king final report

8/13/2019 King Final Report

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iii

CONTENTS

LIST OF FIGURES ........................................................................................................... v

LIST OF TABLES ........................................................................................................... vii

NOMENCLATURE ....................................................................................................... viii

GLOSSARY ..................................................................................................................... ix

ACKNOWLEDGMENT ................................................................................................... x

ABSTRACT ..................................................................................................................... xi

1. Theory .......................................................................................................................... 1

1.1 Coordinate System ............................................................................................. 1

1.2 Twist Parameters ................................................................................................ 4

1.3 Stress, Strain, and Displacement for a Cantilever Beam ................................... 5

1.3.1 Stress Equilibrium .................................................................................. 6

1.3.2 Strain-Displacement Relationships ........................................................ 8

1.3.3 Stress-Strain Relationships .................................................................. 10

1.4 Generic Cantilever Beam Stress, Strain, and Displacement Solutions ............ 11

1.4.1 Functionally Varying Moment of Inertia ............................................. 13

1.4.2 Discontinuous Shear Stress Solution for a Rectangular Beam ............ 18

1.4.3 General Stress Distribution for a Twisted Cantilever Beam ................ 21

1.4.4 General Strain Distribution for a Twisted Cantilever Beam ................ 22

1.4.5 General Vertical Displacement of a Twisted Cantilever Beam ........... 23

1.5 Specific Theoretical Solution ........................................................................... 32

2. Finite Element Analysis (FEA) Model ...................................................................... 38

2.1 Methodology .................................................................................................... 38

2.2 Analytical Results ............................................................................................ 41

2.2.1 Tensile Stress and Strain Values (x-Direction) .................................... 43

2.2.2 Shear Stress and Strain Values (xy-Direction) ..................................... 47

2.2.3 Vertical Displacement (y-Direction) .................................................... 51


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3. Comparison of Beam Theory Formulation Results to FEA Results.......................... 54

3.1 Displacement Values ........................................................................................ 54

3.2 Maximum Tensile Stress Values ...................................................................... 57

3.3 Maximum Shear Stress Values ........................................................................ 58

4. Summary and Conclusion .......................................................................................... 60

5. Appendix ATheoretical Results at SelectedxValues ........................................... 62

6. Appendix BFEA Results at SelectedxValues ...................................................... 67

7. Appendix CABAQUS Input File (.inp) ................................................................. 72

8. References ................................................................................................................ 100


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v

LIST OF FIGURES

Figure 1 - Example of a Twisted Cantilever Beam ........................................................... 1

Figure 290 Twisted Cantilever Beam in thexy-Plane .................................................. 2

Figure 390 Twisted Cantilever Beam in theyz-Plane .................................................. 3

Figure 4 -xyzAxes vs. ABC Axes .................................................................................... 4

Figure 5Arbitrary Beam Cross-Section Subjected to Bending Moment Mz(x)............. 6

Figure 6 - Rotation of Axes and Moment of Inertia ........................................................ 13

Figure 7 - Moment of Inertia Variation within the Twisted Region of a Rectangular

Beam ................................................................................................................................ 17

Figure 8 - Point of Discontinuity in Shear Stress ............................................................ 19

Figure 9 - Width and Thickness Parameters of the Solution ........................................... 32

Figure 10 - Load and Length Parameters of the Solution ................................................ 32

Figure 11 - Midplane Shell of the ABAQUS FEA Model .............................................. 38

Figure 12 - Meshed View of the ABAQUS FEA Model ................................................. 39

Figure 13 - Load and Boundary Conditions of the ABAQUS FEA Model ..................... 40

Figure 14 - FEA Shell Model, Variation of Principal Axes ............................................ 42

Figure 15 - FEA Tensile Stress at the Shells Midplane................................................. 44

Figure 16 - FEA Tensile Stress at the Shell's Positive Face ............................................ 45

Figure 17 - FEA Indicates an Additional Compressive Stress Peak in Twisted Region,

Shells Positive Face........................................................................................................ 45

Figure 18 - FEA Tensile Stress at the Shell's Negative Face .......................................... 46

Figure 19 - FEA Indicates an Additional Tensile Stress Peak in Twisted Region, Shells

Negative Face .................................................................................................................. 46

Figure 20 - FEA Shear Stress Distribution at the Shell's Midplane ................................ 47

Figure 21 - FEA Shear Stress Concentration at Shell's Midplane ................................... 48

Figure 22 - FEA Shear Stress Distribution on Shells Positive Face.............................. 49Figure 23 - FEA Shear Stress Distribution on Shells Negative Face............................. 50

Figure 24 - FEA Displacement Results, Isometric View ................................................ 51

Figure 25 - FEA Displacement Results,xy-Plane ........................................................... 51

Figure 26 - FEA Displacement Results,yz-Plane ............................................................ 52

Figure 27 - FEA Indicates Residual Stiffness and Non-planar Cross-Section atx=x4.. 53


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Figure 28 - Theory vs. FEA, Vertical Displacement ....................................................... 54

Figure 29 - Theoretical Displacement Results vs. Beams of Constant Cross-Sectional

Orientation ....................................................................................................................... 56

Figure 30 - Theory vs. FEA, Maximum Tensile Stress ................................................... 57

Figure 31 - Theory vs. FEA, Maximum Shear Stress ...................................................... 58


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vii

LIST OF TABLES

Table 1 - Summary of FEA Input Parameters ................................................................. 40


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viii

NOMENCLATURE

For the following symbols, ij. Units are in parenthesis.

E - Modulus of Elasticity (psi)

- Strain Tensor (in/in)

ii - Extensional Strain (in/in)

ij - Tensor Shear Strain (in/in)

F - Force Applied to the Free End of the Cantilever Beam (lbs)

G - Shear Modulus (psi)

- Engineering Shear Strain (in/in)

Iii - Moment of Inertia (in )

Iij - Product of Inertia (in )

- Curvature (in-

)

L - Length (in)

M - Moment (in-lb)

- Poissons Ratio (dimensionless)

- Stress Tensor (psi)

ii - Tensile Stress (psi)

t - Thickness of the Rectangular Cross-Section, Less than its Width, w (in)

ij - Shear Stress (psi)

- Angular Measurement Between theBand Y-Axes or the CandZ-Axes (radians or degrees)

u - Longitudinal Displacement (in)

v - Vertical Displacement (in)

w - Transverse Displacement, Only Applicable to Section1.3.2 (in)

w - Width of the Rectangular Cross-Section, Greater than its Thickness, t (in)


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GLOSSARY

ABAQUS -A software application used for both the modeling and analysis of

mechanical components and assemblies (pre-processing) and

visualizing the finite element analysis results (post-processing).

Cantilever Beam-

A projecting beam that is supported at one end and carries a load at

the other end or along its length.

Cartesian Coordinate System -A coordinate system comprised of three mutually orthogonal axes to

represent three-dimensional space.

Deflection - A movement of a structural member resultant of an applied force.

Displacement - See deflection.

Equilibrium - A state in which opposing forces or influences are balanced.

FEA - An acronym for Finite Element Analysis.

Final Orientation - The cross-sectional orientation of the beam in the region beyondL2.

Global Axes -The coordinate system that remains constant throughout the length

of the beam and whose origin is at the fixed end of the beam.

Hooke's Law - A law stating that the strain in a solid is proportional and linearly-related to the applied stress within the elastic limit of the solid.

Initial Orientation - The cross-sectional orientation of the beam from its origin to L1.

Isotropic - Having uniform physical properties in each direction.

Local Axes -The coordinate system that remains perpendicular to the perimeter

of the beams cross-section and whose origin varies to satisfy this

perpendicularity.

Midpoint Rule -Computes an approximation to a definite integral, made by finding

the area of a collection of rectangles whose heights are determined

by the values of the function.

Modulus of Elasticity -The proportionality constant relating a solids extensional strain

value(s) to its longitudinal stress value(s) within the elastic region.

Moment of Inertia - A measurement that quantifies a beams ability to resist bendingabout a particular axis.

Origin -The point in three-dimensional space where all three Cartesian axes

are coincidient, e.g.x =y =z = 0.

Poisson's Ratio - The negative ratio of transverse strain to axial strain.

Product of Inertia -A measurement that quantifies a beams ability to resist shear

deformation about a particular axis.

Shear Modulus -The proportionality constant relating a solids shear strain value(s)

to its shear stress value(s) within the elastic region.

Strain -The ratio of total deformation to the initial dimension of the

material body in which the forces are being applied.

Stress - A measure of the internal forces acting within a deformable body.

Twisted Cantilever Beam - A cantilever beam whose cross-section remains constant, but isrotating about its centroid along the beams length.

Varying Orientation - The cross-sectional orientation of the beam in the region of L1toL2.


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ACKNOWLEDGMENT

I would like to thank my fianc, Jessica Rowe, for her unwavering support,

encouragement, and affection during my graduate studies.


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xi

ABSTRACT

Twisted cantilever beams are defined here as beams whose rectangular cross-sectional

orientation changes along the beams length with respect to global axes. These beams

have a twisted or spiral-type geometric feature somewhere along their length. This

project will formulate solutions for twisted cantilever beams in static bending using

classical beam theory, compare the results to numerical solutions, and discuss

inconsistencies.


