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For any given eccentric load condition P is the only unknown and the equation can be readily solved.

(b) One end fixed, the other free

Consider the strut shown in Fig. 2.11.

Fig. 2.11. Strut with eccentric load (one end fixed, the other free)

The solution of the expression is y = A cos nx + B sin xn + eo At x = 0, y = 0 A + eo = 0 or A = — eo

S = — eo cos nL + eo = eo(l — cos nL) — (S + e)( 1 — cos nL) = 8 — S cos nL + e — e cos nL 8 cos nL = e — e cos nL 8 — e(sec nL — 1) or 5 + e = e sec nL

This is the same form of solution as that obtained previously for pinned ends with L replaced by 2L, i.e. the Smith-Southwell formula will apply in this case provided that the equivalent length of the strut (/ = 2L) is used in place of L. Thus the Smith-Southwell formula can be written in the form

the value of the equivalent length / to be used for any given end condition being given by the diagrams of Fig. 2.6, §2.2.

The exception to this rule, however, is the case of fixed ends where the only effect of eccentricity of loading is to increase the fixing moments within the supports at each end; there will be no effect on the deflection or stress in the strut itself. Thus, eccentricity of loading can be neglected in the case of fixed-ended struts - an important factor since most practical struts can be considered to be of this type.

Mechanics of Materials 2 2.10. Laterally loaded struts

(a) Central concentrated load

With the origin at the centre of the strut as shown in Fig. 2.12, d2y dx2

0 0

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