Abstract
Full Text
THEORY OF ELASTICITY
D. D. IVLEV
APPROXIMATE SOLUTION OF ELASTOPLASTIC PROBLEMS OF THE THEORY OF IDEAL PLASTICITY
(Presented by Academician A. I. Nekrasov, 12 X 1956)
We shall seek the solution of the elastoplastic problem in the form of series in powers of a certain parameter (\delta):
[
\sigma_{ij}=\sum_{n=0}\delta^n\sigma_{ij}^{(n)},\qquad
\sigma_{ii}\equiv\sigma_\rho,\qquad
\sigma_{jj}\equiv\sigma_\theta,\qquad
\sigma_{ij}\equiv\sigma_{ji}\equiv\tau_{\rho\theta}.
\tag{1}
]
We shall confine ourselves to consideration, in the theory of plane strain, of the plasticity conditions of Mises and Saint-Venant ((^1)), which are essentially coincident:
[
\frac14(\sigma_\rho-\sigma_\theta)^2+\tau_{\rho\theta}^2=1;
\tag{2}
]
in the theory of plane stress—the Saint-Venant plasticity condition ((^1)):
[
\frac14(\sigma-\sigma_\theta)^2+\tau_{\rho\theta}^2=
\left[1-\frac12(\sigma_\rho+\sigma_\theta)\right]^2
\quad \text{for } \sigma_\theta>\sigma_\rho>0.
\tag{3}
]
In (1), (2), (3) all quantities are dimensionless. Substituting (1) into (2), with (\tau_{\rho\theta}^0=0), we obtain:
[
\sigma_\rho'-\sigma_\theta'=0,\qquad
(\sigma_\rho''-\sigma_\theta'')\mu+\tau_{\rho\theta}'^{\,2}=0,\qquad
\frac12(\sigma_\rho'''-\sigma_\theta''')\mu+\tau_{\rho\theta}'\tau_{\rho\theta}''=0,
\tag{4}
]
where (\mu=\operatorname{sign}(\sigma_\rho^0-\sigma_\theta^0)).
Substituting (1) into (3) with (\tau_{\rho\theta}^0=0), we obtain:
[
\sigma_\theta'=0,\qquad
(1-\sigma_\rho^0)\sigma_\theta''+\tau_{\rho\theta}'^{\,2}=0,\qquad
(1-\sigma_\rho^0)\sigma_\theta'''-\sigma_\rho'\sigma_\theta''+2\tau_{\rho\theta}'\tau_{\rho\theta}''=0.
\tag{5}
]
If on the boundary (L) of the body under consideration
[
\sigma_{\mathrm n}=P_{\mathrm n},\qquad \tau_{\mathrm n}=P_\tau,
\tag{6}
]
then, assuming the equation of the contour (L) to be given in the form
[
\rho=\sum_{n=0}\delta^n\rho_n(\theta),
]
we obtain the linearized boundary conditions (6) in the form
[
\sigma_\rho' + \frac{d\sigma_\rho^0}{d\rho}\rho_1
=
\frac{dP_{\mathrm n}}{d\rho}\rho_1;\qquad
\tau_{\rho\theta}'-(\sigma_\theta^0-\sigma_\rho^0)\dot R_1
=
\frac{dP_\tau}{d\rho}\rho_1;
]
[
\sigma_\rho''+
\frac{d\sigma_\rho'}{d\rho}\rho_1+
\frac{d^2\sigma_\rho^0}{d\rho^2}\frac{\rho_1^2}{2!}
+
\frac{d\sigma_\rho^0}{d\rho}\rho_2
+
(\sigma_\theta^0-\sigma_\rho^0)\dot R_1^{\,2}
-
2\tau_{\rho\theta}'\dot R_1
=
]
[
\frac{d^2P_{\mathrm n}}{d\rho^2}\frac{\rho_1^2}{2!}
+
\frac{dP_{\mathrm n}}{d\rho}\rho_2;
]
[
\begin{gathered}
\tau_{\rho\theta}^{\prime\prime}
-(\sigma_\theta^0-\sigma_\rho^0)(\dot R_2-R_1\dot R_1)
-(\sigma_\theta'-\sigma_\rho')\dot R_1
+\frac{d}{d\rho}\left[\tau_{\rho\theta}'-(\sigma_\theta^0-\sigma_\rho^0)\dot R_1\right]\rho_1
=\
=\frac{d^2P_\tau}{d\rho^2}\frac{\rho_1^2}{2!}
+\frac{dP_\tau}{d\rho}\rho_2;
\end{gathered}
\tag{7}
]
[
