PLASTIC ANALYSIS OF STRUCTURES References 1 Marshall and Nelson, Structures. 2 Norris, Wilbur and Utku, Elementary Structural Analysis 3 Sir John Baker and J Heyman: Plastic Design of Frames. (Cambridge University Press) In the elastic design methods the assumption is made that the stresses are kept below the elastic limit and that when the load is removed the structure will revert to its original shape. Furthermore one assumes that strain is directly proportional proportional to stress. This theory works well for most building materials under normal loads. It, however, does not work when one tries to determine the failure strength of structures. In the plastic analysis method we are more interested in the strength of the structure. The applied load may not be greater than a predetermined predetermined portion of the load that will cause failure. This failure could be strength or serviceability. Certain materials such as steel have the ability to maintain load even though the yield str ess has been reached.
Figure 1: Graph of stress versus strain. Investigate the stress distribution of a profile that has properties shown in the Figure 1.
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Stress Distribution for Bending Moment Without Axial Load Rectangular Profile
Elastic:
Me
=
1 h 2 b ⋅ ⋅ h ⋅ fy 2 2 3
Plastic:
M
=
b⋅
Shape factor:
F =
p
M p Me
h 1 ⋅
2 2
=
h⋅f
Z p Z e
=
6 4
=
=
1
1 2 ⋅ bh f y 6
⋅
y
4
=
1,5
bh2f y
When the profile is fully plastic, rotation can take place without an increase in the bending moment. A plastic hinge has then been formed. The bending moment will also change as the profile stresses increase fr om elastic to fully plastic.
Elastic: Me
Plastic Analysis
=
qe L2 12
; thus qe
=
12 ⋅ M e 2
L
Plastic: 2 × M p
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=
q p
×
8
L2
; thus q p
=
16 × M p L2
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q p qe
=
16 ⋅ M p 12 ⋅ M e
=
4 3
x Shape factor
Failure will occur when enough plastic hinges have formed to lead to a mechanism. Normally this is one more plastic hinge than the number of redundancies.
Determination of The Maximum Load in Accordance With The Plastic Theory: Static Method Enough assumptions are made so that the bending moment diagram can be solved using the equilibrium equations. Nowhere may the moment be greater than the assumed plastic moment. Example: A uniform beam is loaded as in the sketch. Determine the plastic moment M p that will just cause the beam to fail.
The beam is statically indeterminate to the first degree: Two plastic hinges are required. 1 2
Assume that the negative moment at A is numerically equal to the positive moment at C.
5 VA = 2 MA VA = 0,4 MA (VA – 21) 7 + MA – 18 x 3 = 0 It follows that M A
=
201 3,8
=
of
(0,4 MA – 21)7 + MA = 54
52,89 kN ⋅ m
VA is then = 2,16 kN
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With this assumption MC = 53,53 kN.m. This is greater than the value assumed for the plastic hinge. It is then obvious that the plastic hinges will rather form at A and C.
It follows that
(39 – VA )3 = MA en 9 VA - 2 MA – 21 x 4 = 0 MA = 53,40 kN.m VA = 2,20 kN
The biggest bending moment occurs in two places, namely A and C with a value = 53,4 kN.m. M 53,4 kN.m.
P
=
The load, that is in equilibrium with arbitrary values of the redundancies and where the bending moment is nowhere greater than the maximum plastic moment, is never greater than the failure load.
The Kinematic Method. Positions are assumed for the plastic hinges and virtual work is used to determine the ratio between loads and the bending moment at the plastic hinges. Work done by the hinges = work done by the loads. The previous example is used to illustrate the method.
Mechanism 2 leads to the biggest value of M p and is hence the critical mechanism. A load that is determined by using virtual work based on an arbitrary mechanism is never smaller than the failure load.
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Requirements for determining the failure conditions: 1 2 3
Equilibrium conditions: The moments must be in equilibrium with the applied loads. Failure conditions: no bending moment in the structure may be greater than the plastic moment. Mechanism conditions: enough plastic hinges must form to form a mechanism.
Example:
Mechanism 3 gives the biggest value of Mp = 24,3 kN.m and is thus the critical mechanism. The answer may be checked by calculating the bending moment diagram.
