Matrix Analysis Example Hughes figure 5.12 page 191 ff
6
C
4
4
3
5
FYB
A
3
c
b
1
2
2
3
B
1
a
4
2
1
input data
n_elements := 3
f, δ element; F, ∆ structure; m = element
n_nodes := 3
ie := 1 .. n_elements
1
3
a
n_free := 2 number of degrees of freedom per node
n_dof := n_nodes⋅n_free
input for the class and text problem:
 1 2
elem :=  1 3

 2 3
c
b
4
2
FXB
ORIGIN := 1
nod_el := 2
nodal map of elements
total number of degrees of freedom in structure
n_dof = 6
nodes per element
 0 0
XY :=  1 0

 0 1
in := 1 .. nod_el
location of nodes
 1
A :=  1

 1
 1

E := 1

 1
__________________________________________________________________________________________________________________
element stiffness matrix:
geometry
X
ie , in
:= XY
elemie, in , 1
notes_34_ex_pg_192_2003.mcd
Y
ie , in
:= XY
elemie, in , 2
 0 1
X = 0 0

 1 0
 0 0
Y = 0 1

 0 1
1
L :=
ie
(Xie , 2 − Xie , 1) + (Yie , 2 − Yie , 1)
2
2
 1 
L= 1

 1.414 



angle := if  X
ie
(
ie , 2
angle := if X
ie
ie , 2
−X
ie , 1
−X
ie , 1
 Yie , 2 − Yie , 1  π 
,
 Xie , 2 − Xie , 1 2

 
> 0, atan
< 0, angle + π , angle
ie
ie
)
 0 
=  90

deg
 −45 
don't need angle now but will
later for T
angle
gets angle -π/2 < angle < π/2
 0 
=  90

deg
 135 
angle
gets angle in appropriate quadrant
element stiffness, element coordinates
 1
ie ie  0
⋅
 −1
L
ie
0

A ⋅E
ke :=
ie
0 −1 0 
 1
0
ke = 
1  −1

0
0 0 0
0 1 0
0 0 0
0 −1 0 
 1
0
ke = 
2  −1
0

0 0 0
0 1 0
0 0 0
0 −1 0 
0 0 0
0 1 0
0 0 0
 0.707
0
ke = 
3  −0.707

0

0 −0.707 0 
0
0
0
0 0.707 0 
0
0
0 
transformation matrix
(
λ ie := cos angle
ie
)
(
µ ie := sin angle
ie
 1 
λ= 0

 −0.707 
)
 0 

µ=
1

 0.707 
transform from structure to element; applies at each node of element.
 λ ie

−µ ie
T := 
ie
 0
 0

µ ie
0
λ ie
0
0
λ ie
0
( −µ) ie


0 

µ ie

λ ie

0
1

0
T =
1 0
0

0 0 0
1 0 0
0 1 0
0 0 1
 0
−1
T =
2  0
0

1 0 0
0 0 0
0 0 1
0 −1 0 
0
0 
 −0.707 0.707
−0.707 −0.707
0
0 
T =
3  0
0
−0.707 0.707 
 0
0
−0.707 −0.707 

element stiffness, structure coordinates
Ke := T
ie
T
⋅ke ⋅T
ie
ie ie
 1
0
Ke = 
1  −1
0

notes_34_ex_pg_192_2003.mcd
0 −1 0 
0 0 0
0 1 0
0 0 0
0

0

Ke =
2 0

0

1 0 −1 
0 0 0 
−1 0 1 
 0.354

−0.354

Ke =
3  −0.354

 0.354
0 0 0
2
−0.354 −0.354 0.354
0.354
0.354

0.354 −0.354 
0.354 −0.354 
−0.354 −0.354 0.354

assemble structure stiffness matrix structure coordinates
now we have to deal with total structure model:
 F1 
 F2 
F 
3

F =
 F4 
 
 F5 
 F6 
 ∆1 

 ∆2 
 
 ∆3 
 ∆4 
 
 ∆5 

 ∆6 
∆ =
and
F =
superposing respective element contributions
K⋅∆
convert node number to numbered degree of freedom
j := 1 .. n_free
k := 0 .. n_free − 1
i := 1 .. nod_el n_free
top
ie , 2⋅ j−k
:= n_free elem
−k
j := 1 .. nod_el n_free
K
n_dof , n_dof
K
topie, i , topie, j
:= 0
:= K
topie, i , topie, j
( ie)i, j
+ Ke
set up forces, lhs of F = K* ∆
ii := 1 .. n_dof
ie , j
 1 2 3 4
top =  1 2 5 6

 3 4 5 6
F := 0
ii
notes_34_ex_pg_192_2003.mcd
 F3   3 

:= 
 F4
4
   
1
0

−1
K= 
0
0

0
0
−1
0
0
0
1
0
0
0
−1



0 −0.354 0.354 0.354 −0.354 

0 −0.354 0.354 0.354 −0.354
0
1.354 −0.354 −0.354 0.354
−1 0.354 −0.354 −0.354 1.354
 0
0
 
3
F→ 
 4
 0

 0
3

apply boundary conditions
only degrees of freedom 3 and 4 are unconstrained therefore the reduced equations become
Fred := submatrix(F, 3, 4, 1, 1)
Fred →
 3

 4
Kred := submatrix(K, 3, 4, 3, 4)
Kred =
 1.354 −0.354 

 −0.354 0.354 
 −7 
0
 
3
F=  
4
4

 −4 
solve for ∆and F
∆ ii := 0
and we can solve for ∆3 and ∆4
 ∆3 
−1
:= Kred ⋅Fred

 ∆4 
F := K⋅∆
reverse to calculate element properties
and then the element forces are calculated from the relationships that we began with:
 0 
 0


7 

∆=
 18.314 
 0 

 0 
first get Delta (structure coordinates) of each element
∆eie , i := ∆ top
ie, i
 0 0 7 18.314 
∆e =  0
0
0
0

 7 18.314 0 0 
f := ke ⋅δ ie
ie
ie
apply stress matrix
E
Se :=
ie
L
ie
 −7 
0
f = 
1  7 
0
 
⋅( −1 0 1 0 )
ie
σ ie := Se ⋅δ ie
ie
 0 
0
f = 
2  0
0
 
Se = ( −1 0 1 0 )
1
 0
0
T
∆e = 
 7
 18.314


0 18.314 
0
0 
0
0
7
0

0 

0 
δ1 = 
 7 
 18.314


5.657 

0 
f =
3  −5.657 
 0


Se = ( −1 0 1 0 )
2
 7 
σ= 0

 −5.657 
notes_34_ex_pg_192_2003.mcd
( T)
δ ie := T ⋅ ∆e
ie
⟨ie⟩
4
Se = ( −0.707 0 0.707 0 )
3
0
 
0
δ2 =  
 0
0
 
8


−17.899 
δ3 = 
 0 
 0

