British Standards and Eurocodes, Spon Press.

6. Peter Budgell (ud) What is Non-linearity, website: http://www.2doworld.com, access

date: 2/2010.

7. James R. Clifton and Nicholas J. Carino (1995) Prediction of Cracking in Reinforced

Concrete Structures, NISTIR 5634, Building and Fire Research Laboratory, National

Institute of Standards and Technology, Gaithersburg.

8. R.Santhakumar and et. al (2007) Behaviour of Retrofitted Reinforced Concrete Beams

under Combined Bending and Torsion : A numerical study, Electronic Journal of

Structural Engineering. Accessed 15/3/2010, website:

http://www.ejse.org/Archives/Fulltext/2007/Special/200709.pdf, p3.

9. S. Parvanova, K. Kazakov, I. Kerelezova, G. Gospodinov and M. P. Nielsen (no.date) On a

Diagonal Crack Numerical Model of RC beam with No Shear Reinforcement, University

of Architecture, Civil Engineering and Geodesy, 1 Smirnenski blv., Sofia, Bulgaria.

10. Jonathan Haynes (2009) Concrete Building Design – Level M, University of Salford, p62.

TU NGUYEN 27

Appendix A – Log file without bearing plate

/PREP7

/TITLE, Crack and Cursh Concrete

!* Element type

ET,1,SOLID65

ET,2,BEAM23

R,1, , , , , , ,

RMORE, , , , , , ,

RMORE, ,

KEYOPT,2,2,0

KEYOPT,2,4,0

KEYOPT,2,6,2

KEYOPT,2,10,0

R,2,0.012,

!* Young's Modulus and Poisson's

Ratio of concrete

MPTEMP,,,,,,,,

MPTEMP,1,0

MPDATA,EX,1,,29250E6

MPDATA,PRXY,1,,.2

!* Concrete

TB,CONC,1,1,9,

TBTEMP,0

TBDATA,,1,1,1e6,-1,,

TBDATA,,,,1,,,

!* Young's Modulus and Poisson's

Ratio of steel

MPTEMP,,,,,,,,

MPTEMP,1,0

MPDATA,EX,2,,200e9

MPDATA,PRXY,2,,.3

!* Plastic data - yield stress

TB,BISO,2,1,2,

TBTEMP,0

TBDATA,,460e6,,,,,

!* Input data of Multilinear

TBDE,MISO,1,,,

TB,MISO,1,1,35,0

TBTEMP,0

TBPT,,0.0001,2.925E6

TBPT,,0.0002,5.7E6

TBPT,,0.0003,8.325E6

TBPT,,0.0004,10.8E6

TBPT,,0.0005,13.125E6

TBPT,,0.0006,15.3E6

TBPT,,0.0007,17.325E6

TBPT,,0.0008,19.2E6

TBPT,,0.0009,20.925E6

TBPT,,0.001,22.5E6

TBPT,,0.0011,23.925E6

TBPT,,0.0012,25.2E6

TBPT,,0.0013,26.325E6

TBPT,,0.0014,27.3E6

TBPT,,0.0015,28.125E6

TBPT,,0.0016,28.8E6

TBPT,,0.0017,29.325E6

TBPT,,0.0018,29.7E6

TBPT,,0.0019,29.925E6

TBPT,,0.002,30E6

TBPT,,0.0021,30E6

TBPT,,0.0022,30E6

TBPT,,0.0023,30E6

TBPT,,0.0024,30E6

TBPT,,0.0025,30E6

TBPT,,0.0026,30E6

TBPT,,0.0027,30E6

TBPT,,0.0028,30E6

TBPT,,0.0029,30E6

TBPT,,0.003,30E6

TBPT,,0.0031,30E6

TBPT,,0.0032,30E6

TBPT,,0.0033,30E6

TBPT,,0.0034,30E6

TBPT,,0.0035,30E6

!* Creating Node

