chapter systematic methods of plastic...

CHAPTER 5Systematic Methods

of Plastic Analysis

The examples of combined plastic mechanisms presented in Chapter 4 show that the challenge in analyzing complex struc-tures lies in finding the collapse mechanism that will give

the true (unique) plastic collapse load. Systematic methods of plastic analysis are therefore needed to efficiently identify this correct mech-anism. A number of such methods have been developed in the past, and two are reviewed in this chapter: direct combination of mecha-nisms, which is more suitable for hand calculation, and the method of inequalities, more suitable for linear programming. However, in both these methods, knowledge of the number of basic mechanisms present in a given structure is imperative for a successful solution search, and this topic is therefore presented first.

5.1 Number of Basic MechanismsFour basic mechanism types were identified in the previous chapter:

• Beam

• Panel

• Gable

• Joint

In any given structure, the first task in systematic plastic analysis is to locate and identify those basic mechanisms. To provide some assis-tance in that endeavor, it is worthwhile to formalize the relationship between the degree of indeterminacy of a structure, X, the number of potential plastic hinge locations, N, and the number of possible basic mechanisms, n. This is accomplished by the following equation:

n = N - X (5.1)

Figure 5.1 illustrates how this equation can be used. Note that the same single-bay moment-resisting frame is used throughout the

249

05_Bruneau_Ch05_p249-272.indd 249 6/11/11 2:44:14 PM

S y s t e m a t i c M e t h o d s o f P l a s t i c A n a l y s i s 251

Cas

e 1

Bea

m

Pan

el

Bea

m

Pan

el

Join

t

Join

t

Bea

m

Pan

el

Bea

m

2 Jo

ints

N =

6X

= 1

N =

5 X1

X1

X1

Cas

e 2

Cas

e 3

N =

5X

= 1

N =

4

N =

3X

= 1

N =

2

Fig

ur

e 5

.1

Exa

mpl

es o

f id

entific

atio

n of

bas

ic m

echa

nism

s fo

r a

sim

ple

fram

e in

dete

rmin

ate

to t

he fi

rst

degr

ee.

250



example shown in that figure. Simple observation reveals this frame to be indeterminant to the first degree, X = 1. Using the knowledge developed in Chapter 4, one can identify the number of potential plastic hinge locations, N. Cases 1 and 2 in Figure 5.1 illustrate that it doesn’t matter whether one considers only one plastic hinge at each corner of that frame or two (i.e., one at the end of each member framing into the joint). Case 1 is a simplified version of Case 2 in which the joint mecha-nisms have been implicitly considered. In Case 1, the engineer must determine whether the flexural capacity at the corner joint is equal to the plastic moment of the column or the beam, whereas the more explicit approach used in Case 2 systematically manages this information.

For hand calculations, the approach of Case 1 is recommended when only two members frame into a joint, and the approach of Case 2 must be used at all joints in which more than two members meet (as shown in Figures 5.2 and 5.3). Real hinges (such as at the base of the frames in Figure 5.1) are never included in the value of N because they cannot do plastic work. Note that the number of potential plastic hinge locations is also a function of loading, as shown by a com-parison of Cases 2 and 3 in Figure 5.1.

Once the number of basic mechanisms is known, judgment is often necessary to identify these possible mechanisms. Although it is straightforward to find the 16 basic mechanisms in Figure 5.2, more complex structures, such as the one shown in Figure 5.3, can be more challenging. Note that for the panel mechanisms in those two figures,

N = 34X = 18

N = 34

Beam mech. (typ.)X1 X2

X3

X10 X11X12

X4 X5X6

X13 X14X15

X7 X8X9

X16 X17X18

Joint mech. (typ.)

X = 18

n = 16 = 6 Beams, 7 joints, and 3 panels

Figure 5.2 Examples of identification of basic mechanisms for a three-story frame loaded as shown.


252 C h a p t e r F i v e S y s t e m a t i c M e t h o d s o f P l a s t i c A n a l y s i s 253

the floors above the story displacing laterally move rigidly with the displacing floor; the characteristic of this approach is that a single “panel level” is distorted as a parallelogram for each mechanism. An alternative approach consists of preventing all parallel members from moving except the one being displaced for the panel mechanism con-sidered (as done for the rightmost case in Figure 5.4); the characteristic

Figure 5.3 Examples of identification of basic mechanisms for a three-story frame loaded as shown.

