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March 2014

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Autodesk Robot Structural Analysis Professional - Verification Manual for French Codes

March 2014 page i

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

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

1. CM66 ................................................................................................................................................................................................. 3

VERIFICATION EXAMPLE 1 - AXIAL COMPRESSION I ..................................................................................................................... 4 VERIFICATION EXAMPLE 2 - AXIAL COMPRESSION II .................................................................................................................... 7 VERIFICATION EXAMPLE 3 - BENDING WITH LATERAL/TORSIONAL BUCKLING EFFECT.................................................................. 10 VERIFICATION EXAMPLE 4 - IPE PROFILE LOADED IN UNIAXIAL BENDING AND AXIAL COMPRESSION ............................................ 13 VERIFICATION EXAMPLE 5 - COMPRESSION AND SHEAR FORCE ................................................................................................. 16

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

1. BAEL 91 MOD. 99 - RC COLUMNS ............................................................................................................................................. 23

VERIFICATION EXAMPLE 1 - COLUMN SUBJECTED TO AXIAL LOAD .............................................................................................. 24 VERIFICATION EXAMPLE 2 - COLUMN SUBJECTED TO AXIAL LOAD .............................................................................................. 27 VERIFICATION EXAMPLE 3 - COLUMN SUBJECTED TO AXIAL LOAD AND BIAXIAL BENDING ............................................................ 29 LITERATURE.................................................................................................................................................................................. 35


March 2014 page 1 / 35

INTRODUCTION

This verification manual contains numerical examples for 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. CM66



VERIFICATION EXAMPLE 1 - Axial compression I

Example taken from handbook STRUCTURES METALLIQUES CM66 - additif 80 - Eurocode 3

written by Jean Morel

TITLE: Axial compression (Example 1 page 119).

SPECIFICATION: The column shown below is fully restraint at the bottom end and pinned at the top end about y-y and z-z axes. So the effective length lk along both of axes is 0,5h = 5000 mm. For the design value of the compressive force N=73500 daN check the column made of S.235steel. It has been suggested that a IPE 400 section be considered.

SOLUTION: Define a new type of member. For analysed 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. Then, press Buckling Length coefficient Y icon and select the second icon (value 0.5). Repeat the same operation for Z direction.



In the CALCULATIONS dialog box set Member Verification option for member 1 and switch off Limit State – Serviceability (only Ultimate Limit state will be analysed). 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 member 1 opens the RESULTS dialog box with detailed results for the analysed 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: CM66

ANALYSIS TYPE: Member Verification

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

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

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

LOADS: Governing Load Case: 1 test

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

MATERIAL:

ACIER fy = 23.50 daN/mm2 ---------------------------------------------------------------------------------------------------------------------------------------

SECTION PARAMETERS: IPE 400 ht=40.0 cm

bf=18.0 cm Ay=48.600 cm2 Az=34.400 cm2 Ax=84.464 cm2

tw=0.9 cm Iy=23128.400 cm4 Iz=1317.820 cm4 Ix=46.800 cm4

tf=1.4 cm Wely=1156.420 cm3 Welz=146.424 cm3

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

STRESSES: SigN = 73500.00/84.464 = 8.70 daN/mm2

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

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

BUCKLING PARAMETERS:

About Y axis: About Z axis:

LY=10.00 m MuY=26.09 LZ=10.00 m MuZ=1.49

LfY=5.00 m k0Y=1.03 LfZ=5.00 m k0Z=2.69

Lambda Y=30.22 Lambda Z=126.58

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

VERIFICATION FORMULAS: k0*SigN = 2.69*8.70 = 23.37 < 23.50 daN/mm2 (3.411)

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

Section OK!!!

COMPARISON:

Resistance, interaction expression Robot HANDBOOK

Factored compressive stress in a member k [daN/mm2] Amplification factor for compression stresses k

Check of the formula k /e 1

23,37

2,69

CM66 (3,412)

0,99

23,5

2,71

CM66 (3,411)

1,000



VERIFICATION EXAMPLE 2 - Axial compression II

Example taken from handbook CONCEPTION ET CALCUL DES STRUCTURES METALLIQUES


TITLE: Axial compression (Example 3.2.4.1 page 73).

