l-1 plastic analysis of structures

L-1

PLASTIC ANALYSIS

OF

STRUCTURES

ASSUMPTIONS IN PLASTIC BENDING

1. The material is homogeneous and isotropic.

2. The material obeys Hooke’s law till the stress reaches fy.

3. Member Cross-section is symmetrical about the axis at right angles

to the axis of bending.

4. Cross-section which were plane before bending remain plane after

bending.

5. The value of modulus of Elasticity of the material remains the same

in tension as well as in compression.

6. Effects of temperature, fatigue, shear and axial force are neglected.

7. Idealized bi-linear stress-strain curve applies.

ADVANTAGES OF PLASTIC DESIGN

3

➢ Uniform and realistic factor of safety for all parts.

➢ Saving of material over elastic methods.

➢ No effect due to temperature change, settlement of supports,

imperfections, etc. However instability and elastic deflections

should be Checked.

➢ Idea of collapse mode and strength of the structure.

➢ Plastic design produces a balanced section in a single attempt.

Drawbacks of Plastic Design

4

• Obtaining collapse load is difficult for complicated structure.

•

• Very little savings in column design.

• Difficult to design for fatigue.

• Lateral bracing requirements are more stringent.

• ‘Principle of superposition’ is invalid.

Material Behaviour

A uniaxial tensile stress on a ductile material such as mild

steel typically provides the following graph of stress versus

strain:

As can be seen, the material can sustain strains far in excess

of the strain at which yield occurs before failure. This

property of the material is called its ductility.

Linearly elastic fully plastic

Linearly elastic linearly plastic

IDEALISED STRESS STRAIN CURVE OF

MILD STEEL

(a)

(b) © (d)

ey

Behavior of Steel Plastic Beams

➢ Stress varies linearly from neutral axis to extreme fibers.

➢ When moment increases there will also be linear increase in

moment and stress until yield.

➢ When moment increases beyond yield moment the outer fiber

will have the same stress but will yield.

➢ The process will continue with more and more parts of the

beam x-section stressed to yield point until finally a fully plastic

distribution is approached.

➢ A Plastic Hinge is formed at the maximum stressed location

9

Behavior of Short/Restrained beams

Plastic Hinge

A plastic hinge is a zone of yielding due to flexure in a

structural member.

A plastic hinge (●) is similar to a frictionless hinge ( ) in that

rotation can occur without any change in bending moment.

At the plastic hinge stresses remain constant, but strains and

hence rotations can increase.

However, it should be recognized that the moment is MP at

plastic hinges rather than zero at frictionless hinges.

It will develop in span first at section subjected to greatest

curvature ( least radius of curvature)

13

Plastic Hinges

The effect of plastic hinge is assumed to be concentrated at

one section for analysis purpose.

However, it should be noted that this effect may extend for

some distance along the beam.

Due to formation of plastic hinge one after the other,

redistribution of moment will take place. Structure will

collapse when sufficient no of hinges are formedto render the

structure to unstable state.

If R is the degree of redundancy of the structure than

formation of R+1 hinges will make it unstable ie will lead to

collapse of structure.

St may collapse at less than R+1 hinges but that is called

partial collapse

• Plastic hinge

When the section is completely yielded, the

section is fully plastic.

A fully plastic section behaves like a hinge –

• 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

constant plastic moment

POSITION OF PLASTIC HINGE

It is likely to be formed

(i) at point of application of load

(ii) at section of sudden change in geometry

(iii) at the fixed end

(iv) at the point of zero shear

(v) when two sections with different Mp meet, hinge

shall be formed at a section having smaller Mp

Neutral Axis for Plastic Condition

As long as stress is proportional to strain neutral axis lies at center

CG of section. NA may not remain at CG of section when stress

strain relation becomes non linear.

The neutral axis for plastic condition is different than its counterpart

for elastic condition, unless the section is symmetrical,

The PNA is defined as the axis that splits the cross section into two

equal areas so that the area of compression equals the area of

tension.

The total internal compression must equal the total internal tension.

As all the fibers are considered to have the same stress fy in the

plastic condition, the area above and below the plastic neutral axismust be equal.

17

Neutral Axis for Plastic Condition

So, for a square cross section the plastic and elastic neutral axiscoincide,

The plastic section modulus depends on the location of the

plastic neutral axis, or PNA.

