W nL

(b) Uniformly distributed load

Consider now the uniformly loaded strut of Fig. 2.13 with the origin again selected at the centre but y measured from the maximum deflected position.

The solution of this equation is y = A cos nx + B sin nx w fL2 A e 2w i.e.

w nL

n2P 2

L2 x1

Thus the maximum deflection 8, when y — 0 and x = 0, is given by

and the maximum B.M. is

w nL

In the case of a member carrying a tensile load (i.e. a tie) together with a uniformly distributed load, the above procedure applies with the sign for P reversed. The relevant differential expression then becomes d y 2 w dx2'" y=2EI

The solution of this equation involves hyperbolic functions but remains of identical form to that obtained previously, i.e.


0 0

Post a comment