Actions on Silos According to EN 1991-4

Technical Article

Silos are used as large containers for storing bulk materials such as agricultural products or source materials as well as intermediates of industrial production. Structural engineering of such structures requires a precise knowledge of the stresses due to particulate solids in the building structure. The standard EN 1991‑4 ‘Actions on Silos and Tanks’ [1] provides general principles and requirements for determining these actions.

Scope of Application

The application of the design rules for silos and tanks is subjected to geometrical limitations. In [1], the geometric dimensions are limited to hb / dc < 10 with hb < 100 m and dc < 60 m. In addition, the application limits depend on the cross‑section shape of the silo and on the stored solids.

Figure 01 - Silo Forms Showing Dimensions and Pressure Notation

Properties of Particulate Solids

Annex E of [1] specifies parameters of the most common solids stored in silos, showing the range of particulate solids properties. Furthermore, Section 4 and Annex C of [1] describe test methods for the determination of the stored solids properties.

Wall frictional properties of the particulate solids take into account the roughness of wall surfaces where the solids slide along. Table 4.1 of [1] describes various categories of wall surfaces. The categories of the wall surfaces are shown in the table below. Annex D.2 of [1] also provides information for the evaluation of the wall friction coefficient for the D4 category.

Figure 02 – Categories of Wall Surfaces

You should always determine the loads of a load case for a particular combination of the relevant solid properties. For each of these load cases, the extreme values are reached when the solid properties take different extreme values within discharge flow of the particulate solids. The applicable extreme values of the particulate solid properties are listed in Table 3.1 of [1] for each of the load cases to be examined. The relevant parameters for various load applications are included in the following table.

Figure 03 - Relevant Parameters for Various Load Applications

Action Assessment Class

Silos are divided into three action assessment classes according to their storage capacity and eccentricity in compliance with Table 2.1 of [1].

Figure 04 - Classification of Design Situations

According to the respective action assessment class, various differentiated or simplified loading assessments are adopted.

Loads on Vertical Walls of Silos

Loads on the vertical walls of silos are subjected to a differentiated calculation considering silo slenderness. A distinction is made between:

  • slender silos (hc / dc ≥ 2.0)
  • intermediate slenderness silos (1.0 < hc / dc < 2.0)
  • squat silos (0.4 < hc / dc ≤ 1.0)
  • retaining silos (hc / dc ≤ 0.4 and the silo bottom is flat)

Figure 05 - Pressure Distribution in Silo According to Silo Slenderness

Symmetrical Loads

Symmetrical loads are fixed loads that are uniformly distributed over the silo circumference. Discharge loads arise when the uniform loads in full condition are increased by a load magnifying factor.

Unsymmetrical Loads

Besides the fixed loads, additional free loads are usually to be applied. Distributions of unsymmetrical loads (patch loads) in a silo are caused by actions due to imperfections or eccentricities during filling and discharge of solids.

In the case of thick‑walled circular silos, the patch load applies to two opposite square areas with side length s. In the case of non‑circular silos, the patch loads can be taken into account by increasing the symmetrical loads. The outward patch pressure should be taken to act on a horizontal band on the silo wall at any level, over a vertical height s.

Figure 06 - Patch Load Application

Generally, it is not necessary to apply the patch loads in the case of squat and intermediate slenderness silos.

For silos in Action Assessment Class 2, the patch load method may be approximately used by uniformly increasing horizontal pressures.

Discharge Loads with Large Eccentricities

According to [1], the loads due to large discharge eccentricities should be used as a separate load case. The development of this loading assessment is based on a premise that a flow channel may occur near the wall as a result of large eccentric discharge. A pipe flow channel is assumed, which is constant due to the height of the silo wall, and intersects the silo wall at an opening angle θc.

Figure 07 - Flow Channel and Pressure Distribution for Silos with Large Outlet Eccentricities

However, a theoretical prediction of the geometric shape of a discharge hopper is hardly possible with the tools currently available. Therefore, the flow channel must be specified. The calculation is performed with at least three different flow channel radii rc in order to determine the apparent variations of the flow channel.

In the contact areas of the flowing solid and the silo wall, lower horizontal pressures occur outside the flow channel. In the latter area, the loads of the filling load case apply. Directly next to the flow channel up to the opening angle of 2 θc, the pressure is increased.

