Torque Arm Shape Optimization

Torque Arm Shape OptimizationJay Malaney Graduate Student Department of Mechanical and Aerospace Engineering University of Florida Gainesville, Florida, USA ABSTRACT which can lead to pronounced wheel hop in hard stops. [2] Since the quality of design for automotive With the automotive industry striving towards components is measured by its ability to perform the weight reduction of vehicular components to given function with the least weight, reducing improve fuel efficiency, weight optimization must be machining and material costs. [1] Therefore, we carried out for these components. In order to reduce analyze a Torque Arm model which is a suspension weight without sacrificing its integrity, FEM link used to control wheel motion in the longitudinal Analysis must be carried out. [1] direction. [2] The objective is to optimize the shape of The initial design of torque arm is shown in Fig. 1. the Torque Arm to reduce the weight but on the other The forces acting on the circular cross section on the hand also ensure that the induced maximum stress in right-hand side due to the rotation of the shaft are the component is below the yield stress of 340 MPa converted to an approximate upward force for the material used. The model is optimized for 3 FY=8000N and the horizontal force FX=-4000N variable length constraints between their upper and acting at the center of the circle. lower bound values. Maximum values of stress are studied for the upper, center and lower bound values APPROACH which provides us with 3x3x3=27 data points or I. PRELIMINARY ANALYSIS constraints which are then used to find the optimal design using regression and fmincon function on MATLAB. The optimum weight reduction solution was found which reduced the weight from 2.31 kg in the initial design to 2.08 kg inducing a stress of 334 MPa. INTRODUCTION Fig. 1: Torque Arm Model A torque arm is typically mounted ahead of the wheel. In that position they resist dive Before designing the FEM Model, a preliminary under braking forces and wheel hop analysis is carried out to calculate the approximate under acceleration. On a vehicle with leaf springs, maximum stress in the component and its location such as trucks, the springs themselves provide some using the initial design values of x1=12 cm, x2=1cm longitudinal wheel control. On some leaf spring- and x3=27cm. equipped vehicles, the springs are mounted so that a Assumptions: lesser portion of the spring's length is forward of the 1. The Torque Arm is assumed to be a beam that wheels, improving wheel control on acceleration. A is clamped at the left end with a length of 42cm. side effect of such positioning is that the longer, aft 2. A horizontal force FX = -4000N = -4x105 kg- portion of the springs may not be stiff enough to cm/s2 and a vertical force of FY=8000N = 8x105 control wheel movement under braking forces, kg-cm/s2 is applied at the right end.  3. The cross section of the beam is modelled as shown below with a thickness of 1.0 cm. 4. The 3-cm fillet is neglected in the preliminary analysis. 5. The circles are modelled to be as triangles as shown in the figure below. Fig. 3: Fully Constrained Geometry using Initial Values To ensure the geometry is fully constrained the tangent, symmetry, fixed, horizontal and vertical constrains were used. The fully constrained Fig. 2: Model for Preliminary Analysis geometry allows you to update the sketch without The Bending Stress is given by the following distorting the geometry and the shape of the model. equation, The figure below shows the fully constrained geometry for the test values of x1=8cm, x2=3cm and 𝐹(𝐿 − 𝑥)𝑦 x3=30cm. The figure highlights the simplicity of 𝜎𝑥𝑥 = − (1) 𝐼 updating the model once the geometry has been fully Also, a compressive force of -4000N is applied due constrained. Section VI explains further reasonings to which the stress induced is given by, to opt for a fully constrained geometry. 𝐹 (2) 𝜎𝑥𝑥 = − 𝐴 The bending stress would be maximum at either the top or bottom surface. However, since a compressive force is applied the stress at the top surface would be more negative and the stress and the bottom surface would be less positive since it would be under tension. Hence, the stresses are calculated on the top surface Fig. 4: Fully Constrained Geometry using Test Values which would have the most stress using the above equations. Moment of Inertia, I at any point is given III. MATERIAL PROPERTIES, BOUNDARY by the formula below, CONDITIONS & LOADING CONDITIONS 𝑏ℎ3 𝑏ℎ3 Once the geometry is created, the material properties 𝐼=( )@X − ( ) Hole (3) 12 12 are added to Abaqus. The table below shows the where, b is the thickness of the torque arm. loads applied as well as the material properties. II. GEOMETRY Quantity Given Units Desired Units Using Abaqus a fully constrained geometry is Young’s 206.8 GPa = 206.8x109 206.8x107 modelled such that when the values of x1, x2 and x3 Modulus, E kgm/s2.m2 kg.cm/s2.cm2 can be changed for different mesh trials keeping the Poison’s 0.29 0.29 Ratio,v model symmetric over the horizontal and also, Plane stress 1.0 cm 1.0 cm prevent the geometry being distorted after meshing thickness, t using CST, LST, Q4 or Q8 elements. Density 7850 kg/m3 7850x106 kg/cm3 Shown below is a fully constrained geometry for the Allowable 340 MPa = 340x106 340x104 initial design values of x1=12 cm, x2=1cm and Stress kg.m/s2.m2 kg.cm/s2.cm2 x3=27cm.  FX -4000N = -4000 kg.m/s2 -4x105 partition is done in a manner to create a 4-edge kg.cm/s2 element so that a refined mesh is obtained. FY 8000N = 8000 kg.m/s2 8x105 kg.cm/s2 Table 1: Material Properties and Applied Loads (Unit Conversion to cm) After inputting the data on Abaqus, Reference points are created at the center point of the circles on the left and right side to apply the boundary conditions and loads. MPC constraint is then used which allows us to transfer these conditions from the center point of Fig. 7: Partition the circle to the entire circle. The reference point is called the master node and the circle is called the A Q4 structured mesh was used to arrive at the slave nodes as the mimic the conditions applied at the following result. The figure below shows the mesh reference point i.e. master node. The figure shows obtained. For quadrilateral elements, the element below depicts these conditions. performs best when the shape is a rectangle because the Jacobian matrix is diagonal and constant. When an element is distorted too much, numerical integration becomes inaccurate and the Jacobian is close to zero. Although we cannot make all elements rectangular, the angle between adjacent edges should be close to 90 degrees. Also, a transition from small elements to large elements gives a poor mesh. Keeping these points in mind, a good quality mesh Fig. 5: Reference Points along with MPC Constraint can be produced which is always a recipe for success in finite element analysis. [10] The figure below Reference Point 1 has all its displacement and shows that the mesh quality is good based on the rotations fixed i.e. U1=U2=U3=UR1=UR2=UR3=0. above discussion, help convergence to the exact Reference Point 2 has an upward force FY and the solution. We can see that most of the elements are horizontal force FX, their magnitudes are taken from square with an internal angle of 90o. This is due to Table 1. Figure below shows the loads and boundary the partition that was used previously. conditions applied. Fig 8: Q4 Structured Mesh Fig. 6: Boundary Conditions and Loads The figure below shows the S, Mises for the applied IV. PARTITION, MESH & INITIAL loading for the initial values. The figure also shows DESIGN SIMULATION the location of the maximum stress. A trial and error method were adopted to then arrive at the partitioning for the mesh. The partition adopted is shown in the figure below. Partitioning is essential for complex geometries. A free mesh over complex geometries would