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Structural Analysis - II
Plastic Analysis
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in ...
Module IVModule IV
Plastic Theory
• Introduction-Plastic hinge concept-plastic section modulus-shape
factor-redistribution...
Plastic Analysis Why? What?
• Behaviour beyond elastic limit?
Plastic Analysis - Why? What?
• Behaviour beyond elastic lim...
MaterialsMaterials
• Elastic •Elastic-Perfectly plastic
σ Elastic limitElastic limitσ Elastic limit
εε
Dept. of CE, GCE Ka...
Upper yield
point
A
point
Plasticess
B C
L i ld
range
stre
Lower yield
point
strainO strainO
Idealised stress-strain curve...
stresss
strainO
Idealised stress-strain curve in plastic theory
Dept. of CE, GCE Kannur Dr.RajeshKN
6
• Elastic analysisElastic analysis
- Material is in the elastic state
P f f d i l d- Performance of structures under servi...
AssumptionsAssumptions
• Plane sections remain plane in plastic
condition
S l d l b h• Stress-strain relation is identical...
Process of yielding of a section
• Let M at a cross-section increases gradually.
• Within elastic limit, M = σ.Z
• Z is se...
yσσ
σ yσ σ
Dept. of CE, GCE Kannur Dr.RajeshKN
10
Change in stress distribution duringChange in stress distribution during
yielding
yσ yσ yσσ
yσ yσ yσσ
y
Rectangular cross ...
σy σy
σy σy
σy
σy
Inverted T section
Dept. of CE, GCE Kannur Dr.RajeshKN
12
Plastic hingePlastic hinge
h h l l ld d h• When the section is completely yielded, the
section is fully plastic
A f ll l b...
Mechanical hinge Plastic hinge
Reality ConceptReality Concept
Resists zero Resists a constant
t Mmoment moment MP
Mechanic...
• M – Moment corresponding to working load• M – Moment corresponding to working load
• My – Moment at which the section yi...
Plastic momentPlastic moment
• Moment at which the entire section is
under yield stress
C T
c y t y
C T
A Aσ σ
=
=
2
c t
A...
yσ
C
yc
2
y
A
σ=
yt
yc
A
T
ZSi il
σ
2
yT σ=
•Couple due to
A
σ
yZσSimilar toyσ
( )
A
y y Zσ σ⇒ + =•Couple due to
2
yσ ( )
...
Shape factor for various cross-sections
b
Shape factor for various cross sections
Rectangular cross-section:
d
Section mod...
Circular sectionCircular section
A
d
( )
2
p c t
A
Z y y= +
d 2 3
2 2
8 3 3 6
d d d dπ
π π
⎛ ⎞⎛ ⎞
= + =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠8 3 3 ...
Triangular section
3
bh⎛ ⎞
⎜ ⎟
Triangular section
2
36
2 24
bh
bh
Z
h
⎛ ⎞
⎜ ⎟
⎝ ⎠= = 2
3
h
2 24
3
h
A
3
CG axis h
( )
2
p ...
I sectionI section
20mm
10mm250mm
20mm20mm
200mm
Mp = 259.6 kNmS = 1.132
Dept. of CE, GCE Kannur Dr.RajeshKN
21
Load factor
collapse load M Zσ
Load factor
P y Pcollapse load M
Load factor
working load M Z
Zσ
σ
= = =
Rectangular cross-...
Factor of safetyFactor of safety
Yield Load
Factor of Safety=
WorkingLoad
yW
W
=
YieldStress
WorkingStress
yσ
σ
= =
g
( )
...
Mechanisms of failure
• A statically determinate beam will collapse if one plastic
hinge is developedhinge is developed
• ...
• For a statically indeterminate beam to collapse, more than oney p
plastic hinge should be developed
• The plastic hinge ...
Beam mechanismsBeam mechanisms
D i bDeterminate beams
& frames: Collapse
f fi l i Simple beam
after first plastic
hinge
De...
Indeterminate beams &
frames: More than onea es: Mo e t a o e
plastic hinge
to develop mechanismp
Fixed beam
l h d l h d f...
Indeterminate beam:
More than one plasticMo e t a o e p ast c
hinge to develop
mechanism
Propped cantilever
l h d l h f d ...
Panel mechanism/sway mechanismPanel mechanism/sway mechanism
W
Dept. of CE, GCE Kannur Dr.RajeshKN
29
Gable Mechanism
W
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 assum...
