Let's consider the following figure image of stick to be broken

The grey box contains a blue stick which is fixed. The blue stick has a length of $ a+b+c $ and two diameters $ f,h $. The diameter $ h $ describes the part $ b $ of the stick. The stick is fixed in the plane but the plane is not connected to the grey box. A force $ F $ is pushing against the withe plane like in the picture. How much force is needed to break off the stick in part $ b $?

  • 2
    $\begingroup$ How much effort have you applied to try to obtain a proposed solution? $\endgroup$ Commented Feb 26, 2023 at 20:03
  • $\begingroup$ I don't have an idea how I could solve this because I had never a mechanical problem with a notch. What I also can say is that I see two different ways how this plane could move: One way would be striaght downward if the force is close to the notch or the plane is rotated if the force comes from the outer part of the plane. $\endgroup$
    – hallo007
    Commented Feb 26, 2023 at 20:26
  • $\begingroup$ I just have some knowledge about bending sticks and not about stuff like in my picture. $\endgroup$
    – hallo007
    Commented Feb 26, 2023 at 20:43
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    $\begingroup$ Apply your knowledge about bending sticks to try to solve the problem. We should like to see how far that takes you. $\endgroup$ Commented Feb 26, 2023 at 20:57
  • $\begingroup$ Does the white plane slide against the grey plane or does it tilt ie pivot at the lower left corner? $\endgroup$
    – Solar Mike
    Commented Feb 26, 2023 at 21:18

1 Answer 1


We assume the distance from F to the hinge to be

$$ X_F=a+b+c+d/2$$ We calculate the equivalent I of the cantilever beam, with the parallel axis. When it bends it will rotate about a point at the lower corner of the gray support, call it point A. Let's annotate the thickness of the bar, B.

$$I_{Beam} =I_{stick}+ A_{stick}*Y^2_{stick}$$ $$I_{stick}= bh^3/12$$ $$I_{Beam}=bh^3/12+bh(e+f/2)^2$$

we assume the stick will break at yield stress and ignore 2nd hardening, or if we have it we plug it.

$$\sigma_y=\frac{MC}{I_{Beam}}=\frac{(F*x)(e+f/2)}{bh^3/12+bh(e+f/2)^2}$$ $$F*X=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{e+f/2}$$

$$F=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{(e*f/2)*(a+b+c+d/2)}$$


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