Chapter8c

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CHAPTER 8c. INTRODUCTION TO BEAMS Slide No. 2ENCE 355 Assakkaf

The Collapse Mechanism

Statically Determinate Beam (contd)P

(a)

(b)

Figure 1

Real hinge Real hinge

Plastic hinge

nP



Statically Determinate Beam (contd)

Any further increase in the load will cause

collapse.

Pn

represents the nominal or theoretical

maximum load that the beam can support.

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Statically Indeterminate Beam For statically indeterminate beam to fail, it

is necessary for more than one plastic

hinge to form.

The number of plastic hinges required for

failure of statically indeterminate structure

will be shown to vary from structure to

structure, but never be less than two.

The fixed-end beam of Fig. 2 cannot fail

unless the three hinges shown in the figure

are developed.



Statically Indeterminate Beam (contd)

P

(a)

(b)

Figure 2

Plastic hinge Plastic hinge

Plastic hinge

nP

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Statically Indeterminate Beam (contd) Although a plastic hinge may have formed in a

statically indeterminate structure, the load can

still be increased without causing failure if the

geometry of he structure permits.

The plastic hinge will act like a real hinge as

far as the increased loading is concerned.

As the load is increased, there is a

redistribution of moment because the plastic

hinge can resist no more moment.


The Collapse Mechanism Statically Indeterminate Beam (contd)

As more plastic hinges are formed in the

structure, there will eventually be a

sufficient number of them to cause

collapse.

Actually, some additional load can be

carried after this time before collapse

occurs as the stresses go into the strain

hardening range.

However, deflections that would occur are

too large to be permissible in the design.

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The Mechanism The load may be further increased until the

moment at some point (here it will be at the

concentrated load) reaches the plastic

moment.

Additional load will cause the beam to

collapse.

Therefore, the Mechanism is defined as the

arrangement of plastic hinges and perhaps real

hinges which permit the collapse in a structure

as shown in part (b) of Figs. 1, 2, and 3.


Plastic Analysis of Structure

There are various methods that can be

used to perform plastic analysis for a

given structure.

Two satisfactory method for this type of

analysis are

The virtual-work Method (Energy Method)

Equilibrium Method

In this course, we will focus on the

virtual-work method.

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Example 1Determine the plastic limit (or nominal)

distributed load wn

in terms of the plastic

(or nominal) moment Mn

developed at the

hinges.

ft18=L

(k/ft)n

w


Plastic Analysis of Structure Example 1 (contd)

The collapse mechanism for the beam is

sketched.

ft18=L

(k/ft)n

w

2

L

2

L

2Collapse Mechanism

Lwn

A

B

C

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Example 1 (contd) Because of the symmetry, the rotations

at the end plastic hinges are equal.

The work done by the external load (wnL) is

equal wnL times the average deflection avg

of the mechanism at the center of the

beam.

The deflection is calculated as follows:

2

theory)angle(small2/tan

L

L

=

=



Example 1 (contd)

The internal work absorbed by the hinges

is equal the sum of plastic moments Mn

at

each plastic hinge times the angle through

which it works.

The average deflection avg throughout the

length of the beam is equals one-half thedeflection at the center of the beam, that

is

422

1

2

1avg

LL =

==

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Example 1 (contd)

Applying Eq. 1 (conservation of energy),

yield a relationship between wn

and Mn

as

follows:

( ) ( ) ( ) ( )

nn

nnnn

ML

Lw

MMMLw

WW

44

2

workInternalworkExternal

avg

int.ext.

=

++=

=

=

Left

A

Middle

B

Right

C



Therefore,

For 18-ft span, the plastic limit distributed

load is computed as

2

16

44

L

Mw

ML

Lw

n

n

nn

=

=

( ) 25.20181616

22

nnn

n

MM

L

Mw ===

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Example 2For the propped beam shown, determine

the plastic limit (or nominal) load Pn

in

terms of the plastic (or nominal) moment

Mn

developed at the hinges.

ft20=L

nP

ft10



The collapse mechanism for the beam is sketched.

ft20=L

nP

ft10

2

L

2

L

2Collapse Mechanism

Lwn

A

B

C

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Example 2 (contd) Because of the symmetry, the rotations

at the end plastic hinges are equal.

The work done by the external load (Pn) is

equal Pn

times the deflection of the

mechanism at the center of the beam.

