Steel Structures Details - Column Problems

8/2/2019 Steel Structures Details - Column Problems

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Steel Structures

N-W.F.P University of Engineering &

Technology Peshawar

CE - 411

Instructor: Prof. Dr. Shahzad Rahman

Lecturer: Engr. Kamran Ayub


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Example Problem 1


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Example Problem 1(Contd.)


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Example Problem 1(Contd.)


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Example Problem 2 - LRFD


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Example Problem 2 (Contd.)


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Example Problem 3


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KL is called effective length of column and K

effective length factor.


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Effective length factors (AISC manual)


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EFFECTIVE LENGTH OF

COLUMNS IN FRAMES So far, we have looked at the buckling strength of

individual columns. These columns had various

boundary conditions at the ends, but they werenot connected to other members with moment

(fix) connections.

The effective length factor K for the buckling of an

individual column can be obtained for theappropriate end conditions from Table C-C2.1 of

the AISC Manual


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Columns in Frames

However, when these individual columns

are part of a frame, their ends are

connected to other members (beams etc.).

These frames are sometimes braced and

sometimes unbraced.


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A Braced frame is one in which a sideway(joint translation) is prevented by means of

bracing, shear walls, or lateral support fromadjoining structure.


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A UnBraced does not have any bracingand must depend on stiffness of its ownmembers and rotational rigidity of joints

between frame members to prevent lateralbuckling.


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Conclusions

Critical loads for a column depends on

its stiffness relative to that of beamsframing into it and

Presence or absence of restraint to

lateral displacement of its ends.

Effective length coefficient increases with

decreasing stiffness of the beam and

becomes unity with zero stiffness.


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Braced & Unbraced Frames Similarly you can analyze

multibay, multistory frames.

Assumptions Subjected to vertical loads only

All columns become unstable

simultaneously

All joint rotations at floor are equal Restraining moment distributed in

proportion to stiffnesses.


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Method of Analysis First, you have to determine whether the

column is part of a braced frame or an

unbraced (moment resisting) frame. Then, you have to determine the relative

rigidity factor G for both ends of the column


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G is defined as the ratio of the summationof the rigidity (EI/L) of all columns comingtogether at an end to the summation of therigidity (EI/L) of all beams coming together

at the same end.

G =

It must be calculated for both ends of the

column.

b

b

c

c

L

IE

L

IE


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Then, you can determine the effective lengthfactor K for the column using the calculated value

of G at both ends, i.e., GA

and GB

and the

appropriate alignment chart.

There are two alignment charts provided by the

AISC manual,

One is for columns in braced (sidesway inhibited)

frames. See Figure C-C2.2a on page 16.1-191 ofthe AISC manual. 0 < K 1

The second is for columns in unbraced (sidesway

uninhibited) frames. See Figure C-C2.2b on page

16.1-192 of the AISC manual. 1 < K


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Inelastic Case

G is a measure of the relative flexural rigidityofthe columns (EI

c/L

c) with respect to the beams

(EIb/Lb) However, if column buckling were to occur in the

inelastic range (c

< 1.5), then the flexural rigidity

of the column will be reduced because Ic

will be

the moment of inertia of only the elastic core ofthe entire cross-section.


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rc = 10 ksi

rt = 5 ksi

rt = 5 ksi

rt = 5 ksi

rc = 10 ksi

(a) Initia l state residual st ress (b) Partially y ielded sta te at buckling

Yielded zone

Elastic core, Ic

rc = 10 ksi

rt = 5 ksi

rt = 5 ksi

rt = 5 ksi

rc = 10 ksi

rc = 10 ksi

rt = 5 ksi

rt = 5 ksi

rt = 5 ksi

rc = 10 ksi

(a) Initia l state residual st ress (b) Partially y ielded sta te at buckling

Yielded zone

Elastic core, Ic

Yielded zone

Elastic core, Ic


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Inelastic stiffness

The beams will have greater flexural rigidity

when compared with the reduced rigidity

(EIc) of the inelastic columns. As a result,

the beams will be able to restrain the

columns better, which is good for column

design.


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Inelastic Stiffness

The ratio Fcr/ Fe is called

Stiffness reduction factor

( )22

658.0877.0877.0

)658.0(

,

, 2

2

cc c

Fy

c

Fy

elasticFcr

inelasticFcr

=

=


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Procedure for Column Design

1. Design Load

2. Assume Fcr

Pn = Ag Fcr= Pu

Find Ag

Select a section

3. Find

1. Find

c = E

F

r

LK y


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Column Design

Forc

1.5 Fcr

= Fy

Forc > 1.5 Fcr = Fy

1. Fcr Calculated > Fcr Assumed

2. P > Pu. Check3. IfP > Pu, Design is OK Else Repeat

Steps 2 to 6

2c658.0

2c

877.0


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Column Design Using Design Aid

LRFD manual contains variety of Design aids,helpful in making original trial section.

1. Design load

2. Find section for Corresponding P & KL using

Column Tables in the Manual

3. Calculate an equivalent (KL)eq =

if strong axis buckling needs

checking1. Use the calculated (KL)eq value to find (cPn)

the column strength from column tables

yx

xx

r/r

LK


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Problem 4-11-1


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Solution


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Steel Structures Details - Column Problems

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