Computations on Grain Stored in Various Configurations

Thomas W. Dorn, UNL Extension Educator - Lancaster County, NE

Unconstrained Piles

Unless space is limiting at the site, the least expensive temporary/emergency outside storage scheme will be an unconstrained pile (one that slopes all the way to the ground with no "walls").  All outside storage needs to be placed on a site which has been graded to conduct drainage away from the pile on all sides with plastic sheeting under the pile to prevent moisture migration from the soil into the grain.  When planning such a pile, it would be useful to know the size of pile necessary to store a given quantity of grain or conversely how to estimate the amount of grain in an existing pile.

Estimating the diameter of a conical pile

The volume of a conical shaped pile of grain is a function of the diameter of the base and the height of the peak. The height of a conical pile of grain is likewise dependent on the diameter of the pile and can be computed if the average filling angle (angle of repose) is known for the particular type of grain. One useful result of this relationship is that the diameter of a conical pile of grain can be estimated if the number of bushels is known.

The author has worked out a simple formula which can be used to predict the diameter of a conical pile of grain to hold any given quantity (bushels) of grain. Only two variables are necessary for the computation: the number of bushels and the volume conversion factor (V.F.) from Table 1.

The formula is: D(ft) = (Bu. * V. F.) 1/3
Equation 1

where:

D is the diameter of the pile of grain in ft
Bu is the number of bushels of grain in the pile
V.F. is the volume factor from Table 1.
(   )1/3 is another way to express the mathematical term "cube root"

Hint: Few calculators have a button for cube root.   Extracting the cube root usually requires using the Yx function on a scientific calculator, where y is the result of (Bu. * V.F.) and x is 1/3 for cube root.
Procedure:  First find the result of (Bu. * V.F.) , then press the Y key,  key in 0.3333 and press the =  key.

Example1: One could use Equation 1 to estimate the diameter of a conical corn pile necessary to contain 10,000 bushels of corn.
D(ft) = (10,000 Bu. * 22.52)1/3 =  60.8 feet.

Table 1. Conversion factors for calculations on grain resting at the angle of repose

 Crop Avg. Filling  Angle Height Factor  (H.F.) Volume Factor  (V.F.) Barley 28 0.2659 17.95 Corn (shelled) 23 0.2122 22.52 Oats 28 0.2659 17.95 Grain Sorghum 29 0.2772 17.24 Soybeans 25 0.2332 20.49 Sunflower (non oil) 28 0.2659 17.95 Sunflower (oil) 27 0.2548 18.76 Durum Wheat 23 0.2122 22.52 Wheat 25 0.2332 20.49

The process of solving for the diameter of the pile using Equation 1 also implies the height of the pile necessary to contain the stated number of bushels. However, at times it is useful to estimate the height of a pile of grain once the diameter is known using Equation 2.

H  = D * H.F.
Equation 2

Where:
H is the Height of the pile, ft.
D is the Diameter of the pile, ft.
H.F. is the Height Factor from Table 1

Example 2: The height of a pile of corn 60.8 feet in diameter can be estimated using Equation 2 and the height conversion factor ( H.F.) for from Table 1.

H = D * H.F.
=  60.8 * 0.2122
=  12.9 feet.

Estimating bushels in a pile of grain

Conical Piles

If the diameter and height of a conical pile of grain is known, the bushels in the pile can be estimated using Equation 3.
The formula to calculate the bushels in a cone shaped pile or the cone shaped portion of a windrow style pile is:

Bu. = 0.209 * D2 * H.
Equation 3

Where:
D is the diameter of the base of the pile, ft.
H is the height of the pile, ft.

Windrow Piles

Windrows of grain will consist of three sections, a half cone on each end and a section between the conical ends that is a triangle in cross-section.   The two ends together make a full cone and the bushels in that portion can be estimate using Equation 3.