SPECIFICATION: The column shown aside is fully restraint at the both ends about y-y and z-z axes. The effective length lk along both of axes is 0,5h = 4000 mm. For the design value of the compressive force N=130167 daN check the column made of E24 steel. It has been suggested that a HEB 200 section be considered.

SOLUTION: Define a new type of member. For analysed 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. Then, press Buckling Length coefficient Y icon and select the second icon (value 0.5). Repeat the same operation for Z direction.








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

CODE: CM66


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


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


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

MATERIAL:

ACIER E24 fy = 23.50 daN/mm2 ---------------------------------------------------------------------------------------------------------------------------------------

SECTION PARAMETERS: HEB 200 ht=20.0 cm




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


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




LY=8.00 m MuY=5.67 LZ=8.00 m MuZ=1.99

LfY=4.00 m k0Y=1.10 LfZ=4.00 m k0Z=1.42

Lambda Y=46.83 Lambda Z=78.97

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

VERIFICATION FORMULAS: k0*SigN = 1.42*16.67 = 23.70 > 23.50 daN/mm2 (3.411)

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

Incorrect section!!!

COMPARISON:


1. Factored compressive stress in a member k [daN/mm2]

2. Check of the formula k /e 1

23,7

1,01

24,00

1,02


March 2014 page 10 / 35

VERIFICATION EXAMPLE 3 - Bending with lateral/torsional buckling effect


Written by Jean Morel

TITLE: Bending with lateral-torsional buckling effect (Example 3.3.4.1 page 92).

SPECIFICATION: IS1 simply supported beam over a span of 40.0 m. is laterally restraint at both ends. For the loading shown below check the beam made of IS 1 steel.

SOLUTION: Define a new type of member. For analysed 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 Beam 1 in the Member Type editable field. For defining appropriate load type diagram, press More button. Choose the icon for Load type Y and double-click the first icon (Uniform Loads) in Load Type dialog box. Repeat the same operation for Z direction. Save the newly-created type of member.


March 2014 page 11 / 35





March 2014 page 12 / 35

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

CODE: CM66


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


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

LOADS: Governing Load Case: 1 TEST

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

MATERIAL:

ACIER E24 fy = 23.50 daN/mm2 ---------------------------------------------------------------------------------------------------------------------------------------

SECTION PARAMETERS: IS 1 ht=150.0 cm




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

STRESSES: SigFY = 83600.00/27514.902 = 3.04 daN/mm2

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

LATERAL BUCKLING PARAMETERS: z=0.00 B=1.00 D=2.36 Sig D=1.78 daN/mm2

lD_sup=40.00 m C=1.13 kD=7.63

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


About Y axis: About Z axis: ----------------------------------------------------------------------------------------------------------------------------------

VERIFICATION FORMULAS: kD*SigFY = 7.63*3.04 = 23.20 < 23.50 daN/mm2 (3.611)

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

Section OK!!!

COMPARISON:


1. Factored flexural stress in a member kDfy [daN/mm2] (lateral buckling analysis included)

2. Check of the formula kDfy /e 1

23,20 0,99

24,00 1,02

CONCLUSION: The differences are caused by different way of rounding-off the cross sectional properties (cross sectional area, section modulus, moment of inertia).


March 2014 page 13 / 35

VERIFICATION EXAMPLE 4 - IPE profile loaded in uniaxial bending and axial compression



TITLE: IPE profile loaded in uniaxial bending and axial compression (Example 3.2.4.3 page 77).

SPECIFICATION: The column shown aside is a part of a portal frame. It is fully restrained at the bottom end and is connected by rigid connection with a regel at the top end (lky = 9,53 m.). For the design values of the compressive force N=8000 daN and bending moment My = 12000 daNm check the beam made of E24 steel. It has been suggested that a IPE 360 section be considered.

SOLUTION: Define a new type of member. For analysed 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. Then, select Member length ly - Real radio- button and tape the value 9.53.


