RUPTURE STRESSES IN BEAMS AND CRANE HOOKS. 



213 



ultimate stresses obtained in direct tension and compression. The last two columns 

 of the table give Young's modulus when the beam is placed on its broad or narrow 

 face. 



^ 



~y 



+ 



\ 



General Case for Beams of any Section, but with the Ultimate Distribution 



of Stress a Rectangular One. 



The general case may be deduced as follows : — 

 Let fig. 1 denote the section of the beam, and let 

 A = total sectional area, 



A t = ,, area ultimately subjected to tensile stress, 

 A c = ,, ,, ,, ,, compressive stress. 



t, c, T, C, and M have the same meaning as before. 



Also let NN = ultimate position of neutral axis, the plane 

 of which extends along the length of the 

 beam and divides each cross-section of the 

 beam into areas A { and A e , 

 y t = distance of centroid of area A t from NN, 



yc 11 ii ii -"-e 5 > 



and let the ultimate distribution of stress be a rectangular one. 



From the condition (a) that the tension and compression stress areas are equal 



to each other, we have 



tA t -cA e =0. ..... (la) 



If the ultimate fibre stresses reached at the point of rupture be equal to the 

 ultimate direct stresses T and C in tension and compression respectively, then 



TA«-CA c = 0, 



and 



A,_C 

 A C ~T 



and similarly 



E'&l 



A, + A c T + C 



.-. A, = A 



C 



T + C 



A, = A 



T + C 



We therefore obtain the position of the neutral axis of the beam at fracture point 



by dividing the total area A into areas A { and A„ in the ratio of =. 



Further, since the resisting moment of the section just up to the point of rupture 

 equals the external bending moment, 



C T 



-. M = tk t y t + rA c y c = tA t (y t + y c ) = cA c (y t + y c ) = tA— — (y, + y c ) = cA^—^{y t + y c ) 



T + C v 



T + C x 



t=- 



and c = 



^1 T + C 



Hy t +y c ) C 



M T + C 



Hl/t + Vc) T T~ 



(I) 



(La) 

 (Lb) 



