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Exercise Examples InfoCAD Program System

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Page 1: InfoGraph exercise example

Exercise Examples

InfoCAD Program System

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The description of program functions within this documentation should not be considered a warranty of product features. All warranty and liability claims arising from the use of this documentation are excluded. InfoGraph® is a registered trademark of InfoGraph GmbH, Aachen, Germany. The manufacturer and product names mentioned below are trademarks of their respective owners. This documentation is copyright protected. Reproduction, duplication, translation or electronic storage, of this document or parts thereof, is subject to the written permission of InfoGraph GmbH. © 2010 InfoGraph GmbH, Aachen, Germany. All rights reserved.

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InfoCAD Exercise Examples

© InfoGraph GmbH, July 2008 1

Table of Contents

Introduction 2

Load Combination for Nonlinear Calculations 3

Safety Factors in Nonlinear System Analysis 3 Calculation in the Ultimate Limit State 3

Nonlinear 2D Frame According to DIN 1045-1 4

Nonlinear 3D Frame According to DIN 1045-1 8

Determining Maximum Reinforcement 11

Construction Stages, Creep Redistribution 13 Construction Stages 13 Creep Redistribution as a Result of a System Change 14

Dynamic Train Crossing 16 Track and Train Definition 16 Time Step Calculation 17 Results 18

Prestressed Roof Girder According to DIN 1045-1 20 Static System 22 Entering a Tendon Group 22 Loads 26 Checks According to DIN 1045-1 26 Performing Calculations 27 Results 28

Prestressed Steel Design / Single Design 31

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Introduction This documentation is designed to help experienced users with the advanced calculation and checking methods of the InfoCAD program system. It is limited to describing the various procedures and essential work steps. Basic program operation will not be explained as it is assumed the user is already familiar with the program. For a detailed explanation of the program's theoretical foundations, calculation methods and results, refer to the user manual. The complete program documentation has been integrated into the help system, which means you can view it at any time on your computer. Press the F1 key to open the help section on the command or dialog that you are currently using. Product news, user tips and updated versions of the program are available at www.infograph.de.

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Load Combination for Nonlinear Calculations Extremal internal forces are usually determined by superposing the results of individual load cases according to the superposition principle. This principle, however, is not applicable for the results of nonlinear calculations. You therefore need to determine internal forces for total loads. To allow you to enter individual load cases in such circumstances as well, the InfoCAD program system provides the option to combine load cases for internal forces calculation and then weight them with the requisite partial safety factors and combination coefficients.

Use the Load group load type to group existing load cases and apply load factors to them. At the same time you can select the desired calculation theory, assign a predeformation, if applicable, or activate nonlinear support. This load type is suitable for all nonlinear calculations. The load group is recommended as the only load type to be used within a load case.

Safety Factors in Nonlinear System Analysis Calculation in the Ultimate Limit State

This program module automatically takes the material-dependent safety factors γconcrete,

γreinforcing steel and γconstruction steel into account on the resistance side.

Hence, when performing a nonlinear analysis, you should enter the safety factors and combination

coefficients γG, γQ, und ψ on the load side and activate the deflection theory analysis if necessary

and a predeformation, if applicable. For example: Load case 10: Load group: LC1 * 1.35 LC2 * 1.5 * 0.8 LC3 * 1.5 Predeformation in the x direction Deflection theory Exception: Since DIN 1045 (July 1988 Issue) does not identify a specific safety factor for the

material, calculations are made on the resistance side using the characteristic strengths. This means the safety factor (e.g., 1.75) belongs entirely to the load side.

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Nonlinear 2D Frame According to DIN 1045-1 The reinforced concrete frame depicted below (C20/25, BSt 500/550, 3 cm overlap, requirement class F) will be calculated and checked in accordance with DIN 1045-1. Using the calculated reinforcement, a buckling safety check is then carried out with allowance for any stiffness changes in state II.

5,00 4,009,00

1,00

3,20

4,20

1,20 60 1,203,00

6040

1,00

60

80

Section 1 (Column)

Section 2 (Waler)

Static system and section polygons The beams must be adequately divided into groups for nonlinear calculation. Select five beam elements for each structural component and calculate the load cases shown below. Load

Load case 1: permanent loads

γ = 25 kN/m³

Load case 2: traffic upside

Load case 3: traffic area 1

Load case 4: traffic area 2

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Load case 5: wind from left

Load case 6: wind from right Define the actions and design situations and carry out a reinforced concrete design in accordance with DIN 1045-1:

Design according to DIN 1045-1

Upper and lower longitudinal reinforcement As [cm²]

(maximum from robustness, bending with longitudinal force and crack width)

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Stirrup reinforcement Asb [cm²/m]

Buckling safety check in the ultimate limit state For the buckling safety check, create a new load case 10 consisting of load cases 1 (load factor 1.35) 2, 3, 4 and 5 (load factor 1.5). Use the Load group load type for this load case. Now activate the deflection theory analysis. The check should be performed for this load case and is based on the reinforcement that was calculated in accordance with DIN 1045-1. Choose the settings as shown below.

Now perform the framework analysis. A limit load factor of 1.0 will be reached after the calculated reinforcement distribution is taken into consideration.

If a special reinforcement distribution is to be taken into account during the check, this can beassigned to the steel layers in the section polygon as the base reinforcement. In this case, a design according to DIN 1045-1 is no longer required. In the Start reinforcement field, select NULL.

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Besides the node deformations and internal forces, the Stresses results folder (corresponding load case preceded by 'N') also contains the concrete edge stresses and the strains and curvatures for the system.

The following figures show the analysis results of the buckling and ultimate limit state check.

System deformation [mm] in state II without consideration of concrete tensile strength

Stress distribution on the upper and lower side of the section[MN/m²] To carry out a deformation analysis in the serviceability state, create a load group with 1x loads. Now activate the concrete tensile strength in the nonlinear system analysis. The deformations are calculated as follows:

System deformation [mm] in state II with consideration of concrete tensile strength

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Nonlinear 3D Frame According to DIN 1045-1 This example involves a 3D reinforced concrete frame in state II. The resulting stiffness decrease will be taken into account while an ultimate limit state or buckling safety check is performed with a corresponding reinforcement increase. At first, a linear-elastic internal forces calculation with a DIN 1045-1-compliant design will be performed for the system. The reinforcement determined by this calculation will then be used as the basis for the nonlinear system analysis. The program will increase the reinforcement by the requisite amount.

The following items are described: • Definition of section and material properties and design specifications • Action combinations with partial safety factors and combination coefficients • Bending, shearing and torsional design • Geometrically and physically nonlinear analysis of the entire system

(nonlinear system analysis or buckling safety check) Structure properties Material: C 30/37, BSt 500S Sections:

Shaft section Ring section

Dimensions: ring diameter 10 m shaft height 5 m Support: jointed

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Load:

Load case 1: System dead load and additional loads

Load case 2: Traffic load

Load case 3 and 4: Wind loads in +/- x direction

Tilt in x and y direction α = 1/200

Calculate the above load cases, define a permanent and temporary design situation as per DIN 1045-1 and determine the longitudinal and stirrup reinforcement based on the bending, longitudinal and lateral force and for torsion.

Bending moments My from the fundamental

combination

Bending reinforcement from the design according to DIN 1045-1

Nonlinear internal forces calculation and reinforcement increase Because the superposition principle is not valid for nonlinear calculations, you need to define the actions you want to analyze. When defining actions, be sure to take the safety factors and combination coefficients into account on the load side: Load case 10: LC1 * 1.35 Load case 11: LC1 * 1.35

LC2 * 1.5 * 0.8 LC2 * 1.5 * 0.8 LC3 * 1.5 LC3 * 1.5 Predeformation in x direction Predeformation in y direction Deflection theory Deflection theory

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Create the new load cases 10 and 11. Add the specified load cases with partial safety factors and combination coefficients to these new load cases, activate the deflection theory analysis and select the relevant predeformation. Use the Load group load type for the new load cases.

