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 the

loaded points (w)

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Torsion stiffness requirement:

1. To ensure good handling properties

2. 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

SMC 4133 AUTOMOTIVE STRUCTURES

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

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Load 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 couple

at the front and rear corners

All panels are loaded

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

- 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)

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Example 1

Determine the edge loads for the torsion case

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Example 2

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Example 2

Determine the edge loads for the given torsion load case

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Analysis 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

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Effective 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

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Example 3

Determine torsional stiffness of a box van based on:

a) Given shear rigidity

b) 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/rad

a) K = (2x1400x1250)^2 x (1/(21.9+35+35+31.3+31.3+76553))

= 1.6E+8 Nmm/rad = 2807Nm/degree

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Analysis of body torsional stiffness: Sedan

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

Delta is obtained from

FEA

SMC 4133 AUTOMOTIVE STRUCTURES

DESIGN FOR BODY TORSION

Example 4

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

q = 2678/1250 = 2.1414N/mm

q/T = 2.77E-7 mm^-2

Let 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^2

A6=3120000mm^2, A7=A8=2312500mm^2

Gt 1-6 = 80000 N/mm

Thus, K = 6.55E+ 8 Nmm/rad = 11491 Nm/degree