It is extremely common circular shaft part, under normal circumstances, it is a three-dimensional stress analysis of the problem, but the tension, torsion, pure bending force three common case, the problem can be reduced to two dimensions. Although the capacity of today's computers, deal directly with the three-dimensional problem is not too difficult, but still very significant simplification. Information on elastic mechanics and finite element has been a lot in this article a brief introduction, and then apply it to the analysis of circular shaft.
Elasticity equation in cylindrical coordinates
Geometry axisymmetric circular shaft, using a cylindrical coordinate system is more convenient. In cylindrical coordinates, the strain and displacement following expression
\ [\ begin {pmatrix} \ varepsilon_r \\ \ varepsilon_ \ theta \\ \ varepsilon_z \\ \ gamma_ {r \ theta} \\ \ gamma _ {\ theta z} \\ \ gamma_ {zr} \ end { pmatrix} = \ begin {pmatrix} \ frac {\ partial} {\ partial r} & 0 & 0 \\ \ frac {1} {r} & \ frac {1} {r} \ frac {\ partial} {\ partial \ theta} & 0 \\ 0 & 0 & \ frac {\ partial} {\ partial z} \\ \ frac {1} {r} \ frac {\ partial} {\ partial \ theta} & \ frac {\ partial} {\ partial r} - \ frac {1} {r} & 0 \\ 0 & \ frac {\ partial} {\ partial z} & \ frac {1} {r} \ frac {\ partial} {\ partial \ theta} \\ \ frac {\ partial} {\ partial z} & 0 & \ frac {\ partial} {\ partial r} \ end {pmatrix} \ begin {pmatrix} u_r \\ u_ \ theta \\ u_z \ end
{pmatrix} \] referred to as \ (\ boldsymbol {\ varepsilon} = \ mathbf Su} {\) . Cartesian coordinates the relationship between stress and strain, and similar
\(F_z\)，则圆柱坐标下的平衡方程为
\[ \begin{aligned} & \frac{\partial \sigma_{r}}{\partial r} + \frac{1}{r}\frac{\partial \tau_{r\theta}}{\partial \theta} + \frac{\partial \tau_{zr}}{\partial z} + \frac{1}{r}(\sigma_{r}-\sigma_{\theta}) + F_r= 0 \\ & \frac{\partial \tau_{r\theta}}{\partial r} + \frac{1}{r}\frac{\partial \sigma_{\theta}}{\partial \theta} + \frac{\partial \tau_{\theta z}}{\partial z} + \frac{2}{r}\tau_{r\theta} + F_\theta = 0 \\ & \frac{\partial \tau_{zr}}{\partial r} + \frac{1}{r}\frac{\partial \tau_{\theta z}}{\partial \theta} + \frac{\partial \sigma_{z}}{\partial z} + \frac{1}{r}\tau_{zr} + F_z = 0 \end{aligned} \]
FEM
, The displacement of the respective components of the stress expressed by the above relation, substituting the equilibrium equation to obtain three equations for displacements, displacement component are three unknown, the corresponding solution can be obtained. However, the expression of the equation is very complex, in addition to a very simple case, not the analytical solution, so that applications typically using numerical methods. In the numerical solution, the finite element method is a major way in the field of elastic mechanics is an absolute position. In fact, the finite element method is not to solve partial differential equations, but the use of weighted residuals method into another partial differential equation of the form. This conversion may correspond to the principle of virtual work Mechanics: provided virtual displacement \ (\ Delta \ mathbf {U} \) , the corresponding virtual strain \ (\ delta \ boldsymbol {\ varepsilon} = \ mathbf {S \ delta U} \) , virtual strain energy generated as \ (\ int_ \ Omega \ delta \ boldsymbol {\ varepsilon} ^ T \ boldsymbol {\ sigma} dV = \ int_ \ Omega \ delta \ boldsymbol {\ varepsilon} ^ T \ {D} mathbf \ boldsymbol {\ varepsilon} dV \) , should be equal to the work as an external force on the virtual displacement, i.e. the volume of the work force \ (\ int_ \ Omega \ delta \ mathbf {u ^ TF} dV \) plus the upper boundary of the work force \ (\ int _ {\ partial \ Omega} \ Delta \ mathbf T ^ F {U} dS \) .
