Computer Aided Engineering Design

318 COMPUTER AIDED ENGINEERING DESIGN

For convenience in assembly, the numbers assigned to the degrees of freedom are represented along
the rows and columns. For element 2, θ = 180°,l= 2 m, A = 20 × 10 –4 m^2 and E = 200 × 109 Nm^2.
Hence

12 3 4

ke^27 = 20 10

10 –10

00 00

–1 0 1 0

00 00

1

2

3

4

×

⎡

⎣

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

(11.8c)

For element 3, θ = 135°,l= 2√2 m, A = 50 × 10 –4 m^2 and E = 200 × 109 Nm–2. Thus

125 6

ke^3

8

= 10 10

42

1–1–1 1

–111–1

–111–1

1–1–1 1

1

2

5

6

×

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

(11.8d)

Combining the three local stiffness matrices appropriately to form the global stiffness matrix, we
have the force-displacement relation as

F F F F F F

u

u

x y x y x y

1 x

1

2

2

3

3

8

1

= 10

3.77 – 1.77 – 2 0 –1.77 1.77

–1.77 1.77 0 0 1.77 – 1.77

–202000

0 0 0 0.7 0 – 0.7

• 1.77 1.77 0 0 1.77 – 1.77
  1.77 – 1.77 0 – 0.7 – 1.77 2.47

⎧

⎨

⎪

⎪

⎪⎪

⎩

⎪

⎪

⎪

⎪

⎫

⎬

⎪

⎪

⎪⎪

⎭

⎪

⎪

⎪

⎪

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

11

2

2

3

3

y

x

y

x

y

u

u

u

u

⎧

⎨

⎪

⎪

⎪⎪

⎩

⎪

⎪

⎪

⎪

⎫

⎬

⎪

⎪

⎪⎪

⎭

⎪

⎪

⎪

⎪

(11.8e)

Noting that u 2 x = u 2 y = u 3 x = u 3 y = 0, we can discount the 3rd, 4th, 5th and 6th rows and columns
in the above system to compute only the unknown displacements. Thus

F

F

u

u

x

y

x

y

1

1

8 1

1

= 10

3.77 – 1.77

• 1.77 1.77

⎧

⎨

⎩

⎫

⎬

⎭

⎡

⎣

⎢

⎤

⎦

⎥

⎧

⎨

⎩

⎫
⎬
⎭
Substituting for F 1 x = 100 kN and F 1 y = –200 kN and solving for displacements, we get u 1 x=
–5× 10 –4m and u 1 y = –1.63× 10 –3 m. The reaction forces at the supports, that is, F 2 x,F 2 y,F 3 x and
F 3 y can all be computed by substituting for u 1 x and u 1 y in Eq. (11.8e). The final force vector can be
computed as [100 – 200 100 0 – 200 200]T with the last four entries representing the respective
reaction forces.

11.4 Beam Elements

A beam element shown in Figure 11.6 has four degrees of freedom, two at each node i and j,
respectively, and exhibits bending, that is, transverse deflections (vi and vj) and rotations (θi and θj)
under transverse loads (Fi and Fj) and end moments (Mi and Mj). Treating these degrees of freedom
as unknowns, a cubic interpolation function for transverse displacements v(x) at any point in the
beam may be assumed, that is