give us a distorted element shape which would give us inaccurate results. Since meshing would be done using a Q4 element type (Reason for which is mentioned in Section V). The Fig. 9: Initial Design Stress  V.CONVERGENCE STUDY 𝑓 = 𝑎1 + 𝑎2 𝑥1 + 𝑎3𝑥2 + 𝑎4𝑥3 + 𝑎5𝑥12 For the convergence study the element types + 𝑎6𝑥22 + 𝑎7𝑥32 (4) considered for selection were CST (Constant Strain + 𝑎8𝑥1𝑥2 + 𝑎9𝑥2𝑥3 Triangular), LST (Linear Strain Triangular), Q4 + 𝑎10𝑥1𝑥3 (Quadrilateral Element with 4 nodes) or Q8 The approximate function above has 10 unknown (Quadrilateral Element with 8 nodes). As mentioned coefficients which need to be found using regression in the earlier section the Q4 element type was used (or least squares). for the convergence study. The following reasoning In order to do that, we need to have test functions was adopted for the selection: which will allow us to approximate the values of the 1. The CST (Constant Strain Triangular) unknown coefficients for mass and stress. For that element performs well when strain gradient is each design value i.e. x1, x2 and x3 is sampled in 3 small. In a pure bending problem, σxx in the levels (lower bound, center, and upper bound) giving neutral axis should be zero. Instead, CST a total of 27 combinations for the 3 design variables. elements show oscillating pattern of stress. These 27 sample designs are run on Abaqus which Strain along y-axis is supposed to be linear. would provide us with ample number of data points But, CST elements can only have constant to find a smooth regression curve that would fit these strain in y-direction. CST elements predict values. The mass and the stress are noted down for stress and deflection about ¼ of the exact each of the sample designs. These results have been values. [3] tabulated below. The power of a fully constrained 2. The LST (Linear Strain Triangular) has a geometry is realized here. An unconstrained complete polynomial in its deflection geometry would require the designer to model the equation which would almost exactly mimic same part 27 different times, since updating any one the properties of bending. However, the of the three design parameters would drastically model in not only a bending problem. The modify his initial design. compression force would induce stress concentration near the whole which is not Sr. X1 X2 X3 Max. Stress Mass (kg) accounted for using the LST element. Also, No. (cm) (cm) (cm) (kg.cm/s2.cm2) LST elements have some discontinuities at 1 8 0.5 20 2.41E+00 2.41E+06 the edges. [4] 2 8 0.5 28.5 2.31E+00 3.88E+06 3. The Q8 element would be ideal for solving 3 8 0.5 37 2.21E+00 5.38E+06 the model since it mimics the bending 4 8 1.75 20 2.26E+00 2.74E+06 problem better than the Q4 element due to a 5 8 1.75 28.5 2.08E+00 4.22E+06 quadratic variation stress as compared to the 6 8 1.75 37 1.89E+00 6.25E+06 linear variation of stress in the Q4 element. 7 8 3 20 2.07E+00 3.67E+06 However, limited by 1000 nodes we would 8 8 3 28.5 1.80E+00 5.03E+06 need larger element size that would result in 9 8 3 37 1.53E+00 7.70E+06 a course mesh in which case finite element 10 11.5 0.5 20 2.45E+00 2.30E+06 results would contain large errors. [10] Hence, 11 11.5 0.5 28.5 2.35E+00 3.37E+06 we select Q4 which would allow us to smaller 12 11.5 0.5 37 2.25E+00 4.91E+06 element sizes. 13 11.5 1.75 20 2.34E+00 2.58E+06 Section II under Discussions discusses further about 14 11.5 1.75 28.5 2.15E+00 3.60E+06 the Convergence Study. 15 11.5 1.75 37 1.97E+00 5.82E+06 VI. OPTIMIZATION 16 11.5 3 20 2.17E+00 3.65E+06 17 11.5 3 28.5 2.02E+00 5.05E+06 Data from the convergence study points out that the best approximate solution would be obtained by 18 11.5 3 37 1.64E+00 7.84E+06 using a Q4 element with a global size of 0.6. 