52
3 Investigation of Euler–Bernoulli Beams in the Elastic Range
Fig. 3.11 a Simply supported and b cantilevered Euler–Bernoulli beam loaded by a distributed
load
3.2 Solution
The problem of this example is very similar to the previous example Problem 3.1 and
major parts can be handled in a similar way.
(a) For the case of the simply supported beam we write—as in the case of example Problem 3.1—the finite difference approximation according to Eq. (3.9) at the
inner nodes i = 2, . . . , 4 under consideration of the distributed load as:
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = −q 0 ,
(3.61)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = −q 0 ,
(3.62)
node 4:
E I Y
X 3 (u 6 − 4u 5 + 6u 4 − 4u 3 + u 2 ) = −q 0 .
(3.63)
Consideration of the boundary conditions, i.e. u 1 = u 5 = 0, u 0 = −u 2 and u 6 =
−u 4 , results in the following system of equations:
E I Y
X 3
⎡
⎣
5 −4 1
−4 6 −4
1 −4 5
⎤
⎦
⎡
⎣
u 2
u 3
u 4
⎤
⎦ =
⎡
⎣
−q 0
−q 0
−q 0
⎤
⎦ .
(3.64)
The solution of this linear system of equations gives the unknown nodal values as:
u 2 = −
5q 0
4
2E I Y
, u 3 = −
7q 0
4
2E I Y
, u 4 = −
5q 0
4
2E I Y
,
(3.65)
or with =
L
4
as:
u 2 = −
5q 0 L
4
512E I Y
, u 3 = −
7q 0 L
4
512E I Y
, u 4 = −
5q 0 L
4
512E I Y
.
(3.66)
3 Investigation of Euler–Bernoulli Beams in the Elastic Range
Fig. 3.11 a Simply supported and b cantilevered Euler–Bernoulli beam loaded by a distributed
load
3.2 Solution
The problem of this example is very similar to the previous example Problem 3.1 and
major parts can be handled in a similar way.
(a) For the case of the simply supported beam we write—as in the case of example Problem 3.1—the finite difference approximation according to Eq. (3.9) at the
inner nodes i = 2, . . . , 4 under consideration of the distributed load as:
node 2:
E I Y
X 3 (u 4 − 4u 3 + 6u 2 − 4u 1 + u 0 ) = −q 0 ,
(3.61)
node 3:
E I Y
X 3 (u 5 − 4u 4 + 6u 3 − 4u 2 + u 1 ) = −q 0 ,
(3.62)
node 4:
E I Y
X 3 (u 6 − 4u 5 + 6u 4 − 4u 3 + u 2 ) = −q 0 .
(3.63)
Consideration of the boundary conditions, i.e. u 1 = u 5 = 0, u 0 = −u 2 and u 6 =
−u 4 , results in the following system of equations:
E I Y
X 3
⎡
⎣
5 −4 1
−4 6 −4
1 −4 5
⎤
⎦
⎡
⎣
u 2
u 3
u 4
⎤
⎦ =
⎡
⎣
−q 0
−q 0
−q 0
⎤
⎦ .
(3.64)
The solution of this linear system of equations gives the unknown nodal values as:
u 2 = −
5q 0
4
2E I Y
, u 3 = −
7q 0
4
2E I Y
, u 4 = −
5q 0
4
2E I Y
,
(3.65)
or with =
L
4
as:
u 2 = −
5q 0 L
4
512E I Y
, u 3 = −
7q 0 L
4
512E I Y
, u 4 = −
5q 0 L
4
512E I Y
.
(3.66)
