Practical Mechanics By John Perry

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Practical Mechanics By John Perry
John Perry
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The shear strain in the little prism is B B' divided by p B or o o', so that it is r T, hence the shear stress is NTT (see Art. 87). If a is the area of the end of the little prism in square inches, the shear force acting on it is N r T a, and as this acts in the direction B B' at right angles to the radius, its moment about o o' is N r* T a. But we have a similar moment for every such little area into which the cross section may be divided, and to find the total torque we must take the sum of ...all such terms. Now N and T are the same everywhere, so that in taking such a sum our only difficulty is with the Fig. 45.
96 PRACTICAL MECHANICS. [Chap. IX.
factors r 3 a. But the sum of all such terms as r 2 a is called the moment of inertia of the section about the axis o o', and it has been calculated for us. Thus, if D is the diameter of a round shaft, the moment of inertia of its section about an axis through its centre at right angles to the section is IT D* -r- 32, and for a hollow shaft whose outside diameter is D and inside diameter d, the moment of inertia is f (D 4 d 4 ) -T- 32, and hence we see that the moment M necessary to produce a twist of T radians per inch in a round shaft of diameter D it M = WNT D 4 -i- 32 (1), and for a hollow shaft it is M = 1TNT (D 4 -

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