19 15 0.5 20 2.50E+00 2.16E+06 The optimization can be done by approximating the 20 15 0.5 28.5 2.40E+00 2.93E+06 mass and stress as a quadratic function of 3 design 21 15 0.5 37 2.30E+00 4.53E+06 variables; that is 22 15 1.75 20 2.41E+00 2.38E+06  23 15 1.75 28.5 2.23E+00 3.25E+06 a5 5.20E+03 24 15 1.75 37 2.05E+00 5.42E+06 a6 3.46E+05 25 15 3 20 2.28E+00 4.25E+06 a7 7.20E+03 26 15 3 28.5 2.02E+00 5.04E+06 a8 5.85E+04 27 15 3 37 1.75E+00 8.13E+06 a9 3.25E+04 Table 2: Analysis Results of All Sample Points a10 -3.40E+03 Using the above values of mass and stress, a Table 4: Coefficients for Stress Function using Regression regression model is used to determine the unknown The above values provide a regression curve that is a coefficient, ‘a’ by using the formula given below. very close approximation to the values obtained by 𝑎 = (𝑋T𝑋)-1 𝑋T𝑦 (5) running the sample designs on Abaqus. The table below shows the comparison between these results. Where, y is a 27x1 matrix consisting the values of Abaqus Regression Abaqus Regression mass/stress for the 27 sample designs. y is given by Model Model Model Model the function of the 3 design variables. i.e. Masses Masses Stresses Stresses (kg) (kg) (kg.cm/s2.cm2) (kg.cm/s2.cm2) 𝑦 = 𝑎1 + 𝑎2 𝑥1 + 𝑎3𝑥2 + 𝑎4𝑥3 2.41 2.39 2.41E+06 2.75E+06 + 𝑎5𝑥12 + 𝑎6𝑥22 (6) 2.31 2.29 3.88E+06 3.64E+06 + 𝑎7𝑥32 + 𝑎8𝑥1𝑥2 2.21 2.16 5.38E+06 5.56E+06 + 𝑎9𝑥2𝑥3 + 𝑎10𝑥1𝑥3 2.26 2.24 2.74E+06 2.64E+06 Where, x is a 27x10 matrix consisting of the 2.08 2.06 4.22E+06 3.87E+06 constants x1, x2 and x3 corresponding to equation y for 1.89 1.85 6.25E+06 6.14E+06 the 27 sample designs. A sample row of is shown 2.07 2.05 3.67E+06 3.60E+06 below. 1.80 1.79 5.03E+06 5.18E+06 (7) 1.53 1.50 7.70E+06 7.79E+06 a is a 10x1 matrix consisting of the 10 unknown 2.45 2.44 2.30E+06 2.36E+06 coefficients of the mass/stress for the 27 sample 2.35 2.34 3.37E+06 3.15E+06 designs. 2.25 2.21 4.91E+06 4.97E+06 Regression is performed twice to obtain the values of 2.34 2.32 2.58E+06 2.50E+06 the unknown coefficient, a for mass as well as stress. 2.15 2.14 3.60E+06 3.63E+06 The unknown coefficients, ‘a’ obtained after running 1.97 1.93 5.82E+06 5.80E+06 the equation shown above in MATLAB gives the 2.17 2.17 3.65E+06 3.72E+06 following results. 2.02 1.91 5.05E+06 5.20E+06 a1 2.3106 1.64 1.62 7.84E+06 7.71E+06 a2 0.0302 2.50 2.47 2.16E+06 2.10E+06 a3 -0.0074 2.40 2.37 2.93E+06 2.78E+06 a4 0.0019 2.30 2.24 4.53E+06 4.50E+06 a5 -0.001 2.41 2.38 2.38E+06 2.49E+06 a6 -0.0098 2.23 2.20 3.25E+06 3.52E+06 a7 -0.0002 2.05 1.99 5.42E+06 5.59E+06 a8 0.0077 2.28 2.27 4.25E+06 3.97E+06 a9 -0.0077 2.02 2.00 5.04E+06 5.34E+06 a10 0 1.75 1.71 8.13E+06 7.76E+06 Table 3: Coefficients for Mass Function using Regression Table 5: Comparison between Abaqus Model and Regression Model a1 6.51E+06 Using regression, we have obtained a function of a2 -1.74E+05 mass and stress by substituting the values in Table 3 a3 -1.99E+06 and Table 4 in equation 6. Using these functions, an a4 -2.34E+05 optimization toolbox is required to arrive at the best possible solutions for x1, x2 and x3. To achieve this, we use the function fmincon in Abaqus is 0.221 cm as shown in Fig 10. The location MATLAB. fmincon finds a constrained minimum of for both the displacements are identical which can be a scalar function of several variables starting at an easily proven analytically, since the maximum initial estimate. This is generally referred to as displacement would be obtained the tip of the Torque constrained nonlinear optimization or nonlinear Arm. The error is approximately 10% in these 2 programming. [8] Hence, fmincon minimizes the cost values. function, i.e. mass under the stress constrain of 340 MPa. The optimal values of x1, x2 and x3 obtained from MATLAB are shown in Section V under Discussions. DISCUSSIONS I. PRELIMINARY ANALYSIS The values for the maximum stresses were calculated on Excel. The maximum stresses were obtained at Fig. 10: Maximum Displacement in Initial Design x=0cm where Length = 42cm. The calculations are shown below, II. CONVERGENCE STUDY 𝐹(𝐿 − 𝑥)𝑦 Fig. 10 shows a global size of 0.6 with 991 nodes and 𝜎xx = − = -3.8E+06 kg.cm/s2.cm2 𝐼 833 elements. A convergence study is carried out to find the suitable global size that would be used for 𝐹 the iterations discussed further. 