Applying Classical Beam Theory to Twisted Cantilever Beams and Comparing to the Results of FEA

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1 Applying Classical Beam Theory to Twisted Cantilever Beams and Comparing to the Results of FEA by Mitchell S. King An Engineering Project Submitted to the Graduate Faculty of Rensselaer Polytechnic Institute in Partial Fulfillment of the Requirements for the degree of MASTER OF ENGINEERING IN MECHANICAL ENGINEERING Approved: Professor Ernesto Gutierrez-Miravete, Engineering Project Advisor Rensselaer Polytechnic Institute Hartford, Connecticut December, 2012

2 Copyright 2012 by Mitchell S. King All Rights Reserved ii

3 CONTENTS LIST OF FIGURES... v LIST OF TABLES... vii NOMENCLATURE... viii GLOSSARY... ix ACKNOWLEDGMENT... x ABSTRACT... xi 1. Theory Coordinate System Twist Parameters Stress, Strain, and Displacement for a Cantilever Beam Stress Equilibrium Strain-Displacement Relationships Stress-Strain Relationships Generic Cantilever Beam Stress, Strain, and Displacement Solutions Functionally Varying Moment of Inertia Discontinuous Shear Stress Solution for a Rectangular Beam General Stress Distribution for a Twisted Cantilever Beam General Strain Distribution for a Twisted Cantilever Beam General Vertical Displacement of a Twisted Cantilever Beam Specific Theoretical Solution Finite Element Analysis (FEA) Model Methodology Analytical Results Tensile Stress and Strain Values (x-direction) Shear Stress and Strain Values (xy-direction) Vertical Displacement (y-direction) iii

4 3. Comparison of Beam Theory Formulation Results to FEA Results Displacement Values Maximum Tensile Stress Values Maximum Shear Stress Values Summary and Conclusion Appendix A Theoretical Results at Selected x Values Appendix B FEA Results at Selected x Values Appendix C ABAQUS Input File (.inp) References iv

5 LIST OF FIGURES Figure 1 - Example of a Twisted Cantilever Beam... 1 Figure 2 90 Twisted Cantilever Beam in the xy-plane... 2 Figure 3 90 Twisted Cantilever Beam in the yz-plane... 3 Figure 4 - xyz Axes vs. ABC Axes... 4 Figure 5 Arbitrary Beam Cross-Section Subjected to Bending Moment M z (x)... 6 Figure 6 - Rotation of Axes and Moment of Inertia Figure 7 - Moment of Inertia Variation within the Twisted Region of a Rectangular Beam Figure 8 - Point of Discontinuity in Shear Stress Figure 9 - Width and Thickness Parameters of the Solution Figure 10 - Load and Length Parameters of the Solution Figure 11 - Midplane Shell of the ABAQUS FEA Model Figure 12 - Meshed View of the ABAQUS FEA Model Figure 13 - Load and Boundary Conditions of the ABAQUS FEA Model Figure 14 - FEA Shell Model, Variation of Principal Axes Figure 15 - FEA Tensile Stress at the Shell s Midplane Figure 16 - FEA Tensile Stress at the Shell's Positive Face Figure 17 - FEA Indicates an Additional Compressive Stress Peak in Twisted Region, Shell s Positive Face Figure 18 - FEA Tensile Stress at the Shell's Negative Face Figure 19 - FEA Indicates an Additional Tensile Stress Peak in Twisted Region, Shell s Negative Face Figure 20 - FEA Shear Stress Distribution at the Shell's Midplane Figure 21 - FEA Shear Stress Concentration at Shell's Midplane Figure 22 - FEA Shear Stress Distribution on Shell s Positive Face Figure 23 - FEA Shear Stress Distribution on Shell s Negative Face Figure 24 - FEA Displacement Results, Isometric View Figure 25 - FEA Displacement Results, xy-plane Figure 26 - FEA Displacement Results, yz-plane Figure 27 - FEA Indicates Residual Stiffness and Non-planar Cross-Section at x = x v

6 Figure 28 - Theory vs. FEA, Vertical Displacement Figure 29 - Theoretical Displacement Results vs. Beams of Constant Cross-Sectional Orientation Figure 30 - Theory vs. FEA, Maximum Tensile Stress Figure 31 - Theory vs. FEA, Maximum Shear Stress vi

