Samer Adeeb

Balance Equations: Momentum Balance

Newton and Euler Laws of Motion

Newton’s second law of motions states that the rate of change with respect to the time t of the linear momentum of an object is equal to the net force acting on that object. If F\in\mathbb{R}^3 is the force vector acting on an object with mass m\in\mathbb{R} moving with a velocity vector v\in\mathbb{R}^3, then, by denoting the momentum by p=mv, we have:

    \[ F=\frac{\mathrm{d}p}{\mathrm{d}t}\)=\frac{\mathrm{d}mv}{\mathrm{d}t}\) \]

If the mass m is constant then:

    \[ F=m\frac{\mathrm{d}v}{\mathrm{d}t}\)=ma \]

where a\in\mathbb{R}^3 is the acceleration vector of the object.
Euler extended Newton’s second law of motion to a continuum object and introduced two laws. The first law is the balance of linear momentum while the second is the balance of angular momentum of of a continuum object. These are as follows:

Euler’s Law of the Balance of Linear Momentum

The first equation is a direct extension of Newton’s first law to an arbitrary volume of a continuum body. Given an arbitrary volume of a continuum body represented by the set \Omega\in\mathbb{R}^3, then the net force acting on the arbitrary volume is equal to the rate of change of the linear momentum of that volume. The net force is the resultant of the traction vectors acting on the boundary and the body forces acting on the continuum points. Thus, according to Euler’s Law:

    \[ \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} \! \rho v \, \mathrm{d}x=\int_{\partial\Omega} \! t_n \, \mathrm{d}s+\int_{\Omega} \! \rho b \, \mathrm{d}x \]

where, t is time, \rho:\Omega\times t\rightarrow\mathbb{R} is the mass density distribution, v:\Omega\times t\rightarrow\mathbb{R}^3 is the velocity vector distribution, b:\Omega\times t\rightarrow\mathbb{R}^3 is the body force distribution, t_n:\partial\Omega\times t\rightarrow\mathbb{R}^3 is the traction vector distribution acting on the surface of \Omega, and \mathrm{d}x and \mathrm{d}s are the differential volume and surface elements of the arbitrary volume \Omega.

Euler’s Law of the Balance of Angular Momentum

Euler’s law of angular momentum preservation states that the rate of change of the angular momentum of an arbitrary volume of a continuum body around a fixed origin is equal to the torque (measured around the same fixed point) applied by the traction vectors on the boundary and the body forces acting on the continuum points. Thus, according to Euler’s Law:

    \[ \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} \! r\times\rho v \, \mathrm{d}x=\int_{\partial\Omega} \! r\times t_n \, \mathrm{d}s+\int_{\Omega} \! r\times\rho b \, \mathrm{d}x \]

where, r:\Omega\times t\rightarrow\mathbb{R}^3 is the position vector relative to the fixed origin.
It should be noted that the fixed origin can be arbitrary. If r_2 is the position vector relative to another fixed point such that r=r_2+r_0, then, by replacing r with r_2+r_0 and using the fact that r_0 is a fixed vector:

    \[ \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} \! r_2\times\rho v \, \mathrm{d}x=\int_{\partial\Omega} \! r_2\times t_n \, \mathrm{d}s+\int_{\Omega} \! r_2\times\rho b \, \mathrm{d}x+ r_0\times\left(-\frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} \! \rho v \, \mathrm{d}x+\int_{\partial\Omega} \! t_n \, \mathrm{d}s+\int_{\Omega} \! \rho b \, \mathrm{d}x\right) \]

By using Euler’s law of the balance of linear momentum we reach to the same statement of the balance of angular momentum with r_2 replacing r:

    \[ \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} \! r_2\times\rho v \, \mathrm{d}x=\int_{\partial\Omega} \! r_2\times t_n \, \mathrm{d}s+\int_{\Omega} \! r_2\times\rho b \, \mathrm{d}x \]

