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March 2014

Autodesk Robot Structural Analysis Professional - Verification Manual for Canadian Codes

March 2014 page i

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

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

1. CAN/CSA-S16-09 ........................................................................................................................................................................ 3

VERIFICATION EXAMPLE 1 - AXIAL COMPRESSION .................................................................................................................... 4 VERIFICATION EXAMPLE 2 - FLEXURAL MEMBERS ..................................................................................................................... 7 VERIFICATION EXAMPLE 3 - AXIAL COMPRESSION AND BENDING .............................................................................................. 13 VERIFICATION EXAMPLE 4 - BIAXIAL BENDING AND COMPRESSION ........................................................................................... 16 VERIFICATION EXAMPLE 5 - TENSION MEMBERS ..................................................................................................................... 19

CCOONNCCRREETTEE ................................................................................................................................................................................... 22

1. CSA A23.3-94 – RC BEAMS ..................................................................................................................................................... 23

VERIFICATION EXAMPLE 1 - DETERMINATION OF CAPACITY OF A BEAM .................................................................................... 24 VERIFICATION EXAMPLE 2 - DIMENSIONING OF RECTANGULAR BEAM ....................................................................................... 25 LITERATURE ............................................................................................................................................................................. 26

2. CSA A23.3-94 – RC COLUMNS ................................................................................................................................................ 27

VERIFICATION EXAMPLE 1 - COLUMN SUBJECTED TO AXIAL LOAD AND UNI-AXIAL BENDING – NON-SWAY FRAME ........................ 28 VERIFICATION EXAMPLE 2 - COLUMN SUBJECTED TO AXIAL LOAD AND BIAXIAL BENDING ........................................................... 32 LITERATURE ............................................................................................................................................................................. 39

3. CSA A23.3-04 – RC BEAMS ..................................................................................................................................................... 40

VERIFICATION EXAMPLE 1 - DETERMINATION OF CAPACITY OF A BEAM .................................................................................... 41 VERIFICATION EXAMPLE 2 - DIMENSIONING OF RECTANGULAR BEAM ....................................................................................... 42 LITERATURE ............................................................................................................................................................................. 44

4. CSA A23.3-04 – RC COLUMNS ................................................................................................................................................ 45

VERIFICATION EXAMPLE 1 - COLUMN SUBJECTED TO AXIAL LOAD AND UNI-AXIAL BENDING – NON-SWAY FRAME ........................ 46 LITERATURE ............................................................................................................................................................................. 49


March 2014 page 1 / 49

INTRODUCTION

This verification manual contains numerical examples for elements of structures prepared and originally calculated by Autodesk Robot Structural Analysis Professional version 2013. The comparison of results is still valid for the next versions. Most of the 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. CAN/CSA-S16-09



VERIFICATION EXAMPLE 1 - Axial compression

Example taken from Handbook of Steel Construction (Sixth Edition) Canadian Institute of Steel Construction

TITLE: Axial compression (Example 1 page 4-30).

SPECIFICATION: A W 310x143 column is required to carry a factored load of 3600 kN. The effective length Kz*Lz along the weak axis is 4500 mm. The effective length Ky*Ly along the strong axis is 7600 mm. Use CSA G40.21-M. 300W steel.

SOLUTION: Define a new type of member. For analyzed member pre-defined type of member COLUMN may be initially opened. It can be set in Member type combo-box. Press the Parameters button in DEFINITION-MEMBERS tab, which opens MEMBER DEFINITION – PARAMETERS dialog box. Type a new name Column 1 in the Member Type editable field. For defining an appropriate buckling length about Y axis select Real radio button in Member Length Y area and type 7,60 m in editable field. Save the newly-created type of the member.



In the CALCULATIONS dialog box set Member Verification option for member 1 and switch off Limit State – Serviceability (only Ultimate Limit state will be analyzed). Now, start the 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 window is presented below. Moreover, the printout note containing the same results data as in Simplified results tab of the RESULTS window is added.



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

CODE: CAN/CSA S16-09

ANALYSIS TYPE: Member Verification

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

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

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

LOADS: Governing Load Case: 1 DL1

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

MATERIAL:

STEEL 300W Fy = 300.00 MPa

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

SECTION PARAMETERS: W 310x143

d=32.3 cm

b=30.9 cm Ay=141.522 cm2 Az=45.220 cm2 Ax=182.000 cm2

w=1.4 cm Iy=34800.000 cm4 Iz=11300.000 cm4 J=287.000 cm4

t=2.3 cm Zy=2420.000 cm3 Zz=1110.000 cm3

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

INTERNAL FORCES AND CAPACITIES:

Cf = 3600.00 kN

Cr0 = 4914.00 kN

CLASS: = Plastic

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

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

BUCKLING PARAMETERS:

About Y axis: About Z axis:

Ly = 7.60 m Lamy = 0.678 Lz = 4.50 m Lamz = 0.704

KyLy = 7.60 m KzLz = 4.50 m

KyLy/ry = 54.962 KzLz/rz = 57.110

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

VERIFICATION FORMULAS: Cf/Cr = 3600.00/3842.37 = 0.937 < 1.000 (13.8.2)

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

Section OK!!!