• Collapse load (Wc): Minimum load at whichp c)
collapse will occur – Least value
• Fully plastic moment (MP): Maximum mom...
Determination of collapse load
1. Simple beam
Determination of collapse load
1. Simple beam
Equilibrium method:
W l
Equili...
Virtual work method:
E IW W=
.2
2
u P
l
W Mθ θ
⎛ ⎞
=⎜ ⎟
⎝ ⎠
uW
4 P
u
M
W
l
∴ =
2θl
θ
2
θ
Dept. of CE, GCE Kannur Dr.Rajesh...
2. Fixed beam with UDL 2
l
2
2
.
,
24
CENTRE
w l
M =
2
.
12
ENDS CENTRE
w l
M M>=
Hence plastic hinges will develop at the...
2
.
2 uw l
M =
uw
Equilibrium:
2
8
PM =
16 PM
∴
2θ
θ θ
2
P
uw
l
∴ =
Virtual work: E IW W=
( )
0
22 2
l
l
M
θ
θ θ θ
⎛ ⎞
+⎜ ...
3. Fixed beam with point load3. Fixed beam with point load
uW
2θ
θ θ
u
MP
MP
2θ
Virtual work:
( )2
l
W Mθ θ θ θ
⎛ ⎞
= + +⎜...
4. Fixed beam with eccentric point load4. Fixed beam with eccentric point load
uW
Equilibrium:
u
a b
2 P u
ab
M W
l
=
q
MP...
Virtual work:
1 2a bθ θ=uW
Virtual work:
θ θ+
1θ
a b
2θ
1 2
b
a
θ θ⇒ =
1 2θ θ+ a
⎡ ⎤( ) ( )1 1 1 2 2u PW a Mθ θ θ θ θ= + +...
5. Propped cantilever with point load at
midspanmidspan
MMC2
MP
MP
MP
MC1
MB1
Dept. of CE, GCE Kannur Dr.RajeshKN
40
uW
2θ
Vi t l kVirtual work:
E IW W=
Equilibrium:
E IW W
( ) ( )2
l
W Mθ θ θ
⎛ ⎞
= +⎜ ⎟
.
0.5
4
u
P P
W l
M M+ =
( ) ( )2
2...
6. Propped cantilever with UDL
2
wl Maximum positive BM
8
wl p
x1
MMP
MP
At collapse
x2
E
Required to locate E
Dept. of CE...
2
l ⎛ ⎞
2
2 2 2
2 2
u u
E P P
w lx w x x
M M M
l
⎛ ⎞= − − =⎜ ⎟
⎝ ⎠
( )1
For maximum,
0EdM
=
2
0
dx
=
2 0
2
u P
u
w l M
w x...
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 ...
50 kN 75 kN
1.5 m
1 5 m 4 5 m
50 kN 75 kN
1 5 m1.5 m
θ θ1
4.5 m 1.5 m
Mechanism 1
θ + θ1
1 5 6θ θ
1.5
θ θ⇒ =11.5 6θ θ=
( )...
50 75
1.5 m 1.5 m
θ
θ1
4.5 m
θ1
Mechanism 2
θ + θ1
16 1.5θ θ=
1
1.5
6
θ θ⇒ =
( )1 1 1 1 1
1.5 1.5 1.5
50 1.5 75 1.5
6 6 6
...
Problem 2: A beam of span 6 m is to be designed for an ultimate UDL
f 25 kN/ Th b i i l t d t th d D iof 25 kN/m. The beam...
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 3
2 25 2253EW
θ
θ
+...
Problem 3: Find the collapse load for the frame shownProblem 3: Find the collapse load for the frame shown.
W
F
/ 2 / 2B C...
• Degree of Indeterminacy, N = 5 – 3 = 2
• No. of hinges, n = 5 (at A, B, C, E & F)
• No. of independent mechanisms ,r = n...
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= + + =
/ ...
Beam Mechanism for BC
W
F/ 2 / 2B
C
θ θ
Mp
θ θ
2θ
2
θMp
Mp
2θ
Mp
(2 ) 4I p p p pW M M M Mθ θ θ θ= + + =
2
EW W θ=
8 p
E I ...
Panel Mechanism
2 4I p p p pW M M M Mθ θ θ θ= + + =
Panel Mechanism
W
/ 2Mp Mp/ 2 2 2
E
W
W θ=
F
/ 2
θ
θ
16 p
E I c
M
W W ...
