PLASTIC ANALYSIS Reading – Megson I, Sections 9.10, Megson II, Chapter 18. Neal, B.G., ‘The plastic methods of structural analysis’, 3rd (S.I.) Ed., Chapman & Hall, 1977. Horne, M.R., ‘Plastic theory of structures’, Nelson, London, 1971.
Bending beyond the elastic limit Limit state design of structures requires the prediction of the ultimate strength or collapse load of a structure. Safe loads are then determined as a suitable fraction of the collapse load. As bending of a beam proceeds, strains increase increase steadily, but the corresponding stress values values depend on the material’s stress-strain stress-strain relationship. Some materials materials will fail suddenly suddenly in a brittle fashion fashion when the strain reaches a certain value (e.g. timber, cast iron, glass, etc). Others will yield and flow in a plastic fashion fashion (e.g. many types of steel). Although some structures may fail whilst still in a fully elastic state (by buckling, for example), most will exhibit stresses that exceed the elastic limit before failing. We consider now the behaviour of a beam under steadily increasing bending moment and assume it is made from an elasto-plastic material with an idealised stress-strain relationship as shown in the plot on the right (not to scale). We further assume that the beam crosssection has at least one axis of symmetry which lies in the plane of bending.
σ
yield ‘plateau’ yield stress, σy yield strain -εy εy ε
onset of strain hardening
Yield moment, My
When the stress at the extreme fibre most distant from the neutral axis just reaches yield stress. This defines the maximum moment the beam can resist whilst still fully elastic. It follows that
σ y =
M y y max , or M y = ZE σ y I
where ZE =
I y max
(the elastic section modulus)
Plastic Neutral axis
Under elastic conditions the neutral axis (zero strain locus) passes through the centroid of the crosssection. As parts of the cross-section yield and the stress distribution becomes nonlinear, the need for the tension and compression forces to remain equal causes the position of the neutral axis to move away from the centroid (except in the case of a doubly symmetric symmetric section). σy
A1
_ y1 _ y2
G1
C
plastic neutral axis
MP T
G2 A2
σy
Let the plastic neutral axis divide the section such that the areas above and below are A1 and A2
or
MP = ZP σ y
where ZP is the plastic section modulus.
Plastic section modulus
It follows from the preceding relationship that ZP =
A( y1 + y 2 ) ( 1 + y 2 ) A y or, more strictly ZP = (taking absolute values of y1 and y2 ). 2 2
More generally we could write
∫
ZP = y dA A
where y denotes the distance of each element of area, dA, from the PNA. Shape factor
Maximum elastic moment, M y = ZE σ y , where ZE =
I y max
(the elastic section modulus).
Ultimate (fully plastic) moment, MP = ZP σ y . The ratio of the fully plastic moment to the yield moment depends on the shape of the cross-section and is known as the shape factor, f (Megson’s notation, but also called S and sometimes, v). f
MP
ZP
10 W
plastic neutral axis
yP
140
2m
10 100
σ y = 280 MPa γ steel = 76 kN / m 3 The problem is to determine the load W that will cause collapse of the steel cantilever T-beam. It is obvious that the maximum bending moment occurs at the left hand end. When this peak moment reaches the fully plastic value MP, a plastic hinge will form at the left hand end and the beam will collapse. Plastic neutral axis If yP denotes the position of the PNA, the area above must = A/2. A (140 + 100)10 = 2 2 yP = 120mm
10 yP =
Plastic section modulus
Spread of plasticity in the plastic hinge vicinity [Megson I p.244, Megson II p.602] material at σy
b
c
My
MP σy
σy
yielded elastic
under point load
at b
STRESS DISTRIBUTION at selected points along beam
at c
Plastic Analysis Role in Design
When designing for the strength limit state, the objective is to satisfy the inequality S * ≤ φ Sn .
S* denotes the structural action(s) caused by the characteristic or factored loads, and
φSn denotes the reliable strength of the structure based on nominal strength Sn and strength reduction factor φ. The characteristic loads are normally obtained by reference to Standards (NZS 1170 for example) which present basic loads together with appropriate factors. The intention is to define loadings that typically have a 5% probability of exceedance in 50 years (in other words a value that is unlikely to underestimate the maximum load encountered by the structure during its intended design life). You will be familiar with loadings such as 1.2G + 1.6Q where G and Q represent the intensity of dead and live loads on a timber deck. The reliable strength is intended to represent a value with a 95% probability of exceedance (in other words a value that is unlikely to over-estimate the strength). The structural actions S* and strengths Sn are normally replaced by specific actions such as bending moment in the case of flexural (beam and frame) structures, or axial forces in the case of truss-type structures. Flexural structures:
M * ≤ φ Mn (where Mn may be replaced by MP, the fully plastic moment).