18

Rectangular Beams

Shape Factor= Mp/M= Zp/Ze

= bd2/4/bd2/6 = 1.5

• And so a rectangular section can sustain 50%

more moment than the yield moment, before a

plastic hinge is formed.

• Therefore the shape factor is a good measure of

the efficiency of a cross section in bending.

21

Shape Factor

The ratio of the plastic moment to the yield moment is known as

the shape factor since it depends on the shape of the cross

section.

Significance of Shape Factor

1. It gives an indication of reserve capacity of a section

from on set of yielding at extreme fibres to full

plastification.

2. If My is known,, Mp may be calculated.

3. A section with higher shape factor gives a longer

warning before collapse.

4. A section with higher shape factor is more ductile and

gives greater deflection at collapse.

5. Greater is the shape factor value, greater is collapse

load factor (LF=S*FoS)

Shape Factor v = Mp / My = Zp / Ze

22

•

Plastic Modulus of rolled I section symmetrical Shape

To determine plastic section modulus about Z axis, divide the

section into two area A1 and A2

A1= (h/2)tw = (400/2) ×8.9

= 1780 mm2

A2 = (bf -t)tf = (140-8.9) ×16

= 2097.6 mm2

y1 = D/4= 400/4= 100mm

y2 = ( D/2 – tf/2) = (400/2 – 16/2)

= 100mm

Plastic section Modulus Zp= 2× ( A1y1 + A2y2 ) = 1.1615 ×10 6 mm3

The value of Zp in code is 1.168 ×10 6 mm3 which is 1.27% greater

because of additional material at root

Channel section

This is symmetrical axis therefore equal area axis and centroid shall

coincide.

A= 2(60×10) + 80×10= 2000mm2

Taking moment of area about top flange

2000y = 60×10×5 + 80×10×50 + 60×10×95

Or y=50mm

Let y1=y2=Distance of CG of the area above equal area axis from equal

area axis

y1=y2= A1y’1 +A2y’2 / A1+A2 = (60×10×45 + 40×10×20/ (60×10+

40×10) = 35mm

Plastic sectional modulus= Zp= A(y1+y2)/2= 2000(35+35)/2=70000mm3

Ze = I/ymax= 2.86×106/50= 57333.3

SF= 1.22

There will be two stress blocks, one in tension, theother in compression, both of which will be at yieldstress.

For equilibrium of the cross section, the areas incompression and tension must be equal

Shape factor of Circular sections

R=D/2C

T

fy

4R/3π

4R/3π

Shape factor of Circular sections

Moment of inertia (I)= π D4/ 64

ymax = D /2

Section Modulus (Z) = I/ymax = π D3/ 32

M = fy Z = fy π D3/ 32

Plastic Moment (Mp)= fy.A/2.(d1+d2)

= f y[(π /4 D2 )/2] *(4R/3 π + 4R/3 π )

=f y D3/6

or

Plastic Section Modulus (Zp)= D3/6

S= 16/3. π = 1.697

Load FactorThe plastic load factor at rigid plastic collapse ( λ) is

defined as the lowest multiple of the design loads

which will cause the whole structure, or any part of it

to become a mechanism.

λ = Collapse Load (wult)/ Working Load (wa)

Multiply and divide by L/4

λ = (wult L/4) / (wa L/4)

λ = Mp / Ma ---- --------- -----(1)

=Plastic moment/ Allowable moment

= Mp/Ma

= fy . Zp/ fa. Ze= FOS. Shape Factor

Methods of Plastic Analysis

• Static method or Equilibrium method - Lower bound:

Load determine on the basis of any collapse BMD in

which bending moment at any section is less then the

plastic moment capacity will always be less than or equal

to the actual collapse load.

• Kinematic method or Mechanism method or Virtual

work method –upper bond- Work performed by the

external loads is equated to the internal work absorbed by

plastic hinges

• Load determined by assuming a mechanism will always be

greater than or equal to the actual collapse load. In

otherwords it states that out of various possible

mechanism, the mechanism is one for which loading is

minimum.

• Uniqueness theorem: A bending moment

distribution which satisfies equilibrium,

mechanism and plasticity.

