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Chapter 1

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}
A hollow plastic circular pipe (length Lp, inner and outer diameters d1, d2, respectively; see Fig. 1-45) is inserted as a liner inside a cast iron pipe (length Lc, inner and outer diameters d3, d4, respectively). (a) Derive a formula for the required initial length Lp of the plastic pipe so that

Step-by-Step

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 .