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Determine two values of $\sigma_{22}$ for which the maximum shear stress is 80 MPa

i know that

\begin{equation} \sigma_{ij} = [\sigma]^\top = \begin{bmatrix} \sigma_{11} & \tau_{21} & \tau_{31} \\ \tau_{12} & \sigma_{22} & \tau_{32} \\ \tau_{13} & \tau_{23} & \sigma_{33} \end{bmatrix} = \begin{bmatrix} 90 & 0 & 60 \\ 0 & \sigma_{yy} & 0 \\ 60 & 0 & 0 \end{bmatrix} \end{equation}

and i also know that $\tau_{max}=(\sigma_{I}-\sigma_{III})/2$ <=> $80 = (\sigma_{I}-\sigma_{III})/2$

but in the solutions appears that

Case 1

\begin{equation} \begin{aligned} \sigma_\mathrm{I} &= 120~\mbox{MPa} \\ \sigma_\mathrm{II} &= -30~\mbox{MPa} \\ \sigma_\mathrm{III} &=~? \\ \end{aligned} \end{equation}

\begin{equation} \begin{aligned} \tau_\mathrm{max} &= \frac{1}{2}(\sigma_\mathrm{I}-\sigma_\mathrm{III}) = 80~\mbox{MPa} \Leftrightarrow \sigma_\mathrm{III} = \sigma_\mathrm{I} - 2\tau_\mathrm{max} = -40~\mbox{MPa} \\ \end{aligned} \end{equation}

Case 2

\begin{equation} \begin{aligned} \sigma_\mathrm{I} &= ?~\mbox{MPa} \\ \sigma_\mathrm{II} &= 120~\mbox{MPa} \\ \sigma_\mathrm{III} &= -30~\mbox{MPa} \\ \end{aligned} \end{equation}

\begin{equation} \begin{aligned} \tau_\mathrm{max} &= \frac{1}{2}(\sigma_\mathrm{I}-\sigma_\mathrm{III}) = 80~\mbox{MPa} \Leftrightarrow \sigma_\mathrm{I} = \sigma_\mathrm{III} + 2\tau_\mathrm{max} = 130~\mbox{MPa} \\ \end{aligned} \end{equation}

\begin{equation} \therefore \sigma_{yy} = 130~\mbox{MPa} \vee \sigma_{yy} = -40~\mbox{MPa} \end{equation}

Is the solution right? If its right could someone explain me why is this true? \begin{equation} \begin{aligned} \sigma_\mathrm{I} &= 120~\mbox{MPa} \\ \sigma_\mathrm{II} &= -30~\mbox{MPa} \\ \end{aligned} \end{equation}

peterh
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  • No, you are not allowed to do that. Do not vandalize your own posts. – peterh Jan 14 '21 at 21:35
  • Either you stop, or the mods will suspend you and so you will be stopped. Your post will be rollbacked to its last useful version in both cases. – peterh Jan 14 '21 at 21:37
  • You posted your homework, your teacher found it, and now you have a problem? – peterh Jan 14 '21 at 21:38
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    You can not, you gave it to the community by a CC-BY-SA license. But you can delete your account. Anyways, I think you could profit a lot by keeping these rules, at least if you plan to work and live in well-educated environments. – peterh Jan 14 '21 at 21:42
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    Your questions will be rollbacked even after your account deletion. You have no way to delete them. – peterh Jan 14 '21 at 21:43
  • Btw, playing a cooperative game here would help you a lot to learn. Already you could profit a lot by learning to communicate with educated, literate people. – peterh Jan 14 '21 at 21:44

1 Answers1

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The $x_1x_3$ plane is normal to a principal stress.

That plane has: $\sigma_1=90$,$\sigma_3=0$ and $\tau_{13}=60$. That means that the principal stresses on the $x_1x_3$ are:

$$\sigma_{P1,P2} = \frac{\sigma_1+\sigma_3}{2} \pm \sqrt{\left(\frac{\sigma_1-\sigma_3}{2}\right)^2 + \tau_{13}^2}$$ $$\sigma_{P1,P2} = \frac{90+0}{2} \pm \sqrt{\left(\frac{90-0}{2}\right)^2+ 60^2}$$ $$\sigma_{P1,P2} = 45 \pm 75$$

$$\begin{cases}\sigma_{P1} = 45 + 75 = 120\\\sigma_{P2} = 45 - 75=-30\end{cases}$$

Now, you know that there are two principal axes on plane $x_1x_3$ and you know their values are (120, -30). The third principal stress is $\sigma_{yy}$ but you don't know its magnitude. The third principal stress cannot be between the other two (120, -30), because then $\tau_{max}$ would not involve $\sigma_{yy}$.

Additionally, for two principal stress that I already know ($120,-30$) the maximum shear stress would be $\frac{120-(-30)}{2}=75[MPa]$. Therefore, the constraint that $\tau_{max}\ge 80[MPa]$, will need to be calculated with the use of $\sigma_{yy}$ and one of the other two principal stresses.

So if you order the principal stresses $\sigma_{yy}$ will have to be either

  1. greater than 120[MPa]: in that case the $\tau_{max}$ will be determined by $\sigma_{yy}$ and -30[MPa]
  2. less than 30[MPa]: in that case the $\tau_{max}$ will be determined by $\sigma_{yy}$ and 120[MPa]

Simplistic analogy

I'm writing this simplistic analogy, because of the comments below.

Think that you have a problem where you need to create a set of three numbers which the difference between the largest and the smallest is 16.

You also know that two numbers are 12 and -3.

What numbers would create a set that satisfies that?

  • one set of numbers is [12, -3 , -4], so the unknown that is added to the other known two is -4
  • the other set of numbers is [13, 12, -3 ] so the unknown that is added to the other known two is 13
NMech
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  • Yes, you are right. I mixed the axes. I was just getting ready to change it, although it won't change anything for $\sigma_{P1}$ and $\sigma_{P2}$ . I was also going to reply and explain why two cases. – NMech Dec 07 '20 at 14:44
  • Have a look now to check if its any more clear now. – NMech Dec 07 '20 at 14:59
  • I rewrote a part, but without clearly understanding where you stop following the reasoning, I am afraid it would be futile to rewrite it any further. – NMech Dec 07 '20 at 15:34
  • Principal stresses are by convention ordered by magnitude. Therefore in the first case, $\sigma_{III} = \sigma_{yy}$. In the second case .$\sigma_{I} = \sigma_{yy}$. That is why I use $\sigma_{P1}, \sigma_{P2}$ to avoid the confusion with I, II, II. – NMech Dec 07 '20 at 15:49
  • they are not different. They are reordered, because the first principal stress $\sigma_{I}$ by definition has the greatest magnitude. In the first case $\sigma_{P1}=120MPa$ is the stress with the greatest magnitude. In the second case, the same stress $\sigma_{P1}=120MPa$ is the second largest. – NMech Dec 07 '20 at 16:04