Plastic Analysis Notes

Plastic_Analysis_Notes.doc p1

PLASTIC ANALYSIS Reading – Megson I, Sections 9.10, Megson II, Chapter 18. Neal, B.G., ‘The plastic methods of structural analysis’, 3rd (S.I.) Ed., Chapman & Hall, 1977. Horne, M.R., ‘Plastic theory of structures’, Nelson, London, 1971.

Bending beyond the elastic limit Limit state design of structures requires the prediction of the ultimate strength or collapse load of a structure. Safe loads are then determined as a suitable fraction of the collapse load.

As bending of a beam proceeds, strains increase steadily, but the corresponding stress values depend on the material’s stress-strain relationship. Some materials will fail suddenly in a brittle fashion when the strain reaches a certain value (e.g. timber, cast iron, glass, etc). Others will yield and flow in a plastic fashion (e.g. many types of steel). Although some structures may fail whilst still in a fully elastic state (by buckling, for example), most will exhibit stresses that exceed the elastic limit before failing.

We consider now the behaviour of a beam under steadily increasing bending moment and assume it is made from an elasto-plastic material with an idealised stress-strain relationship as shown in the plot on the right (not to scale).

We further assume that the beam cross-section has at least one axis of symmetry which lies in the plane of bending.

examples of symmetric sections

plane of loading and bendingcoincides with plane of symmetry

M M

neutralaxis

σyεy

σy σy σy

σy σy σy

σε

strain(linear)

elastic neutralaxis

yM = M

‘Yield Moment’y M = M

‘Fully Plastic Moment’p

plasticneutral axis

increasing

maximum stress hasjust reached forthe first time

σy

yield stress penetratesentire cross-section

as strain increases yieldingpenetrates further into the beam

ymax

σ

ε

yield stress, σy

-σy

εy

-εy

yield ‘plateau’

yield strainonset of strainhardening

Plastic_Analysis_Notes.doc p2 Copyright © J.W.Butterworth, July 2005

Yield moment, My

When the stress at the extreme fibre most distant from the neutral axis just reaches yield stress. This defines the maximum moment the beam can resist whilst still fully elastic. It follows that

IyM maxy

y =σ , or yEy ZM σ=

where max

E yIZ = (the elastic section modulus)

Plastic Neutral axis

Under elastic conditions the neutral axis (zero strain locus) passes through the centroid of the cross-section. As parts of the cross-section yield and the stress distribution becomes nonlinear, the need for the tension and compression forces to remain equal causes the position of the neutral axis to move away from the centroid (except in the case of a doubly symmetric section).

σy

σy

y_

1

y_

2

C

T

G1

G2

plastic neutralaxis

A1

A2

MP

Let the plastic neutral axis divide the section such that the areas above and below are A1 and A2 respectively. For zero resultant axial force:

2/AAAAA

21

y2y1

==∴

σ=σ

Thus the plastic neutral axis divides the section into equal areas.

Fully plastic moment, MP

When entire cross-section has reached yield stress, attempting to further increase the applied bending moment will simply result in the beam rotating without further increase in resisting moment. A plastic hinge is said to have formed. A plastic hinge could perhaps be likened to a ‘rusty’ hinge in that it displays a constant resisting moment. MP is the maximum or ‘ultimate’ moment the beam can resist. The condition of full plasticity associated with the fully plastic moment theoretically requires infinite curvature (finite change in slope over a zero length of the beam – i.e. a ‘kink’) implying infinite strain – which is unattainable. In practice strains near the neutral axis will be below yield strain, but will be compensated for by extreme fibre strains reaching strain-hardening levels with consequent small increases in stress above σy.

Let 21 y and y denote the distance of the centroids of areas A1 and A2 from the PNA.

Let C denote the compressive force due to yσ acting on A1, and T the tensile force on A2 –

2/ACT2/AAC

y

y1y

σ==

σ=σ=

Taking moments about the PNA –

2)yy(A

yTyCM

21y

21P

+σ=

+=


or yPP ZM σ=

where ZP is the plastic section modulus.

Plastic section modulus

It follows from the preceding relationship that

( ) ( )2

yyAZ strictly more or,

2yyA

Z 21P

21P

+=

+= (taking absolute values of 21 y and y ).

More generally we could write

∫=A

P dAyZ

where y denotes the distance of each element of area, dA, from the PNA.

Shape factor

Maximum elastic moment, max

EyEy yIZ where ,ZM =σ= (the elastic section modulus).

Ultimate (fully plastic) moment, yPP ZM σ= .

The ratio of the fully plastic moment to the yield moment depends on the shape of the cross-section and is known as the shape factor, f (Megson’s notation, but also called S and sometimes, v).

y

P

y

P

ZZ

MM

f ==

f is a measure of the ‘reserve strength’ in a beam that has reached its maximum elastic moment, My.

Some sample values:

f = 1.5 f 1.7≅ f 1.27≅ f 1.15 to 1.6≅

Example – rectangular beam

4bd

4d

4d

2bdZ

2

P =⎟⎠

⎞⎜⎝

⎛ +=

6bd

2/d12/bd

yIZ

23

maxE === ( - should know this….)

