text:
Elasticity/Equilibrium example 2

Example 2
 Given: The displacement equation of equilibrium for an isotropic inhomogeneous linear elastic material can be written as

\boldsymbol{\nabla} \bullet (\mathbf{C} : \boldsymbol{\nabla} \mathbf{u}) + \mathbf{b} = 0 $$ where

\mathbf{C} = \lambda \mathbf{1}^{(2)}\otimes\mathbf{1}^{(2)} + 2\mu\mathbf{1}^{(4s)} $$ and $$\lambda(\mathbf{x})$$ and $$\mu(\mathbf{x})$$ are the Lamé moduli.

Show:

Show that the displacement equation of equilibrium can be expressed as

\mu \boldsymbol{\nabla}\bullet(\boldsymbol{\nabla}\mathbf{u}) + (\lambda+\mu) \boldsymbol{\nabla}(\boldsymbol{\nabla}\bullet\mathbf{u}) + (\boldsymbol{\nabla}\mathbf{u}+\boldsymbol{\nabla}\mathbf{u}^T) \boldsymbol{\nabla}{\mu} + (\boldsymbol{\nabla}\bullet\mathbf{u}) \boldsymbol{\nabla}{\lambda} + \mathbf{b} = 0 $$

Solution
The skew part of the tensor $$\boldsymbol{\nabla}\mathbf{u}$$ does not affect the stress because it leads to a rigid displacement field. Therefore, the displacement equation of equilibrium may be written as

\boldsymbol{\nabla} \bullet \left[\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u})\right] + \mathbf{b} = 0 $$ where

\text{symm}(\boldsymbol{\nabla}\mathbf{u}) = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) $$ In index notataion,

\text{symm}(\boldsymbol{\nabla}\mathbf{u}) = \boldsymbol{\varepsilon} \equiv \varepsilon_{kl} = \frac{1}{2}(u_{k,l} + u_{l,k}) $$ and

\mathbf{C} \equiv C_{ijkl} = \lambda\delta_{ij}\delta_{kl} + \mu(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) $$ Therefore,
 * $$\begin{align}

\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u}) \equiv C_{ijkl}~\varepsilon_{kl} & = \lambda\delta_{ij}\delta_{kl}~\varepsilon_{kl} + \mu\delta_{ik}\delta_{jl}~\varepsilon_{kl} + \mu\delta_{il}\delta_{jk}~\varepsilon_{kl} \\ & = \lambda~\varepsilon_{mm}\delta_{ij} + \mu~\varepsilon_{ij} + \mu~\varepsilon_{ij} \\ & = \lambda~\varepsilon_{mm}\delta_{ij} + 2\mu~\varepsilon_{ij} \\ & \equiv \lambda~(\text{tr}~\boldsymbol{\varepsilon})\mathbf{1} + 2\mu~\boldsymbol{\varepsilon} \end{align}$$ Now,

\text{tr}~\boldsymbol{\varepsilon} \equiv \varepsilon_{mm} = \frac{1}{2}(u_{m,m} + u_{m,m}) = u_{m,m} \equiv \boldsymbol{\nabla}\bullet\mathbf{u} $$ Hence,

\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u}) = \lambda~(\boldsymbol{\nabla}\bullet\mathbf{u})\mathbf{1} + \mu~(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) $$ Taking the divergence,
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left[\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u})\right]} & = \boldsymbol{\nabla}\bullet{\left[\lambda~(\boldsymbol{\nabla}\bullet\mathbf{u})\mathbf{1} + \mu~(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T)\right]} \\ & = \boldsymbol{\nabla}\bullet{\left[\lambda~(\boldsymbol{\nabla} \bullet\mathbf{u})\mathbf{1}\right]} + \boldsymbol{\nabla}\bullet{\left(\mu~\boldsymbol{\nabla}\mathbf{u}\right)} + \boldsymbol{\nabla}\bullet{\left(\mu~\boldsymbol{\nabla}\mathbf{u}^T\right)} \end{align}$$ Recall that
 * $$\begin{align}

\boldsymbol{\nabla}{\phi} & = \phi_{,j} \\ \boldsymbol{\nabla}{\mathbf{v}} & = v_{i,j} \\ \boldsymbol{\nabla}\bullet{\mathbf{v}} & = v_{j,j} \\ \boldsymbol{\nabla}\bullet{\mathbf{T}} & = T_{ij,j} \end{align}$$ Therefore,
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left[\lambda~(\boldsymbol{\nabla} \bullet\mathbf{u})\mathbf{1}\right]} & \equiv \left(\lambda~u_{k,k}\delta_{ij}\right)_{,j} \\ & = \lambda_{,i}~u_{k,k} + \lambda~u_{k,ki} \\ & \equiv \boldsymbol{\nabla}{\lambda}(\boldsymbol{\nabla}\bullet\mathbf{u}) + \lambda\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} \end{align}$$
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left(\mu~\boldsymbol{\nabla}\mathbf{u}\right)} & \equiv \left(\mu~u_{i,j}\right)_{,j} \\ & = \mu_{,j}~u_{i,j} + \mu~u_{i,jj} \\ & \equiv \boldsymbol{\nabla}{\mu} \boldsymbol{\nabla}\mathbf{u} + \mu\boldsymbol{\nabla}\bullet{(\boldsymbol{\nabla}\mathbf{u})} \end{align}$$
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left(\mu~\boldsymbol{\nabla}\mathbf{u}^{T}\right)} & \equiv \left(\mu~u_{j,i}\right)_{,j} \\ & = \mu_{,j}~u_{j,i} + \mu~u_{j,ij} \\ & \equiv \boldsymbol{\nabla}{\mu} \boldsymbol{\nabla}\mathbf{u}^{T} + \mu\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} \end{align}$$ Hence,
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left[\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u})\right]} & = \boldsymbol{\nabla}{\lambda}(\boldsymbol{\nabla} \bullet\mathbf{u}) + \lambda\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} + \boldsymbol{\nabla}{\mu} \boldsymbol{\nabla}\mathbf{u} + \mu\boldsymbol{\nabla}\bullet{(\boldsymbol{\nabla}\mathbf{u})} + \boldsymbol{\nabla}{\mu} \boldsymbol{\nabla}\mathbf{u}^{T} + \mu\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} \\ & = \mu\boldsymbol{\nabla}\bullet{(\boldsymbol{\nabla}\mathbf{u})} + (\lambda+\mu)\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} + \boldsymbol{\nabla}{\mu}\left(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^{T}\right) + \boldsymbol{\nabla}{\lambda}(\boldsymbol{\nabla}\bullet \mathbf{u}) \end{align}$$ Therefore, the displacement equation of equilibrium can be expressed as required, i.e,

\mu \boldsymbol{\nabla}\bullet(\boldsymbol{\nabla}\mathbf{u}) + (\lambda+\mu) \boldsymbol{\nabla}(\boldsymbol{\nabla}\bullet\mathbf{u}) + (\boldsymbol{\nabla}\mathbf{u}+\boldsymbol{\nabla}\mathbf{u}^T) \boldsymbol{\nabla}{\mu} + (\boldsymbol{\nabla}\bullet\mathbf{u}) \boldsymbol{\nabla}{\lambda} + \mathbf{b} = 0 $$

title: Elasticity/Equilibrium example 2


text: Introduction to Robotics/Visuals/FOR NEXT Loop

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Introduction to Robotics/Visuals/Nested Loops

MyCount VAR Byte FOR MyCount = 0 TO 200 FOR MyCount = 0 TO 50 ...   NEXT NEXT MyCount never reaches 200


text:
Introduction to Robotics/Visuals/Nested Loops 1

MyCount1 VAR Byte MyCount2 VAR Byte FOR MyCount1 = 0 TO 200 FOR MyCount2 = 0 TO 200 ...   NEXT NEXT


text:
Introduction to Robotics/Visuals/FOR NEXT Control Flow

How a FOR / NEXT loop works.


text: Introduction to Robotics/Visuals/Labels and GOTO


text:
Category:Fluid mechanics

Fluid mechanics deals with the motion of liquids and gases.

title: Category:Fluid mechanics


text:
Elasticity/Equilibrium example 3

Example 3
Given:

If a material is incompressible ($$\nu$$ = 0.5), a state of hydrostatic stress ($$\sigma_{11} = \sigma_{22} = \sigma_{33}$$) produces no strain. The corresponding stress-strain relation can be written as
 * $$ \sigma_{ij} = 2\mu\varepsilon_{ij} - p\delta_{ij} $$

where $$p$$ is an unknown hydrostatic pressure which will generally vary with position. Also, the condition of incompressibility requires that the dilatation
 * $$ e = \varepsilon_{kk} = 0~. $$

Show:

Show that the stress components and the hydrostatic pressure $$p$$ must satisfy the equations

\nabla^2{p} = \boldsymbol{\nabla}\bullet{\mathbf{b}} ~; \sigma_{11} + \sigma_{22}  = -2p $$ where $$\mathbf{b}$$ is the body force.

Solution
We have, $$ e = \varepsilon_{kk} = \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33} = 0~.\, $$ Also,

\sigma_{11} = 2\mu\varepsilon_{11} - p ~; \sigma_{22} = 2\mu\varepsilon_{22} - p ~; \sigma_{33} = 2\mu\varepsilon_{33} - p ~.\, $$ Therefore,
 * $$\begin{align}

\sigma_{11} + \sigma_{22} + \sigma_{33} & = 2\mu\left(\varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}\right) - 3p\\ & = -3p \end{align}$$ Since $$\sigma_{11} = \sigma_{22} = \sigma_{33}\,$$, the above relation gives $$ \sigma_{11} = \sigma_{22} = \sigma_{33} = -p \,$$. Therefore,

\sigma_{11} + \sigma_{22} = -2p \, $$ The strain-stress relations are

2\mu\varepsilon_{11} = \sigma_{11} + p ~; 2\mu\varepsilon_{22} = \sigma_{22} + p ~; 2\mu\varepsilon_{12} = \sigma_{12} ~. $$ Differentiating the strains so that they correspond to the compatibilityrelation is two-dimensions, we have

\varepsilon_{11,22} = \frac{1}{2\mu}\left(\sigma_{11,22} + p_{,22}\right) ~; \varepsilon_{22,11} = \frac{1}{2\mu}\left(\sigma_{22,11} + p_{,11}\right) ~; \varepsilon_{12,12} = \frac{1}{2\mu}\left(\sigma_{12,12} \right) ~. $$ In terms of the compatibility equation,
 * $$\begin{align}

& \varepsilon_{11,22} + \varepsilon_{22,11} - 2\varepsilon_{12,12} = \frac{1}{2\mu} \left(\sigma_{11,22} + \sigma_{22,11} - 2\sigma_{12,12} +            p_{,11} + p_{,22}\right) \\ \text{or,}~ & 0 = \sigma_{11,22} + \sigma_{22,11} - 2\sigma_{12,12} + \nabla^2{p} \end{align}$$ From the two-dimensional equilibrium equations,

\sigma_{11,1} + \sigma_{12,2} + b_1 = 0 ~; \sigma_{12,1} + \sigma_{22,2} + b_2 = 0 $$ Therefore, differentiating w.r.t $$x_1$$ and $$x_2$$ respectively,

\sigma_{11,11} + \sigma_{12,21} + b_{1,1} = 0 ~; \sigma_{12,12} + \sigma_{22,22} + b_{2,2} = 0 $$ Adding,

2\sigma_{12,12} + \sigma_{11,11} + \sigma_{22,22} + b_{1,1} + b_{2,2} = 0 $$ Hence,

\sigma_{11,11} + \sigma_{22,22} + b_{1,1} + b_{2,2} = - 2\sigma_{12,12} $$ Substituting back into the compatibility equation,
 * $$\begin{align}

& \sigma_{11,22} + \sigma_{22,11} + \sigma_{11,11} + \sigma_{22,22} + b_{1,1} + b_{2,2} + \nabla^2{p} = 0 \\ \text{or,} ~ &  \nabla^2{\sigma_{11}}+\nabla^2{\sigma_{22}}+\nabla^2{p} + \boldsymbol{\nabla}\bullet{\mathbf{b}}  = 0\\ \text{or,} ~ & \nabla^2{(\sigma_{11}+\sigma_{22}+p)} + \boldsymbol{\nabla}\bullet{\mathbf{b}}  = 0 \\ \text{or,} ~ & \nabla^2{(-2p+p)} + \boldsymbol{\nabla}\bullet{\mathbf{b}}  = 0 \\ \text{or,} ~ & -\nabla^2{p} + \boldsymbol{\nabla}\bullet{\mathbf{b}}  = 0 \end{align}$$ Hence,

{ \nabla^2{p} = \boldsymbol{\nabla}\bullet{\mathbf{b}}} $$

title: Elasticity/Equilibrium example 3


text:
Elasticity/Beam bending example 1

Example 1
Given:

A long rectangular beam with cross section $$ab$$

Find:

A solution for the displacement and stress fields, using strong boundary conditions on the edges $$x_2 = 0$$ and $$x_2 = b$$.

[Hint : Assume that the displacement can be expressed as a second degree polynomial (using the Pascal's triangle to determine the terms) $$u(x,y) = Ax^2 + By^2 + Cxy + Dx + Ey + F$$]

Solution
 Step 1: Boundary conditions

 * $$\begin{align}

\text{at}~ x_1 & = 0 ~; \sigma_{13} = 0 \\ \text{at}~ x_1 & = a ~; u_3 = 0 \\ \text{at}~ x_2 & = 0 ~; \sigma_{23} = 0 \\ \text{at}~ x_2 & = b ~; \sigma_{23} = S \end{align}$$

 Step 2: Assume a solution

Let us assume antiplane strain

u_3(x_1,x_2) = Ax_1^2 + Bx_2^2 + Cx_1x_2 + Dx_1 + Ex_2 + F ~; u_1 = u_2 = 0~. $$

 Step 3: Calculate the stresses

The stresses are given by $$\sigma_{\alpha 3} = \mu u_{3,\alpha}$$, and $$\sigma_{11} = \sigma_{22} = \sigma_{33} = \sigma_{12} = 0$$. Therefore,
 * $$\begin{align}

\sigma_{13} & = \mu u_{3,1} = \mu(2Ax_1 + Cx_2 + D) \\ \sigma_{23} & = \mu u_{3,2} = \mu(2Bx_2 + Cx_1 + E) \end{align}$$

Step 4: Satisfy stress BCs

Thus we have,
 * $$\begin{align}

0 & = \mu(Cx_2 + D) \\ 0 & = \mu(Cx_1 + E) \\ S & = \mu(2bB + Cx_1 + E) \end{align}$$ Since $$x_1$$ and $$x_2$$ can be arbitrary, $$ C = D = E = 0$$.

Hence, $$B = S/2\mu b$$ which gives us
 * $$\begin{align}

u_3 & = Ax_1^2 + \frac{S}{2\mu b}x_2^2 + F \\ \sigma_{13} & = \mu(2Ax_1) \\ \sigma_{23} & = \mu(2\frac{S}{2\mu b}x_2) \end{align}$$ Assume that the body force is zero. Then the equilibrium condition is $$\nabla^2{u_3} = 0$$. Therefore,
 * $$\begin{align}

& u_{3,11} + u_{3,22} = 0 \\ \text{or,} \quad & 2A + 2\frac{S}{2\mu b}  = 0 \\ \text{or,} \quad & A  = - \frac{S}{2\mu b} \end{align}$$ Therefore, the stresses are given by

{ \sigma_{13} = -\frac{S}{b} x_1 ~; \sigma_{23} = \frac{S}{b} x_2 } $$

Step 5: Satisfy displacement BCs

The displacement is given by

u_3 = -\frac{S}{2\mu b}x_1^2 + \frac{S}{2\mu b}x_2^2 + F $$ If we substitute $$x_1 = a$$, we cannot determine the constant $$F$$ uniquely.

