Steel Design
Page 9.1
Theme 9: Plastic Design of Steel Structures
References:
Mahachi: Design of Structural Steelwork to SANS 10162: Chapter 11
SANS 10162-1: The structural use of steel: Part 1: Limit-states design
Study objectives for Theme 9:
To understand the principles behind the plastic analysis and design of steel structures;
To be able to analyze the plastic resistance of steel beams and portal frames; and
To understand the design implications when a structure’s resistance is based on a plastic design.
■
Introduction
Reference:
Mahachi § 11.1
First-order structural analysis methods such as slope-deflection and matrix stiffness methods are
based on linear elasticity. The solutions obtained are valid as long as the resultant flexural stresses
are below the yield stress of the material. [The methods studied thus far ignored the P-δ and P-Δ
(stability) effects.] The relationship between increasing load and the resultant deflection was linear.
The analysis solutions obtained with these first-order linear-elastic methods also depended on
considerations such as movement of supports, lack-of-fit and temperature.
If the material, such as steel, is ductile, the design (say at the ultimate limit state) could be based on
the behaviour of the structure beyond the elastic limit. Such an approach could lead to more
economical structures.
Consider for example a two span uniform continuous beam with a uniformly distributed load:
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓ w (kN/m)
A
1
B
L
C
L
Bending moment diagram:
With an elastic design the moments are as follows:
M1 = 0.070 w L2
MB = - 0.125 w L2
If the design is based on an this elastic analysis the beam would have to resist this maximum
moment at B. The rest of this uniform beam would have reserve strength, significantly more than
the moment between A and B and between B and C.
If the load is increased (say at the ultimate limit state) beyond the elastic moment resistance at B
the beam will start to yield at B. If the material and the flexural behaviour is ductile the beam would
bend plastically at B whist still resisting a moment based on the plastic resistance at B. Further
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Steel Design
Page 9.2
increase in load will then increase the bending moment between A and B and between B and C as
the beam is continuing to yield in bending at B.
It will be shown that a plastic designed beam (meeting the code requirements) may be designed for
the following moments:
M1 = 0.0858 w L2
MB = - 0.0858 w L2
This is a maximum moment which is 31% less.
For elastic design the principle of superposition is valid. This fact facilitates the analyses for
different load combinations in design. Elastic design provides a safe solution for the bending
moments in a structure but cannot accurately predict the collapse strength of a structure. In
addition, an elastic design might not lead to an economic solution.
Plastic design, because of the redistribution of bending moments due to plastic behaviour, could
lead to structures with smaller sections and thus possibly to more economic structural solutions.
However, the requirement of limiting the deflections of the structure under serviceability loads could
dictate the sections. With a plastic design the requirements as far as the bracing of the structure
and the dimensions of the sections are more stringent. This is to ensure that the required amount
of plastic rotation of the sections could take place prior to local or lateral buckling.
(The principles of plastic design are also applied to reinforced concrete flexural members and
composite concrete slab / steel beam flexural members. In these flexural cases the concrete forms
a compression block and the steel reinforcement / section is in tension. The condition is that the
steel should yield before the maximum capacity of the concrete in compression is reached.
Sometimes the term “moment redistribution” is used in the context of plastic design. This term
should not be confused with “moment distribution”; a method used for the elastic analysis of
structures.)
■
Continuous beam
Reference:
Mahachi § 11.2 and Fig 11.1
This example of a two-span beam illustrates the reserve capacity of this structure if the beam is
allowed to form plastic regions under increasing load:
With an elastic analysis (no yielding is considered) the capacity of the beam is P = 57 kN.
With a plastic analysis (yielding and the formation of plastic “hinges” considered) the capacity of the
beam is P = 81 kN.
Techniques for determining the plastic resistance of beams of frames are to be studied in this
theme.
■
Elastic-plastic stress-strain relation
References:
Mahachi § 11.3
The real  -  relationship for steel was discussed before.
The idealized (simplified)  -  relationship is shown in Fig. 11.2.
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Steel Design
■
Page 9.3
Plastic bending without axial force
References:
Mahachi § 11.4
First consider a section subjected to bending moment and thus to bending normal stresses only.
(An axial force will cause additional normal stresses and will influence resulting normal stress
distribution.)
See Mahachi Fig. 11.3: Bending strains, elastic stresses, elastic-plastic stresses and fully plastic
stresses for a symmetrical section.
Yield moment: M Y  Z e f y
Plastic moment: M P  Z pl f y
Ze and Zpl for steel sections are tabulated.
For a symmetrical section the elastic and plastic neutral axes will be at the centre of the section.
However, with an unsymmetrical section such as a T, the neutral axis for elastic bending will be at
the centroid but the plastic neutral axis will divide the section into two equal areas.
