1.1 OVERVIEWBall joints are an important component suspension system found on every modern vehicle. They are ball pin and spherical socket connecting the control arms of the vehicle to steering knuckles. They attach the wheel hub to the rest of suspension system. This connection made the ball joint able to rotate horizontally for steering and vertically for shock absorption, hence the use of ball joints that can move in all directions. Therefore, ball joints must withstand complex loading in vehicle handling.1.

2 PROBLEM STATEMENTSuspension ball joints manage the movement of the vehicle suspension and allow the wheel to move around its vertical axis. This generates a cyclic loading on the ball joints components during its lifetime. It will receive different application and at the same time will rotate around the ball joints axis. Resulting from the load and motion application, the failure of suspension ball joint is located at the spherical contact interface between the ball pin and the ball socket. This interface leads to wear, which is caused by two mechanisms, the friction of the ball socket surface with the ball pin and plastic deformation due to the axial load. Another failure that can occur is fracture of the ball joint.

This fracture can occur when the loads received by the ball joint are more than it can withstand. Since the first failure that will occur is most probably the wear, this project will focus on the wear analysis of the ball joint.1.3 OBJECTIVESThe main objective of this project is to perform the durability analysis of the lower arm ball joint.

The ball joint will be subjected to loads from the road profile and the durability of the ball joint will be analyzed. The main objectives of this project are:1. To predict the reaction forces acting on the ball joint assembly.2. To predict the wear behavior between the ball joint and the cage.3.

To estimate the life cycle of the ball joint assembly.?CHAPTER 2LITERATURE REVIEW2.1 QUARTER CARFor the sake of the vertical dynamics of a vehicle, the well-known 2 DOF quarter car model, shown in Figure 1. The model only includes a corner of a test vehicle, which means it only has one wheel of the vehicle. The model is composed of a sprung mass and an unsprung mass. The damper and the spring of the suspension system are modeled as cs and ks respectively. The tire is modeled as a linear spring without damping, as usually the damping effect of the tire is very small compared to the suspension damping.

The tire spring stiffness is kt. Using Newton’s second law, the differential equations of motion for the sprung mass and unsprung mass are respectively:m_s z ?_1+c_s (z ?_1-z ?_2 )+k_s (z_2-z_2 )=0 (1)m_u z ?_2+c_s (z ?_2-z ?_1 )+k_s (z_2-z_1 )+k_t z_2=k_t z_0 (2) Figure 1: A quarter car modelBy denoting:z={?([email protected]_2 )},u=z_0 (3)the two equations can be combined into a vector equation:M=?(m_s;[email protected];m_u ) C=?(c_s;[email protected]?-c?_s;c_s ) K= k_s-ks-ksks+kt F= 0kt (4)In this model, ms and mu are respectively the mass of the quarter car body and the wheel, ks is the stiffness of the spring incorporated in the suspension system, kt is the tire stiffness, cs is the damping coefficient of the shock absorber, z1 and z2 are the position of the sprung mass and unsprung mass respectively, z0 is the displacement of the road surface 1.2.2 MACPHERSON SUSPENSION SYSTEMSuspension system consists of springs, shock absorbers and linkages that connects a vehicle to its wheels and allows relative motion between the two parts.

Suspension systems serve a dual purpose contributing to the vehicle’s road holding, handling and braking for active safety and driving pleasure. In addition, suspension systems are used for keeping vehicle occupants comfortable and reasonably well isolated from road noise, bumps, and vibrations. compromise.

It is important for the suspension to keep the road wheel in contact with the road surface as much as possible because the road forces are acting on the vehicle through the contact patches of the tires. The main functions of suspension system are as follows:• To protect vehicle from road shocks.• To safeguard passengers from shocks.• To prevent pitching or rolling 2.

