STRESSES IN SUSPENSION BUNKERS. 



309 



and calculate the horizontal pressure PI - 7,680 lb., acting on the plane H-K at \HK above //. 

 Pressure Pi was calculated by the graphic method. Produce Pi until it intersects at O the line 

 of action of the weight of the triangle GHK acting through the center of gravity of the triangle. 

 From O lay off 0-1 W 19,900 lb., acting downwards, and from I lay off 1-2 P\ 7,680 

 lb., acting to the left. Then 0-2 - P t - 21,300 lb. Now P t - area triangle 6'CH-w, and 



% ,c*> 



^^'j* 4000 .'- 5 -!-^ 7 ^ ;< 



* .- I ' 



Surcharge- +.50. 



FIG. 10. 



= areaS'- B-A-s'-w = 1 1,340 \b. Force acts through the center of gravity of area 8-B-A-$. 

 The horizontal pressure on plane C-B = 1,400 lb. = area s'e'n'-w. The vertical pressure on 

 the left-hand side of the bottom A-F is 7,480 lb., acting through the center of gravity of the 

 pressure polygon. The vertical unit pressure at A is 1,412 lb. per sq. ft. 



STRESSES IN SUSPENSION BUNKERS. The suspension bunker shown in (a) Fig. 11, 

 carries a load which varies from zero at the support to a maximum at the center. If the bunker 

 is level full the loading from the supports to the center varies nearly as the ordinatcs to a straight 

 line, while if the bunker is surcharged the straight line assumption for loading is more nearly 

 correct. 



We will, therefore, assume that the loading of the bunker in (a) is represented by the tri- 

 angular loading varying from p = zero at each support to a maximum of p P at the center. 



Let / = one-half the span in feet; 

 S = the sag in feet; 



// = the horizontal component of the stress in the plate in lb. per lineal foot of bin; 

 w = weight of bin filling in lb. per cu. ft. ; 