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The shape (or form factor) for an I-section bent about the x-axis, F, is typically in the order of 1.14.
See Mahachi Fig. 11.4: Moment – curvature relationships for I and rectangular sections.
The moment – curvature relationship for an elastic-plastic analysis is idealized as two straight lines:
the elastic straight line and the horizontal line when M = MP.
See Mahachi Fig. 11.5: Spreading of the plastic region:
In a plastic analysis it is assumed that plasticity is confined to a point on the structure (at the
position of maximum bending moment). After MP has been reached at this point it is assumed that
with a further increase of the load on the structure this position will behave as a plastic hinge.
■
Effects of axial load and shear force on the plastic moment
Reference:
Mahachi § 11.5
See Mahachi Figs. 11.6 and 11.7.
The presence of an axial force (as with a beam-column) could reduce the plastic moment of
resistance. The presence of a significant shear force could also reduce the plastic moment of
resistance. The graphs illustrate that for a relatively small axial or shear force this reduction in the
plastic moment of resistance could be ignored. These effects are thus small for normal beam
structures or low-rise frames and will not be considered for the structures analyzed in this study
theme.
■
Elastic-plastic analyses
When a ductile structure is subjected to increasing load it will form a sequence of plastic hinges
until it reaches a plastic collapse mechanism.
A statically determinate structure (e.g. a simply supported beam or a cantilever beam) could only
form one plastic hinge before becoming a plastic mechanism. The elastic and plastic bending
moment diagrams are the same. (No redistribution of moments are possible.)
A structure statically indeterminate to the first degree, such as a propped cantilever or a two hinged
portal, will form two plastic hinges before becoming a plastic mechanism.
A built-in (fixed-fixed) supported beam is, strictly speaking, indeterminate to the third degree.
However, the redundant related to the axial forces in the beam is solved independently from the
bending action. Considering the flexural action, there are two redundants. I.e. if the two fixed-end
moments (or the reaction and moment at one of the two ends are known), for example, are known
the bending moment diagram can be determined. A built-in beam will form three plastic hinges
before becoming a plastic mechanism.
A continuous beam with multiple spans and general loading will generally form a mechanism in any
one of the spans. If the mechanism is formed in an end span with a hinged end support, two plastic
hinged are required. If the mechanism is formed in an interior span three plastic hinged are
required. The bending moments in the span with the mechanism can be determined. The
remainder of the beam that has not reached the collapse state is still elastic and the bending
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moments in these (elastic) sections of the beam (if required) would have to be determined by elastic
analysis methods.
An elastic-plastic analysis of a structure provides information on the sequence of the formation of
plastic hinges as the load on the structure is increased. If the deflections are also determined
information on the deformation of the structure at the different load levels is also available. This
additional information could be useful to the designer.
In practice, the designer generally does a plastic analysis, only considering the plastic mechanism,
at the ultimate limit state load level. The deflections at the serviceability limit state are then checked
by using an elastic analysis procedure.
Support movements and temperature changes influence an elastic analysis of structure and will
impact on the sequence of the formation of plastic hinges and the deflection of the structure. The
ultimate strength of a ductile structure is, however, independent of support movements and
temperature changes.
Example 1: A two span beam:
A
P
B
C
D
L
L=6m
L/2
L/2
MY = 70 kN.m MP = 81 kN.m EI = 8800 kN.m2 (Approx. a 254 x 146 x 31 I )
Result of elastic plastic analysis: Load vs. deflection at D:
The slope of the load deflection curve is an indication of the stiffness of the structure.
Note the decreasing stiffness of the structure with the formation of plastic hinges.
Also note the overall non-linear behaviour of the structure.
The ultimate load capacity is independent of EI. The amount of deflection depends on EI.
Mechanism load / first plastic hinge load = 81/ 67 = 1.21
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Example 2: A built-in beam:
P
A
B
C
2m
4m
MY = 70 kN.m MP = 81 kN.m EI = 8800 kN.m2 (Say approximately a 254 x 146 x 31 I )
Structural behaviour: Graph of increasing load vs. increasing deflection at B:
Note again the decreasing stiffness of the structure with the formation of plastic hinges and also the
overall non-linear behaviour of the structure.