Hence the displacement boundary conditions have to be applied in a weak sense,
 * $$\begin{align}

& \int_0^b u_3(a, x_2) dx_2 = 0 \\ \text{or,} \quad & \int_0^b \left(-\frac{S}{2\mu b}a^2 + \frac{S}{2\mu b}x_2^2 +                      F\right) dx_2  = 0 \\ \text{or,} \quad & \left. \left[ \frac{S}{2\mu b} \left(-a^2 x_2 + \frac{x_2^3}{3}\right) + F x_2 \right]\right|_0^b = 0\\ \text{or,} \quad & \frac{S}{2\mu b} \left(-a^2 b + \frac{b^3}{3}\right) + F b = 0\\ \text{or,} \quad & \frac{S}{2\mu} \left(-a^2 + \frac{b^2}{3}\right) + Fb = 0\\ \text{or,} \quad & \frac{S}{2\mu} \left(-\frac{a^2}{b}+\frac{b}{3}\right)+ F = 0\\ \text{or,} \quad & F = \frac{S}{2\mu b} \left(a^2 - \frac{b^2}{3}\right) \end{align}$$ Therefore,

{ u_3  = \frac{S}{2\mu b}\left(x_2^2 - x_1^2 + a^2 - \frac{b^2}{3}\right)} $$

title: Elasticity/Beam bending example 1


text: File:Beam bending ex1.png

title: File:Beam bending ex1.png


text:
Elasticity/Plane strain example 1

Example 1
Given:

The plane strain solution for the stresses in a rectangular block with $$0 < x_1 < a$$, $$-b < x_2 < b$$, and $$-c < x_3 < c$$ with a given loading is

\sigma_{11} = \frac{3 F x_1 x_2}{2 b^3}~; \sigma_{12} = \frac{3 F (b^2 - x_2^2)}{4 b^3}~; \sigma_{22} = 0~; \sigma_{33} = -\frac{3 \nu F x_1 x_2}{2 b^3}. $$

Find:

 * 1) Find the tractions on the surfaces of the block and illustrate the results on a sketch of the block.
 * 2) We wish to use this solution to solve the corresponding problem in which the surfaces $$x_3 = \pm c$$ are traction-free. Determine an approximate corrective solution for this problem by offloading the unwanted force and moment results using elementary bending theory.
 * 3) Find the maximum error in the stress $$\sigma_{33}$$ in the corrected solution and compare it with the maximum tensile stress in the plane strain solution.

Solution
The tractions acting on the block are:
 * $$\begin{align}

\text{at}~ x_1 &= 0~,\widehat{\mathbf{n}}{} \equiv (-1,0,0)~,t_i = n_1 \sigma_{1i} \equiv (-\sigma_{11},-\sigma_{12},-\sigma_{13}) \\ \text{at}~ x_1 &= a~,\widehat{\mathbf{n}}{} \equiv (1,0,0)~,t_i = n_1 \sigma_{1i} \equiv (\sigma_{11},\sigma_{12},\sigma_{13}) \\ \text{at}~ x_2 &= -b~,\widehat{\mathbf{n}}{} \equiv (0,-1,0)~,t_i = n_2 \sigma_{2i} \equiv (-\sigma_{21},-\sigma_{22},-\sigma_{23}) \\ \text{at}~ x_2 &= b~,\widehat{\mathbf{n}}{} \equiv (0,1,0)~,t_i = n_2 \sigma_{2i} \equiv (\sigma_{21},\sigma_{22},\sigma_{23}) \\ \text{at}~ x_3 &= -c~,\widehat{\mathbf{n}}{} \equiv (0,0,-1)~,t_i = n_3 \sigma_{3i} \equiv (-\sigma_{31},-\sigma_{32},-\sigma_{33}) \\ \text{at}~ x_3 &= c~,\widehat{\mathbf{n}}{} \equiv (0,0,1)~,t_i = n_3 \sigma_{3i} \equiv (\sigma_{31},\sigma_{32},\sigma_{33}) \end{align}$$ Plugging in the expressions for stress,
 * $$\begin{align}

\text{at}~ x_1 &= 0~,t_i \equiv (0,-\frac{3 F (b^2 - x_2^2)}{4 b^3},0) \\ \text{at}~ x_1 &= a~,t_i \equiv (\frac{3 F a x_2}{2 b^3},\frac{3 F (b^2 - x_2^2)}{4 b^3},0)\\ \text{at}~ x_2 &= -b~,t_i \equiv (0,0,0) \\ \text{at}~ x_2 &= b~,t_i \equiv (0,0,0) \\ \text{at}~ x_3 &= -c~,t_i \equiv (0,0,\frac{3 \nu F x_1 x_2}{2 b^3}) \\ \text{at}~ x_3 &= c~,t_i \equiv (0,0,-\frac{3 \nu F x_1 x_2}{2 b^3}) \end{align}$$ These tractions are illustrated in the following figure

To unload the tractions on the faces $$x_3 = \pm c$$, we have to superpose the solution to a problem with equal and opposite tractions and moments. In order to use elementary bending theory, we have set up a problem with simple boundary conditions.

Let us first consider the force distribution required for the superposed problem. Since the loading is antisymmetric at $$x_3 = \pm c$$, there is no net force is the $$x_3$$ direction. Similarly, there is no net moment about the $$x_2$$ axis.

However, there is a net moment about the $$x_1$$ axis. Hence, the problem to be superposed should have a bending stress distribution $$\sigma_{33} = C x_2$$, where $$C$$ is a constant that is chosen so as to make the total bending moment (original problem + superposed problem) equal to zero. (Note: Think of a beam in the $$x_3 - x_2$$ plane subjected to bending moments at the ends.)

The total stress for the corrected problem is

\sigma_{33} = -\frac{3 \nu F x_1 x_2}{2 b^3} + C x_2 $$ The bending moment for a cross-section of the beam in the $$x_3 - x_2$$ plane about the $$x_1$$ axis is $$ M = -\sigma_{33} I / x_2 \,$$, where $$I = \int_0^a \int_{-b}^b x_2^2 dx_2 dx_1$$.

Since $$\sigma_{33}$$ varies with $$x_1$$, the total bending moment for the beam is given by
 * $$\begin{align}

M & = \int_0^a \int_{-b}^b \left(\frac{-\sigma_{33}}{x_2}\right) x_2^2 dx_2 dx_1 \\ & = \int_0^a \int_{-b}^b \left(\frac{3 \nu F x_1 x_2}{2 b^3} - C x_2 \right) x_2 dx_2 dx_1 \\ & = \int_0^a \left[\frac{3\nu F x_1 x_2^3}{6 b^3} - \frac{C x_2^3}{3} \right]_{-b}^b dx_1 \\ & = \int_0^a \left[\frac{\nu F x_1 b^3}{b^3} - \frac{2C b^3}{3} \right] dx_1 \\ & = \left[\frac{\nu F x_1^2 b^3}{2b^3} - \frac{2C x_1 b^3}{3} \right]_0^a \\ & = \frac{\nu F a^2 b^3}{2b^3} - \frac{2C a b^3}{3} \end{align}$$ Setting the bending moment to zero, we have

C = \frac{3\nu F a}{4 b^3} $$ Therefore, the corrected solution is

{ \sigma_{33} = -\frac{3 \nu F x_1 x_2}{2 b^3} + \frac{3\nu F a x_2}{4 b^3}} $$ Ideally, for a problem with zero tractions on $$x_3 = \pm c$$, we should have $$\sigma_{33} = 0$$. Therefore, the error in our solution is

\sigma_{33}^{\text{err}} = \text{abs}\left(\frac{3 \nu F x_1 x_2}{2 b^3} -                             \frac{3\nu F a x_2}{4 b^3}\right) $$ The error is maximum at $$(0,-b)$$, $$(0,b)$$, $$(a,-b)$$, and $$(a,b)$$. Thus,
 * $$\begin{align}

\left.\sigma_{33}^{\text{err}}\right|_{(0,-b)} & = \text{abs}\left(\frac{3\nu F a b}{4 b^3}\right) = \text{abs}\left(\frac{3\nu F a}{4 b^2}\right) \\ \left.\sigma_{33}^{\text{err}}\right|_{(0,b)} & = \text{abs}\left(\frac{3\nu F a b}{4 b^3}\right) = \text{abs}\left(\frac{3\nu F a}{4 b^2}\right) \\ \left.\sigma_{33}^{\text{err}}\right|_{(a,-b)} & = \text{abs}\left(\frac{3\nu F a b}{4 b^3}\right) = \text{abs}\left(\frac{3\nu F a}{4 b^2}\right) \\ \left.\sigma_{33}^{\text{err}}\right|_{(a,b)} & = \text{abs}\left(\frac{3\nu F a b}{4 b^3}\right) = \text{abs}\left(\frac{3\nu F a}{4 b^2}\right) \end{align}$$ The maximum error is

{\frac{3\nu F a}{4 b^2}} $$ The maximum tensile stress is

\sigma_{11}(a,b) = \frac{3 F a b}{2 b^3} = \frac{3Fa}{2b^2} $$ Therefore, the ratio of the maximum error in $$\sigma_{33}$$ to the maximum tensile stress is

{\text{Ratio} = \frac{\nu}{2}} $$

title: Elasticity/Plane strain example 1


text: File:Plane strain ex1.png

title: File:Plane strain ex1.png


text:
Elasticity/Antiplane shear example 1

Example 1
Given:

The body $$-\alpha < \theta < \alpha$$, $$0 \le r < a$$ is supported at $$r = a$$ and loaded only by a uniform antiplane shear traction $$\sigma_{\theta z} = S$$ on the surface $$\theta = \alpha$$, the other surface being traction-free.

Find:

Find the complete stress field in the body, using strong boundary conditions on $$\theta = \pm\alpha$$ and weak conditions on $$r = a$$.

[Hint: Since the traction $$\sigma_{\theta z}$$ is uniform on the surface $$\theta = \alpha$$, from the expression for antiplane stress we can see that the displacement varies with $$r^1 = r$$. The most general solution for the equilibrium equation for this behavior is $$u(r,\theta) = Ar\cos\theta + Br\sin\theta$$]

Solution
Step 1: Identify boundary conditions

 * $$\begin{align}

\text{at}~ r & = 0 ~; u_r = 0, u_{\theta} = 0 \\ \text{at}~ r & = a ~; u_r = 0, u_{\theta} = 0, u_{z} = 0 \\ \text{at}~ \theta & = -\alpha ~; t_{\theta} = 0, t_{r} = 0, t_{z} = 0 \\ \text{at}~ \theta & = \alpha ~; t_{\theta} = 0, t_{r} = 0, t_{z} = S \end{align}$$ The traction boundary conditions in terms of components of the stress tensor are
 * $$\begin{align}

\text{at}~ \theta & = -\alpha ~; \sigma_{\theta r} = 0, \sigma_{\theta\theta} = 0, \sigma_{\theta z} = 0 \\ \text{at}~ \theta & = \alpha ~; \sigma_{\theta r} = 0, \sigma_{\theta\theta} = 0, \sigma_{\theta z} = S \end{align}$$

Step 2: Assume solution

Assume that the problem satisfies the conditions required for antiplane shear. If $$\sigma_{\theta z}$$ is to be uniform along $$\theta=\alpha$$, then

\sigma_{\theta z} = \frac{\mu}{r} \frac{\partial u_z}{\partial \theta} = C $$ or,

\frac{\partial u_z}{\partial \theta} = \frac{Cr}{\mu} $$ The general form of $$u_z$$ that satisfies the above requirement is

u_z(r,\theta) = Ar\cos\theta + Br\sin\theta + C $$ where $$A$$, $$B$$, $$C$$ are constants.

Step 3: Compute stresses

The stresses are
 * $$\begin{align}

\sigma_{\theta z} & = \frac{\mu}{r} \frac{\partial u_z}{\partial \theta} = \mu \left(-A\sin\theta + B\cos\theta\right) \\ \sigma_{rz} & = \mu \frac{\partial u_z}{\partial r} = \mu \left(A\cos\theta + B\sin\theta\right) \end{align}$$

Step 4: Check if traction BCs are satisfied

The antiplane strain assumption leads to the $$\sigma_{\theta\theta}$$ and $$\sigma_{r\theta}$$ BCs being satisfied. From the boundary conditions on $$\sigma_{\theta z}$$, we have
 * $$\begin{align}

0 & = \mu \left(A\sin\alpha + B\cos\alpha\right) \\ S & = \mu \left(-A\sin\alpha + B\cos\alpha\right) \end{align}$$ Solving,

A = -\frac{S}{2\mu\sin\alpha} ~; B = \frac{S}{2\mu\cos\alpha} $$ This gives us the stress field

\sigma_{\theta z} = \frac{S}{2} \left(\frac{\sin\theta}{\sin\alpha} + \frac{\cos\theta}{\cos\alpha}\right) ~; \sigma_{rz} = \frac{S}{2} \left(-\frac{\cos\theta}{\sin\alpha} + \frac{\sin\theta}{\cos\alpha}\right) $$

Step 5: Compute displacements

The displacement field is

u_z(r,\theta) = \frac{Sr}{2\mu}\left(-\frac{\cos\theta}{\sin\alpha} +                                     \frac{\sin\theta}{\cos\alpha}\right) + C $$ where the constant $$C$$ corresponds to a superposed rigid body displacement.

Step 6: Check if displacement BCs are satisfied

The displacement BCs on $$u_r$$ and $$u_{\theta}$$ are automatically satisfied by the antiplane strain assumption. We will try to satisfy the boundary conditions on $$u_z$$ in a weak sense, i.e, at $$r = a$$,

\int_{-\alpha}^{\alpha} u_z(a, \theta) d\theta = 0~. $$ This weak condition does not affect the stress field. Plugging in $$u_z$$,
 * $$\begin{align}

0 & = \int_{-\alpha}^{\alpha} u_z(a, \theta) d\theta \\ & = \frac{Sa}{2\mu}\int_{-\alpha}^{\alpha} \left(-\frac{\cos\theta}{\sin\alpha} +            \frac{\sin\theta}{\cos\alpha} + C\frac{2\mu}{Sa}\right) d\theta \\ & = \frac{Sa}{2\mu}\int_{-\alpha}^{\alpha} \left(-\frac{\cos\theta}{\sin\alpha} +            \frac{\sin\theta}{\cos\alpha} + C\frac{2\mu}{Sa}\right) d\theta \\ & = \frac{Sa}{2\mu}\left[ \left(-\frac{\sin\theta}{\sin\alpha} -            \frac{\cos\theta}{\cos\alpha} + C\theta\frac{2\mu}{Sa}\right) \right]_{-\alpha}^{\alpha} \\ & = \frac{Sa}{2\mu} \left(-2\frac{\sin\alpha}{\sin\alpha} +           2C\alpha\frac{2\mu}{Sa}\right) \\ & = -\frac{Sa}{\mu} + C\alpha \end{align}$$ Therefore,

C = \frac{Sa}{2\mu\alpha} $$ The approximate displacement field is

u_z(r,\theta) = \frac{S}{2\mu}\left(-r\frac{\cos\theta}{\sin\alpha} +                                    r\frac{\sin\theta}{\cos\alpha} + a\frac{1}{\alpha}\right) $$

title: Elasticity/Antiplane shear example 1


text: File:Antiplane shear ex1.png

title: File:Antiplane shear ex1.png


text:
Elasticity/Energy methods example 1

Example 1
Given:

\Pi^c[\boldsymbol{\sigma}(\mathbf{x}) + \Delta\boldsymbol{\sigma}(\mathbf{x})] - \Pi^c[\boldsymbol{\sigma}(\mathbf{x})] = \int_{\mathcal R} [U^c(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) - U^c(\boldsymbol{\sigma}) - \boldsymbol{\varepsilon}:\Delta\boldsymbol{\sigma}]~dV $$

Show:

\Pi^c[\boldsymbol{\sigma}(\mathbf{x}) + \Delta\boldsymbol{\sigma}(\mathbf{x})] - \Pi^c[\boldsymbol{\sigma}(\mathbf{x})] = \int_{\mathcal R} U^c(\Delta\boldsymbol{\sigma})~dV $$

Solution
For a linear elastic material, the complementary strain energy density is given by

U^c(\boldsymbol{\sigma}) = \frac{1}{2}\boldsymbol{\sigma}:\text{S}:\boldsymbol{\sigma} $$ where $$\text{S}$$ is the compliance tensor.

