Plastic Analysis structural

8/20/2019 Plastic Analysis structural

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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.

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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.

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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)

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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

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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

.

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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⇒ =

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Figure 5(c) (Δ = Lθ/3):

Lθ⎛ ⎞ =3⎝ ⎠

23 3

θ= =

5M θ

2

C2P3 2

∴ =⎜ ⎟⎝ ⎠

C2 p⇒ =

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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

- .

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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

=

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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+⎝ ⎠

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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

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Plastic Analysis structural