7 LIST OF TABLES Table 1 - Summary of FEA Input Parameters vii

8 NOMENCLATURE For the following symbols, i j. Units are in parenthesis. E - Modulus of Elasticity (psi) ε - Strain Tensor (in/in) ε ii - Extensional Strain (in/in) ε ij - Tensor Shear Strain (in/in) F - Force Applied to the Free End of the Cantilever Beam (lbs) G - Shear Modulus (psi) γ - Engineering Shear Strain (in/in) I ii - Moment of Inertia (in 4 ) I ij - Product of Inertia (in 4 ) κ - Curvature (in -1 ) L - Length (in) M - Moment (in-lb) ν - Poisson s Ratio (dimensionless) σ - Stress Tensor (psi) σ ii - Tensile Stress (psi) t - Thickness of the Rectangular Cross-Section, Less than its Width, w (in) τ ij - Shear Stress (psi) ϴ - Angular Measurement Between the B and Y-Axes or the C and Z-Axes (radians or degrees) u - Longitudinal Displacement (in) v - Vertical Displacement (in) w - Transverse Displacement, Only Applicable to Section (in) w - Width of the Rectangular Cross-Section, Greater than its Thickness, t (in) viii

9 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 beyond L 2. 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 linearlyrelated to the applied stress within the elastic limit of the solid. Initial Orientation - The cross-sectional orientation of the beam from its origin to L 1. Isotropic - Having uniform physical properties in each direction. Local Axes - The coordinate system that remains perpendicular to the perimeter of the beam s 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 solid s extensional strain value(s) to its longitudinal stress value(s) within the elastic region. Moment of Inertia - A measurement that quantifies a beam s ability to resist bending about 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 beam s ability to resist shear deformation about a particular axis. Shear Modulus - The proportionality constant relating a solid s 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 is rotating about its centroid along the beam s length. Varying Orientation - The cross-sectional orientation of the beam in the region of L 1 to L 2. ix

10 ACKNOWLEDGMENT I would like to thank my fiancé, Jessica Rowe, for her unwavering support, encouragement, and affection during my graduate studies. x

11 ABSTRACT Twisted cantilever beams are defined here as beams whose rectangular cross-sectional orientation changes along the beam s 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. xi

12 1. Theory 1.1 Coordinate System The analytical case of interest is a rectangular cantilever beam in static bending with an abrupt, 90 twist, like that shown in Figure 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 origin or 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 beam s 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 beam s geometric center, at the fixed end, are defined as the global axes. 1

13 Moving down the beam s length, away from the rigid attachment, in the x-direction, the beam initially has a homogeneous moment of inertia and its initial orientation. At some distance, denoted by L 1, the beam s 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 beam s cross-section continues to rotate until it reaches another distance, denoted by L 2, where the rotation stops but again, the moment of inertia remains continuous. Beyond L 2, the beam s final orientation continues until it reaches its free end, whose distance from the origin is denoted by the length L 3. For the case of a 90 twist, the final orientation is rotated 90 from the initial orientation. See Figure 2 and Figure 3. Y F O X L 1 L 2 L 3 Figure 2 90 Twisted Cantilever Beam in the xy-plane 2

14 Y Z Figure 3 90 Twisted Cantilever Beam in the yz-plane Because the cross-sectional orientation of the beam is varying along its length between L 1 and L 2, 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 beam s length, provided that the A-axis remains collinear with the x-axis. At the beam s 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 beam s cross-section begins to rotate (at L 1 ) 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 always remain collinear, and either can be used interchangeably. See Figure 4. 3

15 Y [ x and A-axes both coming out of the page. ] Y B B ϴ Z Z C C Figure 4 - xyz Axes vs. ABC Axes It is important to describe and define the parameters within the beam s varying orientation (L 1 x L 2 ), because these parameters will contribute to the analytical solutions that are eventually formulated. The difference between the cross-sections at x = L 1 and L 1 x L 2 can be quantified by the angular difference, ϴ, between the y and B-axes or the z and C-axes. The rate of twist can then be defined as the change in angle, ϴ, over the change in length between L 1 and L 2, or. 1.2 Twist Parameters For the problem to be solved, the cross-sectional orientation can be described as: [1] [2] [3] Let ϴ(x) be a linear function of x, so that the rate of twist from L 1 to L 2 is a constant value. Then, by the equalities given by equations [1] through [3]: 4