Figure 08 - Load Application for Silos with Large Outlet Eccentricities

Large Eccentricity Filling Loads

Loads due to eccentric filling must be considered for squat or intermediate slenderness silos.

EN 1991‑4 [1] explains the determination of the additional vertical force (compressive) in the wall per unit length of circumference at any depth zs below the point of highest wall contact. This force per unit circumference should be added to the force arising from wall friction.

Figure 09 - Filling Pressures in Eccentrically Filled Squat or Intermediate Slenderness Silo

Loads on Silo Hoppers and Silo Bottoms

The loads on the walls of silo hoppers should be determined with regard to the steepness of the hopper walls according to [1].

Figure 10 - Filling and Discharge Pressures in Hopper

The standard distinguishes between flat bottoms as well as steep and shallow hoppers. In the case of steep hoppers, there is an additional distinction between the load case of filling and discharge. Kick load at the transition from the vertical walled section to the hopper is already included in the load distributions.

Annex G of [1] provides alternative rules for pressures in hoppers.

Example

The example presents a free standing cylindrical silo for cement with a diameter of 5.00 m and the maximum bulk unit depth of 8.00 m. The silo is made of reinforced concrete with a wall thickness of 0.30 m.

Figure 11 - Layout and Dimensions of Cement Silo

Particulate Solids

The following particulate solid properties of cement are given by Table E.1 of [1].

Unit weight (upper)  γu  =  16.00  kN/m³
Angle of repose  Φr  =  36.00 °
Angle of internal friction (mean)  Φim  =  30.00 °
Factor  aφ  =  1.22  
Lateral pressure ratio (mean)  Κm  =  0.54  
Factor  aΚ  =  1.20  
Wall friction coefficient (wall type D3)  μm  =  0.51  (for concrete)
Factor  aμ  =  1.07  
Patch load solid reference factor  Cop  =  0.50  

Characteristic Particulate Solids Properties

In order to determine the characteristic values of the lateral pressure ratio, wall friction coefficient, and the angle of internal friction, the listed mean values of the particulate solids must be scaled using the conversion factors. The factors ax are specified in Table E.1 of [1] for the available particulate solids.

Upper and lower characteristic value of the lateral pressure ratio:

$$\begin{array}{l}{\mathrm K}_\mathrm u\;=\;{\mathrm a}_\mathrm K\;\cdot\;{\mathrm K}_\mathrm m\;=\;1.20\;\cdot\;0.54\;=\;0.648\\{\mathrm K}_\mathrm l\;=\;\frac{{\mathrm K}_\mathrm m}{{\mathrm a}_\mathrm K}\;=\;\frac{0.54}{1.20}\;=\;0.450\end{array}$$

Upper and lower characteristic value of the wall friction coefficient:

$$\begin{array}{l}{\mathrm\mu}_\mathrm u\;=\;{\mathrm a}_\mathrm\mu\;\cdot\;{\mathrm\mu}_\mathrm m\;=\;1.07\;\cdot\;0.51\;=\;0.546\\{\mathrm\mu}_\mathrm l\;=\;\frac{{\mathrm\mu}_\mathrm m}{{\mathrm a}_\mathrm\mu}\;=\;\frac{0.51}{1.07}\;=\;0.477\end{array}$$

Upper and lower characteristic value of the angle of internal friction:

$$\begin{array}{l}{\mathrm\Phi}_\mathrm{iu}\;=\;{\mathrm a}_\mathrm\Phi\;\cdot\;{\mathrm\Phi}_\mathrm{im}\;=\;1.22\;\cdot\;30.00^\circ\;=\;36.60^\circ\\{\mathrm\Phi}_\mathrm{iu}\;=\;\frac{{\mathrm\Phi}_\mathrm{im}}{{\mathrm a}_\mathrm\Phi}\;=\;\frac{30.00^\circ}{1.22}\;=\;24.59^\circ\end{array}$$

Values of properties to be used for different wall loading assessments

The evaluation of each load case should be made using a single set of consistent values of the solids properties, so that each limit state corresponds to a single defined stored solid condition. The extreme values of solids properties that should be adopted for each load case is given in the following table.