Mechanism load / first plastic hinge load = 122/ 92 = 1.33
Reserve plastic strength:
The ratio of the maximum mechanism load to the load at the first plastic hinge is not a constant for
different structures:
Example
SIN 415_Theme 9_Plastic Design 2018.doc
Load, λ P, to 1st
hinge
(x MP / L)
Load, λ P, to
mechanism
(x MP / L)
Mechanism load /
first plastic hinge
load
4.00
4.00
1.00
5.33
6.00
1.13
Steel Design
Page 9.7
Example
■
Load, λ P, to 1st
hinge
(x MP / L)
Load, λ P, to
mechanism
(x MP / L)
Mechanism load /
first plastic hinge
load
4.92
6.00
1.22
6.75
8.99
1.33
8.00
8.00
1.00
Collapse mechanisms
Reference:
Mahachi § 11.6
Mahachi Example E11.1: Elastic-plastic analysis of a fixed-end beam
The plastic failure load is independent of initial conditions of settlement of supports or temperature
changes. See CSIR Figs. E11.1(b). An elastic bending moment diagram is dependent on
settlement of supports or temperature changes.
An elastic-plastic analysis provides information on the deformation of the structure and the
sequence of the formation of plastic hinges. The last hinge to form requires less ductility than the
preceding plastic hinges.
Elastic-plastic analyses are not part of this module. This module will concentrate on investigating
the ultimate mechanism load.
■
Methods of plastic analysis
Reference:
Mahachi § 11.7
o
Elastic-plastic analysis
o
Kinematic mechanism or virtual work method.
o
Static or equilibrium method
 Equilibrium considering free body diagrams of the structure
 Equilibrium considering a combination of bending moment diagrams: (Also called the “semigraphical” method.)
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Plastic design could be applied to either one of two problem formulations:
 Design problem: Given the loads on the structure determine the plastic resisting moments
required for the various members.
 Analysis problem: Given the plastic bending strengths of the members determine the plastic
collapse load.
An elastic-plastic analysis of a structure provides information on the load-deflection behaviour of the
structure (the EI’s of the various members must be known), the sequence of the formation of plastic
hinges leading to the plastic collapse mechanism and the ultimate load capacity of the structure.
This analysis process provides valuable information on the behaviour of the structure, but this
process is time consuming. This approach will not be studied in this module.
This module will concentrate on the ultimate strength, the load that will cause a plastic collapse
mechanism, of a structure. How this mechanism is reached, or how much the structure will deflect
will not be known. Apart from the loading and the dimensions, only the strength of structure (relative
plastic moment of resistances) will be considered.
The collapse mechanism (apart from simple structures) is generally not known initially. Various
possible mechanisms will be investigated and the critical mechanism found. In the case of a
continuous beams, each span of the beam will be considered and the critical span will be found.
 Kinematic mechanism or virtual work method
Example 1: Propped cantilever (ABC) or end span of continuous beam:
W
L/2
B
W
L/2
C
L/2
L/2
A
Consider a collapse mechanism: Plastic hinges at A and B (and a normal hinge at C). (From the
elastic bending moment diagram it is known that local maximum moments will occur at A and B.)
For a uniform beam the plastic moment of resistances at A and B will be the same = MP
Consider an arbitrary small amount of deformation of the beam as a mechanism: Plastic hinges are
considered to be concentrated at A and B. The segments AB and BC will remain straight.
Assume the hinge at A rotates through a small angle θ.
W
θ
W
θ
θ
θ
L/2
L/2
Δ
2θ
L/2
L/2
B then displaces a vertical distance Δ = [θ (L/2)]
C would then also rotate through an angle θ
At B: The total amount of relative rotation of AB and BC is [θ + θ] = 2θ
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To deform the beam as a mechanism work will be done: External work by the load W and internal
work at the two plastic hinges. (The beam will require work to rotate it at a plastic hinge.)
External work by the point load = W [Δ] = W [θ (L/2)]
Internal work at A and B = MP (θ) + MP (2θ) = MP (3θ)
External work = Internal work
or
W [θ (L/2)]
= MP (3θ)
If the load is known (the normal case), the plastic moment of resistance required: M P  W L
6
If the strength is known, the load that can be supported: W 
6MP
L
Exercise: Fixed-fixed beam or interior span of continuous beam:
W
L/2
B
W
L/2
A
L/2
L/2
C
Show that:
If the load is known, the plastic moment of resistance required: M P  W L
8
If the strength is known, the load that can be supported: W 
8M P
L
Example 2: Propped cantilever (or end span of continuous beam) with two point loads:
A uniform beam is loaded (with ultimate factored loads) as in the sketch. Determine the plastic
moment Mp required if the beam is at the point of forming a plastic collapse mechanism.
(Dimensions in m.)
The collapse mechanism is not known beforehand. From the elastic bending moment diagram it is
known that a maximum negative (hogging) moment will occur at A and that a maximum positive
(sagging) moment will occur at either B or C. Two possible mechanisms will have to be considered:
Mechanism 1: Plastic hinges at A and B and Mechanism 2: Plastic hinges at A and C.
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After investigating all the possible collapse mechanisms the beam has to be designed for the
largest MP found. That is MP = 53.40 kN.m.