W
Combined Mechanism
2 ( ) (2 )
( )
I p pW M M
M
θ θ
θ θ
= +
+ +
W
/ 2 Mp/ 2
θ θ ( )
6
p
p
M
M
θ θ
θ
+ +
=
/ 2
Mp
θ
θ
2
θ
...
Problem 4: A portal frame is loaded upto collapse. Find
the plastic moment capacity required if the frame is of
uniform se...
• Degree of Indeterminacy, N = 4 – 3 = 1
• No. of possible plastic hinges, n = 3
(at B, C and between B&C)
• No. of indepe...
Beam Mechanism for BC B
C
10 kN/m
C
θ θ
2θ
4θMp
Mp
0 4θ+⎛ ⎞ 2θ
Mp
0 4
2 10 1604
2
EW
θ
θ
+⎛ ⎞= × × =×⎜ ⎟
⎝ ⎠
( )2 4I p pW ...
Panel Mechanism
25 kN
4θPanel Mechanism 4θ
Mp Mp
θ θ
E IW W=
( ) 25 4pM θ θ θ⇒ + = ×
50pM kNm⇒ =
Dept. of CE, GCE Kannur D...
Combined Mechanism 10kN/m
4θ 8x θ
25kN 4θ 8 x−
Mθ
Mp
x
xθ
θ+ θ1
θ
θ1
Mp
4m
θ
It is required to locate the
plastic hinge be...
( ) 2
x
W M Mθ θ θ θ θ θ
⎡ ⎤
= + + + = +⎡ ⎤⎣ ⎦ ⎢ ⎥
( )( )5 5 2 8x x+
( )1 1
8
2I p p
x
W M Mθ θ θ θ θ θ= + + + = +⎡ ⎤⎣ ⎦ ⎢...
Problem 5: Determine the Collapse load of the continuous beam.Problem 5: Determine the Collapse load of the continuous bea...
Equilibrium:
Hinges at A, B and D
Equilibrium:
pM>
M
pM
pM
uP
pM
P
E
8MP
4
u
4
uP
8
4
pu
p p u
MP
M M P= + ⇒ =
Moment at E...
Hinges at B and EHinges at B and E
M pM
pM
pM p
4
uP
4
uP
6p pu
M MP
M P= + ⇒ =
4 2
p uM P= + ⇒ =
True Collapse Load,
6 p
...
P PVirtual work:
/ 2 / 2
A B C
D E
θ θ
4 2 2SI = − =
θ
2θ
θ θ
2θ
Hinges at A, B and D
( )
8
2
2
p
u p u
M
P M Pθ θ θ θ
⎛ ⎞...
Problem 6: For the cantilever, determine the collapse load.
W
A
L/2 L/2
2 Mp Mp
A
B
C
• Degree of Indeterminacy N = 0
p
De...
/2 L/2
Wu
L/2
θ Mechanism 1
L/2
Mp
Lθ/2
L
W Mθ θ×
2 pM
W∴ =
2
u pW Mθ θ× = uW
L
∴ =
Wu
L
θ
Mechanism 2
2Mp Lθ2Mp
2 pM
2u p...
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 c...
I t l l ( t f i t )
2 2
03 4
2 yd d y
b y by
σ
σ
⎧ ⎫ −⎛ ⎞
= × + ×⎨ ⎬⎜ ⎟
Internal couple (moment of resistance)
( )0 0
0
2
...
SummarySummary
Plastic Theory
• Introduction-Plastic hinge concept-plastic section modulus-shape
factor-redistribution of ...
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Module4 plastic theory- rajesh sir

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Module4 plastic theory- rajesh sir

  1. 1. Structural Analysis - II Plastic Analysis Dr. Rajesh K. N. Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering Govt. College of Engineering, Kannur Dept. of CE, GCE Kannur Dr.RajeshKN
  2. 2. 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. Dept. of CE, GCE Kannur Dr.RajeshKN 2
  3. 3. 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 Dept. of CE, GCE Kannur Dr.RajeshKN 3
  4. 4. MaterialsMaterials • Elastic •Elastic-Perfectly plastic σ Elastic limitElastic limitσ Elastic limit εε Dept. of CE, GCE Kannur Dr.RajeshKN 4
  5. 5. Upper yield point A point Plasticess B C L i ld range stre Lower yield point strainO strainO Idealised stress-strain curve of mild steel Dept. of CE, GCE Kannur Dr.RajeshKN 5
  6. 6. stresss strainO Idealised stress-strain curve in plastic theory Dept. of CE, GCE Kannur Dr.RajeshKN 6
  7. 7. • 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. Dept. of CE, GCE Kannur Dr.RajeshKN 7 increase in load.
