Automotive Technology - 4.1 Simple Structural Surfaces - continued
4. Example
The most simple type of structure is a closed box, such as a freight container. This
can be modelled by 6 SSSs (for clarity only 1 side panel is shown below) however in the example two additional (equi-spaced)
SSSs are included to represent cross beams to distribute the load over the floor, see diagram below.
The load (P) is assumed to be equally split between the two cross beams which are supported at their ends
by the sidewalls of the container - Equal and opposite reactions, K1, are present in the
container sidewalls acting in their planes and from symmetry: K1 = P/4.
The sidewalls are held in equilibrium by forces acting on their front and rear edges, carried by
the front and rear panels and it is assumed that the container is supported at the lower
edges of these panels by suspension elements carrying loads: Rf and Rr.
Taking moments about the front lower corner: K3 = ( K1 L/3 + K1 2L/3 )/L = K1
Resolving vertically the forces acting on the sidewall gives K2 + K3 - 2 K1 = 0.
Where K2 is the force between the sidewall and the front panel.
It is assumed that the load is symmetrical so: Rf = Rr = K1 = P/4.
In the absence of any external torsional load applied to the container, there are no forces
acting in the plane of the roof or in the plane of the floor.
The next stage of an analysis is to ignore the internal load in the container and assess what
occurs when an external torsional load is applied, such as when one wheel goes over a bump. The
simplest way of doing this is to apply an upward force to one wheel (or to the point in the
structure taking the vertical suspension load) and an equal downward force
to the other wheel on the same axle. This causes most of the panels to be loaded in shear, see
diagram below:
The moment, Mx = r Rr applied to the rear panel, is balanced by
edge loads from the roof, floor and side panels.
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David Grieve, 13th December 2001.