N = 33X = 18

Beam mech. (typ.)

Joint mech. (typ.)

n = 15 = 4 Beams, 6 joints, 1 gable, and 4 panels

N = 16X = 9n = 7

= 4 Joints, 2 beams, and 1 panel = 4 Joints and 3 panels

Panel

Beam

Beam

Panel

Panel

Panel

or

N = 16 X = 9

3 3 3

Figure 5.4 Examples of basic mechanisms identification for a Veerendel truss.



of this approach is that only one line of structural members sways foreachmechanism.Bothapproachesarefine,butmixingtogethercases from both approaches is not recommended, being often a source of confusion (i.e., double-counting some panel mechanisms and missing others).

In some cases, the mechanism identification procedure can be subjective because two (or more) different sets of basic mechanisms can be obtained for the same structure. This is illustrated for the Veerendel truss in Figure 5.4. The basic mechanisms constitute only the building blocks for systematic plastic analysis, so any appropriate set of basic mechanisms will eventually lead to a correct solution (although the length of the “path” toward that goal may vary).

5.2 Direct Combination of MechanismsThe direct-combination-of-mechanisms method is a systematic exten-sion of the upper bound method presented in the previous chapter. In the upper bound method, various basic and combined mechanisms are added or subtracted by trial and error until the mechanism havingthelowestplasticcollapseloadisfound.Byvirtueoftheuniqueness theorem, this calculated collapse load is the true solution if it also produces a statically admissible moment diagram.

Therefore, in this method, the search for the plastic collapse load proceeds through systematic combinations of mechanisms to either increase the external work or decrease the internal work. A tabular procedure is adopted to keep a manageable record of basic mecha-nisms and previously attempted combinations as well as to facilitate the identification of those combinations that can be best combined. This systematic tabular approach is best described through the use of examples.

5.2.1 Example: One-Bay, One-Story FrameThe one-bay, one-story frame previously analyzed in Chapter 4 with a trial-and-error approach (see Figure 4.11) is considered here for the case a = L/2. Obviously, a systematic method of plastic analysis is not necessary for such a simple example, but the objective at this point is to illustrate how to use the tabular procedure; additional complexi-ties at this early stage would only detract from this goal.

First, in this systematic method, an arbitrary sign convention must be adopted. It is expedient to draw a dotted line along the frame members to visually define the adopted sign convention, as shown in Figure 5.5; moments and rotations are positive when they produce tension on the side of structural members adjacent to the dotted lines. Other sign conventions are possible (such as tracking clockwise and counterclockwise plastic hinge rotations) but not used here. Then, one establishes the number of basic mechanisms for the structure using the procedure described in Section 5.1.


254 C h a p t e r F i v e

With this information, the general layout of the table needed for the structure under consideration can be drawn, as shown in Table 5.1. In this example, there are five potential plastic hinge locations and the structure is indeterminant to the third degree, producing two basic mechanisms: one beam and one panel, as shown in Figure 5.5. Note that the sign of the plastic rotations is retained with the plastic hinge data in Table 5.1. Also, for the sake of clarity, basic and combined mechanisms are separated in this table.

The characteristics of the previously sketched basic mechanisms are first entered in this table. Normalized coefficients are used as much as possible for clarity and to expedite the mechanisms combi-nation process. For each mechanism, the internal work that develops ateachplastichingelocationisinventoriedunderPlasticHingeData.Thus, one can directly obtain the total internal work corresponding to any given mechanism by adding the absolute values of the plastic hinge data across a row of the table. For structures composed of more than one section shape, each member would have a different plastic moment. The plastic hinge data therefore simultaneously accounts for the magnitude of the plastic rotation (b q) and plastic moment (a Mp), as shown in the header of Table 5.1.

One should always draw rough sketches of all mechanisms (using the tabulated plastic hinge data) to keep track of the physical mean-ing of each combination. The total external work should always be computed directly from these sketches. For small structures, the rough sketches can be inserted directly into the table.