\begin{gathered}
\sigma_\rho^{\prime\prime\prime}
+\frac{d\sigma_\rho^{\prime\prime}}{d\rho}\rho_1
+\frac{d^2\sigma_\rho'}{d\rho^2}\frac{\rho_1^2}{2!}
+\frac{d^3\sigma_\rho^0}{d\rho^3}\frac{\rho_1^3}{3!}
+2(\sigma_\theta^0-\sigma_\rho^0)(\dot R_1\dot R_2-R_1\dot R_1^2)+\
+\frac{d(\sigma_\theta-\sigma_\rho)}{d\rho}\dot R_1^2\rho_1
-(\sigma_\theta'-\sigma_\rho')\dot R_1^2
-2\tau_{\rho\theta}'(\dot R_2-R_1\dot R_1)-\
-2\frac{d\tau_{\rho\theta}}{d\rho}\dot R_1\rho_1
-2\tau_{\rho\theta}^{\prime\prime}\dot R_1
+\frac{d\sigma_\rho'}{d\rho}\rho_2
+\frac{d^2\sigma_\rho^0}{d\rho^2}\rho_1\rho_2
+\frac{d\sigma_\rho^0}{d\rho}\rho_3=\
=\frac{d^3P_\Pi}{d\rho^3}\frac{\rho_1^3}{3!}
+\frac{d^2P_\Pi}{d\rho^2}\rho_1\rho_2
+\frac{dP_\Pi}{d\rho}\rho_3;
\end{gathered}
]
[
\begin{gathered}
\tau_{\rho\theta}^{\prime\prime\prime}
-2\tau_{\rho\theta}'\dot R_1^2
-(\sigma_\theta^0-\sigma_\rho^0)(\dot R_3-R_1\dot R_2+R_1^2\dot R_1-\dot R_1R_2-\dot R_1^3)-\
-(\sigma_\theta'-\sigma_\rho')(\dot R_2-R_1\dot R_1)
-(\sigma_\theta^{\prime\prime}-\sigma_\rho^{\prime\prime})\dot R_1
+\frac{d}{d\rho}\left[\tau_{\rho\theta}^{\prime\prime}-(\sigma_\theta^0-\sigma_\rho^0)(\dot R_2-R_1\dot R_1)-\right.\
\left.-(\sigma_\theta'-\sigma_\rho')\dot R_1\right]\rho_1
+\frac{d^2}{d\rho^2}\left[\tau_{\rho\theta}'-(\sigma_\theta^0-\sigma_\rho^0)\dot R_1\right]\frac{\rho_1^2}{2!}
+\frac{d}{d\rho}\left[\tau_{\rho\theta}'-(\sigma_\theta^0-\sigma_\rho^0)\dot R_1\right]\rho_2=\
=\frac{d^3P_\tau}{d\rho^3}\frac{\rho_1^3}{3!}
+\frac{d^2P_\tau}{d\rho^2}\rho_1\rho_2
+\frac{dP_\tau}{d\rho}\rho_3,
\end{gathered}
]
where (R_i=\rho_i/\rho_0).
Since on the boundary of the plastic region (L_s) the solutions for the plastic and elastic regions are matched continuously (1):
[
[\sigma_\rho]=[\sigma_\theta]=[\tau_{\rho\theta}]=0
\quad \text{on } L_s,
\tag{8}
]
then, representing the equation of the contour (L_s) in the form
[
\rho_s=\sum_{n=0}\delta^n\rho_{sn}(\theta),
\tag{9}
]
we obtain the linearized matching conditions (8) in the form
[
\begin{gathered}
\left[\sigma_{ij}'+\frac{d\sigma_{ij}^0}{d\rho}\rho_{s1}\right]=0,\qquad
\left[\sigma_{ij}^{\prime\prime}
+\frac{d\sigma_{ij}'}{d\rho}\rho_{s1}
+\frac{d^2\sigma_{ij}^0}{d\rho^2}\frac{\rho_{s1}^2}{2!}
+\frac{d\sigma_{ij}^0}{d\rho}\rho_{s2}\right]=0;\
\left[\sigma_{ij}^{\prime\prime\prime}
+\frac{d\sigma_{ij}^{\prime\prime}}{d\rho}\rho_{s1}
+\frac{d^2\sigma_{ij}'}{d\rho^2}\frac{\rho_{s1}^2}{2!}
+\frac{d^3\sigma_{ij}^0}{d\rho^3}\frac{\rho_{s1}^3}{3!}
+\frac{d\sigma_{ij}^0}{d\rho}\rho_{s3}
+\frac{d\sigma_{ij}'}{d\rho}\rho_{s2}+\right.
\tag{10}\
\left.
+\frac{d^2\sigma_{ij}^0}{d\rho^2}\rho_{s1}\rho_{s2}
\right]=0\ \text{etc.}
\end{gathered}
]
We note that from the equilibrium equations and conditions (8) it follows that
[
\left[\frac{d\sigma_\rho^0}{d\rho}\right]=0
\quad \text{on } L_s .