Use the Nonlinear Analysis option to activate nonlinear analysis in the settings. • Load cases 10 and 11 are to be analyzed. • The analysis is based on the

reinforcement that was calculated during section design in accordance with DIN 1045-1.

• If necessary, the reinforcement will be increased to achieve the required load-bearing safety.

You can now start the framework analysis.

Deformations and increased bending reinforcement from the load-bearing capacity calculation for load case 10

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Determining Maximum Reinforcement The design of the extremal internal forces from the action combinations does not necessarily yield the maximum reinforcing steel reinforcement. Frequently there are combinations with lower internal forces that result in higher reinforcement levels. This is why the InfoCAD program system allows you to design all possible internal force combinations. The basic example below will demonstrate this procedure.

Load case 3: Traffic load

The depicted frame should be designed for the permanent and temporary situation in accordance with DIN 1045-1. LC 1: Dead weight LC 2: Permanent loads LC 3 - LC 10: Traffic loads LC 11 / 12: Wind left / right The extremal internal forces are determined and designed first. All possible combinations are then designed in comparison to the internal forces.

Definition of actions for the permanent and temporary design situation:

G – Dead load: Load case 1 and 2, permanent Gamma.sup / gamma.inf = 1.35 / 1 QN - Working load, traffic load: Load case 3–10, inclusive Gamma.sup / gamma.inf = 1.5 / 0 Combination coefficients psi for: Buildings Working loads - category A – Living quart. and lounges Psi.0 / Psi.1 / Psi.2 = 0.7 / 0.5 / 0.3 QW – Wind load: Load case 11 and 12, exclusively Gamma.sup / gamma.inf = 1.5 / 0 Combination coefficients psi for: Buildings Wind loads Psi.0 / Psi.1 / Psi.2 = 0.6 / 05 / 0.

The program will first calculate the extremal internal forces to determine the shown reinforcement distribution (representation of column reinforcement):

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To design all possible combinations in comparison to the internal forces, the (basic) design of the extreme values will be disabled in the calculation settings and the DIN 1045-1 Design will be restarted.

Based on the partial safety factors and combination coefficients, 5112 possible combinations of internal forces are found for each check location. The design of these combinations leads to the shown reinforcement distribution. The reinforcement was greatly increased at several design points.

As an example, let's analyze the bearing point of the lower left column. The permanent and temporary situation results in the following extremal internal forces and reinforcement (4 designs): Nx [kN] My [kNm] As [cm²] at each location Nx- : -3741.91 801.85 3.23 Nx+ : -1293.00 370.68 1.12 My- : -1332.15 365.28 1.15 My+ : -3212.12 838.62 3.10 As max. at 2 locations on each side: 12.9 cm² The design of all combinations will lead to the maximum reinforcement (5112 designs) given the following internal forces combination: Nx [kN] My [kNm] As [cm²] at each location -1958.45 794.01 5.25 As max. at 2 locations on each side: 21.0 cm² A detailed log records the load cases involved and their weightings: 1.35 * LC 1 + 2 1.05 * LC 6,7,8,9,10 1.50 * LC 12 Hence, when designing all possible combinations, a significantly higher reinforcement is calculated for the column. Due to the extremely high calculation complexity, this method should only be used in exceptional cases. Two additional traffic loads will quadruple, for instance, the number of possible combinations and thus the calculation time.

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Construction Stages, Creep Redistribution This section shows you how to create and process construction stages. For every construction stage a separate project file is created for which all calculation options are available. Using a simple example, the calculation of creep redistribution as a result of a system change is then explained with the help of construction stages. Construction Stages The system 'inherits' all the properties from one construction stage to the next. During this process, the subsequent stage stores all the information about its predecessor. This information prevents the redundant use of load cases, elements and similar items when the subsequent stage is being processed. The program assigns an attribute to the project files to ensure their coherency. Hence there is no need to bypass the system to add or delete construction stages. When calculating a construction stage, the results of the preceding stage are copied and then processed so that they can be combined or superposed with the results from the current file. Procedure:

1. First enter the original system for the first construction stage.

2. Enter a file name ending in (1) and save it. For example, bridge(1).FEM (this activates the Construction Stage... function).

