1

PLASTIC COLLAPSE MECHANISMS

INTRODUCTION In order to develop the concepts of plastic collapse mechanisms, we need first of all to go back to a bit of Theory of structures theory, and remind ourselves of the stress-strain relationship for most construction materials.

Steel is unique in that it has a "strength reserve" in the form of a strain hardening effect. However this effect is ignored in design, and so for all materials we only consider the two zones discussed in the lecture - ie the elastic and plastic zones. We can take the relationship from this curve, and apply it to bending moment stress distributions. Recall that for a given moment applied to a given cross section, the stress distribution looks as follows

At any infinitesimal area δA, a distance y from the Neutral Axis, let the value of stress be σ. The force on this area is then σ δA.

The total force on the section = A

A = A

AyA

M

but net force on the section =0 A

Ay =0, i.e. the N.A. is the centroid of the section

σy

σ

σU

εy ε

δA N.A.

yσ

I

My

2

ELASTICDESIGN The foregoing theory is the basis for elastic design. In this approach to design:

a) the neutral axis passes through the centroid of the section

b) the stress at any point is given by I

My

c) the section is considered to be on the point of failure when the highest stress on the section reaches the yield stress σy.

d) to provide safeguard against failure, a factor of is applied to the yield stress to ensure that in normal circumstances the maximum stress reached does not some working stress σw.

For example, for a rectangular beam, the maximum moment carried in the beam at failure can be assessed as follows

The section Modulus Z, which we shall denote Ze to indicate it is the modulus for the elastic condition, can be obtained in either of two ways:

6

bd

6

bdMyZ

Z

Mcesinor

6

bd

2

d12

bd

Y

IZ

2

y

2y

yE

2

3

E

b

d

σy

22

dby

22

dby

d3

2

6

bd

d3

2.

4

bdM

2y

yy

3

PLASTICDESIGN What happens if we increase the moment on the section above that which produces a maximum stress of ay on the section? The following strain and stress distributions "elastic failure" indicate the response

It is seen that the section does not in fact fail, but that the outermost fibres, which reached a stress level of σy at "elastic failure", continue to strain at the same stress level of σy.

As the moment is increased even further, successive layers of the section will reach yield until, eventually the entire section is at yield. The whole progression can be pictured as follows:- Once the whole section has reached yield the section has reached its maximum moment carrying capacity, and whilst it will continue to carry this amount of moment, it can no longer carry any more. The tiniest increment of moment applied beyond this point will cause the beam to deform continuously at this section.

σyεy

Strain Stress

at failure (first yield) M=My

σy ε >εy

Strain Stress

Partially plastic

εy

εy

M

Φ

σyσw

(2) limit of elastic

design

(1) working

stress

σy

(3) intermediate

stage

section (4)

fully plastic

4

Since free rotation at a point in a structure without taking any moment is associated with a hinge at that point, we say that a "plastic hinge" has formed at this point. Consider the following stress distribution:

At any infinitesimal area dA, a distance y from the Neutral Axis, the value of stress is σy

.The force on this area is then σy δA.

x

0

)xd(

0

x

0

)xd(

0

y

x

0

y

0

)xd(

y

A

y

AA

AA

AAA

0 section the on force net the

section the on force total The

i.e. the neutral (plastic) axis is the axis that divides the section in half by area In this approach to design;

1. the neutral axis passes through the equal area axis of the section 2. the stress at all points is σy 3. the moment of resistance corresponding to this state of stress on the section is

called the ultimate moment of resistance or the plastic moment - at this point a plastic hinge is formed

4. to provide safeguard against failure, a factor of safety in the form of a "load factor" is applied to the failure moment to ensure that in normal circumstances the beam does not collapse.

Note the interpretation of the safety factors for elastic and plastic design with regard to the stress distribution progression above. Elastic design: factor of safety = loads (or moments) giving (2) loads (or moments) giving (1) Plastic design: load factor = loads (or moments) giving (4) loads (or moments) giving (1) The load factor used in plastic theory is usually in the vicinity of 2. A typical value is 1,8. (Modern design codes for specific materials, however, apply a slightly different approach in that they apply factors of safety to both material strengths and to applied loads.)Referring again to the plastic stress distribution above, we can evaluate the moment of resistance for a rectangular beam.

