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001972

Hedi Boukraa, M.Sc.

2025-08-07

Required Reinforcement in Bucket with Rough Inner Walls: Design Approach and Steps

The article describes the complete derivation and design of the required bucket reinforcement. Both horizontal and vertical stirrups are analyzed.

The foundation to be designed is a bucket foundation with rough bucket sides. The connected column has a rectangular cross-section with dimensions 30 cm × 40 cm. The design is performed according to DIN EN 1992-1-1. The materials used are concrete with strength class C 35/45 and reinforcing steel of grade B500S(A). The concrete cover corresponds to the minimum requirement according to the standard, designed for exposure class XC3 for a manufacturing method on prepared subsoil.

Dimensions of the Foundation Slab and Bucket

All dimensions are to be entered in the Geometry tab. The slab has a width of 3.30 m, a length of 2.60 m, and a thickness of 0.36 m. The bucket has a height of 1.31 m; the embedment depth of the column is also 1.31 m. The eccentricity of the column is −0.30 m in the x-direction, related to the center of the foundation slab.

Geometric details of a foundation plate with bucket, focusing on technical aspects of the structure.

Foundation Slab and Bucket Geometry

Top view of bucket foundation shows detailed dimensions for structural design.

Top View of Bucket Foundation with Dimensions

Two variants are possible for the arrangement of the horizontal stirrups in the bucket:

Stirrups enclosing the column
Stirrups lying completely in one bucket wall

Enclosing stirrups are used for this example. In the Reinforcement tab, the reinforcing steel material, the reinforcement type of the slab, and the arrangement type of the horizontal stirrups in the bucket are to be specified.

Detailed view of reinforcement settings for a bucket foundation in a CAD software interface.

Bucket Foundation | Tab Reinforcement

Load Cases

The internal forces of the following load cases are available for the ultimate limit state designs:

Load case	P_Z,d [kN]	P_X,d [kN]	P_Y,d [kN]	M_X,d [kNm]	M_Y,d [kNm]
1	300	-50	20	100	250
2	100	0	0	0	327
3	500	0	0	150	-150

Required Bucket Reinforcement

Required Outer Horizontal Reinforcement

Load combination 1 (LC1+LC2) results in the largest required reinforcement area of the horizontal outer stirrups in the bucket and is therefore governing. The following horizontal forces act on the bucket walls:

\{\begin{cases} H_{t, x} = \frac{6}{5} \cdot \frac{M_{y}}{d} + \frac{6}{5} \cdot H_{x} = \frac{6}{5} \cdot \frac{250 kNm}{1.310 m} + \frac{6}{5} \cdot 50 kN = 289.01 kN \\ H_{t, y} = \frac{6}{5} \cdot \frac{M_{x}}{d} + \frac{6}{5} \cdot H_{y} = \frac{6}{5} \cdot \frac{100 kNm}{1.310 m} + \frac{6}{5} \cdot 20 kN = 115.60 kN \end{cases}

Horizontal Forces on Bucket Walls from Load Combination 1

This stirrup is subjected to tension by two different loads: On the one hand, a tensile force arises due to the bending of the bucket wall, which runs perpendicular to the considered horizontal force. On the other hand, a tensile force acts due to the direct tensile stress of the wall, which is aligned parallel to the considered horizontal force.

From the horizontal force H_t,x, two tensile forces then result in the outer stirrups:

In the wall in the x-direction, the tensile force T_H,x arises due to the tensile stress.
In the wall in the y-direction, the tensile force T_{H,out,x(wall)} results due to the bending of this orthogonal wall.

From the horizontal force H_t,y, the following also results:

In the wall in the y-direction, the tensile force T_H,y arises due to the tensile stress.
In the wall in the x-direction, the tensile force T_{H,out,y(wall)} results due to the bending of this orthogonal wall.

Tensile Forces in Outer Stirrups

To determine the maximum tensile force in the stirrup, both contributions from bending and tension must be added. This is necessary because their causes – the simultaneously acting horizontal components (H_t,x and H_t,y) – originate from the same load case and are therefore effective simultaneously.

Required reinforcement area of the horizontal stirrups in the bucket due to tensile force in the x-direction

The upper horizontal force H_t,x can be split so that half of it acts at each of the quarter points of the longitudinal extension of the column in the y-direction.

Detailed view of the internal moments in a structural model with different load configurations.

