282 



or 



Prof. J. Perry on Struts and 

 m 



cos I 



\/£ 



/« = 



Z cos I 



m F 



V EI 



ft = 



Zcos Z 



v/e 



F 

 EI 



Taking for numerical example a round bar of steel as 

 before, and taking d=l inch, Z = 48 inches, and writing the 

 angle in degrees to simplify calculation, we have 



10-2? 



1-273F, 



f c or f t = 



cos 2-267 VF 



1Jl F 1 cos 2-267 VF / " 



If F = 0, / c or/, = 10'2 



3/i = ^J or 7-821 xl0- 4 m, 



Euler's strength or U for this strut is 1577 lb. It is 

 evidently obtainable by putting the quantity inside the 

 bracket equal to infinity. 



First take F = 1 500 lb. Then 



m 

 ^ 1 = 66' 



2*267 <s/F = 87°-80, 



£or /i=266w±1910. 



m. 



fc 



ft- 



Vi- 



fc OYJt, 



if F=0. 







1910 



-1910 











1 



2176 



- 1644 



•0167 



HV2 



2 



2442 



-1378 



•0333 



20-4 



3 



2708 



-1112 



•0500 



306 



5 



3240 



- 580 



•0833 



510 I 



7-18 



3820 







•1197 



732 



10 



4570 



750 



•1667 



102 



20 



7230 



3410 



•3333 



204 



80 



23190 



19370 



• 



1-333 



816 



So that an endlong load only sufficient by itself to produce 