δAy

σy

σy

x

d-x

5

The maximum moment carried in the beam is

σy

σy

Y2

AC

Y2

AT

d/2yc

yt

4

bd

2

d

2

bd

2

d

2

AMp

2

YYY

6

ShapefactorThe ratio between the plastic and elastic modulus is known as the "shape factor". This is the same ratio as that between plastic moment and maximum elastic moment.

Y

P

e

p

M

Malso

Z

Z where MY = elastic moment at first yield = Ze.σy

for a rectangular beam

6

bd4

bd

2

2

=1.5

For I beams, the shape factor is usually about 1,15.

NoteoncompositebeamsFor uniform beams having a definite material yield point, the plastic moment is always as developed above; ie MP= σy ZP. For composite beams, the value of MP is dependent upon the theory relating to the materials used and the methods of connection of those materials. eg. reinforced concrete section:

x

d

σU (=fcu)

σy (=fy) T= Asσy

CMp=Asσy(d-x/2)

7

PLASTICMODULIFor the symmetrical steel I- section shown below, determine:

1. the elastic moment 2. plastic moment 3. shape factor

flange B= 100 tf= 10 fy= 300 web tw= 6 D= 200

Elastic I= 20982667 Ze= 209827 Me= 62.948

plastic part A y Ay flange 1000 95 95000 y'= 77.46753web 540 45 24300 Zp= 238600 total 1540 119300

Mp= 71.58

Shape factor=Zp/Ze= 1.1371291

10 flange

6 web 200

100

10 flange

8

AxialforceColumns may have to carry significant axial forces in addition to bending moment. The axial force will have the effect of moving the axis of zero strain.

Consider the rectangular section F= y*Z*B Mpc=(B*D2/4 - B*Z2/4)*y

Py=y*B*D Let n=F/Py=y*Z*B / y*B*D =Z/D so that Z=nD Mpc=(B*D2/4 - B*(nD)2/4)*y

Mpc=(B*D2/4 – n2*B* D2/4)*y

Mpc= BD2/4(1-n2)y

Mpc=Mp(1-n2)

For an I section, S.J.Moy (Plastic methods for steel and concrete structures) gives the following “(1)with the zero strain axis in the web: Mp’=Mp-(A2/4t)n2y …(2.17) where A is total area, n P/PP

This equation is valid for P/Pp<=1-2BT/A …(2.18) T is flange thickness (2) With the zero strain axis in the flange: Mp’=[(A2/4B)*(1-n)*(2BD/A-1+n)]y … (2.19)

Shear

Reductionisnegligibleinlowrisestructures(Referto2.6.2BookbyMoy)

B

D

Due to F Due to M

y

y

y

y

y

Total stress

-

-- -

+ + +

Z

Strain

9

PlasticanalysisofstructuresThus far we have been concerned with the development of formulations relating to the Plastic Design of sections. In other words, we have looked at an isolated cross section of a beam which is subjected to a moment, and we have looked at a method of calculating the maximum plastic moment that can be carried by the section. However, we must now turn our attention. to look at the broader context of the whole (or at least the local part of the) structure in which this cross section finds itself, and what effect the development of this plastic moment might have on the overall structure. This is the field of Plastic Analysis.

BehaviourofastaticallydeterminatebeamConsider a simple determinate structure such as a simply supported beam loaded with a point load at some location

Notice two things:

1. Points A and B are pinned supports. Therefore they do not develop moments at any time. The beam is free to rotate at these points, and it does so.

2. The maximum working moment in the beam occurs at the point of application of P. This is due to the nature of the geometry of the structure (the beam) and the loads. Therefore as the load P increases so the moment at all points in the beam increase, and the cross section of the beam at P reaches the value of Mp first.

