robot-manual french codes

8/9/2019 ROBOT-Manual French Codes

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AAuuttooddeesskk

RRoobboottSSttrruuccttuurraallAAnnaallyyssiiss

PPrrooffeessssiioonnaall

VVEERRIIFFIICCAATTIIOONNMMAANNUUAALLFFOORRFFRREENNCCHHCCOODDEESS

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

VERIFICATIONEXAMPLE1 -AXIAL COMPRESSION I ..................................................................................................................... 4VERIFICATIONEXAMPLE2 -AXIAL COMPRESSION II .................................................................................................................... 7

VERIFICATIONEXAMPLE3 -BENDING WITH LATERAL/TORSIONAL BUCKLING EFFECT.................................................................. 10VERIFICATIONEXAMPLE4 -IPEPROFILE LOADED IN UNIAXIAL BENDING AND AXIAL COMPRESSION................. ................ ........... 13

VERIFICATIONEXAMPLE5 -COMPRESSION AND SHEAR FORCE ................................................................................................. 16

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

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

VERIFICATIONEXAMPLE1 -COLUMN SUBJECTED TO AXIAL LOAD .............................................................................................. 24VERIFICATIONEXAMPLE2 -COLUMN SUBJECTED TO AXIAL LOAD .............................................................................................. 27VERIFICATIONEXAMPLE3 -COLUMN SUBJECTED TO AXIAL LOAD AND BIAXIAL BENDING............................................................ 29LITERATURE.................................................................................................................................................................................. 35


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March 2014 page 2/ 35

SSTTEEEELL


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VERIFICATION EXAMPLE 1- Axial compression I

Example taken from handbook STRUCTURES METALLIQUES

CM66 - additif 80 - Eurocode 3written 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 andz-z axes. So the effective length lk along both of axes is 0,5h = 5000 mm. For the design value of thecompressive force N=73500 daN check the column made of S.235steel. It has been suggested that aIPE 400 section be considered.

SOLUTION:Define a new type of member. For analysed member pre-defined type of member COLUMN may beinitially opened. It can be set in Member type combo-box. Press the Parameters button inDEFINITION-MEMBERS tab, which opens MEMBER DEFINITIONPARAMETERS dialog box. Typea new name Column 1in the Member Type editable field. Then, press Buckling Length coefficient Yicon and select the second icon (value 0.5). Repeat the same operation for Z direction.


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In the CALCULATIONS dialog box set Member Verificationoption for member 1 and switch off LimitState Serviceability (only Ultimate Limit state will be analysed). Now, start the calculations bypressing Calculationsbutton.

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 analysedmember.

The view of the RESULTS window is presented below. Moreover, the printout note containing thesame results data as in Simplified resultstab of the RESULTS window is added.


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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 400ht=40.0 cm

bf=18.0 cm Ay=48.600 cm2 Az=34.400 cm2 Ax=84.464 cm2tw=0.9 cm Iy=23128.400 cm4 Iz=1317.820 cm4 Ix=46.800 cm4tf=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=2.69Lambda 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

2,69CM66 (3,412)

0,99

23,5

2,71CM66 (3,411)

1,000


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VERIFICATION EXAMPLE 2- Axial compression II

Example taken from handbook

CONCEPTION ET CALCUL DES STRUCTURES METALLIQUESwritten by Jean Morel

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 lengthlk along both of axes is 0,5h = 4000 mm. For the design value of the compressive force N=130167daN check the column made of E24 steel. It has been suggested that a HEB 200 section beconsidered.

SOLUTION:Define a new type of member. For analysed member pre-defined type of member COLUMN may beinitially opened. It can be set in Member type combo-box. Press the Parameters button in

DEFINITION-MEMBERS tab, which opens MEMBER DEFINITIONPARAMETERS dialog box. Type

a new name Column 1in the Member Type editable field. Then, press Buckling Length coefficient Yicon and select the second icon (value 0.5). Repeat the same operation for Z direction.


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VERIFICATION EXAMPLE 3- Bending with lateral/torsional buckling effect


CONCEPTION ET CALCUL DES STRUCTURES METALLIQUESWritten 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 loadingshown 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 beinitially opened. It can be set in Member type combo-box. Press the Parameters button inDEFINITION-MEMBERS tab, which opens MEMBER DEFINITIONPARAMETERS dialog box. Typea new name Beam 1in the Member Type editable field. For defining appropriate load type diagram,press Morebutton. Choose the icon for Load type Y and double-click the first icon (Uniform Loads) inLoad Type dialog box. Repeat the same operation for Z direction. Save the newly-created type ofmember.


