FEM-Design Useful Examples

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1 FEM-Design Useful Examples 1 Example 1: A simple example Buckling load Design th order theory nd order theory Example 2: A frame type example Buckling load Design th order theory nd order theory Example 3: A truss type example th order theory nd order theory Useful Examples 1

2 2 Useful Examples

3 Info Useful Examples 6.0 Copyright: Structural Design Software in Europe AB Date: For the latest information visit FEM-Design home page at Useful Examples - Info 3

4 4 Useful Examples - Info

1 Example 1: A simple example 1.1 Buckling load Code: Swedish, Material: S275JR, Security class: 3, γ m = 1.0, Section HEA 300 Design values: f yd = 275 / (1.2 1.0) = 229.17 Mpa E d = 2.1 10 5 / (1.

0 for buckling mode 1 meaning flexural buckling in the weak direction. The critical buckling load N cr_z is therefore 756 kn.

5 1 Example 1: A simple example 1.1 Buckling load Code: Swedish, Material: S275JR, Security class: 3, γ m = 1.0, Section HEA 300 Design values: f yd = 275 / ( ) = Mpa E d = / ( ) = Mpa First we would like to find the elastic buckling load. A stability calculation with P = 756 kn will result in a critical factor 1.0 for buckling mode 1 meaning flexural buckling in the weak direction. The critical buckling load N cr_z is therefore 756 kn. The critical load for flexural buckling in the stiff direction is: N cr_y = * 576 = kn The theoretical buckling length is calculated as: l cz = π (E d I z / N cr_z ) = π (1.75 * 10 8 * 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 Design We will perform a design of the column according to the Swedish Code for an axial load P = 510 kn. 1.2.1 1 th order theory Imperfections For a sway structure normally deviation has to be considered.

6 1.2 Design We will perform a design of the column according to the Swedish Code for an axial load P = 510 kn th order theory Imperfections For a sway structure normally deviation has to be considered. For a 1 th order design however deviation as well as initial bow imperfections are considered in the flexural buckling calculation, see K18:55. We start by defining the input necessary for the steel design. In this case we have to define the following: Lateral torsional and shear buckling, flexural buckling stiff direction and flexural buckling weak direction. Lateral torsional and shear buckling This input will influence the result concerning torsional buckling, lateral torsional buckling and shear buckling of the web. As it is a cantilever we define the support conditions as rigid at one end and free at the other. 6 Useful Examples - Example 1: A simple example

We perform a 1 th order analysis for Load Combinations in this case only one called Load and a design

7 Flexural buckling stiff direction and weak direction The 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 1 th order analysis for Load Combinations in this case only one called Load and a design according to the steel code by activating the option Checking. The result Utilization will show the following: Useful Examples - Example 1: A simple example 7

8 The utilization is 103% and the member will be red as the utilization is larger than 100%. A code check will show the utilization for all checks necessary according to the Swedish Code. 8 Useful Examples - Example 1: A simple example

9 Decisive is the Lateral torsional buckling check that in this case without moments 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 2 nd order theory To receive accurate result for the 2 nd order calculation each compressed member should be divided in more than one calculation element that is the default setting.

10 nd order theory To receive accurate result for the 2 nd order calculation each compressed member should be divided in more than one calculation element that is the default setting. An even number should be chosen and we will divide the member in four elements. We go on by defining the input necessary for the steel design. In this case we only have to define input for Lateral torsional and shear buckling where the same input as above will be used. No buckling lengths for flexural buckling are necessary as these effects are considered in the 2 nd order moments. Imperfections For a design based on a 2 nd order analysis it is vital to consider imperfections. For a sway structure normally deviation will be decisive and this effect will be automatically considered based on the buckling shape chosen. We start therefore by making an imperfection calculation where we are interested in the first two buckling shapes. The first buckling shape representing a pure flexural buckling in the weak direction has a critical parameter of 1.78 and a factor 0.03 meaning that the load can be magnified 1.78 times before a theoretical 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 factor is Useful Examples - Example 1: A simple example

We will therefore perform a 2 nd order analysis based on buckling shape one as shown below.

