Analysis of Composite Shafts for Torque Transmission in Automotive Applications by Jeffrey Michael Schurr 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: __/s/_D. Hufner 7/26/2013_ David Hufner, Project Adviser Rensselaer Polytechnic Institute Hartford, Connecticut July, 2013 (For Graduation August, 2013) i © Copyright 2013 by Jeffrey M. Schurr All Rights Reserved ii CONTENTS Analysis of Composite Shafts for Torque Transmission in Automotive Applications ...... i LIST OF TABLES ............................................................................................................. v LIST OF FIGURES .......................................................................................................... vi LIST OF SYMBOLS ....................................................................................................... vii ACKNOWLEDGMENT ................................................................................................ viii ABSTRACT ..................................................................................................................... ix 1. Introduction.................................................................................................................. 1 2. Methodology ................................................................................................................ 3 2.1 Governing Equations .......................................................................................... 3 2.1.1 Torsional Shear Stress Assumption ....................................................... 3 2.1.2 Calculation of the Laminate Stiffness Matrices ..................................... 3 2.1.3 Maximum Stress Failure Criterion ......................................................... 7 3. Problem Parameters ..................................................................................................... 8 3.1 Ply Orientations and Material Properties ........................................................... 8 3.2 Applied Loading............................................................................................... 10 4. Results........................................................................................................................ 11 4.1 Expected Results .............................................................................................. 11 4.2 Analytical Solution .......................................................................................... 11 4.3 Finite Element Solution ................................................................................... 15 4.3.1 Calculation of Principal Stresses Under Loading ................................ 16 5. Discussion .................................................................................................................. 18 5.1 Comparison of Analytical and FEA Results .................................................... 18 5.2 Future Improvements ....................................................................................... 18 6. Conclusions................................................................................................................ 20 6.1 Feasibility of using a Composite Shaft for Torque Transmission.................... 20 7. References.................................................................................................................. 21 iii 8. Appendix A: Matlab Scripts for Analytical Solution ................................................ 22 iv LIST OF TABLES Table 1. Analysis Parameters ............................................................................................ 8 Table 2. Principal Stresses with per material and ply orientation ................................... 14 Table 3. Maximum Allowable Torque at Failure ............................................................ 14 Table 5. Comparison of Analytical and FEA Results for Material 1 ............................. 18 v LIST OF FIGURES Figure 1. Examples of Shafts Used to Transmit Torque ................................................... 1 Figure 2. Example of a car drive shaft used to transmit torque ......................................... 2 Figure 3. Laminated Plate Geometry and Ply Numbering System [6] .............................. 4 Figure 4. Maximum Stress Failure Envelopes for Materials Examined ............................ 9 Figure 5. Finite Element Representation of Shaft ........................................................... 16 Figure 6. Finite Element Results with Material 1 ............................................................ 