Useful Examples 1 FEM-Design Useful Examples 1 Example 1: A simple example ................................................................... 5 1.1 Buckling load ....................................................................................... 5 1.2 Design .................................................................................................. 6 1.2.1 1 th order theory .............................................................................. 6 1.2.2 2 nd order theory ........................................................................... 10 2 Example 2: A frame type example .......................................................... 15 2.1 Buckling load ..................................................................................... 15 2.2 Design ................................................................................................ 18 2.2.1 1 th order theory ............................................................................ 19 2.2.2 2 nd order theory ........................................................................... 23 3 Example 3: A truss type example ............................................................ 31 3.1 1 th order theory .................................................................................. 33 3.2 2 nd order theory .................................................................................. 36
Transcript
FEM-Design Useful Examples1 Example 1: A simple example ................................................................... 5
1.2.1 1th order theory .............................................................................. 61.2.2 2nd order theory ........................................................................... 10
2 Example 2: A frame type example .......................................................... 152.1 Buckling load ..................................................................................... 152.2 Design ................................................................................................ 18
2.2.1 1th order theory ............................................................................ 192.2.2 2nd order theory ........................................................................... 23
3 Example 3: A truss type example ............................................................ 313.1 1th order theory .................................................................................. 333.2 2nd order theory .................................................................................. 36
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Date: 051005
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Useful Examples - Info 3
4 Useful Examples - Info
1 Example 1: A simple example
1.1 Buckling loadCode: Swedish, Material: S275JR,
Security class: 3, γm = 1.0, Section HEA 300
Design values:
fyd = 275 / (1.2 1.0) = 229.17 Mpa
Ed = 2.1 105 / (1.2 1.0) = 1.75 105 Mpa
First we would like to find the elastic buckling load. A stability calculation with P = 756 kN will result in a crit-ical factor 1.0 for buckling mode 1 meaning flexural buckling in the weak direction. The critical buckling load Ncr_z is therefore 756 kN.
The critical load for flexural buckling in the stiff direc-tion is:
Ncr_y = 2.896 * 576 = 2189.4 kN
The theoretical buckling length is calculated as:
lcz = π (Ed Iz / Ncr_z) =π (1.75 * 108 * 6310 * 10-8/ 756)0,5 = 12.0 m
Not surprisingly the factor β is 2.0.
Useful Examples - Example 1: A simple example 5
1.2 DesignWe will perform a design of the column according to the Swedish Code for anaxial load P = 510 kN.
1.2.1 1th order theory
ImperfectionsFor a sway structure normally deviation has to be considered. For a 1th order de-sign however deviation as well as initial bow imperfections are considered in theflexural buckling calculation, see K18:55.
We start by defining the input necessary for the steel design. In this case we haveto define the following: Lateral torsional and shear buckling, flexural bucklingstiff direction and flexural buckling weak direction.
Lateral torsional and shear bucklingThis input will influ-ence the result con-cerning torsional buckling, lateral tor-sional buckling and shear buckling of the web.
As it is a cantilever we define the support con-ditions as rigid at one end and free at the oth-er.
6 Useful Examples - Example 1: A simple example
Flexural buckling stiff direction and weak directionThe theoretical β value is 2.0 as shown above but the Swedish Code predicts β = 2.1 for design, see K18:38.
We perform a 1th order analysis for Load Combi-nations in this case only one called Load and a de-sign according to the steel code by activating the op-tion Checking.
The result Utilization will show the following:
Useful Examples - Example 1: A simple example 7
The utilization is 103% and themember will be red as the utili-zation is larger than 100%.
A code check will show the uti-lization for all checks neces-sary according to the SwedishCode.
8 Useful Examples - Example 1: A simple example
Decisive is the Lateral torsional buckling check that in this case without mo-ments is a pure check for flexural buckling around z-z axis.
Capacities due to instability and used buckling lengths are as shown below.
