Plastic Analysis of
Continuous Beams1
Increasing the applied load until
yielding occurs at some locations
will result in elastic-plastic deformations that will eventually reach
a fully plastic condition.
Fully plastic condition is
defined as one at which a
s fficient number
sufficient
n mber of plastic
hinges are formed to transform
the structure into a mechanism, i.e., the structure is
geometrically unstable.
Additional loading applied to
the fully plastic structure
would lead to collapse.
Design of structures based on
the plastic or limit state
approach is increasingly used
and accepted by various codes of
practice, particularly for steel
construction. Figure 1 shows a
t pical stress
typical
stress-strain
strain curve
c r e for mild
steel and the idealized stressstrain response for performing
plastic analysis.
1
2
1See pages 142 – 152 in your class notes.
ULTIMATE MOMENT
σ
rupture
x
σy
idealized
ε
εy
Figure 1. Mild Steel StressStrain Curve
Consider the beam shown in Fig.
2. Increasing the bending
moment results in going from
elastic cross section behavior
(Fig. 2(a)) to yield of the
outermost fibers (Figs. 2(c) and
(d)) and finally the two yield
zones meet (Fig. 2(e)); the
cross section in this state is
defined to be fully plastic.
σy = yield stress
εy = yield strain
3
4
Also see pages 142 - 152 in your class notes.
1
The ultimate moment is
determined in terms of the yield
stress σ y .
Figure. 2. Stress distribution in a symmetrical cross section subjected to a
bending moment of increasing magnitude: (a) Cross section, (b) Elastic, (c)
Top fibers plastic, (d) Top and bottom
5
fibers plastic, and (e) Fully plastic
The maximum moment which a
section can resist without
exceeding the yield stress
(defined as the yield moment
My) is the smaller of
M y = σ y St
M y = σ y Sc
(2a)
(2b)
St = tension section modulus
( ≡ I / ct )
Sc = compression section
modulus ( ≡ I / cc )
Since the axial force is zero in
this beam case, the neutral axis
in the fully plastic condition
divides the section into two
equal areas, and the resultant
tension and compression are
each equal to σ y A/2, forming a
couple equal to the ultimate
plastic moment Mp
M p = 12 σ y A (yc + y t )
(1)
6
ct = distance from neutral axis
to the extreme tension fiber
cc = distance from neutral
axis to the extreme compression fiber
I = moment of inertia
α = Mp/My > 1 = shape factor
= 1.5 for a rectangular
section
= 1.7 for a solid circular
section
7
= 1.15 – 1.17 for I- or Csection
8
2
PLASTIC BEHAVIOR OF A
SIMPLE BEAM
If a load P at the mid-span of a
simple beam (Fig. 3) is
increased until the maximum
mid-span moment reaches the
fully plastic moment Mp, a plastic
hinge is formed at this section
and collapse will occur under
any further load increase. Since
this structure is statically deter
determinate, the collapse load PC can
easily be calculated to give
PC = 4M p / L
(3)
9
Plastic Hinge Along the
Length of the Simple Beam
11
P
L
2
L
(a) Loaded Beam
Mp
(b) Plastic BMD
PC
θ
θ
2θ
Δ
(c) Plastic Mechanism
Figure 3. Simple Beam
10
The collapse load of the beam
can be calculated by equating
the external and internal work
during a virtual movement of
the collapse mechanism (this
approach is equally applicable
to the collapse analysis of statically indeterminate beams).
Equating the external virtual
work We done by the force PC to
the internal virtual work Wi
done by the moment Mp at the
plastic hinge:
12
3
We = Wi
Lθ
= M p (2θ)
2
PC = 4M p / L
⇒ PC
⇒
which
hich is identical to the res
result
lt
given in (3).
ULTIMATE STRENGTH OF
FIXED-ENDED BEAM
Consider a prismatic fixed-ended
beam subjected to a uniform
load of intensity q (Fig. 4(a)).
