Computer Aided Engineering Design

FINITE ELEMENT METHOD 319

v(x) = a 1 + a 2 x + a 3 x^2 + a 4 x^3 (11.9a)

where constants a 1 ,... , a 4 can be determined using the conditions

v(x)=vi and

dx

dx

v()

= θiatx = 0 ,

v(x)=vj and

dx

dx

v()

= θjatx = l (11.9b)

Solving and rearranging yields

vvv() =^2 –^3 + 1 + –^2 + +^3 –^2 + –

3

3

2

2

3

2

22

2

3

3

3

2

2

x x

l

x

l

x

l

x

l

x x

l

x

l

x

l

x

iijjl

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

θθ

or v(x) = N 1 (x)vi + N 2 (x)θi + N 3 (x)vj + N 4 (x)θj

or v(x) = [N 1 (x) N 2 (x) N 3 (x) N 4 (x)][viθivjθj]T = Nv (11.9c)

withN as the shape function matrix and v the displacement vector. From linear beam theory, plane

y

νi, Fi

θi, Mi

θj, Mj

νj, Fj

x

l

Figure 11.6 A beam element

u

y Neutral axis

∂

∂

ν()x

x

Figure 11.7 Axial displacement u in the beam

cross-sections remain plane after deformation and
hence the axial displacement u due to transverse
displacementv can be expressed as (Figure 11.7)

uy

x

= – ∂

∂

v (11.9d)

wherey is the distance from the neutral axis. The
axial strain is given by

εx u

x

y

x

= = – =

2

2

∂

∂

∂

∂

v Bv (11.9e)

whereB is the strain displacement matrix. From
Eq. (11.9c), we note

∂

∂

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

2

232 = 2 23 2

v (^12) – (^6) vv + (^6) – (^4) + (^6) – (^12) + (^6) – 2
x
x
ll
x
l l l
x
l
x
l l
iiθθj j(11.9f)
and thus comparing with Eq. (11.9e) yields