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C h a p t e r 18 / Plastic Analysis of
Beams and Frames
So far our analysis of the behaviour of structures has assumed that whether the struc-
tures are statically determinate or indeterminate the loads on them cause stresses
which lie within the elastic limit. Design, based on this elastic behaviour, ensures
that the greatest stress in a structure does not exceed the yield stress divided by an
appropriate factor of safety.
An alternative approach is based on plastic analysis in which the loads required to cause
the structure to collapse are calculated. The reasoning behind this method is that, in
most steel structures, particularly redundant ones, the loads required to cause the
structure to collapse are somewhat larger than the ones which cause yielding. Design,
based on this method, calculates the loading required to cause complete collapse and
then ensures that this load is greater than the applied loading; the ratio of collapse load
to the maximum applied load is called the load factor. Generally, plastic, or ultimate
load design, results in more economical structures.
In this chapter we shall investigate the mechanisms of plastic collapse and determine
collapse loads for a variety of beams and frames.
18.1 THEOREMS OF PLASTIC ANALYSIS
Plastic analysis is governed by three fundamental theorems which are valid for elasto-
plastic structures in which the displacements are small such that the geometry of the
displaced structure does not affect the applied loading system.
THE UNIQUENESS THEOREM
The following conditions must be satisfied simultaneously by a structure in its collapsed
state:
The equilibrium condition states that the bending moments must be in equilibrium with
the applied loads.
The yield condition states that the bending moment at any point in the structure must
not exceed the plastic moment at that point.
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18.2 Plastic Analysis of Beams
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The mechanism condition states that sufficient plastic hinges must have formed so that
all, or part of, the structure is a mechanism.
THE LOWER BOUND, OR SAFE, THEOREM
If a distribution of moments can be found which satisfies the above equilibrium and
yield conditions the structure is either safe or just on the point of collapse.
THE UPPER BOUND, OR UNSAFE, THEOREM
If a loading is found which causes a collapse mechanism to form then the loading must
be equal to or greater than the actual collapse load.
Generally, in plastic analysis, the upper bound theorem is used. Possible collapse
mechanisms are formulated and the corresponding collapse loads calculated. From the
upper bound theorem we know that all mechanisms must give a value of collapse load
which is greater than or equal to the true collapse load so that the critical mechanism
is the one giving the lowest load. It is possible that a mechanism, which would give a
lower value of collapse load, has been missed. A check must therefore be carried out
by applying the lower bound theorem.
18.2 PLASTIC ANALYSIS OF BEAMS
Generally plastic behaviour is complex and is governed by the form of the stress–strain
curve in tension and compression of the material of the beam. Fortunately mild steel
beams, which are used extensively in civil engineering construction, possess structural
properties that lend themselves to a relatively simple analysis of plastic bending.
We have seen in Section 8.3, Fig. 8.8, that mild steel obeys Hooke’s law up to a
sharply defined yield stress and then undergoes large strains during yielding until
strain hardening causes an increase in stress. For the purpose of plastic analysis we
shall neglect the upper and lower yield points and idealize the stress–strain curve as
shown in Fig. 18.1. We shall also neglect the effects of strain hardening, but since this
FIGURE 18.1
Idealized
stress–strain curve
for mild steel
sY (Compression)
s
sY
(Tension)

Y
Y
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• Chapter 18 / Plastic Analysis of Beams and Frames
provides an increase in strength of the steel it is on the safe side to do so. Finally we
shall assume that both Young’s modulus, E, and the yield stress, σY, have the same
values in tension and compression, and that plane sections remain plane after bending.
The last assumption may be shown experimentally to be very nearly true.
PLASTIC BENDING OF BEAMS HAVING A SINGLY
SYMMETRICAL CROSS SECTION
This is the most general case we shall discuss since the plastic bending of beams of
arbitrary section is complex and is still being researched.
Consider the length of beam shown in Fig. 18.2(a) subjected to a positive bending
moment, M , and possessing the singly symmetrical cross section shown in Fig. 18.2(b).
If M is sufficiently small the length of beam will bend elastically, producing at any
section mm, the linear direct stress distribution of Fig. 18.2(c) where the stress, σ , at
a distance y from the neutral axis of the beam is given by Eq. (9.9). In this situation
the elastic neutral axis of the beam section passes through the centroid of area of the
section (Eq. (9.5)).
Suppose now that M is increased. A stage will be reached where the maximum direct
stress in the section, i.e. at the point furthest from the elastic neutral axis, is equal
to the yield stress, σY (Fig. 18.3(b)). The corresponding value of M is called the yield
moment, MY, and is given by Eq. (9.9); thus
MY = σYI
y1
(18.1)
FIGURE 18.2
Direct stress due
to bending in a
singly symmetrical
section beam
m
m
(a)
G
(b)
Section mm
Elastic
neutral
axis
G
(c)
M
y
M
x
z
y
FIGURE 18.3
Yielding of a beam
section due to
bending
y
z
y2
y1
G
Elastic
neutral
axis
(c)
(d)
(e)
(b)
(a)
Plastic
neutral
axis
sY
sY
sY
sY
sY
sY
sY
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18.2 Plastic Analysis of Beams
• 595
If the bending moment is further increased, the strain at the extremity y1 of the section
increases and exceeds the yield strain, εY. However, due to plastic yielding the stress
remains constant and equal to σY as shown in the idealized stress–strain curve of
Fig. 18.1. At some further value of M the stress at the lower extremity of the section also
reaches the yield stress, σY (Fig. 18.3(c)). Subsequent increases in bending moment
cause the regions of plasticity at the extremities of the beam section to extend inwards,
producing a situation similar to that shown in Fig. 18.3(d); at this stage the central
portion or ‘core’ of the beam section remains elastic while the outer portions are
plastic. Finally, with further increases in bending moment the elastic core is reduced
to a negligible size and the beam section is more or less completely plastic. Then, for
all practical purposes the beam has reached its ultimate moment resisting capacity;
the value of bending moment at this stage is known as the plastic moment, MP, of the
beam. The stress distribution corresponding to this moment may be idealized into two
rectangular portions as shown in Fig. 18.3(e).
