This example describes the input and functionality of the shear force reduction using the example of a two-span beam.
System, Loading, and Combinatorics
In RFEM 6 or RSTAB 9, we create a two-span beam with a span length of 4.0 m. The rectangular cross-section has the dimensions w/h = 35/50 cm. As material, the concrete grade C30/37 was selected.
The load for this example results from a permanent and a variable load. The permanent loads are entered in Load Case 1. This is the self-weight of the cross-section and a line load of gk = 48.75 kN/m. As a variable load, a line load of qk = 37.5 kN/m and, alternatively, a concentrated load introduction as four single loads with Qk = 37.5 kN (member load, concentrated – nx) is entered.
We enter these two loads separately for the respective field and treat them alternatively in the combinatorics. In RFEM and RSTAB, the automatic combinatorics are activated to create the load combinations for ULS and SLS. To prevent the alternative load cases from being combined with the distributed or concentrated loads, they are each assigned to a group. Download the file at the end of this article to see all the other inputs we have made in RFEM 6 and RSTAB 9.
To explain the following entries or the results of the reduction, the program calculates with individual load combinations that include either a pure uniform load or a combination of an equal and a concentrated load. For this purpose, two load combinations were created manually (CO10 and CO11). The options described below are not available for the result combination design. See also the notes in the last section of this article.
Definition of Design Supports
In RFEM 6, you can assign design supports to a member. To define the supports, use the Design Supports & Deflection tab of the member. You can define design supports at the member start, member end, and internal nodes.
In the settings for the design support, you can decide on a support width and a support type.
Please note that entering the support width without selecting the "Reduction of shear force…" option in the Ultimate Configurations has no influence on the shear force design.
More information about this input window can be found in our online manual about the ultimate configurations for members.
However, it may be necessary to specify supports without activating one of the available options for the shear force reduction if a deformation calculation is performed for the serviceability limit state. In this case, you can use the supports to calculate the reference length l0 to determine the maximum deformation limit value. This article does not go into this option further.
Direct Support
It is necessary to activate the direct support if, according to 6.2.2 (6) or 6.2.3 (8), single loads close to the support are to be reduced with ß = av / 2 d. If there is a secondary beam transferring its load to another beam and not to a "direct support" (column, nodal support, wall, and so on), you should not select the direct support.
Design at Distance d from Support Face
If the design support has been defined correctly and the support width w = 300 mm has been set, you can apply the reduced shear force for the design and determination of the required shear reinforcement by selecting the "Reduction of the shear forces at the support face and distance d according to 6.2.1 (8)" option.
In the following, you can see the shear force distribution Vz from the structural analysis and the shear force distribution Vz,Ed from the concrete design in Load Combination CO10.
The end support and the inner support of Span 1 will be examined.
| Span 1 | End Support | Inner Support |
|---|---|---|
| Effective Depth d on Respective Support Edge | 416.1 mm (bottom reinforcement) | 448 mm (top reinforcement) |
| Distance d from Support Center (0.5w + d) | 566.1 mm | 598.6 mm |
| Governing x-Location at Distance d from Support Edge | x1=0.5661 m | x2 =3.4014 m |
| Vz from Static Analysis in Support Center | Vz (0.00 m) = 192.66 kN |
|
| Vz from Static Analysis in Governing x-Location | Vz(0.5661 m) = 120.22 kN | Vz(3.402 m) = -242.68 kN |
| Vz,Ed from Concrete Design | Vz,Ed(0.5661 m) = Vz,Ed(0.00 m) = 120.22 kN | Vz,Ed(3.402 m) = Vz,Ed(4.00 m) = -242.68 kN |
The distance d from the support edge results in the maximum value of the reduced shear force for the concrete design, Vz,Ed.
Reduction of Concentrated Loads near Support
To explain the reduction of the concentrated loads near the support, the previously discussed two-span beam is now designed for the load combination CO11 with the concentrated loads close to the support and a uniformly distributed load.
In the following, you can see the shear force distribution Vz from the structural analysis and the shear force distribution Vz,Ed from the concrete design in Load Combination CO11. Due to the previously described settings of "Direct Support" and the "Reduction of shear forces with concentrated loads according to 6.2.2 (6) and 6.2.3 (8)" function, the concentrated loads in the range of 0.5d ≤ av < 2d with β = av / 2d are reduced.
The concentrated load F = 56.25 kN at the location x = 4.40 m results in av = 0.25 m. This is thus within the limits of 0.5d ≤ av < 2d ⇔ 0.2243 m ≤ av < 0.8972 m and can be reduced with β = av / 2d = 0.25 m / (2 ⋅ 0.4486 m) = 0.279.
The result of the concrete design indicates a jump of 15.68 kN at the location x = 4.40 m.
ΔVz,Ed = Vz,Ed,left (4.40 m) - Vz,Ed,right (4.40 m) = 249.02 kN - 233.34 kN = 15.68 kN
Since the concentrated load of F = 56.25 kN acts at this location, the distribution of the static analysis Vz shows a jump of 56.25 kN.
ΔVz = Vz,left (4.40 m) - Vz,right (4.40 m) = 289.59 kN - 233.34 kN = 56.25 kN
The quotient of these differences is the reduction factor β.
ΔVz,Ed / ΔVz = 15.68 kN / 56.25 kN = 0.279 = β
According to 6.2.2 (6), when applying the reduced value Vz,ED,red for the design of VRd,c in Equation (6.2a), the applied longitudinal reinforcement (longitudinal reinforcement ratio ρl) must be fully anchored to the support. Furthermore, it is necessary to check the shear force determined without the reduction β with regard to the requirement according to Eq. (6.5).
For the structural components with a calculated shear reinforcement according to [1] 6.2.3, the value Vz,Edmust be applied with ß according to 6.2.3 (8) for the design of VRd,max without reducing the concentrated loads near the support.
Special Cases of Ribs and Result Combinations
In order to reduce the concentrated loads close to the support and to design the uniformly distributed load at the distance d from the support, the add-on module analyzes the distribution of the shear force Vz using the internal forces from RFEM or RSTAB. This shear force distribution analysis allows the program to recognize a uniformly distributed load from a linear distribution of the shear force and the size of the concentrated loads close to the support from the jumps in the shear force distribution.
Therefore, the evaluation of the shear force distribution is the basis of the shear force reduction mentioned here. This also results in the restriction that these options are not available for the design with a result combination (RC), since a uniformly distributed load cannot necessarily be assumed for an RC.
The same applies when designing ribs in the Concrete Design add-on. The rib internal forces are partly composed of the member internal forces of the eccentrically connected T-beam and partly of the integrated surface internal forces of the connected plates. Singularities in the surface internal forces can now ensure that the integrated rib internal force (shear force Vz from RFEM) has no linear distribution in the program. Likewise, jumps in the shear force distribution Vz can result from a possible integration of singular surface internal forces. Therefore, the mentioned options to reduce the shear force are not available for rib member design.