# Confused about shear elasticity and complementary shear stress

I am a self learner of continuum mechanic. I am confused about simple shear stress in situation similar to figure 1, in case $F_\textrm{ext}$ is caused by external perturbation by i.e., human, what are the forces that involve in this scenario? In my mind, there are

1. $F_\textrm{ext}$ is caused by external perturbation

2. $F_\textrm{couple}$ which $F_\textrm{couple}=-F_\textrm{ext}$

3. $F_\textrm{com}$ which are the complementary shear stress multiply with area

4. $F_\textrm{comneg}$ which $F_\textrm{comneg}=-F_\textrm{com}$ however, these three forces could result in figure 2, not figure 1. Therefore, I think there should be constrained forces i.e., from human hand, surrounding material or floor that eliminate the force component in z axis by reaction forces.

5. $F_\textrm{reac}$ which equal to Force components in z axis, $F_\textrm{reac} =-F_\textrm{com}$

Is that right?

in the case of dynamic deformation,it will not necessary that $F_\textrm{couple}$ equal to $-F_\textrm{ext}$ and $F_\textrm{comneg}$ equal to $F_\textrm{com}$. $F_\textrm{reac}$ equal to force components in z axis (in case $F_\textrm{reac}$ is caused from surrounding continuum material) right?

I understand is, in this situation, the object's stiffness reacts to force in x-direction only ($F_\textrm{ext}$ and $F_\textrm{couple}$), not the force in z-direction ($F_\textrm{com}$ and $F_\textrm{comneg}$). Therefore $F_\textrm{reac} =-F_\textrm{com}$. Is this right?

• $F_{com}$ and $F_{comneg}$ aren't exist because of free edges. – lucas May 7 '16 at 18:55
• Ok, I get it now. Anyway, does it have Freac normal to surface AB and CD to stop the object from clockwise rotation? – user3188389 May 8 '16 at 9:47

## 2 Answers

Think of it in this way: If there exists a couple on the center of mass by exerting shear stresses on AB and CD, then shouldn't the block rotate? However, we know from experiment that the block will tend to deform (by a very small amount) as shown in Figure A, so there exists no rotation, which means the system is in rotational equilibrium.

The couple created by the shear stresses is opposed by the couple created by the complementary shear stresses, which prevents turning of the object. The object will not deform to the shape as shown in Figure 2 because $F_{com} = F_{comneg}$ and is actually a reaction couple to the couple produced by the shear stresses, thus causing deformation in only one direction, which is the direction in which the shear stress was applied.

• I still didn't understand why it is not turned out similar to figure 2. If imagine the object as finite group of masses, why isn't a mass at point A goes righty and downward (since the force vectors go that way). Vectors at Point D should move mass at point D toward A. Masses at point B and Point C should move away from each other. I don't think figure 2 is the result of rotation. – user3188389 May 8 '16 at 9:40

In the small deformation limit figure 1 and 2 are equivalent. Figure 1 is only an intuition driven squematic and should not be used to define the shear stress (or strain) since this is not a pure shear stress state (normal stress is present). In order to properly define the shear stress you should consider a differential surface with shear stresses on its faces. This picture, together with the conservation of linear and angular momentum gives as a result a shear stress of equal magnitude in all faces of the differential element.