Analysis Notes

Torsion of flat plates has been analyzed by several, and in practically all cases results assume isotropic material behavior. Below is a typical example. The expression for the angular deflection of one end due to an applied moment while the other end is fixed is shown below:

yP

M L

Wt Gt

3

Where ‘Mt’ is the applied moment, ‘L’ is the length, ‘W’ is the width, ‘t’ is the thickness, ‘G’ is the shear modulus, and ‘‘ is a warping factor relating to the aspect ratio of W/t as shown below by it’s expression:

1

16163

4

43 36 112

.b

a

b

a where a=W/2 and b= t/2.

The table below demonstrates the dependence of on W/t W/t= 1 2 3 4 5 6 7 8 9 10 0.141 0.229 0.263 0.281 0.291 0.298 0.303 0.307 0.31 0.312 .333

For the above model, the boundary conditions are too stiff and as a result the FEA prediction on is roughly 30% too high. In the picture below, the boundary conditions have been optimized such that the worst case error is close to 5%. Table 1 compares theory to FEA predictions in isotropic aluminum :

Table 1W/t= 2 4 6 8 10 0.229 0.281 0.298 0.307 0.312 .333FEA 0.225 0.267 0.289 0.299 0.306 .333% error= 1.75 4.98 3.02 2.61 1.92

The warping factor ‘‘ is determined from mathematical relationships that assume isotropic material behavior. In recent published reports on fracture behavior of single crystal lithium niobate, the above expression for was cited as the basis for the derived stress intensity factor ‘k’. Furthermore, the calculation for ‘k’ uses the term (1+) which is poison’s ratio plus one. In anisotropic material, (like lithium niobate) poison’s ratio in is irrelevant. It is possible to simplify anisotropy for a particular orientation and obtain values like those mentioned above, however a careful, detailed treatise will have to be done to determine the appropriate ‘’, ‘E’ effective tension modulus, and ‘G’ effective shear modulus for that unique orientation.

In this report I have attempted to do just that; to gather the necessary terms to simplify the anisotropy for some orientations of lithium niobate, and tellurium dioxide. The work mentioned earlier using FEA to predict is still necessary for obtaining ‘’. However, two other models are important to construct to obtain effective ‘E’s and ‘G’s. They are mentioned below.

A pure tension mode is used to obtain ‘E’, while a pure torsion model is used to obtain ‘G’. The tricky part is obtaining them with the right orientation. First, an analytical model of the double torsion is constructed and solved for principal stresses at the crack tip. Regardless of isotropic or anisotropic behavior, the largest principal stress is found to occur at right angles to the crack propagation. This is the beauty of the DT specimen, it produces Mode I cracks, the opening type. Since the product of strain and young’s modulus ‘E’ produce stress, than the effective or equivalent young’s modulus for this specimen is computed for the axis at right angles to it. Therefore the tension model must apply tension in this direction and solve for the equivalent tensile modulus, young’s modulus. The boundary conditions are simple and shown below:

For the shear modulus ‘G’, a shaft in torsion, where it’s axis share the same orientation as the DT specimen is constructed. At on end, the nodes will be fixed in ‘r’ and ‘’. At the other end tangential forces at the circumference are to be applied. The resultant angular twist ‘’ for the applied moment will yield ‘G’. Below are the boundary conditions for this model:

Notice that at x=0, u=0, and at x=l, force is applied tangentially to the shaft in intervals.

The double torsion model is now possible to construct. We have determined Eeff, Geff, and for a particular crystal orientation and size. Below is a double torsion model

In the above model, the load applied to both arms, P=.25lb. The maximum displacement, ‘y’ is =DMAX 2.7104110 4 in, incidentally the predicted displacement from the equation at the beginning of this report is =y 6.84610 5 in.

The FEA result is 4x more deflection than the theory of flat plates. The in the

expression was .333 for predicting 6.845 E-05 in. But W/t is only 5.5 therefore is

only .295. This makes the FEA result 3.5x that of the handbook calculation. For this

geometry, a fudge factor should be applied that accounts for the boundary condition, and

the fudge factor for the geometry of the above model is 3.5. The stress intensity

calculation will then be modified accordingly. It looks like if 1/ stays at 3, then 4x3=12.

The new stress intensity equation will replace 3 with 12. Let’s see what stress intensity

does for this model. It now predicts 126.34 psi in**.5.

I believe the torsion arm also bends down like a beam and hence the added deflection. I

could check this out by adding supports left and right of the crack tip or extend the

torsion arm almost all the way to the end. I decided to do just that and made the crack

length term ‘a’ 44mm. Now =DMAX 3.5338410 4 in and =y 3.09310 4 in. There is 14%

difference and that is OK.

The lesson learned from this is to make compliance checks on the test specimens. If

ANSYS predicts 5x the handbook ‘y’ prediction or what ever, check it out.

From my investigation into TeO2 fracture toughness I came across Effective Young’s moduli for various orientations. The published values for [100], [110] are 8735 MPa and 11.2360 MPa respectively. Interestingly, ANSYS predicts 8.66 MPa and 11.2353 MPa respectively. It looks like pretty good agreement to me. I should look at TeO2 samples in the orientations as published and determine what loads will produce stress intensities at the material’s fracture toughness limits. I have all the information! So far I have setup the dbteo3.txt with the correct E, G and . I am going to run it and

then compute the fudge factor as mentioned above. Then the K can be determined.

The work is complete now and I can report on the outcome. A new ratio is being developed that seems intriguing to me; fudge_factor/. The alpha is 1/3, fudge_Factor is the multiplier that equalizes the theory prediction on ‘y;’ (displacement) and the ANSYS model. For TeO2 where the DT specimen dimensions are as shown below, fudge_factor/is:

L=45/25.4 ! total x dimension 39mmW=25/25.4 ! total y dimension 19mmD=2/25.4 ! total z dimension 3.2mmDn=.6/25.4 ! THIN CRACK SECTIONA=11/25.4 ! CRACK LENGTH 'A'DX=.5/25.4 ! CRACK WIDTHC=11/25.4 ! TORSION ARM LENGTHDMAX -1.387897825E-04 SCALAR

For P/2 =.25 lbs, the stress intensity is 65.9 , .072

Another ANSYS run looked at how the compliance is affected by the crack length ‘a’. This is related to the fudge_factor/ ratio. For the TeO2 orientation [110],[110],[001] and the load is applied in the (001) direction on to the [110],[110] plane, the for this is: .152 (w/t=5.5). Isotropic for w/t=5.5 is .295. The fudge_factor/ in the displacement expression to agree with the ANSYS deflection of -.364414685E-02 is 2.234/.295 this ration is 13% higher than 1/.152 (1/110). The significance of this is the way the compliance is affected by the crack length ‘a’. The simple rectangular torsion expression is not too accurate for short crack lengths. As the crack length approaches dimensions close to two independent torsion bars, then the torsion expression more closely predicts the displacement. In the case above, The double torsion specimen’s overall length is 45 mm. The crack length is 44 mm. The torsion expression is within 13% of the ANSYS

prediction. When ‘a’ is 11 mm, it’s within 16%. These values assume that the warping

factor =.152 is used instead of isotropic =.295 for w/t=5.5 in y ....2 P wm2 a

..W t3 G

1

.152

The next step in this analysis is to fabricate some TeO2 samples and test them.