The Differential Equation of Equilibrium: Derivation using a Differential Volume

Balance of Linear Momentum

Newton’s laws of motion were originally formulated for systems of rigid particles mechanically interacting with each other. Newton’s laws can be extended to a continuum by considering a differential volume and the stresses acting on it. Let \Omega\subset\mathbb{R}^3 be a set representing a deformed configuration. Consider a differential cube of mass \Delta m\in\mathbb{R} inside the body with dimensions \Delta x_1,\Delta x_2,\Delta x_3 and whose mass density is \rho=\lim_{\Delta v\rightarrow 0}\frac{\Delta m}{\Delta v}. The differential volume of the cube is given by \Delta v=\Delta x_1 \Delta x_2\Delta x_3. Let F=\{F_1,F_2,F_3\}\in\mathbb{R}^3 be the resultant force acting on the cube which is due to a body forces vector per unit mass b=\{b_1,b_2,b_3\}\in\mathbb{R}^3 acting on the cube and the stresses on the different faces of the cube. To write the differential form of the equilibrium equations, the stresses are assumed to be continuous functions. Such assumption allows the use of the first component in the Taylor expansion for the stresses. Thus, the resultant force component F_1 acting on the cube in the direction of the first basis vector can be evaluated as follows:

    \[ \begin{split} F_1 &= \left(-\sigma_{11}+\sigma_{11}+\frac{\partial \sigma_{11}}{\partial x_1}\Delta x_1\right)\Delta x_2\Delta x_3+\left(-\sigma_{21}+\sigma_{21}+\frac{\partial \sigma_{21}}{\partial x_2}\Delta x_2\right)\Delta x_1\Delta x_3\\ &\qquad+\left(-\sigma_{31}+\sigma_{31}+\frac{\partial \sigma_{31}}{\partial x_3}\Delta x_3\right)\Delta x_1\Delta x_2 +\rho b_1 \Delta x_1\Delta x_2\Delta x_3\\ &=\left(\frac{\partial \sigma_{11}}{\partial x_1}+\frac{\partial \sigma_{21}}{\partial x_2}+\frac{\partial \sigma_{31}}{\partial x_3}+\rho b_1\right)\Delta x_1\Delta x_2\Delta x_3 \end{split} \]

Similarly,

    \[ \begin{split} F_2&=\left(\frac{\partial \sigma_{12}}{\partial x_1}+\frac{\partial \sigma_{22}}{\partial x_2}+\frac{\partial \sigma_{32}}{\partial x_3}+\rho b_2\right)\Delta x_1\Delta x_2\Delta x_3\\ F_3&=\left(\frac{\partial \sigma_{13}}{\partial x_1}+\frac{\partial \sigma_{23}}{\partial x_2}+\frac{\partial \sigma_{33}}{\partial x_3}+\rho b_3\right)\Delta x_1\Delta x_2\Delta x_3 \end{split} \]

If a=\{a_1,a_2,a_3\}\in\mathbb{R}^3 is the acceleration vector of the differential cube, then the statement of Newton’s second law dictates that the resultant force is related to the acceleration via:

    \[ F= \Delta m\,a=\rho \Delta v\,a=\rho \,a\,\Delta x_1\Delta x_2\Delta x_3 \]

Thus, in component form, the equilibrium equations can be written as:

(1)   \begin{equation*} \begin{split} \frac{\partial \sigma_{11}}{\partial x_1}+\frac{\partial \sigma_{21}}{\partial x_2}+\frac{\partial \sigma_{31}}{\partial x_3}+\rho b_1 &=\rho \,a_1\\ \frac{\partial \sigma_{12}}{\partial x_1}+\frac{\partial \sigma_{22}}{\partial x_2}+\frac{\partial \sigma_{32}}{\partial x_3}+\rho b_2&=\rho \,a_2\\ \frac{\partial \sigma_{13}}{\partial x_1}+\frac{\partial \sigma_{23}}{\partial x_2}+\frac{\partial \sigma_{33}}{\partial x_3}+\rho b_3&=\rho \,a_3 \end{split} \end{equation*}

The above equations are usually written in the following compact form with 1\leq i\leq 3:

    \[ \sum_{j=1}^3\frac{\partial \sigma_{ji}}{\partial x_j}+\rho b_i = \rho a_i \]

If the divergence operator is used, the equations of equilibrium can be written in the following vector form:

    \[ \mathrm{div}\sigma+\rho b= \rho a \]

