Problem 3.2. Suspended bar – self weight

Derive the strain function parametrically.

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Draw strain and displacement diagrams.

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Step 1. Give the load and the differential equation of the tensile rod.

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See Eq.(3-26)



$$\begin{gathered} –EA\frac{{\mathrm{d}}^2{u}^{}}{\mathrm{d}{x}^2}=p_x, \\ \mathrm{where} \\ {p}_x=\rho gA \end{gathered}$$

Step 2. Write the general solution of the homogeneous equation.

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See Eqs.(D-29) in the Appendix and Example 3.1.

$u_{\mathrm{hom}}=C_1+C_{2}x$

Step 3. Find a particular solution.

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Eq.(D-30) in the Appendix

$u_{\mathrm{part}}=r_0\frac{x^2}{2}$

where r₀ is determined by substituting the particular solution into the differential equation:

$$\begin{gathered} –EA\frac{{\mathrm{d}}^2\left({\mathrm{r}}_0{\displaystyle \frac{x^2}{2}}\right)}{\mathrm{d}{x}^2}=\rho gA \to –EA{r}_0=\rho gA \\ \to r_0=\mathit{–}\frac{\rho gA}{EA}\mathit{=}\mathit{–}\frac{\rho g}{E} \end{gathered}$$

Step 4. Write the general solution and determine its constants from the boundary conditions.

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Step 5. Express the strain function. Draw the strain and displacement diagrams.

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The differential equation of the tensile rod subjected to its self-weight is

Eq.(3-26)

$$\begin{gathered} –EA\frac{{\mathrm{d}}^2{u}^{}}{\mathrm{d}{x}^2}=p_x, \\ \mathrm{where} \\ {p}_x=\rho gA \end{gathered}$$

The general solution of the homogeneous equation is

$u_{\mathrm{hom}}=C_1+C_{2}x$

See Eqs.D-29) in the Appendix and Example 3.1.

A particular solution is

$u_{\mathrm{part}}=r_0\frac{x^2}{2}$

where r₀ is determined by substituting the particular solution into the differential equation:

$$\begin{gathered} –EA\frac{{\mathrm{d}}^2\left({\mathrm{r}}_0{\displaystyle \frac{x^2}{2}}\right)}{\mathrm{d}{x}^2}=\rho gA \to –EA{r}_0=\rho gA \\ \to r_0=\mathit{–}\frac{\rho gA}{EA}\mathit{=}\mathit{–}\frac{\rho g}{E} \end{gathered}$$

Eq.(D-30) in the Appendix

The constants of the general solution are determined from the boundary conditions. The general solution is 

$u=u_{\mathrm{hom}}+ u_{\mathrm{part}}=C_1+C_{2}x–\frac{\rho g}{E}\frac{x^2}{2}$

The boundary conditions are

$$\begin{gathered} \mathrm{at} x=0 u=0 \to C_1=0 \\ \mathrm{at} x=L N=0 \\ N=EA\epsilon =EA\frac{\mathrm{d}u}{\mathrm{d}x}=EA\left(C_2–\frac{\rho g}{E}x\right) \\ N\left(L\right)=EA\left(C_2–\frac{\rho g}{E}L\right)=0 \to C_2=\frac{\rho g}{E}L \end{gathered}$$

The solution of the differential equation is the displacement function:

$u\mathit{=}\frac{\mathit{\rho }\mathit{g}}{\mathit{E}}\left(\mathit{L}\mathit{x}\mathit{–}\frac{{\mathit{x}}^{\mathit{2}}}{\mathit{2}}\right)$

Its derivative results in the strain function:

$\mathit{\epsilon }=\frac{\mathrm{d}u}{\mathrm{d}x}\mathit{=}\frac{\mathit{\rho }\mathit{g}}{\mathit{E}}\left(\mathit{L}\mathit{–}\mathit{x}\right)$

The strain and displacement functions are given in the Figure.