Plate Yield Line Theory 07-09-2015

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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates

Introduction

Why yield line analysis?

Behaviour of a plate under increasing load

Rules for yield lines

What is analysis by virtual work

External work done by loads: examples

Internal work done by resisting moments: examples

Energy dissipation in a yield line

Rectangular plate with an arbitrary chosen dimension

Four examples

Assignments 3 and 4

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Learning objectives

Understanding the meaning of yield line theory


Using method of virtual work

Finding the decisive yield line pattern

Analysing the failure load of plated structures

variation of:

support conditions: free edge, restraint, simply supported

loadings: uniform, non-uniform, concentrated force, partly

uniform loaded

material behavior: isotropic, orthotropic, moment capacity

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Introduction


We studied already the use of plasticity in case of beam elements (plastic hinges).

Now we extend the theory of plasticity to plates (yield lines).

Regarding plasticity of plates three different solutions techniques can be distinguished:

Incremental (stepwise) elastic-plastic calculation. Discussed last week (also part of assignment 1).

Application of lower-bound theorem based on equilibrium equations.

Application of upper-bound theorem based on a mechanism.

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Introduction Incremental (stepwise) elastic-plastic calculation

This method normally can not be done analytically (by hand) because of its complexity (assignment 1 which is approx. the most simple case - one beam element only – and we concluded that this already resulted into a lot of work to be done). This method in fact can only be carried out by the use of software.

Application of lower-bound theorem based on equilibrium equations

This method will be discussed in week 4.

Application of upper-bound theorem based on a mechanism

This method is expressed by yield-line theory (use of virtual work).It provides an upper-bound solution which forms a restriction for it’s application on arbitrary practical problems.However, from validation of the theory based on experimental research shortcomings has been solved to a large extend. But still precautions should be taken to ensure that calculated failure load at least closely approaches the correct value.

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Introduction


The load increases stepwise and finally the applied moment becomes equal tot the flexural capacity of the cross section of the plate.

Upon yielding, the curvature of the plate at the yield line cross section increases sharply and deflection increases disproportionately.

The elastic curvature along the plate is relatively very small and therefore it is acceptable to consider the plate parts between the yield lines straight (flat).

The resistance at the yield lines is mp. Notice that for some materials there is a difference in mp between positive and negative yield line (like reinforce concrete slab).

A mechanism means there are now other positions possible for the formation of plastic hinges (lines) and the ratio of moment distribution just before collapse is 1:1.

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Why Yield Line analysis


An analysis approach for determining the ultimate load capacity of lateral loaded plates.

It is only applicable to ductile plates (like steel plates) since we assume a certain moment-rotation diagram.

a yield line occurs when the moment capacity has been reached

no additional moment can be taken at the cross section

the cross section can undergo any amount of rotation

The main advantage of this approach over conventional code-based approach to plate analysis and design is it’s ability to cater for irregular geometries, plates with uncommon supports conditions and uncommon loadings (like concentrated point load, partly distributed load, non uniform partly distributed load, plate column supported, etc).

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8



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Consider a rectangular plate which is loaded by a uniformly distributed load q (q is a fixed value and is the load factor.

Starting from unloaded state =0 the load is gradually increased.

In first instance the response of the plate is completely elastic.

At a certain level = e somewhere in the plate the stress state satisfied the yield stress and initial yielding occurs, so-called plastic hinge which for a plate is called a yield line.

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During continuing loading more plastic points appear.

These points chain together and form finally a complete pattern of yield lines and plate zones. For this state = p and the plate deflects unlimited.The unit of the yield bending moment is force and is expressed in Nm/m or shortly N.

The entire increase of plastic deformation is concentrated in a number of yield lines and the plate parts can be considered to be flat.

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Yield lines are straight lines because they represent the

intersection of two planes.

Yield lines represent axes of rotation.

The supported edges of the plate will also establish axes of

rotation. If the edge is fixed, a negative yield line may form

providing constant resistance to rotation. If the edge is simply

supported, the axis of rotation provide zero restraint.

Yield lines form under concentrated loads, radiating outward from

the point of application.

A yield line between two plates must pass through the point of

intersection of the axes of rotation of the adjacent plate segments.

Yield moments are principal moments and therefore twisting

moments are zero along the yield lines and in most cases the

shearing forces are also zero.

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Wrong: yield lines are not straight.

Wrong: points ABCD are no laying is a single plane.

Wrong: yield line of intersect EF is not parallel to AB. Right: yield line is a correct solution.


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Typical yield line patterns.a solid line represents a positive

yield line caused by sagging yield moment.

a broken yield line represents a negative yield line caused by a hogging yield moment

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Once the general pattern of yielding and rotation has been established by applying the rules of yield lines, the specific location and orientation of the axes of rotation and the failure load of the plate can be analyzed by virtual work.

