Chapter 14, Problem 14.15P

Interpretation Introduction

(a)

Interpretation:

The maximum force applied on an aluminum rod needs to be determined.

Concept Introduction:

Expression for stress (S) is given as follows:

  $S=\frac{F}{A}$

Here, Force is F, the cross-sectional area is A.

The expression for Young's Modulus (E) is as follows:

  $E=\frac{S}{e}$

Here, e is the strain.

Length of the rod (L) is expressed as follows:

  $L=L+{\delta }_{\max }$

Strain in the rod is expressed as (e):

  $e=\frac{L-l}{l}$.

Answer to Problem 14.15P

The maximum force applied is 271 N.

Explanation of Solution

  $L=L+{\delta }_{\max }$

Here,

  ${\delta }_{\max }=\text{2 mm}$

l =10 mm

L= 10 + 0.002

L= 10.002 m

Strain,

  $e=\frac{10.002-10}{10}$

e = 0.0002 m

To calculate the area of the rod,

  $\begin{array}{l}A=\frac{\pi }{4}{d}^2\\ $ A=\frac{\pi }{4}{(0.5)}^2$A=0.1963\times {10}^{-4}{m}^2\end{array}$

Selecting the value of Young's modulus as 69×10⁹ N/m² from the table of physical properties of common alloys,

Substitute 69×10⁹ N/m²for E, 0.1963 ×10-4 m² for A and 0.0002 for e in below equations,

  $\begin{array}{l}F=EAe\\ F=(69\times {10}^9)(0.1963\times {10}^{-4})(0.0002)\end{array}$

  $F=271$ N

The maximum force applied is 271 N.

Interpretation Introduction

(b)

Interpretation:

The maximum force applications on magnesium rod needs to be determined.

Concept Introduction:

Expression for stress (S) is given as follows:

  $S=\frac{F}{A}$

Here, Force is F, the cross-sectional area is A.

The expression for Young's Modulus (E) is as follows:

  $E=\frac{S}{e}$

Here, e is the strain.

Length of the rod (L) is expressed as follows:

  $L=L+{\delta }_{\max }$

Strain in the rod is expressed as (e):

  $e=\frac{L-l}{l}$

Answer to Problem 14.15P

The maximum force applied is 176 N.

Explanation of Solution

  $L=L+{\delta }_{\max }$

Here,

  ${\delta }_{\max }=2mm$

l =10 mm

L= 10 + 0.002

L= 10.002 m

Strain,

  $e=\frac{10.002-10}{10}$

e = 0.0002 m

To calculate the area of the rod,

  $\begin{array}{l}A=\frac{\pi }{4}{d}^2\\ $ A=\frac{\pi }{4}{(0.5)}^2$A=0.1963\times {10}^{-4}{m}^2\end{array}$

Selecting the value of Young's modulus as 45×10⁶N/m 2.

Substituting the required values in the below equation,

  $\begin{array}{l}F=EAe\\ F=(45\times {10}^9)(0.1963\times {10}^{-4})(0.0002)\end{array}$

  $F=176$ N

The maximum force applied in the bar is 176 N.

Interpretation Introduction

(c)

Interpretation:

The maximum force applied on beryllium rod needs to be determined.

Concept Introduction:

Expression for stress (S) is given as follows:

  $S=\frac{F}{A}$

Here Force is F, the cross-sectional area is A.

The expression for Young's Modulus (E) is

  $E=\frac{S}{e}$

Where e is the Strain.

Length of the rod (L) is expressed as:

  $L=L+{\delta }_{\max }$

Strain in the rod is expressed as (e):

  $e=\frac{L-l}{l}$

Answer to Problem 14.15P

The maximum force applied on beryllium is 1138 N.

Explanation of Solution

  $L=L+{\delta }_{\max }$

Here,

  ${\delta }_{\max }=2mm$

l =10 mm

L= 10 + 0.002

L= 10.002 m

Strain,

  $e=\frac{10.002-10}{10}$

e = 0.0002 m

To calculate the area of the rod:

  $\begin{array}{l}A=\frac{\pi }{4}{d}^2\\ $ A=\frac{\pi }{4}{(0.5)}^2$A=0.1963\times {10}^{-4}{m}^2\end{array}$

From the table of common metals, select the Young's modulus of beryllium as 287×10⁹ N/m²

Substitute the required values in the below equation,

  $\begin{array}{l}F=EAe\\ F=(287\times {10}^9)(0.1963\times {10}^{-4})(0.0002)\end{array}$

  $F=1138$ N