5003 Lectures - Faculty of Engineering and Applied Science

E5003 - Ship Structures I 111
© C.G. Daley
When using the stiffness method, we always need to find a set of forces and
moments that occur when we impose a movement at a support. The
movement will correct a situation that involved the suppression of a
movement at a support. In our case here, the structure is a beam, and the
supports are at the ends of the beam. The supports prevent the ends of the
beam from moving. There are 3 possible movements at a support for a 2D
problem, and 6 for a 3D problem. Because of this we will define a standard
set of ‘degrees of freedom’ for a beam. A ‘degree of freedom’ can have either a
force or displacement, or a rotation or moment. The standard 2D degrees of
freedom for a beam are shown below;
The degrees of freedom follow the Cartesian system, with the right-hand rule.
These are essentially x, y, rotation (called rz). In general, to impose a unit
movement in one (and only one) of these degrees of freedom, we need to also
impose a set of forces/moments, The forces/moments must be in equilibrium.
These forces/moments will be ‘stiffnesses’.
The mechanics are linear. This means that the set of forces/moments
corresponding to each movement can be added to those of any other
movement. A general solution for any set of movements of the degrees of
freedom can be found by superposition.
For now we will just consider the 2D case and derive the stiffness terms.
There are 6 degrees of freedom. For each degree of freedom, there are
potentially 6 forces or moments that develop. This means that there are a
total of 36 stiffness terms. Any single term would be labeled kij, meaning the
force/moment at i due to a displacement/rotation at j. For example;
k11 = force at 1 due to unit displacement at 1
k41 = moment at 4 due to unit displacement at 1
k26 = force at 2 due to unit rotation at 6
All the terms can be written in matrix form as;
2D beam = 6 degrees of freedom