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March 2014

Autodesk Robot Structural Analysis Professional - Verification Manual for Indian Codes

March 2014 page i

INTRODUCTION .............................................................................................................................................................................. 1

SSTTEEEELL .............................................................................................................................................................................................. 2

1. IS 800:2007 - GENERAL CONSTRUCTION IN STEEL (THIRD REVISION) ............................................................................ 3

GENERAL REMARKS .................................................................................................................................................................. 5 VERIFICATION EXAMPLE 1 - DESIGN OF MEMBERS FOR COMPRESSION .................................................................................... 10 VERIFICATION EXAMPLE 2 - COMBINED COMPRESSION AND BENDING ABOUT BOTH AXES (LATERAL-TORSIONAL BUCKLING) ........ 18

CCOONNCCRREETTEE ................................................................................................................................................................................... 27

1. IS 456:2000 – RC BEAMS ......................................................................................................................................................... 28

VERIFICATION EXAMPLE 1 - DIMENSIONING OF SIMPLY SUPPORTED BEAM ............................................................................... 29 VERIFICATION EXAMPLE 2 - DETERMINATION OF CAPACITY OF A BEAM .................................................................................... 32 LITERATURE ............................................................................................................................................................................. 33

2. IS 456:2000 – RC COLUMNS ................................................................................................................................................... 34

VERIFICATION EXAMPLE 1 - UNIAXIALLY ECCENTRICALLY LOADED BRACED RECTANGULAR COLUMN ......................................... 35 VERIFICATION EXAMPLE 2 - COLUMN SUBJECTED TO AXIAL LOAD AND BIAXIAL BENDING ........................................................... 38 LITERATURE ............................................................................................................................................................................. 41


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INTRODUCTION

This verification manual contains numerical examples for elements of steel structures prepared and originally calculated by Autodesk Robot Structural Analysis Professional version 2013. The comparison of results is still valid for the next versions. All examples have been taken from handbooks that include benchmark tests covering fundamental types of behaviour encountered in structural analysis. Benchmark results (signed as “Handbook”) are recalled, and compared with results of Autodesk Robot Structural Analysis Professional (signed further as “Robot”).

Each example contains the following parts:

- Title of the problem

- Specification of the problem

- Robot solution of the problem

- Outputs with calculation results and calculation notes

- Comparison between Robot results and exact solution

- Conclusions.



SSTTEEEELL



1. IS 800:2007 - General

Construction in Steel

(third revision)



IMPLEMENTED CHAPTERS of IS 800: 2007 (third revision)

List of Indian Standard General Construction in Steel – Code of Practice chapters, implemented to

Autodesk RSA program:

1. Mechanical properties of structural steel (Table 1)

2. Classification of cross-sections - § 3.7 and Table 2

3. Partial Safety Factor for Loads f ( Table 4)

4. Partial Safety Factor for Materials M ( Table 5)

for Limit State Design

5. Design of tension members - § 6

6. Design of compression members - § 7

7. Design of members subjected to bending

Laterally supported beam - § 8.2.1

Laterally unsupported beams - § 8.2.2

Effective length for simply supported beams – Table 15

8. Shear - § 8.4

9. Combined axial force and bending moment - § 9.3

10. Limit State of serviceability - § 5.6 and Table 6

11. Working Stress Design § 11



GENERAL REMARKS

A. Job Preferences If you make first step in RSA program specify your job preferences in JOB PREFERENCES dialog box (click Menu/ Tools/ Job Preferences). Default JOB PREFERENCES dialog box opens:

You can define a new type of Job Preferences to make it easier for future. First of all, make selection of documents and parameters appropriate for India from tabs of the list view in JOB PREFERENCES dialog box. For example to choose code, first click Design codes tab from left list view, then select code from Steel/Aluminum structures combo-box or press More codes button which opens Configuration of Code List:

Put India code into the right list of the box, and then set it as the current code. Press OK.



To choose code combination first click Loads tab from left list view in JOB PREFERENCES dialog box,

Then select code from Code combinations combo-box or press More codes to open the Configuration of Code List. Pick Load combinations from combo box. The new list view appears:

Put IS: 875 (Part5) code into the right list of the box, then set it as the current code. Press OK.



After the job preferences decisions are set, press Save Job Preferences icon in JOB PREFERENCES dialog box. It opens SAVE JOB PREFERENCES dialog box. Type a new name e.g. “ India Limit State” and save it. The new name appears in the combo-box. Press OK button.