Finite element method, the need to solve domain is divided into a number of cells, and then take some of the nodes on the boundary unit, the displacement domain approximately the nodal displacements \ (\ mathbf {u} _i \) interpolation
\ [\ = {U} mathbf \ sum_i n_i \ mathbf {U} _i \]
\ (n_i \) is the interpolation function, only the node is not zero in adjacent cells. This approximate expression is substituted into equation principle of virtual work, can be obtained on \ (\ mathbf {u} _i \) linear equations. In the process of establishing the equation, it is necessary as the form \ (N_iN_j \) , or a derivative thereof integral of the product, and that only when the \ (I \) , \ (J \) if belonging to the same unit is not zero, these integrals of one unit can be calculated, and the obtained equation, most coefficients are zero, it is possible to efficiently build and solve the equations. This advantage is an important reason for the finite element method has been widely applied. Obtained from Equations \ (\ mathbf {u} _i \) , the interpolation formula can be further according to the calculated strain, stress and the like.
Turning next to the subject of this article, about solving three different stress situations. The key is to determine the respective equations, i.e. find the corresponding matrix \ (\ mathbf {S} \ ) and \ (\ mathbf {D} \) .
Twist
In this case, the displacement and \ (\ Theta \) irrelevant, and only \ (u_ \ theta \) is not zero, so that only the strain component \ (\ R & lt gamma_ {\ Theta} \) , \ (\ {gamma_ \ theta z} \) is not zero
\ [\ begin {pmatrix} \ gamma_ {r \ theta} \\ \ gamma _ {\ theta z} \ end {pmatrix} = \ begin {pmatrix} \ frac {\ partial} { \ partial r} - \ frac {
1} {r} \\ \ frac {\ partial} {\ partial z} \ end {pmatrix} u_ \ theta \] stress components accordingly only two
\ [\ begin {pmatrix} \ tau_ {r \ theta} \\ \ tau _ {\ theta z} \ end {pmatrix} = \ frac {E} {2 (1+ \ nu)} \ begin {pmatrix} 1 & 0 \\ 0 & 1 \ end {pmatrix} \ begin {pmatrix
} \ gamma_ {r \ theta} \\ \ gamma _ {\ theta z} \ end {pmatrix} \] coefficients, i.e., where the shear modulus \ (G = \ frac {E } { 2 (1+ \ NU)} \) . Only the original problem into an unknown quantity \ (u_ \ theta \) is a two-dimensional problem. Force the ends of the boundary conditions may be set according to material mechanics, is proportional to \ (R & lt \), The torque dividing ratio of cross-sectional polar moment of inertia.
On this issue but also as a supplement. Noting
\ [\ gamma_ {r \ theta } = r \ frac {\ partial (u_ \ theta / r)} {\ partial r}, \, \ gamma _ {\ theta z} = r \ frac {\ partial (u_ \ theta / r)} {\
partial z} \] If \ (\ psi = u_ \ theta / r \) is unknown, the equation can be written more symmetrical. Virtual strain energy can be expressed as
\ [2 \ pi G \ iint_ \ Omega \ left (\ frac {\ partial \ delta \ psi} {\ partial r} \ frac {\ partial \ psi} {\ partial r} + \ frac {\ partial \ delta \ psi}
{\ partial z} \ frac {\ partial \ psi} {\ partial z} \ right) r ^ 3drdz \] this actually corresponds to the form of conduction equations. If not familiar with such correspondence, can be derived from the balance equation. Since only \ (\ R & lt tau_ {\ Theta} \) , \ (\ of tau _ {\ Theta Z} \) is not zero, three equilibrium equations only consider the second, with the stress (\ PSI \) \ Expression , into equation
\ [\ Frac {\ partial} {\ partial r} \ left (Gr \ frac {\ partial \ psi} {\ partial r} \ right) + \ frac {\ partial} {\ partial z} \ left (Gr \ frac {\ partial \ psi} {
\ partial z} \ right) + \ frac {2} {r} \ left (Gr \ frac {\ partial \ psi} {\ partial r} \ right) = 0 \] sides except in \ (G \) , multiplied by \ (R & lt ^ 2 \)
\ [R & lt ^. 3 \ FRAC {\ partial ^ 2 \ PSI} {\ partial R & lt ^ 2} + 3R ^ 2 \ FRAC {\ partial \ PSI } {\ partial r} + r
^ 3 \ frac {\ partial ^ 2 \ psi} {\ partial z ^ 2} = 0 \] i.e.