Plastic Analysis
HD
=
1, 0 + 1,5 ⋅M p 4
H A
=
20 − HD
=
=
15,179 kN
4,821 kN
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V
=
VD
=
1
( 20 ⋅ 4
A −
6 1 6
(20 ⋅ 4
30 x ⋅ 3
+
+
30 x ⋅ 3
1,5 ⋅ Mp
= −
1,5 ⋅ M p
+
M A
= −
MB
= −
ME
= −
MC
= +
MD
= −
+
−
2 ⋅ 1,5 M ) 2 ⋅ 1,5 Mp )
=
=
1,667 + 0,5 ⋅ M
p
28,333 − 0,5 ⋅ Mp
=
13,810 p
=
16,190
36,43 kN.m
4 ⋅ HA
= −
17,14 + 3 ⋅ VA
= +
24,29 + 3 ⋅ VA
−
17,14 kN.m
24,29 kN.m = Mp ( test )
30 ⋅ 3
24,29 + 20 ⋅ 4 − 4 ⋅ H A
24,29 kN. m = − Mp ( test)
= −
= +
36,43 kN.m = +1,5 ⋅ Mp ( test)
Nowhere in the structure is the bending moment greater than the calculated plastic moment. Enough plastic hinges have formed to create a mechanism. Mechanism 3 may be seen as a combination of mechanisms 1 and 2. The virtual work at the hinges is equal to 4 + 5 – 2 = 7, where the negative 2 is for the hinge that is removed at B (mechanism 1: clockwise, mechanism 2: anti-clockwise). The virtual work done by the loads is equal to 90 + 80 = 170. Mechanisms 1 and 2 can be seen as independent and mechanism 3 as derived from the independent mechanisms.. Number of Independent Mechanisms The bending moment diagram of any structure may be determined if the bending moment at enough possible hinge positions is chosen. The moments at these points are not independent: they are linked by the equilibrium equations. For every possible hinge position we are able to derive an equilibrium equation in terms of the loads and the redundancies. From these equations it is possible to eliminate the redundancies and to derive a set of equilibrium equations that gives the link between the bending moments at possible hinge positions. The number of these independent equilibrium equations is equal to the number of possible hinge positions minus the redundancies. A mechanism is formed when the equilibrium equations are no longer satisfied. The number of independent mechanisms is also equal to the number of independent equilibrium equations. Therefore: Number of independent mechanisms = number of possible hinge positions – number of redundancies.
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Combining of Mechanisms. Combine mechanisms to find big virtual work by the loads and minimum work by the hinges. Example: Determine the value of Q so that failure will just occur.
Number of possible hinge positions Number of redundancies Number of independent mechanisms
= 10 =6 =4
In the following table the independent mechanisms are numbered 1 to 4. Mechanism 4 cannot occur in isolation but may be combined with other mechanisms to eliminate hinges.
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Mechanism 7 is linked to the lowest value of Q and is then the actual failure mechanism. Example
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Determine the value of Mp so that the following structure will just fail. Draw the bending moment diagram that is associated to this failure mechanism.
Number of possible hinge positions Number of redundancies Number of independent mechanisms
=7 =4 =3
Mechanism 5 is critical with Mp = 50,0 kN.m.
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Take moments about the momentary centre of rotation. 25 8 6Q = (20 + 20) ⋅ − 2 ⋅ 50 − 65 ⋅ = 60 therefore Q = 10 kN 3 3 Toets:
Shear force just left of C:
100 + 5 = −75 2 120 – 75 = + 45 + 100 - 45·2 = + 10 kN.m −
Shear force just left of 120 kN load: Bending moment just to the right of B:
The Semi-Graphical Method. The semi-graphical method is based on the static method. Example:
If one uses the XY as the basis of the zero line, we are back to the bending moment diagram of the propped cantilever.
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MC
83,75 −
7
⋅ M A 12 3 MD = 66,75 − ⋅ M A 12
1)
If
−
=
MA
−
for any value of MA
M C then 83,75 −
7 ⋅M A 12
=
M A
leads to
MA = 52,89 kN.m
and
MC = 52,89 kN.m MD = 53,53 kN.m If one had to design the beam for an Mp = 53,3 kN.m, then we would not have a failure mechanism and the beam would be able to carry a greater load. 3 2) If −M A = M D then 66,75 − leads to MA = 53,40 kN.m ⋅M A = M A 12 And MC = 52,60 kN.m MD = 53,40 kN.m If Mp = 53,40 kN.m then we have a failure mechanism. Example The example is statically indeterminate to the first degree. To make the structure statically determinate, remove the horizontal reaction at D as the redundancy. Draw the bending moment diagram of the statically determinate structure for the load as well as for the redundancy.
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The total bending moment diagramme must be read with AB 2C2D as basis.
The maximum positive moment between BC for the statically determinate structure is at X1 at a distance of 4,576 m with a value of 132,28 kN.m. Assume plastic hinges at X 1 and C. Total bending moment: MX = 132,28 – 6H MC = - 6H + 10 Assume that the numerical values at X and C are the same. 132,28 – 6H = 6H – 10 then H = 11,86 en MX = MC = 61,1 and we have a failure mechanism.
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Further Example:
Remove the redundancy at I, and determine the bending moment diagram of the statically determinate structure for the given load as well as the redundancy. Fold the bending moments open and super-impose the bending moments. From the sketch it appears that the biggest moments are at D and H. At D, At H,
252,7 – 8,6667 H 0–6H
= + Mp = - Mp
It follows that 17,229 kN and Mp = 103,377 kN.m. Redraw the bending moment diagramme and ensure that the value of 103,377 is not exceeded.
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