17 Figure 7. Ply Stack Plot of Material 1 ............................................................................. 17 Figure 6. Comparison of predicted and measured biaxial failure surface for unidirectional E-glass/epoxy laminae under combined normal stresses in directions parallel and perpendicular to the fibers [6] ...................................................................... 19 vi LIST OF SYMBOLS Aij Laminate extensional stiffnesses Bij Laminate-coupling stiffnesses Dij Laminate-bending matrix Ei Modulus of elasticity (Pa) FS Factor of safety G12 Shear modulus (GPa) k Ply number Mij Applied moment (N-m) Nxy Shear load per unit length (Pa-mm) Η¬ij Transformed lamina stiffness matrix R Mean radius of shaft (mm) Sij Laminate compliance Sij(bar) Transformed compliance matrix sL(-) Longitudinal compressive strength (MPa) sL(+) Longitudinal tensile strength (MPa) sT(-) Transverse compressive strength (MPa) sT(+) Transverse tensile strength (MPa) T [T] Torque (N-m) Transformation matrix t Wall thickness (m) z Ply thickness (m) ε0 Midplane strain κ Laminate curvature ν12 Poisson's ratio σ Stress (MPa) τxy Torsional shear stress (N/m2 or Pa) vii ACKNOWLEDGMENT I would like to thank my lovely wife for sticking with me through my RPI Master's experience. I could not have made it through the program without her support. viii ABSTRACT The capability of composite shafts to transmit a steady state torque was analyzed by calculating the failure stress utilizing the Maximum Stress failure criterion. Various materials were considered to determine which materials would be able to transmit the required torque assuming the same ply orientation, number of plies, and static loading. The Maximum Stress failure calculations were then compared to a finite element model (FEM) developed in Abaqus/CAE software. The FEM was used to calculate the maximum stresses in the shaft, and were compared against the failure stresses and the stresses calculated utilizing a mathematical model. ix 1. Introduction The purpose of this project is to assess the use of a composite cylindrical shaft for the purposes of transmission of torque, particularly in automotive applications. Shafts are used in many applications to transmit power or torque developed in one location, to another locations. Some examples of where shafts are used are in wind turbines, jet engines, and automobiles. In all of these examples, it is critical that the shaft does not fail while operating within their associated design envelope. The use of composite shafts is becoming more wide spread as the various applications aim to improve efficiency by reducing the amount of rotating mass, thereby reducing the rotating mass moment of inertia. Figure 1 shows various examples of shafts used to transmit power and torque [1] [2] [3]. Figure 1. Examples of Shafts Used to Transmit Torque For this project, the shaft analyzed will be based on the drive shaft of an E90 BMW M3. This shaft is required to transmit approximately 406 N-m (300 lb-ft) of torque under it's 1 highest loading conditions, based on the maximum output of the engine (assuming no increase in torque due to gearing in the transmission). The torque is assumed to be steady state and may be applied in either the positive or negative direction. A 2D mathematical model will be developed using the Maximum Stress failure criterion to determine whether 3 different shaft materials fail under a steady state torque applied by the engine. The objective is to determine whether a composite shaft can be used as an alternative to steel or aluminum drive shafts in this application. Figure 2 depicts a drive shaft from a 2008 BMW M3 [4] used to transmit power from the engine to the rear differential. The length is assumed to be 1,433mm from flange to flange [4]. For this analysis, it was assumed that the shaft is a continuous shaft without universal joints in the center or flanges at the end. A typical automotive drive shaft, made of metal, is approximately 3" in diameter (76.2mm). For this study, the composite drive shaft is assumed to have a wall thickness of 1mm, assuming four plies, nominally 0.25mm thick. Figure 2. Example of a car drive shaft used to transmit torque 2 2. Methodology 2.1 Governing Equations The following sections present the mathematical formulation of the approach used to determine the whether composite shafts will fail under a known loading, as well as the torque required to satisfy the Maximum Stress failure criterion. The equations discussed below represent the process used in solving for the allowable stresses in the composite shafts for use in the Maximum Stress failure criterion. 