MTM1412: Structural Design and Analysis Topic: 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
MTM1412: Structural Design and Analysis Topic: 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 structures variation of: support conditions: free edge, restraint, simply supported
loadings: uniform, uniform, non-uniform, concentrated force, force, partly uniform loaded
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates 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.
MTM1412: Structural Design and Analysis Topic: 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
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates 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
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates 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
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Why Yield Line analysis
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Why Yield Line analysis
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Behaviour Beha viour of a plate plate under under increasing increasing load load Consider a rectangular plate which is loaded by a uniformly distributed load q (q is a fixed value 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.
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Behaviour Beha viour of a plate plate under under increasing increasing load load 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.
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rules for yield lines
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 rotat ion 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.
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rules for yield lines
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rules for yield lines Wrong: points ABCD are no laying is a single plane.
Wrong: yield lines are not straight.
Wrong: yield line of intersect EF is not parallel to AB.
Right: yield line is a correct solution.
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rules for yield lines
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
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rules for yield lines
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rules for yield lines
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
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates 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
( N )
for all regions
(m l )
for all regions
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates What is analysis by Virtual Work? Energy is dissipated in the yield lines only. Energy is dissipated in the yield lines only. Because of equilibrium, the forces and moments per unit length along the yield line are:
Deformations and internal loads in a yield line
mnn = bending moment mns = torsional moment qn = transverse force The plastic deformation equals the difference in rotation of both planes about the s-axis. This angle is small and therefore
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates What is analysis by Virtual Work?
( N )
for all regions
(m l )for all regions
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]
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates What is analysis by Virtual Work? In case of uniform distributed lateral (surface) load, the amount of work equals:
W
q( x , y )w( x , y) dxdy
plate
w(x,y) = increase in displacement during failure
W
plate parts
S = plate area
q S w z
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates What is analysis by Virtual Work? Of all forces along the yield line, only m nn provides a contribution to the energy dissipation
Ed
mnn d ds
along the yield line
The yield line is the intersection between two planes and therefore the value of d is constant.
Ed
m
p
d l s
ls = the length of the yield line
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates External work done by Loads: Examples 1
A square plate carrying a single concentrated load at its center
W P 1 P
2
A plate supported along three sides and free along the fourth. Loaded with a line load “w” per unit length along the free edge
W b w 1 a w 2
1
w (a b ) 1 w (a b )
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates External work done by Loads: Examples 3
A distributed load “w” per unit area on a triangular segment defined by a hinge and yield lines
a b a b 1 1 W w a b 1 w 1 w 2 3 6 6 4
A rectangular plate carrying a distributed load “w” per unit area
1 a b a b 1 w W w a b 1 w 2 2 2
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates External work done by Loads: Examples More complicated trapezoidal shapes may always be subdivided into component triangles and rectangles. 1 2
5
1 2
7,5 3, 75 2 9,375 3
3, 75 3,75 4 9,375 3
5 1,5 3, 75
2
2 5, 625
9,375 9,375 5, 625 24,375
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates External work done by Loads: Examples More complicated trapezoidal shapes may always be E subdivided into component triangles and rectangles.
6
Simply supported plate uniformly loaded by a surface load λq. Displacement at centre = δ
r 1
r
W
q
r
rd dr
plate
R
2 R
W
0 0
q 1
W 2 q
r
rd dr
R
dr
R 2 6
rd
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates 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 one unknown parameter “a”. Total rotation diagonal yield line = θ1 Total rotation horizontal yield line = θ2
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Internal work done by Resisting Moments 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
b
5 25 a 2 a
c
and
a 25 a 2 5
This means 1
1 b
1
a
c
5 25 a 2
1
2
5 a 25 a 2
a 5 25 a 2 5 a 1
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Internal work done by Resisting Moments 1 s
a 5 a 5 25 a 2 2 2 5 a 25 a 2 5 a 1
Similar to
1 a 5 a 2 10 2 5 a 5 a
1
It seems that we can do all this in a simplified way by considering support edges only.
