Plastic Analysis of
1
Continuous Beams
Increasing the applied load until
yielding occurs at some locations
will result in elastic
elastic-plastic
plastic defor
deformations that will eventually reach
a fully plastic condition.
Fully plastic condition is
defined as one at which a
sufficient number of plastic
hinges
g are formed to transform
the structure into a mechanism, i.e., the structure is
geometrically
t i ll unstable.
t bl
1
1See pages 142 – 152 in your class notes.
Additional
Addi
i
l loading
l di applied
li d to
the fully plastic structure
would lead to collapse.
collapse
Design
g of structures based on
the plastic or limit state
approach is increasingly used
and
d accepted
t db
by various
i
codes
d off
practice, particularly for steel
construction Figure 1 shows a
construction.
typical stress-strain curve for mild
steel and the idealized stressstrain response for performing
plastic analysis.
2
σ
rupture
x
σy
idealized
ε
εy
Figure 1. Mild Steel StressStrain Curve
σy = yield stress
εy = yield strain
3
ULTIMATE MOMENT
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.
g 2(c)
( ) and
(d)) and finally the two yield
zones meet (Fig. 2(e)); the
cross section
i in
i this
hi state is
i
defined to be fully plastic.
4
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,
plastic (d) Top and bottom
5
fibers plastic, and (e) Fully plastic
The ultimate moment is
determined in terms of the yield
stress σ y .
Since the axial force is zero in
this beam case,
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
A/2, forming a
couple equal to the ultimate
plastic moment Mp
M p = 12 σ y A (yc + y t )
(1)
6
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
(2a)
M y = σ y Sc
(2b)
St = tension section modulus
(≡ I / ct )
p
section
Sc = compression
modulus ( ≡ I / cc )
7
ct = distance from neutral axis
to the extreme tension fiber
cc = distance from neutral
axis to the extreme compression fiber
p
I = moment of inertia
α = Mp/My > 1 = shape factor
= 1.5
1 5 for a rectangular
section
= 1.7 for a solid circular
section
= 1.15 – 1.17 for I- or Csection
8
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
id
momentt reaches
h the
th
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 determinate, the collapse load PC can
easil be calc
easily
calculated
lated to gi
give
e
PC = 4M p / L
(3)
9
P
L
2
L
(a) Loaded Beam
Mp
(b) Plastic BMD
PC
θ
θ
2θ
Δ
(c) Plastic Mechanism
Figure 3. Simple Beam
10
Plastic Hinge Along the
Length of the Simple Beam
11
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 eq
equally
all applicable
to the collapse analysis of statically indeterminate beams)
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
We = Wi
Lθ
Lθ
⇒ PC
= M p (2θ)
2
⇒ PC = 4M p / L
which is identical to the result
given in (3).
13
ULTIMATE STRENGTH OF
FIXED ENDED BEAM
FIXED-ENDED
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
2
q
L
y
I
M y = σ y S(≡ ) =
c
12
⇒ qy =
12 M y
2
L
through the fully plastic condition
14
in the beam.
q
(a)
L
Mp
(b)
My
My
Mp
Mp
qC
Δ
θ
θ
2θ
Figure 4. Fixed-Fixed Beam
(c)
15
The collapse mechanism is
shown in Fig
Fig. 4(c) and the collapse load is calculated by equating
g the external and internal
virtual works, i.e.
⎛ q CL ⎞ Lθ
2⎜
= M p (θ
θ+ 2θ
θ+ θ)
⎟
⎝ 2 ⎠ 4
16 M p
⇒ qC =
2
L
Sequence off Plastic
S
Pl i Hinge
Hi
Formation:
(1) Fixed-end
Fi d d supports – maxii
mum moment (negative)
(2) Mid-span
Mid
– maximum
i
positive
iti
16
moment
ULTIMATE STRENGTH OF
CONTINUOUS BEAMS
Next consider
N
id the
h three
h
span
continuous beam shown in Fig. 5
with each span having a plastic
moment capacity of Mp. Values
of the collapse
p load correspondp
ing to all possible mechanisms
are determined; the actual
collapse
ll
load
l d is
i the
th smallest
ll t off
the possible mechanism
collapse loads.
loads
17
Mp = constant
P
L
2
A
L
3
E
B
L
C
L
P
(a)
D
F
L
PC1
Δ1
θ
(b)
θ
2θ
PC2
Δ2
θ β
(c)
θ+β
Figure 5. (a) Continuous Beam
(b) Mechanism 1
18
(c) Mechanism 2
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
19
Figure 5(c) (Δ2 = Lθ/3):
⎛ Lθ ⎞
PC2 ⎜
⎟ = M p (θ + θ +β)
⎝ 3 ⎠
2Lβ
L
θ
= Δ2 =
3
3
⇒ β= θ
2
55M pθ
⎛ Lθ ⎞
∴ PC2 ⎜
⎟ =
2
⎝ 3 ⎠
⇒ PC2 = 15M p / 2L
20
The smaller of these two values
i th
is
the ttrue collapse
ll
lload.
d Thus,
Th
PC = 7.5Mp/L and the corresponding bending moment
diagram is shown below.
