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.

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

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,  Image and video hosting by TinyPic

\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):
Image and video hosting by TinyPic

\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

 

 

You can now buy the complete solution manual for Mechanics of Materials: An Integrated Learning System 3rd Edition by Timothy A. Philpot for a very cheap price.

Buy Now button for Mechanics of Materials: An Integrated Learning System 3rd Edition

Advertisements