In the context of plastic analysis done here, MP has a somewhat different meaning. Interpret this
calculated MP as Mu. (Earlier calculations of: Mr = ϕ MP = ϕ fy Zpl. Then MP = fy Zpl was the plastic
resistance of a specific class 1 or 2 steel section.)
Mr = ϕ MP = ϕ fy Zpl ≥ [Mu = 53.40 kN.m]. Considering the ϕ = 0.9 partial safety factor, provide
a steel section with an (fy Zpl) ≥ 59.33 kN.m.
In structures with a more complicated geometry and with more loads it is necessary to calculate the
bending moment diagram for the mechanism that is considered to be critical. The bending moments
at any point, Mi should then be ≤ MP.
Consider the free-bodies indicated. For the mechanism under consideration the bending moments
at A (hogging) and at C (sagging) are known: MP = 53.40 kN.m.
ABC:
VA = 1/9 [ 2(MP) + 21(4) ] = 21.20 kN
and MB = VA (5) - MP = 52.6 kN.m ≤ MP O.K.
The bending moment diagram on tension side (kN.m):
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Example 3: End span of continuous beam, non-uniform section:
Span DA has an MP1
MP1
Span AC has an MP2
D
MP2
C
W
B
A
L/2
L/2
Let MP1 < MP2
The plastic hinge at A will form in the weaker section. I.e. just to the left of the support at A.
Consider the collapse mechanism as before. (See Example 1)
External work by the point load = W [θ (L/2)]
Internal work at A and B = MP1 (θ) + MP2 (2θ)
External work = Internal work
With W and L given, two unknowns remain: MP1 and MP2
If one of MP1 or MP2 is selected, the other one can be calculated.
Alternatively, a ratio of MP1 to MP2 must be selected; say MP1 = 0.80 MP2. Then MP1 and MP2 can
be calculated.
If MP1 > MP2 the calculations would be the same as in Example 1 as the hinge at A will form in AC.
Example 4: Fixed-fixed beam or interior span of continuous beam with uniformly distributed
load and uniform MP :
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
w
w
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
L
L
A
C
From the elastic bending moment diagram it is known that maximum moments (hogging) will occur
at A and C (MA = MC = - w L2 / 12) and a sagging moment at the centre of MB = w L2 / 24.
If the load is increased beyond the elastic limit, the first two plastic hinges will form at A and C at the
same load level. The beam will form a mechanism when the third plastic hinge forms at the centre
of the beam. Due to the symmetry the position of the third hinge is known.
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
θ
B
w
w
θ
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
θ
θ
L/2
L/2
Δ
A
C
2θ
L/2
L/2
Consider the load on AB: The displacement of the centroid of this load = [θ (L/4)]. The same
applies for the load on BC.
External work by the uniformly distributed load = 2 [w L /2] [θ (L/4)] = [w L2 θ] /4
Internal work at A, B and C = MP (4θ)
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External work = Internal work
[w L2 θ] /4 = MP (4θ)
or
2
If the load is known (the normal case), the plastic moment of resistance required: M P  w L
16
If the strength is known, the uniformly distributed load that can be supported:
w
16 M P
L2
 Static or equilibrium method: Equilibrium considering free body diagrams
Example 5: Fixed-hinged beam or end span of continuous beam with uniformly distributed
load and uniform MP :
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
w
w
L
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
L
A
C
From the elastic bending moment diagram it is known that maximum moment (hogging) will occur at
A (MA = - w L2 / 8) and a maximum sagging moment of MB = 9 w L2 / 128. (Redbook Table 5.19)
If the load is increased beyond the elastic limit, the first plastic hinge will form at A. The beam will
form a mechanism when the second plastic hinge forms in sagging bending. The position of the
second hinge is not known by inspection and will have to be calculated.
MP
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓ w
MP
A
B
(L-x)
x
C
Consider the free bodies AB and BC. The moment is a maximum at B, where MB = MP and where
the shear force is zero:
It can be found that: M P 
w x2
2
(a)
Equating the moment at B by considering AB with the moment at B by considering BC:
It can be found that:
x
L MP

2 wL
(b)
Combining the two equations, the roots of a quadratic equation have to be found: Only one root is
acceptable.