Figure 4(b) shows the moment
diagram sequence from the yield
moment My
q y L2
I
M y = σ y S(≡ ) =
c
12
⇒ qy =
13
q
L
Mp
Mp
Mp
qC
Δ
θ
θ
2θ
Figure 4. Fixed-Fixed Beam
through the fully plastic condition
14
in the beam.
⎛ q L ⎞ Lθ
= M p (θ+ 2θ+ θ)
2⎜ C ⎟
⎝ 2 ⎠ 4
16 M p
⇒ qC =
L2
(b)
My
L2
The collapse mechanism is
shown in Fig. 4(c) and the collapse load is calculated by equating the external and internal
virtual works, i.e.
(a)
My
12 M y
Sequence of Plastic Hinge
Formation:
(c)
(1) Fixed-end supports – maximum moment (negative)
15
(2) Mid-span – maximum positive
16
moment
4
ULTIMATE STRENGTH OF
CONTINUOUS BEAMS
Next consider the three span
continuous beam shown in Fig. 5
with each span having a plastic
moment capacity of Mp. Values
of the collapse load corresponding to all possible mechanisms
are determined; the actual
collapse load is the smallest of
the possible mechanism
collapse loads.
17
For this structure, there are two
possible collapse mechanisms
are shown in Figs. 5(b) and (c).
Using the principle of virtual work
(We = Wi) for each mechanism
leads to
Figure 5(b) (Δ1 = Lθ/2):
⎛ Lθ ⎞
PC1 ⎜
⎟ = M p (θ + 2θ + θ)
⎝ 2 ⎠
⇒ PC1 = 8M p / L
Mp = constant
P
L
2
A
E
B
L
P
L
3
(a)
D
F
C
L
L
PC1
Δ1
θ
(b)
θ
2θ
PC2
Δ2
((c))
θ β
θ+β
Figure 5. (a) Continuous Beam
(b) Mechanism 1
18
(c) Mechanism 2
Figure 5(c) (Δ2 = Lθ/3):
⎛ Lθ ⎞
PC2 ⎜
⎟ = M p (θ + θ +β)
⎝ 3 ⎠
2Lβ
= Δ 2 = Lθ
3
3
⇒ β= θ
2
5M pθ
⎛ Lθ ⎞
∴ PC2 ⎜
⎟ =
2
⎝ 3 ⎠
⇒ PC2 = 15M p / 2L
19
20
5
The smaller of these two values
is the true collapse load. Thus,
PC = 7.5Mp/L and the corresponding bending moment
diagram is shown below.
1
2Mp
Mp
L
L
θ
(b)
θ
Δ1
2θ
qC
Δ2
F
D
-Mp
Collapse BMD
(a)
q
2
C
E
-M > -Mp
L
2
Mp
M < Mp
B
qL = P
PC
When collapse occurs, the
part of the beam between A
and C is still in the elastic
range.
A
P
21
θ
(c)
β
θ+β
L1
Figure 6. (a) Continuous Beam
(b) Mechanism 1 22
(c) Mechanism 2
Mechanism 1:
The two span continuous beam
shown in Fig. 6 exhibits some
unique considerations:
P Lθ
We = PC Δ1 = C
2
Wi = 2M pθ+ 2M p (2θ) + M pθ
= 7M pθ
1.the plastic moment capacity of
span 1-2 is different than the
plastic moment capacity of
span 2-3; and
We = Wi
2.the location of the p
positive
moment plastic hinge in span
2-3 is unknown.
M h i
Mechanism
2:
2
23
⇒ PC =
14M p
L
(A)
Δ
Δ
We = q C L1 2 + q C (L − L1) 2
2
2
Δ
= qCL 2
2
24
6
Wi = M pθ+ M p (θ+β)
The problem with this solution
for qCL is that the length L1 is
unknown.
