Question 11.CA.6: A body of mass m moving with a velocity v0 hits the end B of......
A body of mass m moving with a velocity \text{v} _0 hits the end B of the nonuniform rod BCD (Fig. 11.22). Knowing that the diameter of segment BC is twice the diameter of portion CD, determine the maximum value \sigma_m of the stress in the rod.
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)
