98
4 Investigation of Timoshenko Beams in the Elastic Range
Considering 17 domain nodes, a linear system of dimension 30 × 30 can be derived
based on the 15 inner nodes. The vertical displacement is now as
u 9 = u
L
2
= −
512E I Y + 53L
2 k s AG
L
2 q 0
4k s G A
1024E I Y + L 2 k s AG
(4.31)
obtained and the relative error decreases to −66.348%. As can be seen from this
result, a larger number of nodes is required to achieve an acceptable error in the case
of Timoshenko beams.
(b) Single force case
In order to account for single forces which are acting on the structure, the system
of equations given in Eqs. (4.17) and (4.18) must be multiplied by X to obtain on
the right-hand side the expression for the equivalent nodal force. Since this example involves only single forces,
3 it would be sufficient to multiply only the second
equation by X . Thus, the linear system of equations is obtained for this case as:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
−
2E I Y
3X 2 − k s AG −
k s AG
2X
2E I Y
3X 2
0
0
−
2k s AG
X
−
2k s AG
3
k s AG
X
2k s AG
3
0
0
k s AG
2X
E I Y
X 2
0
−
2E I Y
X 2 − k s AG −
k s AG
2X
E I Y
X 2
k s AG
X
−
k s AG
2
−
2k s AG
X
0
k s AG
X
+
k s AG
2
0
0
k s AG
2X
2E I Y
3X 2
0
−
2E I Y
3X 2 − k s AG
0
0
k s AG
X
−
2k s AG
3
−
2k s AG
X
2k s AG
3
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
u 2
φ 2
u 3
φ 3
u 4
φ 4
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0
0
0
F 0
0
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
.
(4.32)
The vertical displacement is for the single load case obtained as
u 3 = u
L
2
= −
32EY Z + 3L
2 k s AG
L F 0
2k s AG
64E I Y + L 2 k s AG
(4.33)
3 In the case of single moments, Eq. (4.17) should be multiplied by X .