L1θ = Δ 2 = (L − L1) β
⇒ β=
L1
θ
L − L1
L1 can be obtained by differentitiating
ti both
b th sides
id off qCL with
ith
respect to L1 and set the result to
zero, i.e.
⎛ 2L − L1 ⎞
∴ Wi = ⎜
⎟ M pθ
⎝ L − L1 ⎠
d(q CL)
−2L1(L − L1)
=
M
2
2 p
dL1
(L1) (L − L1)
∴ We = 1 q C LL1θ
2
We = Wi
⇒ qCL =
−
2 ⎛ 2L − L1 ⎞
M p (B)
L1 ⎜⎝ L − L1 ⎟⎠
2(2L − L1) (L − 2L1)
(L1)2 (L − L1)2
Mp
= 0
(C)
25
Solving (C) for L1:
26
Comparing the result in (A) with
(E) and for qL = P shows that the
failure mechanism for this
beam structure is in span 2-3.
2L21 − 8LL1 + 4L2 = 0
8L ± (8L) 2 − 4(8L2 )
4
= 2L − 2 L
⇒ L1 =
= 0.5858L
M < 2Mp
L1
(D)
Mp
Substituting (D) into (B):
qCL =
-M > -2Mp
11.66 M p
L
(E)
27
-Mp
BMD for Collapse Load qC
28
7
Direct Procedure to
Calculate Positive Moment
Plastic Hinge Location for
Unsymmetrical Plastic
M
Moment
t Diagram
Di
Consider any beam span that is
loaded by a uniform load and the
resulting plastic moment diagram is
unsymmetric. Just as shown
above the location of the maximum
positive moment is unknown. For
example, assume beam span B –
C is subjected to a uniform load
and the plastic moment capacity at
end B is Mp1, the plastic moment29
The location of the positive plastic
moment can be determined using
the bending moment equation
M(x) = ax2 + bx + c
and appropriate boundary
conditions.
(i) x = 0: M = -Mp1 = c
capacity at end C is Mp2 and the
plastic positive moment capacity is
Mp3.
Mp1 ≤ Mp3; Mp2 ≤ Mp3
Mp3
x
-M
Mp11
-Mp2
L1
L
30
Solving for a and b from (ii) and
(iii):
a=
b=
−(M p1 + M p3 )
L21
2(M p1 + M p3 )
L1
(ii) x = L1: M = Mp3 = aL12
+ bL1 + c
⇒ aL12 + bL1 = Mp3 + Mp1
(iii) x = L1: dM/dx = 0 = 2aL1 + b 31
32
8
(iv) x = L:
M = -Mp2 = aL2 + bL + c
= -(Mp1+ Mp3)(L/L1)2
+ 2(Mp1+ Mp3) (L/L1) - Mp1
⎛L⎞
⎜ ⎟ =1
⎝ L1 ⎠
±
4(Mp1 + Mp3)2 − 4(Mp1 − Mp2)(Mp1 + Mp3)
2(Mp1 + Mp3)
⎛ Mp1 −Mp2 ⎞
= 1 ± 1− ⎜
⎟
⎝ Mp1+Mp3 ⎠
0 = -(Mp1+ Mp3)(L/L1)2
+ 2(Mp1+ Mp3) (L/L1)
- Mp1+ Mp2
Solving the quadratic equation:
∴ L1 =
33
L
⎛ M p1 − M p2 ⎞
1+ 1− ⎜ M + M ⎟
p3 ⎠
⎝ p1
34
EPILOGUE
The process described in these
notes and in the example problems uses what is referred to as
an “upper bound” approach;
i e any assumed mechanism can
i.e.,
provide the basis for an analysis.
The resulting collapse load is an
upper bound on the true collapse load. For a number of
trial mechanisms, the lowest
computed
p
load is the best
upper bound. A trial mechanism is the correct one if the
corresponding moment
diagram nowhere exceeds the
plastic moment capacity.
35
9