412 Mr. R. C. Nichols on the Resistances 



directions. Then draw I K and E F parallel to R S, inter- 

 secting the vertical lines through N and P at A and B, 

 C and D, at a distance from each other equal to the depth of 

 the beam, in such manner that the area of the triangle I H K 

 may equal that of the pentagon CLHMD. Then A B C D 

 will represent the section of the beam ; H will be the position 

 of the neutral axis, which will be slightly elevated above the 

 centre of gravity, G, of the section ; I K will represent the 

 compressive strain or stress at the upper side of the beam ; 

 any horizontal lines i k, I m, intersecting the lines of strain or 

 stress IM, KL, will represent the strain or stress at such line, 

 any line £/* intersecting the lines LE, MF, will represent the 

 strain, and the portion of it cd between L C and M D the 

 stress, at such line, and E F and C D respectively the strain 

 and stress at the lower side of the beam. 



And the areas L H M D and IHR will be the areas of 

 maximum tensile stress and of an equal compressive stress 

 respectively, and the sum of these areas each multiplied by 

 the maximum tensile stress and by the distance of its centre of 

 gravity from H will be the total moment of horizontal stress. 



Let AB = 6, AC = <i, e = elongation per unit of length at a 

 depth below the neutral axis equal to half the depth of the 

 beam, H Q, the height of neutral axis above the lower side, = A, 

 MD = c, and h — c=g. 



Then NP _ _S 



RS ~ Be 

 Let this = <r, _ HO . NP do- 



IK= b(d-h) ^ and the A 1 H E = the pentagon CLHMD, 



b ^ J ^=bg + 2bc = b(2h-g). 



Solving this equation with respect to h, we obtain 

 h=d+g— s/2dg 



which is the height of the neutral axis above the lower side 

 of the section ; 



.*. d — 7i = dl s/o-—-\ and c=h— g = d(l— \/<x). 



And the moment of IHK about H is 



1 b(d-hy _ bd* (2s/o— o-) 3 

 3 g ~ 6 2cr ; 