Once the basic mechanisms have been logged, one must search for combinations that minimize the internal work, maximize the external work, or do both. The tabular format presented here facili-tates the identification of combinations that will result in the cancel-lation of plastic hinges (i.e., reduction of internal work). In this simple example, the number of possibilities is limited. The two basic

N = 5X = 3

H = P/2 B

A

P

C

D

E

L/2For signconvention

L/2

L

MPh = L

I. Beam

I + II (Beam + Panel) II – I (Panel – Beam)

II. Panel

2θ

θ

ββ ββ

θ

n = 2

Schematic of combined mechanisms

Figure 5.5 One-bay, one-story frame example.



Mec

hani

smP

last

ic H

inge

Dat

a (`

Mp)

(ap)

/(M

pp)

Inte

rnal

Wor

kExt

erna

l Wor

k

PP

/(M

p/

L)Id

enti

fier

Titl

eA

BC

DE

Wi/

Mpp

We/P

Lp

Bas

ic M

echa

nism

s

IB

eam

0

− 1

+ 2

− 1

0

40.5

8M

p/L

8

IIPan

el− 1

+ 1

0

− 1

+ 1

40.5

8M

p/L

8

Com

bine

d M

echa

nism

s

I +

IIB

eam

+ p

anel

− 1

0

+ 2

− 2

+ 1

61.0

6M

p/L

6 ✔

II –

IPan

el −

bea

m− 1

+ 2

− 2

0

+ 1

6

0•

•

Tab

le 5

.1

Sys

tem

atic

Direct

Com

bina

tion

of M

ech

ani

sms

for

One

-Bay

, O

ne-S

tory

Exa

mple

(Fi

gure

5.5

)

255



mechanisms are first directly combined by simple addition, column by column, of the normalized plastic rotations listed in the respective rows of Table 5.1 corresponding to these two basic mechanisms. Then, these are subtracted to demonstrate that some combinations do not progress the solution. Inspection of the sketches of the mechanism reveals that subtracting the beam mechanism from the panel mecha-nism reduces the amount of external energy.

For each combination, the corresponding plastic collapse load must be calculated from the ratio of the internal work to the external work because these results from separate rows in Table 5.1 are not additive. Whenever one believes that the lowest plastic collapse load has been found, the resulting moment diagram should be drawn. A statically admissible moment diagram will reveal that the true solution has been found, whereas, in the alternative, the resulting diagram will indicate where plastic hinges should develop and thus provide guidance for the most promising subsequent combination.

5.2.2 Example: Two-Story Frame with Overhanging BayThe two-story frame with an overhanging bay, shown in Figure 5.6, is analyzed to illustrate how to combine mechanisms in a more general

I II

III IV

X

V Panel 1 VIIVI

VIII

IX

L/2

L L

L/2 L/2 L/2

XI = V + VI XII = XI + VIII + IX XIII = XII + VII

XIV = XIII + X XV = XIV + beams

1

N = 19X = 9N = 10 = 4 Beams, 3 joints and 3 panels

2

3

4 5 6 7

8

9

1011

12

13 14 15

16

18 1917

Panel 2 Panel 3

3

4

5

7

8

9

10

15

16 17

VIII IXX

XVI

Signconvention

P P

PP

P

PL

L

Figure 5.6 Two-story frame with an overhang example for application of direct combination of mechanism method.



situation.Perthesystematicmethodologypresentedinthepreviousexample, a sign convention is selected, the potential plastic hinge locations are determined, and the basic plastic mechanisms are identified. The layout of Table 5.2 is then established, and the plastic collapse loads for each basic mechanism are calculated. Note that joint mechanisms are used only to eliminate internal work: therefore, no external work or plastic collapse load is calculated for those mech-anisms. Also, when single hinges are used in joints connecting only two members (such as for hinges 12, 13, and 19 in this example), the dotted line indicating the sign convention must extend continuously across such joints to avoid possible confusion (as would have been otherwise the case for hinge 12 here)

There is no single strategy to find the combined mechanism that will produce the lowest plastic collapse load because the particulars of a given problem will suggest different approaches. (In fact, even for a given structure, different engineers may adopt different search sequences.) Here, results for the basic mechanisms indicate that the panel mechanisms produce the largest amount of external work, the lowest plastic collapse load being that of mechanism VI. Therefore, as an arbitrary starting point, one could first try to combine some panel mechanisms to increase the amount of external work. However, as shown by the combination of mechanisms V and VI (noted XI in the table), this also results in significantly more internal work because the plastic hinges of the two mechanisms are additive. This, as well as the sketch of this first combination, suggests that joint mechanisms must be introduced to reduce internal work. A combina-tion of mechanisms V, VI, VIII, and IX is attempted (mechanism XII), but only one plastic hinge is eliminated in the process.

