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?

Answer 1

The answer is "19.735 mpa"

Step-by-step explanation:

Given:

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

`\hat{\wedge} = (1)/(√(2))e_x+(1)/(√(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}(1)/(√(2))&0&(1)/(√(2))\\ \end{array}\right]`

`=(1)/(√(2))\left[\begin{array}{ccc}10* 1&+12 * 0&+ 13 * 1\\12* 1&+11* 0&15* 1\\13* 1&+15* 0&+20* 1\end{array}\right] \\\\\\\\=(1)/(√(2))\left[\begin{array}{c}23&27&33\\ \end{array}\right] \\\\`

The area of acting to surface=
`t^(\hat \wedge) \cdot \hat \wedge= ((23)/(√(2))e_n+(27)/(√(2))e_y+(33)/(√(2))e_z)`

`\sigma_(\wedge)=(23)/(2)+(33)/(2)=(56)/(2) =28 \ mpa`

S hearing component of traction =
`t^(\hat n)-\sigma_\wedge\cdot \hat n\\`

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