Given: σ = ⎡ ⎣ 10 12 13 12 11 15 13 15 20 ⎤ ⎦ MPa at a point. What is the force per unit area at this point acting normal to the surface wit

Question

Given: σ = ⎡ ⎣ 10 12 13 12 11 15 13 15 20 ⎤ ⎦ MPa at a point. What is the force per unit area at this point acting normal to the surface with unit normal vector n = (1/ √ 2)ex + (1/ √ 2)ez? Are there any shear stresses acting on this surface?

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Tryphena 4 years 2021-08-16T21:49:36+00:00 1 Answers 22 views 0

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  1. MichaelMet
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    2021-08-16T21:50:57+00:00
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    Answer:

    The answer is “19.735 mpa“

    Explanation:

    Given:

    [\sigma_{ij}]=\left[\begin{array}{ccc}10&12&13\\\12&11&15\\13&15&20\end{array}\right]

    \hat{\wedge} = \frac{1}{\sqrt{2}}e_x+\frac{1}{\sqrt{2}}e_2\\

    The traction vector acting on this plane=

    \left[\begin{array}{c}t_1^{R}&t_2^{R}&t_3^{R}\\ \end{array}\right] =\left[\begin{array}{ccc}10&12&13\\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{c}\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\ \end{array}\right]

               =\frac{1}{\sqrt{2}}\left[\begin{array}{ccc}10\times 1&+12 \times 0&+ 13 \times 1\\12\times 1&+11\times 0&15\times 1\\13\times 1&+15\times 0&+20\times 1\end{array}\right] \\\\\\\\=\frac{1}{\sqrt{2}}\left[\begin{array}{c}23&27&33\\ \end{array}\right] \\\\

    The area of acting to surface=  t^{\hat \wedge} \cdot \hat \wedge= (\frac{23}{\sqrt{2}}e_n+\frac{27}{\sqrt{2}}e_y+\frac{33}{\sqrt{2}}e_z)

                                              \sigma_{\wedge}=\frac{23}{2}+\frac{33}{2}=\frac{56}{2} =28 \ mpa

    S hearing component of traction = t^{\hat n}-\sigma_\wedge\cdot \hat n\\

    =\frac{23}{2}+\frac{33}{2}=\frac{56}{2} - 28(\frac{1}{\sqrt{2}}e_n+\frac{1}{\sqrt{2}}e_z)\\\\=-\frac{5}{\sqrt{2}}e_n+\frac{27}{\sqrt{2}}e_y+\frac{5}{\sqrt{2}}e_z\\\\\\= \sqrt{-\frac{5}{\sqrt{2}}e_n+\frac{27}{\sqrt{2}}e_y+\frac{5}{\sqrt{2}}e_z}\\\\\\= 19.735 \ mpa

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