Mechanics of Materials 3rd Edition 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.

Mechanics of Materials 3rd Edition by Timothy Philpot Figure P1.3/4


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,

\displaystyle \sum F_y=0

\displaystyle -F_1-15=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_2-30-30-15=0

\displaystyle F_2=-75\:kips

\displaystyle F_2=75\:kips\:\left(C\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

\displaystyle A_{1,\:min}\ge \frac{F_1}{\sigma }

\displaystyle A_{1,\:min}\ge \frac{15\:kips}{40\:ksi}

\displaystyle A_{1,\:min}\ge 0.375\:in.^2

The minimum rod diameter is therefore

\displaystyle A_{1,\:min}=\frac{\pi }{4}\left(d_1\right)^2

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

\displaystyle \therefore d_1=0.691\:in

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

\displaystyle A_{2,\:min}=\frac{F_2}{\sigma }

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

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

The minimum diameter for rod (2) is therefore

\displaystyle A_{2,\:min}=\frac{\pi }{4}\left(d_2\right)^2

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

\displaystyle d_2\ge 1.545\:in.


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