Question

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex + (1/ 2)ez ? Are there any shear stresses acting on this surface?

Answer 1

Complete Question:

Given
`\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]` at a point. What is the force per unit area at this point acting normal to the surface with
`\b n = (1/ √(2) ) \b e_x + (1/ √(2)) \b e_z` ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area,
`\sigma_n = 28 MPa`

There are shear stresses acting on the surface since
`\tau \\eq 0`

Step-by-step explanation:

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

equation of the normal,
`\b n = (1/ √(2) ) \b e_x + (1/ √(2)) \b e_z`

`\b n = \left[\begin{array}{ccc}(1)/(√(2) )\\0\\(1)/(√(2) )\end{array}\right]`

Traction vector on n,
`T_n = \sigma \b n`

`T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}(1)/(√(2) )\\0\\(1)/(√(2) )\end{array}\right]`

`T_n = \left[\begin{array}{ccc}(23)/(√(2) )\\0\\(27)/(√(33) )\end{array}\right]`

`T_n = (23)/(√(2) ) \b e_x + (27)/(√(2) ) \b e_y + (33)/(√(2) ) \b e_z`

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

`\sigma_n = T_n . \b n`

`\sigma \b n = ((23)/(√(2) ) \b e_x + (27)/(√(2) ) \b e_y + (33)/(√(2) ) \b e_z) . ((1/ √(2) ) \b e_x + 0 \b e_y +(1/ √(2)) \b e_z)\\\\\sigma \b n = 28 MPa`

If the shear stress,
`\tau`, is calculated and it is not equal to zero, this means there are shear stresses.

`\tau = T_n - \sigma_n \b n`

`\tau = [(23)/(√(2) ) \b e_x + (27)/(√(2) ) \b e_y + (33)/(√(2) ) \b e_z] - 28( (1/ √(2) ) \b e_x + (1/ √(2)) \b e_z)\\\\\tau = [(23)/(√(2) ) \b e_x + (27)/(√(2) ) \b e_y + (33)/(√(2) ) \b e_z] - [ (28/ √(2) ) \b e_x + (28/ √(2)) \b e_z]\\\\\tau = (-5)/(√(2) ) \b e_x + (27)/(√(2) ) \b e_y + (5)/(√(2) ) \b e_z`

`\tau = \sqrt{(-5/√(2))^2 + (27/√(2))^2 + (5/√(2))^2} \\\\ \tau = 19.74 MPa`

Since
`\tau \\eq 0`, there are shear stresses acting on the surface.