You can check load combination regulations by pressing right button next to Code combinations combo-box in Loads tab JOB PREFERENCES dialog box. It opens proper Editor of code combination regulations dialog box.



B. Calculation method Indian code IS 800: 2007 gives two verification options: Limit State Design and Working Stress Design. In RSA program you must always manually specify:

1. calculation method 2. load code combination -> appropriate for calculation method

ad.1 calculation method

Calculation method (Limit State or Working Stress) can be chosen on Steel /Aluminum Design layout. Press the Configuration button in CALCULATIONS dialog box.

Here you can choose calculation method (not regulation of load combination).



ad.2a load code combination – basic approach

To specify load code combination (Limit State or Working Stress) appropriate for calculation method, click Menu/ Tools/ Job Preferences. JOB PREFERENCES dialog box opens. Select earlier prepared job preferences (as defined in Section A.) by clicking its name from combo-box. In following dialog box India Limit State job preferences should be selected.

By pressing OK button, you accept chosen job preferences for a current task. ad.2b load code combination - alternative (tricky-easy) approach Start in Loads layout. Here, you can prepare load combination for both calculation methods for further using (for member verification). Create manually Limit State Design “LS” load combination and Working Stress Design “WSD” load combination in Load Types dialog box.

In this case, you can use in verification appropriate load combination corresponding to the calculation method.


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VERIFICATION EXAMPLE 1 - Design of members for compression

Example taken from: DESIGNERS’ GUIDE to EN 1993-1-1 Eurocode 3:

Design of steel structures, general rules and rules for buildings

written by L. Gardner, D. A. Nethercot, 2005

TITLE: Example 6.7 – Buckling resistance of a compression member

SPECIFICATION: A circular hollow section CHS member used as an internal column in a building. The column has pinned boundary conditions at each end and the inter-storey height is 4 m. The critical combination of actions results in a design axial force of 1630 kN. Assess the suitability of hot-rolled 244,5x10 CHS in grade S275 steel (fy = 275 MPa) for this application, using Limit State Design.

SOLUTION: Specify appropriate parameters in JOB PREFERENCES dialog box (click Menu/Tools/Job Preferences). Then, choose calculation method Limit State Design in CONFIGURATION dialog box (press Configuration button in CALCULATIONS dialog box). In DEFINITIONS dialog box you can define a new type of member in agreement with structure data It can be set in Member type combo-box. Pre-defined type of member “Column” may be initially opened.


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For chosen member type, press the Parameters button on Members tab, which opens MEMBER DEFINITION – PARAMETERS dialog box.

Type a new name in the Member type editable field, e.g. “column 1”. Change parameters to meet initial data requirements of the structure. In the particular compression case press More button which opens MEMBER DEFINITION – ADDITIONAL PARAMETERS dialog box:


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Switch on Hot-rolled pipes. Click OK. In DEFINITIONS dialog box save the newly-created member type, here “column 1”:

Number of the member must be assigned to appropriate name of Member type. (It is very important when you verify different member types.)


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In the CALCULATIONS dialog box set: -> Verification option; here Member Verification, -> Loads cases ; here for n

o 1 for Limit State Design

-> Limit state ; here only Ultimate Limit state will be analyzed, so switch off Limit stat –Serviceability. Now, start calculations by pressing Calculations button.

MEMBER VERIFICATION dialog box with most significant results data will appear on screen.

Pressing the line with results for the member 1 opens the RESULTS dialog box with detailed results for the analyzed member. The view of the RESULTS windows are presented below.


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Simplified results tab

Detailed results tab

Pressing the Calc.Note button in “RESULTS –Code” dialog box opens the printout note for the analyzed member. You can obtain Simplified results printout or Detailed results printout. It depends on which tab is active. The printout note view of Simplified results is presented below.


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RESULTS for Limit State Design method:

a) In the first step CHS 244,5x10 section was considered. The results are presented below.

