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FUNDAMENTAL STRUCTURAL ANALYSISJaroon Rungamornrat Method of Consistent Deformation

Copyright 2011 J. Rungamornrat

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CHAPTER 10

METHOD OF CONSISTENT DEFORMATION

A method of consistent deformation, also known as aflexibility methodor aforce method, is a well-

known technique widely used in the analysis of statically indeterminate structures. In this method, a

set of independent static quantities (e.g. support reactions and internal forces at certain locations)

that cannot be determined from static equilibrium equations is chosen as a set of primary unknowns

called redundants. An additional set of equations called compatibility equations is formed to

solve all such unknown redundants. After all redundants are resolved, the structure becomes

statically determinate and, as a consequence, all remaining static quantities can readily be

determined from static equilibrium equations and the displacement and rotation at any point within

the structure can be computed using several techniques discussed previously (e.g. method curvature

area, conjugate structure analogy, energy methods).

10.1 Basic Concept

To clearly demonstrate the basic concept underlying the method of consistent deformation, let

consider a statically indeterminate frame with the degree of static indeterminacy equal to 2, i.e. DI

= 2, and subjected to external loads as shown schematically in Figure 10.1.

Figure 10.1: Schematic of a statically indeterminate frame with DI = 2

For this particular structure, there are two extra static unknowns or two redundants that cannot be

determined by static equilibrium. Let RCX and RCY be support reactions at a pinned support at point

C. If both the support reactions RCX and RCY are known, the given structure now becomes statically

determinate and all other static quantities (remaining support reactions and all internal forces) can

readily be calculated from static equilibrium.

Now, let consider the following thought process. First, we remove the pinned support at the

point C from the original structure or, equivalently, release both the support reactions RCX and RCY.

The resulting structure, shown in Figure 10.2(a), is therefore statically determinate and is generally

termed as the primary structure. Next, we apply two forces R1 and R2, in addition to existing

external loads acting to the original structure, to the primary structure at point C in the direction of

the support reactions RCX and RCY, respectively. The resulting structure, shown in Figure 10.2(b), is

still statically determinate and is termed as the admissible structure. It is important to note that thetwo forces R1 and R2 up to this state can take any value. Since both the primary structure and the

X

Y

P

q

RCY

RCX

A

B C


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admissible structure are statically determinate, all support reactions and the internal forces can be

determined from static equilibrium and the displacement and rotation at any point within both

structures can readily be calculated from various techniques such as the method of curvature area,

the conjugate structure analogy, the principle of complementary virtual work, etc.

Figure 10.2: (a) Schematic of primary structure and (b) admissible structure of the original structure

shown in Figure 10.1

If the structure under consideration is linear, the method of superposition can be employed to

simplify the analysis procedure. For instance, any response of the admissible structure can be

obtained by superposing the response of the following structures: (i) the primary structure shown in

Figure 10.2(a), (ii) the structure subjected to the force R1 as shown in Figure 10.3(a), and (iii) the

structure subjected to the force R2 as shown in Figure 10.4(a). Let u1 and u2 be the displacement at

the point C in the direction of R1 and R2 of the admissible structure. From the method of

superposition, we then obtain

1211o11 uuuu (10.1)

2221o22 uuuu (10.2)

Figure 10.3: (a) Schematic of the released structure subjected to the force R1 and (b) schematic ofthe released structure subjected to a unit force in the direction of R1.

X

Y

P

q

(a)

X

Y

q

R2

R1

(b)

A

B C B C

u1o

u2o

A

X

Y

(a)

R1

(b)

A

B C

u11

u21

X

Y

1

A

B C

f11

f21


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Figure 10.4: (a) Schematic of the released structure subjected to the force R2 and (b) schematic of

the released structure subjected to a unit force in the direction of R2.

where u1o and u2o are the displacements at the point C of the primary structure shown in Figure

10.2(a) in the direction of the force R1 and the force R2, respectively; u11 and u12 are the

displacements at the point C of the structure shown in Figure 10.3(a) in the direction of the force R1

and the force R2, respectively; and u21 and u22 are the displacements at the point C of the structure

shown in Figure 10.4(a) in the direction of the force R1 and the force R2, respectively. It is worth

noting that calculation of uij (i, j = 1, 2) involves analyses of a statically determinate structure (a

statically stable released structure of the original structure shown in Figure 10.1) subjected either to

a force R1 or a force R2 one at a time. To further simplify such analysis, we exploit the linearity of

the structure and then analyze the structures subjected to a unit force shown in Figure 10.3(b) and