• Collapse load (Wc):

• Minimum load at which collapse will occur

• Fully plastic moment (MP): Maximum moment capacity

for design – Highest value capacity for design

METHOD OF ANALYSIS

STATIC METHOD:-

The procedure for application of static theorem is as follows

:

1. Convert the structure into statically determine structure

by removing the redundant forces.

2. Draw free bending moment diagram for the structure.

3. Draw the bending moment diagram for the redundant

forces.

4. Draw the composite bending moment diagram in

such a way that a mechanism is obtained

5. Find out the value of collapse load by solving

equilibrium equations.

6. Check the moments to ensure that . If it is so, correct

value of collapse load is obtained.

This method is suitable only for simple structures. For

complicated frames, the method becomes very difficult and,

therefore, kinematic method is preferred.

Kinematic method

It is based on the kinematic or upper bound theorem according to

which a load computed on the basis of an assumed mechanism will

always be greater than or at best equal to the true ultimate load.

For the application of this method, it is very essential to know the

possible types and number of mechanisms.

There are four types of independent mechanisms : (i) beam

mechanism, (ii) sway mechanism, (iii) gable mechanism, and (iv)

joint mechanism.

Various combinations of the independent mechanism may be made

to obtain certain number of composite mechanisms.

Plastic Collapse Mechanism-Beams

Plastic Design of Portal Frames

37

DETERMINATION OF COLLAPSE LOAD FOR SOME STANDARD CASES

OF BEAMS

W

Real Hinge

Plastic hinge

No of independent mechanism

• No of indeterminacy = j

• No of possible plastic hinge locations = k

• No of independent mechanism= k- j

No of indeterminacy = j =0

No of possible plastic hinge locations = k=1

No of independent mechanism= k- j= 1

Plastic Hinge will be formed under load W

SIMPLY SUPPORTED BEAM HAVING Mp as PLASTIC

MOMENT CAPACITY

Kinematic Method

Δ = θ L/2

External work =

Load × Deflection= Wu× θL/2

Internal Work = Moment ×

Rotation = Mp (θ + θ)

By principle of virtual work

External Work done =

Internal work done

Wu (L/2 θ) = Mp (θ+θ)

Wu = 4 Mp/L

Wu represents the nominal or

theoretical maximum load that

the beam can support.

SIMPLY SUPPORTED BEAM CARRYING A CONCENTRATED

LOAD HAVING Mp AS PLASTIC MOMENT CAPACITY

• No of indeterminacy = j =0

• No of possible plastic hinge locations = k=1

• No of independent mechanism= k- j= 1

• Plastic Hinge will be formed under load W

Kinematic Method

Δ = aθ=b θ1 or θ1=a/b θ

External work =

Load × Deflection= Wu×a θ


Rotation

= Mp (θ + θ1)



Internal work done

Wu aθ = Mp (θ+θ1)

Wu a θ = Mp (θ+a/bθ)

Wu a θ = Mp θ(b+a)/b

Wu = MpL/ab

No of indeterminacy = j =2

No of possible plastic hinge

locations = k=3

No of independent mechanism=

k- j= 1

Plastic Hinge will be formed

under load W and at both

support

FIXED BEAM HAVING Mp AS PLASTIC MOMENT

CAPACITY

Kinematic Method

Δ = θ L/2

External work =

Load × Deflection= Wu× θL/2


Rotation

= Mpθ+ Mp (θ + θ) + Mpθ



Internal work done

Wu (L/2 θ) = 4Mpθ

Wu = 8 Mp/L

Wu represents the nominal or

theoretical maximum load that

the beam can support.

Static Method

WuL/8 = Mp

Or Wu= 8Mp/L

Kinematic Method

Δ = θ L/2

External work =

Load × AVERAGE Deflection


Rotation

Wu1/2 (L/2 θ) = Mp (θ+θ)

Wu = 8 Mp/L

SIMPLY SUPPORTED BEAM HAVING Mp AS PLASTIC

MOMENT CAPACITY

FIXED BEAM WITH udl OVER SPAN HAVING Mp AS

PLASTIC MOMENT CAPACITY

Static Method

WuL/8 = 2Mp

Or Wu= 16Mp/L

Kinematic Method

Δ = θ L/2

External work =

Load × average Deflection

= Wu×1/2 × θL/2

Internal Work = Moment × Rotation

= Mpθ+ Mp (θ + θ) + Mpθ


External Work done = Internal work

done

Wu × 1/2 ×(L/2 θ) = 4Mpθ

Wu =16Mp/L

Propped Cantilever• The propped beam of is an example of a structure that

will fail after two plastic hinges develop.