3. Perform the necessary calculations and checks for this project.

4. Select the Construction Stage... command and click New... The second construction stage (here: Bridge(2).fem) is automatically generated and activated.

5. Now perform the desired system modifications or additions.

6. When the internal force calculation begins, the results from the previous system will be copied and made available for load case combinations and superpositions.

7. Using this method, you can create any number of additional construction stages.

You can switch between the individual construction stages by simply double-clicking them. Click New... to attach or add a subsequent stage after the currently selected construction stage.

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Wait until the preceding stage has been fully processed before generating the next construction stage. Later changes to a construction stage concerning information inherited by thesubsequent stage must be manually added to the successor using the Edit... option.

Exceptions:

1. Changes within a load case. To update the results, you need to recalculate the construction stage and the subsequent stage in sequential order.

2. Changes to the element properties. Creep Redistribution as a Result of a System Change The original system, construction stage 1, consists of 2 single-span beams with a length of 4.00 m and 6.00 m. Material: C20/25, Section: T-beam.

For the original system the following is calculated: Load case 1: Dead weight Load case 2: Creep t=10 to t=50 days, with ϕt=0.66, creep-generating continuous load LC 1

Now the construction stage 2 is generated and the system changes are implemented here. By removing the joint, a continuous model is created. The additional load is applied and calculated: Load case 3: Additional load 10 kN/m The internal forces and deformations after 50 days result from the dead load + creep 10/50 + the additional load: Load case 4: Superposition LC 1, 2 and 3 (superposition of the results) The creep redistribution up to the final state is calculated by: Load case 5: Creep t=50 to final state, ϕt =2.04, creep-generating continuous load: LC 4

The final state results from load case 4 and 5: Load case 6: Final state (superposition of the results LC4+LC5) The superposition load type is especially suitable for adding the results of different load cases (which can derive from different construction stages).

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Construction

stage Internal forces My Deformations

LC 1: Dead load

LC 2: Creep t=10-

50 d

Bridge (1).fem

LC 3: Additional

load

LC 4: LC 1+2+3

t=50 d

LC 5: Creep t=50-∞

d

LC 6: Final state

Bridge (2).fem

For comparison:

Monolithic model

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Dynamic Train Crossing The 'FEM Dynamics' program module offers an easy way to analyze the dynamic stress on any beam and shell structure based on predefined trains such as the ICE or Thalys or user-defined trains. To use this feature, the following steps are required: • Define the track • Specify the train properties • Set the specifications for the time step calculation • Perform the calculation • Display and process the results

Track and Train Definition The track is defined by entering a continuous line on the structure. Open a new load case, choose the Train crossing load type and enter the track of the train. This can be done using area elements and beams with polygon sections:

The only requirement is that the track runs within the section dimensions of the elements.Eccentricities of the beam axes are automatically taken into account.

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After having defined the continuous line, the Train crossing dialog will appear:

Select a train and the speed you want to have analyzed. If you want to analyze multiple trains or tracks simultaneously (e.g., to simulate a train meeting), use the start time to control the time offset. If you want to use a customized train, edit the axle distance and load on the Train tab. You can also paste data into this tab from the clipboard.

You have now completed the definition of a track.

In the calculation, the train is treated as a group of individual loads traveling along the track. Theaxle loads always act along the global z axis.

Time Step Calculation Dynamic train loads are only taken into consideration for dynamic calculations. Dynamic train loads are specified on the Dynamics tab in the FEM analysis Settings dialog.

The train crossing can be analyzed using direct or modal time step integration. Select the number and duration of the time steps to produce the desired analysis period and the tracks of the train. For modal time step integration, the number of eigenvalues to be considered for the system must be defined additionally.

The duration assigned to the selected time steps mainly depends on the vibration behavior of thestructure. The duration should be less than 1/10 of the period of oscillation of the determinantresonant frequency.

The resonant frequencies, eigenmodes and all time steps will be calculated as part of the FEM analysis. The deformations, speeds, accelerations and system internal forces for each time step will also be determined and then stored.