5.146

ZZ

fE

P ===

If b=40mm, d=120mm and =σ y 250,000 kPa, the fully plastic moment will be

kNm36000,2504

12.004.0ZM2

yPP =××

=σ=

Example – collapse load of a steel T-beam

d/2

d/2

b


W

2m

yP

100

140

10

10

plasticneutralaxis

3steel

y

m/kN76MPa280

=γ

=σ

The problem is to determine the load W that will cause collapse of the steel cantilever T-beam.

It is obvious that the maximum bending moment occurs at the left hand end. When this peak moment reaches the fully plastic value MP, a plastic hinge will form at the left hand end and the beam will collapse.

Plastic neutral axis

If yP denotes the position of the PNA, the area above must = A/2.

mm120y2

10)100140(2Ay10

P

P

=

+==

Plastic section modulus

36

3

iiP

m1099mm000,99

25101001010206010120PNA) the about rectangle each of moments area the (summing yAZ

−×=

=

××+××+××=

= ∑

Fully plastic moment

mkN72.27000,2801099

ZM6

yPP

−=

××=

σ=−

Collapse load

Ignoring self-weight, the maximum bm is 2W kN-m at the left hand end. Equating this to MP :

kN86.13W72.27W2

collapse

collapse

=

=

Show also that PNA is 18.75 mm from the elastic neutral axis, 36E m101.55Z −×= , f = 1.798,

My = 15.42 kNm, and load at first yield, Wy = 7.71 kN. What value of W would cause collapse if self-weight was included? (13.68 kN)

Wcollapse

yielded material

plastic hinge

Collapse mechanism


Spread of plasticity in the plastic hinge vicinity [Megson I p.244, Megson II p.602]

cb

material at σy

My

MP

σy

under point load

σy

at b

yielded

elastic

at c

STRESS DISTRIBUTIONat selected points along beam

COLLAPSE MECHANISM

The form of the yielded zone will vary according to the bending moment diagram and the shape of the cross-section.

Once sufficient plastic hinges have formed to create a mechanism there will be no further increase in stress or load. However, strains and displacements will continue to increase as plastic flow proceeds in the plastic hinge zones.

Suitability for plastic analysis

Note that not all structural members are capable the ductile behaviour needed to form stable plastic hinges. Timber, for example tends to fail in a brittle fashion but can achieve ductile performance by means of steel connectors (but ductility is necessarily confined to the connector locations). Similarly unreinforced concrete fails in a brittle manner, but when suitable reinforced with steel becomes satisfactorily ductile. Structural steel is probably the pre-eminent ductile material, but care is still needed to ensure that undesirable buckling doesn’t occur prior to the establishment of to be suitably proportioned and intervene before the formation of plastic hinges.


Plastic Analysis Role in Design

When designing for the strength limit state, the objective is to satisfy the inequality

n* SS φ≤ .

S* denotes the structural action(s) caused by the characteristic or factored loads, and

φSn denotes the reliable strength of the structure based on nominal strength Sn and strength reduction factor φ.

The characteristic loads are normally obtained by reference to Standards (NZS 1170 for example) which present basic loads together with appropriate factors. The intention is to define loadings that typically have a 5% probability of exceedance in 50 years (in other words a value that is unlikely to underestimate the maximum load encountered by the structure during its intended design life). You will be familiar with loadings such as 1.2G + 1.6Q where G and Q represent the intensity of dead and live loads on a timber deck.

The reliable strength is intended to represent a value with a 95% probability of exceedance (in other words a value that is unlikely to over-estimate the strength).

The structural actions S* and strengths Sn are normally replaced by specific actions such as bending moment in the case of flexural (beam and frame) structures, or axial forces in the case of truss-type structures.

Flexural structures: n* MM φ≤ (where Mn may be replaced by MP, the fully plastic moment).

Truss structures: n* NN φ≤ (where N* denotes the axial force in a truss member)

Plastic analysis provides the means of determining the design actions M*, throughout the structure resulting from the application of the factored loads.

Plastic analysis can also be used to determine the magnitude of applied loads that would bring about the collapse of a given structure.

Example – Required strength of a simply supported beam

A simply supported beam of 8m span is to carry a uniform spread load of 12.5kN/m (factored). Determine the required fully plastic strength of the beam.

w

wL /82

collapsemechanism

Maximum bm, wL2/8 occurs at mid-span with a value = 12.5 x 82/8 = 100kNm.

Collapse will ensue if the beam has a fully plastic moment of φMp = 100kNm.

Hence the required beam strength, Mp = 100/φ = 100/0.9 = 111kNm.

(taking φ = 0.9, the value for a normal steel beam)


Example – Collapse load of a simply supported (determinate) beam with overhang (from Megson)

The beam ABC has span AB = L and overhang BC = L/2. Point loads of 4W and W act at mid-span AB and C respectively.

L/2 L/2 L/2

4W W

3WL/4

A BC

D

WL/2

collapse mechanism The bm diagram is readily obtained revealing a maximum bm of 3WL/4 at D. If is gradually increased a plastic hinge will eventually form at D creating the collapse mechanism shown.

The value of W is determined from the knowledge that

L3M4W

M4WL3

Pcollapse

P

=

=

The formation of a plastic hinge immediately created a mechanism, allowing collapse to occur. This will always be the case for determinate structures, but not necessarily for indeterminate structures as the next example shows.

