I have a half cylinder which is significantly thick but not very long. In the half cylinder, a force $F_{app}$ is being applied over a certain area as shown in the figure below:

enter image description here

I have to find the hoop stress in the free edges.

I understand how the hoop stress is calculated over cylindrical vessels when there is internal pressure all over the cylindrical surface. Can anyone provide any insight how to start when there is force acting over a partial area. And finally how will the results change when the thickness is significantly large.


1 Answer 1


enter image description here

The sketch above forms the base for solving reactions at "a" and "b". Note that if the loads are acting on the near edge, then there is a twisting need to be considered.

Without twisting, find the reactions that contribute to normal stress at "a" and "b":

$\sum Ma = 0$

$Rb = \dfrac{Fx*x + Fy*y}{D}$, pointing down. "D" is the diameter measured from center to center.

$\sum Fy = 0$

$Fa = Fy - Fb$, assume "positive force" pointing up

  • $\begingroup$ The force is acting all along the longitudinal direction of the cylinder. Also, since the thickness is very small, we can neglect the twisting effect. $\endgroup$
    – s6292_1997
    Nov 12, 2022 at 5:31
  • $\begingroup$ Then you can get the reactions on the holding edges "a" and "b" by equilibrium, can't you? $\endgroup$
    – r13
    Nov 12, 2022 at 13:03
  • $\begingroup$ Right. The vertical component of the force, $F_y$ will give its contribution to the hoop stress $\sigma_h$. How will the horizontal component, $F_x$ help? GIve rise to uniform motion if the system is free? $\endgroup$
    – s6292_1997
    Nov 12, 2022 at 13:57

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