ABSTRACT
For an extremely 'long' beam — we will need to assess what 'long' actually means — the boundary conditions are:
• slope dy/dx = 0 at x = 0 from symmetry
• moment M = EId2y/dx2 — » 0 as x — » oo
• shear force F = EId3y/dx3 — > 0 as x — > oo
These give the solution in the section of the beam for positive x, the boundary conditions for negative x are the same but with the sign of the shear force at the origin reversed, from symmetry. The first form of the solution (8.2) is the more convenient with the final result:
PLL y = ^-^e ^ (cos fix + sin fix) (8.5)
and this profile of deflections, together with the corresponding variations of slope (normalised with Pfi2 /kB\ bending moment (normalised with P/4//) and shear force (normalised with P/2), is plotted in Fig 8.2 as a function of the normalised position on the beam fix.