March 2014 page 14 / 35





March 2014 page 15 / 35

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

CODE: CM66


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


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


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

MATERIAL:

ACIER E24 fy = 23.50 daN/mm2 ---------------------------------------------------------------------------------------------------------------------------------------

SECTION PARAMETERS: IPE 360 ht=36.0 cm




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


SigFY = 12000.00/903.644 = 13.28 daN/mm2

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




LY=9.53 m MuY=46.400 LZ=6.00 m MuZ=7.509

LfY=9.53 m k1Y=1.007 LfZ=6.00 m k1Z=1.048

Lambda Y=63.725 kFY=1.034 Lambda Z=158.405

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

VERIFICATION FORMULAS: k1*SigN + kFY*SigFY = 1.048*1.10 + 1.034*13.28 = 14.89 < 23.50 daN/mm2 (3.521)

1.54*TauZ = 1.540*1.39 = 2.14 < 23.50 daN/mm2 (1.313)

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

Section OK!!!

COMPARISON:


1. Factored compressive stress in a member n [daN/mm2]

2. Factored flexural stress in a member fy [daN/mm2] (lateral buckling analysis not included)

Check of the formula (k1n + kff )/ e 1,0 ( 3,521 from the CM66 code)

1,10

13,28

0,634

1,10

13,3

0,613



March 2014 page 16 / 35

VERIFICATION EXAMPLE 5 - Compression and shear force

Example taken from handbook STRUCTURES METALLIQUES CM66 - additif 80 - Eurocode 3


TITLE: Beam-column (Example 2 page 119).

SPECIFICATION: The column is fully restrained at the bottom end and pinned at the top end about y-y and z-z axes (lky = lkz = 0,7). For the loading conditions (3 load cases) as shown below check the beam-column made of S.235 steel. It has been suggested that HEA 200 shape might be used.

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. Then, press Buckling Length coefficient Y icon and select the third icon (value 0.7). Repeat the same operation for Z direction. To define an appropriate load type diagram, press More. Choose the icon for Load type Y and double-click the third icon (Concentrated Force in Center) in Load Type dialog box. Repeat the same operation for Z direction. Save the newly-created type of member.


March 2014 page 17 / 35

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 member 1 opens the RESULTS dialog box with detailed results for the analyzed member. Analyze the member separately for each previously defined load case. 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.


March 2014 page 18 / 35

RESULTS: 1. Load case no. 1

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

CODE: CM66


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


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

LOADS: Governing Load Case: 1 test1

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

MATERIAL: ACIER fy = 23.50 daN/mm2

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

SECTION PARAMETERS: HEA 200 ht=19.0 cm




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


SigFZ = 2250.00/133.551 = 16.85 daN/mm2 ---------------------------------------------------------------------------------------------------------------------------------------




LY=4.00 m MuY=32.536 LZ=4.00 m MuZ=11.769

LfY=2.80 m k1Y=1.010 LfZ=2.80 m k1Z=1.029

Lambda Y=33.809 Lambda Z=56.215 kFZ=1.107


March 2014 page 19 / 35

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

VERIFICATION FORMULAS: k1*SigN + kFZ*SigFZ = 1.029*5.57 + 1.107*16.85 = 24.38 > 23.50 daN/mm2 (3.521)

1.54*TauY = |0.000*-0.52| = |0.00| < 23.50 daN/mm2 (1.313) ---------------------------------------------------------------------------------------------------------------------------------------


2. Load case no. 2

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

CODE: CM66 ANALYSIS TYPE: Member Verification

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


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


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

MATERIAL:

ACIER fy = 23.50 daN/mm2

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





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


SigFY = 2250.00/388.647 = 5.79 daN/mm2

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


March 2014 page 20 / 35


lD_inf=4.00 m C=1.560 kD=1.002

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


About Y axis: About Z axis: LY=4.00 m MuY=32.536 LZ=4.00 m MuZ=11.769

LfY=2.80 m k1Y=1.010 LfZ=2.80 m k1Z=1.029

Lambda Y=33.809 kFY=1.036 Lambda Z=56.215

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

VERIFICATION FORMULAS: k1*SigN + kD*kFY*SigFY = 1.029*5.57 + 1.002*1.036*5.79 = 11.74 < 23.50 daN/mm2 (3.731)