10.4(b) instead of the structures in Figure 10.3(a) and 10.4(a), respectively. The displacements u ij at

the point C can then be obtained, in terms of response of the structure subjected to the unit force, by

a linearity rule as

1212111111 Rfu;Rfu (10.3)

2222221212 Rfu;Rfu (10.4)

where f11 and f21 are the displacements at the point C of the released structure subjected to a unit

load in the direction of the force R1 (as shown in Figure 10.3(b)) in the direction of the force R1 and

the force R2, respectively; and f12 and f22 are the displacements at the point C of the released

structure subjected to a unit load in the direction of the force R2 (as shown in Figure 10.4(b)) in thedirection of the force R1 and the force R2, respectively. Again, fij (i, j = 1, 2) can be calculated from

any convenient method such as the method of curvature area, conjugate structure analogy, the

principle of complementary virtual work, etc.

By substituting (10.3) and (10.4) into the relations (10.1) and (10.2), we obtain the

displacements u1 and u2 at the point C of the admissible structure in the direction of the force R1

and the force R2, respectively, as

212111o11 RfRfuu (10.5)

222121o22 RfRfuu (10.6)

These two equations can, alternatively, be expressed in a matrix form as

X

Y

(a)

R1

(b)

A

B C

u12

u22

X

Y

1

A

B C

f12

f22


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Rfu R

R

ff

ff

u

u

u

uo

2

1

2221

1211

o1

o1

2

1

(10.7)

where uo is a vector containing components of the displacement at the point C of the primary

structure given by

o1

o1

ou

uu (10.8)

R is a vector containing arbitrary loads applied to the admissible structure at the locations and

directions of releases given by

2

1

R

RR (10.9)

and fis termed as aflexibility matrix of the released structure defined by

2221

1211

ff

fff (10.10)

with fij (i, j = 1, 2) denoting the flexibility coefficients. It is evident from equation (10.7) that the

displacement at the point C of the admissible structure shown in Figure 10.2(b) can equivalently be

obtained by performing analysis of a set of simpler structures (structures shown in Figure 10.2(a),

Figure 10.3(b), and Figure 10.4(b)) and following by employing linearity of the structure via the

method of superposition. Equivalence between the admissible structure and the set of simpler

structures is clearly illustrated by Figure 10.5. It is important to remark that other quantities or

responses of the admissible structure can also be determined using the method of superposition in

the same fashion.

As a final step in our thought process, we aim to make a connection between the statically

determinate admissible structure (shown in Figure 10.2(b)) and the statically indeterminate original

structure (shown in Figure 10.1). The key difference between the admissible structure and the

original structure is that the kinematical conditions at the released locations of the admissible

structure (the point C for this particular case) are, in general, not the same as those of the original

structure. This results from that the additional applied loads R1 and R2 in the admissible structure

can vary arbitrarily. For instance, the vertical and horizontal components of the displacement at thepoint C of the admissible structure are, in general, not equal to zero. For the admissible structure to

become the original structure, the applied loads R1 and R2 must be chosen to be identical to the

support reactions RCX and RCY of the original structure. Since both RCX and RCY are unknown a

priori, R1 and R2 must be chosen such that the kinematical conditions at the released locations of

the admissible structure are identical to those of the original structure. Let ur be a vector containing

the (prescribed) displacement components of the original structure at the released point. The

kinematical conditions or compatibility conditions for determining R1 and R2 are

2

1

2221

1211

o2

o1

r2

r1

orR

R

ff

ff

u

u

u

uRfuu (10.11)


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Figure 10.5: Equivalence between the admissible structure and a set of simpler structures

Equation (10.11) is known as the compatibility equation or the continuity equation. For the

special case when there is no support settlement taking place at the point C, i.e. ur = 0, and the

compatibility equation (10.11) reduces to

1o 11 12 1

o

2o 21 22 2

u f f R 0

u f f R 0

u f R 0 (10.12)

The forces R1 and R2 obtained by solving the compatibility equation (10.11) are therefore the

support reactions RCX and RCY of the original structure. The admissible structure with the known or

solved R1 and R2 is now identical to the original structure and, as a consequence, all quantities of

interest of the original structure can equivalently be obtained from the admissible structure or from

superposition of responses of a set of simpler structures as indicated by the correspondence shown

in Figure 10.5.

10.2 Choice of Released Structures

While the basic concept of the method of consistent deformation has already been demonstrated in

the previous section, two very important issues remain to be addressed. These issues are related to

the following questions: how many redundants to be removed from the original structure to obtain

a statically determinate released structure? and what is the necessary condition for the releasedstructure?

