User:Eml5526.s11.team01.roark/Mtg05

[[media: Fe1.s11.mtg5.djvu| Mtg 5]]: Fri, 14 Jan 11 [[media: Fe1.s11.mtg5.djvu| Page 5-1]]
 * Team, wiki user account, HW1

Q: “A” in (1) [[media: Fe1.s11.mtg4.djvu| p. 4-2]], no “A” in [[media: Fe1.s11.mtg4.djvu|(2) p. 4-2]] }-heat pb

“A” in [[media: Fe1.s11.mtg2.djvu|(3) P.2-3]], no “A” in [[media: Fe1.s11.mtg4.djvu|(3)-(6) p. 4-3]] }-Elasticity (E in $$\displaystyle \mathbf D$$ [[media: Fe1.s11.mtg4.djvu|(4) p 4-3]])

Answer: Elasticity

Dimensional Analysis (part 1)

Elastic Bar (dynamic): [[media: Fe1.s11.mtg2.djvu|(3) P.2-3]]

$$\displaystyle \left[ f \right]=\frac{F}{L}$$

(f=force per unite length)

$$\displaystyle \left[ m\frac{{{\partial }^{2}}u}{\partial {{t}^{2}}} \right]=\underbrace{\left[ m \right]}_{{}^{M}\!\!\diagup\!\!{}_{L}\;}\underbrace{\left[ \frac{{{\partial }^{2}}u}{\partial {{t}^{2}}} \right]}_{{}^{L}\!\!\diagup\!\!{}_\;}=\frac{F}{L}$$

$$\displaystyle \left[ \frac{{{\partial }^{2}}u}{\partial {{t}^{2}}} \right]=\frac{\left[ {{\partial }^{2}}u \right]}{\left[ \partial {{t}^{2}} \right]}$$

$$\displaystyle \left[ {{\partial }^{2}}u \right]=L$$

$$\displaystyle \left[ \partial {{t}^{2}} \right]={{T}^{2}}$$

F=Force (Dimension)

L=Length(Dimension)

$$\displaystyle m=\rho A$$= mass per unit length, M =Mass (dimension)

[[media: Fe1.s11.mtg5.djvu| Page 5-2]]

$$\displaystyle F=\frac{ML}$$

$$\displaystyle (\underbrace{f}_{Force}=ma)$$

$$\displaystyle \left[ f \right]=F$$

$$\displaystyle \left[ \frac{\partial }{\partial x}\left( EA\frac{\partial u}{\partial x} \right) \right]=\frac{F}{L}$$

With

$$\displaystyle A\sigma =A\left( E\varepsilon \right)=EA\frac{\partial u}{\partial x}$$

$$\displaystyle \left[ \varepsilon \right]=\frac{\left[ \partial u \right]}{\left[ \partial x \right]}=\frac{L}{L}=1\to \left[ \sigma  \right]=\left[ E \right]$$

$$\displaystyle \left[ \sigma \right]=\frac{F}\to \left[ A\sigma  \right]={{L}^{2}}\frac{F}=F$$

$$\displaystyle \left[ \frac{\partial }{\partial x} \right]=\frac{1}{\left[ \partial x \right]}=\frac{1}{L}$$

=>Must have “EA” for consistency in dimension.

Derivation (formulation <-quotation by Einstein) 

Let’s look @ dimension of 2-D or 3-D elastodynamic [[media: Fe1.s11.mtg4.djvu| (3)-(6) p. 4-3.]]

[[media: Fe1.s11.mtg5.djvu| Page 5-3]]

$$\displaystyle \left[ \right]=\frac{F}$$

Where $$\displaystyle $$is body tone, i.e., force per unit volume

$$\displaystyle \left[ \rho \frac{\partial {{t}^{2}}} \right]=\underbrace{\left[ \rho \right]}_{Mass\_Density={}^{M}\!\!\diagup\!\!{}_\;}\underbrace{\left[ \frac{\partial {{t}^{2}}} \right]}_{Accel={}^{L}\!\!\diagup\!\!{}_\;}=\left( \frac{1} \right)\underbrace{\left( \frac{ML} \right)}_{F}=\frac{F}$$

$$\displaystyle \left[{\rm div}{\mathbf \sigma } \right]$$

$$\displaystyle div{\mathbf \sigma }=\frac{\partial {{\sigma }_{ij}}}{\partial {{x}_{i}}}{{{\mathbf e}}_{j}}=\sum\limits_{i}{\frac{\partial {{\sigma }_{ij}}}{\partial {{x}_{i}}}}{{{\mathbf e}}_{j}}$$

