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Steel Structures
SE-505
Plastic Analysis and Design of Structures
M.Sc. Structural Engineering
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Inelastic or Ductile Design of Structures(Plastic Analysis and Design)
Steel
fy
fu
Strain
Stress
0.1 to 0.2%
εy
1.5% 20 to 25%
Strain Hardening
Typical Stress Strain Diagram
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Inelastic or Ductile Design of Structures
fy
Strain
Stress
εy
20 to 25%
Assumed Bilinear Stress Strain Diagram
We neglect strain hardening and keep it as extra margin available in some cases.
Before fracture, the strain is about 100 to 250 times the strain at first yield.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Load Deformation Curve For Flexure
Full Plastic Moment DistributionP
L/2 L/2
Rotation becomes free at the section of full plastic moment (Mp).
Py
∆c
εy
PuPH Formed
Shaded Portion yields
Py = Load at which only outer fibers yield
PH
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Load Deflection Curve (contd…)
When load is less, stresses are less than Fy, load and deflection are linearly related. For the simply supported beam with a point load at center:
48EIPL∆
3
= P∆ ∝
When P > Py, yielding will penetrate inside at the maximum moment section
IMyf = S
MIyM
F yyy ==At start of yielding:
SFM yy ×=or
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresWhen yielding penetrates inside, the moment of inertia of the elastic portion reduces and hence the deflections starts increasing at a greater rate.
After complete yielding and formation of the hinge, the deformations continue to increase until the rotation at any plastic hinge exceeds the available capacity.
Pu depends on the cross-sectional properties and also on the end conditions.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Full Plastic Moment Capacity, MP
At the section of MP moment becomes constant and rotation becomes free. Section can rotate without further increase of load. This particular section is called “Plastic Hinge”.
PH
Mechanism, three hinges in a span
After mechanism, deformation will become very large at constant load until rotation capacity of section is exhausted and after this final failure will take place.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Residual StressesHot rolled sections have residual stresses due to differential cooling during manufacture.
CC
T
T
C
T
Parts smaller in thickness will cool first.
Maximum residual stress in hot rolled section is up-to 30 to 40% of Fy
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresBending TheoryAssumptions For Elastic Theory
1. Material is homogeneous and isotropic.
2. Member is subjected to bending moment only.
3. The ordinary bending formula is developed for objects symmetrical at least about one axis.
4. Material obeys hooks law (Not applicable for inelastic bending).
5. Plane section remains plane, even after bending i.e. warping is not there.
y εRyε ∝⇒=
y = distance of fiber where strain is to be calculated.R = Radius of curvature
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Inelastic Bending (For Different Stages of Loading)
section
M1=My ε = εy
ε < εy
Fy
f < Fy
ε > εy
εy
Fy
Fy
M3 >Myε > εy
ε > εy
εy
εy
Fy
Fy
M4 =MPε > εy
ε > εy
Fy
Fy Fy
Fy
Assumed
εy
M2=My ε > εy
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Inelastic Bending (contd…)
When section is fully plastic N.A. may not pass through centroid of cross section
Fy
FyC
T
AC
AT
C
T
yCyT
Total tension = Total Compression
Equal Area Axis
yCyT FAFA ×=×2AAA CT ==
The axis dividing the area into two equal halves is called the equal area axis. The stress changes its sense at this axis.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Inelastic Bending (contd…)
TcP yTyCM ×+×=
TycyP y2AFy
2AFM ××+××=
( )TcyP yy2AFM +××=
( )Tc yy2A
+×=First moment of area about equal area axis, called Plastic Section Modulus, Z
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Inelastic Bending (contd…)
ZFM yP ×=
SFM yy ×=
My = Moment at which yielding starts at the outer edge
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Shape Factor (F)
0.1>==y
P
MM
SZF
Shape factor depends upon shape of cross section.
Rectangular section, F = 1.5 MP = 1.5 My
Circular section, F = 1.698
Diamond section, F = 2
W-Section, F = 1.1 to 1.18 (average = 1.15)
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Ductility is the property of any material under which it shows excessive deformations before fracture / failure.
