Take the simplest example of some uniform bar being stressed by a force N, I understand that we need to adjust an angle to determine the stress in terms of shear and principal stresses on that plane, but what are the physical reasons for this? Is it that at a point there will be different internal forces between particles that are positioned around that point depending on their relative position to the point?

Why is shearing experienced at an angle to an axial force? What causes there to be a shearing effect on a plane that is at an angle to the direction of N?


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What helped me understand this was drawing a square on the side of the bar and then watch what happens when you simply stretch the bar in axial direction.

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The physical reason is for this is a force equilibrium. Each square can in general experience normal and shear stresses on its edges. Combination of the normal and shear stress on the edge has to result in a force in axial direction in case of the stretched bar.

  • $\begingroup$ Is it better then to see plane stress as what happens to a stress element at different angles then to try and see it as a different plane in the actual object, i.e. a plane of atoms or a 1-d plane of mass. $\endgroup$ Mar 3, 2023 at 9:26
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    $\begingroup$ The important thing is different orientation of a plane on which you want to calculate the stresses. That is why in general you need 6 components to fully describe stress state at a point; either there are 3 normal and 3 shear components in global coordinate system or there are just 3 normal (principal stresses) which require specific orientation defined by 3 angles. Even in case of the plane, 3 numbers would be required for plane orientation (lets say $z$ is in normal direction) and then you can have 1 normal stress $\sigma_z$ and 2 shear stresses $\tau_x$ and $\tau_y$, so again, 6 in total. $\endgroup$ Mar 3, 2023 at 17:12

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