By P. R. Lancaster, D. Mitchell
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Additional resources for Advanced Solid Mechanics: Theory, worked examples and problems
708, P3 = -6. CP2 - p3), ~CP3- Pl) and hence calculate the normal stresses on these shear planes. 6 the resultant shear stress and its direction cosines in relation to the given axes. What are the principal stresses and the direction cosines of the greatest of these? 6 A resistance strain gauge rosette has arms a, b and c. The following readings are obtained -3e a Show that the maximum shear stress at the point is given by ~[28)~ 1 + \} 3 assuming that the measurements are taken on the surface of a thin plate subjected to a two-dimensional stress ~stem.
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).
Thus ar + a6 --:2,--- + R cos (26 - a) where R is the radius of ar + a6 ----2--- cry and + ~1ohr circle. R cos 26 cos a + R sin 26 sin a + T R cos a = FD or AF R sin a a y 'r6 a6 - a r cos 26 2 = ar + a6 --2-- = r -y(l - cos 26) + a a a y a sin 2 6 a r + a6 2 X a T r xy r + a6 2(1 + a 6 cos 2 6 + r6 cos 26) + T r6 sin 26 + T r6 sin 26 sin 26 (1. 22b) - R cos(26 - a) cos 2 6 -R sin(26 a 6 sin 2 6 - + - T r6 a) a6 - ar sin 26 2 + T r6 cos 26 (1. 22 are the transform equations for rectangular coordinates to polar coordinates.
Advanced Solid Mechanics: Theory, worked examples and problems by P. R. Lancaster, D. Mitchell