+ +R1 R2

X

Y

q

R2

R1B C

A

X

Y

P

q

A

B C

u1o

X

Y

1

A

B C

f11

f21

X

Y

1

A

B C

f12

f22

u2o

P


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To answer the first question, the degree of static indeterminacy of the given structure must

be determined and this number is in fact the number of all unknown redundants to be removed from

the original structure to render the resulting released structure statically determinate. By recalling

discussion in Chapter 1, the degree of static indeterminacy of a structure, denoted by DI, can readily

be computed from

cjma nnnrDI (10.13)

where ra is the number of all components of the support reactions, nm is the number of components

of the internal force of all members (e.g. the number of components of the internal force for an

individual truss, beam, and frame members is 1, 2, and 3, respectively), nj is the number of

independent equilibrium equations at all nodes (e.g. the number of independent equilibrium

equations for an individual node of truss, beam, and frame structures is 2, 2, and 3, respectively),

and nc is the number of static conditions associated with all internal releases present within the

structure. For instance, the degree of static indeterminacy of the frame structure shown in Figure

10.1 is DI = (3 + 2) + (3 2) (3 3) (0) = 2.

Figure 10.6: Examples of (a) primary structure and (b) admissible structures of original structure

shown in Figure 10.1

(a) (b)

R2

R1

X

Y q

A

B

C

R2R1 X

Y q

A

B

C

X

Y q

A

B

C

X

Y q

A

BC

R1X

Yq

A

B C

X

Yq

A

BC

R2

P P

PP

P P


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Response to the second question is somewhat more difficult. The technique requires only

that the original structure must be released to obtain a primary structure that is statically

determinate and, of course, must also be statically stable. Note that while the degree of static

indeterminacy or the number of redundants of a given structure can readily be determined, the

process of constructing the primary structure is nontrivial since a choice of redundants is not unique

and, at the same time, is not fully arbitrary. In fact they must be chosen such that the releasedstructure is statically stable; i.e. there is no development of rigid body motion of the entire structure

or any portion of the structure. Examples of valid sets of redundants, primary structures and

admissible structures of the original structure shown in Figure 10.1 are given, in addition to those

indicated in Figure 10.2, in Figure 10.6.

It is noted that a valid set of redundants is not necessary consisting of only components of

support reactions, the internal force such as bending moment, shear force, and axial force at certain

locations can also be chosen as redundants. For instance, the last primary structure shown in Figure

10.6 is obtained by removing the moment reaction at the point A and the bending moment at the

point B. Examples of invalid sets of redundants that produce statically unstable, released structures

are shown in Figure 10.7; these released structures cannot be used as the primary structures in the

analysis by the method of consistent deformation.

Figure 10.7: Examples of statically unstable released structures

10.3 Compatibility Equations for General Case

Consider a statically indeterminate structure with the degree of static indeterminacy DI = N (i.e. the

number of redundants is equal to N). Let R = {R1, R2, R3, , RN} be a proper set of redundants

chosen to construct the primary structure and let ur = {u1r, u2r, u3r, , uNr} be a vector of

kinematical conditions of the original structure where uir

(i = 1, 2, , N) denotes the prescribed

displacement (or prescribed rotation or prescribed internal discontinuity depending on the type of

the redundant) of the original structure at a location and in the direction of the released redundant

Ri. For the constructed primary structure (a structure resulting from releasing all the redundants R

and subjected to the same set of external applied loads as that for the original structure), let define

uo = {u1o, u2o, u3o, , uNo} as a vector of kinematical quantities of the primary structure where u io (i

= 1, 2, , N) denotes the displacement (or the rotation or the discontinuity condition) of the

primary structure at a location and in the direction associated with the released redundant Ri. Next,

consider a released structure subjected only to a unit value of the redundant Ri and let define the

flexibility coefficient fij (j = 1, 2, , N) as the displacement (or rotation or the discontinuity

condition) of this released structure at the location and in the direction associated with the released

redundant Rj. A set of compatibility equations necessary and sufficient for solving all the unknownredundants R is given by

X

Yq

A

B C

P

X

Yq

A

BC

P


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N

2

1

NN2N1N

N22221

N11211

No

o2

o1

Nr

r2

r1

R

R

R

fff

fff

fff

u

u

u

u

u

u

(10.14)

or, in a matrix form, by

or Rfuu (10.15)

where fis the flexibility matrix of the released structure given by

NN2N1N

N22221

N11211

fff

fff

fff

f (10.16)

10.3.1 Vector of kinematical conditions ur

A vector of kinematical conditions ur for a given statically indeterminate structure is known a

priori. Determination of this vector requires only the knowledge of a choice of redundants, support

conditions at locations and in directions where support reactions are chosen as redundants, and

continuity conditions at locations where the internal forces are chosen as redundants.