COMPARISON:

Resistance, interaction expression Robot HANDBOOK

1. Factored compressive resistance of a member Cr [kN]

2. Check of the formula Cf/Cr 1

3842.37

0.937

3840

0.938



VERIFICATION EXAMPLE 2 - Flexural members


TITLE: Flexural members (Examples 1, 2, 3, page 5-91).

SPECIFICATION: 1. Design a simply supported beam spanning 8 meters to carry a uniformly distributed load of 15

kN/m. Specified live load and 7 kN/m. Specified dead load. The dead load includes an assumed beam dead load of 0.7 kN/m. Live load deflection is limited to L/300. Assume the beam frames into supporting members and that the beam is laterally supported. Use CSA G40.21-M. 300W steel.

2. Same as in (1), except assume that beam is laterally supported at quarter points, mid span and ends of beam.

3. Same as in (1) except assume that beam is laterally supported at mid point and ends of beam only.

SOLUTION: Create, for simplicity, an example consists of 3 separate members. Each member should have the same length, support conditions and load type. Then define a new type of member 1. For analyzed member pre-defined type of member BEAM may be initially opened. It can be set in Member type combo-box. Press the Parameters button in DEFINITION-MEMBERS tab, which opens MEMBER DEFINITION – PARAMETERS dialog box. Type a new name Beam1 in the Member Type editable field. Double click an Lateral Buckling icon that opens Lateral Buckling Type dialog box. Select the last icon (No Lateral Buckling) and click OK Save the newly-created type of member1.



For member 2 also pre-defined type of member BEAM may be initially opened. Change the name to Beam2. Select the icon Lateral Buckling Coefficient – Upper Flange that opens Effective Length of Beams between Supports dialog box. Select the icon Intermediate Bracing. In Internal bracing dialog go to Define manually coordinates of the existing bracings field and type relative coordinates of bracing at quarter points and in mid span 0,25; 0.5; 0,75, then click OK. Set also om2 coefficient on 1.0 (manually or by choosing appropriate icon) and save the newly created type of member 2.

Similarly define the set of parameters for member 3. Name the set of parameters as Beam 3 and define internal bracing in the middle of the member 3. In the CALCULATIONS dialog box set Member Verification option for members 1to3 and switch off Limit State – Serviceability (only Ultimate Limit state will be analyzed). Now, start the calculations by pressing Calculations button.



Member Verification dialog box with most significant results data will appear on screen. Pressing the lines with results for the members 1 2 3 opens the RESULTS dialog box with detailed results for the analyzed member.


RESULTS:

1. Member 1. In the first step W410x54 was considered.

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



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

CODE GROUP:

MEMBER: 1 POINT: 2 COORDINATE: x = 0.50 L = 4.00 m

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

LOADS:

Governing Load Case: 1 DL1

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

MATERIAL:



March 2014 page 10 / 49

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

SECTION PARAMETERS: W 410x54 d=40.3 cm


w=0.8 cm Iy=18600.00 cm4 Iz=1010.00 cm4 J=22.60 cm4

t=1.1 cm Zy=1050.00 cm3 Zz=177.00 cm3

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


Mfy = 249.60 kN*m

Mry = 283.50 kN*m

CLASS: = Plastic

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




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

VERIFICATION FORMULAS: Mfy/Mry = 249.60/283.50 = 0.88 < 1.00 (13.8.2(a))

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

Section OK!!!


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



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


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

LOADS:


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

MATERIAL:


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


March 2014 page 11 / 49


d=40.3 cm


w=0.8 cm Iy=18600.00 cm4 Iz=1010.00 cm4 J=22.60 cm4

t=1.1 cm Zy=1050.00 cm3 Zz=177.00 cm3

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

INTERNAL FORCES AND CAPACITIES: Mfy = 249.60 kN*m

Mry = 283.50 kN*m

CLASS: = Plastic

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

LATERAL BUCKLING PARAMETERS: Le = 2.00 m om2 = 1.00 Mre = 283.50 kN*m

k = 0.75 Mu = 1020.52 kN*m

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



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

VERIFICATION FORMULAS:

Mfy/Mry = 249.60/283.50 = 0.88 < 1.00 (13.8.2(a))

Mfy/Mre = 249.60/283.50 = 0.88 < 1.00 (13.8.2(c))

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

Section OK!!!


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



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


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

LOADS:


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

MATERIAL:



March 2014 page 12 / 49

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



w=0.8 cm Iy=21600.00 cm4 Iz=1200.00 cm4 J=32.80 cm4

t=1.3 cm Zy=1190.00 cm3 Zz=209.00 cm3

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


Mfy = 249.60 kN*m

Mry = 321.30 kN*m

CLASS: = Plastic

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


k = 0.00 Mu = 350.00 kN*m

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



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

VERIFICATION FORMULAS: Mfy/Mry = 249.60/321.30 = 0.78 < 1.00 (13.8.2(a))

Mfy/Mre = 249.60/263.97 = 0.95 < 1.00 (13.8.2(c))

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

Section OK !!! COMPARISON:


(1) results for element No1 (W 410x54) 1. Factored moment resistance of a member Mry [kNm]

2. Check formula Mfy/Mry 1 (2) results for element No2 (W 410x54) 1. Factored moment resistance of a member Mry [kNm]

2. Check formula Mfy/Mry 1 (3) results for element No1 (W 410x60) 1. Factored moment resistance of a member Mry [kNm] (as for laterally supported member) 2. Factored moment resistance of a member Mre [kNm] (as for laterally unsupported member)

Check formula Mfy/Mry 1

Check formula Mfy/Mre 1

283,5 0,880

283,5 0,880

321,3

263,97 0,777 0,946

284 0,880

284 0,880

321

264 0,779 0,947

.


