Dynamic Responses - Mechanics of Structures: Variational and Computational Methods - page 587

2

t 2

where p is a vector of applied loadings. The inertia force would be − ρ(∂ w/∂ ) =− ρ w ¨ ,

where is the deflection, so that the principle of virtual

work for a beam model with no shear deformation becomes

ρ is the mass per unit length and w

x δwρ x δw

2 EI 2

x

] a = − δ W =

x δw x ∂

∂ w dx + w ¨ dx − p z dx − [ M δθ + V δw 0 (10.7)

Introduce the shape functions for the static response of a beam [Chapter 4, Eq. (4.47)]
w = Nv i , where v i θ b ] T = [ w a θ a w b and

T

2 3

1 − 3 ξ + 2 ξ

2 3

( − ξ + 2 ξ − ξ )

N = (10.8)

2 3

3 ξ − 2 ξ

2 3

(ξ − ξ )

giving the principle of virtual work of Eq. (10.7) for M elements in the form

v iT N T p z dx

M

N ,xx EI N ,xx dx v i N T N dx v i

− δ W =

1 δ

+

x ρ

− = 0

(10.9)

x x

i =

k i v i m i v i p i

+ − = 0

x N 2 N

x 2 , and the boundary terms have been ignored. Substitute N where N ,xx = ∂ = ∂ /∂

= x ρ of Eq. (10.8) into m i N T N dx and carry out the integration. Then, the consistent mass

matrix for the beam element of Fig. 10.1 becomes

−

156 22 54 13

0 ρ

−

2

− −

2

= ρ
420

22 4 13 3

m i N T N dx

= (10.10a)

54 − 13 156 22

13 − 3

2

22 4

2

Note that this consistent mass matrix is a full symmetric matrix.
If axial motion of the beam element is considered, the nodal displacement vector is

b ] T and the shape function is developed in Chapter 4, Example 4.1.
The consistent mass matrix is

v = [ u a w θ

a u b

w θ

a b

140 0 0 70 0 0

0 156 − 22 0 54 13

0 ρ

2 2

0 − 22 4 0 − 13 − 3

= ρ
420

m i N T N dx = (10.10b)

70 0 0 140 0 0

0 54 − 13 0 156 22

2 2

0 3 − 3 0 2 4

FIGURE 10.1
A uniform beam segment with in-plane bending about the y axis. Sign Convention 2.