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).
σ
yield ‘plateau’
yield stress, σy
yield strain
onset of strain
hardening
-εy
We further assume that the beam crosssection has at least one axis of symmetry
which lies in the plane of bending.
εy
ε
-σy
plane of loading and bending
coincides with plane of symmetry
examples of symmetric sections
M
M
neutral
axis
εy
σy
elastic neutral
axis
ε
σy
strain
(linear)
plastic
neutral axis
σ
σy
σy
σy
M = Mp
‘Fully Plastic Moment’
M = My
‘Yield Moment’
maximum stress has
just reached σ y for
the first time
Plastic_Analysis_Notes.doc
σy
increasing
ymax
y
σy
p1
as strain increases yielding
penetrates further into the beam
yield stress penetrates
entire cross-section
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
σy =
My y max
I
where ZE =
, or My = ZE σ y
I
y max
(the elastic section modulus)
Plastic Neutral axis
Under elastic conditions the neutral axis (zero strain locus) passes through the centroid of the crosssection. 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
A1
G1
_
y1
_
y2
C
plastic neutral
axis
MP
G2
A2
σy
T
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:
A1 σ y = A2 σ y
∴ A1 = A2 = A / 2
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 y1 and y2 denote the distance of the centroids of areas A1 and A2 from the PNA.
Let C denote the compressive force due to σ y acting on A1, and T the tensile force on A2 –
C = σ y A1 = σ y A / 2
T = C = σ yA / 2
Taking moments about the PNA –
MP = Cy1 + Ty2
=
σ yA( y1 + y2 )
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or
MP = Z P σ y
where ZP is the plastic section modulus.
Plastic section modulus
It follows from the preceding relationship that
ZP =
A(y1 + y 2 )
2
or, more strictly Z P =
(
A y1 + y 2
)
2
(taking absolute values of y1 and y2 ).
More generally we could write
ZP =
∫ y dA
A
where y denotes the distance of each element of area, dA, from the PNA.
Shape factor
Maximum elastic moment, My = ZE σ y , where ZE =
I
y max
(the elastic section modulus).
Ultimate (fully plastic) moment, MP = Z P σ y .
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).
f=
MP
My
=
ZP
Zy
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
b
bd ⎛ d d ⎞ bd 2
ZP =
⎜ + ⎟=
2 ⎝4 4⎠
4
ZE =
f=
I
bd / 12 bd
=
=
y max
d/2
6
ZP
ZE
3
=
d/2
2
( - should know this….)
d/2
6
= 1.5
4
If b=40mm, d=120mm and σ y = 250,000 kPa, the fully plastic moment will be
MP = Z P σ y =
0.04 × 0.12 2
× 250,000 = 36 kNm
4
Example – collapse load of a steel T-beam
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10
W
yP
plastic
neutral
axis
140
2m
10
100
σ y = 280 MPa
γ steel = 76 kN / m 3
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.
A (140 + 100)10
=
2
2
yP = 120mm
10 yP =
Plastic section modulus
ZP =
∑Ay
i
i
(summing the area moments of each rectangle about the PNA)
= 120 × 10 × 60 + 20 × 10 × 10 + 100 × 10 × 25
= 99,000 mm3
= 99 × 10 −6 m3
Fully plastic moment
MP = ZP σ y
yielded material
Wcollapse
= 99 × 10 × 280,000
−6
= 27.72 kN − m
Collapse load
Ignoring self-weight, the maximum bm is 2W
kN-m at the left hand end. Equating this to
MP :
plastic hinge
Collapse mechanism
2Wcollapse = 27.72
Wcollapse = 13.86 kN
Show also that PNA is 18.75 mm from the elastic neutral axis, ZE = 55.1 × 10 −6 m3 , 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)
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Spread of plasticity in the plastic hinge vicinity [Megson I p.244, Megson II p.602]
material at σy
b
c
My
MP
σy
σy
yielded
elastic
under point load
at b
at c
STRESS DISTRIBUTION
at 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 crosssection.
