## 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

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

$\displaystyle \sigma _2=\frac{-75\:kips}{2.4053\:in^2}$

$\displaystyle \sigma _2=-15.2789\:ksi$

$\displaystyle \sigma _2=15.28\:ksi\:\left(Compression\right)$

## 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

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

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

$\displaystyle d_2\ge 1.545\:in.$

## Solution:

From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi

$\displaystyle \sigma =\frac{P}{A}$

$\displaystyle A_{min}=\frac{P}{\sigma }$

$\displaystyle A_{min}=\frac{27\:kips}{18\:ksi}$

$\displaystyle A_{min}=1.500\:in.^2$

The cross-sectional area of the aluminum tube is given by

$\displaystyle A=\frac{\pi }{4}\left(D^2-d^2\right)$

Set this expression equal to the minimum area and solve for the maximum inside diameter

$\displaystyle \frac{\pi }{4}\left[\left(2.50\:in\right)^2-d^2\right]=1.500\:in^2$

$\displaystyle \left(2.50\:in\right)^2-d^2=\frac{4}{\pi }\left(1.500\:in^2\right)$

$\displaystyle \left(2.50\:in\right)^2-\frac{4}{\pi }\left(1.500\:in^2\right)=d^2$

$\displaystyle d_{max}=2.08330\:in$

The outside diameter D, the inside diameter d, and the wall thickness t are related by

$D=d+2t$

Therefore, the minimum wall thickness required for the aluminum tube is

$\displaystyle t_{min}=\frac{D-d}{2}$

$\displaystyle t_{min}=\frac{2.50\:in-2.08330\:in}{2}$

$\displaystyle t_{min}=0.20835\:in$

$\displaystyle t_{min}=0.208\:in$

## Maximum Load of a Member | Stress | Mechanics of Materials | 3rd Edition | Timothy Philpot | Problem P1.1

### SOLUTION:

The cross-sectional area of the stainless steel tube is

$\displaystyle A=\frac{\pi }{4}\left(D^2-d^2\right)$

$\displaystyle A=\frac{\pi }{4}\left[\left(60\:mm\right)^2-\left(50\:mm\right)^2\right]$

$\displaystyle A=863.938\:mm^2$

The normal stress in the tube can be expressed as

$\displaystyle \sigma =\frac{P}{A}$

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.

$\displaystyle P_{max}=\sigma _{max}A$

$\displaystyle P_{max}=\left(200\:MPa\right)\left(863.938\:mm^2\right)$

$\displaystyle P_{max}=\left(200\:\frac{N}{mm^2}\right)\left(863.938\:mm^2\right)$

$\displaystyle P_{max}=172\:788\:N$

$\displaystyle P_{max}=172.8\:kN$