Computer Aided Engineering Design

320 COMPUTER AIDED ENGINEERING DESIGN

B = – 3

y

l

[12x – 6ll(6x – 4l) – (12x – 6l) l(6x – 2l)]

From truss analysis, the local stiffness matrix for a beam element may directly be written as

kBB =

V

TEdV

∫

=

12 – 6

(6 – 4 )

–(12 – 6 )

(6 – 2 )

2

6

0

E

ydA

l

xl

lx l

xl

lx l

A

l

∫∫

⎡

⎣

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

[12x – 6ll(6x – 4l) –(12x – 6l) l(6x – 2l)]dx

=

36 (4 – 4 + ) 12 (6 – 7 + 2 ) – 36 (4 – 4 + ) 12 (6 – 5 + )

12 (6 – 7 + 2 ) 4 (9 – 12 + 4 ) – 12 (6 – 7 + 2 )

6

0

22 2 2 22 22

EI 2222 2 22

l

x xl l l x xl l x xl l l x xl l

l l x xl l l x xl l l x xl l

∫

4 4(9 – 9 + 2)

• 36 (4 – 4 + ) – 12 (6 – 7 + 2 ) 36 (4 – 4 + ) – 12 (6 – 5 + )
  12 (6 – 5 + ) 4 (9 – 9 + 2 ) – 12 (6

22 2

22 2 2 22 22

2222 2 2

l x xl l

x xl l l x xl l x xl l l x xl l

lx xl l l x xl l l x–– 5 + ) 4 (9xl l^222 l x – 6 + )xl l^2

dx

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ =

12 6 –12 6

64 –62

–12 – 6 12 –6

62 –64

3

22

22

EI

l

ll

ll ll

ll

ll ll

⎡

⎣

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

(11.9g)

whereI is the second moment of area of the cross-section given as IydA = ∫A^2. The assembly
procedure for beam elements is similar to that in truss elements with the difference that two degrees
of freedom are considered in the local coordinate system per node. Note that the stiffness assembly
is based on: (a) the interelement continuity of primary variables (deflection and slope) and (b) the
interelement equilibrium of secondary variables (shear forces and bending moments) at nodes common
to the elements.

Example 11.2. Given is a composite beam with varying cross sections as shown in Figure 11.8 with
external loads and displacement boundary conditions. Solve for transverse deflections and slopes and
compare with the analytical result.
For the three elements, there are four nodes and two degrees of freedom per node. The global
displacement vector U is such that for node j,U(2j – 1) = vj and U(2j) = θj. The global stiffness
matrix is of size 8 × 8 which is determined as follows:
For elements 1, 2 and 3, using Eq. (11.9g), the stiffness matrices are

1 2 3 4 34 56 5 6 7 8

kkk 1

6

2

6

3

=^106

1

12 6 –12 6

64–62

–12–6 12–6

62–64

1

2

3

4

= 10

63–63

32–3 1

–6 –3 6 –3

31–32

3

4

5

6

= 10

12 3 – 12 3

3 1 – 3 0.5

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

• –12–3 12–3

3 0.5 – 3 1

5

6

7

8

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