General Method for Deriving an Element Stiffness Matrix
step I: select suitable displacement function
beam likely to be polynomial with one unknown coefficient for each (of four) degrees of freedom
 v1 
 v'
1
dof = δ =   in matrix notation:
 v2 

 v'2 
say ...
2
 C1 
C
2
C :=  
 C3 

 C4 
2
(
2
H( x) := 1 x x
3
 C1 + x⋅ C2 + x2⋅C3 + x3⋅C4 
 C1 
δ ( x) → 
δ ( 0) → 

2
 C2 
 C2 + 2⋅ x⋅C3 + 3⋅ x ⋅C4 
in matrix form:
3
x
)
v( x) := H( x)⋅C
C
5.3.6
2
d
v( x) → C2 + 2⋅ x⋅C3 + 3⋅ x ⋅C4
dx
v( x) → C1 + x⋅ C2 + x ⋅C3 + x ⋅C4
v x) 
 v(

δ ( x) := d
 v( x)
 dx

3
v( x) = C1 + x⋅ C2 + x ⋅C3 + x ⋅C4
 C1 + L⋅ C2 + L2⋅C3 + L3⋅C4 
δ ( L) → 

2
 C2 + 2⋅ L⋅C3 + 3⋅ L ⋅C4 
for manipulation
 C1 
 1 x x2 x3   C2
 1 x x2 x3 



δ ( x) =
⋅
δ_over_C( x) := 
5.3.7


2  C3 
2
 0 1 2x 3x  
 0 1 2x 3x 
 C4 
step II: relate general displacements within element to its nodal displacement
 1 L L2 L3 
δ_over_C( L) → 

2
 0 1 2⋅ L 3⋅ L 
 v1 
v
1_p 
define A such that
δ_nodes = A⋅C
δ_nodes = 
 v2 

 v2_p 
 1 0 0 0 
0 1 0 0 
A := stack( δ_over_C( 0) , δ_over_C( L) )
A→
2
3 
1 L L L 
 0 1 2⋅ L 3⋅ L2


 1 0 0 0
δ_over_C( 0) → 
 0 1 0 0
in single matrix form:
form by stacking
node 1 with node 2
δ_nodes = A⋅C
=>
0
0
1
0
L L
2


3 
L

0
0
1 L 3⋅ L
2

4

0
0
0 
 L


4
L
0
0 
1  0
−1
5.3.8b
A
→
⋅
4 
2
3
2
3
L
 −3⋅L −2⋅L 3⋅ L −L 
2
2
 2⋅ L
L
−2⋅L L 

−1
C := A
 1
0
A=
1
0

δ_nodes
⋅δ_nodes
1
notes_36_element_stiffness_general_2.mcd
v( x) := H( x)⋅C
C
v( x)
δ_nodes
(
3
x
)
5.3.9a
−1
v( x) := H( x)⋅A
⋅δ_nodes
δ_nodes
 L3 − 3⋅ x2⋅L + 2⋅ x3 L2 − 2⋅ x⋅L + x2
2 −3⋅L + 2⋅ x 2 −L + x 

x⋅
−x ⋅
x ⋅

3
2
3
2
L
L
L
L


simplify →
−1
shape function defined
(
2
H( x) → 1 x x
2
N( x) := H( x)⋅A
A
N( x) → 1 − 3⋅ξ + 2⋅ξ
3
2
with
3
2
ξ=
ξ⋅L − 2⋅ξ ⋅L + ξ ⋅L 3⋅ξ − 2⋅ξ
3
x
=>
L
2
3
−ξ ⋅L + ξ ⋅L
)
x := ξ⋅L
L
5.3.9b although text has mix of ξ and x
Step III: express the internal deformation in terms of the nodal displacement
area resets x, redefines C, H and v
our problem is one of solid mechanics
deformation is strain: du/dx,
; plane elasticity
bending curvature d2 u/dx2 . v_2pr = d2 u/dx2.
2
3
2
v( x) → C1 + x⋅ C2 + x ⋅C3 + x ⋅C4
d
2
dx
−1
v_2pr(x) := ( 0 0 2 6⋅x )⋅A
B( x) →
⋅δ_nodes
δ_nodes
v( x) → 2⋅ C3 + 6⋅ x⋅C4
v_2pr(x) := ( 0 0 2 6⋅x )⋅C
C
−1
B( x) := ( 0 0 2 6⋅x )⋅A
A
v_2pr( x) := B( x)⋅δ_nodes
δ_nodes
 −6 + 12⋅ x −4 + 6⋅ x 6 − 12⋅ x −2 + 6⋅ x 
 2
3 L
2
2
3 L
2
L
L L
L
L 
L
5.3.10
step IV: express the internal force in terms of the nodal displacement
the "internal force" is the bending moment and as with internal deformation,
this is a problem in bending so the relationship is
we just developed
2
d
2
v( x) = E⋅ I⋅v_2pr( x
) dx
v_2pr( x) := B( x)⋅δ_nodes
δ_nodes
copy here for use elsewhere:
v_2pr( x) →
M ( x) = E⋅I⋅
B( x) :=
 −6 + 12⋅ x −4 + 6⋅ x 6 − 12⋅ x −2 + 6⋅ x 
 2
3 L
2
2
3 L
2
L
L L
L
L 
L
  −6 + 12⋅ x  ⋅δ_nodes  −4 + 6⋅ x  ⋅δ_nodes  6 − 12⋅ x  ⋅δ_nodes  −2 + 6⋅ x  ⋅δ_nodes 
 2
L
 2
L