The formula to calculate the bushels of grain in the interior of the windrow, (between the conical ends of the windrow) is:

Bu = 0.4 * W * H * L
Equation 4

Where:
W is the width of the pile, ft.
H is the height of the pile, ft.
L is the length of the pile, ft.

Example 3:  Calculate the bushels of corn in a windrow type pile 50 feet wide and 100 feet long.   This pile would contain three sections. 1/2 of a 50 foot diameter cone on each end and a triangular shaped section 50 feet wide by 50 feet long.

Using Eq. 2, the height can be calculated : H = (50 * 0.2122) = 10.61 feet.
Using Eq. 3, the two half cones together contain: 0.209 * 50 * 50 * 10.61 = 5,544 bu.
Using Eq. 4, the interior of the pile contains: 0.4 * 50 * 10.61 = 212.2 bu/ft * 50 ft = 10,610 bu
The pile would therefore contain 16,154 bushels.

Constrained Piles

Estimating the bushels of grain in a round bin

The amount of grain in a circular bin is computed using Equation 5 for the portion that is cylindrical in nature and using Equation 3 for any portion that is conical in nature such as the grain in a hopper bottom or grain that is left peaked in the the top of the bin.

For cylindrical volumes of grain,

Bu. = 0.628 * D2 * H
Equation 5

Where:

D is the diameter of the bin, ft.
H is the height of the grain, ft.
0.628 is a conversion constant

Example 4: Calculate the number of bushels of corn in a 30' diameter bin 18' high with a drying floor 1.0' above the concrete pad and peaked 6 feet above the level of the eaves.

Calculate the height of the grain at the bin wall. 18.0' - 1.0' = 17.0'
Calculate the bushels of grain to the eaves using Equation 5.
Bu. = 0.628 * D2 * H
Bu. = 0.628 * 30 * 30 * 17
= 9,608

Calculate the bushels in the peaked grain at the top of the bin using Equation 3.
Bu. = 0.209 * D2 * H.
= 0.209 * 30 * 30 * 6
= 1,129

The total quantity of corn in the bin is 9,608 + 1,129 = 10,737 bushels

Example 5: Calculate the bushels of corn in a temporary containment structure made with large round bales. The bale structure is 40 feet in diameter (inside), 5 feet high at the wall, with grain flowing over the bales an additional 2 feet to promote better drainage (44 foot diameter cone) .

Using Equation 2, the peak height would be: H = 44 * 0.2123 = 9.34 feet higher than the bales.
Using Equation 3, the 44' by 9.34' peak would contain:
Bu. = 0.209 * D2 * H
Bu. = 0.209 * 44 * 44 * 9.34 = 3779 bu

Using Equation 5, the 40' by 5'  "bin" would contain:
Bu. = 0.628 * D2 * H
Bu. = 0.628 * 40 * 40 * 5 = 5,024 bu.

The total grain contained by this structure would therefore be:  5,024 + 3,779 = 8,803 bushels.

Estimating the bushels in flat storage (rectangular building).

Existing buildings such as machine sheds are sometimes used to store grain in emergency situations. Ordinary machine sheds are not designed to withstand the loads that grain will exert on the wall.  (Grain exerts a lateral pressure on a wall of about 23 pounds per square foot per foot of grain depth).  This pressure tries to push the bottom of the wall out and also causes increased forces at the top of the wall.  Posts, trusses, and post-to-truss connections can fail when buildings are inadequately reinforced for grain storage.

University of Nebraska engineers recommend that grain placed in non-reinforced buildings not be piled more than 2 feet deep against the side walls.   However, grain that is dry and cool may be peaked (piled higher in the middle of the building than at the walls) to increase the storage capacity of the building.  This subject explained further in Temporary/ Emergency Grain Storage Options by Dorn, etal. on this web site.

Multi-purpose buildings, those which have been specifically designed to withstand the lateral pressure imposed by grain on the walls, may be filled normally with grain either leveled on top or allowed to remain peaked for short term storage.  If uncertain about the capabilities of a building, check with the manufacturer.  They often can provide building strength information, especially for steel frame structures.