The deflection is calculated as follows:

2

theory)angle(small

2/

tan

LL

=

=



The internal work absorbed by the hinges


at


which it works.

Note that in example, we have only two

plastic hinges at points A and B of the

mechanism. Point C is a real hinge, and nomoment occurs at that point.

Also note that the external work is

calculated using and not avg. because of

the concentrated load Pn

in that location.

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Example 2 (contd)


yield a relationship between Pn

and Mn

as

follows:

( ) ( ) ( )

nn

nnn

ML

P

MMP

WW

32

2


int.ext.

=

+=

=

=

Left

A

Middle

B



Therefore,

For 20-ft span, the plastic limit load Pn

is

computed as

L

MP

ML

P

n

n

nn

6

32

=

=

n

nnn

n M

MM

L

MP 3.0

10

3

20

66====

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Example 3 (contd)

Because of the unsymmetry, the rotations

1 and 2 at the end plastic hinges are not

equal.

We need to find all rotations in terms, say

1

The work done by the external load (Pn) is

equal Pn

times the deflection of the

mechanism at the center of the beam.



From triangles ABE and BCE:

3

2L

3

L

12

)( 21 +

nP

A

B

CE

1221

222

111

2or33

2

:3and2Eqs.fromThus,

33/tan:BCE

3

2

3/2tan:ABE

==

==

==

LL

L

L

L

L

11121

12

1

32:BAt

2:AAt

:CAt

Therefore,

=+=+=

==

=

B

A

C

(2)

(3)

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Example 3 (contd) The internal work absorbed by the hinges


at


which it works.

Note that in example, we have three plastic

hinges at points A, B, and C of the

mechanism. Also there is no real hinge.

Also note that the external work is

calculated using and not avg. because ofthe concentrated load P

nin that location.




yield a relationship between Pn

and Mn

as

follows:

( ) ( ) ( ) ( )

nn

nnnn

ML

P

MMMP

WW

11

111

int.ext.

63

2

32


=

++=

=

=

Left

A

Middle

BRight

C2Eq.from

3

21

L=

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Example 3 (contd) Therefore,

For 30-ft span, the plastic limit load Pn

is

computed as

L

MP

ML

P

n

n

nn

9

63

2

=

=

nnn

n MM

L

MP 3.0

3099 ===



Complex Structures

If a structure (beam) has more than one

distributed or concentrated loads, there

would be different ways in which this

structure will collapse.

To illustrate this, consider the propped

beam of Fig. 4.

The virtual-work method can be applied to

this beam with various collapse

mechanisms.

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Complex Structures (contd)

ft30=L

nP

ft10

nP6.0

ft10ft10

Figure 4




The beam with its two concentrated loads

is shown in Fig. 5 together with four

possible collapse mechanisms and the

necessary calculations.

It is true that the mechanisms of parts (a),

(c), and (d) of Fig. 5 do not control, butsuch a fact is not obvious for those taking

an introductory course in plastic analysis.

Therefore, it is necessary to consider all

cases.

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Figure 5a. Various Cases of Collapse Mechanism

ft30=L

nP

ft10

nP6.0

ft10ft10

2

3

20 10

( ) ( ) ( )

nn

nn

nnn

MP

PM

PPM

227.0

4.4

10206.05

==

+=

Real hinge


Plastic Analysis of StructureFigure 5b. Various Cases of Collapse Mechanism

ft30=L

nP

ft10

nP6.0

ft10ft10

3

10 20

2

( ) ( ) ( )

nn

nn

nnn

MP

PM

PPM

154.0

5.6

20106.04

=

=

+=

Real hinge

Controls

Controls

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Figure 5c. Various Cases of Collapse Mechanism

ft30=L

nP

ft10

nP6.0

ft10ft10

2

10

( ) ( )

nn

nn

nn

MP

PM

PM

3.0

33.3

103

==

=

Real hinge


Plastic Analysis of StructureFigure 5d. Various Cases of Collapse Mechanism

ft30=L

nP

ft10

nP6.0

ft10ft10

10

( ) ( ) ( )

nn

nn

nnn

MP

PM

PPM

1875.0

33.5

10106.03

=

=

+=

Real hinge

10

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The value for which the collapse load Pn

is

the sm ll st in terms of Mn

is the correct

value.

or the value where Mn

is the gr t st in

terms of Pn.

For this beam, the second plastic hinge

forms at the concentrated load Pn, and P

n

equals 0.154 Mn.

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