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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 n1 Cu n1 Rn1 Fn1 f n1 Mu n1 Cu n1 g n1 g n1 f n1 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 n1 / 2 u n1 u n t u (4) from（4）we get un1 un t u n u n1 / 2 n u n1 / 2 u n1 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 kt u u cn 1 ktH n / 2 1 ktH n / 2 (11) Then， we get f n1 f n f k d u f rn (12 where kd k 1 ktH n / 2 frn kt ucn 1 ktH 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 un1 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 Mun 1 Cu n 1 (16) From equation (5), we get 2 un 1 t 4 2 un 1 t u n 1 un 1 (17) From equation（14）we get f K d un1 (18) Substituting from equations (17) and (18) into equation (16), we have K tn1un1 g n1,old Thus we get an improvement u n 1 to un 1 (19)