Chapter 2, Problem 2.82P

To determine

The expression for force versus piston travel d up to d = 10 mm.

Answer to Problem 2.82P

The expression for Force F versus piston travel d is

  $F=8835.732d+294522\ln {\left(1+\frac{d}{25}\right)}^{0.4}$

And the value of force F at d =10 mm is 127996.71 N

Explanation of Solution

Given data:

  $\begin{array}{l}\text{Diamter of cylindrical slug, }{D}_1=25\text{mm}\\ \text{Height of cylindrical slug, }d=25\text{mm,}\\ \text{Diamterof cavity in a rigid die, }{D}_2=25\text{mm,}\\ \text{Pressure}\text{acting of matrix is, }{p}_m=150\frac{\Delta V}{V_{\text{om}}}\text{MPa,}\\ \text{Stress of the slug material, }\sigma =600{\epsilon }^{0.4}.\end{array}$

A cylindrical is surrounded by a compressible matrix. Cylindrical slug is compressed vertically by a force F which is to be determined. When the force F is acted then there will be the generation of stress inside the cylindrical slug and pressure is also generated inside the compressible matrix. The force acted will be equal to the force generated inside the cylindrical slug plus the force generated inside the matrix.

Calculation:

Drawing the free body diagram of the given system,

  

From the free body diagram, it can be written as,

  $F=F_{\text{matrix}}+F_{\text{cylinder}}\mathrm{...........}(1)$

Area of the matrix and cylinder can be given as −

  $\begin{array}{l}\text{Area of the matrix,}\\ A_m=\frac{\pi }{4}\left(D_1^2-D_2^2\right)\\ A_m=\frac{\pi }{4}\left({50}^2-{25}^2\right)\\ A_m=1472.622{\text{mm}}^2\end{array}$

  $\begin{array}{l}\text{Area of the cylinder,}\\ A_c=\frac{\pi }{4}{D}_2^2\\ A_c=\frac{\pi }{4}\times {25}^2\\ A_c=490.87{\text{mm}}^2\end{array}$

  $\begin{array}{l}\text{Change in the volume for matrix can be given as,}\\ \Delta V=A_m\left[25-\left(25-d\right)\right]\\ \Delta V=A_m\times d\\ \Delta V=1472.622\times d\end{array}$

Calculating the force exerted by the matrix:-

  $\begin{array}{l}{F}_{\text{matrix}}=p_{\text{m}}\times Area\\ F_{\text{matrix}}=150\times \frac{\Delta V}{V_{\text{om}}}\times Area\\ \text{Volume of the matrix will be, }{V}_{\text{om}}=A_m\times 25\\ F_{\text{matrix}}=150\times \frac{\Delta V}{A_m\times 25}\times A_m\\ F_{\text{matrix}}=150\times \frac{\Delta V}{25}\\ F_{\text{matrix}}=150\times \frac{1472.622\times d}{25}\\ F_{\text{matrix}}=8835.732d\mathrm{...............}(2)\end{array}$

The strain developed inside the cylinder for deformation of height d is −

  $\text{Strain,}e=\frac{d}{25}$

True strain inside the cylinder would be −

  $\begin{array}{l}\text{True strain, }\epsilon =\ln \left(1+\text{e}\right)\\ \epsilon =\ln \left(1+\frac{d}{25}\right)\end{array}$

Writing true stress-true strain relation,

  $\begin{array}{l}\sigma =600{\epsilon }^{0.4}\\ \sigma =600\ln {\left(1+\frac{d}{25}\right)}^{0.4}\end{array}$

The force will be given as,

  $\begin{array}{l}{F}_c=\sigma A_c\\ F_c=600\ln {\left(1+\frac{d}{25}\right)}^{0.4}\times 490.87\\ F_c=294522\ln {\left(1+\frac{d}{25}\right)}^{0.4}\mathrm{...................}(3)\end{array}$

From equation (1), (2) and (3),

  $\begin{array}{l}F=F_{\text{matrix}}+F_{\text{cylinder}}\\ \text{But,}\\ F_{\text{matrix}}=8835.732d\\ F_c=294522\ln {\left(1+\frac{d}{25}\right)}^{0.4}\\ \text{Therefore,}\\ F=8835.732d+294522\ln {\left(1+\frac{d}{25}\right)}^{0.4}\end{array}$

The above is the required expression for force versus d .

Putting d = 10 mm we get,

  $\begin{array}{l}F=8835.732d+294522\ln {\left(1+\frac{d}{25}\right)}^{0.4}\\ F=8835.732\times 10+294522\ln {\left(1+\frac{10}{25}\right)}^{0.4}\\ F=88357.32+39639.39\\ F=127996.71\text{N}\end{array}$

Hence, the expression for Force F versus piston travel d is

  $F=8835.732d+294522\ln {\left(1+\frac{d}{25}\right)}^{0.4}$

And the value of force F at d =10 mm is 127996.71 N