Determination of Internal Moment

Half of the half horizontal tensile force H_t,x is distributed to each of the two stirrup levels. The proportional tensile force in the outer stirrups is then:

T_{h, x} = \frac{|H_{t, x}|}{4} = \frac{|289.01 kN|}{4} = 72.25 kN

Proportional Tension Force T_h,x

The required reinforcement area of the horizontal stirrups in the bucket due to the tensile force in the x-direction is therefore:

A_{sw, h, req} (H_{t, x}) = \frac{T_{h, x}}{f_{yd}} = \frac{72.25 kN}{43.47 kN / {cm}^{2}} = 1.66 cm ²

Required Reinforcement Area of Horizontal Stirrups in Bucket Due to Tensile Force in x-Direction

Required reinforcement area of the horizontal stirrups in the bucket due to tensile force in the y-direction

The required reinforcement area of the horizontal stirrups in the bucket due to the tensile force in the y-direction H_t,y is determined analogously to the determination from H_t,x:

T_{h, y} = \frac{|H_{t, y}|}{4} = \frac{|115.60 kN|}{4} = 28.90 kN

A_{sw, h, req} (H_{t, y}) = \frac{T_{h, y}}{f_{yd}} = \frac{28.90 kN}{43.47 kN / cm ²}

Required Reinforcement Area of Horizontal Stirrups in Bucket Due to Tensile Force in y-Direction

Required reinforcement area of the horizontal outer stirrups in the bucket due to bending of H_t,x

If the upper right corner is considered on its own (see image above), it must be in equilibrium of forces. The horizontal internal forces must be balanced with the external, loading horizontal force. Since no vertical external forces act in the section considered here, only a rotational equilibrium now has to be found. First, the sum of the moments about point P is formed. For this, the lever arms are calculated:

a_{2, y} = \frac{c_{y}}{4} = \frac{0.30 m}{4} = 0.08 m

a_{3, y} = \frac{c_{y}}{2} + a_{ty} + c_{(k)} + \frac{d_{s, h}}{2} = \frac{0.30 m}{2} + 0.10 m + 0.05 m + \frac{0.014 m}{2} = 0.31 m

a_{4, y} = \frac{d_{b, y}}{2} - c_{(k)} - \frac{d_{s, h}}{2} = \frac{1.24 m}{2} - 0.05 m - \frac{0.014 m}{2} = 0.56 m

Determination of Lever Arms

The moment about point P is therefore calculated as:

M_{Ed} = \frac{|H_{t, x}|}{4} \cdot (a_{3, y} + a_{4, y}) - \frac{|H_{t, x}|}{2} \cdot a_{2, y} = \frac{|289.01 kN|}{4} \cdot (0.31 m + 0.56 m) - \frac{|289.01 kN|}{2} \cdot 0.08 = 52.02 kNm

Bending Design Moment of Bucket Wall in y-Direction

This design bending moment M_Ed must be balanced by an opposite moment. This is created by the concrete compressive force on the compressed inner side of the bucket wall in the y-direction with the lever arm z to the pivot point P. To determine the necessary concrete strain, proceed as follows:

It is assumed that the concrete stress is distributed uniformly over the compression zone.
Starting from a concrete strain of 0.0 ‰, this is increased step by step until the resulting internal moment corresponds to the design bending moment M_Ed.

At the same time, it is assumed that the steel on the outer side of the bucket wall has already reached its maximum strain. As soon as the calculated internal moment is greater than the design bending moment M_Ed, the iteration is terminated.

The following values were reached at the end of the iteration:

Designation	Value
Required reinforcement area A_s,w,erf	5.60 cm²
Design bending moment M_Ed	52.02 kNm
Lever arm of internal forces z	0.21 m
Used neutral axis depth x	0.04 m
Reinforcement strain on tension side ε_s	18.90 ‰
Reinforcement stress on tension side σ_s	449.899 N/mm²
Reinforcement strain on compression side ε_s,c	1.2 ‰
Reinforcement stress on compression side σ_s,c	247.515 N/mm²
Concrete strain on compression side ε_c	-3.5 ‰
Concrete stress on compression side σ_c	-19.833 N/mm²
Concrete compressive force F_c	252.07 kN
Tensile force F_s	252.07 kN

The required reinforcement area in the bucket due to bending is 5.60 cm². This bending tensile force is not distributed equally to both horizontal stirrups, but follows the compression strut mechanics. For more information, see the Concrete Foundations manual.