In this project, a MacPherson suspension system as shown in Figure 2 is used as part of the quarter car design, which offer the possibility to choose the ideal settings from existing solutions which satisfies all the imposed constraints 3. Figure 2: MacPherson Suspension System.The roles of a suspension system are to support the weight of the vehicle to isolate the vibrations from the road and to maintain the traction between the tire and the road.2.3 ROAD PROFILEA profile of a road is an intrinsic part of the interaction between the pavement and a vehicle suspension, and it influences ride, handling, fatigue, fuel consumption, tire wear, maintenance costs, and vehicle delay costs. It is a two-dimensional cross-sectional shape of the road surface, taken along an imaginary line for the purpose of ride quality analysis. As shown in Figure 3, a profile consists of longitudinal and lateral profiles.

Profiles taken along the lateral direction show the super elevation and crown of the road design 4, while this research focuses on the longitudinal profile, as it shows the design grade, roughness and texture, and is the primary source of vehicle vibrations. Figure 3: Road ProfileIn physical testing and simulations, the road profile can be classified such as: isolated bump, sinusoidal, fixed waveform, and stochastic roads 5. An isolated bump is a nominally smooth basic road with a local disturbance, which results in a solitary wave.

It includes ramps or true bumps. A ramp is caused by a disturbance with a residual change of road height, while a true bump returns to its original elevation after a disturbance.2.3 RIGID BODY ANALYSISMultibody simulation deals with study and analysis of dynamic behavior of system of flexible and/or rigid interconnected bodies.

These bodies are subjected to constrain with respect to one another through a kinematic constraint modeled as joints. These systems can represent an automobile, a robot with manipulator arms, an aircraft and so on as an assembly of rigid and flexible parts. The components may be subjected to large displacement, large rotation, and effects of finite strain.

Multibody systems have conventionally been modeled as rigid body systems with superimposed elastic effects of one or more components. A major limitation of these methods is that non-linear large-deformation, finite strain effects or non-linear material cannot be incorporated completely into model.The FE method used in ANSYS offers an attractive approach to modeling a multibody system. The ANSYS multibody analysis method may require more computational resources and modeling time compared to standard analysis; it has the following advantages: The finite element mesh automatically represents the geometry while the large deformation/rotation effects are built into the finite element formulation. Inertial effects are greatly simplified by the consistent mass formulation or even point mass representations. Interconnection of parts via joints is greatly simplified by considering the finite motions at the two nodes forming the joint element.

A general step for FEM for non-linear analysis is as follows: Build the model: A flexible mechanism usually comprises of flexible and/or rigid body parts connected via joint elements. The modeling the flexible parts with any of the 3-D solid, shell, or beam elements. The flexible and/or rigid parts are connected using joint elements.

Two parts may be simply connected to ensure that the displacements at the joints are identical. But the two connected parts may involve joint such as the universal joint or a planar joint. While modeling these joints, a suitable kinematic constraint is implemented on the relative motion (displacement and rotation) between the two nodes that form the joint. Define element types: Simulation of a flexible multibody involving flexible and rigid components joined together subjected to some form of kinematic constraints, using appropriate joint and contact element types. Define materials: Defining the linear and non-linear material properties for each components of multibody system. Mesh the model: Mesh the all flexible components of multibody system. Two nodes define joint elements and no special meshing is required to define them.

Solve the model: Multibody analyses generally involve large rotations in static or transient dynamics analysis, so non-linear geometric effects must be accounted for. Review the results of model: Results from a flexible multibody analysis consist mainly of displacements, velocities, accelerations, stresses, strains, and reaction forces in structural components. Constraint forces, current relative positions, relative velocities, and relative accelerations in joint elements are also available 6.2.4 FATIGUE ANALYSISFatigue is a failure under a repeated or varying load which never reaches a level of sufficient to cause failure in a single application. It can also be thought of as the initiation and growth of a crack, or growth from a pre-existing defect, until it reaches a critical size, such as separation into two or more parts.Fatigue analysis itself usually refers to one of two methodologies: either Stress-Life (S-N) method, or the Local Strain or Strain-Life (e-N) method.

The S-N method commonly referred to as Total Life since it makes no distinction between initiating or growing a crack and the e-N method is commonly known as the Crack Initiation method which concerns itself only with the initiation of a crack 7.2.4.