Balance of Angular Momentum

The three angular momentum balance equations can be used to derive the symmetry of the stress tensor. Consider a differential rectangular volume oriented with the coordinate system with volume \Delta v=\Delta x_1 \Delta x_2\Delta x_3. Let the stresses on the sides with the negative normals have stresses \sigma_{ij} and the stresses on the sides with positive normals have stresses \sigma_{ij}+\Delta \sigma_{ij}. The first two equations for the balance of linear momentum dictate the following:

    \[\begin{split} \frac{\mathrm{d}}{\mathrm{d}t}(\rho v_1 \Delta v) &=\Delta \sigma_{11}\Delta x_2\Delta x_3+\Delta \sigma_{21}\Delta x_1\Delta x_3+\Delta \sigma_{31}\Delta x_1\Delta x_2+\rho b_1 \Delta v\\ \frac{\mathrm{d}}{\mathrm{d}t}(\rho v_2 \Delta v) &=\Delta \sigma_{22}\Delta x_1\Delta x_3+\Delta \sigma_{12}\Delta x_2\Delta x_3+\Delta \sigma_{32}\Delta x_1\Delta x_2+\rho b_2 \Delta v \end{split} \]

The moment of the external forces (stresses on the boundary and body forces) acting on the rectangular volume around the axis e_3 have the following form:

    \[\begin{split} M_{external}= & r_1\rho b_2\Delta v - r_2\rho b_1\Delta v + (\sigma_{12}-\sigma_{21})\Delta v \\ &+ r_1 \Delta \sigma_{22}\Delta x_1 \Delta x_3-r_2\Delta \sigma_{11}\Delta x_2 \Delta x_3\\ &+(r_1+\Delta x_1/2)\Delta \sigma_{12}\Delta x_2 \Delta x_3-(r_2+\Delta x_2/2)\Delta \sigma_{21}\Delta x_1 \Delta x_3\\ &+r_1\Delta\sigma_{32}\Delta x_1 \Delta x_2-r_2\Delta \sigma_{31}\Delta x_1 \Delta x_2 \end{split} \]

The rate of change of angular momentum around the axis e_3 is denoted T_3 and has the form:

    \[ T_3= r_1 \frac{\mathrm{d}\rho v_2 \Delta v}{\mathrm{d}t}-r_2 \frac{\mathrm{d}\rho v_1 \Delta v}{\mathrm{d}t} \]

By equating M_{external}=T_3, using the equilibrium equations, and neglecting the smaller terms we get:

    \[ (\sigma_{12}-\sigma_{21})\Delta v=0 \]

Therefore, \sigma_{12}=\sigma_{21}. By repeating the procedure for the moment around other axes we reach:

    \[ \sigma_{ij}=\sigma_{ji} \]

The Differential Equation of Equilibrium: Derivation using Integrals over Arbitrary Volumes

The same differential equations of motion can be obtained using an equivalent but simpler derivation technique that relies on the integration over an arbitrary volume \Omega\in\mathbb{R}^3.

Balance of Linear Momentum

Using Euler’s law of the balance of linear momentum we have:

    \[ \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} \! \rho v \, \mathrm{d}x=\int_{\partial\Omega} \! t_n \, \mathrm{d}s+\int_{\Omega} \! \rho b \, \mathrm{d}x \]

The differential operator \frac{\mathrm{d}}{\mathrm{d}t} on the left hand side cannot be interchanged with the integration operator unless the volume of integration is independent of time. Therefore, we can replace the differential volume \mathrm{d}x with J\mathrm{d}X where J is the determinant of the deformation gradient and \mathrm{d}X is the corresponding differential volume in the reference configuration, which is independent of time. We will also utilize the relationship between the traction vector and the stress tensor: t_n=\sigma^Tn, and the divergence theorem which lead to:

    \[\begin{split} \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega_0} \! \rho vJ \, \mathrm{d}X &=\int_{\partial\Omega} \! \sigma^Tn \, \mathrm{d}s+\int_{\Omega} \! \rho b \, \mathrm{d}x\\ \int_{\Omega_0} \! \frac{\mathrm{d}\rho vJ}{\mathrm{d}t} \, \mathrm{d}X &=\int_{\Omega} \! \mathrm{div}\sigma \, \mathrm{d}x+\int_{\Omega} \! \rho b \, \mathrm{d}x \end{split} \]

