Chapter 4, Problem 4.59P

(a)

To determine

The reduction in the height of the specimen.

Explanation of Solution

Formula used:

The expressionfor the volume of the original specimen is given as,

  $V=\frac{\pi }{4}\left(d_o{}^2-d_i{}^2\right)\cdot h$

Here, $d_o$ is outer diameter of the specimen, $d_i$ is inner diameter of the specimen and $h$ is the height of specimen.

The expression for the volume during the plastic deformation of the original specimen is given as,

  $V_1=\frac{\pi }{4}\left(d_o{}^2-d_i{}^2\right)\cdot h$

Here, $d_o$ is outer diameter of the specimen and $d_i$ is inner diameter of the specimen.

Calculation:

Refer figure 4.8(a) “Charts to determine friction in ring compression test” to obtain the values as,

The outer diameter of the specimen is $d_o=0.75\text{in}$ .

The outer diameter of the specimen is $d_i=0.375\text{in}$ .

Height of the specimen is $h=0.25\text{in}$ .

The value for the volume of original specimen can be calculated by substituting the values of $d_0$ , $d_i$ and $h$ as,

  $\begin{array}{c}V=\frac{\pi }{4}\left(d_o{}^2-d_i{}^2\right)\cdot h\\ V=\frac{\pi }{4}\left(d_o{}^2-d_i{}^2\right)\times h\\ V=\frac{\pi }{4}\left({0.75}^2{\text{in}}^2-{0.375}^2{\text{in}}^2\right)\times 0.25\text{in}\\ V=0.0828{\text{in}}^3\end{array}$

The value for the height of specimen is calculated as,

  $\begin{array}{c}V=V_1\\ 0.0828{\text{in}}^3=\frac{\pi }{4}\left(d_0^2-d_i^2\right)\cdot h\\ h=\frac{0.105}{\left(d_0^2-d_i^2\right)}\end{array}$

The reduction in height of specimen is calculated by taking different values of $d_i$ and $d_0$ from the figure 4.8(a) as,

Specimen (Number)	$d_i\left(\text{in}\right)$	$d_0\left(\text{in}\right)$	Formula	Calculation	$h$
$1.$	$0.375$	$0.75$	$h=\frac{0.105}{\left(d_0^2-d_i^2\right)}$	$h=\frac{0.105{\text{in}}^{\text{3}}}{\left({0.75}^2{\text{in}}^2-{0.375}^2{\text{in}}^2\right)}$	$0.25\text{in}$
$2.$	$0.477$	$0.97$	$h=\frac{0.105}{\left(d_0^2-d_i^2\right)}$	$h=\frac{0.105{\text{in}}^{\text{3}}}{\left({0.97}^2{\text{in}}^2-{0.477}^2{\text{in}}^2\right)}$	$0.147\text{in}$
$3.$	$0.282$	$1.04$	$h=\frac{0.105}{\left(d_0^2-d_i^2\right)}$	$h=\frac{0.105{\text{in}}^{\text{3}}}{\left({1.04}^2{\text{in}}^2-{0.282}^2{\text{in}}^2\right)}$	$0.104\text{in}$

(b)

To determine

The coefficient of the friction for the each specimen.

Explanation of Solution

The value of coefficient is calculated from the figure 4.8(a) corresponding to the different values of percentage reduction in height and reduction in internal diameter is given as,

Specimen (Number)	% Reduction in height	% Reduction in Internal diameter	Coefficient of Friction   $\left(\mu \right)$
1.	41.2	-27.2	0.01
2.	58.4	24.8	0.10
3.	60	53.1	0.20