\ [\ frac {\ partial} {\ partial r} \ left (r ^ 3 \ frac {\ partial \ psi } {\ partial r} \ right) + \ frac {\ partial} {\ partial z} \ left (r ^ 3 \ frac {\ partial \ psi} {\ partial z} \ right ) = 0 \]
is a two-dimensional conduction equation, but conductivity is not constant, but equal to \ (R & lt ^. 3 \) . Because of this relationship, the thickness of the plate is taken as \ (R & lt ^. 3 \) , via electrical conduction simulation, numerical methods in mature before the torsional shaft can be accurately analyzed.
Tensile
Quite the stretched and twisted in front of the case with a complementary. Also the displacement \ (\ Theta \) independent, but \ (u_ \ theta \) is zero, and \ (diode with U_r \) , \ (u_z \) is not zero. In this case, only shear strain \ (\ gamma_ {zr} \ ) is not zero
\ [\ begin {pmatrix} \ varepsilon_r \\ \ varepsilon_ \ theta \\ \ varepsilon_z \\ \ gamma_ {zr} \ end {pmatrix} = \ begin {pmatrix} \ frac {\ partial} {\ partial r} & 0 \\ \ frac {1} {r} & 0 \\ 0 & \ frac {\ partial} {\ partial z} \\ \ frac {\ partial} {\ partial z
} & \ frac {\ partial} {\ partial r} \ end {pmatrix} \ begin {pmatrix} u_r \\ u_z \ end {pmatrix} \] stress correspondingly simplified
\ [\ Begin {pmatrix} \ sigma_r \\ \ sigma_ \ theta \\ \ sigma_z \\ \ tau_ {zr} \ end {pmatrix} = \ frac {E} {(1+ \ nu) (1-2 \ nu )} \ begin {pmatrix} 1- \ nu & \ nu & \ nu & 0 \\ \ nu & 1- \ nu & \ nu & 0 \\ \ nu & \ nu & 1- \ nu & 0 \\ 0 & 0 & 0 & \ frac { 1-2 \ nu} {2} \ end {pmatrix} \ begin {pmatrix} \ varepsilon_r \\ \ varepsilon_ \ theta \\ \ varepsilon_z \\ \ gamma_ {zr} \ end {pmatrix } \]
original question turned on \ (diode with U_r \) , \ (u_z \) is a two-dimensional problem. Force the ends of the boundary conditions is a uniform force, equal to the total force divided by the cross sectional area. Commercial finite element software provides such a method and geometric forces are axisymmetric case special processing.
bending
In this case two complex than many of the above, the displacement component is not zero, and with \ (\ Theta \) changes. Looking for ideas, look at one kind of situation has been simple analytical solutions. Polar moment of inertia of cross section of a cylindrical rod \ (I_P = \ FRAC {\ PI. 4 D ^ {64}} \) , the torque (M \) \ curvature of the curved under \ (\ kappa = \ frac { M} {EI_p} \) . Provided in the bending \ (an xz \) plane, bent \ (X \) the negative direction of the shaft, the displacement of solution as \ (u_x = - \ frac12 \ kappa (z ^ 2 + \ nu (x ^ 2-y ^ 2)) \) , \ (u_y = - \ Kappa \ NU XY \) , \ (u_z = \ Kappa an xz \) . Is converted to cylindrical coordinates \ (diode with U_r = - \ frac12 \ Kappa (Z ^ 2 + \ R & lt NU ^ 2) \ COS \ Theta \) , \ (U_ \ Theta = \ frac12 \ Kappa (Z ^ 2- \ NU R & lt ^ 2) \ SiN \ Theta \) , \ (u_z = \ Kappa Zr \ COS \ Theta \) , and the stress \ (\ sigma_z = \ frac { M} {I_p} x = \ frac {M} {I_p} R & lt \ COS \ Theta \) , other stresses are zero.