It can be shown that MP = 0.0858 w L2
SIN 415_Theme 9_Plastic Design 2018.doc
and
x = 0.414 L
Steel Design
Page 9.13
Elastic Analysis
MA = - 0.1250 q L2
Plastic Analysis
Mpos,max = 0.0703 q L2
MP = 0.0858 q L2
 Static or equilibrium method: Equilibrium considering a combination of bending
moment diagrams
Example 6: Propped cantilever (ABC) or end span of continuous beam:
W
L/2
B
W
L/2
C
L/2
L/2
A
The beam is statically indeterminate to the first degree. Consider the load W and the moment at A
separately:
+
WL/4
MA/2
MA
-
Consider the mechanism as shown in Example 1:
At A hogging:
- MA = - MP
(a)
At B sagging:
+ MB = +MP = WL/4 – MP/2
(b)
Solve the two equations and find:
SIN 415_Theme 9_Plastic Design 2018.doc
MP 
WL
6
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Page 9.14
Example 7: Propped cantilever (or end span of continuous beam) with two point loads:
The beam is statically indeterminate to the first degree. Consider the point loads and the moment at
A separately: (Bending moments kN.m)
Consider the first mechanism as shown in Example 2:
At A hogging:
- MA = - MP
(a)
At B sagging:
+ MB = +MP = 83.75 – 7MP/12
(b)
Solve and find:
MP = 52.89 kN.m
Consider the second mechanism as shown in Example 2:
At A hogging:
- MA = - MP
At C sagging:
+ MC = +MP = 66.75 – 3MP/12 (d)
Solve and find:
MP = 53.40 kN.m
(c)
Complete the problem as in Example 2.
Mahachi Example E11.2: Plastic design of a continuous beam
With a continuous beam consider each mechanism in each span separately. Find the span with the
critical mechanism, i.e. the mechanism with largest MP.
DE is the span with the critical mechanism MP = 41.7 kN.m
Read MP as Mu, i.e. Mu = 41.7 kN.m
Find a section with Mr = ϕ MP = ϕ fy Zpl ≥ [Mu = 41.7 kN.m].
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Example 8
The ultimate factored distributed load on a continuous beam is shown. The beam has a constant
cross section of Class 1. The beam is provided with continuous lateral support such that lateral
torsional buckling is not possible. Determine Mr required.
Because the section is uniform the formulae of Examples 4 and 5 may be used.
Ignoring the effect of shear force on MP:
Read MP as Mu, i.e. Mu = 462 kN.m
Find a section with Mr = ϕ MP = ϕ fy Zpl ≥ [Mu = 462 kN.m].
Example 9
The plastic moments of resistance (Class 1) for the sections of the continuous beam are given and
shown. Determine the load factor, λ, which would lead to a plastic collapse mechanism in the centre
span of the beam. (The beam is provided with continuous lateral support such that lateral torsional
buckling is not possible and the webs have stiffeners under the point loads).
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Check the mechanism method with the equilibrium method (combining bending moment diagrams.)
■
Stability considerations
Reference:
Mahachi § 11.7.1
In the case of certain (unusual) structures with slender columns and high axial compressive loads
stability considerations (P-  and P-∆-effects, see Theme 6) would influence the deflections and
bending moments and such structures would require a second-order elastic-plastic analysis.
■
Incremental collapse
Reference:
Mahachi § 11.7.2
If significant repeated reversing loads are applied to a structure, as in the case of cranes and
bridges, and the structure reaches regions of plasticity under the action of these loads, the repeated
yielding and unloading could lead to increasing deflections and eventually incremental collapse.
Plastic design would thus not be used with heavy crane buildings and road and rail bridges.
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SANS 10162 Clause 8.6: Plastic analysis: Extract from the code:
Under a particular loading combination, the forces and moments throughout all or part of the
structure may be determined by a plastic analysis, provided that
a) the steel used has fy  0,85 fu and exhibits the load-strain characteristics necessary to achieve
moment redistribution,
b) the width-to-thickness ratios meet the requirements for class 1 sections, as given in 11.2,
c) the members are braced laterally in accordance with 13.7,
d) web stiffeners are supplied on a member at a point of load application where a plastic hinge
would form,
e) splices in beams or columns are designed to transmit the greater of 1,1 times the maximum
calculated moment under ultimate load at the splice location, or 0,25 Mp, whichever is greater,
f)
members are not subject to repeated heavy impact or fatigue loading, and
g) the influence of inelastic deformation on the strength of the structure is taken into account. (See also
8.7.)
■
Continuous beams
It is possible for a structure to form a partial mechanism where a part of the structure remains
elastic. This is the case for continuous beams. Only one span would form the mechanism (unless
there are identical spans with identical loads). Each span is to be investigated and the critical span
identified. See CSIR Example E11.2.
■
Application of plastic design to portal frames
Reference:
Mahachi § 11.8
The application of plastic design may well lead to lighter sections than an elastic designed portal
frame, providing deflection is not a governing point. However, additional bracing may be required.
■
Fundamental theorems of plastic collapse
Upper bound (or unsafe) theorem
 A bending moment diagram obtained from equilibrium (statics or virtual work) together with
 A mechanism
Will provide an upper bound on the load carrying capacity of the structure:
i.e. the structure will be only be able to support this or a smaller load;
i.e. the MP obtained could provide an unsafe design.