Useful Examples - Example 1: A simple example 9
1.2.2 2nd order theoryTo receive accurate result for the 2nd order calculation each compressed membershould be divided in more than one calculation element that is the default set-ting. An even number should be chosen and we will divide the member in fourelements. We go on by defining the input necessary for the steel design. In thiscase we only have to define input for Lateral torsional and shear bucklingwhere the same input as above will be used. No buckling lengths for flexuralbuckling are necessary as these effects are considered in the 2nd order moments.
ImperfectionsFor a design based on a 2nd order analysis it is vital to consider imperfections.For a sway structure normally deviation will be decisive and this effect will beautomatically considered based on the buckling shape chosen.
We start therefore by making an imperfection calculation where we are interest-ed in the first two buckling shapes.
The first buckling shape representing a pure flexu-ral buckling in the weak direction has a critical pa-rameter of 1.78 and a fac-tor 0.03 meaning that the load can be magnified 1.78 times before a theo-retical buckling will occur and the deviation in the column top is 0.03 m.
For buckling shape 2, pure flexural buckling in the stiff direction the critical fac-tor is 5.151.
10 Useful Examples - Example 1: A simple example
To be able to make a design the analysis must be connected to one of the available buckling shapes. Normally the first buckling shape will be decisive as in this case. We will therefore per-form a 2nd order analysis based on buckling shape one as shown be-low.
The deviation will result in a moment distribution Mz as shown above.
The design will result in the following:
Useful Examples - Example 1: A simple example 11
The utilization is 104% and the member will be red as the utilization is larger than 100%.
A code check will show the utilization for all checks nec-essary according to the code.
The first stress check is also checking flexural buckling around y-y and z-z axis. The lateral buckling is decisive in this case as apart from the 2nd order moment Mz also the re-duction because of torsional buckling that is not consid-ered in the analysis is consid-ered here.
If the design instead had been based on the second buckling mode the result would have been the follow-ing:
12 Useful Examples - Example 1: A simple example
In this case the utilization will only be 52% as the moment Mz is zero. Insteadthe moment My is calculated with regard to the deviation. In the above resultflexural buckling around the y-y axis is considered but not the more dangerousbuckling around the z-z axis. This tells us that it is important to consider themost dangerous buckling shape in the design.
Useful Examples - Example 1: A simple example 13
14 Useful Examples - Example 1: A simple example
2 Example 2: A frame type exampleWe will design a space frame structure according to a 1th and a 2nd order analysisfor the Swedish Code.
Useful Examples - Example 2: A frame type example 15
Sections: Columns, Rectangular hollow sections VKR 300 x 300 x 10,Beams, I-sections, HEB 260,Diagonals, Circular hollow sections CHS 101.6 x 3.6.
Design values:
fyd = 275 / (1.2 1.0) = 229.17 Mpa
Ed = 2.1 105 / (1.2 1.0) = 1.75 105 Mpa
16 Useful Examples - Example 2: A frame type example
To be able to perform a design based on a 1th order analysis we need to estimatethe buckling length of the columns. To accomplish this we try to calculate theelastic buckling load for the structure. A stability calculation with P = 1360 kNapplied at all columns will result in a critical factor 1.0 for buckling mode 1. Thecritical buckling load Ncr_z is therefore 1360 kN.
The theoretical buckling length for the columns are calculated as:
lcz = γ (Ed Iz / Ncr_z) = π (1.75 108 * 16026 10-8 / 1360)0.5 = 14.3 m.
This means that the factor β is 2.4.
Useful Examples - Example 2: A frame type example 17
2.2 DesignWe will design the frame for the following loads.
All line loads 8 kN/m and all point loads 500 kN.
18 Useful Examples - Example 2: A frame type example
ImperfectionsFor a sway structure normally deviation has to be considered. For a 1th order de-sign however deviation as well as initial bow imperfections are considered in theflexural buckling calculation if certain tolerance rules are fulfilled which we as-sume in this case, see K18:55.
We start by defining the input necessary for the steel design.
Useful Examples - Example 2: A frame type example 19
Lateral torsional and shear bucklingThis input will influence the result concerning torsional buckling, lateral tor-sional buckling and shear buckling of the web.
We will define all members as hinged at both ends. The load level influencingthe lateral torsional buckling capacity is set to top of the beam. No end stiffenerswill be defined meaning weak end stiffening according to BSK99.
