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