CHAP. XI.] CAUSING MAXIMUM STRAINS. 16 



* (? .4- ). Join A H and L. Make B G = C S = 

 o V *' 



Join A G and S. The maximum vertical ordinates between 

 the two parabolas and the lines AH, A G, H L, and G S, as ' 

 shown in the figure, give the maximum moments. 



The points I, K and O, P or M, N show the limits of devia- 

 tion of the points of inflection. 



If ^-~ is less than QO > tne beam will require to be 



held down at the abutments. 



If the beam be continuous for three spans only, I in the 



F/20 m\ 

 expression for B H = (~- + I must have a value given to 



_i 



2 ' 



Shearing Forces. 



The maximum shear at any pier (B or C) will obtain simul- 

 taneously with the maximum moment over that pier. 

 Construction. PI. 18, Fig. 76. 

 Let A B C be part of the beam. First, for any inner span as 



I. At B and C erect BG = CH = + Make BD 



\ 2i o / 



and C E each = - (p + m). Join D and E to midspan F, and 



a 



draw G K and H K parallel to D F and F E respectively. 

 Second, for either end span as 4 At B erect a perpen- 



2 7 3 



dicular = ~ (l?+ m )i which, if ^ = - I, will coincide with B D. 



At A make AL = - B D. Join D and L to M distant - ^ from A. 

 2 o 



Make A O = - 1 (p+m) -~-j , and draw O N parallel to L M. 



2i ' o2i (?[ 



Sketch in curves as shown by the dotted curves in the figure, 



giving additional depth to the ordinates there of ^ and - 



o o 



respectively. Then the vertical distances between O a b D and 

 A B give the maximum shearing forces for either end span, and 

 those between G c d H and B C, the shearing forces for the re- 

 maining spans. 



