# Mechanics of Materials 3rd Edition by Timothy A. Philpot, P1.4

## Solution:

Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, $\displaystyle \sum F_y=0$ $\displaystyle -F_1-15\:kips=0$ $\displaystyle F_1=-15\:kips$ $\displaystyle F_1=15\:kips\:\left(C\right)$

Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): $\displaystyle \sum F_y=0$ $\displaystyle -F_2-30\:kips-30\:kips-15\:kips=0$ $\displaystyle F_2=-75\:kips$ $\displaystyle F_2=75\:kips\:\left(Compression\right)$

From the given diameter of rod (1), the cross-sectional area of rod (1) is $\displaystyle A_1=\frac{\pi }{4}\left(1.75\:in.\right)^2$ $\displaystyle A_1=2.4053\:in^2$

and thus, the normal stress in rod (1) is $\displaystyle \sigma _1=\frac{F_1}{A_1}$ $\displaystyle \sigma _1=\frac{-15\:kips}{2.4053\:in^2}$ $\displaystyle \sigma _1=-6.23627\:ksi$ $\displaystyle \sigma _1=6.24\:ksi\:\left(Compression\right)$

From the given diameter of rod (2), the cross-sectional area of rod (2) is $\displaystyle A_2=\frac{\pi }{4}\left(2.50\:in.\right)^2$ $\displaystyle A_2=4.9087\:in^2$

Accordingly, the normal stress in rod (2) is $\displaystyle \sigma _2=\frac{F_2}{A_2}$ $\displaystyle \sigma _2=\frac{-75\:kips}{2.4053\:in^2}$ $\displaystyle \sigma _2=-15.2789\:ksi$ $\displaystyle \sigma _2=15.28\:ksi\:\left(Compression\right)$