1.54*TauZ = 0.000*1.67 = 0.00 < 23.50 daN/mm2 (1.313)

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

Section OK!!! 3. Load case no. 3

STEEL DESIGN --------------------------------------------------------------------------------------------------------------------------------------- CODE: CM66


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


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


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

MATERIAL:

ACIER fy = 23.50 daN/mm2

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


March 2014 page 21 / 35


bf=20.0 cm Ay=40.000 cm2 Az=12.350 cm2 Ax=53.831 cm2 tw=0.7 cm Iy=3692.150 cm4 Iz=1335.510 cm4 Ix=18.600 cm4


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


SigFY = 2250.00/388.647 = 5.79 daN/mm2

SigFZ = 2250.00/133.551 = 16.85 daN/mm2

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


lD_inf=4.00 m C=1.560 kD=1.002

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


About Y axis: About Z axis: LY=4.00 m MuY=32.536 LZ=4.00 m MuZ=11.769

LfY=2.80 m k1Y=1.010 LfZ=2.80 m k1Z=1.029

Lambda Y=33.809 kFY=1.036 Lambda Z=56.215 kFZ=1.107

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

VERIFICATION FORMULAS: k1*SigN + kD*kFY*SigFY + kFZ*SigFZ = 1.029*5.57 + 1.002*1.036*5.79 + 1.107*16.85 = 30.39 > 23.50

daN/mm2 (3.731)

1.54*TauY = |0.000*-0.52| = |0.00| < 23.50 daN/mm2 (1.313)

1.54*TauZ = 0.000*1.67 = 0.00 < 23.50 daN/mm2 (1.313)

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


COMPARISON:


Case 1


2. Factored flexural stress in a member fz [daN/mm2] (lateral buckling analysis included)

Check of the formula (k1n + kfzfz )/ e 1,0 ( 3,521 from the CM66 code) Case 2


2. Factored flexural stress in a member fy [daN/mm2] (lateral buckling analysis included)

Check of the formula (k1n + kfyfy )/ e 1,0 ( 3,521 from the CM66 code) Case 3


2. Factored flexural stress in a member fy [daN/mm2] (lateral buckling analysis included)

3. Factored flexural stress in a member fz [daN/mm2] (lateral buckling analysis included)

4. Check of the formula (k1n + kfyfy + kfzfz)/ e 1,0 ( 3,521 from the CM66 code)

5,57

16,85

1,038

5,57

5,79

0,500

5,57

5,79

16,85

1,293

5,6

16,8

1,030

5,6

5,8

0,498

5,6

5,8

16,8

1,285



March 2014 page 22 / 35

CCOONNCCRREETTEE


March 2014 page 23 / 35

1. BAEL 91 mod. 99 - RC columns


March 2014 page 24 / 35

VERIFICATION EXAMPLE 1 - Column subjected to axial load

Example based on: [2] J. Perchat, “Pratique du BAEL 91”, Deuxième èdition, Eyrolles, 1998, Example 2, pp. 98

DESCRIPTION OF THE EXAMPLE: The capacity of the column is determined with the Robot program and verified against the example solved in [2]. In [2], the reinforcement is assumed a priori, and the capacity is checked for that reinforcement. For our purpose, we define the axial force equal to the capacity determined in [2], and then compare the results.

LOADS:

Nu = 1310 (kN)

GEOMETRY:

lf = 2.80 (ft)

cross section: 30x30 (cm)

MATERIAL:

Concrete : fc28 = 25.00 (MPa) Unit weight = 2501.36 (kG/m3) Longitudinal reinforcement : type HA fe = 500.00 (MPa)

Fig.1. Cross section with reinforcement determined in [2] (4HA16). IMPORTANT STEPS: In the dialog box Buckling length set the length lf in both directions (Fig.1.2.). The program calculates the slenderness for the most unfavorable direction.