Examination of the results so far reveals that adding panel 3 would reduce the amount of internal work (trading hinges 10, 17, and 12 for 19) and increase the amount of external work. This is tried with mechanism XIII, which gives the lowest value of the plastic collapse load at this point. As examination of that mechanism in Figure 5.6 suggests, adding mechanism X removes one further plastic hinge (mechanism XIV) and gives a lower value for the collapse load. Mechanism XV investigates if the additional external work produced by the beam mechanisms exceeds the extra internal work thus intro-duced; the resulting plastic collapse load is found to be higher than for mechanism XIV. Although one could attempt some additional trial combinations if deemed necessary (such as mechanism XVI in Figure 5.6), it appears that no other combination of mechanisms would provide a lower value for the plastic collapse load. The moment diagram for mechanism XIV (not shown here) is found to be statically admissible confirming that the true solution has been found.

As demonstrated by the above examples, without engineering judgment, the direct combination method, even using a systematic procedure, can rapidly become excruciatingly laborious as structural



Mec

hani

smP

last

ic H

inge

Dat

a (`

Mp )(ap

)/(M

Pp

)In

tern

al

Wor

kExt

erna

l W

ork

PP/

(Mp/L)

Iden

tifie

rN

ame/

sket

ch1

23

45

67

89

10

11

12

13

14

15

16

17

18

19

Wi/

Mpp

We/P

Lp

Bas

ic M

echa

nism

s

IB

eam

1− 1

+ 2

− 1

40.5

8 M

p/L

8

IIB

eam

2− 1

+ 2

− 1

40.5

8 M

p/L

8

IIIB

eam

3− 1

+ 2

− 1

40.5

8 M

p/L

8

IVB

eam

4− 1

+ 2

− 1

40.5

8 M

p/L

8

VPan

el 1

− 1

+ 1

+ 1

+ 1

− 1

− 1

61.0

6 M

p/L

6

VI

Pan

el 2

− 1

+ 1

+ 1

− 1

42.0

2 M

p/L

2

VII

Pan

el 3

− 1

− 1

− 1

+ 1

41.0

4 M

p/L

4

VIII

Join

t 1

− 1

+ 1

+ 1

3−

−−

IXJo

int

2− 1

+ 1

− 1

+ 1

4−

−−

XJo

int

3− 1

+1+

13

−−

−

Com

bine

d M

echa

nism

s

XIV +

VI

− 1

+ 1

+ 1

− 1

− 1

+ 1

+ 1

+ 1

− 1

− 1

10

33.3

3 M

p /L

3.3

3

XII

XI +

VIII

+ IX

− 1

+ 1

+ 1

− 1

+ 1

+ 1

+ 1

− 1

− 1

93

3 M

p/L

3.0

XIII

XII

+ VII

− 1

+ 1

+ 1

− 1

+ 1

− 1

− 1

74

1.7

5 M

p /L

1.7

5

XIV

XIII

+ X

− 1

+ 1

+ 1

− 1

+ 1

− 1

64

1.5

Mp /L

1.5

✔

XVXI

V +

I +

III

− 1

+ 1

+ 2

− 2

+ 2

− 2

10

52.0

Mp /L

2.0

Tab

le 5

.2

Sys

tem

atic

Direct

Com

bina

tion

of M

ech

ani

sms

for

Two-S

tory

Fra

me

with

an

Ove

rhangi

ng

Bay

(Fi

gure

5.6

)

258



complexitygrows.Partlyforthatreason,themethodpresentedinthefollowing section, although nearly suitable only for implementation in computer programs, is preferable when one is dealing with extremely large problems (or for small problems whenever one is ready to trade control against expediency, reserving engineering judgment for the task of checking the validity of the generated computer results).

