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Chapter 10

Q. 10.4

A prismatic linearly elastic rod of cross-sectional area A and length L of negligible mass is hanging freely when a rigid weight W is allowed to fall freely from a height on to the rod as shown in Figure 10.20. The end of the rod contains flange (whose weight is neglected). Assuming no energy loss, calculate the longitudinal extension of the rod due to the impact.

10.20

Step-by-Step

Verified Solution

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]