# Analysis of Dynamic Forces Due to Heavy Mechanisms Mounted on a aVehicle in Motion by FEM (IECTET 2009)

Location: Nagpur, Maharashtra, India

 Event Date/Time: Dec 16, 2009 End Date/Time: Dec 18, 2009 Registration Date: Jun 25, 2009 Early Registration Date: Jun 25, 2009

#### Description

A.B. Totey (M.Tech IV Semester CAD/CAM, GHRCE)
arvindb_totey@yahoo.com
Prof. I.A. Khan (Lecturer, Dept. of Mech. Engg, GHRCE)
iak20041978@yahoo.co.in
Analysis Of Dynamic Forces Due To Heavy Mechanisms Mounted On A Vehicle In Motion By FEA

Abstract
Trailer bed forms the structural backbone of a commercial vehicle When heavy mechanisms which are to be transported from one place to another, are directly mounted on the trailer bed. When vehicle travels along the road the trailer bed is subjected to dynamic forces induced by different road conditions and different modes of movements. These dynamic forces from the mechanism are transmitted to the trailer bed through contact points, causing deformation, shear, stress etc. in trailer bed and also toppling and skidding of the vehicle may occur.
Motion analysis determines variation in acceleration points, reactions and inertial forces acting on the trailer bed. For specified range of velocity considering different road conditions, different modes of movements and combination of both.
CAD geometry is the staring point for FEA. The loads are determined by analytical treatments the load consists of inertial forces which are then transferred to the components of trailer bed which are selected for analysis.
FEA can work on the components of trailer bed to find its deformation, shear and stresses. Which locates the critical points having highest stress value. The best and worst zone on the trailer bed are determined.

Key Words: Trailer bed, Modes of movements, Road conditions, Motion analysis, Stress.
1. Introduction

Heavy mechanisms which are to be transported are mounted on trailer bed. Due to this dynamic forces are induced into the system during transportation.
These dynamic forces are at the joints in the form of impacts. This is caused mainly due to acceleration and the clearance available. The magnitude of these forces increases as the speed increases and are transmitted to the vehicle frame which disturbs the smooth movements of the vehicle and some time may lead to instability or toppling of the vehicle. For reducing such dynamic forces only stoppers are provided to restrict mobility of the links during transportation. But at high speeds even small clearance may give large forces. Therefore the transportation of such heavy mechanisms becomes unsafe and restricts the speed of the vehicle.

2. Objectives

â€¢ To determine the velocity and acceleration at different points in vehicle volume by motion analysis.
â€¢ To determine the dynamic forces at these points considering different road conditions and modes of movements.
â€¢ To determine the effect of these dynamic forces by FEA.

3. Planning

By motion analysis values of velocity and acceleration at different points are determined according to these values of velocity and acceleration dynamic forces are determined analytically using basic principles. Finite Element Analysis is done using dynamic forces as inputs.

4. Motion Analysis using basic principles

The velocity and acceleration of points in the given vehicle volume are determined for different cases. These values are then plotted in the given volume, which will give clear pattern of the distribution of the acceleration. This distribution will help in identifying the worst and suitable zones. These values of acceleration are determined by the motion analysis. The motion analysis is done considering following cases -

4.1. Different modes of movements
4.1.1. Acceleration mode
4.1.2. Braking mode
4.1.3. Turning mode
4.2.1. Over pit holes
4.2.2. Over the bumps

4.1.1) Acceleration while accelerating

When vehicle is moving on a straight path & accelerating in the same direction the inertial forces will be in opposite direction.
Therefore, the acceleration of a point in the vehicle volume due to the acceleration of vehicle can be explained as follows.
AA p (x, y, z,)x = Ft/M

Fig-4.1.1.1.

4.1.2) Acceleration while Braking

When the vehicle is moving on a straight path there is relative acceleration due to change in velocity of the vehicle. And the inertial force is acting to the direction of the motion of vehicle. When the brakes are applied, the acceleration due to braking is opposite to the direction of the motion of the vehicle.

The acceleration of a point due to the braking is in the x direction i.e. direction along the length of the vehicle can be determined as follows.
AB p (x, y, z,)x = Fb/M

Fig-4.1.2.1.