N, ,0,0,0,,,,

N, ,0,0,0.05,,,,

N, ,0,0,0.1,,,,

N, ,0,0,0.2,,,,

N, ,0,0,0.25,,,,

FLST,4,6,1,ORDE,2

FITEM,4,1

FITEM,4,-6

NGEN,2,6,P51X, , , ,0.05, ,1,

FLST,4,6,1,ORDE,2

FITEM,4,7

FITEM,4,-12

NGEN,4,6,P51X, , , ,0.1, ,1,

FLST,4,6,1,ORDE,2

FITEM,4,25

FITEM,4,-30

NGEN,2,6,P51X, , , ,0.05, ,1,

FLST,4,36,1,ORDE,2

FITEM,4,1

FITEM,4,-36

NGEN,26,36,P51X, , ,0.11, , ,1,

!* Creating SOLID65 element

E,1,2,38,37,7,8,44,43

E,2,3,39,38,8,9,45,44

E,3,4,40,39,9,10,46,45

E,4,5,41,40,10,11,47,46

E,5,6,42,41,11,12,48,47

FLST,4,5,2,ORDE,2

FITEM,4,1

FITEM,4,-5

EGEN,5,6,P51X, , , , , , , , , , ,

FLST,4,1,2,ORDE,1

FITEM,4,21

FLST,4,25,2,ORDE,2

FITEM,4,1

FITEM,4,-25

EGEN,25,36,P51X, , , , , , , , , , ,

NPLOT

!* Element Type 2

TYPE, 2

MAT, 2

REAL, 2

ESYS, 0

SECNUM,

TSHAP,LINE

TU NGUYEN 28

!* Creating BEAM23 element

E,8,44

E,9,45

E,10,46

E,11,47

FLST,4,4,2,ORDE,2

FITEM,4,626

FITEM,4,-629

EGEN,25,36,P51X, , , , , , , , , , ,

N, ,0,0,0.15,,,,

!* Apply boundary condition

FLST,2,36,1,ORDE,2

FITEM,2,901

FITEM,2,-936

D,P51X, , , , , ,UX, , , , ,

FLST,2,6,1,ORDE,2

FITEM,2,1

FITEM,2,-6

D,P51X, , , , , ,UY,UZ, , , ,

FLST,5,5,2,ORDE,2

FITEM,5,396

FITEM,5,-400

CM,_Y,ELEM

ESEL, , , ,P51X

CM,_Y1,ELEM

CMSEL,S,_Y

CMDELE,_Y

SFE,_Y1,6,PRES, ,1.5e6, , ,

!* Analysis type

TIME,1

AUTOTS,1

NSUBST,0,0,0

OUTRES,ALL,ALL

EPLOT

/VIEW,1,1,1,1

/REPLOT

FINISH

/SOL

! /STATUS,SOLU

SOLVE

!* DONE

Appendix B – Log file with bearing plate

!* Brick plate ET,3,SOLID45 KEYOPT,3,1,0 KEYOPT,3,2,1 KEYOPT,3,4,0 KEYOPT,3,5,0 KEYOPT,3,6,0 R,3,0.15, !* Young's modulus and Poisson's ration of load pad MPTEMP,,,,,,,, MPTEMP,1,0 MPDATA,EX,3,,66e9 MPDATA,PRXY,3,,0.3

!* Creating plate Node N, ,0,-0.05,0,,,, N, ,0,-0.05,0.05,,,, N, ,0,-0.05,0.1,,,, N, ,0,-0.05,0.2,,,, N,940,0,-0.05,0.15,,,, N, ,0,-0.05,0.2,,,, N, ,0,-0.05,0.25,,,, FLST,4,6,1,ORDE,2 FITEM,4,937 FITEM,4,-942 NGEN,2,36,P51X, , ,0.11, , ,1,

!* Creating element TYPE, 3 MAT, 3 REAL, 3 ESYS, 0 SECNUM, TSHAP,LINE e,1,2,38,37,937,938,974,973 FLST,4,1,2,ORDE,1 FITEM,4,726 EGEN,5,1,P51X, , , , , , , , , , , !* Boundary condition FLST,2,36,1,ORDE,2 FITEM,2,901 FITEM,2,-936 D,P51X, , , , , ,UX, , , , , FLST,2,6,1,ORDE,2 FLST,2,12,1,ORDE,4 FITEM,2,937 FITEM,2,-942 FITEM,2,973 FITEM,2,-978 D,P51X, , , , , ,UY, , , , ,

Appendix C – LINK8 (3D Spar)

ET,2,LINK8

R,2,0.0001131, ,

!* please replace to BEAM23 element

Documents

Modeling of Reinforced Concrete Beam