Example – collapse load of an indeterminate beam

Consider the propped cantilever with central point load, W. An elastic analysis (e.g. by using integration, moment-area or similar) gives the bms shown, with maximum moment of 3WL/16 at the support.

As W increases the 1st plastic hinge forms when 3WL/16 = MP (i.e. when W = 16MP/3L). However, this does not create a mechanism and W may be further increased until eventually the bm at mid-span also reaches MP. At this stage a 2nd plastic hinge forms and this time a mechanism is created and collapse ensues.

Finally we carry out a static equilibrium analysis of the beam at the instant of collapse to determine the value of W.

But hold on, this is an indeterminate structure isn’t it? How can we use just static equilibrium to analyse it?

Answer is that each plastic hinge provides the value of

W

L/2 L/2

3WL/16 = MP

5WL/32

bm, elastic analysis

1st plastic hinge forms here

bm increases with W

MP

MP

bm at collapse

Wcollapse

collapse mechanism


the bm at its location. This additional information allows the analysis to be completed.

VA VC

W

A

B

CMP

MP

Taking moments about B for segment BC:

LM2

V

,2LVM

PC

PCP

=

=

Taking moments about A for AC:

LM6

LLM2

ML2W

,LV2LWM

P

PP

CP

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

−=

Giving the required collapse load in terms of the beam’s fully plastic strength.

Alternative equilibrium calculation using virtual work principle

Virtual work (Megson Section 15.2) provides powerful alternative principles that are widely used in theoretical mechanics. Here we will apply the principle of virtual displacements which is an alternative equilibrium criterion to Newton’s Laws. It states that if a structure in equilibrium is given a virtual displacement the sum of the internal and external virtual work done will be zero.

To illustrate the principle we apply it first to the simply supported beam ABC:

Apply a vertical virtual displacement ∆ to whole:

WVV,0VVWdone VW

CA

CA

=+

=∆−∆−∆=

Next apply a rotational virtual displacement δθ about A:

2WV

,0LV2LWdone VW

C

C

=

=δθ−δθ=

Giving us the expected results.

VC

B

CMP

VA VC

W

A B C

∆

δθ

VC

Lδθ


In applying the principle the displacements are referred to as virtual to distinguish them from real displacements that result from things such as applied loads. The application of a virtual displacement is not permitted to alter any external or internal forces (or stresses) that may be present. Forces are effectively “frozen” during the process.

Virtual work is computed as the product of the real forces acting through virtual displacements, or as real moments acting through virtual rotations.

We now apply the principle to the beam ABC at the instant of collapse, using the collapse mechanism as the virtual displacement:

Denoting the rotations of the beam segments AB and BC by θ (assumed small, since we need only consider the initial movement of collapse), we deduce from the geometry that the rotation at the mid-span hinge is 2θ.

θ+θ=

θ=

2MMVW Internal2LWVW External

PP

where the internal work is obtained as the product of moment x rotation.

Equating the internal and external VW gives

LM6

W

M32LW

P

P

=

θ=θ

The same result as before. However, the virtual work approach avoids the need to calculate the intermediate result VC and provides a consistent (and very simple) approach to the equilibrium calculation. Note that it was not necessary to carry out an elastic analysis. Plastic analysis requires us to consider only the final plastic collapse state, not the elastic state that precedes it.

Example – Guessing collapse mechanisms W

L/3 2L/3

For the fixed end beam shown we guess a collapse mechanism (A) as shown below: W

L/3 L/2

θθ

2θ

Denoting the rotations as shown, we proceed directly to the calculation of the collapse load using virtual work principle:

W

collapse mechanism as virtual displacement

θ θ

2θ

MP

MP


)2(MVW Internal

,3LWVW External

P θ+θ+θ=

θ=

and equating,

L12MW P=

Repeating for the mechanism (B) below: W

L/3 L/3

θ 2θ

3θ

)32(MVW Internal

,3LWVW External

P θ+θ+θ=

θ=

and equating,

L18MW P=

Not surprisingly we see that the result depends on the choice of the collapse mechanism.

Trying one more, (C): W

L/3 2L/3

θ2θ

3θ

)32(MVW Internal

,23LWVW External

P θ+θ+θ=

θ=

and equating,

L9MW P=

Each mechanism results in a different value of collapse load, W. Which one, if any, is correct?

If we draw the bm diagram for each collapse mechanism we obtain the results shown in the next figure.

Note that for mechanisms (A) and (B) the resulting bms (that satisfy equilibrium) exceed MP over the shaded part of the diagram. This is impossible of course, as MP is by definition the maximum bm that the beam can resist, and indicates that the resulting value of W must be wrong.

Mechanism (C) however gives a distribution of bm that nowhere exceeds MP. It also gives the lowest value of collapse load, W = 9MP/L. Obviously if the beam was able to collapse at this load, it would not continue on to carry the higher loads calculated for mechanisms (A) and (B).

We conclude that the mechanism which gives the lowest collapse load is probably the correct one.


W=12M /LPMP

MP

MP

MP

MP

MP

W=18M /LP

W=9M /LP

MP

MP

MP

(A)

(B)

(C)

Cases (A) and (B) are said to violate the Yield Condition, i.e. they have M > MP.

