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} and d_{2}, respectively; see Fig.1-48) is inserted as a liner inside a cast iron pipe (length L_{c} , inner and outer diameters d_{3} and 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 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, find the 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 the initial and final volumes (mm³ ) for the plastic pipe.
Numerical data and pipe cross-section properties are
L_{c}=0.25 m \quad E_{c}=170 GPa \quad E_{p}=2.1 GPa \quad v_{c}=0.3 \quad v_{p}=0.4d_{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
Use the four-step problem-solving approach to find the dimensions and force required to fit a plastic liner into a cast iron pipe.
1. Conceptualize: Application of a compressive force P results in compressive normal strains and extensional lateral strains in the plastic pipe, while the cast iron pipe is stress-free. The initial length of the plastic pipe (L_{p} ) is greater than that of the cast iron pipe (L_{c} ). With full application of force P, the lengths are made equal.
The initial cross-sectional areas of the plastic and cast iron pipes are
2. Categorize : The two requirements are (a) 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 and (b) the final lengths of the two pipes are the same. The first requirement depends on lateral strain and the second on normal strain. Each requirement leads to an expression for shortening of the plastic pipe. Equating the two expressions (i.e., enforcing compatibility of displacements) leads to a solution for the required length of the plastic pipe.
3- Analyze:
Part (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 \left(d_{3}-d_{2}\right) between the plastic pipe and the inner surface of the cast iron pipe. The required extensional lateral strain is positive (here,\left.\varepsilon_{\text {lat }}=\varepsilon^{\prime}\right):
\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.1-14 \varepsilon^{\prime}=-\nu \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}Use the compressive normal strain \varepsilon_{p} to compute the shortening \delta_{p 1} of the plastic pipe as
\delta_{p 1}=\varepsilon_{p} L_{p}The required shortening of the plastic pipe (so that it has the same final length as that of the cast iron pipe) is
\delta_{p 2}=-\left(L_{p}-L_{c}\right)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}} \quad \text { or } \quad L_{p}=\frac{L_{c}}{1-\frac{d_{3}-d_{2}}{v_{p} d_{2}}}Part (b): Now substitute the numerical data to find the initial length L_{p}, change in thickness \Delta t_{p}, and final thickness t_{p f} 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}:
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 \quad \text { so } \quad t_{p f}=t_{p}+\Delta t_{p}=0.10018 mm
Part (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.1-12), shows that it is well below the ultimate stress for selected plastics (see Table I-3, Appendix I); this is also the final normal stress in the plastic pipe:
\sigma=E \varepsilon Eq.1-12
\sigma_{p}=E_{p} \varepsilon_{p}=-9.55 MPaThe required downward force to compress the plastic pipe is
P_{ 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.
| Table I-3 | |||||
| Mechanical Properties | |||||
| Material | Yield Stress \sigma_{Y} | Ultimate Stress \sigma_{U} | percent Elongation (2-in. gage length) |
||
| ksi | MPa | ksi | MPa | ||
| Aluminum alloys 2014-T6 6061-T6 7075-T6 |
5–70 60 40 70 |
35–500 410 270 480 |
15–80 70 45 80 |
100–550 480 310 550 |
1–45 13 17 11 |
| Brass | 10–80 | 70–550 | 30–90 | 200–620 | 4–60 |
| Bronze | 12–100 | 82–690 | 30–120 | 200–830 | 5–60 |
| Cast iron (tension) | 17–42 | 120–290 | 10–70 | 69–480 | 0–1 |
| Cast iron (compression) | 50–200 | 340–1400 | |||
| Concrete (compression) | 1.5-10 | 10-70 | |||
| Copper and copper alloys | 8–110 | 55–760 | 33–120 | 230–830 | 4–50 |
| Glass Plate glass Glass fibers |
5–150 10 1000–3000 |
30–1000 70 7000–20,000 |
0 | ||
| Magnesium alloys | 12–40 | 80–280 | 20–50 | 140–340 | 2–20 |
| Monel (67% Ni, 30% Cu) | 25–160 | 170–1100 | 65–170 | 450–1200 | 2–50 |
| Nickel | 15–90 | 100–620 | 45–110 | 310–760 | 2–50 |
| Plastics
Nylon |
6–12 |
40–80 |
20–100 |
||
| Rock (compression)
Granite, marble, quartz |
8–40 |
50–280 |
|||
| Rubber | 0.2–1.0 | 1–7 | 1–3 | 7–20 | 100–800 |
| Steel
High-strength |
50–150 |
340–1000 |
80–180 |
550–1200 |
5–25 |
| Steel, structural ASTM-A36 ASTM-A572 ASTM-A514 |
30–100 36 50 100 |
200–700 250 340 700 |
50–120 60 70 120 |
340–830 400 500 830 |
10–40 30 20 15 |
| Steel wire | 40–150 | 280–1000 | 80–200 | 550–1400 | 5–40 |
| Titanium alloys | 110–150 | 760–1000 | 130–170 | 900–1200 | 10 |
| Tungsten | 200-600 | 1400-4000 | 0-4 | ||
| Wood (bending)
Douglas fir |
5–8 |
30–50 |
8–12 |
50–80 |
|
| Wood (compression parallel to grain)
Douglas fir |
4–8 |
30–50 |
6–10 |
40–70 |
|
Part (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_{\text { pfinal }}=L_{c} A_{p f} \text { or } V_{\text { pfinal }}=8663 mm ^{3}4. Finalize: The ratio of final to initial volume reveals little change in the volume of the plastic pipe:
\frac{V_{\text {pfinal }}}{V_{\text {pinit }}}=0.99908The 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.