examination paper for tmr4205 buckling and … of marine technology page 1 of 10 examination paper...

Department of Marine Technology

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Examination paper for TMR4205 Buckling and collapse of marine structures Academic contact during examination: Jørgen Amdahl Phone: 73595544 / 95745663 Examination date: 21.05.2013 Examination time (from-to): 09.00-13.00 Permitted examination support material: (D) – Neither printed nor

handwritten note as are permitted. Approved, simple calculator is permitted.

Other information: Language: English Number of pages: 10 Number of pages enclosed: 0

Checked by:

____________________________

Date

Signature


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PROBLEM 1

Figure 1

a) Figure 1shows the cross-section of a beam subjected to bending. Determine the

position of the plastic neutral axis and calculate the plastic section modulus for bending about the strong axis.

b) Sketch the stress distribution for pure plastic bending. Sketch in principle the stress distribution for combined plastic bending and axial force, when the axial force is relatively small (compared to the plastic axial force). How much does the presence of the axial force reduce the plastic bending capacity when the cross-section is subjected to an axial force of 0.9 MN? The yield strength of the material is fy = 300 MPa.


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Figure 2

c) Figure 2 shows a beam with two concentrated loads, with a magnitude of P and

2/3P, respectively. The plastic moment capacity of the beam is MP. The beam is fully clamped at A and simply supported at B. Sketch the potential collapse mechanisms. Determine the plastic collapse load for the mechanisms.

d) Select the critical mechanism (of those analysed in pt. c) and determine whether the associated collapse load is the true collapse load. If it is not the true collapse load, how is the calculated critical load compared to the true collapse load?

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PROBLEM 2 a) Sketch the principle force-end shortening relationship (or alternatively force-mid

point lateral displacement relationship) for a perfect, elastic column subjected to axial compression. In the same sketch plot also the corresponding relationship for an elastic column with imperfect geometry. What is the major challenge in estimating the Euler buckling stress for a compression member that is part of a structural system?

b) Sketch the principle force-end shortening relationship (or alternatively force-mid point lateral displacement relationship) for imperfect, elasto-plastic columns subjected to axial compression, one curve for a slender column and one curve for a stocky (small slenderness) column. Explain the reason for the difference between the two curves

c) A brace member of a jacket has been subjected to a relatively small ship impact. The

brace cross-section is not damaged, but the brace has got an out-of-straightness (lateral deformation) of 10% of the brace diameter. This is significantly larger than tolerance limit for fabrication of the member, and conventional (Ultimate Limit State) code checking does not apply. On the basis of linear elastic analysis it is found that the brace, which has an Euler buckling stress of 430 MPa, is subjected to axial stress of 190 MPa in damaged condition. Formulate a buckling/failure criterion that can be used for the bent brace, and check whether the brace satisfies this criterion. As this is a damaged condition, all partial safety factors can be taken equal to unity. For simplicity the radius of gyration can be taken equal to i ≈ 0.35D , where D = brace diameter. The yield strength of the brace material is fy = 300 MPa.

d) For non-flooded braces located in deep water there will be two constant stress components that are not necessarily included in a linear analysis of load effects. Which are they? How do these stress components affect the column buckling capacity of slender braces? In what way are they accounted for in column buckling checks of stocky (small slenderness) braces according to the ISO/NORSOK code? (A brief discussion is requested; no calculations shall be carried out)

e) Describe the reasoning behind the ISO/NORSOK code formulation for checking of deep water braces subjected to combined axial force and bending


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PROBLEM 3

Figure 3

Figure 3 shows a plate girder in a platform deck. It is fabricated in steel with yield stress fy = 300 MPa. The elastic modulus is E = 2.1�105 MPa. The plate girder is subjected to a shear force V = 2MN. The girder has vertical stiffeners with a spacing of a = 2000 mm. a) Calculate the utilisation with respect to buckling of the web.

b) Buckling of the web may not necessarily mean that the capacity of the girder is fully

utilised. Describe briefly the possible redistribution of load carrying that may take place in the post-buckling range. What condition must be fulfilled for this to be realised? (So far weak flanges shall be assumed). Calculate the capacity of the girder on the basis of weak flanges (Basler´s method) .