Hypothesis

A collapse load calculated on the basis of an assumed mechanism is greater than or equal to the true collapse load.

(Later we show that this is correct.)

Example – collapse? load based on satisfaction of equilibrium and yield

We attempt to salvage some useful information from the previous analyses using guessed mechanisms.

Taking mechanism (A) we observe that the calculated collapse load is 12MP/L. The maximum apparent bm occurs under the point load and from the bm diagram can be seen to have a value of Mmax = 5MP/3.

If load W is reduced, internal actions such as bms will be reduced proportionately.

To reduce the largest bm to MP, we need to reduce W by the ratio of MP/Mmax = 3/5.

This will give the following load and bms: W=36M /5LP

3M /5P(A)

MP

3M /5P

original bm

no plastic hingesat supports now

This leaves just a single plastic hinge under the point load.


We now have a load, 36MP/5L, that is in equilibrium with a set of internal moments that nowhere exceed MP. However, there are not enough plastic hinges left to create a mechanism, so we conclude that the applied load is less than the collapse load.

Hypothesis

If a bm distribution can be found that is (1) in equilibrium with the applied load(s), and (2) ≤ MP everywhere, then the applied load is ≤ true collapse load.

(Later we show that this is correct.)

Thus by considering just one mechanism, (A), we have been able to show that the true collapse load lies in the range

LM12W

LM2.7

LM12

WL5M36

PP

PP

≤≤

≤≤

With the true collapse load of LM9 P lying close to the mean of the upper and lower bounds.

Example – beam with distributed loading

Examples so far have all carried point loads, making it easy to guess the likely location of plastic hinges (since the bm diagram consists of straight line segments peak values must occur at the loads or at fixed ends). With distributed loads it is not so easy to identify the plastic hinge locations.

The approach used in this example is to treat the hinge location as a variable, calculate the collapse load and then change the variable until the collapse load is minimised.

For the propped cantilever we assume plastic hinges form at the fixed end and at a point distant x from the fixed end.

The geometry of the resulting mechanism has the angles shown in the figure.

)LxL(MVW Internal2

xLx)xL(w2x)xL(wxVW External

P +−θ=

−θ−+θ−=

Equating:

)xL(Lx)xL2(M2

w P

−−

=

Correct value of x will minimise the collapse load. Hence seek x such that 0dxdW

= , leading to

2P

2P

22

LM

657.11

LM

)246(w

L586.0L)22(x

0L2Lx4x

=

+=

=−=

=+−

w

(L-x)θ xθ

Lθ

x L-x


Reduction of MP due to axial load (Megson I p.250Megson II p.611)

When a structural member is subjected to axial load (tension or compression) in addition to moment, the fully plastic moment will be reduced (since some strength is used up in resisting the axial force). The reduced plastic moment is known as MRP or, (Megson) MP,R.

The amount by which MP is reduced depends on the shape of the cross-section and the magnitude of the axial force. We examine the case of a beam with a rectangular cross-section.

d/2

d/2

b

PM

σ = P/AM = 0 small M

σy

compressiveyield

σy

tensileyield

σy

σy

σy

fullyplastic

a

a

d/2+ad/2-a

stressresisting

P

stressresisting

M

stressresisting

M

cross-section

Assume the axial force is applied first followed by gradually increasing moment until the cross-section reaches its fully yielded state as illustrated above. Yield stress is eventually reached over the entire cross-section, with equal areas of tensile and compressive stress at the top and bottom resisting the moment, and a central area of compressive stress resisting the axial force.

)a4d(b)a

2d)(a

2d(bM 2

2

yyRP −σ=+−σ= (1)

)a2(bP yσ= (2)

substituting for a in equation 1:

)b4

P4d(bM 22

y

22

yRP σ−σ=

b4P

4bdM

y

2

y

2

RP σ−σ= (3)

Now let Py = squash load (axial load that will fully yield cross-section).

Thus yy bdP σ=

also y

2

P 4bdM σ= , and since yab2P σ= ,

we get ,P4dPMMy

2

PRP −= or

)n1(M

PPMMM

2P

2

yPPRP

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

where yPPn =

MM

RP

P

PPy

1.0

1.0


Reduction of MP due to shear force

The presence of shear force will also cause a reduction in MP, although the effect is less serious than in the case of axial force. As detailed analysis is more complicated discussion is deferred until later courses.

Moment-curvature relationship (Megson II, p.600)

Consider a beam of rectangular cross-section under the action of a bm M (<MP). Let de denote the depth of the elastic core, then

yee

y

2e

2dd

2ddb

6bdM σ⋅

+⋅

−⋅+σ⋅=

where

1st term = bm resisted by the elastic core, and

2nd term = bm resisted by the yielded (plastic) zones.

i.e. y2e

22

e )dd(4b

6bdM σ−+=

y

2e

2

12d

4dbM σ⎟⎟

⎠

⎞⎜⎜⎝

⎛−= (1)

The curvature of the elastic core (and of the beam),

e

e

EIM

R1

=κ=

where Me = resisting moment of elastic core yeZ σ=

Ie = moment of inertia of elastic core

12bdE

6bd

EIZ

3e

y

2e

e

yeσ

=σ

=κ∴

i.e. e

y

Ed2σ

=κ (2)

Equations (1) and (2) define bm and curvature in terms of de, where de ranges from d down to zero.

when yy

2

e M6

bdM,dd =σ== , the yield moment.

when Py

2

e M4

bdM,0d =σ== , the fully plastic moment

and ∞→κ .