Brittleness is the property of any material under which sudden fracture occurs and deformations are comparatively very less before fracture.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Advantages of Plastic Analysis and Design1. Difference between load analysis and strength evaluation is
removed (unlike LRFD). (Analysis and design both are carried out in in-elastic range.)
2. Reserve strength of most heavily stresses section and other less stressed sections in case of indeterminate structures is utilized. This makes the structure 10 to 15% economical in case of indeterminate structures.
12
2wL12
2wL
24
2wL
Still Stable
Redistribution of moments is only there in plastic analysis
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Advantages of Plastic Analysis and Design (contd…)
3. True collapse mechanism is predicted more accurately.
4. Used for Shakedown analysis, analysis for blast and seismic loading.
5. For smaller structures, plastic method is simple than elastic method using hand computations.
6. Overall F.O.S is same as in other methods.
7. At service stage the structure is still elastic.
8. At working load deflections will be less.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Disadvantages of Plastic Analysis and Design1. Gives upper bound solution. Can give unsafe design if all
failure possibilities are not considered.
2. Principle of supper position is not valid. So we have to make separate analysis for all the load combinations.
3. Deflections may be critical in some cases.
4. The local and lateral stability becomes much more important. Where PH is formed, the rotation capacity is important to reach to PHs at other points. Every section and connection has a particular rotation capacity.
5. The solution procedure can become very length for large size frames.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresDuctility
“Measure of deformation capacity of a member before final failure”
Deformation is mostly in in-elastic range. For flexural section ductility is usually measured in terms of rotation capacity.
Compact Section“A section which does not show stability problem before
reaching the plastic moment and still provides rotation capacityat a constant moment”
For plastic analysis and design section must be compact because we need rotation capacity.
Pdb LL ≤ For rotation capacity See condition of compactness in specifications
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresTypes of SteelMild steel and high strength steel can be used for plastic design.
Plastic HingeWhen a section of a structural member reaches a maximum value of moment (Full Plastic Moment) and free rotation can occur at this constant moment, we can say that a plastic hinge is developed at this section. yielding
MP
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresPossible Locations of Plastic Hinge
1. Under the point loads
2. At connection of members
3. At change of geometry
4. At the point where shear force changes sign.(in case of UDL)
Change in geometry
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresLength of Plastic Hinge P
L/2 L/2
x
MyMy
MP
B.M.D
x2L2L
MM
y
P
−=
15.1FMM
y
P == For W-section
x2L2L15.1−
=
L0652.0x =
Length of PH = 2 × xL1304.0=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresLength of Plastic Hinge
6.7Lx2 =
For rectangular section, F = 1.5
3Lx2 =
For W-section, F = 1.15
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Flexural Strength For Partially Plastic Section
ε = f/E
Strain Diagram
Fy
Fy Fy Fy
Stress Diagram
+ -
Fy
yoΦ Φ. E
1 2 3
Se Z Ze
Partially Plastic Section
b
d y
Two sections are considered at unit distance apart
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Flexural Strength For Partially Plastic Section (contd…)
φ = Curvature (Rotation per unit length)
Original stress diagram = 1 + 2 + 3
Se = Elastic section modulus of the part which is still elastic
Ze = Plastic section modulus of the inner part that is still elastic.
Z = Plastic section modulus of the entire cross-section
From strain diagram
o
y
o
y
EyF
yε
tan ==φ
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Flexural Strength For Partially Plastic Section (contd…)
From strain diagram
o
y
EyF
=φ
For larger angle in radians
= −
o
y1
EyF
tanφ
For smaller angle in radians
o
y
o
y
EyF
yε
tan ==φ
1
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Flexural Strength For Partially Plastic Section (contd…)
( )62 2
oyb×=
For rectangular section
( )eey ZZSFM −+= 2
2yo
b2
32
oby=
6
2bhSe =
d/4
b
d/2⇒×
×= 2
42ddbZ
4
2bdZ =
Using Eq:1, 2 we can draw moment curvature relationship
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Flexural Strength For Partially Plastic Section (contd…)
2yo
( ) 22
42
oo
e byybZ =×
=
b
−+= 2
22
432
ooy bybdbyF
−×= 2
22
341
4 dybdF o
y
( )eey ZZSFM −+=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Moment Curvature Relationship For a Particular Section, (M-φ Curve)Benefits of M-φ Curve