The problem now, therefore, is to determine the plastic moment, MP. First, however,
we must investigate the position of the neutral axis of the beam section when the latter
is in its fully plastic state. One of the conditions used in establishing that the elastic
neutral axis coincides with the centroid of a beam section was that stress is directly
proportional to strain (Eq. (9.2)). It is clear that this is no longer the case for the stress
distributions of Figs 18.3(c), (d) and (e). In Fig. 18.3(e) the beam section above the
plastic neutral axis is subjected to a uniform compressive stress, σY, while below the neu-
tral axis the stress is tensile and also equal to σY. Suppose that the area of the beam
section below the plastic neutral axis is A2, and that above, A1 (Fig. 18.4(a)). Since
MP is a pure bending moment the total direct load on the beam section must be zero.
Thus from Fig. 18.4
σYA1 = σYA2
so that
A1 = A2
(18.2)
FIGURE 18.4
Position of the
plastic neutral axis
in a beam section
Area, A1
Area, A2
(a)
(b)
Plastic neutral
axis
C2
C1
sY
sY
y1
y2
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• Chapter 18 / Plastic Analysis of Beams and Frames
Therefore if the total cross-sectional area of the beam section is A
A1 = A2 = A2
(18.3)
and we see that the plastic neutral axis divides the beam section into two equal areas.
Clearly for doubly symmetrical sections or for singly symmetrical sections in which the
plane of the bending moment is perpendicular to the axis of symmetry, the elastic and
plastic neutral axes coincide.
The plastic moment, MP, can now be found by taking moments of the resultants of the
tensile and compressive stresses about the neutral axis. These stress resultants act at
the centroids C1 and C2 of the areas A1 and A2, respectively. Thus from Fig. 18.4
MP = σYA1ȳ1 + σYA2ȳ2
or, using Eq. (18.3)
MP = σY A2 (ȳ1 +ȳ2)
(18.4)
Equation (18.4) may be written in a similar form to Eq. (9.13); thus
MP = σYZP
(18.5)
where
ZP = A(ȳ1 +ȳ2)
2
(18.6)
ZP is known as the plastic modulus of the cross section. Note that the elastic modulus,
Ze, has two values for a beam of singly symmetrical cross section (Eq. (9.12)) whereas
the plastic modulus is single-valued.
SHAPE FACTOR
The ratio of the plastic moment of a beam to its yield moment is known as the shape
factor, f. Thus
f = MP
MY
= σYZP
σYZe
= ZP
Ze
(18.7)
where ZP is given by Eq. (18.6) and Ze is the minimum elastic section modulus, I/y1.
It can be seen from Eq. (18.7) that f is solely a function of the geometry of the beam
cross section.
EXAMPLE 18.1 Determine the yield moment, the plastic moment and the shape
factor for a rectangular section beam of breadth b and depth d.
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18.2 Plastic Analysis of Beams
• 597
The elastic and plastic neutral axes of a rectangular cross section coincide (Eq. (18.3))
and pass through the centroid of area of the section. Thus, from Eq. (18.1)
MY = σYbd
3/12
d/2
= σY bd
2
6
(i)
and from Eq. (18.4)
MP = σY bd2
(
d
4
+ d
4
)
= σY bd
2
4
(ii)
Substituting for MP and MY in Eq. (18.7) we obtain
f = MP
MY
= 3
2
(iii)
Note that the plastic collapse of a rectangular section beam occurs at a bending
moment that is 50% greater than the moment at initial yielding of the beam.
EXAMPLE 18.2 Determine the shape factor for the I-section beam shown in
Fig. 18.5(a).
Again, as in Ex. 18.1, the elastic and plastic neutral axes coincide with the centroid,
G, of the section.
In the fully plastic condition the stress distribution in the beam is that shown in
Fig. 18.5(b). The total direct force in the upper flange is
σYbtf
(compression)
FIGURE 18.5 Beam
section of Ex. 18.2
(a)
(b)
G
Elastic and
plastic
neutral
axes
b = 150 mm
d = 300 mm
tw = 8 mm
tf = 12 mm
z
y
sY
sY
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• Chapter 18 / Plastic Analysis of Beams and Frames
and its moment about Gz is
σYbtf
(
d
2
− tf
2
)
≡ σYbtf
2
(d − tf )
(i)
Similarly the total direct force in the web above Gz is
σYtw
(
d
2
− tf
)
(compression)
and its moment about Gz is
σYtw
(
d
2
− tf
)
1
2
(
d
2
− tf
)
≡ σYtw
8
(d − 2tf )2
(ii)
The lower half of the section is in tension and contributes the same moment about Gz
so that the total plastic moment, MP, of the complete section is given by
MP = σY
[
btf (d − tf ) + 14 tw(d − 2tf )
2]
(iii)
Comparing Eqs (18.5) and (iii) we see that ZP is given by
ZP = btf (d − tf ) + 14 tw(d − 2tf )
2
(iv)
Alternatively we could have obtained ZP from Eq. (18.6).
The second moment of area, I, of the section about the common neutral axis is
I = bd
3
12
− (b − tw)(d − 2tf )
3
12
so that the elastic modulus Ze is given by
Ze =
I
d/2
= 2
d
[
bd3
12
− (b − tw)(d − 2tf )
3
12
]
(v)
Substituting the actual values of the dimensions of the section in Eqs (iv) and (v) we
obtain
ZP = 150 × 12(300 − 12) + 14 × 8(300 − 2 × 12)
2 = 6.7 × 105 mm3
and
Ze = 2
300
[
150 × 3003
12
− (150 − 8)(300 − 24)
3
12
]
= 5.9 × 105 mm3
Therefore from Eq. (18.7)
f = MP
MY
= ZP
Ze
= 6.7 × 10
5
5.9 × 105 = 1.14
and we see that the fully plastic moment is only 14% greater than the moment at initial
yielding.
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18.2 Plastic Analysis of Beams
• 599
EXAMPLE 18.3
Determine the shape factor of the T-section shown in
Fig. 18.6.
