Composites

Modeling

Assignment

Saibudeen, Babar Khan (2839130)M.Sc., COMMAS

Contents

1 Introduction 1

2 Elastic modulus and Failure properties of a ply 12.1 Longitudinal modulus, E1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Transverse modulus, E2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.3 Inplane shear modulus, G12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.4 Poisson’ ratio, ν12 and ν21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.5 Failure properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.6 Longitudinal tensile strength, F1t . . . . . . . . . . . . . . . . . . . . . . . . . 42.7 Longitudinal compressive strength, F1c . . . . . . . . . . . . . . . . . . . . . . 42.8 Transverse tensile and compressive strength, F2t and F2c . . . . . . . . . . . . 52.9 Inplane shear strength, F6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.10 Mechanical properties of a ply . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Maximum deflection from classical beam bending formula 5

4 Deflection from FEA 7

5 Failure analysis 9

6 Resin infusion simulation in the beam 10

7 Conclusion 12

List of Figures

1 Mechanical properties of Resin and Fibre . . . . . . . . . . . . . . . . . . . . 12 Actual stress distribution in the RVE [1] . . . . . . . . . . . . . . . . . . . . . 33 Screenshot of the results from LAP software . . . . . . . . . . . . . . . . . . . 74 Deflection of the laminate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Failure analysis of the beam with hole . . . . . . . . . . . . . . . . . . . . . . 96 First ply failure of the beam with hole . . . . . . . . . . . . . . . . . . . . . . 107 Resin filling time for the given BC . . . . . . . . . . . . . . . . . . . . . . . . 118 Resin filling time under 1 hour . . . . . . . . . . . . . . . . . . . . . . . . . . 11

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Figure 1: Mechanical properties of Resin and Fibre

1 Introduction

The problem has a hollow beam with the dimensions of 40 mm width, 80 mm height and400 mm length. The material is made of composites with symmetric lay-up, which differin their angle of orientation, made out of T300 carbon fibre and Hexcel 8551-7 epoxy resin.The lay-up of the plies are [0, 90, 45,−45]s. Each ply has a thickness of 0.25 mm. It is acantilever beam which is one end being fixed and the other end being free. A pressure loadof 0.000625 kN/mm2 is applied to the upper beam surface. The questions of the problem areaddressed in the following sections.

2 Elastic modulus and Failure properties of a ply

The material properties for the fibre and the resin are given in the figure 1. Using these values,the elastic modulus and failure properties for the ply are calculated by using micromechan-ics. There are different models, namely Rule of Mixtures (ROM), Hopkins-Chamis model andHalpin-Tsai relation, available to calculate these properties. Each model has a different way ofmodeling the Representative Volume Element (RVE) for the composite and so the mechanicalproperties calculated are different for each modeling.

2.1 Longitudinal modulus, E1

The longitudinal modulus E1 is calculated from the Rule of Mixtures. The modeling of themodulus is good and give a reasonable result. This gives the formula,

E1 = EfVf + EmVm (1)

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where,

Ef : Longitudinal modulus of fibreEm : Modulus of matrixVf : Fibre volume ratioVm : Matrix volume ratio

2.2 Transverse modulus, E2

The transverse modulus E2 is calculated from the Hopkins-Chamis model with the equation 2.

E2 = Em

[(1−

√Vf

)+

√Vf

1−√

Vf (1− Em/Ef2)

](2)

where,

Ef2 : Transverse modulus of fibre

The modeling of the fibre and the matrix by Rule of Mixture method does not represent theexact model of RVE. The stress distributions in both models is very different due to the stressconcentrations introduced by the stiff cylindrical fibre. The greater proportion of stiff fibrematerial at location A (see figure 2), leads to lesser deformation here and a greater estimate forE2. This is not represented in Rule of Mixtures and so it is a limitation.

The Hopkins-Chamis model considers the representation of the fibre as a simple squarewhich is a reasonable assumption compared to the assumption of Rule of mixture’.

The Halpin-Tsai relation also give a good estimate for the transverse modulus and it is asemi-empirical formula. Therefore, it requires an empirical curved fitting parameter.

