Module4 rajesh sir

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

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.
2

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
3

MaterialsMaterials
• Elastic •Elastic-Perfectly plastic
σ Elastic limitElastic limitσ Elastic limit
εε
4

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
5

stresss
strainO
Idealised stress-strain curve in plastic theory
6

• 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.
7
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
8

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
9

yσσ
σ yσ σ
10

Change in stress distribution duringChange in stress distribution during
yielding
yσ yσ yσσ
yσ yσ yσσ
y
Rectangular cross section
11

σy σy
σy σy
σy
σy
Inverted T section
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 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
13

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
14

• 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
15

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= =
1616

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
>
17
p( )
Z

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
1818

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
19

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
20
b

I sectionI section
20mm
10mm250mm
20mm20mm
200mm
Mp = 259.6 kNmS = 1.132
21

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∴ = = ÷ =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2222

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
23

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
24

• 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.
25

Beam mechanismsBeam mechanisms
D i bDeterminate beams
& frames: Collapse
f fi l i Simple beam
after first plastic
hinge
2626

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
27

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

Panel mechanism/sway mechanismPanel mechanism/sway mechanism
W
29

Gable Mechanism
W
Gable Mechanism
Composite (combined) MechanismComposite (combined) Mechanism
- Combination of the above
30

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.
31

• 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
32

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
3333

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
34

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
35

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
∴ =
36
⎝ ⎠

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
∴ =
37

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
∴ =
38

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
=
2

5. Propped cantilever with point load at
midspanmidspan
MMC2
MP
MP
MP
MC1
MB1
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
u PW Mθ θ θ= +⎜ ⎟
⎝ ⎠
4
6 PM
6 P
u
M
W
l
∴ =
6 P
u
M
W
l
∴ =
41

6. Propped cantilever with UDL
2
wl Maximum positive BM
8
wl p
x1
MMP
MP
At collapse
x2
E
Required to locate E
42
q

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),
l

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
44

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∴ =
45
45.83pM∴

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)
g p ( g )

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
47
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 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)
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.
W
F
/ 2 / 2B C
/ 2
W/2
Mp
E
2Mp
/ 2
W/2
2Mp
/ 2
D
AA
49

• No. of hinges, n = 5 (at A, B, C, E & F)
•Beam Mechanisms for members AB & BC•Beam Mechanisms for members AB & BC
•Panel MechanismPanel Mechanism
50

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
51

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= ⇒ =
52

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
53

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 =
54

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
55

• No. of possible plastic hinges, n = 3
(at B, C and between B&C)
•Beam Mechanism for BC
•Panel Mechanism
56

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∴ =
57

Panel Mechanism
25 kN
4θPanel Mechanism 4θ
Mp Mp
θ θ
E IW W=
( ) 25 4pM θ θ θ⇒ + = ×
50pM kNm⇒ =
58

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
−⎛ ⎞⎛ ⎞= × + × + × − ×⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
59

( ) 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
60

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
61

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.

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 =
63

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
⎛ ⎞
= + ⇒ =⎜ ⎟
⎝ ⎠
64

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
65

/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
66
( )TrueCollapseLoad, , p
cWLowest of the above
L
=

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
=
−
67

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
68
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.
69

Module4 rajesh sir

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Module4 rajesh sir