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1.Theory1.1 Coordinate SystemThe analytical case of interest is a rectangular cantilever beam in static bending with an

abrupt, 90 twist, like that shown inFigure 1. The beam is rigidly affixed at one end,

preventing any displacements or rotations at that end. The geometric center of the beam

at the affixed end will be defined as the originor the point at which all coordinates

(x,y,z) are equal to zero. Cartesian coordinates will be used in these solutions because

traditional cantilever beams in bending have documented analytical solutions in

Cartesian coordinates as well. Note that the placement of the axes shown in Figure 1

does not place the vertex at the analytical origin.

Figure 1 - Example of a Twisted Cantilever Beam

Moving away from the rigid attachment at (0,0,0), parallel to the length of the beam will

be defined as: moving in the x-direction, defining the x-axis. The direction of the

y-axis is orthogonal to the x-axis and is chosen to be parallel to one side of the beams

cross-section at the origin. The z-axis is orthogonal to both the x and y-axes. These

three axes whose origin occurs at the beams geometric center, at the fixed end, are

defined as the global axes.


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Moving down the beams length, away from the rigid attachment, in thex-direction, the

beam initially has a homogeneous moment of inertia and its initial orientation. At

some distance, denoted by L1, the beams cross-section begins to rotate about its x-axis,

but its moment of inertia remains continuous, where it can be described as having a

varying orientation. The beams cross-section continues to rotate until it reaches

another distance, denoted by L2, where the rotation stops but again, the moment of

inertia remains continuous. BeyondL2, the beams final orientation continues until it

reaches its free end, whose distance from the origin is denoted by the length L3. For the

case of a 90 twist, the final orientation is rotated 90 from the initial orientation. See

Figure 2 andFigure 3.

Figure 290 Twisted Cantilever Beam in the xy-Plane

Y

XO

L3

L2

L1

F


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Because the cross-sectional orientation of the beam is varying along its length between

L1andL2, another set of axes are defined in order more easily express twist parameters

and cross-sectional orientations. These local axes are also Cartesian, but are denoted

(A,B,C) instead of (x,y,z). The origin of the local axes is not necessarily at the same

location as that of the global axes, but can be anywhere along the beams length,

provided that the A-axis remains collinear with thex-axis. At the beams fixed end, at

its geometric center, the A-axis is collinear with the x-axis, the B-axis is collinear with

the y-axis, and the C-axis is collinear with the z-axis. Once the beams cross-section

begins to rotate (at L1) the B and C-axes develop an angular measurement greater than

zero between their global counterparts (y and z, respectively). The A-axis and x-axis

will alwaysremain collinear, and either can be used interchangeably. SeeFigure 4.

Figure 390 Twisted Cantilever Beam in the yz-Plane

Y

Z


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It is important to describe and define the parameters within the beams varying

orientation (L1 x L2), because these parameters will contribute to the analytical

solutions that are eventually formulated. The difference between the cross-sections at

x =L1andL1xL2can be quantified by the angular difference, , between theyand

B-axes or thezand C-axes. The rate of twist can then be defined as the change in angle,

, over the change in length betweenL1andL2, or .1.2 Twist ParametersFor the problem to be solved, the cross-sectional orientation can be described as: [1] [2]

[3]

Let (x)be a linear function of x, so that the rate of twist from L1 to L2 is a constant

value. Then, by the equalities given by equations[1] through[3]:

Figure 4 - xyzAxes vs. ABC Axes

Z

Y

B

C

Z

Y

B

C

[xand A-axes both coming out of the page. ]


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Substituting the boundary conditions in the expression for yields: Solving for the constants M and N, and subtracting these equalities gives:

The function (x)can then be written as:

[4]So, the rate of twist is: [5]Rewriting equation[1] through[3] with equation[4]: [6]

[7] [8]Together, equations [6] through [8] fully describe the beams twist parameters and can

be derived using the valuesL1andL2.

1.3 Stress, Strain, and Displacement for a Cantilever BeamBecause the cantilever beam of interest has a functionally varying cross-sectional

orientation along its length, the stress distribution throughout the beam must be solved

for a generic case first so that the twist parameters developed in section 1.2 can be


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applied. Let a beam of arbitrary cross-section be subjected to a bending moment,Mz(x),

about thez-axis, which is a function of longitudinal (x) position only. The origin of the

cross section is at the beams centroid, and the yandz-axes are the principal axes. See

Figure 5.

1.3.1 Stress EquilibriumAs a result of the applied load, F, and subsequent moment,Mz(x), the beam experiences

displacements, strains, and a state of stress. Each can be represented by a tensor at each

point throughout its volume. The generic stress tensor, [], is given by:

[9]

In order to satisfy equilibrium, the stress tensor must be symmetric such that:

Figure 5Arbitrary Beam Cross-Section Subjected to Bending Moment Mz(x)

Z

Y

[Xand A-axes both coming out of the page. ]

Arbitrary

Cross-Section

Mz(x)


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Because stresses can be related to displacements (as will be shown in following

discussion), and because the displaced shape of an end-loaded cantilever beam is similar

to that of a beam in pure bending, it is reasonable to assume that the cantilever beam has

a stress tensor similar to that of a beam in pure bending. Beginning with the bending

moment,Mz(x):

[10] [11]

[12]

That is, the normal stresses, ii, and thex-zandy-zshear stresses,xzand yz, are identical

to those for a beam in pure bending, but no specific assumptions are made about the

other shear stress, xy, only that it is some function ofx,y, andz.

The equations of 3-D stress equilibrium with no body forces are now noted to augment

the discussion and simplify the equations above:

[13]

[14] [15]

Based on equations[10] through [12], equations [13] and [14] reduce to:

[16] [17]


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Equations [16] and [17] make up the 3-D stress equilibrium equations for a cantilever

beam experiencing plane stress. Upon inspection, one can see that equation [17] is only

satisfied if xyis constant in thex-domain. Rewriting equations[10] and[11] for clarity,

equation [12] can be simplified to:

[18] [19]

[20]

1.3.2 Strain-Displacement RelationshipsAs mentioned, the beam also experiences a state of strain. The generic stress tensor, [],

is given by:

[21]where iiare normal strains and ijare shear strains.

The strain tensor is symmetric, such that:


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For small deflections (where sin ), the following expressions relate strain to

displacement:

[22]

[23] [24]

[25]

[26]

[27]Engineering shear strain is related to tensor shear strain by:

Rewriting equations [22] through [27], the components of the strain tensor, [], are:

[28] [29] [30]

[31] [32] [33]


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1.3.3 Stress-Strain RelationshipsFor an isotropic, linear-elastic material (obeying Hookes Law), the stress-strain

relationships are as follows:

[34] [35]whereEis the modulus of elasticity and G is the shear modulus. The two modulii are

related through Poissons Ratio, , by the following:

[36]

For each component of the strain tensor, the relationships are:

[37]

[38]

[39] [40] [41]

[42]


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1.4 Generic Cantilever Beam Stress, Strain, and DisplacementSolutions

Equation [18] indicates that the normal stress in the x-direction is dependent on the

bending moment, Mz(x), which is a function of x. This bending moment is simply the

force,F, multiplied by the distance from the origin, and can be written as:

[43]So, the normal stress in thex-direction is:

[44]Substituting equation [44] in thex-direction stress equilibrium equation ([16]) gives: [45]Integrating equation [45]:

Pis a constant of integration and can be determined on the basis that in order to satisfy

boundary equilibrium around the perimeter of the beam, the shear force must be equal to

zero at these points.

The total stress distribution is now repeated for clarity:

[46] [47] [48]


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The strains at each point can be found directly from the resultant stresses, by inserting

equations [46] through [48] into equations [37] through [42]. Inserting equation [46]

and [47] into the stress-strain relationships (equation [37]), and that into the strain-

displacement relationship (equation [22]) leads to a noteworthy:

[49]Equation [49] can be integrated in x to solve for the longitudinal displacement of the

beam, once the function ofIzzis known.

For small strains and displacements in the elastic range, and assuming that plane sections

remain plane, the curvature of the beams neutral surface can be expressed in the

following form: where is the curvature. Inserting equation [43] gives: [50]Equation [50] is a second-order linear differential equation, and is the governing

equation for the elastic curve. The product EI is the flexural rigidity of the beam.

Because the moment of inertia,Izz, varies with respect tox, it must be first formulated in

order to integrate equation [50] and solve for the vertical displacement.

Up to this point, each parameter of the stress, strain, and displacement components is

known except the moment of inertia about the neutral axis, Izz, and the constant of

integration, P, for the shear term. This moment of inertia and P are derived in the

following discussion.


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1.4.1 Functionally Varying Moment of InertiaConsider the plane area shown inFigure 6below. The moments and product of inertia

with respect to the local BC-axes are:

[51] [52] [53]

The same forms of expressions exist for the global coordinate system, in xyz-

coordinates.