1 Stress-Life MethodTo determine the strength of materials under the action of fatigue loads, specimens are subjected to repeated or varying forces of specified magnitudes while the cycles or stress reversals are counted to destruction. The most widely used fatigue-testing device is the R. R.Moore high-speed rotating-beam machine. This machine subjects the specimen to pure bending (no transverse shear) by means of weights. Other fatigue-testing machines are available for applying fluctuating or reversed axial stresses, torsional stresses, or combined stresses to the test specimens.

To establish the fatigue strength of a material, quite several tests are necessary because of the statistical nature of fatigue. For the rotating-beam test, a constant bending load is applied, and the number of revolutions (stress reversals) of the beam required for failure is recorded. The first test is made at a stress that is somewhat under the ultimate strength of the material. The second test is made at a stress that is less than that used in the first. This process is continued, and the results are plotted as an S-N diagram (Figure 4). This chart may be plotted on semi log paper or on log-log paper. In the case of ferrous metals and alloys, the graph becomes horizontal after the material has been stressed for a certain number of cycles.

Plotting on log paper emphasizes the bend in the curve, which might not be apparent if the results were plotted by using Cartesian coordinates. Figure 4: A S-N diagram of completely reversed axial fatigue tests.The ordinate of the S-N diagram is called the fatigue strength Sf; a statement of this strength value must always be accompanied by a statement of the number of cycles N to which it corresponds. In the case of the steels, a knee occurs in the graph, and beyond this knee failure will not occur, no matter how great the number of cycles.

The strength corresponding to the knee is called the endurance limit Se, or the fatigue limit. The graph of Figure 4 never does become horizontal for nonferrous metals and alloys, and hence these materials do not have an endurance limit.The body of knowledge available on fatigue failure from N = 1 to N = 1000 cycles is generally classified as low-cycle fatigue, as indicated in Figure 4. High-cycle fatigue, then, is concerned with failure corresponding to stress cycles greater than 103 cycles. A finite-life region and an infinite-life region are also distinguished in Figure 4. The boundary between these regions cannot be clearly defined except for a specific material; but it lies somewhere between 106 and 107 cycles for steels, as shown in Figure 4 8.

2.4.2 Strain-Life MethodAccording to Budynas and Nisbett (2006) 8, the best approach yet advanced to explain the nature of fatigue failure is called by some the strain-life method. The approach can be used to estimate fatigue strengths, but when it is so used it is necessary to compound several idealizations, and so some uncertainties will exist in the results. For this reason, the method is presented here only because of its value in explaining the nature of fatigue.A fatigue failure almost always begins at a local discontinuity such as a notch, crack, or other area of stress concentration. When the stress at the discontinuity exceeds the elastic limit, plastic strain occurs.

If a fatigue fracture is to occur, there must exist cyclic plastic strains. Thus, the behavior of materials subject to cyclic deformation shall be investigated.In 1910, Bairstow verified by experiment Bauschinger’s theory that the elastic limits of iron and steel can be changed, either up or down, by the cyclic variations of stress 9. In general, the elastic limits of annealed steels are likely to increase when subjected to cycles of stress reversals, while cold-drawn steels exhibit a decreasing elastic limit.In general, the elastic limits of annealed steels are likely to increase when subjected to cycles of stress reversals, while cold-drawn steels exhibit a decreasing elastic limit.

R. W. Landgraf has investigated the low-cycle fatigue behavior of many very high-strength steels, and during his research he made many cyclic stress-strain plots 10.

Figure 5 has been constructed to show the general appearance of these plots for the first few cycles of controlled cyclic strain. In this case the strength decreases with stress repetitions, as evidenced by the fact that the reversals occur at ever-smaller stress levels. As previously noted, other materials may be strengthened, instead, by cyclic stress reversals. Figure 5: True stress–true strain hysteresisThe SAE Fatigue Design and Evaluation Steering Committee released a report in 1975 in which the life in reversals to failure is related to the strain amplitude ??/2 11. The report contains a plot of this relationship for SAE 1020 hot-rolled steel; the graph has been reproduced as Figure 6.