2.3 Inplane shear modulus, G12

The inplane shear modulus G12 calculation is underestimated by the Rule of Mixture. TheHopkins-Chamis and Halpin-Tsai models give better results. For a UD ply, G12 = G13. Here,it is calculated using Hopkins-Chamis method which is given by

G12 = Gm

[(1−

√Vf

)+

√Vf

1−√

Vf (1−Gm/Gf12)

](3)

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Figure 2: Actual stress distribution in the RVE [1]

where,

Gf12 : Inplane shear modulus of fibre

2.4 Poisson’ ratio, ν12 and ν21

The Poisson’ ratio ν12 and ν21 are calculated from Rule of Mixture and it is reasonable fordesign purposes.

ν12 = νf12Vf + νmVm (4)1

ν21=

Vf

ν21f+

Vm

νm(5)

2.5 Failure properties

For the failure properties, theories have been developed for UD plies using simple mechanics ofmaterials approaches. This has not been as successful as micromechanics stiffness models sinceaccurate modeling of stress and strain distributions are not possible. The other problems arecomplex interactions of fibre and matrix failure (which occur at different strains) and statisticalvariations in materials properties. Therefore, they are not very reliable and usually macro-mechanical failure models with coupon test data are used. This is the important limitation ofmicromechanics in calculation of failure properties.

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2.6 Longitudinal tensile strength, F1t

For the longitudinal tensile strength F1t with the fibre dominated (high Vf ) failure case, theequation used is

F1t = σfVf + σ∗m(1− Vf ) (6)

where,

σf : tensile strength in fibreσm : tensile strength in matrix

2.7 Longitudinal compressive strength, F1c

For the longitudinal compressive strength F1c, three criteria are considered and the least failureload is used.

• Fiber micro-bucklingIt is triggered by fibre misallignment from manufacturing. The buckling load is given bythe first equation (eq. 7) for in-phase (shear) mode and the second equation (eq. 8) for outof phase (extension) mode

F1c =Gm

1− Vf

(7)

F1c = 2Vf

√VfEmEf

3(1− Vf )(8)

• Transverse tensile rupture due to Poisson’ effectAssuming a maximum strain failure criteria, the failure stress F1c is given by the equation9

F1c =E1ϵ

T2ult

ν12(9)

• Shear failure without bucklingThis is the minimum of the three cases presented and so it can be directly used as thefailure value of E2 and the strength is calculated from the equation 10

F1c = 2 (τfultVf + τmultVm) (10)

where,

τfult : ultimate shear strength of fibreτmult : ultimate shear strength of matrix

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2.8 Transverse tensile and compressive strength, F2t and F2c

This method utilises Stress Concentration Factor (SCF), F to calculate the ultimate transversetensile strength. It is given by

F =1

ds

[Em

Ef2− 1

]+ 1

(11)

F2t = E2ϵmult

F(12)

(13)

The same is done for F2c with a different value for ϵmult. The SCF is the same for both tensionand compression cases. The modeling is not so good because the true stress distribution is farmore complex than in the idealised unit cell model which is a limitation.

2.9 Inplane shear strength, F6

This uses an identical approach and formulae as those used to derive F2t. The equation is givenas

Fs =1

ds

[Gm12

Gf12− 1

]+ 1

(14)

F2t = G12γm12ult

Fs

(15)

(16)

2.10 Mechanical properties of a ply

The formulae given above were used to find out the mechanical properties of the complete plywith the properties of fibre and resin from the figure 1. They are given in the table 2.1.

3 Maximum deflection from classical beam bending formula

The classical beam bending formula for the cantilever is given as

δ =ql4

8D+

ql2

2F(17)

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Parameters of the composite Notation Value UnitLongitudinal elastic modulus E11 128.89 kN/mm2

Transverse elastic modulus E22 7.63 kN/mm2

Shear modulus G12 3.69 kN/mm2

Poisson’ ratio ν12 0.28Poisson’ ratio ν21 0.33Longitudinal tensile strength F11t 1.395 kN/mm2