2.1.1 Torsional Shear Stress Assumption This project assumes a pure torsional load on a composite shaft. From Reference [5], we know that the torsional shear stress in a thin walled cylinder, τxy, is represented by Equation [1]. ππ₯π¦ = π 2ππ 2 π‘ [1] Where T is the applied torque, R is the mean radius, and t is the wall thickness. However, since this will be applied to a composite material analysis, the shear loading per unit length, Nxy, is required. This is calculated by rearranging Equation [1] to solve for Nxy in terms of applied torque, shown in Equation [2]. ππ₯π¦ = ππ₯π¦ π‘ = π 2ππ 2 [2] The shear loading per unit length, Nxy in Equation [2], was then used to perform the composite material failure analysis described in detail below. 2.1.2 Calculation of the Laminate Stiffness Matrices The shearloading, Nxy, calculated in Section 2.1.1 above can then be used in conjunction with the laminate extensional matrix [Aij], laminate-coupling stiffness matrix [Bij], and 3 laminate-bending matrix [Dij] to calculate the midplane strains (ε0) and the laminate curvatures (κ), as defined in Reference [6]. Equations [3], [4], and [5] show the equations used to calculate the [Aij], [Bij], and [Dij] matrices. π‘⁄ 2 π΄ππ = ∫ π Μ Μ Μ Μ Μ Μ Μ Μ (π ππ )π ππ§ = ∑(πππ )π (π§π − π§π−1 ) −π‘⁄ 2 π‘⁄ 2 π΅ππ = ∫ π=1 π 1 2 2 Μ Μ Μ Μ Μ Μ Μ Μ (π ππ )π π§ππ§ = ∑(πππ )π (π§π§ − π§π−1 ) 2 −π‘⁄ 2 π·ππ = ∫ π‘⁄ 2 [3] [4] π=1 π 1 3 2 3 Μ Μ Μ Μ Μ Μ Μ Μ π§ ππ§ = (π ) ∑(π ππ π ππ )π (π§π§ − π§π−1 ) 3 −π‘⁄ 2 [5] π=1 Where t is the overall thickness of the laminate, Η¬ij are the components of the transformed lamina stiffness matrix, k is the ply number, and z is the ply thickness. Figure 3 below depicts the ply numbering system used in this analysis, and is derived from Figure 7.9 of Reference [6]. zo Middle Surface z1 t zk-1 t/2 zk Figure 3. Laminated Plate Geometry and Ply Numbering System [6] Expanding the matrices calculated in Equations [3], [4], and [5] results in the stress resultant matrix, shown in Equation [6] below. 4 ππ₯ π΄11 ππ¦ π΄12 ππ₯π¦ π΄ = 16 ππ₯ π΅11 ππ¦ π΅12 {ππ₯π¦ } [π΅16 π΄12 π΄22 π΄26 π΅12 π΅22 π΅26 π΄16 π΄26 π΄66 π΅16 π΅26 π΅66 π΅11 π΅12 π΅16 π·11 π·12 π·16 π΅12 π΅22 π΅26 π·12 π·22 π·26 0 ππ₯ π΅16 ππ¦0 π΅26 0 π΅66 πΎπ₯π¦ π·16 π π₯ π·26 π π¦ π·66 ] {π π₯π¦ } [6] However, in this analysis, only pure shear loading is assumed, which reduces the contents of the matrix shown in Equation [6]. The laminate considered in this problem is symmetric which reduces the Bij matrix to zero, indicating that there is no crosscoupling of the stress within the laminate. Additionally, since there are no applied moments, Mij, the bending matrix, and therefore the curvatures, reduce to zero. 2.1.2.1 Components of the Transformed Lamina Stiffness Matrix, Η¬ij The components of the Η¬ij matrix, defined above, are related to the compliances, Sij, and the engineering constants for each material and ply orientation. The engineering constants for the various materials used in this analysis are presented in Table 1. The engineering constants are the modulus of elasticity for each ply direction (E1,2) and the Poisson's ratio (ν12,21). The engineering constants are then used in Equations [7] through [10] below. π11 = 1 πΈ1 1 πΈ2 π21 π12 =− =− πΈ2 πΈ1 π22 = π12 = π21 π66 = 5 1 πΊ12 [7] [8] [9] [10] and combine to for the [Sij] compliance matrix shown in Equation [11] below. π11 πππ = [π21 0 π12 π22 0 0 0] π66 [11] It is then necessary to transform the compliance matrix [Sij] into the various ply orientations utilizing the transformation matrix, [T], shown in Equation [12] in order to solve for the transformed lamina stiffness matrix [Η¬ij]. πππ 2 π [π] = [ π ππ2 π −πππ ππ πππ π ππ2 π πππ 2 π πππ ππ πππ 2πππ ππ πππ −2πππ ππ πππ ] πππ 2 π − π ππ2 π [12] where θ represents the orientation of the ply being analyzed. The Sij(bar) matrix, or transformed compliance matrix, is equal to the transpose of the transformation matrix multiplied by the compliance matrix and the transformation matrix. Sij(bar) is dependent on the ply orientation due to the transformation matrix associated with each ply. The equation used to calculate the Sij(bar) matrix is shown in Equation [13] below. Sij(bar)=[T]T*Sij*[T] [13] The transformed lamina stiffness matrix is then simply calculated by inverting the transformed compliance matrix, and is shown in Equation [14] below. Η¬ij = Sij-1 6 [14] 2.1.3 Maximum Stress Failure Criterion The Maximum Stress Criterion is an analysis technique used to