Problem 4.12. Elastic and plastic resistance of circular cross sections

Problem a)

Ratio of plastic and elastic resistance, T_R,pl/T_R,el =

Problem b)

Ratio of plastic and elastic resistance, T_R,pl/T_R,el =

Problem a)

Step 1. Draw stress diagram which belongs to elastic failure. Determine elastic torsional resistance of the ring cross section.

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See Figure 3.51 and Eq.(3-79)

$$\begin{gathered} T_{\mathrm{R},\mathrm{el}}={\int }_{A}r\tau \mathrm{d}A={\int }_{A}r\vartheta Gr\mathrm{d}A=\vartheta G{\int }_{\alpha R}^R{\int }_0^{2r\pi }{r}^{2}ds\mathrm{d}r=\vartheta G\frac{\pi }{2}\left(R^4–{\left(\alpha R\right)}^4\right)=f\frac{\pi }{2}{R}^3\left(1–{\alpha }^4\right) \\ \mathrm{where} f=\vartheta GR \to \vartheta =\frac{f}{GR} \mathrm{and} \alpha =\frac{R_i}{R}=0.8 \end{gathered}$$

Step 2. Draw stress diagram which belongs to plastic failure. Determine plastic torsional resistance of the ring cross section.

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Step 3. Give the ratio of the plastic and elastic resistances.

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Problem b)

Step 1. Draw stress diagram which belongs to elastic failure. Determine elastic torsional resistance of the solid circular cross section.

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See Figure 3.51 and Equation (3-79)

$$\begin{gathered} T_{\mathrm{R},\mathrm{el}}={\int }_{A}r\tau \mathrm{d}A={\int }_{A}r\vartheta Gr\mathrm{d}A=\vartheta G{\int }_0^R{\int }_0^{2r\pi }{r}^{2}ds\mathrm{d}r=\vartheta G\frac{\pi }{2}{R}^4=f\frac{\pi }{2}{R}^3 \\ \mathrm{where} f=\vartheta GR \to \vartheta =\frac{f}{GR} \end{gathered}$$

Step 2. Draw stress diagram which belongs to plastic failure. Determine plastic torsional resistance of the solid circular cross section.

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Step 3. Give the ratio of the plastic and elastic resistances.

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Problem a)

Stress diagram which belongs to elastic failure is given below.

See Figure 3.51 and Equation (3-79)

Elastic torsional resistance of the ring cross section is

$$\begin{gathered} T_{\mathrm{R},\mathrm{el}}={\int }_{A}r\tau \mathrm{d}A={\int }_{A}r\vartheta Gr\mathrm{d}A=\vartheta G{\int }_{\alpha R}^R{\int }_0^{2r\pi }{r}^{2}ds\mathrm{d}r=\vartheta G\frac{\pi }{2}\left(R^4–{\left(\alpha R\right)}^4\right)=f\frac{\pi }{2}{R}^3\left(1–{\alpha }^4\right) \\ \mathrm{where} f=\vartheta GR \to \vartheta =\frac{f}{GR} \mathrm{and} \alpha =\frac{R_i}{R}=0.8 \end{gathered}$$

Stress diagram which belongs to plastic failure is given in the next Figure.

Plastic torsional resistance of the ring cross section is$$\begin{gathered} T_{\mathrm{R},\mathrm{el}}={\int }_{A}r\tau \mathrm{d}A={\int }_{A}rf\mathrm{d}A=f{\int }_{\alpha R}^R{\int }_0^{2r\pi }rds\mathrm{d}r=f\frac{2\pi }{3}{R}^3\left(1–{\alpha }^3\right) \\ \mathrm{where} \tau =f \end{gathered}$$

The ratio of the plastic and elastic resistances:$\frac{{\mathit{T}}_{\mathbf{R}\mathbf{,}\mathbf{pl}}}{{\mathit{T}}_{\mathbf{R}\mathbf{,}\mathbf{el}}}=\frac{f\frac{2\pi }{3}{R}^3(1–{\alpha }^3)}{f\frac{\pi }{2}{R}^3(1–{\alpha }^4)}=\frac{4}{3}\frac{\left(1–{\mathrm{\alpha }}^3\right)}{\left(1–{\mathrm{\alpha }}^4\right)}=\frac{4}{3}\frac{\left(1–0.8^3\right)}{\left(1–0.8^4\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{102}$

Problem b)

Stress diagram which belongs to elastic failure is shown in the Figure.

See Figure 3.51 and Equation (3-79)

Elastic torsional resistance of the solid circular cross section is

$$\begin{gathered} T_{\mathrm{R},\mathrm{el}}={\int }_{A}r\tau \mathrm{d}A={\int }_{A}r\vartheta Gr\mathrm{d}A=\vartheta G{\int }_0^R{\int }_0^{2r\pi }{r}^{2}ds\mathrm{d}r=\vartheta G\frac{\pi }{2}{R}^4=f\frac{\pi }{2}{R}^3 \\ \mathrm{where} f=\vartheta GR \to \vartheta =\frac{f}{GR} \end{gathered}$$

Stress diagram which belongs to plastic failure is given in the next Figure.

Plastic torsional resistance of the solid circular cross section is

$$\begin{gathered} T_{\mathrm{R},\mathrm{el}}={\int }_{A}r\tau \mathrm{d}A={\int }_{A}rf\mathrm{d}A=f{\int }_0^R{\int }_0^{2r\pi }rds\mathrm{d}r=f\frac{2\pi }{3}{R}^3 \\ \mathrm{where} \tau =f \end{gathered}$$

The ratio of the plastic and elastic resistances:

$\frac{{\mathit{T}}_{\mathbf{R}\mathbf{,}\mathbf{pl}}}{{\mathit{T}}_{\mathbf{R}\mathbf{,}\mathbf{el}}}=\frac{f\frac{2\pi }{3}{R}^3}{f\frac{\pi }{2}{R}^3}=\frac{4}{3}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{33}$