The moments and product of inertia in the BC-plane are constant values, equal to those

of the yz-plane at the origin. However, as the angular measurement increases, the

moments and product of inertia in the yz-plane change. To obtain these quantities, the

coordinates of the differential element dA are expressed in terms of theyz-coordinates as

follows:

Figure 6 - Rotation of Axes and Moment of Inertia

Arbitrary

Cross-Section

C

B

dA

BdA

CdA

y

z

Z

Y[xand A-axes both coming out of the page. ]


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[54]

[55]

Substituting these values in equations [51] through [53] gives:

Using the following trigonometric identities, the form of IYY, IZZ, and IYZ can be

simplified.


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These are complicated expressions in their most reduced form, unlike the simple and

familiar (for example). However, if the general expressions for momentsand product of inertia in the localcoordinate system (equations [51], [52], and [53]) are

substituted in the above integrals, they take on a more practical form.

[56]

[57]

[58]IYY,IZZ, andIYZare the moments and product of inertia in the global coordinate system at

any point along thex-axis, andIBB,ICC, andIBCare the moments and product of inertia in

the local coordinate system.

With the moments and products of inertia defined as such, the reader is referred back to

Figure 2 to be reminded of the cross-sectional orientation at the origin. Specifically, for

this problem, the local moments and product of inertia are the same as those of a

rectangular cross-section in bending, whereICCis the strong axis of bending andIBBis

the weak axis of bending at the origin. The limits of integration are defined by the

beams perimeter. Thus:


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[59]

[60]

[61]where wis the width of the beam, and tis the thickness.

For any symmetric cross-section whose centroid is at the origin,IBCwill be zero, so anyterms containing the beams localproduct of inertia will drop out of equations [56], [57],

and [58]. Substituting equations [59] through [61] into these equations gives:

By substituting in the relationship between andx(see equations[6], [7],and[8]),

these equations yield the full form of the functionally varying moments of inertia.

[62]


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[63]

[64]

Taking (for example) values of t =0.25 and w = 1.00, the moments and product of

inertia change in the twisted region of the beam as shown inFigure 7below:

Figure 7 - Moment of Inertia Variation within the Twisted Region of a Rectangular Beam


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1.4.2 Discontinuous Shear Stress Solution for a Rectangular BeamThe stress distributions will be solved to eventually formulate the strain and

displacement solutions. However, the shear stress, xy, still has an undefined constant,P,

which must be found to fully formulate the stress in the beam. As stated in section1.4.1,

Pis a constant of integration that can be determined on the basis that in order to satisfy

boundary equilibrium around the perimeter of the beam, the shear force must be equal to

zero at these points. But because the beams cross-section is rotating, this constant is

also related to the twist parameters forL1xL2.

For the beams initial orientation:

Substituting this value into equation [48] gives:

[65]

For the beams final orientation:

Substituting this value into equation [48] gives:

[66]


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For the beams varying orientation, the y-coordinate of the beams outermost fiber is

constantly changing. This coordinate is needed to solve the shear stress distribution in

this region. The perimeter of the beam can be represented by four straight lines whose

orientation varies with the x-position, or . Because the perimeter edges intersect one

another at right angles, however, the function of the outermost y-coordinate will not be a

continuous function. The function changes when they-axis coincides with the corner of

the beams cross-section. For any rectangular beam, this occurs at an angle of . SeeFigure 8.

For the varying orientation, where 0arctan(t/w), they-coordinate of the outermost

fiber of the beams cross-section is:

Z

Y

w

t

= arctan (t/w)

Figure 8 - Point of Discontinuity in Shear Stress


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Substituting this value into equation [48] and using the relationship between and x

(equation [7]) gives:

[67]

Similarly, for the varying orientation, where arctan(t/w)/2, they-coordinate of

the outermost fiber of the beams cross-section is:

Substituting this value into equation [48] and using the relationship between andx

(equation [7]) gives:

[68]


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1.4.4 General Strain Distribution for a Twisted Cantilever BeamInserting equations [69] through [74] into the stress-strain relationships given by

equations [37] through [42]:

AnyOrientation: [75]

Any

Orientation: [76]

Any

Orientation: [77]

Initial

Orientation:

[78]

Varying

Orientation:

[79]

Varying

Orientation:

[80]

Final

Orientation:

[81]


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1.4.5 General Vertical Displacement of a Twisted Cantilever BeamNow that the moment of inertia,Izz, has been derived, equation [50] can be integrated to

obtain the y-displacement function. However, because Izz is dependent upon the

longitudinal position, x, the vertical displacement function must be dissected into three

conditional equations, depending on the magnitude of x. The first solution presented

will be for the case of . Because this x-location is within the initialorientation and the moment of inertia is constant in this region, equation [50] is

integrated twice inx, as would be done for a normal cantilever beam with constant cross-

section:

where C1is a constant of integration. Integrating again inxgives:

The constants of integration, C1 and C2 can be determined by applying the boundary

conditions of the beam. At the fixed end,

, the displacement and the slope of the

beam are equal to 0. Therefore, . So, for the region of : [82]The displacement function for x values in the varying orientation becomes more

complicated, however. For values ofxwhere , the moment of inertia cannotbe considered constant and excluded from the integrals, as was done above. Inserting

the equality forIzzinto equation [50], for values ofxwhere , and integratingonce with respect toxgives:


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The second term of they-displacement slope solution (above) indicates an inherent flaw

in the application of classical beam theory to this problem. Even a simplified version of

the function does not stay within the real domain; the varying moment ofinertia functions presence in the denominator complicates the solution beyond any

reasonable point.

Because the solution is beyond the scope of this document, a piecewise approximation is

instead presented using the midpoint rule. If the second term above is broken into many

separate intervals, the integral can be carried out by approximating the value of x for

each separate interval. That is, the distance from L1to L2will be divided byNnumber

of divisions, and the average longitudinal value between each point will be used in place

ofx. The midpoint rule is defined as:

The midpoint of each interval is equal to the variable xn, where N is the number of

predefined intervals chosen:

Because the midpoint rule is only needed for values of xwhere , the valueof a is already known and equal to L1. Likewise, the value of b will be set to x.

Therefore, the second term of the exact y-displacement slope solution can be

approximated as:


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In this form, the first term of the exact solution can be solved directly, and the second

term can be approximated. Of course, as , the function will converge to the exactsolution, but in this form it is much easier to deal with. Thus, the slope of the beam,, for can be written as:

The constant of integration, C3, can be determined by applying the boundary condition

provided by equation [82]: at ,


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Rewriting the expression for the slope of the beam in the region of :

To solve for the displacements, the equality above must again be integrated in x.

However, the second term again creates problems and yields solutions of non-real

numbers. Thus, the midpoint rule must be applied a second time. Let the second term

above be defined as an arbitrary function ofx,(x), so that:

Integrating the second term using the midpoint rule gives:


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There have certainly been more elegant expressions derived in engineering. Writing the

total solution fory-displacement when

gives:

The constant of integration, C4, can be found by using equation [82] at :


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The constant of integration, C5, can be determined by recalling the function derived for

the beams slope between , and inserting :


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The slope of the beam for can then be written as:

Notice that in this region only one term of the beams slope is dependent on x. This

simplifies things much more than before when deriving the beams deflection in the

varying orientation. That is, all integrals henceforth can be computed directly, and no

more approximations are needed. To solve for they-displacement, the equality above is

integrated inxto give:


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The final constant of integration can be determined by the boundary condition:

at

. Therefore,

(from equation [83]).

[84]

Equations [83] and [84] are very complicated and tedious to carry out in real

applications. A less cumbersome method is to use the equation of the beams slope,, and integrate the numerical value(s) over x, rather than carrying through to aclosed-form solution. This method will be used to obtain values in section 1.5 and in

Appendix A.


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1.5 Specific Theoretical SolutionAdvancing from the generic case, the specifics of the problem are now defined and solved.

Figure 9 shows the beams cross-section at two distinct points and assigns its width (w)

and thickness (t). Let these values be 1.0 and 0.25, respectively. Figure 10 is arepetition ofFigure 2,but assigns values to the applied force,F, and lengthsL1,L2, andL3.

Let the material properties of the beam be those of mild steel (ms):Figure 10 - Load and Length Parameters of the Solution

Y

XO

7.5

4.5

3

10 lbsFigure 9 - Width and Thickness Parameters of the Solution

Z

Y

w

t

Z

Y

w

t

[Initial Orientation]

[Final Orientation]

w= 1.0

t= 0.25


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For comparison to numerical results, the stress, strain, and displacement values at five

distinct values ofxwill be found. These fivex-locations are as follows:

corresponds to the start of the twist,

corresponds to the midpoint of the beam and

its twist feature, corresponds to the end of the twist, and corresponds to the end ofthe beam. In addition, at eachxvalue the stress, strain, and vertical displacement valueswill be formulated for three values ofy, corresponding to the positive/negative outermost

beam fiber y value, and at , the neutral axis. The following solution is that of . The analytical results for all other points can be found in Appendix A.In thebeams local coordinates, the moments and products of inertia are:

For and the moment of inertia,IZZ, is:

(See equation [55]).