The figure shows the relationship for a rectangular section. Similar plots can be obtained for sections of other shapes, such as I-beams.

Simplifying assumptions about the moment-curvature relationship are often made. Three of these are shown in the figure above.

MMP

κκy

1.0

1/3

2/3

0

a b c

1 2 3

Behaviour Model

oac =

obc =

oc =

rigid plastic

elastic-perfectly plastic

elasto-plastic

σy

σy

de

(d-d )/2e

(d+d )/2e

yielded zone

elastic core

cross-section stress distribution


Residual stress resulting from removal of load after MP achieved

We consider what happens if, after applying increasing load to a beam until the maximum moment reaches MP, we then remove the load, causing the bending moment to return to zero.

σ

ε

σy

-σy

εy

σy

unloading(elastic)

stress under MP

a

b c

d

MP MP

When the moment reaches MP, the entire cross-section is assumed to have reached either the tensile or compressive yield stress.

Reducing the moment to zero is equivalent to applying a bm of MP in the reverse direction.

Referring to the stress-strain diagram, note that as strain is reduced, stress reduces according to the straight line cd – i.e. a linear elastic response. Thus applying reverse MP results in a stress distribution of the form:

σmax

change in stressdue to reverse MP

MP MP

σmax

Where E

Pmax Z

M=σ .

But as ,ZM yPP σ=

section rrectangula a for 5.1

fZZ

y

yyE

Pmax

σ=

σ=σ=σ

Superimposing the stress distributions reveals the final state of residual stress following the removal of the load. Note that the final stresses are self-equilibrating (i.e. zero bending moment and axial force).

M=0

σy

M=0

1.5σy

1.5σyσy

+ =

0.5σy

0.5σy

σyσy

MP -MP residual stresswhen M=0


PLASTIC ANALYSIS THEOREMS (Megson II, 18.1)

Three conditions that must be satisfied by a structure on the point of collapse are:

1. EQUILIBRIUM CONDITION At collapse, the bending moments must correspond to a state of equilibrium between the external loads and the internal actions.

2. MECHANISM CONDITION At collapse there must be sufficient plastic hinges to create a partial or complete collapse mechanism.

3. YIELD CONDITION At collapse the bending moments must everywhere be ≤ MP.

Using these conditions we can now state three fundamental theorems of plastic analysis –

LOWER BOUND THEOREM

If the bending moments are in equilibrium with the external load and M ≤ MP everywhere, the load is a lower bound (i.e. load is ≤ collapse load).

UPPER BOUND THEOREM

For an assumed mechanism in which the virtual work done in the plastic hinges equals the virtual work done by the external loads, the load is an upper bound (i.e. load is ≥ collapse load).

UNIQUENESS THEOREM

If a bending moment distribution can be found that satisfies the three conditions of equilibrium, mechanism and yield, then the corresponding load is the collapse load.

Proofs of the theorems are straightforward, but rather unexciting. See Horne, 1971 or Neal, 1977.

The three theorems are summarised diagrammatically below:

UNIQUENESS load = true collapse load

MECHANISM

EQUILIBRIUM

YIELD

UPPER BOUNDload true collapse load≥

LOWER BOUNDload true collapse load≤

.

There are a number of corollaries (easily proved consequences) of the theorems:

1. The collapse load of a structure cannot be decreased by increasing the strength of any part of it (corollary of lower bound theorem).

2. If the collapse loads are determined for all possible mechanisms, the actual collapse load will be the smallest of these (corollary of upper bound theorem).

3. The collapse load of a structure cannot be increased by decreasing the strength of any part of it (corollary of upper bound theorem).

4. The initial state of stress has no effect on the collapse load (excluding elastic stability effects) (from uniqueness theorem).


5. If a structure is subjected to any programme of proportional or non-proportional loading, collapse will occur at the first combination of loads for which a bm distribution satisfying the conditions of equilibrium, mechanism and yield can be found (from uniqueness theorem).

Note

1. The uniqueness theorem does not assert that the bm distribution at collapse is unique. The bm distribution at collapse may depend on factors such as initial state of stress and loading history. Nor does it assert that the collapse mechanism is unique. There may be alternative mechanisms, but they will lead to the same collapse load.

2. An assumed plastic mechanism leading to a collapse load need not imply that a bm distribution in equilibrium with the external loads can exist for such a mechanism – as shown below:

Example – assumed mechanism, equilibrium not satisfied W

L/2 L/2 L/4

W/4

WW/4

W W/4

Correct mechanism Assumed mechanism

θθ θ θ θ

7M48W

4L

4W

2LW3M

P

P

=

θ−θ=θ

LM8W

2LW4M

P

P

=

θ=θ

True collapse bm diagram(satisfies equilibrium)

7W/48

7W/48

WL/16WL/8 WL/8

WL/8 not in equilibrium

Assumed mechanism - bm diagram(doesn’t satisfy equilibrium, but thecalculated collapse load is still a validupper bound)


Mechanism Method of Analysis This method of analysis is based directly on the upper bound theorem. The basic idea is to try all the likely collapse mechanisms and select the one which gives the lowest collapse load. The steps involved are as follows:

1. Identify the likely plastic hinge locations (under point loads, at supports, at joints, at zero shear positions under spread loads).

