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Suppose a solid, right-circular, cylindrical body has a long axis of length $L$ and end-faces each having a diameter $D$ was placed fully submerged into a fluid having a hydrostatic pressure of $p$, as shown in the figures below.

enter image description here

enter image description here

If one wanted to know the stress distribution within the cylinder's body one might bisect the cylinder in two ways as shown in the figure. One way would be to bisect along the cylinder's long axis, denoted as Surface 1, where the surface $A_1$ is equal to $LD$. Another way would be to bisect the cylinder diametrically, denoted as Surface 2 in the figure, where its surface area $A_2$ is equal to $\frac{\pi D^2}{4}$.

At first thought, if the confining fluid pressure acting on the cylinder's outer surfaces is hydrostatic, then I would think the resulting stress induced within the cylinder's body would be isostatic. However, would this be true for any length $L$ to end-face diameter $D$ relationship? How might one show the resulting axial (along the cylinder's long axis) and lateral (or radial-direction) internal stress using a free body diagram?


My attempt at this is as follows. For Surface 1 I can imagine the resulting internal forces having to counter act the normal forces $F_n$ acting on the plane caused by the external pressure. I might show this using the figure below.

enter image description here

Assuming some infinitesimal area for the external pressure to act on, I would think I could find the force acting on Surface 1 by first integrating the pressure force acting on the quarter-circle and then multiplying by 2 and then by the length of the cylinder $L$, i.e.,

$$F_n=\int_0^{\frac{\pi}{2}}F_R \cos\theta ~\mathrm d\theta$$

evaluating the integral,

$$F_n = F_R\left[\sin\left(\frac{\pi}{2}\right)-\sin(0)\right]=F_R$$

multiplying by 2 and by $L$,

$$F_n = 2F_RL$$

Since pressure is force over area, $F_R$ is equated to pressure in the following manner

$$F_R=pA=p\left(\frac{\pi D}{2}\right)L$$

where $\left(\frac{\pi D}{2}\right)$ is the arc-length of the half-circle. Substituting, we solve for the stress acting on surface 1,

$$\sigma_1=\frac{F_{n_1}}{A_1}=\frac{2p(\frac{\pi D}{2})L^2}{DL}=p\pi L$$

The stress for surface 2 due to the pressure acting on the end face would be

$$\sigma_2=p$$

Therefore, the cylinder's length to diameter stress relationship might be said to be equal to

$$\frac{p\pi L}{p}=\pi L$$

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  • $\begingroup$ "Check my work" is not an acceptable question on this site. Do you have some reason for doubting the result? A conceptual difficulty? Otherwise, we expect you to make use of available resources, such as classmates or a teacher, to "check your work". $\endgroup$ – sammy gerbil Nov 5 '16 at 16:16
  • $\begingroup$ @sammygerbil you are right. I doubt the result because I doubt my mathematical and problem solving abilities. I would go to a teacher or classmates if I was in school, but, I am not in school. How might I rephrase my question? $\endgroup$ – Armadillo Nov 5 '16 at 16:29
  • $\begingroup$ Also, I do not know how the tag "homework" got added to my question. $\endgroup$ – Armadillo Nov 5 '16 at 16:31
  • $\begingroup$ I added the tag is "homework-and-EXERCISES". The latter includes calculations of a desired result, such as here, especially where the question is "check my work". ... Your paragraph "At first thought ..." looks like the start of a conceptual doubt, but you don't give any reason for it. Your intuition tells you the internal stress is isostatic, but you don't seem to have a reason for doubting this - eg some contrary result or principle. $\endgroup$ – sammy gerbil Nov 5 '16 at 16:40
  • $\begingroup$ @sammygerbil Well, I showed my work and found that the stress values are not the same. If my work is right, I do not know. I will update my question to explicitly state why I doubt the isostatic condition. From that point forward what more can I do to improve this question? If this is still an unacceptable question I will delete it. Just let me know $\endgroup$ – Armadillo Nov 5 '16 at 16:50
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Based on @Sammygerbil's suggestion, I continued to work on this problem, checking my calculation. With more internet & resource searching I found the correspondence and answer by user @Jared (here) to be most helpful. Following his answer, we can determine the amount of force acting on Surface 1 due to the confining pressure. Using the cartoon below we can more easily visualize the problem.

enter image description here

This cartoon attempts to show the cylinder in a perspective view, illustrating the cylinder's end face, the differential area on which the confining pressure acts at the cylinder's lateral surface, and the angles and component force vectors as a result of the applied pressure in relation to the long-axis of the cylinder. Noting the problem's symmetry, and that the problem concerns the static condition, the forces acting vertically downward are equal to the forces acting vertically upward. Therefore, we can analyze half of the cylinder.

Again, following Jared's explanation, we note that the differential force due to the confining pressure is proportional to the differential area on which the pressure acts, i.e.,

$$\tag{1} dF=p*dA$$

Since the fluid's pressure force acts normal to the surface on which it acts and the surface it acts on is curved, there are two directional components of the resulting force in relation to Surface 1: a force parallel to Surface 1, denoted as $F_h$ (h for horizontal), and a force orthogonal to Surface 1, denoted as $F_v$ (v for vertical). The vertical component of the differential force is proportional to the $\sin$ of the angle $(\theta)$ between the centerline axis of Surface 1 and the direction of the differential force. This can be written mathematically as,

$$\tag{2} dF_v(\theta)=p*dA*\sin (\theta)$$

All the vertical force components due to the pressure about the circumference of the cylinder can be summed by integration of the differential areas, from $0$ radians to $\pi$ radians and along the length of the cylinder from $0$ to L. The differential area at each value of $\theta$ is equal to

$$\tag{3} dA=R\cdot d\theta * dx$$

where $R$ is the radius of the cylinder and $dx$ is the differential length along the cylinder's long axis. Substituting equation 3 into equation 2,

$$\tag{4} dF_v(\theta,x)=p*R*\sin (\theta) d\theta * dx$$

we now have an equation of differential force as a function of location on the cylinder's lateral surface. Integrating from $0$ to $\pi$ and from $0 \leq x \leq L$ gives:

$$\tag{5} F_v= p*R \int\limits_0^\pi\sin(\theta)d\theta \int\limits_0^L dx$$

$$\tag{6} F_v= p*R [-\cos(\pi)+\cos(0)] L$$

$$\tag{7} F_v= p*R [-(-1)+(1)] L$$

$$\tag{8} F_v= 2pRL$$

The area of Surface 1 is $DL$ which is equivalent to $2RL$. Therefore, the stress at Surface 1 is

$$\tag{9} \sigma_1= \frac{2pRL}{2RL}=p$$

Thus, the resulting stress induced within the cylinder's body is indeed isostatic and this is true for any length to end-face diameter relationship.

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