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

#### Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Figure P1.3/4. If the normal stress in each rod must be limited to 40 ksi, determine the minimum diameter required for each rod.

SOLUTION:

Cut a FBD through rod (1). The FBD should include the free end of the rod at A.  As a matter of course, 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):

$\sum F_y=0$

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

$F_2=-75\:kips$

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

Notice that rods (1) and (2) are in compression. In this situation, we are concerned only with the stress magnitude; therefore, we will use the force magnitudes to determine the minimum required cross-sectional areas. If the normal stress in rod (1) must be limited to 40 ksi, then the minimum cross-sectional area that can be used for rod (1) is

$A_{1,min}=\frac{F_1}{\sigma }$

$A_{1,min}=\frac{15\:kips}{40\:ksi}$

$A_{1,min}=0.375\:in^2$

The minimum rod diameter is therefore

$A_{1,min}=\frac{\pi }{4}\cdot d_1^2$

$\frac{\pi }{4}d_1^2=0.375\:in^2$

$d_1=\sqrt{\frac{4\cdot \left(0.375\:in^2\right)}{\pi }}$

$d_1=0.691\:in$

Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of

$A_{2,min}=\frac{F_2}{\sigma }$

$A_{2,min}=\frac{75\:kips}{40\:ksi}$

$A_{2,min}=1.875\:in^2$

The minimum diameter for rod (2) is therefore

$A_{2,min}=\frac{\pi }{4}\cdot d_2^2$

$\frac{\pi }{4}d_{_2^2}=1.875\:in^2$

$d_2=\sqrt{\frac{4\cdot \left(1.875\:in^2\right)}{\pi }}$

$d_2=1.545\:in$

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