6.3 Answers for Problems from Chap. 3
125
6.3 Answers for Problems from Chap. 3
3.7 Finite difference approximation of a simply supported and cantilevered
beam based on three domain nodes
The single force F 0 is understood as the integral value of a distributed load q 0 , which
is acting over a length of (a) X or (b) X/2.
(a) Simply supported beam:
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = −F 0 .
(6.56)
Boundary conditions: u(X = 0) = u 1 = 0, u(X =L)=u 3 = 0, M Y (X = 0) = 0 →
u 0 = −u 2 , M Y (X = L) = 0 → u 4 = −u 2 .
u 2 = −
X
3 F 0
4E I Y
= −
1
32
0.03125
F 0 L
3
E I Y
.
(6.57)
Relative error to analytical solution: 50%.
(b) Cantilevered beam:
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = 0 ,
(6.58)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = −F 0 .
(6.59)
Boundary conditions: u(X = 0)=u 1 =0,
du 1
dX
= 0 → u 0 = u 2 , E I Y
d
2 u
dX 2
3
= −M Y =
0 → u 4 = −u 2 + 2u 3 , E I Y
d
3 u
dX 3
3
= 0 → u 5 = 4u 3 − 4u 2 + 0.
Linear system of equations:
E I Y
X 3
3 −1
−4 2
u 2
u 3
=
0
−F 0
.
(6.60)
Solution:
u 2 = −
1
2
X
3 F 0
E I Y
= −
1
16
F 0 L
3
E I Y
,
(6.61)
u 3 = −
3
2
X
3 F 0
E I Y
= −
3
16
0.1875
F 0 L
3
E I Y
.
(6.62)
Relative error at node 3 compared to analytical solution: 43.75%.