A u t o

Transcription

A u t o

®
Autodesk
™
Robot Structural Analysis
Professional
VERIFICATION MANUAL
FOR FRENCH CODES
March 2014
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Autodesk Robot Structural Analysis Professional - Verification Manual for French Codes
INTRODUCTION .................................................................................................................................................................................. 1
STEEL ................................................................................................................................................................................................... 2
1. CM66 ................................................................................................................................................................................................. 3
VERIFICATION EXAMPLE 1
- AXIAL COMPRESSION I ..................................................................................................................... 4
- AXIAL COMPRESSION II .................................................................................................................... 7
- BENDING WITH LATERAL/TORSIONAL BUCKLING EFFECT.................................................................. 10
- IPE PROFILE LOADED IN UNIAXIAL BENDING AND AXIAL COMPRESSION ............................................ 13
- COMPRESSION AND SHEAR FORCE ................................................................................................. 16
CONCRETE ........................................................................................................................................................................................ 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 i
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.
March 2014
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STEEL
March 2014
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1. CM66
March 2014
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- 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.
March 2014
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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.
March 2014
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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 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
Factored compressive stress in a member k [daN/mm2]
Amplification factor for compression stresses k
Robot
HANDBOOK
23,37
23,5
2,69
CM66 (3,412)
2,71
CM66 (3,411)
0,99
1,000
Check of the formula k /e  1
March 2014
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- 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:
icon and select the second icon (value 0.5). Repeat the same operation for Z direction.
March 2014
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member.
March 2014
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STEEL DESIGN
--------------------------------------------------------------------------------------------------------------------------------------CODE: CM66
--------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1
POINT: 1
--------------------------------------------------------------------------------------------------------------------------------------LOADS:
--------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
ACIER E24
fy = 23.50 daN/mm2
--------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: HEB 200
ht=20.0 cm
bf=20.0 cm
Ay=60.000 cm2
Az=18.000 cm2
Ax=78.081 cm2
tw=0.9 cm
Iy=5696.180 cm4
Iz=2003.370 cm4
Ix=61.400 cm4
tf=1.5 cm
Wely=569.618 cm3
Welz=200.337 cm3
--------------------------------------------------------------------------------------------------------------------------------------STRESSES:
SigN = 130167.00/78.081 = 16.67 daN/mm2
About Y axis:
About Z axis:
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
k0*SigN = 1.42*16.67 = 23.70 > 23.50 daN/mm2 (3.411)
---------------------------------------------------------------------------------------------------------------------------------------
Incorrect section!!!
COMPARISON:
Robot
HANDBOOK
1. Factored compressive stress in a member k [daN/mm2]
23,7
24,00
2. Check of the formula k /e  1
1,01
1,02
March 2014
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- 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
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
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member.
March 2014
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STEEL DESIGN
--------------------------------------------------------------------------------------------------------------------------------------CODE: CM66
--------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1
POINT: 2
--------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 1 TEST
--------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
ACIER E24
fy = 23.50 daN/mm2
--------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: IS 1
ht=150.0 cm
bf=40.0 cm
Ay=320.000 cm2
Az=213.000 cm2
Ax=533.000 cm2
tw=1.5 cm
Iy=2063617.667 cm4
Iz=42706.604 cm4
Ix=1757.794 cm4
tf=4.0 cm
Wely=27514.902 cm3
Welz=2135.330 cm3
--------------------------------------------------------------------------------------------------------------------------------------STRESSES:
SigFY = 83600.00/27514.902 = 3.04 daN/mm2
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 kDfy [daN/mm2]
(lateral buckling analysis included)
2. Check of the formula kDfy /e  1
Robot
HANDBOOK
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 12 / 35
- 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:
a new name Column 1 in the Member Type editable field. Then, select Member length ly - Real radiobutton and tape the value 9.53.
March 2014
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member.
March 2014
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STEEL DESIGN
--------------------------------------------------------------------------------------------------------------------------------------CODE: CM66
--------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1
POINT: 1
--------------------------------------------------------------------------------------------------------------------------------------LOADS:
--------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
ACIER E24
fy = 23.50 daN/mm2
--------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: IPE 360
ht=36.0 cm
bf=17.0 cm
Ay=43.180 cm2
Az=28.800 cm2
Ax=72.729 cm2
tw=0.8 cm
Iy=16265.600 cm4
Iz=1043.450 cm4
Ix=36.200 cm4
tf=1.3 cm
Wely=903.644 cm3
Welz=122.759 cm3
--------------------------------------------------------------------------------------------------------------------------------------STRESSES:
SigN = 8000.00/72.729 = 1.10 daN/mm2
SigFY = 12000.00/903.644 = 13.28 daN/mm2
About Y axis:
About Z axis:
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
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 (k1n + kff )/ e  1,0
( 3,521 from the CM66 code)
Robot
1,10
HANDBOOK
1,10
13,28
13,3
0,634
0,613
CONCLUSION:
March 2014
page 15 / 35
- 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 beamcolumn 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
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 16 / 35
State – Serviceability (only Ultimate Limit state will be analyzed). Now, start the calculations by
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
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RESULTS:
1. Load case no. 1
STEEL DESIGN
--------------------------------------------------------------------------------------------------------------------------------------CODE: CM66
--------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1
POINT: 1
--------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 1 test1
--------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
ACIER
fy = 23.50 daN/mm2
--------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: HEA 200
ht=19.0 cm
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
tf=1.0 cm
Wely=388.647 cm3
Welz=133.551 cm3
--------------------------------------------------------------------------------------------------------------------------------------STRESSES:
SigN = 30000.00/53.831 = 5.57 daN/mm2
SigFZ = 2250.00/133.551 = 16.85 daN/mm2
About Y axis:
LY=4.00 m
MuY=32.536
LfY=2.80 m
k1Y=1.010
Lambda Y=33.809
March 2014
About Z axis:
LZ=4.00 m
MuZ=11.769
LfZ=2.80 m
k1Z=1.029
Lambda Z=56.215
kFZ=1.107
page 18 / 35
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
--------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1
POINT: 1
--------------------------------------------------------------------------------------------------------------------------------------LOADS:
--------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
ACIER
fy = 23.50 daN/mm2
--------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: HEA 200
ht=19.0 cm
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
tf=1.0 cm
Wely=388.647 cm3
Welz=133.551 cm3
--------------------------------------------------------------------------------------------------------------------------------------STRESSES:
SigN = 30000.00/53.831 = 5.57 daN/mm2
SigFY = 2250.00/388.647 = 5.79 daN/mm2
---------------------------------------------------------------------------------------------------------------------------------------
March 2014
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LATERAL BUCKLING PARAMETERS:
z=0.000
B=1.000
D=1.401
Sig D=20.34 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
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
--------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1
POINT: 1
--------------------------------------------------------------------------------------------------------------------------------------LOADS:
--------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
ACIER
fy = 23.50 daN/mm2
---------------------------------------------------------------------------------------------------------------------------------------
March 2014
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SECTION PARAMETERS: HEA 200
ht=19.0 cm
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
tf=1.0 cm
Wely=388.647 cm3
Welz=133.551 cm3
--------------------------------------------------------------------------------------------------------------------------------------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
z=0.000
B=1.000
D=1.401
Sig D=20.34 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
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]
Check of the formula (k1n + kfzfz )/ e  1,0
Case 2
Check of the formula (k1n + kfyfy )/ e  1,0
Case 3
3. Factored flexural stress in a member fz [daN/mm2]
4. Check of the formula (k1n + kfyfy + kfzfz)/ e  1,0
Robot
HANDBOOK
5,57
5,6
16,85
16,8
1,038
1,030
5,57
5,6
5,79
5,8
0,500
0,498
5,57
5,6
5,79
5,8
16,85
16,8
1,293
1,285
CONCLUSION:
March 2014
page 21 / 35
CONCRETE
March 2014
page 22 / 35
1. BAEL 91 mod. 99 - RC columns
March 2014
page 23 / 35
- 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
cross section: 30x30
(ft)
(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 24 / 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 25 / 35
2.5.1 Slenderness analysis
Lu (m)
2.80
2.80
Direction Y:
Direction Z:
K
1.00
1.00