• Three hinges are required for collapse, but there is a

real hinge on the right end.

• In this beam the largest elastic moment caused by the

design concentrated load is at the fixed end.

• As the magnitude of the load is increased a plastic hinge

will form at that point.

Two hinges required for

mechanism

Kinematic Method

Δ = aθ=b θ1 or θ1=a/b θ

External work =

Load × Deflection


Rotation

Propped cantilever BEAM HAVING Mp AS PLASTIC

MOMENT CAPACITY

Wu aθ = Mp (θ+θ1) + Mp θ

Wu a θ = Mp (θ+a/bθ) + Mp θ

Wu a θ = Mp[ θ(b+a)/b + θ]

Wu a θ = Mp[ θL/b + θ]

Wu= Mp(L+b)/ab

Propped cantilever BEAM HAVING Mp AS PLASTIC

MOMENT CAPACITY

Find the collapse load for a propped cantilever subjected to

udl – w/unit length as shown below. Plastic moment capacity

is Mp. Find maxm Mp.

No of indeterminacy =2-1= 1

No of plastic hinge formed= 1+1=2

One hinge will be formed at the fixed end and second at a

distance x from the propped end.

External work done in the case of uniformly distributed load

is computed multiplying the load and the average

deflection or by multiplying the load length and the area of

the deflection shape (mechanism).

From the mechanism,

∆ = (L – x) θ= xθ1

θ 1 = L – x

x

External work =

Load × average Deflection

= (wuL) × 1/2 (L-x)θ

= 1/2 × wuL.(L-x)θ


= Mpθ+ Mp (θ + θ1)

= Mpθ+ Mp θ(1 + (L-x)/x)

= Mp(1+1+ (L-x)/x) θ

= Mp(L+x)/x. θ


External Work done = Internal work done

1/2 × wuL.(L-x)θ = Mp(L+x)/x. θ

Mp = wuL /2. [x(L-x)]/(L+x) ----(1)

For maximum value of Mp

dMp/dx=0

dMp/dx=0

=[ wuL /2.(L+x)(1×(L-x) + x ×(-1) + x ×(L-x) × ]/ (L+x)2

(L+x)(L-2x)- x (L-x) =0

x2 +2xL- L2 =0

x=0.414L

Put value of x in eq 1 ie

Mp = wuL /2. [x(L-x)]/(L+x) ----(1)

Mp (max) = wuL /2. [0.414L(L-0.414L)]/(L+0.414L)

Mp= wu L2/ 11.56

W=wu L

Mp= W L/ 11.56

A fixed ended beam is subjected to a load W at 1/3rd span.

Estimate collapse load.

•

To convert the beam into mechanism three plastic hinges are

required. There can be two cases

Case-I

• Two plastic hinges at support and one at B where cross section

changes ie in limb BC

Δ= L/3. θ

External work = Load × Deflection




Wu× L/3θ = 5 Mpθ

Wu= 15Mp/L

Case II

• Two plastic hinge will develop at the support and one below

the load

• Plastic Hinge will be formed at B and E

Δ= L/3. θ1 = 2/3 Lθ or θ1 = 2θ

External work = Load × Deflection= Wu× 2/3 Lθ


= 2Mpθ1 + 2Mp (θ + θ1) + Mpθ



Wu× 2/3 Lθ = 11 Mpθ

Wu= 16.5Mp/L

Collapse load shall be minimum of above two loads ie 15Mp/L

Continuous Beam

• Note: In case of continuous beam, the

mechanism is formed for its various

spans and collapse load found for each

span. Collapse load for the beam will be

minimum of the spans

Find out the collapse load for a continuous beam of

uniform cross section

Span AB

• The end A is simple support and end B is intermediate support

Therefore AB acts as propped cantilever with B as fixed end.