Truss structures:
N * ≤ φ Nn (where N* denotes the axial force in a truss member)
Plastic analysis provides the means of determining the design actions M*, throughout the structure resulting
Example – Collapse load of a simply supported (determinate) beam with overhang (from Megson)
The beam ABC has span AB = L and overhang BC = L/2. Point loads of 4W and W act at mid-span AB and C respectively. 4W
W B
A
C
D L/2
L/2
L/2 WL/2
3WL/4
collapse mechanism
The bm diagram is readily obtained revealing a maximum bm of 3WL/4 at D. If is gradually increased a plastic hinge will eventually form at D creating the collapse mechanism shown. The value of W is determined from the knowledge that 3WL = MP 4
the bm at its location. This additional information allows the analysis to be completed. W A
C
MP
B MP V A
VC
Taking moments about B for segment BC:
L , 2P 2MP VC = L
C
MP = VC
MP
B VC
Taking moments about A for AC:
L MP = W − VC L, 2 2MP 2 ⎛ W = ⎜⎜ MP + L ⎝ L 6MP = L
⎞ L ⎟⎟ ⎠
Giving the required collapse load in terms of the beam’s fully plastic strength.
In applying the principle the displacements are referred to as virtual to distinguish them from real displacements that result from things such as applied loads. The application of a virtual displacement is not permitted to alter any external or internal forces (or stresses) that may be present. Forces are effectively “frozen” during the process. Virtual work is computed as the product of the real forces acting through virtual displacements, or as real moments acting through virtual rotations.
We now apply the principle to the beam ABC at the instant of collapse, using the collapse mechanism as the virtual displacement:
Denoting the rotations of the beam segments AB and BC by θ (assumed small, since we need only consider the initial movement of collapse), we deduce from the geometry that the rotation at the mid-span hinge is 2θ.
L External VW = W θ 2 Internal VW = MP θ + MP 2θ
W MP
L W θ = 3M θ
θ
2θ
MP collapse mechanism as virtual displacement
where the internal work is obtained as the product of moment x rotation. Equating the internal and external VW gives
θ
L External VW = W θ, 3 Internal VW = MP (θ + 2θ + θ) and equating,
W=
12MP L
Repeating for the mechanism (B) below: W 2θ
θ 3θ
L/3
L External VW = W θ, 3 Internal VW = MP (θ + 2θ + 3θ) and equating,
W=
18MP L
L/3
MP
W=12MP/L
MP
(A) MP
MP
W=18MP/L
MP
(B) MP
MP
W=9MP/L MP
(C) MP
Cases (A) and (B) are said to violate the Yield Condition, i.e. they have M > MP.
We now have a load, 36MP/5L, that is in equilibrium with a set of internal moments that nowhere exceed MP. However, there are not enough plastic hinges left to create a mechanism, so we conclude that the applied load is less than the collapse load. Hypothesis
If a bm distribution can be found that is (1) in equilibrium with the applied load(s), and (2) ≤ MP everywhere, then the applied load is ≤ true collapse load. (Later we show that this is correct.) Thus by considering just one mechanism, (A), we have been able to show that the true collapse load lies in the range
36MP 12MP ≤W≤ 5L L 7.2MP 12MP ≤W≤ L L With the true collapse load of
9MP lying close to the mean of the upper and lower bounds. L
Example – beam with distributed loading
Examples so far have all carried point loads, making it easy to guess the likely location of plastic hinges (since the bm diagram consists of straight line segments peak values must occur at the loads or at fixed ends). With distributed loads it is not so easy to identify the plastic hinge locations. The approach used in this example is to treat the hinge location as a variable, calculate the collapse load and
Reduction of MP due to axial load (Megson I p.250Megson II p.611)
When a structural member is subjected to axial load (tension or compression) in addition to moment, the fully plastic moment will be reduced (since some strength is used up in resisting the axial force). The reduced plastic moment is known as MRP or, (Megson) MP,R. The amount by which MP is reduced depends on the shape of the cross-section and the magnitude of the axial force. We examine the case of a beam with a rectangular cross-section.
σy
σy
b
σy stress resisting M
M
d/2-a
d/2+a
P
a
stress resisting P
a stress resisting M
σy
σ = P/A
M=0
small M
compressive yield
tensile yield
d/2
d/2
σy
fully plastic
cross-section
Assume the axial force is applied first followed by gradually increasing moment until the cross-section reaches its fully yielded state as illustrated above. Yield stress is eventually reached over the entire crosssection, with equal areas of tensile and compressive stress at the top and bottom resisting the moment, and a central area of compressive stress resisting the axial force. d d d2 MRP = σ y b( − a)( + a) = σ y b( − a 2 ) 2 2 4
(1)
Reduction of MP due to shear force
The presence of shear force will also cause a reduction in MP, although the effect is less serious than in the case of axial force. As detailed analysis is more complicated discussion is deferred until later courses.