It’s a must to investigate all possible mechanisms for any plate to confirm that the correct solution, giving the lowest failure load, has been found.

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What is analysis by Virtual Work?


The external work done by a load to cause a small arbitrary virtual deflection must equal the internal work done as the plate rotates at the yield lines to accommodate this deflection.

Elastic rotations and deflections are not considered when writing the work equations, as they are very small compared to plastic deformations.

External energy

Internal energy

Expended by loads moving Dissipated by rotations about yield lines

Expended Dissipated

W Ed

for all regions( )N for all regions( )m l

dW E

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Energy is dissipated in the yield lines only.

Deformations and internal loads in a yield line

Energy is dissipated in the yield lines only.

Because of equilibrium, the forces and moments per unit length along the yield line are:

mnn = bending momentmns = torsional momentqn = transverse force

The plastic deformation equals the difference in rotation of both planes about the s-axis.This angle is small and therefore

tan sind d d

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for all regions for all regions( ) ( )N m l

N Load(s) acting within a particular region [kN]

δ Vertical displacement of the load(s) N on each region expressed as a fraction of unity [m]

m The moment in or the moment of resistance of the plate per metre run [kNm/m]

l The length of yield line or its projected length onto the axis of rotation for that region [m]

The rotation of the region about its axis of rotation [m/m]

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In case of uniform distributed lateral (surface) load, the amount of work equals:

w(x,y) = increase in displacement during failure



( , ) ( , )plate

W q x y w x y dxdy

z

plate parts

W q S w

S = plate areaw = vertical displacement (gravity) at plate centre

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Of all forces along the yield line, only mnn provides a contribution to the energy dissipation

The yield line is the intersection between two planes and therefore the value of is constant.


d nn d

along the yield line

E m ds

ls = the length of the yield line

d

d p d sE m l

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External work done by Loads: Examples


A square plate carrying a single concentrated load at its center

A plate supported along three sides and free along the fourth. Loaded with a line load “w” per unit length along the free edge

1W P P

11 2 ( ) 1 ( )

2W b w a w w a b w a b

1

2

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A distributed load “w” per unit area on a triangular segment defined by a hinge and yield lines

A rectangular plate carrying a distributed load “w” per unit area

3

4

1 11 1

2 3 6 6

a b a bW w a b w w

11 1

2 2 2

a b a bW w a b w w

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More complicated trapezoidal shapes may always be subdivided into component triangles and rectangles.

5

17,5 3,75 2 9,375

2 3

13,75 3,75 4 9,375

2 3

1,5 3,75 2 5,6252

9,375 9,375 5,625 24,375

5

24,375W w

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More complicated trapezoidal shapes may always be subdivided into component triangles and rectangles.

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Simply supported plate uniformly loaded by a surface load λq.Displacement at centre = δ

E

1r

r

R

2

0 0

2

1

26

r

plate

R

W q rd dr

rW q rd dr

R

RW q

rd

dr

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Internal work done by Resisting Moments


Summing the products of yield moment “m” per unit length of plastic hinges times the plastic rotation at the respective yield lines.

Consider a rectangular plate simply supported.

Assume a most logic yield pattern, with oneunknown parameter “a”.

Total rotation diagonal yield line = θ1

Total rotation horizontal yield line = θ2

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The correct value of “a” will be such as to maximize the moment resistance required to support the uniform distributed load “w”.

From comparison of triangular shapes it is found

This means

2 25 25 25 and

5

a a ab c

a

12 2 2

1 1 5 1 5

55 25 25 25

a a

b c aa a a a

2

1 22

5 5

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2

12

1 5 525 2 2

5 525

a as a

a aa

Similar to

It seems that we can do all this in a simplified way by considering support edges only.

1 1 52 10 2

5 5

aa

a a

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The total internal work done by resisting moments is

For a number of different values of “a” results in the following data

2

1 225 4 20 2dE m a m a

“a” Ed

6 11.36 m

6.5 11.08 m

7 10.87 m

7.5 10.69 m

And we know already that W=Ed

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The external work done by uniform distributed load is


External work done by Loads=

1 1

20 2 5 2 5 4 10 22 2 3 2 3

W q a a a

“a” W

6 80,0qδ

6.5 78,4qδ

7 76,6qδ

7.5 75,0qδ

“a” m

6 80,0q/11.36=7,07q

6.5 78,4q/11.08=7,08q

7 76,6q/10.87=7,04q

7.5 75,0q/10.69=7,02q

With δ=unit=1. gives

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The largest values a=6,5 is most critical and the required resisting moment for the plate is 7.08 q.

