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)
Ǭij
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].
𝜏𝑥𝑦 =
𝑇
2𝜋𝑅2 𝑡
[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].
𝑁𝑥𝑦 = 𝜏𝑥𝑦 𝑡 =
𝑇
2𝜋𝑅2
[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.
𝑡⁄
2
𝐴𝑖𝑗 = ∫
𝑁
̅̅̅̅
̅̅̅̅
(𝑄
𝑖𝑗 )𝑘 𝑑𝑧 = ∑(𝑄𝑖𝑗 )𝑘 (𝑧𝑘 − 𝑧𝑘−1 )
−𝑡⁄
2
𝑡⁄
2
𝐵𝑖𝑗 = ∫
𝑘=1
𝑁
1
2
2
̅̅̅̅
̅̅̅̅
(𝑄
𝑖𝑗 )𝑘 𝑧𝑑𝑧 = ∑(𝑄𝑖𝑗 )𝑘 (𝑧𝑧 − 𝑧𝑘−1 )
2
−𝑡⁄
2
𝐷𝑖𝑗 = ∫
𝑡⁄
2
[3]
[4]
𝑘=1
𝑁
1
3
2
3
̅̅̅̅
̅̅̅̅
𝑧
𝑑𝑧
=
(𝑄
)
∑(𝑄
𝑖𝑗 𝑘
𝑖𝑗 )𝑘 (𝑧𝑧 − 𝑧𝑘−1 )
3
−𝑡⁄
2
[5]
𝑘=1
Where t is the overall thickness of the laminate, Ǭij 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
𝑁𝑥
𝐴11
𝑁𝑦
𝐴12
𝑁𝑥𝑦
𝐴
= 16
𝑀𝑥
𝐵11
𝑀𝑦
𝐵12
{𝑀𝑥𝑦 } [𝐵16
𝐴12
𝐴22
𝐴26
𝐵12
𝐵22
𝐵26
𝐴16
𝐴26
𝐴66
𝐵16
𝐵26
𝐵66
𝐵11
𝐵12
𝐵16
𝐷11
𝐷12
𝐷16
𝐵12
𝐵22
𝐵26
𝐷12
𝐷22
𝐷26
0
𝜀𝑥
𝐵16
𝜀𝑦0
𝐵26
0
𝐵66 𝛾𝑥𝑦
𝐷16 𝜅𝑥
𝐷26 𝜅𝑦
𝐷66 ] {𝜅𝑥𝑦 }
[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, Ǭij
The components of the Ǭij 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.
𝑆11 =
1
𝐸1
1
𝐸2
𝜐21
𝜐12
=−
=−
𝐸2
𝐸1
𝑆22 =
𝑆12 = 𝑆21
𝑆66 =
5
1
𝐺12
[7]
[8]
[9]
[10]
and combine to for the [Sij] compliance matrix shown in Equation [11] below.
𝑆11
𝑆𝑖𝑗 = [𝑆21
0
𝑆12
𝑆22
0
0
0]
𝑆66
[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 [Ǭij].
𝑐𝑜𝑠 2 𝜃
[𝑇] = [ 𝑠𝑖𝑛2 𝜃
−𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
𝑠𝑖𝑛2 𝜃
𝑐𝑜𝑠 2 𝜃
𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
2𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
−2𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 ]
𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛2 𝜃
[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.
Ǭij = 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.
(−)
< 𝜎1 < 𝑠𝐿
(−)
< 𝜎2 < 𝑠𝑇
−𝑠𝐿
−𝑠𝑇
(+)
(+)
|𝜏12 | < 𝑠𝐿𝑇
[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.
𝑁𝑥𝑦 =
𝑁𝑥𝑦 =
𝑇
2𝜋𝑅2
𝑇
= 111.1 𝑁 − 𝑚 = 1.11𝑥10−4 𝑇 𝐺𝑃𝑎 − 𝑚𝑚
2
2𝜋0.03785𝑚
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.
𝜀𝑥𝑜
0
0
′
{ 𝜀𝑦 } = [𝐴 ] { 0 }
𝑜
𝑁𝑥𝑦
𝛾𝑥𝑦
0.4501 −0.3125
= [−0.3125 0.4501
0
0
0
={
}
0
−15
3.345𝑥10 𝑇
0
0
−10
{
}
0
0 ] 𝑥10
1.11𝑥10−4 𝑇
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.
𝜀𝑥𝑜
𝜎𝑥
𝑜
{ 𝜎𝑦 }
= [Ǭ]+45º { 𝜀𝑦 }
𝑜
𝜏𝑥𝑦 +45º
𝛾𝑥𝑦
42.89 29.78 29.16
0
9
= [29.78 42.89 29.16] 𝑥10 {
}
0
29.16 29.16 33.21
3.345𝑥10−12
𝜎𝑥
0.0975𝑇
{ 𝜎𝑦 }
= {0.0975𝑇 } 𝑀𝑃𝑎
𝜏𝑥𝑦 +45º
0.1111𝑇
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 Ǭ 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.
𝜎𝑥
−0.0975𝑇
{ 𝜎𝑦 }
= {−0.0975𝑇 } 𝑀𝑃𝑎
𝜏𝑥𝑦 −45º
0.1111𝑇
[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.
𝜎𝑥
𝜎1
0.5 0.5 1.0 0.0975𝑇
{ 𝜎2 }
= [𝑇]+45º { 𝜎𝑦 } = [ 0.5 0.5 −1.0] {0.0975𝑇}
𝜏𝑥𝑦
𝜏12 +45º
−0.5 0.5
0
0.1111𝑇
0.2086𝑇
= {−0.0135𝑇} 𝑀𝑃𝑎
0𝑇
𝜎𝑥
𝜎1
−0.2086𝑇
𝜎
𝜎
{ 2}
= [𝑇]−45º { 𝑦 } = { 0.0135𝑇 } MPa
𝜏𝑥𝑦
𝜏12 −45º
0𝑇
[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.
𝐹𝑆 =
𝑀𝑎𝑥 𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑇𝑜𝑟𝑞𝑢𝑒
𝐷𝑒𝑠𝑖𝑔𝑛 𝑇𝑜𝑟𝑞𝑢𝑒
[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