To explain the graph, we first define the following terms: Fatigue ductility coefficient ??’?_F is the true strain corresponding to fracture in one reversal (point A in Figure 5). The plastic-strain line begins at this point in Figure 6 Fatigue strength coefficient ??’?_F is the true stress corresponding to fracture in one reversal (point A in Figure 5). Note in Figure 6 that the elastic-strain line begins at ??’?_F/E.

Fatigue ductility exponent c is the slope of the plastic-strain line in Figure 6 and is the power to which the life 2N must be raised to be proportional to the true plastic strain amplitude. If the number of stress reversals is 2N, then N is the number of cycles. Fatigue strength exponent b is the slope of the elastic-strain line and is the power to which the life 2N must be raised to be proportional to the true-stress amplitude. Figure 6: A log-log plot showing how the fatigue life is related to the true-strain amplitudeNow, from Figure 5, we see that the total strain is the sum of the elastic and plastic components.

Therefore, the total strain amplitude is half the total strain range??/2=????_e/2+????_p/2 (5)The equation of the plastic-strain line in Figure 6 is????_p/2=??^’?_F (2N)^c (6)The equation of the elastic strain line is????_e/2=??^’?_F/E (2N)^b (7)Therefore, from Eq. (5), we have for the total-strain amplitude??/2=??^’?_F/E (2N)^b+??^’?_F (2N)^c (8)which is the Manson-Coffin relationship between fatigue life and total strain.Though Eq. (8) is a perfectly legitimate equation for obtaining the fatigue life of a part when the strain and other cyclic characteristics are given, it appears to be of little use to the designer.

The question of how to determine the total strain at the bottom of a notch or discontinuity has not been answered. There are no tables or charts of strain concentration factors in the literature.2.5 LIFE CYCLEThe total number of cycles for which a specimen sustains before failure is called “fatigue (cyclic) life”, denoted by N.

The graph by plotting values of Sa and N is called “S-N curve” or “Wöhler diagram”. When a specimen does not fail even if the specified cycle is reached, test is stopped, and the corresponding stress value is marked on the curve as “runout” (given by an arrow as in Figure 4). The fatigue limit in such case is assumed as 50×108 cycles for design purposes.

Cycle of 100 refers to ultimate tensile strength (Sut) while endurance strength (Se ? 0.5Sut) is obtained at 106 cycles. The stress amplitude is about 0.8Sut by which high cycle fatigue starts (103 cycles). The correlation between S and N in high cycle region (103 ? N ? 106) can be obtained based on equation of the line (i.e y=bx+c) 12:log??S_f=b log??N+c? ? (9)Rearrange Eq.

(9), the fatigue life cycle isN=?10?^(-c/b) S_f^(1/b) (10)whereb=-1/3 log?((0.8S_ut)/S_e ) (11)c=log??(0.8S_ut )^2/S_e ? (12)?CHAPTER 3METHODOLOGY3.1 LITERATURE SURVEYLiterature survey is done to study all criteria and standard specification used when analyzing the durability of lower arm ball joint. By the end of the surveys, the standard quarter car used, and road profile are determined. The material used for the ball joint is also decided.

3.2 QUARTER CAR MODELINGAs discussed in Chapter 2.1, this model is then made as a 3D model using CATIA as shown in Figure 7 and imported to ANSYS for rigid body analysis.

The parameters for the quarter car is as below: Table 1 Parametes: Quarter-car suspension system and controlParameter Value UnitsSprung mass (m_s) 208 kgUnsprung mass (m_u) 28 kgSpring stiffness ?(k?_s) 18,709 N/mTire stiffness (k_t) 127,200 N/m Figure 7: Quarter Car 3D Model TEMP3.3 ROAD PROFILE SELECTIONIn this work, the shape filter method (Feng Tyan et al 2008) is used for generating the road profiles. The road profile can be represented by Power Spectral Density (PSD) function. Power spectral densities of roads show a characteristic drop in magnitude with wave number. The road surface profile is measured with respect to reference plane. Random road profile can be approximated by a PSD in the form of?(?)=??(?/?_0 )^(-?) (13)Or?(n)=?n?(n/n_0 )?^(-?) (14)?=2?/L in rad/m denotes the angular spatial frequencyL is wave length?_0??(?_0) in m²/ (rad/m) describes the values of the PSD at the reference wave number(?_0 )=1rad/m.

