Tie-rods with Lateral Loads. 275 



to see that the solution becomes 



M M 



^=T 



FcosZ.. 



EI 



— 7 ==cos«a / — + — ^ cos- 7 #. 17 



/F V El FI r 21 K 



VET EI 4^" F 



We can at once work out from this, cases in which with such 

 a lateral loading we have any equal couples M applied at the 

 ends, including the case in which the end is hinged, but the 

 resultant push at the ends is applied with the inaccuracy A, in 

 which case — M = F/<. In case the strut is fixed at the ends 



we apply the condition -^ = 0, when x = l. This gives us 

 ax 



8(U-F) V EI 



It is not instructive to pursue this example unless numerical 

 values are taken for the dimensions of the strut and the 

 loading W. If the strut is hinged at the ends and h = 0, then 

 M = 0, and 



^f7^V 0S ^ (18) 



As this is a simple case, common in practice, I will work 

 out the stresses. 



(18) is evidently true even when F = 0. The deflexion in 

 the middle is 



*--±^- s a»> 



and the greatest bending-moment fi is 



^F^ + iWl, 

 or 



If W = and if fu, has any value whatever, the denominator 

 must be 0. Putting it equal to 0, we have Euler's law for 

 the strength of struts which are so long that they bend before 



