Intro to Plastic Analysis

7/28/2019 Intro to Plastic Analysis

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School of the Built Environment

Structural Analysis Notes

Introduction to Plastic Analysis

These notes are designed to complement the lecture material, not replace it. They

should serve as a reminder of what is already in your mind and are not intended as a

self-teaching aid.

Plastic vs Elastic methods

In elastic analysis, we assume that all the materials are behaving elastically (i.e. stress

is proportional to strain) so if we double the values of the loads on a structure then we

will double the stresses in the members. This allows us to specify a permissible stress

by dividing the breaking stress by a factor of safety.

In plastic analysis, which is only used for certain types of structure, we allow the

material (usually steel) to pass its yield stress in certain places. There is therefore nolonger a linear relationship between stress and load, and we therefore apply a load

factor (symbol )to ensure safety of the structure.

Formation of Plastic Hinges

The basis of plastic analysis is the formation of a plastic hinge. Figure 1 shows the

cross-section of a beam which is subject to a bending moment, together with the

corresponding strain distribution and elastic stress distribution.

Figure 1:Elastic stress distribution in a beam.

1

D

Bstrain, stress,


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Remember from your previous work that the strain distribution is dependent solely

upon the deformed shape of the beam (assumed circular) and the stress distribution is

derived from this strain distribution and the relationship = E. This will work for

any values of strain until we reach the limit of proportionality, where the stress in the

outer fibres of the beam reaches the yield stress, y.

If we make some simplifying assumptions about the behaviour of the material after it

yields (we assume that no work hardening occurs, figure 2) then we can predict what

will happen when this point is reached.

Figure 2:Simplified stress-strain diagram

This means that, although the value of strain can continue to increase in line with the

curvature of the beam, the value of y cannot be exceeded anywhere in the cross-

section of the beam. The resulting stress distributions are shown in figure 3.

Figure 3: Stress distributions after yield.

As more and more of the cross-section reaches the yield stress, there comes a point

where essentially all of the cross-section is acting in a plastic manner. At this point,

no more bending moment can be sustained and the beam will therefore continue to

bend at a constant value of bending moment. The value of this is called the Plastic

Moment of Resistance and is denoted by the symbol MP. It is very important to note

that, for a particular beam cross-section, the bending moment cannot exceed the

value MP.

This is a highly localised phenomenon, and once this happens we say that a plastic

hinge has formed.

2

stress,

y

yy

yy

y

y

strain,

D

B Fully elastic elastic-plastic Fully plastic

increasing bending moment


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So, how can we calculate the value of MP? Recall first that the elastic section

modulus, which we will denote by ZE, is related to the bending moment by:

ME = y ZE

This is the maximum bending moment which can be sustained before yield occurs.Remember also that ZE is related to the I-value of the section. We can write a similar

equation for the fully-plastic situation:

MP = y ZP

Now, as with the equation of circular bending (elastic) we can calculate the bending

moment from the stress profile:

Figure 4:Bending moment in the fully-plastic condition.

From figure 4, we can write:

MP = FC (yc + yt)

Now, for equilibrium, FC = FT and since the stress is constant throughout the section,

this must mean that the plastic neutral axis is an equal area axis. There is the same

area of cross-section above the plastic neutral axis as there is below it, i.e.

AC = AT = A/2.

and, since FC = AC y (and FT = AT y)

MP = A(yc + yt) y2

from which we can infer that the plastic section modulus can be calculated as follows:

ZP = A(yc + yt)

2

Note that the values yc and yt are the distances from the plastic neutral axis to the

centroids of the section areas in compression and tension, respectively.

Check your class notes for examples of how to calculate the plastic section modulus.

3

FT

y

D

B

y

FC

yc

yt


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Plastic Collapse

Steel frames do not normally break when they fail, but form plastic hinges which

then rotate, allowing the frame to collapse. In order to do this, there must be

sufficient hinges to form a mechanism. A mechanism is not stable because it willcontinue to deflect without any increase in load, hence leading to collapse of all or

part of the structure.

We have covered simple collapse mechanisms in class. These are the beam

mechanism, which can occur in continuous beams and in portal frames, and the sway

mechanism, which occurs only in portal frames:

Figure 5:Simple collapse mechanisms. .

Possible hinge positions: Joints or supports

Concentrated loads

Towards centre of distributed loads

(i.e. anywhere where you might expect a maximum value in the bending moment

diagram).

