Lammps 金刚石弹性模量的计算

第一种方法

#计算 弹性模量

设置：



运行lammps：

提取出每次修正的初始能量：（附上 python 数据提取代码）

def read(path):

b=2

f=open(path)

fw=open(path+'.output','w')

for line in f.readlines():

b=b+1

if 'Energy initial' in str(line):

b=0

if b==1:

print(line)

fw.write(line)

提取的数据如下:

-471563.715475 -471563.715475 -471563.715475

-471559.573965 -471559.573965 -471559.573967

-471549.996058 -471549.996058 -471549.996058

-471535.007192 -471535.007192 -471535.007192

-471514.632964 -471514.632964 -471514.632964

-471488.899114 -471488.899114 -471488.899114

-471457.831529 -471457.831529 -471457.831529

-471421.456236 -471421.456236 -471421.456236

-471379.799399 -471379.799399 -471379.799399

-471332.887319 -471332.887319 -471332.887319

-471280.746425 -471280.746425 -471280.746425

-471223.403279 -471223.403279 -471223.403279

-471160.884564 -471160.884564 -471160.884564

-471093.217089 -471093.217089 -471093.217089

-471020.427778 -471020.427778 -471020.427778

-470942.543674 -470942.543674 -470942.543674

-470859.59193 -470859.59193 -470859.59193

-470771.599811 -470771.599811 -470771.599811

-470678.594686 -470678.594686 -470678.594686

-470580.604026 -470580.604026 -470580.604026

-470477.655406 -470477.655406 -470477.655406

只取第一列数据：

并绘制 能量-应变 曲线：





Matlab 五次多项式拟合：

Linear model Poly5:

f(x) = p1*x^5 + p2*x^4 + p3*x^3 + p4*x^2 + p5*x + p6

Coefficients (with 95% confidence bounds):

p1 = 3.509e+06 (3.507e+06, 3.51e+06)

p2 = -1.294e+06 (-1.294e+06, -1.294e+06)

p3 = -1.245e+06 (-1.245e+06, -1.245e+06)

p4 = 1.214e+06 (1.214e+06, 1.214e+06)

p5 = 943.2 (943.2, 943.2)

p6 = -4.716e+05 (-4.716e+05, -4.716e+05)

Goodness of fit:

SSE: 2.548e-10

R-square: 1

Adjusted R-square: 1

RMSE: 4.122e-06

二次项系数：1.214e+06

>>> Vc=3.567**3*20**3

>>> C2=1.214e+06

>>> C11=2*C2/Vc/6.2415e-3

>>> C11

1071.4215876370854

同样也可以计算C12=101.7246



第二种方法：

……http://files.cnblogs.com/bacazy/ELASTIC.zip