If the support reaction Ri is chosen as one of redundants, the corresponding kinematical

condition uir

vanishes if there is no support settlement in the direction of Ri

and uir

= if thesupport settlement takes place in the direction of Ri with an amount of. If the internal force Ri ischosen as one of redundants, the corresponding kinematical condition uir vanishes if there is no

discontinuity (e.g. relative rotation, gap, overlapping) taking place in the direction of Ri and uir = if the discontinuity takes place in the direction of Ri with an amount of . Note that suchdiscontinuity occurs only at a point where the internal constraint cannot completely be developed,

e.g. a flexible rotational point and an extensible point.

10.3.2 Vector of kinematical quantities uo

As evident from the definition described above, a vectoruo contains the displacement components,

rotations, or internal discontinuity (e.g. relative rotation, gap, overlapping) of the primary structureat the locations and in the directions of released redundants. Since the primary structure is statically

determinate, computation of the vectoruo can readily be achieved by using various techniques such

as the method of curvature area, the conjugate structure analogy, and the principle of

complementary virtual work or the unit load method once the static analysis for the internal forces

of such structure is completed.

10.3.3 Flexibility matrix f

The flexibility coefficient fij can readily be computed as follow: (i) obtain the released structure by

releasing all the redundants and removing all external applied loads applied from the original

structure, (ii) apply a unit value of the redundant Ri to the released structure, (iii) compute thedisplacement (or rotation or internal discontinuity) at the location and in the direction of the


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released redundant Rj by using various techniques similar to those used to compute uo, and (iv) set

fij equal to the quantity computed from (iii). This process is repeated until all coefficients fij are

obtained.

The flexibility matrix f of the released structure possesses following two important

properties: symmetry and positive definiteness. The former significantly reduces the computational

effort and storage while the latter implies that the flexibility matrix f is nonsingular or invertibleand, as a consequence, the system of linear algebraic equations (10.14) possesses a unique solution.

To prove these two properties, let consider a released structure obtained by releasing all the

redundants and removing all external applied loads from the original structure. For convenience, let

define the location and the direction of the released redundant Ri of the released structure as the ith

degree of freedom. Next, let the released structure be subjected to a set of loads P = {P1, P2, P3, ,

PN} where Pi is a concentrated load acting at the ith degree of freedom. This process is assumed to

be continuous in the sense that the loads P increases continuously from zero to their final values

{P1, P2, P3, , PN}. The corresponding kinematical quantity (e.g. displacement, rotation, internal

discontinuity) at the ith degree of freedom due to the loads P is denoted by ui and it can readily be

computed by a method of superposition, i.e.

Pfu (10.17)

where f is the flexibility matrix of the released structure given by (10.16) and u = {u1, u2, u3, ,

uN}. Note that the flexibility coefficient fij can be viewed as the value of kinematical quantity at the

ith

degree of freedom due to a unit load acting only at the jth

degree of freedom. The total work done

W due to the loads P for the entire process is given, for a linearly elastic structure, by

PfPuP 2

1

2

1W

TT (10.18)

Since the work done W is a scalar quantity, taking the transpose of (10.18) yields the identical

result, i.e.

PfPPuTTT

2

1

2

1W (10.19)

Combining (10.18) and (10.19) leads to

0PffP TT (10.20)

Since (10.20) is valid for an arbitrary set of loads P, it can be deduced that

jiij

T ff ff (10.21)

or the flexibility matrix is essentially symmetric. The property (10.21) is also known as the

Maxwells reciprocal theorem. This theorem simply states that the value of kinematic quantity at

the ith

degree of freedom due to a unit load acting at the jth

degree of freedom is equal to the value

of kinematic quantity at the jth

degree of freedom due to a unit load acting at the ith

degree of

freedom. To clearly demonstrate the theorem, let consider two identical simply-supported beams as

shown in Figure 9.8. One of the beams is subjected to a unit (vertical) load at a point 1 and the

other is subjected to a unit (vertical) load at a point 2. From Maxwells reciprocal theorem, it can bededuced that the (vertical) deflection at the point 2 due to the unit (vertical) load applied at the


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point 1, f21, is equal to the (vertical) deflection at the point 1 due to the unit (vertical) load applied

at the point 2, f12.

Figure 10.8: Schematic of two identical beams subjected to a unit load applied at different locations

Next, lets consider two identical rigid frames as shown in Figure 10.9. One of the frames is

subjected to a unit horizontal load at a point 1 and the other is subjected to a unit vertical load at a

point 2. From Maxwells reciprocal theorem, it can be deduced that the vertical displacement at the

point 2 due to the unit horizontal load applied at the point 1, f21, is equal to the horizontal deflection

at the point 1 due to the unit vertical load applied at the point 2, f12.

