User:Eml4500.f08.bottle.butler/HW 5

Principle of Virtual Work
We derive $$ k^{(e)}_{4X4}=T^{(e)^{T}}_{4X2}\hat{k}^{(e)}_{2X2}T^{(e)}_{2X4} $$ Recall the FD relationship with axial dof's $$q^{(e)}: \! $$ $$ \hat{k}^{(e)}_{2X4}q^{(e)}_{2X1}=p^{(e)}_{2X1} $$ $$ \Rightarrow equation (1) \hat{k}^{(e)}q^{(e)}-p^{(e)}= 0_{2X1}\! $$ Therefore: $$ \hat{W}_{2X1}\circ \underbrace{(\hat{k}^{(e)}q^{(e)}-p^{(e)})}_{2X1}=0_{1X1}$$ for all $$ \hat{W}_{2X1} \! $$

Recall that $$ q^{(e)}_{2X1}=T^{(e)}_{2X4}d^{(e)}_{4X1} $$ similarly, $$ \hat{w_{2X1}}=T^{(e)}_{2X4}w_{4X1}$$ where $$\hat{w}_{2X1}$$ is the virtual displacement corresponding to $$q^{(e)}_{2X1}$$ and $$w_{4X1} \!$$ is the virtual displacement in the global coordinate system corresponding to $$d^{(e)}_{4X1}$$

If we place equation $$ (3) \! $$ and $$ (4) \! $$ in equation $$ (2) \! $$ then we get $$ equation(5) = \! $$$$ (T^{(e)}w)\circ [\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}] =0 $$ for all $$ w_{4X1} \!$$

Recall: $$ equation (6) = \! $$$$ (AB)^{T}=B^{T}A^{T}\! $$  (see below for proof) and $$ equation (7) = \!$$ $$a\circ b=a^{T}b$$ Substituting equations $$ (6) \!$$ and $$ (7) \!$$ into $$ (5) \!$$ we get $$(T^{(e)}w)^{T}[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}]=0_{1X1} $$ for all $$ w_{4X1} \! $$ $$ \Rightarrow w^{T}T^{(e)}[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}]-0 $$ $$ \Rightarrow w\circ [\underbrace {(T^{(e)^{T}}\hat{k}^{(e)}T^{(e)})}_{k^{(e)}}d^{(e)}-\underbrace {(T^{(e)^{T}}P^{(e)})}_{f^{(e)}}]$$ for all $$ w_{4X1} \! $$  $$ \Rightarrow w\circ [k^{(e)}d^{(e)}-f^{(e)}]=0 $$ for all $$ w \! $$  therefore $$ k^{(e)}d^{(e)}=f^{(e)} \!$$

Continuous Case with PDE's
Motivational Model Problem: Elastic Bar with varying $$ A(x) \! $$ and $$ E(x) \! $$, subject to a varying axial load (distributed load), concentrated loading, and inertial forces (damping).

image $$ \sum{F_{x}}=0=-N(x,t)+N(x +dx, t)+f(x,t)dx-m(x)\ddot{u} \! $$ $$ equation (1) \! $$$$=\frac{\partial N}{\partial x}+h.o.t.+f(x,t)-m(x)\ddot{u}dx \! $$ neglecting the higher order terms, we recall the Taylor series expansion: $$f(x+dx)=f(x)+\frac{df(x)}{dx}+ \underbrace {\frac{1}{2}\frac{d^{2}f(x)}{dx^{2}}dx^{2}+\cdots}_{h.o.t.} $$

Equation $$ (1) \! $$ becomes $$ equation (2) = \! $$$$\frac{\partial N}{\partial x}+f=m\ddot{u}\! $$ the equation for the motion of the elastic bar $$ equation (3)= \! $$ $$ N(x,t)=A(x)\underbrace {\sigma (x)}_{E(x)\underbrace {\varepsilon (x,t)}_{\frac{\partial u (x,t)}{\partial x}}}$$ equation $$ (3) \! $$ into equation $$ (2) \! $$ yields: $$ \frac{\partial }{\partial x}[A(x)E(x)\frac{\partial u}{\partial x}]+f(x,t)=m(x)\underbrace {\ddot{u}}_{\frac{\partial ^{2}u}{\partial t^{2}}} $$ In order to solve this problem, we need 2 boundary conditions and 2 initial conditions.