March 2014 page 13 / 49

VERIFICATION EXAMPLE 3

- Axial compression and bending


TITLE: Axial compression and bending (Example 1, page 4-108).

SPECIFICATION: Design a steel column for the third storey of a six storey building for the loading conditions shown below. Moments are due to rigidly framed beams and gravity loading, and cause bending about the Y-

Y axis of the column. The P effects have been included in the analysis. Beams framing to the minor axis have flexible connections. It has been suggested that W 310x129 shape might be used CSA G40.21-M. 300W steel.

SOLUTION: Define a new type of member. For analyzed member pre-defined type of member COLUMN may be initially opened. It can be set in Member type combo-box. Press the Parameters button in DEFINITION-MEMBERS tab, which opens MEMBER DEFINITION – PARAMETERS dialog box. Type a new name Column 1 in the Member Type editable field. Double click an Lateral Buckling icon that opens Lateral Buckling Type dialog box. Select the first icon (Element loaded Symmetrically) and click OK. Select the icon Lateral Buckling Coefficient – Upper Flange that opens Effective Length of Beams between Supports dialog box. Choose third radio button that set the lateral buckling coefficient to 1.0. Repeat the previous procedure for the lower flange. Save the newly-created type of the member.


March 2014 page 14 / 49





March 2014 page 15 / 49

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



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


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

LOADS: Governing Load Case: 1 TEST

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

MATERIAL:


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


d=31.8 cm


w=1.3 cm Iy=30800.000 cm4 Iz=10000.000 cm4 J=213.000 cm4

t=2.1 cm Zy=2160.000 cm3 Zz=991.000 cm3

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


Cf = 2000.00 kN Mfy = 300.00 kN*m

Cr0 = 4455.00 kN Mry = 583.20 kN*m

Vfz = -135.14 kN

CLASS: = Plastic Vrz = 742.35 kN

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


k = -0.667 Mu = 6595.19 kN*m

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



Ly = 3.70 m Lamy = 0.334 Lz = 3.70 m Lamz = 0.586

KyLy = 3.70 m om1y = 0.400 KzLz = 3.70 m

KyLy/ry = 27.081 U1y = 0.419 KzLz/rz = 47.527

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


Cf/Cr0 + 0.85*Mfy/Mry = 0.886 < 1.000 (13.8.2(a))

Cf/Cry1 + 0.85*U1y*Mfy/Mry = 0.650 < 1.000 (13.8.2(b))

Cf/Crz + 0.85*U1y*Mfy/Mre = 0.964 < 1.000 (13.8.2(c))

Vfz/Vrz = 0.182 < 1.000 (13.4.1)

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

Section OK !!!

COMPARISON:


1. Factored compressive resistance of a member Cr [kN] 2. Factored moment resistance of a member Mry [kNm] (as for laterally supported member) 3. Factored moment resistance of a member Mre [kNm] (as for laterally unsupported member) 4. Cross sectional strength check Overal member strength check Lateral-torsional buckling strength check

3797.31 583.20

583.20

0.886 0.650 0.964

3800 583

583

0.885 0.650 0.963


March 2014 page 16 / 49

VERIFICATION EXAMPLE 4 - Biaxial bending and compression


TITLE: Biaxial bending and compression (Example 2, page 4-110).

SPECIFICATION: Design a steel column for the third story of a six story building for the loading conditions shown. Moments are due to rigidly framed beams and gravity loading, and cause bending about the Y-Y and Z-Z axis of the column. The direction of the moments is such that double curvature is induced in the

column. The P effects have been included in the analysis. Beams framing to the minor axis have flexible connections. It has been suggested that W 310x143 shape might be used CSA G40.21-M. 300W steel.

SOLUTION: Define a new type of member. For analyzed member pre-defined type of member COLUMN may be initially opened. It can be set in Member type combo-box. Press the Parameters button in DEFINITION-MEMBERS tab, which opens MEMBER DEFINITION – PARAMETERS dialog box. Type a new name Column 1 in the Member Type editable field. Double click an Lateral Buckling icon that opens Lateral Buckling Type dialog box. Select the first icon (Element loaded Symmetrically) and click OK. Select the icon Lateral Buckling Coefficient – Upper Flange that opens Effective Length of Beams between Supports dialog box. Choose third radio button that set the lateral buckling coefficient to 1.0. Repeat the previous procedure for the lower flange. Save the newly-created type of the member.


March 2014 page 17 / 49

In the CALCULATIONS dialog box set Member Verification option for member 1 and switch off Limit State – Serviceability (only Ultimate Limit state will be analyzed). Now, start the calculations by pressing Calculation button.