Therefore,

U^c(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) = \frac{1}{2}(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}):\text{S}:(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) = \frac{1}{2}(\sigma_{ij}+\Delta\sigma_{ij})S_{ijkl}(\sigma_{kl}+\Delta\sigma_{kl}) $$ or (using the symmetry of the compliance tensor),
 * $$\begin{align}

U^c(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) & = \frac{1}{2}\left[\sigma_{ij}\sigma_{kl}+\sigma_{ij}\Delta\sigma_{kl}+ \sigma_{kl}\Delta\sigma_{ij}+\Delta\sigma_{ij}\Delta\sigma_{kl}\right] S_{ijkl} \\ &= \frac{1}{2}\left[\sigma_{ij}S_{ijkl}\sigma_{kl}+ \sigma_{ij}S_{ijkl}\Delta\sigma_{kl}+ \sigma_{kl}S_{ijkl}\Delta\sigma_{ij}+ \Delta\sigma_{ij}S_{ijkl}\Delta\sigma_{kl}\right] \\ &= \frac{1}{2}\left[\sigma_{ij}S_{ijkl}\sigma_{kl}+ \varepsilon_{kl}\Delta\sigma_{kl}+ \varepsilon_{ij}\Delta\sigma_{ij}+ \Delta\sigma_{ij}S_{ijkl}\Delta\sigma_{kl}\right] \\ &= \frac{1}{2}\left[\sigma_{ij}S_{ijkl}\sigma_{kl}+ 2\varepsilon_{kl}\Delta\sigma_{kl}+ \Delta\sigma_{ij}S_{ijkl}\Delta\sigma_{kl}\right] \\ &= \frac{1}{2}\boldsymbol{\sigma}:\text{S}:\boldsymbol{\sigma} + \boldsymbol{\varepsilon}:\Delta\boldsymbol{\sigma} + \frac{1}{2}\Delta\boldsymbol{\sigma}:\text{S}: \Delta\boldsymbol{\sigma} \\ &= U^c(\boldsymbol{\sigma}) + \boldsymbol{\varepsilon}:\Delta\boldsymbol{\sigma} + U^c(\Delta\boldsymbol{\sigma}) \end{align}$$ Therefore,

U^c(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) = U^c(\boldsymbol{\sigma}) + \boldsymbol{\varepsilon}:\Delta\boldsymbol{\sigma} + U^c(\Delta\boldsymbol{\sigma}) $$ Plugging into the given equation

\Pi^c[\boldsymbol{\sigma}(\mathbf{x}) + \Delta\boldsymbol{\sigma}(\mathbf{x})] - \Pi^c[\boldsymbol{\sigma}(\mathbf{x})] = \int_{\mathcal R} [U^c(\boldsymbol{\sigma}) + \boldsymbol{\varepsilon}:\Delta\boldsymbol{\sigma} + U^c(\Delta\boldsymbol{\sigma}) - U^c(\boldsymbol{\sigma}) - \boldsymbol{\varepsilon}:\Delta\boldsymbol{\sigma}]~dV $$ or,

{ \Pi^c[\boldsymbol{\sigma}(\mathbf{x}) + \Delta\boldsymbol{\sigma}(\mathbf{x})] - \Pi^c[\boldsymbol{\sigma}(\mathbf{x})] = \int_{\mathcal R} U^c(\Delta\boldsymbol{\sigma})~dV } $$ Hence shown.

title: Elasticity/Energy methods example 1


text:
Elasticity/Energy methods example 2

Example 2
Given:

The potential energy functional for a membrane stretched over a simply connected region $$\mathcal{S}$$ of the $$x_1-x_2$$ plane can be expressed as

\Pi[w(x_1,x_2)] = \frac{1}{2}\int_{\mathcal{S}} \eta\left[(w_{,1})^2 + (w_{,2})^2\right] ~dA - \int_{\mathcal S} pw~dA $$ where $$w(x_1,x_2)$$ is the deflection of the membrane, $$p(x_1,x_2)$$ is the prescribed transverse pressure distribution, and $$\eta$$ is the membrane stiffness.

Find:

 * 1) The governing differential equation (Euler equation) for $$w(x_1,x_2)$$ on $${\mathcal S}$$.
 * 2) The permissible boundary conditions at the boundary $$\partial{\mathcal S}$$ of $${\mathcal S}$$.

Solution
The principle of minimum potential energy requires that the functional $$\Pi$$ be stationary for the actual displacement field $$w(x_1,x_2)$$. Taking the first variation of $$\Pi$$, we get

\delta\Pi = \frac{\eta}{2}\int_{\mathcal S} \left[(2 w_{,1})\delta w_{,1} + (2 w_{,2})\delta w_{,2}\right] ~dA - \int_{\mathcal S} p~\delta w~dA $$ or,

\delta\Pi = \eta\int_{\mathcal S} \left[w_{,1}~\delta w_{,1} + w_{,2}~\delta w_{,2}\right] ~dA - \int_{\mathcal S} p~\delta w~dA $$ Now,
 * $$\begin{align}

(w_{,1}~\delta w)_{,1} & = w_{,11}~\delta w + w_{,1}~\delta w_{,1} \\ (w_{,2}~\delta w)_{,2} & = w_{,22}~\delta w + w_{,2}~\delta w_{,2} \end{align}$$ Therefore,

w_{,1}~\delta w_{,1} + w_{,2}~\delta w_{,2} = (w_{,1}~\delta w)_{,1} - w_{,11}~\delta w + (w_{,2}~\delta w)_{,2} - w_{,22}~\delta w $$ Plugging into the expression for $$\delta\Pi$$,

\delta\Pi = \eta\int_{\mathcal S}  \left[(w_{,1}~\delta w)_{,1} + (w_{,2}~\delta w)_{,2} - (w_{,11} + w_{,22})~\delta w\right]~dA - \int_{\mathcal S} p~\delta w~dA $$ or,

\delta\Pi = \eta\int_{\mathcal S}  \left[(w_{,1}~\delta w)_{,1} + (w_{,2}~\delta w)_{,2}\right]~dA - \eta\int_{\mathcal S} \nabla^2{w}~\delta w~dA - \int_{\mathcal S} p~\delta w~dA $$ Now, the Green-Riemann theorem states that

\int_{\mathcal S} (Q_{,1} - P_{,2})~dA = \oint_{\partial{\mathcal S}} (P~dx_1 + Q~dx_2) $$ Therefore,

\delta\Pi = \eta\oint_{\partial{\mathcal S}} \left[(w_{,1}~\delta w)~dx_2 - (w_{,2}~\delta w)~dx_1\right] - \int_{\mathcal S}\left[\eta\nabla^2{w}+p\right]~\delta w~dA $$ or,

\delta\Pi = \eta\oint_{\partial{\mathcal S}} \left[w_{,1}~\frac{dx_2}{ds} - w_{,2}~\frac{dx_1}{ds}\right]\delta w~ds - \int_{\mathcal S}\left[\eta\nabla^2{w}+p\right]~\delta w~dA $$ where $$s$$ is the arc length around $$\partial{\mathcal S}$$.

The potential energy function is rendered stationary if $$\delta\Pi = 0$$. Since $$\delta w$$ is arbitrary, the condition of stationarity is satisfied only if the governing differential equation for $$w(x_1,x_2)$$ on $${\mathcal S}$$ is

{      \eta\nabla^2{w}+p = 0   \forall(x_1,x_2)~\in~{\mathcal S}  } $$ The associated boundary conditions are

{ w_{,1}~\frac{dx_2}{ds} - w_{,2}~\frac{dx_1}{ds} = 0 \forall(x_1,x_2)~\in~\partial{\mathcal S}  } $$

title: Elasticity/Energy methods example 2


text:
Elasticity/Minimizing a functional

Minimizing a functional in 1-D
In 1-D, the minimization problem can be stated as

Find $$u(x)$$ such that

U[u(x)] = \int_{x_0}^{x_1} F(x,u,u^{'}) dx $$ is a minimum.

We have seen that the minimization problem can be reduced down to the solution of an Euler equation

\frac{\partial F}{\partial u} - \frac{d}{dx}\left(\frac{\partial F }{\partial u^{'}} \right) = 0 $$

with the associated boundary conditions

\eta(x_0) = 0 ~\text{and}~ \eta(x_1) = 0 $$ or,

\left.\frac{\partial F}{\partial u^{'}} \right|_{x_0} = 0 ~\text{and} \left.\frac{\partial F}{\partial u^{'}} \right|_{x_1} = 0 $$

Minimizing a Functional in 3-D
In 3-D, the equivalent minimization problem can be stated as

Find $$\mathbf{u}(\mathbf{x})$$ such that

U[\mathbf{u}(\mathbf{x})] = \int_{\mathcal{R}} F(\mathbf{x},\mathbf{u},\boldsymbol{\nabla}\mathbf{u})~dV $$ is a minimum.

We would like to find the Euler equation for this problem and the associated boundary conditions required to minimize $$U$$.

Let us define all our quantities with respect to an orthonormal basis $$(\widehat{\mathbf{e}}_{i})$$.

Then,

\mathbf{x} = x_i\widehat{\mathbf{e}}_{i} ; \mathbf{u} = u_i\widehat{\mathbf{e}}_{i} ; \boldsymbol{\nabla} \mathbf{u} = u_{i,j} \widehat{\mathbf{e}}_{i}\otimes\widehat{\mathbf{e}}_{j} $$ and

U[\mathbf{u}(\mathbf{x})] = \int_{\mathcal{R}} \tilde{F}(x_i, u_i, u_{i,j})~dV $$

Taking the first variation of $$U$$, we get

\delta U = \int_{\mathcal{R}} \left(\frac{\partial\tilde{F} }{\partial u_i} \delta u_i +    \frac{\partial \tilde{F}}{\partial u_{i,j}} \delta u_{i,j}\right) dV $$

All the nine components of $$\delta u_{i,j}$$ are not independent. Why ?

The variation of the functional $$U$$ needs to be expressed entirely in terms of $$\delta u_i$$. We do this using the 3-D equivalent of integration by parts - the divergence theorem.

Thus,
 * $$\begin{align}

\int_{\mathcal{R}} \frac{\partial \tilde{F}}{\partial u_{i,j}} \delta u_{i,j}~ dV &= \int_{\mathcal{R}} \frac{\partial }{\partial x_j} \left(\frac{\partial \tilde{F}}{\partial u_{i,j}}                 \delta u_i\right) dV - \int_{\mathcal{R}} \frac{\partial }{\partial x_j} \left(\frac{\partial \tilde{F}}{\partial u_{i,j}} \right) \delta u_i~ dV \\ & =   \int_{\partial\mathcal{R}} \frac{\partial \tilde{F}}{\partial u_{i,j}} \delta u_i~n_j~dA - \int_{\mathcal{R}} \frac{\partial }{\partial x_j} {}{}\left(\frac{\partial \tilde{F}}{\partial u_{i,j}} \right) \delta u_i~ dV \end{align}$$ Substituting in the expression for $$\delta U$$, we have,
 * $$\begin{align}

\delta U &= \int_{\mathcal{R}} \frac{\partial \tilde{F}}{\partial u_i} \delta u_i~dV + \int_{\partial\mathcal{R}} \frac{\partial \tilde{F}}{\partial u_{i,j}} \delta u_i~n_j~dA - \int_{\mathcal{R}} \frac{\partial }{\partial x_j} \left(\frac{\partial \tilde{F}}{\partial u_{i,j}} \right) \delta u_i~ dV \\ &= \int_{\mathcal{R}}\left[\frac{\partial \tilde{F}}{\partial u_i} - \frac{\partial }{\partial x_j}\left(\frac{\partial \tilde{F}}{\partial u_{i,j}} \right)\right] \delta u_i~ dV + \int_{\partial\mathcal{R}} \frac{\partial \tilde{F} }{\partial u_{i,j}} \delta u_i~n_j~dA \end{align}$$ For $$U$$ to be minimum, a necessary condition is that $$\delta U = 0$$ for all variations $$\delta\mathbf{u}$$.

Therefore, the Euler equation for this problem is

\frac{\partial \tilde{F}}{\partial u_i} - \frac{\partial }{\partial x_j} \left(\frac{\partial \tilde{F}}{\partial u_{i,j}} \right) = 0 ~\forall\mathbf{x} \in \mathcal{R} $$ and the associated boundary conditions are

\frac{\partial \tilde{F}}{\partial u_{i,j}} = 0 \text{or,} \delta u_i = 0 ~\forall\mathbf{x} \in \partial\mathcal{R} $$

title: Elasticity/Minimizing a functional


text:
Elasticity/Kinematic admissibility

Kinematically admissible displacement field
Consider a body $$B$$ with a boundary $$\partial B$$ with an applied body force field $$\tilde{\mathbf{f}}$$.

Suppose that displacement BCs $$\mathbf{u} = \tilde{\mathbf{u}}$$ are prescribed on the part of the boundary $$\partial B^u$$.

Suppose also that traction BCs $$\widehat{\mathbf{n}}{}\bullet\boldsymbol{\sigma} = \tilde{\mathbf{t}}$$ are applied on the portion of the boundary $$\partial B^t$$.

A displacement field $$(\mathbf{u})$$ is kinematically admissible if
 * $$\mathbf{u}$$ satisfies the displacement boundary conditions  $$\mathbf{u} = \tilde{\mathbf{u}}$$ on $$\partial B^u$$.
 * $$\mathbf{u}$$ is continuously differentiable, i.e. $$\mathbf{u} \in C^3(\mathcal{R})$$.
 * $$|\boldsymbol{\nabla}{\mathbf{u}}| \ll 1$$.

A kinematically admissible displacement field needs only to satisfy compatibility condition and the displacement boundary conditions - but not the traction boundary conditions or equilibrium.

Recall that a kinematically admissible displacement field is used to define the principle of minimum potential energy.

title: Elasticity/Kinematic admissibility


text:
Elasticity/Principle of minimum potential energy

Strain Energy Density
The strain energy density is defined as

U(\boldsymbol{\varepsilon}) = \int_0^{\varepsilon} \boldsymbol{\sigma}:d\boldsymbol{\varepsilon} $$ If the strain energy density is path independent, then it acts as a potential for stress, i.e.,

\sigma_{ij} = \frac{\partial U(\varepsilon) }{\partial \varepsilon_{ij}} $$ For adiabatic processes, $$U$$ is equal to the change in internal energy per unit volume.

For isothermal processes, $$U$$ is equal to the Helmholtz free energy per unit volume.

The natural state of a body is defined as the state in which the body is in stable thermal equilibrium with no external loads and zero stress and strain.

When we apply energy methods in linear elasticity, we implicitly assume that a body returns to its natural state after loads are removed. This implies that the Gibb's condition is satisfied :

U(\boldsymbol{\varepsilon}) \ge 0 \text{with} U(\boldsymbol{\varepsilon}) = 0 \text{iff} \boldsymbol{\varepsilon} = 0 $$

Principle of Minimum Potential Energy
This principle states that
 * If the prescribed traction and body force fields are independent of the deformation
 * then the actual displacement field makes the potential energy functional an absolute minimum.

In other words, the principle of minimum potential energy states that the potential energy functional

\Pi[\mathbf{u}] = \frac{1}{2}\int_{\mathcal{B}} \boldsymbol{\sigma}:\boldsymbol{\varepsilon}~dV - \int_{\mathcal{B}} \mathbf{f}\bullet\mathbf{u}~dV - \int_{\partial{\mathcal{B}}^t} \mathbf{t}\bullet\mathbf{u}~dA $$ is minimized by the actual displacement field.

Proof
The first step in the proof is to show that the actual displacements make the function $$\Pi\,$$ stationary. The second step is to show that the stationary point is actually the minimum.