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.
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Plastic Analysis
Role in Design
When designing for the strength limit state, the objective is to satisfy the inequality
S * ≤ φ Sn .
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:
M * ≤ φ Mn (where Mn may be replaced by MP, the fully plastic moment).
Truss structures:
N * ≤ φ Nn (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
collapse
mechanism
2
wL /8
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)
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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.
4W
W
B
A
C
D
L/2
L/2
L/2
WL/2
3WL/4
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
3WL
= MP
4
4MP
Wcollapse =
3L
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.
W
Example – collapse load of an indeterminate beam
Consider the propped cantilever with central point load,
W. An elastic analysis (e.g. by using integration, momentarea 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.
L/2
L/2
bm, elastic analysis
3WL/16 = MP
5WL/32
1st plastic hinge
forms here
bm increases with W
MP
bm at collapse
Finally we carry out a static equilibrium analysis of the
beam at the instant of collapse to determine the value
of W.
MP
But hold on, this is an indeterminate structure isn’t it?
How can we use just static equilibrium to analyse it?
Wcollapse
Answer is that each plastic hinge provides the value of
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collapse mechanism
Copyright © J.W.Butterworth, July 2005
the bm at its location. This additional information allows the analysis to be completed.
W
A
C
MP
B
MP
VA
VC
Taking moments about B for segment BC:
L
,
2P
2MP
VC =
L
MP = VC
C
MP
B
VC
Taking moments about A for AC:
L
− VC L,
2
2MP ⎞
2⎛
W = ⎜⎜ MP +
L ⎟⎟
L⎝
L
⎠
6MP
=
L
MP = W
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:
W
Apply a vertical virtual displacement ∆ to whole:
A
B
VW done = W∆ − VA ∆ − VC ∆ = 0,
VA + VC = W
∆
Next apply a rotational virtual displacement δθ about
A:
VW done = W
VC =
W
2
C
VA
VC
δθ
L
δθ − VC Lδθ = 0,
2
Lδθ
VC
Giving us the expected results.
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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.
W
θ
MP
θ
2θ
MP
L
θ
2
Internal VW = MP θ + MP 2θ
External VW = W
collapse mechanism as virtual displacement
where the internal work is obtained as the product of moment x rotation.
Equating the internal and external VW gives
L
θ = 3MP θ
2
6MP
W=
L
W
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
θ
θ
2θ
L/3
L/2
Denoting the rotations as shown, we proceed directly to the calculation of the collapse load using virtual
work principle:
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L
θ,
3
Internal VW = MP (θ + 2θ + θ)
External VW = W
and equating,
W=
12MP
L
Repeating for the mechanism (B) below:
W
θ
2θ
3θ
L/3
L/3
L
θ,
3
Internal VW = MP (θ + 2θ + 3θ)
External VW = W
and equating,
W=
18MP
L
Not surprisingly we see that the result depends on the choice of the collapse mechanism.
Trying one more, (C):
W
2θ
θ
3θ
L/3
2L/3
L
External VW = W 2θ,
3
Internal VW = MP (θ + 2θ + 3θ)
and equating,
W=
9MP
L
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.
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W=12MP/L
MP
MP
(A)
MP
W=18MP/L
MP
MP
(B)
MP
W=9MP/L
MP
MP
(C)
MP
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=36MP/5L
(A)
3MP/5
3MP /5
no plastic hinges
at supports now
MP
original bm
This leaves just a single plastic hinge under the point load.
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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
12MP
36MP
≤W≤
L
5L
7.2MP
12MP
≤W≤
L
L
With the true collapse load of
9MP
lying close to the mean of the upper and lower bounds.
L
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.
w
The geometry of the resulting mechanism has the angles shown in
the figure.
x
L−x
+ w( L − x ) x θ
2
2
Internal VW = MP θ(L − x + L)
External VW = wx(L − x)θ
Equating:
(L-x)θ
xθ
Lθ
x
2MP (2L − x)
w=
Lx(L − x)
Correct value of x will minimise the collapse load. Hence seek x such that
L-x
dW
= 0 , leading to
dx
x 2 − 4Lx + 2L2 = 0
x = (2 − 2 )L = 0.586L
w = (6 + 4 2 )
= 11.657
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MP
L2
MP
L2
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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.