  8. 8. AssumptionsAssumptions • Plane sections remain plane in plastic condition S l d l b h• Stress-strain relation is identical both in compression and tensionp Dept. of CE, GCE Kannur Dr.RajeshKN 8
  9. 9. 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 innerWhen 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 Dept. of CE, GCE Kannur Dr.RajeshKN 9
  10. 10. yσσ σ yσ σ Dept. of CE, GCE Kannur Dr.RajeshKN 10
  11. 11. Change in stress distribution duringChange in stress distribution during yielding yσ yσ yσσ yσ yσ yσσ y Rectangular cross section Dept. of CE, GCE Kannur Dr.RajeshKN 11
  12. 12. σy σy σy σy σy σy Inverted T section Dept. of CE, GCE Kannur Dr.RajeshKN 12
  13. 13. Plastic hingePlastic hinge h h l l ld d h• When the section is completely yielded, the section is fully plastic A 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 duePlastic hinge is defined as an yielded zone due to bending in a structural member, at which large rotations can occur at a section atlarge rotations can occur at a section at constant plastic moment, MP Dept. of CE, GCE Kannur Dr.RajeshKN 13
  14. 14. Mechanical hinge Plastic hinge Reality ConceptReality Concept Resists zero Resists a constant t Mmoment moment MP Mechanical Hinge Plastic Hinge with M = 0Mechanical Hinge Plastic Hinge with MP= 0 Dept. of CE, GCE Kannur Dr.RajeshKN 14
  15. 15. • 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 yσ MP Dept. of CE, GCE Kannur Dr.RajeshKN 15
  16. 16. Plastic momentPlastic moment • Moment at which the entire section is under yield stress C T c y t y C T A Aσ σ = = 2 c t A A A⇒ = = •NA divides cross-section into 2 equal parts 2 A 2 y A C T σ= = Dept. of CE, GCE Kannur Dr.RajeshKN 1616
  17. 17. yσ C yc 2 y A σ= yt yc A T ZSi il σ 2 yT σ= •Couple due to A σ yZσSimilar toyσ ( ) A y y Zσ σ⇒ + =•Couple due to 2 yσ ( ) 2 y c t y py y Zσ σ⇒ + = Plastic modulus is the shape factor( )1pZ > Dept. of CE, GCE Kannur Dr.RajeshKN 17 p( ) Z
  18. 18. Shape factor for various cross-sections b Shape factor for various cross sections Rectangular cross-section: d Section modulus ( )3 212bdI bd( ) ( ) 12 2 6 bdI bd Z y d = = = 2 A bd d d bd⎛ ⎞ ( ) 2 2 2 4 4 4 p c t A bd d d bd Z y y ⎛ ⎞ = + = + =⎜ ⎟ ⎝ ⎠ Plastic modulus Z Shape factor = 1.5pZ Z Dept. of CE, GCE Kannur Dr.RajeshKN 1818
  19. 19. Circular sectionCircular section A d ( ) 2 p c t A Z y y= + d 2 3 2 2 8 3 3 6 d d d dπ π π ⎛ ⎞⎛ ⎞ = + =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠8 3 3 6π π⎝ ⎠⎝ ⎠ 1 7pZ S = = ( )4 364d d Z π π = = 1.7S Z = = 2 32 Z d Dept. of CE, GCE Kannur Dr.RajeshKN 19
  20. 20. Triangular section 3 bh⎛ ⎞ ⎜ ⎟ Triangular section 2 36 2 24 bh bh Z h ⎛ ⎞ ⎜ ⎟ ⎝ ⎠= = 2 3 h 2 24 3 h A 3 CG axis h ( ) 2 p c t A Z y y= + b E l i cy S = 2.346 Equal area axis ty Dept. of CE, GCE Kannur Dr.RajeshKN 20 b