The mass balance equations can be used to simplify the left hand side:

    \[\begin{split} \int_{\Omega_0} \! \frac{\mathrm{d}\rho J}{\mathrm{d}t}v+\frac{\mathrm{d}x}{\mathrm{d}t}\rho J \, \mathrm{d}X &=\int_{\Omega} \! \mathrm{div}\sigma \, \mathrm{d}x+\int_{\Omega} \! \rho b \, \mathrm{d}x\\ \int_{\Omega_0} \! \frac{\mathrm{d}x}{\mathrm{d}t}\rho J \, \mathrm{d}X &=\int_{\Omega} \! \mathrm{div}\sigma \, \mathrm{d}x+\int_{\Omega} \! \rho b \, \mathrm{d}x\\ \int_{\Omega} \! \rho a \, \mathrm{d}x &=\int_{\Omega} \! \mathrm{div}\sigma \, \mathrm{d}x+\int_{\Omega} \! \rho b \, \mathrm{d}x\\ 0 &=\int_{\Omega} \!\mathrm{div}\sigma +\rho b - \rho a\, \mathrm{d}x \end{split} \]

Since the integration domain is arbitrary, then the integrand is equal to zero. Therefore, the linear momentum balance equation is:

    \[ \mathrm{div}\sigma +\rho b = \rho a \]

Balance of Angular Momentum

Using Euler’s law of the balance of angular momentum we have:

    \[ \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} \! r\times \rho v \, \mathrm{d}x=\int_{\partial\Omega} \! r\times t_n \, \mathrm{d}s+\int_{\Omega} \! r\times \rho b \, \mathrm{d}x \]

The differential operator \frac{\mathrm{d}}{\mathrm{d}t} on the left hand side cannot be interchanged with the integration operator unless the volume of integration is independent of time. Therefore, we can replace the differential volume \mathrm{d}x with J\mathrm{d}X where J is the determinant of the deformation gradient and \mathrm{d}X is the corresponding differential volume in the reference configuration, which is independent of time. We also utilize the mass balance equations:

    \[\begin{split} \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega_0} \! r\times \rho v J \, \mathrm{d}X &=\int_{\partial\Omega} \! r\times t_n \, \mathrm{d}s+\int_{\Omega} \! r\times \rho b \, \mathrm{d}x\\ \int_{\Omega_0} \! \frac{\mathrm{d}r\times\rho vJ}{\mathrm{d}t} \, \mathrm{d}X &=\int_{\partial\Omega} \! r\times t_n \, \mathrm{d}s+\int_{\Omega} \! r\times \rho b \, \mathrm{d}x\\ \int_{\Omega_0} \! r\times\frac{\mathrm{d}\rho vJ}{\mathrm{d}t} \, \mathrm{d}X &=\int_{\partial\Omega} \! r\times t_n \, \mathrm{d}s+\int_{\Omega} \! r\times \rho b \, \mathrm{d}x\\ \int_{\Omega_0} \! r\times\left(\frac{\mathrm{d}\rho J}{\mathrm{d}t}v+\frac{\mathrm{d} v}{\mathrm{d}t}\rho J\right) \, \mathrm{d}X &=\int_{\partial\Omega} \! r\times t_n \, \mathrm{d}s+\int_{\Omega} \! r\times \rho b \, \mathrm{d}x\\ \int_{\Omega_0} \! r\times\frac{\mathrm{d} v}{\mathrm{d}t}\rho J \, \mathrm{d}X &=\int_{\partial\Omega} \! r\times t_n \, \mathrm{d}s+\int_{\Omega} \! r\times \rho b \, \mathrm{d}x \end{split} \]

Therefore, we have:

    \[ \int_{\Omega} \! r\times \left(\rho a-\rho b\right) \, \mathrm{d}x =\int_{\partial\Omega} \! r\times t_n \, \mathrm{d}s \]

Using the balance of linear momentum equation and replacing t_n with \sigma^T n we get:

    \[ \int_{\Omega} \! r\times \mathrm{div}\sigma \, \mathrm{d}x =\int_{\partial\Omega} \! r\times \sigma^T n \, \mathrm{d}s \]

