SMC 4133 AUTOMOTIVE STRUCTURES DESIGN FOR BODY TORSION Body torsion strength requirement The body has to recover its shape with little to no permanent

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Body torsion strength requirement•The body has to recover its shape with little to no permanent deformation during twist ditch maneuver•The twist ditch torque can be obtained by multiplying axle load (W) by half of the wheel track (t).•The angle of twist can be determined by 2 x deflection divided by width of theloaded points (w)

All materials in this slide are taken from Donald E Malen. 2011. Fundamentals of Automobile Body Structure Design, SAE International.



Torsion stiffness requirement:1.To ensure good handling properties2.To ensure a solid structural feel and minimize relative deformations – squeaks & rattles

-As a vehicle turns a corner, it will roll and causes a weight transfer. It then can affect steering characteristics-High body torsional stiffness is required to ensure good vehicle handling-Typical roll stiffness is 1000 Nm/deg while ride spring rate = 23.4 N/mm



DESIGN FOR BODY TORSION- Let’s view the stiffness system as a

series connection of springs- Keff/Kroll = 1.0- Kbody = 10 Kroll - Kbody = 10000 Nm/deg for good handling

For good solid structure feel:-Vehicle torsional frequency from 22-25 Hz-Torsional stiffness = 12000 Nm/deg-Torsion strength = 6250 Nm



DESIGN FOR BODY TORSIONLoad Path Analysis-To determine loads on individual structure elements-With these loads those elements can be designed

Let’s begin with a simple structure i.e.a closed box.

The box is loaded by a twisting coupleat the front and rear corners

All panels are loaded




- Edge loads & shear flows can be calculated

-AQ = T-A is a coefficient matrix-Q is an edge load matrix-T is an applied torque matrix

Shear flow, q = Q/L (N/m)




Example 1Determine the edge loads for the torsion case



DESIGN FOR BODY TORSIONExample 2




Example 2Determine the edge loads for the given torsion load case



DESIGN FOR BODY TORSIONAnalysis of body torsional stiffness: Closed box

- Energy method will be used to predict torsional stiffness by taking into account panel dimensions, thicknesses and material properties



DESIGN FOR BODY TORSIONEffective shear rigidity

- to predict realistic torsional stiffness where in reality the body panels differ considerably from an ideal flat plate

- Typically, the body panels are crown shape, have holes, cut-outs and framework with flexible joints




Example 3Determine torsional stiffness of a box van based on:a)Given shear rigidityb)Effective shear rigidity: rear hatch opening

Data: w = 1400mm, h = 1250mm, L = 2000mm, G = 80000N/mm^2, t = 1mm

Solution:a)K = (2x1400x1250)^2 x (1/(2x(21.9+35+31.3)) = 6.95E+10 Nmm/rad = 1.22E+6 Nm/degree

b) Work done = Energy in the joints ½ x F x delta = 4 x ½ x Kj x theta^2 theta = delta/b, S = 4Kj/b^2, Gt = 4Kj /ab Given Kj = 0.1E+8Nmm/rada)K = (2x1400x1250)^2 x (1/(21.9+35+35+31.3+31.3+76553)) = 1.6E+8 Nmm/rad = 2807Nm/degree



DESIGN FOR BODY TORSIONAnalysis of body torsional stiffness: Sedan

Gt = (Q/delta) x (H/L)

Delta is obtained from FEA



DESIGN FOR BODY TORSIONExample 4

From Example 2, determine the cabin torsional stiffness with side-frame.

q = 2678/1250 = 2.1414N/mmq/T = 2.77E-7 mm^-2Let Q/delta = 374.5 N/mm, Gt7-8 = 374.5x1250/2000 =234N/mm (side frame)A1=A5=1170000mm^2, A2=1103087mm^2, A3=1950000mm^2, A4=872067mm^2A6=3120000mm^2, A7=A8=2312500mm^2Gt 1-6 = 80000 N/mmThus, K = 6.55E+ 8 Nmm/rad = 11491 Nm/degree


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SMC 4133 AUTOMOTIVE STRUCTURES DESIGN FOR BODY TORSION Body torsion strength requirement The body has to recover its shape with little to no permanent