16 Substituting the boundary conditions in the expression for ϴ yields: Solving for the constants M and N, and subtracting these equalities gives: The function ϴ(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 beam s twist parameters and can be derived using the values L 1 and L Stress, Strain, and Displacement for a Cantilever Beam Because 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 5

17 applied. Let a beam of arbitrary cross-section be subjected to a bending moment, M z (x), about the z-axis, which is a function of longitudinal (x) position only. The origin of the cross section is at the beam s centroid, and the y and z-axes are the principal axes. See Figure 5. [ X and A-axes both coming out of the page. ] Y Arbitrary Cross-Section Z M z (x) Figure 5 Arbitrary Beam Cross-Section Subjected to Bending Moment M z (x) Stress Equilibrium As a result of the applied load, F, and subsequent moment, M z (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, [σ], is given by: [9] In order to satisfy equilibrium, the stress tensor must be symmetric such that: 6

18 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, M z (x): [10] [11] [12] That is, the normal stresses, σ ii, and the x-z and y-z shear stresses, τ xz and τ yz, are identical to those for a beam in pure bending, but no specific assumptions are made about the other shear stress, τ xy, only that it is some function of x, y, and z. 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] 7

19 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 τ xy is constant in the x-domain. Rewriting equations [10] and [11] for clarity, equation [12] can be simplified to: [18] [19] [20] Strain-Displacement Relationships As mentioned, the beam also experiences a state of strain. The generic stress tensor, [ε], is given by: [21] where ε ii are normal strains and ε ij are shear strains. The strain tensor is symmetric, such that: 8

20 For small deflections (where sin α α), 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, [ε], are: [28] [29] [30] [31] [32] [33] 9

21 1.3.3 Stress-Strain Relationships For an isotropic, linear-elastic material (obeying Hooke s Law), the stress-strain relationships are as follows: [34] [35] where E is the modulus of elasticity and G is the shear modulus. The two modulii are related through Poisson s Ratio, ν, by the following: [36] For each component of the strain tensor, the relationships are: [37] [38] [39] [40] [41] [42] 10

22 1.4 Generic Cantilever Beam Stress, Strain, and Displacement Solutions Equation [18] indicates that the normal stress in the x-direction is dependent on the bending moment, M z (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: So, the normal stress in the x-direction is: [43] [44] Substituting equation [44] in the x-direction stress equilibrium equation ([16]) gives: [45] Integrating equation [45]: P is 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] 11

23 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 straindisplacement 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 of I zz is known. For small strains and displacements in the elastic range, and assuming that plane sections remain plane, the curvature of the beam s neutral surface can be expressed in the following form: where κ 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, I zz, varies with respect to x, 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, I zz, and the constant of integration, P, for the shear term. This moment of inertia and P are derived in the following discussion. 12

24 1.4.1 Functionally Varying Moment of Inertia Consider the plane area shown in Figure 6 below. The moments and product of inertia with respect to the local BC-axes are: [51] [52] The same forms of expressions exist for the global coordinate system, in xyz- coordinates. [53] [ x and A-axes both coming out of the page. ] Y B Arbitrary Cross-Section da C da z ϴ Z y B da C Figure 6 - Rotation of Axes and Moment of Inertia 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 θ 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 the yz-coordinates as follows: 13

25 [54] [55] Substituting these values in equations [51] through [53] gives: Using the following trigonometric identities, the form of I YY, I ZZ, and I YZ can be simplified. 14

26 These are complicated expressions in their most reduced form, unlike the simple and familiar (for example). However, if the general expressions for moments and product of inertia in the local coordinate system (equations [51], [52], and [53]) are substituted in the above integrals, they take on a more practical form. [56] [57] [58] I YY, I ZZ, and I YZ are the moments and product of inertia in the global coordinate system at any point along the x-axis, and I BB, I CC, and I BC are 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, where I CC is the strong axis of bending and I BB is the weak axis of bending at the origin. The limits of integration are defined by the beam s perimeter. Thus: 15

27 [59] [60] [61] where w is the width of the beam, and t is the thickness. For any symmetric cross-section whose centroid is at the origin, I BC will be zero, so any terms containing the beam s local product 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 ϴ and x (see equations [6], [7], and [8]), these equations yield the full form of the functionally varying moments of inertia. [62] 16

28 [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 in Figure 7 below: Figure 7 - Moment of Inertia Variation within the Twisted Region of a Rectangular Beam 17