n=?/2? is the spatial frequency.n=0.1 cycle/m, ? is the waviness, for most of the road surface, ?=2The following guide lines are taken from the ISO 8608. New roadway layers such as asphalt or concrete layers can be assumed to have a good or even a very good roughness quality. Old roadway layers which are not maintained may be classified as having a medium roughness. Road way layers consisting of cobble stones or similar material may be classified as medium (average) or bad (poor, very poor).

It is well known that the amount of road excitation imposed at the vehicle tire depends on the road roughness which is a function of the road roughness coefficient and the velocity (v).Let s be the path variable by introducing the wave length ?=2?/?And assuming that s=0 at t=0 the term ?_s can be written as?_s=2?/L s=2? V/? t=?t where ? (rad/s) is the angular velocity in time domain. Then ?V=? hence in the time domain the excitation frequency is given by f=?/2?=v/?For most of the vehicles, the rigid body vibration is in between f=0.5Hz to f=15Hz. This range is covered by waves which satisfy the condition0.5Hz?V/??15Hz.

The random road profile is generated by shaping filter method. The profile can be approximated by PSD distribution?(?)=(2?V?^2)/(?^2+?^2 V^2 ) (15)where; ?^2 denotes the road roughness variance (m²)V the aircraft speed (m/s)? depends on the type of road surface (rad/m)Hence if the vehicle runs with the constant velocity V, the PSD is given by the above equation (15) and the road profile signal may be obtained as the output of a linear filter expressed by the differential Equation (16),d/dt z_R (t)=-?Vz_R (t)+?(t) (16)where ?(t) is a white noise process with the spectral density ?(?).The road roughness standard deviations for various types of roads are as given in Table 2.Table 2: Road roughness standard deviatonRoad class ?(?10?^(-3) m)Roughness variance ?(?_0 )(?10?^(-6) m^3 ),?_0=1Power spectral density u ((rad)?(m))A (very good) 2 1 0.127B (good) 4 4 0.127C (average) 8 16 0.

127D (poor) 16 64 0.127E (very poor) 32 256 0.127The road profiles are generated by using the values given in the Table 2 in the random road generator as shown in Figure 8. The random road generation is done through MATLAB/SIMULINK 13. Figure 8: Simulink model of random road generator3.4 RIGID BODY ANALYSISThe procedure for rigid and dynamic analysis of mechanism in ANSYS Workbench software is as follows 6:3.

4.1 Attach geometryTo import the geometry, following step can be performed. From the analysis system subroutine, select the geometry cell. Browse to the CAD file from the following access points: Right-click on the geometry cell in the project schematic and choose import geometry. The model cell in the project schematic can be selected via double click. Subsequently the mechanical application displays the geometry.

The dialogue box related to geometry import as shown in Figure 9. Figure 9: Step for attaching the geometry3.4.

2 Define engineering dataAfter importing the assembly to the workbench, all engineering data used for this project is defined in the table below: attached later Figure 10: Step for defining the engineering data3.5 BALL JOINT MODELINGThe measurements of main parts of ball joint have been made in order to create a three-dimensional geometry by CATIA software. Then, the model (geometry) has been imported to ANSYS workbench software. Initially the 3D drawing of ball joint is done by using CATIA according to dimensions specified in the Table 3 14.Table 3 Dimension of Suspension Ball JointDiameter Dimension in mmD1 30D2 20D3 15 Figure 11: 3D Model of Ball Joint3.5.1 MaterialAccording to Jitendra Shinde and Sunil Kadam (2016), the Spectro test results the ball joint was manufactured using an EN 18D steel joy.

