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

The frame consists of four beam elements which are interconnected in five nodes. The element
stiffness matrices and vectors of nodal reactions due to the span loading are found in the local coordinates
for each element
1 2 1 1
1
1
1
x ~
2
y ~ x ~
3
y ~
4
y ~ x ~
y ~ x ~
2
2
are transformed to the global co-ordinates
1 2 2 3
1 x
2
y
y
and they can be represented as
x
⎡K11,1
K12,1
⎤ ⎡K11,2
K12,2
⎤ ⎡K11,3
K12,3
⎤ ⎡K11,3
K12,3
⎤
K 1 = ⎢ ⎥ K 2 = ⎢
⎥ K 3 = ⎢
⎥ K 1 = ⎢
⎥
⎣K
21,1 K22,1⎦
⎣K
21,2 K22,2
⎦ ⎣K
21,3 K22,3
⎦ ⎣K
21,3 K22,3
⎦
Note, that the local element numbers of element nodes can be replaced with the global numbers:
Element 1: 1 – 1, 2 – 2, element 2: 1 – 2, 2 – 3, element 3: 1 – 2, 2 – 4, element 4: 1 – 3, 2 – 5.
These relations enable a proper “addressing” of the subsequent submatrices of the element
stiffness matrices K e in the global stiffness matrix of the frame K. The assembly scheme is
2
4
3
y
x
3
y
4
5
x
K =
K 11,1 K 12,1
K 21,1
K 22,1+K 11,2+
+K 11,3
K 12,2
K 12,3
K 21,2 K 22,2 +K 11,4 K 12,4
K 22,3
K 21,3
K 21,4 K 22,4
Note, that in this scheme each box corresponds to a node and it represents 3×3 elements of the
global stiffness matrix corresponding to three displacements at the given node.
Let us also consider the assembly of the global force vector P.
Out of four beams constituting the frame only one – No. 4 has a span loading and consequently
the non-zero vector of reactions due to the loading. These vectors for the other beams are zero:
~ ~ ~
R R = R = R = R = R = 0
01 = 02 03 01 02 03
To calculate the vector for the fourth element the loading can be split into uniformly distributed load
acting along the beam and transverse to the beam
q
q 1 l
q 2
2 l
2
12
q q 2 l
1 = qsinϕcosϕ
q 2 qcos 2 ϕ
l
=
+ 2
l
ϕ
l 2 l
q 1 l
2 2
2 l
2
12