]
Consequently, the matching conditions (10) for (\sigma_\rho) and (\tau^{\rho\theta}) play the role of boundary conditions for determining the stresses in the elastic region, while the matching condition for (\sigma_\theta) serves to determine (\rho_{sn}). The greatest difficulty and interest in such problems is the determination of the equation of the boundary of the plastic zone (L_s). We shall present several approximate solutions.
- Biaxial tension of a thick plate with a circular hole of radius (a) by forces (P_1) and (P_2). Denoting
[
\delta=\frac{|P_1-P_2|}{2k},
\tag{11}
]
where (k=\frac14\sigma_s) according to Saint-Venant and (k=\frac13\sigma_s) according to Mises, one may obtain:
[
\rho_s=1+\delta\cos 2\theta-\frac34\delta^2(1-\cos 4\theta)+\frac58\delta^3(-\cos 2\theta+\cos 6\theta)+
]
[
+\frac{7}{64}\delta^4(-1-4\cos 4\theta+5\cos 8\theta)+\cdots
\tag{12}
]
Expansion (12) coincides exactly with the expansion of the equation of an ellipse with semiaxes ((1+\delta)), ((1-\delta)), which, as shown in ((^2)), is the exact solution of this problem.
- Biaxial tension of a thin plate with a circular hole of radius (a) by forces (P_1) and (P_2). Introducing (\delta) (11), one may obtain:
[
\rho_s=1+4\delta^\cos 2\theta-8\delta^{2}(1-2\cos 4\theta)-80\delta^{3}(\cos 2\theta-\cos 6\theta)+
]
[
+32\delta^{4}(1-16\cos 4\theta+14\cos 8\theta)+\cdots,
\tag{13}
]
where (\delta^*=\delta/\alpha;\ \alpha=a/r_s^0;\ r_s^0) is the size of the radius of the plastic zone for (\delta=0). The first approximation (13) was obtained in ((^3)).
- Biaxial tension of a thin plate with an elliptical hole by forces (P_1d_2) and (P_2d_2), directed at an angle (\theta_0) to the principal central axes of the ellipse. Introducing (\delta) (11), one may obtain:
[
\rho_s=1+\delta^(4d_2\cos 2(\theta-\theta_0)+3\alpha d_1\cos 2\theta)+\delta^{2}{d_1^2(\alpha^2/4-8\alpha^4)-
]
[
-(18d_1d_2\alpha\cos 2\theta_0+8d_2^2)+[-d_1^2(15/4\,\alpha^2-8\alpha^3-3/4\,\alpha^4)+
]
[
+(18d_1d_2\alpha\cos 2\theta_0+16d_2^2\cos 4\theta_0)]\cos 4\theta+
]
[
+[18d_1d_2\alpha\sin 2\theta_0+16d_2^2\sin 4\theta_0]\sin 4\theta}+\cdots,
]
where the equation of the ellipse of the hole is represented in the form
[
\rho=\alpha+\delta\alpha d_1\cos 2\theta-\delta^2\frac{3\alpha d_1^2}{4}(1-\cos 4\theta)+\cdots .
]
For (d_1=0,\ \theta_0=0,\ d_2=1) there is the case of biaxial tension of a thin plate with a circular hole; for (d_2=0,\ d_1=1), the case of uniform tension of a thin plate with an elliptical hole.
- An eccentric tube under the action of internal pressure (p_0). Referring all linear quantities to the outer radius of the tube (b), denote (\delta=c/b), where (c) is the eccentricity of the tube. One may obtain:
[
\rho_s=\beta_0-\delta\frac{2\beta_0^4}{1-\beta_0^4}\cos\theta+\delta^2\left{\frac{\beta_0^3(2-\beta_0^4-\beta_0^6)}{(1-\beta_0^2)(1-\beta_0^4)^2}+\frac{2\beta_0^7}{(1-\beta_0^4)^2}+\right.
]
[
+\frac{1}{N}\left[-\frac{(1-\beta_0^2)(1-3\beta_0^4)}{\beta_0}
+\frac{(1-\beta_0^2)^2(5+3\beta_0^4)\beta_0}{(1-\beta_0^4)}-\right.
]
[
\left.\left.-\frac{\beta_0^3}{(1-\beta_0^4)^2}\bigl[(1+\beta_0^2)^4+4\beta_0^4(2-\beta_0^2)^2-4(1+4\beta_0^4)\bigr]
+\frac{2\beta_0^3(1-\beta_0^2)^2}{(1+\beta_0^2)^2}\right]\cos 2\theta\right},
]
where (\beta_0=r_s^0/b,\ N=(\beta_0-1/\beta_0)^4).
Moscow State University
named after M. V. Lomonosov
Received
9 X 1956
References
- V. V. Sokolovskii, Theory of Plasticity, Moscow, 1950.
- L. A. Galin, Applied Mathematics and Mechanics, vol. 3 (1946).
- A. P. Sokolov, DAN, vol. 60, No. 1 (1948).
- G. N. Savin, Stress Concentration around Holes, Moscow, 1951.