𝜎xx = − = -2E+05 kg.cm/s2.cm2 𝐴 A convergence study is done to find the convergence Adding the values in the equation above we obtain, rate. For this, we need displacements for different σmax = -4 x 106 kg.cm/s2.cm2 global sizes that would alter the number of nodes to The units for calculations are referenced in Table 1. see which global size provides the best results. The The maximum deflection can be calculated by beam table below shows the desired no. of nodes, actual theory by the equation given below, [5] number of nodes and elements along with the 𝐹Y𝐿3 displacement at the point of application of force. 𝛿max = (8) Fig. 11 shows plot for the convergence study. 3𝐸𝐼 Here we only consider the bending load since the Desired Nodes No. of Global Displacement axial load does not contribute to the net deformation. # of Elements Size (cm) The maximum deflection using the Excel Calculator Nodes was at L=42cm where the value of deflection was, 1000 991 833 0.6 1.93E-01 𝐹Y𝐿3 500 478 368 0.905 1.90E-01 𝛿max = = 0.245 𝑐𝑚 250 246 174 1.5 1.87E-01 3𝐸𝐼 Table 6: Convergence Study Results From Fig. 9, we can see that the maximum stress from Abaqus shows a maximum stress of -2.87E+06. Convergence Study We get an error of approximately 28.5%. This error 0.194 is due to the design simplification considered in the Displacement (cm) preliminary. Although, it helps us see that the figures 0.192 are in the same ballpark which gives us some idea of 0.19 the stresses to expect while running the simulations 0.188 on Abaqus. Fig. 9, shows the location of maximum 0.186 stress at the top right-hand corner of the center cut 0 200 400 600 800 1000 1200 out. Another region of high stress concentration is on the top surface close to the fixed end. This location No. of Nodes is shown by the preliminary analysis as well, where Fig. 11: Convergence Study Graph the maximum stress is calculated at the fixed end. The maximum displacement that is obtained from We now use Richardson’s Extrapolation to find out the convergence rate using the formula given below. [6] ||𝑢3 − 𝑢2|| ℎ2 𝛼 =( ) (9) ||𝑢2 − 𝑢1|| ℎ1 Where h1, h2 and h3 is the representative length of the element (global size) such that h1>h2>h3. Thus, h3 Fig. 12: Design Sample 1 (x1=8cm, x2=0.5cm and x3=20cm) corresponds to the element with the least global size. Using the formula, we obtain α=1.11. The true displacement can be obtained by using the set of equations given below. [6] 𝑢ℎ = 𝑢0 + 𝑔ℎ𝛼 + 𝑂(ℎ𝑎+1 ) (10) 𝑢ℎ3 − 𝑢ℎ2 = 𝑔𝑝𝛼 (𝑝𝛼 − 1)ℎ1 𝛼 (11) Fig. 13: Design Sample 2 (x1=8cm, x2=0.5cm and x3=28.5cm) Or 𝑢ℎ2 − 𝑢ℎ1 = 𝑔(𝑝𝛼 − 1)ℎ1 𝛼 (12) Using the above equation, we get, g=-0.005681. Using the value of g in equation for uh we get the value of true displacement uo=0.195cm. The value of true displacement determines that element size that should be used to arrive close to the exact results. Since, the displacement using an element size of 0.6 provides the most accurate Fig. 14: Design Sample 3 (x1=8cm, x2=0.5cm and x3=37cm) results, we use those values to carry out the 27 design samples explained in Section VI Under Approach. While performing the 27 iterations, we might need to make a few changes to the partitions to the initial III. ANALYSIS OF ABAQUS REULTS design shown in Fig. 7 to ensure the generation of a From the preliminary analysis, the maximum stress good mesh. It is also essential to note the structure of should be obtained at the top surface as explained in the element around the point of maximum stress. Too Section I under Approach. However, Fig. 9 showed distorted elements near the area of maximum stress the location of maximum stress at the top right corner would provide inaccurate results. [10] of the center cut out. This is due to less material at IV. ANALYSIS OF REGRESSION that point, which would result in a small moment on FUNCTION & FMINCON inertia. From the bending equations referenced in Section I