Using Newton’s 2nd law for the masses in Fig. 1.7(c) we find:
Àf 1 ¼ m 1 € x 1 ; Àf 2 ¼ m 2 € x 2 :
ð1:122Þ
Inserting (1.121)–(1.122) into (1.120) and rearranging, the equations of motion
take the form of (1.118):
l
3
486EI
8m 1 7m 2
7m 1 8m 2
! € x 1
€ x 2
& '
þ
x 1
x 2
& '
¼
0
0
& '
:
ð1:123Þ
1.8.3 The Stiffness Method
The stiffness method will produce at system in the form (1.117) with M = diag(m 1
m 2 ÁÁÁ m n ). The elements k ij of the stiffness matrix K are called stiffness influence
coefficients or just stiffness coefficients. The coefficients are determined this way:
k ij is the value of the force f i required to keep the system in a position where x j
= 1 while all the other x-values are zero.
To see this we write out the components of relation (1.115) for a system with
three degrees of freedom:
f 1
f 2
f 3
8
<
:
9
=
;
¼
k 11 k 12 k 13
k 21 k 22 k 23
k 31 k 32 k 33
2
4
3
5
x 1
x 2
x 3
8
<
:
9
=
;
¼
k 11 x 1 þ k 12 x 2 þ k 13 x 3
k 21 x 1 þ k 22 x 2 þ k 23 x 3
k 31 x 1 þ k 32 x 2 þ k 33 x 3
8
<
:
9
=
;
:
ð1:124Þ
As appears, a coefficient such as k 23 can be computed as the value of f 2 obtained
when letting x 1 = x 2 = 0 and x 3 = 1, because in that case the equation for f 2
becomes f 2 = k 23 , as the rule says.
The computation of stiffness coefficients is often convenient in cases where
flexible elements are connected in series, as in the below example.
Example The system in Fig. 1.8(a) has two degrees of freedom. Figure 1.8(b)
shows the system stripped from its masses, whose influence are replaced by forces
f 1 and f 2 . These same forces appear as reaction forces in Fig. 1.8(c), showing the
free-body diagrams of the masses. Using the stiffness method we start by writing
the linear force relation f = Kx component-wise:
f 1
f 2
& '
¼
k 11 x 1 þ k 12 x 2
k 21 x 1 þ k 22 x 2
&
'
:
ð1:125Þ
40
1 Vibration Basics
Àf 1 ¼ m 1 € x 1 ; Àf 2 ¼ m 2 € x 2 :
ð1:122Þ
Inserting (1.121)–(1.122) into (1.120) and rearranging, the equations of motion
take the form of (1.118):
l
3
486EI
8m 1 7m 2
7m 1 8m 2
! € x 1
€ x 2
& '
þ
x 1
x 2
& '
¼
0
0
& '
:
ð1:123Þ
1.8.3 The Stiffness Method
The stiffness method will produce at system in the form (1.117) with M = diag(m 1
m 2 ÁÁÁ m n ). The elements k ij of the stiffness matrix K are called stiffness influence
coefficients or just stiffness coefficients. The coefficients are determined this way:
k ij is the value of the force f i required to keep the system in a position where x j
= 1 while all the other x-values are zero.
To see this we write out the components of relation (1.115) for a system with
three degrees of freedom:
f 1
f 2
f 3
8
<
:
9
=
;
¼
k 11 k 12 k 13
k 21 k 22 k 23
k 31 k 32 k 33
2
4
3
5
x 1
x 2
x 3
8
<
:
9
=
;
¼
k 11 x 1 þ k 12 x 2 þ k 13 x 3
k 21 x 1 þ k 22 x 2 þ k 23 x 3
k 31 x 1 þ k 32 x 2 þ k 33 x 3
8
<
:
9
=
;
:
ð1:124Þ
As appears, a coefficient such as k 23 can be computed as the value of f 2 obtained
when letting x 1 = x 2 = 0 and x 3 = 1, because in that case the equation for f 2
becomes f 2 = k 23 , as the rule says.
The computation of stiffness coefficients is often convenient in cases where
flexible elements are connected in series, as in the below example.
Example The system in Fig. 1.8(a) has two degrees of freedom. Figure 1.8(b)
shows the system stripped from its masses, whose influence are replaced by forces
f 1 and f 2 . These same forces appear as reaction forces in Fig. 1.8(c), showing the
free-body diagrams of the masses. Using the stiffness method we start by writing
the linear force relation f = Kx component-wise:
f 1
f 2
& '
¼
k 11 x 1 þ k 12 x 2
k 21 x 1 þ k 22 x 2
&
'
:
ð1:125Þ
40
1 Vibration Basics