1. For any value of M we can calculate φ and rotation capacity.
2. We can develop load-deflection curves.
3. We can calculate section ductility.
E.φ (N/mm3)
Moment (N-mm)
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Example: Draw M-φ relationship for W 250 x 70
bf = 254 mm
tf = 14.2 mm
d = 253 mm
tw = 8.6 mm
Ix = 11,300x104 mm4
Zx = 990 x 103 mm3
A = 9290 mm2
Sx = 895 x 103 mm3
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution:Let’s take first point at yo = d/2
mm5.1262d=
126.5250
yF
Eo
y ==φ
3/97.1E mmN=φ
( )eey ZZSFM −+=
For yo = d/2
ZZSS exe == ,
mkN75.22310/10895250M 63 −=××=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution:2nd point: yo = d/2 – tf/2
mm4.1192t
2d f =−
3
o
y /094.2119.4250
yF
E mmN===φ
( )eey ZZSFM −+=Calculations for Se
( ) ( )12
6.2246.825412
8.238254I33
e×−
−×
=
44105654 mm×=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Solution: (contd…)
33
o
ee mm106.473
yIS ×==
46.2246.8
21.74.1191.72542
2×+
−××=eZ
33103.526 mmZe ×=
( ) 6333 10/103.52610990106.473250M ×−×+×=
mkN3.234M −=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution:3rd point: yo = d/2 – tf
mm3.112t2d
f =−
3
o
y /23.2112.3250
yF
E mmN===φ
33e mm103.72S ×=
mkN5.238M −=
33105.108 mmZe ×=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution:4th point: yo = (d/2 – tf)/2
mm15.562
t2d
f
=
−
3
o
y /45.456.15250
yF
E mmN===φ
33e mm1008.18S ×=
mkN2.245M −=
331011.27 mmZe ×=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Solution:5th point: yo = 20 mm
3
o
y /5.1220250
yF
E mmN===φ
33e mm10293.2S ×=
mkN2.247M −=
331044.3 mmZe ×=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Solution:
EφEφy = 1.97
M223.75
247.5
Point corresponding to EφP
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Solution:
From curve
y
P
MM
=y
P
EEφφ
y
P
MM
×= yP φφ
75.2235.247
000,20097.1
P ×=φ
mmrad /1009.1 5−×=
Maximum value of section ductility (µ = φu/ φ P ) is 3 for ordinary structures and 22 for special earthquake resistant structures
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Simplification of M-φ Curve
φφP
My ≈ MP
M
Bi-Linear Curve
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Section Ductility is defined as the rotation at ultimate (φu) divided by the rotation at first yield extended linearly to Mp (φy).
y
u
φφµ =
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresLoad Deflection Curve
Example: Using the section of previous example and simplified M-φ curve plot the load deflection curve for the beam shown and hence estimate the member ductility. Assume the following:
1. Section ductility, µ = 3
2. Length of plastic hinge is d/2 on each side of maximum moment section.
3. My ≈ MP
8m
w (kN/m)
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution:
L=8m
w (kN/m)
wy = Value of “w” that causes first yielding anywhere in the beam.
12Lw
M2
yy =
24Lw
2M 2
yy =
yM
yww =
CASE-A : Before the development of end hinges or elastic range
CASE-B : Formation of central hinge.
CASE-C : Final failure
B.M.D
B
A
pφ pφ2pφ
2pφ
Curvature Diagram
C
y1∆y2∆
y3∆
Elastic Curve
d/2 d/2
d
Length of Plastic Hinge
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
∆y1 = Deflection at the stage of yielding at the ends
∆y2 = Deflection at the stage of yielding at the center
∆y3 = Final failure
Final failure is the stage when the rotation capacity at the ends or at the center exhausts.