under Approach, we notice that that both From Table 6, we notice that the values of the mass the moment of inertia as well as the area is inversely and stress are not exactly equal to their respective proportional with the stress, resulting in high stresses results for all the 27 design simulations. at that point. The main reason for this discrepancy is that This was an interesting result. While simulating the regression is a statistical process for estimating the 27 sample designs on Abaqus, the location of the relationships among variables. [7] This is not always maximum stress shifted to different loads. Most of exact. Regression tries its best to establish a close them conform with the preliminary analysis. relationship between the dependent (mass) and However, some had it at different locations like the independent variables (x1, x2 and x3). Initial Design shown in Fig. 13 and Fig. 14. The Also, fmincon is a powerful optimization tool but different locations of stress are shown in the figures depending on the problem you are trying to solve below. there could be better optimization tools that could be leveraged to come to a global minimum for a The final parameters for the weight optimization are solution. shown in Table 11. V. ANALYSIS OF OPTIMIZATION TOOL X1 X2 X3 Mass Stress RESULTS 2.08 3.34E+6 8 cm 2.48 cm 22.6 cm kg (kg.cm/s2.cm2) Simulating the initial design on Abaqus yields the Table 10: Final Optimization Results following results: We notice some difference in the values of weight Mass Max. Stress Displacement (kg) (kg.cm/s2.cm2) (cm) for the optimization as shown by MATLAB in Table 2.31E+00 2.87E+06 1.93E-01 10 and the actual mass shown by Abaqus. This is due Table 7: Initial Design Results to approximation of the function coefficients, ‘a’ found by regression. However, this error is Studying the values of mass and stress of the 27 approximately 0.5% which is negligible. Similarly, sample designs in Table 2, we try to analyze at which due to the same reasons, the final design also has values of x1, x2 and x3 would the mass be the least at some scope to drop the weight further by tweaking which point the maximum stress induced would be the parameters x1, x2 and x3. under 340 MPa. The most suitable designs are highlighted in bold in Table 2 and tabulated below REFERENCES for reference. [1] Weight Optimization; Retrieved From: Sr. X1 X2 X3 Max. Stress https://elenoenergy.com/weight- Mass (kg) No. (cm) (cm) (cm) (kg.cm/s2.cm2) optimization-using-cae-techniques.html 4 8 1.75 20 2.26E+00 2.74E+06 [2] Radius Rod; Retrieved From: 23 15 1.75 28.5 2.23E+00 3.25E+06 https://en.wikipedia.org/wiki/Radius_rod Table 8: Extract from Table 2 Highlighting Suitable Design [3] Nam Ho Kim; Chapter 6 Finite Elements for Results Plane Solids; EML 4500 Finite Element The results that are obtained using MATLAB Analysis and Design. Optimization function fmincon are shown below. [4] Amit H. Verma; CE 595: Course Part 2; Finite Element Modeling and Analysis. X1 X2 X3 Mass [5] Handbook of Engineering Mechanics; W. 8 cm 2.48 cm 22.6 cm 2.071 kg Table 9: fmincon Optimization Results Flugge: McGraw-Hill 1962 [6] Nam Ho Kim; Chapter 9 Finite Element We notice that the values of x1, x2 and x3 are close to Procedure and Modeling; EML 5526 Finite the to Sample Design 4 shown in the table above. Element Analysis. CONCLUSIONS [7] Regression Analysis; Retrieved From: https://en.wikipedia.org/wiki/Regression_an These results are simulated on Abaqus to check if the alysis#Underlying_assumptions fmincon optimization satisfied the constraints and [8] UC Berkeley Wireless Research Center; IC gave us an optimal value of the weight. The results Design Classes Project 1 Solutions. from Abaqus are shown below. [9] MATLAB Nonlinear Optimization with fmincon; Retrieved From: https://www.youtube.com/watch?v=_Il7GQ dL3Sk&t=750s [10] Nam-Ho Kim, Bhavani V. Sankar- Introduction to Finite Element Analysis and Design-Wiley (2008) Fig. 15: Final Design Stress