Presentation Lecture 1

7/28/2019 Presentation Lecture 1

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Plastic Design Lecture 1

Associate Professor Bill Wong

Department of Civil Engineering


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Introduction The plastic method usually results in a more

economical design, especially for structures with high

degree of indeterminacy.

The first structure designed plastically in the U.S. wasin 1957.

Examples are .......


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Plastic design example First high-rise building designed by plastic

method:Bladensburg, Maryland, USA


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Introduction Material behaviour for a beam

Maximum moment at mid-span

Stress variation as load increases

Compression

Tension

Plastic

neutral axis

fy

fy

fe fy

fe fy

fy

fy

Elastic Elastic-plastic Plastic

Mp

= Ms

curvature


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Plastic analysis for indeterminate

structures When the bending moment at a section reaches its

plastic moment Mp (=Ms), the section behaves like

a hinge. This section is called a plastic hinge (within avery small length).

As load increases, more plastic hinges occur until the

structure collapses. For design, Design load = Collapse load.

Load

Collapse

load

Cross-section in fully plastic state

Collapse

Deflection

Slope Stiffness of structureElastic

state

Elastic-plastic state

(reserve strength)


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Plastic vs. elastic design

Elastic design: the first plastic hinge should occurat or above the design load level

Plastic design: the last plastic hinge should occurat or above the design load level

Load Design

load

Plastic design

Elastic design

Deflection

Reserve

strength


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Limitations when using plastic method Sections showing no sign of local buckling: use

compact sections - Mp =Ms =Zefyand Ze = lesser

of S and 1.5Z

Ductile enough to undergo plastic rotation

Hot-formed, doubly-symmetric I-sections with fy 450 MPa

No fatigue requirements e

fy

0.15

0.2 fy

y 6e


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Limitations when using plastic method No lateral-torsional buckling:

Ms

=Mb

=s

m

Ms

and s

m

1.0

Provide adequate lateral restraints:

L

r fym

y +( )80 50

250


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Plastic bending:

Local buckling:


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Spreadsheet method for plastic analysis Example

100 kN

6m 6m

AB

C

Mp =

270 kNm

225 kNm 187.5 kNm

AB

Mp =

270 kNm

225 kNm

A B

Cup filling analogy:

Stage 1:

For P = 100 kN,

MA

= 3PL/16 = 225 kNm,

MB = 5PL/32 = 187.5 kNm

For A, A

= 270/225 = 1.2

For B, B

= 270/187.5 = 1.44

load factor1 = 1.2, hinge formed at A.

Total MA

= 1.2x225 = 270 kNm,

Total MB

= 1.2x187.5 = 225 kNm.


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Spreadsheet method for plastic analysis Stage 2:

Total c =1+2 =1.2 + 0.15 = 1.35

Collapse load Pw =cP = 1.35x100 = 135 kN

For P = 100 kN,

MA

= 0, MB

= PL/4 = 300 kNm

For B, remaining plastic moment capacity

= 270-225 = 45kNmload factor

2= 45/300 = 0.15,

Total MA

= 270 kNm,

Total MB = 225 + 0.15x300 = 270 kNm.

Mp =

270 kNm

225 kNm

A B

45 kNm

Fig. 8.8


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Spreadsheet implementation


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Notes to computer analysis1. Perform a linear analysis by computer for the

structure subject to original loading.

2. Scale the loading and calculate the load factor foreach member; the smallest is the critical cr.

3. Calculate the residual plastic moments for all othersections.

4. Insert hinge at the section with cr and repeat theabove (1) to (4) until collapse.

5. Theory: Determinant of [K] = 0. When using

computer: Run time error due to zero determinant,or dramatic increase in displacements.

6. Collapse load = design load X . cr

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Presentation Lecture 1