29 1.4.2 Discontinuous Shear Stress Solution for a Rectangular Beam The stress distributions will be solved to eventually formulate the strain and displacement solutions. However, the shear stress, τ xy, still has an undefined constant, P, which must be found to fully formulate the stress in the beam. As stated in section 1.4.1, P is 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 beam s cross-section is rotating, this constant is also related to the twist parameters for L 1 x L 2. For the beam s initial orientation: Substituting this value into equation [48] gives: [65] For the beam s final orientation: Substituting this value into equation [48] gives: [66] 18

30 For the beam s varying orientation, the y-coordinate of the beam s 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 ϴ. 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 the y-axis coincides with the corner of the beam s cross-section. For any rectangular beam, this occurs at an angle of. See Figure 8. Y ϴ = arctan (t/w) Z w t Figure 8 - Point of Discontinuity in Shear Stress For the varying orientation, where 0 ϴ arctan (t/w), the y-coordinate of the outermost fiber of the beam s cross-section is: 19

31 Substituting this value into equation [48] and using the relationship between ϴ and x (equation [7]) gives: [67] Similarly, for the varying orientation, where arctan (t/w) ϴ π/2, the y-coordinate of the outermost fiber of the beam s cross-section is: Substituting this value into equation [48] and using the relationship between ϴ and x (equation [7]) gives: [68] 20

32 1.4.3 General Stress Distribution for a Twisted Cantilever Beam Consolidating equations [6], [7], [8], [46], and [65] through [68]: Any Orientation: [69] Any Orientation: [70] Initial Orientation: [71] Varying Orientation: [72] Varying Orientation: [73] Final Orientation: [74] 21

33 1.4.4 General Strain Distribution for a Twisted Cantilever Beam Inserting equations [69] through [74] into the stress-strain relationships given by equations [37] through [42]: Any Orientation: [75] Any Orientation: [76] Any Orientation: [77] Initial Orientation: [78] Varying Orientation: [79] Varying Orientation: [80] Final Orientation: [81] 22

34 1.4.5 General Vertical Displacement of a Twisted Cantilever Beam Now that the moment of inertia, I zz, has been derived, equation [50] can be integrated to obtain the y-displacement function. However, because I zz 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 initial orientation and the moment of inertia is constant in this region, equation [50] is integrated twice in x, as would be done for a normal cantilever beam with constant crosssection: where C 1 is a constant of integration. Integrating again in x gives: The constants of integration, C 1 and C 2 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 of x where, the moment of inertia cannot be considered constant and excluded from the integrals, as was done above. Inserting the equality for I zz into equation [50], for values of x where, and integrating once with respect to x gives: 23

35 The second term of the y-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 of inertia function s 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 L 1 to L 2 will be divided by N number of divisions, and the average longitudinal value between each point will be used in place of x. The midpoint rule is defined as: The midpoint of each interval is equal to the variable x n, where N is the number of predefined intervals chosen: Because the midpoint rule is only needed for values of x where, the value of a is already known and equal to L 1. 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: 24

36 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 exact solution, 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, C 3, can be determined by applying the boundary condition provided by equation [82]: at, 25

37 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 of x, β(x), so that: Integrating the second term using the midpoint rule gives: 26

38 There have certainly been more elegant expressions derived in engineering. Writing the total solution for y-displacement when gives: The constant of integration, C 4, can be found by using equation [82] at : 27

39 The y-displacement solution for becomes: [83] Finally, the displacement solution for values of x where can be formulated using the same process as the other intervals. Because this x-location is within the final orientation and the moment of inertia is constant in this region, equation [50] is integrated twice in x, as was done to derive equations [82] and [83], only the contribution to the displacement from the varying orientation is given the limits of L 1 and L 2. 28

40 The constant of integration, C 5, can be determined by recalling the function derived for the beam s slope between, and inserting : 29

41 The slope of the beam for can then be written as: Notice that in this region only one term of the beam s slope is dependent on x. This simplifies things much more than before when deriving the beam s deflection in the varying orientation. That is, all integrals henceforth can be computed directly, and no more approximations are needed. To solve for the y-displacement, the equality above is integrated in x to give: 30

42 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 beam s slope,, and integrate the numerical value(s) over x, rather than carrying through to a closed-form solution. This method will be used to obtain values in section 1.5 and in Appendix A. 31

