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 = π × - Moment of inertia of area of bar: I = π ×
0.012 2 = 1.13x10-4 m2 4
0.012 4 = 1x10-9 m4 64
- 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 TU NGUYEN
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σ cu = σ 'cu (2(
ε ε ) − ( ) 2 ) (Design stress-strains for concrete Clause 3.1.7) ε0 ε0
σ cu = fcu when ε 0 ≤ ε c ≤ ε cu 2 Where: σ cu : stress, σ 'cu : maximum stress, ε : strains, ε 0 : maximum strain. Where: ε 0 =0.002, ε cu 2 = 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)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.001 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 0.0018 0.0019 0.002 0.0021 0.0022 0.0023 0.0024 0.0025 0.0026 0.0027 0.0028 0.0029 0.003 0.0031 0.0032 0.0033 0.0034 0.0035
2.925 5.7 8.325 10.8 13.125 15.3 17.325 19.2 20.925 22.5 23.925 25.2 26.325 27.3 28.125 28.8 29.325 29.7 29.925 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30
Young’s Modulus (N/mm2) 29250 28500 27750 27000 26250 25500 24750 24000 23250 22500 21750 21000 20250 19500 18750 18000 17250 16500 15750 15000 15000 15000 15000 15000 15000 15000 15000 15000 15000 15000 15000 15000 15000 15000 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 =
1
γs
fyAs ∴ 0.459x30x250x x = 0.87x460x452
The area of tension reinforcement: As = 4x π × r 2 = 4x π ×
∴ x = 53 mm ≤ 0.617h = 0.617x400 = 250mm
12 2 = 452 mm2 4
This neutral axis is at ultimate moment of resistance of the cross-section. Where:
γ c 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. γ s 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-
s 42.4 ) = 0.87x460x452x(400) x10-6= 68500 Nm 2 2
(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 TU NGUYEN
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Maximum moment: M =
PL 3× 68500 = 68500 Nm ∴ P = = 37400N 3 5.5
[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 longterm 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
∴ b× x
x = ae As (d − x) 2
(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 =
ES 200 = = 6.82 EC , eff 29.3
Where: effective modulus: Ec,eff = 29.3/(1+0) = 29.3 kN/mm2 (Assume that creep ignored therefore φ (∞, t 0 ) = 0, Ec,eff = Ecm/(1+ φ (∞, t 0 ) ) Therefore: 250x
x2 = 6.82x452(350-x) 2
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: TU NGUYEN
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400 h bh × + ae As d 250 × 400 × + 6.82 × 452 × 350 2 2 = = 204mm xSLS = bh + ae As 250 × 400 + 6.82 × 452 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 = b × = 250 ×
x3 + ae As (d − x) 2 3
82 3 + 6.82 × 452(350 − 82) 2 = 2.9x109 mm4 3
Calculate the curvature of the cracked section:
68.5 × 10 6 1 M = = 8.06x10-6 (mm)x = r EC ,eff I Cr 29.3 × 10 3 × 2.9 × 10 9 Calculate deflection of serviceability limit state due to cracked section:
Δ = f1 L2 (1 / r ) = 0.1065 × 5500 2 × 8.06 × 10 −6 = 26mm Where: f1 = (0.125 −
α2 6
) 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. TU NGUYEN
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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 TU NGUYEN
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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
o
PRXY (Poisson’s ratio): 0.2
2
N/m
(Please see previous calculation)
+ Rebar (Material Properties 2): TU NGUYEN
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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. Modelling -> Create -> Copy -> Nodes -> Copy - ITEM NUMBER OF COPIES: 4 - DX (X-offset in active CS): 0 - DY (X-offset in active CS): 0.1 - DZ (X-offset in active CS): 0 - INC (Node number increment): 6 + Generating node in Y direction for the rest of concrete. Picking node from 31 to 36. 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 - INC (Node number increment): 6 + Generating node in X direction - ITEM NUMBER OF COPIES: 26 - 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. TU NGUYEN
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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 - ITEM NUMBER OF COPIES: 5 - NINC (Node number increment): 6 + Generating elements in X direction - ITEM NUMBER OF COPIES: 25 - 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
To simply create element in 3D, at command-line: e,8,44 for
Rebar 2
9
45
Rebar 1.