Moment-curvature relationship (Megson II, p.600)
Consider a beam of rectangular cross-section under the action of a bm M (
σy
bd d − de d + de ⋅ ⋅ σ y M = e ⋅ σ y + b ⋅ 6 2 2 2
(d-de)/2
where
(d+de)/2
de
1st term = bm resisted by the elastic core, and 2nd term = bm resisted by the yielded (plastic) zones. i.e.
M=
elastic core
bde 2 b 2 + (d − de2 )σ y 6 4
⎛ d2 de2 ⎞ M = b⎜⎜ − ⎟⎟σ y 4 12 ⎠ ⎝ The curvature of the elastic core (and of the beam),
cross-section
σy
stress distribution
(1)
Residual stress resulting from removal of load after MP achieved
We consider what happens if, after applying increasing load to a beam until the maximum moment reaches MP, we then remove the load, causing the bending moment to return to zero. σ σy
MP
σy
b
c unloading (elastic)
MP
a
ε
εy
-σy
d
stress under MP
When the moment reaches MP, the entire cross-section is assumed to have reached either the tensile or compressive yield stress. Reducing the moment to zero is equivalent to applying a bm of MP in the reverse direction. Referring to the stress-strain diagram, note that as strain is reduced , stress reduces according to the straight line cd – i.e. a linear elastic response. Thus applying reverse MP results in a stress distribution of the form: σmax
MP
MP
PLASTIC ANALYSIS THEOREMS (Megson II, 18.1) Three conditions that must be satisfied by a structure on the point of collapse are: 1.
EQUILIBRIUM CONDITION At collapse, the bending moments must correspond to a state of equilibrium between the external loads and the internal actions.
2.
MECHANISM CONDITION At collapse there must be sufficient plastic hinges to create a partial or complete collapse mechanism.
3.
YIELD CONDITION At collapse the bending moments must everywhere be ≤ MP.
Using these conditions we can now state three fundamental theorems of plastic analysis – LOWER BOUND THEOREM If the bending moments are in equilibrium with the external load and M ≤ M P everywhere, the load is a lower bound (i.e. load is ≤ collapse load). UPPER BOUND THEOREM For an assumed mechanism in which the virtual work done in the plastic hinges equals the virtual work done by the external loads, the load is an upper bound (i.e. load is ≥ collapse load). UNIQUENESS THEOREM If a bending moment distribution can be found that satisfies the three conditions of equilibrium, mechanism
5.
If a structure is subjected to any programme of proportional or non-proportional loading, collapse will occur at the first combination of loads for which a bm distribution satisfying the conditions of equilibrium, mechanism and yield can be found (from uniqueness theorem).
Note 1.
The uniqueness theorem does not assert that the bm distribution at collapse is unique . The bm distribution at collapse may depend on factors such as initial state of stress and loading history. Nor does it assert that the collapse mechanism is unique . There may be alternative mechanisms, but they will lead to the same collapse load .
2.
An assumed plastic mechanism leading to a collapse load need not imply that a bm distribution in equilibrium with the external loads can exist for such a mechanism – as shown below:
Example – assumed mechanism, equilibrium not satisfied W
L/2
W/4
L/2
L/4
W/4 W θ
W θ
θ
W/4 θ
θ
Mechanism Method of Analysis This method of analysis is based directly on the upper bound theorem. The basic idea is to try all the likely collapse mechanisms and select the one which gives the lowest collapse load. The steps involved are as follows: 1.
Identify the likely plastic hinge locations (under point loads, at supports, at joints, at zero shear positions under spread loads).
2.
Sketch all the likely collapse mechanisms.
3.
For each mechanism use virtual work to calculate the collapse load factor.
4.
Select the mechanism which gives the lowest load.
5.
For this chosen case check that M ≤ MP (this is just to check that the selected mechanism is indeed the correct one). If this condition is not satisfied the correct mechanism has been overlooked.
Example – Fixed end beam with point loads 2W
A B 2m
3.5W
MP = 120kNm
D C
3m
1m 3.5W
Yield check on mechanism giving lowest load: We only need to find the moment at C, MC, since we know the bm at A, B and D is MP. Using virtual work again, we take the known bms and load from the chosen mechanism (1) and use mechanism (2) as a virtual displacement to determine MC: 2(62.61) θ
5θ
MP 3.5(62.61) 6θ
MC (unknown)
MP (θ + 5θ) + MC (6θ) = 2(62.61)(θ)2 + 3.5(62.61)(5θ)1
MC = 104.35kNm (≤ MP , OK) 120
giving the collapse bm diagram below:
120
MP
Example – Fixed base rectangular frame, varying M P, point loads (examples of frame analyses can be found in Megson II from p.613) 8W kN 6W kN 4
3
2
MP (beam) = 60kNm
4m
MP (both columns) = 40kNm
1
5
6m
1.