STEEL DESIGN ----------------------------------------------------------------------------------------------------------------------------------------

CODE: IS800: 2007 Indian Standard - General Construction in Steel - Code of Practice (Third Revision)

ANALYSIS TYPE: Member Verification

----------------------------------------------------------------------------------------------------------------------------------------

CODE GROUP:

MEMBER: 1 POINT: 1 COORDINATE: x = 0.00 L = 0.00 m

----------------------------------------------------------------------------------------------------------------------------------------

LOADS:

Governing Load Case: 1 STA1

----------------------------------------------------------------------------------------------------------------------------------------

MATERIAL:

S275 fy = 275.0 MPa fu = 430.0 MPa

E = 210000.0 MPa gM0=1.100 gM1=1.250

----------------------------------------------------------------------------------------------------------------------------------------

SECTION PARAMETERS: CHS 244.5x10

D=24.4 cm Avy=46.919 cm2 Avz=46.919 cm2 Ag=73.700 cm2

B=24.4 cm Iy=5073.000 cm4 Iz=5073.000 cm4 It=10150.000 cm4

tw=1.0 cm Zey=414.969 cm3 Zez=414.969 cm3

tf=1.0 cm Zpy=550.236 cm3 Zpz=550.236 cm3

----------------------------------------------------------------------------------------------------------------------------------------

INTERNAL FORCES AND CAPACITIES: N = 1630.0 kN

Nd = 1842.5 kN

Pd = 1669.6 kN

Class of section = 1

----------------------------------------------------------------------------------------------------------------------------------------

LATERAL BUCKLING PARAMETERS: ----------------------------------------------------------------------------------------------------------------------------------------

BUCKLING PARAMETERS:

About Y axis: About Z axis:

Ly = 4.00 m Xy = 0.906 Lz = 4.00 m Xz = 0.906

ky*Ly = 4.00 m Pdy = 1669.6 kN kz*Lz = 4.00 m Pdz = 1669.6 kN

ky*Ly/ry = 48.213 kz*Lz/rz = 48.213

Lamy = 0.555 Lamz = 0.555

----------------------------------------------------------------------------------------------------------------------------------------

VERIFICATION FORMULAS:

Section strength check: N/Nd = 0.885 < 1.000 (9.3.1.1) OK!

Global stability check of member:

ky*Ly/ry = 48.213 < (k*L/r),max = 200.000 kz*Lz/rz = 48.213 < (k*L/r),max = 200.000 STABLE

P/min(Pdy,Pdz) = 0.976 < 1.000 (7.1.2) OK!

----------------------------------------------------------------------------------------------------------------------------------------

Section OK !!!


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b) From economical reason try to check the lighter CHS section.

Being still in RESULTS- CODE dialog box, type CHS 273x8 in the editable field below drawing of section and press ENTER. Calculations (and results) are refreshed instantly.

The results for new selected section are presented below.


March 2014 page 17 / 41

COMPARISON:

Resistance, interaction expression RSA

IS 800: 2007 Limit State Design

Handbook EC3: 2005

For CHS 244,5x10: 1. Cross-section compression resistance 2. Member buckling resistance in compression

ϪM0 = 1,1

N/ Nd = 1630/ 1842,5 = 0,885

N/ Pd = 1630/ 1669,6

= 0,976

ϪM0 = 1,0

NEd/ Nc,Rd = 1630/2026,8 = 0,804

NEd/ Nb,Rd = 1630/1836,5 = 0,888

CONCLUSIONS: Underlined values results from handbook (EC3 example) are different then values calculated by RSA

program for Indian code IS 800 because partial safety factors M0 are different in both codes. If the design resistances from handbook are divided by partial safety factor from IS 800 code

( M0 = 1,1 ): 1. NEd/ (Nc,Rd :1,1) = 1630 / (2026,8 : 1,1 ) = 1630 / 1842,5 = 0,8846 = 0,885 2. NEd/ (Nb,Rd :1,1) = 1630 / (1836,5 : 1,1 ) = 1630 / 1669,5 = 0,9763 = 0,976 .


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VERIFICATION EXAMPLE 2 - combined compression and bending about both axes

(lateral-torsional buckling)

Example taken from: DESIGNERS’ GUIDE to EN 1993-1-1 Eurocode 3:

Design of steel structures, general rules and rules for buildings

written by L. Gardner, D. A. Nethercot, 2005

TITLE: Example 6.10 -- Member resistance under combined bi-axial bending and axial compression SPECIFICATION: Verify if H 305x305x240 section in grade S275 has sufficient available strength to support the axial forces and moments listed below for Limit State Design:

Material Properties: S275 fy = 275 MPa , E= 210GPa

The member is to be designed as a ground floor column in a multi-storey building. The column of length 4,2 m is fixed in-plane (for buckling about major axis bending) and pinned out-of-plane, with diagonal bracing provided in both directions ( ky =0,8 ; kz=1,0). The column is laterally and torsionally unrestrained.

