Stresses in Bins and Hoppers, by Gabriel I. Tardos

-- 8. Solved Problems --

Computing Stresses using Janssen's Equations Fine sand with a bulk density of 1,560 kg/m3 and angle of internal friction of = 24.1o is stored in a tall cylindrical bin of diameter D = 2 b = 1.8 m and an angle of wall friction of W = 20.2o. Take g10 m/sec2. a) Calculate the maximum vertical stress in the bin assuming active state of stress b) Use Janssen's equations to determine the depth at which the vertical stress reaches 99% of its maximum value; also calculate the wall pressure at this point. c) Compare the hydrostatic pressure at the point calculated in question b) to the actual stresses present in the powder. d) Recalculate questions a)-c) for the case of passive stress in the bin. Solution: a) The maximum average vertical stress in the bin is given by Eq. 7 with m = 2 and Janssen's constant for the active stress from Eq. 19       K = [(1 - sin ) / (1 + sin )] tan W = [(1 - sin 24.1o) / (1 + sin 24.1o)] tan 20.2o = 0.153 With this value the vertical stress becomes       y,max = B g b / m K = 1,560 x 10 x 0.9 / (2 x 0.153) = 45.9 kPa b) To calculate the depth in the bin where the stress reaches 99% of its maximum value, again Eq. 7 is used in the form       y,max (1 - exp [-mKy/b]) = 0.99 y,max This yields for the distance y the expression       y99% =[ b ln (0.01)] / (m K) = 0.9 x 4.6 / (2 x 0.153) = 13.55 m The normal average wall stress is now given by Eq. 9       n = 0.99 y,max K cot W = 0.99 x 45.9 x 0.153 x cot 20.2o = 19.1 kPa c) The hydrostatic pressure is given by       y,h = B g y99% = 1,560 x 10 x 13.55 = 211 kPa This value is significantly higher then the average vertical stress (45.5 kPa) and the normal stress on the wall (19.1 kPa) at the same point in the silo. d) Janssen's constant for the passive state of stress is from Eq. 19       K = [(1 + sin ) / (1 - sin )] tan W = [(1 + sin 24.1o) / (1 - sin 24.1o)] tan 20.2o = 0.876 With this value, y,max = 8 kPa, the depth at which the stress reaches 99% of its maximum value becomes y = 2.36 m, n =18.8 kPa and the hydrostatic pressure is y,h = 36.8 kPa. It can be seen that the stresses in this case are significantly smaller compared with the active state. Design of a Hopper Using Jenike's Method It is desired to design a silo with a hopper from which a material with bulk density 1700 kg/m3 and angle of internal friction of = 40o can be emptied safely. Four sets of shear tests have been conducted on the material and the results for the instantaneous unconfined shear strength and the corresponding consolidating stresses are listed below fc [kN/2] 0.9 1.35 1.75 1.95 1 [kN/2] 1.1 1.75 2.75 3.25 Using a hopper half-angle = 25o and an angle of wall friction W=20o, calculate the minimum span of the exit opening, B, for either a 2D plane or a 3D conical hopper. Solution Data from the table in the problem is represented graphically in Fig. 14.

Fig. 14 - Design of a hopper using Jenike's method

Using the Jenike charts in Fig. 12 for a material with an internal angle of friction of = 40o, a hopper half-angle = 25o and with W = 20o, one finds ff2D = 1.4 and ff3D = 1.5. Lines through the origin with slopes 1/1.4 and 1/1.5 are also shown in Fig. 14. At the intersection, one finds the critical values of the unconfined yield strength to be fc, 2D = 1.45 kN/m2 and fc,3D = 1.6 kN/m2. Values of the correction function H() are obtained from Fig'. 13 as H()2D = 1.15 and H()3D = 2.4 for = 25o. From Eq. 30, with H() replacing the geometric factor m, the critical value of the span becomes for the 2D hopper       Bcrit,2D = H()2D fc,2D / (B g) = 1.15 x 1.45 x 10 3 / (1,700 x 10) = 0.098 m while for the 3D hopper Bcrit,D = 0.226 m. It can be seen that for identical conditions and the same hopper half angle, the conical hopper has a critical opening diameter that is about double the span of the 2D hopper.