11 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 perform a 2 nd order analysis based on buckling shape one as shown below. The deviation will result in a moment distribution M z as shown above. The design will result in the following: Useful Examples - Example 1: A simple example 11

The first stress check is also checking flexural buckling around y-y and z-z axis.

12 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 necessary 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 2 nd order moment M z also the reduction because of torsional buckling that is not considered in the analysis is considered here. If the design instead had been based on the second buckling mode the result would have been the following: 12 Useful Examples - Example 1: A simple example

13 In this case the utilization will only be 52% as the moment M z is zero. Instead the moment M y is calculated with regard to the deviation. In the above result flexural buckling around the y-y axis is considered but not the more dangerous buckling around the z-z axis. This tells us that it is important to consider the most dangerous buckling shape in the design. Useful Examples - Example 1: A simple example 13

14 14 Useful Examples - Example 1: A simple example

15 2 Example 2: A frame type example We will design a space frame structure according to a 1 th and a 2 nd order analysis for the Swedish Code. 2.1 Buckling load Code: Swedish, Material: S275JR, Security class: 3, γ m = 1.0 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: f yd = 275 / (1.

16 Sections: Columns, Rectangular hollow sections VKR 300 x 300 x 10, Beams, I-sections, HEB 260, Diagonals, Circular hollow sections CHS x 3.6. Design values: f yd = 275 / ( ) = Mpa E d = / ( ) = Mpa 16 Useful Examples - Example 2: A frame type example

17 To be able to perform a design based on a 1 th order analysis we need to estimate the buckling length of the columns. To accomplish this we try to calculate the elastic buckling load for the structure. A stability calculation with P = 1360 kn applied at all columns will result in a critical factor 1.0 for buckling mode 1. The critical buckling load N cr_z is therefore 1360 kn. The theoretical buckling length for the columns are calculated as: l cz = γ (E d I z / N cr_z ) = π ( * / 1360) 0.5 = 14.3 m. This means that the factor β is 2.4. Useful Examples - Example 2: A frame type example 17

18 2.2 Design We 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

Line loads 1 kn/m and point loads 5 kn. Load combination: 1.3 Vertical load + 1.0 Horizontal load. 2.2.1 1 th order theory Imperfections For a sway structure normally deviation has to be considered.

19 Line loads 1 kn/m and point loads 5 kn. Load combination: 1.3 Vertical load Horizontal load th order theory Imperfections For a sway structure normally deviation has to be considered. For a 1 th order design however deviation as well as initial bow imperfections are considered in the flexural buckling calculation if certain tolerance rules are fulfilled which we assume 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

The load level influencing the lateral torsional buckling capacity is set to top of the beam.

20 Lateral torsional and shear buckling This input will influence the result concerning torsional buckling, lateral torsional buckling and shear buckling of the web. We will define all members as hinged at both ends. The load level influencing the lateral torsional buckling capacity is set to top of the beam. No end stiffeners will be defined meaning weak end stiffening according to BSK99. Flexural buckling stiff and weak direction 20 Useful Examples - Example 2: A frame type example

21 The flexural buckling factor will be chosen as 1.0 for all beams and 2.4 for all columns in accordance with the calculation above. We perform a 1 th order analysis for Load Combinations in this case only one called Load and a design according to the steel code by activating the option Checking. The result Utilization will show the following: Four columns have utilization over 100% and will be shown in red colour. The bottom columns C3 and C6 have a utilization of 107%. A code check will show the utilization for all checks necessary according to the code. Useful Examples - Example 2: A frame type example 21

22 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 2 nd order theory For a 2 nd order design all compressed members ought to be divided into more than one finite element to get accurate results.

