Post on 07-Aug-2018
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Plastic Analysis oflastic Analysis of
Continuous Beams
Increasing the applied load until
yielding occurs at some locations
- -mations that will eventually reach
Fully plastic condition is
sufficient number of plastic
hin es are formed to transform
the structure into a mecha-nism, i.e., the structure is
1
geome r ca y uns a e.
1See pages 142 – 152 in your class notes.
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Additional loading applied to
the fully plastic structure.
Desi n of structures based on
the plastic or limit state
approach is increasingly used
an accep e y var ous co es o
practice, particularly for steel.
typical stress-strain curve for mil
steel and the idealized stress-
strain response for performingplastic analysis.
2
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σ rupture
xσy
idealized
y
Figure 1. Mild Steel Stress-
σy = yield stress
3
y = yield strain
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Consider the beam shown in Fig
2. Increasing the bending
moment results in going from
(Fig. 2(a)) to yield of the
outermost fibers Fi s. 2 c and
(d)) and finally the two yield
zones meet (Fig. 2(e)); thecross section in this state is
defined to be fully plastic.
4
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determined in terms of the yieldstress .σ
Since the axial force is zero in
,in the fully plastic condition
divides the section into two
equal areas, and the resultant
tension and compression are,
couple equal to the ultimateyσ
1 p y c t2
M A (y y )= σ + (1)
6
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The maximum moment which a
section can resist without
exceeding the yield stress
(defined as the yield moment
My) is the smaller of
y y tM S= σ (2a
y y c
S = tension section modulu
( )
S = com ression section
tI / c≡
modulus ( )cI / c≡
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=
to the extreme tension fiber c
axis to the extreme com-
ression fiber
I = moment of inertia
= Mp/M
y> 1 = shape factor
= .
section
=
section
= – - -
8
section
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PLASTIC BEHAVIOR OF A
If a load P at the mid-span of a
simple beam (Fig. 3) is
increased until the maximum
m -span momen reac es efully plastic moment Mp, a plasti
and collapse will occur under
any further load increase. Since
this structure is statically deter-
minate, the collapse load PC ca
y u v
9
C p
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P
2
(a) Loaded Beam
p
(b) Plastic BMD
θθ
C
2θ(c) Plastic Mechanism
Δ
10Figure 3. Simple Beam
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Length of the Simple Beam
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can be calculated by equatingthe external and internal work
during a virtual movement of
the collapse mechanism (this
u yto the collapse analysis of sta
.
Equating the external virtual
work We done by the force PC to
the internal virtual work Widone by the moment Mp at the
12
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e i C p
W W P M (22
= ⇒ = θ
C p⇒ =
which is identical to the result
given in (3).
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ULTIMATE STRENGTH OF
-
Consider a prismatic fixed-ended
load of intensity q (Fig. 4(a)).
diagram sequence from the yield
2yq LI= =c 12
y12 M
y 2L
14
in the beam.
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q(a)
LM
M M
Mp Mp
qC
(c)θ θΔ
15
Figure 4. Fixed-Fixed Beam
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The collapse mechanism is
- .
lapse load is calculated by equa-ting the external and internal
L Lθ
virtual works, i.e.
p2 4 = + +⎝ ⎠
pC
2
q
L
⇒ =
equence of Plastic Hinge
Formation:
(1) Fixed-end supports – maxi-mum moment (negative)
16
-span – max mum pos v
moment
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CONTINUOUS BEAMSNext consider the three span
continuous beam shown in Fig. 5
moment capacity of Mp. Values
of the colla se load corres ond-
ing to all possible mechanisms
are determined; the actualco apse oa s e sma es o
the possible mechanism
.
17
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Mp = constant
L
23 A
DE F
L L L
PC1(b
P
θθ2θ
1Δ
(c
2Δ θ β
Figure 5. (a) Continuous Beam
18
(c) Mechanism 2
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For this structure, there are two
possible collapse mechanisms
are shown in Figs. 5(b) and (c).