3
2
3
2
L 
L 
L 
L 
 L

L


define:
M ( x) := E⋅I⋅v_2pr(
v_2pr x)
bb :=
E⋅ I
L
3
M ( x)
bb⋅δ_nodes
simplify → [ −6⋅L + 12⋅x 2⋅(−2⋅L + 3⋅ x)⋅L 6⋅ L − 12⋅x 2⋅(−L + 3⋅ x)⋅L ]
5.3.12a
2
notes_36_element_stiffness_general_2.mcd
since M(x) is linear, we can calculate M(x) at the nodes (N.B. M is the internal moment)
define S
 M( 0)  = S⋅δ_nodes

 M ( L) 
5.3.12b
M ( 0)
 M( 0) 
bb⋅δ_nodes 
S_over_bb = 
 M ( L) 
 bb⋅δ_nodes 
bb⋅δ_nodes
M ( L)
bb⋅δ_nodes
substituting bb = E*I/L^3
S=
(

E⋅ I  −6⋅L −4⋅L
L
3
(
2
6⋅ L −2⋅L
2
)
(
2
→ −6⋅L −4⋅L
(
→ 6⋅ L 2⋅ L
2
2
6⋅ L −2⋅L
−6⋅L 4⋅ L
2
)
)
note that this is similar to M1 and M2 with sign reversal in top element
)
2
2 
 6⋅ L 2⋅ L −6⋅L 4⋅ L 
step V: obtain the element stiffness matrix ke by relating nodal forces to nodal
displacements
 v1_star 
v1'_star 
δ_star := 
 v2_star 
 v2'_star


T
we will do this by the principle of virtual work:
assume an arbitrary virtual nodal displacement:
external virtual work is force * virtual deflection:
actual nodal forces are:
Wext := δ_star ⋅ff
 f1 
M1 
f := 
 f2 
 M2
 
Wext → v1_star⋅f1 + v1'_star⋅M1 + v2_star⋅f2 + v2'_star⋅M2
L
internal work = work done in imposing curvature on the beam:
for an arbitrary virtual curvature v''_star(x)
⌠
T
Wint :=  v_2pr_star( x) ⋅M ( x) dx
⌡
0
M ( x) = internal_moment
using transpose as v''_star(x) is a scalar but will involve 4 x 1 vectors to multiply the scalar M(x) with vector components later.
if arbitrary virtual curvature v''_star(x) is imposed indirectly by virtual nodal displacement
v''_star(x) is related to the δ_star by B(x)
v_2pr( x) := B( x)⋅δ_nodes
δ_nodes
from above
and ...
now using
is understood to be nodal
δ_star
v_2pr_star( x) := B( x)⋅δ_star
δ_star
v_2pr_star( x) = ( B( x)⋅δ_star) = δ_star ⋅ B( x)
T
T
M ( x) = E⋅I⋅
2
d
2
v( x) = E⋅ I⋅v_2pr( x)
T
T
v_2pr( x) := B( x)⋅δ_nodes
δ_nodes
M ( x) = E⋅I⋅ B( x)⋅δ_nodes
dx
L
L
⌠
⌠
T
T
T
Wint =  v_2pr_star( x) ⋅M ( x) dx =  δ_star ⋅ B( x) ⋅E⋅I⋅ B( x)⋅δ_nodes dx
⌡
⌡
0
0
3
notes_36_element_stiffness_general_2.mcd
taking the constants outside the integral and equating internal to external work the constants have to come
out of the correct side to maintain matrix math
L
⌠
T
Wext = δ_star ⋅f = Wint = δ_star ⋅E⋅I  B( x) ⋅ B( x) dx⋅δ_nodes
⌡
T
T
cancelling δ_star =>
0
L
⌠
T
f = E⋅I  B( x) ⋅ B( x) dx⋅δ_nodes = ke⋅δ_nodes
⌡
what we came for
0
 −6 + 12⋅ x 
3
 L2
L 
 −4

x

+ 6⋅

2 
L

L
T
B( x) → 

 6 − 12⋅ x 
3
 L2
L 


 −2 + 6⋅ x 
2
 L
L 

result;
copied from
rhs
L
⌠
T
ke = E⋅ I⋅  B( x) ⋅ B( x) dx
⌡
0
B( x) →
 −6 + 12⋅ x −4 + 6⋅ x 6 − 12⋅ x −2 + 6⋅ x 
 2
3 L
2
2
3 L
2
L
L L
L
L 
L
L
⌠
T
all we need is  B( x) ⋅ B( x) dx
⌡
0
(it won't compute symbolicly so I wrote it out in
the collapsed area)
 12
 L3
 6

L
 L2
⌠
T
 B( x) ⋅ B( x) dx = 
⌡
0
 −12
 L3

 6
 L2

6
−12
2
3
L
4
L
−6
L
2
2
L
L
−6
L
2
12
L
3
−6
L
2

2
L 

2

L 

−6

2
L

4 
 12 6

 L3 L2
 6 4

L
 L2 L
⌠
T
ke = E⋅ I⋅  B( x) ⋅ B( x) dx = E⋅ I⋅ 
⌡
0
 −12 −6
 L3 L2

 6 2
 2 L
 L
6
L

−12
L
3
−6
L
2
12
L
3
−6
L
2

L 

2

L 

−6

2
L

4 
6
2
L

so ...
 12
 fy1 

 M1  = E⋅ I ⋅  6⋅ L
 fy2 
3 −12
L 

 6⋅ L
 M2 

6⋅ L
−12
6⋅ L
  v1 
2
2
4⋅ L −6⋅L 2⋅ L   θ 1 
 ⋅ 
−6⋅L 12 −6⋅L
  v2 
2
2 
2⋅ L −6⋅L 4⋅ L   θ 2 
we now have
4
f = ke⋅δ
notes_36_element_stiffness_general_2.mcd