assess the stress required to make a given composite lamina fail. The Maximum Stress Criterion for orthotropic laminae was first suggested in 1920 by Jenkins as an extension of the Maximum Normal Stress Theory (or Rankine's Theory) for isotropic materials [6]. The Maximum Stress Criterion is defined as one having no interactions between the stress components. This criteria compares the individual stress components with the corresponding material allowable strength values. The failure surface for the Maximum Stress Criteria is rectangular in stress space [7]. In particular, the Maximum Stress Criterion is effective at predicting failure in material with uniaxial stress in the principal material directions. The Maximum Stress criterion predicts failure when any principal material axis stress component exceeds the corresponding strength. The Maximum Stress failure theory, which embodies a very simple, but carefully structured, set of non-interactive criteria to identify failure mechanisms and to take appropriate post-initial failure action [8]. This is the primary case being assessed in this project. Equations [15], [16], and [17] show the inequalities which must be satisfied for the Maximum Stress criterion. (−) < π1 < π πΏ (−) < π2 < π π −π πΏ −π π (+) (+) |π12 | < π πΏπ [15] [16] [17] where the numerical values of sL(-) and sT(-) are assumed to be positive, and are defined by the material properties. It is assumed that shear failure along the principal material axes is independent of the sign of the shear stress, τ12. All of the stress shown in Equations [15], [16], and [17] are in the principal direction for the material being analyzed. The principal direction is based on fiber orientation within each ply of the laminate. 7 3. Problem Parameters 3.1 Ply Orientations and Material Properties This analysis assumed the use of three separate materials used to form a shaft for the transmission of torque (i.e., pure shear loading). The materials chosen for this analysis are AS4-3501-6 carbon/epoxy (Material 1), T300-976 carbon/epoxy (Material 2), and EGlass/epoxy (Material 3). The material properties for the three materials were obtained from References [9], [10], and [11] respectively. The key properties assumed for this analysis are compiled in Table 1 below. Table 1. Analysis Parameters Material Property Material 1 Material 2 Material 3 127 135 44.8 11.15 9.27 12.4 Shear Modulus (G12) (GPa) 6.56 6.15 5.52 Poisson's Ratio (ν12) 0.28 0.31 0.28 Young's Modulus in the fiber direction (E1) (GPa) Young's Modulus orthogonal to the fiber direction (E2) (GPa) Thickness per ply (mm) 0.25 SL(+) (MPa) 1950 1455 1035 SL(-)(MPa) 1480 1296 620 ST(+)(MPa) 48.0 39.0 48.3 ST(-)(MPa) 200 206.8 137.9 SLT(MPa) 79 76.5 68.9 It was also assumed that all shafts had the same ply orientations, number of plies, and thickness per ply. All materials were assumed to have a ply orientation of [+45/-45/45/+45]. The 45º angle was selected because of the fundamental assumption of the problem that the torque would be placed on the shaft as a pure torsional input. Therefore, by using the ply orientations selected, the shaft would have the first principal axis in the direction of loading. In any composite material, it is desirable to have stresses, after axis transformation, oriented in the same direction as the reinforcing 8 fibers.. This is because, in general, composites are weaker in the shear, or material transverse, direction. In this problem, the +/- 45º orientation ensures that the shaft can transmit both a positive or negative applied torque. If it was known that the torque would only be applied in one continuous direction, the plies could have all been aligned to one direction to ensure that the highest strengths would be in the loading, or principal direction. The Maximum Stress failure envelopes were plotted using the material properties shown in Table [1]. The failure envelopes are shown on the same figure, Figure [4], below to illustrate the difference between the various materials for their expected failure properties based on the information provided in Table [1] above. 