Rebar 3
10
46
Similarly to creating node, the rebar 1 should be copy to the
Rebar 4
11
47
end of the beam: ITEM NUMBER OF COPIES: 25, and NODE NUMBER INCREMENT: 36
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1 ELEMENTS FEB 26 2010 12:01:25
Reinforcement
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.
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1 ELEMENTS FEB 26 2010 12:02:51
U PRES-NORM .150E+07
Y
Z
X
Pressure
Beam support
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 =
40000 = 1.5x106 N/m2 = 0.15x107N/m2 0.11× 0.25
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,
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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 CRACKS AND CRUSHING FEB 17 2010 22:51:17
STEP=1 SUB =1 TIME=.45
Region of crack and crush
Y Z
X
Figure 14 – Crack and Crush in half-beam However, crack and crush start occurring in the head of concrete block at the last step: CRACKS AND CRUSHING FEB 17 2010 22:47:14
STEP=1 SUB =33 TIME=.940747
Region of crack and crush
Y Z
X
Figure 15 – More crack and crush at the head of block
Crack and crush appear more the last step as shown in figure below:
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1 CRACKS AND CRUSHING FEB 26 2010 12:24:44
STEP=1 SUB =37 TIME=.980747
More cracks and crushes
Y Z
X
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 DISPLACEMENT FEB 17 2010 22:53:17
STEP=1 SUB =1 TIME=.45 RSYS=0 DMX =.013588
Y Z
X
Figure 17: Deformed shape at 0.45
At the last step, the deflection is now 0.043747 (m) = 44 mm
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1 DISPLACEMENT FEB 26 2010 12:26:08
STEP=1 SUB =37 TIME=.980747 RSYS=0 DMX =.043747
Y Z
X
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 Δ mm. 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%.
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ANSYS
Hand-calculation
13.6mm
26 mm
21
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 Bending moment Shear force
68.5 kNm 40 kN
ANSYS without Brick plate 69.6kNm 40.8kN
ANSYS with Brick plate 58.9kNm 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 ELEMENTS MAR
U
1 2010 21:00:01
PRES-NORM .150E+07
Y Z
X
Bearing plate modeled with SOLID45 as a brick (Please see appendix: Clear log file)
Crack and Crush Concrete
Figure 20.b – Structural model in ANSYS
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1 CRACKS AND CRUSHING MAR
STEP=1 SUB =58 TIME=.827031
1 2010 20:48:15
Y Z
X
Figure 21.a – Model of bearing plate at 83%
1 CRACKS AND CRUSHING MAR
STEP=1 SUB =23 TIME=.830747
1 2010 20:56:25
Y Z
X
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
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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.
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Appendix A – Log file without bearing plate /PREP7
