# Navier ground pressure

The analysis is based on a ground pressure calculated according to the Naviers equation. For centric load this means a uniform ground pressure and for eccentric a linearly varying pressure.

With eccentricity with regard to one axis the stress distribution will be as above.

# Design forces

The design sections when calculating moments is at the face of the column/wall and for shear forces at a distance d from the column/wall according to EN 1992-1-1 6.2.1 (8).

# Bending reinforcement design

From the design moments the required bending reinforcement in x- and y-direction are calculated according to EN 1992-1-1 3.1.7 and 6.1 with a stress-strain condition according to the below figure.

λ = 0,8 and η = 1 for fck ≤ 50 MPa
λ = 0,8 - (fck / 50) / 400 and η = 1,0 – (fck - 50) / 200 for 50 ≤ fck ≤ 90 MPa

The value of the design compressive strength is defined as:

• fcd = αcc fck / γc
• αcc = 1,0
• γc is the partial safety factor.

Finnish Annex

αcc = 0,85

UK Annex

αcc = 0,85

If the design moment is larger than the moment corresponding to balanced reinforcement required compression reinforcement is calculated.

# Choice of bar diameter

The bar diameter choice is made in the following way:

The program demands a minimum and maximum limit for the interval where the diameter is chosen. The calculation starts with the minimum diameter and then the following three situations can occur:

The distance between the bars is less than the minimum spacing, min cc, defined by the user. The program will then choose the next larger diameter inside the allowed interval and the check is repeated. If the largest allowed diameter has been reached and the bar distance still is less than mincc a check is made that the smallest distance between parallel reinforcement units is fulfilled according to the code if code check has been activated. If this condition cannot be fulfilled the program will choose a diameter larger than the largest allowed diameter.

The distance between the bars is larger than the maximum allowed which is 2*plate thickness. The program will then choose the nearest allowed smaller diameter and repeats the check. If the smallest allowed diameter has been reached and the bar distance still is to large the program will choose a diameter smaller than the smallest allowed diameter. If for the current reinforcement quality the smallest available diameter has been reached and the bar distance still is to large this reinforcement will be chosen with the distance 2*plate thickness.

The bar distance is between the chosen mincc and 2*plate thickness. The current diameter and bar distance will be chosen.

Minimum reinforcement area

Ultimate Limit State

Minimum tensile reinforcement in both directions should according to EN 1992-1-1 9.3.1.1 not be less than:

• As,min ≥ 0,26 (fctm / fyk) b d > 0,0013 b d

where

• b is the width of the slab perpendicular to the reinforcement direction
and
• d is the effective depth of the slab in this direction.

The distance between bars are also not to exceed 2h ≤ 250 mm.

• h is the plate thickness.

Swedish annex

As,min > (0,26 fctm / fyk b d)/4 > (0,0013 b d)/4

Serviceability Limit State

With regard to demands for crack width reduction a minimum reinforcement
according to EN 1992-1-1 7.3.2 will be required in both directions.

• As,min = kc k fct,eff Act / fyk
• Act = Ac/2 for pure bending
• k = 1.0 for h < 300 mm
• k = 0.65 for h > 800 mm
For values in between interpolation is used.
• fct,eff = fctm
• kc = 0.4 for pure bending.

If both ULS and SLS load cases are defined the largest minimum reinforcement area from above will be used.

# Shear check

The capacity with regard to shear is calculated according to EN 1992-1-1 6.2.2. If the shear capacity is inadequate the reinforcement could be increased if the user so wishes. Maximum reinforcement to be used is then ρ ≤ 0,02.

# Punching check

The punching check is made according to EN 1992-1-1 6.4. If the reinforcement area differs in the two directions then a medium value will be used in the calculations. The checks will be made at the column perimeter and for four control perimeters situated 0,5d, d, 1,5d and 2d from the column face.

The punching force is the column load reduced with the ground pressure within the control perimeter according to EN 1992-1-1 6.4.3 (8). The eccentricity is for interior columns calculated according to EN 1992-1-1 6.4.3 (6.43) and for edge- and corner columns are the values β = 1,4 and 1,5 chosen. If the capacity is inadequate the reinforcement will be increased in both directions until the capacity is adequate or the maximum allowed reinforcement ρ ≤ 0,02 is reached. This reinforcement is placed within a distance equal to the column width plus 3d on each side.

Finnish annex

Without shear reinforcement

The capacity is calculated according to EN 1992-1-1 6.4.4 with the factor CRd,c calculated in the following way:

• CRd,c = 0.3 (D/d + 1.5) /(γc (D/d + 4))
• D is the diameter for a circular column and for rectangular columns D = (c1 c2)0.5
• where c1 and c2 are the column measurements.
• d is the effective thickness of the slab
• vmin = 0
• k1 = 0.1

With shear reinforcement

The capacity with shear reinforcement is calculated according to EC2 6.4.5 with VRd,c calculated with the following CRd,c value:

• CRd,c = 0.3 (D/d + 1.5) /(4.5 γc (D/d + 4))
• kmax = 1.6

The formula 6.4.5 (6.52) has been derived for radial reinforcement. In the formula the total reinforcement area inside a region defined by the circumference u1 is used, wherein the term 1.5 (d/sr) Asw is replaced by the total reinforcement area. When calculating the reinforcement area it must be secured that that the reinforcement is adequately anchored at both sides of the punching crack. Normally reinforcement situated within maximum the distance 1.5 d from the column is considered.

# Anchorage lengths

The available and required anchorage lengths are calculated according to EN 1992-1-1 8.4 and 9.8.2.2.

If the reinforcement area has been increased with regard to shear, punching or crack widths it is considered in the calculation.

The distance to the most outer crack x = h/2

Swedish annex

[7] 6.9:233, EN 1992-1-1 12.3.1

• x = max [( fctd / (3 qd))0.5 h, h/2]
• x ≤ available physical length
• fctd = αct,pl fctk,0.05 / γc
• αct,pl = 0.5

# Crack widths

The crack width calculation is performed according to EN 1992-1-1 7.3.4 for all serviceability limit load cases. A crack width design can be performed meaning that the reinforcement area will be increased to reach the crack width criteria defined by the user.