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Let us consider a rod having a young's modulus $Y$. Let it be of length $l$, and suppose it is suspended from a point P. Let us pull the rod with a force $F$ at a point Q which is at a distance $2/3l$ from the point P. Next let us apply the same force at the other end point of the rod. Is the length of the rod in both the cases same?

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It is not clear over what length you apply $F$ the second time. Same stress, cross-section and stiffness will however always lead to the same strain, so in the frist case the rod will be $l+\frac{2}{3}l\varepsilon$, in the second (whole length loaded??) $l+l\varepsilon$, where strain $\varepsilon=\frac{F/A}{Y}$, where $A$ is cross-section. –  eudoxos Sep 27 '11 at 15:41
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2 Answers

The base length for the purpose of computing the strain is the length over which the tension/compression acts.

So in the first case the strain is $\Delta l/(2/3 L) = 3 \Delta l/ 2 L$.

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Hello Dmckee can you provide a reason for your answer? –  Primeczar Aug 27 '11 at 6:51
    
Because that is the definition of strain. Because the whole point of the measure is understanding how materials deform under loads you don't care about the unloaded parts of the rod. –  dmckee Aug 27 '11 at 6:58
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Deflection is $$\delta = \frac{F\,L}{Y\,A}$$ where $F$ is force, $L$ is the loaded length, $Y$ is modulus of elasticity and $A$ is the cross sectional area.

So the deflection $\delta$ is proportional to the loaded length. In your case a) that is $L=\frac{2}{3}l$, but in case b) I cannot figure out from the wording where points P and Q are. Figure out what $L$ is since the rest is the same, you will have your answer.

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