SOLUTION: Specify appropriate parameters in JOB PREFERENCES dialog box (click Menu/Tools/Job Preferences). Then, choose calculation method Limit State Design in CONFIGURATION dialog box ( press Configuration button in CALCULATIONS dialog box ).

section A-A, x=0 section B-B, x=L

NEd = 3440 kN

My,Ed = 420 kNm

Mz,Ed = 0

NEd = 3440 kN

My,Ed = 420 kNm

Mz,Ed = 110 kNm


March 2014 page 19 / 41

In DEFINITIONS dialog box you can define a new type of member in agreement with structure data It can be set in Member type combo-box. Pre-defined member type “Column” may be initially opened.

For chosen member type, press the Parameters button on Members tab, which opens MEMBER DEFINITION – PARAMETERS dialog box.

Type a new name in the Member type editable field, e.g. “My Mz Nc”. Change parameters to meet initial data requirements of the structure. In this particular combined compression and bending about both axes case set the following parameters: 1. for Y buckling define appropriate value of ky by manually entering 0,8 value in editable field or by pressing ky icon which opens BUCKLING DIAGRAM dialog box:

press third icon “0.8” and switch on Non-sway structure radio button.


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2. for lateral-buckling define:

lateral buckling type by pressing Lat. buckling type icon which opens LATERAL BUCKLING TYPE dialog box

select first icon and press OK.

lateral buckling length coefficient by pressing Upper flange or Lower flange button. It opens EFFECTIVE LENGTH FOR BEAMS, BETWEEN SUPPORTS dialog box:

Change conditions at supports. Click the first option for unrestrained and free to rotate on plane, press OK.


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3. for additional code parameters press More… button which opens MEMBER DEFINITION – ADDITIONAL PARAMETERS dialog box:

3.1. Click Load type My icon. It opens LOAD TYPE dialog box;

Select first option, press OK. Do the same for Load type Mz icon.


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3.2. Switch on Not weakened in Parameters of a weakened section. Click OK. 3.3. Type for gM0= 1,0 and for gM1= 1,0 ; in this particular case it will be helped in results comparison to EC3 ;

After changes dialog box looks like:

Now, you can SAVE changes of the newly-created member type named e.g. “ My Mz Nc “ in MEMBER DEFINITION- PARAMETERS dialog box.


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In DEFINITIONS dialog box number of the member must be assigned to appropriate name of Member type.

(very important when you verify different member types.)

You must press SAVE button to attribute the member type to number of member, here “My Mz Nc” to n

o 1.

In the CALCULATIONS dialog box set: -> Verification option; here Member Verification, -> Loads cases ; here for n

o 1 for Limit State Design

-> Limit state ; here only Ultimate Limit state will be analyzed, so switch off Limit stat –Serviceability. Check CONFIGURATION dialog box.

Now, start verificiations by pressing Calculations button.

MEMBER VERIFICATION dialog box with most significant results data will appear on screen.

Pressing the line with results for the member 1 opens the RESULTS dialog box with detailed results for the analyzed member. The view of the RESULTS windows are presented below.


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Simplified results tab

Detailed results tab

Pressing the Calc.Note button in “RESULTS –Code” dialog box opens the printout note for the analyzed member. You can obtain Simplified results printout or Detailed results printout. It depends on which tab is active. The printout note view of Simplified results is presented below.


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RESULTS for Limit State Design method:

STEEL DESIGN ----------------------------------------------------------------------------------------------------------------------------------------

CODE: IS800: 2007 Indian Standard-General Construction in Steel-Code of Practice (Third Revision) ANALYSIS TYPE: Member Verification

----------------------------------------------------------------------------------------------------------------------------------------

CODE GROUP:

MEMBER: 1 column POINT: COORDINATE: x = 1.00 L = 4.20 m

----------------------------------------------------------------------------------------------------------------------------------------

LOADS: Governing Load Case: 1 Limit State Design

----------------------------------------------------------------------------------------------------------------------------------------

MATERIAL: S 275 fy = 275.00 MPa fu = 430.00 MPa

E = 210000.00 MPa gM0=1.000 gM1=1.000

----------------------------------------------------------------------------------------------------------------------------------------