Longitudinal compressive strength (Transverse splitting) F11c 2.648 kN/mm2

Longitudinal compressive strength (shear) F11c 1.151 kN/mm2

Transverse tensile strength (SCF=2.56) F22t 0.131 kN/mm2

Transverse compressive strength (SCF=2.56) F22c 0.268 kN/mm2

Shear strength (SCF=4.07) F12 0.046 kN/mm2

Table 2.1: Mechanical properties of the composite

where,

q : distributed loadD : Flexural rigidity, EI

F : Shear rigidity, GA

The elastic modulus of the laminate (with various lay-ups) is calculated from the LAP soft-ware. The elastic properties and failure properties of the laminate is given with the angles offibre orientation. Then the program is run and the elastic modulus and failure properties areobtained (figure 3). Here, the Membrane stiffnesses are used instead of Flexural’ since all thesections are in compression or tension eventhough the beam undergoes bending. The secondmoment of inertia for the box cross section is calculated as

D =BD3 − bd3

12= 18.90GNmm2

where,

B : Width of the beam’ Cross SectionD : Height of the beam’ CSb : Width of the inner empty beam’ CSd : Height of the inner empty beam’ CS

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Figure 3: Screenshot of the results from LAP software

The shear rigidity (only web is considered to carry the shear) is calculated as

F = 2(Gxy)wtwbw

= 2× 18.453× 2× 76

= 5609.712kN

Thus from the formula 17, the maximum deflection at the free end for the cantilever box beamis turned out to be δ = 4.58mm

4 Deflection from FEA

The cantilever box laminate beam is modeled in Visual CRASH-PAM and is run to obtain thedeflection from Finite Element Analysis. The following procedure is followed.

1. Initially in Visual Crash MESH, the corner nodes for the entire structure is defined byusing Node ⟩ By XYZ locate and then by using 2D⟩3/4 point mesh, a mesh is createdwith the element size of 2 mm. A fine mesh is chosen so as to capture the failure, which isdone later, correctly in the fixed edge and around the hole. Here, no thickness is consideredfor the structure as it is a shell structure will be given later.

2. Then the model is used in Visual Crash PAM. All DOFs of the free edge in the origin isfixed and the other edge is left free. This is applied by the menu Crash ⟩ Loads ⟩ Dis-placement BC. Then the pressure load is applied over the top surface of the beam by using

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the option Crash ⟩ Loads ⟩ Pressure Face where the pressure load of 0.000625 kN/mm2

is given.

3. Then the ply cards for the UD is defined using Crash ⟩ Materials ⟩ Composites ⟩ Ply. Theply type is chosen as ITYP=1. The max Stress criteria is selected by these values, IFAILNP= 1 and FAILTYP= 5. Then the option FAILDAM= 0 is selected to just visualise thedamage values. Also the elastic modulus and failure parameters are entered which wasfound out in the Question 1. The shear modulus in 12, 23 and 13 planes are considered thesame. Similarly, the positive and negative shear strength in 12, 23 and 13 are consideredto be the same.

4. Then the Material Editor is opened by the option Crash ⟩ Materials ⟩ Structural andthen the type 131 Multilayered Orthotropic BiPhasic is selected. The ILAY is selected as1 and then NOPER is given as 8. Each ply is given a thickness, a fibre angle and an id forthe ply. Here, 8 plies are given with the thickness and angles. To get the contour of PlyMaximum Stress Criterion, an auxillary variable, each ply id was given the number 26,which is the auxiliary variable for Ply Maximum Stress Criterion.

5. Now the ply is converted into layer by the command Convert Ply to Layer

6. Now the part is linked with the material id, thickness is given and the reference fibredirection is given in the axis direction of the beam.

7. The control parameters are set starting with RUNEND as 1.0.

8. The model should be in IMPLICIT method. Then add the controls ICTRL with ER-FOUTPUT as 3 for storing both curve and contour data and ICOMPRES as 0 for turningoff the compression of the file. In the control ECTRL, the SHELL FORMULATION isgiven as 6, which is the element formulation. Under the ICTRL, the ANALYSIS TYPEis given as STATIC and LINEAR.

9. Then the model is run in PAM-CRASH and the results are then post-processed.

The deflection which was obtained is shown in the figure 4. The maximum deflection ob-tained is 4.69 mm. Therefore, the deflection is different from the one calculated by the clas-sical bending cantilever hollow beam equation, which 4.58 mm. The increase in FEA couldbe attributed to the 3D nature of the problem in FEA whereas the classical bending formulaconsiders the beam to be a line (ie it gives deflection of the neutral axis) and it calculates thedeflection based on the material properties and load. Thus, the poisson’ ratio and edges mayinfluence the result in the FEA. Further, each cross section of the box beam may act locally asa beam owing to the applied pressure. This can be seen in the figure 4 where the centre of theCS of the beam has more deflection than the places nearer to the ’local’ edge.