!* Input data of Multilinear
N, ,0,0,0.2,,,,
/TITLE, Crack and Cursh Concrete
TBDE,MISO,1,,,
N, ,0,0,0.25,,,,
!* Element type
TB,MISO,1,1,35,0
FLST,4,6,1,ORDE,2
ET,1,SOLID65
TBTEMP,0
FITEM,4,1
ET,2,BEAM23
TBPT,,0.0001,2.925E6
FITEM,4,-6
R,1, , , , , , ,
TBPT,,0.0002,5.7E6
NGEN,2,6,P51X, , , ,0.05, ,1,
RMORE, , , , , , ,
TBPT,,0.0003,8.325E6
FLST,4,6,1,ORDE,2
RMORE, ,
TBPT,,0.0004,10.8E6
FITEM,4,7
TBPT,,0.0005,13.125E6
FITEM,4,-12
KEYOPT,2,2,0
TBPT,,0.0006,15.3E6
NGEN,4,6,P51X, , , ,0.1, ,1,
KEYOPT,2,4,0
TBPT,,0.0007,17.325E6
FLST,4,6,1,ORDE,2
KEYOPT,2,6,2
TBPT,,0.0008,19.2E6
FITEM,4,25
KEYOPT,2,10,0
TBPT,,0.0009,20.925E6
FITEM,4,-30
TBPT,,0.001,22.5E6
NGEN,2,6,P51X, , , ,0.05, ,1,
TBPT,,0.0011,23.925E6
FLST,4,36,1,ORDE,2
R,2,0.012,
TBPT,,0.0012,25.2E6
FITEM,4,1
!* Young's Modulus and Poisson's
TBPT,,0.0013,26.325E6
FITEM,4,-36
Ratio of concrete
TBPT,,0.0014,27.3E6
NGEN,26,36,P51X, , ,0.11, , ,1,
MPTEMP,,,,,,,,
TBPT,,0.0015,28.125E6
MPTEMP,1,0
TBPT,,0.0016,28.8E6
!* Creating SOLID65 element
MPDATA,EX,1,,29250E6
TBPT,,0.0017,29.325E6
E,1,2,38,37,7,8,44,43
MPDATA,PRXY,1,,.2
TBPT,,0.0018,29.7E6
E,2,3,39,38,8,9,45,44
!* Concrete
TBPT,,0.0019,29.925E6
E,3,4,40,39,9,10,46,45
TB,CONC,1,1,9,
TBPT,,0.002,30E6
E,4,5,41,40,10,11,47,46
TBTEMP,0
TBPT,,0.0021,30E6
E,5,6,42,41,11,12,48,47
TBDATA,,1,1,1e6,-1,,
TBPT,,0.0022,30E6
FLST,4,5,2,ORDE,2
TBDATA,,,,1,,,
TBPT,,0.0023,30E6
FITEM,4,1
TBPT,,0.0024,30E6
FITEM,4,-5
!* Young's Modulus and Poisson's
TBPT,,0.0025,30E6
EGEN,5,6,P51X, , , , , , , , , , ,
Ratio of steel
TBPT,,0.0026,30E6
FLST,4,1,2,ORDE,1
MPTEMP,,,,,,,,
TBPT,,0.0027,30E6
FITEM,4,21
MPTEMP,1,0
TBPT,,0.0028,30E6
FLST,4,25,2,ORDE,2
MPDATA,EX,2,,200e9
TBPT,,0.0029,30E6
FITEM,4,1
MPDATA,PRXY,2,,.3
TBPT,,0.003,30E6
FITEM,4,-25
TBPT,,0.0031,30E6
EGEN,25,36,P51X, , , , , , , , , , ,
!* Plastic data - yield stress
TBPT,,0.0032,30E6
NPLOT
TB,BISO,2,1,2,
TBPT,,0.0033,30E6
!* Element Type 2
TBTEMP,0
TBPT,,0.0034,30E6
TYPE, 2
TBDATA,,460e6,,,,,
TBPT,,0.0035,30E6
MAT,
!* Creating Node
REAL,
2
N, ,0,0,0,,,,
ESYS,
0
N, ,0,0,0.05,,,,
SECNUM,
N, ,0,0,0.1,,,,
TSHAP,LINE
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!* Creating BEAM23 element
D,P51X, , , , , ,UX, , , , ,
!* Analysis type
E,8,44
FLST,2,6,1,ORDE,2
TIME,1
E,9,45
FITEM,2,1
AUTOTS,1
E,10,46
FITEM,2,-6
NSUBST,0,0,0
E,11,47
D,P51X, , , , , ,UY,UZ, , , ,
OUTRES,ALL,ALL
FLST,4,4,2,ORDE,2
EPLOT
FITEM,4,626
FLST,5,5,2,ORDE,2
/VIEW,1,1,1,1
FITEM,4,-629
FITEM,5,396
/REPLOT
EGEN,25,36,P51X, , , , , , , , , , ,
FITEM,5,-400
FINISH
N, ,0,0,0.15,,,,
CM,_Y,ELEM
/SOL
!* Apply boundary condition
ESEL, , , ,P51X
! /STATUS,SOLU
FLST,2,36,1,ORDE,2
CM,_Y1,ELEM
SOLVE
FITEM,2,901
CMSEL,S,_Y
!* DONE
FITEM,2,-936
CMDELE,_Y SFE,_Y1,6,PRES, ,1.5e6, , ,
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
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