Page 336 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

Sec. 10.5 Vibrations Involving Beam Elements 323

Solution: Antisymmetric Mode. The deflection and slope of stations 1 and 2 are identical,
i.e., W = ¿¡2 ^1 = ^2- These conditions are imposed on the preceding equation by
j
adding column 3 to column 1 and column 4 to column 2. This results in identical
equations for { - and

(o^ml '366 221' El '12 6/1
420 22/ 5/^ . 6/ 10/q.

By letting A = w^ml^/A20EI, the determinant^ of this equation
(12 - 366A) (6 - 22A)/
= 0
(6 - 22A)/ (10 - 5A)/2
results in two roots:

A, = 0.0245 = 3.21

2.543 = 32.68

The lowest natural frequency corresponds to the simple shape displayed in Fig.
10.5-4(a) and is of acceptable accuracy. However, the second antisymmetric mode
would be of more complex shape and a>2 computed with the few stations used in this
example would not be accurate. Several more stations would be necessary to ade­
quately represent the higher modes.
j
Symmetric Mode: For the symmetric mode, we have W = «2 = 0 and 62 = —
Deleting columns 1 and 3 and subtracting column 4 from column 2, we obtain just one
equation for 6{.
^ml (11/2) ^ :^((,/2)
420 = 0

A and o) are then
15.14i

Example 10.5-5
Figure 10.5-5 shows external loads acting on the portal frame. Examine the boundary
condition and determine the stiffness matrix in terms of the coordinates given.
Solution: The condition of no extension of members = Uj is satisfied by adding column
3 to column 1 in Eq. (c). Example 10.5-3. This eliminates the extension term R. We

^When the determinant is multiplied out, E becomes a factor that cancels out. Thus, we can let
^1.0 in the matrices of the frequency equation without altering the values of A, and A2.