The last equation can be written in component form to show the symmetry of the stress matrix. The i^{th} component of the above vector equation can be written as:

    \[ \int_{\Omega} \! \sum_{j,k,l=1}^3\varepsilon_{ijk} r_j\frac{\partial\sigma_{lk}}{\partial x_l} \, \mathrm{d}x =\int_{\partial\Omega} \! \sum_{j,k,l=1}^3 \varepsilon_{ijk} r_j \sigma_{lk} n_l \, \mathrm{d}s \]

Using the divergence theorem, the right hand side can be converted into a volume integral as follows:

    \[ \int_{\Omega} \! \sum_{j,k,l=1}^3\varepsilon_{ijk} r_j\frac{\partial\sigma_{lk}}{\partial x_l} \, \mathrm{d}x =\int_{\Omega} \! \sum_{j,k,l=1}^3 \frac{\partial \varepsilon_{ijk}r_j\sigma_{lk}}{\partial x_l} \, \mathrm{d}x \]

The right hand side can be further simplified as follows:

    \[ \begin{split} \int_{\Omega} \! \sum_{j,k,l=1}^3\varepsilon_{ijk} r_j\frac{\partial\sigma_{lk}}{\partial x_l} \, \mathrm{d}x &=\int_{\Omega} \! \sum_{j,k,l=1}^3 \varepsilon_{ijk}\left(\frac{\partial r_j}{\partial x_l}\sigma_{lk}+r_j\frac{\partial \sigma_{lk}}{\partial x_l}\right) \, \mathrm{d}x\\ \int_{\Omega} \! \sum_{j,k,l=1}^3\varepsilon_{ijk} r_j\frac{\partial\sigma_{lk}}{\partial x_l} \, \mathrm{d}x &=\int_{\Omega} \! \sum_{j,k,l=1}^3 \varepsilon_{ijk}\left(\delta_{jl}\sigma_{lk}+r_j\frac{\partial \sigma_{lk}}{\partial x_l}\right) \, \mathrm{d}x\\ 0 &=\int_{\Omega} \! \sum_{j,k,l=1}^3 \varepsilon_{ijk}\delta_{jl}\sigma_{lk} \, \mathrm{d}x\\ 0 &=\int_{\Omega} \! \sum_{j,k=1}^3 \varepsilon_{ijk}\sigma_{jk} \, \mathrm{d}x \end{split} \]

Therefore, the integrand is equal to zero. Setting i=1,2 and 3, the three components of the above vector equation have the form:

    \[\begin{split} 0 & =\varepsilon_{123}\sigma_{23}+\varepsilon_{132}\sigma_{32} \\ 0 & =\varepsilon_{213}\sigma_{13}+\varepsilon_{231}\sigma_{31} \\ 0 & =\varepsilon_{312}\sigma_{12}+\varepsilon_{321}\sigma_{21} \end{split} \]

Therefore:

    \[ \sigma=\sigma^T \]

It should be noted that the book by P. Chadwick has another proof showing the same result without relying on the component form of the vector equation.

Solution of the Equilibrium Equations in Static Problems

A continuum problem is solved if a stress field (the stress at every point inside the body) and an acceleration field are found such that the three equations of force equilibrium (conservation of linear momentum equations) are satisfied at every point. On the other hand, the restriction that \sigma=\sigma^T ensures the conservation of angular momentum and no additional information can be obtained from considering the three equations of moment equilibrium. The unknown stress and velocity fields are required to also satisfy certain boundary conditions, which are traditionally given as either displacements or external traction forces on the exterior of the body and initial conditions for velocities. In static equilibrium, which is the purpose of many engineering applications, the acceleration vector is assumed to be zero and, thus, the equilibrium equation is simplified and the stress field becomes the only unknown. In that case, the equations become:

(2)   \begin{equation*} \begin{split} \frac{\partial \sigma_{11}}{\partial x_1}+\frac{\partial \sigma_{21}}{\partial x_2}+\frac{\partial \sigma_{31}}{\partial x_3}+\rho b_1 &=0\\ \frac{\partial \sigma_{12}}{\partial x_1}+\frac{\partial \sigma_{22}}{\partial x_2}+\frac{\partial \sigma_{32}}{\partial x_3}+\rho b_2&=0\\ \frac{\partial \sigma_{13}}{\partial x_1}+\frac{\partial \sigma_{23}}{\partial x_2}+\frac{\partial \sigma_{33}}{\partial x_3}+\rho b_3&=0 \end{split} \end{equation*}

and \sigma_{ji}=\sigma_{ij}. The problem is mathematically formulated as follows:
Let \Omega_0\subset \mathbb{R}^3 be a set representing a reference configuration of a continuum body. Given a constant or variable body forces vector b\in\mathbb{R}^3, find the distribution of the stresses inside the continuum body that would satisfy the equations of equilibrium (Eq. 2) above. The boundary conditions for the equations of equilibrium are given on two parts of the boundary of \Omega_0. On the first part, \Omega_n, the external traction vectors t_n are known so we have the boundary conditions for \sigma since \sigma^Tn=t_n. On the second part, \Omega_u, the displacement of the continuum is given. The boundary of \Omega_0 is \partial \Omega_0=\Omega_n\cup \Omega_u.

Difficulties Associated with Obtaining a Solution to the Equilibrium Equations

There are two difficulties associated with solving Eq. 2. First, these are three equations of static equilibrium. However, there are six unknowns (six stress variables). Thus, in their current form, many solutions could possibly satisfy the equilibrium equations! The second issue is that some of the boundary conditions contain expressions of displacements, while the equations themselves in this form do not have displacements as variables! These two major issues impel replacing the six unknown stress variables in the above equations with three unknown variables (usually the three displacements u_1, u_2, and u_3. This can be performed by using a “constitutive equation” that describes the relationship between the stress and the strains inside the material. By replacing the stresses with the strains, and by using the relationship between the strains and the displacements described in the strain measures section, the problem becomes well posed, i.e., a solution can be obtained. The following few problems illustrate the applications and the solutions of the equilibrium equations given simplified assumptions allowing such solutions to exist.

Examples and Problems

Example 1

A stress field over a body that is in static equilibrium has the following form:

    \[ \sigma=\left(\begin{matrix}5x_1^2+3x_2+x_3& -x_2x_1& 0 \\ -x_2 x_1 & 5 x_2 & 0\\ 0 & 0 & 9x_3^2\end{matrix}\right) \]

where x_1, x_2, and x_3 are the coordinates inside the body. Find the body forces vector field that is in equilibrium with this stress field.

Solution

Since the given stress matrix is symmetric, it automatically satisfies the angular momentum balance equations. Using Eq. 2 above, the body forces vector field has the following form:

    \[ \rho b=\left(\begin{array}{c}-9x_1\\x_2-5\\-18x_3\end{array}\right) \]

View Mathematica Code
x={x1,x2,x3};
s={{5*x1^2+3x2+x3,-x2*x1,0},{-x2*x1,5x2,0},{0,0,9x3^2}};
rhob1=-Sum[D[s[[i,1]],x[[i]]],{i,1,3}]
rhob2=-Sum[D[s[[i,2]],x[[i]]],{i,1,3}]
rhob3=-Sum[D[s[[i,3]],x[[i]]],{i,1,3}]

Example 2

The shown vertical beam has a varying circular cross sectional area with a radius r=2 metres at the top varying linearly to r=1 metres at the bottom. The beam is used to carry a concentrated load of P=50N, has a density \rho=20 kg/m^3, and is subjected to a gravity field with g = 10 m/sec^2. Assuming that (a) the only nonzero stress component is \sigma_{22}, (b) \sigma_{22} is constant on every cross section perpendicular to e_2, and (c) the undeformed and deformed coordinates coincide, find the distribution of the stress component \sigma_{22} along the length of the beam.
eqproblem1

Solution

The radius r and the area A vary with X_2 according to the equations:

    \[\begin{split} r&=\left(2-\frac{X_2}{L}\right)\\ A&=\pi\left(2-\frac{X_2}{L}\right)^2 \end{split} \]