43 1.5 Specific Theoretical Solution Advancing from the generic case, the specifics of the problem are now defined and solved. Figure 9 shows the beam s 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 a repetition of Figure 2, but assigns values to the applied force, F, and lengths L 1, L 2, and L 3. Y Y Z w Z t w [Final Orientation] t [Initial Orientation] w = 1.0 t = 0.25 Y Figure 9 - Width and Thickness Parameters of the Solution 10 lbs O X Figure 10 - Load and Length Parameters of the Solution Let the material properties of the beam be those of mild steel (ms): 32

44 For comparison to numerical results, the stress, strain, and displacement values at five distinct values of x will be found. These five x-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 of the beam. In addition, at each x value the stress, strain, and vertical displacement values will be formulated for three values of y, 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 the beam s local coordinates, the moments and products of inertia are: For and the moment of inertia, I ZZ, is: (See equation [55]). Therefore, the stress distribution for and is: 33

45 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 is found using the slope equation that, when integrated in x, gives equation [83]. Because the value of x = 3.75 lies in the region of. For the following solution, let N = 5. 34

46 To extract a vertical displacement from the beam s slope at this point, integrated again in x with the appropriate limits: is simply The constant of integration, C 7, is determined by the boundary condition provided by the beam s vertical displacement in its initial orientation. At (from equation [82]), 35

47 The very outer-most beam fiber that lies on the y-axis is derived in section 1.4.2, and is given by the equality: Since at, the outer-most beam fiber on the y-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. 36

48 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. 37

49 2. Finite Element Analysis (FEA) Model 2.1 Methodology To 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 (preprocessing) 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 chosen because 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 the x-values chosen for examination of the theoretical solutions. See Table 1 for the values of x that 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 38

50 After the geometry was defined, the element was assigned a shell thickness (t = 0.25 ) and material properties (E = 30x10 6 psi, ν = 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 at x = 0 equal to zero (shown as orange cones in Figure 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 39

51 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 L 1 3 L L w 1.00 t 0.25 E 30 x 10 6 psi ν 0.3 F -10 lbs 40

52 2.2 Analytical Results Explicit 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 stresses cannot be directly extracted from ABAQUS in the traditional manner. As shown in Figure 14 below, the model s 2 and 3 axes (corresponding to the B and 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 Results Options Transformation). 41

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

54 NOTE: The following graphical representations extracted from ABAQUS give stress, strain, and displacement values in index notation. Therefore, Tensile Stress and Strain Values (x-direction) The midplane of the FEA model shows tensile stress values (σ 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. See Figure 15 for a view of the tensile stress results at the midplane. At the shell model s positive face, however, ABAQUS shows a severe increase in compressive stress as the beam approaches its final orientation. At the model s 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 and Figure 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. 43

55 Figure 15 - FEA Tensile Stress at the Shell s Midplane 44

56 Figure 16 - FEA Tensile Stress at the Shell's Positive Face Area of compressive stress peak Figure 17 - FEA Indicates an Additional Compressive Stress Peak in Twisted Region, Shell s Positive Face 45

57 Figure 18 - FEA Tensile Stress at the Shell's Negative Face Area of tensile stress peak Figure 19 - FEA Indicates an Additional Tensile Stress Peak in Twisted Region, Shell s Negative Face 46

58 2.2.2 Shear Stress and Strain Values (xy-direction) Figure 20 shows the shear stress distribution of the shell s 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 - FEA Shear Stress Distribution at the Shell's Midplane Figure 20 and Figure 21 also show that at the beam s midpoint, in the middle of its twist feature, 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. 47

59 Negative Shear Stress Concentration Figure 21 - FEA Shear Stress Concentration at Shell's Midplane The results of the positive and negative faces of the shell model indicate severe shear stress concentrations after the twist midpoint and before the beam s 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. 48

60 Shear Stress Concentrations Figure 22 - FEA Shear Stress Distribution on Shell s Positive Face 49

61 Shear Stress Concentrations Figure 23 - FEA Shear Stress Distribution on Shell s Negative Face 50

62 2.2.3 Vertical Displacement (y-direction) Figure 24 through Figure 26 below show the twisted cantilever beam s displacement distribution. Note that in Figure 26, there is a positive z displacement 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 51

63 Δz Figure 26 - FEA Displacement Results, yz-plane 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 = x 4. The displacements across the beam s 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 (x 4 ), the rotations about the x-axis disappear, corresponding to traditional theory. At 52

