why dm=m/A *dA i made it dm= (den)*2*PI*r*R*d0 because dm=(den)*dv , dv=2*PI*r*dL , dL=Rd0 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 this case, I am treating the spherical shell as being made up of “infinite” hoops. I have understood this solution but why don’t you use the same method in finding the moment of inertia of the solid sphere. 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 … because what used here is a full size ring, not half. for the dm=msin(theta)d(theta)/2, where did the 2 come from and how does it get into the denominator? Hence, $$r = R \, \text{sin} \: \theta \tag{4}$$. I don’t know it But (adsbygoogle = window.adsbygoogle || []).push({}); The problem is envisioned as dividing an infinitesemally thin spherical shell of density σ per unit area into circular strips of infinitesemal width. If you integrate from 0 to 2 pi, you’ll be double counting as I’ve used a complete hoop in the integration. Engineering Book Store The summation of volumes occupied by all such elements present from radius 0 to R will yield the total volume. The total mass of the shell is M and its radius is R. Figure %: A thin spherical shell. Engineering Calculators 8.3.1.1 Membrane Stresses in Simple Thin Shells of Revolution. This is the condition that we must have to note it to consider a cylindrical or spherical shell as thin cylindrical or spherical shell. 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” I do no understand why does the change in area, dA = R dθ * 2(pi)r, how is the circumference of the hoop(cross section) related to the dA? ], $$\begin{aligned} dA &= \text{length} \times \text{breadth} \\ &= \text{circumference} \times \text{arc length} \\ &= 2 \pi r \times R \: d \theta \end{aligned} \tag{3}$$. { Dm = Mean Diameter (Outside diameter - t). You can’t just multiply Rdø by 2pi*r can you? Hence, dm is the mass of one of those “infinite” hoops. Why is there a difference and when is it made? Point lying inside the shell: Is my whole idea wrong or is it something else? A spherical mass can be thought of as built up of many infinitely thin spherical shells, each one nested inside the other. Advertising Center If one of these shells can be treated as a point mass, then a system of shells (i.e. For example in a sphere of radius $R$ and $n=4$ shells, each of thickness $h=R/4$, the total sum of the minor term represents 1/64th of the shell volume. | Feedback Hello, This is a great content and I love it. t/d < 1/20. The last integration gives 4/3. Disclaimer We can derive the formula for volume of sphere in a number of ways. $$\begin{aligned} dm &= \frac{M}{A} \, dA \end{aligned} \tag{2} $$, ,where $A$ is the total surface area of the shell – $4 \pi R^{2}$, If $A$ is the total surface area of the shell, $dA$ is the area of one of the many thin circular hoops. Condition for thin cylindrical or spherical shell. In order to continue, we will need to find an expression for $dm$ in Equation 1. A classic problem in mechanics is the calculation of the gravity force that would be experienced by a mass m that was attracted by a uniform spherical shell of mass M. The law of gravity applies, but calculus must be used to account for the fact that the mass is distributed over the surface of a sphere. This is known as the axial or longitudinal stress and is usually less than the hoop stress. Units for t, and d are inches (in). Electric field due to a uniformly charged thin spherical shell: Consider a spherical shell having surface charge density σ and radius R. The electric field resulting from such a spherical shell is radial and hence electric field intensity is calculated for a point lying inside and outside the spherical shell. I did similar integration but instead of using angle I did the integration over x, and then it holds r=sqrt(R^2-x^2), and dA=2rpi*dx, but i got 3pi/16mR^2 instead of 2/3mR^2. document.write(''); Rdtheta is so small that we can consider it perpendicular. 1 Introduction Thin shell structures find wide applications in many branches of engineering. t pr Max 4 τ = Summary for Spherical pressure vessel with r/t large: t pr 1 2 2 σ σ = = very thin cylindrical shell under uniform compression or a very thin spherical shell under uniform external pressure. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Notice that the thin spherical shell is made up of nothing more than lots of thin circular hoops. (from the pitfallsnasa.ppt file) 7.3.3. For those who needs a little bit more help, read on. so, dx=-Rsin(theta)d(theta) 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. The equation for rotational inertia of an object is, Here, is mass of the object and is distance from the axis. so the area of this hoop is just the circumference of a circle x thickness, Here x=Rcos(theta) The circumference of the hoop multiplied by the “thickness” of the hoop gives the area of the hoop. (In this example the shell radii are $\frac{h}{2}$, $\frac{3h}{2}$, $\frac{5h}{2}$ and $\frac{7h}{2}$). 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