Problem 3.4. Linearly distributed tensile load

Derive the displacement function parametrically.

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

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See Eq.(3-26)



$$\begin{gathered} –\stackrel{}{EA}\frac{{\mathrm{d}}^2{u}^{}}{\mathrm{d}{x}^2}=p_x, \\ \mathrm{where} \\ {p}_x=p_0\frac{x}{L} \end{gathered}$$

Step 2. Write the solution of the homogeneous equation.

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Step 3. Find a particular solution.

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

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The differential equation of the rod subjected to linearly varying tensile load is

See Eq.(3-26)

The solution of the homogeneous equation is

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

The following particular solution satisfies the inhomogeneous equation

$u_{\mathrm{part}}=–\frac{p_0}{EAL}\frac{x^3}{6}$

$–\stackrel{}{EA}\frac{d^2{u}_{\mathrm{part}}}{d{x}^2}=–\stackrel{}{EA}\frac{d^2\left(–{\displaystyle \frac{p_0}{EAL}}{\displaystyle \frac{x^3}{6}}\right)}{d{x}^2}=–\stackrel{}{EA}\left(–\frac{p_0}{EAL}x\right)=p_0\frac{x}{L}$

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

$u=u_{\mathrm{hom}}+ u_{\mathrm{part}}=C_1+C_{2}x–\frac{{\rho }_0}{EAL}\frac{x^3}{6}$

At both supports the displacement is zero.

$$\begin{gathered} \mathrm{at} x=0 u=0 \to C_1=0 \\ \mathrm{at} x=L u=0 \\ u\left(L\right)=C_{2}L–\frac{{\rho }_0}{EAL}\frac{L^3}{6}=0 \to C_2=\frac{{\rho }_{0}L}{6EA} \end{gathered}$$

The solution of the differential equation, i.e. the displacement function is

$\mathit{u}\mathit{=}\frac{{\mathit{p}}_{\mathbf{0}}}{\mathbf{6}\mathit{E}\mathit{A}\mathit{L}}\left(\mathit{x}{\mathit{L}}^{\mathbf{2}}\mathbf{–}{\mathit{x}}^{\mathit{3}}\right)$

$\epsilon =\frac{p_0}{6EAL}\left(L^2–3x^2\right)$