Question 1.8: A hollow plastic circular pipe (length Lp, inner and outer d...
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 :
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} ):
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}:
(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 :
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 |
The required downward force to compress the plastic pipe is
P_{\text {reqd }}=\sigma_{p} A_{p}=-330 NBoth 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
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 }}}=0.99908Note: 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 .