At this point a plastic hinge forms at P and the structure becomes unstable and collapses

a b A B

Өa Өb

e=2, r=2 and D=r-e =0 determinate

bending moment diagram

Mmax=Pab/L deflected shape

bending moment diagram

Mmax=Pab/L

structure r=2, e=2 and c=1 D=r-e-c =-1 unstable, collapses

10

Behaviourofastaticallyindeterminatebeam

w

wl2

12wl2

12

24

wl

12

wl

8

wl 222

From elastic theory we know the value of the bending moment at the fixed ends, and we can calculate the value of the moment at midspan. The maximum moments in the beam occur at the ends

w

w

Since the plastic moment has been reached at the supports, plastic hinges form at these points. The beam does not collapse because a beam on hinged supports at its ends is a simply supported beam, which is a stable and determinate structure.

D=0

Mp

24

wl2

As the value of w is increased so the moments along the beam also do. Since the maximum moments occur at the ends these will be the sections where the plastic moment capacity of the beam, Mp will be reached first

Mp

=Mp/2

Now consider the following simple indeterminate beam loaded with a uniform load

Mp≠wl2 12

)MM(M8

wlM PP

2

If the load w is further increased, the beam now acts as a simply supported beam, but with a non-increasing moment at the hinges of MP

Mp

Mp At some higher value of w, the moment at the centre also reaches the plastic value MP

Mp

Mp

We now have the situation of a simply supported beam with an internal hinge - an unstable condition - and the beam collapses. r=2, e=3and c=1 → D=-1

11

ThecollapsemechanismWe have seen that for a statically determinate structure (D=0), the insertion of a hinge leads to the formation of a mechanism and the collapse of the structure (ie D = -1, an unstable structure). We have also seen that for an indeterminate structure (D=2), the insertion of two hinges leads to a determinate one (D=0), and the insertion of a further hinge leads to collapse (D= -1). In general, if a structure is indeterminate to the n'th degree, then a mechanism is formed when n+l plastic hinges are developed. This rule must be applied not only to the whole structure, however, but also to elements of it, because local collapse of the structure is also considered to be a failure of the structure.

D= n= (hinges)

D=

n= n=

n= n=

or

or or

D= n= n= n=

for frames: D=3m-3j+r-c

12

From the above we can see that, for a given structure, a number of different collapse mechanisms are possible depending on the applied loading system. For each loading system only one of the possible mechanisms can be the first to form. Plastic analysis is concerned with establishing

i) given the structural strength, which of these possible mechanisms is the true collapse mechanism, and what are the associated collapse loads

or, conversely ii) given the actual loads, what are the required collapse moment capacities of the structure Consider the previous statically indeterminate beam:

1) Given the structural strength MP, the collapse occurs at a load of wu=L

M16 P

or 2) Given the applied load to the structure of wu, the required structural strength to

avoid collapse is MP=16

Lw 2u

[compare beam capacity by elastic analysis (but plastic depth MP not MY)- limit of

elastic analysis occurs when support moment reaches MP 2

Pe

2

e L

M12w.e.i

12

wLM

eue

u w33.1w.e.i33.112

16

w

w ]

MP MP

MP

wL2/8 = 2MP

13

CollapsemechanismsforsimplebeamsSince we have established that for collapse mechanism to form we need a sufficient number of plastic hinges to develop, which in turn means that at each of these hinge locations the bending moment diagram has peaks of value MP, we can carry out the analysis by a graphical trial-and-error procedure, or by inspection.