• Plastic Hinge will be formed at B and D

Δ= L/2. θ



Wu× L/2θ = 3 Mpθ

Wu= 6Mp/L

Collapse Load for Beam = 6Mp/L

Span BC

The end B is intermediate support and end C simple

support Therefore BC acts as propped cantilever with B as

fixed end.

Plastic Hinge will be formed at B and E

Δ= L/3. θ1 = 2/3 Lθ or θ1 = 2θ



Wu× 2/3 Lθ = 4 Mpθ

Wu= 6Mp/L

Collapse Load for Beam = 6Mp/L

• A continuous beam ABC simply supported at A and C is made by

rigidly connecting at B two sections ISMB250 and ISMB 200 as

shown in Figure below. Using plastic method of design determine

the value of K so that both spans may collapse simultaneously.

No of indeterminacy = 1

No of plastic hinge formation locations= 4

No of independent mechanisms= 4-1=3

(2 for beam AB and 1 for BC)

Beam Mechanism-1 Span AB

(hinge under load and at B)

From Fig-

2θ= 4ϕ or ϕ = θ/2



Wu× 2θ + 2Wu × 2ϕ = Mpθ + Mp ϕ + Mp ϕ

4Wuθ = 2Mp θ

Wu= Mp/2

Collapse Load for Beam = Mp/2

• Beam Mechanism II for span AB

Hinge at B and under load 2W

From Fig

4θ= 2ϕ or ϕ = 2θ



Wu× 2θ + 2Wu × 2ϕ = Mpθ + Mp ϕ + Mp ϕ

10Wuθ = 5Mp θ

Wu= Mp/2

For span AB, least value of load= Collapse Load for Beam

= Mp/2

• Beam MechanismIII for span BC

Hinge at B and under load KW

From Fig

2θ= 2ϕ or ϕ = θ



KWu 2θ = Mpθ + Mp θ + Mp θ

2KWuθ = 3Mp θ

Wu= 3Mp/2K

Since both spans collapse simultaneously

Mp/2= 3Mp/2K

or K= 3

PORTAL FRAMES

Degree of redundancy of Portal Frame

Degree of redundancy ,

T= 3a+R-3

where

a= No of areas completely enclosed by the members

R= Total no of reaction component

For a single storey portal frame fixed at ends

a=0 ; R= 3+3=6 ; T=3

No of hinges required for mechanism=3+1=4

FIND OUT COLLAPSE LOAD FOR A PORTAL FRAME

SHOWN BELOW. BEAM COLUMN OF SAME CROSS

SECTION. AASUME H=L/2, FIXED AT A AND PINNED AT E

Beam Mechanism

End B and D of the Beam BD are fixed joints.

So the possible location of hinges are B,D and C(below load)

Δ= L/2. θ

External work =

Load × Deflection

Internal Work =

Moment × Rotation



Wu× L/2θ = Mpθ +Mp(θ + θ) + Mpθ

Wu= 8Mp/L

Column AB and ED are of same length and deflect by same

amount and rotations at the end same

Hinges can be foemed at A,B and D

Δ= h. θ





Wu× L/2θ = Mpθ +Mpθ + Mp θ

Wu= 6Mp/L

SWAY MECHANISM

Possible location of plastic hinges are A, C D and E

A plastic hinge does not form at B

column AB and ED are of same length and deflect by same

amount and rotations at the end same

Hinges can be foemed at A,C E and D





Wu× L/2θ + Wu× L/2θ = Mpθ +Mp(θ + θ) + Mp(θ + θ)

Wu= 5Mp/L

Thus collapse load= 5Mp/L (least of three)

COMBINED MECHANISM

Determine the collapse load for the portal frame shown below.

Assume the Beam has plastic moment 2Mp and column Mp.

Solution :

T = 3a + R – 3

a = 0 ; R = 3 + 3 = 6 ;

T = 6 – 3 = 3

Thus the frame is statically indeterminate to third degree. The total

number of independent mechanisms are given by

N = n – T

where n = number of possible hinges

= 5 (one each at points A, B, C, D and E)

N = 5 – 3 =2

Thus there are two independent mechanisms : (i) beam mechanism,

and (2) panel mechanism. In addition to these, a combined

mechanism, consisting of beam and panel mechanism is possible.

All the three mechanisms are shown in Fig. ………..

3.13Find Mp ; assume frame has same crossection

l-1 plastic analysis of structures

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