Lecture 12 Plasticity of structures Load-carrying capacity Print version Lecture on Theory of Elasticity and Plasticity of Dr. D. Dinev, Department of Structural Mechanics, UACEG 12.1 Contents 1 Introduction 1 2 Material modeling 2 3 Load-carrying capacity 3 4 Plastic hinges 7 5 Plastic analysis 9 6 Yield line theory of slabs 10 6.1 Introduction ..................................... 10 6.2 Yield lines ...................................... 11 6.3 Ultimate moment of resistance ........................... 12 6.4 Analysis by virtual work principle ......................... 13 6.5 Minimum load principle .............................. 15 12.2 1 Introduction Introduction Plastic behavior • Plastic deformations 12.3 1
Transcript
Lecture 12Plasticity of structuresLoad-carrying capacityPrint version Lecture on Theory of Elasticity and Plasticity of
Dr. D. Dinev, Department of Structural Mechanics, UACEG
• Plastic behavior of the material describes the deformation of the body undergoing non-reversible changes of shape in response to applied forces
12.5
Material modeling
Results from a tensile test
• The stress-strain relation is non-linear• The transition from elastic to plastic behavior is called yielding
12.6
Material modeling
Mild steel
• σy = 200−400 MPa• σu = 400−600 MPa
12.7
2
Material modeling
Mild steel
• εy ≈ 0.1%• εu = 30−50%• The most important zone is Oab
12.8
Material modeling
Idealized curve
• Elastic-perfectly plastic model12.9
3 Load-carrying capacity
Load-carrying capacity
Plastic bending
• Load-carrying capacity- the maximum bending moment which can bear a section beforefailure• Assumptions- The Bernoulli hypothesis is valid• Euler-Bernoulli beam
– ε = zκ- kinematic equations
– σ = Eε- constitutive equations
– M =∫
A zσdA = EIκ- equilibrium equations12.10
Load-carrying capacity
Stage 1- elastic behavior
• Consider a rectangular cross-section with dimensions of b and h• Maximum stress is σmax = σy
3
• Moment equilibrium gives
My = σybh2
612.11
Load-carrying capacity
Stage 1- elastic behavior• The curvature is
κy = σy2
Eh• The yielding point is reached
12.12
Load-carrying capacity
Stage 2- partial plastification• When the load is increased beyond the yielding the result is a partial plastification
12.13
Load-carrying capacity
Stage 2- partial plastification• The curvature is 2κy and moment capacity of section is
M2 =1148
σybh2
12.14
Load-carrying capacity
4
Stage 3- almost full plastification
• The load is increased and the section is almost fully plastic12.15
Load-carrying capacity
Stage 4- filly plastic section
• All fibers of the section are filly plastic12.16
Load-carrying capacity
Stage 4- filly plastic section
• The curvature is κp = ∞ and the plastic moment is
Mp = σybh2
4
12.17
Load-carrying capacity
Stage 4- filly plastic section
• Moment - curvature relationship12.18
5
Load-carrying capacity
Stage 4- filly plastic section
• The comparison of the bending moments
η =Mp
My= 1.5
• The above ratio defines so called shape factor• Circular section- η = 1.7• Pipe section- η = 1.273• I-section- η = 1+d/2
1+d/3 , where d = 12
htwbt f
12.19
Load-carrying capacity
RC section
• RC section with σ − ε- curves of concrete and reinforcing steel• Assumption for a tension failure
12.20
Load-carrying capacity
RC section
• Moment-curvature diagram• The presence of composite section require a definition of an ultimate moment capacity of
section Mu
12.21
6
Load-carrying capacity
RC section
• Moment capacity of section12.22
Load-carrying capacity
RC section
• ∑H = 0- height of the compression zone
As fy = 0.85 f ′cab → a =As fy
0.85 f ′cb
where As is the area of the reinforcement, fy is yield strength of the reinforcement, f ′c isthe compressive cylinder strength of the concrete, b is the beam width
12.23
Load-carrying capacity
RC section
• ∑M = 0 - moment capacity- Mu
M = T(
d− a2
)→ Mu = φM = φAs fy
(d− As
1.7bfy
f ′c
)where d is the distance from the centroid of the reinforced steel to the extreme concretecompression fiber, φ = 0.9 is a strength reduction factor to obtain the design strength
12.24
4 Plastic hinges
Plastic hinges
7
Hinge formation
• The presence of unrestricted plastic flow at section leads us to the concepts of the formationof plastic hinges in beams
12.25
Plastic hinges
Hinge formation
• When the load is increased and Mp is reached the mid-span section is fully plastic12.26
Plastic hinges
Hinge formation
• No plasticity occurs in sections where the bending moment is < My
12.27
Plastic hinges
Hinge formation
• The beam behaves as two rigid bodies connected by a plastic hinge which allows them torotate relatively to each other
12.28
Plastic hinges
Hinge formation
• The value of Fu is given by Fu =4Mp` is the ultimate (collapse) load