c) To improve the capacity of the girder, the following measures have been proposed in

prioritized order: 1. Apply a horizontal stiffener in the middle of the web 2. Reduce the spacing of the vertical stiffeners to 1000 mm 3. Introduce a diagonal flat bar (100 x 15 mm) stiffener in 1-direction ( see

Figure 3) between the vertical stiffeners 4. Introduce a diagonal flat bar (100 x 15 mm) stiffener in 2-direction ( see

Figure 3) between the vertical stiffeners Do you agree with this prioritization? I you disagree, you may propose a new prioritized order based on simple considerations without detailed calculations.

d) If the flanges are strong enough, they will contribute by increasing the capacity with

respect to shear loading. Explain this with a simple sketch. The girder is subjected to simultaneous bending moment, which induces axial stresses of 100 MPa in the flanges. Will these stresses affect the shear capacity substantially? (State the reasoning behind your answer)


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Given information: Elastic section modulus

W = 4 4( 2 )

32D D t

Dπ ⎡ ⎤− −⎣ ⎦

Plastic section modulus

Z = 3 31 ( 2 )6D D t⎡ ⎤− −⎣ ⎦

Formulas related to buckling of beam-columns: Local buckling axial compression

2

1 0 0 412

1 047 0 274 0 412 1 382

1 382

cl

y

cl

y

cl ce

f . .ff . . . .f

f f .

= λ ≤

= − λ ≤ λ ≤

= ≤ λ

0 3xetf . Er

= , y

xe

ff

λ =

Column buckling

c2

y

f 0.9 for 1.34f

λλ

= >

21.0 0.28 for 1.34c

y

ff

λ λ= − ≤

( )2

2,y EE

f EIff k A

πλ = =l

Beam-columns - hydrostatic pressure included

0.52 2

1 1.01 1

my mya mz mz

a ach mh

Ey Ez

C Cf f

f f

σσ σσ σ

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟+ + =⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟− −⎢ ⎥⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

use Cmz = Cmy = 1.0


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Local buckling - hydrostatic pressure included

2 2

1.0my mza x

cl mhf fσ σσ σ ++ + =

The ECCS design curve

1

1

x iox Y

x

E

AwW

σσ σσσ

+ =−

BV = τYbt crτ

Yτ+ 3

21− crτ

Yτ⎛⎝⎜

⎞⎠⎟

ba

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Density of sea water: 1000 kg/m3 Acceleration of gravity: 10 m/s2

Maximum shear stress: Qrt

τπ

=

Stresses in closed cylinder under hydrostatic pressure:

2xpr prt t

= =θσ σ

Eσ =2π E

12 1− 2ν( )tb

⎛⎝⎜

⎞⎠⎟2

⋅ k λ = YσEσ

crσ =σY

1+ λ4

( )22

E 2

E t kb12 1

πτν

⎛ ⎞= ⋅⎜ ⎟− ⎝ ⎠ k = 5.34 + 4 b

a⎛⎝⎜

⎞⎠⎟2

λ = Yσ / 3

Eτ

crτ =

σY / 3

1+ λ4

Effective width:

ebb=

1.8β

−0.8

2β β ≥ 1

ebb= 1 β ≤ 1

β =bt

σY

E


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0.9 1.9 0.91e2

aa , a b

ααββ β

⎛ ⎞= + − =⎜ ⎟

⎝ ⎠

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⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡ −−−

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

θ

θ

φ

φφ

φφφ

φφφφ

B

B

A

A

3

2253

4223

22532253

B

B

A

A

w

w

4EI

6EI12EI

2EI6EI4EI

6EI 12EI 6EI 12EI

=

M

Q

M

Q

Stiffness matrix with stability functions ( φ-factors

Compression Tension

( )

φφφ

φφφ

φφφ

φβφ

ββφ

215

214

213

1

2

2

1

23

21

43

4131

=

+−=

+=

−=

=

1

tan

( )

φφφ

φφφ

φφφ

φβφ

ββφ

215

214

213

1

2

2

1

23

21

43

4131

=

+−=

+=

−=

⎭⎬⎫=

1

tanh

lEI

NNN

2 2

2

EE

ππβ == ,


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Stability functions

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5

f-val

ue

Axial force r = N/NE

f4

f2

f3

f5

f1

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