SECTION PARAMETERS: UC 305x305x240 D=35.26 cm Avy=239.70 cm2 Avz=81.10 cm2 Ag=306.00 cm2

B=31.79 cm Iy=64200.00 cm4 Iz=20310.00 cm4 It=1271.00 cm4

tw=2.30 cm Zey=3641.52 cm3 Zez=1277.76 cm3

tf=3.77 cm Zpy=4243.00 cm3 Zpz=1945.00 cm3

----------------------------------------------------------------------------------------------------------------------------------------

INTERNAL FORCES AND CAPACITIES: N = 3440.0 kN My = -420.0 kN*m Mz = 110.0 kN*m Vy = -26.2 kN

Nd = 8415.0 kN Mdy = 1166.8 kN*m Mdz = 534.9 kN*m Vdy = 3805.7 kN

Pd = 6639.7 kN Mndy = 773.7 kN*m Mndz = 502.7 kN*m Vz = -200.0 kN

Vdz = 1287.6 kN

Class of section = 1 Mymax = 420.0 kN*m Mzmax = 110.0 kN*m

----------------------------------------------------------------------------------------------------------------------------------------

LATERAL BUCKLING PARAMETERS: L,LT,low=5.04 m Mcr = 11963.5 kN*m fbd = 268.02 MPa X,LT = 0.975

fcr,b = 2819.59 MPa Lam,LT = 0.312 Mbdy = 1137.2 kN*m KLT = 0.892

----------------------------------------------------------------------------------------------------------------------------------------

BUCKLING PARAMETERS:

About Y axis: About Z axis: Ly = 4.20 m Xy = 0.976 Lz = 4.20 m Xz = 0.789

ky*Ly = 3.36 m Pdy = 8213.3 kN kz*Lz = 4.20 m Pdz = 6639.7 kN

ky*Ly/ry = 23.197 Ky = 1.028 kz*Lz/rz = 51.553 Kz = 1.204

Lamy = 0.267 Cmy = 0.400 Lamz = 0.594 Cmz = 0.600

----------------------------------------------------------------------------------------------------------------------------------------

VERIFICATION FORMULAS: Section strength check:

N/Nd = 0.409 < 1.000 (9.3.1.1) OK!

My/Mdy = 0.360 < 1.000 (9.3.1.1) OK!

Mz/Mdz = 0.206 < 1.000 (9.3.1.1) OK!

(My/Mndy)^ 2.000 + (Mz/Mndz)^2.044 = 0.340 < 1.000 (9.3.1.1) OK!

Vy/Vdy = 0.007 < 1.000 (8.4) OK!

Vz/Vdz = 0.155 < 1.000 (8.4) OK!

Global stability check of member: ky*Ly/ry = 23.197 < (k*L/r),max = 200.000 kz*Lz/rz = 51.553 < (k*L/r),max = 200.000 STABLE

P/Pdy + Ky*Cmy*Mymax/Mbdy + 0.6*Kz*Cmz*Mzmax/Mdz = 0.656 < 1.000 (9.3.2.2) OK!

P/Pdz + KLT*Mymax/Mbdy + Kz*Cmz*Mzmax/Mdz = 0.988 < 1.000 (9.3.2.2) OK!

----------------------------------------------------------------------------------------------------------------------------------------

Section OK !!!


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COMPARISON:

Resistance, interaction expression RSA

IS 800: 2007 Limit State Design

Handbook EC3: 2005

(Annex B, method 2)

For UB 305x305x240: 1. Cross-section compression resistance 2. Cross-section bending resistance 4. Cross-section resistance under

My+Mz+V+N 5. Member buckling resistance in compression

ky - buckling length coefficient Y 6. Member buckling resistance in bending

-lateral buckling length -critical M for lateral-torsional buckling

7. Member buckling resistance in combined bending and axial compression

ϪM0 = 1,0

N/ Nd = 3440/ 8415 = 0,409

My/Mdy =420/ 1166,8

= 0,360

Mz/Mdz =110/ 534,9 = 0,206

(9.3.1.1) = 0,340

N/ Pdy= 3440/8213,3 = 0,419 ;(ky=0,8)

