Module4 plastic theory- rajesh sir

Structural Analysis - II

Plastic Analysis

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKN

Module IVModule IV

Plastic Theory

• Introduction-Plastic hinge concept-plastic section modulus-shape factor-redistribution of moments-collapse mechanism-

• Theorems of plastic analysis - Static/lower bound theorem; Kinematic/upper bound theorem-Plastic analysis of beams and portal frames b equilibrium and mechanism methodsportal frames by equilibrium and mechanism methods.

Plastic Analysis Why? What?• Behaviour beyond elastic limit?

Plastic Analysis - Why? What?• Behaviour beyond elastic limit?

• Plastic deformation collapse load• Plastic deformation - collapse load

• Safe load load factor• Safe load – load factor

• Design based on collapse (ultimate) load limit • Design based on collapse (ultimate) load – limit design

• Economic - Optimum use of material

MaterialsMaterials

• Elastic •Elastic-Perfectly plastic

σ Elastic limitElastic limitσ Elastic limit

Upper yield point

Apoint

Plastic es

s B CL i ld

rangest

re Lower yield point

strainO strainO

Idealised stress-strain curve of mild steel

strainO

Idealised stress-strain curve in plastic theory

• Elastic analysisElastic analysis- Material is in the elastic state

P f f d i l d - Performance of structures under service loads - Deformation increases with increasing load

• Plastic analysis• Plastic analysis– Material is in the plastic state– Performance of structures under

ultimate/collapse loads / p– Deformation/Curvature increases without an

increase in load.

AssumptionsAssumptions

• Plane sections remain plane in plastic condition

S l d l b h• Stress-strain relation is identical both in compression and tensionp

Process of yielding of a section

• Let M at a cross-section increases gradually.

• Within elastic limit, M = σ.Z• Z is section modulus, I/y

• Elastic limit – yield stresses reached M ZMy = σy.Z

• When moment is increased, yield spreads into inner When moment is increased, yield spreads into inner fibres. Remaining portion still elastic

Fi ll th ti ti i ld• Finally, the entire cross-section yields

σ yσ σ

Change in stress distribution during Change in stress distribution during yielding

yσ yσyσσ

yσ yσyσσy

Rectangular cross section

σy σyσy σy

σyσy

Inverted T section

Plastic hingePlastic hinge

h h l l ld d h• When the section is completely yielded, the section is fully plasticA f ll l b h l k h • A fully plastic section behaves like a hinge –Plastic hinge

Plastic hinge is defined as an yielded zone due Plastic hinge is defined as an yielded zone due to bending in a structural member, at which

large rotations can occur at a section at large rotations can occur at a section at constant plastic moment, MP

Mechanical hinge Plastic hingeReality ConceptReality Concept

Resists zero Resists a constant t Mmoment moment MP

Mechanical Hinge Plastic Hinge with M = 0Mechanical Hinge Plastic Hinge with MP= 0

• M – Moment corresponding to working load• M – Moment corresponding to working load

• My – Moment at which the section yieldsy

• MP – Moment at which entire section is under yield stress

Plastic momentPlastic moment• Moment at which the entire section is

under yield stressC T

c y t y

C TA Aσ σ=

=2c tAA A⇒ = =

•NA divides cross-section into 2 equal parts

A2 yAC T σ= =

Cyc 2 y

ZSi il σ2 yT σ=

•Couple due to Aσ

yZσSimilar toyσ

( )A y y Zσ σ⇒ + =•Couple due to2 yσ ( )

2 y c t y py y Zσ σ⇒ + =

Plastic modulusis the shape factor( )1pZ

Shape factor for various cross-sectionsb

Shape factor for various cross sections

Rectangular cross-section:d

Section modulus ( )3 212bdI bd( )( )

bdI bdZy d

2A bd d d bd⎛ ⎞( )2

2 2 4 4 4p c tA bd d d bdZ y y ⎛ ⎞= + = + =⎜ ⎟

⎝ ⎠Plastic modulus

ZShape factor = 1.5 pZZ

Circular sectionCircular section

( )2p c tAZ y y= +

d 2 32 28 3 3 6d d d dπ

π π⎛ ⎞⎛ ⎞= + =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠8 3 3 6π π⎝ ⎠⎝ ⎠

