1
$\begingroup$

I have a problem in a solids course about mohrs circle and its principal forces. I have solved to its last part and it all checks up when putting the right angle theta which makes the shear stresses zero(61.87) but something weird happens when i pull arctan. Do come to that answer numerically. EX

For the last part i have that $\tan(2\theta) = \dfrac{20 \cdot 2}{60 - 86.6}$.

When you put in the known angle for theta they are practically the same answers. Next step is naturally to do arctan to get to theta.

$2\theta = \arctan\left(\dfrac{20\cdot2}{60-86.717}\right)$ here is the answer supposed to be $123.7$ for the right hand equation instead i get $-56$. What is going wrong? ¨ I have checked with degrees and radian.\ Theta will be 61.87. i know that angle from approximation and website http://www.jnovy.com/jnovy/calcs/MohrsCircle2d/mohrsCircle2d.html where sigmax is 60 sigmay is 86.717 and tau is 20

$\endgroup$
4
  • 1
    $\begingroup$ $\tan$ is a periodic function with period 180, i.e. $\tan \alpha = \tan (\alpha+180)$. So when you take the inverse, the answer is only accurate up to a multiple of 180. The calculator does not know which period of the function you want, and probably only gives answers in the range $(-90,90]$. $\endgroup$
    – Jaap Scherphuis
    Commented Sep 22, 2023 at 14:35
  • 1
    $\begingroup$ The range of angles that arctan returns is from -90 to 90 degrees. In this case you have to add 180 degrees to the angle returned by arctan to get the correct angle. $\endgroup$
    – Quadrics
    Commented Sep 22, 2023 at 14:36
  • $\begingroup$ Do you know if your calculator supports the two-argument $\arctan$? It also gives the expected result. $\endgroup$
    – user170231
    Commented Sep 22, 2023 at 15:18
  • $\begingroup$ The reason that the one-argument arctan function has a range of $~-90^\circ < \theta < 90^\circ~$ is that the arctan function is a function. This means that given any real number $~r,~$ the expression arctan$(r)~$ must return exactly one specific value. Note that (for example) $~\tan(-56^\circ) = \tan(124^\circ).$ $\endgroup$
    – user2661923
    Commented Sep 22, 2023 at 15:22

1 Answer 1

Reset to default
0
$\begingroup$

Calculators are programmed to return inverse trig functions only in a given range:

For $\arcsin x= \theta$, $[-90 \leq \theta \leq 90]$ (Right half of the unit circle. $\sin \theta = \sin(180 - \theta)$)

For $\arccos x = \theta$, $[-180 \leq \theta \leq 180]$ (The top half of the unit circle. $\cos \theta = \cos (-\theta)$)

For $\arctan x = \theta$, $[-90 \leq \theta \leq 90]$ (The right half of the unit circle. $\tan \theta = \tan (\theta + 180)$)

If you got $\arctan x = -56$, the other angles are:

  1. $-56 + 180 = 124$
  2. $-56 + 360k$, where $k$ is an integer
  3. $124 + 360k$, where $k$ is an integer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .