A u t o

Transcription

A u t o

®
Autodesk
™
Robot Structural Analysis
Professional
VERIFICATION MANUAL
FOR AUSTRALIAN CODES
March 2014
© 2014 Autodesk, Inc. All Rights Reserved. Except as otherwise permitted by Autodesk, Inc., this
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Autodesk Robot Structural Analysis Professional - Verification Manual for Australian Codes
INTRODUCTION ..............................................................................................................................................................................1
STEEL ..............................................................................................................................................................................................2
1. AS 4100-1998 - STEEL................................................................................................................................................................3
VERIFICATION EXAMPLE 1
March 2014
- AXIAL COMPRESSION ....................................................................................................................4
- AXIAL COMPRESSION SLENDER SECTION ........................................................................................8
- BENDING – SIMPLY SUPPORTED BEAM ..........................................................................................11
- BENDING – BRACED BEAM ...........................................................................................................15
- SHEAR STIFFENERS ....................................................................................................................19
- BIAXIAL BENDING ........................................................................................................................23
page i
INTRODUCTION
This verification manual contains numerical examples for elements of steel structures prepared and
originally calculated by Autodesk Robot Structural Analysis Professional version 2013. The
comparison of results is still valid for the next versions.
All examples have been taken from handbooks that include benchmark tests covering fundamental
types of behaviour encountered in structural analysis. Benchmark results (signed as “Handbook”) are
recalled, and compared with results of Autodesk Robot Structural Analysis Professional (signed further
as “Robot”).
Each example contains the following parts:
- title of the problem
- specification of the problem
- Robot solution of the problem
- outputs with calculation results and calculation notes
- comparison between Robot results and exact solution
- conclusions.
March 2014
page 1 / 28
STEEL
March 2014
page 2 / 28
1. AS 4100-1998
March 2014
page 3 / 28
- Axial compression
Example taken from The Behavior and Design of Steel Structures to AS 4100
N.S.Trahair and M.A.Bradford, Taylor & Francis, 3rd edition 2008
TITLE:
Checking a UB compression member. (Example 3.12.1 page 89).
SPECIFICATION:
A compression member is simply supported at each end and has a central brace of lateral deflection
in the minor principal plane. The member is modeled with a cantilever, the bracing conditions are
created by means of buckling length coefficients. The design axial load N = 562,5 kN is checked for
the assumed section 460UB82.1, steel Grade 300.
SOLUTION:
Define a new type of member. For analysed member pre-defined type of member COLUMN may be
initially opened. Press the Parameters button in DEFINITIONS/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 Z icon and select the last icon (Intermediate bracing) that
opens Internal bracing dialog window. Define the coordinates of the existing bracing, type in: 0.50
(relative) in the Coordinate of the existing bracing edit box. Close dialog box by pressing OK.
March 2014
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Save the newly-created type of member Column1.
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.
March 2014
page 5 / 28
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: AS4100 - 1998 Australian Standard - Steel Structures
ANALYSIS TYPE: Member Verification
---------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1 BAR_1
POINT: 1
COORDINATE: x = 0.00 L =
0.00 m
---------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 1 STA1
---------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
STEEL 300
fy = 300.00 MPa
fu = 440.00 MPa
E = 200000.00 MPa
G = 80000.00 MPa
Fi=0.90
---------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: UB 460x82.1
d=46.0 cm
Ag=105.000 cm2
Iy=37200.000 cm4
Iz=1860.000 cm4
b=19.1 cm
An=105.000 cm2
Zy=1617.391 cm3
Zz=194.764 cm3
tw=1.0 cm
Awy=61.120 cm2
Sy=1840.000 cm3
Zz=303.000 cm3
tf=1.6 cm
Awz=45.540 cm2
J=70.100 cm4
kf=0.97
Ae=101.611 cm2
Zey=1840.000 cm3
Zez=292.147 cm3
---------------------------------------------------------------------------------------------------------------------------------------INTERNAL FORCES AND CAPACITIES:
N = 562.50 kN
N* = 562.50 kN
Fi*Ns = 2743.50 kN
Fi*Nc = 772.91 kN
CLASS: = Slender
March 2014
page 6 / 28
---------------------------------------------------------------------------------------------------------------------------------------LATERAL BUCKLING PARAMETERS:
---------------------------------------------------------------------------------------------------------------------------------------BUCKLING PARAMETERS:
About Y axis:
About Z axis:
ly = 12.00 m
alfa,cy = 0.76
lz = 6.00 m
alfa,cz = 0.28
l,ey = 12.00 m
Ncy = 2306.14 kN
l,ez = 6.00 m
Ncz = 858.79 kN
lambda,ny = 68.70
lambda,nz = 153.62
---------------------------------------------------------------------------------------------------------------------------------------VERIFICATION FORMULAS:
Section strength check:
N/(Fi*Ns) = 0.21 < 1.00 (6.1) OK!,
Global stability check of member:
key*ly/ry = 68.70 < (ke*l/r),max = 200.00
kez*lz/rz = 153.62 < (ke*L/r),max = 200.00 STABLE
N*/(Fi*min(Ncy,Ncz)) = 0.73 < 1.00 (6.1) OK!,
----------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
COMPARISON:
Resistance, interaction expression
1. Section capacity ΦNs
2. Member capacity ΦNc
March 2014
Robot
2743,50
772,91
Handbook
2777
777
page 7 / 28
- Axial compression slender section
TITLE:
Section capacity of slender compression member. (Example 4.9.1 page 139).