1 7pZS = =

( )4 364d dZπ π

= = 1.7SZ

= =2 32Z

Triangular section3bh⎛ ⎞

⎜ ⎟

Triangular section

2362 24

bhbhZ h

⎛ ⎞⎜ ⎟⎝ ⎠= = 2

2 243h

3CG axis h

( )2p c tAZ y y= + b

E l icy

S = 2.346Equal area axis

I sectionI section

10mm250mm

20mm20mm200mm

Mp = 259.6 kNmS = 1.132

Load factor

collapse load M Zσ

Load factor

P y Pcollapse load MLoad factorworking load M Z

Rectangular cross-section:2

4P y P ybdM Zσ σ= =

6 1 6ybd bdM Z

σσ σ= = =4y y

2 25yPM bd bdLFσ

σ⎛ ⎞ ⎛ ⎞

∴ = = ÷ =⎜ ⎟ ⎜ ⎟

6 1.5 6

2.254 1.5 6yLF

Mσ∴ = = ÷ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Factor of safetyFactor of safety

Yield LoadFactor of Safety =Working Load

YieldStressWorkingStress

( )= 1.5/1 5

=( )/ 1.5yσ

El ti A l i F t f S f tElastic Analysis - Factor of Safety

Plastic Analysis - Load Factor

Mechanisms of failure• A statically determinate beam will collapse if one plastic

hinge is developedhinge is developed

• Consider a simply supported beam with constant cross • Consider a simply supported beam with constant cross section loaded with a point load P at midspan

• If P is increased until a plastic hinge is developed at the point of maximum moment (just underneath P) an unstable t t ill b t d structure will be created.

• Any further increase in load will cause collapse• Any further increase in load will cause collapse

• For a statically indeterminate beam to collapse, more than one y pplastic hinge should be developed

• The plastic hinge will act as real hinge for further increase of load (until sufficient plastic hinges are developed for collapse )collapse.)

• As the load is increased, there is a redistribution of moment, , ,as the plastic hinge cannot carry any additional moment.

Beam mechanismsBeam mechanisms

D i b Determinate beams & frames: Collapse

f fi l i Simple beamafter first plastic hinge

Indeterminate beams & frames: More than one a es: Mo e t a o e plastic hingeto develop mechanismp

Fixed beam

l h d l h d fPlastic hinges develop at the ends first

Beam becomes a simple beamBeam becomes a simple beam

Plastic hinge develops at the centreg p

Beam collapses

Indeterminate beam: More than one plastic Mo e t a o e p ast c hinge to develop mechanism

Propped cantilever

l h d l h f d fPlastic hinge develops at the fixed support first

Beam becomes a simple beamBeam becomes a simple beam

Plastic hinge develops at the centreg p

Beam collapses

Panel mechanism/sway mechanism Panel mechanism/sway mechanism

Gable MechanismW

Gable Mechanism

Composite (combined) MechanismComposite (combined) Mechanism

- Combination of the above

Methods of Plastic Analysisy

• Static method or Equilibrium method - Lower bound: A load computed on the basis of an assumed - Lower bound: A load computed on the basis of an assumed

equilibrium BM diagram in which the moments are not greater than MP is always less than (or at the worst equal to) the true ultimate l d load.

• Kinematic method or Mechanism method or Virtual work • Kinematic method or Mechanism method or Virtual work method - Work performed by the external loads is equated to the internal

work absorbed by plastic hinges

Upper bound: A load computed on the basis of an assumed - Upper bound: A load computed on the basis of an assumed mechanism is always greater than (or at the best equal to) the true ultimate load.

• Collapse load (Wc): Minimum load at which p c)collapse will occur – Least value

• Fully plastic moment (MP): Maximum moment capacity for design – Highest valuecapacity for design – Highest value

Determination of collapse load

1. Simple beam

Determination of collapse load

1. Simple beam

Equilibrium method:

PW lM =

MPM 4 P

∴ =u l

Virtual work method:E IW W=

.22u PlW Mθ θ⎛ ⎞ =⎜ ⎟

⎝ ⎠

∴ =2θl θ2θ

2. Fixed beam with UDL2l

2. ,24CENTRE

w lM =

2.12ENDS CENTREw lM M>=

Hence plastic hinges will develop at the ends first.