FIGURE 18.6 Beam
section of Ex. 18.3
200 mm
G
z
yP
y
ye
sY
sY
(a)
(b)
150 mm
10 mm
7 mm
Plastic
neutral axis
Elastic
neutral axis
In this case the elastic and plastic neutral axes are not coincident. Suppose that the
former is a depth ye from the upper surface of the flange and the latter a depth yP. The
elastic neutral axis passes through the centroid of the section, the location of which is
found in the usual way. Hence, taking moments of areas about the upper surface of
the flange
(150 × 10 + 190 × 7)ye = 150 × 10 × 5 + 190 × 7 × 105
which gives
ye = 52.0mm
The second moment of area of the section about the elastic neutral axis is then, using
Eq. (9.38)
I = 150 × 52
3
3
− 143 × 42
3
3
+ 7 × 148
3
3
= 11.1 × 106 mm4
Therefore
Ze = 11.1 × 10
6
148
= 75 000 mm3
Note that we choose the least value for Ze since the stress will be a maximum at a point
furthest from the elastic neutral axis.
The plastic neutral axis divides the section into equal areas (see Eq. (18.3)). Inspection
of Fig. 18.6 shows that the flange area is greater than the web area so that the plastic
neutral axis must lie within the flange. Hence
150yP = 150(10 − yP) + 190 × 7
from which
yP = 9.4mm
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• Chapter 18 / Plastic Analysis of Beams and Frames
Equation (18.6) may be interpreted as the first moment, about the plastic neutral axis,
of the area above the plastic neutral axis plus the first moment of the area below the
plastic neutral axis. Hence
ZP = 150 × 9.4 × 4.7 + 150 × 0.6 × 0.3 + 190 × 7 × 95.6 = 133 800 mm3
The shape factor f is, from Eq. (18.7)
f = MP
MY
= ZP
Ze
= 133 800
75 000
= 1.78
MOMENT–CURVATURE RELATIONSHIPS
From Eq. (9.8) we see that the curvature k of a beam subjected to elastic bending is
given by
k = 1
R
= M
EI
(18.8)
At yield, when M is equal to the yield moment, MY
kY = MY
EI
(18.9)
The moment–curvature relationship for a beam in the linear elastic range may
therefore be expressed in non-dimensional form by combining Eqs (18.8) and
(18.9), i.e.
M
MY
= k
kY
(18.10)
This relationship is represented by the linear portion of the moment–curvature dia-
gram shown in Fig. 18.7. When the bending moment is greater than MY part of the
beam becomes fully plastic and the moment–curvature relationship is non-linear. As
the plastic region in the beam section extends inwards towards the neutral axis the
curve becomes flatter as rapid increases in curvature are produced by small increases in
moment. Finally, the moment–curvature curve approaches the horizontal line M =MP
M/MY
1
M
MP
M
MY
f
k /kY
1
FIGURE 18.7 Moment–curvature diagram
for a beam
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18.2 Plastic Analysis of Beams
• 601
Fully plastic
zone
(a)
(b)
sY
sY
d
de
b
Elastic and
plastic neutral axis
FIGURE 18.8 Plastic
bending of a
rectangular-section beam
as an asymptote when, theoretically, the curvature is infinite at the collapse load. From
Eq. (18.7) we see that when M =MP, the ratio M/MY = f , the shape factor. Clearly the
equation of the non-linear portion of the moment–curvature diagram depends upon
the particular cross section being considered.
Suppose a beam of rectangular cross section is subjected to a bending moment which
produces fully plastic zones in the outer portions of the section (Fig. 18.8(a)); the
depth of the elastic core is de. The total bending moment, M , corresponding to the
stress distribution of Fig. 18.8(b) is given by
M = 2σYb 12 (d − de)
1
2
(
d
2
+ de
2
)
+ 2σY
2
b
de
2
2
3
de
2
which simplifies to
M = σYbd
2
12
(
3 − d
2
e
d2
)
= MY
2
(
3 − d
2
e
d2
)
(18.11)
Note that when de = d, M =MY and when de = 0, M = 3MY/2=MP as derived in
Ex. 18.1.
The curvature of the beam at the section shown may be found using Eq. (9.2) and
applying it to a point on the outer edge of the elastic core. Thus
σY = E de
2R
or
k = 1
R
= 2σY
Ede
(18.12)
The curvature of the beam at yield is obtained from Eq. (18.9), i.e.
kY = MY
EI
= 2σY
Ed
(18.13)
Combining Eqs (18.12) and (18.13) we obtain
k
kY
= d
de
(18.14)
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• Chapter 18 / Plastic Analysis of Beams and Frames
Substituting for de/d in Eq. (18.11) from Eq. (18.14) we have
M = MY
2
(
3 − k
2
Y
k2
)
so that
k
kY
=
1
√
3 − 2M/MY
(18.15)
Equation (18.15) gives the moment–curvature relationship for a rectangular section
beam for MY ≤M ≤MP, i.e. for the non-linear portion of the moment–curvature dia-
gram of Fig. 18.7 for the particular case of a rectangular section beam. Corresponding
relationships for beams of different section are found in a similar manner.
We have seen that for bending moments in the range MY ≤M ≤MP a beam sec-
tion comprises fully plastic regions and a central elastic core. Thus yielding occurs in
the plastic regions with no increase in stress whereas in the elastic core increases in
deformation are accompanied by increases in stress. The deformation of the beam is
therefore controlled by the elastic core, a state sometimes termed contained plastic
flow. As M approaches MP the moment–curvature diagram is asymptotic to the line
M =MP so that large increases in deformation occur without any increase in moment,
a condition known as unrestricted plastic flow.
PLASTIC HINGES
The presence of unrestricted plastic flow at a section of a beam leads us to the concept
of the formation of plastic hinges in beams and other structures.
Consider the simply supported beam shown in Fig. 18.9(a); the beam carries a concen-
trated load, W , at mid-span. The bending moment diagram (Fig. 18.9(b)) is triangular
in shape with a maximum moment equal to WL/4. If W is increased in value until
WL/4=MP, the mid-span section of the beam will be fully plastic with regions of
plasticity extending towards the supports as the bending moment decreases; no plas-
ticity occurs in beam sections for which the bending moment is less than MY. Clearly,
unrestricted plastic flow now occurs at the mid-span section where large increases in
deformation take place with no increase in load. The beam therefore behaves as two
rigid beams connected by a plastic hinge which allows them to rotate relative to each
other. The value of W given by W = 4MP/L is the collapse load for the beam.
