Now, at this point, we will use the substitution: $u = \cos{\theta}$. Your situation is the same as his: https://www.miniphysics.com/uy1-calculation-of-moment-of-inertia-of-thin-spherical-shell.html/comment-page-1#comment-217, Hi I think instead of “Recall: Moment of inertia for a hoop: I = r2 dm” This is why I use $ dm = \frac{m}{A} dA $, and not dV. so, dx=-Rsin(theta)d(theta) Have a closer look at the figure. My doubt is that for thin spherical shell if we put the value of E= 0 , I.e. Use the equation for rotational inertia of an object to calculate the rotational inertia of a thin spherical of inertia about an axis tangent to any point on its surface. Engineering Videos The hoop stress equation for thin shells is also approximately valid for spherical vessels, including plant cells and bacteria in which the internal turgor pressure may reach several atmospheres. . For application of the law of gravity inside a uniform spherical shell of mass M, a point is chosen on the axis of a circular strip of mass. Sub. | Contact | Privacy Policy, Home But I think the formula for spherical cap area is usually obtained by integration, so this is some sort of circular reasoning. for a point inside the shell , then we will get V= 0 but … When you use integration by substitution, u = cos theta becomes du = -sin theta d(theta). $R d(\theta)$ is the arc length. I have understood this solution but why don’t you use the same method in finding the moment of inertia of the solid sphere. Disclaimer Notice that the thin spherical shell is made up of nothing more than lots of thin circular hoops. Home University Year 1 Mechanics UY1: Calculation of moment of inertia of a thin spherical shell. Rdtheta is so small that we can consider it perpendicular. so the area of this hoop is just the circumference of a circle x thickness, Here x=Rcos(theta) Japanese gas companies added a touch of cha-racter to giant spherical gas tanks. Consider a thin cylinder of internal diameter d and wall thickness t, subject to internal gauge pressure P. The following stresses are induced in the cylinder-(a) Circumferential tensile stress (or hoop stress) σ … Now, we split the $\sin^{3}{\theta}$ into two, $$I = \frac{M R^{2}}{2} \int\limits_{0}^{\pi} \sin^{2}{\theta} \: \sin{\theta} \: d \theta$$, $$I = \frac{M R^{2}}{2} \int\limits_{0}^{\pi} (1 – \cos^{2}{\theta}) \: \sin{\theta} \: d \theta$$. I’m pretty sure you can handle the simple integration in Equation 7 by yourself. DFM DFA Training Hence, we have to find a way to relate r r with θ θ. . 2πr is area. } A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections. In addition, a case study of internal stresses developed in a soda can will be presented and discussed. 8.3.1.1 Membrane Stresses in Simple Thin Shells of Revolution. t/d < 1/20. TANGENTIAL STRESS, σt (Circumferential Stress) Consider the tank shown being subjected to an internal pressure p. The length of the tank is L and the wall thickness is t. Therefore, Now, in Equation 3, notice that you will have different r r for different hoops. The circumference of the hoop multiplied by the “thickness” of the hoop gives the area of the hoop. You can see the calculations for that of a sphere by following this link: https://www.miniphysics.com/uy1-calculation-of-moment-of-inertia-of.html. The problem is envisioned as dividing an infinitesemally thin spherical shell of … else In this post, we will derive the following formula for the volume of a ball: \begin{equation} V = \frac{4}{3}\pi r^3, 7.3.3. A solid, spherically symmetric body can be modeled as an infinite number of concentric, infinitesimally thin spherical shells. Please help. This is obtained by spinning the ring in the horizontal plane (around the z-axis). dA = length×breadth = circumference×arc length = 2πr×R dθ (3) (3) d A = length × breadth = circumference × arc length = 2 π r × R d θ. Dm = Mean Diameter (Outside diameter - t). Because of the symmetry of the sphere and of the pressure loading, the circumferential (or tangential or hoop) stress t at any location and in any tangential orientation must be the same (and there will be zero shear stresses). 0. In order to continue, we will need to find an expression for $dm$ in Equation 1. ... We can also choose a thin spherical shell element of radius r (such that 0 < r < R) and infinitesimal thickness dr. I don’t quite understand why are we taking the limits from 0 to pi. I checked the calculation several times and it seems to be correct. They are not perpendicular. Hello. Hence, using Equation 4 in Equation 3, $dA$ can be expressed by: $$dA = 2 \pi R^{2} \text{sin} \: \theta \: d \theta \tag{5}$$. very thin cylindrical shell under uniform compression or a very thin spherical shell under uniform external pressure. Downloads document.write('