Figure 12 – Values of Properties to be Used for Different Wall Loading Assessments

The angle of wall friction must always be less than or equal to the angle of internal friction of the stored solid, that is Φwh ≤ Φi. Otherwise, the material will rupture internally if slip at the wall contact demands a greater shear stress than the internal friction can sustain. This means that, in all cases, the wall friction coefficient should not be taken as greater than tanΦi (i.e. μ = tanΦw ≤ tanΦi). This is considered in the table above where the relevant values are in bold.

Actions

Actions are determined according to [1]. Only the filling loads on vertical walls and vertical pressures on flat bottoms of the silo should be calculated here.

Silo Classification

The classification of the silo is based on the slenderness and the action assessment class.

Slenderness
$$1.0\;<\;\frac{{\mathrm h}_\mathrm c}{{\mathrm d}_\mathrm c}\;=\;\frac{8.00}{5.00}\;=\;1.6\;<\;2$$

The silo is classified as an intermediate slenderness silo in accordance with 1.5.21 of [1].

Action Assessment Class
$$\mathrm{Capacity}\;=\;\mathrm V\;\cdot\;{\mathrm\gamma}_\mathrm u\;=\;157.08\;\cdot\;16.00\;=\;2,513.27\;\cong\;\frac{2,513.27}{9.80665}\;=\;256.28\;\mathrm t$$

According to Table 2.1 of [1], at least Action Assessment Class 2 must be selected.

Construction form
$$\frac{{\mathrm d}_\mathrm c}{\mathrm t}\;=\;\frac{5.00}{0.20}\;=\;25\;<\;200$$

The silo is classified as a thick‑walled silo in accordance with 1.5.43 of EN 1991‑4 [1].

Symmetrical Filling Loads on Vertical Walls

Horizontal Pressure
Janssen characteristic depth zo
$$\begin{array}{l}{\mathrm z}_o\;=\;\frac1{\mathrm K\;\cdot\;\mathrm\mu}\;\cdot\;\frac{\mathrm A}{\mathrm U}\;\;\;\;\;(5.75)\\{\mathrm z}_o\;=\;\frac1{0.648\;\cdot\;0.458}\;\cdot\;\frac{19.63}{15.71}\;=\;4.22\;\mathrm m\\\end{array}$$
Vertical distance ho

For a symmetrically filled circular silo, the vertical distance ho between the equivalent surface of the solid and the highest solid‑wall contact is calculated as follows:

$$\begin{array}{l}{\mathrm h}_o\;=\;\frac{{\mathrm d}_\mathrm c}{6\;\cdot\;\tan\;{\mathrm\Phi}_\mathrm r}\;\;\;\;\;(5.77)\\{\mathrm h}_o\;=\;\frac{5.00}{6\;\cdot\;\tan\;36.00^\circ}\;=\;0.61\;\mathrm m\\\end{array}$$
Parameter n
$$\begin{array}{l}\mathrm n\;=\;-(1\;+\;\tan\;{\mathrm\Phi}_\mathrm r)\;\cdot\;\frac{1\;-\;{\mathrm h}_o}{{\mathrm z}_o}\;\;\;\;\;(5.76)\\\mathrm n\;=\;-(1\;+\;\tan\;36.00^\circ)\;\cdot\;\frac{1\;-\;0.61}{4.22}\;=\;-1.48\\\end{array}$$
Asymptotic horizontal pressure at great depth due to stored particulate solid pho
$$\begin{array}{l}{\mathrm p}_\mathrm{ho}\;=\;\mathrm\gamma\;\cdot\;\mathrm K\;\cdot\;{\mathrm z}_o\;\;\;\;\;(5.73)\\{\mathrm p}_\mathrm{ho}\;=\;16.00\;\cdot\;0.648\;\cdot\;4.22\;=\;43.70\;\mathrm{kN}/\mathrm m^2\\\end{array}$$
Horizontal pressure phf(z)
$$\begin{array}{l}{\mathrm p}_\mathrm{hf}(\mathrm z)\;=\;{\mathrm p}_\mathrm{ho}\;\cdot\;{\mathrm Y}_\mathrm R(\mathrm z)\;=\;{\mathrm p}_\mathrm{ho}\;\cdot\;\left(1\;-\;\left(\frac{\mathrm z\;-\;{\mathrm h}_\mathrm o}{{\mathrm z}_\mathrm o\;-\;{\mathrm h}_\mathrm o}\;+\;1\right)^\mathrm n\right)\;\;\;\;\;(5.71)\\{\mathrm p}_\mathrm{hf}(0.61)\;=\;0\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{hf}(1.61)\;=\;43.70\;\cdot\;\left(1\;-\;\left(\frac{1.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;13.26\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{hf}(2.61)\;=\;43.70\;\cdot\;\left(1\;-\;\left(\frac{2.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;20.93\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{hf}(3.61)\;=\;43.70\;\cdot\;\left(1\;-\;\left(\frac{3.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;25.83\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{hf}(4.61)\;=\;43.70\;\cdot\;\left(1\;-\;\left(\frac{4.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;29.19\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{hf}(5.61)\;=\;43.70\;\cdot\;\left(1\;-\;\left(\frac{5.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;31.62\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{hf}(6.61)\;=\;43.70\;\cdot\;\left(1\;-\;\left(\frac{6.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;33.43\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{hf}(7.61)\;=\;43.70\;\cdot\;\left(1\;-\;\left(\frac{7.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;34.83\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{hf}(8.00)\;=\;43.70\;\cdot\;\left(1\;-\;\left(\frac{8.00\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;35.29\;\mathrm{kN}/\mathrm m^2\end{array}$$
Wall Frictional Traction
Janssen characteristic depth zo
$$\begin{array}{l}{\mathrm z}_o\;=\;\frac1{\mathrm K\;\cdot\;\mathrm\mu}\;\cdot\;\frac{\mathrm A}{\mathrm U}\;\;\;\;\;(5.75)\\{\mathrm z}_o\;=\;\frac1{0.648\;\cdot\;0.458}\;\cdot\;\frac{19.63}{15.71}\;=\;4.22\;\mathrm m\\\end{array}$$
Vertical distance ho

For a symmetrically filled circular silo, the vertical distance ho between the equivalent surface of the solid and the highest solid‑wall contact is calculated as follows:

$$\begin{array}{l}{\mathrm h}_o\;=\;\frac{{\mathrm d}_\mathrm c}{6\;\cdot\;\tan\;{\mathrm\Phi}_\mathrm r}\;\;\;\;\;(5.77)\\{\mathrm h}_o\;=\;\frac{5.00}{6\;\cdot\;\tan\;36.00^\circ}\;=\;0.61\;\mathrm m\\\end{array}$$
Parameter n
$$\begin{array}{l}\mathrm n\;=\;-(1\;+\;\tan\;{\mathrm\Phi}_\mathrm r)\;\cdot\;\frac{1\;-\;{\mathrm h}_o}{{\mathrm z}_o}\;\;\;\;\;(5.76)\\\mathrm n\;=\;-(1\;+\;\tan\;36.00^\circ)\;\cdot\;\frac{1\;-\;0.61}{4.22}\;=\;-1.48\\\end{array}$$
Asymptotic horizontal pressure at great depth due to stored particulate solid pho
$$\begin{array}{l}{\mathrm p}_\mathrm{ho}\;=\;\mathrm\gamma\;\cdot\;\mathrm K\;\cdot\;{\mathrm z}_o\;\;\;\;\;(5.73)\\{\mathrm p}_\mathrm{ho}\;=\;16.00\;\cdot\;0.648\;\cdot\;4.22\;=\;43.70\;\mathrm{kN}/\mathrm m^2\\\end{array}$$
Wall frictional traction pwf(z)
$$\begin{array}{l}{\mathrm p}_\mathrm{wf}(\mathrm z)\;=\;\mathrm\mu\;\cdot\;{\mathrm p}_\mathrm{ho}\;\cdot\;{\mathrm Y}_\mathrm R(\mathrm z)\;=\;\mathrm\mu\;\cdot\;{\mathrm p}_\mathrm{ho}\;\cdot\;\left(1\;-\;\left(\frac{\mathrm z\;-\;{\mathrm h}_\mathrm o}{{\mathrm z}_\mathrm o\;-\;{\mathrm h}_\mathrm o}\;+\;1\right)^\mathrm n\right)\;\;\;\;\;(5.72)\\\;{\mathrm p}_\mathrm{wf}(0.61)\;=\;0\;\mathrm{kN}/\mathrm m^2\\\;{\mathrm p}_\mathrm{wf}(1.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{1.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;6.07\;\mathrm{kN}/\mathrm m^2\\\;{\mathrm p}_\mathrm{wf}(2.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{2.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;9.58\;\mathrm{kN}/\mathrm m^2\\\;{\mathrm p}_\mathrm{wf}(3.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{3.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;11.82\;\mathrm{kN}/\mathrm m^2\\\;{\mathrm p}_\mathrm{wf}(4.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{4.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;13.36\;\mathrm{kN}/\mathrm m^2\\\;{\mathrm p}_\mathrm{wf}(5.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{5.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;14.47\;\mathrm{kN}/\mathrm m^2\\\;{\mathrm p}_\mathrm{wf}(6.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{6.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;15.30\;\mathrm{kN}/\mathrm m^2\\\;{\mathrm p}_\mathrm{wf}(7.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{7.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;15.94\;\mathrm{kN}/\mathrm m^2\\\;{\mathrm p}_\mathrm{wf}(8.00)\;=\;0.458\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{8.00\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;16.15\;\mathrm{kN}/\mathrm m^2\end{array}$$
Vertical Pressure
Janssen characteristic depth zo
$$\begin{array}{l}{\mathrm z}_\mathrm o\;=\;\frac1{\mathrm K\;\cdot\;\mathrm\mu}\;\cdot\;\frac{\mathrm A}{\mathrm U}\;\;\;\;\;(5.75)\\\;z_\mathrm o\;=\;\frac1{0.450\;\cdot\;0.477}\;\cdot\;\frac{19.63}{15.71}\;=\;5.83\;\mathrm m\\\end{array}$$
Parameter n
$$\begin{array}{l}\mathrm n\;=\;-(1\;+\;\tan\;{\mathrm\Phi}_\mathrm r)\;\cdot\;\frac{1\;-\;{\mathrm h}_\mathrm o}{{\mathrm z}_\mathrm o}\;\;\;\;\;(5.76)\\\mathrm n\;=\;-(1\;+\;\tan\;36.00^\circ)\;\cdot\;\frac{1\;-\;0.61}{5.83}\;=\;-1.55\\\end{array}$$
Vertical pressure pvf(z)