  21. 21. I sectionI section 20mm 10mm250mm 20mm20mm 200mm Mp = 259.6 kNmS = 1.132 Dept. of CE, GCE Kannur Dr.RajeshKN 21
  22. 22. Load factor collapse load M Zσ Load factor P y Pcollapse load M Load factor working load M Z Zσ σ = = = Rectangular cross-section: 2 4 P y P y bd M Zσ σ= = 2 2 6 1 6 ybd bd M Z σ σ σ= = =4 y y 2 2 2 25yPM bd bd LF σ σ ⎛ ⎞ ⎛ ⎞ ∴ = = ÷ =⎜ ⎟ ⎜ ⎟ 6 1.5 6 2.25 4 1.5 6 yLF M σ∴ = = ÷ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN 2222
  23. 23. Factor of safetyFactor of safety Yield Load Factor of Safety= WorkingLoad yW W = YieldStress WorkingStress yσ σ = = g ( ) = 1.5 /1 5 yσ σ = ( )/1.5yσ El ti A l i F t f S f tElastic Analysis - Factor of Safety Plastic Analysis - Load Factor Dept. of CE, GCE Kannur Dr.RajeshKN 23
  24. 24. 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 dstructure will be created. • Any further increase in load will cause collapse• Any further increase in load will cause collapse Dept. of CE, GCE Kannur Dr.RajeshKN 24
  25. 25. • For a statically indeterminate beam to collapse, more than oney p plastic 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. Dept. of CE, GCE Kannur Dr.RajeshKN 25
  26. 26. Beam mechanismsBeam mechanisms D i bDeterminate beams & frames: Collapse f fi l i Simple beam after first plastic hinge Dept. of CE, GCE Kannur Dr.RajeshKN 2626
  27. 27. Indeterminate beams & frames: More than onea es: Mo e t a o e plastic hinge to 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 Dept. of CE, GCE Kannur Dr.RajeshKN 27
  28. 28. Indeterminate beam: More than one plasticMo 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 Dept. of CE, GCE Kannur Dr.RajeshKN
  29. 29. Panel mechanism/sway mechanismPanel mechanism/sway mechanism W Dept. of CE, GCE Kannur Dr.RajeshKN 29
  30. 30. Gable Mechanism W Gable Mechanism Composite (combined) MechanismComposite (combined) Mechanism - Combination of the above Dept. of CE, GCE Kannur Dr.RajeshKN 30
  31. 31. 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 dload. • 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. Dept. of CE, GCE Kannur Dr.RajeshKN 31
  32. 32. • Collapse load (Wc): Minimum load at whichp c) collapse will occur – Least value • Fully plastic moment (MP): Maximum moment capacity for design – Highest valuecapacity for design – Highest value Dept. of CE, GCE Kannur Dr.RajeshKN 32
  33. 33. Determination of collapse load 1. Simple beam Determination of collapse load 1. Simple beam Equilibrium method: W l Equilibrium method: . 4 u P W l M = MP M 4 P u M W l ∴ =u l Dept. of CE, GCE Kannur Dr.RajeshKN 3333
  34. 34. Virtual work method: E IW W= .2 2 u P l W Mθ θ ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ uW 4 P u M W l ∴ = 2θl θ 2 θ Dept. of CE, GCE Kannur Dr.RajeshKN 34
