CHAPTER IX. 

 STEEL GRAIN ELEVATORS. 



Introduction. Grain elevators, or "silos," as they arc called in Europe, may be divided into 

 two classes according to the arrangement of the bins and elevating machinery: (a) elevators 

 which are self contained, with all the storage bins in the main elevator or working house; and 

 (6) elevators having a working house containing the elevating machinery, while the storage is in 

 bins connected with the working house by conveyors. The working house is usually rectangular 

 in shape, with square or circular bins; while the independent storage bins are usually circular. 



With reference to the materials of which they are constructed, elevators may be divided 

 into (i) timber; (2) steel; (3) concrete; (4) tile, and (5) brick. Steel grain elevators, only, will 

 be considered in this chapter. For a complete treatise on the design of grain elevators, see the 

 author's "The Design of Walls, Bins and Grain Elevators." 



STRESSES IN GRAIN BINS. The problem of calculating the pressure of grain on bin 

 walls is somewhat similar to the problem of the retaining wall, but is not so simple. The theory 

 of Rankine will apply in the case of shallow bins with smooth walls where the plane of rupture 

 cuts the grain surface, but will not apply to deep bins or bins with rough walls. (It should be 

 remembered that Rankine assumes a granular mass of unlimited extent.) 



Stresses in Deep Bins. Where the plane of rupture cuts the sides of the bin the solution for 

 shallow bins does not apply. 



Nomenclature. The following nomenclature will be used: 

 <f> = angle of repose of the filling; 



<f>' = the angle of friction of the filling on the bin walls; 

 It, = tan = coefficient of friction of filling on filling; 

 n' = tan <(>' = coefficient of friction of filling on the bin walls; 

 x = angle of rupture; 

 w = weight of filling in Ib. per cu. ft. ; 

 V = vertical pressure of the filling in Ib. per sq. ft.; 

 L = lateral pressure of the filling in Ib. per sq. ft. ; 

 A = area of bin in sq. ft.; 

 U = circumference of bin in ft.; 

 R = A/U = hydraulic radius of bin. 



Janssen's Solution. The bin in (a) Fig. i, has a uniform area A, a constant circumference U, 

 and is filled with a granular material weighing w per unit of volume, and having an angle of repose 

 <t>. Let V be the vertical pressure, and L be the lateral pressure at any point, both V and L 

 being assumed as constant for all points on the horizontal plane. (More correctly V and L will 

 be constant on the surface of a dome as in (6).) 



The weight of the granular material between the sections of y and y + dy = A-w-dy; the 

 total frictional force acting upwards at the circumference will be = L- U'tan 4>'-dy\ the total 

 perpendicular pressure on the upper surface will be = V-A; and the total pressure on the lower 

 surface will be = ( V + d V)A. 



Now these vertical pressures are in equilibrium, and 



V-A - (V + dV)A + A-wdy - L-U-ten+'-dy = o 

 and 



= (w- 



(i) 



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