A rectangular element in a linearly elastic, isotropic material is subjected to tensile stresses of 83 and 65 N/mm^2 on mutually perpendicular planes. Determine the strain in the direction of each stress and in the direction perpendicular to both stresses. Find also the principal strains, the

Question

A rectangular element in a linearly elastic, isotropic material is subjected to tensile stresses of 83 and [latex]65 N / mm ^{2}[/latex] on mutually perpendicular planes. Determine the strain in the direction of each stress and in the direction perpendicular to both stresses. Find also the principal strains, the maximum shear stress, the maximum shear strain, and their directions at the point. Take [latex]E=200,000 N / mm ^{2} ext { and } v=0.3[/latex].

Accepted Answer

If we assume that [latex]\sigma_{x}=83 N / mm ^{2} ext { and } \sigma_{y}=65 N / mm ^{2}[/latex], then from Eqs (1.52),

[latex]\left.\begin{array}{l} \varepsilon_{x}=\frac{1}{E}\left(\sigma_{x}-v \sigma_{y} ight) \ \varepsilon_{y}=\frac{1}{E}\left(\sigma_{y}-v \sigma_{x} ight) \ \varepsilon_{z}=\frac{-v}{E}\left(\sigma_{x}-\sigma_{y} ight) \ \gamma_{x y}=\frac{ au_{x y}}{G} \end{array} ight\}[/latex] (1.52)

[latex]\varepsilon_{x}=\frac{1}{200,000}(83-0.3 imes 65)=3.175 imes 10^{-4}[/latex]

[latex]\varepsilon_{y}=\frac{1}{200,000}(65-0.3 imes 83)=2.005 imes 10^{-4}[/latex]

[latex]\varepsilon_{z}=\frac{-0.3}{200,000}(83+65)=-2.220 imes 10^{-4}[/latex]

In this case, since there are no shear stresses on the given planes, [latex]\sigma_{x}[/latex] and [latex]\sigma_{y}[/latex] are principal stresses, so that [latex]\varepsilon_{x}[/latex] and [latex]\varepsilon_{y}[/latex] are the principal strains and are in the directions of [latex]\sigma_{x}[/latex] and [latex]\sigma_{y}[/latex]. It follows from Eq. (1.15) that the maximum shearstress (in the plane of the stresses) is

[latex] au_{\max }=\frac{\sigma_{ I }-\sigma_{ II }}{2}[/latex] (1.15)

[latex] au_{\max }=\frac{83-65}{2}=9 N / mm ^{2}[/latex]

acting on planes at [latex]45^{\circ}[/latex] to the principal planes. Further, using Eq. (1.50), the maximum shear strain is

[latex]\gamma=\frac{2(1+v)}{E} au[/latex] (1.50)

[latex]\gamma_{\max }=\frac{2 imes(1+0.3) imes 9}{200,000}[/latex]

so that [latex]\gamma_{\max }=1.17 imes 10^{-4}[/latex] on the planes of maximum shear stress.

Question 1.4: A rectangular element in a linearly elastic, isotropic mater...

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