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Member Verification dialog box with most significant results data will appear on screen. Pressing theline 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 resultstab of the RESULTS window is added.


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

ANALYSIS TYPE: Member Verification

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

CODE GROUP:MEMBER: 1 POINT: 2 COORDINATE: x = 0.50 L = 20.00 m

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

LOADS:Governing Load Case: 1 TEST---------------------------------------------------------------------------------------------------------------------------------------

MATERIAL:

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

SECTION PARAMETERS: IS 1ht=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/mm2lD_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

23,200,99

24,001,02

CONCLUSION:

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


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


CONCEPTION ET CALCUL DES STRUCTURES METALLIQUESwritten by Jean Morel

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 isconnected by rigid connection with a regel at the top end (lky = 9,53 m.). For the design values of thecompressive force N=8000 daN and bending moment My = 12000 daNm check the beam made ofE24 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 beinitially opened. It can be set in Member type combo-box. Press the Parameters button inDEFINITION-MEMBERS tab, which opens MEMBER DEFINITIONPARAMETERS dialog box. Type

a new name Column 1in the Member Type editable field. Then, select Member length ly - Real radio-button and tape the value 9.53.


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Member Verification dialog box with most significant results data will appear on screen. Pressing theline with results for member 1 opens the RESULTS dialog box with detailed results for the analysedmember.

The view of the RESULTS window is presented below. Moreover, the printout note containing thesame results data as in Simplified resultstab of the RESULTS window is added.


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VERIFICATION EXAMPLE 5- Compression and shear force

Example taken from handbook STRUCTURES METALLIQUES

CM66 - additif 80 - Eurocode 3written by Jean Morel

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 andz-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 beinitially opened. It can be set in Member type combo-box. Press the Parameters button inDEFINITION-MEMBERS tab, which opens MEMBER DEFINITIONPARAMETERS dialog box. Typea new name Column 1in the Member Type editable field. Then, press Buckling Length coefficient Yicon and select the third icon (value 0.7). Repeat the same operation for Z direction. To define anappropriate load type diagram, press More. Choose the icon for Load type Y and double-click the thirdicon (Concentrated Force in Center) in Load Type dialog box. Repeat the same operation for Zdirection. Save the newly-created type of member.


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RESULTS:1. Load case no. 1

STEEL DESIGN---------------------------------------------------------------------------------------------------------------------------------------CODE: CM66ANALYSIS TYPE: Member Verification---------------------------------------------------------------------------------------------------------------------------------------

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

LOADS:Governing Load Case: 1 test1---------------------------------------------------------------------------------------------------------------------------------------

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

SECTION PARAMETERS: HEA 200ht=19.0 cm

bf=20.0 cm Ay=40.000 cm2 Az=12.350 cm2 Ax=53.831 cm2tw=0.7 cm Iy=3692.150 cm4 Iz=1335.510 cm4 Ix=18.600 cm4tf=1.0 cm Wely=388.647 cm3 Welz=133.551 cm3---------------------------------------------------------------------------------------------------------------------------------------

STRESSES: SigN = 30000.00/53.831 = 5.57 daN/mm2SigFZ = 2250.00/133.551 = 16.85 daN/mm2

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

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


About Y axis: About Z axis:

LY=4.00 m MuY=32.536 LZ=4.00 m MuZ=11.769LfY=2.80 m k1Y=1.010 LfZ=2.80 m k1Z=1.029

Lambda Y=33.809 Lambda Z=56.215 kFZ=1.107


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

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)---------------------------------------------------------------------------------------------------------------------------------------

I ncorr ect section! ! !