Proof of the Principle of Stationary Potential Energy
The first variation of the potential energy functional $$\Pi\,$$ is

\delta\Pi[\mathbf{u},\delta\mathbf{u}] = \int_B C_{ijkl}~u_{i,j}~\delta u_{k,l}~dV - \int_B \tilde{f}_i~\delta u_i~dV - \int_{\partial B^t} \tilde{t}_i~\delta u_i~dA $$ or,
 * $$\begin{align}

\delta\Pi[\mathbf{u},\delta\mathbf{u}] & = \int_B \sigma_{ji}~\delta u_{i,j}~dV - \int_B \tilde{f}_i~\delta u_i~dV - \int_{\partial B^t} \tilde{t}_i~\delta u_i~dA \\ & = \int_B (\sigma_{ji}~\delta u_i)_{,j}~dV - \int_B \sigma_{ji,j}~\delta u_i~dV - \int_B \tilde{f}_i~\delta u_i~dV - \int_{\partial B^t} \tilde{t}_i~\delta u_i~dA \\ & = \int_{\partial B} \sigma_{ji}~\delta u_i~n_j~dA - \int_B \sigma_{ji,j}~\delta u_i~dV - \int_B \tilde{f}_i~\delta u_i~dV - \int_{\partial B^t} \tilde{t}_i~\delta u_i~dA \\ & = \int_{\partial B^t} \sigma_{ji}~n_j~\delta u_i~dA - \int_{\partial B^t} \tilde{t}_i~\delta u_i~dA - \int_B \sigma_{ji,j}~\delta u_i~dV - \int_B \tilde{f}_i~\delta u_i~dV \\ & = \int_{\partial B^t} \left[\sigma_{ji}~n_j-\tilde{t}_i\right]~\delta u_i~dA - \int_B \left[\sigma_{ji,j} + \tilde{f}_i\right]~\delta u_i~dV \end{align}$$ or,

\delta\Pi[\mathbf{u},\delta\mathbf{u}]= \int_{\partial B^t} \left[\widehat{\mathbf{n}}{}\bullet\boldsymbol{\sigma} -\tilde{\mathbf{t}}\right]\bullet\delta\mathbf{u}~dA - \int_B \left[\boldsymbol{\nabla}\bullet{\boldsymbol{\sigma}} + \tilde{\mathbf{f}}\right]\bullet\delta\mathbf{u}~dV $$ Therefore, $$\delta\Pi[\mathbf{u},\delta\mathbf{u}] = 0$$ (i.e. $$\Pi$$ is stationary) only if
 * $$\begin{align}

\widehat{\mathbf{n}}{}\bullet\boldsymbol{\sigma} & = \tilde{\mathbf{t}} \\ \boldsymbol{\nabla}\bullet{\boldsymbol{\sigma}} + \tilde{\mathbf{f}} & = 0 \end{align}$$ which are the conditions that only the actual displacement field satisfies.

Proof of the Principle of Minimum Potential Energy
To prove that $$\Pi[\mathbf{u},\delta\mathbf{u}]$$ is not only stationary, but also the global minimum all we now need to show is that

\delta^2\Pi[\mathbf{u},\delta\mathbf{u}] > 0 $$

Now,

\delta^2\Pi[\mathbf{u},\delta\mathbf{u}]= \int_B C_{ijkl}~\delta u_{i,j}~\delta u_{k,l}~dV $$

If the displacement field is a pure rigid body motion, then the strain energy density

U = \frac{1}{2} C_{ijkl} w_{ij} w_{kl} = 0 $$ where $$\mathbf{w}$$ is the spin tensor given by

\mathbf{w} = \frac{1}{2} \left(\boldsymbol{\nabla}u - \boldsymbol{\nabla}u^T\right) ; w_{ij} = \frac{1}{2} \left(u_{i,j} - u_{j,i}\right) $$

If the displacement field does not contain any rigid body motion, then the strain energy density is given by

U = \frac{1}{2} C_{ijkl} \varepsilon_{ij} \varepsilon_{kl} > 0 $$ where $$\boldsymbol{\varepsilon}$$ is the strain tensor given by

\boldsymbol{\varepsilon} = \frac{1}{2} \left(\boldsymbol{\nabla}u + \boldsymbol{\nabla}u^T\right) ; \varepsilon_{ij} = \frac{1}{2} \left(u_{i,j} + u_{j,i}\right) $$

Therefore, for a displacement field containing both spin and strain

U = \frac{1}{2} C_{ijkl}~\delta u_{i,j}~\delta u_{k,l} \ge 0 $$ or,

\delta^2\Pi[\mathbf{u},\delta\mathbf{u}]= \int_B C_{ijkl}~\delta u_{i,j}~\delta u_{k,l}~dV \ge 0 $$

This means that $$\delta^2\Pi > 0$$ for all values of $$\delta\mathbf{u}$$ other than rigid body motion, in which case $$\delta^2\Pi = 0$$.

Hence, for mixed boundary value problems $$\delta^2\Pi > 0$$ for all $$\delta\mathbf{u}$$, as long as the displacement BCs prevent rigid body motions. Therefore, the potential energy functional is minimized by the actual displacement field.


text:
Elasticity/Static admissibility

Statically admissible stress field
This type of field is used in the definition of the principle of minimum complementary energy.

A statically admissible stress field is one that satisfies equilibrium (with the prescribed body force) and traction boundary conditions but not compatibility or the displacement boundary conditions.

title: Elasticity/Static admissibility


text:
Elasticity/Principle of minimum complementary energy

Complementary strain energy
The complementary strain energy density is given by

U^c(\boldsymbol{\sigma}) = \int_0^{\boldsymbol{\sigma}} \varepsilon_{ij}~d\sigma_{ij} $$

For linear elastic materials

U^c = \frac{1}{2}\boldsymbol{\sigma}:\boldsymbol{\varepsilon} = U $$

Principle of minimum complementary energy
Let $$\mathcal{A}$$ be the set of all statically admissible states of stress.

Let the complementary energy functional be

\Pi^c[\boldsymbol{\sigma}] = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma})~dV - \int_{\partial{\mathcal{B}}^u} \mathbf{t}\bullet\widehat{\mathbf{u}}~dA $$

Then the Principle of Stationary Complementary Energy states that: Among all stress fields $$\boldsymbol{\sigma}$$ in $$\mathcal{A}$$, the functional           $$\Pi^c$$ is rendered stationary only by actual            stress fields which satisfy compatibility and the displacement BCs.

The Principle of Minimum Complementary Energy states that: For linear elastic materials, the complementary energy functional is rendered an absolute minimum by the actual stress field.

Note that the complementary energy corresponding to the actual stress field is the negative of the potential energy corresponding to the actual displacement field.

Proof
Let $$[\mathbf{u},\boldsymbol{\varepsilon},\boldsymbol{\sigma}]$$ be a solution of a the mixed boundary value problem of linear elasticity.

Let $$\tilde{\boldsymbol{\sigma}} \in \mathcal{A}$$.

Define

\boldsymbol{\sigma}^{'} = \tilde{\boldsymbol{\sigma}} - \boldsymbol{\sigma} $$

Then $$\boldsymbol{\sigma}^{'}$$ satisfies equilibrium and the traction BCs, i.e.,
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\boldsymbol{\sigma}^{'}} = 0 &\forall \mathbf{x}\in\mathcal{B} \\ \mathbf{t}^{'} = \widehat{\mathbf{n}}{}\bullet\boldsymbol{\sigma}^{'} = 0 &\forall \mathbf{x}\in\partial{\mathcal{B}}^t \end{align}$$

Since

\boldsymbol{\varepsilon} = \text{S}:\boldsymbol{\sigma} $$ and

\int_{\mathcal{B}} U^c(\boldsymbol{\sigma}^{'}+\boldsymbol{\sigma})~dV = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma}^{'})~dV + \int_{\mathcal{B}} U^c(\boldsymbol{\sigma})~dV + \int_{\mathcal{B}} \boldsymbol{\sigma}^{'}:(\text{S}:\boldsymbol{\sigma})~ dV $$ we have,
 * $$\text{(1)} \qquad

\int_{\mathcal{B}} \left[U^c(\tilde{\boldsymbol{\sigma}}) - U^c(\boldsymbol{\sigma})\right]~dV = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma}^{'})~dV + \int_{\mathcal{B}} \boldsymbol{\sigma}^{'}:\boldsymbol{\varepsilon}~ dV $$

We can also show that

\int_{\partial\mathcal{B}} \mathbf{t}\bullet\mathbf{u}~dA = \int_{\mathcal{B}} \mathbf{u}\bullet(\boldsymbol{\nabla}\bullet{\boldsymbol{\sigma}})~dV + \frac{1}{2}\int_{\mathcal{B}} \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{u}+\boldsymbol{\nabla}\mathbf{u}^T)~dV $$

Therefore,
 * $$\text{(2)} \qquad

\int_{\partial{\mathcal{B}}^{u}} \mathbf{t}^{'}\bullet\mathbf{u}~dA = \int_{\mathcal{B}} \boldsymbol{\sigma}^{'}:\boldsymbol{\varepsilon}~dV $$

Now,

\Pi^c[\boldsymbol{\sigma}] = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma})~dV - \int_{\partial{\mathcal{B}}^u} \mathbf{t}\bullet\widehat{\mathbf{u}}~dA $$

Hence,

\Pi^c[\tilde{\boldsymbol{\sigma}}] = \int_{\mathcal{B}} U^c(\tilde{\boldsymbol{\sigma}})~dV - \int_{\partial{\mathcal{B}}^u} (\mathbf{t}+\mathbf{t'})\bullet\widehat{\mathbf{u}}~dA $$

Therefore,
 * $$\text{(3)} \qquad

\Pi^c[\tilde{\boldsymbol{\sigma}}] - \Pi^c[\boldsymbol{\sigma}] = \int_{\mathcal{B}} \left[U^c(\tilde{\boldsymbol{\sigma}})-U^c(\boldsymbol{\sigma})\right]~dV - \int_{\partial{\mathcal{B}}^u} \mathbf{t}^{'}\bullet\widehat{\mathbf{u}}~dA $$

From equations (1), (2), and (3), we have,

\Pi^c[\tilde{\boldsymbol{\sigma}}] - \Pi^c[\boldsymbol{\sigma}] = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma}^{'})~dV + \int_{\partial{\mathcal{B}}^{u}} \mathbf{t}^{'}\bullet\mathbf{u}~dA - \int_{\partial{\mathcal{B}}^u} \mathbf{t}^{'}\bullet\widehat{\mathbf{u}}~dA $$

Since $$\mathbf{u} = \widehat{\mathbf{u}}$$ on $$\partial{\mathcal{B}}^u$$, we have,

\Pi^c[\tilde{\boldsymbol{\sigma}}] - \Pi^c[\boldsymbol{\sigma}] = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma}^{'})~dV > 0 $$

Hence, proved.


text:
Elasticity/More variational principles

More variational principles
Recall that the admissible states appropriate to the minimum       principles are required to meet certain field equations and the         appropriate boundary conditions. In some cases, we would like to use variational principles in       which the admissible states satisfy as few constraints as possible. Such variational principles are useful for symmetric elastic fields.

Let $$I$$ be a scalar-valued functional. Let $$\mathcal{A}$$ be the set of all admissible states.

Let $$s$$ and $$\tilde{s}$$ be two admissible states $$\in\mathcal{A}$$. Let $$\lambda$$ be a Lagrange multiplier such that $$s+\lambda\tilde{s}$$ is also an admissible state $$\in\mathcal{A}$$.

Let

\delta_{\tilde{s}} I[s] = \left.\frac{d}{d\lambda} I[s+\lambda\tilde{s}]\right|_{\lambda=0} $$

Then,

\delta I[s] = 0 $$ only if $$\delta_{\tilde{s}} I[s]$$ exists and equals zero for all $$\tilde{s}$$ that satisfy the above requirements.

There is an infinite number of possible functional $$I[s]$$ that satisfy these requirements. Two examples are:
 * The Hellinger-Reissner functional
 * The Hu-Washizu functional

title: Elasticity/More variational principles


text:
File:Mok That was a great movie.ogg

Used with the WikiU Film School Lesson on how to do Animatic Dialog Recording.

title: File:Mok That was a great movie.ogg


text:
Elasticity/Hu-Washizu principle

Hu-Washizu Variational Principle
In this case, the admissible states are not required to meet any of the field equations or boundary conditions.

Let $$\mathcal{A}$$ denote the set of all admissible states and let $$\mathcal{W}$$ be a functional on $$\mathcal{A}$$ defined by

{\mathcal W}[s] = \int_{\mathcal{B}} U(\boldsymbol{\varepsilon}) - \int_{\mathcal{B}} \boldsymbol{\sigma}:\boldsymbol{\varepsilon}~dV - \int_{\mathcal{B}} (\boldsymbol{\nabla}\bullet{\boldsymbol{\sigma}} + \mathbf{f})\bullet\mathbf{u}~dV + \int_{\partial{\mathcal{B}}^{u}} \mathbf{t}\bullet\widehat{\mathbf{u}}~dA + \int_{\partial{\mathcal{B}}^{t}} (\mathbf{t}-\widehat{\mathbf{t}})\bullet\mathbf{u}~dA $$ for every $$s = [\mathbf{u},\boldsymbol{\varepsilon},\boldsymbol{\sigma}] \in \mathcal{A}$$.

Then,

\delta {\mathcal W}[s] = 0 $$ at an admissible state $$s$$ if and only if $$s$$ is a solution of the mixed problem.

title: Elasticity/Hu-Washizu principle


text:
Elasticity/Hellinger-Reissner principle

Hellinger-Prange-Reissner Variational Principle
In this case, we assume that the elasticity field is invertible and $$s$$ is smooth on $$\mathcal{B}$$. We also assume that $$\mathcal{A}$$ is the set of all admissible states that satisfy the strain-displacement relations, the traction-stress relations and the balance of angular momentum.

Let $$\mathcal{A}$$ denote the set of all admissible states and let $$\mathcal{H}$$ be a functional on $$\mathcal{A}$$ defined by

{\mathcal H}[s] = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma}) - \int_{\mathcal{B}} \boldsymbol{\sigma}:\boldsymbol{\varepsilon}~dV + \int_{\mathcal{B}} \mathbf{f}\bullet\mathbf{u}~dV + \int_{\partial{\mathcal{B}}^{u}} \mathbf{t}\bullet(\mathbf{u}-\widehat{\mathbf{u}})~dA + \int_{\partial{\mathcal{B}}^{t}} \widehat{\mathbf{t}}\bullet\mathbf{u}~dA $$ for every $$s = [\mathbf{u},\boldsymbol{\sigma}] \in \mathcal{A}$$.

Then,

\delta {\mathcal H}[s] = 0 $$ at an admissible state $$s\in\mathcal{A}$$ if and only if $$s$$ is a solution of the mixed problem.

title: Elasticity/Hellinger-Reissner principle


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Elasticity/Rayleigh-Ritz method

The Rayleigh-Ritz method
The potential energy functional has the form

\Pi[\mathbf{u}] = \frac{1}{2}\int_{\mathcal B} \boldsymbol{\nabla}\mathbf{u}:(\text{C}:\boldsymbol{\nabla}\mathbf{u})~dV - \int_{\mathcal B} \mathbf{f}\bullet\mathbf{u}~dV - \int_{\partial{\mathcal B}} \widehat{\mathbf{t}}\bullet\mathbf{u}~dV $$

The standard method of finding an approximate solution to the mixed boundary value problem is to minimize $$\Pi$$ over a restricted class of functions (the Rayleigh-Ritz method), by assuming that

\mathbf{u}_{\text{approx}} = \mathbf{w}_0 + \sum_{n=1}^{N} a_n \mathbf{w}_n $$ where $$\mathbf{w}_n$$ are functions that are chosen so that they vanish on $$\partial{\mathcal B}^{u}$$ and $$\mathbf{w}_0$$ is a function that approximates the boundary displacements on $$\partial{\mathcal B}^{u}$$. The constants $$a_n$$ are then chosen so that they make $$\Pi[\mathbf{u}_{\text{approx}}]$$ a minimum.

