10.4 Dynamic Stiffness Matrix in Bending
407
Assembling two element stiffness matrices, the overall stiffness matrix is
[K (λ)] = 8.28 × 10
7
⎡
⎣
0.6094 0.3634 0
0.3634 0.8448 0.3815
0 0.3815 0.2354
⎤
⎦
Now, the determinant of the overall stiffness matrix is
|K (λ)| = 1.166 × 10
5
Next, choose b 1 = 2.32 and following identical steps as above, we obtain the
element stiffness matrix of the two spans as
[K (λ)] 1 =
8.28 × 10
7
6.1
3.7094 2.2221
2.2221 3.7094
and
[K (λ)] 2 =
8.28 × 10
7
9.15
2.1093 3.5289
3.5289 2.1093
The overall stiffness matrix resulting from the assembly of the beam element
becomes
[K (λ)] = 8.28 × 10
7
⎡
⎣
0.6081 0.3643 0
0.3643 0.8386 0.3857
0 0.3857 0.2305
⎤
⎦
The determinant of the overall stiffness matrix is
|K (λ)| = −2.9065 × 10
5
The value of b 1 for which the determinant will be zero can be obtained by linear
interpolation, that is
b 1 = 2.31 + 0.01
1.166
1.166 + 2.9065
= 2.313
or
L 1
4
mp 2
E I
= 2.313
or
407
Assembling two element stiffness matrices, the overall stiffness matrix is
[K (λ)] = 8.28 × 10
7
⎡
⎣
0.6094 0.3634 0
0.3634 0.8448 0.3815
0 0.3815 0.2354
⎤
⎦
Now, the determinant of the overall stiffness matrix is
|K (λ)| = 1.166 × 10
5
Next, choose b 1 = 2.32 and following identical steps as above, we obtain the
element stiffness matrix of the two spans as
[K (λ)] 1 =
8.28 × 10
7
6.1
3.7094 2.2221
2.2221 3.7094
and
[K (λ)] 2 =
8.28 × 10
7
9.15
2.1093 3.5289
3.5289 2.1093
The overall stiffness matrix resulting from the assembly of the beam element
becomes
[K (λ)] = 8.28 × 10
7
⎡
⎣
0.6081 0.3643 0
0.3643 0.8386 0.3857
0 0.3857 0.2305
⎤
⎦
The determinant of the overall stiffness matrix is
|K (λ)| = −2.9065 × 10
5
The value of b 1 for which the determinant will be zero can be obtained by linear
interpolation, that is
b 1 = 2.31 + 0.01
1.166
1.166 + 2.9065
= 2.313
or
L 1
4
mp 2
E I
= 2.313
or