SPECIFICATION:
Determine the design axial compression section capacity of the lightly welded I-section of steel
strength fy=310 MPa. The member is modeled with a cantilever, the checked section is shown in
figure below.
SOLUTION:
Define a new type of member. For analysed member pre-defined type of member SIMPLE BAR may
be initially opened. Press the Parameters button in DEFINITIONS/MEMBERS tab, which opens
MEMBER DEFINITION – PARAMETERS dialog box. Type a new name Simple bar 1 in the Member
Type editable field.
Then, select LW – lightly welded longitudinally type in the combo list of General parameters / Residual
stresses.
Save the newly-created type of member Simple bar1.
March 2014
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member.
March 2014
page 9 / 28
STEEL DESIGN
---------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1 BAR_1
POINT: 1
0.00 m
---------------------------------------------------------------------------------------------------------------------------------------LOADS:
---------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
STEEL C310
fy = 310.00 MPa
fu = 430.00 MPa
E = 200000.00 MPa
G = 80000.00 MPa
Fi=0.9000
---------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: I-BISYM_1
d=42.0 cm
Ag=120.000 cm2
Iy=38960.000 cm4
Iz=10670.000 cm4
b=40.0 cm
An=120.000 cm2
Zy=1855.238 cm3
Zz=533.500 cm3
tw=1.0 cm
Awy=80.000 cm2
Sy=2040.000 cm3
Zz=1000000.000 cm3
tf=1.0 cm
Awz=40.000 cm2
J=39.370 cm4
kf=0.7650
Ae=91.803 cm2
Zey=2040.000 cm3
Zez=800.250 cm3
N = 105.00 kN
N* = 105.00 kN
Fi*Ns = 2561.29 kN
CLASS: = Slender
About Y axis:
About Z axis:
N/(Fi*Ns) = 0.0410 < 1.0000 (6.1) OK!,
----------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
COMPARISON:
1. Section capacity ΦNs
March 2014
Robot
2561,29
Handbook
2561
page 10 / 28
- Bending – simply supported beam
TITLE:
Checking a beam supported at both ends. (Example 6.13.1 page 267).
SPECIFICATION:
Check the simply supported beam, 12 m long 610UB125 of grade 300 steel shown in the figure below.
Lateral deflections are effectively prevented and twist rotations are partially restrained. Design load is
a central concentrated top flange load combination (1.25x40)+(1.5x50) = 125 kN.
SOLUTION:
Define a new type of member. For analysed member pre-defined type of member BEAM may be
DEFINITION – PARAMETERS dialog box. Type a new name Beam1 in the Member Type editable
field.
Then, press upper flange button in the Lateral buckling parameters.
March 2014
page 11 / 28
Select P-Partially restrained for
member beginning and end. Close
dialog box by pressing OK.
Then, press the icon by the alpha,m in the main dialog Lateral buckling parameters.
Select Concentrated force distant
from the axis icon and let the
distant equal to 0.0. Close dialog
box by pressing OK.
Save the newly-created type of member Beam1.
March 2014
page 12 / 28
member.