RFEM 6 / RSTAB 9 | Concrete Foundations

Angle of Dispersal

To determine the portion of the tensile force that is absorbed by the outer bucket stirrup, you can follow these steps:

Determination of the vertical and horizontal legs of the concrete compression strut within the bucket wall:

\begin{array}{l} a_{5, x} = d_{x (wall)} - k_{x (wall)} - \frac{X_{x (wall)}}{4} = 0.223 m - 0.832 \cdot \frac{0.04 m}{4} = 0.21 m \\ a_{6, y} = \frac{d_{b, y}}{2} - \frac{3}{8} \cdot c_{y} - c_{(k)} - \frac{d_{s, h}}{2} = \frac{1.24 m}{2} - \frac{3}{8} \cdot 0.30 m - 0.05 m - \frac{0.014 m}{2} = 0.45 m \end{array}

Vertical and Horizontal Lever Arms of Concrete Compression Strut Within Bucket Wall

Load distribution angle within the bucket wall:

θ_{1, x} = \arctan (\frac{a_{5, x}}{a_{6, y}}) = \arctan (\frac{0.21 m}{0.45 m}) = 25.39 °

Angle of Dispersal Within Bucket Wall

Proportional compressive force in outer stirrups due to bending of the bucket wall:

P_{1, x (wall)} = \frac{|H_{t, x}|}{4 \cdot \sin (θ_{1, x})} = \frac{|289.01 kN|}{4 \cdot \sin (25.39 °)} = 168.48 kN

Proportional Compressive Force in Outer Stirrup Due to Bending of Bucket Wall

Proportional tensile force in outer stirrups due to bending of the bucket wall:

T_{h, out, x (wall)} = P_{1, x (wall)} \cdot \cos (θ_{1, x}) = 168.48 kN \cdot \cos (25.39 °) = 152.20 kN

Proportional Tensile Force in Outer Stirrups Due to Bending of Bucket Wall

The required reinforcement area of the horizontal outer stirrups in the bucket due to bending of Ht,x is:

A_{sw, h, out, req} (M_{Ed} | H_{t, x}) = \frac{T_{h, out, x (wall)}}{f_{yd}} = \frac{152.20 kN}{43.47 kN / cm ²} = 3.50 cm ²

Required Reinforcement Area of Horizontal Outer Stirrups in Bucket Due to Bending of Ht,x

Analogous to the bucket wall in the y-direction, the bending failure of the bucket wall in the x-direction due to H_T,y is calculated and the required reinforcement area of the horizontal outer stirrups in the bucket due to bending of H_t,y A_sw,h,out,erf (M_Ed|H_t,y) is determined. It amounts to 0.76 cm².

As explained above, the contributions from bending and tension must be added to determine the required reinforcement area of the horizontal outer stirrups in the bucket:

The required reinforcement area of the horizontal outer stirrups in the bucket due to bending of H_t,x A_sw,h,out,erf(M_Ed|H_t,x) is added to the required reinforcement area of the horizontal stirrups in the bucket due to the tensile force in the y-direction A_sw,h,erf(H_t,y).
The required reinforcement area of the horizontal outer stirrups in the bucket due to bending of H_t,y A_sw,h,out,erf(M_Ed|H_t,y) is added to the required reinforcement area of the horizontal stirrups in the bucket due to the tensile force in the x-direction A_sw,h,erf(H_t,x).

A_{sw, \erf, B, h} = \{\begin{array}{l} (A_{sw, h, \erf} (H_{t, x}) + A_{sw, h, out, req} (M_{Ed} | H_{t, y}) = 1.66 cm ² + 0.76 cm ² = 2.42 cm ² \\ (A_{sw, h, \erf} (H_{t, y}) + A_{sw, h, out, req} (M_{Ed} | H_{t, x}) = 0.66 cm ² + 3.50 cm ² = 4.17 cm ² \end{array} = 4.17 cm ²

Required Reinforcement Area of Horizontal Outer Stirrups in Bucket

Required Horizontal Reinforcement in the y-direction

First, the horizontal stirrup lying on the outside of the bucket wall in the y-direction is considered. Load combination 2 (LC1+LC3) results in the largest required reinforcement area of the horizontal stirrups in the y-direction and is therefore governing.

3D structural model display: reinforcement analysis in the y-direction with detailed element views.

Determination of Reinforcement in y-Direction

From the figure shown above, it can be seen that only horizontal forces in the x-direction are assumed to cause tensile forces in this stirrup. For more information, see the Concrete Foundations manual.

