HOOKS, LINKS, AND SPRINGS 



151 



In the present case p = CE = 4.4", and the distance from the line of action of 

 the load to the neutral axis of the section is 



c + OC = 1.79" + 2.22" = 4.01"; 



consequently the external moment M acting on the section is 

 M = 20,000 x 4.01 = 80,200 in. Ib. 



The fiber most distant from the neutral axis is at the back of the hook. The 

 quantity y in formula (98) 

 is the distance of this ex- 

 treme fiber from the gravity 

 axis of the original section; 

 hence y = 2.8". Also the dis- 

 tance d between the gravity 

 axis of the original cross sec- 

 tion 1 and that of the trans- 

 formed section 2 is d = .40". 

 Substituting these values of 

 .V. /', />, d, and y in formula 

 (98), Article 125, the maxi- 

 mum bending stress p in the 

 hook is found to be 



p = 11,123 lb./in.2. 



To this must be added the 

 direct stress on the section. 

 which is found by dividing 

 the load by the area of the 

 section. Consequently, the 

 iiKixiuiuni total stress is 



ll,123+5102=10,2251b./inA 



Since the ultimate 

 strength of wrought iron 

 in tension may be taken as 

 64,000 lb./in. 2 , the factor of 

 safety is about 3J. 



Problem 149. By the 

 formula given in Article 123, 



calculate the maximum bend- Fi<;. 105 



in- stress and the maximum 



total stress on the hook shown in Fig. 104, and compare the results with those of 

 the preceding problem. 



NOTE. The lever arm of the force in this formula is assumed to be the distance from 

 tlit- axis <>f tin- hook to the center of gravity of the original cross section, or the distance 

 CE iu Fig. 105. Hence, in this case, 



M = 20,000 X 4.4 = 88,000 in. Ib. 