5.3 Method of InequalitiesAlthough the systematic method of plastic analysis formulated in the previous section can also be computerized, the most computationally efficient solution procedure for plastic analysis is the method of inequalities, based on a matrix formulation of the lower bound method. However, because formal presentation of this mathematical procedure would be abstract, an example is presented instead to illustrate how a systematic lower bound method can proceed from the method of inequalities.

Considering the small size of this example problem, the corre-sponding amount of computations necessary for the solution pre-sented hereafter may appear excessive, particularly when one can find the correct answer within minutes and with a minimum of calculations by using the graphical approach presented earlier. Nonetheless, the principal objective here is to illustrate how a systematic computerized procedure can be designed to automatically converge to the correct solution. In the process, for further clarity, expanded equations instead of matrix notation have been used (the matrix methods implemented in computer programs being too numerically intensive for hand calculations).

The frame considered in this example is shown in Figure 5.7, along with the four potential plastic hinge locations corresponding to theloadingcondition.Giventhatthisframeisindeterminatetothesecond degree (x = 2), it is possible to express the moments at each of the potential plastic hinge locations as a function of any two arbitrarily selected redundant forces. The chosen sign convention, indicated by the dotted line, is drawn along the frame members in Figure 5.7. In this problem, the two reactions at the right support are selected as the redundant forces, and a set of equations can be written for the moment at the potential plastic hinge locations as a function of these forces.

M4 = -2RL (5.2a)

M3 = -2RL + VL (5.2b)

M2 = -2RL + 2VL - 2FL (5.2c)

M1 = 2VL - 2FL - 6FL (5.2d)



This set of four equations cannot be solved because it contains six unknowns (four moments and two redundant forces). However, one can eliminate the two redundant forces using algebraic manipula-tions and substitutions. As a result, the above system of equations is reduced to the following two equations and four unknowns:

-M2 + 2M3 -M4 = 2FL (5.3a)

-M1 + M2 -M4 = 6FL (5.3b)

Note that these two equations represent two equilibrium equa-tions for this structure. Incidentally, one can obtain the same two equations, with much less difficulty, by writing the virtual-work equations for the two basic mechanisms of this structure. Hence, for the basic beam and the panel mechanisms shown in Figure 5.7, equat-ing internal and external work (Chapter 4) gives:

-M2q + 2M3q - M4q = 2FLq (5.4a)

M2q - M1q - M4q = 6FLq (5.4b)

from which the qs can be eliminated. This should not be surprising because, in essence, work equations are also equilibrium equations.

However, for these basic mechanisms, the hinge moments, Mi, are expressed not in terms of Mp , but rather as unknowns whose values remain to be determined. For large structural systems, this approach to obtain a set of equilibrium equations is preferred because it is considerably easier to write virtual-work equations for a number of basic mechanisms than equilibrium equations in terms of redundants.

N = 4X = 2

H = 3F

2F

2L

2L

Frame properties and loading

L L

1

2 3 4

For signconvention

X = 2

R

VPlastic hingelocations andsign convention Redundancies

MP

MP

MP

MP

MP

MP

MP/2

BeamPanel

Basic mechanisms Resulting moment diagram

2θ

θ

ββ ββ

θ

n = 2

Figure 5.7 Frame example of the method of inequalities.



Nonetheless, at this point, the above remains a system with more unknowns (four) than equations (two). However, within the context of plastic design, a solution is possible because additional constraints exist that limit the magnitude of moment at each potential plastic hinge as follows:

-Mp ≤ M1 ≤ MP (5.5a)

-Mp ≤ M2 ≤ MP (5.5b)

-Mp ≤ M3 ≤ MP (5.5c)

-Mp ≤ M4 ≤ MP (5.5d)

In practical situations, the plastic moments would likely differ at the different plastic hinge locations; different positive and negative values of plastic moment are even possible at a given location for composite structures. However, for this example, a single value of plastic moment is assumed throughout the entire frame for the sake of simplicity.

The above equilibrium equations and inequality relationships, coupled with the knowledge that a lower bound approach is a search for the largest possible applied loads (while respecting all above equalities and inequalities), make possible a solution to this problem. The algorithm to achieve such a solution typically uses the equilib-rium equations and the inequality conditions to systematically eliminate the unknowns, one by one, performing all possible cross-comparisons of inequality equations in the process, until all remain-ing equations are expressed in terms of known quantities. One can then find the largest possible value for the applied loads (often expressed as a function of a common load, F) simply by scanning all resulting inequalities. The largest value of F that satisfies all resulting constraints gives the collapse load.

