10.3 Methods Analogous to Classical Methods in Statical Analysis
401
Equation (10.33) is the three-moment equation for the freely vibrating beam. In
addition to the three bending moments at the three supports, it contains additional
unknown λL associated with α and β terms.
Example 10.1 Determine the frequency equation of the continuous beam by using
three–moment theorem.
Three supports of Fig. 10.1 are labelled as 1, 2 and 3 from the left.
We have
M 1 = M 3 = 0
( a )
Three-moment equation for this problem reduces to
M 2 α 2 + M 2 α 3 = 0
( b )
Substituting α 2 and α 3 into Eq. (b), we get
M 2 L 2 θ 2
3E 2 I 2
+
M 2 L 3 θ 3
3E 3 I 3
= 0
( c )
Now, E 2 I 2 = E 3 I 3 ; therefore, Eq. (c) becomes
L 2 θ 2 + L 3 θ 3 = 0
( d )
Substituting θ 2 and θ 3 from Eqs. (10.25)–(d), we get
cot λL 2 − coth λL 2 = coth λL 3 − cot λL 3
(e)
Equation (e) agrees with Eq. (10.18).
Indeterminate frame problems can be formulated by other classical methods as
well. However, we pass on to more formalised approach in the next section.
10.4 Dynamic Stiffness Matrix in Bending
From the discussion of two-span continuous beam in the previous section, it is
evident that as the number of spans goes on increasing, the number of constants
to be evaluated also increases. As such, the problem tends to become more cumbersome. Dynamic stiffness matrix approach, which has a close resemblance with static
direct stiffness matrix method, forms a more elegant method to the solution of these
problems.
A beam segment with the prescribed displacements at its ends is shown in
Fig. 10.4. The end displacements vary harmonically with time, frequency and phase
difference, for all are same. As such, the end moments and shears also vary at the
401
Equation (10.33) is the three-moment equation for the freely vibrating beam. In
addition to the three bending moments at the three supports, it contains additional
unknown λL associated with α and β terms.
Example 10.1 Determine the frequency equation of the continuous beam by using
three–moment theorem.
Three supports of Fig. 10.1 are labelled as 1, 2 and 3 from the left.
We have
M 1 = M 3 = 0
( a )
Three-moment equation for this problem reduces to
M 2 α 2 + M 2 α 3 = 0
( b )
Substituting α 2 and α 3 into Eq. (b), we get
M 2 L 2 θ 2
3E 2 I 2
+
M 2 L 3 θ 3
3E 3 I 3
= 0
( c )
Now, E 2 I 2 = E 3 I 3 ; therefore, Eq. (c) becomes
L 2 θ 2 + L 3 θ 3 = 0
( d )
Substituting θ 2 and θ 3 from Eqs. (10.25)–(d), we get
cot λL 2 − coth λL 2 = coth λL 3 − cot λL 3
(e)
Equation (e) agrees with Eq. (10.18).
Indeterminate frame problems can be formulated by other classical methods as
well. However, we pass on to more formalised approach in the next section.
10.4 Dynamic Stiffness Matrix in Bending
From the discussion of two-span continuous beam in the previous section, it is
evident that as the number of spans goes on increasing, the number of constants
to be evaluated also increases. As such, the problem tends to become more cumbersome. Dynamic stiffness matrix approach, which has a close resemblance with static
direct stiffness matrix method, forms a more elegant method to the solution of these
problems.
A beam segment with the prescribed displacements at its ends is shown in
Fig. 10.4. The end displacements vary harmonically with time, frequency and phase
difference, for all are same. As such, the end moments and shears also vary at the