Design and strength assessment of a welded connection of a plane frame.

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Design and strength assessment of a welded connection of a plane frame
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Transcript of Design and strength assessment of a welded connection of a plane frame.

Design and strength assessment of a welded connection of a plane frame

Advanced Design for Mechanical System Pract04

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Structural connections

• Structural connections of a plane frame must be able to transfer

1) internal forces between beam and beam

2) internal forces between beam and column

3) reaction forces between column and ground

• These are typical permanent connections and can be riveted, bolted or welded

• The basic criterion in the design of connections include- assessment of their static strength and endurance- assessment of the right transfer of the internal forces

The welded connection at point B must be designed

Advanced Design for Mechanical System Pract04

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Reaction and Internal moments

•Reaction forces and internal moments can be evaluate:

1) using handbook formulae for a similar structure loaded with distributed load or concentrated load, and then applying the superposition principle.

2) applying the Principle of Virtual Work

3) by means FEM model of the frame

The suggestion is to evaluate reaction forces and internal moments by means of handbook formulae and to compare results with results obtained

by Principle of Virtual Work or FEM analysis

Advanced Design for Mechanical System Pract04

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Moment for distributed load: handbook formulae*

*Manuale for mechanical engineer, Hoepli edition 1994, (in italian).

Advanced Design for Mechanical System Pract04

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Moment for concentrated load: handbook formulae

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Principle of virtual work for frames

structure real on the nsdeformatio related theare ,,,

structureauxiliary on the forces internal theare ,,,

applied is load theepoint wher theof structure, real on the nt,displaceme theis

structureauxiliary on the load applied theis P

beameach of axis thealong coordinater curvilinea theis s

where

1,111

workinternal

1,beams all

111

workexternal

loads applied all1

dddd

MTMN

dMdTdMdNP

t

s

t

s s s

For plane frame Mt=0 and the deformations due to the axial and the shear forces are negligible, only the internal bending must be taken

into account

Advanced Design for Mechanical System Pract04

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The examined plane frame

A

B E

p

Q/2

l/2

h

E

p

Q/2

B

A

RE

ME

• The plane frame is symmetric only half of the frame have to be considered

The structure is two times hyperstatic

The internal moment M(x) on the real structure is

M(x)=M0(x)+RE*M1(x)+ME*M2(x)

RE and ME: hyperstatic unknown

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B

Internal moment M(x) on the real structure

M(x)=M0(x)+RE*M1(x)+ME*M2(x)

E

p

Q/2

B

A 1

E EB

1

Isostatic structure

A A

Auxiliary structure n.1

Auxiliarystructure n. 2

11

RA

MA 1*h

M0(x) M1(x) M2(x)

Advanced Design for Mechanical System Pract04

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PVW for the auxiliary structure n. 1

structure real on the rotations

structureauxiliary on the moments)( and

structure real on the Epoint ofnt displaceme0

structureauxiliary on the force applied1

where

)()(01

,1,1

0

2/

0

,1,1

EJ

(x)Md

EJ

(x)Md

xM(x)M

dxMdxM

BEBE

ABAB

BEAB

h l

BEBEABAB

Advanced Design for Mechanical System Pract04

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PVW for the auxiliary structure n. 2

structure real on the rotations

structureauxiliary on the moments)( and

structure real on the Epoint ofrotation 0

structureauxiliary on themoment applied1

where

)()(01

,2,2

0

2/

0

,2,2

EJ

(x)Md

EJ

(x)Md

xM(x)M

dxMdxM

BEBE

ABAB

BEAB

h l

BEBEABAB

Advanced Design for Mechanical System Pract04

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Hyperstatic unknown

h l

BEBEABAB

h l

BEBEABAB

dxMdxM

dxMdxM

0

2/

0

,2,2

0

2/

0

,1,1

)()(01

)()(01

• The system

allows the calculation of RE and ME

Advanced Design for Mechanical System Pract04

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FEM Analysis

20000 N/m

20000 N

Constrains:Point A U1=U2=UR3=0Point E U1=UR3=0

Deformed shapeModel

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Moment at nodes

31420 Nm

10370 Nm

Example of results

The same cross section IPE 330 has been used for the beam and for the column

Advanced Design for Mechanical System Pract04

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The cross section of the beam and of the column

