Stress tensor and traction vector

Question

Hey all. I have this problem with the above figure. The lengths of the three sides are $AB=a, AG=b, GB=c$. Also, P,Q,R are forces perpendicular to each side as you can see and P,Q,R are their magnitudes.. Considering the stress field homogeneous and that $t$ is the thickness, I need to identify the stress tensor.

What I've done is to take the normal unit vectors of sides AG and AB, which are $$\mathbfη_{AG} = (-1,0), \mathbfη_{AB}=(0,-1)$$ We know that $$\boldsymbol{t} = σ\cdot \mathbf{n} (1)$$ where $\mathbfσ$ is the stress tensor and $\mathbf n$ is the unit vector normal to the side.

In AG side we have that $\mathbfσ_{ΑG}=\mathbf{R}/A=-{R}/bt \mathbf i$ and in AB we have that $\mathbfσ_{ΑB}=\mathbf{Q}/A=-{Q}/at \mathbf j$. By $\mathbf σ$ here I mean the traction vector.

So from (1) we have that, for $η_{AG}$ $$\boldsymbol{t}=\sigma\boldsymbol{n} = \left(\begin{array}{ccc} σ_{xx} & σ_{xy}\\ σ_{yx} & σ_{yy} \end{array}\right) \left(\begin{array}{c} -1\\ 0\\\end{array}\right)= \left(\begin{array}{c} -R/bt\\ 0\end{array}\right).$$

Which shows us that $σ_{xx}=R/bt$ and that $σ_{yx}=0$. Similarly for the other unit vector we get that $σ_{yy}=Q/at$ and $σ_{xy}=0$.

My question is : The solution of this problem that I have, doesn't take the $σ_{xy}$ as 0. It proceeds and finds it as non zero using the same technique with the 3rd side of our triangle. Can you explain to me why it's not $0$ even though we just found that $σ_{xy}=σ_{yx}=0$?

Sorry for any mistakes, English is not my main language. In the figure, whereever you see $Γ$ it's $G$, $Δ$ is $D$. Thanks in advance.

EDIT : I'm gonna add what the solution does in order to get the value for $σ_{xy}$. The normal unit vector of GB is $$\mathbfη_{GB}=(sinθ,cosθ)=(b/c , a/c)$$ where $θ$ is the angle between the sides AB and BG. The traction vector of BG side can be written as $$\mathbfσ_{BG}=(Pb/c^2 t)\mathbf i + (Pa/c^2 t)\mathbf j$$ Now we do the following one more time $$\boldsymbol{t} = σ\cdot \mathbf{n}$$ and we get $$ \left(\begin{array}{ccc} R/bt & σ_{xy}\\ σ_{yx} & Q/at \end{array}\right) \left(\begin{array}{c} b/c\\ a/c\\\end{array}\right)= \left(\begin{array}{c} Pb/c^2 t\\ Pa/c^2 t\end{array}\right)$$

From here, if you continue you'll get that $$σ_{xy} = \frac{P}{ct}-\frac{R+Q}{t(a+b)}$$ Is this wrong somehow?

Answer 1

I get $\sigma_{xx}=R/(lt\cos{\omega})$and $\sigma_{yy}=Q/(lt\sin{\omega})$, where l is the length of GB. I also get $P\cos{\omega}=R$ and $P\sin{\omega}=Q$. From this, it immediately follows that $$\sigma_{xx}=\sigma_{yy}=\frac{P}{lt}$$ That is, the state of stress in the material is isotropic. Of course, we should immediately have realized this because the only way the traction on the arbitrary side of the prism could be orthogonal to the side, while, at the same time, the tractions on the faces of constant x and y are normal to those faces, is if the stress tensor is isotropic.