Suppose,

\Pi[\mathbf{u}_{\text{approx}}] = \Pi_{\text{approx}} = \Pi[a_1,a_2,,a_n] $$ Then,

\Pi_{\text{approx}} = A + \frac{1}{2}\sum_{m,n=1}^{N} B_{mn}~a_m~a_n + \sum_{n=1}^N D_n~a_n $$ where,
 * $$\begin{align}

A & = \int_{\mathcal B} U(\mathbf{w}_0)~dV - \int_{\mathcal B} \mathbf{f}\bullet\mathbf{w}_0~dV - \int_{\partial{\mathcal B}^{t}} \widehat{\mathbf{t}}\bullet\mathbf{w}_0~dA \\ B_{mn} & = \int_{\mathcal B} \boldsymbol{\nabla}{\mathbf{w}_m}:(\text{C}:\boldsymbol{\nabla}{\mathbf{w}_n})~dV\\ D_n & = \int_{\mathcal B} \boldsymbol{\nabla}{\mathbf{w}_0}:(\text{C}:\boldsymbol{\nabla}{\mathbf{w}_n})~dV - \int_{\mathcal B} \mathbf{f}\bullet\mathbf{w}_n~dV - \int_{\partial{\mathcal B}^{t}} \widehat{\mathbf{t}}\bullet\mathbf{w}_n~dA \end{align}$$ To minimize $$\Pi_{\text{approx}}$$ we use the relations

\frac{\partial \Pi}{\partial a_i} = 0 (i=1,2,,n) $$ to get a set of $$N$$ equations which provide us with the values of $$a_i$$.

This is the approach taken for the displacement-based finite element method. If, instead, we choose to start with the complementary energy functional, we arrive at the stress-based finite element method.

title: Elasticity/Rayleigh-Ritz method


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Elasticity/Energy methods example 3

Example 3 : Torsion of a cylinder
Suppose that the cross-section of the cylinder is bounded by the curves:
 * $$\begin{align}

x_1 & = c \\ x_1 & = d \\ x_2 & = g(x_1) \\ x_2 & = -g(x_1) \end{align}$$

A statically admissible Prandtl stress function for this cross-section is

\phi = [x_2^2 - g^2(x_1)] f(x_1) $$ with the restrictions that $$f(x_1)\,$$ is twice continuously differentiable and $$f(c) = f(d) = 0\,$$. We seek to derive the best approximate Prandtl stress function of this form by minimizing the complementary energy.

You can show that the complementary energy per unit length of the cylinder can be expressed as

\Pi^c[\phi(x_1,x_2)] = \int_{\mathcal S} \left[\frac{1}{2\mu}(\phi_{1}^2   + \phi_{,2}^2) - 2\alpha\phi\right]~dA $$ Plugging in the given form of $$\phi$$ and after some algebra, we get

\Pi^c[f(x_1)] = \frac{15}{16\mu} \int_c^d \left[\frac{1}{2} g^5(f^{'})^2 + \frac{5}{2}g^4 g^{'} f f^{'} + \frac{15}{4}g^3 (g^{'} f)^2 + \frac{5}{4}g^3 f^2 + \frac{5}{2}\mu\alpha g^3 f\right]~dx_1 $$

Taking the variation of $$\Pi$$, and after considerable manipulation, we get

\delta\Pi^c = \frac{15}{16\mu} \int_c^d \frac{g^5}{4}\left[f^{} - \frac{5}{g^2}(g^2 f)^{} + \frac{10}{g^2} f + \frac{10\mu\alpha}{g^2}\right]\delta f~dx_1 $$

Now, if we consider the cross-section to be rectangular, then we have $$c = -a$$, $$d = a$$, and $$g = b$$. Therefore, the above equation reduces to

\delta\Pi^c = \frac{15}{16\mu} \int_{-a}^a \frac{b^5}{4}\left[f^{} - 5f^{} + \frac{10}{b^2} f + \frac{10\mu\alpha}{b^2}\right]\delta f~dx_1 $$ Therefore, the function $$f$$ that minimizes $$\Pi^c$$ satisfies the equation

- 4f^{''} + \frac{10}{b^2} f + \frac{10\mu\alpha}{b^2} = 0 $$ or,

f^{''} - \frac{5}{2b^2} f - \frac{5\mu\alpha}{2b^2} = 0 $$ with the static admissibility conditions $$f(-a) = f(a) = 0\,$$.

The general solution of the above equation is

f = A\sinh\left(\sqrt{2.5}~\frac{x_1}{b}\right) + B\cosh\left(\sqrt{2.5}~\frac{x_1}{b}\right) - \mu\alpha $$ Therefore, from the BCs,

A = 0; B = \frac{\mu\alpha}{\cosh{\sqrt{2.5}~a/b}} $$

We can substitute back into $$\phi$$ to get the approximate Prandtl stress function for this problem. The error between the exact and this approximate solution is generally less than 1%.

title: Elasticity/Energy methods example 3


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Elasticity/Energy methods example 4

Application of the Principle of Virtual Work
The virtual work done by the external applied forces in moving through the virtual displacement $$\delta w(x)\,$$ is given by

\delta W_{\text{ext}} = \int_0^L q~\delta w~dx + P~\delta w(L) $$ The work done by the internal forces are,
 * $$\begin{align}

\delta W_{\text{int}} & = \int_{\mathcal R} \delta U~dV \\ & = \int_0^L\int_{\mathcal S}\delta\left(\frac{1}{2}\sigma_{ij}\varepsilon_{ij}\right)~dA~dx\\ & = \int_0^L\int_{\mathcal S}\sigma_{ij} \delta\varepsilon_{ij}~dA~dx \end{align}$$ From beam theory, the displacement field at a point in the beam is given by

u = -z\frac{dw}{dx} ; v = 0 ; w = w(x) $$ The strains are, neglecting Poisson effects,

\varepsilon_{xx} = \frac{\partial u}{\partial x} = -z\frac{d^2 w}{dx^2} ; \varepsilon_{yy} = 0 ; \varepsilon_{zz} = 0 $$ and the corresponding stresses are

\sigma_{xx} = E\varepsilon_{xx} = -Ez\frac{d^2 w}{dx^2} ; \sigma_{yy} = 0 ; \sigma_{zz} = 0 $$ If we also neglect the shear strains and stresses, we get
 * $$\begin{align}

\delta W_{\text{int}} = & = \int_0^L\int_{\mathcal S}\sigma_{xx} \delta\varepsilon_{xx}~dA~dx \\ & = \int_0^L\int_{\mathcal S} E~\varepsilon_{xx} \delta\varepsilon_{xx}~dA~dx \\ & = \int_0^L E\frac{d^2w}{dx^2}\frac{d^2(\delta w)}{dx^2} \left(\int_{\mathcal S} z^2~dA\right)~dx \\ & = \int_0^L E~I\frac{d^2w}{dx^2}\frac{d^2(\delta w)}{dx^2} dx \end{align}$$ Therefore, from the principle of virtual work,

\delta W = \int_0^L \left(E~I\frac{d^2w}{dx^2}\frac{d^2(\delta w)}{dx^2}    + q~\delta w\right)~dx + P~\delta w(L) = 0 $$ Integrating by parts and after some manipulation, we get,
 * $$\begin{align}

0 &= \int_0^L \left[\frac{d^2}{dx^2}\left(E~I\frac{d^2w}{dx^2}\right) + q + P\delta(L-x)\right]\delta w~dx + \left.\left[\left(E~I\frac{d^2w}{dx^2}\right)\delta\left(\frac{dw}{dx}\right) + \frac{d}{dx}\left(E~I\frac{d^2w}{dx^2}\right)\delta w\right]\right|_0^L \end{align}$$ where $$\delta(L-x)$$ is the Dirac delta function,

\int_{-\infty}^{\infty} \delta(L-x)~f(x)~dx = f(L) $$ The Euler equation for the beam is, therefore,

\frac{d^2}{dx^2}\left(E~I\frac{d^2w}{dx^2}\right) + q + P\delta(L-x) = 0 $$ and the boundary conditions are
 * $$\begin{align}

E~I\frac{d^2w}{dx^2}(L) & = 0 \\ \left.\frac{d}{dx}\left(E~I\frac{d^2w}{dx^2}\right)\right|_{x=L} & = 0 \end{align}$$

Application of the Hellinger-Prange-Reissner variational principle
The governing equations of the cantilever beam can be written as

Kinematics

\kappa = \frac{d^2 w}{dx^2} ; w(0) = 0 ; \left.\frac{dw}{dx}\right|_{x=0} = 0 $$

Constitutive Equation

M = EI\kappa $$

Equilibrium (kinetics)

\frac{d^2M}{dx^2} + q + P\delta(L-x) = 0 ; M(L) = 0 ; \left.\frac{dM}{dx}\right|_{x=L} = 0 $$

Recall that the Hellinger-Prange-Reissner functional is given by

{\mathcal H}[s] = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma}) - \int_{\mathcal{B}} \boldsymbol{\sigma}:\boldsymbol{\varepsilon}~dV - \int_{\mathcal{B}} \mathbf{f}\bullet\mathbf{u}~dV + \int_{\partial{\mathcal{B}}^{u}} \mathbf{t}\bullet(\mathbf{u}-\widehat{\mathbf{u}})~dA + \int_{\partial{\mathcal{B}}^{t}} \widehat{\mathbf{t}}\bullet\mathbf{u}~dA $$

If we apply the strain-displacement constraints using the Lagrange multipliers $$\boldsymbol{\lambda}\,$$ and the displacement boundary conditions using the Lagrange multipliers $$\boldsymbol{\mu}\,$$, we get a modified functional

\bar{\mathcal H}[\mathbf{u},\boldsymbol{\varepsilon},\boldsymbol{\lambda},\boldsymbol{\mu}] = \int_{\mathcal{B}} \left[U(\boldsymbol{\varepsilon}) + \boldsymbol{\lambda}:[\frac{1}{2}(\boldsymbol{\nabla}\mathbf{u}-\boldsymbol{\nabla}\mathbf{u}^T) - \boldsymbol{\varepsilon}] - \mathbf{f}\bullet\mathbf{u}\right]~dV - \int_{\partial{\mathcal{B}}^{u}} \boldsymbol{\mu}\bullet(\mathbf{u}-\widehat{\mathbf{u}})~dA - \int_{\partial{\mathcal{B}}^{t}} \widehat{\mathbf{t}}\bullet\mathbf{u}~dA $$

For the cantilevered beam, the above functional becomes
 * $$\begin{align}

\bar{\mathcal H}[w,\kappa,\lambda,\mu_1,\mu_2] = & \int_0^L \left[\frac{EI}{2}\kappa^2 + \left(\frac{d^2w}{dx^2}-\kappa\right)\lambda + [q + P\delta(L-x)]~w\right]~dx \\ & - M(L)\frac{dw}{dx}(L) - \frac{dM}{dx}(L)~w(L) + \mu_1[w(0) - 0] + \mu_2[\frac{dw}{dx}(0) - 0] \end{align}$$

Taking the first variation of the functional, we can easily derive the Euler equations and the associated BCs.
 * $$\begin{align}

\delta\kappa : & EI\kappa - \lambda = 0 \\ \delta w : & \frac{d^2\lambda}{dx^2} + q + P\delta(L-x) = 0 \\ \delta\lambda : & \frac{d^2 w}{dx^2} -\kappa = 0 \end{align}$$ and
 * $$\begin{align}

\frac{d}{dx}(\delta w) : & \lambda(0) = \mu_2, \lambda(L) = M(L)\\ \delta w : & \frac{d\lambda}{dx}(0) = -\mu_1 , \frac{d\lambda}{dx}(L) = \frac{dM}{dx}(L)\\ \delta\mu_1 : & w(0) = 0 \\ \delta\mu_1 : & \frac{dw}{dx}(0) = 0 \end{align}$$

The same process can be used to derive Euler equations using the Hu-Washizu variational principle.

title: Elasticity/Energy methods example 4


text: File:Cantilever.png

title: File:Cantilever.png


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User:Johnkecops

Hai...I'm John Kecops, Biotechnologist from University of Indonesia. I'm interested in studying Marxism philosophy, but only as a tool in observing social-politics phenomenon. My favorite link is the marxists internet archive []

title: User:Johnkecops


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User talk:Johnkecops

Hello Johnkecops, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 00:11, 7 August 2007 (UTC)

title: User talk:Johnkecops


text:
Elasticity/Sample quiz1

 * 1) Show the directions of normal and shear stresses on a cube of material.  Also label the arrows using the standard symbols and subscripts.
 * 2) Since you are using arrows, you are suggesting that the stresses you have shown are vectors. Are they ?
 * 3) Write down the strain-displacement relations.
 * 4) Write down the stress-strain relations for an isotropic, linear elastic material in matrix form.

title: Elasticity/Sample quiz1


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Elasticity/Sample quiz 2

 * 1) Write down the equation of equilibrium in the absence of body forces.  Use index notation.
 * 2) Write down the relationship between the traction vector acting on an arbitrary surface of a body and the stress tensor in the body.
 * 3) Write down the basis transformation rule for the stress tensor in both index notation and matrix notation.
 * 4) Consider a square block under pure shear as shown below.  Draw the directions of the maximum and minimum principal stresses on the figure.  Ignore the third dimension.

title: Elasticity/Sample quiz 2


text: File:PureShear.jpg

title: File:PureShear.jpg


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Elasticity/Sample quiz 3

Given the displacement vector
 * $$     \mathbf{u} = X_1^2 X_2 \widehat{\mathbf{e}}_{1} + X_3 \widehat{\mathbf{e}}_{2} + X_3^2 \widehat{\mathbf{e}}_{3}

$$
 * 1) Find the strain tensor $$\boldsymbol{\varepsilon}$$ and the rotation tensor $$\mathbf{W}\,$$.
 * 2) Write down the compatibility equation in the $$\widehat{\mathbf{e}}_{1}$$- $$\widehat{\mathbf{e}}_{2}$$ plane. Use index notation with commas representing partial derivatives with respect to Lagrangian material co-orinates.

title: Elasticity/Sample quiz 3


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Elasticity/Sample quiz4

 * 1) What is the number of governing equations in the linear theory of elasticity ?  What is the number of unknowns ?
 * 2) The Lame-Navier equations represent the equilibrium equations in terms of displacements.  Why are these equations rarely used ?
 * 3) What are Michell's equations (use words) ? How are they different from Beltrami's equations?
 * 4) Why do we need boundary conditions ? Can I apply full traction and displacement boundary conditions at the same point ? Do boundary conditions have to be applied at all points on the surface of a body ?
 * 5) What is St. Venant's principle ?
 * 6) What is the superposition principle ?
 * 7) I have done a lot of algebra and come up with two different solutions for the same linear elasticity problem.  How do I know which solution is correct ?
 * 8) Give an example of a problem where the Betti-Rayleigh reciprocity relations can be used.

title: Elasticity/Sample quiz4


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Elasticity/Sample quiz5

 * An infinite body has a uniform state stress of pure hydrostatic pressure $$-p$$. This stress state is perturbed by a spherical void of radius $$a$$ which leads to a new stress state

\sigma_{rr} = -p\left(1 - \frac{a^3}{r^3}\right) ~; \sigma_{\theta\theta} = \sigma_{\phi\phi} = -p\left(1 + \frac{a^3}{r^3}\right) $$
 * What do you expect the values of $$\sigma_{r\theta}\,$$, $$\sigma_{\theta\phi}\,$$, and $$\sigma_{\phi r}\,$$ to be? What is the stress concentration factor at the void?

 * What are the displacement conditions for antiplane strain in rectangular co- ordinates? Give an example of a problem that can be approximated by antiplane strain.

 * What are the strain conditions for plane strain in rectangular co-ordinates? Give an example of a problem that can be approximated by plane strain.  What is the difference between plane strain and generalized plane strain?