Figure 10.9: Schematic of two identical rigid frames subjected to a unit load applied at different

locations

Similarly, lets consider the same frames as shown in Figure 10.10; one of them is subjected to a

unit horizontal load at a point 1 and the other is subjected to a unit moment at a point 2. From

Maxwells reciprocal theorem, it implies that the rotation at the point 2 due to the unit horizontal

load applied at the point 1, f21, is equal to the horizontal deflection at the point 1 due to the unit

moment applied at the point 2, f12.

Figure 10.10: Schematic of two identical identical frames subjected to a unit load applied atdifferent locations

1

2

f21

P = 1

P = 1

1 f12

2

P = 1

1

2 1

2f21 f12

P = 1

1

2

f21

P = 1

21 f12

M = 1


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Since the primary structure is statically stable, any set of non-zero loads P produces a set of

non-zero displacements u such that the external work done is always greater than zero, i.e.

02

1W T uP (10.22)

By employing the relation (10.17) along with (10.22), we then obtain

0T PfP (10.23)

for every set of non-zero loads P. This completes the proof of positive definiteness of the flexibility

matrix. The positive definiteness of f implies that all its eigen values are positive. Now, by

choosing a particular set of applied loads P = {P1 = 0, P2 = 0, Pi-1 = 0, Pi = 1, Pi+1 = 0, , PN =

0}, the condition (10.23) implies that

0fiiT

PfP (10.24)

That is all diagonal entries must be positive.

Example 10.1 Lets consider a statically indeterminate truss shown below. Properties of each

member (e.g. cross sectional area, Young modulus, and coefficient of thermal expansion) are given

in a table below. Analyze this structure for all support reactions and member forces when it is

subjected to the following four effects: (i) applied loads P and 3P at joints 2 and 4, respectively, (ii)

increase in temperature T in a member 35, (iii) the length of a member 13 is eo longer than L dueto fabrication error, and (iv) a pinned support at a joint 1 is subjected to the leftward settlement s.In addition, determine the vertical displacement at a joint 4.

Members Area, iA Young modulus, iE Coefficient of thermal expansion, i

13, 35 2A E

24, 46 2A E

12, 34, 56 A E

23, 36 A E

L

L

2

L

1

3P

P4

3

6

5


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Solution First, we determine the degree of static indeterminacy or the number of redundants, DI =

(2 + 2) + (9 1) (62) = 13 12 = 1. Thus, the structure is statically indeterminate with degree ofstatic indeterminacy equal to 1. Next, we choose a horizontal reaction at a joint 5, R1, as a redundant

and the corresponding primary structure, denoted by structure 0, is given in the figure below. It is

worth noting that the primary structure is subjected to the same loading conditions, temperature

change, error from fabrication, and support settlement as those for the original structure except thatthe redundant is released. Consider also the other system used for computing the flexibility matrix

f, a released structure subjected only to a unit load at the location and in the direction of the

redundant denoted by structure I, as shown in the figure below.

Since both the structure 0 and the structure I are statically determinate, all support reactions and

member forces can readily be calculated from static equilibrium and the method of joints or the

method of sections. Results are reported in above figures. To determine the redundant R1, it is

required to set up the following compatibility equation

111o1r1 Rfuu (10.25)

Since there is no settlement in the horizontal direction of the support 5 of the original structure, then

u1r which is the horizontal displacement at the joint 5 of the original structure vanishes, i.e. u1r = 0.

The horizontal displacement u1o at the joint 5 of the primary structure can readily be computed by

using the principle of complementary virtual work. By choosing the structure I as the virtual

structure, u1o can be obtained as follows:

MemberiA iE i iL

0

iF iT ieI

iF

ii

i

I

i

0

i

EA

LFF

0

iiii FTL 0

iiFe

13 2A E L P 0 oe 1

2AE

PL

0oe

35 2A E L 0 T 0 1 0 LT 0

24 2A E L -2P 0 0 0 0 0 0

46 2A E L -2P 0 0 0 0 0 0

2

1

3P

P4

3

6

5

2PP

P(0)(+P)

(-2P)

( 22 P)

(-3P)

( 2 P)

(-P)

(-2P)(-2P)

Structure 0

Primary structure

2

11

4

3

6

5

00

1(+1)(+1)

(0)(0)(0)(0)(0)

(0)(0)

Structure I

Released structure loaded by unit redundant


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12 A E L -P 0 0 0 0 0 0

23 A E L2 P2 0 0 0 0 0 0

34 A E L -3P 0 0 0 0 0 0

36 A E L2 P22 0 0 0 0 0 056 A E L -2P 0 0 0 0 0 0

2AE

PL LT oe

s1os1oC u1u1W

m

1i

ii

m

1i

iiii

m

1i ii

iiiC FeFTL

EA

LFFU

2AE

PL+ TL oe

CC UW 1ou2AEPL + TL oe s

The flexibility coefficient f11 or the horizontal displacement at the joint 5 of the structure I can also

be computed from the principle of complementary virtual work by choosing the structure I itself as

the virtual structure. Details of calculations are given below.