2. Sketch all the likely collapse mechanisms.

3. For each mechanism use virtual work to calculate the collapse load factor.

4. Select the mechanism which gives the lowest load.

5. For this chosen case check that M ≤ MP (this is just to check that the selected mechanism is indeed the correct one). If this condition is not satisfied the correct mechanism has been overlooked.

Example – Fixed end beam with point loads 2W

2m

4θ

3.5W

3m 1m

AB C

DM = 120kNmP

2θ

6θ

2W

θ 5θ

6θ

3.5W

1)

2)2W

3.5W

Mechanism 1:

θ++=θ+θ )264(M1)2(W5.32)4(W2 P

lowestkN61.6223M12W P ←==

Mechanism 2:

θ++=+θ+θ )561(M1)5(W5.32)(W2 P

kN98.665.21

M12W P ==


Yield check on mechanism giving lowest load:

We only need to find the moment at C, MC, since we know the bm at A, B and D is MP. Using virtual work again, we take the known bms and load from the chosen mechanism (1) and use mechanism (2) as a virtual displacement to determine MC:

θ 5θ

6θ

3.5(62.61)

2(62.61)

M (unknown)C

MP MP

1)5)(61.62(5.32))(61.62(2)6(M)5(M CP θ+θ=θ+θ+θ

)OK,M(kNm35.104M PC ≤= giving the collapse bm diagram below:

120

120

120

104.35

The same result could be found by ‘conventional’ methods:

Consider segment BD and take moments about B:

0V4M361.625.3M DPP =−+××+

kN35.224VD =

Summing moments about C for segment CD:

1VMM DPC ×=+

)OK,M(.kNm35.104120135.224M PC ≤=−×=

3.5W

3m 1m

B C

DMP

VD

1m

C DMC

VD


Example – Fixed base rectangular frame, varying MP, point loads (examples of frame analyses can be found in Megson II from p.613)

4m

6m 6m

M (both columns) = 40kNmP

6W kN

8W kN

M (beam) = 60kNmP

1

2 3 4

5

1. Likely plastic hinge positions identified at points 1, 2, 3, 4, and 5.

Note: At the joints between beam and columns, a plastic hinge will form in the weaker member (the column) as soon as the moment reaches 40kNM.

2. Sketch candidate collapse mechanisms:

6W8W

θ

θ

θ

θ

6W 8Wθ θ

2θ 90°

8W6W θ θ

θ

θ

θ

‘SWAY’ mechanism ‘BEAM’ mechanism(example of partial collapse mechanism)

‘COMBINED’ mechanism

Note that the ‘combined’ mechanism is obtained by combining (adding) the ‘sway’ and ‘beam’ mechanisms. In the process the plastic hinge rotations at hinge location 2 cancel each other, removing the plastic hinge and leaving the beam and column meeting at right angles.

3. Calculate collapse load factor (W) for each mechanism using virtual work equation:

SWAY

67.6W.4.W6)4(M )column(P

=

θ=θ

BEAM

167.468

260240W

.6.W8)2(M)2(M )beam(P)column(P

=×

×+×=

θ=θ+θ


COMBINED

889.34824

260440W

.6.W8.4.W6)2(M)4(M )beam(P)column(P

=+

×+×=

θ+θ=θ+θ

4. Conclude collapse load, W = 3.889 (upper bound theorem)

5. Check yield condition to ensure M ≤ MP for selected mechanism: Need to find the moments at points 1 to 5. However, we know that the moments at 1, 3, 4 and 5 are the MP values for those locations. This leaves just M2 to find.

Using the beam mechanism as a virtual displacement to find M2:

kNm7.26M611.31)(40)2(60)(M

2

2

=

θ××=θ+θ+θ

Hence collapse bm diagram as shown with M ≤ MP everywhere proving solution is correct (uniqueness theorem).

Example – Frame with sloping member

Sloping members complicate the mechanism geometry, and we introduce the concept of instantaneous centre of rotation. Otherwise it is business as usual.

L

L L

3W2W

1

23 4

5

all membershave same MP

1. Diagram above shows frame details and 5 possible plastic hinge locations.

23.33 kN

31.11kN

1

2 3 4

5

60

4026.7

40 40

31.11θ θ

2θ

M2

M = 60(M )

3

P(beam)

M = 40(M )

4

P(column)


2Wθ θ

2θ

3W

ic

θ

θ

θ

θ

θ

2θ

2θ

θ θ

4θ

θθ

4θ

plastic hingecancels

Mechanism A Mechanism C= 2A + B

Mechanism B

3W3W

2W2W

2. Sketches above show three likely mechanisms.

Mechanism A – familiar beam mechanism.

Mechanism B – a ‘sway’ type mechanism.

Note:

Joint 2 can only rotate about joint 1 and so moves in a direction perpendicular to 1-2.