MMPMB1MP

2.2 uw lM =uw

Equilibrium:

28PM =

16 PM∴

2θθ θ

Virtual work: E IW W=

θθ θ θ

⎛ ⎞+⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟

16 PMw∴ =( )22 22 2u Pw M θ θ θ⎛ ⎞ = + +⎜ ⎟⎜ ⎟

⎝ ⎠⎜ ⎟⎝ ⎠

⎝ ⎠

3. Fixed beam with point load3. Fixed beam with point load

2θθ θ

Virtual work:

( )2lW Mθ θ θ θ⎛ ⎞ = + +⎜ ⎟Equilibrium:

Virtual work:

( )22u PW Mθ θ θ θ= + +⎜ ⎟

⎝ ⎠ 24P ulM W=

∴ = 8 Pu

4. Fixed beam with eccentric point load4. Fixed beam with eccentric point load

Equilibrium:

2 P uabM Wl

2 PM lW∴ =MP

Virtual work:

1 2a bθ θ=uW

Virtual work:

θ θ+

1θa b

2θ1 2

θ θ⇒ =1 2θ θ+ a

⎡ ⎤( ) ( )1 1 1 2 2u PW a Mθ θ θ θ θ= + + +⎡ ⎤⎣ ⎦

b⎡ ⎤( )2 2 22 2u PbW b Ma

θ θ θ⎡ ⎤= +⎢ ⎥⎣ ⎦

( )2 2

22 2P P

uM Mb a bWb aba

θ θθ

+⎡ ⎤∴ = =+⎢ ⎥⎣ ⎦

M lWab

5. Propped cantilever with point load at midspanmidspan

Vi t l kVirtual work:

E IW W=Equilibrium:

E IW W

( ) ( )2lW Mθ θ θ⎛ ⎞ = +⎜ ⎟

.0.54u

P PW lM M+ =

( ) ( )22u PW Mθ θ θ= +⎜ ⎟

⎝ ⎠4

6 PM6 P

∴ =6 P

6. Propped cantilever with UDL

2wl Maximum positive BM

MPAt collapse

Required to locate E

Dept. of CE, GCE Kannur Dr.RajeshKN42

2l ⎛ ⎞22 2 2

2 2u u

E P Pw lx w x xM M M

l⎛ ⎞= − − =⎜ ⎟⎝ ⎠

For maximum, 0EdM=

2 02u P

uw l Mw x

l− − = ( )2

2 0.414x l=From (1) and (2),

211.656 Pu

=From (2),

Problem 1: For the beam, determine the design plastic moment icapacity.

50 kN 75 kN

1.5 m 1.5 m

D f d 3 2 1• Degree of Indeterminacy, N = 3 – 2 = 1• No. of hinges, n = 3• No. of independent mechanisms ,r = n - N = 2

50 kN 75 kN

1 5 m 4 5 m

50 kN 75 kN

1 5 m1.5 mθ θ1

4.5 m 1.5 m

Mechanism 1θ + θ1

1 5 6θ θ1.5θ θ⇒ =11.5 6θ θ=

( ) 1.5 1.550 1.5 75 1.5 Mθ θ θ θ θ⎛ ⎞ ⎛ ⎞+ × = + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 6θ θ⇒ =

( )50 1.5 75 1.56 6pMθ θ θ θ θ+ + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

45 83M∴ =

45.83pM∴

1.5 m 1.5 m

Mechanism 2θ + θ1

16 1.5θ θ=

θ θ⇒ =

( )1 1 1 1 11.5 1.5 1.550 1.5 75 1.56 6 6pMθ θ θ θ θ⎛ ⎞ ⎛ ⎞× + = + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠6 6 6⎝ ⎠ ⎝ ⎠

87.5pM kNm∴ =

87.5 kNm=Design plastic moment (Highest of the above)

g p ( g )

Problem 2: A beam of span 6 m is to be designed for an ultimate UDLf 25 kN/ Th b i i l t d t th d D iof 25 kN/m. The beam is simply supported at the ends. Design a

suitable I section using plastic theory, assuming σy= 250 MPa.