The length, LP, of the plastic region of the beam may be found using the fact that at
each section bounding the region the bending moment is equal to MY. Thus
MY = W2
(
L − LP
2
)
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18.2 Plastic Analysis of Beams
• 603
Region of plasticity
(a)
(b)
W
LP
L
MY
MP
2
W
2
W
FIGURE 18.9 Formation of a
plastic hinge in a simply
supported beam
Substituting for W (=4MP/L) we obtain
MY = MPL (L − LP)
from which
LP = L
(
1 − MY
MP
)
or, from Eq. (18.7)
LP = L
(
1 − 1
f
)
(18.16)
For a rectangular section beam f = 1.5 (see Ex. 18.1), giving LP =L/3. For the
I-section beam of Ex. 18.2, f = 1.14 and LP = 0.12L so that the plastic region in this
case is much smaller than that of a rectangular section beam; this is generally true for
I-section beams.
It is clear from the above that plastic hinges form at sections of maximum bending
moment.
PLASTIC ANALYSIS OF BEAMS
We can now use the concept of plastic hinges to determine the collapse or ultimate
load of beams in terms of their individual yield moment, MP, which may be found for
a particular beam section using Eq. (18.5).
For the case of the simply supported beam of Fig. 18.9 we have seen that the formation
of a single plastic hinge is sufficient to produce failure; this is true for all statically
determinate systems. Having located the position of the plastic hinge, at which the
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• Chapter 18 / Plastic Analysis of Beams and Frames
moment is equal to MP, the collapse load is found from simple statics. Thus for the
beam of Fig. 18.9, taking moments about the mid-span section, we have
WU
2
L
2
= MP
or
WU = 4MPL
(as deduced before)
where WU is the ultimate value of the load W .
EXAMPLE 18.4 Determine the ultimate load for a simply supported, rectangular
section beam, breadth b, depth d, having a span L and subjected to a uniformly
distributed load of intensity w.
The maximum bending moment occurs at mid-span and is equal to wL2/8 (see Section
3.4). The plastic hinge therefore forms at mid-span when this bending moment is equal
to MP, the corresponding ultimate load intensity being wU. Thus
wUL2
8
= MP
(i)
From Ex. 18.1, Eq. (ii)
MP = σY bd
2
4
so that
wU = 8MP
L2
= 2σYbd
2
L2
where σY is the yield stress of the material of the beam.
EXAMPLE 18.5 The simply supported beam ABC shown in Fig. 18.10(a) has a
cantilever overhang and supports loads of 4W and W . Determine the value of W at
collapse in terms of the plastic moment, MP, of the beam.
The bending moment diagram for the beam is constructed using the method of Section
3.4 and is shown in Fig. 18.10(b). Clearly as W is increased a plastic hinge will form
first at D, the point of application of the 4W load. Thus, at collapse
3
4
WUL = MP
so that
WU = 4MP
3L
where WU is the value of W that causes collapse.
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18.2 Plastic Analysis of Beams
• 605
(a)
(b)
C
A
D
B
ve
ve
4W
W
3WL
L 2
L 2
L 2
4
2
WL
FIGURE 18.10 Beam of Ex. 18.5
The formation of a plastic hinge in a statically determinate beam produces large,
increasing deformations which ultimately result in failure with no increase in load.
In this condition the beam behaves as a mechanism with different lengths of beam
rotating relative to each other about the plastic hinge. The terms failure mechanism
or collapse mechanism are often used to describe this state.
In a statically indeterminate system the formation of a single plastic hinge does not
necessarily mean collapse. Consider the propped cantilever shown in Fig. 18.11(a). The
bending moment diagram may be drawn after the reaction at C has been determined
by any suitable method of analysis of statically indeterminate beams (see Chapter 16)
and is shown in Fig. 18.11(b).
As the value of W is increased a plastic hinge will form first at A where the bending
moment is greatest. However, this does not mean that the beam will collapse. Instead
it behaves as a statically determinate beam with a point load at B and a moment MP at
A. Further increases in W eventually result in the formation of a second plastic hinge
at B (Fig. 18.11(c)) when the bending moment at B reaches the value MP. The beam
now behaves as a mechanism and failure occurs with no further increase in load. The
bending moment diagram for the beam is now as shown in Fig. 18.11(d) with values of
bending moment of−MP at A and MP at B. Comparing the bending moment diagram at
collapse with that corresponding to the elastic deformation of the beam (Fig. 18.11(b))
we see that a redistribution of bending moment has occurred. This is generally the
case in statically indeterminate systems whereas in statically determinate systems the
bending moment diagrams in the elastic range and at collapse have identical shapes
(see Figs 18.9(b) and 18.10(b)). In the beam of Fig. 18.11 the elastic bending moment
diagram has a maximum at A. After the formation of the plastic hinge at A the bending
moment remains constant while the bending moment at B increases until the second
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• Chapter 18 / Plastic Analysis of Beams and Frames
FIGURE 18.11
Plastic hinges in a
propped cantilever
A
B
C
(a)
(b)
W
L /2
L /2
3WL /16
5WL/32
ve
ve
Elastic bending moment diagram
A
B
WU
C
Collapse mechanism
Bending moment diagram at collapse
MP
ve
ve
MP
(c)
(d)
plastic hinge forms. Thus this redistribution of moments tends to increase the ultimate
strength of statically indeterminate structures since failure at one section leads to other
portions of the structure supporting additional load.
Having located the positions of the plastic hinges and using the fact that the moment
at these hinges is MP, we may determine the ultimate load, WU, by statics. Therefore
taking moments about A we have
MP = WU L2 − RCL
(18.17)
where RC is the vertical reaction at the support C. Now considering the equilibrium
of the length BC we obtain
RC
L
2
= MP
(18.18)
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18.2 Plastic Analysis of Beams
• 607
Eliminating RC from Eqs (18.17) and (18.18) gives
WU = 6MPL
(18.19)
Note that in this particular problem it is unnecessary to determine the elastic bending
moment diagram to solve for the ultimate load which is obtained using statics alone.
This is a convenient feature of plastic analysis and leads to a much simpler solution of
statically indeterminate structures than an elastic analysis. Furthermore, the magni-
tude of the ultimate load is not affected by structural imperfections such as a sinking
support, whereas the same kind of imperfection would have an appreciable effect on
the elastic behaviour of a structure. Note also that the principle of superposition (Sec-
tion 3.7), which is based on the linearly elastic behaviour of a structure, does not hold
for plastic analysis. In fact the plastic behaviour of a structure depends upon the order
in which the loads are applied as well as their final values. We therefore assume in
plastic analysis that all loads are applied simultaneously and that the ratio of the loads
remains constant during loading.