Joint 4 likewise moves perpendicular to 4-5.

As joints 2 and 4 also belong to member 2-4 they must rotate about a common instantaneous centre of rotation.

Since 2 moves perpendicular to 1-2 its centre of rotation must lie on 1-2, and similarly the centre of rotation of 4 must lie on 4-5. Member 4-5 must therefore rotate about the intersection of 1-2 and 4-5, marked as ‘ic’ in the diagram. From the geometry of the diagram it can be seen that 2-4 rotates through an angle θ.

Mechanism C – a ‘combined’ mechanism.

We seek to combine mechanisms A and B such that plastic hinge cancellation occurs. Noting that the hinge at 2 has a negative rotation of θ in mechanism A (causing tensile stress on outside of frame) and a positive rotation of 2θ in mechanism B, we use the combination 2A + B to eliminate the plastic hinge at 2.

3. Collapse loads for each mechanism.

A:

L3M8

W

Gq4.M.2L.W3

P

P

=

=θ

B:

L7M12

W

6.M.L.W2.2L.W3

P

P

=

θ=θ+θ

C: lowest

L13M20

W

10.M.L.W23.2L.W3

P

P

←=

θ=θ+θ


4. Mechanism C gives the lowest collapse load, providing us with our answer (according to the lower bound theorem).

5. To prove that our solution is correct we need to demonstrate that for the chosen collapse mechanism M ≤ MP throughout the frame.

Since there are plastic hinges at locations 1, 3, 4 and 5, we only need to show that M2 ≤ MP.

Using mechanism A as a virtual displacement in a similar way to the previous example we obtain the virtual work (equilibrium) equation as:

θ+θ=θ 3.M.M.2L.W3 P2

subst. L13M20

W P=

P

P2

M692.0

M)32660(M

−=

−=

The negative value for M2 simply indicates that it acts in the opposite direction to that assumed in the diagram.

Thus M2 ≤ MP and the final bm diagram will be as shown.

Example – Beam design

5m 8m

2kPa (dead) 5kPa (live)

3m 3m

A reinforced concrete floor in an industrial building is to be supported by steel universal beams spaced 3m apart. Each beam is 13m long and supported at three points creating spans of 5m and 8m. The floor is 150mm thick and is attached to the top surface of the beam, providing restraint against lateral buckling. Partitions, etc, result in an additional dead load of 2kPa over the floor area. A live load of 5kPa is also specified.

Determine the required size of UB for strength limit state loading of 1.2G+1.5Q. Use grade 300 steel and take the weight of reinforced concrete as 24kN/m3.

θ θ

2θ

3WM2 MP

MP

MP

MP

MP

MP0.69MP


Objective

To select a beam size such that M* < φMn, where φ = 0.9, M* is the design bending moment for the factored (strength limit state) loading and Mn is the nominal (in this case fully plastic) moment capacity of the selected beam.

Loading

Dead: Concrete slab: 24 x 0.15 x 3 10.8kN/m

Partitions, etc. 2 x 3 6.0

Self weight UB 60kg/m 0.6 (guess)

Total G = 17.4kN/m

Live: Specified 5 x 3 Q = 15kN/m

Factored 1.2G+1.5Q 43.4kN/m

Analysis

Longest span will collapse first with a mechanism identical to a propped cantilever (see p.12 of these notes).

5m 0.586L(4.7m)

3.3m

We’ve already worked out the collapse load for this case (see p.12) and obtained 2P

LM

657.11w = .

Hence the maximum design action (bending moment) imposed by the factored loads will be

kNM238657.11

84.43657.11L.wM

2

2*

=

×=

=

Selection of beam with adequate strength

We require a beam with φMn ≥ M*, and since Mn = ZPσy,

33

3

yP

mm10881m00088.0

000,3009.0238.

9.0238Z

×≥

≥

×≥

σ×≥

Select 360UB51 from table of UB section properties (next page). It has a plastic section modulus of 897x103 mm3, giving φZPσy = 242kNM (>238).

Finally check that self-weight of chosen beam is in agreement with initially guessed value.

360UB51: self weight = 51kg/m, compared with guessed value of 60kg/m – OK.


Alternative design based on linear elastic analysis

Designers frequently opt to determine the design actions by means of a linear elastic analysis – e.g. by moment distribution or more likely by computer analysis, rather than by plastic analysis. The loading, 43.4kN/m is the same but the analysis will give the bms shown below. The beam section is then selected such that M* < φMn as in the design based on plastic analysis. The maximum design bm M*, now occurs only at the interior support.

Thus, kNm266M* = , requiring

33

yP

mm10985

266Z

×≥

φσ≥

Select 410UB53.7 from table of UB section properties (next page). It has a plastic section modulus of 1060x103 mm3, giving φZPσy = 286kNM (>266).

The design is effectively based on the lower bound theorem in that only one plastic hinge is allowed to form (at the point of maximum bm) and so there is no mechanism.

The approach is more conservative and leads to a less economical choice of beam. However, there are advantages in that the structure does not have to satisfy such stringent ductility conditions as that designed using plastic analysis. Providing sufficient restraint to ensure satisfactory plastic hinge rotation can increase the cost of a plastic-based design making it less economic.

