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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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Learning objectives
Understanding the meaning of yield line theory
Rules for yield lines
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
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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Why Yield Line analysis
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
8
Why Yield Line analysis
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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Behaviour of a plate under increasing load
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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Behaviour of a plate under increasing load
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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Rules for yield lines
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Rules for yield lines
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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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.
Rules for yield lines
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Rules for yield lines
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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Rules for yield lines
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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Rules for yield lines
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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?
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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What is analysis by Virtual Work?
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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What is analysis by Virtual Work?
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
What is analysis by Virtual Work?
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
( , ) ( , )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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What is analysis by Virtual Work?
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.
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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External work done by Loads: Examples
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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External work done by Loads: Examples
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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External work done by Loads: Examples
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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Internal work done by Resisting Moments
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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Internal work done by Resisting Moments
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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Internal work done by Resisting Moments
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
The external work done by uniform distributed load is
Internal work done by Resisting Moments
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Internal work done by Resisting Moments
External work done by Loads=
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Internal work done by Resisting Moments
External work done by Loads=
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Internal work done by Resisting Moments
External work done by Loads=
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Rectangular plate with arbitrary chosen dimensions
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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Rectangular plate with arbitrary chosen dimensions
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
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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Rectangular plate with arbitrary chosen dimensions
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
14
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Rectangular plate with arbitrary chosen dimensions
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Rectangular plate with arbitrary chosen dimensions
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Rectangular plate with arbitrary chosen dimensions
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Rectangular plate with arbitrary chosen dimensions
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
51
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
52
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Example 3Equilibrium
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
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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Example 3Equilibrium
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Example 3
22
2 2
36 2
6 2 2
x xq am
x b x a
58
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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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MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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.
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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
61
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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: a.romeijn@tudelft.nl
Deadline 14 sept. 10.00 a.m.
62
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.
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Assignment 4
D
C
A
B
E
a-x1 x1
a
a
D
C
A
B
a
aPositive yield moment Negative yield moment
63
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.
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
Assignment 4
D
C
A
B
E
a-x1 x1
a
a
D
C
A
B
a
aPositive yield moment Negative yield moment
64
MTM1412: Structural Design and AnalysisTopic: Yield Line Analysis of plates
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: a.romeijn@tudelft.nl
Deadline 14 sept. 10.00 a.m.