An alternative and powerful method of analysis uses the principle of virtual work (see
Section 15.2), which states that for a structure that is in equilibrium and which is given
a small virtual displacement, the sum of the work done by the internal forces is equal
to the work done by the external forces.
Consider the propped cantilever of Fig. 18.11(a); its collapse mechanism is shown in
Fig. 18.11(c). At the instant of collapse the cantilever is in equilibrium with plastic
hinges at A and B where the moments are each MP as shown in Fig. 18.11(d). Suppose
that AB is given a small rotation, θ . From geometry, BC also rotates through an angle θ
as shown in Fig. 18.12; the vertical displacement of B is then θL/2. The external forces
A
B
C
L /2
L /2
WU
u
u
2u
FIGURE 18.12 Virtual displacements
in propped cantilever of Fig. 18.11
on the cantilever which do work during the virtual displacement are comprised solely of
WU since the vertical reactions at A and C are not displaced. The internal forces which
do work consist of the plastic moments, MP, at A and B and which resist rotation. Hence
WUθ
L
2
= (MP)Aθ + (MP)B2θ
(see Section 15.1)
from which WU = 6MP/L as before.
We have seen that the plastic hinges form at beam sections where the bending moment
diagram attains a peak value. It follows that for beams carrying a series of point loads,
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• Chapter 18 / Plastic Analysis of Beams and Frames
plastic hinges are located at the load positions. However, in some instances several
collapse mechanisms are possible, each giving different values of ultimate load. For
example, if the propped cantilever of Fig. 18.11(a) supports two point loads as shown in
Fig. 18.13(a), three possible collapse mechanisms are possible (Fig. 18.13(b–d). Each
possible collapse mechanism should be analysed and the lowest ultimate load selected.
A
A
A
A
B
B
B
B
C
C
C
C
D
D
D
D
(a)
(b)
(c)
(d)
W1
W2
FIGURE 18.13 Possible collapse
mechanisms in a propped
cantilever supporting two
concentrated loads
The beams we have considered so far have carried concentrated loads only so that the
positions of the plastic hinges, and therefore the form of the collapse mechanisms, are
easily determined. This is not the case when distributed loads are involved.
EXAMPLE 18.6 The propped cantilever AB shown in Fig. 18.14(a) carries a uni-
formly distributed load of intensity w. If the plastic moment of the cantilever is MP
calculate the minimum value of w required to cause collapse.
Peak values of bending moment occur at A and at some point between A and B so that
plastic hinges will form at A and at a point C a distance x, say, from A; the collapse
mechanism is then as shown in Fig. 18.14(b) where the rotations of AC and CB are
θ and φ respectively. Then, the vertical deflection of C is given by
δ = θx = φ(L − x)
(i)
so that
φ = θ
x
L − x
(ii)
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18.2 Plastic Analysis of Beams
• 609
FIGURE 18.14
Collapse
mechanism for a
propped cantilever
A
u
RA
RB
C
x
L  x
B

f
u + f
(a)
(b)
A
B
w
L
The total load on AC is wx and its centroid (at x/2 from A) will be displaced a vertical
distance δ/2. The total load on CB is w(L− x) and its centroid will suffer the same
vertical displacement δ/2. Then, from the principle of virtual work
wx
δ
2
+ w(L − x) δ
2
= MPθ + MP(θ + φ)
Note that the beam at B is free to rotate so that there is no plastic hinge at B.
Substituting for δ from Eq. (i) and φ from Eq. (ii) we obtain
wL
θx
2
= MPθ + MP
(
θ + θ
x
L − x
)
or
wL
θx
2
= MPθ
(
2 +
x
L − x
)
Rearranging
w = 2MP
Lx
(
2L − x
L − x
)
(iii)
For a minimum value of w, (dw/dx)= 0. Then
dw
dx
= 2MP
L
[−x(L − x) − (2L − x)(L − 2x)
x2(L − x)2
]
= 0
which reduces to
x2 − 4Lx + 2L2 = 0
Solving gives
x = 0.586L
(the positive root is ignored)
Then substituting for x in Eq. (iii)
w (at collapse) = 11.66MP
L2
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• Chapter 18 / Plastic Analysis of Beams and Frames
We can now use the lower bound theorem to check that we have obtained the critical
mechanism and thereby the critical load. The internal moment at A at collapse is
hogging and equal to MP. Then, taking moments about A
RB L − wL
2
2
= −MP
which gives
RB = 4.83MP
L
Similarly, taking moments about B gives
RA = 6.83MP
L
Summation of RA and RB gives 11.66MP/L=wL so that vertical equilibrium is
satisfied. Further, considering moments of forces to the right of C about C we have
MC = RB(0.414L) − w0.414L
2
2
Substituting for RB and w from the above gives MC =MP. The same result is obtained
by considering moments about C of forces to the left of C. The load therefore satisfies
both vertical and moment equilibrium.
The bending moment at any distance x1, say, from B is given by
M = RBx1 − wx
2
1
2
Then
dM
dx1
= RB − wx1 = 0
so that a maximum occurs when x1 =RB/w. Substituting for RB, x1 and w in the expres-
sion for M gives M =MP so that the yield criterion is satisfied. We conclude, therefore,
that the mechanism of Fig. 18.14(b) is the critical mechanism.
PLASTIC DESIGN OF BEAMS
It is now clear that the essential difference between the plastic and elastic methods of
design is that the former produces a structure having a more or less uniform factor of
safety against collapse of all its components, whereas the latter produces a uniform
factor of safety against yielding. The former method in fact gives an indication of
the true factor of safety against collapse of the structure which may occur at loads
only marginally greater than the yield load, depending on the cross sections used. For
example, a rectangular section mild steel beam has an ultimate strength 50% greater
than its yield strength (see Ex. 18.1), whereas for an I-section beam the margin is in the
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18.2 Plastic Analysis of Beams
• 611
range 10–20% (see Ex. 18.2). It is also clear that each method of design will produce
a different section for a given structural component. This distinction may be more
readily understood by referring to the redistribution of bending moment produced by
the plastic collapse of a statically indeterminate beam.
Two approaches to the plastic design of beams are indicated by the previous analysis.
