Advanced Structural Analysis
Finite Element Method
Beam Element
Beam element:
Dy1
1
Dy2
q2
q1
2
We will go through the 7 steps for a bar element that is subject to bending and
shear only – that is a member with the stiffness equation,
é V1 ù é 12 EI / L3
êM ú ê
2
6
EI
/
L
1
ê ú=ê
ê V2 ú ê- 12EI / L3
ê ú ê
2
ëM 2 û êë 6EI / L
6EI / L2
4EI / L
- 6EI / L2
2EI / L
- 12EI / L3
- 6EI / L2
12 EI / L3
- 6EI / L2
6EI / L2 ù é D1 ù
úê ú
2EI / L ú ê q1 ú
- 6EI / L2 ú êD 2 ú
úê ú
4EI / L úû ë q 2 û
FEM for beam element
The steps in the process
1. Identify the problem geometry
2. Select a displacement function
3. Relate member displacements to the node displacements – N matrix
4. Write the strain-displacement relationships – B matrix
5. Write the stress-strain relationships
6. Relate node loads to node displacements – K matrix and force vector
7. Post- processing
Before we start,
Equation relating the forces to the displacements was derived as:
2
d y
M = EI K = EI 2
dx
Before we start,
Equation relating the forces to the displacements was derived in the ENG 1020 as:
differentiate
load
differentiate
shear force
integrate
bending moment
integrate
From the principles of geometry we know that the relationship between curvature, slope and deflection is:
differentiate
deflection
integrate
differentiate
slope
curvature
integrate
load
deflection
shear force
slope
bending moment
M = EIK
integrate integrate
integrate integrate
Linking these two together enables us to calculate the deflection of a beam when we know the load,
curvature
Step 1 - Identify the problem geometry
Dy1
1
The force matrix
Dy2
q2
q1
é Fy 1 ù
êM ú
is, [F ] = êê 1 úú
Fy 2
ê
ú
ëM2 û
2
and the displacement matrix is
é Δy 1 ù
êθ ú
[U] = ê 1 ú
êΔy 2 ú
ê
ú
ë θ2 û
Using the local coordinates we can write the member stiffness equation in
general form as,
[F] = [K][U]
(A)
Step 2 - Select a displacement function
To describe the displacement of the member between nodes choose the
following polynomial displacement function:
v = a1 + a2 x + a3 x 2 + a4 x 3
1
v
x
2
v = α1 + α 2 x + α 3 x 2 + α 4 x 3
θ=
dv
= α 2 + 2α 3 x + 3α 4 x 2
dx
The displacement function can be written in more general terms as,
[u(x,y)] = [f(x,y)][a]
or in matrix form (for this example) as,
é α1 ù
ê ú
é v ù é 1 x x 2 x 3 ù êα 2 ú
êθ ú = ê
2ú
ë û ë0 1 2x 3x û êα 3 ú
ê ú
ëα 4 û
(B)
Step 3 - Relate member displacements to the node displacements
The coefficients a will vary with changes in the node displacements, so better that
we express the displacements in the member directly in terms of the node
displacements.
