Stress tensor at a point Q in a body has components:
| 1 -1 0 |
|-1 2 1 |
| 0 1 3 |
(i) Calculate components of the stress force f across a small area of surface at Q normal to n = (2,1,-1).
(ii) The component of f in the direction of n is called the normal stress while the component of f tangential to the surface is called the shear stress. Find the normal stress and the shear stress at Q.
f = fi = pijnj
The Attempt at a Solution
(i) I take the stress force at Q across the surface normal to n to be the product of the stress tensor pij and the normal vector n and arrive at:
f = (1x2 + -1x1 + 0x-1)i + (-1x2 + 2x1 + 1x-1)j + (0x2 + 1x1 +3x-1)k
= ( 1, -1, -2 )
(ii) I think this is a simple algebraic exercise - ie what components of ( 1, -1, -2) project onto n to give the normal stress, and which components project onto the plane normal to n to give the shear stress.
Am I reading this correctly, and what is the most efficient way of computing these components? I can't help but think the answer might be staring at me from within the tensor pij.