The most direct method would calculate the working loads, determine the required
strength of the beam by the application of a suitable load factor, obtain by a suitable
analysis the required plastic moment in terms of the ultimate load and finally, knowing
the yield stress of the material of the beam, determine the required plastic section
modulus. An appropriate beam section is then selected from a handbook of structural
sections. The alternative method would assume a beam section, calculate the plastic
moment of the section and hence the ultimate load for the beam. This value of ultimate
load is then compared with the working loads to determine the actual load factor, which
would then be checked against the prescribed value.
EXAMPLE 18.7 The propped cantilever of Fig. 18.11(a) is 10 m long and is required
to carry a load of 100 kN at mid-span. If the yield stress of mild steel is 300 N/mm2,
suggest a suitable section using a load factor against failure of 1.5.
The required ultimate load of the beam is 1.5× 100= 150 kN. Then from Eq. (18.19)
the required plastic moment MP is given by
MP = 150 × 10
6
= 250 kN m
From Eq. (18.5) the minimum plastic modulus of the beam section is
ZP = 250 × 10
6
300
= 833 333 mm3
Referring to an appropriate handbook we see that a Universal Beam,
406 mm× 140 mm× 46 kg/m, has a plastic modulus of 886.3cm3. This section there-
fore possesses the required ultimate strength and includes a margin to allow for its
self-weight. Note that unless some allowance has been made for self-weight in the
estimate of the working loads the design should be rechecked to include this effect.
EFFECT OF AXIAL LOAD ON PLASTIC MOMENT
We shall investigate the effect of axial load on plastic moment with particular reference
to an I-section beam, one of the most common structural shapes, which is subjected
to a positive bending moment and a compressive axial load, P, Fig. 18.15(a)).
If the beam section were subjected to its plastic moment only, the stress distribution
shown in Fig. 18.15(b) would result. However, the presence of the axial load causes
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• Chapter 18 / Plastic Analysis of Beams and Frames
FIGURE 18.15
Combined bending
and axial
compression
Plastic neutral axis
due to bending
Plastic neutral axis
due to bending
and axial load
(a)
(b)
Stress due to
axial load
tw
G
a
a
z
y
a
a
(c)
(d)
sY
sY
sY
sY
sY
sY
additional stresses which cannot, obviously, be greater than σY. Thus the region of
the beam section supporting compressive stresses is increased in area while the region
subjected to tensile stresses is decreased in area. Clearly some of the compressive
stresses are due to bending and some due to axial load so that the modified stress
distribution is as shown in Fig. 18.15(c).
Since the beam section is doubly symmetrical it is reasonable to assume that the area
supporting the compressive stress due to bending is equal to the area supporting
the tensile stress due to bending, both areas being symmetrically arranged about the
original plastic neutral axis. Thus from Fig. 18.15(d) the reduced plastic moment,
MP,R, is given by
MP,R = σY(ZP − Za)
(18.20)
where Za is the plastic section modulus for the area on which the axial load is assumed
to act. From Eq. (18.6)
Za = 2atw
2
(a
2
+ a
2
)
= a2tw
also
P = 2atwσY
so that
a = P
2twσY
Substituting for Za, in Eq. (18.20) and then for a, we obtain
MP,R = σY
(
ZP − P
2
4twσ 2Y
)
(18.21)
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18.3 Plastic Analysis of Frames
• 613
Let σa be the mean axial stress due to P taken over the complete area, A, of the beam
section. Then
P = σa A
Substituting for P in Eq. (18.21)
MP,R = σY
(
ZP − A
2
4tw
σ 2a
σ 2Y
)
(18.22)
Thus the reduced plastic section modulus may be expressed in the form
ZP,R = ZP − Kn2
(18.23)
where K is a constant that depends upon the geometry of the beam section and n is
the ratio of the mean axial stress to the yield stress of the material of the beam.
Equations (18.22) and (18.23) are applicable as long as the neutral axis lies in the web
of the beam section. In the rare case when this is not so, reference should be made
to advanced texts on structural steel design. In addition the design of beams carrying
compressive loads is influenced by considerations of local and overall instability, as we
shall see in Chapter 21.
18.3 PLASTIC ANALYSIS OF FRAMES
The plastic analysis of frames is carried out in a very similar manner to that for beams
in that possible collapse mechanisms are identified and the principle of virtual work
used to determine the collapse loads. A complication does arise, however, in that
frames, even though two-dimensional, can possess collapse mechanisms which involve
both beam and sway mechanisms since, as we saw in Section 16.10 in the moment
distribution analysis of portal frames, sway is produced by any asymmetry of the loading
or frame. Initially we shall illustrate the method by a comparatively simple example.
EXAMPLE 18.8 Determine the value of the load W required to cause collapse of
the frame shown in Fig. 18.16(a) if the plastic moment of all members of the frame is
200 kN m. Calculate also the support reactions at collapse.
We note that the frame and loading are unsymmetrical so that sway occurs. The bending
moment diagram for the frame takes the form shown in Fig. 18.16(b) so that there are
three possible collapse mechanisms as shown in Fig. 18.17.
In Fig. 18.17(a) the horizontal member BCD has collapsed with plastic hinges forming
at B, C and D; this is termed a beam mechanism. In Fig. 18.17(b) the frame has
swayed with hinges forming at A, B, D and E; this, for obvious reasons, is called a
sway mechanism. Fig. 18.17(c) shows a combined mechanism which incorporates both
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• Chapter 18 / Plastic Analysis of Beams and Frames
FIGURE 18.16
Portal frame of
Ex. 18.8
(b)
(a)
2 m
2 m
2 m
4 m
A
E
B
W
C
D
W
(c)
A
B
E
D
C
3u
2u
2u
2u
4u
u
u
u
(a)
A
B
E
D
C
2u
2u
u
u
(b)
A
B
E
D
2u
4u
4u
u
FIGURE 18.17 Collapse mechanisms for the frame of Ex. 18.8
the beam and sway mechanisms. However, in this case, the moments at B due to the
vertical load at C and the horizontal load at B oppose each other so that the moment
at B will be the smallest of the five peak moments and plastic hinges will form at the
other locations. We say, therefore, that there is a hinge cancellation at B; the angle
ABC then remains a right angle. We shall now examine each mechanism in turn to
determine the value of W required to cause collapse. We shall designate the plastic
moment of the frame as MP.