32.33
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
Reinforcement
[2]
4 HA16
Robot
8 HA12
Nu
1310 kN*
1340 kN
* According to accurate calculations should be 1308.5 kN
March 2014
page 26 / 35
- 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
Determine the reinforcement of three columns. The data is given below.
LOADS:
N1 = 1650
N2 = 2150
N3 = 2770
(kN)
(kN)
(kN)
GEOMETRY:
lf = 3.2
cross section: 30x60
(ft)
(cm)
MATERIAL:
Concrete
: fc28 = 25.00 (MPa)
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 27 / 35
RESULTS OF THE CALCULATIONS:
Column
Nu
RESULTS:
As (cm2)
Reinforcement
March 2014
1
1650
[3] / Robot
7.7
4HA14 +
2HA10
7.7
4HA12 +
4HA10
2
2150
[3] / Robot
7.7
4HA14 +
2HA10
7.7
4HA12 +
4HA10
3
2770
[3] / Robot
29.5
6HA25
23.6
4HA20 +
14HA10
page 28 / 35
- Column subjected to axial load and biaxial bending
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
Longitudinal reinforcement
Transversal reinforcement
3.
: fc28 = 25.00 (MPa)
: type
HA
: type
HA
fe = 500.00 (MPa)
fe = 500.00 (MPa)
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 29 / 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
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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:
DL1
LL1
N
(kN)
500
200
MyA
(kN*m)
100
50
MyB
(kN*m)
30
30
MyC
(kN*m)
72
42
MzA
(kN*m)
20
10
MzB
(kN*m)
30
20
MzC
(kN*m)
40*
30*
1.35DL1+1.5LL1
975
210
85.5
160.2
42
70.5
99
Case
1
2
Dimensioning
combination
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 31 / 35
7.2. THE INFLUENCE OF SLENDERNESS
Two independent calculations of the total moment for both directions are carried out.
Y DIRECTION
Slenderness analysis:

l0
= 40.41
r
  lim according to 4.3.5.3.5(2)
Check if




lim  max 25;
u 
15 


 u 
= 30.38
N Sd
= 0.244
Ac  f cd
  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).
crit  25  (2 
e01
)
e02
= 39.82
85.5
= 0.088 (m)
975
210
e01 
= 0.215 (m)
975
e01 
  crit
- slenderness effects have to be taken into account.
The abovementioned requirement means that the total eccentricity of the axial force in Z
( e z  M y / N ) direction will be:
etot  e0  ea  e2
(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
etot  e0  ea .
Calculation of initial eccentricity e0
For the mid-height section, we have:
e0  ee  0.4  e01  0.6  e02 = 0.164 (m)
Calculation of additional eccentricity ea
March 2014
page 32 / 35
ea  
l0
= 0.018 (m),
2
1

= 0.0022
100 h
h = 21.0 (m)
  1 / 200
We assume  = 0.005
Calculation of second-order eccentricity e2
2
l0 1
= 0.043 (m)
10 r
K1 = 1 (   35)
e2  K 1 
f yd
1
= 0.009
 2K 2
r
0.9d  E s
K2 = 1
d = 0.552 (m)
Es = 200 (GPa)
f yd = 435 (MPa)
The total eccentricity in Z direction:
etot  e0  ea  e2 = 0.225 (m)
The total moment My:
M y  N  etot  219 (kNm)
Z DIRECTION
Slenderness analysis:

l0
= 41.57
r
Check if
  lim according to 4.3.5.3.5(2)




lim  max 25;
u 
  lim
15 


 u 
= 30.38
N Sd
= 0.244
Ac  f cd
- column is slender
Since the column is sway in Z direction, the slenderness effects have to be taken into account.
March 2014
page 33 / 35
The abovementioned requirement means that the total eccentricity of the axial force in Y
( e y  M z / N ) direction will be:
etot  e0  e2
Calculation of initial eccentricity e0
For the mid-height section, we have:
e0 
M sd
= 0.102 (m)
N sd
(the moment in mid-height section is given directly by the user)
Calculation of second-order eccentricity e2
2
l 1
= 0.057 (m)
e2  K 1  0
10 r
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:
l0  n l0 = 6.79 (m)
N sd

n  1 

N sd





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:
N sd
N sd

1.35  500
1.5  200
1.0 
0.3 = 0.785
975
975
Thus, the influence of creep is taken into account by means of n =1.3359
The other parameters are:
K1 = 1 (   35)
f yd
1
= 0.014
 2K 2
r
0.9d  E s
K2 = 1
d = 0.351 (m)
Es = 200 (GPa)
f yd = 435 (MPa)
The total eccentricity in Z direction:
etot  e0  e2 = 0.158 (m)
The total moment My:
M y  N  etot  154 (kNm)
March 2014
page 34 / 35
7.3. FINAL RESULT
M y = 219 (kNm)
M z = 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
March 2014
page 35 / 35

A u t o

Transcription

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