Tie-rods with Lateral Loads, 



211 



Now it might be supposed that in the second direction the 

 strut is as if fixed at the ends ; but the pins are very short, 

 and the resultant load is certainly not applied axially, and it 

 is safer to assume that in this direction also the strut is as if 

 hinged at the ends. I assume this, although perhaps from 

 the possibility of accidental untrue loading we ought to go 

 farther in this safe direction and assume that the strength of the 

 strut is less. Assuming it hinged also in the second direction, 

 however, the thrust per square inch which it can receive is 



'Elijj^/bdy and in this case I,= j~~ 



co = 6'll~ xlO 6 



so that, if E = 3xl0' 



This is the value to use in formula (23). 



Taking / for the steel used as 20,000 lb. per square inch, 

 being the proof stress which the material will stand for an 

 infinite number of reversals of stress, we find for equal 

 strength in the two directions, applying (23), that 



8*4 x 10 8 (l-308^Vl- j 2 ) = rin 2 r/d. 



Thus, for example, if b = l, and 1 = 30, and r = 12, we have 

 the following tables : — 



If 5=1. 



If b = 1-5. 



d. 



n. 



1 







1-5 



205 



2 



277 



2-5 



327 



3 



368 



4 



437 



6 



545 



d. 



n. 



1-5 







2 



125 



25 



170 



3 



202 



4 



249 



6 



318 



10 



440 



To illustrate these results. On a certain locomotive engine 

 the coupling-rods are 68 inches long or Z=34", d=o%, b = l& 

 r=12. 



Now it will be found that this rod is equally strong in the 

 two directions if n — 29S revolutions per minute. The limit 

 of speed which has practically been settled for the engine bv 

 the breaking of similar coupling-rods is 258 revolutions 

 minute. 



per 