23 nd order theory For a 2 nd order design all compressed members ought to be divided into more than one finite element to get accurate results. For this example we will divide all members in 4 elements. We go on by defining the input necessary for the steel design. In this case we only have to define input for lateral torsional and shear buckling where the same input as above will be used. As mentioned before no buckling lengths for flexural buckling are necessary as these effects are considered in the 2 nd order moments. Useful Examples - Example 2: A frame type example 23

24 Imperfections For a design based on a 2 nd order analysis it is vital to consider imperfections. For a sway structure normally deviation will be decisive and this effect will be automatically considered based on the buckling shape chosen. We start therefore by making an imperfection calculation where we will calculate the first five buckling shapes. The first buckling shape representing a pure side sway in the load direction has a critical parameter of 1.87 and a factor meaning that the load can be magnified 1.87 times before a theoretical buckling will occur and the deviation in the column top is m. For buckling shape 2, pure side sway in the perpendicular direction the critical factor is and for buckling shape 3, a twisting of the entire structure the critical factor is Useful Examples - Example 2: A frame type example

25 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 but as the first three shapes have about the same critical factor we will check all three buckling shapes. We start with a 2 nd order analysis based on buckling shape one. The moment distribution M z for column C3 will differ as shown above. Useful Examples - Example 2: A frame type example 25

26 The design based on an analysis with imperfection according to buckling shape 1 will result in the following: The same four columns as for the 1 th order design above have utilization larger than 100%. The maximum utilization is 122% for member C6 and 120% for member C3. 26 Useful Examples - Example 2: A frame type example

27 The first stress check that in this case is also a check with regard to flexural buckling 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 in the following utilizations. Useful Examples - Example 2: A frame type example 27

28 28 Useful Examples - Example 2: A frame type example

29 As can bee seen some of the members C5, B3, B6, B13 and B16 have a somewhat larger utilization with regard to buckling shape 2 and B15 for buckling shape 3. To reach a structure that fulfil the requirements of the code the column sections should be increased. Useful Examples - Example 2: A frame type example 29

30 30 Useful Examples - Example 2: A frame type example

31 3 Example 3: A truss type example We will design a truss structure according to a 1 th and a 2 nd order analysis for the Swedish Code. Code: Swedish, Material: S275JR, Security class: 3, γ m = 1.0 Design values: f yd = 275 / ( ) = Mpa E d = / ( ) = Mpa Useful Examples - Example 3: A truss type example 31

32 Sections: Beams, I-sections HEA 500 and HEA 300, Diagonals, Circular hollow sections CHS x 5.0 All members have a stiff connection to each other except the diagonals that are hinged at both ends. We will design the truss for the following loads: 32 Useful Examples - Example 3: A truss type example

33 All line loads 15 kn/m, all vertical point loads 50 kn and all horizontal point loads 100 kn. Load combination: 1.0 Vertical load Horizontal load th order theory Imperfections As mentioned above, for a 1 th order design initial bow imperfections are considered in the flexural buckling calculation. We start by defining the input necessary for the steel design. Lateral torsional and shear buckling This input will influence the result concerning torsional buckling, lateral torsional buckling and shear buckling of the web. Useful Examples - Example 3: A truss type example 33

No end stiffeners will be defined meaning weak end stiffening according to BSK99.

34 We will define all members as hinged at both ends. The load level influencing the lateral torsional buckling capacity is set to top of the beam. No end stiffeners will 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 members. 34 Useful Examples - Example 3: A truss type example

35 A design based on a 1 th order analysis will result in the following utilization: All members have utilization under 100% so the structure fulfils the requirements of the code. The diagonal B30 have the highest utilization 97%. Useful Examples - Example 3: A truss type example 35