“a” m

6 80,0q/11.36=7,07q

6.5 78,4q/11.08=7,08q

7 76,6q/10.87=7,04q

7.5 75,0q/10.69=7,02q

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The largest values a=6,5 is most critical and the required resisting moment for the plate is m=7.08 q.

This method is fantastic because for each plate configuration, boundary conditions and loading we can analyse:- Max loading, or- Needed minimum plastic moment capacity

Plate

thickness

“t”

12

p y

tF f

2

2 1 12 2 2 4

p p y y

t t t tM F f f

21

4pW b t

21

6eW b t

Shape factor = 1.5

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Example for unequal values of mp.

Values of mp are 5, 5, and 7.5 at points A, B and C respectively (kNm/m).

We consider a sag yield line at point B, unit vertical displacement ”1” and position “x”.

1 1

10 52 2

W w x w x w

1 1 1 1

10 10

1 1 1 15 5 5 7.5

10 10

A B B CE m m m mx x x x

Ex x x x

W kNm/m

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2

1 1 1 15 5 5 7.5

10 10

5 200

2 20

Ex x x x

xE

x x

2

2

5 2005

2 20

40

2 20

W E

xw

x x

xw

x x

The minimum value of w is found by differentiating to x and set this equal to zero

W kNm/m

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2

22

1 2 20 4 20 400

2 20

4.72

x x x xdw

dx x

x

2

40

2 20

xw

x x

Substitution of this value for w results in

This is super because now we can:- analyse the position of yield lines- analyse the plate capacity (failure load)

2 2

40 4.72 400.89 kNm/m

2 20 2 4.72 20 4.72

xw

x x

W kNm/m

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Energy Dissipation in a Yield Line

The yield line ab divides the plate into two rigid portions A and B.

Part A rotates θA about support axis eg.Part B rotates θB about support axis df.

The rotations are represented by vector following the right-hand-corkscrew rule.

The energy dissipation per unit length of the yield line is

cos cos

i nA nB

i A A B B

E m

E m

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Energy Dissipation in a Yield Line

For the length “l” of the yield line

Therefore

cos cosi A A B BE m l l

i A A B BE m l l

(projection of l on an axis) (rotation of rigid region about that axis)iE m

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Rectangular plate with arbitrary chosen dimensions

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Plate part Area Displacement centre

of gravity

ABE

EFB

BDF

1 1

2 2b a

1

2a a

1

3w

1

3w

2

3w

Yield line lx ly

FE 0 0

EC

2b a

a1

2a

x y

4w

a

2w

a

w

a

1 1

22 2

a b a

40



The work done by external load q yields:

The internal work yields

21 1 2 12 2

4 3 4 3 2 3

1 2

2 3

w wW q ba a b a w a

W qa b a w

2 1

2 4 42

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2

d p p

d p

w w wE m b a m a a

a a a

bE m w

a

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2

2

1 2

2 3

1 2 14

2 3 2

1

282

3

p

p

b a

W qa a a w

qa w m w

m

qa

The variable α has to be determined such that is minimised.

The following condition exists1

02

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2

22

2

2

2

2 1 2 1

3 2 3 28

2

3

2 1 2 10

3 2 3 2

4 4 3 0

1 3 1

2

pmd

d qa

A number of examples.

The choice of a mechanism whichdoes not give the lowest value of not necessarily leads to large mistakes in the load factor

Type of plate β α

Square plate 1 0.5 24.00

Length twice the width 2 0.5 14.40

Length twice the width: optimised 2 14.14

Infinetely long plate 8.00

1

13 1 0.6514

13 0.866

2

2

p

qa

m

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As shown the load carryingcapacity pf the plate increase withdecrasing span in x-direction.

The maximum is reached for a square plate, which can resist a three times higher failure load thanthe infinitely long plate.

The value of 8 is logic because

21

8M q a

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Comparison elastic versus plastic

Shape factor plate

0.752.7

0.28

2

2

11

4 1.51

16

t

t

13.0

8

24

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Example 1A simply supported square steel plate is loaded by a uniform distributed load. Calculate the load carrying capacity.

Assuming point “O” is displaced by δ

Internal work

2tan

/ 2

2

x

x

a a

a

20 4 8

d x x y y

d

E m a m a

E m a ma

46


Example 1External work

Equilibrium

2

14

2 2 3

3

d

d

d

E qdxdy w

aE q a

q aE

2

2

83

24

q am

q am

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Example 2 A simply supported rectangular steel plate with length 1.5 x width. The plate is loaded by a uniform distributed load.Calculate the load carrying capacity.