For the example at hand, the equilibrium equations are first used to express two of the remaining unknowns in terms of the others:

M FL M M3 2 412

= + +( ) (5.6a)

M1 = - 6FL + M2 - M4 (5.6b)

These results from the equilibrium equations are then substituted in the appropriate inequalities:

- ≤ - + - ≤M FL M M Mp p6 2 4 (5.7a)

- ≤ ≤M M Mp p2 (5.7b)

- ≤ + + ≤M FL M M Mp p12 2 4( ) (5.7c)

- ≤ ≤M M Mp p4 (5.7d)



From this point on, further elimination of unknowns through the inequality relationships proceeds by construction of all possible infer-ences from the inequality set. For example, to eliminate M2 from the inequality set, all inequality equations that contain M2 must be rear-ranged to isolate that term within the inequalities. This gives:

- ≤ ≤

- - - ≤ ≤ - -

- + +

M M M

M FL M M M FL M

M FL M

p p

p p

p

2

4 2 42 2 2 2

6 44 2 46≤ ≤ + +M M FL Mp

(5.8a)

(5.8b)

(5.8c)

Then the left side of the inequalities given in Eq. (5.8) is system-atically compared with the right side of the same inequalities. More explicitly, the left-side inequality of Eq. (5.8a) is compared with the right side inequalities of Eqs. (5.8a), (5.8b), and (5.8c); the left-side inequality of Eq. (5.8b) is compared with the same three right-side inequalities; and finally, the left side of Eq. (5.8c) is similarly com-pared. This gives:

- ≤M Mp p (5.9a)

- ≤ - -M M FL Mp p2 2 4 (5.9b)

- ≤ + +M M FL Mp p 6 4 (5.9c)

- - - ≤2 2 4M FL M Mp p (5.9d)

- - - ≤ - -2 2 2 24 4M FL M M FL Mp p (5.9e)

- - - ≤ + +2 2 64 4M FL M M FL Mp p (5.9f)

- + + ≤M FL M Mp p6 4 (5.9g)

- + + ≤ - -M FL M M FL Mp p6 2 24 4 (5.9h)

- + + ≤ + +M FL M M FL Mp p6 64 4 (5.9i)

Equations (5.9a), (5.9e), and (5.9i) are automatically satisfied. Then, by finding matching pairs, one can write Eq. (5.10a) [from Eqs. (5.9c) and (5.9g)], Eq. (5.10b) [from Eqs. (5.9b) and (5.9d)], and Eq. (5.10c) [from Eqs. (5.9f) and (5.9h)] as follows:

- - ≤ ≤ -2 6 2 64M FL M M FLp p (5.10a)

- - ≤ ≤ -3 2 3 24M FL M M FLp p (5.10b)

- - ≤ ≤ -32

4 32

44M FL M M FLp p (5.10c)

- ≤ ≤M M Mp p4 (5.10d)



ByrepeatingtheaboveprocesstoeliminateM4, that is, by system-atically comparing each left side of the inequalities given in Eq. (5.10) with the corresponding four right sides of the same set of inequality equations, one obtains the following set of 16 inequalities:

- - ≤ -2 6 2 6M FL M FLp p (5.11a)

- - ≤ -2 6 3 2M FL M FLp p (5.11b)

- - ≤ -2 6 32

4M FL M FLp p

(5.11c)

- - ≤2 6M FL Mp p (5.11d)

- - ≤ -3 2 2 6M FL M FLp p (5.11e)

- - ≤ -3 2 3 2M FL M FLp p (5.11f)

- - ≤ -3 2

3

24M FL

MFLp

p

(5.11g)

- - ≤3 2M FL Mp p (5.11h)

- - ≤ -

3

24 2 6

MFL M FLp

p

(5.11i)

- - ≤ -

3

24 3 2

MFL M FLp

p

(5.11j)

- - ≤ -

3

24

3

24

MFL

MFLp p

(5.11k)

- - ≤

3

24

MFL Mp

p

(5.11l)

- ≤ -M M FLp p2 6 (5.11m)

- ≤ -M M FLp p3 2 (5.11n)

