Exercise 4

NTNUNorwegian University of Science and Technology.

Faculty of Marine Technology

Department of Marine Structures. TMR4205 Buckling and Collapse of StructuresBuckling of truss-works________________________________________________________________________________________________Date: February, 2011 Signature: EKim Distributed 18.02.11 Due Date: 26.02.11PROBLEM 1

Figure 1 shows a column subjected to axial compression. The ends of the column are elastically restrained against rotation. The restraints are conveniently modeled with a rotational spring with stiffness kA and kB, [Nm/rad] respectively. The elastic buckling load shall be calculated by means of the procedure described in Chapter 2.8. Consider especially Example 1. (Hint:The stiffness of the spring must be added to the stiffness matrix)

Figure 1 An end Truss work of an Offshore Compression Module..

a) Calculate the critical load for different values of the rotational spring stiffnesses kA and kB ( for example with reference to Figure 2.38 (A =4, B =0), (A =4, B =2), (A =25, B =0), (A =25, B =2)). Compare the results with those of Figure 2.38.

The solution of the determinant singularity is conveniently found by an iterative procedure, e.g. by the bisection method. A simple algorithm may be programmed in Visual basic if Excel is used or programmed in MATLAB

b) Use the input files for the plane frame to calculate the buckling load for element 4. c) Determine the rotational rigidity of each end of element 4 by applying unit moments at the end. For this purpose element 4 should be removed from the analysis (Element 4 is most easily removed using the NONSTRU command). Adopting these rotational flexibilities use the algorithm developed in pt a) to determine the elastic buckling load for element 3. Use the ECCS column curve a to assess the critical stress for elasto-plastic buckling. Compare with USFOS predictions. Comment upon sources of deviation between the two results.

.

Figure 2 Plane frame

PROBLEM 2 (Voluntary)Figure 1 shows the end trusswork of an offshore compression module. The outer, vertical column is to be designed for buckling. The column is subjected to a centrally applied compressive design load of 390 kN.

Figure 1 An end Trusswork of an Offshore Compression Module.A hot-rolled IPE profile is to be used. In Table 1, geometry data are listed for such profiles. It can be assumed that the column is simply supported at both ends for strong-axis bending. For weak axis bending, it is restrained against displacements at 1/3- and 2/3-lengths.

d) Determine the required profile using the ECCS buckling design curves. The material factor can be set equal to 1.15. The yield stress is (Y = 240 MPa.

It is recommended that a simple spreadsheet algorithm (Excel) be developed in order to solve this problem. The input parameters should be profile dimensions, material parameters, buckling curve parameters etc. For later exercises it is beneficial that the algorithm is able to handle unequal flanges in calculating cross-sectional moment of inertia.

Plot the acting stress versus critical stress as a function of cross-sectional area (for a few cross-sections close to the one selected for design). Indicate which buckling mode is governing.

b) As an alternative to steel, aluminium can be considered. What is the required cross-section if an I-profile listed in Table 2 is to be used.

The yield strengths can be assumed to be (02 = (Y = 240 MPa. According to Eurocode 9 the buckling curve for aluminium can be determined with ( = 0.2 and

= 0.1.

Elastic modulus of Aluminium:E = 7 ( 104 MPa.

Density

:( = 2700 kg/m3

Material factor

:

=1.10

What is the weight ratio between sections in aluminium and steel that satisfy the design criterion?

Table 1.CONTINENTAL IPE SECTION *see key below

sizeDIMENSIONSWT per metre M KgSection A cm sqr.

hmm bmm amm emm rmm hmm

8080463.85.25606.07.64

100100554.15.77758.110.3

120120644.46.379310.413.2

140140734.76.9711212.916.4

160160825.07.4912715.820.1

180180915.38.0914618.823.9

2002001005.68.51215922.428.5

2202201105.99.21217826.233.4

2402401206.29.81519030.739.1

2702701356.610.21522036.145.9

3003001507.110.71524942.253.8

3303301607.511.51827149.162.6

3603601708.012.71829957.172.7

4004001808.613.52133166.384.5

4504501909.414.62137977.698.8

50050020010.216.02142690.7116.0

55055021011.117.224468106.0134.0

60060022012.019.024514122.0156.0

CONTINENTAL IPE SECTION - KEY

Table 2.

Aluminium profiles

h x b x s x t

120 x 58 x 5 x 8

140 x 66 x 6 x 9

160 x 74 x 7 x 10

180 x 82 x 7 x 10

200 x 90 x 8 x 11

220 x 100 x 9 x 12

240 x 110 x 10 x 13

260 x 120 x 11 x 14

280 x 130 x 12 x 15

300 x 140 x 13 x 16

340 x 160 x 15 x 17

380 x 180 x 17 x 19

PROBLEM 3 (voluntary)

Figure 1

Figure 1 shows a vertical section of a platform deck. The deck is built as a truss-work.

The braces are all oriented 450 relative to vertical. The stiffness in bending is given by

the factor . At the end of the deck a derrick is located with weight P=7.

a) Use a truss-work approach to calculate the forces in the various members, in particular for the members with circles.

b) Use the energy method to calculate the critical axial force for the diagonal marked with*. The following displacement function is proposed

where u x/ l is a non-dimensional axial co-ordinate og w is the lateral deflection of the brace. Discuss the goodness of this displacement function with respect to boundary conditions. In what end will you locate the origin of the co-ordinate system?

c) Calculate the critical force wrt. to buckling. What is the effective buckling length?

d) Use the G-method to obtain a better estimate of the critical force. Select a reasonable displacement pattern for the members when determining the appropriate coefficients. What do you obtain for the elastic buckling force now? Calculate the critical force, selecting a column curve at your liking. The yield stress is 300.

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390KN

9m

390KN

Column

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N

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(B

EI

kA

kB

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B

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