The required reinforcement area of the horizontal stirrups in the y-direction in the bucket due to the bending caused by H_t,x results from the total required bending reinforcement of the wall in the y-direction, minus the portion that is already absorbed by the outer stirrups. For load combination 2 (LC2), the total required reinforcement area due to bending in the y-direction is 5.83 cm². Of this, 3.63 cm² is already absorbed by the outer horizontal stirrups. Thus, for the horizontal stirrups in the y-direction in the bucket due to the bending caused by H_t,x, a still required reinforcement area of:

A_{sw, h, in, req} (M_{Ed} | H_{t, x}) = A_{sw, h, req} (M_{Ed} | H_{t, x}) - A_{sw, h, out, req} (M_{Ed} | H_{t, x}) = 5.83 cm ² - 3.63 cm ² = 2.20 cm ²

Required Reinforcement Area of Horizontal Stirrups in y-Direction in Bucket Due to Bending of H_t,x

The required reinforcement area of the horizontal stirrups in the bucket due to the tensile force in the x-direction must also be taken into account. It is:

A_{sw, h, req} (H_{t, x}) = \frac{T_{h, x}}{f_{yd}} = \frac{74.89 kN}{43.47 kN / {cm}^{2}} = 1.72 cm ²

Required Reinforcement Area of Horizontal Stirrups in Bucket Due to Tension in x-Direction

The required reinforcement area of the horizontal stirrups in the bucket in the y-direction is the maximum of A_sw,h,erf(H_t,x) and A_sw,h,in,erf (M_Ed|H_t,x):

A_{sw, req, B, h, y} = \max \{\begin{array}{l} A_{sw, h, req} (H_{t, x}) = 1.72 cm ² \\ A_{sw, h, in, req} (M_{Ed} | H_{t, x}) = 2.20 cm ² \end{array} = 2.20 cm ²

Required Reinforcement Area of Horizontal Stirrups in Bucket in y-Direction

Required Horizontal Reinforcement in the x-direction

Analogous to the reinforcement in the y-direction, only horizontal forces acting perpendicular to the direction of the stirrup leg cause a tensile force in the stirrup. LC3 (LC1+LC4) is governing here.

The reinforcement in the x-direction of a concrete structure is displayed.

Determining Reinforcement in x-Direction

The required reinforcement area of the horizontal stirrups in the bucket in the x-direction is the maximum of A_sw,h,erf(H_t,y) and A_sw,h,in,erf(M_Ed|H_t,y):

Required Vertical Reinforcement in the x-direction

To determine the vertical edge reinforcement of the bucket wall in the x-direction, the load case that leads to the maximum horizontal force in the x-direction is considered (LC2). The horizontal force H_t,x = 299.54 kN is distributed equally to both bucket wall panels.

The slope of the concrete compression strut, which forms diagonally across the bucket wall panel in the x-direction, is determined as follows:

Force model for determining the vertical edge pull force

Force Model for Determining Vertical Edge Tension Force

Thus, the edge tensile force can be determined: Only half of the edge tensile force is considered, since the stirrup is designed with two legs. The resulting required reinforcement is:

A_{sw, req, B, v, x} = 0.5 \cdot \frac{T_{v, x}}{f_{yd}} = 0.5 \cdot \frac{183.50 kN}{43.347 kN / cm ²} = 2.11 cm ²

Required Reinforcement Area of Vertical Stirrups in Bucket in x-Direction

The required reinforcement area of the vertical stirrups in the bucket in the y-direction is calculated in the same way. The result is a required reinforcement area A_sw,erf,B,v,y of 0.93 cm².

The required bucket reinforcement is listed in the result table "Concrete Foundations" in the section "Reinforcement at the Foundation". In addition, the reinforcement can be displayed graphically via the results navigator.

Input of the Bucket Reinforcement

Two distribution areas are to be applied for the outer stirrups. The first distribution area extends over one third of the bucket height. The second distribution area corresponds to the remaining portion of the bucket height. Four stirrups with a diameter of 14 mm are used per distribution area. The arrangement is identical for the two remaining stirrup groups in the x- and y-directions.

Two stirrups are selected for each edge of the bucket wall panels in the x- and y-directions. In addition, the bucket walls are reinforced constructively by vertical stirrups at a distance of 20 cm.

Bucket Reinforcement Input

The following two images show the layout and the description of the bucket reinforcement:

Horizontal Stirrups | Layout and Description

Vertical reinforcement in the y-direction | Layout and description

Vertical Reinforcement in y-Direction | Layout and Description

Author

Hedi works in Product Engineering, specializing in reinforced concrete, and is also involved in quality assurance.

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