Therefore, the stress distribution for and is:


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In tensor form:

The strains at and can be found by inserting these values into thestress-strain relationships, given by equations [37] through [42]:

The vertical displacement at this point is found using the slope equation that, when

integrated inx, gives equation [83]. Because the value ofx= 3.75 lies in the region of . For the following solution, letN= 5.


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To extract a vertical displacement from the beams slope at this point, is simplyintegrated again inxwith the appropriate limits:

The constant of integration, C7, is determined by the boundary condition provided by the

beams vertical displacement in its initial orientation.

At (from equation [82]),


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The very outer-most beam fiber that lies on the y-axis is derived in section1.4.2,and is

given by the equality:

Since at , the outer-most beam fiber on they-axis is: For and the stress distribution is:

In tensor form:

The strains at

and

can be found by inserting these values into

the stress-strain relationships, given by equations [37] through [42]:

The vertical displacement at this point does not depend on y, so it is the same as

previously calculated.


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For and the stress distribution is:

In tensor form:

The strains at and can be found by inserting these values intothe stress-strain relationships, given by equations [37] through [42]:

The vertical displacement at this point does not depend on y, so it is the same as

previously calculated.


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2.Finite Element Analysis (FEA) Model2.1 MethodologyTo compare and validate the theoretical results, the finite element analysis (FEA)

software ABAQUS was used. ABAQUS is a powerful software application used for

both the modeling and analysis of mechanical components and assemblies (pre-

processing) and visualizing the finite element analysis results (post-processing).

ABAQUS offers a variety of different modeling approaches for the formulation of a

cantilever beam problem. The user has the option of using 3D-continuum parts, shell

assemblies, axisymmetric models, planar parts, and many other choices. For this

particular problem, a shell assembly was created and meshed to generate S4R elements

(4-sided shell elements using reduced integration methods). This option was chosenbecause it most closely matches the theoretical results of a traditional cantilever beam,

without a twist feature.

First, the midplane shell geometry was created by extruding line-connectors to specific

datum planes. These datum planes represent thex-values chosen for examination of the

theoretical solutions. SeeTable 1 for the values ofxthat represent these datum planes.

The figure below shows the shell geometry created in ABAQUS to create an FEA model

of the twisted cantilever beam.

Figure 11 - Midplane Shell of the ABAQUS FEA Model


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After the geometry was defined, the element was assigned a shell thickness ( t= 0.25)

and material properties (E = 30x106 psi, = 0.3) identical to those of the theoretical

solution in Section 1.5. The shell geometry was then meshed to form S4R elements,

with an approximate global seed size of 0.1. The seed size of the model determines the

coarseness of the mesh. ABAQUS generally selects an appropriate seed size based on

the number of elements it is able to process; with the student version of ABAQUS (used

herein), the number of elements is limited to 1,000.

It is possible to extrude the twisted region of the cantilever beam as one element,

however,Figure 11 above shows that it was created with two separate elements. This

was done to force the meshing to generate element points at the x = 3.75 point. This is

useful in that point results at the middle of the twist feature can later be extracted,

instead of using points nearby and approximating. Figure 12,below, shows the meshed

model of the twisted cantilever beam.

Figure 12 - Meshed View of the ABAQUS FEA Model

Finally, the root of the beam was fixed by creating an initial job step and setting all

displacement and rotations atx= 0 equal to zero (shown as orange cones inFigure 13).

A second job step was created to define the end-load on the beam. To avoid point

effects, a transverse shear, line-load on the far edge was used (shown as red arrows in


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the negative y direction in Figure 13). This model was submitted for analysis and

completed successfully.

Figure 13 - Load and Boundary Conditions of the ABAQUS FEA Model

Table 1 - Summary of FEA Input Parameters

Variable Value

L1 3

L2 4.5

L3 7.5

w 1.00

t 0.25

E 30 x 10 psi

0.3

F -10 lbs


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2.2 Analytical ResultsExplicit numerical results for x values corresponding to those analyzed in Section 1.5

and Appendix A can be found in Appendix B. The model used for analysis is a shell

model whose thickness or normal direction defines the z-axis, therefore shear stressescannot be directly extracted from ABAQUS in the traditional manner. As shown in

Figure 14below, the models 2 and 3 axes (corresponding to the Band C-axes) evolve

over the length of the beam. Thus, the numerical results provided by ABAQUS

correspond to the stress states in the ABC-coordinate system (see Figure 4). A direct

comparison can be made to any x-direction values, but to compare shearing components

or any direction other than x, the values must be properly transformed into the global

coordinate system (xyz). To apply this transformation of results, the user must specify a

field output transformation option corresponding to the global axes (under

ResultsOptionsTransformation).


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Figure 14 - FEA Shell Model, Variation of Principal Axes


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NOTE:

The following graphical representations extracted from ABAQUS give stress, strain,

and displacement values in index notation. Therefore,

2.2.1 Tensile Stress and Strain Values (x-Direction)The midplane of the FEA model shows tensile stress values (xx) peaking at the fixed

end of the beam, as is the case for a traditional cantilever beam. It does not indicate any

specific stress-increasing effect induced by the twist feature. SeeFigure 15 for a view of

the tensile stress results at the midplane.

At the shell models positive face, however, ABAQUS shows a severe increase in

compressive stress as the beam approaches its final orientation. At the models negative

face, the same region indicates a severe increase in tensile stress. See Figure 16 and

Figure 17 for the tensile stresses at the positive face of the shell model andFigure 18 and

Figure 19 for the tensile stresses at the negative face of the shell model.

Because the relationship between stress and strain is linear (see equations [34] and [35])the graphical results of only stress are given. The strain distributions predicted by

ABAQUS are similar to their corresponding stress distributions, shown below.


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Figure 15 - FEA Tensile Stress at the Shells Midplane


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Area of compressive stress peak

Figure 16 - FEA Tensile Stress at the Shell's Positive Face

Figure 17 - FEA Indicates an Additional Compressive Stress Peak in Twisted Region, Shells Positive Face


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Area of tensile stress peak

Figure 18 - FEA Tensile Stress at the Shell's Negative Face

Figure 19 - FEA Indicates an Additional Tensile Stress Peak in Twisted Region, Shells Negative Face


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2.2.2 Shear Stress and Strain Values (xy-Direction)Figure 20 shows the shear stress distribution of the shells midplane. Notice that the

initial orientation shows a parabolic distribution, as is the case of classical beam theory.

However, Figure 20 also shows that the shear stress has a negative shear stress

concentration at the origin, near the top and bottom fibers (the blue corners). This may

be attributed to point effects and discontinuity extrapolations generated by the numerical

integration routine of ABAQUS. Notice that just adjacent to these points, the shear

stress is positive, and more close to a zero value.

Figure 20 andFigure 21 also show that at the beams midpoint, in the middle of its twistfeature, there is a negative shear stress concentration. Again, note that there is a linear

relationship between stress and strain. Thus, only the distributions for stress are shown.

Figure 20 - FEA Shear Stress Distribution at the Shell's Midplane


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The results of the positive and negative faces of the shell model indicate severe shear

stress concentrations after the twist midpoint and before the beams final orientation

begins. Most notably, there is a large increase in positive shear in the xy-plane. This

rebounds the color spectrum of the post-processor, and results in the rest of the beam

appearing as a constant shear value. This is not the case, however. The shear stress still

follows a parabolic distribution over the height of the beam, up to the twisted region.

See Figure 22 and Figure 23 for the shear stress distributions on the positive and

negative faces of the shell model.

Negative Shear Stress Concentration

Figure 21 - FEA Shear Stress Concentration at Shell's Midplane


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Shear Stress Concentrations

Figure 22 - FEA Shear Stress Distribution on Shells Positive Face


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Shear Stress Concentrations

Figure 23 - FEA Shear Stress Distribution on Shells Negative Face


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2.2.3 Vertical Displacement (y-Direction)Figure 24 through Figure 26 below show the twisted cantilever beams displacement

distribution. Note that inFigure 26,there is a positive zdisplacement at the end of the

beam. This feature indicates shear coupling within the layers of the shell, which is not

addressed by classical beam theory.

Figure 24 - FEA Displacement Results, Isometric View

Figure 25 - FEA Displacement Results, xy-Plane


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When viewing the rotational displacements about the x-axis, an interesting point arises:

the twist feature seems to have imposed a structural anomaly in the beam at the end of its

twist, at x = x4. The displacements across the beams width at the end of the twist

(z-direction) follow a parabolic distribution. Classical beam theory suggests that vertical

displacements are independent of this direction. Looking at regions far from this point

(x4), the rotations about the x-axis disappear, corresponding to traditional theory. At

z

Figure 26 - FEA Displacement Results, yz-Plane


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x=x4, however, the outer edges of the beam have a higher residual stiffness than the

centerline, and experience smaller displacements. SeeFigure 27.

x4= 4.5

Figure 27 - FEA Indicates Residual Stiffness and Non-planar Cross-Section at x= x4


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3.Comparison of Beam Theory Formulation Results to FEAResults

3.1 Displacement ValuesThe vertical displacement functions derived in section1.4.5 show the closest correlation

to FEA results. Equation [82] matches the displacements calculated by ABAQUS

almost exactly. Equation [83] does a reasonable job up to the end of the twist but

beyond this (Equation [84]), predicts a slightly stiffer beam. Figure 28 shows the

displacements of the beam as calculated in this document and by FEA. Notice that each

method plots a smooth, cubic displacement function along the beams length.

