Advanced Solid Mechanics: Theory, worked examples and by P. R. Lancaster, D. Mitchell

By P. R. Lancaster, D. Mitchell

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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. De) Both these equations give compatible results for both strain and displacement.

Fig. nr) 53 ~ r2 Tre Also u = 2C + B(3 + 2~nr) + (2. 20c) % [-~(1 + v) + 2Cr(l - v) + B[Cl - 3v)r + 2r(l - v) (~nr- 1~ + Mcose v 4Bre = -E- Msine + Ncose + Lr - + Nsine (2. 22a) (2. 22b) However, now B, L, M and N are not necessarily zero since the bar will be displaced by the couples M. The boundary conditions are: Radial stresses are zero at the inside and outside surfaces a r = 0 for r = a and r = b (a) There can be no resultant force over the end of the bar (b) By taking moments about the axis 0 (c) From (a) -+ A a2 2Cr + B(l + A b2 2C + B(l + -+ (b) gives Ja 6dr = 1 _cp(dr) d2 dr2 0 2~na) 0 2~nb) r = ~d~ dr b and this is already satisfied by (a).

P = Hence xJfl(y)dy Now the constants L, M and N have to be determined. There can be no resultant force over the end of the plate so for unit thickness or Ja(Lxy + Mx)dy + Nxy 0 -a 2Max 0 giving M and q, =0 = ~xy3 6 (2. ) Thus that is - L 3P 2a 3 and the complete stress function is cp = _ 3P 2a 3 xy 3 + 3P 4a xy If I is the second moment of area for the cross-sectio n I Hence 2a 3; 3 L and a a (2. 9a) X y 0 (2. 9 can be found. Note that ox = -(Px)y/I = -My/I where M is the bending moment at a section x, and when y is negative·, stresses are tensile as expected.

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