Stress Tensor - simple problem (I think)

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SUMMARY

The discussion focuses on calculating the stress force components at point Q using the given stress tensor pij and the normal vector n = (2,1,-1). The stress force f is computed as f = (1, -1, -2) through the product of the stress tensor and the normal vector. The participants confirm that the normal stress is derived from the projection of f onto n, while the shear stress is obtained by subtracting this projection from f. The conversation emphasizes the importance of understanding vector projections for efficient computation of these stress components.

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rarkup
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Homework Statement



Stress tensor at a point Q in a body has components:

pij:
| 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.

Homework Equations



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.
 
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Welcome to PF!

Hi rarkup! Welcome to PF! :smile:
rarkup said:
(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 )

Yup, that's fine! :biggrin:
(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.

Just find the component of ( 1, -1, -2) in the direction of ( 2, 1, -1) in the usual way (and then subtract that from the whole stress to get the shear stress) …

this has nothing to do with the tensor any more. :wink:
 
Thanks for the welcome and your response tiny-tim.

Now it's time to dust off the Algebra books and relearn how to compute components of a vector in the direction of another.
 

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