Figure 28 - Theory vs. FEA, Vertical Displacement


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Overall, it is shown that for vertical displacements, classical beam theory can be applied

to cantilever beams with functionally varying moments of inertia to obtain reasonable

results. The percent difference of the beams vertical deflection at the end of the beam

(x= 7.5) using beam theory is only 4.5% different than the value predicted by FEA.

Classical beam theory (as applied in the derivation of theoretical values within),

however, does not account for out-of-plane displacements that may occur, as these are

resultant of shear-coupling and non-planar cross-sectional deformations. Figure 26

shows that FEA results predict out-of-plane displacements. Abiding by the key

assumptions of traditional beam mechanics (see section 1.3.1) forces one to discount

out-of-plane effects. In the case of a twisted cantilever beam, this may disregard key

aspects of the beams internal mechanics. Beam theory does not leaveroom for out-of-

plane displacements but does predict fairly accurate in-plane displacements when

compared to the results of FEA.

When compared to the analytical solutions for similar rectangular beams with constant

cross-sections (seeFigure 29), the vertical displacement curve of this particular twisted

cantilever beam seems to relate more closely to the beam experiencing strong-axis

bending, though this correlation will be a function of the length and twist parameters. If

the twist was closer to the fixed end of the beam (near x= 0), the displacement curve of

the twisted cantilever beam would be expected to more closely follow that of the

constant cross-section beam in weak-axis bending. Conversely, the closer the twist is to

the applied load (nearx=L3), the more closely the displacement curve would follow that

of the constant cross-section beam in strong-axis bending.


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Figure 29 - Theoretical Displacement Results vs. Beams of Constant Cross-Sectional Orientation


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3.2 Maximum Tensile Stress ValuesThe theoretical values of tensile stress (xx) match the numerical results very well outside

of the beams twisted region. Within the twisted region,however, the ABAQUS model

generally predicts much higher values, and does not indicate maximum or minimums atexactly the same value of x. The FEA model also does not follow the smooth curve

based on trigonometric modifications of the moment of inertia in this region. At each

tensile stress maximum point, the theoretical value is 9% less than that of the numerical

solution. See Figure 30 for the maximum tensile stresses predicted by theoretical

methods and by numerical methods. The theoretical method, as presented herein, should

not be used to calculate maximum tensile stress within the varying-orientation region of

a twisted cantilever beam.

Figure 30 - Theory vs. FEA, Maximum Tensile Stress


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3.3 Maximum Shear Stress ValuesAgain, beyond the twisted region, the theoretical stress values closely match the

numerical values. However, for the shear stress (xy), the FEA results vary greatly from

the theoretical results in the twisted region. Both methods allude to a maximum orminimum at the twists midpoint (x = 3.75), but the theoretical results appear to have an

inverted shape compared to those of the FEA shell model. SeeFigure 31.

The FEA model shows the shear stress peaking at the midpoint of the twist, but this

effect is not predicted by the application of classical beam theory. A possible

explanation is that, contrary to classical beam theory, plane sections are not remaining

plane under deformation (as shown by Figure 27). This effect would cause the shear

deformations to have a much greater impact on the distribution of flexural stresses

Figure 31 - Theory vs. FEA, Maximum Shear Stress


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within the beam than calculated with traditional methods. Figure 31 shows the severity

of error in using traditional beam theory to calculate shear stresses with in the beams

twisted region.

For the application of failure criterion (such as von Mises, Tresca, maximum stress, etc),

the shear stress equations developed by this document would not accurately demonstrate

maximum shear stress values, and could lead to errors. However, both theoretical and

FEA models indicate a peak of sorts at the twists midpoint. To mitigate gross errors

and propagate conservatism, a scalar factor could be applied to the constant shear stress

values of the initial or final orientations formulated by traditional methods to estimate

the maximum shear stress at the midpoint of the twist. In this particular problem, the

scalar factor is approximately 8/3. Additional case studies of beam geometries should be

performed to confirm the validity of this value and method. Refining the tensile and

shear stress equations could be possible by using the methodology of Timoshenko,

where plane sections do not necessarily remain plane, as demonstrated in Chapter 12 of

Reference [d].


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4.Summary and ConclusionBased on comparison of theoretical results to FEA results, application of classical beam

theory to twisted cantilever beams of rectangular cross-section is only useful in

calculating vertical displacement values, derived from the curvature of the beam at its

neutral surface. This approach is the beginning of the direct method. The vertical

displacement function developed correlates very well with FEA results. The theoretical

curve follows a smooth, negative-cubic shape along its length as would be expected with

a cantilever beam. However, classical beam theory does not leave room for deformation

coupling, non-planar cross-sections, and gives no method of solving for the out-of-plane

displacements that FEA predicts.

When the indirect method is used to calculate stresses and strains in the twisted region of

the cantilever beam, the results of this document demonstrate that classical beam theory

is not applicable. If the exact stress distribution was needed within the twisted region,

using the indirect method and classical beam theory would not predict reliable values.

The margin of error within the twisted region is too great to be safely applied to real

applications of twisted cantilever beams.

A possible explanation for the stress calculation error is that cross-sections of the beam

that begin plane are not remaining plane under load and subsequent displacement, which

is a critical assumption of classical beam theory. A non-planar section becomes evident

by studying the FEA rotational and displacement results at the end of the twist, presented

in Figure 27. Because the beams cross-section is being distorted as a result of the

applied load, a key stipulation of beam theory is obviously not being satisfied.

The theoretical stress and strain functions derived by this document are notrecommended for detailed analyses. However, based on the applied load and beam

geometry, the results presented herein suggest that the method of calculating vertical

displacement gives reasonable results. These displacement results could be applied to

the strain-displacement equations (section1.3.2), and then to the stress-strain equations

(section1.3.3)to approximatestress and strain values within the beams twisted region.


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Note that although the vertical displacement functions derived are very close to FEA

results, full application of the direct method to find stresses and strains will still be an

approximation. Under an applied load, FEA shows that the beam experiences out-of-

plane displacement. This z-displacement will contribute to the stress and strain

distribution throughout the beam, though its magnitude is much less than the vertical

displacements. Therefore, applying the direct method to calculate stresses will notyield

exact results, but could be used to estimate stress and strain values within the beams

twisted region.


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5.Appendix ATheoretical Results at Selected xValuesx1

x= 0, y = 0

Tensile Stress, xx(Equation [69]) 0 psi

Shear Stress, xy(Equation [71]) -60 psi

Longitudinal Strain, xx(Equation [75]) 0 in/in

Shear Strain, xy(Equation [78]) -2.600 x 10-6

in/in

Vertical Displacement, v(Equation [82]) 0.000 in

x= 0, y = 0.5

Tensile Stress, xx(Equation [69]) 1,800 psi

Shear Stress, xy(Equation [71]) 0 psi

Longitudinal Strain, xx(Equation [75]) 6.000 x 10-5

in/in

Shear Strain, xy(Equation [78]) 0 in/in


x= 0, y = -0.5

Tensile Stress, xx(Equation [69]) -1,800 psi


Longitudinal Strain, xx(Equation [75]) -6.000 x 10-5in/in




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x2x = 3.00, y = 0





in/in

Vertical Displacement, v (Equation [82]) -4.680 x 10-4

in

x = 3.00, y = 0.5



Longitudinal Strain, xx(Equation [75]) 3.600 x 10-5in/in


Vertical Displacement, v (Equation [82]) -4.680 x 10-4 in

x = 3.00, y = -0.5



Longitudinal Strain, xx(Equation [75]) -3.600 x 10-5

in/in



in


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x3x = 3.75, y = 0


Shear Stress, xy(Equation [73]) -14.12 psi



in/in

Vertical Displacement, v (Equation [83]) -7.297x 10-4

in

x = 3.75, y =

= 0.1768

Tensile Stress, xx(Equation [69]) 599.0 psi




Vertical Displacement, v (Equation [83]) -7.297x 10-4 in

x = 3.75, y = - = -0.1768Tensile Stress, xx(Equation [69]) -599.0 psi



in/in


Vertical Displacement, v (Equation [83]) -7.297x 10-4

in


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x4x = 4.50, y = 0





in/in


in

x = 4.50, y = 0.125






x = 4.50, y = -0.125




in/in



in


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x5x = 7.50, y = 0





in/in


in

x = 7.50, y = 0.125






x = 7.50, y = -0.125






in


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6.Appendix BFEA Results at Selected xValuesx1