v1 = α 1
Substitute for x = 0 and x = L in the displacement function:
é v1 ù é1
ê θ ú ê0
ê 1ú = ê
êv 2 ú ê1
ê ú ê
ëθ 2 û ë0
or in matrix form,
or [U]=[A][a]
inverting, [a]=[A]-1[U] or,
0 0
0 ù é α1 ù
1 0
0 úú êêα 2 úú
L L2 L3 ú êα 3 ú
úê ú
1 2L 3L2 û ëα 4 û
é 1
é α1 ù ê
êα ú ê 0
ê 2 ú = ê- 3
êα 3 ú ê L2
ê ú ê 2
ëα 4 û ê 3
ë L
0
1
2
L
1
L2
0
0
3
L
2
- 3
L
θ1 = α 2
v 2 = α1 + α 2 L + α 3 L2 + α 4 L3
θ 2 = α 2 + 2α 3 L + 3α 4 L
0 ù
é v1 ù
0 úú ê ú
1 θ1
- úê ú
L ú êv2 ú
1 ú êθ ú
ë 2û
L2 úû
Combining this with equation (B) we can write the displacement at any point on the
member in terms of the displacements at the nodes as,
[u(x,y)] = [f(x,y)] [A]-1[U]
(C)
Step 4 - Write the strain-displacement relationships
Strain is the rate of change of displacement, so strains can be obtained from the
displacement function by differentiation
e(x,y)
= {differential of u(x,y)}
= {differential of [f(x,y)] [A]-1[U]}
= {differential of [f(x,y)]} [A]-1[U]
Let
[B] = {differential of [f(x,y)]} [A]-1, then
[e(x,y)]
= [B][U]
For our example of the beam subject to bending and shear, the beam
displacements are due to the curvature caused by bending,
v(x) = α1 + α2 x + α3 x 2 + α4 x 3
d 2v
ε(x,y) = - y 2 = - y (2α3 - 6α4 x )
dx
(D)
Step 4 - Write the strain-displacement relationships
(continued)
v(x) = α1 + α2 x + α3 x 2 + α4 x 3
d 2v
ε(x , y) = - y 2 = - y (2α3 - 6α4 x )
dx
and from before,
therefore,
é α1 ù
êα ú
[ε(x,y)] = y[0 0 - 2 - 6 x]ê 2 ú
ê α3 ú
ê ú
ëα 4 û
or in matrix form,
é 1
é α1 ù ê
êα ú ê 0
ê 2 ú = ê- 3
êα 3 ú ê L2
ê ú ê 2
ëα 4 û ê 3
ë L
0
1
2
L
1
L2
0
0
3
L
2
- 3
L
é 1
ê 0
ê
[ε(x,y)] = y[0 0 - 2 - 6 x]ê- 32
ê L
ê 2
êë L3
0 ù
é v1 ù
0 úú ê ú
1 θ1
- úê ú
L ú êv2 ú
1 ú êθ ú
ë 2û
L2 úû
0
1
2
L
1
L2
0
0
3
L
2
- 3
L
0 ù
é v1 ù
0 úú ê ú
1 θ1
- úê ú
L ú ê v2 ú
1 ú êθ ú
ë 2û
L2 úû
é v1 ù
ê ú
éæ 6 12 x ö æ 4 6 x ö æ 6 12 x ö æ 2 6 x öù ê θ1 ú
= y êç 2 - 3 ÷ ç - 2 ÷ ç - 2 + 3 ÷ ç - 2 ÷ú
L ø èL L ø è L
L ø è L L ø û ê v2 ú
ëè L
ê ú
ëθ2 û
Step 4 - Write the strain-displacement relationships
(continued)
so for the particular case of a member which carries the load by bending, the
strains in the member are related to the displacements at the nodes by
[e(x,y)] = [B][U] where,
[B] = y éêæç 62 - 123x ö÷
ëè L
æ 4 6 x ö æ 6 12 x ö æ 2 6 x öù
ç - 2 ÷ ç - 2 + 3 ÷ ç - 2 ÷ú
L ø èL L ø è L
L ø è L L øû
Step 5 - Write the stress - strain relationships
For the finite element member, the stress-strain relationship can be written
generally in terms of the strain matrix derived in Step 4 as,
[s(x,y)] = [D][e(x,y)] where [D] is called the elasticity matrix.
Substituting [e(x,y)]= [B][U] from before,
[s(x,y)] = [D][B][U]
(E)
For our beam example, the stress [s(x,y)] is represented by the bending
moment.
