Module4 plastic theory- rajesh sir
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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
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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.
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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
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MaterialsMaterials
• Elastic •Elastic-Perfectly plastic
σ Elastic limitElastic limitσ Elastic limit
εε
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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
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stre
sss
strainO
Idealised stress-strain curve in plastic theory
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• 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.
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increase in load.
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AssumptionsAssumptions
• Plane sections remain plane in plastic condition
S l d l b h• Stress-strain relation is identical both in compression and tensionp
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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
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yσσ
σ yσ σ
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Change in stress distribution during Change in stress distribution during yielding
yσ yσyσσ
yσ yσyσσy
Rectangular cross section
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σy σyσy σy
σyσy
Inverted T section
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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
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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
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• 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
yσ
CC
T
yσMP
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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
2
A2 yAC T σ= =
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yσ
Cyc 2 y
Aσ=
yt
yc
AT
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
>
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p( )Z
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Shape factor for various cross-sectionsb
Shape factor for various cross sections
Rectangular cross-section:d
Section modulus ( )3 212bdI bd( )( )
122 6
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
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Circular sectionCircular section
A
d
( )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
d
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Triangular section3bh⎛ ⎞
⎜ ⎟
Triangular section
2362 24
bhbhZ h
⎛ ⎞⎜ ⎟⎝ ⎠= = 2
3h
2 243h
A
3CG axis h
( )2p c tAZ y y= + b
E l icy
S = 2.346Equal area axis
ty
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I sectionI section
20mm
10mm250mm
20mm20mm200mm
Mp = 259.6 kNmS = 1.132
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Load factor
collapse load M Zσ
Load factor
P y Pcollapse load MLoad factorworking load M Z
Zσσ
= = =
Rectangular cross-section:2
4P y P ybdM Zσ σ= =
2 2
6 1 6ybd bdM Z
σσ σ= = =4y y
2 2
2 25yPM bd bdLFσ
σ⎛ ⎞ ⎛ ⎞
∴ = = ÷ =⎜ ⎟ ⎜ ⎟
6 1.5 6
2.254 1.5 6yLF
Mσ∴ = = ÷ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
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Factor of safetyFactor of safety
Yield LoadFactor of Safety =Working Load
yWW
=
YieldStressWorkingStress
yσσ
= =g
( )= 1.5/1 5
yσσ
=( )/ 1.5yσ
El ti A l i F t f S f tElastic Analysis - Factor of Safety
Plastic Analysis - Load Factor
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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
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• 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.
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Beam mechanismsBeam mechanisms
D i b Determinate beams & frames: Collapse
f fi l i Simple beamafter first plastic hinge
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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
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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
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Panel mechanism/sway mechanism Panel mechanism/sway mechanism
W
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Gable MechanismW
Gable Mechanism
Composite (combined) MechanismComposite (combined) Mechanism
- Combination of the above
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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.
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• 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
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Determination of collapse load
1. Simple beam
Determination of collapse load
1. Simple beam
Equilibrium method:
W l
Equilibrium method:
.4u
PW lM =
MPM 4 P
uMWl
∴ =u l
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Virtual work method:E IW W=
.22u PlW Mθ θ⎛ ⎞ =⎜ ⎟
⎝ ⎠
uW4 P
uMWl
∴ =2θl θ2θ
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2. Fixed beam with UDL2l
2
2. ,24CENTRE
w lM =
2.12ENDS CENTREw lM M>=
Hence plastic hinges will develop at the ends first.
MP
M
MC1
MB1
MC2
MMPMB1MP
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2.2 uw lM =uw
Equilibrium:
28PM =
16 PM∴
2θθ θ
2P
uwl
∴ =
Virtual work: E IW W=
( )0
22 2
ll M
θθ θ θ
⎛ ⎞+⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟
16 PMw∴ =( )22 22 2u Pw M θ θ θ⎛ ⎞ = + +⎜ ⎟⎜ ⎟
⎝ ⎠⎜ ⎟⎝ ⎠
2uwl
∴ =
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⎝ ⎠
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3. Fixed beam with point load3. Fixed beam with point load
uW
2θθ θ
uMP
MP2θ
Virtual work:
( )2lW Mθ θ θ θ⎛ ⎞ = + +⎜ ⎟Equilibrium:
Virtual work:
( )22u PW Mθ θ θ θ= + +⎜ ⎟
⎝ ⎠ 24P ulM W=
8 Pu
MWl
∴ = 8 Pu
MWl
∴ =
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4. Fixed beam with eccentric point load4. Fixed beam with eccentric point load
uW
Equilibrium:
u
a b
2 P uabM Wl
=
q
MP
l
2 PM lW∴ =MP
uWab
∴ =
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Virtual work:
1 2a bθ θ=uW
Virtual work:
θ θ+
1θa b
2θ1 2
ba
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
2
22 2P P
uM Mb a bWb aba
θ θθ
+⎡ ⎤∴ = =+⎢ ⎥⎣ ⎦
2 Pu
M lWab
=
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5. Propped cantilever with point load at midspanmidspan
MMC2
MP
MP
MP
MC1
MB1
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uW
2θ
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
uMWl
∴ =6 P
uMWl
∴ =
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6. Propped cantilever with UDL
2wl Maximum positive BM
8wl p
x1
MMP
MPAt collapse
x2
E
Required to locate E
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2l ⎛ ⎞22 2 2
2 2u u
E P Pw lx w x xM M M
l⎛ ⎞= − − =⎜ ⎟⎝ ⎠
( )1
For maximum, 0EdM=
2
0dx
=
2 02u P
uw l Mw x
l− − = ( )2
2 0.414x l=From (1) and (2),
211.656 Pu
Mwl
=From (2),
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Problem 1: For the beam, determine the design plastic moment icapacity.