A horizontal slice of the beam can be analyzed to find the equilibrium equation as follows:
eqproblem2
Equating the sum of the vertical forces acting on the slice to zero yields:

    \[ -\sigma_{22}A+\rho g A (dX_2)+\left(\sigma_{22}+\frac{\partial \sigma_{22}}{\partial X_2}(dX_2)\right)\left(A+\frac{\partial A}{\partial X_2}(dX_2)\right)=0 \]

Simplifying and rearranging:

    \[ \frac{\partial \sigma_{22}}{\partial X_2}A+\sigma_{22}\frac{\partial A}{\partial X_2}+\rho g A=0 \]

A boundary condition for the stress is given at the free end as follows:

    \[ @X_2=L:\sigma_{22}=\frac{P}{A|_{X_2=L}}=\frac{P}{\pi} \]

Even with the very simplified assumption that \sigma_{22} is the only nonzero component of the stress matrix, the differential equation is relatively complicated. The command “DSolve” in Mathematica can be used to solve differential equations and was used to find the following distribution of the stress:

    \[ \sigma_{22}=\frac{1}{3}\left(2\rho g L-\rho gX_2+\left(\frac{L^2(3P-\rho g L\pi)}{\pi (X_2-2L)^2}\right)\right)=\frac{1}{3}\left(2000-\frac{23806.3}{(X_2-10)^2}-200X_2\right) N/m^2 \]

View Mathematica Code
Clear[s,x,L,g]
A=Pi*(2-x/L)^2;
a=FullSimplify[DSolve[{s'[x]*A+D[A,x]*s[x]+ro*g*A==0,(s[L]==P/A/.x->L)},s[x],x]]
ss=s[x]/.a[[1]]
N[ss/.{L->5,ro->20,g->10,P->50}]

Example 3

The shown horizontal beam has a density \rho=1 kg/m^3 and is under a constant horizontal body force component b_1=10 N/kg. Assuming that (a) \sigma_{11} is the only nonzero stress component, (b) \sigma_{11} is only a function of x_1, and (c) the undeformed and deformed coordinates coincide, find the distribution of \sigma_{11} along the beam.

eqproblem3

Solution

It should be noted that the first two assumptions are very strong. For example, closer to the fixed boundary, there could be other nonzero stress components which would cause stress concentrations. Additionally, \sigma_{11} could be higher in the middle and lower closer to the boundaries. However, to easily achieve a closed form solution to the equilibrium equations, such assumptions are required. Otherwise, a numerical solution (for example, using finite element analysis) would be required to find all the components of the stress.

The equation of equilibrium in the horizontal direction is given by:

    \[ \frac{\partial \sigma_{11}}{\partial x_1}+\rho b_1=0 \]

Assuming that \sigma_{11} is a function of only x_1 then:

    \[ \frac{\mathrm{d} \sigma_{11}}{\mathrm{d} x_1}+\rho b_1=0 \]

Integrating the above equation yields:

    \[ \sigma_{11}=-10x_1+C \]

where C is a constant that can be obtained from the boundary condition given for the stress:

    \[ @x_1=5:\sigma_{11}=20\Rightarrow -10(5)+C=20\Rightarrow C=70 N/m^2 \]

Therefore, the stress distribution is:

    \[ \sigma_{11}=-10x_1+70 N/m^2 \]

Notice that if the end of the beam x_1=5 metres was also fixed, then the given boundary conditions would not be sufficient to solve the differential equation of equilibrium and a constitutive law would be required to replace the stresses with expressions of the displacement in the differential equation of equilibrium.

Example 4

The following is the stress field inside a body in static equilibrium

    \[ \sigma=\left(\begin{matrix}\alpha x_1^2+\delta x_1+3x_2+x_3 & -\beta x_2 x_1 & 0 \\ -\beta x_2 x_1 & 5 x_2^2 & 0\\ 0 & 0 & \gamma x_3^2\end{matrix}\right) \]

If the body is subjected to the following body forces vector:

    \[ \rho b=\left(\begin{array}{c} 10 x_1\\10x_2\\10x_3\end{array}\right) \]

then, find the values of \alpha, \beta, \delta, and \gamma such that static equilibrium is achieved.

