User:Eml4500.f08.gravy.jad/Notes6

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Lecture 29

HW6

Continuing Case: PVW of Dynamics of Elastic Bars

PDE: $$\ \frac{d}{dx} \left \lbrack (EA) \frac{du}{dx} \right \rbrack + f = m \ddot{u} \,$$ Equation(1)

$$\ \Rightarrow \mathbf{m} \mathbf{\ddot{d}} + \mathbf{k} \mathbf{d} = \mathbf{F} \,$$ Equation(2)

Derive (2) from (1): $$\ \int_{x=0}^{x=L} W(x) \left \lbrace \frac{d}{dx} \left \lbrack EA \frac{du}{dx} \right \rbrack + f - m \ddot{u} \right \rbrace \, dx = 0 \,$$ for all possible W(x). Where w(x) is the weighting function.

(1) $$\ \Rightarrow \,$$(3) trivial (3) $$\ \Rightarrow \,$$(1) not trivial (3) rewritten as: $$\ \int W(x)g(x)dx =0 \,$$for all W(x) Since (3) holds for all W(x), select W(x)=g(x), then (3) becomes $$\ \int g^{2}(x)dx = 0 \,$$

Integration by Parts: r(x), s(x)

$$\ (rs)' = r's + rs' \,$$ where $$\ r'=frac{dr}{dx} \,$$ and $$\ s'=frac{ds}{dx} \,$$

$$\ \int (rs)' = rs - \int rs' \,$$ $$\ \Rightarrow \int r's = rs - \int rs' \,$$

Recall cont. PVW (equation 3)

1st term: $$\ r(x)=(E,A) \frac{du}{dx}, s(x)=w(x) \,$$

By integrating by parts: $$\ \int_{x=0}^{x=L} \underbrace{w(x)}_{s} \frac{d}{dx} \underbrace{\left \lbrack (EA) \frac{du}{dx} \right \rbrack}_{r} dx = \left \lbrack W(EA) \frac{du}{dx} \right \rbrack |_{x=0}^{x=L} \,$$

$$\ = W(L) \left \lbrack (EA)(L) \right \rbrack \frac{du}{dx}(L,t) - W(0) \left \lbrack (EA)(O) \right \rbrack \frac{du}{dx} (0,t) \,$$

Now consider model prob:



At x=0, select W(x) such that W(0)=0 (ie. kinetically dismiss-able)

Motivation: Discrete PVW applied to Eq $$\ \mathbf{W}_{6x1} \cdot \underbrace{ \mathbf{k}_{6x1} \mathbf{d}_{2x1}-\mathbf{F}_{6x1} }_{6x1} = 0_{1x1} \,$$

$$\ F^{T} = \left \lfloor F_{1} F_{2} F_{3} F_{4} F_{5} F_{6} \right \rfloor \,$$

Where $$\ F_{1}, F_{2}, F_{5}, F_{6} \,$$ are unknown reactions. Since $$\ \mathbf{W} \,$$ can be selected arbitraily, select $$\ \mathbf{W} \,$$ such that $$\ W_{1}=W_{2}=W_{5}=W_{6}=0 \,$$, so to eliminate equations involving unknown reactions, thus eliminating 1,2,5,6; $$\ \mathbf{\overline{k}} \mathbf{\overline{d}}=\mathbf{\overline{F}} \,$$

Please note: $$\ \overline{\mathbf{W}} \cdot (\mathbf{\overline{kd}} - \mathbf{\overline{F}})=0 \,$$ for all $$\ \mathbf{W} \,$$

This concludes the motivation.

Unknown reaction term $$\ N(0,t) = \left \lbrack (EA)(0) \right \rbrack \frac{du}{dx} (0,t) \,$$

Continuing PVW: $$\ W(L)F(L) - \int_{0}^{L} \frac{dW}{dx}(EA) \frac{du}{dx} dx + \int_{0}^{L} W(x) \left \lbrack f-m \ddot{u} \right \rbrack dx=0 \,$$ for all W(x) such that W(0)=0.

$$\ \Rightarrow \int_{0}^{L} W(m \ddot{u}) dx + \int_{0}^{L} \frac{dW}{dx} (EA) \frac{du}{dx} dx = W(L)F(t) + \int_{0}^{L} Wf dx \,$$ for all W(x) such that W(0)=0.

Constant Setting (PVW)

Inertia : $$\ \int_{0}^{L} Wm \ddot{u} dx = \alpha \,$$

Stiffness : $$\ \int_{0}^{L} \frac{dW}{dx} (EA) \frac{du}{dx} dx = \beta \,$$

Applied force : $$\ W(L)F(t) + \int_{0}^{L} Wf dx = \gamma \,$$

For all W(x) such that: $$\ \underbrace{W(0)=0}_{b.c.} \,$$

Discrete setting (PVW)

Inertia : $$\ \overline{\mathbf{W}} \cdot ( \overline{\mathbf{md}}) (1) \,$$

Stiffness : $$\ \overline{\mathbf{W}} \cdot ( \overline{\mathbf{kd}}) (2) \,$$

Applied force : $$\ \overline{\mathbf{W}} \cdot \overline{\mathbf{F}} (3) \,$$

For all W(x) such that: b.c.'s eliminated

Stiffness term:



Assume displacement u(x) for $$\ x_{i} \le x \le x_{i+1} \,$$

Motivation for linear interpolation of u(x)



Deformed shape is a straight line. ie. there was an implicit assumption of linear interpolation of displacement between 2 nodes

consider the case where there are only axial-displacement (ie. zero transverse displacement)

Question: Express V(x) in terms of $$\ d_{i} = u(x_{i}) \,$$ and $$\ d_{i+1} = u_(x_{i+1}) \,$$ as a linear function in x (ie. linear interpolation

$$\ u(x) = N_{i}(x) d_{i} + N_{i+1}(x)d_{i+1} \,$$ $$\ where N_{i}(x) and N_{i+1}(x) are linear functions in x \,$$

insert HW

$$\ N_{i}(x)= \frac{x-x_{i}}{x){1+1}-x_{i}} \,$$