For general shaft bending problem solution in front of inspired us assume: \ (diode with U_r = \ bar diode with U_r \ COS \ Theta \) , \ (U_ \ Theta = \ bar U_ \ Theta \ SiN \ Theta \) , \ (u_z = \ bar u_z \ COS \ Theta \) , and the overline term \ (\ Theta \) regardless of the strain can be obtained
\[ \begin{aligned} \varepsilon_{r} & = \frac{\partial \bar u_r}{\partial r}\cos\theta \\ \varepsilon_{\theta} &= \frac{1}{r}\left(\bar u_\theta\cos\theta + \bar u_r\cos\theta\right) \\ \varepsilon_{z} &= \frac{\partial \bar u_z}{\partial z}\cos\theta \\ \gamma_{r\theta} & = -\frac{\bar u_r}{r}\sin\theta + \frac{\partial \bar u_\theta}{\partial r}\sin\theta - \frac{\bar u_\theta}{r}\sin\theta \\ \gamma_{\theta z} & = \frac{\partial \bar u_\theta}{\partial z}\sin\theta - \frac{1}{r}\bar u_z\sin\theta \\ \gamma_{zr} &= \ Frac {\ partial \ bar u_r} {\ partial z} \ cos \ theta + \ frac {\ partial \ bar u_z} {\ partial r} \ cos \ theta \ end {aligned} \] \ (\ cos \ theta \)with only\ (\ sigma_z \)correlation, and further a\ (\ sin \ theta \)or only with\ (\ cos \ theta \)
As can be seen, each and only oneRelated, and boundary conditions are also the same, so the assumption is feasible. In calculating the virtual work, for \ (\ Theta \) integral appears \ (\ int ^ {2 \ pi} _ {0} \ cos ^ 2 \ theta d \ theta \) or \ (\ int ^ {2 \ {0}} _ PI \ SiN ^ 2 \ Theta D \ Theta \) , their values are \ (\ PI / 2 \) , can be eliminated from the equation, the resulting equation and \ (\ Theta \ ) has nothing to do. The original problem into about \ (\ bar diode with U_r \) , \ (\ bar U_ \ Theta \) , \ (\ bar u_z \) is a two-dimensional problem. Force the ends of the same boundary conditions may be set according to material mechanics, is proportional to \ (R & lt \) , divided by the cross-sectional moment of inertia moment ratio.
Implementation
Finite element software capable of self-defined equations, there are many, as used herein, is open source FreeFem ++. It's a more detailed document, the old version of the document has a Chinese version, the basic usage and the current version is the same. In the manual also gives examples of elasticity, but does not use a matrix to define the problem. Personally I feel that the use of a matrix will be more easy to understand, but also very simple to achieve in the FreeFem ++. The key point is to use the func instead matrix matrix equations to define the required declaration. This article is the most complex to bend the Case
func D = E/(1+nu)/(1-2*nu)*
[[1-nu, nu, nu, 0, 0, 0],
[nu, 1-nu, nu, 0, 0, 0],
[nu, nu, 1-nu, 0, 0, 0],
[0, 0, 0, (1-2*nu)/2, 0, 0],
[0, 0, 0, 0, (1-2*nu)/2, 0],
[0, 0, 0, 0, 0, (1-2*nu)/2]];Then using macros to define the strain
macro epsilon(uz, ur, ut) [dx(uz), dy(ur), (ut+ur)/y, dy(ut)-(ur+ut)/y, dx(ut)-uz/y, dx(ur)+dy(uz)] //Virtual Work integral equations in the time domain definitions can be written as
int2d(Th) (epsilon(uz,ur,ut)'*D*epsilon(vz,vr,vt)*y)Here is \ (X \) axis of the column coordinates \ (Z \) axis, \ (Y \) axis of the column coordinates \ (R & lt \) axis. Plane infinitesimal \ (dxdy \) corresponds to a cylindrical coordinate microvolume \ (2 \ PI ydxdy \) , so the integral equation of the integral is multiplied by the \ (Y \) , \ (2 \ PI \) This constants can be eliminated from both sides of the equation.
Further, the strain formulation, some components appear in the \ (. 1 / R & lt \) , in \ (r = 0 \) occurs when the singular. Therefore, when the geometric definition as is best to avoid \ (r = 0 \) at, say, even if the solid shaft, may leave a tiny hole in the center.