By P. R. Lancaster, D. Mitchell

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**Extra info for Advanced Solid Mechanics: Theory, worked examples and problems**

**Example text**

Hence the principal stresses can be found for 23 and so and p2 = _E__ (e2 1 - v2 + vel) It is also possible to find other direct and shear strains for the given point. Solution of equations (i), (ii) and (iii) can be tedious and if many results are to be analysed a small desk computer will be found useful. It will be appreciated that e1, e 2 , p 1 and p 2 are the strains and stresses at the surface of the material. Given eb 1 X 10- 4 3 X 10- 4 = ec with A. 6667 The signs are determined by the loading situation, for example, on a beam the rosette can be on the tensile or compressive side.

20b) o (2. 21 giving u = ~[- ~( 1 + v) + 2 Cr ( 1 - v) + B [ (1 - J 2 ( 1 - v) ( r Q. 2ld directly f 2 (r) = Lr (2. 2lf) (L. 22b give the displacements for any radius r at any inclination e. 2le is independent of r. M and N can therefore only describe translatory movements of the whole body. Consequently M and N have no effect on the strain and can be eliminated from the displacement equation. 22b, B must be zero otherwise shearing would occur on radial planes due to the tangential displacement of a point given by (ra, e = 0, 2~, 4~, etc).

10c that a constant -Pa 2/2GI occurs. 10b). It is usual in such instances to let A+ B = -Pa2/2GI. This does not increase the constants of integration, which decide the boundary conditions because if A is known, B can be found. Separating the variables· for X -Px2 dfz(x) A=--+--2EI dx giving Px 3 f 2 (x) = 6EI + Ax + F and for y ~ - vPy2 dfl (y) B + 2GI - 2E I + ~ giving _ ~ vPy 3 fl (y) - 6GI - 6EI + By + H F and H both constants of integration to be determined from the boundary conditions. lOd) (2.