σy
σy
b
σy
stress
resisting
M
M
d/2-a
d/2+a
P
a
stress
resisting
M
σy
σ = P/A
M=0
small M
compressive
yield
tensile
yield
d/2
a
stress
resisting
P
d/2
σy
fully
plastic
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 crosssection, 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.
MRP = σ y b(
d
d
d2
− a)( + a) = σ y b(
− a2 )
2
2
4
(1)
(2)
P = σ y b(2a)
substituting for a in equation 1:
MRP = σ y b(
MRP =
d2
P2
)
−
4
4σ 2y b 2
bd2
P2
σy −
4
4σ y b
(3)
Now let Py = squash load (axial load that will fully yield cross-section).
Thus
Py = bdσ y
also
MP =
we get
MRP = MP −
bd2
σ y , and since P = 2abσ y ,
4
P2d
, or
4Py
⎛P
MRP = MP − MP ⎜
⎜ Py
⎝
= MP (1 − n 2 )
where n =
⎞
⎟
⎟
⎠
MRP
MP
2
P
Py
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1.0
P
Py
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1.0
Copyright © J.W.Butterworth, July 2005
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
yielded zone
σy
2
bde
d − de d + de
M=
⋅ σy + b ⋅
⋅
⋅ σy
6
2
2
(d-de)/2
where
(d+de)/2
de
1st term = bm resisted by the elastic core, and
2nd term = bm resisted by the yielded (plastic)
zones.
i.e.
M=
σy
elastic core
bde 2 b 2
+ (d − de2 )σ y
6
4
cross-section
stress distribution
⎛ d2 de2 ⎞
⎟σ y
M = b⎜⎜
−
12 ⎟⎠
⎝ 4
(1)
The curvature of the elastic core (and of the beam),
M
1
=κ= e
R
EIe
where Me = resisting moment of elastic core = Ze σ y
Ie = moment of inertia of elastic core
bde2
σy
Ze σ y
∴κ =
= 6 3
EIe
bd
E e
12
i.e.
κ=
M 1.0
MP
a b
Behaviour Model
2/3
oac = rigid plastic
2σ y
(2)plastic
obc = elastic-perfectly
Ede
Equations (1) and (2) define bm and curvature in terms of
de, where de ranges from d down to zero.
when de = d, M =
bd2
σ y = My , the yield moment.
6
2
bd
σ y = MP , the fully plastic moment
4
and κ → ∞ .
when de = 0, M =
c
oc = elasto-plastic
1/3
0
1 2 3
κ
κy
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.
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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
σy
c
σy b
unloading
MP
MP
(elastic)
a
ε
εy
-σy
d
When the moment
reaches MP, the entire cross-section is assumed to have reached either the tensile or compressive yield
stress.
stress under MP
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
MP
MP
σmax
change in stress
due to reverse MP
MP
.
ZE
Where
σ max =
But as
MP = ZP σ y ,
σ max =
ZP
σ y = fσ y
ZE
= 1.5σ y for a rectangular section
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).
σy
M=0
1.5σy
0.5σy
M=0
+
σy
1.5σy
MP
Plastic_Analysis_Notes.doc
=
p15
-MP
σy
σy
0.5σy
residual stress
when M=0
Copyright © J.W.Butterworth, July 2005
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
2.
MECHANISM CONDITION
3.
YIELD CONDITION
At collapse the bending moments must everywhere be ≤ MP.
At collapse, the bending moments must correspond to a state of equilibrium between the external
loads and the internal actions.
At collapse there must be sufficient plastic hinges to create a partial or complete collapse
mechanism.
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:
.