$$\begin{array}{l}{\mathrm p}_\mathrm{vf}(\mathrm z)\;=\;\mathrm\gamma\;\cdot\;{\mathrm z}_\mathrm v(\mathrm z)\;=\;\mathrm\gamma\;\cdot\;\left({\mathrm h}_\mathrm o\;-\;\frac1{\mathrm n\;+\;1}\;\cdot\;\left({\mathrm z}_\mathrm o\;-\;{\mathrm h}_\mathrm o\;-\;\frac{(\mathrm z\;+\;{\mathrm z}_\mathrm o\;-\;2\;\cdot\;{\mathrm h}_\mathrm o)^{\mathrm n+1}}{({\mathrm z}_\mathrm o\;-\;{\mathrm h}_\mathrm o)^\mathrm n}\right)\right)\;\;\;\;\;\;(5.79)\\{\mathrm p}_\mathrm{vf}(0.61)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(0.61\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;9.69\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{vf}(1.61)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(1.61\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;23.65\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{vf}(2.61)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(2.61\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;34.51\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{vf}(3.61)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(3.61\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;43.27\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{vf}(4.61)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(4.61\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;50.52\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{vf}(5.61)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(5.61\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;56.65\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{vf}(6.61)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(6.61\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;61.92\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{vf}(7.61)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(7.61\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;66.50\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{vf}(8.00)\;=\;16.00\;\cdot\;\left(0.61\;-\;\frac1{-1.55\;+\;1}\;\cdot\;\left(5.83\;-\;0.61\;-\;\frac{(8.00\;+\;5.83\;-\;2\;\cdot\;0.61)^{-1.55+1}}{(5.83\;-\;0.61)^{-1.55}}\right)\right)\;=\;68.15\;\mathrm{kN}/\mathrm m^2\end{array}$$
Vertical forces (compressive) in the wall nsk(z)
$$\begin{array}{l}{\mathrm n}_\mathrm{zSk}(\mathrm z)\;=\;\mathrm\mu\;\cdot\;{\mathrm p}_\mathrm{ho}(\mathrm z)\;\cdot\;(\mathrm z\;-\;{\mathrm z}_\mathrm v)\;\;\;\;\;(5.81)\\{\mathrm n}_\mathrm{zSk}(0.61)\;=\;0.00\;\mathrm{kN}/\mathrm m\\{\mathrm n}_\mathrm{zSk}(1.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;(1.61\;-\;1.48)\;=\;2.55\;\mathrm{kN}/\mathrm m\\{\mathrm n}_\mathrm{zSk}(2.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;(2.61\;-\;2.16)\;=\;8.97\;\mathrm{kN}/\mathrm m\\{\mathrm n}_\mathrm{zSk}(3.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;(3.61\;-\;2.70)\;=\;18.02\;\mathrm{kN}/\mathrm m\\{\mathrm n}_\mathrm{zSk}(4.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;(4.61\;-\;3.16)\;=\;28.96\;\mathrm{kN}/\mathrm m\\{\mathrm n}_\mathrm{zSk}(5.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;(5.61\;-\;3.54)\;=\;41.30\;\mathrm{kN}/\mathrm m\\{\mathrm n}_\mathrm{zSk}(6.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;(6.61\;-\;3.87)\;=\;54.72\;\mathrm{kN}/\mathrm m\\{\mathrm n}_\mathrm{zSk}(7.61)\;=\;0.458\;\cdot\;43.70\;\cdot\;(7.61\;-\;4.16)\;=\;68.98\;\mathrm{kN}/\mathrm m\\{\mathrm n}_\mathrm{zSk}(8.00)\;=\;0.458\;\cdot\;43.70\;\cdot\;(8.00\;-\;4.26)\;=\;74.81\;\mathrm{kN}/\mathrm m\end{array}$$