March 2014 page 25 / 35

Fig. 1.2. Buckling parameters of the column. In the Loads dialog box put the axial force N, equal to the capacity determined in [2].

Fig. 1.3. Loads. RESULTS OF THE CALCULATIONS: The reinforcement generated by the program (Fig 1.4.) is different than that determined in [2]. The greater reinforcement generated by the program is the result of the fact, that 4HA16 is not actually sufficient reinforcement. Although the calculations with small accuracy carried out in [2] show, that the capacity is 1310 kN, after the accurate calculations it can be proven that it is in fact 1308 kN, thus 4HA16 reinforcement is not correct. Robot finds the sufficient reinforcement 8HA12. The calculation of capacity is illustrated in the calculation note:

Fig. 1.4. Reinforcement generated by the program (8HA12).


March 2014 page 26 / 35

2.5.1 Slenderness analysis

Lu (m) K

Direction Y: 2.80 1.00 32.33 Direction Z: 2.80 1.00 32.33

2.5.2 Detailed analysis

= max (y ; z)

= 32.33

< 50

= 0,85/(1+0,2*(/35)^2) = 0.73 Br = 0.08 (m2) A= 9.05 (cm2)

Nulim = [Br*fc28/(0,9*b)+A*Fe/s] = 1339.79 (kN)

FINAL VERIFICATION:

Quantity [2] Robot

Reinforcement 4 HA16 8 HA12

uN 1310 kN* 1340 kN

* According to accurate calculations should be 1308.5 kN


March 2014 page 27 / 35

VERIFICATION EXAMPLE 2 - Column subjected to axial load

Example based on: [3] J-P. Mougin, “Béton Armé. BAEL 91 et DTU associés”, Eyrolles, 1995, Example 1, pp. 113

DESCRIPTION OF THE EXAMPLE: Determine the reinforcement of three columns. The data is given below.

LOADS:

N1 = 1650 (kN)

N2 = 2150 (kN)

N3 = 2770 (kN)

GEOMETRY:

lf = 3.2 (ft)

cross section: 30x60 (cm)

MATERIAL:

Concrete : fc28 = 25.00 (MPa) Unit weight = 2501.36 (kG/m3) Longitudinal reinforcement : type HA fe = 400.00 (MPa)

IMPORTANT STEPS: Define three columns separately. For each column, follow the steps as below: In the dialog box Buckling length set the length lf in both directions (Fig.2.1.). The program calculates the slenderness for the most unfavorable direction.

Fig. 2.1. Buckling parameters of the column. In the Loads dialog box put the axial force N.


March 2014 page 28 / 35

RESULTS OF THE CALCULATIONS:

Column 1 2 3

Nu 1650 2150 2770

RESULTS: [3] / Robot [3] / Robot [3] / Robot

As (cm2) 7.7 7.7 7.7 7.7 29.5 23.6

Reinforcement 4HA14 +

2HA10

4HA12 +

4HA10

4HA14 +

2HA10

4HA12 +

4HA10 6HA25

4HA20 +

14HA10


March 2014 page 29 / 35

VERIFICATION EXAMPLE 3 - 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.

NOTE: In Robot the calculations of compression with bending are carried out using EC2 code [4].

1. SECTION DIMENSIONS

2. MATERIALS

Concrete : fc28 = 25.00 (MPa) Unit weight = 2447.32 (kG/m3) Longitudinal reinforcement : type HA fe = 500.00 (MPa) Transversal reinforcement : type HA fe = 500.00 (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 30 / 35

4. LOADS

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: Program generates the reinforcement 6 HA 20.

6. RESULTS OF THE SECTION CALCULATIONS: The dimensioning combination is 1.35DL1+1.50LL1 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 (marked as (C)).