  35. 35. 2. Fixed beam with UDL 2 l 2 2 . , 24 CENTRE w l M = 2 . 12 ENDS CENTRE w l M M>= Hence plastic hinges will develop at the ends first. MP M MC1 MB1 MC2 MMP MB1 MP Dept. of CE, GCE Kannur Dr.RajeshKN 35
  36. 36. 2 . 2 uw l M = uw Equilibrium: 2 8 PM = 16 PM ∴ 2θ θ θ 2 P uw l ∴ = Virtual work: E IW W= ( ) 0 22 2 l l M θ θ θ θ ⎛ ⎞ +⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟ 16 PM w∴ =( )22 2 2 2 u Pw M θ θ θ⎛ ⎞ = + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟ ⎝ ⎠ 2uw l ∴ = Dept. of CE, GCE Kannur Dr.RajeshKN 36 ⎝ ⎠
  37. 37. 3. Fixed beam with point load3. Fixed beam with point load uW 2θ θ θ u MP MP 2θ Virtual work: ( )2 l W Mθ θ θ θ ⎛ ⎞ = + +⎜ ⎟ Equilibrium: Virtual work: ( )2 2 u PW Mθ θ θ θ= + +⎜ ⎟ ⎝ ⎠ 2 4 P u l M W= 8 P u M W l ∴ = 8 P u M W l ∴ = Dept. of CE, GCE Kannur Dr.RajeshKN 37
  38. 38. 4. Fixed beam with eccentric point load4. Fixed beam with eccentric point load uW Equilibrium: u a b 2 P u ab M W l = q MP l 2 PM l W∴ = MP uW ab ∴ = Dept. of CE, GCE Kannur Dr.RajeshKN 38
  39. 39. Virtual work: 1 2a bθ θ=uW Virtual work: θ θ+ 1θ a b 2θ 1 2 b a θ θ⇒ = 1 2θ θ+ a ⎡ ⎤( ) ( )1 1 1 2 2u PW a Mθ θ θ θ θ= + + +⎡ ⎤⎣ ⎦ b⎡ ⎤ ( )2 2 22 2u P b W b M a θ θ θ ⎡ ⎤= +⎢ ⎥⎣ ⎦ ( ) 2 2 2 2 2 2P P u M Mb a bW b aba θ θ θ +⎡ ⎤∴ = =+⎢ ⎥⎣ ⎦ 2 P u M l W ab = Dept. of CE, GCE Kannur Dr.RajeshKN 2
  40. 40. 5. Propped cantilever with point load at midspanmidspan MMC2 MP MP MP MC1 MB1 Dept. of CE, GCE Kannur Dr.RajeshKN 40
  41. 41. uW 2θ Vi t l kVirtual work: E IW W= Equilibrium: E IW W ( ) ( )2 l W Mθ θ θ ⎛ ⎞ = +⎜ ⎟ . 0.5 4 u P P W l M M+ = ( ) ( )2 2 u PW Mθ θ θ= +⎜ ⎟ ⎝ ⎠ 4 6 PM 6 P u M W l ∴ = 6 P u M W l ∴ = Dept. of CE, GCE Kannur Dr.RajeshKN 41
  42. 42. 6. Propped cantilever with UDL 2 wl Maximum positive BM 8 wl p x1 MMP MP At collapse x2 E Required to locate E Dept. of CE, GCE Kannur Dr.RajeshKN 42 q
  43. 43. 2 l ⎛ ⎞ 2 2 2 2 2 2 u u E P P w lx w x x M M M l ⎛ ⎞= − − =⎜ ⎟ ⎝ ⎠ ( )1 For maximum, 0EdM = 2 0 dx = 2 0 2 u P u w l M w x l − − = ( )2 2 0.414x l=From (1) and (2), 2 11.656 P u M w l =From (2), Dept. of CE, GCE Kannur Dr.RajeshKN l
  44. 44. 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 Dept. of CE, GCE Kannur Dr.RajeshKN 44
  45. 45. 50 kN 75 kN 1.5 m 1 5 m 4 5 m 50 kN 75 kN 1 5 m1.5 m θ θ1 4.5 m 1.5 m Mechanism 1 θ + θ1 1 5 6θ θ 1.5 θ θ⇒ =11.5 6θ θ= ( ) 1.5 1.5 50 1.5 75 1.5 Mθ θ θ θ θ ⎛ ⎞ ⎛ ⎞ + × = + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 6 θ θ⇒ = ( )50 1.5 75 1.5 6 6 pMθ θ θ θ θ+ + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 45 83M∴ = Dept. of CE, GCE Kannur Dr.RajeshKN 45 45.83pM∴
  46. 46. 50 75 1.5 m 1.5 m θ θ1 4.5 m θ1 Mechanism 2 θ + θ1 16 1.5θ θ= 1 1.5 6 θ θ⇒ = ( )1 1 1 1 1 1.5 1.5 1.5 50 1.5 75 1.5 6 6 6 pMθ θ θ θ θ ⎛ ⎞ ⎛ ⎞ × + = + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠6 6 6⎝ ⎠ ⎝ ⎠ 87.5pM kNm∴ = 87.5 kNm=Design plastic moment (Highest of the above) Dept. of CE, GCE Kannur Dr.RajeshKN g p ( g )