12.29
Plastic hinges
Hinge formation
• The length of the plastic hinge is
Lp = L(1−η)
12.30
8
5 Plastic analysis
Plastic analysis
Ultimate limit state
• The main task of the structural engineering is to design the structural members so they cancarry the loads under all possible conditions including ultimate limit states• The elastic distribution of stresses can be obtained by solution of the elasticity problem• However the structural elements do not behave elastically near ultimate load and bending
capacity of section is based on a plastic analysis• It is reasonable to use methods of analysis and design which recognize plasticity
12.31
Plastic analysis
Theorems of plasticity- Hillerborg 1975
• Lower-bound theorem- If there is a load qu for which it is possible to find a momentfield that fulfills all equilibrium conditions and the moment at no point is higher than theyield moment, then qu is a lower-bound value of the carrying capacity. The structure cancertainly carry the load qu• Upper-bound theorem- If for a small virtual increment of deformation, the internal energy
taken up by the structure on the assumption that the moment in every point where thecurvature is changed equals the yielding moment and this energy is found to equal thework performed by the load qu for the same increment of deformation then qu is an upper-bound value of the carrying capacity. Loads greater than qu are certainly high enough tocause moment failure of the structure
12.32
Plastic analysis
Theorems of plasticity
• Values of the ultimate load according to the upper-bound and lower-bound theorems12.33
Plastic analysis
9
Example
• Consider a propped cantilever beam with a force at the mid-span• The section starts to yield when the maximum moment reach to My• A force F2 can be added and produces a collapse mechanism• The ultimate load Fu = F1 +F2 and plastic hinges are formed at maximum moments• The equilibrium gives the magnitude of the ultimate load based on the lower-bound theo-
rem12.34
Plastic analysis
Example
• The virtual work principle gives the value of the ultimate load based on the upper-boundtheorem
12.35
6 Yield line theory of slabs
6.1 Introduction
Yield line theory of slabs
Introduction• The method for a limit analysis of RC slabs known as yield line theory was initiated by
Ingerslev (1921) and extended by Johanson (1932)
10
• The ultimate load is estimated by postulating a collapse mechanism that is compatible withthe BCs (upper-bound approach)• The moment at the plastic hinge lines are the ultimate moment of resistance of the section• The ultimate load is determined using the principle of virtual work and it is either correct
or too high
Note
• Thus all possible collapse mechanisms must be examined to ensure that the load-carryingcapacity is not overestimated
12.36
6.2 Yield lines
Yield line theory of slabs
Yield lines
• When a collapse mechanism has developed, the plastic deformations along the yield linesare much greater than the elastic deformations of the segments between yield lines• The theory assumes that the segments are plane• The geometry of deformations gives the basic rules for the determination of the pattern of
yield lines
– To act as plastic hinges of a collapse mechanism made up of plane segments, yieldlines must be straight lines forming axis of rotation for the movements of the seg-ments
– The supports of the slab will act as axes of rotation. If the edge is fixed, a yield linemay form along the support. An axis of rotation will pass over a column
– For compatibility of deformation, a yield line must pass through the intersection ofthe axes of rotation of the adjacent segments
12.37
Yield line theory of slabs
Yield lines
• Crack pattern in a RC slab (Mörsch, 1922)12.38
11
Yield line theory of slabs
Yield lines
• Boundary conditions, axes of rotation and yield lines12.39
Yield line theory of slabs
Yield lines
• Yield line patterns for uniformly loaded slabs12.40
6.3 Ultimate moment of resistance
Yield line theory of slabs
Ultimate moment of resistance
• For a yield line that runs orthogonal to the reinforcement the ultimate moment of resistanceper unit width is
Mu = φAs fy
(d− As
1.7fy
f ′c
)where As is area of the reinforcement per unit width
12.41
12
Yield line theory of slabs
Ultimate moment of resistance• In the usual case the reinforcement bars are orthogonal to each other and not coincide with
a general yield line, then we may apply the Johansen’s yield criterion• ∑ t = 0 gives
Mun = Mux cos2α +Muy sin2
α
12.42
Yield line theory of slabs
Ultimate moment of resistance• ∑n = 0 gives
Munt = (Mux−Muy)sinα cosα
• When Mux = Muy, thus Munt = 0 the slab is isotropically reinforced• What is the slab with Mux 6= Muy?!?