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Results The results of the train crossing can be plotted over the analysis period for each node. The representation area is split into two panes for this purpose. In the left pane, select the nodes for which you want to generate a results curve.

You can also access individual system deformations for each time step.

The load case combination can be used to determine the 'envelopes' of all time steps for the deformations and internal forces and, if necessary, they can be superposed with static loadcases. As the individual time steps mutually exclude one another, add the time step calculation to theload case combination as an exclusive action. Note that no soil reactions are available for the time step calculation, meaning they might bemissing in the superposition.

Extremal internal forces My of a main girder

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You can use the InfoGraph Systemviewer to animate the deformations of the train crossing. To do so, open the file with Systemviewer and start the calculated time step integration from within the Results folder.

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Prestressed Roof Girder According to DIN 1045-1 This example involves a wide-spanned roof construction that is represented as a continuous girder over two spans with a double-sided cantilever. A T-beam is selected as the section. The figure below shows the system in longitudinal and lateral section view. Prestressing with subsequent bond is applied to the roof construction in the longitudinal direction. The lateral direction is not analyzed. To edit this system, perform the following work steps: • Enter a static system • Enter a tendon group • Enter the loads • Carry out the calculations and checks • Display the results

Static system and dimensions Material Concrete: C45/55 Reinforcing steel: BSt 500/550, axis distance from edge 5 cm Section

Tendon groups Four bundled tendons are arranged in this example. The tendon group guide is shown in the next figure. The depicted tendon group ordinates zv of the spline points refer to the upper edge of the

section.

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Prestressing steel and prestressing system Prestressing steel quality St 1500/1770 Certification of the prestressing system DIN 1045-1, Cona 1206 Number of tendons in the bundle 4 Section surface area Az 1800 mm²

E-modulus of the prestressing steel 195,000 MN/m² 0.1% strain limit (yield strength) of the prestressing steel fp0.1k 1500 MN/m²

Tensile strength of the prestressing steel fpk 1770 MN/m²

Permitted prestressing force of the tendon Vzul = Pm0 2295 kN

Friction coefficients when increasing or releasing strain 0,2 Unintentional angle displacement of a tendon ß’ 0.3 °/m Slippage at prestressed tie bolt 6 mm Duct diameter 82 mm

Allowance value for ensuring an over-stressing reserve κ 1.5 Scattering coefficients of the internal prestressing as per DIN 1045-1, Equ. 52/53 Construction stage according to DAfStb (German Committee of Reinforced Concrete) Book 525 (rsup/rinf) 1.0 / 1.0

Final state(rsup/rinf) 1.1 / 0.9

Prestressing procedure Double-sided overstressing by applying the allowance value κ to the maximum permitted force P0,max as specified in DAfStb Book 525. No release occurs.

Loads Load case 1: Dead load (G1) Load case 2: Additional loads q=11.06 kN/m (G2) Load case 3: Traffic load (snow load) q=7.90 kN/m (Q) Load case 10: Prestressing (P) Load case 15: Creep-generating continuous load: G1+P+G2 Load case 20: Creep and shrinkage (CSR)

Coefficients: ϕt∞ = 2.55; ρ = 0.8; εt∞ = -24.8 · 10-5

Creep-generating continuous load case: 15 The redistribution of internal forces between concrete and prestressing steel are

taken into account.

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Static System To accurately measure the prestressing and the creep and shrinkage behavior, first divide the structure into beams with a length of 4 meters. Specify the supports based on a fork support of the system. Entering a Tendon Group The definition of the tendon groups does not depend on the elements. The assignment and force introduction onto the structure occur while the Prestressing load case is being calculated. To avoid any conflicts, a distinction is made between beam and area/solid prestressing. The tendon curve on which the calculation approach is based is represented by a 3D cubical spline function. The spline function is the curve that runs through all specified spline points with the least amount of curvature. The basis for the prestressing analysis is provided by a tendon group force curve that allows for strain increase, release and slippage. To include prestressing in the FEM analysis, you need to define a load case with the Prestress load type.

Use the Prestressing function to enter and edit tendon groups. Click Define and then specify the desired tendon group properties in the resulting dialog. The spline points are defined graphically.

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The Prestressing System tab contains all the properties that are assigned to the tendon groups. They apply to each individual tendon of the group.

If the allowance value κ is used, then the factor for the first over-stressing refers to the maximum permitted force P0,max as specified in DAfStb

Book 525. Using the factor specified for the release, the maximum prestressing force remaining in the tendon group is defined with respect to Vzul. A factor of '0' means no release.

Exit the dialog and enter the spline points of the tendon group (start, supports, midspans) at the nodes of the beam array.

When you are done, click OK. The tendon group no. 1 is now positioned along the centroidal axis of the beam array.

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Additional editing should take place in a section view. To access a section view, select the entire beam series with the View/Beam option and define the view plane from the first node to the last node in the global z direction.

Click the View option again and add tendon group 1 to the Visible tendon groups list.

The tendon group is now displayed in the selected section. Next, use the Zero point option to move the position of the reference system from the centroid to the upper edge of the section and then in the Base points option select Spline points.

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If the tendon group is selected, you can reposition it by first clicking the existing spline point and then the Spline points option. Only change the respective Zv position. The Insert option allows you to add additional spline points behind the active spline point.

Use this method to assign the desired geometry to the tendon group.

In the tendon group view, the coordinates always refer to the reference system.

Click the Representation option to activate the resulting prestressing force curve in the tendon group view:

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Loads Enter the following loads: Load case 1: Dead weight Load case 2: Additional loads q = 11.06 kN/m Load case 3: Snow load p = 7.90 kN/m Load case 10: Prestressing Load case 15: Creep-generating continuous load case.

Load cases 1, 2 and 10 are grouped into this load case with the Insert load type. Load case 20: Creep and shrinkage.

The Creep and Shrinkage load type lets you calculate the redistribution of internal forces between the concrete and prestressed steel. Specify load case 15 as a creep-generating continuous load case.

Checks According to DIN 1045-1 The following checks are carried out as a part of this example: Checks at the ultimate limit state • Minimum reinforcement for securing ductile member behavior • Bending with or without longitudinal force or longitudinal force only • Lateral force under consideration of the minimum level of reinforcement

Checks at the serviceability limit state • Limiting the concrete compressive stresses • Limiting the reinforcing steel stresses • Limiting the prestressing steel stresses • Minimum reinforcement for the crack width limitation • Limiting the crack width via direct calculation

The following input is required for these checks: • Activate the checks in the settings • Select and adjust the settings for the checks in the element properties • Define the actions

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To take different (construction) stages into account, three situations are defined for both the permanent and temporary and the rare design combination: Construction stage: Dead load G and prestressing P (tendon not grouted) t0: Dead load G, P, additional load and snow load (tendon grouted)

t∞: Dead load G, P, additional load, snow load and CSR (tendon grouted)

Performing Calculations Perform the following calculations after all system specifications have been made: • Finite Elements • DIN 1045-1 Design

The necessary design situations are used for all checks depending on the requirement class. Each situation is checked independently and the maximum reinforcement for each steel layer is stored.

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Results A number of different calculation results are shown below: Internal forces

Bending moment My from the third permanent and temporary design situation

Normal force Nx from the first permanent and temporary design situation

Reinforcement

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Shear reinforcement

Excerpt from the 'DIN 1045-1 Design' log (short version)

