[Physics] Confused about shear elasticity and complementary shear stress

classical-mechanicscontinuum-mechanicsnewtonian-mechanicsstress-strain

I am a self learner of continuum mechanic. I am confused about simple shear stress in situation similar to figure 1,
enter image description here

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.
    enter image description here
    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?

Best Answer

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.

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