Filling Patch Loads on Vertical Walls

Dimension of the patch load zone
$$\begin{array}{l}\mathrm s\;=\;\frac{\mathrm\pi\;\cdot\;{\mathrm d}_\mathrm c}{16}\;\;\;\;\;\;\;(5.12)\\\mathrm s\;=\;\frac{\mathrm\pi\;\cdot\;5.00}{16}\;=\;0.98\;\mathrm m\end{array}$$
Janssen characteristic depth zo
$$\begin{array}{l}{\mathrm z}_\mathrm o\;=\;\frac1{\mathrm K\;\cdot\;\mathrm\mu}\;\cdot\;\frac{\mathrm A}{\mathrm U}\;\;\;\;\;(5.75)\\\;{\mathrm z}_\mathrm o\;=\;\frac1{0.648\;\cdot\;0.458}\;\cdot\;\frac{19.63}{15.71}\;=\;4.22\;\mathrm m\\\end{array}$$
Vertical distance ho

For a symmetrically filled circular silo, the vertical distance between the equivalent surface of the solid and the highest solid‑wall contact is calculated as follows:

$$\begin{array}{l}{\mathrm h}_\mathrm o\;=\;\frac{{\mathrm d}_\mathrm c}{6\;\cdot\;\tan\;{\mathrm\Phi}_\mathrm r}\;\;\;\;\;(5.77)\\\;{\mathrm h}_\mathrm o\;=\;\frac{5.00}{6\;\cdot\;\tan\;36.00^\circ}\;=\;0.61\;\mathrm m\\\end{array}$$
Parameter n
$$\begin{array}{l}\mathrm n\;=\;-(1\;+\;\tan\;{\mathrm\Phi}_\mathrm r)\;\cdot\;\frac{1\;-\;{\mathrm h}_\mathrm o}{{\mathrm z}_\mathrm o}\;\;\;\;\;(5.76)\\\mathrm n\;=\;-(1\;+\;\tan\;36.00^\circ)\;\cdot\;\frac{1\;-\;0.61}{4.22}\;=\;-1.48\\\end{array}$$
Asymptotic horizontal pressure at great depth due to stored particulate solid pho
$$\begin{array}{l}{\mathrm p}_\mathrm{ho}\;=\;\mathrm\gamma\;\cdot\;\mathrm K\;\cdot\;{\mathrm z}_\mathrm o\;\;\;\;\;(5.73)\\\;{\mathrm p}_\mathrm{ho}\;=\;16.00\;\cdot\;0.648\;\cdot\;4.22\;=\;43.70\;\mathrm{kN}/\mathrm m^2\\\end{array}$$
Filling patch load factor (load magnification factor) Cpf
$$\begin{array}{l}\mathrm E\;=\;2\;\cdot\;\frac{{\mathrm e}_\mathrm f}{{\mathrm d}_\mathrm c}\;\;\;\;\;(5.10)\\\mathrm E\;=\;2\;\cdot\;\frac{0.00}{5.00}\;=\;0.00\end{array}$$ $$\begin{array}{l}{\mathrm C}_\mathrm{pf}\;=\;0.21\;\cdot\;{\mathrm C}_\mathrm{op}\;\cdot\;(1\;+\;2\;\cdot\;\mathrm E²)\;\cdot\;\left(1\;-\;\mathrm e^{-1.5\cdot(\frac{{\mathrm h}_\mathrm c}{{\mathrm d}_\mathrm c}\;-\;1)}\right)\;\;\;\;\;\;\;(5.9)\\{\mathrm C}_\mathrm{pf}\;=\;0.21\;\cdot\;0.50\;\cdot\;(1\;+\;2\;\cdot\;0.002)\;\cdot\;\left(1\;-\;\mathrm e^{-1,5\cdot(\frac{8.00}{5.00}\;-\;1)}\right)\;=\;0.06\;\geq\;0\end{array}$$
Filling patch load
$$\begin{array}{l}{\mathrm p}_\mathrm{pf}(\mathrm z)\;=\;{\mathrm C}_\mathrm{pf}\;\cdot\;{\mathrm p}_\mathrm{hf}(\mathrm z)\;=\;{\mathrm C}_\mathrm{pf}\;\cdot\;{\mathrm p}_\mathrm{ho}\;\cdot\;{\mathrm Y}_\mathrm R(\mathrm z)\;=\;{\mathrm C}_\mathrm{pf}\;\cdot\;{\mathrm p}_\mathrm{ho}\;\cdot\;\left(1\;-\;\left(\frac{\mathrm z\;-\;{\mathrm h}_\mathrm o}{{\mathrm z}_\mathrm o\;-\;{\mathrm h}_\mathrm o}\;+\;1\right)^\mathrm n\right)\;\;\;\;\;\;(5.8)\\{\mathrm p}_\mathrm{pf}(0.61)\;=\;0\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pf}(1.61)\;=\;0.06\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{1.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;0.83\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pf}(2.61)\;=\;0.06\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{2.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;1.30\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pf}(3.61)\;=\;0.06\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{3.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;1.61\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pf}(4.61)\;=\;0.06\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{4.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;1.82\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pf}(5.61)\;=\;0.06\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{5.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;1.97\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pf}(6.61)\;=\;0.06\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{6.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;2.08\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pf}(7.61)\;=\;0.06\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{7.61\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;2.17\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pf}(8.00)\;=\;0.06\;\cdot\;43.70\;\cdot\;\left(1\;-\;\left(\frac{8.00\;-\;0.61}{4.22\;-\;0.61}\;+\;1\right)^{-1.48}\right)\;=\;2.20\;\mathrm{kN}/\mathrm m^2\end{array}$$

 

$$\begin{array}{l}{\mathrm p}_\mathrm{pfi}(\mathrm z)\;=\;\frac{{\mathrm p}_\mathrm{pf}(\mathrm z)}7\;\;\;\;\;(5.13)\\{\mathrm p}_\mathrm{pfi}(0.61)\;=\;0\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pfi}(1.61)\;=\;\frac{0.83}7\;=\;0.12\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pfi}(2.61)\;=\;\frac{1.30}7\;=\;0.19\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pfi}(3.61)\;=\;\frac{1.61}7\;=\;0.23\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pfi}(4.61)\;=\;\frac{1.82}7\;=\;0.26\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pfi}(5.61)\;=\;\frac{1.97}7\;=\;0.28\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pfi}(6.61)\;=\;\frac{2.08}7\;=\;0.30\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pfi}(7.61)\;=\;\frac{2.17}7\;=\;0.31\;\mathrm{kN}/\mathrm m^2\\{\mathrm p}_\mathrm{pfi}(8.00)\;=\;\frac{2.20}7\;=\;0.31\;\mathrm{kN}/\mathrm m^2\end{array}$$