16.00 48.00 48.00 16.00128.00

Beam 16

Design of longitudinal reinforcement (M) Nominal reinforcement to guarantee robustness (Charact. C.). (R) Nominal or required reinforecement for crack width limitation. Increase of reinforcement due to crack width check is marked by "!". (B) Design of reinforement at ultimate limit state. In case of dominant bending, compression reinforcement is marked with "*". Beam Reinforcement Nx My Mz Xi1*Ap req.As No. Se. Lo. Lay. Type [kN] [kNm] [kNm] [cm²] [cm²] 16 1 1 1 M -80.36 -3907.96 0.00 . 44.91 R -5909.70 -4315.09 0.00 . 59.87 B -7385.39 -275.25 0.00 . 0.00 2 M -80.36 -3907.96 0.00 . 44.91 R -5909.70 -4315.09 0.00 . 59.87 B -7385.39 -275.25 0.00 . 0.00 3 M -80.36 -3907.96 0.00 . 0.00 R -5909.70 -4315.09 0.00 . 0.00 B -7385.39 -275.25 0.00 . 0.00 4 M -80.36 -3907.96 0.00 . 0.00 R -5909.70 -4315.09 0.00 . 0.00 B -7385.39 -275.25 0.00 . 0.00 16 1 2 1 M 122.97 -12038.40 0.00 . 44.91 R -5798.38 -10888.74 0.00 . 59.91 B -6442.64 -17292.95 0.00 . 21.97 2 M 122.97 -12038.40 0.00 . 44.91 R -5798.38 -10888.74 0.00 . 59.91 B -6442.64 -17292.95 0.00 . 21.97 3 M 122.97 -12038.40 0.00 . 0.00 R -5798.38 -10888.74 0.00 . 0.00 B -6442.64 -17292.95 0.00 . 8.98* 4 M 122.97 -12038.40 0.00 . 0.00 R -5798.38 -10888.74 0.00 . 0.00 B -6442.64 -17292.95 0.00 . 8.98*

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Shear reinforcement from ultimate limit state design The percentage of nominal reinforcement acc. to 13.2.3 (5) is considered. VRd, TRd Resisting lateral force and torsional moment Angle Angle cot Theta between the compressive strut and the beam axis Asb,Asl.T Req. stirrup reinf. from lateral force and torsion, torsional long. reinf. Asl Req. longitudinal reinf. acc. to Pic. 32 for req. Asb. Beam Qy/ Asb.y Qz/ Asb.z Asl Q/VRd+ Asb.T, Asl.T No. Loc. VRd Angle[cm²/m] VRd Angle[cm²/m] [cm²] Mx/TRd [cm²/m, cm²] 16 1 0.00 3.00 0.00 0.33 2.35 9.76 . . . . 2 0.00 3.00 0.00 0.49 1.76 19.82 . . . . Check of crack widths The check calculates the crack width directly. (CC) Charact. (rare), (TC) Frequent, (QC) Quasi-continuous combination Beam Rei. Nx My Mz Sigma.x wk per. No. Se. C. Lo. Lay. [kN] [kNm] [kNm] [MN/m²] [mm] [mm] 16 1 TC 1 . -8123.93 -782.14 0.00 -2.45 0.00 0.20 2 1 -5798.38 -9473.06 0.00 1.98 0.03 0.20 Check of concrete compressive stress For the check, a cracked concrete section (II) is assumed if the tensile stress from the char. comb. exceeds the value of fctm. Otherwise, a non-cracked section (I) is used. If the strain is not treatable on cracked section, (I*) is marked. (CC) Characteristic (rare) combination, (QC) Quasi-continuous combination (t,b) Position of the edge point: above, below of centre Beam min Sigma.x per. Sigma.x Se.- Side Period Situation No. Se. Loc. [MN/m²] [MN/m²] Pnt. t b 16 1 1 (I) -7.17 -27.00 8 . x Final CC.3 (I) -5.74 -20.25 8 . x Final QC.1 2 (I) -16.05 -27.00 8 . x Final CC.3 (I) -13.55 -20.25 8 . x Final QC.1 Check of steel stress For the check, a cracked concrete section is assumed. For tendon groups without bond and/or for situations before grouting, the prestressing steel stress is checked acc. to Eq. (49). Type B Bending reinf., layer number, Characteristic (rare) combination (CC) Type P Prestressing steel, Tendon number, Quasi-continuous combination (QC) and Characteristic (rare) combination (CC) Beam Steel As Sigma.s per. No. Se. Lo. Type No. [cm²] [MN/m²] [MN/m²] 16 1 1 B 1 59.87 -4.90 400.00 B 2 59.87 -4.90 400.00 B 3 0.00 . 400.00 B 4 0.00 . 400.00 P 1 72.00 896.21 1150.50 (QC) P 1 72.00 1032.93 1275.00 (CC) 16 1 2 B 1 59.91 53.69 400.00 B 2 59.91 53.69 400.00 B 3 8.98 -48.91 400.00 B 4 8.98 -48.91 400.00 P 1 72.00 924.12 1150.50 (QC) P 1 72.00 1007.77 1275.00 (CC)