March 2014
page 13 / 28
STEEL DESIGN
---------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1 BAR_1
POINT: 3
6.00 m
---------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 3 KOMB1 1*1.25+2*1.50
---------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
STEEL 300
fy = 280.00 MPa
fu = 440.00 MPa
E = 200000.00 MPa
G = 80000.00 MPa
Fi=0.90
---------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: UB 610x125
d=61.2 cm
Ag=160.000 cm2
Iy=98600.000 cm4
Iz=3930.000 cm4
b=22.9 cm
An=160.000 cm2
Zy=3222.222 cm3
Zz=343.231 cm3
tw=1.2 cm
Awy=89.768 cm2
Sy=3680.000 cm3
Zz=536.000 cm3
tf=2.0 cm
Awz=72.828 cm2
J=156.000 cm4
kf=1.00
Ae=160.000 cm2
Zey=3680.000 cm3
Zez=514.847 cm3
My = 375.00 kN*m
M*y = 375.00 kN*m
Fi*Msy = 927.36 kN*m
Fi*Mry = 927.36 kN*m
Vz = -62.50 kN
CLASS: = Compact
Fi*Vvz = 1179.81 kN
kt = 1.05
kr = 1.00
alfa,m = 1.35
alfa,s = 0.16
kl = 1.40
le = 17.70 m
Moa = 190.53 kN*m
Fi*Mby = 203.26 kN*m
About Y axis:
About Z axis:
My/(Fi*Msy) = 0.40 < 1.00 (5.1) OK!,
Vz/(Fi*Vvz) = 0.05 < 1.00 (5.11.1) OK!,
M*y/(Fi*Mby) = 1.84 > 1.00 (5.1) Not OK!,
----------------------------------------------------------------------------------------------------------------------------------------
Incorrect section !!!
COMPARISON:
1. Section capacity ΦMs
2. Member capacity ΦMb
March 2014
Robot
927,36
203,26
Handbook
927,4
202,8
page 14 / 28
- Bending – braced beam
Example taken from Steel Designers’ Handbook to AS 4100
B.Gorenc, R.Tinyou, A.Syam, UNSW Press, 7th edition 2007
TITLE:
Checking a braced beam. (Example 5.3.2 page 113).
SPECIFICATION:
Check the simply supported beam, 8 m long 360UB50.7 of grade 300 steel shown in the figure below.
Top flange has one lateral restraint at the midspan, ends are partially restrained. Nominal loads are
permanent action WG=38.4 kN/m and imposed action PQ=10kN, WQ=38.4 kN/m. Design load is a
combination of (1.2xG)+(1.5xQ).
SOLUTION:
field.
March 2014
page 15 / 28
Select Intermediate bracings in the
Types of flange restraints group.
Select P-Partially restrained for
member beginning and end.
Select L-Laterally restrained for
intermediate bracing.
Select User-defined definition for
Twist restraint factor kt and type in
kt = 1,1 value.
Press the icon next to Intermediate
bracing, which opens Internal
bracing definition window...
March 2014
page 16 / 28
Define the coordinates of the existing lateral bracing, type in: 0.50 (relative) in the Coordinate of the
existing bracing edit box. Close dialog box by pressing OK.
Close Lateral buckling coefficients dialog box by pressing OK.
Save the newly-created type of member Beam1.
member.
March 2014
page 17 / 28
STEEL DESIGN
---------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1 BAR_1
POINT: 1
4.00 m
---------------------------------------------------------------------------------------------------------------------------------------LOADS:
Governing Load Case: 3 KOMB1 1*1.200+2*1.500
---------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
STEEL 300
fy = 300.00 MPa
fu = 440.00 MPa
E = 200000.00 MPa
G = 80000.00 MPa
Fi=0.900
d=35.6 cm
Ag=64.700 cm2
Iy=14200.000 cm4
Iz=960.000 cm4
b=17.1 cm
An=64.700 cm2
Zy=797.753 cm3
Zz=112.281 cm3
tw=0.7 cm
Awy=39.330 cm2
Sy=897.000 cm3
Zz=173.000 cm3
tf=1.1 cm
Awz=25.988 cm2
J=24.100 cm4
kf=1.000
Ae=64.700 cm2
Zey=897.000 cm3
Zez=168.421 cm3
My = 124.68 kN*m
M*y = 124.68 kN*m
Vz = -7.50 kN
CLASS: = Compact
Fi*Vvz = 449.07 kN
kt = 1.100
kr = 1.000
alfa,m = 1.424
alfa,s = 0.379
kl = 1.400
le = 6.16 m
Moa = 130.49 kN*m
Fi*Mby = 130.50 kN*m
About Y axis:
About Z axis:
My/(Fi*Msy) = 0.515 < 1.000 (5.1) OK!,
Vz/(Fi*Vvz) = 0.017 < 1.000 (5.11.1) OK!,
M*y/(Fi*Mby) = 0.955 < 1.000 (5.1) OK!,
----------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
COMPARISON:
1. Section capacity ΦMs
2. Member capacity ΦMb
Robot
242,19
130,50
Handbook
242
129
.
