Chapter 9, Problem 17P

To determine

How much does a point which is exerted by a load of force $8500\text{N}$ move down due to the application of the force.

Answer to Problem 17P

Solution: The point which is exerted by a load of force $8500\text{N}$ move down by $\text{0}\text{.86}\text{mm}$.

Explanation of Solution

Given info: The diameter of the steel cable $1.27\text{cm}$ and it is $5.75\text{m}$ long. The aluminium cylinder has an inner inside diameter of $16.14\text{cm}$. The outer diameter is $16.24\text{cm}$. The unloaded length of the column is $3.25\text{m}$.

The upward forces which support the load of $8500\text{N}$ is the result of the tension force of the cable and the compression force of the column. Hence the sum of the tension force and the compression force of cable and column respectively will be equal to the load.

$F_L=F_T+F_C$ (I)

• $F_L$ is the load
• $F_T$ is the tension force
• $F_C$ is the compression force

Since the cable and the column are supporting the walkway from up and down, the change in length of cable and column is same.

Formula to relate tensile stress and tensile strain is given by,

$\frac{F}{A}=Y\frac{\Delta L}{L_0}$

• $F$ is the force
• $A$ is cross-sectional area
• $Y$ is young’s modulus
• $\Delta L$ is change in length
• $L_0$ is the initial length

Re-writing the above equation to find an expression for force,

$F=AY\frac{\Delta L}{L_0}$

The tension force of the steel cable will be,

$F_T=A_{cab}{Y}_s\frac{\Delta L}{L_{o,cab}}$ (II)

• $A_{cab}$ is cross-sectional area of the cable
• $Y_s$ is young’s modulus of steel
• $\Delta L$ is change in length
• $L_{0,cab}$ is the initial length of the cable

The compression force of the aluminium column will be,

$F_C=A_{col}{Y}_a\frac{\Delta L}{L_{o,col}}$ (III)

• $A_{col}$ is cross-sectional area of the column
• $Y_a$ is young’s modulus of aluminium
• $\Delta L$ is change in length
• $L_{0,col}$ is the initial length of the column

Substitute equation (II) and (III) in (I),

$F_L=A_{cab}{Y}_s\frac{\Delta L}{L_{o,cab}}+A_{col}{Y}_a\frac{\Delta L}{L_{o,col}}$

Re-write the above equation to get an expression for the change in length,

$\Delta L=\frac{F_L}{\frac{Y_s}{L_{o,cab}}{A}_{cab}+\frac{Y_a}{L_{o,col}}{A}_{col}}$ (IV)

The area of the column in contact will be,

$A_{col}=\frac{\pi d_{out}^2-\pi d_{in}^2}{4}$ (V)

The area of cable in contact will be,

$A_{cab}=\frac{\pi d_{cab}^2}{4}$ (VI)

Substitute (V) and (VI) in (IV),

$\Delta L=\frac{F_L}{\frac{Y_s}{L_{o,cab}}\left(\frac{\pi d_{cab}^2}{4}\right)+\frac{Y_a}{L_{o,col}}\left(\frac{\pi d_{out}^2-\pi d_{in}^2}{4}\right)}$

Substitute $1.27\text{cm}$ for $d_{cab}$, $5.75\text{m}$ for $L_{0,cab}$, $16.14\text{cm}$ for $d_{in}$, $16.24\text{cm}$ for $d_{out}$, $7.0\times {10}^{10}\text{Pa}$ for $Y_a$, $20\times {10}^{10}\text{Pa}$ for $Y_s$, $3.25\text{m}$ for $L_{0,col}$, $8500\text{N}$ for $F_L$ to determine the change in length,

$\begin{array}{c}\Delta L=\frac{8500\text{N}}{\frac{20\times {10}^{10}\text{Pa}}{5.75\text{m}}\left(\frac{\pi {\left(1.27\text{cm}\right)}^2}{4}\right)+\frac{7.0\times {10}^{10}\text{Pa}}{3.25\text{m}}\left(\frac{\pi {\left(16.24\text{cm}\right)}^2-\pi {\left(16.14\text{cm}\right)}^2}{4}\right)}\\ =8.6\times {10}^{-4}\text{m}\\ \text{=}\text{0}\text{.86}\text{mm}\end{array}$

Conclusion:

The point which is exerted by a load of force $8500\text{N}$ move down by $\text{0}\text{.86}\text{mm}$