146
6 Answers to Supplementary Problems
0 0 0 0
F 0 X
3
2F
3
T .
(6.188)
The value of the deflection at the right-hand end can be obtained from the solution
of the linear system of equations as:
u(X = L) = −
u 3
3
+
4u 4
3
−
2X φ 3
9
+
8X φ 4
9
+
2X Q 5
3k s AG
.
(6.189)
(b) Solutions for dim. = 2 × (domain nodes − 1).
The fictitious node at the right-hand boundary can be eliminated for example in the
case of five domain nodes by the kinematic relationship
φ 5 = −
du Z (L)
dX
+ γ X Z (L) ≈ −
u 6 − u 4
2X
+
τ X Z (L)
G
= −
u 6 − u 4
2X
+
Q Z (L)
k s AG
,
(6.190)
or
u 6 = u 4 − φ 5 2X +
Q Z (L)
k s AG
.
(6.191)
The condition for the bending moment at the free boundary gives: φ 6 = φ 4 .
In the case of this approach, the system of equations given in Eq. (6.187) is extended
by two columns and rows whereas the last two rows reads as:
⎡
⎢
⎣
. . . 0
E I Y
X
0
−
2E I Y
X
. . .
2k s AG
X
0 −
2k s AG
X
−2k s AG
⎤
⎥
⎦
⎡
⎢
⎢
⎣
u 5
φ 5
⎤
⎥
⎥
⎦ =
⎡
⎢
⎣
0
q 0 X
2
⎤
⎥
⎦ ,
(6.192)
or in the case of the single force as:
⎡
⎢
⎣
. . . 0
E I Y
X
0
−
2E I Y
X
. . .
2k s AG
X
0 −
2k s AG
X
−2k s AG
⎤
⎥
⎦
⎡
⎢
⎢
⎣
u 5
φ 5
⎤
⎥
⎥
⎦ =
⎡
⎢
⎢
⎣
−Q Z (L))X
F 0 − 2Q Z (L)
⎤
⎥
⎥
⎦ .
(6.193)
The convergence rate is shown in Fig. 6.13.
Let us finally remind that the equilibrium between the internal reaction (shear
force) and external load at the right boundary gives: Q Z (L) = −F 0 .
u Z (L) = −
224E I Y + 55k s AG L
2
L
2 q 0
8k s AG
64E I Y + k s AG L 2
(5 nodes) ,
(6.194)