  47. 47. Problem 2: A beam of span 6 m is to be designed for an ultimate UDL f 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 Mechanism25 kN/m θ θ 3 m 2θ Dept. of CE, GCE Kannur Dr.RajeshKN 47 3 m 3 m
  48. 48. 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 3 2 25 2253EW θ θ +⎛ ⎞= × × =×⎜ ⎟ ⎝ ⎠2 E ⎜ ⎟ ⎝ ⎠ 2 225I EW W M θ θ= ⇒ = 112.5pM kNm∴ =2 225I E pW W M θ θ⇒ p Plastic modulus P P M Z = 6 112.5 10× = 5 3 4.5 10 mm= ×Plastic modulus P y Z σ 250 Z 5 4.5 10× 5 3 3 913 10PZ Z S = .5 0 1.15 = 5 3 3.913 10 mm= ×Section modulus Assuming shape factor S = 1.15 Adopt ISLB 275@330 N/m (from Steel Tables – SP 6) Dept. of CE, GCE Kannur Dr.RajeshKN Adopt ISLB 275@330 N/m (from Steel Tables SP 6)
  49. 49. 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/2 2Mp / 2 D AA Dept. of CE, GCE Kannur Dr.RajeshKN 49
  50. 50. • 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 MechanismPanel Mechanism Dept. of CE, GCE Kannur Dr.RajeshKN 50
  51. 51. 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 2 E W W θ= W/2 E / 2 θ 2θ 2Mp 28 p E I c M W W W= ⇒ = W/2 / 2 p θ 2Mp Dept. of CE, GCE Kannur Dr.RajeshKN 51
  52. 52. Beam Mechanism for BC W F/ 2 / 2B C θ θ Mp θ θ 2θ 2 θMp Mp 2θ Mp (2 ) 4I p p p pW M M M Mθ θ θ θ= + + = 2 EW W θ= 8 p E I c M W W W= ⇒ = Dept. of CE, GCE Kannur Dr.RajeshKN 52
  53. 53. Panel Mechanism 2 4I p p p pW M M M Mθ θ θ θ= + + = Panel Mechanism W / 2Mp Mp/ 2 2 2 E W W θ= F / 2 θ θ 16 p E I c M W W W= ⇒ = W/2 2 θ E / 2 E 2Mp Dept. of CE, GCE Kannur Dr.RajeshKN 53
  54. 54. W Combined Mechanism 2 ( ) (2 ) ( ) I p pW M M M θ θ θ θ = + + + W / 2 Mp/ 2 θ θ ( ) 6 p p M M θ θ θ + + = / 2 Mp θ θ 2 θ 2θ θ θ 3 2 2 2 4 E W W W Wθ θ θ= + = W/2 2 θ p E 2 2 2 4 8M / 2 8 p E I c M W W W= ⇒ =2Mp ( ) 8 TrueCollapseLoad, , p c M WLowest of the above = Dept. of CE, GCE Kannur Dr.RajeshKN 54
  55. 55. Problem 4: A portal frame is loaded upto collapse. Find the plastic moment capacity required if the frame is of uniform section throughout. 10 kN/m 25 kN B C 25 kN Mp 8m Mp 4 m Mp DA Dept. of CE, GCE Kannur Dr.RajeshKN 55
  56. 56. • 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 Dept. of CE, GCE Kannur Dr.RajeshKN 56
  57. 57. Beam Mechanism for BC B C 10 kN/m C θ θ 2θ 4θMp Mp 0 4θ+⎛ ⎞ 2θ Mp 0 4 2 10 1604 2 EW θ θ +⎛ ⎞= × × =×⎜ ⎟ ⎝ ⎠ ( )2 4I p pW M Mθ θ θ θ= + + = 40pM kNm∴ = Dept. of CE, GCE Kannur Dr.RajeshKN 57
  58. 58. Panel Mechanism 25 kN 4θPanel Mechanism 4θ Mp Mp θ θ E IW W= ( ) 25 4pM θ θ θ⇒ + = × 50pM kNm⇒ = Dept. of CE, GCE Kannur Dr.RajeshKN 58
  59. 59. Combined Mechanism 10kN/m 4θ 8x θ 25kN 4θ 8 x− Mθ Mp x xθ θ+ θ1 θ θ1 Mp 4m θ It 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⎛ ⎞⎛ ⎞ ( ) ( ) 18 25 4 10 10 8 2 2 E x x W x x θ θ θ −⎛ ⎞⎛ ⎞= × + × + × − ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN 59