- ≤ -M

MFLp

p3

24

(5.11o)

- ≤M Mp p (5.11p)

ByeliminatingEqs.(5.11a),(5.11f),(5.11k),and(5.11p),whichpro-vide no useful information, one obtains the following 12 inequalities for F, respectively:

- ≤5

4

M

LFp

(5.12a)

- ≤7

4

M

LFp

(5.12b)



- ≤M

LFp

2

(5.12c)

FM

Lp≤

5

4

(5.12d)

FM

Lp≤

9

4

(5.12e)

- ≤2M

LFp

(5.12f)

FM

Lp≤

7

4

(5.12g)

- ≤9

4

M

LFp

(5.12h)

- ≤5

8

M

LFp

(5.12i)

FM

Lp≤

2

(5.12j)

FM

Lp≤

2

(5.12k)

FM

Lp≤

5

8

(5.12l)

The range of values of F that can satisfy all the above inequalities is − Mp/2 ≤ F ≤ Mp/2, or in absolute value, F = Mp/2. From this result, one can calculate the actual value of each previously eliminated unknown by going in reverse through the elimination order. Hence, one calculates M4 first by substituting the value for F into Eq. (5.10). This gives, for the positive value of F (negative value would simply give a reversed moment diagram):

( ) ( )- - = - ≤ ≤ - = -2 3 5 2 34M M M M M M Mp p p p p p (5.13a)

( ) ( )- - = - ≤ ≤ = -3 4 2 34M M M M M M Mp p p p p p (5.13b)

( . . ) ( . . )- - = - ≤ ≤ - = -1 5 2 3 5 0 5 1 5 24M M M M M M Mp p p p p p (5.13c)

- ≤ ≤M M Mp p4 (5.13d)

These inequalities, in particular Eqs. (5.13a) and (5.13d), constrain M4 to be equal to -Mp , which indicates presence of a plastic hinge at



that location. Then, substituting this result together with the value of F into Eq. (5.8) gives:

- ≤ ≤M M Mp p2 (5.14a)

- -

- - = -

≤ ≤ =2 2

12

2 2 22M M M M M M Mp p p p p( ) pp p pM M-

- -

2

12

( )

(5.14b)

- +

+ - =

≤ ≤ = +M M M M M M Mp p p p p p612

3 61

2( )22

+ -

M Mp p( )

(5.14c)

From those inequalities it follows that M2 must be equal to Mp. Finally, when one substitutes all previously obtained results into the equilibrium equations, Eq. (5.6), the remaining two unknowns can be calculated:

M M M M Mp p p p312

12

12

= + - =( ) (5.15a)

M M M M Mp p p p1 6 12

= -

+ - - = -( ) (5.15b)

These results indicate, according to the previously defined sign convention, that the plastic collapse mechanism for this structure for the given loading condition is the basic panel mechanism. The result-ing statically admissible moment diagram is shown in Figure 5.7.

Formally, for any given structure, a computerized solution is pos-sible provided that the following two mathematical constructions can be formulated:

• Asetofn = N − X equilibrium equations, obtained directly from the basic independent mechanisms expressed in the following matrix format:

c c c c cc c c c cc c c c

N

N

11 12 13 14

21

31

1

22 23 24 2

32 33 344 3

2 3 4

1

2

c

c c c c c

MMM

N

n n n n nN1

33

4

1

2

3

M

M

kkkk

kN N

=

4

P

(5.16)



• AseriesofN inequalities expressing the plastic moment con-straints at each potential plastic hinge location, such that:

- ≤ ≤

- ≤ ≤

- ≤ ≤

M M M

M M M

M M M

p p

p p

pN N pN

1 1 1

2 2 2

(5.17)

The matrix solution of this problem per the philosophy expressed in the above example can be accomplished through use of the stan-dard simplex method algorithm developed for multidimensional optimizationinlinearprogramming(Cambridge1992,Dantzig1963,Livesley 1975).

5.4 Self-Study ProblemsProblem 5.1 Redo Problem 4.12 using the systematic approach of Section 5.2.

Problem 5.2 Forallthestructuresshownhere(CasesAtoG),useanupperbound approach to find the maximum load, P, that can be applied. Make sure to:

(a) Determinethenumberofbasicplasticcollapsemechanismsandsketch each one of them.