36 Decisive is the flexural buckling check. Capacities due to instability and used buckling lengths are as shown below nd order theory We start by defining the input necessary for the steel design. As mentioned above we only have to define input for lateral torsional and shear buckling where the same input as above will be used. Imperfections For a design based on a 2 nd order analysis it is vital to consider imperfections. In a truss structure normally some of the members are hinged at both ends and only affected by a compressive or tensile force. For compressed members it is vital that a moment will be present as the buckling effect lies in the magnitude of the 36 Useful Examples - Example 3: A truss type example

37 2 nd order moment. Therefore no design will be possible if the moment is less than the moment due to initial bow imperfection and this effect will be automatically considered if a buckling shape representing this effect will be chosen. We start therefore by making an imperfection calculation. Finite element division For a structure it is important to decide if we are interested in a global buckling or local buckling mode. For the present structure the three diagonals B30, B32 and B34 will be compressed and are likely to be exposed to local buckling. If this should be possible these members has to be divided into at least 2 finite elements. We start by making an imperfection calculation with the default setting 1 finite element for each member. The critical factor of the first buckling shape is 49 times the applied load and the buckling shape represents a global buckling as seen below. A design based on buckling shape 1 will not be possible for the three members B30, B32 and B34 as the moment is zero in these members. A message as shown below will be displayed. Useful Examples - Example 3: A truss type example 37

Either another buckling shape should be chosen or these members could be designed according to 1 th order theory. If we divide all members in e.g. 4 finite elements the following imperfection result will be calculated.

We can also see that the second buckling shape has a negative critical factor. This is due to the fact that we also have divided members in tension.

38 Either another buckling shape should be chosen or these members could be designed according to 1 th order theory. If we divide all members in e.g. 4 finite elements the following imperfection result will be calculated. As can bee seen the critical perimeters are now significantly smaller as also local buckling is considered. We can also see that the second buckling shape has a negative critical factor. This is due to the fact that we also have divided members in tension. (See also the theory manual Applied Theory and Design). If we restore the division 38 Useful Examples - Example 3: A truss type example

39 to the default setting for the three diagonals in tension B29, B31 and B33 the following imperfection result will be calculated. It is therefore recommended only to divide members in compression but as in reality members can be in compression for one load combination and in tension for another this is often not possible. Buckling shapes with negative critical factors should then be ignored and only shapes with a positive critical factor are to be used in the design. Above the first 6 shapes have been calculated. They all have similar critical factors representing local buckling around y-y and z-z direction for the three diagonals in compression as seen below. Useful Examples - Example 3: A truss type example 39

40 The design based on an analysis with imperfection according to buckling shape 1 will result in the following: 40 Useful Examples - Example 3: A truss type example

41 The maximum utilization is 84% for member B30 according to the initial bow imperfection for buckling shape 1. Members B32 and B34 are not affected by any 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 buckling check will be decisive. Useful Examples - Example 3: A truss type example 41

42 A design with regard to buckling shape 2 will display the same result. A design with regard to buckling shape 3 will show: Not surprisingly member B34 will be designed and the maximum utilization is 81%. In this case there will be no design for members B30 and B32. A design 42 Useful Examples - Example 3: A truss type example

43 with regard to buckling shape 5 or 6 will accordingly mean design for member B32 but not for members B30 and B34. The above emphasizes a problem when using 2 nd order design for truss type structures and the imperfection is calculated according to one of the buckling shapes. This is due to the fact that a truss type structure normally contains a lot of members that are only subjected to an axial force and as the flexural buckling effect lies in the magnification of the bending moment and consequently each member must have a moment, a large number of buckling shapes must be checked. For structures like this maybe a 1 th order design is preferable as the buckling lengths normally are easy to estimate. Usually complemented with a global stability analysis to guarantee that the critical factor is acceptable. Useful Examples - Example 3: A truss type example 43

Question Paper Code : 11410

Reg. No. : Question Paper Code : 11410 B.E./B.Tech. DEGREE EXAMINATION, APRIL/MAY 2011 Fourth Semester Mechanical Engineering ME 2254 STRENGTH OF MATERIALS (Common to Automobile Engineering and Production