Question 10.4: A prismatic linearly elastic rod of cross-sectional area A a...

Let us assume that the rod deflects by an amount δ. Considering conservation of energy, we can say;

Initial potential energy of the weight = Strain energy stored within the elastic rod If datum is assumed to be at the final elongated position of the rod, then the above equation is written as:

W(h +δ ) =U               (1)

where for linearly elastic rod, we have

U=\underset{\sout{V}}{\iiint} \tilde{u} d \sout{V} \quad \text { and } \quad \tilde{u}=\frac{\sigma^2}{2 E}=\frac{E ∈^2}{2}

Obviously, ∈= δ /L, so

U=\underset{\sout{V}}{\iiint}\left\lgroup \frac{E \delta^2}{2 L^2} \right\rgroup d \sout{V}=\frac{E \delta^2}{2 L^2} \times A L=\frac{A E \delta^2}{2 L}

From Eq. (1), we get

\left\lgroup \frac{A E}{2 L} \right\rgroup \delta^2=W(h+\delta)

or            \left\lgroup \frac{A E}{2 L} \right\rgroup \delta^2-W \delta-W h=0

Therefore,              \delta=\frac{W \pm \sqrt{W^2+(2 A E W h / L)}}{(A E / L)}

Neglecting physically impossible negative root, we obtain the deflection of the rod as

\delta=\left\lgroup \frac{W L}{A E} \right\rgroup\left[1+\sqrt{1+\frac{2 A E h}{W L}}\right]