RESULTS:

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




March 2014 page 18 / 49

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

CODE GROUP:


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

LOADS:

Governing Load Case: 1 test

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

MATERIAL:


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



w=1.4 cm Iy=34800.000 cm4 Iz=11300.000 cm4 J=287.000 cm4

t=2.3 cm Zy=2420.000 cm3 Zz=1110.000 cm3

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


Cf = 2000.00 kN Mfy = 300.00 kN*m Mfz = -100.00 kN*m

Cr0 = 4914.00 kN Mry = 653.40 kN*m Mrz = 299.70 kN*m

Vfy = -54.05 kN Vfz = -135.14 kN

CLASS: = Plastic Vry = 2521.92 kN Vrz = 805.82 kN

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


k = -0.667 Mu = 7735.47 kN*m

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



Ly = 3.70 m Lamy = 0.330 Lz = 3.70 m Lamz = 0.579

KyLy = 3.70 m om1y = 0.400 KzLz = 3.70 m om1z = 0.400

KyLy/ry = 26.758 U1y = 0.417 KzLz/rz = 46.957 U1z = 0.456

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

VERIFICATION FORMULAS: Cf/Cr0 + 0.85*Mfy/Mry + 0.60*Mfz/Mrz = 0.997 < 1.000 (13.8.2(a))

Mfy/Mry + Mfz/Mrz = 0.793 < 1.000 (13.8.2(b))

Cf/Crz + 0.85*U1y*Mfy/Mre + 0.832*U1z*Mfz/Mrz = 0.992 < 1.000 (13.8.2(c))

Vfy/Vry = 0.021 < 1.000 (13.4.1) Vfz/Vrz = 0.168 < 1.000 (13.4.1)

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

Section OK !!!

COMPARISON:


1. Factored compressive resistance of a member Cr [kN] 2. Factored moment resistance of a member Mry [kNm] (as for laterally supported member) 3. Factored moment resistance of a member Mre [kNm] (as for laterally unsupported member) 4. Cross sectional strength check Additional criterion in 13.8.2 b Lateral-torsional buckling strength check

4207.8 653.4

653.4

0.997 0.793 0.992

4200 653

653

0.998 0.793 0.959

CONCLUSION: The differences are caused by different way of rounding-off the cross-sectional properties (cross-sectional area, section modulus, moment of inertia). A small difference in formula 13.8.2.c is caused by changes between version S16.1-94 and S 16-09.


March 2014 page 19 / 49

VERIFICATION EXAMPLE 5 - Tension members

Example taken from handbook Limit States Design in Structural Steel, by Kulak, Adams, & Gilmor

Canadian Institute of Steel Construction

TITLE: Tension members.

SPECIFICATION: Design the tension diagonal of an all-welded Pratt roof truss in which the chords are made from DLL 3x2.5x10.375 sections. The factored load Tf in the member under consideration is 630 kN and its length is 4 m. It has been suggested that 2 - 76x64x9.5 angles might be used. Use CSA G40.21-M. 300W steel.

SOLUTION: For analyzed member pre-defined type of member SIMPLY BAR may be used.



March 2014 page 20 / 49



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



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

CODE GROUP:


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

LOADS:

Governing Load Case: 1 test

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

MATERIAL:


March 2014 page 21 / 49


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

SECTION PARAMETERS: DLL 3x2.5x10.375 d=7.6 cm


w=1.0 cm Iy=137.773 cm4 Iz=215.071 cm4 J=7.825 cm4

t=1.0 cm Zy=48.014 cm3 Zz=56.266 cm3

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


Tf = -630.00 kN

Tr = 668.90 kN

CLASS: = Plastic

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




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


Tf/Tr = 630.00/668.90 = 0.942 < 1.000 (13.9(a))

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

Section OK!!!

COMPARISON:


1. Factored tensile resistance of a member Tr [kN]

2. Check of the formula Tf/Tr 1

668.9

0.942

670

0.940


March 2014 page 22 / 49

CCOONNCCRREETTEE


March 2014 page 23 / 49

1. CSA A23.3-94 – RC beams


March 2014 page 24 / 49

VERIFICATION EXAMPLE 1 - Determination of capacity of a beam

Example based on: [3] Canadian Portland Cement Association, “Concrete Design Handbook”, Second Edition, 1998,

Example 2.1, pp. 2-7

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:

Longitudinal reinforcement: 4 No. 25

Stirrups: No.10

GEOMETRY:

Cross-section: 40x60 [cm]

clear cover: 4 [cm]

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: fc’ = 30 [MPa]

Steel: fy = 400 [MPa]

RESULTS OF THE CALCULATION:

Quantity [1] ROBOT

Resisting moment 206.8kNm 205.6 kNm


March 2014 page 25 / 49

VERIFICATION EXAMPLE 2 - Dimensioning of rectangular beam


Example 2.2 pp. 2-7

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

LOADS:

In the reference example [1] only the ultimate moment is given. Thus, we assume the geometry of the beam and determine the load in order to obtain the moment specified in [1].

uniformly distributed: p=81 [kN/m]

GEOMETRY:

clear span: l0=6 [m]

support width: a=40 [cm]

cross section: 40x60 [cm]

MATERIAL:



IMPORTANT STEPS: Define the geometry of the beam (Fig.1.1) and the loads (Fig.1.2). Select support type in dialog box. Set proper materials (Calculation Options).

Fig. 1.1 Beam geometry


March 2014 page 26 / 49

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 24.9 cm2 24.7 cm2

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

In order to compare the real (provided) reinforcement, set the same diameter of bars as assumed in [1] in Reinforcement pattern/ Bottom Reinf. As presented in the table below, ROBOT generates the same number of bars as in [1].