  60. 60. ( ) 2 x W M Mθ θ θ θ θ θ ⎡ ⎤ = + + + = +⎡ ⎤⎣ ⎦ ⎢ ⎥ ( )( )5 5 2 8x x+ ( )1 1 8 2I p p x W M Mθ θ θ θ θ θ= + + + = +⎡ ⎤⎣ ⎦ ⎢⎣ − ⎥⎦ ( )( )5 5 2 8 4 E I p x x W W M + − = ⇒ = For maximum, 0PdM dx = 2.75x m⇒ = ( )( )5 5 2 8 68.91 4 p x x M kNm + − ∴ = = ( )Design plastic moment of resistance, ,largest of the ab 68.o 91ve pM kNm= 4 Dept. of CE, GCE Kannur Dr.RajeshKN 60
  61. 61. Problem 5: Determine the Collapse load of the continuous beam.Problem 5: Determine the Collapse load of the continuous beam. P / 2 / 2 P A B C/ 2 / 2A B C D 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 Dept. of CE, GCE Kannur Dr.RajeshKN 61
  62. 62. Equilibrium: Hinges at A, B and D Equilibrium: pM> M pM pM uP pM P E 8MP 4 u 4 uP 8 4 pu p p u MP M M P= + ⇒ = Moment at E is greater than Mp. Hence this mechanism is not possible. Dept. of CE, GCE Kannur Dr.RajeshKN
  63. 63. Hinges at B and EHinges at B and E M pM pM pM p 4 uP 4 uP 6p pu M MP M P= + ⇒ = 4 2 p uM P= + ⇒ = True Collapse Load, 6 p u M P = Dept. of CE, GCE Kannur Dr.RajeshKN 63
  64. 64. P PVirtual work: / 2 / 2 A B C D E θ θ 4 2 2SI = − = θ 2θ θ θ 2θ Hinges at A, B and D ( ) 8 2 2 p u p u M P M Pθ θ θ θ ⎛ ⎞ = + + ⇒ =⎜ ⎟ ⎝ ⎠ Hinges at B and E ( ) 6 2 2 p u p u M P M Pθ θ θ ⎛ ⎞ = + ⇒ =⎜ ⎟ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN 64
  65. 65. Problem 6: For the cantilever, determine the collapse load. W A L/2 L/2 2 Mp Mp A B C • 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 Dept. of CE, GCE Kannur Dr.RajeshKN 65
  66. 66. /2 L/2 Wu L/2 θ Mechanism 1 L/2 Mp Lθ/2 L W Mθ θ× 2 pM W∴ = 2 u pW Mθ θ× = uW L ∴ = Wu L θ Mechanism 2 2Mp Lθ2Mp 2 pM 2u pW L Mθ θ× = p uW L ∴ = ( ) 2 T C ll L d pM WL t f th b Dept. of CE, GCE Kannur Dr.RajeshKN 66 ( )TrueCollapseLoad, , p cWLowest of the above L =
  67. 67. 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. yσ Distance of CG from NA, y 0 0 0 0 0 1 2 2 2 2 2 3 y y d d b y y y by y y σ σ ⎡ ⎤⎛ ⎞ ⎛ ⎞ − × × + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦′ 2 2 03 4d y− 0 0 2 2 y y y d b y by σ σ ⎣ ⎦= ⎛ ⎞ − +⎜ ⎟ ⎝ ⎠ ( ) 0 012 y d y = − Dept. of CE, GCE Kannur Dr.RajeshKN 67
  68. 68. I t l l ( t f i t ) 2 2 03 4 2 yd d y b y by σ σ ⎧ ⎫ −⎛ ⎞ = × + ×⎨ ⎬⎜ ⎟ Internal couple (moment of resistance) ( )0 0 0 2 2 2 12 yb y by d y σ= × − + ×⎨ ⎬⎜ ⎟ −⎝ ⎠⎩ ⎭ 2 2 3 4d y03 4 12 y d y bσ − = 2 bd External bending moment = 0.9 0.9 0.9 4 p p y y bd M Z σ σ= × = × 2 2 2 3 4d bd Equating the above, 2 2 2 03 4 0.9 12 4 y y d y bd bσ σ − = × 0 0.274y d⇒ = 02 0.548y d= =Hence, depth of elastic core Dept. of CE, GCE Kannur Dr.RajeshKN 68 02 0.548y dHence, depth of elastic core
  69. 69. 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. Dept. of CE, GCE Kannur Dr.RajeshKN 69

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