Since
d2y
M = EI K = EI 2
dx
we write, [D ] = [E ]
[B] = y éêæç 62 - 123x ö÷
ëè L
æ 4 6 x ö æ 6 12 x ö æ 2 6 x öù
ç - ÷ ç- +
÷ ç - ÷
L ø è L L2 ø è L2 L3 ø è L L2 øúû
therefore,
é v1 ù
ê ú
éæ 6 12 x ö æ 4 6 x ö æ 6 12 x ö æ 2 6 x öù ê θ1 ú
[σ(x,y)] = [D][B][U ] = [E ]y êç 2 - 3 ÷ ç - 2 ÷ ç - 2 + 3 ÷ ç - 2 ÷ú
L ø èL L ø è L
L ø è L L ø û ê v2 ú
ëè L
ê ú
ëθ2 û
Step 6 - Relate node loads to node displacements
Arbitrary nodal displacements
éu *1 ù
êu * ú
[U *] = ê 2 ú
ê ... ú
ê
ú
ëu *n û
result in a nodal loads
é F *1 ù
êF * ú
[F *] = ê 2 ú
ê ... ú
ê
ú
ë F *n û
external work: Wexternal = u*1 x F*1 + u*2 x F*2 + ....... + u*n x F*n = [U*]T[F*]
internal work: Winternal =
Substituting
vol
ò [ε * (x, y)] [σ * (x, y ]dV
T
[e*(x,y)]= [B][U*] and
[s*(x,y)] = [D][B][U],
vol
Winternal =
T
[
]
B
ò [U *][D][B][U ]dV
Equating the external and internal work,
é vol T
ù
[F ] = ê ò [B] [D][B]dV ú [U ]
êë
úû
(F)
vol
This is the member stiffness equation, [F] = [K][U] where [K ] = ò [B]T [D][B]dV
Step 6 - Relate node loads to node displacements
[D] = [E ]
[B] = y éêæç 62 - 123x ö÷
ëè L
æ 4 6 x ö æ 6 12 x ö æ 2 6 x öù
ç - ÷ ç- +
÷ ç - ÷
L ø è L L2 ø è L2 L3 ø è L L2 øúû
vol
[K ] = ò [B]T [D][B]dV, so
é 6 12x ù
ê L2 - L3 ú
ê 4 6x ú
Lê
- 2 ú
L
[K ] = ò ê 6 L12x ú[EI]éêæç 62 - 12x3 ö÷ æç 4 - 6x2 ö÷ æç - 62 + 12x3 ö÷ æç 2 - 6x2 ö÷ùú
L ø èL L ø è L
L ø è L L øû
ëè L
0 ê+ 3 ú
2
ê L
L ú
ê 2 6x ú
ê
ú
ë L L2 û
which multiplies out as,
Step 6 - Relate node loads to node displacements
é æ 36 144x 144x 2 ö
÷÷
ê çç 4 - 5 +
6
L
L
L
è
ø
ê
2
ê æ 24 84 x 72 x ö
ç
+ 5 ÷÷
Lê ç 3 4
L
L
L ø
[K ] = [EI]ò êê è
æ 36 144x 144x 2 ö
0
÷÷
êçç - 4 + 5 6
L
L
L
ø
êè
2
ê æ 12 60 x 72 x ö
ê çç 3 - 4 + 5 ÷÷
L
L ø
ë èL
æ 24 84 x 72 x 2 ö
çç 3 - 4 + 5 ÷÷
L
L ø
èL
æ 16 48 x 36 x 2 ö
çç 2 - 3 + 4 ÷÷
L
L ø
èL
æ 24 84 x 72 x 2 ö
çç - 3 + 4 - 5 ÷÷
L
L ø
è L
æ 8 36 x 36 x 2 ö
çç 2 - 3 + 4 ÷÷
L
L ø
èL
æ 36 144x 144x 2 ö
çç - 4 + 5 ÷
L
L6 ÷ø
è L
æ 24 84 x 72 x 2 ö
çç - 3 + 4 - 5 ÷÷
L
L ø
è L
æ 36 144x 144x 2 ö
çç 4 - 5 +
÷÷
6
L
L
L
è
ø
2
æ 12 60 x 72 x ö
çç - 3 + 4 - 5 ÷÷
L
L ø
è L
æ 12 60 x 72 x 2 ö ù
çç 3 - 4 + 5 ÷÷ ú
L
L øú
èL
æ 8 36 x 36 x 2 ö ú
çç 2 - 3 + 4 ÷÷ ú
L
L øú
èL
dx
æ 12 60 x 72 x 2 öú
çç - 3 + 4 - 5 ÷÷ú
L
L øú
è L
æ 4 24 x 36 x 2 ö ú
çç 2 - 3 + 4 ÷÷ ú
L
L øû
èL
and performing the integration,
é 12 EI
ê L3
ê 6EI
ê 2
L
[K ] = ê 12
ê- EI
ê L3
ê 6EI
ê 2
ë L
6EI
L2
4EI
L
6EI
- 2
L
2EI
L
12 EI
L3
6EI
- 2
L
12 EI
L3
6EI
- 2
L
-
6EI ù
L2 ú