Real reinforcement [1] ROBOT

tension reinf. Ast 5 No.25 5 No.25

LITERATURE

[1] Canadian Portland Cement Association, “Concrete Design Handbook”, Second Edition, 1998.


March 2014 page 27 / 49

2. CSA A23.3-94 – RC columns


March 2014 page 28 / 49

VERIFICATION EXAMPLE 1 - Column subjected to axial load and uni-axial bending –

non-sway frame



DESCRIPTION OF THE EXAMPLE: Design the rectangular column in a non-sway frame.

In the following example, the results of the program, concerning the calculations of reinforcement and buckling analysis are compared to the results of [2].

LOADS:

Unfactored Loads: Dead Load Live Load Factored Loads

(1.25 DL + 1.5 LL)

Axial Load (kN) 1776 1320 4200

Top Moment (kNm) 112 66 239

Bottom Moment (kNm) 12 7 26

GEOMETRY:

lu=8.1 [m]


MATERIAL:

Concrete : CONCRETE 40 fc' = 40.00 (MPa) Longitudinal reinforcement : Grade 400 fy = 400.00 (MPa)

Fig.1. Cross section with longitudinal reinforcement determined in [2] (12 No.25). IMPORTANT STEPS: In the dialog box Buckling length set buckling parameters (Fig.1.2.).


March 2014 page 29 / 49

Fig. 1.2. Buckling parameters of the column. In the Loads dialog box put the forces and the ratio of long-term to total load for each load case.

Fig. 1.3. Loads.

In the Calculation Option/ General dialog box check: Design – unidirectional bending: My direction (Fig. 1.4.).

Fig. 1.4. Unidirectional bending RESULTS OF REINFORCEMENT CALCULATION: The reinforcement generated by the program (Fig 1.5.) is different than that calculated in [2]. The authors of [2] find the reinforcement of 12 No. 25 bars, thus the total area is equal to 60.0 cm

2. The

calculations with the program result in reinforcement with 10 No. 25 bars, thus the total area is equal to 50 cm

2. The reinforcement determined by the program is more optimal solution. As shown by the

verification of two reinforcement patterns, carried out in the program, the capacity coefficients are equal to 1.03 and 1.01 for 12 No. 25 and 10 No. 25 respectively.


March 2014 page 30 / 49

Fig. 1.5. Reinforcement generated by the program (10 No.25).

In order to verify the results of buckling analysis, after the modification of reinforcement to the form as in [2] (see Fig. 1), the verification is carried out. RESULTS OF BUCKLING ANALYSIS:

Quality (Unit) [2] Robot

r

klu (-) 45.42 47.1

gI (mm4) 5.2 x 109 5.2 x 109

seI (mm4) 1.6 x 108 1.5 x 108

gc

d

sesgc

IE

IEIE

EI

25.0

1

)2,0(

max (Nmm2) 4.1 x 1013 4.1 x 1013

2

2

)( u

ckl

EIP

(N) 8741 x 103 8743 x 103

2

14.06.0M

MCm (-)

0.56

0.56

)03.015(2 hPMC um (kNm) 126 126

c

u

m

c

P

P

MCM

75.01

2

(kNm) 373 371


March 2014 page 31 / 49

FINAL VERIFICATION:

NOTE: the results of reinforcement calculation concern the automatic calculation of reinforcement. In the case of buckling analysis, the reinforcement obtain in the program is modified to a form as in the reference example, in order to enable the comparison of total moments.

Quantity [2] Robot

sA

SdRd /

60.0 cm2

1.03

50.0 cm2

1.01

cM 373 kNm 371 kNm


March 2014 page 32 / 49

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

DESCRIPTION OF THE EXAMPLE: Following example illustrates the procedure of dimensioning of biaxial bending of column, which is non-sway in one direction, whereas sway in the other. The results of the program are accompanied by the „manual” calculations.

1. SECTION DIMENSIONS

2. MATERIALS Concrete : CONCRETE 30 fc' = 30.00 (MPa) Longitudinal reinforcement : Grade 500 fy = 500.00 (MPa) Transversal reinforcement : Grade 400 fy = 400.00 (MPa)

3. BUCKLING MODEL

As can be seen the sway column is assumed for Z direction, and the non-sway column for Y direction.


March 2014 page 33 / 49

4. LOADS

NOTE: The column is sway for Z direction, thus the ratio of non-sway moment to total moment should be defined in the load table.

NOTE: Let us assume, the moments in Y direction are linearly distributed along the height of the column. Thus, we define only the ends’ moments for Y direction. In Z direction however, we assume the mid-height moment is not a result of the linear distribution. For such a case, Robot let the user define the moments in the mid-section explicitly.

5. CALCULATED REINFORCEMENT: The program generates the reinforcement 16 No.30.

6. RESULTS OF THE SECTION CALCULATIONS: The dimensioning combination is 1.25 DL1+1.5 LL1 The dimensioning section (where the most unfavorable set of forces is found) is for that combination the section in the mid-height of the column.