Assuming a displacement at “O” = δ

Rotations of the parts are:

1,3 xx

2,4

2

/ 2y

l l

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Example 2

Internal work:

1, 2 : 2

2 3,4 : 1,5 2

1 3 : 2

d s

d

d

d

E m l

Part E m lx

Part E m ll

Total E mlx l

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Example 2

External work:

2

1 1,2 : 2

2 3

1 3,4 : (1,5 2 ) 2 4

2 2 2 2 3

3 :

4 3

W qdxdy w

Part W q l x

l lPart W q l x q x

l lxTotal W q

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Example 2Equilibrium

This means

The optimised m is found by

2

2 2

1 3 32

4 3

39 48 6

24 33

l lxml q

x l

l xql

l x l xqm

l l x

x

( )m f x

0dm

dx

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Example 2

2 2

2 2 2

2

2 2 2

3 2 2

22

9 4

24 3

3 9 8 3 9 40

24 3

024

3 9 8 3 9 4 0

9 8 12 0

1.26 and 0.595

9 0.595 4 0.5950.0584

24 3 0.595

l x l xqm

l x

l x l l x l x l xdm q

dx l x

q

l x l l x l x l x

l l x lx

x l x l

l l l lqm q

l l

2l

2 1: 0.042example ql

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Example 3 A rectangular steel plate with length ‘b’ and width ‘a’. The plate have three different support conditions: clamped, simply supported, freed edge. The steel plate is loaded by a uniform distributed load.Calculate the load carrying capacity.

Assuming a displacement at “O” = δ

Rotations of the parts are:

1,3 xx

2, ya

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Example 3Internal work:

'

'

2 2

1, 2 : 2

3: 2

Assume m=m

: 2 2

d s

d

d

d

E m l

Part E m ax

Part E m x m ba a

mTotal E x bx a

ax

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Example 3External work:

1 1,2 : 2

2 3

1 3 : 2 2

2 2 3

: 3 26

W qdxdy w

Part W q a x

Part W q b x a q x a

q aTotal W b x

55



2 2

22

2 2

2

2 2 3 2

36 2

6 2 2

mx bx a qa b x

ax

x xq am

x b x a

um

v

du dvv u

dm dv dx

dx v

56



2 2 22

22 2

2 2 2

(36 4 ) 2 2 (4 ) 36 2

6 2 2

(36 4 ) 2 2 (4 ) 36 2 0

x x b x a x b x xdm qa

dx x b x a

x x b x a x b x x

solver equation

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Example 3

22

2 2

36 2

6 2 2

x xq am

x b x a

58


Example 4

D

C

A

B

a

aCapacity negative yield moment = m-

Capacity positive yield moment = m+

12

2 3

d

d

E m a m a m aa a a

W q a a

E W

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Example 4

Capacity negative yield moment = m-

Capacity positive yield moment = m+

D

C

A

B

aa

This way we can also handle orthotropic material behaviour

2

23

23

30

60

9

dE W

am a m a q

q am m

mm q

a

mm q

a

mm m m q

a

60

A steel plate is simply supported at two sides only. The minimum length of side b = length of side a. The other two sides are free edges. The plate is uniform loaded by “q”. A yield pattern (line AE) is considered as shown in figure below.


Assignment 3

Based on upper-bound theory (virtual work) determine:

1. Internal work.

2. External work.

3. Optimised value of m by solving dm/d.

4. Plot the relation between and β starting with β=1.

5. Plot the relation between and β.

A

B

D

C

E

a

b=βa

a 2

p

qa

m

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Report no specific conditions: hand written, word editor, etc. It should

clearly show answers obtained.

Focus should be right answering the questions: 2 students per

assignment allowed.

Name, study number and assignment number on front page

Digital using e-mail: [email protected]

Deadline 14 sept. 10.00 a.m.

mailto:[email protected]

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A square steel plate is simply supported at two sides only. The other two sides are free edges. The plate is at location D loaded by a concentrated force F. Two alternative yield pattern are shown. The material is orthotropic, m+ and m-, which are not the same.


Assignment 4

D

C

A

B

E

a-x1 x1

a

a

D

C

A

B

a

aPositive yield moment Negative yield moment

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Determine the value for decisive F expressed by m+ and expressed by m- and for which ratio of m+/m- are the results the same.


Assignment 4

D

C

A

B

E

a-x1 x1

a

a

D

C

A

B

a

aPositive yield moment Negative yield moment

64


Report no specific conditions: hand written, word editor, etc. It should

clearly show answers obtained.

Focus should be right answering the questions: 2 students per

assignment allowed.

Name, study number and assigment number on front page

Digital using e-mail: [email protected]

Deadline 14 sept. 10.00 a.m.

mailto:[email protected]

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Plate Yield Line Theory 07-09-2015

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