MemberiA iE i iL

I

iFI

iF

ii

i

I

i

I

i

EA

LFF

13 2A E L 1 1

2AE

L

35 2A E L 1 1

2AE

L

24 2A E L 0 0 0

46 2A E L 0 0 0

12 A E L 0 0 0

23 A E L2 0 0 0

34 A E L 0 0 0

36 A E L2 0 0 056 A E L 0 0 0

AE

L

1111C ff1W

m

1i ii

iiiC

EA

LFFU

AE

L

CC UW 11fAE

L


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By substituting u1r, u1o, and f11 into the compatibility equation (10.25), the redundant R1 can be

solved as follows:

1so RAE

LeTL

2AE

PL0

L

AE

L

eAETAE

2

PR so1

After the redundant R1 was solved, the response of the original structure can be obtained by

superposing the response of the primary structure and the response of the structure I multiplied by

the computed redundant R1. Thus, the remaining support reactions of the original structure, in

addition to R1, are given by

L

AE

L

eAETAE

2

PR)1(PR so1x1

; PR)0(PR 1y1 ; P2R)0(P2R 1y5

All member forces are given by

Member Member forcesPrimary structure Member forcesStructure I Member forcesOriginal structure

13 P 1L

AE

L

eAETAE

2

P so

35 0 1L

AE

L

eAETAE

2

P so

24 -2P 0 -2P

46 -2P 0 -2P

12 -P 0 -P23 P2 0 P2

34 -3P 0 -3P

36 P22 0 P22

56 -2P 0 -2P

Similarly, the vertical displacement at the joint 4 of the original structure, u4y, is equal to the sum of

the vertical displacement at the joint 4 of the primary structure, u4yo, and the vertical displacement

at the joint 4 of the structure I, u4yI, multiplied by the redundant R1. To proceed, we again employ

the principle of complementary virtual work (or the unit load method) with a virtual structure

shown below.

2

1

1

4

3

6

5

1/21/2

0

(0)(0)

(-1/2)

( 2/2 )

(-1)(-1/2)

(-1/2)(-1/2)

( 2/2 )


15/42



447

MemberiA iE

i

iL

0

iF iT

ie iF

ii

ii

0

i

EA

LFF

iiii FTL

iiFe

13 2A E L P 0oe 0 0 0 0

35 2A E L 0 T 0 0 0 0 0

24 2A E L -2P 0 0 -1/22AE

PL 0 0

46 2A E L -2P 0 0 -1/22AE

PL 0 0

12 A E L -P 0 0 -1/22AE

PL 0 0

23 A E L2 P2 0 0 /22 AE

PL2 0 0

34 A E L -3P 0 0 -1AE

PL3 0 0

36 A E L2 P22 0 0/22

AE

PL22 0 0

56 A E L -2P 0 0 -1/2AE

PL 0 0

AE

PL23

2

11

0 0

The displacement u4yo is obtained as follow

4yo4yoC u0u1W

m

1i

ii

m

1i

iiii

m

1i ii

iiiC FeFTL

EA

LFFU

AE

PL23

2

11

CC UW 4you AEPL23

211

The displacement u4yI is obtained is obtained in a similar fashion as follows:

4yI4yIC uu1W

m

1i ii

iiiC

EA

LFFU 0

CCU

W

4yIu 0


16/42



448

MemberiA iE i iL

I

iF iF

ii

ii

I

i

EA

LFF

13 2A E L 1 0 0

35 2A E L 1 0 0

24 2A E L 0 -1/2 0

46 2A E L 0 -1/2 0

12 A E L 0 -1/2 0

23 A E L2 0 /22 0

34 A E L 0 -1 0

36 A E L2 0 /22 0

56 A E L 0 -1/2 0

0

Therefore, the vertical displacement at the joint 4 of the original structure, u4y, is equal to

4y 4yo 4yI 1

11 PL 11 PLu u u R 3 2 0 3 2

2 AE 2 AE

Downward

It is important to emphasize that the virtual structure chose for computing the displacement of a

statically indeterminate truss is not necessary to be of identical geometry to the original truss but

can be its (statically determinate and statically stable) released structure. Use of this statically

determinate structure in the analysis significantly reduces the computational effort and avoids

solving another statically indeterminate structure.

