Question 11.CA.1: A rod consists of two portions BC and CD of the same materia......

Use Eq. (11.14) for the strain energy of each of the two portions, and add the expressions obtained:

U=\frac{P^2 L}{2 A E}            (11.14)

U_n=\frac{P^2\left(\frac{1}{2} L\right)}{2 A E}+\frac{P^2\left(\frac{1}{2} L\right)}{2\left(n^2 A\right) E}=\frac{P^2 L}{4 A E}\left(1+\frac{1}{n^2}\right)

or

U_n=\frac{1+n^2}{2 n^2} \frac{P^2 L}{2 A E}               (1)

Check that, for n = 1,

U_1=\frac{P^2 L}{2 A E}

which is the same as Eq. (11.14) for a rod of length L and uniform cross section of area A. Also note that for n>1, U_n<U_1 . As an example, when n=2, U_2=\left(\frac{5}{8}\right) U_1 . Since the maximum stress occurs in portion CD of the rod and is equal to \sigma_{\max }=P / A , then for a given allowable stress, increasing the diameter of portion BC of the rod results in a decrease of the overall energy-absorbing capacity. Unnecessary changes in cross-sectional area should be avoided in the design of members subjected to loads (such as impact loadings) where the energy-absorbing capacity of the member is critical.