64 x = x 4, however, the outer edges of the beam have a higher residual stiffness than the centerline, and experience smaller displacements. See Figure 27. x 4 = 4.5 Figure 27 - FEA Indicates Residual Stiffness and Non-planar Cross-Section at x = x 4 53

65 3. Comparison of Beam Theory Formulation Results to FEA Results 3.1 Displacement Values The vertical displacement functions derived in section 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 beam s length. Figure 28 - Theory vs. FEA, Vertical Displacement 54

66 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 beam s 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 beam s internal mechanics. Beam theory does not leave room for out-ofplane 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 (see Figure 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 (near x = L 3 ), the more closely the displacement curve would follow that of the constant cross-section beam in strong-axis bending. 55

67 Figure 29 - Theoretical Displacement Results vs. Beams of Constant Cross-Sectional Orientation 56

68 3.2 Maximum Tensile Stress Values The theoretical values of tensile stress (σ xx ) match the numerical results very well outside of the beam s twisted region. Within the twisted region, however, the ABAQUS model generally predicts much higher values, and does not indicate maximum or minimums at exactly 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 57

69 3.3 Maximum Shear Stress Values Again, beyond the twisted region, the theoretical stress values closely match the numerical values. However, for the shear stress (τ xy ), the FEA results vary greatly from the theoretical results in the twisted region. Both methods allude to a maximum or minimum at the twist s midpoint (x = 3.75 ), but the theoretical results appear to have an inverted shape compared to those of the FEA shell model. See Figure 31. Figure 31 - Theory vs. FEA, Maximum Shear Stress 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 58

70 within the beam than calculated with traditional methods. Figure 31 shows the severity of error in using traditional beam theory to calculate shear stresses within the beam s 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 twist s 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]. 59

71 4. Summary and Conclusion Based 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 beam s 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 not recommended 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 (section 1.3.2), and then to the stress-strain equations (section 1.3.3) to approximate stress and strain values within the beam s twisted region. 60

72 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-ofplane 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 not yield exact results, but could be used to estimate stress and strain values within the beam s twisted region. 61

73 5. Appendix A Theoretical Results at Selected x Values x 1 x = 0, y = 0 Tensile Stress, σ xx (Equation [69]) Shear Stress, τ xy (Equation [71]) Longitudinal Strain, ε xx (Equation [75]) Shear Strain, ε xy (Equation [78]) Vertical Displacement, v (Equation [82]) 0 psi -60 psi 0 in/in x 10-6 in/in in x = 0, y = 0.5 Tensile Stress, σ xx (Equation [69]) Shear Stress, τ xy (Equation [71]) Longitudinal Strain, ε xx (Equation [75]) Shear Strain, ε xy (Equation [78]) Vertical Displacement, v (Equation [82]) 1,800 psi 0 psi x 10-5 in/in 0 in/in in x = 0, y = -0.5 Tensile Stress, σ xx (Equation [69]) Shear Stress, τ xy (Equation [71]) Longitudinal Strain, ε xx (Equation [75]) Shear Strain, ε xy (Equation [78]) Vertical Displacement, v (Equation [82]) -1,800 psi 0 psi x 10-5 in/in 0 in/in in 62

74 x 2 x = 3.00, y = 0 Tensile Stress, σ xx (Equation [69]) 0 psi Shear Stress, τ xy (Equation [71]) -60 psi Longitudinal Strain, ε xx (Equation [75]) 0 in/in Shear Strain, ε xy (Equation [78]) x 10-6 in/in Vertical Displacement, v (Equation [82]) x 10-4 in x = 3.00, y = 0.5 Tensile Stress, σ xx (Equation [69]) 1,080 psi Shear Stress, τ xy (Equation [71]) 0 psi Longitudinal Strain, ε xx (Equation [75]) x 10-5 in/in Shear Strain, ε xy (Equation [78]) 0 in/in Vertical Displacement, v (Equation [82]) x 10-4 in x = 3.00, y = -0.5 Tensile Stress, σ xx (Equation [69]) -1,080 psi Shear Stress, τ xy (Equation [71]) 0 psi Longitudinal Strain, ε xx (Equation [75]) x 10-5 in/in Shear Strain, ε xy (Equation [78]) 0 in/in Vertical Displacement, v (Equation [82]) x 10-4 in 63

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