addition of shearing forces and stresses I would like to ask for help regarding the summation of non parallel shearing forces acting on an elastic body. In the image force F (Left to right direction,in yellow) acts on the top horizontal surface "A" of an elastic cuboid body that is attached to the table only at its bottom surface. I know how to calculate the stress and strain. But what happen if a second force "F_vertical" (in black) acts on the top portion of the vertical surface facing the reader, again direction left to right. how do I add the stress and strain imposed by both forces. (I am a physician not an an engineer or physicist ).

thank you Yehuda Raveh

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    $\begingroup$ If the deformations are small you can just use linear superposition, i.e., compute the stresses for each loading condition separately and add them up. $\endgroup$ – Biswajit Banerjee Dec 11 '18 at 2:17
  • $\begingroup$ Yes. deformations are very small. I do know that if force F acted alone, the shear stress would be F/A and strain would be delta-X /L. I don't know how to calculate stress and strain for F_vertical alone. thank you. Yehuda $\endgroup$ – Yehuda Raveh Dec 13 '18 at 11:59
  • $\begingroup$ For a shear traction $t_v = F_{v}/A_v$, you will have to solve a boundary value problem. That's probably hard to do by hand and a computational tool such as finite elements will be easier. The stresses and strains in the block will not be homogeneous. However, if you just want a rough estimate, just assume that the force $F_v$ acts over the entire area and do the calculation as you would for $F$ alone. $\endgroup$ – Biswajit Banerjee Dec 14 '18 at 20:24

There's some ambiguity in your diagram and description because you have one force acting on a point and another force acting over a portion of a surface. Also, although you say, "I know how to calculate the stress and strain [for the first loading configuration]," it's not clear which idealizations you're applying to get these answers.

For example, one possible idealization is that the force $F$ acts over the entire top surface, thus producing a state of pure shear throughout the body. (An equal and opposite force would act over the bottom surface due to its attachment to the table.) Assuming that the $z$-axis points in the upward vertical direction, the original state of stress would be $$\sigma = \begin{pmatrix} 0 & 0 & F/A_\mathrm{top} \\ 0 & 0 & 0 \\ F/A_\mathrm{top} & 0 & 0 \end{pmatrix},$$

where $A_\mathrm{top}$ is the cross-sectional area of the top surface.

(If you don't apply this idealization of $F$ acting across the entire top, then you'll get a stress variation around the point where $F$ is drawn as acting. Think about pushing at a single location on a very compliant material such as a sponge; most likely, you'll see a lot more deformation around the point your finger touches than elsewhere.)

Inside the material, any small element would be subjected to the following stress state:

enter image description here

These four forces are all necessary to apply a state of pure shear upon an internal element; any fewer would result in translation and/or rotation. This stress state becomes poorly defined at the surface because you're not applying any upward or downward forces; the material is effectively generating these forces internally.

So that's just the simple case of a single force! The revised state of stress is more complex. Just slightly inside the material where the force $F_\mathrm{vertical}$ is applied, the revised stress state is

$$\sigma^\prime = \begin{pmatrix} 0 & F_\mathrm{vertical}/A_\mathrm{side} & F/A_\mathrm{top} \\ F_\mathrm{vertical}/A_\mathrm{side} & 0 & 0 \\ F/A_\mathrm{top} & 0 & 0 \end{pmatrix},$$

where $A_\mathrm{side}$ includes only the area upon which $F_\mathrm{vertical}$ acts.

Elsewhere in the block, however, the stress will vary. The force on the bottom, which originally needed to resist only $F$, now must resist a torque generated from the off-center application of $F_\mathrm{vertical}$. I don't know of a solution for the general state of stress; you may need to use finite element analysis for that.

  • $\begingroup$ sorry for the ambiguity. Indeed force F applies to all top surface "A". I do know that if force F acted alone, the shear stress would be F/A and strain would be delta-X /L. I am trying to learn the matrices you wrote. If there is simpler way to express the sum of the two shearing forces, please let me know. $\endgroup$ – Yehuda Raveh Dec 13 '18 at 11:50
  • $\begingroup$ There is the von Mises stress, which is a coordinate-system-invariant number that expresses the stress state. For a couple of shear stresses, the von Mises stress would be $\sigma_\mathrm{vM}=\sqrt{3(\sigma_{12}^2+\sigma_{13}^2)}$. But again, note that this will vary throughout the material because $F_\mathrm{vertical}$ is asymmetrically applied. $\endgroup$ – Chemomechanics Dec 13 '18 at 15:54

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