Computer (matrix) version of the stiffness method 1. The computer ...

Reactions number 1 and 4 created by the action of displacements number 1 and 4 are obtained in
the same way as in the case of the plane truss element. Note, that these displacements do not
yield any other reactions. Also, these reactions are zero under the action of all the other
displacements. Hence, in the classical plane beam element the action of axial forces and
displacements is fully decoupled from the combined action of shear and bending. The reactions
due to the transverse displacements and cross-section rotations can be found using the slopedeflection
formulae from the classical stiffness method. This yields:
~
R
3
~
R
6
~
R
2
~
R
5
2EI
= M =
ψ
l
2EI
⎛
l ⎝
6EI
4EI
l
6EI
5 2
( 2ϕ
)
~ ~
~ ~ ~ ~
i + ϕk
− 3 ik = ⎜2q3
+ q6
− 3 ⎟ = q2
+ q3
− q5
q6
ik +
2
2
2EI
= M =
ψ
l
2EI
⎛
l ⎝
q
~
− q
~
⎞
l ⎠
l
6EI
2EI
l
l
6EI
2EI
l
5 2
( 2ϕ
)
~ ~
~ ~ ~ ~
k + ϕi
− 3 ik = ⎜2q6
+ q3
− 3 ⎟ = q2
+ q3
− q5
q6
ki +
2
2
6EI
= −T
=
ψ
l
6EI
⎛
q
~
− q
~
⎞
l ⎠
l
12EI
6EI
l
12EI
4EI
l
5 2
( ϕ
)
~ ~
~ ~ ~ ~
i + ϕk
− 2 ik = ⎜q3
+ q6
− 2 ⎟ = q2
+ q3
− q5
q6
ik +
2
2
3
2
3
6EI
6EI
⎛
= T ( )
~
ki = − ϕ
2 i + ϕk
− 2ψ
ik = − ⎜q
2
l
l ⎝
12EI
~ 6EI
~ 12EI
~ 6EI
= − q
~
3 2 − q
2 3 + q
3 5 − q
2
l l l l
l
⎝
3
6
+ q
~
6
q
~
q
~
− 2
− q
~
⎞
l ⎠
Summarising these calculations yield the following stiffness matrix for the plane beam element
~
K
e
⎡
⎢
⎢
⎢ 0
⎢
⎢
⎢ 0
= ⎢
⎢−
⎢
⎢
⎢
⎢
⎢
⎣
EA
l
EA
l
0
0
12EI
l
6EI
l
12EI
−
3
l
6EI
l
0
3
2
0
2
0
6EI
2
l
4EI
l
0
6EI
−
2
l
2EI
l
5
− q
~
l
−
2
EA
l
0
0
EA
l
0
0
⎞
⎟
⎠
=
l
0
12EI
−
3
l
6EI
−
2
l
0
12EI
3
l
6EI
−
2
l
l
⎤
0 ⎥
6EI
⎥
⎥
2
l ⎥
2EI
⎥
l ⎥
⎥
0 ⎥
⎥
6EI
⎥
−
2
l
⎥
4EI
⎥
⎥
l ⎦
It is worth to note, that the element stiffness matrices are symmetric. This is the consequence of
the reactions reciprocity theorem (Rayleigh’s theorem) stating that
k ij = k ji
The physical interpretation of the entries in this matrix is unchanged, i.e. for instance the second
column collects the values of reactions created by the action of the displacement q ~
1 .
After the derivation of these two examples of the element stiffness matrices let us now deal with
the loading. Generally, the loading is split into actions directly at the nodes and the actions along
the elements. The former one will be considered at the later stages. The latter one must be
transferred to the nodes in the form of appropriate reactions. They are assembled into a vector of
reactions due to the span loading
~
R
~
[ R0
i
] 1,2,..., 6
0 e =
6× 1
i =
This vector enters the equation of element equilibrium as an additional term
l
2 =
6EI
l
2