Example1Simply supported Bending Moment: (notes P10) Mmax=PxL/3x2L/3/L=2PL/9

MP+2/3 MP=2PL/9 5/3 MP=2PL/9 P=7.5 MP/L or MP=2PL/15 Alternatively: Graphically Method:

1. Draw the s.s. bmd first to scale 2. Superimpose end moment bmd graphically so that 1 = 2 3. Measure MP to scale

P

L/3 2L/3

MP

MP

2Mp/3

2PL/9

MP

+- + 2PL/9

14

Example2 Free bending moment diagram: There are 2 possibilities: a) MP+2/3 MP=288 MP=172.8 kNm b) MP+1/3 MP=252 MP=189 kNm From the previous example where more than one mechanism is a possibility, we can see that having considered all the candidate mechanisms, the one that requires the largest MP to resist it is the critical one. In other words, given a certain load system, the mechanism

288 252

96 kN 84 kN

MP

MP

MP

MP Graphically:

MP

MP

Graphically:

MP

MP

A B C D

108 72

15

which requires the highest structural strength to resist failure, ie the highest MP, will be the correct mechanism. Conversely, given the member capacity MP, the mechanism requiring the lowest load to cause failure will occur first, and is the correct mechanism. We can see from the above that one way to find the correct mechanism is to exhaustively try every possibility. But what if the system has many possible mechanisms (this would create a huge amount of work), or what if we overlooked some possibilities (if we overlooked the correct mechanism we would get an erroneous result)? It would be useful to have a set of conditions, which, if satisfied, would confirm whether or not a selected mechanism is the correct one. General collapse conditions A structure is just on the point of collapse when the following three conditions apply a) Equilibrium condition The system of bending moments must be in equilibrium with the external loads. b) Yield condition The bending moments may nowhere exceed the plastic moment values of the members. c) Mechanism condition There must be sufficient plastic hinges to form a collapse mechanism. If a system of bending moments can be found which satisfies these three conditions, then that system defines the true collapse load.

Example2continueda) MC= 252-MP/3 = 252-172.8/3 = 194.4 > MPthis is not the correct solution ,

yield condition is not satisfied b) MB= 288-2MP/3 = 288-2*189/3 = 162 < MPall collapse conditions satisfied

→ this is the correct mechanism

16

EquilibriumMethod(alsocalledStaticalmethod)Examples 1 & 2 which were presented earlier in this course (P14), made use of Equilibrium. We saw how equilibrium must be satisfied at every stage of loading (P10 Behavior of a statically indeterminate beam). In this method, a moment diagram is drawn that satisfies equilibrium conditions. “When a sufficient no of plastic hinges have been developed to allow instantaneous hinge rotations without developing increased resistance, a mechanism is said to have occurred”1. If a mechanism is obtained and the moment at any point does not exceed Mp then the load calculated is the true load. The aim of this method is to find an equilibrium moment diagram in which no moment exceeds Mp. The following approach may be used:

1. Select redundant forces / moments 2. Draw the bending moment diagram for the determinate structure 3. Draw the bending moment diagram at failure due to the redundancies 4. Superimpose the two moment diagrams so that a mechanism forms. Sketch the

mechanism 5. Calculate the value of the ultimate load by solving the equilibrium equation 6. Check that M Mp at any point, if this is the case then the true ultimate load is

obtained. If M Mp at any point then repeat steps 3 to 6 so that a plastic hinge is formed at the section of maximum moment, and continue until the three basic conditions are satisfied (p15)

Example 1 Pin based rectangular frame:

1 C.J.Salmon and J.E.Johnson

+

H

WW

h

L

17

Mechanismmethod‐applicationofvirtualWorkThe graphical method, while convenient for simple beams, becomes complicated and unwieldy when applied to more complex structures. The mechanism method provides a much simpler alternative in these circumstances. In this approach the system is assumed to act as a rigid bar mechanism. Any flexure of the members is ignored. Use is made of the principle that; External work done by applied forces = Internal work absorbed in plastic hinge rotation Remember that work is done by an action moving through a certain displacement i.e. Work done = Force x distance travelled through or Work done = Moment x angle rotated through in radians A number of examples will be given in lectures to demonstrate this approach. Reading study chapter 9 of Solving Problems in Structures volume 2 (PCL Croxton & LH Martin) Tutorial exercises Do problems 1, 4, 5, 8 and 9 from the text.

Plasticanalysis–Mechanismmethod

Example1

W

L/2 L/2

18

Example2

Example3

W

3L/4 L/4

W

3L/4 L/4

19

Example4

Example5

w

x L-x

w

L/2 L/2