34.3

265.8

225.8


UNIVERSAL BEAMS – Dimensions and Properties

Flange Overalldepth

Width Thick-ness

Web thick-ness

Rootradius

Depthbet-

weenflanges

Web slend-erness

Flangeout-

stand

Gross area

About x-axis About y-axis Torsion

con-stant

Warp-ing

con-stant Designation

d B T t r d1 td1 2

tB − Ag Ix Zx ZPx rx Iy Zy ZPy ry J Iw

kg/m mm mm mm mm mm mm mm2 106mm4 103mm3 103mm3 mm 106mm4 103mm3 103mm3 mm 103mm4 109mm6

125 612 229 19.6 11.9 14. 572 48.1 5.54 16000 986 3230 3680 249 39.3 343 536 49.6 1560 3450 113 607 228 17.3 11.2 14.0 572 51.1 6.27 14500 875 2880 3290 246 34.3 300 469 48.7 1140 2980

610UB

101 602 228 14.8 10.6 14.0 572 54.0 7.34 13000 761 2530 2900 242 29.3 257 402 47.5 790 2530 92.4 533 209 15.6 10.2 14.0 502 49.2 6.37 11800 554 2080 2370 217 23.8 228 355 44.9 775 1590 530UB

82 528 209 13.2 9.6 14.0 502 52.3 7.55 10500 477 1810 2070 213 20.1 193 301 43.8 526 1330 82.1 460 191 16.0 9.9 14.0 428 43.3 5.66 10500 372 1610 1840 188 18.6 195 303 42.2 701 919 74.6 457 190 14.5 9.1 11.4 428 47.1 6.24 9520 335 1460 1660 188 16.6 175 271 41.8 530 815

460UB

67.1 454 190 12.7 8.5 11.4 428 50.4 7.15 8580 296 1300 1480 186 14.5 153 238 41.2 378 708 59.7 406 178 12.8 7.8 11.4 381 48.8 6.65 7640 216 1060 1200 168 12.1 135 209 39.7 337 467 410UB 53.7 403 178 10.9 7.6 11.4 381 50.1 7.82 6890 188 933 1060 165 10.3 115 179 38.6 234 394 56.7 359 172 13.0 8.0 11.4 333 41.6 6.31 7240 161 899 1010 149 11.0 128 198 39.0 338 330 50.7 356 171 11.5 7.3 11.4 333 45.6 7.12 6470 142 798 897 148 9.60 112 173 38.5 241 284

360UB

44.7 352 171 9.7 6.9 11.4 333 48.2 8.46 5720 121 689 777 146 8.10 94.7 146 37.6 161 237 46.2 307 166 11.8 6.7 11.4 284 42.3 6.75 5930 100 654 729 130 9.01 109 166 39.0 233 197 40.4 304 165 10.2 6.1 11.4 284 46.5 7.79 5210 86.4 569 633 129 7.65 92.7 142 38.3 157 165

310UB

32.0 298 149 8.0 5.5 13.0 282 51.3 8.97 4080 63.2 424 475 124 4.42 59.3 91.8 32.9 86.5 92.9 37.3 256 146 10.9 6.4 8.9 234 36.6 6.40 4750 55.7 435 486 108 5.66 77.5 119 34.5 158 85.2 31.4 252 146 8.6 6.1 8.9 234 38.4 8.13 4010 44.5 354 397 105 4.47 61.2 94.2 33.4 89.3 65.9

250UB

25.7 248 124 8.0 5.0 12 232 46.4 7.44 3270 35.4 285 319 104 2.55 41.1 63.6 27.9 67.4 36.7 29.8 207 134 9.6 6.3 8.9 188 29.8 6.65 3820 29.1 281 316 87.3 3.86 57.5 88.4 31.8 105 37.6 25.4 203 133 7.8 5.8 8.9 188 32.3 8.15 3230 23.6 232 260 85.4 3.06 46.1 70.9 30.8 62.7 29.2 22.3 202 133 7.0 5.0 8.9 188 37.5 9.14 2870 21.0 208 231 85.5 2.75 41.3 63.4 31.0 45.0 26.0

200UB

18.2 198 99 7.0 4.5 11.0 184 40.9 6.75 2320 15.8 160 180 82.6 1.14 23.0 35.7 22.1 38.6 10.4 22.2 179 90 10.0 6.0 8.9 159 26.5 4.20 2820 15.3 171 195 73.6 1.22 27.1 42.3 20.8 81.6 8.71 18.1 175 90 8.0 5.0 8.9 159 31.8 5.31 2300 12.1 139 157 72.6 0.975 21.7 33.7 20.6 44.8 6.80

180UB

16.1 173 90 7.0 4.5 8.9 159 35.3 6.11 2040 10.6 123 138 72.0 0.853 19.0 29.4 20.4 31.5 5.88 18.0 155 75 9.5 6.0 8.0 136 22.7 3.63 2300 9.05 117 135 62.8 0.672 17.9 28.2 17.1 60.5 3.56 150UB 14.0 150 75 7.0 5.0 8.0 136 27.2 5.00 1780 6.66 88.8 102 61.1. 0.495 13.2 20.8 16.6 28.1 2.53

Documents

Plastic Analysis Notes