To calculate the bushels in a rectangular structure, one needs to calculate the volume.  If grain is peaked in the building, it will be necessary to calculate the volume of  the three-dimensional rectangular portion and the peaked portion separately.

The bushel volume of the rectangular portion is calculated using Equation 6.

Bu. = L * W * H / 1.25
Equation 6.

Example 6:  A flat storage structure 80 feet long by 50 feet wide is filled and leveled with a grain depth of 10 feet.

Using Equation 6 this building contains Bu. =  80 * 50 * 10 / 1.25  =  32,000 Bushels

When the grain is peaked in a rectangular structure, the shape of the peak will depend on the relative dimensions of the building.  The grain in a square building will slope from the side walls to a common peak in the middle of the building.  A rectangular building will have a peak height that is dependent on the width of the building.  Most people will move the filling auger as the building is being filled, resulting in a ridge that runs down the length of the building.   The shape of the ridge also depends on whether the grain is allowed to rest against the gable ends of the building.   In the examples given in this discussion, it will be assumed that grain is not allowed to rest against the gable ends on either end of the building.

The slope of the peak will conform to the angle of repose consistent with the type of grain.  (See Table 1).  The height of the peak can be estimated using Equation 2, substituting the width of the building for the diameter of the cone.

Example 7:  A farmer has a machine shed that is 50' wide and 80' long with a workshop near the front end.   The farmer wants to store corn in this building but needs to reserve the final 25' or so open so he can continue to use the workshop   A temporary divider "wall" constructed of large square bales of hay is placed across the machine shed, 50 feet from the the back wall.  The auger spout is placed 25 feet from the back of the building, centered between the side walls (25 feet from each side wall).   The grain is allowed to peak in the middle of the building resembling a pyramid.  Filling is stopped when the grain reaches a height of 2 feet on all four walls.

The bushels in the 50' by 50' by 2' high portion can be computed using Equation 6.

Bu. =  50 * 50 * 2 / 1.25  =  4,000 Bushels

The  height of the peak in a square building is calculated using Equation 2, substituting the width of the building for the diameter of the cone in the equation.

H = W * H.F.
=  50 * 0.2122
=  10.6  feet.

The volume of the peak in a square building is calculated using Equation 3, substituting the width of the square for the diameter of the cone in the equation.

Bu. = 0.209 * W2 * H.
= 0.209 *  50 * 50 * 10.6
=  5,539 Bushels

The total volume of this temporary grain storage structure is therefore 9,539 bushels.

If grain is piled in a flat storage structure with the auger moved forward each time the grain has reached the proper height so that a ridge is formed in the peaked grain in the middle of the building.  The volume is calculated similar to Example 5 for a windrow of grain in an unconstrained pile except the volume contained in the base must be added.

Example 8: Assume a structure like the one discussed in Example 7, except the entire 80' by 50' structure is filled 2' high on all side walls and peaked in the middle.

The bushels in the 80' by 50' by 2' high portion can be computed using Equation 6.

Bu. =  80 * 50 * 2 / 1.25
=  6,400 Bushels

The peak consists of a 1/2 pyramid on each end, 50' wide by 25' deep.  The remaining peak has a triangular cross section 50' wide by (80' - 50' = 30') long.

The height of the peak is calculated using Equation 2,  based on the shortest dimension as the width of the building.

H = W * H.F.
=  50 * 0.2122
=  10.6  feet.

The volume of the two 1/2 pyramidal shaped sections together has been calculated in Example 7.

Bu. = 0.209 * W2 * H.
= 0.209 *  50 * 50 * 10.6
=  5,539 Bushels

The volume of the triangular shaped section between the two ends is calculated using Equation 4.

Bu = 0.4 * W * H * L
= 0.4 * 50 * 10.6 * 30
= 6,360 Bushels

The total capacity of this structure therefore is 6,400 + 5,539 + 6,360 = 18,299 bushels