弧长法的Python实现

弧长法点击这里:

非线性 | 弧长法(Arc-Length Methods)

改进弧长法点击这里：

非线性|弧长法改进

对于一个非线性有限元模型，只有一个自由度

u

,外荷载

f_0^{ext}=1

,内力为

f_{int}=-4(u-1)^2 4

切线刚度矩阵

K_t = frac {partial f_{int}}{partial u}=-8(u-1)

假设某一荷载步迭代收敛时荷载因子,

lambda_0 = 3.84,u_0 =0.8

。接下来以

Delta lambda_1 = 0.26

开始。以下是改进弧长法手算过程

Python代码：

代码语言：javascript复制

import math

def stiff(u):
    f_int = -4*(u-1)**2   4
    Kt = -8*(u-1)
    return f_int, Kt



f_ext = 1

lamd_0 = 3.84
u0 = 0.8
delta_lamd = 0.26

tol = 1e-7
conv = 2e11

max_iter = 200
niter = 0
sum_u = 0
sum_lamd = 0
print('niter      desplacement                   lamda                   conv')
while ( conv > tol and niter < max_iter ):
    niter  = 1

    if niter == 1:
        f_int, Kt = stiff( u0 )
        lamd_1 = lamd_0   delta_lamd
        f_resid = lamd_1 * f_ext - f_int

        delta_u = f_resid / Kt

        u1 = u0   delta_u
        sum_u = sum_u   delta_u
        sum_lamd = sum_lamd   delta_lamd

        u0 = u1
        lamd_0 = lamd_1
    else:
        f_int, Kt = stiff( u0 )
        f_resid =  f_int - lamd_0 * f_ext 

        delta_u_ext = f_ext / Kt
        delta_u_resid = f_resid / Kt

        delta_lamd1 = delta_u_resid * sum_u / (delta_u_ext * sum_u   sum_lamd )

        lamd_1 = lamd_0   delta_lamd1

        delta_u = delta_lamd1 *  delta_u_ext - delta_u_resid 

        sum_u = sum_u   delta_u
        sum_lamd = sum_lamd   delta_lamd1
        u1 = u0   delta_u

        u0 = u1
        lamd_0 = lamd_1
        delta_lamd = delta_lamd1

    conv = math.sqrt( f_resid**2 )
    print(format(niter, '>3x'), format(u1, '>20.12f'), 
                    format(lamd_1, '>26.14f'), format(f_resid, '>28.16e') )

输出结果：

python methods 模型

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