RUPTURE STRESSES IN BEAMS AND CRANE HOOKS. 



219 



divide up the section of the hook in the above ratio as shown in fig. 5. Let y t and 

 y c be the distances of centroids of A T and A c from the neutral axis NN. 



Then we have, since the external moment equals the internal resisting moment, 



= M TTc 2/ ' + cA fTc 2/ " 

 and at the fracture point, when t and c equal T and C respectively, 



WL = *Aj^(y, + y c ) = cA_I_G/ r : „.) 



(II.) 



T + C v '" •"' T + C7 

 Now, when the resultant tensile force W becomes important and is taken into 



consideration, we have 



tA t -cA a = W. 



We may imagine that this is the previous case, with L diminishing as W increases. 

 Since the stress limits at fracture are T and C as before, these cannot be altered 



Fig. 5. 



to meet the resultant W, which can only be balanced by a transfer of part of the 

 compression -area to the tension side. Of course, the moment of these rearranged 

 areas must be equal to the external moment. Let the shaded area in fig. 5 be the part 

 so transferred from compression to tension, and let the actual final position of the 

 neutral axis be N 2 at a distance x from N, and z t and z c be the respective distances of the 

 centroids of the areas A t and A c from those of the areas A T and A c of the special case. 

 From the equation of moments we have 



W(L + x) = tk t (y t + x - z t ) + cA e (y c -x + z L ). 



Now 



C 



Aj = A T + a = A 

 A f = A c - a = A 



r+c 



T 



TTc 



+ a 



y t + 



A 



and 



(». 



C 



. W(L + x) = tk^L-# t + cA^-^ + x(tA t - cA c ) - a?(t + c). 

 TRANS. ROY. SOC. EDIN., VOL. L. PART I. (NO. 7). 



30 