N/ Pdz= 3440/6639,7 =

0,518

My/Mdy =420/1137,2 = 0,369

LLT = 5,04 m

Mcr = 11 963,5 kNm

0,656 (9.3.2.2) 0,988 (9.3.2.2)

ϪM0 = 1,0

NEd/ Nc,Rd = 3440/ 8415 = 0,409

My,Ed/My,c,Rd = 420/1168

= 0,360

Mz,Ed/Mz,c,Rd= 110/536,5

= 0,205

(6.2.9.1.(6)) = 0,340

NEd/ Ny,b,Rd =3440/8314,0 = 0,414 ; (ky=0,7)

NEd/ Ny,b,Rd =3440/6640

= 0,518

My,Ed/Mb,Rd =420/1152

= 0,36

Lcr = 4,2 m

Mcr = 17 114 kNm

0,66 (6.61)

0,97 (6.62)

CONCLUSIONS: The calculation example was made according to:

- IS:800 code using RSA program and - EC3:2005 code taken from handbook.

To make comparison more adequate, the same value of the partial safety factors M0=1,0 were assumed in both methods. The small differences in results are caused generally by:

small differences in values for elastic and plastic section modulus about Y & Z axis;

different values for buckling length coefficients in both codes;

different way of lateral buckling length coefficient description in both codes (effective length for beams between supports are different); in Indian code there is very precision method to define conditions at support;

different accuracy of parameters (e.g. the cross-sectional properties) in calculations.


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CCOONNCCRREETTEE


March 2014 page 28 / 41

1. IS 456:2000 – RC beams


March 2014 page 29 / 41

VERIFICATION EXAMPLE 1 - Dimensioning of simply supported beam

Example based on: [1] S. N. Sinha, “Reinforced Concrete Design”, Second Edition, 2002, Example 6.1, pp. 239

DESCRIPTION OF THE EXAMPLE: Design a simply supported beam. In this example, the results of the program are compared against [1]. The comparison concerns the amount of longitudinal reinforcement and shear reinforcement.

LOADS:

uniformly distributed: p=50 [kN/m]

GEOMETRY:

clear span: l0=6 [m]

support width: a=30 [cm]

cross section: 30x60 [cm]

MATERIAL:

Concrete: M20

Steel: Fe 415

IMPORTANT STEPS: Define the geometry of the beam (Fig.1.1) and the loads (Fig.1.2). Select support type in dialog box Dimension definition/Span geometry as Masonry (shear will be calculated from the axis of the support). Set proper materials (Calculation Options).

Fig. 1.1 Beam geometry


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Fig. 1.2 Loads and the calculation model

RESULTS OF LONGITUDINAL REINFORCEMENT (REINFORCEMENT FOR BENDING) CALCULATION: The theoretical areas of reinforcement determined by the program are presented on the graph in Fig.1.3. The values in the midspan, compared with [1], are presented in the table below.

Theoretical areas [1] ROBOT

tension reinf. Ast 23.65 cm2 23.77 cm2

tension reinf. Asc 7.87 cm2 7.97 cm2

Fig. 1.3. Theoretical (required) areas of reinforcement in beam.

The real reinforcement generated by ROBOT is different than in [1], but it fulfills the capacity requirements as well. The real reinforcement in the midspan is compared in the table below.

Real reinforcement [1] ROBOT

tension reinf. Ast 3 28 + 2 20

(24.75 cm2)

3 32

(24.13 cm2)

tension reinf. Asc 3 20

(9.42 cm2)

4 16

(8.04 cm2)

RESULTS OF TRANSVERSAL REINFORCEMENT (SHEAR REINFORCEMENT) CALCULATION: In [1] the beam is divided into three segments with idealised diagrams of shear force assumed. In the first and last segment (close to supports) the redistribution of sheaf force is assumed. To enable the distribution in ROBOT select Calculation Options/Advanced/Redistribution of a shear force near supports. The shear force on the distance equal to effective height d from the support is assumed as at the end of this distance. This way, the reduced shear force is equal to 211.57 kN, while the maximum shear force is 257.51 kN, Fig.1.4.


March 2014 page 31 / 41

Fig. 1.4. Diagram of shear force in beam (blue) and reduced shear force (green).

The shear reinforcement generated automatically by the program is bigger than this determined in [1]. The reason is following: the authors of [1] assume the full area of longitudinal bars influencing the

shear strength of concrete c. This is not a case in ROBOT, where the area of steel is assumed taking into account the development length. The area of bars develops from zero in point where the bar starts and reaches the maximum (full) area at the distance equal to development length (see Fig.1.5).