March 2014 page 31 / 35

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

(kN)

MyA

(kN*m)

MyB

(kN*m)

MyC

(kN*m)

MzA

(kN*m)

MzB

(kN*m)

MzC

(kN*m)

1 DL1 500 100 30 72 20 30 40*

2 LL1 200 50 30 42 10 20 30*

Dimensioning

combination 1.35DL1+1.5LL1 975 210 85.5 160.2 42 70.5 99

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 32 / 35

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

Slenderness analysis:

r

l0 = 40.41

Check if lim according to 4.3.5.3.5(2)

u

15

;25maxlim = 30.38

cdc

Sd

ufA

N

= 0.244

lim - column is slender

Check, if the slenderness effects have to be taken into account. For this, we check if crit

according to (4.62).

)2(2502

01

e

ecrit = 39.82

975

5.8501 e = 0.088 (m)

975

21001 e = 0.215 (m)

crit - slenderness effects have to be taken into account.

The abovementioned requirement means that the total eccentricity of the axial force in Z

( NMe yz / ) direction will be:

20 eeee atot (it is decided to consider the additional eccentricity to act in Z direction, thus

increasing moment My)

NOTE: If the requirement 4.62 is fulfilled ( crit ), the total eccentricity would be calculated as

atot eee 0 .

Calculation of initial eccentricity e0

For the mid-height section, we have:

02010 6.04.0 eeee e = 0.164 (m)

Calculation of additional eccentricity ea


March 2014 page 33 / 35

2

0lea = 0.018 (m),

h100

1 = 0.0022

h = 21.0 (m)

200/1

We assume = 0.005

Calculation of second-order eccentricity e2

r

lKe

1

10

2

0

12 = 0.043 (m)

K1 = 1 ( )35

s

yd

Ed

fK

r

9.02

12 = 0.009

K2 = 1

d = 0.552 (m)

Es = 200 (GPa)

ydf = 435 (MPa)

The total eccentricity in Z direction:

20 eeee atot = 0.225 (m)

The total moment My:

toty eNM 219 (kNm)

Z DIRECTION

Slenderness analysis:

r

l0 = 41.57

Check if lim according to 4.3.5.3.5(2)

u

15

;25maxlim = 30.38

cdc

Sd

ufA

N

= 0.244

lim - column is slender

Since the column is sway in Z direction, the slenderness effects have to be taken into account.


March 2014 page 34 / 35

The abovementioned requirement means that the total eccentricity of the axial force in Y

( NMe zy / ) direction will be:

20 eeetot

Calculation of initial eccentricity e0

For the mid-height section, we have:

sd

sd

N

Me 0 = 0.102 (m) (the moment in mid-height section is given directly by the user)

Calculation of second-order eccentricity e2

r

lKe

1

10

2

0

12 = 0.057 (m)

Since the column is sway, and the slenderness effects have to be considered, we take into account the influence of long-term effects (creep), increasing the buckling length of the column:

00 lnl = 6.79 (m)

sd

sd

N

Nn

1

The ratio of long-term load to total load is calculated as a weighted average from the load cases. The weight factors are assumed according to the axial forces. Thus:

3.0975

2005.10.1

975

50035.1

sd

sd

N

N = 0.785

Thus, the influence of creep is taken into account by means of n =1.3359

The other parameters are:

K1 = 1 ( )35

s

yd

Ed

fK

r

9.02

12 = 0.014

K2 = 1

d = 0.351 (m)

Es = 200 (GPa)

ydf = 435 (MPa)

The total eccentricity in Z direction:

20 eeetot = 0.158 (m)

The total moment My:

toty eNM 154 (kNm)


March 2014 page 35 / 35

7.3. FINAL RESULT

yM = 219 (kNm)

zM = 154 (kNm)

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] B.A.E.L. 91. Règles techniques de conception et de calcul des ouvrages et constructions en béton armé suivant la méthode des états-limites. Mod. 99. [2] J. Perchat, “Pratique du BAEL 91”, Deuxième èdition, Eyrolles, 1998, Example 2, pp. 98 [3] J-P. Mougin, “Béton Armé. BAEL 91 et DTU associés”, Eyrolles, 1995, Example 1, pp. 113

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AAuuttooddeesskk RRoobboott SSttrruuccttuurraall … · 2014. 4. 11. · Autodesk Robot Structural Analysis Professional - Verification Manual for French Codes March 2014 page 5