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Prestressed Steel Design / Single Design This example shows how the result of a prestressed steel design according to DIN technical report with a single design according to DIN 1045-1 is derived. Since the checking program, in accordance with the DIN technical report, automatically considers the prestressed steel as a resistance, you need to add the prestressed steel to the single design as an additional steel layer. Internal forces at the check location:

Action My [kNm] Nx [kN] G1 937.5 0.0 G2 2500.0 0.0 max. Q 4500.0 0.0 P -637.2 -1.062.0 CSR 63.7 106.2 Permanent and temporary design situation 10,817.1 -955.8

Design according to DIN technical report, detailed calculation log:

Section 1, Polygon - C45/55, 1 Tendon groups with supplemental bond Steel 1; Design mode: Standard (B) fck=45 Tendon groups with bond No. E-Modul fp0,1k fpk y z Ap Prestress Inclin. [MN/m²] [MN/m²] [MN/m²] [m] [m] [cm²] [kN] [°] 1 195000 1570 1770 0.250 1.350 10.00 1062.00 0.00 1. Permanent and temporary comb. (PC.1): G1+P+CSR1+QK, Final state grouted Loss of prestress by CSR in tendon groups No. CSR[%] No. CSR[%] No. CSR[%] No. CSR[%] No. CSR[%] No. CSR[%] 1 10.00 -.- -.- -.- -.- -.- Stat. determ. part (P+CSR): Nx0=-955.80 kN; My0=-573.48; Mz0=0.00 kNm Concrete section Bond section Nx[kN] My[kNm] Mz[kNm] Nx[kN] My[kNm] Mz[kNm] Nx- : -955.80 4067.15 0.00 0.00 4640.63 0.00 Nx+ : -955.80 4067.15 0.00 0.00 4640.63 0.00 My- : -955.80 2864.02 0.00 0.00 3437.50 0.00 My+ : -955.80 10817.14 0.00 0.00 11390.62 0.00 Mz- : -955.80 4067.15 0.00 0.00 4640.63 0.00 Mz+ : -955.80 4067.15 0.00 0.00 4640.63 0.00 Design of longitudinal reinforcement Reinforcement Nx My Mz max Sx kc Xi1*Ap req.As Situation Lay. Type [kN] [kNm] [kNm] [MN/m²] [cm²] [cm²] 1 B -955.80 10817.14 0.00 . . . 28.74* PC.1.My+ 2 B -955.80 10817.14 0.00 . . . 28.74* PC.1.My+ 3 B -955.80 10817.14 0.00 . . . 89.53 PC.1.My+ 4 B -955.80 10817.14 0.00 . . . 89.53 PC.1.My+

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Calculation process in the checking program:

1. The statically determined share from the prestressing and creep & shrinkage

((P + CSR) * cos α * distance from center of mass) is derived from the internal forces on the concrete section.

2. This yields the internal forces on the composite section (statically undetermined share from P + CSR and the internal forces from the external load).

3. The internal forces acting on the composite section are designed. In the design, P + CSR is applied to the resistance side as a prestressed steel layer.

Single design according to DIN 1045-1

For the single design, the internal forces from the external load plus the statically undetermined share from P + CSR (statically determined system, hence = 0) are required. This information can be obtained from the log.

Internal forces at the composite section: Action My: 1.35 * (G1+G2) +1.5 * Q = 11,390.6 kNm

To take the statically determined effect from P + CSR into account on the resistance side, the prestressed steel layer in the section is defined (layer 5): Zv0 = P + CSR = 0.9 * P = 0.9 * 1062.0 = 955.8 kN

Specifications for the section:

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Result of the single design:

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