MECHANISM
UNIQUENESS
UPPER BOUND
load ≥ true collapse load
EQUILIBRIUM
LOWER BOUND
load ≤ true collapse load
load = true collapse load
YIELD
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).
Plastic_Analysis_Notes.doc
p16
Copyright © J.W.Butterworth, July 2005
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
W/4
L/2
L/4
W/4
W
θ
W
θ
MP 3θ = W
W=
48MP
7
θ
θ
Correct mechanism
W/4
θ
Assumed mechanism
L
W L
θ−
θ
2
4 4
MP 4θ = W
W=
7W/48
8MP
L
L
θ
2
WL/8
WL/8
WL/16
7W/48
WL/8
Assumed mechanism - bm diagram
(doesn’t satisfy equilibrium, but the
calculated collapse load is still a valid
upper bound)
True collapse bm diagram
(satisfies equilibrium)
Plastic_Analysis_Notes.doc
not in equilibrium
p17
Copyright © J.W.Butterworth, July 2005
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
A
3.5W
MP = 120kNm
B
2m
D
C
3m
1m
3.5W
1)
4θ
2θ
2W
6θ
2)
θ
2W
5θ
3.5W
6θ
Mechanism 1:
2W(4 θ)2 + 3.5W(2θ)1 = MP (4 + 6 + 2) θ
W=
12MP
= 62.61kN
23
← lowest
Mechanism 2:
2W(θ)2 + 3.5W(5θ)1+ = MP (1 + 6 + 5) θ
W=
12MP
= 66.98kN
21.5
Plastic_Analysis_Notes.doc
p18
Copyright © J.W.Butterworth, July 2005
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:
MP
2(62.61)
θ
5θ
MP
3.5(62.61)
6θ
MC (unknown)
MP (θ + 5θ) + MC (6θ) = 2(62.61)(θ)2 + 3.5(62.61)(5θ)1
MC = 104.35kNm (≤ MP , OK)
giving the collapse bm diagram below:
120
120
104.35
120
The same result could be found by ‘conventional’ methods:
3.5W
Consider segment BD and take moments about B:
MP + 3.5 × 62.61 × 3 + MP − 4VD = 0
VD = 224.35kN
MP
D
B
C
3m
VD
1m
Summing moments about C for segment CD:
MC + MP = VD × 1
MC = 224.35 × 1 − 120 = 104.35kNm. (≤ MP , OK )
MC
C
D
VD
1m
Plastic_Analysis_Notes.doc
p19
Copyright © J.W.Butterworth, July 2005
Example – Fixed base rectangular frame, varying MP, point loads
(examples of frame analyses can be found in Megson II from p.613)
8W kN
6W kN
4
3
2
MP (beam) = 60kNm
4m
MP (both columns) = 40kNm
1
5
6m
1.
6m
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:
8W
6W
6W
θ
8W
6W
θ
2θ
θ
θ
θ
θ
θ
‘SWAY’ mechanism
‘BEAM’ mechanism
(example of partial collapse mechanism)
θ
90°
θ
θ
8W
θ
‘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
MP (column) (4θ) = 6W.4.θ
W = 6.67
BEAM
MP ( column) (2θ) + MP (beam) (2θ) = 8W.6.θ
40 × 2 + 60 × 2
8×6
= 4.167
W=
Plastic_Analysis_Notes.doc
p20
Copyright © J.W.Butterworth, July 2005
COMBINED
MP ( column) (4θ) + MP (beam) (2θ) = 6W.4.θ + 8W.6.θ
31.11kN
23.33 kN
40 × 4 + 60 × 2
24 + 48
= 3.889
W=
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.
3
2
4
1
5
M 4 = 40
(M P(column) )
M2
31.11
θ
θ
2θ
M 3 = 60
(M P(beam) )
Using the beam mechanism as a virtual displacement to find
M2:
M2 (θ) + 60(2θ) + 40(θ) = 31.11 × 6 × θ
M2 = 26.7kNm
Hence collapse bm diagram as shown with M ≤ MP
everywhere proving solution is correct (uniqueness theorem).