100 Material 1 (AS4-3501-6) Material 2 (T300-976) Material 3 (E-Glass Epoxy) 50 0 σy MPa -50 -100 -150 -200 -250 -2,500 -2,000 -1,500 -1,000 -500 0 σx MPa 500 1,000 1,500 2,000 Figure 4. Maximum Stress Failure Envelopes for Materials Examined 9 2,500 3.2 Applied Loading It was assumed that the load applied to the shaft was a steady state torque. This is an important assumption in that it can be assumed that the shaft will never see a load greater than the steady state load. The assumed steady state load on the shaft was 406 N-m. 10 4. Results 4.1 Expected Results It is expected that the composite shafts will be capable of transmitting the required torque of 406 N-m. The limiting directions will most likely in the transverse tension direction because it is associated with the lowest strength. The shaft is expected to be successful in transmitting the applied load because the ply orientation of composite materials can be optimized for a particular stress orientation, such that the applied torque loading will result in normal stresses in the transformed ply. 4.2 Analytical Solution As discussed in Section 2 above, the Maximum Stress failure criterion was used to determine whether the three chosen composite materials could transmit the required torques. This was accomplished by calculating the required stress to satisfy the Maximum Stress failure criterion for each material assessed. In addition, the factor of safety was calculated for each material assessed based on the 406 N-m steady state torque load. For illustrative purposes, the following equations (Equations [18] through [25]) show the calculations using the properties from Material 1. To begin the analytical solution, Equations [1] and [2] were solved in terms of an unknown torque (T) to determine the largest torque each shaft could transmit without failure. This resulted Equations [18] and [19] below. ππ₯π¦ = ππ₯π¦ = π 2ππ 2 π = 111.1 π − π = 1.11π₯10−4 π πΊππ − ππ 2 2π0.03785π 11 [18] [19] Additionally, as discussed above, it is assumed that there is only a pure torsional loading placed on the shaft. Therefore, Equation [20] is valid for this analysis. Nx=Ny=Mx=My=Mxy=0 [20] Now, the strains are solved for in the x-y coordinate system. The strains are solved for utilizing Equation [6], taking advantage of the fact that the laminate is symmetric ([Bij] = 0) and that there is only a pure torsional load applied ([Mij]=0). Equation [21] shows the reduced version of Equation [6] with the assumptions mentioned above. ππ₯π 0 0 ′ { ππ¦ } = [π΄ ] { 0 } π ππ₯π¦ πΎπ₯π¦ 0.4501 −0.3125 = [−0.3125 0.4501 0 0 0 ={ } 0 −15 3.345π₯10 π 0 0 −10 { } 0 0 ] π₯10 1.11π₯10−4 π 0.3011 [21] It is now possible to utilize the transformed lamina stiffness matrices, calculated using the process detailed in Section 2.1.2.1, for both the +45º and -45º plies. The transformed lamina stiffness matrices can be multiplied by the strains calculated in Equation [21] to determine the stresses in each ply, in the x-y coordinate system. An example of this calculation is shown in Equation [22] for the + 45º plies. ππ₯π ππ₯ π { ππ¦ } = [Η¬]+45º { ππ¦ } π ππ₯π¦ +45º πΎπ₯π¦ 42.89 29.78 29.16 0 9 = [29.78 42.89 29.16] π₯10 { } 0 29.16 29.16 33.21 3.345π₯10−12 ππ₯ 0.0975π { ππ¦ } = {0.0975π } πππ ππ₯π¦ +45º 0.1111π 12 [22] The stresses in the -45º plies are calculated using the same equations as the +45º plies shown above. However, for the -45º plies, the Η¬ matrix associated with the -45º plies is used. The resulting stresses in the x-y coordinates for the -45º plies are shown in Equation [23] below. ππ₯ −0.0975π { ππ¦ } = {−0.0975π } πππ ππ₯π¦ −45º 0.1111π [23] Now, in order to apply the Maximum Stress failure criterion, the stresses in each ply must be transformed into the principal axes of the materials. This is done by multiplying the transformation matrix [T], for each ply direction, by the stresses calculated in Equations [22] and [23] above. Equations [24] and [25] show this for both the +45º plies and -45º plies respectively. ππ₯ π1 0.5 0.5 1.0 0.0975π { π2 } = [π]+45º { ππ¦ } = [ 0.5 0.5 −1.0] {0.0975π} ππ₯π¦ π12 +45º −0.5 0.5 0 0.1111π 0.2086π = {−0.0135π} πππ 0π ππ₯ π1 −0.2086π π π { 2} = [π]−45º { π¦ } = { 0.0135π } MPa ππ₯π¦ π12 −45º 0π [24] [25] At this point, the applied steady state loading of 406 N-m can be applied to determine what the stresses are in the principal direction. Table 2 below shows the calculated principal stresses relative to their respective failure criteria as defined by Equations [15], [16], and [17]. 