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Internal work done by Resisting Moments The total internal work done by resisting moments is
Ed m 25 a 2 1 4 m 20 2 a 2 For a number of different values of “a” results in the following data “a”
Ed
6
11.36 m
6.5 7 7.5
11.08 m 10.87 m
10.69 m
And we know already that W=E d
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Internal work done by Resisting Moments
=
External work done by Loads
The external work done by uniform distributed load is
W q 20 2 a 5
“a”
6
6.5 7
W
“a”
80,0qδ
6
78,4qδ
1 1 2 a 5 4 10 a 2 2 2 3 2 3
76,6qδ
With δ=unit=1. gives
m
6.5 7
80,0q/11.36=7,07q
78,4q/11.08=7,08q
76,6q/10.87=7,04q
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates
“a”
6
m
6.5 7
7.5
80,0q/11.36=7,07q
78,4q/11.08=7,08q
76,6q/10.87=7,04q
75,0q/10.69=7,02q
The largest values a=6,5 is most critical and the required resisting moment for the plate is 7.08 q.
MTM1412: Structural Design and Analysis Topic: 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 t F p f y 1 2
Plate thickness “t”
M p Fp 2
t 2
2
t
t
t
2
2
4
f y 1 f y
1
W p
1 4 1
b t 2 Shape factor = 1.5
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Internal work done by Resisting Moments
=
External work done by Loads Example for unequal values of m p. Values of m p are 5, 5, and 7.5 at points A, B and C respectively (kNm/m).
W kNm/m
We consider a sag yield line at point B, unit vertical displacement ”1” and position “x”. W wx E m A E 5
1
1 x
1 2
w 10 x
2
5w
1
1
x
10 x
mB m B 1
1
5 5
1 10
mC
7.5
1 10
1 10 x
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Internal work done by Resisting Moments
=
External work done by Loads E 5 E
W kNm/m
1
1
5 5
x x 5 x 200
1 10 x
7.5
1 10 x
2 x 20 x 2
W E 5w w
5 x 200 2 x 2 20 x x 40
2 x 20 x 2
The minimum value of w is found by differentiating to x and set this equal to zero
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Internal work done by Resisting Moments
=
External work done by Loads w
W kNm/m
dw dx
x 40 2 x 20 x 2
2 1 2 x 20 x 4x 20 x 40
2 x
2
20
2
0
x 4.72
Substitution of this value for w results in w
x 40 2 x 2 20 x
4.72 40 2 4.72 2 20 4.72
0.89 kNm/m
This is super because now we can: - analyse the position of yield lines
MTM1412: Structural Design and Analysis Topic: 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 θ A 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
Ei m nA nB
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Energy Dissipation in a Yield Line For the length “l” of the yield line
Ei m A l cos A B l cos B Ei m A lA B l B Therefore
Ei m
(projection of l on an axis) (rotation of rigid region about that axis)
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rectangular plate with arbitrary chosen dimensions
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rectangular plate with arbitrary chosen dimensions Plate part
Area
ABE EFB
1 b a 2 2
1
1
2
1 1 2
BDF
a
1 2
Yield line FE EC
Displacement centre of gravity
2
3
b 2 a
3
1
a a
lx
b 2 a
3 ly 0
1
w
w
w
x 4w
y 0
a 2
w
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rectangular plate with arbitrary chosen dimensions The work done by external load q yields: w 1 2 1 w 1 W 2 q ba a b 2 a w a 2 3 4 3 2 3 4 W
2 qa b a w 2 3
1
The internal work yields
Ed m p b 2 a 4
w 2w 1 w 4 m p a a a a 2 a
b 1 w a 2
Ed 4 m p
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rectangular plate with arbitrary chosen dimensions b a W
2 qa a a w 2 3
1
2 1 qa 2 w 4 m p w 2 3 2
1
1 2 m p 8 2 2 qa 3
The variable α has to be determined such that is minimised. 1
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rectangular plate with arbitrary chosen dimensions 2 1 2 1 m p d 3 2 2 3 2 8 2 2 d qa 2 3 2 1 2 1 2 0 3 2 3 2 Type of plate 4 2 4 3 0
1 3 2 1 2
A number of examples.