4

collapse load

beam

mechanism

sway

mechanism

collapse load

Small circlesrepresent hinge

positions.

Dotted lines

represent the

original shape

of the structure.


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Plastic analysis procedure

When carrying out a plastic analysis by hand, it is easiest to proceed by using the

unsafe method i.e. we look at as many mechanisms as we can find, choose the

most likely and then check that it is the correct one. The advantage with working this

way is that we are always dealing with determinate problems. However, because weare working from the unsafe side, it is important to check the answer to make sure that

the chosen mechanism is indeed the unique solution. Recall the requirements of the

uniqueness theorem:

mechanism sufficient hinges must form to produce a collapse mechanism.

yield the bending moment cannot exceed the value of MP relevant to

a particular part of the structure.

equilibrium the system of forces (including the reactions) must be in

equilibrium with the bending moments.

It is the yield condition which we have to check when we have chosen a mechanism.

The calculations done for each mechanism are actually very straightforward. They

involve the principle of virtual work the virtual bit simply means that the work

is not really done, but this is what would happen if it did!

Work is done in two ways:

Work is done by (collapse) forces moving along their lines of action

Work is done by plastic moments rotating at the positions of plastic hinges.

These two are of opposite sign and, since energy can neither be created nor destroyed,

then they must be equal.

The steps in the procedure are as follows:

1. Identify possible hinge positions.

2. Multiply all working loads by to obtain collapse loads.

3. Using combinations of hinge positions identified in step 1, identify possible

collapse mechanisms.

a. Independent mechanisms (beam, sway)

b. Combined mechanisms.

4. Choose the most critical mechanism (lowest load factor or highest MP) and

check that the appropriate value of MP has not been exceeded at the remaining

hinge positions.

5


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Examples

The beam is to have a constant value of MP. We want to find the minimum value of

MP which will ensure a value of of 1.75 (the usual load factor for steel).

There are possible plastic hinge positions at A, B and C and a free hinge at D. There

are also hinge positions under the centre of the udl and under each of the twoconcentrated loads.

Each span is treated separately and a value of MP obtained. The highest value is

chosen and the other spans checked to make sure this value is not exceeded.

Span AB:

Plastic hinges are required at both ends for this mechanism. We can infer that the

third hinge is in the centre because the mechanism is symmetrical (this would not

have been the case for span CD).

One of the angles is given an arbitrary value of and other angles and distances

calculated from it. Note that we use the small-angle approximation, so the distancedropped by the centre of the beam is multiplied by the 3m distance, i.e. 3.

Because the load is uniformly distributed along the length of this span, not all of it

drops through the distance of 3 during collapse. The average distance dropped will

be half of this.

We can therefore write expressions for the external work done by the load and the

internal work done by the plastic hinges and equate them:

17.5 6 3/2 = MP( + 2 + )

i.e. MP = 39.4 kNm

6

3

Audl 10 kN/m

B C D

30 kN 20 kN

6m

2m

5m 8m

3m

A Budl = 10 1.75 = 17.5kN/m

23m 3m


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Note that the (arbitrary) value of drops out of the equation.

Span BC:

Take a little time to ensure you understand how the relative angle and distance values

have been obtained.

The work equation is written in the same way (note that all of the concentrated loaddrops through 2)

52.5 2 = MP( + 5/3 + 2/3)

i.e. MP = 31.5 kNm

Span CD:

Note that the hinge at D is free to rotate, as it is a pinned end.

35 5 = MP( + 8/3)

i.e. MP = 47.7 kNm

The highest value of MP has been calculated for span CD and this is therefore likely

to be the unique value. All that remains is to draw the bending moment diagram,

assuming that plastic hinges occur at all of the support points but making sure that

mid-span hinges do not occur in spans other than CD.

7

D

C

2 2/3

5/33m2m

30 1.75 = 52.5kN

B

55/3

8/3

3m5m

20 1.75 = 35kN

C


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Figure 6: Collapse bending moment diagram.

I hope you can see from the collapse BMD that MP is not exceeded in spans AB or

BC.

Please note two important points about the frame above. Firstly, members have

different relative values of MP. This means that, whatever value of MP is calculated,

then a steel section with (in this case) twice this value will need to be chosen for thelegs. Secondly, note that the foundation at D is incapable of supporting a moment

(i.e. it is a pin).