March 2014 page 34 / 49

Since the column is found as slender, the second-order effects are taken into account in both directions. In parallel the other sections (at the ends of the column) are checked for all combinations of loads. In the top and bottom ends’ sections of the column in Y direction, the influence of buckling has not been taken into account, since the structure is non-sway in this direction. In Z direction however, the influence of slenderness is taken into account for all three sections of the column. All the results of total forces for each combination and each section of the column may be seen in the table “Intersection” at the Column-results layout. 7. CALCULATIONS OF TOTAL MOMENT: 7.1. LOADS For the dimensioning combination, the loads are:

Case

N

(kip)

MyA

(kip-ft)

MyB

(kip-ft)

MyC

(kip-ft)

MzA

(kip-ft)

MzB

(kip-ft)

MzC

(kip-ft)

1 DL1 600 100 80 92 50 20 60*

2 LL1 200 40 30 36 40 10 45*

Dimensioning

combination 1.25G1+1.5Q1 1050 185 145 169 122.5 40 142.5

where A, B and C denote upper, lower and mid-height sections of the column respectively. * - the values are written “by hand” by the user (see point 4 – Loads)


March 2014 page 35 / 49

7.2. THE INFLUENCE OF SLENDERNESS Two independent calculations of the total moment for both directions are carried out. Y DIRECTION

Slenderness:

r

lk u = 56.12

ulk = 8.1 (m)

r = 0.144 (m)

)'/(

2

11025

gcf AfP

M

M

= 41.03

2

1

M

M = 0.78

r

lk u>

)'/(

2

11025

gcf AfP

M

M

column is slender

The initial moments at the end of the column:

M1 = 145.00 (kNm) M2 = 185.00 (kNm)

Calculation of critical force:

2

2

)( u

ckl

EIP

gc

d

sesgc

IE

IEIE

EI

25.0

1

)2,0(

max (10-11), (10-12)

The moment of inertia of steel is calculated according to the scheme:


March 2014 page 36 / 49

i

ii

s zAsI = 22554 (cm4)

gI 416667 (cm4)

cE = 28325 (MPa)

sE = 200000 (MPa)

d is calculated as a weighted average from the load cases. The weight factors are assumed

according to the axial forces. Thus:

d = 01050

2005.11

1050

60025.1

=0.714

mkPa 9505225,0

mkPa 400821

)2,0(

max4

4

gc

d

sesgc

IE

IEIE

EI

Thus,

EI 40082 kPa m4

cP 6030 kN

Check if )03,015(2 hPMC um - OK

4.0

4.06.0max 2

1

M

M

Cm = 0.91

)03,015( hPu = 31.5 kNm

2MCm = 169 kNM (note, that it is the same value as MyC in table in point 7.1 – Loads)

The dimensioning moment in Y direction is equal to:

c

u

m

c

P

P

MCM

75,01

2

= 220 (kNm)

Z DIRECTION

NOTE: In most cases, the sway column is calculated for the end moment M2, taking into account the effects of slenderness. In this case however, the presence of moment in Y direction (which is also increased due to slenderness), causes the mid-height section to be the most unfavorable (even though the end moment in Z direction is greater). Slenderness: Since the column is sway, the slenderness effects are taken into account.


March 2014 page 37 / 49

The following table illustrates how the division of moments into sway and non-sway is carried out for the particular load combination. We consider here directly the mid-height moment (its values have been given directly in the load table).

Case MzC Load factors

MzC

(dimensioning

combination)

Mzns/Mz Mzns =

(Mzns/Mz)*Mz

Mzs =

(1-Mzns/Mz)*Mz

G1 60 1.25 75 1 75 0

Q1 45 1.5 67.5 0.5 33.75

33.75

SUM - 142.5 - 108.75 33.75

nsM 108.75 (kNm)

sM 33.75 (kNm)

The magnification factor for sway column is equal to:

25.1)1/(1 Qs (Q coefficient is defined in the Story parameters dialog box)

nsss

I

c MMM = 150.94 kNm

NOTE: Since the values of moment do not result from the linear distribution, but were defined directly

by the user, we do not deal here with 1M and 2M , but use directly the value of the mid-height

moment).

Check, if further magnification of moment is required

gc

u

u

Af

Pr

l

'

35

r

lu = 86.6.97

gc

u

Af

P

'

35= 83.67.

gc

u

u

Af

Pr

l

'

35 the magnification of the mid-height moment is necessary

NOTE: The magnification below is carried out only if we consider the mid-height section).


March 2014 page 38 / 49

c

u

I

c

c

P

P

MM

75,01

NOTE: Since the values of moment do not result from the linear distribution but were defined directly

by the user, we do not deal here with 2MCm but use directly the value of the mid-height moment).