4.1.3) Acceleration when vehicle is taking a turn

There can be a situation when the vehicle is moving on curved track. The center of turn will be rear wheel axle as shown in figure
When the vehicle is moving on a curved track of radius Rc, the vehicle will experience an angular speed with respect to the radius of curve.
ï· = Vv / Rc

Fig-4.1.3.1.

Variation of Acceleration along the length during turn about â€œY axisâ€
( R+Z ) m X (m) At m/sec2
Vv = 8.33 m/s Vv = 11.11 m/s Vv = 13.88 m/s
3.75 2 0.74 1.33 2.08
3.75 1 0.67 1.22 1.90
3.75 0 (Front Axle) 0.65 1.18 1.83
3.75 1 0.67 1.22 1.90
3.75 2 0.74 1.33 2.08
3.75 3 0.83 1.51 2.35
3.75 4 0.95 1.72 2.68
3.75 5 1.08 1.96 3.06
3.75 6 1.22 2.22 3.47
3.75 7 1.38 2.49 3.89
3.75 8 1.53 2.77 4.32
3.75 9 (Rear Axle) 1.69 3.06 4.78
3.75 10 1.85 3.35 5.23
3.75 11 2.02 3.64 5.7

Variation of Acceleration along the width during turn about â€œY axisâ€
X (m) ( R+Z) m At m/sec2
Vv = 8.33 m/s Vv = 11.11 m/s Vv = 13.88 m/s
0 3.75 0.65 1.18 1.83
0 4.1 0.71 1.28 2.00
0 4.5 0.78 1.41 2.20
0 4.9 0.85 1.53 2.40
0 5.3 0.91 1.66 2.60
0 5.5 0.94 1.72 2.70
0 5.9 1.02 1.85 2.89
0 6.25 1.08 1.89 3.06
Graph showing variation of Acceleration along the length during turn about â€œY axisâ€

Graph-4.1.3.1.

Graph showing variation of Acceleration along the width during turn about â€œY axisâ€

Graph-4.1.3.2.

4.2.1) Vehicle through a pit

Considering that only one wheel is going through the pit. When the wheel is passing through the pit, it will be having a downward motion i.e. it will have a downward velocity. This downward velocity will be causing a downward acceleration. The downward velocity will depend upon the rotation of the wheel in the pit. If the rotation of the wheel is 1800 then the diameter of the wheel will be equal to the depth of the pit. At this point the vertical velocity is infinity theoretically.

Fig-4.2.1.1.
Variation in acceleration due to Rotation about Diagonally opposite Corner when a vehicle is moving through pit
Vv = 8.33 m/sec Vv = 11.11 m/sec Vv = 13.88 m/sec
Vy = 5.89 Vy = 7.86 Vy = 9.81
D = (x2+z2)1/2 D = (x2+z2)1/2 D = (x2+z2)1/2
D Ay D Ay D Ay
1.04 0.41 1.04 0.73 1.04 1.21
0.00 (Left) 0.00 0.00 (Left) 0.00 0.00 (Left) 0.00
1.04 0.41 1.04 0.73 1.04 1.21
2.08 0.83 2.08 1.47 2.08 2.43
3.12 1.24 3.12 2.20 3.12 3.64
4.16 1.65 4.16 2.94 4.16 4.85
5.20 2.06 5.20 3.67 5.20 6.07
6.24 2.48 6.24 4.40 6.24 7.28
7.28 2.89 7.28 5.14 7.28 8.49
8.32 3.30 8.32 5.87 8.32 9.70
9.36 (Right) 3.71 9.36 (Right) 6.60 9.36 (Right) 10.92
10.40 4.13 10.40 7.34 10.40 12.13
Tab-4.2.1.1.

Graph showing Variation of Acceleration when the vehicle is rotating about its diagonally opposite corner passing thorough pit

Graph-4.2.1.1.

4.2.2) Vehicle over a bump.

It is considered that only one wheel is passing over the bump. The vehicle will experience a velocity in the vertical direction. Thus the frame of the vehicle will experience angular velocities about two axes. The angular speed about the respective axes will depend on the vehicle speed, height and width of the bump.