(We = Wi) for each mechanism
leads to
Figure 5(b) (Δ1 = Lθ/2):
LP M 2θ⎛ ⎞ = θ+ θ+ θ
C1P 8M / L⇒ =
19
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Figure 5(c) (Δ = Lθ/3):
Lθ⎛ ⎞ =3⎝ ⎠
23 3
θ= =
5M θ
2
C2P3 2
∴ =⎜ ⎟⎝ ⎠
C2 p⇒ =
20
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The smaller of these two values
s e rue co apse oa . us,
PC = 7.5Mp/L and the corres-
diagram is shown below.
,part of the beam between A
and C is still in the elastic
range.
M < p
C A B
-M-M > -M
DE F
21Collapse BMD
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P qL = P
q2
2M M1 2
L L
P
(b
θ θ
2 θ
θ β
C
β2Δ
L1+
22
(b) Mechanism 1
(c) Mechanism 2
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The two span continuous beam.
unique considerations:
1.the plastic moment capacity of
span 1-2 is different than the
p as c momen capac y o
span 2-3; and2.the location of the positive
moment plastic hinge in span
- .
23
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Mechanism 1:
Ce C 1P L
W P 2
θ= Δ =
i p p pW 2M 2M (2 ) M= θ+ θ + θ
p= p14M==
L
Mechanism 2:
e C 1 C 1W q L q (L L )2 2= + −
24Cq L
2
=
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W M M= θ+ θ+
1 2 1L (L L )θ = Δ = − β1
1
L
L Lβ =
−⇒ θ
1i
2L LW M
⎛ ⎞−∴ = θ
1
1W q LL
−
∴ = θ
e iW W=
C 1
p2 2L L
q L M⎛ ⎞−
= ⎜ ⎟−⇒ (B
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The problem with this solution
for qC
L is that the length L1
is
unknown.
L1 can be obtained by differen-
tiating both sides of qCL with
respect to L1 and set the result tzero, i.e.
C 1 1 p2 2
d(q L) 2L (L L )
MdL
− −
= −
1 12(2L L ) (L 2L ) M− −
−1 1(L ) (L L )
0−
= (C26
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Solving (C) for L1:
2 21 12L 8LL 4L 0− + =2 2
18L (8L) 4(8L )
L4
± −⇒ =
2L 2 L
0.5858L
= −
= D
Substituting (D) into (B):
pC .q LL
=
27
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Comparing the result in (A) with
an or q = s ows a e
failure mechanism for this- .
L1M < 2Mp
p
-M- > - p
BMD for Collapse Load q
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Direct Procedure to
Calculate Positive Momen
Plastic Hinge Location for
Unsymmetrical Plastic
Moment Dia ram
Consider any beam span that is
resulting plastic moment diagram i
unsymmetric. Just as shownabove the location of the maximum
positive moment is unknown. For
, –C is subjected to a uniform load
and the lastic moment ca acit a
29end B is Mp1, the plastic moment
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capacity at end C is M 2 and the
plastic positive moment capacity is
Mp3.
Mp1 ≤ Mp3; Mp2 ≤ Mp3
Mp3
x
-Mp1
-Mp2
1
L
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Solving for a and b from (ii) and
(iii):
p1 p3
2a
−=
p1 p32(M M )+
1
L
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(iv) x = L:
M = -Mp2 = aL2 + bL + c
= -(Mp1+ Mp3)(L/L1)2
+ 2(Mp1+ Mp3) (L/L1) - Mp1
= - p1 p3 1
+ 2(Mp1+ Mp3) (L/L1)
- Mp1+ Mp2
Solving the quadratic equation:
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L⎛ ⎞
1
2
L
4 M M 4 M M M M
⎝ ⎠
+ − − +
p1 p32(M M )±
+
p1 p2
p1 p3M M1 1 −
+= ± − ⎜ ⎟⎝ ⎠
L
p1 p2
1M M−
∴ =⎛ ⎞−
p1 p3M M+⎝ ⎠
34
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EPILOGUEEPILOGUE
notes and in the example pro-blems uses what is referred to as
an “ upper bound” approach;
i.e., any assumed mechanism ca
v y .The resulting collapse load is an
-
lapse load. For a number of
trial mechanisms, the lowest
computed load is the best
upper bound. A trial mecha-
corresponding moment