 * What are the stress conditions for plane stress in rectangular co-ordinates? Give an example of a problem that can be approximated by plane stress.

 * Can the Airy stress function be used for three-dimensional problems? Write down the relation between the Airy stress function and stress in rectangular co-ordinates.  How does this relation change when you solve problems that involve a body force?  What additional restriction on the Airy stress function must be checked to see if compatibility is satisfied?

 * In the absence of body forces, the displacements (plane strain/stress) can be expressed as

2\mu u_1 = - \varphi_{,1} + \alpha \psi_{,2} ~; 2\mu u_2 = - \varphi_{,2} + \alpha \psi_{,1} $$
 * What is $$\psi\,$$ and what are the restrictions on it?

title: Elasticity/Sample quiz5


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Lesson:Thumbnail Storyboard

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File:Thumbnail story2.jpg

Lesson:Thumbnail Storyboard

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Urantia Book/Science vs Religion

The Urantia Book goes to great lengths to show how science and religion are not incompatible. Science is like the art of measuring: man's study of the finite. Religion is man's study of the infinite.

Before this lesson gets developed, I'd like to share the experience that prompted it's creation. The Times has been publishing newer editions of the Complete World History, or some title like that. A reviewer on Amazon.com is irritated that evolution is the only theory offered to explain man's origins. The review suggests both creation and intelligent design as alternatives. This lack of recognition of the beliefs of millions of people around the world led the reviewer to feel as though the Times authors are intentionally misleading readers.

"Never attribute to malice that which can be explained by stupidity."

It pains me to see scientists and religionists butting heads in such a futile manner. They both think they know more than the other. The scientist has hundreds of experiments to point to as his authority (although Philosophy can render sensory input meaningless). The religionist can only point to a subjective set of morals, ostensibly given by God. The scientist questions the religionist's veracity, while the religionist looks down at the seemingly confused, immoral scientist.

In searching for answers, the scientist works from here and now, from the things he can see, and works backwards, in a series of carefully measured steps. The religionist starts at God and works his way to the here and now. Science will never prove or disprove God. Neither will any possible science in the future be able to prove or disprove God.

Science and religion need to leave each other alone. Our species was once without science. But while science has given us a long list of horrors, it has also given us unimagined opportunities. Science eases our lives, religion motivates us.

title: Urantia Book/Science vs Religion


text:
World War II/Important People

People in WWII (World War Two) who were of importance, divided into Allies and Axis

ALLIES

 * 1) Franklin D Roosevelt of United States of America; died during World War II.
 * 2) Joseph Stalin of Soviet Union
 * 3) Harry S. Truman of United States of America; came after FDR
 * 4) Winston Churchill of Great Britain
 * 5) Chiang Kai-shek of China
 * 6) Charles de Gaulle of France

Secondary:

 * 1) Neville Chamberlain of Great Britain; Succeeded by Churchill
 * 2) Władysław Raczkiewicz President and head of state of Poland 1939-47

AXIS

 * 1) Adolf Hitler of Germany
 * 2) Benito Mussolini of Italy
 * 3) Hirohito of Japan

And a recognition to all those military who died as casualties. As well as a recognition to the thousands of civilians that died to the hands of the Allies and Axis

title: World War II/Important People


text:
File:Fred Sample thumbnail Storyboard.png

Lesson:Thumbnail Storyboard


text:
User talk:Dreadatticus

Hello Dreadatticus, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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text:
User:Tomiko~enwikiversity

Thomas Hardiyanto's account

title: User:Tomiko~enwikiversity


text:
Elasticity/Sample midterm 1

Sample Midterm Problem 1
Given:

The vectors $$\mathbf{a}\,$$, $$\mathbf{b}\,$$, and $$\mathbf{c}\,$$ are given, with respect to an orthonormal basis $$(\widehat{\mathbf{e}}_{1},\widehat{\mathbf{e}}_{2},\widehat{\mathbf{e}}_{3})$$, by

\mathbf{a} = 5~\widehat{\mathbf{e}}_{1} - 3~\widehat{\mathbf{e}}_{2} + 10~\widehat{\mathbf{e}}{3}~; \mathbf{b} = 4~\widehat{\mathbf{e}}_{1} + 6~\widehat{\mathbf{e}}_{2} - 2~\widehat{\mathbf{e}}_{3}~; \mathbf{c} = 10~\widehat{\mathbf{e}}_{1} + 6~\widehat{\mathbf{e}}_{2} $$

Find:

 * (a) Evaluate $$ d = a_m~c_m~b_1$$.
 * (b) Evaluate $$ \mathbf{D} = \mathbf{a}\otimes\mathbf{c}$$. Is $$\mathbf{D}\,$$ a tensor?  If not, why not? If yes, what is the order of the tensor?
 * (c) Name and define $$\delta_{ij}\,$$ and $$e_{ijk}\,$$.
 * (d) Evaluate $$ g = D_{ij} \delta_{ij} \,$$.
 * (e) Show that $$ \delta_{ik} e_{ikm} = 0 \,$$.
 * (f) Rotate the basis $$(\widehat{\mathbf{e}}_{1},\widehat{\mathbf{e}}_{2},\widehat{\mathbf{e}}_{3})$$ by 30 degrees in the counterclockwise direction around $$\widehat{\mathbf{e}}_{3}$$ to obtain a new basis $$(\mathbf{e}_1^{'},\mathbf{e}_2^{'},\mathbf{e}_3^{'})$$. Find the components of the vector $$\mathbf{b}\,$$ in the new basis $$(\mathbf{e}_1^{'},\mathbf{e}_2^{'},\mathbf{e}_3^{'})$$.
 * (g) Find the component $$D_{12}\,$$ of $$\mathbf{D}\,$$ in the new basis $$(\mathbf{e}_1^{'},\mathbf{e}_2^{'},\mathbf{e}_3^{'})$$.

Part (a)

d = [(5)(10) + (-3)(6) + (10)(0)](4) = 128 $$

{d = 128} $$

Part (b)

\mathbf{D} = a_i~c_j = \begin{bmatrix} (5)(10) & (5)(6) & (5)(0) \\              (-3)(10) & (-3)(6) & (-3)(0) \\               (10)(10) & (10)(6) & (10)(0)              \end{bmatrix} $$

{\mathbf{D}   = \begin{bmatrix} 50 & 30 & 0 \\              -30 & -18 & 0 \\               100 & 60 & 0              \end{bmatrix} } $$

{\mathbf{D}~\text{is a second-order tensor}.} $$

Part (c)

{\delta_{ij} = \text{Kronecker delta}} $$

{e_{ijk} = \text{Permutation symbol}} $$

{        \delta_{ij} = \begin{cases} 1 & \rm{if}~ i=j \\ 0 & \rm{otherwise} \end{cases} }   $$

{        e_{ijk} = \begin{cases} 1 & \rm{if}~ ijk = 123,~231,~312 \\ -1 & \rm{if}~ ijk = 321,~213,~132 \\ 0 & \rm{otherwise} \end{cases} }   $$

Part (d)

g = D_{kk} = D_{11} + D_{22} + D_{33} = 50 - 18 + 0 = 32 \, $$

{g = 32}\, $$

Part (e)

{ \delta_{ik} e_{ikm} = e_{jjm} = 0} $$ Because $$jjm$$ cannot be an even or odd permutation of $$1,2,3$$.

Part (f)
The basis transformation rule for vectors is

v_i^{'} = l_{ij} v_j $$ where

l_{ij} = \widehat{\mathbf{e}}{i}^{'}\bullet\widehat{\mathbf{e}}{j} = \cos(\widehat{\mathbf{e}}{i}^{'},\widehat{\mathbf{e}}{j}) $$ Therefore,
 * $$\begin{align}

\left[L\right] &= \begin{bmatrix} \cos(30^o) & \cos(90^o-30^o) & \cos(90^o) \\ \cos(90^o+30^o) & \cos(30^o) & \cos(90^o) \\ \cos(90^o) & \cos(90^o) & \cos(0^o) \end{bmatrix} \\ &= \begin{bmatrix} \cos(30^o) & \sin(30^o) & \cos(90^o) \\ -\sin(30^o) & \cos(30^o) & \cos(90^o) \\ \cos(90^o) & \cos(90^o) & \cos(0^o) \end{bmatrix} \\ &= \begin{bmatrix} \sqrt{3}/2 & 1/2 & 0 \\ -1/2 & \sqrt{3}/2 & 0 \\ 0 & 0 & 1             \end{bmatrix} \end{align}$$ Hence,
 * $$\begin{align}

b_1^{'} & = l_{11} b_1 + l_{12} b_2 + l_{13} b_3 = (\sqrt{3}/2)(4) + (1/2)(6) + (0)(-2) = 2\sqrt{3} + 3 = 6.46 \\ b_2^{'} & = l_{21} b_1 + l_{22} b_2 + l_{23} b_3 = (-1/2)(4) + (\sqrt{3}/2)(6) + (0)(-2) = -2 + 3\sqrt{3} = 3.2\\ b_3^{'} & = l_{31} b_1 + l_{32} b_2 + l_{33} b_3 = (0)(4) + (0)(6) + (1)(-2) = -2     \end{align}$$ Thus,

{\mathbf{b}^{'} = 6.46~\mathbf{e}_1^{'}~ +~ 3.2~\mathbf{e}_2^{'}~ -~ 2\mathbf{e}_3^{'} } $$

Part (g)
The basis transformation rule for second-order tensors is

D_{ij}^{'} = l_{ip} l_{jq} D_{pq} \, $$ Therefore,
 * $$\begin{align}

D_{12}^{'} = & l_{11} l_{21} D_{11} + l_{12} l_{21} D_{21} + l_{13} l_{21} D_{31} + l_{11} l_{22} D_{12} + l_{12} l_{22} D_{22} + l_{13} l_{22} D_{32} +\\ & l_{11} l_{23} D_{13} + l_{12} l_{23} D_{23} + l_{13} l_{23} D_{33} \\ = & l_{11} (l_{21} D_{11} + l_{22} D_{12} + l_{23} D_{13}) + l_{12} (l_{21} D_{21} + l_{22} D_{22} + l_{23} D_{23}) +\\ & l_{13} (l_{21} D_{31} + l_{22} D_{32} + l_{23} D_{33}) \\ = & (\frac{\sqrt{3}}{2})\left[(-\frac{1}{2})(50)+(\frac{\sqrt{3}}{2})(30)+(0)(0)\right] + (\frac{1}{2})\left[(-\frac{1}{2})(-30)+(\frac{\sqrt{3}}{2})(-18)+(0)(0)\right] + \\ & (0)\left[(-\frac{1}{2})(100)+(\frac{\sqrt{3}}{2})(60)+(0)(0)\right]\\ = & (\frac{\sqrt{3}}{2})\left[-25+ 15\sqrt{3}\right] + (\frac{1}{2})\left[15 - 9\sqrt{3}\right] \\ = & -25\frac{\sqrt{3}}{2} + \frac{45}{2} + \frac{15}{2} - 9\frac{\sqrt{3}}{2} \\ = & -17\sqrt{3} + 30 \end{align}$$

{D_{12}^{'} = -17\sqrt{3} + 30 = 0.55} $$

title: Elasticity/Sample midterm 1


text:
Elasticity/Sample midterm 2

Sample Midterm Problem 2
Given:

A strain gage rosette provides the following data

\varepsilon_{1} = 0.01; \varepsilon_{2} = 0.02; \varepsilon_{30^o} = 0 $$ where the $$X_1$$ and $$X_2$$ directions are perpendicular to each other and $$\varepsilon_{30^o}$$ is the extensional strain of a line element at an angle of $$30^o$$ to the $$X_1$$ axis (in the counterclockwise direction).

Find:
 * (a) Compute $$\varepsilon_{60^o}$$.
 * (b) Is the result valid if the material is anisotropic ?

Part (a)
From the previous problem, for an angle of rotation of 30$$^o$$, the rotation matrix $$\left[L\right]$$ is

l_{ij} = \left[L\right] = \begin{bmatrix} \sqrt{3}/2 & 1/2 & 0 \\ -1/2 & \sqrt{3}/2 & 0 \\ 0 & 0 & 1           \end{bmatrix} $$ Therefore, the components of strain in the rotated co-ordinate system are given by

\left[\boldsymbol{\varepsilon}\right]^{'} = \left[L\right] \left[\boldsymbol{\varepsilon}\right] \left[L\right]^T \text{or,} \varepsilon^{'}_{ij} = l_{ip} l_{jq} \varepsilon_{pq} $$ Since we are given $$\varepsilon_{30^o} = \varepsilon^{'}_{11}$$, we will calculate the value of this strain in terms of the original components of strain. Thus,
 * $$\begin{align}

\varepsilon^{'}_{11} = & l_{1p} l_{1q} \varepsilon_{pq} \\ = & l_{11}l_{11}\varepsilon_{11} + l_{12}l_{11}\varepsilon_{21} + l_{13}l_{11}\varepsilon_{31} + l_{11}l_{12}\varepsilon_{12} + l_{12}l_{12}\varepsilon_{22} + l_{13}l_{12}\varepsilon_{32} +\\ & l_{11}l_{13}\varepsilon_{13} + l_{12}l_{13}\varepsilon_{23} + l_{13}l_{13}\varepsilon_{33} \\ = & l_{11}(l_{11}\varepsilon_{11} + l_{12}\varepsilon_{12} + l_{13}\varepsilon_{13}) + l_{12}(l_{11}\varepsilon_{21} + l_{12}\varepsilon_{22} + l_{13}\varepsilon_{23}) + \\ & l_{13}(l_{11}\varepsilon_{31} + l_{12}\varepsilon_{32} + l_{13}\varepsilon_{33}) \\ = & (\frac{\sqrt{3}}{2})\left[(\frac{\sqrt{3}}{2})(0.01) + (\frac{1}{2})\varepsilon_{12}\right] + (\frac{1}{2})\left[(\frac{\sqrt{3}}{2})\varepsilon_{12} + (\frac{1}{2})(0.02)\right] \\ = & (3/4)(0.01) + (\sqrt{3}/2)\varepsilon_{12} + (1/4)(0.02) \\ = & (5/4)(0.01) + (\sqrt{3}/2)\varepsilon_{12} \end{align}$$ Therefore,

(5/4)(0.01) + (\sqrt{3}/2)\varepsilon_{12} = \varepsilon_{30^o} = 0 $$ Hence,

\varepsilon_{12} = -(2.5)(0.01)/\sqrt{3} $$ Next, for an angle of rotation of 60$$^o$$, the matrix $$\left[L\right]$$ is
 * $$\begin{align}

\left[L\right]  &= \begin{bmatrix} \cos(60^o) & \sin(60^o) & \cos(90^o) \\ -\sin(60^o) & \cos(60^o) & \cos(90^o) \\ \cos(90^o) & \cos(90^o) & \cos(0^o) \end{bmatrix} \\ &= \begin{bmatrix} 1/2 & \sqrt{3}/2 & 0 \\ -\sqrt{3}/2 & 1/2 & 0 \\ 0 & 0 & 1             \end{bmatrix} \end{align}$$ Therefore, $$\varepsilon_{60^o} = \varepsilon^{'}_{11}$$, is given by
 * $$\begin{align}

