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## Q. 1.8

A hollow plastic circular pipe (length $L_{p}$, inner and outer diameters $d_{1}$, $d_{2}$, respectively; see Fig. 1-45) is inserted as a liner inside a cast iron pipe (length $L_{c}$, inner and outer diameters $d_{3}$, $d_{4}$, respectively).
(a) Derive a formula for the required initial length $L_{p}$ of the plastic pipe so that when it is compressed by some force P, the final length of both pipes is the same and also, at the same time, the final outer diameter of the plastic pipe is equal to the inner diameter of the cast iron pipe.
(b) Using the numerical data given below, find initial length $L_{p}$ (m) and final thickness $t_{p}$ (mm) for the plastic pipe.
(c) What is the required compressive force P (N)? What are the final normal stresses (MPa) in both pipes?
(d) Compare initial and final volumes (mm$^{3}$) for the plastic pipe. Numerical data and pipe cross-section properties :

$\begin{gathered}L_{c}=0.25 {~m} \quad E_{c}=170 {GPa} \quad E_{p}=2.1 {GPa} \quad v_{c}=0.3 \quad v_{p}=0.4 \\d_{1}=109.8 {mm} \quad d_{2}=110 {mm} \quad d_{3}=110.2 {mm} \\d_{4}=115 {mm} \quad t_{p}=\frac{d_{2}-d_{1}}{2}=0.1 {mm}\end{gathered}$

## Verified Solution

The initial cross-sectional areas of the plastic and cast iron pipes are

$A_{p}=\frac{\pi}{4}\left(d_{2}^{2}-d_{1}^{2}\right)=34.526 {mm}^{2} \quad A_{c}=\frac{\pi}{4}\left(d_{4}^{2}-d_{3}^{2}\right)=848.984 {mm}^{2}$

(a) Derive a formula for the required initial length $L_{p}$ of the plastic pipe.
The lateral strain resulting from compression of the plastic pipe must close the gap ($d_{3}$$d_{2}$) between the plastic pipe and the inner surface of the cast iron pipe. The required extensional lateral strain is positive (here, $\varepsilon_{\text {lat }}=\varepsilon^{\prime}$ ):

$\varepsilon_{\text {lat }}=\frac{d_{3}-d_{2}}{d_{2}}=1.818 \times 10^{-3}$

The accompanying compressive normal strain in the plastic pipe is obtained using Eq. (1-17) $\varepsilon^{\prime}=-v \varepsilon$ , which requires Poisson’s ratio for the plastic pipe and also the required lateral strain:

$\varepsilon_{p}=\frac{-\varepsilon_{\text {lat }}}{v_{p}} \text { or } \varepsilon_{p}=\frac{-1}{v_{p}}\left(\frac{d_{3}-d_{2}}{d_{2}}\right)=-4.545 \times 10^{-3}$

We can now use the compressive normal strain $\varepsilon_{p}$ to compute the short-ening $\delta_{p 1}$ of the plastic pipe as

$\delta_{p 1}=\varepsilon_{p} L_{p}$

At the same time, the required shortening of the plastic pipe (so that it will have the same final length as that of the cast iron pipe) is

$\delta_{p 2}=-\left(L_{p}-L_{c}\right)$

Now, equating $\delta_{p 1}$ and $\delta_{p 2}$ leads to a formula for the required initial length $L_{p}$ of the plastic pipe:

$L_{p}=\frac{L_{c}}{1+\varepsilon_{p}} \text { or } L_{p}=\frac{L_{c}}{1-\frac{d_{3}-d_{2}}{v_{p} d_{2}}}$

(b) Now substitute the numerical data to find the initial length $L_{p}$, change in thickness $\Delta t_{p}$, and final thickness $t_{pf}$for the plastic pipe.
As expected, $L_{p}$ is greater than the length of the cast iron pipe, $L_{c}$= 0.25 m, and the thickness of the compressed plastic pipe increases by $\Delta t_{p}$:

$\begin{gathered}L_{p}=\frac{L_{c}}{1-\left(\frac{d_{3}-d_{2}}{v_{p} d_{2}}\right)}=0.25114 {m} \\\\\Delta t_{p}=\varepsilon_{\text {lat }} t_{p}=1.818 \times 10^{-4} {mm} \text { so } t_{p f}=t_{p}+ \Delta t_{p}=0.10018 {mm}\end{gathered}$

(c) Next find the required compressive force P and the final normal stresses in both pipes.
A check on the normal compressive stress in the plastic pipe, computed using Hooke’s Law [Eq. (1-15) $\sigma=E \varepsilon$] shows that it is well below the ultimate stress for selected plastics (see Table H-3, Appendix H); this is also the final normal stress in the plastic pipe :

Table H-3
Mechanical Properties

 Material Yield stress $\sigma_{Y}$ Ultimate stress $\sigma_{U}$ Percent elongation (25 mm gage length) MPa MPa Aluminum alloys 2014-T6 6061-T6 7075-T6 35–500 410 270 480 100–550 480 310 550 1–45 13 17 11 Brass 70–550 200–620 4–60 Bronze 82–690 200–830 5–60 Cast iron (tension) 120–290 69–480 0–1 Cast iron (compression) 340–1,400 Concrete (compression) 10–70 Copper and copper alloys 55–760 230–830 4–50 Glass Plate glass Glass fibers 30–1,000 70 7,000-20,000 0 Magnesium alloys 80–280 140–340 2–20 Monel (67% Ni, 30% Cu) 179-1,100 450–1,200 2–50 Nickel 100–620 310–760 2–50 Plastics Nylon Polyethylene 40–80 7–28 20–100 15–300 Rock (compression) Granite, marble, quartz Limestone, sandstone 50–280 20–200 Rubber 1–7 7–20 100-800 Steel High-strength Machine Spring Stainless Tool 340–1,000 340–700 400–1,600 280–700 520 550–1,200 550–860 700–1,900 400–1,000 900 5–25 5–25 3–15 5–40 8 Steel, structural ASTM-A36 ASTM-A572 ASTM-A514 200–700 250 340 700 340–830 400 500 830 10–40 30 20 15 Steel wire 280–1,000 550–1,400 5–40 Titanium alloys 760–1,000 900–1,200 10 Tungsten 1,400–4,000 0–4 Wood (bending) Douglas fir Oak Southern pine 30–50 40–60 40–60 50–80 50–100 50–100 Wood (compression parallel to grain) Douglas fir Oak Southern pine 30–50 30–40 30–50 40–70 30–50 40–70
$\sigma_{p}=E_{p} \varepsilon_{p}=-9.55 {MPa}$

The required downward force to compress the plastic pipe is

$P_{\text {reqd }}=\sigma_{p} A_{p}=-330 N$

Both the initial and final stresses in the cast iron pipe are zero because no force is applied to the cast iron pipe.

(d) Lastly, compare the initial and final volumes of the plastic pipe.
The initial cross-sectional area of the plastic pipe is

$A_{p}=34.526 {mm}^{2}$

The final cross-sectional area of the plastic pipe is

$A_{p f}=\frac{\pi}{4}\left[d_{3}^{2}-\left(d_{3}-2 t_{p f}\right)^{2}\right]=34.652 {mm}^{2}$

The initial volume of the plastic pipe is

$V_{\text {pinit }}=L_{p} A_{p}=8671 {mm}^{3}$

and the final volume of the plastic pipe is

$V_{p \text { final }}=L_{c} A_{p f} \text { or } V_{p \text { final }}=8663 {mm}^{3}$

The ratio of final to initial volume reveals little change:

$\frac{V_{\text {pfinal }}}{V_{\text {pinit }}}=0.99908$

Note: The numerical results obtained in this example illustrate that the dimensional changes in structural materials under normal loading conditions are extremely small. In spite of their smallness, changes in dimensions can be important in certain kinds of analysis (such as the analysis of statically indeterminate structures) and in the experimental determination of stresses and strains .