BEAM MECHANISM
Suppose that BC is given a small rotation θ . Since CD=CB then CD also rotates
through the angle θ and the relative angle between CD and the extension of BC is 2θ .
Then, from the principle of virtual work
W 2θ = MPθ + MP2θ + MPθ
(i)
which gives
W = 2MP
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18.3 Plastic Analysis of Frames
• 615
In the virtual work equation 2θ is the vertical distance through which W moves and
the first, second and third terms on the right hand side represent the internal work
done by the plastic moments at B, C and D respectively.
SWAY MECHANISM
The vertical member AB is given a small rotation θ , ED then rotates through 2θ .
Again, from the principle of virtual work
W 4θ = MPθ + MPθ + MP2θ + MP2θ
(ii)
i.e.
W = 3
2
MP
COMBINED MECHANISM
Since, now, there is no plastic hinge at B there is no plastic moment at B. Then, the
principle of virtual work gives
W 4θ + W 2θ = MPθ + MP2θ + MP3θ + MP2θ
(iii)
from which
W = 4
3
MP
We could have obtained Eq. (iii) directly by adding Eqs (i) and (ii) and anticipating
the hinge cancellation at B. Eq. (i) would then be written
W 2θ = {MPθ} + MP2θ + MPθ
(iv)
where the term in curly brackets is the internal work done by the plastic moment at B.
Similarly Eq. (ii) would be written
W 4θ = MPθ + {MPθ} + MP2θ + MP2θ
(v)
Adding Eqs (iv) and (v) and dropping the term in curly brackets gives
W 6θ = 8MPθ
as before.
From Eqs (i), (ii) and (iii) we see that the critical mechanism is the combined
mechanism and the lowest value of W is 4MP/3 so that
W = 4 × 200
3
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• Chapter 18 / Plastic Analysis of Beams and Frames
D
C
B
2 m
2 m
2 m
E
A
4 m
RA,H
RA,V
RE,V
RE,H
W
266.7 kN
W
266.7 kN
FIGURE 18.18 Support
reactions at collapse in
the frame of Ex. 18.8
i.e.
W = 266.7kN
Figure 18.18 shows the support reactions corresponding to the collapse mode. The
internal moment at D is MP (D is a plastic hinge) so that, taking moments about D for
the forces acting on the member ED
RE,H × 2 = MP = 200 kN m
so that
RE,H = 100 kN
Resolving horizontally
RA,H + 266.7 − 100 = 0
from which
RA,H = −166.7 kN
(to the left)
Taking moments about A
RE,V × 4 + RE,H × 2 − 266.7 × 2 − 266.7 × 4 = 0
which gives
RE,V = 350.1kN
Finally, resolving vertically
RA,V + RE,V − 266.7 = 0
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18.3 Plastic Analysis of Frames
• 617
i.e.
RA,V = −83.4 kN (downwards)
In the portal frame of Ex. 18.8 each member has the same plastic moment MP. In
cases where the members have different plastic moments a slightly different approach
is necessary.
EXAMPLE 18.9
In the portal frame of Ex. 18.8 the plastic moment of the member
BCD is 2MP. Calculate the critical value of the load W .
Since the vertical members are the weaker members plastic hinges will form at B in
AB and at D in ED as shown, for all three possible collapse mechanisms, in Fig. 18.19.
This has implications for the virtual work equation because in Fig. 18.19(a) the plastic
FIGURE 18.19
Collapse
mechanisms for
the frame of
Ex. 18.9
(b)
A
B
E
4u
2u

D
A
B
D
C
(a)
E
2u
2u
u
u
(c)
A
B
C
D
E
2u
2u
u
3u
4u
moment at B and D is MP while that at C is 2MP. The virtual work equation then
becomes
W 2θ = MPθ + 2MP2θ + MPθ
which gives
W = 3MP
For the sway mechanism
W 4θ = MPθ + MPθ + MP2θ + MP2θ
so that
W = 3
2
MP
and for the combined mechanism
W 4θ + W 2θ = MPθ + 2MP2θ + MP3θ + MP2θ
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618
• Chapter 18 / Plastic Analysis of Beams and Frames
from which
W = 5
3
MP
Here we see that the minimum value of W which would cause collapse is 3MP/2 and
that the sway mechanism is the critical mechanism.
We shall now examine a portal frame having a pitched roof in which the determination
of displacements is more complicated.
EXAMPLE 18.10 The portal frame shown in Fig. 18.20(a) has members which have
the same plastic moment MP. Determine the minimum value of the load W required
to cause collapse if the collapse mechanism is that shown in Fig. 18.20(b).
FIGURE 18.20
Collapse
mechanism for
the frame of
Ex. 18.10
A
B
C
D
B
A
C
D
a
u
E
C
D
W
E
5 m
(a)
(b)
5 m
2 m
5 m
In Exs 18.8 and 18.9 the displacements of the joints of the frame were relatively simple
to determine since all the members were perpendicular to each other. For a pitched
roof frame the calculation is more difficult; one method is to use the concept of
instantaneous centres.
In Fig. 18.21 the member BC is given a small rotation θ . Since θ is small C can be
assumed to move at right angles to BC to C′. Similarly the member DE rotates about
E so that D moves horizontally to D′. Further, since C moves at right angles to BC
and D moves at right angles to DE it follows that CD rotates about the instantaneous
centre, I, which is the point of intersection of BC and ED produced; the lines IC and
ID then rotate through the same angle φ.
From the triangles BCC′ and ICC′
CC′ = BCθ = ICφ
so that
φ = BC
IC
= θ
(i)
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18.3 Plastic Analysis of Frames
• 619
A
B
C
C
D
D
I
F
E
f
a
f
u
FIGURE 18.21 Method of
instantaneous centres for
the frame of Ex. 18.10
From the triangles EDD′ and IDD′
DD′ = EDα = IDφ
Therefore
α = ID
ED
φ = ID
ED
BC
IC
θ
(ii)
Now we drop a perpendicular from C to meet the horizontal through B and D at F.
Then, from the similar triangles BCF and BID
BC
CI
= BF
FD
= 5
5
= 1
so that BC=CI and, from Eq. (i), φ= θ . Also
CF
ID
= BF
BD
= 5
10
= 1
2
from which ID= 2CF= 4 m. Then, from Eq. (ii)
α = 4
5
θ
Finally, the vertical displacement of C to C′ is BFθ (=5θ).