Lower bound (or safe) theorem
 A bending moment diagram obtained from equilibrium (statics or virtual work) together with
 The yield requirement, i.e that bending moment at any point, Mi  MP
Will provide a lower bound on the load carrying capacity of the structure:
i.e. the structure will be able to support this or possibly a larger load;
i.e. the MP obtained would provide an safe design, but the structure might be uneconomical.
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Uniqueness theorem
 A bending moment diagram obtained from equilibrium (statics or virtual work) together with
 A mechanism together with
 The yield requirement, i.e. that bending moment at any point, Mi  MP
Will provide the load carrying capacity of the structure:
i.e. the structure will be able to support this load;
i.e. the MP obtained would provide the economical design.
Example 2 and 7: Refer to problem statement:
The first attempt at a solution (hinges @ A and B) with equilibrium and a mechanism provides an
MP = 52,89 kN.m.
i.e. MP required  52,89 kN.m. Providing this strength might thus produce an unsafe solution. The
only way to check is to draw the bending moment diagram for this assumed mechanism.
From this bending moment diagram a maximum moment of 53,53 kN.m is found. If a moment of
resistance Mr = 53,53 kN.m is provided, two conditions are met: equilibrium and Mi  Mr.
Thus if the beam is designed for MP = 53,53 kN.m, the design is safe but (in this case only slightly)
uneconomical.
Thus
52,89 kN.m  MP-economical  53,53 kN.m
The second attempt at a solution (hinges @ A and C), with equilibrium and a mechanism and
Mi  MP (the bending moment diagram) provides an MP = 53,40 kN.m.
Providing this strength produces a safe and economical solution.
Elastic bending moment diagram:
If we employ the elastic bending moment diagram and design for the largest bending moment we
have a case of the lower bound or safe theorem because we have equilibrium and Mi  MP
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Example 10(a): Hinged supported rectangular portal frame
A rectangular frame with hinged supports and the same section for the columns and beam is
considered. [The frame is braced (by purlins and sheeting rails) in order that out-of-plane buckling is
not possible.] A first-order elastic analysis, with a = 2 m and P = 22 kN, provides the following nonbracketed values for the bending moment diagram:
(b) Elastic bending moment diagram
If a 254 x 146 x 37 I-section is chosen, a second-order elastic analysis provides the values in
brackets.
Plastic design: Consider a similar frame, with the MP for the columns and beam the same. The
collapse load P in terms of MP or, alternatively, the value required for MP in terms of P is required.
A class 1 section will be used and the other plastic design requirements are to be met.

Kinematic, mechanism or work method:
The beam mechanism: Fig. (b):
Virtual work by loads = virtual work by plastic hinges
2 P (3 a ) = (1 + 2 + 1)  MP
giving:
SIN 415_Theme 9_Plastic Design 2018.doc
P 
2MP
3a
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Sway mechanism: Fig. (c):
Virtual work by loads = virtual work by plastic hinges
P (4 a ) = (1 + 1)  MP
giving:
P
MP
2a
Combined mechanism: Fig. (d):
Eliminate plastic rotation at B by adding the beam and sway mechanisms.
Virtual work by loads
= virtual work by plastic hinges
P (4 a ) + 2 P (3 a ) = (2 + 2)  MP
P 
4MP
10 a
or
MP 
giving:
10 a P
4
The combined mechanism resists the smallest load and is the critical mechanism of the
mechanisms considered. It is important to draw the bending moment diagram for the critical
mechanism to ensure that the final bending moments nowhere exceed MP.
Consider Fig. (f): Equilibrium:
Moments about D for DE gives: R E 
MP
and
4a
Horizontal equilibrium gives: R A 
Now draw the bending moment diagram for collapse mechanism:
SIN 415_Theme 9_Plastic Design 2018.doc
3M P
20 a
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With a = 2 m and P = 22 kN
At all positions on the structure the M ≤ MP (yield criterion) and the bending moment diagram is
based on equilibrium. Thus it is safe to design according to this bending moment diagram.
To design according to this bending moment diagram (satisfying the yield and equilibrium criteria) is
also economic because this bending moment diagram corresponds to a mechanism. The
uniqueness theorem is thus satisfied.

Statical, equilibrium or semi-graphical method:
The portal supported by two hinges is statically indeterminate to the first degree. Consider the
horizontal reaction at E as the redundant reaction.
2P
4aH
P
4aP
+
+
5aP
O
P
-
O E
oooooo
P/3
H
-
O
O
oooooo
H
5P/3
Free bending moment diagram
Redundant H bending moment diagram
The portal requires 4 hinges for a mechanism: Thus two plastic hinges (in addition to the two hinges
at the supports). A possible mechanism is shown in Fig. (d) (“Combined Mechanism”) with plastic
hinges at C and D.