50 kN 75 kN
1.5 m 1.5 m
7.5 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
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50 kN 75 kN
1.5 m
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∴ =
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50 75
1.5 m 1.5 m
θθ1
4.5 m
θ1
Mechanism 2θ + θ1
16 1.5θ θ=
11.56
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)
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g p ( g )
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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
66 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
θ θ
3 m2θ
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3 m 3 m
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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
S=
.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)
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Adopt ISLB 275@330 N/m (from Steel Tables SP 6)
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Problem 3: Find the collapse load for the frame shownProblem 3: Find the collapse load for the frame shown.
W
F
/ 2 / 2B C
/ 2
W/2
Mp
E
2Mp
/ 2
W/22Mp
/ 2
DAA
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• 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
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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
B
θ
Mp
2 2EWW θ=
W/2E
/ 2 θ2θ
2Mp
28 pE I c
MW W W= ⇒ =
W/2
/ 2
p
θ
2Mp
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Beam Mechanism for BCW
F/ 2 / 2BC
θ θMp
θ θ
2θ
2θMp
Mp
2θMp
(2 ) 4I p p p pW M M M Mθ θ θ θ= + + =
2EW W θ=
8 pE I c
MW W W= ⇒ =
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Panel Mechanism2 4I p p p pW M M M Mθ θ θ θ= + + =
Panel Mechanism
W
/ 2Mp Mp/ 2 2 2EWW θ=
F
/ 2θ
θ
16 pE I c
MW W W= ⇒ =
W/2 2θ
E
/ 2
E
2Mp
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WCombined Mechanism
2 ( ) (2 )
( )I p pW M M
M
θ θ
θ θ
= +
+ +
W
/ 2 Mp/ 2θ θ ( )
6p
p
M
M
θ θ
θ
+ +
=/ 2
Mp
θθ
2θ
2θ
θ θ
32 2 2 4E
WW W Wθ θ θ= + =W/2 2
θ
p
E2 2 2 4
8M/ 2
8 pE I c
MW W W= ⇒ =2Mp
( )8
TrueCollapse Load, , pc
MWLowest of the above =
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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
25 kN
Mp
8m
Mp4 m Mp
DA
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• 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
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Beam Mechanism for BC B C
10 kN/m
Cθ θ
2θ
4θMpMp
0 4θ+⎛ ⎞ 2θMp
0 42 10 16042EW θ θ+⎛ ⎞= × × =×⎜ ⎟
⎝ ⎠
( )2 4I p pW M Mθ θ θ θ= + + =
40pM kNm∴ =
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Panel Mechanism 25 kN4θPanel Mechanism 4θ
Mp Mp
θ θ
E IW W=
( ) 25 4pM θ θ θ⇒ + = ×
50pM kNm⇒ =
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Combined Mechanism 10kN/m
4θ 8x θ25kN 4θ 8 x−
Mθ
Mp
xxθ
θ+ θ1
θθ1
Mp
4m
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 −⎛ ⎞⎛ ⎞= × + × + × − ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
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( ) 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
4p
x xM kNm
+ −∴ = =
( )Design plastic moment of resistance, ,largest of the ab 68.o 91ve pM kNm=
4
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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
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Equilibrium:
Hinges at A, B and D
Equilibrium:
pM>M
pMpM
uP
pM
PE
8MP
4u
4uP
84
pup p u
MP M M P= + ⇒ =
Moment at E is greater than Mp. Hence this mechanism is not possible.
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Hinges at B and EHinges at B and E
M pM
pMpM p
4uP
4uP
6p pu M MP M P= + ⇒ =4 2p uM P= + ⇒ =
True Collapse Load,6 p
u
MP =
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P PVirtual work:
/ 2 / 2A B CD E
θ θ
4 2 2SI = − =θ
2θ
θ θ
2θHinges at A, B and D
( )8
22
pu p u
MP M Pθ θ θ θ⎛ ⎞ = + + ⇒ =⎜ ⎟⎝ ⎠ Hinges at B and E
( )6
22
pu p u
MP M Pθ θ θ⎛ ⎞ = + ⇒ =⎜ ⎟⎝ ⎠
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Problem 6: For the cantilever, determine the collapse load.
W
A L/2 L/2
2Mp Mp
A
BC
• Degree of Indeterminacy N = 0
p
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
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/2 L/2Wu
L/2θ Mechanism 1
L/2
MpLθ/2
LW Mθ θ×2 pM
W∴ =2u pW Mθ θ× = uW
L∴ =
Wu
Lθ Mechanism 2
2Mp Lθ2Mp
2 pM2u pW L Mθ θ× = p
uWL
∴ =
( )2
T C ll L d pMWL t f th b
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( )TrueCollapse Load, , pcWLowest of the above
L=
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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.
yσ
Let the elastic core be of depth 2y0
02yExternal bending moment must be resisted by the internal couple.
yσDistance of CG from NA,
y
0 0 0 0 01 2
2 2 2 2 3y
yd db y y y by y
y
σσ ⎡ ⎤⎛ ⎞ ⎛ ⎞− × × + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦′
2 203 4d y−
0 02 2y
y
ydb y by
σσ
⎣ ⎦=⎛ ⎞− +⎜ ⎟⎝ ⎠
( )0
012y
d y=
−
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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
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02 0.548y dHence, depth of elastic core
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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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