x= 0, y = 0

Tensile Stress, xx -3.691 psi

Shear Stress, xy -62.26 psi

Longitudinal Strain, xx -1.096 x 10-11

in/in

Shear Strain, xy -3.676 x 10-6

in/in

Vertical Displacement, v 0.000 in

x= 0, y = 0.5

Tensile Stress, xx 1,717 psi


Longitudinal Strain, xx 5.527 x 10-5

in/in


in/in


x= 0, y = -0.5

Tensile Stress, xx -1,719 psi


Longitudinal Strain, xx -5.533 x 10-5in/in

Shear Strain, xy -1.300 x 10-5 in/in



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x2x = 3.00, y = 0




in/in


in/in

Vertical Displacement, v -4.812 x 10-6

in

x = 3.00, y = 0.5

Tensile Stress, xx 818.1 psi

Shear Stress, xy 35.16 psi

Longitudinal Strain, xx 3.225 x 10-5in/in

Shear Strain, xy -1.502 x 10-6in/in

Vertical Displacement, v -4.840 x 10-6 in

x = 3.00, y = -0.5

Tensile Stress, xx

-850.3 psi



in/in


in/in


in


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x3x = 3.75, y = 0




in/in


in/in


in

x = 3.75, y =

= 0.1768



Longitudinal Strain, xx 8.139 x 10-5in/in

Shear Strain, xy -7.471 x 10-5 in/in


x = 3.75, y = - = -0.1768Tensile Stress,

xx

-2,576 psi



in/in


in/in


in


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x4x = 4.50, y = 0




in/in


in/in


in

x = 4.50, y = 0.125



Longitudinal Strain, xx 9.758 x 10-5 in/in

Shear Strain, xy -6.052 x 10-6in/in


x = 4.50, y = -0.125

Tensile Stress, xx

-2,930 psi



in/in


in/in


in


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x5x = 7.50, y = 0

Tensile Stress, xx 8.282 x 10-6

psi



in/in

Shear Strain, xy x 10-6

in/in


in

x = 7.50, y = 0.125

Tensile Stress, xx 47.86 psi

Shear Stress, xy 0.3965 psi

Longitudinal Strain, xx 2.559 x 10-6 in/in

Shear Strain, xy 3.436 x 10-8in/in


x = 7.50, y = -0.125

Tensile Stress, xx

-47.86 psi



in/in


in/in


in


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7.Appendix CABAQUS Input File (.inp)*Heading

ProjectLoad

** Job name: TFBshell Model name: TFBshell*Preprint, echo=YES, model=YES, history=NO, contact=NO

**

** PARTS**

*Part, name=TFBshell

*Node1, 0.5, 0., 7.5

2, -0.5, 0., 7.5

3, -0.5, 0., 4.54, 0., 0., 4.5

5, 0.5, 0., 4.5

6, -0.353553385, 0.353553385, 3.757, 0.353553385, -0.353553385, 3.75

8, 0., 0.5, 3.

9, 0., 0., 3.10, 0., -0.5, 3.

11, 0., 0.5, 0.

12, 0., -0.5, 0.13, 0.400000006, 0., 7.5

14, 0.300000012, 0., 7.5

15, 0.200000003, 0., 7.516, 0.100000001, 0., 7.5

17, 0., 0., 7.5

18, -0.100000001, 0., 7.5

19, -0.200000003, 0., 7.5

20, -0.300000012, 0., 7.521, -0.400000006, 0., 7.5

22, -0.5, 0., 7.4000001

23, -0.5, 0., 7.3000001924, -0.5, 0., 7.19999981

25, -0.5, 0., 7.0999999

26, -0.5, 0., 7.

27, -0.5, 0., 6.9000001

28, -0.5, 0., 6.80000019

29, -0.5, 0., 6.69999981

30, -0.5, 0., 6.599999931, -0.5, 0., 6.5

32, -0.5, 0., 6.4000001

33, -0.5, 0., 6.30000019

34, -0.5, 0., 6.19999981

35, -0.5, 0., 6.099999936, -0.5, 0., 6.

37, -0.5, 0., 5.9000001

38, -0.5, 0., 5.8000001939, -0.5, 0., 5.69999981

40, -0.5, 0., 5.5999999

41, -0.5, 0., 5.542, -0.5, 0., 5.4000001

43, -0.5, 0., 5.30000019

44, -0.5, 0., 5.1999998145, -0.5, 0., 5.0999999

46, -0.5, 0., 5.

47, -0.5, 0., 4.900000148, -0.5, 0., 4.80000019

49, -0.5, 0., 4.69999981

50, -0.5, 0., 4.599999951, -0.400000006, 0., 4.5

52, -0.300000012, 0., 4.5

53, -0.200000003, 0., 4.5

54, -0.100000001, 0., 4.5

55, 0.100000001, 0., 4.5

56, 0.200000003, 0., 4.557, 0.300000012, 0., 4.5

58, 0.400000006, 0., 4.5


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59, 0.5, 0., 4.5999999

60, 0.5, 0., 4.6999998161, 0.5, 0., 4.80000019

62, 0.5, 0., 4.9000001

63, 0.5, 0., 5.64, 0.5, 0., 5.0999999

65, 0.5, 0., 5.19999981

66, 0.5, 0., 5.30000019

67, 0.5, 0., 5.400000168, 0.5, 0., 5.5

69, 0.5, 0., 5.5999999

70, 0.5, 0., 5.69999981

71, 0.5, 0., 5.80000019

72, 0.5, 0., 5.900000173, 0.5, 0., 6.

74, 0.5, 0., 6.0999999

75, 0.5, 0., 6.1999998176, 0.5, 0., 6.30000019

77, 0.5, 0., 6.4000001

78, 0.5, 0., 6.579, 0.5, 0., 6.5999999

80, 0.5, 0., 6.69999981

81, 0.5, 0., 6.8000001982, 0.5, 0., 6.9000001

83, 0.5, 0., 7.

84, 0.5, 0., 7.099999985, 0.5, 0., 7.19999981

86, 0.5, 0., 7.3000001987, 0.5, 0., 7.4000001

88, -0.49759236, 0.0490085706, 4.40625

89, -0.490392804, 0.0975444019, 4.3125014390, -0.478470445, 0.145141393, 4.21875191

91, -0.461939752, 0.191341713, 4.125

92, -0.44096002, 0.235699534, 4.0312476293, -0.415734202, 0.277786046, 3.93749785

94, -0.386505216, 0.317196637, 3.84375

95, -0.282842726, 0.282842726, 3.7596, -0.212132037, 0.212132037, 3.75

97, -0.141421363, 0.141421363, 3.75

98, -0.0707106814, 0.0707106814, 3.7599, 0., 0., 3.75

100, 0.0707106814, -0.0707106814, 3.75

101, 0.141421363, -0.141421363, 3.75102, 0.212132037, -0.212132037, 3.75

103, 0.282842726, -0.282842726, 3.75

104, 0.386505216, -0.317196637, 3.84375

105, 0.41573438, -0.277785748, 3.93749857

106, 0.440960169, -0.235699236, 4.03124809

107, 0.461939752, -0.191341713, 4.125108, 0.478470564, -0.14514108, 4.21875238

109, 0.490392864, -0.0975440741, 4.31250191

110, 0.49759236, -0.0490085706, 4.40625111, -0.317196637, 0.386505216, 3.65625

112, -0.277785748, 0.41573438, 3.56250143

113, -0.235699236, 0.440960169, 3.46875191114, -0.191341713, 0.461939752, 3.375

115, -0.14514108, 0.478470564, 3.28124738

116, -0.0975440741, 0.490392864, 3.18749785

117, -0.0490085706, 0.49759236, 3.09375118, 0., 0.400000006, 3.

119, 0., 0.300000012, 3.120, 0., 0.200000003, 3.

121, 0., 0.100000001, 3.

122, 0., -0.100000001, 3.123, 0., -0.200000003, 3.

124, 0., -0.300000012, 3.

125, 0., -0.400000006, 3.

126, 0.0490085706, -0.49759236, 3.09375

127, 0.0975444019, -0.490392804, 3.18749857

128, 0.145141393, -0.478470445, 3.28124809


85/111

74

129, 0.191341713, -0.461939752, 3.375

130, 0.235699534, -0.44096002, 3.46875262131, 0.277786046, -0.415734202, 3.56250215

132, 0.317196637, -0.386505216, 3.65625

133, 0., 0.5, 2.9000001134, 0., 0.5, 2.79999995

135, 0., 0.5, 2.70000005

136, 0., 0.5, 2.5999999

137, 0., 0.5, 2.5138, 0., 0.5, 2.4000001

139, 0., 0.5, 2.29999995

140, 0., 0.5, 2.20000005

141, 0., 0.5, 2.0999999

142, 0., 0.5, 2.143, 0., 0.5, 1.89999998

144, 0., 0.5, 1.79999995

145, 0., 0.5, 1.70000005146, 0., 0.5, 1.60000002

147, 0., 0.5, 1.5

148, 0., 0.5, 1.39999998149, 0., 0.5, 1.29999995

150, 0., 0.5, 1.20000005

151, 0., 0.5, 1.10000002152, 0., 0.5, 1.