\varepsilon^{'}_{11} = & l_{1p} l_{1q} \varepsilon_{pq} \\ = & l_{11}l_{11}\varepsilon_{11} + l_{12}l_{11}\varepsilon_{21} + l_{13}l_{11}\varepsilon_{31} + l_{11}l_{12}\varepsilon_{12} + l_{12}l_{12}\varepsilon_{22} + l_{13}l_{12}\varepsilon_{32} +\\ & l_{11}l_{13}\varepsilon_{13} + l_{12}l_{13}\varepsilon_{23} + l_{13}l_{13}\varepsilon_{33} \\ = & l_{11}(l_{11}\varepsilon_{11} + l_{12}\varepsilon_{12} + l_{13}\varepsilon_{13}) + l_{12}(l_{11}\varepsilon_{21} + l_{12}\varepsilon_{22} + l_{13}\varepsilon_{23}) + \\ & l_{13}(l_{11}\varepsilon_{31} + l_{12}\varepsilon_{32} + l_{13}\varepsilon_{33}) \\ = & (\frac{1}{2})\left[(\frac{1}{2})(0.01) + (\frac{\sqrt{3}}{2})\varepsilon_{12}\right] + (\frac{\sqrt{3}}{2})\left[(\frac{1}{2})\varepsilon_{12} + (\frac{\sqrt{3}}{2})(0.02)\right] \\ = & (1/4)(0.01) + (\sqrt{3}/2)\varepsilon_{12} + (3/4)(0.02) \\ = & (7/4)(0.01) + (\sqrt{3}/2)(-(2.5)(0.01)/\sqrt{3} ) \\ = & (7/4)(0.01) - (5/4)(0.01) = (1/2)(0.01) = 0.005 \\ \end{align}$$ Therefore,

{\varepsilon_{60^o} = 0.005} $$

Part (b)

{\text{The result is valid for all materials.}} $$

title: Elasticity/Sample midterm 2


text:
Elasticity/Sample midterm3

Sample Homework Problem 3
For an isotropic material with $$E = 100$$ GPa and $$\nu = 0.25$$, find the stress tensor and strain energy density at a point in a body if the components of the strain tensor are given by

\varepsilon_{ij} = \begin{bmatrix} 200 & 100 & 0 \\                      100 & 200 & 100 \\                       0 & 100 & 0                      \end{bmatrix} \times 10^{-6}. $$

Solution
The shear modulus ($$\mu$$) is given by

\mu = \frac{E}{2(1+\nu)} = \frac{100}{2.5} = 40~\text{GPa} = 40\times 10^6 ~\text{KPa} $$ The Lamé modulus ($$\lambda$$) is given by

\lambda = \frac{E\nu}{(1+\nu)(1-2\nu)} = \frac{25}{(1.25)(0.5)} = 40~\text{GPa} = 40\times 10^6 ~\text{KPa} $$ The stress-strain relation for isotropic materials is
 * $$\begin{align}

\sigma_{ij} & = 2\mu\varepsilon_{ij} + \lambda\varepsilon_{kk}\delta_{ij} \\ & = 40(2\varepsilon_{ij} + \varepsilon_{kk}\delta_{ij}) \end{align}$$ Therefore, (after converting $$\mu$$ and $$\lambda$$ into KPa so that the $$10^{-6}$$ term in the strain cancels out),
 * $$\begin{align}

\sigma_{11} & = 40(2\varepsilon_{11} + \varepsilon_{11} + \varepsilon_{22}                    + \varepsilon_{33}) \\ & = 40\left[(3)(200) + 200 + 0\right] = (40)(800) = 32000 \\ \sigma_{22} & = 40(2\varepsilon_{22} + \varepsilon_{11} + \varepsilon_{22}                    + \varepsilon_{33}) \\ & = 40\left[(3)(200) + 200 + 0\right] = (40)(800) = 32000 \\ \sigma_{33} & = 40(2\varepsilon_{33} + \varepsilon_{11} + \varepsilon_{22}                    + \varepsilon_{33}) \\ & = 40\left[(3)(0) + 200 + 200\right] = (40)(400) = 16000 \\ \sigma_{23} & = 40(2\varepsilon_{23}) = (40)(200) = 8000\\ \sigma_{31} & = 40(2\varepsilon_{31}) = (40)(0) = 0\\ \sigma_{12} & = 40(2\varepsilon_{12}) = (40)(200) = 8000 \end{align}$$ In 3$$\times$$3 matrix form (after converting into MPa from KPa)

{ \sigma_{ij} = \begin{bmatrix} 32 & 8 & 0 \\               8 & 32 & 8 \\                0 & 8 & 16                 \end{bmatrix} ~\text{MPa} } $$ The strain energy density is given by

U(\boldsymbol{\varepsilon}) = \frac{1}{2} \sigma_{ij}\varepsilon_{ij} $$ Therefore,
 * $$\begin{align}

U(\boldsymbol{\varepsilon}) & = \frac{1}{2}\left[ \sigma_{11} \varepsilon_{11} + \sigma_{22} \varepsilon_{22} + \sigma_{33} \varepsilon_{33} + 2 \sigma_{23} \varepsilon_{23} + 2 \sigma_{31} \varepsilon_{31} + 2 \sigma_{12} \varepsilon_{12} \right] \\ & = \frac{1}{2}\left[(32)(200) + (32)(200) + (16)(0) + (2)(8)(100) + (2)(0)(0) + (2)(8)(100)\right]~\text{Pa}\\ & = \frac{1}{2}\left[6400 + 6400 + 1600 + 1600\right]~\text{Pa}\\ & = 8000~\text{Pa} = 8 ~\text{KPa} \end{align}$$ The strain energy density is

{ U = 8~\text{KPa}} $$

title: Elasticity/Sample midterm3


text:
User:Teacherbrook

Hello - I need to create this userpage soon.

Right now, I will let you know:

Brook Walter Snyder www.omaisa.com

Hope you see my website, and you can find my email!

Thank you....

title: User:Teacherbrook


text:
Elasticity/Sample midterm 4

Part (a)
A solid is subjected to stresses as shown by the arrows in the figure below. Indicate the indices for each of the stress components, and whether the stresses should be positive or negative.

Part (b)
Suppose that the stress tensor field in a body is given by

\sigma_{ij} = \begin{bmatrix} x_1^2 + x_2 & - 2 x_3^2 &  x_1^2 + x_2 \\ - 2 x_3^2  &  x_1 + x_3 & -x_3^2 + x_2 \\ x_1^2 + x_2 & -x_3^2 + x_2 & x_1^2 + x_2 \end{bmatrix} (\text{MPa}) $$ Find the body force distribution required to maintain equilibrium. ($$x_1$$, $$x_2$$, and $$x_3$$ are in meters). Show units.

Solution
The equation of equilibrium is

\boldsymbol{\nabla}\bullet\boldsymbol{\sigma} + \mathbf{b} = 0\text{or,}\sigma_{ij,i} + b_j = 0 $$ Therefore,
 * $$\begin{align}

\frac{\partial (x_1^2 + x_2)}{\partial x_1} + \frac{\partial (-2 x_3^2)}{\partial x_2} + \frac{\partial (x_1^2 + x_2)}{\partial x_3} + b_1 & = 0 \\ \frac{\partial (-2 x_3^2)}{\partial x_1} + \frac{\partial (x_1 + x_3)}{\partial x_2} + \frac{\partial (-x_3^2 + x_2)}{\partial x_3} + b_2 & = 0 \\ \frac{\partial (x_1^2 + x_2)}{\partial x_1} + \frac{\partial (-x_3^2 + x_2)}{\partial x_2} + \frac{\partial (x_1^2 + x_2)}{\partial x_3} + b_3 & = 0 \end{align}$$ or,
 * $$\begin{align}

2~x_1 + 0 + 0 + b_1 & = 0 \\ 0 + 0 - 2~x_3 + b_2 & = 0 \\ 2~x_1 + 1 + 0 + b_3 & = 0 \end{align}$$ The required body forces are (in MN/m$$^3$$)

{ b_1 = -2~x_1~;b_2 = 2~x_3~;b_3 = - (1 + 2~x_1)} $$

Part (c)
Find the surface tractions at the internal point $$\mathbf{r} = (1,-1,2)$$ in the body on an internal surface with a surface normal $$\widehat{\mathbf{n}} = (-1,-1,1)$$.

Solution
The surface traction is given by

\mathbf{t} = \widehat{\mathbf{n}}\bullet\boldsymbol{\sigma} \text{or,} t_j = n_i \sigma_{ij} $$ The stress at point $$\mathbf{r}$$ is

\left[\boldsymbol{\sigma}\right] = \begin{bmatrix} 0 & -8 & 0 \\ -8 & 3 & -5 \\ 0 & -5 & 0              \end{bmatrix}~(\text{MPa}) $$ Therefore,
 * $$\begin{align}

t_1 & = n_1 \sigma_{11} + n_2 \sigma_{21} + n_3 \sigma_{31} \\ & = (-1)(0) + (-1)(-8) + (1)(0) = 8 ~\text{MPa}\\ t_2 & = n_1 \sigma_{12} + n_2 \sigma_{22} + n_3 \sigma_{32} \\ & = (-1)(-8) + (-1)(3) + (1)(-5) = 0 ~\text{MPa}\\ t_3 & = n_1 \sigma_{13} + n_2 \sigma_{23} + n_3 \sigma_{33} \\ & = (-1)(0) + (-1)(-5) + (1)(0) = 5 ~\text{MPa}\\ \end{align}$$ The traction vector is (after converting $$\widehat{\mathbf{n}}{}$$ into a unit normal)

{\mathbf{t} = (1/\sqrt{3})(8,~0,~5) (\text{MPa})} $$

Part (d)
Find the hydrostatic and deviatoric stress at the point $$\mathbf{r}=(1,-1,2)$$.

Solution
The hydrostatic stress is given by

\boldsymbol{\sigma}_h = \frac{\text{tr}~\boldsymbol{\sigma}}{3} \mathbf{I} $$ In this case,

\text{tr}\boldsymbol{\sigma} = \sigma_{11} + \sigma_{22} + \sigma_{33} = 0 + 3 + 0 = 3   $$ Therefore,

{       \boldsymbol{\sigma}_h = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1                 \end{bmatrix}~(\text{MPa}) }   $$ The deviatoric stress is given by

\boldsymbol{\sigma}_d = \boldsymbol{\sigma} - \boldsymbol{\sigma}_h $$ Therefore,

{       \boldsymbol{\sigma}_d = \begin{bmatrix} -1 & -8 & 0 \\ -8 & 2 & -5 \\ 0 & -5 & -1                 \end{bmatrix}~(\text{MPa}) }   $$

Part (e)
Find the principal stresses at this point.

Solution
The principal stresses can be found using the equation

\text{det}(\boldsymbol{\sigma} - \lambda\mathbf{I}) = 0 $$ where $$\lambda$$ is a principal stress. In expanded form,

\text{det} \begin{bmatrix} \sigma_{11} - \lambda & \sigma_{12} & \sigma_{13} \\ \sigma_{12} & \sigma_{22} - \lambda & \sigma_{23} \\ \sigma_{13} & \sigma_{23} & \sigma_{33} - \lambda \end{bmatrix} = 0 $$ Substituting the values of stress into the above equation,

\text{det} \begin{bmatrix} - \lambda & -8 & 0 \\ -8 & 3 - \lambda & -5 \\ 0 & -5 & - \lambda \end{bmatrix} = 0 $$ Expanding out,

-\lambda\left[(3-\lambda)(-\lambda) - (-5)(-5)\right] - (-8)\left[(-8)(-\lambda) - (-5)(0)\right] = 0 $$ or,

-\lambda(-3\lambda+\lambda^2-25) + 8(8\lambda) = 0 $$ or,

3\lambda^2 - \lambda^3 + 25\lambda + 64\lambda = 0 $$ or,

\lambda^3 - 3\lambda^2 - 89\lambda = 0 $$ Thus, the first possible value of $$\lambda = 0$$ MPa. Also,

\lambda^2 - 3\lambda - 89 = 0 $$ Therefore,

\lambda = \frac{3 \pm \sqrt{9+356}}{2} = (11.05, -8.05) \text{MPa} $$ The principal stresses are (in MPa)

{\sigma_1 = 11.05~;\sigma_2 = 0~;\sigma_3 = -8.05} $$

Part (f)
Find the principal direction corresponding to the intermediate principal stress.

Solution
The directions of the principal stresses can be found using the equation

(\boldsymbol{\sigma} - \lambda\mathbf{I})\widehat{\mathbf{n}}{} = 0 $$ For the principal direction $$\widehat{\mathbf{n}}{2}$$ (corresponding to the principal stress $$\sigma_2$$, we have,

\begin{bmatrix} 0 & -8 & 0 \\ -8 & 3 & -5 \\ 0 & -5 & 0        \end{bmatrix} \begin{bmatrix} n^1 \\ n^2 \\ n^3 \end{bmatrix} = 0 $$ Hence,
 * $$\begin{align}

(0)(n^1) - (8)(n^2) + (0)(n^3) & = 0 \\ (-8)(n^1) + (3)(n^2) + (-5)(n^3) & = 0 \end{align}$$ gives us $$n^2 = 0$$ and

8~n^1 = - 5~n^3 $$ Now, $$n^3 = \sqrt{1 - (n^1)^2 - (n^2)^2} = \sqrt{1-(n^1)^2}$$. Therefore,

8~n^1 = - 5~\sqrt{1-(n^1)^2} $$ Taking squares of both sides,

64(n^1)^2 = 25\left[1-(n^1)^2\right] $$ So we get,

(n^1)^2 = 25/89,\text{or}n^1 = 5/\sqrt{89} $$ Therefore,

n^3 = \sqrt{1-(n^1)^2} = 8/\sqrt{89} $$ The direction corresponding to the intermediate principal stress is

{\widehat{\mathbf{n}}{2} = (5/\sqrt{89},~0,~8/\sqrt{89})} $$

Part (g)
The symmetry of the stress tensor can be derived from a certain balance principle. Name the principle and write it down in index notation.

Solution
The balance principle is

{\text{Conservation of angular momentum}} $$ In index notation

{      \int_{\partial\Omega} e_{ijk}~x_j~n_l~\sigma_{lk}~ds + \int_{\Omega} e_{ijk}~x_j~b_k~dV = \frac{d}{dt} \int_{\Omega} \rho~e_{ijk}~x_j~v_k~dV }   $$

title: Elasticity/Sample midterm 4


text:
Elasticity/Sample midterm5

Sample Midterm Problem 5
Suppose that, under the action of external forces, a material point $$\mathbf{p} = (X_1,X_2,X_3)$$ in a body is displaced to a new location $$\mathbf{q} = (x_1,x_2,x_3)$$ where

x_1 = A~X_1 + \kappa~X_2~; x_2 = A~X_2 + \kappa~X_1~; x_3 = X_3 $$ and $$A$$ and $$\kappa$$ are constants.

Part (a)
A displacement field is called proper and admissible if the Jacobian ($$J$$) is greater than zero. If a displacement field is proper and admissible, then the deformation of the body is continuous.

Indicate the restrictions that must be imposed upon $$A$$ so that the deformation represented by the above displacement is continuous.

Solution
The deformation gradient $$(F)$$ is given by
 * $$\begin{align}

F_{ij} & = \frac{\partial x_i }{\partial X_j} \\ & = \begin{bmatrix} A & \kappa & 0 \\ \kappa & A & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{align}$$ Therefore, the requirement is that $$J = \text{det}(F) > 0$$ where

J = A^2 - \kappa^2 $$ The restriction is

{|A| > |\kappa|} $$

Part (b)
Suppose that $$A = 0$$. Calculate the components of the infinitesimal strain tensor $$\boldsymbol{\varepsilon}$$ for the above displacement field.

Solution
The displacement is given by $$\mathbf{u} = \mathbf{x} - \mathbf{X}$$. Therefore,

\mathbf{u} = \begin{bmatrix} \kappa~X_2-X_1 \\ \kappa~X_1-X_2 \\ 0 \end{bmatrix} $$ The infinitesimal strain tensor is given by

\boldsymbol{\varepsilon} = \frac{1}{2}(\boldsymbol{\nabla}u + \boldsymbol{\nabla}u^T) $$ The gradient of $$\mathbf{u}$$ is given by

\boldsymbol{\nabla}u = \begin{bmatrix} -1 & \kappa & 0 \\ \kappa & -1 & 0 \\ 0 & 0 & 0              \end{bmatrix} $$ Therefore,

{     \boldsymbol{\varepsilon} = \begin{bmatrix} -1 & \kappa & 0 \\ \kappa & -1 & 0 \\ 0 & 0 & 0             \end{bmatrix} }   $$

Part (c)
Calculate the components of the infinitesimal rotation tensor $$\mathbf{W}$$ for the above displacement field and find the rotation vector $$\boldsymbol{\omega}$$.