Flexural buckling stiff and weak direction
20 Useful Examples - Example 2: A frame type example
The flexural buckling factor will be chosen as 1.0 for all beams and 2.4 for allcolumns in accordance with the calculation above.
We perform a 1th order analysis for Load Combinations in this case only onecalled Load and a design according to the steel code by activating the optionChecking.
The result Utilization will show the following:
Four columns have utilization over 100% and will be shown in red colour. Thebottom columns C3 and C6 have a utilization of 107%. A code check will showthe utilization for all checks necessary according to the code.
Useful Examples - Example 2: A frame type example 21
Decisive is the flexural buckling check.
Capacities due to instability and used buckling lengths are as shown below.
22 Useful Examples - Example 2: A frame type example
2.2.2 2nd order theoryFor a 2nd order design all compressed members ought to be divided into morethan one finite element to get accurate results. For this example we will divideall members in 4 elements. We go on by defining the input necessary for thesteel design. In this case we only have to define input for lateral torsional andshear buckling where the same input as above will be used. As mentioned beforeno buckling lengths for flexural buckling are necessary as these effects are con-sidered in the 2nd order moments.
Useful Examples - Example 2: A frame type example 23
ImperfectionsFor a design based on a 2nd order analysis it is vital to consider imperfections.For a sway structure normally deviation will be decisive and this effect will beautomatically considered based on the buckling shape chosen.
We start therefore by making an imperfection calculation where we will calcu-late the first five buckling shapes.
The first buckling shape representing a pure side sway in the load direction has acritical parameter of 1.87 and a factor 0.026 meaning that the load can be magni-fied 1.87 times before a theoretical buckling will occur and the deviation in thecolumn top is 0.026 m.
For buckling shape 2, pure side sway in the perpendicular direction the criticalfactor is 1.949 and for buckling shape 3, a twisting of the entire structure thecritical factor is 2.029.
24 Useful Examples - Example 2: A frame type example
To be able to make a design the analysis must be connected to one of the availa-ble buckling shapes. Normally the first buckling shape will be decisive but asthe first three shapes have about the same critical factor we will check all threebuckling shapes. We start with a 2nd order analysis based on buckling shape one.
The moment distribution Mz for column C3 will differ as shown above.
Useful Examples - Example 2: A frame type example 25
The design based on an analysis with imperfection according to buckling shape1 will result in the following:
The same four columns as for the 1th order design above have utilization largerthan 100%. The maximum utilization is 122% for member C6 and 120% formember C3.
26 Useful Examples - Example 2: A frame type example
The first stress check that in this case is also a check with regard to flexuralbuckling around y-y and z-z axis will be decisive in this case.
A design based on the second buckling and third buckling mode will result inthe following utilizations.
Useful Examples - Example 2: A frame type example 27
28 Useful Examples - Example 2: A frame type example
As can bee seen some of the members C5, B3, B6, B13 and B16 have a some-what larger utilization with regard to buckling shape 2 and B15 for bucklingshape 3.
To reach a structure that fulfil the requirements of the code the column sectionsshould be increased.
Useful Examples - Example 2: A frame type example 29
30 Useful Examples - Example 2: A frame type example
3 Example 3: A truss type exampleWe will design a truss structure according to a 1th and a 2nd order analysis for theSwedish Code.
ImperfectionsAs mentioned above, for a 1th order design initial bow imperfections are consid-ered in the flexural buckling calculation.
We start by defining the input necessary for the steel design.
Lateral torsional and shear bucklingThis input will influence the result concerning torsional buckling, lateral tor-sional buckling and shear buckling of the web.
Useful Examples - Example 3: A truss type example 33
We will define all members as hinged at both ends. The load level influencingthe lateral torsional buckling capacity is set to top of the beam. No end stiffenerswill be defined meaning weak end stiffening according to BSK99.
Flexural buckling stiff and weak direction
The flexural buckling factor will be chosen as 1.0 in both directions for all mem-bers.