13 Table 2. Principal Stresses with per material and ply orientation -SL(-) -1480 -1480 -1296 -1296 -620 -620 Material Angle +45 AS4-3501-6 carbon/epoxy -45 +45 T300-976 carbon/epoxy -45 +45 EGlass/epoxy -45 < σ1 < 84.7 -84.7 86.0 -86.0 74.2 -74.2 SL(+) 1950 1950 1455 1455 1035 1035 < σ2 < -5.5 5.5 -4.2 4.2 -16.0 16.0 -ST(-) -200 -200 -206.8 -206.8 -137.9 -137.9 ST(+) 48.0 48.0 39.0 39.0 48.3 48.3 * All units in MPa In addition to calculating the principal stresses to determine whether the materials fail under the applied load, the Maximum Stress failure criterion can also be applied to the results of Equations [24] and [25], utilizing the inequalities in Equations [15], [16], and [17] to determine the torque required to make the ply fail. The maximum torque loading that each ply and orientation can take before failure is shown in Table 3 below. Allowable Torque Maximum Table 3. Maximum Allowable Torque at Failure AS4-3501-6 Carbon Epoxy +45º -45º T300-976 Carbon/Epoxy +45º -45º E-Glass/Epoxy +45º -45º σ1 9,346.2 7,093.5 6,865.6 6,115.3 5,664.6 3,393.3 σ2 14,765.0 3,543.6 20,157.0 3,801.4 3,493.7 1,223.7 τ12 Inf Inf Inf Inf Inf Inf * All units in N-m As the Maximum Stress failure criterion states, the limiting torque for the shaft is defined by the minimum allowable stress in any ply and principal orientation. For all of the materials analyzed, the allowable shear stress was infinite. This is because shear stress along the 1,2 direction is zero, meaning that the plies will not fail due to shear loading. 14 The failure of Material 1 will be due to transverse tensile failure of the -45º ply. The maximum allowable transverse tension allowed by Material 1 was 3,543 N-m. This is still significantly more than the design load of 406 N-m expected during normal operation. This results in a Factor of Safety (FS), calculated by Equation [26], of approximately 8.7. πΉπ = πππ₯ π΄ππππ€ππππ πππππ’π π·ππ πππ πππππ’π [26] The failure of Material 2 will, again, be due to transverse tension in the -45ºply. The maximum allowable transverse tension allowed by Material 2 was 3,8015 N-m. This too is more than the design load of 406 N-m, resulting in a factor of safety of approximately 9.3. The failure of Material 3 will, again, be due to transverse tension in the -45ºply. The maximum allowable transverse tension allowed by Material 3 was 1223 N-m. This too is more than the design load of 406 N-m, resulting in a factor of safety of approximately 3. 4.3 Finite Element Solution A finite element model (FEM) was created as an independent assessment of the results obtained utilizing the analytical model. Material 1 was used to define the material properties of the FEM. The FEM was created using utilizing Abaqus/CAE software. Figure 5 shows a graphic representation of the mesh used on the FEA model developed in Abaqus. 15 Figure 5. Finite Element Representation of Shaft The shaft was assumed to be one continuous length. All shaft and material properties were assigned to the modeled shaft as defined in Table 1 based on the material properties of Material 1. The shaft was modeled as a continuous shell, with 4 plies of composite material, in the orientation of [+45/-45]s. The FEA model utilized standard S4R linear elements in a quad mesh. 4.3.1 Calculation of Principal Stresses Under Loading The principal stresses were calculated for each of the 4 plies utilizing the FEA model for a 406 N-m applied load. The load was applied to a reference node at the center of the shaft, which was connected to the edge nodes by a beam constraints. The end without the load applied had a boundary condition restraining translations and rotations in the three axes, again utilizing a reference node and beam constraints. The maximum principal stress was output from the resulting model. The maximum calculated was 83.23 MPa. An image of the FEA model with the results is shown in Figure [6]. 16 Figure 6. Finite Element Results with Material 1 A representation of the plies and the fiber orientation in the FEA model was also extracted to help visualize the composite layup. This ply stack plot is shown in Figure [7] below. Figure 7. Ply Stack Plot of Material 1 17 5. Discussion 5.1 Comparison of Analytical and FEA Results The maximum principal stress and minimum principal stresses were compared between the analytical approach and the FEA model results. The calculated stress at the applied load in each ply of Material 1 is shown in Table 4 below, as well as the percent difference between the analytical calculation and the FEM. Table 4. Comparison of Analytical and FEA Results for Material 1 Stresses (magnitude) Analytical Solution FEA (Outer Layer) Percent Difference σ1 (S11) σ2 (S22) 84.7 MPa 5.5 MPa 83.2 MPa 5.4 MPa 1.8% 1.8% As shown in Table 5, the calculated stresses are similar between the two analyses. Since the two methods used produce a similar result, it can be assumed that the result is accurate for either type of solution methodology, and that the analytical approach does a sufficient job of capturing the mechanics of the system, which allows for solutions in a relatively small amount of computation time as opposed to developing a finite element model. Additionally, it can be assumed that the analytical solutions obtained for Materials 2 and 3 are accurate by extension, since they utilize the same MATLAB code used to develop the solution for Material 1, which is verified by the FEA analysis. 