40
26.7
60
40
40
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.
3W
2W
2
3
L
4
all members
have same MP
1
5
L
1.
L
Diagram above shows frame details and 5 possible plastic hinge locations.
Plastic_Analysis_Notes.doc
p21
Copyright © J.W.Butterworth, July 2005
ic
θ
θ
3W
θ
3W
θ
2W
2θ
2W
3W
Mechanism A
Mechanism B
θ
4θ
θ
θ
4θ
θ
2θ
2θ
2.
θ
plastic hinge
cancels
θ
θ
Mechanism C
= 2A + B
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:
L
3W. .θ = MP .4Gq
2
8MP
W=
3L
B:
L
3W. .θ + 2W.L.θ = MP .6θ
2
12MP
W=
7L
C:
L
3W. .3θ + 2W.L.θ = MP .10θ
2
20MP
W=
← lowest
13L
Plastic_Analysis_Notes.doc
p22
Copyright © J.W.Butterworth, July 2005
2W
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:
L
3W. .θ = M2 .θ + MP .3θ
2
M2
3W
θ
20MP
subst. W =
13L
θ
2θ
MP
MP
60
− 3)MP
26
= − 0.692MP
M2 = (
MP
0.69M P
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.
MP
MP
MP
Example – Beam design
5m
8m
2kPa (dead)
3m
5kPa (live)
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.
Plastic_Analysis_Notes.doc
p23
Copyright © J.W.Butterworth, July 2005
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.
2x3
6.0
Self weight UB
60kg/m
0.6 (guess)
Total
Live:
G = 17.4kN/m
Specified
Factored
5x3
Q = 15kN/m
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).
0.586L
(4.7m)
5m
3.3m
We’ve already worked out the collapse load for this case (see p.12) and obtained w = 11.657
MP
.
L2
Hence the maximum design action (bending moment) imposed by the factored loads will be
w.L2
11.657
43.4 × 82
=
11.657
= 238 kNM
M* =
Selection of beam with adequate strength
We require a beam with φMn ≥ M*, and since Mn = ZPσy,
ZP ≥
238
0.9 × σ y
≥.
238
0.9 × 300,000
≥ 0.00088 m 3
≥ 881 × 10 3 mm 3
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.
Plastic_Analysis_Notes.doc
p24
Copyright © J.W.Butterworth, July 2005
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.
265.8
Thus,
M * = 266kNm , requiring
ZP ≥
266
φσ y
34.3
≥ 985 × 10 3 mm 3
225.8
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.
Plastic_Analysis_Notes.doc
p25
Copyright © J.W.Butterworth, July 2005
Flange
Overall
depth
Root
radius
Depth
between
flanges
Web
slenderness
Flange
outstand
Gross
area
d1
t
B−t
2
Ag
Width
Thickness
d
B
T
t
r
d1
mm
mm
mm
mm
mm
mm
Designation
kg/m
Web
thickness
About x-axis
Ix
2
mm
6
10 mm
Zx
4
3
10 mm
3
Warping
constant
ry
J
Iw
mm
10 mm
About y-axis
ZPx
3
Torsion
constant
3
10 mm
rx
Iy
mm
10 mm
6
Zy
4
3
10 mm
ZPy
3
3
10 mm
3
3
4
9
10 mm
6
610UB
125
113
101
612
607
602
229
228
228
19.6
17.3
14.8
11.9
11.2
10.6
14.