2EI ú
ú
L ú
6EI
- 2 úú
L
4EI ú
ú
L û
which is the same result as previously derived using the direct stiffness method
Step 7 - Write the stress-displacement relationships
In Step 5 we had the relationship, [s(x,y)] = [D][B][U]. Putting,
[H] = [D][B]
(G)
we get [s(x,y)] = [H][U] relating the stress to the displacements.
For our beam example,
[D] = [E ]
[B] = y éêæç 62 - 123x ö÷
ëè L
Therefore,
æ 4 6 x ö æ 6 12 x ö æ 2 6 x öù
ç - ÷ ç- +
÷ ç - ÷
L ø è L L2 ø è L2 L3 ø è L L2 øúû
[H ] = [E ]y éêæç 62 - 123x ö÷
ëè L
æ 4 6 x ö æ 6 12 x ö æ 2 6 x öù
ç - ÷ ç- +
÷ ç - ÷
L ø è L L2 ø è L2 L3 ø è L L2 øúû
4L - 6 2 L ù
ú
L ë - 6 - 2L 6 - 4 L û
and substituting for x = 0 and x = L at the two nodes, éê H1 ùú = EI2 éê 6
ëH 2 û
Hence the stresses (which for this element
are represented by the moments) are:
é v1 ù
4L - 6 2L ù êêθ1 úú
é M1 ù
EI é 6
êM ú = [H][U ] = 2 ê- 6 - 2L 6 - 4Lú êv ú
L ë
û 1
ë 2û
ê ú
ëθ1 û
(which are the same equations for the end moments due to sway and rotation as we used
for the direct stiffness method)
Summary of steps and equations
The table below summarises the steps in formulating the stiffness matrix for the member,
using the general process for finite elements.
Step
1
2
3
4
5
6
7
[e(x,y)]
[s(x,y)]
[a]
[A]
[B]
[D]
[F]
[H]
[K]
[u(x,y)]
[U]
Task
Identify the problem geometry
Choose the coordinate system
Number the nodes
Select a displacement function
Relate the displacements at any point in
the member to the node displacements
Relate the strains at any point in the
member to the node displacements
Relate the stresses at any point in the
member to the node displacements
Relate node loads to the node
displacements
Key Equation
(A) [F] = [K][U]
Relate the stresses at any point in the
member to the node displacements
[s(x,y)] = [H][U], where,
(G) [H] = [D][B]
(B) [u(x,y)] = [f(x,y)][a]
(C) [u(x,y)] = [f(x,y)] [A]-1[U]
(D) [e(x,y)] = [B][U]
(E) [s(x,y)] = [D][B][U]
é vol T
ù
(F) [F ] = ê ò [B] [D][B]dV ú [U ]
ëê
ûú
strain in the member
stress in the member
constants in the displacement function
coefficients in the displacement function
matrix relating strains to the node displacements
elasticity matrix for relating stress to strain
forces at the member nodes
matrix relating stress to the node displacements
member stiffness matrix
displacement function - displacements within the member (as a function of x and y, the location relative to the nodes)
displacements at the member nodes