5.2 Future Improvements Future improvements could be to utilize the results of test data for each of the material type to improve the accuracy of the failure criterion used. An example of test data compared to various analytical failure criterion is shown in Figure [6]. Specifically, Figure 8 depicts a comparison of a predicted and measured biaxial failure surface on a E- 18 Glass/epoxy laminate under combined normal stresses in directions parallel (σx) and perpendicular (σy) to the fibers [6]. Figure 8. Comparison of predicted and measured biaxial failure surface for unidirectional E-glass/epoxy laminae under combined normal stresses in directions parallel and perpendicular to the fibers [6] Additionally, future design iterations could consider techniques to optimize global shaft properties. An example of a design modifications could be adding stiffeners to the shaft to better optimize the design. 19 6. Conclusions 6.1 Feasibility of using a Composite Shaft for Torque Transmission Based on the analyses performed in this project, a composite shaft could be a suitable replacement for a typical metal shaft used in automobiles. This conclusion is based on the stresses present in the analyzed shaft assuming a stead state applied loading of 406 N-m relative to the stresses required for failure according to the Maximum Stress failure criterion. This was verified by calculating the principal stresses in shafts assuming 3 different materials and comparing them to the Maximum Stress failure criterion. Factors of safety were also calculated for each material based on the minimum torque required to satisfy one of the Maximum Stress failure criterion, and the design load of 406 N-m. For assessing a shaft with the ply orientation and loading presented in this project, it can be concluded that utilizing the Maximum Stress failure criterion is relatively good approximation in determining the failure of a shaft, due to the applied load. This is because the transverse stress is only a small percentage of the overall longitudinal stress, which therefore has a minimal impact when utilizing the Maximum Stress failure criterion. This conclusion is supported by the plotted Maximum Stress failure envelop as well as the test data for a bi-axial loading shown in Figure [6]. However, for more complex material layups and applied loading, it may be required to utilize additional failure criterion and test data to accurately predict the failure of the material due to cross coupling of the stresses. 20 7. References [1] "Wind Turbines." - Kinetic Wind Energy Generator Technology. N.p., n.d. Web. 25 Oct. 2012. <http://www.alternative-energy-news.info/technology/windpower/wind-turbines/>. [2] "Innovative Lightweight Construction." BMW M3 Sedan: Innovative Lightweight Construction. N.p., n.d. Web. 25 Sept. 2012. <http://www.bmw.co.za/products/automobiles/m/m3sedan/m3sedan_lightweight. asp>. [3] Wikipedia. Wikimedia Foundation, n.d. Web. 25 October 2012 <http://en.wikipedia.org/w/index.php?title=File:Turbofan_operation_lbp.svg>. [4] "RealOEM.com." RealOEM.com. N.p., n.d. Web. 25 Sept. 2012. <http://www.realoem.com/bmw/showparts.do?model=VA93>. [5] Shames, Irving Herman, and Francis A. Cozzarelli. Elastic and Inelastic Stress Analysis. Washington, DC: Taylor and Francis, 1997. Print. [6] Gibson, Ronald F. Principles of Composite Material Mechanics. Boca Raton, FL: CRC, 2007. Print. [7] Beer, Ferdinand P., E. Russell Johnston, and John T. DeWolf. Mechanics of Materials. Boston: McGraw-Hill Higher Education, 2006. Print. [8] NASA. Langley Research Center. Progressive Failure Analysis Methodology For Laminated Composite Structures, NASA/TP-1999-209107, August 6, 1999. By David W. Sleight. S.l.: S.n., 1999. Print. [9] Soden, P.D., Hinton, M.J. and Kaddour, A.S., “Lamina Properties, Lay-up Configurations and Loading Conditions for a Range of Fibre-Reinforced Composite Laminates.” Composites Science and Technology, Vol. 58, 1011-1022, 1998. [10] MIL-HDBK-17-2F, “Department of Defense Handbook: Composite Materials Handbook.” U.S. Army Research Laboratory, Materials Directorate, Aberdeen Proving Ground, MD. [11] Zwben, C., “Static Strength & Elastic Properties”, Mechanical Behavior & Properties of Composite Materials, Delaware Composites Design Encyclopedia, Vol. 1, Technomic Publishing, Lancaster PA, pp. 66-69, (1989). 