Fig. 1.5. Diagram of theoretical (blue) and real (red) area of reinforcement.

Taking this effect into account, the c near support is equal to 0.288 MPa and not 0.639 MPa as in [1]. This leads to the spacing of stirrups equal to 12cm within the end segment. In order to verify the calculations of shear in ROBOT based on [1], the spacings will be manually defined and the anchorage of main bars will be increased (in order to increase the area of still taken

for determination of c ). This allows us to check the calculations of shear capacity carried out in ROBOT. First, define the spacing of stirrups as in [1]. Then, increase the length of hooks of main bottom bars. Next, select Results/Freeze reinforcement. In Reinforcement Pattern/General dialog box enable Consider hook length in anchorage length. The capacity for shear is presnted in Fig.1.6. This proves, that taking into consideration the need of proper anchorage of longitudinal bars, the calculation of shear leads to a similar results as in [1].

Fig. 1.6. Diagram of distribution of shear force (green) and shear capacity (red) for the spacing of

stirrups as in [1]


March 2014 page 32 / 41

VERIFICATION EXAMPLE 2 - Determination of capacity of a beam


DESCRIPTION OF THE EXAMPLE: Determine the bending moment capacity of the beam with the assumed reinforcement. Two cases of reinforcement given in [1] are analysed here. The reinforcement is defined and the results concerning capacity are compared.

REINFORCEMENT:

GEOMETRY:

Cross-section:

case I: 30x53.6 [cm]

case II: 30x53.8 [cm]*

* In example, there is no height of the section given, instead the effective depth is assumed and the clear cover of bars. This means, that for two cases of reinforcement, where different diameters of bars are used, different height of the beam must be defined.

The length of the beam and other geometrical parameters are variables that have no influence on the analysed results since we analyze the capacity of the section.

MATERIAL:

Concrete: M20

Steel: Fe 415

RESULTS OF THE CALCULATION:

The capacity determined by ROBOT (Fig. 2.1) for the reinforcement assumed in [1] is found to be in agreement with the results in [1].

Case [1] ROBOT

I 250.69 kNm 250.78 kNm

II 309.32 kNm 309.32 kNm

Case Tension steel Compression steel

I 4 22 4 16

II 4 25 4 16


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Fig. 2.1. Diagrams of the moment capacities (red) for the two cases of reinforcement.

LITERATURE

[1] S. N. Sinha, “Reinforced Concrete Design”, Second Edition, 2002


March 2014 page 34 / 41

2. IS 456:2000 – RC columns


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VERIFICATION EXAMPLE 1 - Uniaxially eccentrically loaded braced rectangular column


DESCRIPTION OF THE EXAMPLE: Design the rectangular column in a braced (non-sway) frame. Column is subjected to uniaxial eccentrical load.

In the following example, the results of the program, concerning the calculations of longitudinal reinforcement and capacity are compared to the results of [1].

LOADS:

Ultimate axial load: P = 2000 [kN]

Ultimate axial moment: Mx = 400 [kNm]

GEOMETRY:

Unsupported length lu=3.5 [m]

Effective lengths: lx = 3 [m]

ly = 2.75 [m]

Cross section: 40x60 [cm]

MATERIAL:

Concrete : M 25 Longitudinal reinforcement : Fe 415

IMPORTANT STEPS:

In the dialog box Buckling length set buckling parameters (Fig.1.1.).

Fig. 1.1. Buckling parameters of the column.


March 2014 page 36 / 41

In the Calculation Option/ General dialog box select the proper method of calculation (Fig. 1.2.).

Fig. 1.2. Selection of calculation method.

In order to compare the results of calculation with the final solution assumed in [1], select the same diameter of main bars as in [1]. The diameter of bars may be set in Reinforcement pattern/Longitudinal bars dialog box (Fig. 1.3.).

Fig. 1.3. Parameters of main bars.