Where $$\displaystyle {{{\mathbf e}}_{j}}$$is the basis vector (see fig [[media: Fe1.s11.mtg4.djvu| p 4-2]]) and where $$\displaystyle \frac{\partial {{\sigma }_{ij}}}{\partial {{x}_{i}}}$$is the Einstein summation convention on repeated indices.

$$\displaystyle \left[ div{\mathbf \sigma } \right]=\left[ \frac{\partial {{\sigma }_{ij}}}{\partial {{x}_{i}}} \right]=\frac{\left[ \partial {{\sigma }_{ij}} \right]}{\left[ \partial {{x}_{i}} \right]}=\frac{\left[ \sigma \right]}{L}=\frac{{}^{F}\!\!\diagup\!\!{}_\;}{L}=\frac{F}$$

Again dimensions consistent (no “EA”)

[[media: Fe1.s11.mtg5.djvu| Page 5-4]]

[[media: Fe1.s11.mtg4.djvu| (3-6) p. 4-3]]: 3 scalar coupled PDEs, Compact or tensor form.

$$\displaystyle \underbrace{\frac{\partial {{\sigma }_{ij}}}{\partial {{x}_{i}}}}_{\begin{smallmatrix}

3Terms, \\

i=1,2,3

\end{smallmatrix}}+{{b}_{j}}=\rho \frac{\partial {{t}^{2}}},\underbrace{j=1,2,3}_{3eqs}$$

$$\displaystyle \underbrace{{{\sigma }_{11}},}_{\begin{smallmatrix}

Normal \\

Stress

\end{smallmatrix}}\underbrace{{{\sigma }_{21}},{{\sigma }_{31}}}_{=0}$$

(Assuming no shear in the above equation)

$$\displaystyle \frac{\partial {{\sigma }_{11}}}{\partial {{x}_{1}}}+{{b}_{1}}=\rho \frac{\partial {{t}^{2}}}$$

$$\displaystyle \displaystyle {{x}_{1}}=x,{{\sigma }_{11}}=\sigma ,{{u}_{1}}=u$$

[[media: Fe1.s11.mtg5.djvu| Page 5-5]]

$$\displaystyle \frac{\partial \sigma }{\partial x}+\underbrace{b}_{\begin{smallmatrix}

Body \\

force

\\

\left( {F}/\; \right)

\end{smallmatrix}}=\frac{{{\partial }^{2}}u}{\partial {{t}^{2}}}$$

Consider $$\displaystyle \int\limits_{A}{b(x,t)\underbrace{dA}_{dydz}}=f(x,t)$$

Assume $$\displaystyle \displaystyle b(x,t)$$ is dependent on $$\displaystyle \displaystyle x$$ only, not $$\displaystyle \displaystyle (x,y,z)$$

$$\displaystyle \int{\rho }\frac{{{\partial }^{2}}u}{\partial {{t}^{2}}}dA=\rho \frac{{{\partial }^{2}}u}{\partial {{t}^{2}}}\underbrace{\int_{A}^ – {dA}}_{A}$$

Assume $$\displaystyle \rho (x,t)\And u(x,t)$$

Consider

$$\displaystyle \int\limits_{A}{\underbrace{\frac{\partial \sigma }{\partial x}}_{\frac{\partial }{\partial x}\left( E(x)\frac{\partial u}{\partial x} \right)}}dA=\underbrace{\left( \int\limits_{A}{dA} \right)}_{A(x)}\left( \frac{\partial }{\partial x}\left( E(x)\frac{\partial u}{\partial x} \right) \right)$$

The above is WRONG. Miss term w/ A’ (change in area), $$\displaystyle \frac{dA}{dx}={A}'$$

Derivation

[[media: Fe1.s11.mtg5.djvu| Page 5-6]]

HW 1.1: Derive [[media: Fe1.s11.mtg2.djvu| (3) p. 2-3]]



$$\displaystyle N(x)=\sigma (x)n(x)A(x)=-\sigma (x)A(x)$$

$$\displaystyle N(x+dx)=\sigma (x+dx)n(x+dx)A(x+dx)= \underbrace{\color{red}{+}}_{NOTE} \sigma (x+dx)A(x+dx)$$

End HW 1.1

Why $$\displaystyle n(x)$$and $$\displaystyle n(x+dx)$$? Mimic 3-D Case