153, 0., 0.5, 0.899999976

154, 0., 0.5, 0.800000012155, 0., 0.5, 0.699999988

156, 0., 0.5, 0.600000024157, 0., 0.5, 0.5

158, 0., 0.5, 0.400000006

159, 0., 0.5, 0.300000012160, 0., 0.5, 0.200000003

161, 0., 0.5, 0.100000001

162, 0., 0.400000006, 0.163, 0., 0.300000012, 0.

164, 0., 0.200000003, 0.

165, 0., 0.100000001, 0.166, 0., 0., 0.

167, 0., -0.100000001, 0.

168, 0., -0.200000003, 0.169, 0., -0.300000012, 0.

170, 0., -0.400000006, 0.

171, 0., -0.5, 0.100000001172, 0., -0.5, 0.200000003

173, 0., -0.5, 0.300000012

174, 0., -0.5, 0.400000006

175, 0., -0.5, 0.5

176, 0., -0.5, 0.600000024

177, 0., -0.5, 0.699999988178, 0., -0.5, 0.800000012

179, 0., -0.5, 0.899999976

180, 0., -0.5, 1.181, 0., -0.5, 1.10000002

182, 0., -0.5, 1.20000005

183, 0., -0.5, 1.29999995184, 0., -0.5, 1.39999998

185, 0., -0.5, 1.5

186, 0., -0.5, 1.60000002

187, 0., -0.5, 1.70000005188, 0., -0.5, 1.79999995

189, 0., -0.5, 1.89999998190, 0., -0.5, 2.