2. Load case no. 2

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

CODE: CM66ANALYSIS TYPE: Member Verification---------------------------------------------------------------------------------------------------------------------------------------



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

SECTION PARAMETERS: HEA 200ht=19.0 cmbf=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 cm4tf=1.0 cm Wely=388.647 cm3 Welz=133.551 cm3---------------------------------------------------------------------------------------------------------------------------------------

STRESSES: SigN = 30000.00/53.831 = 5.57 daN/mm2SigFY = 2250.00/388.647 = 5.79 daN/mm2

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


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LATERAL BUCKLING PARAMETERS:z=0.000 B=1.000 D=1.401 Sig D=20.34 daN/mm2lD_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=11.769LfY=2.80 m k1Y=1.010 LfZ=2.80 m k1Z=1.029Lambda 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: CM66ANALYSIS TYPE: Member Verification---------------------------------------------------------------------------------------------------------------------------------------



MATERIAL:

ACIER fy = 23.50 daN/mm2

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


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SECTION PARAMETERS: HEA 200ht=19.0 cm


---------------------------------------------------------------------------------------------------------------------------------------STRESSES: SigN = 30000.00/53.831 = 5.57 daN/mm2

SigFY = 2250.00/388.647 = 5.79 daN/mm2

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

LATERAL BUCKLING PARAMETERS:z=0.000 B=1.000 D=1.401 Sig D=20.34 daN/mm2lD_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=11.769LfY=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 > 23.50daN/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)

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

I ncorr ect section! ! !

COMPARISON:


Case 1

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

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

Check of the formula (k1n + kfzfz)/ e1,0( 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)/ e1,0

( 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)/ e1,0( 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

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


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CCOONNCCRREETTEE


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1. BAEL 91 mod. 99 - RC columns


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VERIFICATION EXAMPLE 1- Column subjected to axial load

Example based on:

[2] J. Perchat, Pratique du BAEL 91, Deuxime dition, 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 examplesolved in [2]. In [2], the reinforcement is assumed a priori, and the capacity is checked for thatreinforcement. For our purpose, we define the axial force equal to the capacity determined in [2], andthen 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 lengthset the length l fin both directions (Fig.1.2.).The program calculatesthe slenderness for the most unfavorable direction.


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Fig. 1.2. Buckling parameters of the column.

In the Loadsdialog 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]. Thegreater reinforcement generated by the program is the result of the fact, that 4HA16 is not actuallysufficient reinforcement. Although the calculations with small accuracy carried out in [2] show, that thecapacity is 1310 kN, after the accurate calculations it can be proven that it is in fact 1308 kN, thus4HA16 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).


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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) = 0.73Br = 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


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RESULTS OF THE CALCULATIONS:

Column 1 2 3

Nu 1650 2150 2770

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

As (cm

2

) 7.7 7.7 7.7 7.7 29.5 23.6

Reinforcement4HA14 +2HA10

4HA12 +4HA10

4HA14 +2HA10

4HA12 +4HA10

6HA254HA20 +14HA10


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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 isnon-sway in one direction, whereas sway in the other. The results of the program are accompanied bythe 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.


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4. LOADS

NOTE: Let us assume, the moments in Y direction are linearly distributed along the height of thecolumn. Thus, we define only the ends moments for Y direction. In Z direction however, we assumethe mid-height moment is not a result of the linear distribution. For such a case, Robot let the userdefine 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+1.50LL1The dimensioning section (where the most unfavorable set of forces is found) is for that combinationthe section in the mid-height of the column (marked as (C)).


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Since the column is found as slender, the second-order effects are taken into account in bothdirections.

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 notbeen taken into account, since the structure is non-sway in this direction. In Z direction however, theinfluence 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 thetable Intersection at the Column-results layout.

7. CALCULATIONS OF TOTAL MOMENT:

7.1. LOADS

For the dimensioning combination, the loads are:

CaseN(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*

Dimensioningcombination

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)


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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

u fA

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, thusincreasing 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


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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

012 = 0.043 (m)

K1 = 1 ( )35

s

yd

Ed

f

Kr 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.


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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

012 = 0.057 (m)

Since the column is sway, and the slenderness effects have to be considered, we take into accountthe 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 theload 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)


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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 columnhas been presented. The results obtained with the program (see point 6Results of the SectionCalculations) are in agreement with the manual calculations (see point 7.3 Final Result).

LITERATURE

[1] B.A.E.L. 91. Rgles techniques de conception et de calcul des ouvrages et constructions en btonarm suivant la mthode des tats-limites. Mod. 99.[2] J. Perchat, Pratique du BAEL 91, Deuxime dition, Eyrolles, 1998, Example 2, pp. 98[3] J-P. Mougin, Bton Arm. BAEL 91 et DTU associs, Eyrolles, 1995, Example 1, pp. 113

robot-manual french codes

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