Fig-4.2.2.1.
Variation in acceleration due to Rotation about Diagonally opposite Corner when a vehicle is moving over a bump

Vv = 8.33 m/sec Vv = 11.11 m/sec Vv = 13.88 m/sec
Vy = 7.19 Vy = 9.59 Vy = 11.97
D = (x2+z2)1/2 D = (x2+z2)1/2 D = (x2+z2)1/2
D Ay D Ay D Ay
1.04 0.62 1.04 1.10 1.04 1.70
0.00 (Left) 0.00 0.00 (Left) 0.00 0.00 (Left) 0.00
1.04 0.62 1.04 1.10 1.04 1.70
2.08 1.23 2.08 2.21 2.08 3.41
3.12 1.85 3.12 3.31 3.12 5.11
4.16 2.47 4.16 4.41 4.16 6.82
5.20 3.08 5.20 5.52 5.20 8.52
6.24 3.70 6.24 6.62 6.24 10.22
7.28 4.32 7.28 7.72 7.28 11.93
8.32 4.93 8.32 8.83 8.32 13.63
9.36 (Right) 5.55 9.36 (Right) 9.93 9.36 (Right) 15.34
10.40 6.17 10.40 11.03 10.40 17.04
Tab-4.2.2.1.

Graph showing Variation of Acceleration when the vehicle is rotating about its diagonally opposite corner passing over bump

Graph-4.2.2.1.
5. Calculations of dynamic forces

Total external work done by weight (W) = Internal stored energy
W x (h + = ( /2E) volume
Kinetic energy of moving system is stored as strain energy, therefore ,
Â½ mv2 = ( /2E) volume F is the equivalent static load which will produce the stress
= F/A
V2 = 2as
The kinetic energy body can be stored in resisting member in form of strain energy. Hence, equation 4.2.1 can be reduced as
m.a.s = ( volume
Therefore,
Using the above formula the dynamitic forces are calculated for different cases.

6. Finite Element Analysis

CAD geometry is the first step for FEA. The forces determined are transmitted to the component of the trailer bed which are selected for analysis. The boundary conditions are applied on the trailer bed and the stresses, deformations are identified.

6.1. Mesh model of trailer bed

6.2. Deformation of trailer bed through pit at
30 Kmph

6.3 Von-Mises stresses in a trailer bed through pit at 30 Kmph

7. Conclusion

Calculations based on analytical work gives us the following results.
â€¢ No variation of acceleration in x direction i.e. along the length of the vehicle is observed.
â€¢ The acceleration in vertical upward direction is highest at the front right wheel through pit.
â€¢ The acceleration in the horizontal direction is highest at the back left wheel during turning.
â€¢ The center area is of minimum acceleration value which is called the best zone.
â€¢ The front right wheel area through pit and back left wheel during turning is the worst zone having maximum acceleration.
FEA gives the deformation and stresses of the trailer bed components for these values. The stress depends upon the force and the cross section area. The stresses will be more where the cross section area is less or the forces induced are more. The total deflection under the load is 0 to 11.35 mm. Which can be minimized by leaf spring which are not considered here. The maximum value of Von-Mises stresses are found to be 27.22 mm.

8. References

[1] R. Chakravaty, M.Tech. Project, Dynamic Forces
Due to Heavy Linkages Mounted on Vehicle in
Motion, VNIT Nagpur, 2005
DesCommVehChassBodyPart1990
[3] www.dot.state.mn.us/design/rdm/english/3e.pdf
[4] R.S. Khurmi & J.K. Gupta â€œTheory of Machineâ€,
S chand & company LTD, New Delhi, 1976,
p.p.226-229
[5] Teo Han Fui, Roslan Abd. Rehman, Static & Dynamic Structural Analysis of a 4.5 Ton Truck Chassis Dec-2007, No.24,56-57.
[6] S.S.Sane, Ghanashyam Jadhav and Anandaraj.H, Stress analysis of light commercial vehicle chassis by FEM.
[7] Paul Dvorak, Paul Kurowski, a closer look at motion analysis Jul-7-2005, London Ontario, Canada.

#### Venue

Digdao Hill, Hingna, MIDC Road, Nagpur