SimWe仿真论坛 » A60：有限元理论及CAX二次开发 » 最简单的Newton-Raphson法求解程序

molen 发表于 2007-12-21 07:26

最简单的Newton-Raphson法求解程序

Newton-Raphson法是有限元非线性分析的基础。_"GZSm+P i8C&Ka
下面的这个FORTRAN程序是最简单的非线性求解演示，一个单自由度的非线性材料桁架用N－R法求解。
ac@;U,fOp1a 很有参考价值。+k+PhBT r1Uf2^
Y:Ylp4] ~7YH_m(S*NZ
PROGRAM truss_box_1_2DR!O5H3U9}
! -------------------------------------------------------------------------$S| pAn1i^(U*C0H ]
! Newton-Raphson solver for 1 d.o.f. nonlinear truss examplel"CzbO1aa+f
! Input: r*h6n8_#p9\j.@,G
!  d        = horizontal span
S DE1I$cV)NF !  x_0      = initial height!glWIw
!  area_0   = initial area
x!yx(U|Y !  E        = Young's modulus KX/s @A^K?7d"]d
!  n_incr   = number of load increments

mF ~!lvfz}D !  f_incr   = force increment
Ok(KpJ5o !  c_norm   = residual force convergence norm{a jk1B)K
!  m_iter   = max number of Newton-Raphson iterations(qx*T,F ]%q4l]
! -------------------------------------------------------------------------9wk/|!P(J,Y
IMPLICIT none~dUOTb7I9V
INTEGER,   PARAMETER :: D_P = kind (1.d0) ! machine portable,S\D.W[!S+Y
INTEGER,   PARAMETER :: n_incr = 5, m_iter = 20
/D$fo~2x3@1_6d/v
c REAL(D_P), PARAMETER :: d = 2.5d3, x_0 = 2.5d3, area_0 = 1.d2,     &
:mt j}P.}*qbN8{(y#e                          E = 5.d5, f_incr = 0.3d7, c_norm = 1.d-20, &
umad0yEu+L                          tolerence = TINY (1.d0)
RJbT@,__z INTEGER   :: incrm            ! current increment
)D*f0fk2p-?S*p,\/a INTEGER   :: n_iter           ! current N-R iteration number
:M}.X;K^/l5t${ REAL(D_P) :: force = 0.d0     ! current force}-LVy0K@1Z
REAL(D_P) :: vertical         ! current vertical truss force'{6kU M6X2]&a
REAL(D_P) :: residual = 0.d0  ! current residual
;` V@*ZcB REAL(D_P) :: r_norm           ! current norm of residual }$O t7^ tP
REAL(D_P) :: length, length_0 ! current and initial length
Jl`U;GD.Ys REAL(D_P) :: stiff            ! current stiffness
V1{'H(S7OA` REAL(D_P) :: stress           ! current stressp?@(qf
REAL(D_P) :: x, area          ! current height and area'`&pIQl'wq V'V
REAL(D_P) :: volume           ! current volume
)_7Tn"i!P REAL(D_P) :: u                ! current displacement increment
\$J\%K*D !  initialize geometry and stiffness a.Q2P0oV vB)L'N)e
  x = x_0 ; area = area_01bWZ.td9]
  length_0 = sqrt ( x **2 + d **2 )

_-eX%ok v   volume   = area_0 * length_0
#T*@2Qrh"k,^6v3P   stiff    = (area / length_0) * E * (x / length_0 ) **2
`"I O l6Yg@0vel !  start load increments*[1c [3[/}JS
  DO incrm = 1, n_incrK@'I$j:mi!?
    force    = force + f_incrg$oX/e BNK
    residual = residual - f_incr
c)a!d(Kr,T2cE6BL !  Newton-Raphson iteration5l8D`h7aK v
    r_norm = c_norm * 2 ; n_iter = 0  ! initialize'Vx+@MR
Q
    DO WHILE ( (r_norm > c_norm) .and. &*ZJ"[ xGX
               (n_iter < m_iter)       );e5^_Rr9_ eD0w
      n_iter = n_iter + 1
vsAUHS/D6E9d !     Find geometry increment5DZN/Mc$yR&{`nh%M
      u      = -residual / stiff      ! displacement increment Ty{
@6{
      x      = x + u                  ! update height
B-S%{b9i-A       length = sqrt ( x **2 + d **2 ) ! update lengthj P8R5N-dD+SY
      area   = volume / length        ! update area'R\3D W+{.k2B%e
      Xm6x}S$[
!     Find stresses and residual forces
zaJr'H1~$kD9_       stress   = E * log (length / length_0)  ! axial stressz(D5Oa5W
      vertical = stress * area * x / length   ! vertical force`+I T&tK%D_ @D$u
      residual = vertical - force
Yr;fH[}c O-u       r_norm   = abs (residual / force)
_G!v*h`(Z       print "('increment = ', i3, ' iteration = ', i3,     &6u8\3BQ-S`r,x
      &       ' residual = ', g10.4, ' x = ', g10.4,       &
kIf d3P h9H       &       ' force = ', g10.4)", incrm, n_iter, r_norm, &:N8Q-_&p1cJ
F
      &       x, force
0{7a4I7qDP*e !     Find stiffness and check stability
J]c)Q"C d.AW s|!d       stiff = (area/length) * (E - 2 * stress) * (x/length) **2 &$jJ'EWu.hSr
            + (stress * area / length)
g`_.?8w|,}miD       IF ( abs ( stiff) < tolerence )  THEN
k J9FL c4XA         print *, 'Near zero stiffness'-m1^lC.W6r\UB
        STOP     'Near zero stiffness'
\`#\:z3Ik       END IF ! unstable(V!d2OG*[NQ
    END DO ! Newton-Raphson iteration$i(~P:Vq$Q\['hhu.E7cV
  END DO ! incrmT&L0[ Ft$Bb
END PROGRAM truss_box_1_2

RS9C%bv1F%a !

{*QG#y2` a.wn ! Running with parameters E = 5.d5, f_incr = 1.5d7, c_norm = 1.d-20:,H'? O~!x!z{9J
! increment = 1 iteration = 1 residual = 0.2184     x = 4621. force = 0.1500E+08d1@)cmqN#I
! increment = 1 iteration = 2 residual = 0.4239E-01 x = 5540. force = 0.1500E+08BJ#H!kF C
! increment = 1 iteration = 3 residual = 0.2973E-02 x = 5822. force = 0.1500E+08
avJ3N/jI"KY:f9dN ! increment = 1 iteration = 4 residual = 0.1806E-04 x = 5844. force = 0.1500E+08IU\ ji[%z
! increment = 1 iteration = 5 residual = 0.6780E-09 x = 5845. force = 0.1500E+08)n~G
n.Pq/F
! increment = 1 iteration = 6 residual = 0.1242E-15 x = 5845. force = 0.1500E+08
]XOzbF |ai ! increment = 1 iteration = 7 residual =  0.000     x = 5845. force = 0.1500E+08
4K:A e8?
l1X
C$R*T/A.T\ Y }
U [[i] 本帖最后由 molen 于 2007-12-21 07:28 编辑 [/i]]

查看完整版本: 最简单的Newton-Raphson法求解程序