Instability in torsion of a thin-walled tube Show that necking cannot take place in the torsion of a thin-walled tube made of a material whose true stress–true strain curve is given by σ= Kϵ^n.

Question

Instability in torsion of a thin-walled tube Show that necking cannot take place in the torsion of a thin-walled tube made of a material whose true stress–true strain curve is given by σ= K[latex]ε^n[/latex].

Accepted Answer

According to Eq. (2.21), the expression for torque, T, is

[latex] au=\frac{T}{2 \pi r^{2} t}[/latex] (2.21)

[latex]T=2 \pi r^{2} t au[/latex]

where for this case the shear stress, τ, can be related to the normal stress, σ using Eq. (2.56) for the distortion-energy criterion as [latex] au=\sigma / \sqrt{3}[/latex] (see Section 2.11.7 for details). The criterion for instability in torsion would be

[latex]\bar{\sigma}=\frac{1}{\sqrt{2}}\left[\left(\sigma_{1} -0 ight)^{2}+\left(0+\sigma_{1} ight)^{2}+\left(-\sigma_{1}-\sigma_{1} ight)^{2} ight]^{1 / 2}=\sqrt{3 \sigma_{1}}=\sqrt{3 au_{1}}[/latex] (2.56)

[latex]\frac{d T}{d \epsilon}=0[/latex]

Because r and t are constant, we have

[latex]\frac{d T}{d \epsilon}=\left(\frac{2}{\sqrt{3}} ight) \pi r^{2} au \frac{d \sigma}{d \epsilon}[/latex]

For a material represented by [latex]\sigma=K \epsilon^{n}[/latex],

[latex]\frac{d \sigma}{d \epsilon}=n K \epsilon^{n-1}[/latex]

Therefore,

[latex]\frac{d T}{d \epsilon}=\left(\frac{2}{\sqrt{3}} ight) \pi r^{2} au n K \epsilon^{n-1}[/latex]

Since none of the quantities in this expression are zero, dT/dε cannot be zero. Consequently, a tube in torsion does not undergo instability.

Question 2.2: Instability in torsion of a thin-walled tube Show that necki...

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