Calculation of critical force:

2

2

)( u

ckl

EIP

gc

d

sesgc

IE

IEIE

EI

25.0

1

)2,0(

max (10-11), (10-12)

i

ii

s yAsI = 15482 (cm4)

gI 266667 (cm4)

cE = 28325 (MPa)

sE = 200000 (MPa)

d is calculated as a weighted average from the load cases. The weight factors are assumed

according to the axial forces. Thus:

d = 01050

2005.11

1050

60025.1

=0.714

mkPa 1888325,0

mkPa 268751

)2,0(

max4

4

gc

d

sesgc

IE

IEIE

EI

Thus,

EI 26875 (kPa m4)

cP 2192 (kN)

The dimensioning moment in Y direction is equal to:

c

u

I

c

c

P

P

MM

75,01

= 418 (kNm)

7.3. FINAL RESULT

cyM = 220 (kNm)

czM = 418 (kNm)


March 2014 page 39 / 49

8. CONCLUSIONS

The algorithm of calculations of the total moments (i.e. slenderness effects) in non-sway/sway column has been presented. The results obtained with the program (see point 6 – Results of the Section Calculations) are in agreement with the manual calculations (see point 7.3 – Final Result)

LITERATURE

[1] CSA Standard A23.3-94. Structures (Design). [2] Canadian Portland Cement Association, “Concrete Design Handbook”, Second Edition, 1998, Example 8.1, pp. 8-6


March 2014 page 40 / 49

3. CSA A23.3-04 – RC beams


March 2014 page 41 / 49

VERIFICATION EXAMPLE 1 - Determination of capacity of a beam



DESCRIPTION OF THE EXAMPLE: In this paragraph, the same example as in chapter 2 - verification example 1, is analyzed. The example derived from [3] was created for the old edition of the code CSA A23.3-94 [1]. Now, the example is solved based on the new edition CSA A23.3-04 [2], and the differences that occured in the new version respect to the previous one are specified and disucussed.

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:

Longitudinal reinforcement: 4 No. 25

Stirrups: No.10

GEOMETRY:

Cross-section: 40x60 [cm]

clear cover: 4 [cm]

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:



RESULTS OF THE CALCULATION:

The capacity determined by ROBOT (Fig. 1.1) for the reinforcement assumed in [1] is found to be in very good agreement with the results in [1].

Quantity [1] ROBOT

Resisting moment 206.8 kNm 206.7 kNm

DIFFERENCES IN THE NEW VERSION RESPECT TO THE PREVIOUS ONE: The calculations based on new edition of the code give similar results to the reference [1] obtained for

the previous edition of the code. In fact, the only change that will impact the result is the change in c coefficient introduced by the new code. The coefficient changes from 0.6 up to 0.65.

For the c = 0.6, the resistance factor is Kr = 1.7797, and the resisting moment is: Mr = Kr bd

2 = 206.8kNm

For the c = 0.65, the resistance factor is Kr = 1.7892, and the resisting moment is: Mr = Kr bd

2 = 207.9kNm


March 2014 page 42 / 49

VERIFICATION EXAMPLE 2 - Dimensioning of rectangular beam


Example 2.2 pp. 2-7


Design reinforcement of a beam beam. In this example, the results of the program are compared against [1]. The comparison concerns the amount of longitudinal tension reinforcement.

LOADS:

In the reference example [1] only the ultimate moment is given. Thus, we assume the geometry of the beam and determine the load in order to obtain the moment specified in [1].

uniformly distributed: p=81 [kN/m]

GEOMETRY:

clear span: l0=6 [m]

support width: a=40 [cm]


MATERIAL:



IMPORTANT STEPS: Define the geometry of the beam (Fig.1.1) and the loads (Fig.1.2). Select support type in dialog box. Set proper materials (Calculation Options).

Fig. 1.1 Beam geometry


March 2014 page 43 / 49

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 24.9 cm2 24.4 cm2

As can be seen, very good agreement of the results is obtained.

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

DIFFERENCES IN THE NEW VERSION RESPECT TO THE PREVIOUS ONE: The calculations based on new edition of the code give similar results to the reference [1] obtained for

the previous edition of the code. In fact, the only change that will impact the result is the change in c coefficient introduced by the new code. The coefficient changes from 0.6 up to 0.65.

For the c = 0.6, and the resistance factor is Kr = 3.48, the reinforcement ratio and the area are:

= 1.14 %, Ast = 24.9 cm2

For the c = 0.65, and the resistance factor is Kr = 3.48, the reinforcement ratio is:

= 1.13 %, Ast = 24.6 cm2

The value calculated in ROBOT is in very good agreement with the value presented above. In order to compare the real (provided) reinforcement, set the same diameter of bars as assumed in [1] in Reinforcement pattern/ Bottom Reinf. As presented in the table below, ROBOT generates the same number of bars as in [1].

Real reinforcement [1] ROBOT

tension reinf. Ast 5 No.25 5 No.25


March 2014 page 44 / 49

LITERATURE

[1] CSA Standard A23.3-94. Structures (Design). [2] CSA Standard A23.3-04. Design of concrete structures. [3] Canadian Portland Cement Association, “Concrete Design Handbook”, Second Edition, 1998.


March 2014 page 45 / 49

4. CSA A23.3-04 – RC columns


March 2014 page 46 / 49

VERIFICATION EXAMPLE 1 - Column subjected to axial load and uni-axial bending –

non-sway frame




Design the rectangular column in a non-sway frame.

In the following example, the results of the program, concerning the calculations of reinforcement and buckling analysis are compared to the results of [3].

LOADS:

Unfactored Loads: Dead Load Live Load Factored Loads

(1.25 DL + 1.5 LL)

Axial Load (kN) 1776 1320 4200

Top Moment (kNm) 112 66 239

Bottom Moment (kNm) 12 7 26

GEOMETRY:

lu=8.1 [m]


MATERIAL:

Concrete : CONCRETE 40 fc' = 40.00 (MPa) Longitudinal reinforcement : Grade 400 fy = 400.00 (MPa)

Fig.1. Cross section with longitudinal reinforcement determined in [3] (12 No.25).


