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What pressure can a spherical steel container withstand, if its internal radius is R and the wall thickness is d? Do calculations for R = 50 cm, d = 5 mm. |
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Answer» elastic force `Delta T = sigma DeltaS = sigma d Deltal` ACTS on an element of area `Delta S = d Delta l` on the periphery of this segment. The normal COMPONENT of the elementary force is `DeltaT_n = DeltaT sin alpha = sigma d Deltal alpha`. SUMMING over the completo circumference of the segment, we obtain the total force of the normal pressure: `T_n = sigma d . 2 pi alpha sin alpha = 2 pi sigma Rd sin^2 alpha` This force compensates the force of the gaseous pressure `F = pS` ACTING on the segment. For a small angle the segment.s area is `S = pi a^2 = pi R^2 sin^2 alpha`, and the force of pressure is `F = pi p R^2 sin^2 alpha.` It follows from the balance of forces that `p = 2 sigma d//R`. 
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