Chapter 1.2, Problem 6P

Two brass rods AB and BC, each of uniform diameter, will be brazed together at B to form a nonuniform rod of total length 100 m that will be suspended from a support at A as shown. Knowing that the density of brass is 8470 kg/m³, determine (a) the length of rod AB for which the maximum normal stress in ABC is minimum, (b) the corresponding value of the maximum normal stress.

Fig. P1.6

To determine

The length of brass rod AB such that the maximum normal stress in ABC is minimum.

Answer to Problem 6P

The length of brass rod AB such that the maximum normal stress in ABC is minimum as $\underset{\_}{35.7\text{mm}}$.

Explanation of Solution

Given information:

The total length of the rod is $100\text{m}$.

The length of rod AB is a.

The length of rod BC is $\left(100-a\right)$.

The density of the brass is $8,470\text{kg}/{\text{m}}^{\text{3}}$.

The diameter of the rod AB is $15\text{mm}$.

The diameter of the rod BC is $10\text{mm}$.

Calculation:

Calculate the area $\left(A_{AB}\right)$ of rod AB using the relation:

$A_{AB}=\frac{\pi }{4}{d}_{AB}^2$ (1)

Here, $d_{AB}$ is diameter of rod AB.

Substitute $15\text{mm}$ for AB in Equation (1).

$\begin{array}{c}{A}_{AB}=\frac{\pi }{4}\times {\left(15\right)}^2\\ =\frac{\pi }{4}\times 225\\ =176.715{\text{mm}}^{\text{2}}\times \left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^{\text{2}}}\right)\\ =176.715\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Calculate the area $\left(A_{BC}\right)$ of rod BC using the relation:

$A_{BC}=\frac{\pi }{4}{d}_{BC}^2$ (2)

Here, $d_{BC}$ is diameter of rod BC.

Substitute $10\text{mm}$ for BC in Equation (2).

$\begin{array}{c}{A}_{BC}=\frac{\pi }{4}\times {\left(10\right)}^2\\ =\frac{\pi }{4}\times 100\\ =78.54{\text{mm}}^{\text{2}}\times \left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^{\text{2}}}\right)\\ =78.54\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Calculate the weights $\left(W_{AB}\right)$ of rod AB using the formula:

$W_{AB}=\rho g{A}_{AB}{l}_{AB}$ (3)

Here, $\rho$ is density of brass rod, g is acceleration due to gravity, and $l_{AB}$ is length of rod AB.

Substitute $8,470\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81{\text{m}/\text{s}}^2$ for g, $176.715\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{AB}$, and a for $l_{AB}$ in Equation (3).

$\begin{array}{c}{W}_{AB}=8,470\times 9.81\times 176.715\times {10}^{-6}\times a\\ =14.683a\end{array}$

Calculate the weights $\left(W_{BC}\right)$ of rod BC using the formula:

$W_{BC}=\rho g{A}_{BC}{l}_{BC}$ (4)

Here, $l_{BC}$ is length of rod BC.

Substitute $8,470\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81{\text{m}/\text{s}}^2$ for g, $78.54\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{BC}$, and $\left(100-a\right)$ for $l_{BC}$ in Equation (4).

$\begin{array}{c}{W}_{BC}=8,470\times 9.81\times 78.54\times {10}^{-6}\times \left(100-a\right)\\ =6.5259\times \left(100-a\right)\\ =652.59-6.5259a\end{array}$

Find the force $\left(P_A\right)$ acting at point A using the relation:

$P_A=W_{AB}+W_{BC}$ (5)

Here, $W_{AB}$ is a weight of rod AB and $W_{BC}$ is a weight of rod BC.

Substitute $14.683a$ for $W_{AB}$ and $\left(652.59-6.5259a\right)$ for $W_{BC}$ in Equation (5).

$\begin{array}{c}{P}_A=14.683a+652.59-6.5259a\\ =8.157a+652.59\\ =652.59+8.157a\end{array}$

Find the normal stress acting at point A using the formula:

${\sigma }_A=\frac{P_A}{A_{AB}}$ (6)

Here, ${\sigma }_A$ is normal stress acting at point A and $P_A$ is the force acting at point A.

Substitute $\left(652.59+8.157a\right)$ for $P_A$ and $176.715\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{AB}$ in Equation (6).

$\begin{array}{c}{\sigma }_A=\frac{652.59+8.157a}{176.715\times {10}^{-6}{\text{m}}^{\text{2}}}\\ =\left(3.6930\times {10}^6\right)+\left(46.160\times {10}^{3}a\right)\end{array}$

Find the normal stress acting at point B using the formula:

${\sigma }_B=\frac{P_B}{A_{BC}}$ (7)

Here, ${\sigma }_B$ is normal stress acting at point B and $P_B$ is the force acting at point B.

Consider $P_B$ is equal to $W_{BC}$. So, the value of $P_B$ is $\left(652.59-6.5259a\right)$.

Substitute $\left(652.59-6.5259a\right)$ for $P_B$ and $78.54\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{BC}$ in Equation (7).

$\begin{array}{c}{\sigma }_B=\frac{652.59-6.5259a}{78.54\times {10}^{-6}{\text{m}}^{\text{2}}}\\ =8.3090\times {10}^6-83.090\times {10}^{3}a\end{array}$

Find the length of brass rod AB for the maximum stress in ABC as minimum:

${\sigma }_A={\sigma }_B$ (8)

Here, ${\sigma }_A$ is normal stress acting at point A and ${\sigma }_B$ is normal stress acting at point B.

Substitute $\left(8.3090\times {10}^6-83.090\times {10}^{3}a\right)$ for ${\sigma }_B$ and $\left[\left(3.6930\times {10}^6\right)+\left(46.160\times {10}^{3}a\right)\right]$ for ${\sigma }_A$ in Equation (8).

$\begin{array}{l}3.6930\times {10}^6+46.160\times {10}^{3}a=8.3090\times {10}^6-83.090\times {10}^{3}a\\ 129,250a=4,616,000\\ a=35.71\text{m}\end{array}$

Thus, the length of brass rod AB is $\underset{\_}{35.7\text{mm}}$.

To determine

The value of maximum normal stress of brass rods.

Answer to Problem 6P

The maximum normal stress of brass rods is $\underset{\_}{5.34\text{MPa}}$.

Explanation of Solution

Calculation:

Find the maximum normal stress of brass rod using the relation:

${\sigma }_A=3.6930\times {10}^6+46.160\times {10}^{3}a$

Substitute $35.71\text{m}$ for a.

$\begin{array}{c}{\sigma }_A=3.6930\times {10}^6+46.160\times {10}^3\left(35.71\right)\\ =3.6930\times {10}^6+1,648,373.6\\ =5.34\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =5.34\text{MPa}\end{array}$

${\sigma }_B=8.3090\times {10}^6-83.090\times {10}^{3}a$.

Substitute $35.71\text{m}$ for a in.

$\begin{array}{c}{\sigma }_B=8.3090\times {10}^6-83.090\times {10}^3\left(35.71\right)\\ =8.3090\times {10}^6-2,967,143.9\\ =5.34\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =5.34\text{MPa}\end{array}$

Thus, the maximum normal stress of brass rods is $\underset{\_}{5.34\text{MPa}}$.