β
2 qa
α
m p
Square plate
1
0.5
24.00
Length twice the width
2
0.5
14.40
Length twice the width: optimised
2
1
1
The choice of a mechanism which Infinetely long plate does not give the lowest value of t il lead to la
4 2
13 1 0.651
3 0.866
14.14 8.00
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rectangular plate with arbitrary chosen dimensions
As shown the load carrying capacity pf the plate increase with decrasing span in x-direction. The maximum is reached for a square plate, which can resist a three times higher failure load than the infinitely long plate. The value of 8 is logic because
M
1 8
q a 2
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Rectangular plate with arbitrary chosen dimensions
Comparison elastic versus plastic
0.75 0.28
2.7
1 3.0 8 24
1 Shape factor plate
4 1 6
1 t 2 1 t
2
1.5
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 1
A simply supported square steel plate is loaded by a uniform distributed load. Calculate the load carrying capacity. Assuming point “O” is displaced by δ tan x x
a/2
2 a
2 a
Internal work Ed
m a m a x
Ed m
x
2
y
y
a 0 4 8m
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 1
External work Ed
qdxdy w
Ed q a
a 1
q a 2
E d
3
Equilibrium 8m m
q a 2
q a2 24
3
4
2 2 3
MTM1412: Structural Design and Analysis Topic: 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 x 2,4 y
x
l/2
2 l
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 2 Internal work:
Ed
m l
s
Part 1,2 : Ed m l
x
2
Part 3,4 : Ed m 1, 5l
2 l
1 3 x l
Total : Ed 2 ml
2
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 2 External work:
W
qdxdy w
1 Part 1,2 : W q l x 2 2 3 l 1 l Part 3,4 : W q (1, 5l 2 x) 2 q x 4 2 2 2 2 3
3 l 2 lx Total : W q 4 3
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 2
Equilibrium
3 l 2 lx 1 3 2 ml q 3 x l 4 3l x ql 2 2 l x l x 9 4 8 6 q m 24 l l 3 x 3 x This means
m f ( x)
The optimised m is found by
dm dx
0
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 2
m
2
x 4 l x2
l 3 x
q
dx 24
24
9l
l 3 x 9 l 2 8 l x 3 9 l 2 x 4 l x2 0 2 24 l 3 x
dm q
q
0
l 3 x 9 l 2 8 l x 3 9 l 2 x 4 l x2 0 9l 8l x 12lx 0 3
2
2
x 1.26l m
q 24
and
x 0.595l 2
0.0584ql
9 l 0.595l 4 l 0.595l 2
l 3 0.595l
2
MTM1412: Structural Design and Analysis Topic: 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 x
2, y
x
a
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 3
Internal work:
Ed
m l
s
Part 1, 2 : Ed m a Part 3: Ed m x
Assume m=m Total : Ed
x
2 '
a
2 m b
a
'
m ax
2
2 x bx 2a
2
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 3
External work: W
qdxdy w
Part 1,2 : W q
1
2
a x 2 3
Part 3 : W q b 2 x a Total : W
q a 6
2
3b 2 x
1
2
3
q x a 2
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 3
Equilibrium
m ax
2 x 2 bx 2a 2 qa 3b 2 x
m m
dm dx
qa 6
36 x 2 x 2
2
2 x
2
b x 2a 2
u v v
du
dv
u v2
dv dx
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 3
Equilibrium 2 2 2 qa 2 (36 4 x) 2 x b x 2a (4 x b) 36 x 2 x 2 dx 6 2 x 2 b x 2a2
dm
(36 4 x ) 2 x 2 b x 2a 2 (4 x b) 36 x 2 x2 0 solver equation
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 3 m
q a2 6
36 x 2 x 2
2 x
2
b x 2a 2
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 4 Capacity negative yield moment = m-
a
A
D
Capacity positive yield moment = m+
Ed m a W q
a
1
2 Ed W B
C
a
a
a
m a m a
a a 2 3
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Example 4
Ed a
A
D
2m
2m
m
m
m
W
am
m
a
B
Capacity negative yield
C
0
m
q
3
qa
q
0
aq
a2
3 3m
a 6m
a
m
Capacity positive yield
q
9m
a
This way we can also handle orthotropic material behaviour
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Assignment 3 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. b=βa
Based on upper-bound theory (virtual work) determine:
a
A
a
E
D
1.
Internal work.
2.
External work.
3.
Optimised value of m by solving dm/d.
qa
2
4.
Plot the relation between β=1.
5.
Plot the relation between and β
m p
and β starting with
MTM1412: Structural Design and Analysis Topic: 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:
[email protected]
Deadline 14 sept. 10.00 a.m.
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Assignment 4 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. a
a
a-x1 A
x1 E
D
a
A
D
a
C
C
MTM1412: Structural Design and Analysis Topic: Yield Line Analysis of plates Assignment 4 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. a
a x1
a-x1 A
E
D
a
A
D
a
C
C