When collapse occurs,
occurs the
part of the beam between A
and C is still in the elastic
range.
Mp
M < Mp
A
B
C
E
-M > -Mp
F
D
-Mp
Collapse BMD
21
P
qL = P
L
2
(a)
q
2
1
2Mp
Mp
L
L
PC
θ
(b)
θ
Δ1
2θ
qC
Δ2
θ
(c)
β
θ+β
L1
Figure 6. (a) Continuous Beam
(b) Mechanism 1 22
(c) Mechanism 2
The two span continuous beam
shown in Fig
Fig. 6 exhibits some
unique considerations:
1.the plastic moment capacity of
span 1-2 is different than the
plastic
l ti momentt capacity
it off
span 2-3; and
2.the location of the positive
moment plastic hinge in span
2 3 is unknown
2-3
unknown.
23
Mechanism 1:
PC Lθ
We = PC Δ1 =
2
Wi = 2M pθ + 2M p (2θ) + M pθ
= 7M pθ
We = Wi
⇒ PC =
14M p
L
(A)
Mechanism 2:
Δ2
Δ2
We = q C L1
+ q C (L − L1)
2
2
Δ2
= qCL
2
24
Wi = M pθ + M p (θ +β
β)
L1θ = Δ 2 = (L − L1) β
L1
θ
⇒ β=
L − L1
⎛ 2L − L1 ⎞
M pθ
∴ Wi = ⎜
⎟
⎝ L − L1 ⎠
∴ We = 1 q C LL1θ
2
We = Wi
2 ⎛ 2L − L1 ⎞
⇒ qCL = ⎜
M
(B)
p
⎟
L1 ⎝ L − L1 ⎠
25
The problem with this solution
for qCL is that the length L1 is
unknown.
L1 can be obtained by differentiating both sides of qCL with
respect to L1 and set the result to
zero, i.e.
d(q C L)
−2L1(L − L1)
=
Mp
2
2
dL1
(L1) (L − L1)
−
2(2L − L1)(L − 2L1)
= 0
2
(L1) (L − L1)
2
Mp
(C)
26
Solving (C) for L1:
2
2
2L1 − 8LL1 + 4L = 0
8L ± (8L) 2 − 4(8L2 )
⇒ L1 =
4
= 2L − 2 L
= 0.5858L
( )
(D)
Substituting (D) into (B):
qCL =
11 66 M p
11.66
L
(E)
27
Comparing the result in (A) with
(E) and
d ffor qL
L = P shows
h
th t the
that
th
failure mechanism for this
beam structure is in span 2
2-3
3.
M < 2Mp
L1
Mp
-M
M > -2M
2Mp
-Mp
BMD for Collapse Load qC
28
Direct Procedure to
Calculate Positive Moment
Plastic Hinge Location for
Unsymmetrical Plastic
Moment Diagram
g
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 –
example,
C is subjected to a uniform load
and
a
d tthe
ep
plastic
ast c moment
o e t capac
capacity
ty at
end B is Mp1, the plastic moment29
capacity at end C is Mpp2 and the
plastic positive moment capacity is
Mp3.
Mp1 ≤ Mp3; Mp2 ≤ Mp3
Mp3
x
-Mp1
L1
-Mp2
L
30
The location of the positive plastic
momentt can be
b determined
d t
i d using
i
the bending moment equation
M(x) = ax2 + bx + c
and appropriate boundary
conditions.
(i) x = 0: M = -Mp1 = c
(ii) x = L1: M = Mp3 = aL12
+ bL1 + c
⇒ aL12 + bL1 = Mp3 + Mp1
(iii) x = L1: dM/dx = 0 = 2aL1 + b 31
Solving for a and b from (ii) and
(iii):
a=
b=
−(M p1 + M p3 )
2
L1
2(M p1 + M p3 )
L1
32
(iv) x = L:
M = -Mp2 = aL2 + bL + c
= -(Mp1+ Mp3)(L/L1)2
+ 2(Mp1+ Mp3) (L/L1) - Mp1
0 = -(M
(Mp1+ Mp3)(L/L1)2
+ 2(Mp1+ Mp3) (L/L1)
- Mp1+ Mp2
Solving the quadratic equation:
33
⎛L⎞
⎜ ⎟ =1
⎝ L1 ⎠
±
4(Mp1 + Mp3)2 − 4(Mp1 − Mp2)(Mp1 + Mp3)
2(Mp1 + Mp3)
⎛ Mp1 −Mp2 ⎞
= 1 ± 1− ⎜
⎟
M
+
M
⎝ p1 p3 ⎠
∴ L1 =
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
pro ide the basis for an anal
provide
analysis.
sis
The resulting collapse load is an
upper bound on the true col
collapse load. For a number of
trial mechanisms, the lowest
computed 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