191, 0., -0.5, 2.0999999

192, 0., -0.5, 2.20000005193, 0., -0.5, 2.29999995

194, 0., -0.5, 2.4000001

195, 0., -0.5, 2.5

196, 0., -0.5, 2.5999999

197, 0., -0.5, 2.70000005

198, 0., -0.5, 2.79999995


86/111

75

199, 0., -0.5, 2.9000001

200, 0.40867731, 0., 6.19405651201, -0.402668566, 0., 6.39775133

202, 0.405933738, 0., 5.29558468

203, 0.408387423, 0., 6.49111652204, 0.401252568, 0., 6.69411993

205, 0.396833479, 0., 6.90104437

206, 0.00177802215, 0., 7.40288019

207, -0.197118372, 0., 7.39842939208, -0.298042893, 0., 7.3983407

209, -0.400736928, 0., 4.99964094

210, 0.000279046595, 0., 4.59903193

211, 0.201234028, 0., 4.59916306

212, 0.403696954, 0., 4.99643135213, -0.402956039, 0., 5.09924746

214, 0.404304177, 0., 5.09595919

215, -0.405808479, 0., 5.19816256216, 0.40518558, 0., 5.19583607

217, -0.407385081, 0., 5.29740095

218, 0.406625658, 0., 5.39631796219, -0.406762481, 0., 5.59426737

220, 0.406653494, 0., 5.69659758

221, -0.40634349, 0., 5.89342594222, 0.407420099, 0., 5.99554634

223, -0.40429756, 0., 6.19505644

224, 0.409388989, 0., 6.29291248225, -0.403280675, 0., 6.49861288

226, -0.406411737, 0., 6.69895744227, -0.405461013, 0., 6.89886618

228, -0.400929481, 0., 7.0986414

229, -0.399379015, 0., 7.19837999230, -0.398801684, 0., 7.29853868

231, -0.100126587, 0., 4.59898329

232, -0.200560495, 0., 4.59896088233, -0.300716281, 0., 4.59911108

234, -0.400495023, 0., 4.59947872

235, 0.301283509, 0., 4.59913683236, 0.402510345, 0., 4.79805994

237, -0.407594681, 0., 5.3964119

238, -0.407321811, 0., 5.49507999239, 0.406856358, 0., 5.49654388

240, 0.406758994, 0., 5.59697676

241, -0.406829447, 0., 5.69340563242, -0.406897724, 0., 5.79300213

243, 0.406882733, 0., 5.79643965

244, 0.407251507, 0., 5.89637852

245, -0.405940652, 0., 5.99382353

246, -0.405343413, 0., 6.09392929

247, 0.408475012, 0., 6.09557915248, -0.403198481, 0., 6.29597473

249, 0.410051048, 0., 6.39194059

250, -0.404992044, 0., 6.59876156251, 0.406056046, 0., 6.59184217

252, -0.406782091, 0., 6.79884672

253, 0.397338152, 0., 6.7973218254, -0.403321207, 0., 6.99862671

255, 0.398649305, 0., 7.00376129

256, 0.102255292, 0., 7.40408039

257, 0.400675923, 0., 7.10461521258, 0.202405795, 0., 7.40403891

259, 0.302304924, 0., 7.40319681260, 0.40184623, 0., 7.20414352

261, 0.401590854, 0., 7.40185547

262, -0.0971012264, 0., 7.39980888263, 0.295431882, 0., 6.7917347

264, 0.100899935, 0., 4.59913445

265, -0.200336963, 0., 4.69831038

266, -0.400373787, 0., 4.79929972

267, -0.400186002, 0., 4.89960384

268, 0.302564979, 0., 4.69811058


87/111

76

269, -0.301119387, 0., 5.00126266

270, 0.314894289, 0., 5.4961009271, 0.314381897, 0., 5.79639339

272, 0.30728507, 0., 6.6842556

273, -0.301167935, 0., 7.09433603274, -0.296987951, 0., 7.1945982

275, -0.31428501, 0., 5.48812866

276, -0.313560694, 0., 5.98459959

277, 0.321501642, 0., 6.28869247278, 0.322397113, 0., 6.48214102

279, 0.298777401, 0., 7.00802946

280, 0.00669529289, 0., 7.30547333

281, 0.204572856, 0., 7.30841064

282, 0.303512573, 0., 7.3060894283, -0.399060756, 0., 7.39900494

284, 0.000929877104, 0., 4.69793987

285, 0.403036028, 0., 4.89737177286, 0.305870652, 0., 4.99487877

287, -0.306456804, 0., 5.10053301

288, 0.307883441, 0., 5.09456158289, -0.313620389, 0., 5.19682837

290, 0.310606509, 0., 5.19448519

291, 0.313224137, 0., 5.2947917292, -0.315620482, 0., 5.39094973

293, -0.312159777, 0., 6.08479261

294, 0.319978833, 0., 6.1921401295, -0.310105145, 0., 6.18633223

296, -0.307616264, 0., 6.2891984297, -0.30550006, 0., 6.39351463

298, -0.315976441, 0., 6.79735613

299, -0.307581723, 0., 6.99457979300, -0.296388656, 0., 7.29612589

301, -0.0996720865, 0., 4.69822311

302, -0.300613523, 0., 4.69879389303, -0.400424659, 0., 4.69941044

304, 0.401785314, 0., 4.6987381

305, -0.316448867, 0., 5.29398108306, 0.314912826, 0., 5.39621305

307, 0.314139575, 0., 5.59668589

308, -0.314390153, 0., 5.58593273309, -0.315074533, 0., 5.68460941

310, 0.314020574, 0., 5.69721174

311, -0.315427691, 0., 5.78401661312, 0.31517747, 0., 5.89609241

313, -0.315054536, 0., 5.88357019

314, 0.316314787, 0., 5.99555731

315, 0.318404227, 0., 6.09523869

316, 0.324144006, 0., 6.38569736

317, -0.306122333, 0., 6.49719858318, -0.310529053, 0., 6.59844065

319, 0.317961097, 0., 6.58154917

320, -0.314852864, 0., 6.69842958321, -0.31351608, 0., 6.89608908

322, 0.295243055, 0., 6.90154171

323, 0.105890259, 0., 7.30880499324, 0.301904917, 0., 7.10959387

325, 0.303527385, 0., 7.20823908

326, 0.402051926, 0., 7.30309677

327, -0.299800634, 0., 4.90010214328, -0.0902894214, 0., 7.29561043

329, -0.0986664221, 0., 4.79851437330, 0.0018088948, 0., 4.79800415

331, 0.222399965, 0., 5.29814672

332, 0.216678962, 0., 5.19791412333, -0.223152682, 0., 5.97148514

334, -0.220869288, 0., 6.0717802

335, -0.192721233, 0., 7.29373407

336, -0.299998164, 0., 4.79916191

337, 0.210820511, 0., 5.0974164

338, -0.220917225, 0., 5.47843266


88/111

77

339, 0.243320033, 0., 6.47432995

340, -0.218023121, 0., 6.1738553341, 0.186521545, 0., 6.78018904

342, 0.1101702, 0., 7.21433544

343, 0.204664022, 0., 7.1148324344, -0.200195268, 0., 7.08509207

345, -0.225005433, 0., 6.89159298

346, 0.202432424, 0., 4.69789648

347, 0.303624541, 0., 4.7969141348, 0.22168529, 0., 5.59982252

349, -0.190822065, 0., 7.18799162

350, 0.236797214, 0., 6.56837702

351, 0.2008463, 0., 7.01311731

352, -0.21349068, 0., 6.98613358353, 0.101697974, 0., 4.69795609

354, 0.304460764, 0., 4.89592218

355, -0.208076537, 0., 5.10710716356, -0.199524254, 0., 5.00600624

357, -0.223547459, 0., 5.38253593

358, 0.225628793, 0., 5.39937782359, 0.236854687, 0., 6.29055977

360, 0.242798716, 0., 6.38344049

361, -0.213846207, 0., 6.27833939362, -0.206707627, 0., 6.49696875

363, -0.199223369, 0., 4.79917574

364, -0.226176366, 0., 5.19723749365, -0.227950245, 0., 5.28914165

366, 0.224521562, 0., 5.49932766367, 0.221601427, 0., 5.70128012

368, -0.22309956, 0., 5.57635784

369, 0.22423631, 0., 5.90105867370, -0.22571373, 0., 5.67425919

371, 0.22282511, 0., 5.80190802

372, -0.22619018, 0., 5.77294731373, -0.225415528, 0., 5.87130213

374, 0.226192445, 0., 6.00080109

375, 0.232350484, 0., 6.19606352376, 0.229102418, 0., 6.10029364

377, -0.208279058, 0., 6.38662672

378, -0.226509944, 0., 6.70077181379, -0.217268422, 0., 6.60135412

380, 0.220396414, 0., 6.66632652

381, -0.229614273, 0., 6.79647446382, 0.196128488, 0., 6.90512657

383, 0.0182983484, 0., 7.2118907

384, 0.206072107, 0., 7.21169806

385, -0.198362261, 0., 4.90161896

386, 0.102622084, 0., 4.79722118

387, 0.203470454, 0., 4.79676676388, -0.0974442586, 0., 5.13574553

389, -0.098619014, 0., 7.06667614

390, -0.119545288, 0., 6.96879148391, -0.128163159, 0., 6.1582818

392, 0.109173544, 0., 7.12110043

393, 0.133301482, 0., 5.30742502394, 0.131907314, 0., 5.81121492

395, -0.139652088, 0., 5.66371918

396, -0.122907341, 0., 6.26307774

397, 0.138223693, 0., 6.63145018398, 0.122799806, 0., 5.20717192

399, -0.0742361471, 0., 7.17870903400, 0.204644904, 0., 4.89598274

401, -0.125714183, 0., 5.46752167

402, -0.131364495, 0., 6.05617237403, -0.143076956, 0., 6.88430548

404, 0.206699133, 0., 4.99664974

405, -0.0973391682, 0., 4.90246487

406, 0.136469558, 0., 5.5020051

407, 0.163831994, 0., 6.38763094

408, 0.171233878, 0., 6.46854591


89/111

78

409, 0.159717858, 0., 6.54961777

410, -0.143010616, 0., 6.70844793411, -0.156049699, 0., 5.20429611

412, 0.139882848, 0., 5.40635729

413, -0.130619228, 0., 5.369452414, 0.127579197, 0., 5.60603809

415, 0.1291866, 0., 5.71005011

416, -0.134329766, 0., 5.56901455

417, 0.136376977, 0., 6.0110755418, -0.138612881, 0., 5.76024199

419, 0.134122372, 0., 5.91191626

420, -0.13651605, 0., 5.85700512

421, 0.144195423, 0., 6.20690966

422, -0.134087384, 0., 5.95578814423, 0.152796224, 0., 6.30109358

424, 0.139660269, 0., 6.1107173

425, -0.113268718, 0., 6.37452126426, -0.0975690931, 0., 6.5010314

427, -0.148312464, 0., 6.79710388

428, -0.12851657, 0., 6.61295128429, 0.0305773336, 0., 6.66694832

430, 0.0714686736, 0., 6.75505114

431, 0.0954950154, 0., 6.91913986432, 0.0903138742, 0., 6.82106161

433, 0.103952818, 0., 4.897964

434, -0.0581134595, 0., 5.24749279435, 0.0296035167, 0., 5.2230773

436, -0.0235007312, 0., 5.45896435437, -0.0394197069, 0., 6.14110708

438, 0.0842929482, 0., 6.40796709

439, -0.0695208237, 0., 6.80322504440, 0.0592189096, 0., 5.41695166

441, 0.0543499924, 0., 5.49868059

442, 0.0251491554, 0., 5.61175299443, -0.0447129048, 0., 5.93976307

444, -0.0469984338, 0., 5.8409977

445, -0.033807084, 0., 6.24395227446, 0.0498633273, 0., 6.12507343

447, 0.112137116, 0., 6.465343

448, 0.0102930805, 0., 6.74083328449, -0.000287796778, 0., 7.03742981

450, 0.0422273129, 0., 5.82510805

451, -0.0423180871, 0., 6.04010296452, 0.106647633, 0., 5.00105667

453, 0.00298698479, 0., 4.90101719

454, -0.0636882186, 0., 6.72270203

455, 0.0052099661, 0., 6.46184301

456, 0.0445605069, 0., 5.32480288

457, -0.0420513712, 0., 5.35015059458, -0.0589031056, 0., 5.65282202

459, 0.038012702, 0., 5.72390032

460, -0.0526096337, 0., 5.57187128461, -0.0504724607, 0., 5.74457073

462, 0.0445855856, 0., 5.92536259

463, 0.0469786003, 0., 6.0254817464, 0.0651964545, 0., 6.32155228

465, 0.0554221906, 0., 6.22453547

466, -0.021974273, 0., 6.35031176

467, -0.0482959114, 0., 6.63777971468, -0.0203869492, 0., 6.55431509

469, -0.0783420578, 0., 6.87100172470, -0.0177761577, 0., 6.92647791

471, 0.00460041501, 0., 6.82224131

472, 0.00513232173, 0., 5.00768137473, 0.0167262461, 0., 7.13218832

474, -0.0966288298, 0., 5.01241636

475, 0.0115827657, 0., 5.11680508

476, 0.101649359, 0., 7.02168989

477, 0.400879443, 0., 4.59934616

478, 0.108288564, 0., 6.70781231


90/111

79

479, 0.112396933, 0., 5.10517406

480, 0.0559308752, 0., 6.59166718481, -0.140449628, 0., 5.27679873

482, 0.0822723955, 0., 6.51833534

483, -0.39453128, 0.0404181406, 4.4025116484, -0.294307023, 0.0310180821, 4.39972687

485, -0.195317686, 0.0211791564, 4.39685583

486, -0.0981834605, 0.0110054938, 4.39340591

487, 0.0974388197, -0.0110404696, 4.3922596488, 0.196074381, -0.0214761999, 4.39582109

489, 0.29577893, -0.0310798828, 4.40002489

490, -0.389100462, 0.0795408636, 4.3074441

491, -0.291077256, 0.0610979162, 4.30242586

492, -0.194597781, 0.0423649475, 4.29530096493, -0.100957528, 0.0233076978, 4.28333521

494, -0.00332703814, 0.000786339573, 4.27837133

495, 0.095376201, -0.0224310998, 4.27942276496, 0.292506605, -0.0619109347, 4.30082226

497, -0.38214317, 0.11793372, 4.21415281

498, -0.289467514, 0.090895541, 4.20945358499, -0.202075586, 0.0650578141, 4.20256901

500, -0.11561846, 0.0423492603, 4.16471672

501, 0.0976936817, -0.0368712172, 4.15537786502, 0.198754832, -0.067814976, 4.18600655

503, 0.291765392, -0.0932855457, 4.20449114

504, 0.385729671, -0.118881494, 4.21451283505, -0.370616794, 0.154640973, 4.12252522

506, -0.281834334, 0.117777593, 4.12200212507, -0.0968274847, 0.0492477678, 4.0506916

508, -0.0110924458, 0.00555647956, 4.05654287

509, 0.139335796, -0.0686392635, 4.06290531510, 0.215747297, -0.0993557125, 4.08788395

511, 0.294349134, -0.125744611, 4.11446762

512, -0.353612244, 0.189381301, 4.03046989513, -0.16724059, 0.110600732, 3.94203758

514, -0.0869640633, 0.0568458512, 3.94714236

515, -0.00753175607, 0.00488119386, 3.95089102516, 0.0717929453, -0.046249371, 3.95350266

517, 0.152061835, -0.0962316468, 3.96120858

518, 0.243869111, -0.148629859, 3.97731876519, -0.332861006, 0.222312957, 3.9376936

520, -0.231885433, 0.189902529, 3.84473753

521, -0.154716372, 0.12640582, 3.84584355522, -0.0786142051, 0.0640072823, 3.8474617

523, 0.0728859231, -0.0588177741, 3.8516171

524, 0.149941236, -0.120037109, 3.85534334

525, 0.227919579, -0.181463853, 3.85790133

526, 0.39161098, -0.0797342956, 4.30819273

527, 0.333999455, -0.21827659, 3.94724059528, 0.378465265, -0.157423183, 4.12358427

529, 0.396312237, -0.0401718915, 4.40353394

530, 0.364216506, -0.1934174, 4.03382397531, 0.307291895, -0.248437062, 3.85075688

532, -0.309068471, 0.253469706, 3.84407544

533, -0.000869911979, 9.93500

king final report

Documents