under endlong compression. 



195 



siderable. A Table is given showing the values of P and W 

 measured at the breakdown point : — 



Table I. 



TFlbs. 



Plbs. 



Mean P lbs. 



TFlbs. 



Plbs. 



Mean P lbs. 







108 I 



104-5 



12 



70 "I 



70 







101 J 





12 



70 J 





5 



88 "1 

 90 J 





20 



46 "1 







89 





[ 



47 



5 





20 



48 J 





10 



73 1 



74 



30 



27 "] 



26-5 



10 



75 J 





30 



26 J 





From this table it will be seen that when the value of W was 

 30 lbs., a vertical load equal to 26 lbs. was sufficient to break the 

 strut-beam ; with this lateral loading however an endlong load of 

 25 lbs. was withstood, and it was then noted that the central 

 deflection was about 2 inches, though the calculated value on 

 the usual elastic hypothesis was only 1'7 inch. The formula used 

 to calculate the central deflection is 1 



# 



w 



2P Wp 



a r^r I I 



where y 1 = central deflection, I = length of strut, a = *JEI, 

 E = Young's Modulus and / = Moment of Inertia of section about 

 the neutral axis. 



1 The differential equation to be solved is of the form 



W 



x = 0, 



<Py P 



dx 2 + El y + 2EI ' 



where y is deflection at any point and x is distance measured from one end of the 

 strut. 



This, on solution, gives 



maximum 



and maximum 



where z is half the thickness of the rod 

 VOL. XI. PT. III. 



deflection = g . {^j^l y/ ~ - || . 



Wz l~F . i IT 



siiess =-2r\/ip- t&n 2\/Ei> 



15 



