Problem 3.1. Tension rod

Reaction force at the left support, A [kN]=

Reaction force at the right support, B [kN]=

Displacement at the middle, u_m [mm]=

Maximum tensile stress, σ^max [N/mm²]=

Draw normal force diagram.

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Draw displacement diagram.

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Step 1. Calculate the displacement removing the left support.

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See solution of a cantilever subjected to tension in Example 3.1.

Left half of the rod is unloaded, the right half is loaded by a force, P = 300×10³ N, thus (E = 200×10³ N/mm²)
$u_0=\frac{P}{EA}\frac{L}{2}=\frac{300\times {10}^3}{200\times {10}^3\times {20}^2\pi /4}300=1.43 \mathrm{mm}$

Step 2. The displacement is restrained by support A. Determine the force which would cause the restrained displacement at the end of the cantilever (the difference between the displacement of the free end and the distance to the support is: u₀ – 0.2).

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Step 3. Give the support reactions, the internal forces and displacements with the superposition of the above cases.

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Step 4. Calculate stresses and strains along the length of the rod.

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See Eqs.(3-23) and (3-28)

$$\begin{gathered} \epsilon =\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{N}{EA} \\ \sigma =E\epsilon =\frac{N}{A} \\ {\mathit{\sigma }}^{\mathbf{max}}=\frac{N^{\max }}{A}=\frac{170.9\times {10}^3}{{20}^2\times \pi /4}\mathbf{=}\mathbf{544}\mathbf{.}\mathbf{1}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$

First  we remove the left support, and we calculate the displacement of a cantilever loaded at the middle:

See solution of a cantilever subjected to tension in Example 3.1.

Left half of the rod is unloaded, the right half is loaded by a force, P = 300×10³ N, thus (E = 200×10³ N/mm²)
$u_0=\frac{P}{EA}\frac{L}{2}=\frac{300\times {10}^3}{200\times {10}^3\times {20}^2\pi /4}300=1.43 \mathrm{mm}$

The displacement is restrained by support A. Now we determine the force which would cause the restrained displacement at the end of the cantilever (the difference between the displacement of the free end and the distance to the support is: $u_0–0.2$).  This compressive normal force acts on the enitre length of the beam:

$u_0–0.2=\frac{A}{EA}L \to A=\left(u_0–0.2\right)\frac{EA}{L}=\left(1.43–0.2\right)\frac{200\times 1000\times {20}^2\pi /4}{2\times 300}=129.5\times {10}^3 \mathrm{N}$

The support reactions, the internal forces and displacements are obtained by the superposition of the above cases. Superposition results in 0.2 mm displacement at the end of the rod, the reaction forces and the middle displacement are the following

$$\begin{gathered} \mathit{A}\mathbf{=}\mathbf{129}\mathbf{.}\mathbf{1}\mathbf{ }\mathbf{kN}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)} \\ \mathit{B}=P–A=300–129.1\mathbf{=}\mathbf{170}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{kN}\mathbf{ }\mathbf{\left(}\mathbf{tension}\mathbf{\right)} \\ {\mathit{u}}_{\mathbf{m}}=1.43–1.23/2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{82}\mathbf{ }\mathbf{mm} \end{gathered}$$

Left half of the rod is under compression, right half is under tension.

Stesses and strains along the length of the rod are

See Eqs.(3-23) and (3-28)

$$\begin{gathered} \epsilon =\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{N}{EA} \\ \sigma =E\epsilon =\frac{N}{A} \\ {\mathit{\sigma }}^{\mathbf{max}}=\frac{N^{\max }}{A}=\frac{170.9\times {10}^3}{{20}^2\times \pi /4}\mathbf{=}\mathbf{544}\mathbf{.}\mathbf{1}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$