# A Maxwell spring-damper element consists of... spring, see fig 1.

```A Maxwell spring-damper element consists of a nonlinear damper in series with a
spring, see fig 1.
Fig 1
Consider the uniaxial case, the force deformation relationship is
f  kuk  c sgn( u c )(abs(u c )) 
(1)
u  u j  ui  u k  uc
(2)
Where the subscript k denotes quantites associated with the spring, and c denontes quantites
associated with the nonlinear damper. Quantities with no subscript denote the total element
deformation, force, etc.
Suppose the dynamic equilibrium equation at the end of time step n+1 be expressed as
n1  Cu n1   Rn1  Fn1  f n1  Mu
n1  Cu n1
g n1  g n1  f n1  Mu
(3)
Where the gn 1 are the dynamic unbalanced force and gn 1 are static out of balance force，
fn+1 are the equivalent nodal forces corresponding to the internal forces of the
Maxwell-spring-damper element, Fn+1 are the nodal forces corresponding to stresses of other
elements, we set Fn+1=0 since we only consider spring-damper element here.
Adopt Newmark time integration method and let   1 / 2,   1 / 4 ，we have
n  u
n 1  / 4
un 1  un  tu n  t 2 u
n  u
n1  / 2
u n1  u n  t u
(4)
from（4）we get
un1  un  t u n  u n1  / 2
n  u
n1  / 2
u n1  u n  t u
from（5）we get
(5)
2
u  u n
t
4
4
n
 2 u  u n  u
t
t
u n 1 
n 1
u
(6)
In the following, we will derive the relationship between the incremental element force  f and
the incremental total element displacement u from step n to n+1. Obviously, the following
relationship exists
f  k (u  u c )
(7)
Let’s suppose the incremental displacement of damper between time step n and n+1 satisfy
u c  t (u cn  u cn 1 ) / 2
(8)
where subscript n denotes quantities at time step n and so .
in equation (8), u cn 1 can be expressed as truncated Taylor series at time tn
u cn 1  u cn 
where H n 
u c
f
f
 u cn 
t t n
1
1 
u cn f  u cn  H n f
c
(9)
1
( u cn )1  , in deriving equation (9), we have used equation (1).
c
Substituting from（9）into equation（8）, we have
u c  tu cn 
1
tH n f
2
(10)
Substituting from equation（10）into equation 7）, we have
f 
k
kt
u 
u cn
1  ktH n / 2
1  ktH n / 2
(11)
Then， we get
f n1  f n  f  k d u  f rn
(12
where
kd 
k
1  ktH n / 2
frn  
kt
ucn
1  ktH n / 2
(13
Substituting equations (6)（12） into equation（3）, we get the incremental equation of motion
(
4
2
 4
n   Cu n
M  C  K d）u  Rn 1  frn  f n  M  u n  u
2
t
t
 t

(14)
Solving equation（15）, we get u ，then we have
un1  un  u
(15)
Substituting from equation (15) into equation (3) usually leads to a residual g n 1 which is not
zero.
We can correct un 1 by the modified Newton-Raphson method，using the truncated Taylor series,
we have
g n 1,new  g n 1,old  f n 1  Mun 1  Cu n 1
(16)
From equation (5), we get
2
un 1
t
4
 2 un 1
t
u n 1 
un 1
(17)
From equation（14）we get
f  K d un1
(18)
Substituting from equations (17) and (18) into equation (16), we have
K tn1un1   g n1,old
Thus we get an improvement u n 1 to un 1
(19)
```