How is shear stress different from normal stress?

Question

I feel like I have not understood the difference between these two concepts as well as I thought. My prior understanding of shear stress was that it is stress generated after force is applied parallel (or coplanar) to the surface or cross section of an object, while normal stress is generated after force is applied perpendicular to the surface or cross-section of object.

Now with this definition, I just realized that there is a hole in these two "distinct" definitions. If we take the foregoing definition of shear stress to be correct, we can say that an object of certain volume has innumerable cross-sections (whether it is vertical, diagonal or horizontal part of the object). No matter where or what position I am exerting force on an object, I am SIMULTANEOUSLY exerting it parallel or coplanar to one of the surface/cross section of an object. With this reasoning in mind, then wouldn't shear stress and normal stress go together all the time.

For example, imagine a cube box that has a certain 3D volume. This cube is made of infinite horizontal, vertical surface/ layers (like layers of solid molecules arranged in a lattice) in the xyz plane that come together to give this cube box a macroscopic length, width and height.

Now if I use my hand or finger to push the cube box at top ( horizontal) surface/outer perimeter downward, I see that I am using a normal force to those specific infinite horizontal surface/layer making up the length aspect of the object, because it is force exerted 90 degree to them. But when I do that, am I not exerting shear force at the same time to the specific vertical surface/layer ( out of again the infinite vertical layers making up the cube) that up the vertical aspect of the cube box, because ,to those specific layer, the same normal force applied to the infinite horizontal surface layers is appearing as shear force because the force is parallel to those vertical surface/layers?

If someone can clarify on these concepts, that would be much appreciated. Thanks.

Answer 1

Normal stresses cause changes in a differential element's side lengths, whereas shear stresses cause changes in the element's corner angles:

In other words, the differential element contains no spatial extent that would allow us to talk about internal stress variation.

However, in finite-sized real objects, the orientation of a differential element is arbitrary, and an axial load can absolutely induce shear:

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(images from my website)

The interplay between normal and shear stresses in real objects can be addressed through coordinate system transformations. An intuitive graphical approach is Mohr's circle, from which we obtain certain interesting and useful implications (for instance, pure uniaxial or biaxial tension can produce internal shear, but pure shear can't produce tension—at least not for isotropic materials).

EDIT: Thank you for clarifying. Yes, a uniaxial normal force produces shear within a material. This is what the shearing diamond on the rubber band is meant to show. Mohr's circle can be used to show that a uniaxial normal stress of σ produces a shear stress of σ/2 within the material. However, that shear stress is oriented not perpendicularly to the surface but at an angle of 45°.

In addition, you mention pushing on the surface with your finger. Again, yes, that localized load would produce both normal compressive stress under your finger and shear stress around the edge of your finger where the load drops to zero:

This is a more complex stress state that has been investigated theoretically and experimentally in the context of soil mechanics near heavy buildings and nanoindentation, for example:

The stress state around a point load on a semi-infinite medium was first studied in detail by Boussinesq in Applications des Potentiels à l’Étude de l’Équilibre et du Mouvement des Solides Élastiques (Paris: Gauthier-Villars, 1885). See equations 8-2e-g here, for example, which express the shear stresses.

EDIT 2: I saw that you were interested specifically in vertical shear directly below your finger. Directly below your (assumed-to-be-symmetric) finger, there is no vertical shear. There are two ways to see this: first, $\tau_{rz}$ in Eq. 8-2e in the link is zero for $r=0$, corresponding to the vertical axis under the contact point. But an equation alone may not be satisfying. Here's a graphical interpretation or symmetry argument: As shown in the first image above, a shear stress state requires four forces to be applied to a differential element—any fewer and the element would accelerate away. So any downward shear force we might posit must include a leftward or rightward force on the top of the differential element. But how would Nature choose a leftward or rightward direction if everything is rotationally symmetric? The only conclusion is that the magnitude of this shear stress in this orientation must be zero directly below your finger. (However, as I indicated above and as predicted by Mohr's circle, there is a shear stress oriented at 45° vertically that's induced by the normal compressive stress.)

Answer 2

It you consider any tiny imaginary surface oriented in an arbitrary direction within a deforming fluid, the contact force per unit surface area that the molecules on one side of the surface exert on the molecules on the other side of the surface can be resolved into two components: one in the direction normal to the surface and the other in a direction tangent to the surface. The one normal to the surface is a tension or compression stress, and the one tangent to the surface is a shear stress. How these stresses are related to how the fluid is deforming (i.e., the velocity gradients within the fluid) is laid out in detail in Chapter 1 of Transport Phenomena by Bird, Stewart, and Lightfoot.

Answer 3

My prior understanding of shear stress was that it is stress generated after force is applied parallel (or coplanar) to the surface or cross section of an object, while tension stress is generated after a pulling force is applied to an object and compression stress is generated after a push/squishing force is applied to the object.

That is essentially correct. Shear stress at a cross section of an object equals the force applied parallel to the section (shear force) divided by the cross sectional area. Tensile or compressive stress at a cross section of an object equals the force applied perpendicular to the surface (normal force) divided by the cross sectional area.

To help visualize the difference, consider the difference between pulling a rope apart, which is an example of a tensile stress failure, and cutting the rope with a pair of scissors (or "shears"), which is an example of a shear stress failure.

Hope this helps.

Answer 4

My prior understanding of shear stress was that it is stress generated after force is applied parallel (or coplanar) to the surface or cross section of an object, while normal stress is generated after force is applied perpendicular to the surface or cross-section of object.

The forces applied to the surfaces of the object, as well restriction to displacement (like normal forces and/or friction) are better understood as boundary conditions for the elastic field inside the solid.

Comparing the position $(x,y,z)$ of the point inside the object under load, and the same position unloaded, there is a field of displacements $u(x,y,z)$ corresponding to the difference of location of each point.

For homogeneous and isotropic objects, and small displacements, it is intuitive to think of small cubes of any orientation around each point. Depending on how this cubes are deformed, we can say that for that point and orientation all 3 pairs of faces have only normal stresses or also shear stresses.

There is only normal stresses if the cube preserve 90 degrees of all angles, and only the length of the edges change. Shear stresses are a measure of the angular distortion of the cube.

Answer 5

See below the Figure 31-9 from Feynman Volume 2. The stress tensor describes the forces acting on the faces of a small cube of material. By squashing the cube between your thumb and finger in the center of the cube's faces you are exerting forces $-S_{yy}$ on the top and bottom surfaces, but you are not causing any shear forces $S_{yx}$ on the left and right faces (unless they were glued to two vertical walls). You could move your thumb to be above the right face and your finger to above the he left face, but then the two $S_{yx}$ forces would cause the cube to rotate, until you used the thumb and finger of your other hand to exert $S_{xy}=S_{yx}$ forces to stop the rotation. Now you are exerting equal shear forces on the vertical and horizontal surfaces (ie: the stress tensor is symmetric so the little cube doesn't rotate), and you have strained the cube to become a parallelepiped.