We consider a frame element with loads applied between the ends (Fig. 1.11).
Here, we rewrite Eq. (1.40) in the form
Note that Eq. (1.42) is a second-order equation and requires two boundary conditions, whereas Eq. (1.14) is a fourth-order equation requiring four boundary conditions. These (2+4=6) boundary conditions should be well posed. They are divided into static and kinematic boundary conditions.
The fixed-end displacements and rotation
at the end and at the beginning of the element are respectively
The fixed-end forces and the moments , at the end and at the beginning of the element, respectively, are shown in Eq. (1.52).
At unloaded joints (contacts) of elements, the sum of the fixed-end forces and moments is zero.
Sometimes the fixed-end forces and moments at joints are called internal reactions1.8 1.9 or joint contact forces1.10 [GN12]. We shall call them contact forces and contact moments.
A boundary value problem is a differential equation together with a set of boundary conditions. Solutions to the differential equations of a frame element are given with the basic system of equations (1.47). The boundary value problem should be well posed. The kinematic and static boundary conditions [KW90] of a frame element are shown in Figs. (1.10) and (1.12).