Pressures on Flat Bottoms

The vertical pressure acting on flat bottoms of the intermediate slenderness silos cannot be taken as uniform and the calculation is based on the following load assessments:

$$\begin{array}{l}{\mathrm C}_\mathrm b\;=\;1.0\;\;\;\;\;\;(6.3)\\{\mathrm p}_\mathrm{vb}\;=\;{\mathrm C}_\mathrm b\;\cdot\;{\mathrm p}_\mathrm{vf}(\mathrm{hc})\;=\;1.0\;\cdot\;68.15\;=\;68.15\;\mathrm{kN}/\mathrm m²\;\;\;\;\;\;(6.2)\\{\mathrm h}_\mathrm{tp}\;=\;\tan\;{\mathrm\Phi}_\mathrm r\;\cdot\;\frac{{\mathrm d}_\mathrm c}2\;=\;\tan\;36.00^\circ\;\cdot\;\frac{5.00}2\;=\;1.82\;\mathrm m\;\;\;\;\;\;(\mathrm{Figure}\;6.3)\\{\mathrm p}_\mathrm{vtp}\;=\;\mathrm\gamma\;\cdot\;{\mathrm h}_\mathrm{tp}\;=\;16.00\;\cdot\;1.82\;=\;29.06\;\mathrm{kN}/\mathrm m²\;\;\;\;\;\;(6.15)\\{\mathrm p}_\mathrm{vho}\;=\;\mathrm\gamma\;\cdot\;{\mathrm z}_\mathrm v\;=\;16.00\;\cdot\;0.61\;=\;9.69\;\mathrm{kN}/\mathrm m²\;\;\;\;\;\;(5.79)\\{\mathrm{Δp}}_\mathrm{sq}\;=\;{\mathrm p}_\mathrm{vtp}\;-\;{\mathrm p}_\mathrm{vho}\;=\;29.06\;-\;9.69\;=\;19.37\;\mathrm{kN}/\mathrm m²\;\;\;\;\;\;(6.14)\\{\mathrm p}_\mathrm{vsq}\;=\;{\mathrm p}_\mathrm{vb}\;+\;{\mathrm{Δp}}_\mathrm{sq}\;\cdot\;\frac{2.0\;-\;{\displaystyle\frac{{\mathrm h}_\mathrm c}{{\mathrm d}_\mathrm c}}}{2.0\;-\;{\displaystyle\frac{{\mathrm h}_\mathrm{tp}}{{\mathrm d}_\mathrm c}}}\;=\;68.15\;+\;19.37\;\cdot\;\frac{2.0\;-\;{\displaystyle\frac{8.00}{5.00}}}{2.0\;-\;{\displaystyle\frac{1.82}{5.00}}}\;=\;72.89\;\mathrm{kN}/\mathrm m²\;\;\;\;\;\;(6.13)\end{array}$$

The bottom load magnifying factor Cb applies to silos of Action Assessment Class 2 under the condition that the stored solids do not tend toward dynamic behaviour during the discharge process.

The vertical pressure pvsq on the bottom of a silo may be taken to act both after filling and during discharge.

Entering Loads in RFEM

The defined load can be entered in RFEM. Figure 13 shows the exemplary filling patch load for z = 4.61 m. This load can be entered in RFEM as a free variable load. The load input is displayed in Figure 14.

Figure 13 – Filling Patch Load (z = 4.61 m)

Figure 14 – Entering Filling Patch Load (z = 4.61 m) in RFEM

Reference

[1]   Eurocode 1 - Actions on structures - Part 4: Silos and tanks; EN 1991‑4:2010‑12

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Structural engineering software for finite element analysis (FEA) of planar and spatial structural systems consisting of plates, walls, shells, members (beams), solids and contact elements