March 2014
page 18 / 28
- Shear stiffeners
TITLE:
Shear capacity of a stiffened plate girder web. (Example 4.9.6 page 142).
SPECIFICATION:
Determine the design shear capacity of the plate girder web of grade 300 steel shown in figure below
with intermediate stiffeners spaced at 1800 mm. The member is modeled with a 9.0 m simple beam.
SOLUTION:
field.
Then, select LW – lightly welded longitudinally type in the combo list of General parameters / Residual
stresses.
March 2014
page 19 / 28
Switch off lateral buckling check by choosing the last icon in
the Lateral Buckling Type dialog opened from the main
parameters dialog box.
Close this dialog box by pressing OK.
In order to define stiffeners positions press Stiffeners button in the main parameters dialog box.
Define stiffeners coordinates along the beam
length. Type in the coordinates 0.0 0.2 0.4 0.6 0.8
1.0 (Relative), which gives stiffeners spacing at
1800 mm.
member.
March 2014
page 20 / 28
STEEL DESIGN
---------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1 BAR_1
POINT:
COORDINATE: x = 0.25 L = 2.25 m
---------------------------------------------------------------------------------------------------------------------------------------LOADS:
---------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
STEEL C310
fy = 310.00 MPa
fu = 430.00 MPa
E = 200000.00 MPa
G = 80000.00 MPa
Fi=0.90
---------------------------------------------------------------------------------------------------------------------------------------SECTION PARAMETERS: I-BISYM_2
d=154.0 cm
Ag=310.000 cm2
Iy=1205463.333 cm4
Iz=21345.833 cm4
b=40.0 cm
An=310.000 cm2
Zy=15655.368 cm3
Zz=1067.292 cm3
tw=1.0 cm
Awy=160.000 cm2
Sy=17785.000 cm3
Zz=1000000.000 cm3
tf=2.0 cm
Awz=150.000 cm2
J=256.401 cm4
kf=1.00
Ae=310.000 cm2
Zey=10778.517 cm3
Zez=1600.938 cm3
My = 759.38 kN*m
M*y = 1012.50 kN*m
Fi*Msy = 3007.21 kN*m
Fi*Mry = 3007.21 kN*m
Vz = 225.00 kN
CLASS: = Slender
Fi*Vvz = 1875.53 kN
About Y axis:
About Z axis:
----------------------------------------------------------------------------------------------------------------------------------------
March 2014
page 21 / 28
VERIFICATION FORMULAS:
My/(Fi*Msy) = 0.25 < 1.00 (5.1) OK!,
Vz/(Fi*Vvz) = 0.12 < 1.00 (5.11.1) OK!,
------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
COMPARISON:
1. Section capacity ΦVu
March 2014
Robot
1875,5
Handbook
1880
page 22 / 28
- Biaxial bending
TITLE:
Checking the biaxial bending capacity. (Example 7.7.5 page 318).
SPECIFICATION:
Check the adequacy of the beam-column which is the 250UB37.3 of grade 300 steel. Lateral
deflections and twist rotations are prevented at the ends and at mid-span. The member has the static
model and design load values shown in the figure below.
SOLUTION:
field.
March 2014
page 23 / 28
Press the buckling model icon next to the ke,y edit box in
the Buckling length coefficient Y frame and select the third
model (1.0).
Press the load model icon next to the beta,my edit box in
the Buckling length coefficient Y frame and select the
second model (-1.0).
Press the buckling model icon next to the ke,z edit box in the
Buckling length coefficient Z frame and select the last model
(intermediate bracing).
That opens Internal bracing dialog window. Define the
coordinates of the existing bracing, type in: 0.50 (relative) in
the Coordinate of the existing bracing edit box. Close dialog
box by pressing OK.
March 2014
page 24 / 28
Press the buckling model icon next to the ke,x edit box in the
Torsional buckling frame and select the first model (0.7).
Close dialog box by pressing OK.
Select Intermediate bracings in the
Types of flange restraints group.
Select F-Fully restrained for
member beginning and end.
Select L-Laterally restrained for
intermediate bracing.
Press the icon next to Intermediate
bracing, which opens Internal
bracing definition window.