Equations at C and D:
5aP - 4aH = MP
- - - - - - - - - (i)
- 4aH = - MP
- - - - - - - - - (ii)
Solve: MP = (5/2) aP
or
P
2MP
5a
Calculating the bending moment at B ( = 0.6 MP ) and inspecting the diagram indicate that at all
positions on the structure the M ≤ MP (yield criterion). This diagram is equivalent to the diagram on
the previous page.
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Example 10(b): Fixed supports rectangular portal frame (Coates et al Example 14.6-4)
A rectangular frame with fixed supports and the same section for the columns and beam is
considered. A first-order elastic analysis, with a = 2 m and P = 22 kN, provides the following nonbracketed values for the bending moment diagram:
(b) Elastic bending moment diagram
If a 254 x 146 x 37 I-section is chosen, a second-order elastic analysis provides the values in
brackets.
Plastic design: Consider a similar frame, with the MP for the columns and beam the same.

Kinematic, mechanism or work method:
Refer to the solution for example 10(b):
The beam mechanism: The result is the same: P 
2MP
3a
Sway mechanism: There are now two additional plastic hinges giving:
P
MP
a
Combined mechanism: Again eliminate the plastic rotation at B by adding the beam and sway
mechanisms giving:
P
3M P
5a
or
MP 
5a P
3
The combined mechanism resists the smallest load and is the critical mechanism of the
mechanisms considered. It is important to draw the bending moment diagram for the critical
mechanism to ensure that the final bending moments nowhere exceed MP.
The bending moment diagram for collapse mechanism:
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Page 9.23
With a = 2 m and P = 22 kN

Statical or equilibrium method:
The fixed portal is statically indeterminate to the 3rd degree. Consider the three reactions at A as
the three redundants. The redundants may be chosen as acting in any direction. If the direction
chosen was not the correct direction the solution of the equilibrium equations will indicate a
negative sign.
A sign convention for the bending moment diagrams is required. Choose tension on the inside of
the portal as positive. The plastic hinges will also have the same sign convention associated with
their sense of bending.
Let a = 2 m and P = 22 kN. The four bending moment diagrams are shown: The free (or released)
bending moment diagram and the three bending moment diagrams associated with each of the
three redundants.
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The four bending moment diagrams are added to give the final moment diagram. A plastic collapse
mechanism for the frame is postulated. The sums of the above moments are equated to the plastic
moment at the point of each hinge.
Sum of moments at A:
Sum of moments at C:
Sum of moments at D:
Sum of moments at E:
0
0
-264
-88
+
0 + 0 - N
+ 6V - 8H - N
+ 12V - 8H - N
+ 12V + 0 - N
= - MP
= + MP
= - MP
= + MP
Rearrange the equations:
Sum of moments at A:
0 + 0 - N + MP =
0
Sum of moments at C:
6V - 8H - N - MP =
0
Sum of moments at D:
12V - 8H - N + MP = 264
Sum of moments at E:
12V + 0 - N - MP = 88
Solve the equations: V= 19.56 kN; H = -3.67 kN and N = MP = 73.33 kN.m giving the b.m.d. earlier.
Example 11: Rectangular portal frame: Non-uniform MP
A frame with fixed supports and different sections for the beam and columns is shown. The value
required for MP (where the columns will have a 50% larger bending resistance than the beam),
based on a first-order plastic analysis, is required. A class 1 section will be used and the other
plastic design requirements are to be met.
If the sections are not uniform and a plastic hinge is to form at a certain position the plastic hinge
will form in the weaker of the two sections meeting at that position.
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Statical or equilibrium method:
This structure is indeterminate to the 3rd degree. Three redundants will have to be chosen and released to
produce a free or statically determinate frame. Four bending moment diagrams for the free frame are to be
drawn: One for the given loads and one each for the three redundants.
A mechanism is postulated. Four plastic hinges are required to form an overall mechanism. The bending
moment at each plastic moment is known to be equal to the plastic moment of resistance at that point. This
process leads to four equations which can be solved for the three redundants and MP. The bending moment
diagram for the frame is calculated. Design for the bending moments, or alternatively, postulate a new
mechanism if the bending moment at any position is larger than the calculated MP (or larger than 1,5 MP).
If the third mechanism (the “combined mechanism”) above was chosen, and the same redundants are chosen
as in Example 10, the four equations are found to be:
Sum of moments at A:
Sum of moments at E:
Sum of moments at C:
Sum of moments at D:
0
0
-90
-10
+
+
+
+
0 + 0 - N = - 1,5 MP
3V - 4H - N =
+ MP
6V - 4H - N =
- MP
6V + 0 - N = + 1,5 MP
Solve the equations: V= 13,81 kN; H = -4,82 kN; N = 36,43 kN.m and MP = 24,29 kN.m - giving the
b.m.d. for the chosen plastic mechanism (shown earlier).
