Ryd

The maximum deflection will be at the centre and again equivalent to that obtained when r = R, i.e. from eqn. (7.29),

FR2 FR2

maximum deflection = — —— [log R — 1 ] + ——

SnD 4nD

FR2 16 nD FR2 16 nD

Substituting for D,

- 4jtEt3

Again substituting for dd/dr and 6/r from eqn. (7.28) into eqns (7.14) and (7.15) yields

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