M sc steel-01-prof-zahid_siddiq

57
Steel Structures SE-505 Plastic Analysis and Design of Structures M.Sc. Structural Engineering

Transcript of M sc steel-01-prof-zahid_siddiq

Page 1: M sc steel-01-prof-zahid_siddiq

Steel Structures

SE-505

Plastic Analysis and Design of Structures

M.Sc. Structural Engineering

Page 2: M sc steel-01-prof-zahid_siddiq

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

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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.

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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

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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

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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.

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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.

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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

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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

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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

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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.

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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

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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

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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)

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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.

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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

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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.

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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.

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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

Page 20: M sc steel-01-prof-zahid_siddiq

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

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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

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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=

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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

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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

Page 25: M sc steel-01-prof-zahid_siddiq

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

tan ==φ

Page 26: M sc steel-01-prof-zahid_siddiq

Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem

Steel Structures

Flexural Strength For Partially Plastic Section (contd…)

From strain diagram

o

y

EyF

For larger angle in radians

= −

o

y1

EyF

tanφ

For smaller angle in radians

o

y

o

y

EyF

tan ==φ

1

Page 27: M sc steel-01-prof-zahid_siddiq

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

Page 28: M sc steel-01-prof-zahid_siddiq

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 −+=

Page 29: M sc steel-01-prof-zahid_siddiq

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)

Page 30: M sc steel-01-prof-zahid_siddiq

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

Page 31: M sc steel-01-prof-zahid_siddiq

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 −=××=

Page 32: M sc steel-01-prof-zahid_siddiq

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×=

Page 33: M sc steel-01-prof-zahid_siddiq

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 −=

Page 34: M sc steel-01-prof-zahid_siddiq

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 ×=

Page 35: M sc steel-01-prof-zahid_siddiq

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 ×=

Page 36: M sc steel-01-prof-zahid_siddiq

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 ×=

Page 37: M sc steel-01-prof-zahid_siddiq

Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem

Steel Structures

Solution:

EφEφy = 1.97

M223.75

247.5

Point corresponding to EφP

Page 38: M sc steel-01-prof-zahid_siddiq

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

Page 39: M sc steel-01-prof-zahid_siddiq

Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem

Steel Structures

Simplification of M-φ Curve

φφP

My ≈ MP

M

Bi-Linear Curve

Page 40: M sc steel-01-prof-zahid_siddiq

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 =

Page 41: M sc steel-01-prof-zahid_siddiq

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)

Page 42: M sc steel-01-prof-zahid_siddiq

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

Page 43: M sc steel-01-prof-zahid_siddiq

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 =×

=

Page 44: M sc steel-01-prof-zahid_siddiq

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.

Page 45: M sc steel-01-prof-zahid_siddiq

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∆ −=

Page 46: M sc steel-01-prof-zahid_siddiq

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 =

Page 47: M sc steel-01-prof-zahid_siddiq

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

Page 48: M sc steel-01-prof-zahid_siddiq

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.

Page 49: M sc steel-01-prof-zahid_siddiq

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 =

Page 50: M sc steel-01-prof-zahid_siddiq

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

Page 51: M sc steel-01-prof-zahid_siddiq

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.

Page 52: M sc steel-01-prof-zahid_siddiq

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

Page 53: M sc steel-01-prof-zahid_siddiq

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 =

Page 54: M sc steel-01-prof-zahid_siddiq

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 =×=

Page 55: M sc steel-01-prof-zahid_siddiq

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

Page 56: M sc steel-01-prof-zahid_siddiq

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

Page 57: M sc steel-01-prof-zahid_siddiq

Prof. Zahid Ahmad Siddiqi and Dr. Azhar Saleem

Concluded