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Discussion:
The frames in Examples 10 and 11 have simple geometries and have only a few point loads. The
amount of calculation effort required to do a first-order plastic analysis was about the same for the
two methods (the mechanism or equilibrium methods). However, if the frame does not have a
rectangular form or if the frame has many point loads, the equilibrium becomes the chosen method.
Example 12: Pitched portal frame: Uniform section.
The figure shows a two hinged pitched portal frame. The section is uniform. Determine the plastic
moment of resistance required based on a first-order plastic collapse mechanism.
The dashed lines indicate a possible plastic collapse mechanism.
The following sign convention is employed: Tension on inside of the portal is a positive bending
moment. (Tension on outside of portal is negative bending moment.)
The structure is statically indeterminate to the first degree. Remove the horizontal reaction at J
(= H) as redundant. The bending moment diagram for the free (redundant released) structure
subjected to the given loads is (values given on a next diagram):
The bending moment diagram for the free structure subjected to the redundant is:
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Alternatively, both diagrams are now drawn on the same horizontal base line; the bending moment
diagram for the redundant is inverted:
The magnitudes between the bending moment diagrams are the bending moments in the frame.
(For an elastic analysis the value of H could be determined by superimposing the two load cases; calculating
the horizontal deflections at I; adding these deflections and setting the sum = 0.)
For a plastic analysis a collapse mechanism is postulated. The bending moments at the plastic hinges are
known. From the above diagram it appears that the biggest moments (the largest distances between the two
lines) are at D and I. Inspecting the postulated mechanism (the first figure) it is seen that at one hinge (at D)
tension is on the inside and at another hinge (at I) tension is on the outside: at the first hinge positive bending
moment and at the second hinge negative bending moment.
Write equations for the bending moments at D and I:
252,7 – 8,667 H = + MP
0
– 6,000 H = - MP
Solving these equations: H = 17,23 kN and MP = 103,38 kN.m.
The bending moments at the other positions are calculated (with the know H) and it is found that
Mi ≤ MP. The structure may now be designed according to this bending moment diagram.
The statical or equilibrium method was used above. The latter way the bending moment diagrams were drawn
is called the semi-graphical method. If preferred, the bending moment diagrams could have been drawn in the
conventional manner. The mechanism is shown in dotted lines.
To do the same problem with the kinematic (virtual work or mechanism) method would be quite onerous: The
calculation of the plastic rotation angles requires careful geometrical considerations. For a pitched portal frame
with one degree of indeterminacy (a practical type of structure) the statical method is recommended.
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Example 13: Portal frame with u.d.l.: Uniform section.
The figure shows a two hinged pitched portal frame. The section is uniform. Determine the plastic
moment of resistance required based on a first-order plastic collapse mechanism. A possible plastic
collapse mechanism is shown. The position of the plastic hinge in the beam is not yet known.
The following sign convention is employed: Tension on inside of portal is a positive bending
moment.
The structure is statically indeterminate to the first degree. Remove the horizontal reaction at D
(= H) as redundant.
The bending moment diagram for the free (redundant released) structure subjected to the given
loads is:
The value of the maximum moment between B and C: MB = 82 kN.m and MC = 10 kN.m provides:
VBC = 22,8 kN. Position where Vx = 0 gives x = 4,75 m and Mmax = 136,15 kN.m
The bending moment diagram for the free structure subjected to the redundant is:
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Both diagrams are now drawn on the same horizontal base line. The bending moment diagram for
the free (redundant released) structure subjected to the given loads is A B1 X1 C1 D.
The bending moment diagram for the redundant, inverted, is A B2 X2 C2 D:
The magnitudes between the bending moment diagrams, e.g. B2B1 ; X2X1 and C2C1 , are the
bending moments in the frame.
For a plastic analysis a collapse mechanism is postulated. The bending moments at the plastic
hinges are known. From the above diagram it appears that the biggest moments (the largest
distances between the two lines) are at X and C. Inspecting the postulated mechanism (the first
figure) it is seen that at one hinge (at X) tension is on the inside and at another hinge (at C) tension
is on the outside: at the first hinge positive bending moment and at the second hinge negative
bending moment.
Write equations for the bending moments at X and C:
136,15 – 6,000 H = + MP
10,00 – 6,000 H = - MP
Solving these equations: H = 12,179 kN and MP = 63,08 kN.m.
The bending moments at the other positions are inspected and it is found that Mi ≤ MP.
The structure may now be designed according to this bending moment diagram.
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Example 14: Frame statically indeterminate to second degree
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