Solution
The infinitesimal rotation tensor is given by

\mathbf{W} = \frac{1}{2}(\boldsymbol{\nabla}u - \boldsymbol{\nabla}u^T) $$ Therefore,

{     \mathbf{W} = \begin{bmatrix} 0 & 0 & 0 \\                    0 & 0 & 0 \\                     0 & 0 & 0                     \end{bmatrix} }   $$ The rotation vector $$\boldsymbol{\omega}$$ is

{       \boldsymbol{\omega} = \begin{bmatrix} 0 \\ 0 \\ 0                    \end{bmatrix} }   $$

Part (d)
Do the strains satisfy compatibility ?

Solution
The compatibility equations are
 * $$\begin{align}

\varepsilon_{11,22} + \varepsilon_{22,11} - 2\varepsilon_{12,12} & = 0 \\ \varepsilon_{22,33} + \varepsilon_{33,22} - 2\varepsilon_{23,23} & = 0 \\ \varepsilon_{33,11} + \varepsilon_{11,33} - 2\varepsilon_{13,13} & = 0 \\ (\varepsilon_{12,3} - \varepsilon_{23,1} + \varepsilon_{31,2})_{,1} - \varepsilon_{11,23} & = 0 \\ (\varepsilon_{23,1} - \varepsilon_{31,2} + \varepsilon_{12,3})_{,2} - \varepsilon_{22,31} & = 0 \\ (\varepsilon_{31,2} - \varepsilon_{12,3} + \varepsilon_{23,1})_{,3} - \varepsilon_{33,12} & = 0 \end{align}$$ All the equations are trivially satisfied because there is no dependence on $$X_1$$, $$X_2$$, and $$X_3$$.

{\text{Compatibility is satisfied.}} $$

Part (e)
Calculate the dilatation and the deviatoric strains from the strain tensor.

Solution
The dilatation is given by

e = \text{tr} \boldsymbol{\varepsilon} $$ Therefore,

{e = -2~\text{(Note: Looks like shear only but not really.)}} $$ The deviatoric strain is given by

\boldsymbol{\varepsilon}_d = \boldsymbol{\varepsilon} - \frac{\text{tr}~\boldsymbol{\varepsilon}}{3} \mathbf{I} $$ Hence,

{        \boldsymbol{\varepsilon}_d = \begin{bmatrix} -\cfrac{1}{3} & \kappa & 0 \\ \kappa & -\cfrac{1}{3} & 0 \\ 0 & 0 & -\cfrac{2}{3} \\ \end{bmatrix} }   $$

Part (f)
What is the difference between tensorial shear strain and engineering shear strain (for infinitesimal strains)?

Solution
The tensorial shear strains are $$\varepsilon_{12}$$, $$\varepsilon_{23}$$, $$\varepsilon_{31}$$. The engineering shear strains are $$\gamma_{12}$$, $$\gamma_{23}$$, $$\gamma_{31}$$.

The engineering shear strains are twice the tensorial shear strains.

Part (g)
Briefly describe the process which you would use to calculate the principal stretches and their directions.

Solution

 * Compute the deformation gradient ($$\mathbf{F}$$).
 * Compute the right Cauchy-Green deformation tensor ($$\mathbf{C} = \mathbf{F}^{T}\bullet\mathbf{F}$$).
 * Calculate the eigenvalues and eigenvectors of $$\mathbf{C}$$.
 * The principal stretches are the square roots of the eigenvalues of $$\mathbf{C}$$.
 * The directions of the principal stretches are the eigenvectors of $$\mathbf{C}$$.

title: Elasticity/Sample midterm5


text:
Elasticity/Sample final 1

Sample Final Exam Problem 1
A spherical shell with an outer radius of 5 cm has a spherical inclusion of radius 4 cm embedded in it. The shell is made of AISI-4340 steel which has a Young's modulus of 200 GPa, a Poisson's ratio of 0.3 and a tensile yield stress of 860 MPa. The inclusion is made of a rubbery material with a Young's modulus of 1 GPa, a Poisson's ratio of 0.49 and a tensile yield stress of 10 MPa. Due to a reaction in the rubbery material, the radius of the inclusion increases by 0.4 mm. Assume linear elastic behavior.
 * (a) What is the change in the outer radius of the shell caused by this expansion of the inclusion?
 * (b) Will the shell yield in shear due to this expansion ?

Solution
Calculate bulk modulus ($$K$$) and shear modulus ($$\mu$$) of the shell.
 * $$\begin{align}

K & = \frac{E}{3(1-2\nu)} = \frac{200}{3(1-2\times 0.3)} = 167~\text{GPa} \\ \mu & = \frac{E}{2(1+\nu)} = \frac{200}{2(1+0.3)} = 77~\text{GPa} \end{align}$$ Assume that there is perfect bonding at the interface. The increase in the radius of the inclusion causes an internal pressure in the shell.

The radial displacement of the inside of the shell is $$ka$$ where

k = \frac{0.4 \text{(mm)}}{40 \text{(mm)}} = 0.01 $$ From the equation for the radial displacement in a hollow spherical shell, we have (assuming that the internal pressure is $$p_i$$ and the external pressure is zero),

u_r = \frac{r}{2\mu+3\lambda}\left(\frac{a^3p_i}{b^3-a^3}\right) + \frac{a^3b^3}{4\mu r^2}\left(\frac{p_i}{b^3-a^3}\right) = \frac{r}{3K}\left(\frac{a^3p_i}{b^3-a^3}\right) + \frac{a^3b^3}{4\mu r^2}\left(\frac{p_i}{b^3-a^3}\right) $$ At $$r = a$$,

\left.u_r\right|_{r=a} = ka = \left[\frac{a^4}{3K}+\frac{ab^3}{4\mu}\right] \left(\frac{p_i}{b^3-a^3}\right) \Rightarrow \left[\frac{a^3}{3K}+\frac{b^3}{4\mu}\right] \left(\frac{p_i}{b^3-a^3}\right) = k $$ Therefore,

p_i = (12 K\mu k)\left(\frac{b^3-a^3}{4\mu a^3 + 3K b^3}\right) = (4\mu k)\left(        \cfrac{\cfrac{b^3}{a^3}-1}{\cfrac{b^3}{a^3}+\cfrac{4\mu}{3K}}\right) $$ Plugging in values (with $$b/a = 1.25$$)

p_i = (4)(77)(0.01)\left(        \cfrac{1.25^3-1}{1.25^3+\cfrac{(4)(77)}{(3)(167)}}         \right) = (3.08)\left( \cfrac{1.953-1}{1.953+0.615} \right) = (3.08)(0.371) = 1.14~\text{GPa} = 1140~\text{MPa} $$ Therefore, the displacement of the shell at $$r = b$$ is

\left.u_r\right|_{r=b} = \frac{b}{3K}\left(\frac{a^3p_i}{b^3-a^3}\right) + \frac{a^3b^3}{4\mu b^2}\left(\frac{p_i}{b^3-a^3}\right) = \left[\frac{1}{3K}+\frac{1}{4\mu}\right] \left(\cfrac{p_i b}{\cfrac{b^3}{a^3}-1}\right) $$ Plugging in values,

\left.u_r\right|_{r=b} = \left[\frac{1}{(3)(167)}+\frac{1}{(4)(77)}\right] \left(\cfrac{(1.14)(5)}{(1.25)^3-1}\right) = [0.002 + 0.0032] \left(\cfrac{5.7}{0.953}\right) = 0.031~\text{cm} = 0.31~\text{mm} $$

(a) The change in the outer radius of the shell is 0.31 mm.

The non-zero stresses in the shell are given by

\sigma_{rr} = \left[\frac{p_i}{(b/a)^3 - 1}\right] - \frac{b^3}{r^3} \left[\frac{p_i}{(b/a)^3 - 1}\right] = \left(1- \frac{b^3}{r^3}\right)\left[\frac{p_i}{(b/a)^3 - 1}\right] $$ and

\sigma_{\theta\theta} = \sigma_{\phi\phi} = \left[\frac{p_i}{(b/a)^3 - 1}\right] + \frac{b^3}{2 r^3} \left[\frac{p_i}{(b/a)^3 - 1}\right] = \left(1+ \frac{b^3}{2r^3}\right)\left[\frac{p_i}{(b/a)^3 - 1}\right] $$ Therefore, the maximum shear stress is given by

\tau_{\text{max}} = \frac{1}{2}\left|\sigma_{rr}-\sigma_{\theta\theta}\right| = \frac{1}{2} \left(\frac{3b^3}{2r^3}\right)\left[\frac{p_i}{(b/a)^3 - 1}\right] $$ The magnitude of the maximum shear stress is largest at $$r = a$$. Hence,

\left.\tau_{\text{max}}\right|_{r=a} = \left(\frac{3}{4}\right)\left[\frac{p_i}{1-(a/b)^3}\right] = \left(\frac{3}{4}\right)\left[\frac{1.14}{1-(0.8)^3}\right] = 1.75~\text{GPa} = 1750~\text{MPa} $$ The yield stress in shear is $$\tau_{Y} = \sigma_{Y}/2 = 430$$ MPa which is much lower than the maximum shear stress.

(b) The shell will yield in shear due to the given expansion.

title: Elasticity/Sample final 1


text:
Elasticity/Sample final 2

Sample Final Exam Problem 2
A long elastic cylinder with flat end caps is constructed from 7075 T6 Aluminum alloy. The cylinder has an outer radius of 10 cm and a thickness of 5 mm. The material has a Young's modulus of 72 GPa, a Poisson's ratio of 0.33 and a yield stress under uniaxial tension of 500 MPa. A circular hole of diameter 10 mm is drilled through the thickness of the cylinder so that the contents of the inside can be viewed. The ends of the drilled hole are covered with a strong, flexible and transparent membrane to allow for visibility. The inside of the cylinder is now pressurized uniformly to a pressure of 20 MPa.
 * (a) Give estimates of the magnitudes of the maximum and minimum hoop stress at the drilled hole (using a coordinate system that is centered on the hole).
 * (b) Does the stress state at the hole lead to the yielding of the Aluminum alloy? Use the von Mises yield criterion

\sigma^2_{Y} = \frac{1}{2}\left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2+ (\sigma_3 - \sigma_1)^2\right]~. $$ Clearly state all the non-standard assumptions that you make.

Solution
The stress field in a cylinder with end caps under internal pressure can be approximated (using plane-strain) by
 * $$\begin{align}

\sigma_{rr} & = \frac{p_i}{(b/a)^2-1} - \frac{b^2}{r^2}\frac{p_i}{(b/a)^2-1} = \left(1 - \frac{b^2}{r^2}\right)\frac{p_i}{(b/a)^2-1} \\ \sigma_{\theta\theta} & = \frac{p_i}{(b/a)^2-1} + \frac{b^2}{r^2}\frac{p_i}{(b/a)^2-1} = \left(1 + \frac{b^2}{r^2}\right)\frac{p_i}{(b/a)^2-1} \\ \sigma_{zz} & = \frac{p_i}{(b/a)^2-1} = 185~\text{MPa} \\ \sigma_{r\theta} & = 0 \end{align}$$ At the inner radius of the cylinder $$r = a$$. Therefore,
 * $$\begin{align}

\left.\sigma_{rr}\right|_{r=a} & = \left(1 - \frac{b^2}{a^2}\right)\frac{p_i}{(b/a)^2-1} = -p_i = - 20\text{MPa} \\ \left.\sigma_{\theta\theta}\right|_{r=a} & = \left(1 + \frac{b^2}{a^2}\right)\frac{p_i}{(b/a)^2-1} = \left(1+(10/9.5)^2\right)(185) = 390~\text{MPa} \end{align}$$ At the outer radius of the cylinder $$r = b$$. Therefore,
 * $$\begin{align}

\left.\sigma_{rr}\right|_{r=b} & = 0 \\ \left.\sigma_{\theta\theta}\right|_{r=b} & = \frac{2p_i}{(b/a)^2-1} = 370~\text{MPa} \end{align}$$ At this stage, we make a few assumptions The problem is then converted into one of a circular hole in an infinite thin plate under biaxial tension with $$S_1 = \sigma_{\theta\theta}^{\text{cyl}} = 390$$ MPa and $$S_2 = \sigma_{zz}^{\text{cyl}} = 185$$ MPa.\\
 * Because the cylinder is thin, we can assume that the stresses at $$r = a$$ represent the state of stress in the cylinder.
 * Because the radius of the cylinder is large compared to the radius of the drilled hole, we can consider the cylinder to be approximated by an infinite plate.
 * Because the diameter of the drilled hole is larger than the thickness of the cylinder, we can assume plane stress conditions.
 * We neglect the radial stress in the cylinder because it is small compared to the hoop and axial stresses.

For a plate with a circular under uniaxial tension ($$S_1$$), the stresses are
 * $$\begin{align}

\sigma_{rr} & = \frac{S_1}{2}\left(1-\frac{a^2}{r^2}\right) + \frac{S_1\cos(2\theta)}{2}\left(\frac{3a^4}{r^4}-\frac{4a^2}{r^2}+1\right)\\ \sigma_{\theta\theta} & = \frac{S_1}{2}\left(1+\frac{a^2}{r^2}\right) - \frac{S_1\cos(2\theta)}{2}\left(\frac{3a^4}{r^4}+1\right)\\ \sigma_{r\theta} & = \frac{S_1\sin(2\theta)}{2}\left(\frac{3a^4}{r^4}-\frac{2a^2}{r^2}-1\right) \end{align}$$ At the drilled hole $$r = a$$. Therefore,
 * $$\begin{align}

\sigma_{rr} & = 0 \\ \sigma_{\theta\theta} & = S_1[1 - 2\cos(2\theta)]\\ \sigma_{r\theta} & = 0 \end{align}$$ To get the solution for a biaxial stress field, we superpose the solution in a coordinate system where $$\theta = \theta+\pi/2$$, which is,
 * $$\begin{align}

\sigma_{rr} & = 0 \\ \sigma_{\theta\theta} & = S_2[1 + 2\cos(2\theta)]\\ \sigma_{r\theta} & = 0 \end{align}$$ Therefore, the stress field at the drilled hole is
 * $$\begin{align}

\sigma_{rr} & = 0 \\ \sigma_{\theta\theta} & = (S_1+S_2) + 2(S_2-S_1)\cos(2\theta)\\ \sigma_{r\theta} & = 0 \end{align}$$ The stress $$\sigma_{\theta\theta}$$ at the hole is a maximum or minimum when $$d\sigma_{\theta\theta}/d\theta = 0$$, i.e.,

4(S_2-S_1)\sin(2\theta) = 0 \Rightarrow \theta = 0 ~\text{or}~ \theta = \frac{\pi}{2} $$ The hoop stresses at the drilled hole for these values of $$\theta$$ are
 * $$\begin{align}

\sigma_{\theta\theta} & = (S_1+S_2) + 2(S_2-S_1)\cos(2\theta) = (390+185)+2(185-390)\cos(0) = 165~\text{MPa}\\ \sigma_{\theta\theta} & = (S_1+S_2) + 2(S_2-S_1)\cos(2\theta) = (390+185)+2(185-390)\cos(\pi) = 985~\text{MPa} \end{align}$$

The maximum and minimum hoop stresses at the hole are 985 MPa and 165 MPa, respectively.

At the hole, the maximum principal stress is $$\sigma_1 = 985$$ MPa. The remaining principal stresses are $$\sigma_2 = 0$$, $$\sigma_3 = 0$$. Therefore, the von Mises yield stress is given by

\sigma_{Y} = \sqrt{\frac{1}{2}\left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2+ (\sigma_3 - \sigma_1)^2\right]} = \sigma_1 $$ The uniaxial yield strength of the cylinder is $$500$$ MPa, which is considerably lower than the von Mises yield stress.

The material around the hole will yield under the applied pressure.

title: Elasticity/Sample final 2