材料的名义应力、应变与真实应力、应变转换公式的推导

fem178

发布于 2021-05-18 07:16:53

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代码可运行

文章被收录于专栏：数值分析与有限元编程数值分析与有限元编程

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代码可运行

材料的名义(Nominal)应力、应变是基于变形前的数据计算得到，

$\sigma_N = \frac {F}{A_0} , \epsilon_N = \frac {\Delta L}{L_0}$

其中

$A_0$

为试件初始截面面积，

$L_0$

为试件初始长度。名义应力、应变也叫工程(Engineering)应力、应变。

CAE软件需要采用基于变形后的应力、应变，即真实的应力、应变。

$\sigma_T = \frac {F}{A} , \epsilon_T = \frac {\Delta L}{L}$

其中

$A$

为试件当前截面面积，

$L$

为试件当前长度。

两种应力、应变的转化公式为：

$\sigma_T = \sigma_N(1+\epsilon_N) , \epsilon_T = ln(1+\epsilon_N)$

下面来推导这两个公式。

一) 了解定积分的精确定义。点击这里：

定积分的精确定义(重排版)

二) 根据试件的体积不变的原则可得

$A_0L_0=AL$

，即

$\frac {A_0}{A} = \frac {L}{L_0}$

于是

$\begin{split} \sigma_T &= \frac {F}{A} \\ &=\frac {F}{A_0} \frac {A_0}{A}\\ &=\frac {F}{A_0} \frac {L}{L_0}\\ &=\frac {F}{A_0} \frac {L-L_0+L_0}{L_0}\\ & =\sigma_N(1+\epsilon_N) \end{split}$

三) 假设荷载

$F$

分为

$n$

个增量步，且每个增量步产生相同的伸长量

$\delta L$

,如图所示

$\delta L = \frac {L-L_0}{n}$

总应变

$\begin{split} \epsilon_T&=\epsilon_1+ \epsilon_2+\epsilon_3+\cdots +\epsilon_n \\ &= \frac {\delta L}{L_0+\delta L}+\frac {\delta L}{L_0+2\delta L}+\frac {\delta L}{L_0+3\delta L}+\cdots +\frac {\delta L}{L_0+n \delta L}\\ &=[\frac {1}{L_0+\frac {L-L_0}{n}\times1}+\frac {1}{L_0+\frac {L-L_0}{n}\times2}+\frac {1}{L_0+\frac {L-L_0}{n}\times3}+\cdots+\frac {1}{L_0+\frac {L-L_0}{n}\times n}]\frac {L-L_0}{n}\\ \end{split}$

再和定积分的精确定义比较

$\int_{a}^b = \sum_{i=1}^n f(a+\frac {b-a}{n}\times i) \frac {b-a}{n}$

故

$\begin{split} \epsilon_T &= \int_{L_0}^L \frac {1}{x} dx\\ &=ln(L-L_0)\\ &=ln(\frac {L-L_0+L_0}{L_0})\\ & =ln(1+\epsilon_N) \end{split}$


import math
import matplotlib.pyplot as plt
import numpy as np

plt.rcParams['font.sans-serif'] = ['SimHei']  # 正常显示中文标签
plt.rcParams['axes.unicode_minus'] = False    # 正常显示负号


# 名义应力应变
eps_N   = np.array([0, 0.00060117, 0.0010815,0.0017279,0.0022288,0.0028608,0.00348034,
                0.00404227,0.0045566,0.0051164,0.0058212,0.00674012,0.0078904,0.0092845,
                0.010934,0.012828,0.014924,0.017178])
sigma_N = np.array([490.393,503.8289, 512.22608,520.62322,529.020377,537.41752,545.81467,554.21182,
            562.60897,571.00612,579.40327,587.800419,596.197567,604.594716,612.99186,
            621.3890,629.78616,638.18331])

n = len(sigma_N)
print(n)

#真实应力应变
sigma_T = np.zeros((n))
eps_T   = np.zeros((n))

for i in range(n):
    sigma_T[i] = sigma_N[i] *( 1 + eps_N[i] )
    eps_T[i]   = math.log( 1 + eps_N[i] )

v1 = np.array( [ 0 ] )
eps_N1 = np.hstack( (v1, eps_N) )
sigma_N1 = np.hstack( (v1, sigma_N) )
eps_T1 = np.hstack( (v1, eps_T) )
sigma_T1 = np.hstack( (v1, sigma_T) )

fig, axs = plt.subplots(1, 1,  figsize=(14,6) )

axs.plot(eps_N1, sigma_N1, label="M1", linewidth = 2) 
axs.plot(eps_T, sigma_T, label="M1", linewidth = 2, color = "dimgrey")

plt.legend(["Nominal","True"]) 
plt.xlim(-0.0001, 0.019)
plt.ylim(0, 760)
axs.set_xlabel('Strain', fontsize = 18)
axs.set_ylabel('Stress', fontsize = 18)
fig.savefig('./f118.png', dpi = 400) #保存图片 
plt.show()