There are two independent mechanisms, a beam mechanism in member BC and a

sway mechanism in the frame as a whole. There will also be a combined mechanism

which, as the name suggests, combines both of these independent mechanisms.

8

2MP

63.047.7

A B C D

47.7

78.8

A

B C

D

2MP

MP

30 kN

50 kN

10m

4m

10m

8m


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Beam BC:

Note that all hinges

form at the same

value of MP.

Once again, make sure you are happy with the relationships between angles and

distances.

52.5 4 = MP( + 5/3 + 2/3)

i.e. MP = 31.5 kNm

Sway mechanism:

Note the following:

1. the rotation at top and bottom of a leg is always the same,

2. the relationship between the hinge rotations on the two legs depends upon

their relative lengths,

3. we ignore axial effects, so the horizontal displacement at the top is the same

on both sides.

Note also that there is no work done by the vertical force because it has not moved

along its line of action. Remember also that there is a free hinge at D.

Another point worth noting is that at B and C, the hinge always forms on the weaker

side, in the roof beam, at a value of MP, whereas the hinge at A, because it forms in

the leg rotates at 2MP.

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MPMP

4/5

30 1.75 = 52.5 kN

42/3

B C30 1.75 = 52.5 kN

4m 6m

5/3

A

B C

D

50 kN 1.75 = 87.5 kN8

8m

4/5

8

10m

2MP


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load A B C D

87.5 8 = (2MP 4/5) + (MP 4/5) + (MP ) + 0

i.e. MP = 205.9 kNm

Combined mechanism:

Remember that, when we are looking for combinations of mechanisms, we are

looking forhinge cancellation, i.e. joints which, in the two independent mechanisms,

are rotating in opposite directions. This occurs in this frame at joint C. In the roof

beam mechanism, the angle between DC and CB is closing, whereas in the sway

mechanism it is opening. When we combine the two, this rotation cancels out and the

joint remains at the same angle (i.e. a right angle)

Note that the total hinge rotation at B is 4/5 + = 9/5.

V load H load A B C D

52.5 4 + 87.5 8 = (2MP 4/5) + (MP 9/5) + (MP 5/3) + 0

i.e. MP = 179.6 kNm

The highest calculated value of MP is 205.9 kNm for the sway mechanism, and this is

therefore the critical mechanism. All that remains to do is to check that, for this

mechanism, the final bending moment at the roof-beam hinge position under the load

is less than MP.

10

MP

4/5

30 1.75 = 52.5 kN

A

B

C

D

50 kN 1.75 = 87.5 kN

8

8m

4/5

8

2MP

4

5/3MP


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To do this, we draw free-body diagrams of the separate members and calculate the

horizontal and vertical forces on them. Because we are dealing with a mechanism,

this means that the structure will be determinate and we can therefore calculate

bending moments anywhere we choose. Note that we use the collapse loads, not the

working loads.

HAB 10 = 411.8 + 205.9 i.e. HAB = 61.7 kN

HCD 8 = 205.9 i.e. HCD = 25.7 kN

Check H=0: 61.7 + 25.7 = 87.4 87.5 i.e.OK

Taking moments about B for the roof beam:

VC 10 + 52.5 4 = 205.9 + 205.9 i.e. VC = 20.2 kN

Taking moments about C for the roof beam:

VB 10 = 205.9 + 205.9 + 52.5 6 i.e. VB = 72.7 kN

Check V=0: 72.7 20.2 = 52.5 i.e.OK

Hence, the bending moment under the roof load can be calculated:

MX = 72.7 4 205.9 = 84.9 kNm

which is less that the MP value for the roof beam, which tells us that our choice ofmechanism is correct because the yield criterion has not been violated.

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52.5 kN

87.5 kN

205.9 kNm 205.9 kNm

411.8 kNm

HAB

HAB

HCD

VC

HCD

VB

X


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An alternative approach would have been to sketch the bending moment diagram

we know the bending moments at each of the corners and superimpose the simply-

supported BMD for the roof beam.

Figure 7:Collapse BMD for the roof beam.

Try the calculations for yourself and check that you get the same answer as the free-

body diagram calculations. Do not be put off by the odd shape of the BMD and donot forget that this is not a built-in beam!

Generally, for the sway mechanism you can use either free-body calculations or

sketch the BMD. For the beam and combined mechanisms it is usually easier to do

the calculations.

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205.9 kNm

205.9 kNm

Pab/L

Documents

Intro to Plastic Analysis