Introduction to Elasticity/Rotating rectangular beam

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[edit] Example : Rotating Rectangular Beam

A rotating rectangular beam

The body force potential is given by

\text{(52)}\qquad V = -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right)

Hence,

\text{(53)}\qquad \nabla^2{V} = V_{,11} + V_{,22} = -\cfrac{\rho\dot{\theta}^2}{2} \left(2 + 2\right)

or,

\text{(54)}\qquad \nabla^2{V} = -2\rho\dot{\theta}^2

The compatibility condition (in terms of stress) is

\text{(55)} \qquad \nabla^4{\varphi} + \left(2 - \cfrac{1}{\alpha}\right) \nabla^2{V} = 0

Plug V in to get

\text{(56)} \qquad \nabla^4{\varphi} - \left(2-\cfrac{1}{\alpha}\right) 2\rho\dot{\theta}^2 = 0

Since V is even in x1 and x2 and BCs are homogeneous, assume

\text{(57)} \qquad \varphi = A x_1^4 + B x_1^2 x_2^2 + C x_2^4 + D x_1^2 + E x_2^2

Hence,

\begin{align} \text{(58)}\qquad \sigma_{11} & = \varphi_{,22} + V = 2 B x_1^2 + 12 C x_2^2 + 2 E -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right)\\ \text{(59)}\qquad\sigma_{22} & = \varphi_{,11} + V = 12 A x_1^2 + 2 B x_2^2 + 2 D -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right)\\ \text{(60)}\qquad\sigma_{12} & = -\varphi_{,12} = - 4 B x_1 x_2 \end{align}

The traction BCs are

\begin{align} \text{(61)} \qquad \text{at}~ x_1 = \pm a & & t_1 = t_2 = 0 \Rightarrow \sigma_{11} = \sigma_{12} = 0 \\ \text{(62)} \qquad \text{at}~ x_2 = \pm b & & t_1 = t_2 = 0 \Rightarrow \sigma_{12} = \sigma_{22} = 0 \end{align}

Apply BCs at x_2 = \pm b.

\begin{align} \text{(63}) \qquad \sigma_{22} = 0 & = 12 A x_1^2 + 2 B b^2 + 2 D -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + b^2\right)\\ \text{(64}) \qquad \sigma_{12} = 0 & = - 4 B b x_1 \end{align}

Therefore,

\begin{align} \text{(65)} \qquad B & = 0 \\ \text{(66)} \qquad A & = \cfrac{\rho\dot{\theta}^2}{24} \\ \text{(67)} \qquad D & = \cfrac{\rho\dot{\theta}^2b^2}{4} \end{align}

We then have,

\text{(68)} \qquad \varphi = \cfrac{\rho\dot{\theta}^2}{24} x_1^4 + C x_2^4 + \cfrac{\rho\dot{\theta}^2b^2}{4} x_1^2 + E x_2^2

Plug into compatibility equation

\text{(69)} \qquad \varphi_{,1111} + 2\varphi_{,1122} + \varphi_{,2222} - \left(2-\cfrac{1}{\alpha}\right) 2\rho\dot{\theta}^2 = 0

to get

\text{(70)} \qquad 24 \left(\cfrac{\rho\dot{\theta}^2}{24} + C\right) - \left(2-\cfrac{1}{\alpha}\right) 2\rho\dot{\theta}^2 = 0

or,

\begin{align} C & = \left(2-\cfrac{1}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{12} - \cfrac{\rho\dot{\theta}^2}{24} \\ & = \left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{24} \qquad \text{(71)} \end{align}

Apply BCs at x_1 = \pm a.

\begin{align} \text{(72)} \qquad \sigma_{11} = 0 & = 2 B a^2 + 12 C x_2^2 + 2 E -\cfrac{\rho\dot{\theta}^2}{2} \left(a^2 + x_2^2\right)\\ & = \left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{2} x_2^2 + 2 E -\cfrac{\rho\dot{\theta}^2}{2} \left(a^2 + x_2^2\right) \qquad \text{(73)} \end{align}

Strong BCs imply that

\text{(74)} \qquad \left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{2} -\cfrac{\rho\dot{\theta}^2}{2} = 0

which cannot be true. So weak BCs on σ11 need to be applied at x_1 = \pm a.

\begin{align} \text{(75)} \qquad \text{at}~ x_1 = \pm a & & \int_{-b}^{b} \sigma_{11} dx_2 = 0 \end{align}

Hence,

\text{(76)} \qquad \left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{2}\cfrac{2b^3}{3} + 4 Eb -\cfrac{\rho\dot{\theta}^2}{2} \left(2a^2b + \cfrac{2b^3}{3}\right)=0

or,

\text{(77)} \qquad \left(2-\cfrac{2}{\alpha}\right)\cfrac{\rho\dot{\theta}^2b^2}{3} + 4 E -\rho\dot{\theta}^2a^2 = 0

Hence,

\text{(78)} \qquad E = \cfrac{\rho\dot{\theta}^2}{4}\left[a^2- \left(2-\cfrac{2}{\alpha}\right)\cfrac{b^2}{3} \right]

The stress field is, therefore,

\begin{align} \text{(79}) \qquad \sigma_{11} & = \left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{2} x_2^2 + \cfrac{\rho\dot{\theta}^2}{2}\left[a^2- \left(2-\cfrac{2}{\alpha}\right)\cfrac{b^2}{3} \right] -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right)\\ \text{(80}) \qquad \sigma_{22} & = \cfrac{\rho\dot{\theta}^2}{2} x_1^2 + \cfrac{\rho\dot{\theta}^2b^2}{2} -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right) \\ \qquad \sigma_{12} & = 0 \end{align}

or,

\begin{align} \text{(81)} \qquad \sigma_{11} & = \cfrac{\rho\dot{\theta}^2}{2}\left[\left(a^2 - x_1^2\right)+ 2\left(1-\cfrac{1}{\alpha}\right)\left(x_2^2-\cfrac{b^2}{3}\right)\right] \\ \text{(82)} \qquad \sigma_{22} & = \cfrac{\rho\dot{\theta}^2}{2}\left(b^2 - x_2^2\right) \\ \sigma_{12} & = 0 \end{align}

The displacements can be found in the standard manner.

Stresses (σ11) in a rotating rectangular beam
Stresses (σ22) in a rotating rectangular beam
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