User:Eml5526.s11.team5.srv/hw3

Problem
 Show that the equivalent stiffness of a  spring aligned in the x direction for the bar of thickness t with a centered square hole shown in the figure is :  

 where E is the Young's Modulus and t is the width of the bar  



Solution
To find the equivalent stiffness of the bar, we have to subdivide the bar into 4 parts and each part behaves like an individual spring with stiffness K1, K2, K3 and K4 respectively. It can be shown from the figure below.

The Equivalent spring - mass system can be shown from the figure below.



From the Hooke's Law,


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$$\sigma = E \varepsilon$$ $$
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 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.5.2)
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$$\frac{F}{A} = E \frac{\delta }{L}$$ $$
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 *  $$ \displaystyle (Eq. 3.5.3 )
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$$\frac{F}{\delta} = E \frac{A}{L}$$ $$ From the definition of Stiffness(K),
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 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.5.4 )
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$$K = \frac{F}{\delta}$$ $$
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 *  $$ \displaystyle\!
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Therefore $$ \displaystyle (Eq. 3.5.4 ) $$ becomes


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$$K = E \frac{A}{L}$$ $$
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 *  $$ \displaystyle (Eq. 3.5.5   )
 * }

Hence the stiffness of the springs K1, K2, K3 and K4 can be determined as,


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$$K_{1} = \frac{EA_{1}}{L_{1}} = \frac{Eat}{\frac{l}{10}}$$                   (From figure 2) $$
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 *  $$ \displaystyle\!
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$$ K_{1}= \frac{10Eat}{l}$$ $$
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 *  $$ \displaystyle (Eq. 3.5.6  )
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Similarly


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$$K_{4}= \frac{10Eat}{l}$$ $$
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 *  $$ \displaystyle (Eq.3.5.7   )
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$$K_{2} = \frac{EA_{2}}{L_{2}} = \frac{E. 2b. t}{\frac{4l}{5}}$$                  (From figure 2) $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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$$K_{2}= \frac{5Ebt}{2l}$$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.8  )
 * }

Similarly

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$$K_{3}= \frac{5Ebt}{2l}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.9  )
 * }

From Figure 2 and 3, The total stiffness of the bar (K) can be derived as follows,

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$$\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{(K_{2} + K_{3})} + \frac{1}{K_{4}}$$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.10)
 * }

Due to symmetry,$$K_{1} = K_{4}\quad$$ and $$K_{2} = K_{3},\quad$$ therefore,

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$$\frac{1}{K} = \frac{2}{K_{1}} + \frac{1}{2K_{2}}$$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.11)
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$$K = \frac{2K_{1}K_{2}}{K_{1}+4K_{2}} = \frac{2(\frac{10Eat}{l})(\frac{5Ebt}{2l})}{(\frac{10Eat}{l} + \frac{20Ebt}{2l})}\quad$$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
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$$K = \frac{(\frac{10Eat}{l})(\frac{5Ebt}{l})}{\frac{10Eat}{l}(a+b)}$$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

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$$K = \frac{5Etab}{(a+b)l}$$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 3.5.12)
 * }