Question 11.CA.6: A body of mass m moving with a velocity v0 hits the end B of......

Making n = 2 in Eq. (1) from Concept Application 11.1, when rod BCD is subjected to a static load $P_m$, its strain energy is

$U_m=\frac{5 P_m^2 L}{16 A E}$            (1)

where A is the cross-sectional area of segment CD. Solving Eq. (1) for $P_m$, the static load that produces the same strain energy as the given impact load is

$P_m=\sqrt{\frac{16}{5} \frac{U_m A E}{L}}$

where $U_m$ is given by Eq. (11.33). The largest stress occurs in segment CD. Dividing $P_m$ by the area A of that portion,

$U_m=\frac{1}{2} m ν_0^2$             (11.33)

$\sigma_m=\frac{P_m}{A}=\sqrt{\frac{16}{5} \frac{U_m E}{A L}}$        (2)

or substituting for $U_m$ from Eq. (11.33) gives

$\sigma_m=\sqrt{\frac{8}{5} \frac{m ν_0^2 E}{A L}}=1.265 \sqrt{\frac{m ν_0^2 E}{A L}}$

Comparing this with the value obtained for $σ_m$ in the uniform rod of Fig. 11.21 and making V = AL in Eq. (11.35), note that the maximum stress in the rod of variable cross section is 26.5% larger than in the lighter uniform rod. Thus, as in our discussion of Concept Application 11.1, increasing the diameter of segment BC results in a decrease of the energy-absorbing capacity of the rod.

$\sigma_m=\sqrt{\frac{2 U_m E}{V}}=\sqrt{\frac{m ν_0^2 E}{V}}$          (11.35)