EN 18D Steel Alloy properties: Yield stress = 565 MPa Ultimate tensile strength = 887 MPa Density = 7.85 g/cm Poisson’s ratio = 0.29The second component is the ball socket which is made with polymer-based material 15.3.6 FATIGUE ANALYSISAfter importing the 3D model of ball joint to ANSYS Workbench, the analysis will be done with the following steps 16: Make contact between the ball pin and the ball socket under connections feature. Customize the connections so that it is appropriate for the analysis.

Apply fixed support at the ball socket and loading corresponding to the maximum developed load of 1000 pounds. Insert fatigue tool. Specify fatigue loading as coming from a scale history and select scale history file containing strain gauge results over time (ex. Common FilesAnsys IncEngineering DataLoad HistoriesSAEBracketHistory.dat). Define the scale factor to be .005.

(We must normalize the load history so that the FEM load matches the scale factors in the load history file). ((1 FEM load)/1000lbs)×(1000lbs/(200 strain gauge))=((1 FEM load)/(200 strain gauge))=needed load scale factor Specify a bin size of 32 (Rainflow and damage matrices will be of dimension 32×32). Specify Goodman theory to account for mean-stress effects. (The chosen theory will be illustrated graphically in the graphics window). Specify that a signed Von-Mises stress will be used to compare against fatigue material data. (Use signed since Goodman theory treats negative and positive mean stresses differently.) Perform fatigue calculations (Solve command in context menu).

View rainflow and damage matrix. Plot life, damage, and factor of safety contours over the model at a design life of 1000. (The fatigue damage and FS if this loading history was experienced 1000 times). Thus, if the loading history corresponded to the loading experienced by the part over a month’s time, the damage and FS will be at design life of 1000 months. Note that although a life of only 88 loading blocks is calculated, the needed scale factor (since [email protected]=.61) is only .

61 to reach a life of 1000 blocks. Plot factor of safety as a function of the base load (fatigue sensitivity plot, a 2-D XY plot). Copy and paste to create another fatigue tool and specify that mean stress effects will be ignored (SN-None theory) This will be done to ascertain to what extent mean stress is affecting fatigue life.

Perform fatigue calculations. View damage and factor of safety and compare results obtained when using Goodman theory to get the extent of any possible mean stress effect. Change bin size to 50, rerun analysis, and compare fatigue results to verify that the bin size of 32 was of adequate size to get desired precision for alternating and mean stress bins. 3.7 LIFE CYCLE ESTIMATIONThe life cycle of the ball joint is estimated using the fatigue life cycle as discussed in Chapter 2.

5 by using Eq. (10) until Eq. (12).?CHAPTER 4RESULTS AND DISCUSSIONThe results obtained will be discussed based on the ball joint wear behavior and its life cycle.

4.1 WEAR BEHAVIOURFrom the figure below, it is observed that the ball socket and the ball joint does not made any contact or have little contact in the middle since there is no wear damage in that area for the ball socket. From this result, we can say that the ball joint also may not have any wear damage in the middle but slightly further than the mid-point. This is because the ball joint is tilting when it is under cyclic loading. Figure 14: Analysis of the ball socket contact interface4.2 LIFE CYLEIt is observed that from the results, the equivalent Von-Mises stress goes beyond ultimate tensile strength of the material and hence there is crack initiation in the ball joint at the specified loading conditions and that crack leads towards the fatigue failure of the ball joint as the ball joint is under cyclic loading as described in the problem statement.

Figure 12: Equivalent Von-Mises stressThe minimum life cycle of the ball joint is 264.9 cycles while the maximum is 1(106) cycles as shown in figure below: Figure 13: Fatigue lifeCHAPTER 5CONCLUSION AND FUTURE WORKBased from the results of the simulation, it can be concluded that when the ball joint is under cyclic loading, the ball pin and socket will have wear damage after a certain amount of time. When the ball joint has reached its life cycle limit, it will crack which is most likely to occur first before the ball pin fracture. In order to validate the simulation results, the simulation must be performed with the same parameters as in previous papers conducted by other researchers regarding ball joint analysis or related material.For Final Year Project 2, the result from the simulation should be obtained using the correct command and parameters so it can be compared to the previous researched.