March 2014 page 47 / 49

IMPORTANT STEPS: In the dialog box Buckling length set buckling parameters (Fig.1.2.).

Fig. 1.2. Buckling parameters of the column. In the Loads dialog box put the forces and the ratio of long-term to total load for each load case.

Fig. 1.3. Loads.

In the Calculation Option/ General dialog box check: Design – unidirectional bending: My direction (Fig. 1.4.).

Fig. 1.4. Unidirectional bending RESULTS OF REINFORCEMENT CALCULATION: The reinforcement generated by the program (Fig 1.5.) is different than that calculated in [3]. The authors of [3] find the reinforcement of 12 No. 25 bars, thus the total area is equal to 60.0 cm

2. The

calculations with the program for the new edition of code [2] result in reinforcement with 8 No. 25 bars, thus the total area is equal to 40 cm

2. Calculations based on the previous edition of code [1] resulted

in reinforcement 10 No. 25 bars (50 cm2).

DIFFERENCES IN THE NEW VERSION RESPECT TO THE PREVIOUS ONE: The calculations based on new edition of the code give smaller reinforcement than for the previous

edition of the code while the capacity coefficiet is similar. This is because the change in c coefficient introduced by the new code. The coefficient changes from 0.6 up to 0.65, thus in all calculations the reinforcement should be smaller (concrete has greater strength).


March 2014 page 48 / 49

Fig. 1.5. Reinforcement generated by the program (8 No.25).

Quantity [2] Robot

Reinforcement

sA

12 No.25

60.0 cm2

8 No.25

40.0 cm2

In order to verify the results of buckling analysis, after the modification of reinforcement to the form as in [2] (see Fig. 1), the verification is carried out. RESULTS OF BUCKLING ANALYSIS:

Quality (Unit) [3] Robot

r

klu (-) 45.42 47.1

gI (mm4) 5.2 x 109 5.2 x 109

seI (mm4) 1.6 x 108 1.5 x 108

gc

d

sesgc

IE

IEIE

EI

25.0

1

)2,0(

max (Nmm2) 4.1 x 1013 4.1 x 1013

2

2

)( u

ckl

EIP

(N) 8741 x 103 8743 x 103

2

14.06.0M

MCm (-)

0.56

0.56

)03.015(min hPM u (kNm) 126 126

c

u

m

c

P

P

MCM

75.01

2

(kNm) 373 371


March 2014 page 49 / 49

DIFFERENCES IN THE NEW VERSION RESPECT TO THE PREVIOUS ONE:

The only difference in buckling analysis between the new [1] and the previous edition [2] of the code concerns the way the minimum eccentricity is taken into account. In previous version, the moment in the mid-height section (Cm*M2) was expected to exceed the minimum moment. In the current version, the end moment should be greater than the minimum one. This changes the scheme of the calculations presented in the calculation note. Below, the fragments of the calculation notes for both codes are presented, showing the discussed difference.

New edition of code [2]: 2.5.1.1.3 Buckling analysis

M1 = -25.50 (kN*m) M2 = 239.00 (kN*m) Mmid = 133.20 (kN*m) Case: Cross-section in the middle of the column, Slenderness taken into account

ns = Cm / [1-(Pu / 0.75Pc)] = 1.55 Cm = 0.6 + 0.4(M1/M2) = 0.56 Pc = 8743.30 (kN) M2 = 239.00 (kN*m) Mmin = 126.00 (kN*m) Mo = max (M2 ; Mmin) = 239.00 (kN*m)

Mc = ns * Mo = 370.50 (kN*m)

Previous edition of code [1]:

2.5.1.1.3 Buckling analysis

M1 = -25.50 (kN*m) M2 = 239.00 (kN*m) Mmid = 133.20 (kN*m) Case: Cross-section in the middle of the column, Slenderness taken into account M = Cm*M2 = 133.20 (kN*m) Cm = 0.6 + 0.4(M1/M2) = 0.56

Mmin = 126.00 (kN*m) Mo = max (M ; Mmin) = 133.20 (kN*m)

1 / [1-(Pu / 0.75Pc)] = 2.78 Pc = 8743.30 (kN) Mc = 1 / [1-(Pu / 0.75Pc)] * Mo = 370.50 (kN*m)

CONCLUSIONS:

The results of reinforcement calculation concern the automatic calculation of reinforcement. The calculations carried out based on the new edition of the code [2] give smaller reinforcement than for

the old edition [1] and for the reference [3], because of the change in c coefficient introduced by the new code, which is supposed to lead to the decrease of the amount of reinforcement. In the case of buckling analysis, the reinforcement obtained in the program is modified to a form as in the reference example, in order to enable the comparison of total moments. As can be seen, the results of the calculation according to the new edition of code [2] are the same as according to the previous edition [1], and athe same time similar to those in the reference [3], except the change in considering the minimal moment introduced in new edition of the code.

LITERATURE

[1] CSA Standard A23.3-94. Structures (Design). [2] CSA Standard A23.3-04. Design of concrete structures. [3] Canadian Portland Cement Association, “Concrete Design Handbook”, Second Edition, 1998, Example 8.1, pp. 8-6.

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