25 kN/m25 kN/m

• Degree of Indeterminacy, N = 2 – 2 = 0• No. of hinges, n = 1• No. of independent mechanisms, r = n-N = 1

Mechanism 25 kN/m

3 m2θ

3 m 3 m

Internal work done 0 2 0 2IW M Mθ θ= + × + =te a o k do e 0 2 0 2I p pW M Mθ θ+ +

External work done 0 32 25 2253EW θ θ+⎛ ⎞= × × =×⎜ ⎟⎝ ⎠2E ⎜ ⎟⎝ ⎠

2 225I EW W M θ θ= ⇒ = 112.5pM kNm∴ =2 225I E pW W M θ θ⇒ p

Plastic modulus PP

MZ =6112.5 10×

= 5 34.5 10 mm= ×Plastic modulus Py

Zσ 250

Z54.5 10× 5 33 913 10PZZ

.5 01.15

= 5 33.913 10 mm= ×Section modulus

Assuming shape factor S = 1.15

Adopt ISLB 275@330 N/m (from Steel Tables – SP 6)

Adopt ISLB 275@330 N/m (from Steel Tables SP 6)

Problem 3: Find the collapse load for the frame shownProblem 3: Find the collapse load for the frame shown.

/ 2 / 2B C

W/22Mp

• Degree of Indeterminacy, N = 5 – 3 = 2

• No. of hinges, n = 5 (at A, B, C, E & F)

• No. of independent mechanisms ,r = n - N = 3

• Beam Mechanisms for members AB & BC• Beam Mechanisms for members AB & BC

• Panel Mechanism Panel Mechanism

Beam Mechanism for AB

2 2 (2 ) 7W M M M Mθ θ θ θ+ +B M

Beam Mechanism for AB

2 2 (2 ) 7I p p p pW M M M Mθ θ θ θ= + + =

2 2EWW θ=

/ 2 θ2θ

28 pE I c

MW W W= ⇒ =

Beam Mechanism for BCW

F/ 2 / 2BC

θ θMp

(2 ) 4I p p p pW M M M Mθ θ θ θ= + + =

2EW W θ=

8 pE I c

MW W W= ⇒ =

Panel Mechanism2 4I p p p pW M M M Mθ θ θ θ= + + =

Panel Mechanism

/ 2Mp Mp/ 2 2 2EWW θ=

16 pE I c

MW W W= ⇒ =

W/2 2θ

WCombined Mechanism

2 ( ) (2 )

( )I p pW M M

/ 2 Mp/ 2θ θ ( )

32 2 2 4E

WW W Wθ θ θ= + =W/2 2

E2 2 2 4

8 pE I c

MW W W= ⇒ =2Mp

TrueCollapse Load, , pc

MWLowest of the above =

Problem 4: A portal frame is loaded upto collapse. Findthe plastic moment capacity required if the frame is ofuniform section throughout.

10 kN/m

25 kNB C

Mp4 m Mp

• Degree of Indeterminacy, N = 4 – 3 = 1

• No. of possible plastic hinges, n = 3 (at B, C and between B&C)

• No. of independent mechanisms ,r = n - N = 2

• Beam Mechanism for BC

• Panel Mechanism

Beam Mechanism for BC B C

10 kN/m

Cθ θ

4θMpMp

0 4θ+⎛ ⎞ 2θMp

0 42 10 16042EW θ θ+⎛ ⎞= × × =×⎜ ⎟

⎝ ⎠

( )2 4I p pW M Mθ θ θ θ= + + =

40pM kNm∴ =

Panel Mechanism 25 kN4θPanel Mechanism 4θ

E IW W=

( ) 25 4pM θ θ θ⇒ + = ×

50pM kNm⇒ =

Combined Mechanism 10kN/m

4θ 8x θ25kN 4θ 8 x−

θ+ θ1

θIt is required to locate the plastic hinge between B & C

Assume plastic hinge is formed at x from B

( ) 18x xθ θ= −( ) 1

( )8θ θ⎛ ⎞⎛ ⎞ ( ) ( ) 1825 4 10 10 82 2Ex xW x xθ θθ −⎛ ⎞⎛ ⎞= × + × + × − ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) 2 xW M Mθ θ θ θ θ θ⎡ ⎤= + + + = +⎡ ⎤⎣ ⎦ ⎢ ⎥