Load at the First Yield: wy1
12Lw
MM2
1yPy ==
m/kN41.468
125.247L
12Mw 22P
1y =×
=×
=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
Deflection at the First Yield: ∆ y1
CD
tDC
Rotation between two points C & D ∫ ×=D
C
dxφ
=×∫D
C
dxφ
First moment of area of curvature diagram between C & D about point D.
tDC = Tangential deviation of any point D on the elastic curve formtangent drawn at point C on the elastic curve.
== ∫C
D
dx..xtCD φ
Area of curvature diagram between C & D.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
Deflection at the First Yield: ∆ y1
∆y1 = First moment of curvature diagram between A & C about A
ACt
A BC
c∆
cACt ∆=
∫=C
AAC dxxt ..φ
yφ2211y1 xAxA∆ −=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
Deflection at the First Yield: ∆ y1
ph φ23
=2211y1 xAxA∆ −=
pφ=h
2Lb =
b85
bh32A1 =
bhA2 =
42285
223
32∆y1
LLLLpp ×
×−
×
×= φφ
32L∆
2
py1 φ=
3280001009.1∆
25
y1××
=−
mm8.21∆y1 =
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
Load at the Second Yield: wy2
P
22y
P M8
LwM =+−
Due to end moment, at the center
Due to applied load, at the center
Must be Mp at the center to produce PH
8Lw
M22
2yP =
mkN87.618
5.24716LM16w 22
P2y −=
×==
Assuming sufficient rotation capacity is available at the ends. Otherwise final failure will take place before the formation of second hinge
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
Deflection at the Second Yield: wy2
××−
×
×=
4L
2L
2L
85
2L2
32∆ ppy2 φφ
ph φ2=
pφ=h
2Lb =
b85
bh32A1 =
bhA2 =
12L∆
2
py2 ×=φ1200081009.1
25 ××= −
mm13.58∆y2 =
Area of additional curvature diagram due to end hinge rotations may be ignored because the moment arm about the end is very small.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
Rotation Capacity At The End Hinges:
Pavailable 3 φφ ×=
23θ Pavailable
d××= φ
22531009.13 5 ×××= −
rad310136.4 −×=
rad004136.0θavailable =
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
Rotation Capacity At The End Hinges:
22531009.1004136.0θ 5
balance ××−= −
After the formation of hinge, remaining rotation capacity
rad00276.0θbalance =
Rotation capacity used up-to the formation of central hinge = Difference of area between the two curvature diagrams.
62L
32
32
2L2
32θ P
PPLφφφ =××−××=
0145.0θ =
P2φ
P32φ
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel StructuresSolution: (contd…)
Check For Rotation Capacity At The End Hinges:
This is the rotation capacity required for the formation of central hinge but the capacity available is only 0.00276rad. So before the formation of central hinge the rotation capacity at the ends will exhaust and failure will occur.
0145.0θ req =
Concrete frames may have such situation.
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Example: Same as previous example but the specially designed end connection provides a total rotation capacity of 0.03 rad.
Solution: Calculations upto wy1 and wy2 are the same.
Check For the Rotation Capacity
0.03θavailable =After the formation of end hinge
2253101.09-0.03θ 5-
balance ××=
0145.00286.0θbalance >= So central hinge will form
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Solution: Rotation capacity after the formation of second hinge
0145.00286.0θbalance −=
rad0141.0θbalance =θbalance for central hinge
( ) d13θ Pbalance ×−= φ
5321009.12θ 5balance ×××= −
rad0055.0θbalance =
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Solution: Failure Stage:
m/kN9.61ww 2y3y ==As after the formation of second hinge beam can’t take more load because of the formation of mechanism
θ θ∆'2θ
2L
2L
2Lθ'∆' ×=
For the rotation capacity of central hinge
0.00552θ =
20.0055θ =
mm03.112
80002
0055.0∆y3 =×=
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Solution: Failure Stage:
y3y2 ∆∆ +=Total deflection
mm16.6903.1113.58 =+=
∆(mm)
w
46.41w y1 =
87.61
8.21 18.58 16.69∆u =y∆Pseudo
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Steel Structures
Solution:
yPseudo∆∆
=Member ductility
mm06.2941.4687.618.21o∆Pseud y =×=
29.0669.16µ =
38.2µ = Less then section ductility
Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
Concluded