21 8. Appendix A: Matlab Scripts for Analytical Solution % This script is used to calculate the largest torque (T) that can be % transmitted by a composite power transmission shaft without failure % according to the Maximum Stress Criterion. Additionally, it calculates % the stresses in the principal material directions based on a known % applied load. % Author: Jeffrey M. Schurr % Revision: (-) %dated: 04/17/2013 clear; close all; clc % %% Laminate Definition % % Laminate #1: AS4/3501-6 carbon epoxy [(45/-45)]s theta(1)= 45; E1(1)=127e9; E2(1)=11.15e9; G12(1)=6.557e9; v12(1)=0.2786; t(1)=0.25; theta(2)=-45; E1(2)=127e9; E2(2)=11.15e9; G12(2)=6.557e9; v12(2)=0.2786; t(2)=0.25; theta(3)=-45; E1(3)=127e9; E2(3)=11.15e9; G12(3)=6.557e9; v12(3)=0.2786; t(3)=0.25; theta(4)= 45; E1(4)=127e9; E2(4)=11.15e9; G12(4)=6.557e9; v12(4)=0.2786; t(4)=0.25; SLp=1950; SLn=-1480; STp=48.0; STn=-200; SLT=79; % % Laminate #2: T300-976 carbon/epoxy % theta(1)= 45; E1(1)=135e9; E2(1)=9.27e9; G12(1)=6.15e9; v12(1)=0.31; t(1)=0.25; % theta(2)=-45; E1(2)=135e9; E2(2)=9.27e9; G12(2)=6.15e9; v12(2)=0.31; t(2)=0.25; % theta(3)=-45; E1(3)=135e9; E2(3)=9.27e9; G12(3)=6.15e9; v12(3)=0.31; t(3)=0.25; % theta(4)= 45; E1(4)=135e9; E2(4)=9.27e9; G12(4)=6.15e9; v12(4)=0.31; t(4)=0.25; % SLp=1455; SLn=-1296; STp=39; STn=-206.8; SLT=76.5; % % Laminate #3: E-Glass/epoxy % theta(1)= 45; E1(1)=44.8e9; E2(1)=12.4e9; G12(1)=5.52e9; v12(1)=0.28; t(1)=0.25; % theta(2)=-45; E1(2)=44.8e9; E2(2)=12.4e9; G12(2)=5.52e9; v12(2)=0.28; t(2)=0.25; % theta(3)=-45; E1(3)=44.8e9; E2(3)=12.4e9; G12(3)=5.52e9; v12(3)=0.28; t(3)=0.25; % theta(4)= 45; E1(4)=44.8e9; E2(4)=12.4e9; G12(4)=5.52e9; v12(4)=0.28; t(4)=0.25; % SLp=1035; SLn=-620; STp=48.3; STn=-137.9; SLT=68.9; % % Define cell arrays to store the transformed stiffness matrices of each % individual ply 22 % Qbar=cell(length(theta),1); % % Loop through all plies, calculate Qbar for each ply % v21=zeros(1,4); T=cell(length(theta),1); for i=1:length(theta) v21(i)=v12(i)*(E2(i)/E1(i)); S=[1/E1(i) -v21(i)/E2(i) 0; -v12(i)/E1(i) 1/E2(i) 0; 0 0 1/G12(i)]; c=cosd(theta(i)); s=sind(theta(i)); T{i,1}=[c^2 s^2 2*c*s; s^2 c^2 -2*c*s; -c*s c*s (c^2 - s^2)]; Sbar=T{i,1}'*S*T{i,1}; Qbar{i,1}=inv(Sbar); end % % Calulate the z distances for each ply interface % z=zeros(1,5); total_thick=sum(t); z(1)=-total_thick/2; for k=2:length(theta)+1 z(k)=z(k-1)+t(k-1); end % % Calculate the A, B, and D matrices % A = [0 0 0; 0 0 0; 0 0 0]; B = [0 0 0; 0 0 0; 0 0 0]; D = [0 0 0; 0 0 0; 0 0 0]; for i = 1:length(theta) k = i+1; Aply = Qbar{i,1}*(z(k)-z(k-1)); A = A+Aply; Bply = (1/2)*Qbar{i,1}*(z(k)^2-z(k-1)^2); B = B+Bply; Dply = (1/3)*Qbar{i,1}*(z(k)^3-z(k-1)^3); D = D+Dply; end %====================================================================== ==== % R = 0.03785; % mean wall radius in m (76.2mm shaft diameter) Nxy = 1/(2*pi*R^2); % N/m Nxy = Nxy*10^-6; % GPa-mm % % Assume Nx=Ny=Mx=My=Mxy=0 (pure torsional loading) % strain = A\[0; 0; Nxy]; % % For in the x-y coordinate system stress.Pos = Qbar{1,1}*strain*1e3; %Now in MPa stress.Neg = Qbar{2,1}*strain*1e3; %Now in MPa % % Principal stresses stress.principalPos = T{1,1}*stress.Pos; %Results in Mpa 23 stress.principalNeg = T{2,1}*stress.Neg; %Results in Mpa % % Torque for the +45 plys Torque1.Pos.Nm(1,1) = SLp/stress.principalPos(1,1); %Result in N-m Torque1.Pos.Nm(2,1) = STn/stress.principalPos(2,1); %Result in N-m Torque1.Pos.Nm(3,1) = SLT/stress.principalPos(3,1); %Result in N-m % Torque for the -45 plys Torque1.Neg.Nm(1,1) = SLn/stress.principalNeg(1,1); %Result in N-m Torque1.Neg.Nm(2,1) = STp/stress.principalNeg(2,1); %Result in N-m Torque1.Neg.Nm(3,1) = SLT/stress.principalNeg(3,1); %Result in N-m % Torque1.Pos.ftlb = Torque1.Pos.Nm*0.737; % Converting into ft-lbs from N-m Torque1.Neg.ftlb = Torque1.Neg.Nm*0.737; % Converting into ft-lbs from N-m % % Maximum torque before failure in the laminate Smallest.ftlb = min(Torque1.Pos.ftlb,Torque1.Neg.ftlb); Smallest.Nm = min(Torque1.Pos.Nm,Torque1.Neg.Nm); trueMin.ftlb = min(Smallest.ftlb); trueMin.Nm = min(Smallest.Nm); % % Factor of Safety re: 406 N-m (300 ft-lb) requirement FS.Nm=trueMin.Nm/406; % N-m FS.ftlb=trueMin.ftlb/300; % ft-lbs % % Saves to a .mat file all of the max Torques for use in MS Excel % save('Failure Torque','-append','Torque1'); %% Principal Stresses at load % Applied Loading appLoad = 406; %N-m % Stresses in +/- 45º plies due to load loadStress.plus = stress.principalPos*appLoad; loadStress.neg = stress.principalNeg*appLoad; 24