34 Useful Examples - Example 3: A truss type example
A design based on a 1th order analysis will result in the following utilization:
All members have utilization under 100% so the structure fulfils the require-ments of the code. The diagonal B30 have the highest utilization 97%.
Useful Examples - Example 3: A truss type example 35
Decisive is the flexural buckling check.
Capacities due to instability and used buckling lengths are as shown below.
3.2 2nd order theoryWe start by defining the input necessary for the steel design. As mentionedabove we only have to define input for lateral torsional and shear bucklingwhere the same input as above will be used.
ImperfectionsFor a design based on a 2nd order analysis it is vital to consider imperfections. Ina truss structure normally some of the members are hinged at both ends and onlyaffected by a compressive or tensile force. For compressed members it is vitalthat a moment will be present as the buckling effect lies in the magnitude of the
36 Useful Examples - Example 3: A truss type example
2nd order moment. Therefore no design will be possible if the moment is lessthan the moment due to initial bow imperfection and this effect will be automat-ically considered if a buckling shape representing this effect will be chosen.
We start therefore by making an imperfection calculation.
Finite element divisionFor a structure it is important to decide if we are interested in a global bucklingor local buckling mode. For the present structure the three diagonals B30, B32and B34 will be compressed and are likely to be exposed to local buckling. Ifthis should be possible these members has to be divided into at least 2 finite ele-ments.
We start by making an imperfection calculation with the default setting 1 finiteelement for each member.
The critical factor of the first buckling shape is 49 times the applied load and thebuckling shape represents a global buckling as seen below.
A design based on buckling shape 1 will not be possible for the three membersB30, B32 and B34 as the moment is zero in these members. A message asshown below will be displayed.
Useful Examples - Example 3: A truss type example 37
Either another buckling shape should be cho-sen or these members could be designed ac-cording to 1th order theory.
If we divide all members in e.g. 4 finite ele-ments the following imperfection result willbe calculated.
As can bee seen the critical perimeters arenow significantly smaller as also local buck-ling is considered. We can also see that thesecond buckling shape has a negative critical
factor. This is due to the fact that we also have divided members in tension. (Seealso the theory manual Applied Theory and Design). If we restore the division
38 Useful Examples - Example 3: A truss type example
to the default setting for the three diagonals in tension B29, B31 and B33 thefollowing imperfection result will be calculated.
It is therefore recommended only to dividemembers in compression but as in realitymembers can be in compression for one loadcombination and in tension for another this isoften not possible. Buckling shapes with neg-ative critical factors should then be ignoredand only shapes with a positive critical factorare to be used in the design.
Above the first 6 shapes have been calculat-ed. They all have similar critical factors rep-resenting local buckling around y-y and z-zdirection for the three diagonals in compres-sion as seen below.
Useful Examples - Example 3: A truss type example 39
The design based on an analysis with imperfection according to buckling shape1 will result in the following:
40 Useful Examples - Example 3: A truss type example
The maximum utilization is 84% for member B30 according to the initial bowimperfection for buckling shape 1. Members B32 and B34 are not affected byany moment according to this shape and a design will not be possible.
For member B30 the stress check that in this case also is the flexural bucklingcheck will be decisive.
Useful Examples - Example 3: A truss type example 41
A design with regard to buckling shape 2 will display the same result. A designwith regard to buckling shape 3 will show:
Not surprisingly member B34 will be designed and the maximum utilization is81%. In this case there will be no design for members B30 and B32. A design
42 Useful Examples - Example 3: A truss type example
with regard to buckling shape 5 or 6 will accordingly mean design for memberB32 but not for members B30 and B34.
The above emphasizes a problem when using 2nd order design for truss typestructures and the imperfection is calculated according to one of the bucklingshapes. This is due to the fact that a truss type structure normally contains a lotof members that are only subjected to an axial force and as the flexural bucklingeffect lies in the magnification of the bending moment and consequently eachmember must have a moment, a large number of buckling shapes must bechecked.
For structures like this maybe a 1th order design is preferable as the bucklinglengths normally are easy to estimate. Usually complemented with a global sta-bility analysis to guarantee that the critical factor is acceptable.
Useful Examples - Example 3: A truss type example 43
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