• Cross sections can be choose on the basis of the bending moment only

• On each cross section act the bending moment due to the distributed load constant and the bending moment due to the concentrated load Q varying sinusoidally with time

xy

z

xy

z

Mb,p Mb,Qsint Maximum of Mb,p

and Mb,Qsint

E

tMMM Qppbtotb sin,,,

Advanced Design for Mechanical System Pract04

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Bending stress on the cross section at point E

• We consider the cross section at point E where both Mb,p and Mb,Qsint are maximum.

• The bending stresses result linearly varying with the distance from the neutral axis:

• and sinusoidally varying with time

yJ

tMy

J

M

a

xx

Qb

m

xx

pbtotb

sin,,

,

A

a

aAa

a

aa

A

a

a

max,,totb

max,,totb

t

m atotb,

Advanced Design for Mechanical System Pract04

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• It is maximum at the points that are most distant from the neutral axis of the section

• The condition

where is the safety factor and lim can be obtained from the Haigh diagram of the material

allows the calculation of Jxx of the beam section.

2

sin

2

max,

,

max,

,max,,

h

J

tMh

J

M

a

xx

Qb

m

xx

pbtotb

limmax,, totb

Advanced Design for Mechanical System Pract04

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Bending Haigh diagram

m

a

UTS

a,f

m,max

a,max

lim

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Design of the structural node

• The node between the column and the beam, realized with a double T section, must be designed in order to realize a clamped constrain.

• The aim is to transfer the boundary moment M1, from the transverse beam to the vertical column

M2

M1

M3

ht

hc

0321 MMM

Advanced Design for Mechanical System Pract04

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The welded connection

• The end of the horizontal beam, upper plate, lower plate and web are welded to the upper plate of the column

• The weld is a fillet weld type

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Moment transfer

• An overview of the deformations of the node is given in the figure, as result of a finite element analysis.

• The level of deformation, in absence of any reinforcement, is quite high, and not acceptable.

• In the double T sections, if subjected to flexural moment in the plane of the web, the axial forces that originate from the flexural moment are transmitted by the upper and lower plate.

• As a consequence the upper plate of the column receives the normal forces of the flexural moment from the transverse beam, and deflects, except close to the web.

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• From the previous considerations, it is intuitive that local reinforcements are needed, to correctly transfer the flexural moment to the upper and lower plate of the column.

Reinforcements

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• In the adopted solution, the node is considered as a group of four beams, plus a diagonal member, all hinged at their ends.

Adopted solution

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• Axial load on the upper and lower plate of the beam, transferred to the reinforcement results

• Axial load on the upper and lower plate of the column

• Let M be the moment to be transmitted to the column.

cc

tt

h

MS

h

MS

D

C

E

A

St

Sd

Sc

Sc

Sc

Sc• On the diagonal AD acts the force:

• If the contribution of the web of the column is take into account, by means of the coefficient :

c

tctd h

hhSS

22

122

c

tctd h

hhSS

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• The comparison between the reinforced node (a) and the one without reinforcement (b) allow to visualize their different behavior.

(a) (b)

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Verification of the beam-column welded joint

• In the following the verification of the welding is reported. Let the two profiles be a IPExxx for the beam and for the column.

• The reference sections of the fillet of the welding are place as shown in figure below.

• J is the moment of inertia of the resistant section of the welding• MB and TB are the bending moment and the shear that must be

transmitted by the welded joint

TB

MB

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• So that the reference stress results:

• The corresponding safety coefficients then results:

• At point A (top of the horizontal fillet) only T due to bending is present and he corresponding safety coefficients results:

• At the edges of the fillets welding along the web, where bending and shear are present, the stresses are

222*TLTT

LTT and

* LIMK

T

LIMK