Chapter 2, Problem 2.3.26P

A uniformly tapered lube AB of circular cross section and length L is shown in the figure. The average diameters at the ends are d_Aand d_£= 2d _t. Assume E is constant. Find the elongation S of the tube when it is subjected to loads P acting at the ends. Use the following numerical data:^ = 35 mm, L = WO mm, E = 2.1 GPa. and P = 25 tN. Consider the following cases.

(a) A hole of constant diameter d_Ais drilled from B toward A to form a hollow section of length x - U2.

(b) A hole of variable diameter a\.x) is drilled, from B toward A to form a hollow section of length x = L/2 and constant thickness t = d_A/20.

  

To determine

The elongation of the tube when a hole of constant diameter d_A is drilled from B to A to form a hollow section.

Answer to Problem 2.3.26P

The elongation of the tube when a hole of constant diameter d_A is drilled from B to A to form a hollow section is 2.18 mm.

Explanation of Solution

Given:

  $\begin{array}{l}P=25\text{ kN}\\ L=300\text{ mm}\\ d_A=35\text{ mm}\\ d_B=2d_A\\ E=2.1\text{ GPa}\end{array}$

  $d\left(\zeta \right)=d_A\left(1+\frac{\zeta }{L}\right)$

From the given information, the area is calculated.

For the length $L-x$

  $A\left(\zeta \right)=\frac{\pi d{\left(\zeta \right)}^2}{4}$

For the length x .

  $A\left(\zeta \right)=\frac{\pi }{4}\left[d{\left(\zeta \right)}^2-d_A^2\right]$

The elongation is calculated.

  $\begin{array}{l}\delta =\frac{P}{E}\left[{\displaystyle \int \frac{1}{A\left(\zeta \right)}d\zeta }\right]\\ \delta =\frac{P}{E}\left[{\displaystyle \underset{0}{\overset{L-x}{\int }}\frac{4}{\pi d{\left(\zeta \right)}^2}d\zeta +{\displaystyle \underset{L-x}{\overset{L}{\int }}\frac{4}{\pi \left(d{\left(\zeta \right)}^2-d_A^2\right)}d\zeta }}\right]\\ \delta =\frac{P}{E}\left[{\displaystyle \underset{0}{\overset{L-x}{\int }}\frac{1}{\left[\frac{\pi }{4}{\left[d_A\left(1+\frac{\zeta }{L}\right)\right]}^2\right]}}d\zeta +{\displaystyle \underset{L-x}{\overset{L}{\int }}\frac{1}{\left[\frac{\pi }{4}\left[{\left[d_A\left(1+\frac{\zeta }{L}\right)\right]}^2-d_A^2\right]\right]}d\zeta }\right]\\ \delta =\frac{P}{E}\left[4\frac{L^2}{\left(x-2\right)\pi d_A^2}+\left(4\frac{L}{\pi d_A^2}+{\displaystyle \underset{L-x}{\overset{L}{\int }}\frac{1}{\left[\frac{\pi }{4}\left[{\left[d_A\left(1+\frac{\zeta }{L}\right)\right]}^2-d_A^2\right]\right]}}d\zeta \right)\right]\\ \delta =\frac{P}{E}\left[4\frac{L^2}{\left(x-2\right)\pi d_A^2}+\left(4\frac{L}{\pi d_A^2}-2L\frac{\ln 3}{\pi d_A^2}+2L\frac{-\ln \left(L-x\right)+\ln \left(3L-x\right)}{\pi d_A^2}\right)\right]\end{array}$

Given $x=\frac{L}{2}$

  $\delta =\frac{P}{E}\left(\frac{4}{3}\frac{L}{\pi d_A^2}-2L\frac{\ln 3}{\pi d_A^2}+2L\frac{-\ln \left(\frac{1}{2}L\right)+\ln \left(\frac{5}{2}L\right)}{\pi d_A^2}\right)$

Substituting the values:

  $\begin{array}{l}\delta =\frac{P}{E}\left(\frac{4}{3}\frac{L}{\pi d_A^2}-2L\frac{\ln 3}{\pi d_A^2}+2L\frac{-\ln \left(\frac{1}{2}L\right)+\ln \left(\frac{5}{2}L\right)}{\pi d_A^2}\right)\\ \delta =\frac{25}{2.1}\left(\frac{4}{3}\frac{300}{\pi {\left(35\right)}^2}-2\left(300\right)\frac{\ln 3}{\pi {\left(35\right)}^2}+2\left(300\right)\frac{-\ln \left(\frac{1}{2}300\right)+\ln \left(\frac{5}{2}300\right)}{\pi {\left(35\right)}^2}\right)\\ \delta =2.18\text{ mm}\end{array}$

Conclusion:

The elongation of the tube when a hole of constant diameter d_A is drilled from B to A to form a hollow section is 2.18 mm.