( )( )5 5 2 8x x+

( )1 1 82I p p x

W M Mθ θ θ θ θ θ= + + + = +⎡ ⎤⎣ ⎦ ⎢⎣ − ⎥⎦

( )( )5 5 2 84E I p

x xW W M

+ −= ⇒ =

For maximum, 0PdMdx

2.75x m⇒ =

( )( )5 5 2 868.91

x xM kNm

+ −∴ = =

( )Design plastic moment of resistance, ,largest of the ab 68.o 91ve pM kNm=

Problem 5: Determine the Collapse load of the continuous beam.Problem 5: Determine the Collapse load of the continuous beam.P

/ 2 / 2P

A B C/ 2 / 2A B CD E

4 2 2SI = − =A collapse can happen in two ways:

1 D t hi d l i t A B d D1. Due to hinges developing at A, B and D

2. Due to hinges developing at B and E

Equilibrium:

Hinges at A, B and D

Equilibrium:

pup p u

MP M M P= + ⇒ =

Moment at E is greater than Mp. Hence this mechanism is not possible.

Hinges at B and EHinges at B and E

pMpM p

6p pu M MP M P= + ⇒ =4 2p uM P= + ⇒ =

True Collapse Load,6 p

P PVirtual work:

/ 2 / 2A B CD E

4 2 2SI = − =θ

2θHinges at A, B and D

pu p u

MP M Pθ θ θ θ⎛ ⎞ = + + ⇒ =⎜ ⎟⎝ ⎠ Hinges at B and E

pu p u

MP M Pθ θ θ⎛ ⎞ = + ⇒ =⎜ ⎟⎝ ⎠

Problem 6: For the cantilever, determine the collapse load.

A L/2 L/2

2Mp Mp

• Degree of Indeterminacy N = 0

Degree of Indeterminacy, N 0

• No of possible plastic hinges n = 2 (at A&B)No. of possible plastic hinges, n 2 (at A&B)

• No of independent mechanisms r = n - N = 2• No. of independent mechanisms ,r = n - N = 2

/2 L/2Wu

L/2θ Mechanism 1

MpLθ/2

LW Mθ θ×2 pM

W∴ =2u pW Mθ θ× = uW

L∴ =

Lθ Mechanism 2

2Mp Lθ2Mp

2 pM2u pW L Mθ θ× = p

T C ll L d pMWL t f th b

( )TrueCollapse Load, , pcWLowest of the above

Problem 7: A beam of rectangular section b x d is subjected to a bending moment of 0.9 Mp. Find out the depth of elastic core.

Let the elastic core be of depth 2y0

02yExternal bending moment must be resisted by the internal couple.

yσDistance of CG from NA,

0 0 0 0 01 2

2 2 2 2 3y

yd db y y y by y

σσ ⎡ ⎤⎛ ⎞ ⎛ ⎞− × × + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦′

2 203 4d y−

0 02 2y

ydb y by

⎣ ⎦=⎛ ⎞− +⎜ ⎟⎝ ⎠

I t l l ( t f i t ) 2 2

03 42 yd d yb y byσ

σ⎧ ⎫ −⎛ ⎞= × + ×⎨ ⎬⎜ ⎟

Internal couple (moment of resistance)

( )0 00

22 2 12yb y by

d yσ= × − + ×⎨ ⎬⎜ ⎟ −⎝ ⎠⎩ ⎭

2 23 4d y03 412 y

d y bσ−=

2bdExternal bending moment = 0.9 0.9 0.9

4p p y ybdM Z σ σ= × = ×

2 2 23 4d bdEquating the above,

2 2 203 4 0.9

12 4y yd y bdbσ σ−

0 0.274y d⇒ =

02 0.548y d= =Hence, depth of elastic core

02 0.548y dHence, depth of elastic core

SummarySummary

Plastic Theory

• Introduction-Plastic hinge concept-plastic section modulus-shape factor-redistribution of moments-collapse mechanism-

• Theorems of plastic analysis - Static/lower bound theorem; Kinematic/upper bound theorem-Plastic analysis of beams and portal frames b equilibrium and mechanism methodsportal frames by equilibrium and mechanism methods.

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