280 



Prof. J. Perry on Struts and 



A numerical example will illustrate these formulse. I will 

 take a round rod of mild steel lying in a horizontal position, 

 so that it is loaded laterally by its own weight, dimensions 

 being in inches. If d is its diameter, the weight of 1 cubic 

 inch of material is 0'28 lb. and E = 3x 10 7 , 



64 



and Z = 



ird 3 



fc or ft= 



32 ; 

 4-03 x 10 6 , 



U: 



3-6x10^, 



+ 



3-6xl0 6 4-F ~d 2 

 I 2 4 



the plus sign being taken in calculating f c and the minus in 

 calculating f t . Also 



'lldH 2 



yi= d* — * 



3-6xl0 6 ^-F 

 I 2 



Take d=l inch, 1=60 inches, 

 4-03 x 10 



396 



/cOr ^ = 1000-F ±1,273F and ^ 1= l000^F- 



From this I have calculated the following table. The 

 negative values of F indicate that instead of a strut we have 

 a tie. 



F. 



A 



ft. 



Vi> 



1000 



GO 



CO 



CO 



900 



41446 



39154 



396 



800 



21168 



19132 



1-98 



700 



14324 



12542 



1-32 



500 



8697 



7423 



0-792 



300 



6648 



4866 



0-566 



100 



4605 



4351 



0440 







4030 



4030 



0-396 



- 100 



3537 



3791 



0360 



- 300 



2718 



3482 



0-305 



- 500 



2031 



3323 



0264 



- 700 



1479 



3261 



0-233 



- 900 



975 



3267 



0-208 



-1200 



304 



3360 



0-180 



-1348 







3432 



0-169 



We see, then, that a comparatively small thrust F produces 

 excessive stress in the strut and great deflexion y x . 



We see also that a tensile force F x of 900 lb. halves the 



