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

#### Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Figure P1.3/4. The diameter of rod (1) is 1.75 in. and the diameter of rod (2) is 2.50 in. Determine the normal stresses in rods (1) and (2).

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,

$\sum F_y=0$

$-F_1-15\:kips=0$

$F_1=-15\:kips$

$F_1=15\:kips\:\left(Compression\right)$

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):

$\sum F_y=0$

$-F_2-30\:kips-30\:kips-15\:kips=0$

$F_2=-75\:kips$

$F_2=75\:kips\:\left(Compression\right)$

From the given diameter of rod (1), the cross-sectional area of rod (1) is

$A_1=\frac{\pi }{4}\left(1.75\:in\right)^2$

$A_1=2.4053\:in^2$

and thus, the normal stress in rod (1) is

$\sigma _1=\frac{F_1}{A_1}$

$\sigma _1=\frac{-15\:kips}{2.4053\:in^2}$

$\sigma _1=-6.23627\:ksi$

$\sigma _1=6.23627\:ksi\:\left(Compression\right)$

From the given diameter of rod (2), the cross-sectional area of rod (2) is

$A_2=\frac{\pi }{4}\left(2.50\:in\right)^2$

$A_2=4.9087\:in^2$

Accordingly, the normal stress in rod (2) is

$\sigma _2=\frac{F_2}{A_2}$

$\sigma _2=\frac{-75\:kips}{4.9087\:in^2}$

$\sigma _2=-15.2790\:ksi$

$\sigma _2=15.2790\:ksi\:\left(Compression\right)$