By P. R. Lancaster, D. Mitchell
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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, for a cylindrical body where the stresses are distributed symmetrically about an axis there is only a radial displacement possible.
1 Stress Boundaries To solve the problem of an end loaded cantilever plate the direct approach will be given in detail ~ ~ ~ ,..... p __,. X L y Fig. 2 The force P acts over the free end of the plate and Ox represents the centroidal axis for the undeformed plate. 2. Take origin of coordinates at 0. n boundary conditions At y ± a T xy and at y ± a a y and at X = 0 a X that the bending moment for all values of x for all values of x 0 for all values of y, which also means is zero at X = 0. 0 0 35 Note that the shear force is constant along the plate length and consequently there can be no variation of shear stress from one section to another, that is T will vary with y but not with x, so xy (2.
4) (-az- + -az-) (cr + a ) = ax2 ayz x Y 1 ) [ax [r-::-"V Lax + a'(! 3a. The stress or strain compatibility conditions are not solvable except for the simplest of problems. 6) which is a biharmonic differential equation representing the compatability equation for stress. also difficult to solve for any but the simplest of problems. However, numerical methods can be used as will be seen later for quite complex problems. 6a is that functions of x and y can be devised which represent particular stressing conditions.
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