Define the coordinates of the
existing lateral bracing, type in:
0.50 (relative) in the Coordinate of
the existing bracing edit box.
Close Lateral buckling coefficients dialog box by pressing OK.
March 2014
page 25 / 28
Define the same settings for for the lower flange in the Lateral buckling parameters.
Press Load level button in the Lateral buckling parameters. Select
the first icon, load in the central axis.
Then, press the icon by the alpha,m in the
main dialog Lateral buckling parameters.
Select Concentrated force distant from the
axis icon and let the distant equal to 0.0.
member.
March 2014
page 26 / 28
STEEL DESIGN
---------------------------------------------------------------------------------------------------------------------------------------CODE GROUP:
MEMBER: 1 BAR_1
POINT:
4.50 m
---------------------------------------------------------------------------------------------------------------------------------------LOADS:
---------------------------------------------------------------------------------------------------------------------------------------MATERIAL:
STEEL 300
fy = 320.00 MPa
fu = 440.00 MPa
E = 200000.00 MPa
G = 80000.00 MPa
Fi=0.900
d=25.6 cm
Ag=47.500 cm2
Iy=5570.000 cm4
Iz=566.000 cm4
b=14.6 cm
An=47.500 cm2
Zy=435.156 cm3
Zz=77.534 cm3
tw=0.6 cm
Awy=31.828 cm2
Sy=486.000 cm3
Zz=119.000 cm3
tf=1.1 cm
Awz=16.384 cm2
J=15.800 cm4
kf=1.000
Ae=47.500 cm2
Zey=486.000 cm3
Zez=116.301 cm3
N = 200.00 kN
My = 52.79 kN*m
Mz = -8.10 kN*m
Vy = 9.00 kN
N* = 200.00 kN
M*y = 52.79 kN*m
M*z = -8.10 kN*m
Fi*Vvy = 549.99 kN
Fi*Ns = 1368.00 kN
Fi*Msz = 33.49 kN*m
Fi*Nc = 413.13 kN
Fi*Mrz = 33.49 kN*m
Vz = -11.73 kN
CLASS: = Compact
Fi*Miy = 104.79 kN*m
Fi*Miz = 19.68 kN*m
Fi*Vvz = 283.12 kN
Fi*Moy = 83.00 kN*m
Fi*Mcy = 83.00 kN*m
kt = 1.000
kr = 1.000
alfa,m = 1.350
alfa,s = 0.486
kl = 1.000
le = 4.50 m
Moa = 107.45 kN*m
Fi*Mby = 91.81 kN*m
----------------------------------------------------------------------------------------------------------------------------------------
March 2014
page 27 / 28
BUCKLING PARAMETERS:
About Y axis:
About Z axis:
ly = 9.00 m
alfa,cy = 0.582
lz = 4.50 m
alfa,cz = 0.302
l,ey = 9.00 m
Ncy = 884.16 kN
l,ez = 4.50 m
Ncz = 459.03 kN
lambda,ny = 94.03
beta,my = -1.00
lambda,nz = 147.49
beta,mz = 0.20
N/(Fi*Ns) = 0.146 < 1.000 (6.1) OK!,
(My/(Fi*Mry))^1.546 + (Mz/(Fi*Mrz))^1.546 = 0.333 < 1.000 (8.3.4) OK!,
Vy/(Fi*Vvy) = 0.016 < 1.000 (5.11.1) OK!,
Vz/(Fi*Vvz) = 0.041 < 1.000 (5.11.1) OK!,
key*ly/ry = 94.03 < (ke*l/r),max = 200.000
kez*lz/rz = 147.49 < (ke*L/r),max = 200.000 STABLE
N*/(Fi*min(Ncy,Ncz)) = 0.484 < 1.000 (6.1) OK!,
(M*y/(Fi*Mcy))^1.4 + (M*z/(Fi*Miz))^1.4 = 0.819 < 1.000 (8.4.5.1) OK!,
----------------------------------------------------------------------------------------------------------------------------------------
Section OK !!!
COMPARISON:
1.
2.
3.
4.
5.
6.
Section capacity ΦNs
Section capacity ΦMsy
Section capacity ΦMsz
Member capacity ΦMiy
Member capacity ΦMiz
Member capacity ΦMoy
Robot
1368,00
139,97
33,49
104,79
19,68
83,00
Handbook
1368,0
140,0
33,5
104,7
19,7
84,6
.
March 2014
page 28 / 28

A u t o

Transcription

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