6m
Likely plastic hinge positions identified at points 1, 2, 3, 4, and 5.
Note: At the joints between beam and columns, a plastic hinge will form in the weaker member (the column) as soon as the moment reaches 40kNM. 2.
Sketch candidate collapse mechanisms:
COMBINED MP (column) (4θ) + MP (beam) (2θ) = 6W.4.θ + 8W.6.θ 40 × 4 + 60 × 2 24 + 48 = 3.889
W=
4.
Conclude collapse load, W = 3.889 (upper bound theorem)
5.
Check yield condition to ensure M ≤ MP for selected mechanism: Need to find the moments at points 1 to 5. However, we know that the moments at 1, 3, 4 and 5 are the MP values for those locations. This leaves just M2 to find.
31.11kN 23.33 kN 3
2
4
1
5
M4 = 40 (MP(column))
M2 θ
31.11
θ
2θ
M 3 = 60 (M P(beam) )
Using the beam mechanism as a virtual displacement to find M2: M2 (θ) + 60(2θ) + 40(θ) = 31.11 × 6 × θ M2 = 26.7kNm
Hence collapse bm diagram as shown with M ≤ MP everywhere proving solution is correct (uniqueness theorem).
40
26.7
60
ic θ θ
θ
3W
3W θ
θ
2W
2θ
2W
3W
4θ
2W
θ θ
2θ
2θ
4θ
plastic hinge cancels θ
θ
Mechanism A
2.
Mechanism B
θ
θ
Mechanism C = 2A + B
Sketches above show three likely mechanisms. Mechanism A – familiar beam mechanism. Mechanism B – a ‘sway’ type mechanism. Note: Joint 2 can only rotate about joint 1 and so moves in a direction perpendicular to 1-2. Joint 4 likewise moves perpendicular to 4-5. As joints 2 and 4 also belong to member 2-4 they must rotate about a common instantaneous centre of
4.
Mechanism C gives the lowest collapse load, providing us with our answer (according to the lower bound theorem).
5.
To prove that our solution is correct we need to demonstrate that for the chosen collapse mechanism M ≤ MP throughout the frame. Since there are plastic hinges at locations 1, 3, 4 and 5, we only need to show that M2 ≤ MP. Using mechanism A as a virtual displacement in a similar way to the previous example we obtain the virtual work (equilibrium) equation as:
L 3W. .θ = M2 .θ + MP .3θ 2 subst. W =
3W M2
20MP 13L
θ
θ
MP 2θ
MP
60 − 3)MP 26 = − 0.692MP
M2 = (
The negative value for M2 simply indicates that it acts in the opposite direction to that assumed in the diagram. Thus M2 ≤ MP and the final bm diagram will be as shown.
MP
0.69MP
MP MP MP
Objective To select a beam size such that M* < φMn, where φ = 0.9, M* is the design bending moment for the factored (strength limit state) loading and Mn is the nominal (in this case fully plastic) moment capacity of the selected beam. Loading Dead:
Concrete slab:
24 x 0.15 x 3
10.8kN/m
Partitions, etc.
2x3
6.0
Self weight UB
60kg/m
0.6 (guess)
Total Live:
Specified
Factored
G = 17.4kN/m 5x3
Q = 15kN/m
1.2G+1.5Q
43.4kN/m
Analysis Longest span will collapse first with a mechanism identical to a propped cantilever (see p.12 of these notes).
5m
0.586L (4.7m)
3.3m
We’ve already worked out the collapse load for this case (see p.12) and obtained w = 11.657
MP . L2
Alternative design based on linear elastic analysis
Designers frequently opt to determine the design actions by means of a linear elastic analysis – e.g. by moment distribution or more likely by computer analysis, rather than by plastic analysis. The loading, 43.4kN/m is the same but the analysis will give the bms shown below. The beam section is then selected such that M* < φMn as in the design based on plastic analysis. The maximum design bm M*, now occurs only at the interior support. 265.8
Thus,
M * = 266kNm , requiring
ZP ≥
266 φσ y
≥ 985 × 10 3 mm3
34.3 225.8
Select 410UB53.7 from table of UB section properties (next page). It has a plastic section modulus of 1060x10 3 mm3, giving φZPσ y = 286kNM (>266). The design is effectively based on the lower bound theorem in that only one plastic hinge is allowed to form (at the point of maximum bm) and so there is no mechanism . The approach is more conservative and leads to a less economical choice of beam. However, there are advantages in that the structure does not have to satisfy such stringent ductility conditions as that designed using plastic analysis. Providing sufficient restraint to ensure satisfactory plastic hinge rotation can increase the cost of a plastic-based design making it less economic.