Example 10.2 Determine all support reactions and member forces of a statically indeterminate truss

subjected to external applied loads as shown in the figure below. Given that the axial rigidity EA is

constant throughout.

L

L

2

L

1

3P

P4

3

6

5


17/42



449


(2 + 1) + (111) (62) = 14 12 = 2. Thus, the structure is statically indeterminate with degreeof static indeterminacy equal to 2. Next, we choose the internal force of the member 23, denoted by

F1, and the internal force of the member 36, denoted by F2, as two redundants and the corresponding

primary structure obtained by releasing the two redundants via cutting the two members 23 and 36

and denoted by a structure 0, is given in the figure below. Consider also the other two systemsused for computing the flexibility matrix f: one associated with a released structure subjected to a

pair of unit and opposite forces at the cut of the member 23, denoted by a structure I, and the

other associated with a released structure subjected to a pair of unit and opposite forces at the cut of

the member 36, denoted by a structure II.

Since all three structures, the structure 0, the structure I and the structure II, are statically

determinate, all support reactions and member forces can readily be calculated from static

equilibrium and the method of joints or the method of sections. Results are reported in the figures.

To determine the redundants F1 and F2, we need to set up the following compatibility equations

2

1

2221

1211

o2

o1

r2

r1

F

F

ff

ff

u

u

u

u(e10.2.1)

2PP

P(2P)

( 2 P)

(0)

(0)

(0)

(0)

(0)(-P)

(2P)

(0)

( 22 P)

2

1

3P

P4

3

6

5

Structure 0

Primary structure

( 2/1 )

(1)

( 2/1 )

(1)

(0)

(0)

(0)( 2/1 )

(0)

(0)11

( 2/1 )

00

0

2

1

4

3

6

5

Structure I


18/42



450

Since there is no gap and overlapping of the members 23 and 36 in the original structure, then u 1r=

u2r = 0. The overlapping generated at the member 23, u1o, and the overlapping generated at themember 36, u2o, of the primary structure (the structure 0) can be computed using the principle of

complementary virtual work with the structure I and structure II be chosen as the virtual structures,

respectively.

MemberiA

iE iL

0

iF I

iF II

iF

ii

i

I

i

0

i

EA

LFF

ii

i

II

i

0

i

EA

LFF

13 A E L 2P 2/1 0AE

PL2 0

35 A E L 2P 0 2/1 0AE

PL2

24 A E L -P 2/1 02AE

PL2 0

46 A E L 0 0 2/1 0 0

12 A E L 0 2/1 0 0 0

14 A E L2 P2 1 0AE

2PL 0

23 A E L2 0 1 0 0 0

34 A E L 0 2/1 2/1 0 0

45 A E L2 P2 0 1 0AE

2PL

36 A E L2 0 0 1 0 0

56 A E L 0 0 2/1 0 0

AEPL222

AEPL22

( 2/1 )

(1)

( 2/1

)

(1)

(0)

(0)

(0) ( 2/1 )

(0)

(0)

11

( 2/1

)

00

0

2

1

4

3

6

5

Structure II


19/42



451

CC UW AE

PL

2

22

EA

LFFu

11

1i ii

i

I

i

0

i1o

CC UW AE

PL22

EA

LFFu

11

1i ii

i

II

i

0

i2o

Note that the minus sign of u1o and u2o indicates that a gap occurs for both the member 23 and the

member 36 in the original structure.

The flexibility coefficient fij can also be computed from the principle of complementary

virtual work by properly choosing a pair of actual and virtual structure from the structure I and the

structure II as shown below. By using the symmetric property of the flexibility matrix, only three

flexibility coefficients f11, f12 and f22 needs to be computed. In particular, the entry f11 is obtained by

choosing the structure I as the actual and virtual structures, the entry f22 is obtained by choosing the

structure II as the actual and virtual structures, and the entry f12 is obtained by choosing the

structure I as the actual structure and the structure II as the virtual structure. Details of calculation

are shown below.