14.0
14.0
572
572
572
48.1
51.1
54.0
5.54
6.27
7.34
16000
14500
13000
986
875
761
3230
2880
2530
3680
3290
2900
249
246
242
39.3
34.3
29.3
343
300
257
536
469
402
49.6
48.7
47.5
1560
1140
790
3450
2980
2530
530UB
92.4
82
533
528
209
209
15.6
13.2
10.2
9.6
14.0
14.0
502
502
49.2
52.3
6.37
7.55
11800
10500
554
477
2080
1810
2370
2070
217
213
23.8
20.1
228
193
355
301
44.9
43.8
775
526
1590
1330
460UB
82.1
74.6
67.1
460
457
454
191
190
190
16.0
14.5
12.7
9.9
9.1
8.5
14.0
11.4
11.4
428
428
428
43.3
47.1
50.4
5.66
6.24
7.15
10500
9520
8580
372
335
296
1610
1460
1300
1840
1660
1480
188
188
186
18.6
16.6
14.5
195
175
153
303
271
238
42.2
41.8
41.2
701
530
378
919
815
708
410UB
59.7
53.7
406
403
178
178
12.8
10.9
7.8
7.6
11.4
11.4
381
381
48.8
50.1
6.65
7.82
7640
6890
216
188
1060
933
1200
1060
168
165
12.1
10.3
135
115
209
179
39.7
38.6
337
234
467
394
360UB
56.7
50.7
44.7
359
356
352
172
171
171
13.0
11.5
9.7
8.0
7.3
6.9
11.4
11.4
11.4
333
333
333
41.6
45.6
48.2
6.31
7.12
8.46
7240
6470
5720
161
142
121
899
798
689
1010
897
777
149
148
146
11.0
9.60
8.10
128
112
94.7
198
173
146
39.0
38.5
37.6
338
241
161
330
284
237
310UB
46.2
40.4
32.0
307
304
298
166
165
149
11.8
10.2
8.0
6.7
6.1
5.5
11.4
11.4
13.0
284
284
282
42.3
46.5
51.3
6.75
7.79
8.97
5930
5210
4080
100
86.4
63.2
654
569
424
729
633
475
130
129
124
9.01
7.65
4.42
109
92.7
59.3
166
142
91.8
39.0
38.3
32.9
233
157
86.5
197
165
92.9
250UB
37.3
31.4
25.7
256
252
248
146
146
124
10.9
8.6
8.0
6.4
6.1
5.0
8.9
8.9
12
234
234
232
36.6
38.4
46.4
6.40
8.13
7.44
4750
4010
3270
55.7
44.5
35.4
435
354
285
486
397
319
108
105
104
5.66
4.47
2.55
77.5
61.2
41.1
119
94.2
63.6
34.5
33.4
27.9
158
89.3
67.4
85.2
65.9
36.7
200UB
29.8
25.4
22.3
18.2
207
203
202
198
134
133
133
99
9.6
7.8
7.0
7.0
6.3
5.8
5.0
4.5
8.9
8.9
8.9
11.0
188
188
188
184
29.8
32.3
37.5
40.9
6.65
8.15
9.14
6.75
3820
3230
2870
2320
29.1
23.6
21.0
15.8
281
232
208
160
316
260
231
180
87.3
85.4
85.5
82.6
3.86
3.06
2.75
1.14
57.5
46.1
41.3
23.0
88.4
70.9
63.4
35.7
31.8
30.8
31.0
22.1
105
62.7
45.0
38.6
37.6
29.2
26.0
10.4
180UB
22.2
18.1
16.1
179
175
173
90
90
90
10.0
8.0
7.0
6.0
5.0
4.5
8.9
8.9
8.9
159
159
159
26.5
31.8
35.3
4.20
5.31
6.11
2820
2300
2040
15.3
12.1
10.6
171
139
123
195
157
138
73.6
72.6
72.0
1.22
0.975
0.853
27.1
21.7
19.0
42.3
33.7
29.4
20.8
20.6
20.4
81.6
44.8
31.5
8.71
6.80
5.88
150UB
18.0
14.0
155
150
75
75
9.5
7.0
6.0
5.0
8.0
8.0
136
136
22.7
27.2
3.63
5.00
2300
1780
9.05
6.66
117
88.8
135
102
62.8
61.1.
0.672
0.495
17.9
13.2
28.2
20.8
17.1
16.6
60.5
28.1
3.56
2.53
UNIVERSAL BEAMS – Dimensions and Properties
Plastic_Analysis_Notes.doc
p26
Copyright © J.W.Butterworth, July 2005