12.43
6.4 Analysis by virtual work principle
Yield line theory of slabs
Virtual work principle• Suppose that a rigid body is in equilibrium under the action of a system of forces• If the body is given a small arbitrary displacements, consistent with the BCs the sum of the
work done by the forces (force times its corresponding displacement) will be zero becausethe resultant force is zero• Hence the principle states:
– If a body that is in static equilibrium under the system of forces is given a virtualdisplacement, the sum of the virtual work by the forces is zero
12.44
Yield line theory of slabs
Analysis by virtual work principle• To analyze the slab by the virtual work principle, a yield line pattern is postulated for the
slab at the ultimate load• The segments of the slab may be regarded as rigid bodies because the slab deformation
occurs only at the yield lines• The segments of the slab are in equilibrium under external loading and the bending and
twisting moments and shear forces along the yield lines• A convenient point within the slab is chosen and given a small displacement δ in the
direction of the load• Then the resulting displacements of all points of the slab δ (x,y) and rotations of the slab
segments about the yield lines may be expressed by δ and the dimensions of the segments12.45
13
Yield line theory of slabs
Analysis by virtual work principle• The work W done by a ultimate uniform load qu is
W =∫
Aquδ (x,y)dA = ∑Qu∆
where Qu is load resultant on the segment and ∆ is the displacement of its centroid• The work done by internal forces U is due only to the bending moments
U = ∑Munθn`0
where Mun is the ultimate moment normal to the yield line, θn is the rotation and `0 is theyield line length• The twisting moments and shear forces do not produce work because they cancel each
other at the opposite sides of the yield line12.46
Yield line theory of slabs
Analysis by virtual work principle• The equilibrium is U =W or
∑Munθn`0 = ∑Qu∆
• Since most slabs are rectangular and the reinforcement is orthogonal we know the Mux andMuy and it is easier to deal with the directional components of the internal work
12.47
Yield line theory of slabs
Analysis by virtual work principle• For an arbitrary yield line we have
∑Munθu`0 = ∑Muxθxy0
+∑Muyθyx0
where θx and θy are components of θu; x0 and y0 are components of `0• Therefore
∑Qu∆ = ∑Muxθxy0
+∑Muyθyx0
12.48
14
6.5 Minimum load principle
Yield line theory of slabs
Minimum load principle
• In most cases a yield line pattern cannot be drown without unknown dimensions locatingthe yield line position• The unknown dimensions must be included in the virtual work equation• The equation for ultimate load has the form of Qu = f (x1,x2, . . . ,xn)
12.49
Yield line theory of slabs
Minimum load principle
• Since the upper-bound approach is used the values for xn required those values that givethe minimum value for Qu and may be found by solving a set of equations
∂Qu
∂x1= 0, . . .
∂Qu
∂xn= 0
• The values for x1,x2, . . . ,xn are substituted back into the ultimate load equation to obtainthe minimum Qu
12.50
Yield line theory of slabs
15
Example 1
• Consider a rectangular slab with an orthotropic reinforcement• Determine the ultimate load for the given yield line pattern
12.51
Yield line theory of slabs
Example 1
• The yield line pattern12.52
Yield line theory of slabs
16
Example 2
• For the given pattern find the optimum position of the yield lines and the ultimate load12.53