Solution

The first step is to check that the stress matrix is symmetric to satisfy balance of angular momentum. Indeed, the given stress matrix is symmetric.
The equations of static equilibrium are given as:

    \[ \begin{split} \frac{\partial \sigma_{11}}{\partial x_1}+\frac{\partial \sigma_{21}}{\partial x_2}+\frac{\partial \sigma_{31}}{\partial x_3}+\rho b_1 &=2\alpha x_1+\delta  -\beta x_1+10x_1=0\\ \frac{\partial \sigma_{12}}{\partial x_1}+\frac{\partial \sigma_{22}}{\partial x_2}+\frac{\partial \sigma_{32}}{\partial x_3}+\rho b_2&=-\beta x_2+10x_2+10x_2=0\\ \frac{\partial \sigma_{13}}{\partial x_1}+\frac{\partial \sigma_{23}}{\partial x_2}+\frac{\partial \sigma_{33}}{\partial x_3}+\rho b_3&=2\gamma x_3 + 10x_3=0 \end{split} \]

Rearranging:

    \[ \begin{split} (2\alpha-\beta+10) x_1+\delta &=0\\ (-\beta+20) x_2&=0\\ (2\gamma+10) x_3 &=0 \end{split} \]

These equations have to be satisfied at every point inside the object, i.e., they have to be satisfied for all possible values of x_1, x_2, and x_3. Therefore \alpha=5, \beta=20, \delta =0, and \gamma=-5.

View Mathematica Code
x= {x1, x2, x3};
s = {{alpha*x1^2 + delta*x1 + 3*x2 + x3, -beta*x2*x1, 0}, {-beta*x2*x1, 5 x2^2, 0}, {0, 0, gamma*x3^2}};
pb = {10 x1, 10 x2, 10 x3};
Print["Equilibrium Equations:"]
Eq1 = Sum[D[s[[i, 1]], x[[i]]], {i, 1, 3}] + pb[[1]] == 0
Eq2 = Sum[D[s[[i, 2]], x[[i]]], {i, 1, 3}] + pb[[2]] == 0
Eq3 = Sum[D[s[[i, 3]], x[[i]]], {i, 1, 3}] + pb[[3]] == 0
Print["Solving Equations:"]
ss1 = ForAll[{x1, x2, x3}, Eq1 && Eq2 && Eq3]
Solve[Resolve[ss1], {alpha, beta, delta, gamma}]

Problems

  1. A stress field inside a continuum has the following form:

        \[ \sigma=\left(\begin{matrix}\alpha x_1^2+\delta x_1+3x_2+x_3 & -\beta x_2 x_1 & 0 \\ -\beta x_2 x_1 & 5 x_2^2 & 0\\ 0 & 0 & \gamma x_3^2\end{matrix}\right) \]

    The body forces vector applied is given by:

        \[ \rho b=\left(\begin{array}{c}10\\0\\0\end{array}\right) \]

    Show that the values of the constants \alpha,\beta,\delta, and \gamma so that the stress field is in equilibrium with the body forces vector field are 5, 10, -10, and 0, respectively.

  2. A stress field inside a continuum has the following form:

        \[ \sigma=\left(\begin{matrix}\alpha x_1(x_1-5)+\gamma& \beta x_1 x_2 (x_1-5)(x_2-5)& 0 \\ \beta x_1 x_2 (x_1-5)(x_2-5)& 0 & 0\\ 0 & 0 & 0\end{matrix}\right) \]

    If the stress field is in equilibrium with a zero body forces vector field and the values of \sigma_{11} is equal to 5 when x_1=0, show that the values of the constants \alpha, \beta, and \gamma are 0, 0, and 5, respectively.

  3. A two dimensional stress field on a rectangular plate of dimensions 1 and 5 units and 0.01 units of thickness has the following form:

        \[ \sigma=\left(\begin{matrix}5x_1x_2+B & A & 0 \\ A & 0 & 0\\ 0 & 0 & 0\end{matrix}\right) \]

    Find the equilibrium body forces vector and the constants A and B given that for x_1=0, \sigma_{11}=2 units and \sigma_{12}=2 units. Then, drawn the contour plot of the von Mises stress on the plate and identify the critical location (maximum von Mises stress) if the bottom left corner of the rectangular plate has coordinates (0,0).

Leave a Reply

Your email address will not be published. Required fields are marked *