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Figure 4: Deflection of the laminate

Figure 5: Failure analysis of the beam with hole

5 Failure analysis

Now, the model used in the above section is changed by including a central hole of diameter 60mm. It is done by using the option 2D ⟩ Hole on 2D mesh. The holes have their centre at thecoordinates (400, 0, 40)and(400, 40, 40). In the dialog box, No. of Rings is given as 6, TotalRing Width as 7 and No. of Elems. in Each Ring as 116 with no biasing. Then, the model isrun and analysed.

From the figure 5, we can see the maximum factor of Ply Maximum Stress Criterion forthe ply which attained the maximum factor out of the remaining plies. Therefore, the ultimatepressure which causes the first ply failure is obtained by

Ultimate pressure to cause first ply failure = 0.654× 0.000625

= 0.00095566 kN/mm2

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Figure 6: First ply failure of the beam with hole

Thus, the first ply failure will occur if this load is applied to the structure and it is shown in thefigure 6. The first ply failure occurs in the third ply of the lay-ups. Here, the factor 1 is reachedwhich shows that the failure has onsetted.

6 Resin infusion simulation in the beam

The model which was used in the previous section is used to simulate the resin infusion inPAM-RTM. The following steps were followed.

1. After importing the file in PAM RTM, the quad mesh is converted into triangular shellmesh by the option Mesh ⟩ Transform ⟩ Split Quads. Then the mesh is scaled to operatein metres instead of mm, by the option Mesh ⟩ Transform ⟩ Scale where a value of 0.001is given for the axes.

2. Now, the Groups are defined to define the pressure boundary conditions later. First, a freeedge of the beam is selected by using the Selection (plus) button and then by the menupath Selection ⟩ Pick Boundary. After the edge is highlighted, the menu path Groups ⟩Create is used to classify it as a separate Group. The same is done for the other edge andthe boundaries of the two holes.

3. Under the Numerical heading, Save filling factor and Save pressure are checked.

4. Under the Materials ⟩ Resins section, Resin viscosity at 23 ◦C is given as 0.35 Pa s underthe option Value orFunction.

5. Then under the Materials ⟩ Reinforcements ⟩ Default fabric section, permeability isgiven as 1e− 11m2 for all directions since the fabric is considered isotropic.

6. Then under the Zones section, the porosity and the thickness are given as 0.5 and 0.002mm respectively.

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Figure 7: Resin filling time for the given BC

Figure 8: Resin filling time under 1 hour

7. Now, under the Boundary conditions section, new pressure boundary condition is createdby giving the left edge’ group ID and a value of 1 bar = 1e5N/m2. The same is done forthe right edge with a pressure value of 0 bar ie., vacuum. The simulation is run and theresults are seen.

From the figure 7, we can see that filling time for the resin over the complete structure took15348/3600 = 4.263 hours, which is the infusion time. Therefore, if a resin solidifies into gelafter 2 hours, the infusion will not be successful and it will go beyond the hole but it will notreach the other edge as seen from the figure 7 with 6.14e3− 7.67e3 seconds.

To have the infusion time under 1 hour, the pressure boundary conditions are changed. Theedges of the holes are given 1 bar and the free edges of the beam in vacuum state. Now, theinfusion time of resin is reduced to 2909/3600 = 0.808 hour (from figure 8), which is lessthan an hour. If the same BC’s are reversed for this problem, the infusion time came around4162/3600 = 1.1567 hours

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7 Conclusion

The method to calculate the ply properties from the fibre and resin properties are learned andthen calculated. Then the properties for the composite material is calculated from the LAPsoftware to calculate the deflection analytically. Later it is compared with the values from theFEA. Then a hole is inserted and then the ultimate pressure for the structure is found out. Thena RTM simulation is done to find out the resin infusion time and the boundary condition withthe infusion time under 1 hour.

References

[1] Pickett, Anthony., Composites Modelling, IFB 2013-2014.

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