RESULTS OF CAPACITY CALCULATION:

Quantity [1]

Robot

(results presented

in calculation note)

1.4765 1.4645*

Muy,l 421.2 436.6 **

Muz,l 247.2 264.5 **

Capacity coefficient 1/0.996=1.004 1.026


March 2014 page 37 / 41

NOTES:

* - In [1] the equation scckuz fyAfP 75.0446.0 is used, while Robot uses the equation

given in code [2], which is scckuz fyAfP 75.045.0

** - In [1] the simplification takes place due to the use of approximated interaction curves. Robot calculates the capacity based on the equilibrium of forces in section. RESULTS OF LONGITUDINAL REINFORCEMENT CALCULATION: Reinforcement generated by the program (Fig.1.4), after assuming that the diameter 28mm should be used, is the same as determined in [1].

Fig. 1.4. Reinforcement generated by the program (8 28).

Quantity [1] Robot

Reinforcement 8 28 8 28


March 2014 page 38 / 41

VERIFICATION EXAMPLE 2 - Column subjected to axial load and biaxial bending


DESCRIPTION OF THE EXAMPLE: Design the rectangular column in a braced (non-sway) frame. Column is subjected to biaxial eccentrical load. In the following example, the results of the program, concerning the calculations of longitudinal reinforcement and buckling analysis are compared to the results of [1].

LOADS:

Ultimate axial load: P = 2000 [kN]

Ultimate axial moments: Mx1 = 225 [kNm] My1 = 125 [kNm]

Mx2 = 175 [kNm] My2 = 75 [kNm]

GEOMETRY:

Unsupported length lu=9.0 [m]

Effective lengths: lx = 8 [m]

ly = 6 [m]

Cross section: 40x60 [cm]

MATERIAL:

Concrete : M 25 Longitudinal reinforcement : Fe 415

IMPORTANT STEPS:

In the dialog box Buckling length set buckling parameters (Fig.2.1.).

Fig. 2.1. Buckling parameters of the column.


March 2014 page 39 / 41

In the Calculation Option/ General dialog box select the proper method of calculation (Fig. 2.2.).

Fig. 2.2. Selection of calculation method.

In the Calculation Option/ General dialog box enable the use of k coefficient for calculation of the additional moments (Fig. 2.3.).

Fig. 2.3. Parameter k for calculation of additional moments.

In order to compare the results of calculation with the final solution assumed in [1], select the same diameter of corner and intermediate bars as in [1]. The diameter of bars may be set in Reinforcement pattern/Longitudinal bars dialog box (Fig. 2.4.).

Fig. 2.4. Parameters of main bars.

Define loads (Fig. 2.5.).

Fig. 2.5. Loads definition.


March 2014 page 40 / 41

RESULTS OF LONGITUDINAL REINFORCEMENT CALCULATION: Reinforcement generated by the program (Fig.2.6.) is the same as determined in [1] (after the same diamteres were assumed).

Fig. 2.6. Reinforcement generated by the program (4 28 + 8 25).

Quantity [1] Robot

Reinforcement 4 28 + 8 25 4 28 + 8 25

RESULTS OF BUCKLING ANALYSIS:

Quantity [1]

Robot

(results presented


ky 0.8223 0.8590*

kz 0.7896 0.7899

Muy 292.71 296.63

Muz 176.06 176.10

NOTES: * - This difference is caused by different Pub and Puz values. In [1] Pub are calculated based on approximated tables, while in Robot the exact calculations are carried out. Moreover, in [1] the

equation scckuz fyAfP 75.0446.0 is used, while Robot uses the euqtion given in code [2], which

is scckuz fyAfP 75.045.0 .

RESULTS OF CAPACITY CALCULATION:

Quantity [1]

Robot

(results presented


1.4118 1.3886*

Muy,l 493.2 527.7**

Muz,l 292.8 316.5**

Capacity coefficient 1/0.964=1.037 1.085


March 2014 page 41 / 41

NOTES:

* In [1] the equation scckuz fyAfP 75.0446.0 is used, while Robot uses the euqtion given

in code [2], which is scckuz fyAfP 75.045.0 .

** - In [1] the simplification takes place due to the use of approximated interaction curves. Robot calculates the capacity based on the equilibrium of forces in section. Moreover the capacity is in [1] found for the assumed percentage of reinforcement. At the end however, the greater reinforcement is assumed, thus capacities are in [1] underestimated. In Robot, capacities are calculated for the real reinforcement.

LITERATURE

[1] S. N. Sinha, “Reinforced Concrete Design”, Second Edition, 2002 [2] Indian Standard IS 456:2000 Plain and Reinforced Concrete – Code of Practice (Fourth Revision), 2003

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