To determine

The elongation of the tube when a hole of variable diameter $d\left(x\right)$ is drilled from B to A to form a hollow section.

Answer to Problem 2.3.26P

The elongation of the tube when a hole of variable diameter $d\left(x\right)$ is drilled from B to A to form a hollow section is 6.74 mm.

Explanation of Solution

  $d\left(\zeta \right)=d_A\left(1+\frac{\zeta }{L}\right)$

From the given information, the area is calculated.

For the length $L-x$

  $A\left(\zeta \right)=\frac{\pi d{\left(\zeta \right)}^2}{4}$

For the length x .

  $A\left(\zeta \right)=\frac{\pi }{4}\left[d{\left(\zeta \right)}^2-{\left(d\left(\zeta \right)-2\frac{d_A}{20}\right)}^2\right]$

The elongation is calculated.

  $\begin{array}{l}\delta =\frac{P}{E}\left[{\displaystyle \int \frac{1}{A\left(\zeta \right)}d\zeta }\right]\\ \delta =\frac{P}{E}\left[{\displaystyle \underset{0}{\overset{L-x}{\int }}\frac{4}{\pi d{\left(\zeta \right)}^2}d\zeta +{\displaystyle \underset{L-x}{\overset{L}{\int }}\frac{4}{\pi \left[d{\left(\zeta \right)}^2-{\left(d\left(\zeta \right)-2\frac{d_A}{20}\right)}^2\right]}d\zeta }}\right]\\ \delta =\frac{P}{E}\left[{\displaystyle \underset{0}{\overset{L-x}{\int }}\frac{4}{\left[\pi {\left[d_A\left(1+\frac{\zeta }{L}\right)\right]}^2\right]}}d\zeta +{\displaystyle \underset{L-x}{\overset{L}{\int }}\frac{4}{\left[\pi \left[{\left[d_A\left(1+\frac{\zeta }{L}\right)\right]}^2-{\left[d_A\left(1+\frac{\zeta }{L}\right)-2\frac{d_A}{20}\right]}^2\right]\right]}d\zeta }\right]\\ \delta =\frac{P}{E}\left[4\frac{L^2}{\left(x-2L\right)\pi d_A^2}+4\frac{L}{\pi d_A^2}+\left(20L\frac{\ln 3+\ln 13+2\ln d_A+\ln L}{\pi d_A^2}\right)-20L\frac{2\ln d_A+\ln \left(39L-20x\right)}{\pi d_A^2}\right]\end{array}$

Given $x=\frac{L}{2}$

  $\delta =\frac{P}{E}\left(\frac{4}{3}\frac{L}{\pi d_A^2}+20L\frac{\ln 3+\ln 13+2\ln d_A+\ln L}{\pi d_A^2}-20L\frac{2\ln d_A+\ln \left(29L\right)}{\pi d_A^2}\right)$

Substituting the values:

  $\begin{array}{l}\delta =\frac{P}{E}\left(\frac{4}{3}\frac{L}{\pi d_A^2}-2L\frac{\ln 3}{\pi d_A^2}+2L\frac{-\ln \left(\frac{1}{2}L\right)+\ln \left(\frac{5}{2}L\right)}{\pi d_A^2}\right)\\ \delta =\frac{25}{2.1}\left(\frac{4}{3}\frac{300}{\pi {\left(35\right)}^2}-2\left(300\right)\frac{\ln 3}{\pi {\left(35\right)}^2}+2\left(300\right)\frac{-\ln \left(\frac{1}{2}300\right)+\ln \left(\frac{5}{2}300\right)}{\pi {\left(35\right)}^2}\right)\\ \delta =6.74\text{ mm}\end{array}$

Conclusion:

The elongation of the tube when a hole of variable diameter $d\left(x\right)$ is drilled from B to A to form a hollow section is 6.74 mm.