136 



W. H. WARREN. SOME APPLICATIONS OF THE 



f \y d y 

 > 



and since the beam is rectangular in section b is constant 



dM / 



.\S = dx ( y y'a y but the shearing stress F = dM . •. H = 



I Jy ° <** 



The shearing stress at the neutral axis will be found by integ- 

 rating between the limits y 1 =~ , and u = o. : H = S- d " = 3 — 



° ° 2 a I 8 2b d 



It therefore follows that the intensity of horizontal shearing 

 stress at any point in the cross section of a beam can be determined. 

 For example take the case of a compound beam formed with two 

 plain beams each 12 inches by 12 inches. 



The intensity of shearing stress 3 inches 

 from the neutral axis is — 



ydy 



F 

 I 



«!_.*!> - 1 lvi«d* 



8 128 $ ~ I X 128 



At 6" from the neutral axis it is- 



«?1 



32 J 



F 12 



I X 128 a 



At 9" from the neutral axis it is- 



128*'" 5 ~ i x 128 d " 



At 12" from the neutral axis the horizontal 

 shearing stress is zero. 



These stresses are represented graphically in the diagram fig. 4, 

 but they may also be determined by setting out the central ordin- 

 of a parabola = ^- d with a base equal to the depth of the beam d 

 and drawing the curve in the usual way. 



In order that the compound beam referred to formed with two 

 plain beams each 12 inches by 12 inches should be equal in strength 

 to that of abeam 24 inches deep, it is necessary that means should 

 be provided for resisting the horizontal shearing stresses which 

 have been investigated, and also the vertical shearing stresses 

 which are of equal intensity. The method of doing this is illus- 

 trated in Plate 7, which shows a design proposed by the author 

 for a timber viaduct carrying a single line of railway. The func- 

 tion of the bolts and wedges is to resist the shearing stresses 

 referred to, and it is clear that the bolts can be screwed up and 

 the wedges driven in further from time to time as the timber 

 shrinks. Again the timbers are separated so that they may season 