(b) Also clearly sketch all the combined plastic collapse mechanisms considered in solving this problem.

(c) ExceptforCaseG,verifythatthesolutionisstaticallyadmissibleand draw the corresponding moment diagram. Also show the value and direction for all the reactions obtained in the process.

P

Case A

2MP

0.5L

P

0.5L

0.5L

MP

0.5L



0.5MP

2MP

0.5L

0.5L

L/3 1.5L

MP

MP

2L/3

2P

P

Case B

2MP

4MP

8

5

5 5 7.5 7.5

MP MP

2PP

P

Case C

P

2P3L

1.5L 1.5L

L

2MP

4MP

0.5MPMP

MP

Case D


S y s t e m a t i c M e t h o d s o f P l a s t i c A n a l y s i s 2692P

2MP

2MP MP

3L 2L

2.5L

2.5L

L

Case E

10

Case F

5

6

23

MP

MP/2

P3

P2

P1

Note: For Case F, P1 = P, P2 = 2P and P3 = 1.5P:

P

P

6

Case G

8

2

2

44MP

2MP

2MP

MP 1.5MP

268



Problem 5.3 For all the structures shown below (Cases A to C), assuming that loads could be applied at any ends of any member and acting in any direc-tion, but not within the span of any member, find the number of basic plastic mechanisms, and sketch them. Here, it is not required to calculate the plastic collapse load, but only to show the correct number of basic plastic mechanism and sketch how these mechanisms would deform.

Case A

Case B

Case C



Problem 5.4 For all the structures shown (Cases A to C), determine the number of basic plastic collapse mechanisms, and sketch each one of them. Here, it is not required to calculate the plastic collapse load, but only to show the cor-rect number of basic plastic mechanism and sketch how these mechanisms would deform.

Case A

P

P

P

Case B

P

Case C

Problem 5.5 For the structure shown below, use the upper bound theorem to find the maximum load V that can be applied to this structure. First, draw a sketch showing the plastic mechanism for each of the plastic mechanism that wouldneedtobeconsideredforgenericvaluesofVandP.Then,solveforthe



case P = 0. Note that all beams have a strength of Mp, and all columns have a strength of 3Mp.

For this problem, it is not required to check whether the solution is stati-cally admissible.

V

Mp

6L

P

3L

5L

3Mp

3Mp

Problem 5.6 For the structures shown below, all structural members have a strength of Mp, except the rightmost column which has a strength of 0.5Mp.

(a) Calculate the number of basic plastic mechanisms for the structure showninpart(a)ofthefigurebelow.Drawasketchshowingtheplas-tic mechanism for each of the basic plastic mechanism identified.

(b) Calculate the number of basic plastic mechanisms for the structure showninpart(b)ofthefigurebelow.Drawasketchshowingtheplastic mechanism for each of the basic plastic mechanism identi-fied. Note that the only difference between the structures in parts AandBofthefigurebelowisthatthetwomiddlecolumnsforthestructureinpartBhavehingesattheirtopandbottom.

(c) For the structure shown in part (b) of the following figure, use the upper bound theorem to find the maximum load V that can be applied to this structure. For this problem, check that the solution is a statically admissible solution.

V

All members Mp except

(a)

3L

3L

3L

3L3L

4.5L 4.5L

0.5Mp V

All members Mp except

(b)

3L

3L

3L

3L3L

4.5L 4.5L

0.5Mp



Problem 5.7 Forthestructuresshown(CasesAandB),only:

(a) Determinethedegreeofindeterminacyofthestructureshown.(b) Determinethenumberofbasicmechanisms,andillustratethemby

sketches.

P

P

Case A Case B

Problem 5.8 Determinetheplasticstrengthof theVierendeel trussshownbelow. All connections are moment resisting.

3@L = 3L

L

3P5P

1.5Mp

2Mp 2MpMp

1.5Mp 1.5Mp 1.5Mp

Mp

ReferencesCambridgeUniversityPress.1992.Numerical Recipes in Fortran. Cambridge, MA.Dantzig,G.B.1963.Linear Programming and Extensions.Princeton,NJ:Princeton

UniversityPress.Livesley, R. K. 1975. Matrix Methods of Structural Analysis, 2nd ed. Oxford, UK:

PergamonPress.


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