The equation of virtual work is then
W 5θ = MPθ + MP(θ + α) + MP(φ + α) + MPα
Substituting for φ and α in terms of θ from the above gives
W = 1.12MP
The failure mechanism shown in Fig. 18.20(b) does not involve sway. If, however,
a horizontal load were applied at B, say, then sway would occur and other possible
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• Chapter 18 / Plastic Analysis of Beams and Frames
FIGURE 18.22
Possible collapse
mechanisms
for the frame
of Ex. 18.10
with sway
A
A
B
B
C
C
D
D
E
E
(a)
(b)
failure mechanisms would have to be investigated; two such mechanisms are shown
in Fig. 18.22. Note that in Fig. 18.22(a) there is a hinge cancellation at C and in
Fig. 18.22(b) there is a hinge cancellation at B. In determining the collapse loads of
such frames the method of instantaneous centres still applies.
PROBLEMS
P.18.1 Determine the plastic moment and shape factor of a beam of solid circular cross
section having a radius r and yield stress σY.
Ans. MP = 1.33σYr3, f = 1.69.
P.18.2 Determine the plastic moment and shape factor for a thin-walled box girder
whose cross section has a breadth b, depth d and a constant wall thickness t. Calculate
f for b= 200 mm, d= 300 mm.
Ans. MP = σYtd(2b+ d)/2, f = 1.17.
P.18.3 A beam having the cross section shown in Fig. P.18.3 is fabricated from mild
steel which has a yield stress of 300 N/mm2. Determine the plastic moment of the
section and its shape factor.
Ans. 256.5kNm, 1.52.
15 mm
250 mm
300 mm
15 mm
15 mm
75 mm
FIGURE P.18.3
P.18.4 A cantilever beam of length 6 m has an additional support at a distance of 2 m
from its free end as shown in Fig. P.18.4. Determine the minimum value of W at which
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Problems
• 621
collapse occurs if the section of the beam is identical to that of Fig. P.18.3. State clearly
the form of the collapse mechanism corresponding to this ultimate load.
Ans. 128.3 kN, plastic hinge at C.
2 m
2 m
2 m
D
C
A
B
2W
W
FIGURE P.18.4
P.18.5 A beam of length L is rigidly built-in at each end and carries a uniformly dis-
tributed load of intensity w along its complete span. Determine the ultimate strength
of the beam in terms of the plastic moment, MP, of its cross section.
Ans. 16MP/L2.
P.18.6 A simply supported beam has a cantilever overhang and supports loads as
shown in Fig. P.18.6. Determine the collapse load of the beam, stating the position of
the corresponding plastic hinge.
Ans. 2MP/L, plastic hinge at D.
W
W
L /2
L /3
L /3
L /3
2W
A
B
C
D
E
FIGURE P.18.6
P.18.7 Determine the ultimate strength of the propped cantilever shown in Fig. P.18.7
and specify the corresponding collapse mechanism.
Ans. W = 4MP/L, plastic hinges at A and C.
A
B
C
D
L/3
L/3
L/3
W
W
FIGURE P.18.7
P.18.8 The working loads, W , on the propped cantilever of Fig. P.18.7 are each 150 kN
and its span is 6 m. If the yield stress of mild steel is 300 N/mm2, suggest a suitable
section for the beam using a load factor of 1.75 against collapse.
Ans. Universal Beam, 406 mm× 152 mm× 67 kg/m.
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• Chapter 18 / Plastic Analysis of Beams and Frames
P.18.9 The members of a steel portal frame have the relative plastic moments shown
in Fig. P.18.9. Calculate the required value of M for the ultimate loads shown.
Ans. 36.2 kN m.
A
M
B
C
D
E
M
2M
3 m
3 m
25 kN
30 kN
30 kN
3 m
4 m
F
FIGURE P.18.9
P.18.10 The frame shown in Fig. P.18.10 is pinned to its foundation and has relative
plastic moments of resistance as shown. If M has the value 108 kN m calculate the
value of W that will just cause the frame to collapse.
Ans. 60 kN.
6 m
2M
0.75W
3 m
3 m
3 m
4 m
A
M
W
W
C
2M
B
D
F
E
FIGURE P.18.10
P.18.11 Fig. P.18.11 shows a portal frame which is pinned to its foundation and which
carries vertical and horizontal loads as shown. If the relative values of the plastic
moments of resistance are those given determine the relationship between the load
W and the plastic moment parameter M . Calculate also the foundation reactions at
collapse.
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Problems
• 623
Ans. W = 0.3M . Horizontal: 0.44W at A, 0.56W at G. Vertical: 0.89W at A,
2.11W at G.
9 m
2m
3 m
3 m
3 m
3 m
6 m
G
M
F
W
W
W
C
D
E
2M
W
B
A
FIGURE P.18.11
P.18.12 The steel frame shown in Fig. P.18.12 collapses under the loading shown.
Calculate the value of the plastic moment parameter M if the relative plastic moments
of resistance of the members are as shown. Calculate also the support reactions at
collapse.
Ans. M = 56 kN m. Vertical: 32 kN at A, 48 kN at D. Horizontal: 13.3 kN at A,
33.3 kN at D.
40 kN
20 kN B
A
M
M
E
F
2M
C
D
6 m
3 m
3 m
3 m
3 m
3 m
40 kN
FIGURE P.18.12
P.18.13 The pitched roof portal frame shown in Fig. P.18.13 has columns with a plastic
moment of resistance equal to M and rafters which have a plastic moment of resistance
equal to 1.3M . Calculate the smallest value of M that can be used so that the frame
will not collapse under the given loading.
Ans. M = 24 kN m.
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• Chapter 18 / Plastic Analysis of Beams and Frames
30 kN
5 kN
M
10 kN
C
D
E
B
1.3M
1.3M
3 m
5 m
3 m
6 m
M
A
FIGURE P.18.13
P.18.14 The frame shown in Fig. P.18.14 is pinned to the foundation at D and to a wall
at A. The plastic moment of resistance of the column CD is 200 kN m while that of
the rafters AB and BC is 240 kN m. For the loading shown calculate the value of P at
which collapse will take place.
Ans. P = 106.3 kN.
B
C
P
D
P
A
6 m
6 m
M
2 m
3 m
3 m
4 m
P/5
4 m
FIGURE P.18.14