MemberiA

iE iL

I

iF II

iF

ii

i

I

i

I

i

EA

LFF

ii

i

II

i

I

i

EA

LFF

ii

i

II

i

II

i

EA

LFF

13 A E L 2/1 02AE

L 0 0

35 A E L 0 2/1 0 02AE

L

24 A E L 2/1 0

2AE

L 0 0

46 A E L 0 2/1 0 02AE

L

12 A E L 2/1 02AE

L 0 0

14 A E L2 1 0AE

L2 0 0

23 A E L2 1 0AE

L2 0 0

34 A E L 2/1 2/12AE

L

2AE

L

2AE

L

45 A E L2 0 1 0 0AE

L2

36 A E L2 0 1 0 0AE

L2

56 A E L 0 2/1 0 02AE

L

AEL222

2AEL AEL222


20/42



452

CC UW AE

L222

EA

LFFf

11

1i ii

i

I

i

I

i11

CC UW 2111

1i ii

i

II

i

I

i12 f

2AE

L

EA

LFFf

CC UW AE

L222

EA

LFFf

11

1i ii

i

II

i

II

i22

By substituting u1r, u1r, u1o, u2o, and finto the compatibility equations (e10.2.1), the two redundants

F1 and F2 can be solved as follows:

2

1

F

F

2222/1

2/1222

AE

L

22

2/22

AE

PL

0

0

1

2

F 20 18 2P

F 47 32 2 28 31 2

0.4927P

0.7787

Once the redundants F1 and F2 were solved, the responses of the original structure can be obtained

by superposing the responses of the primary structure, the responses of the structure I multiplied by

F1 and the responses of the structure I multiplied by F2. All support reactions of the original

structure are given by

PF)0(F)0(PR 21x1 ; PF)0(F)0(PR 21y1 ; P2F)0(F)0(P2R 21y5

Similarly, all member forces are given in the table below.

Member 0iF

I

iF II

iF )P7787.0(F)P4927.0(FFFII

i

I

i

0

ii

132P 2/1 0 1.6516P

352P 0 2/1 1.4494P

24 -P 2/1 0 -1.3484P

46 0 0 2/1 -0.5506P

12 0 2/1 0 -0.3484P

14 P2 1 0 -0.9215P

23 0 1 0 0.4927P

34 0 2/1 2/1 -0.8990P

45 P2 0 1 -0.6355P

36 0 0 1 0.7787P

56 0 0 2/1 -0.5506P


21/42



453

Example 10.3 Consider a statically indeterminate beam as shown in the figure below. The flexural

rigidity EI is assumed to be constant throughout. The beam is subjected to a uniformly distributed

load 2q on the segment AB and a concentrated load qL at a point C and, during load application, the

support at B settles downward with an amount of o. Determine all support reactions, sketch theSFD and BMD for the case that o = 0, and compute the (vertical) deflection at point C. Consider

only the bending effect in the analysis.


(2 + 1) + (22) (32) = 7 6 = 1. Thus, the structure is statically indeterminate with degree ofstatic indeterminacy 1. Next, we choose a moment reaction at point A, M1, as a redundant and the

corresponding primary structure, denoted by structure 0, is given in the figure below. Consider

also a released structure (used for computing the flexibility matrix f) subjected only to a unit

moment at point A, denoted by structure I, as shown in the figure below.

Since both the structure 0 and the structure I are statically determinate, all support reactions and the

bending moment diagram can readily be obtained from static equilibrium with results given in

above figures. To determine the redundant M1, we need to set up the following compatibility

equation

111o1r1 Mfuu (e10.3.1)

Since there is no support rotation at the fixed support, then u1r = 0. The rotation at the point A, u1o,

of the primary structure (structure 0) can readily be computed using the principle of complementary

virtual work with the structure I being chosen as the virtual structure as follows:

2L L

A

qL

B

C

2q

A

qL

B

C

2q

3qL/2 7qL/2

M0

-qL2

3qL2

B

C

1/2L 1/2L

-4qL2

1

MI

-1

Structure 0 Structure I


22/42



454

L2/u)(1/2Lu1W o1oo1oC

BC

C

AB

CC UUU dxEI

MMdx

EI

MMBCAB L

0

I0L

0

I0

3EIqL

04

12LEI

4qL

3

1

3

12LEI

3qL

2

1 322

CC UW 1ouL23EI

qL o3

CW

The flexibility coefficient f11 can also be computed from the principle of complementary virtualwork by choosing the structure I as both the actual and virtual structures as shown below.

1111C ff1W

BC

C

AB

CCU

U

U dxEI

MM

dxEI

MM BCABL

0

IIL

0

II

3EI2L

3

2

2LEI

1

2

1

CC UW 11f3EI

2LCCW

By substituting u1r, u1o, and f11 into the compatibility equation (e10.3.1), the redundant M1 can be

solved as follows:

1o

3

M3EI

2L

L23EI

qL0

2

o

2

1L4

EI3

2

qLM

CCW

After the redundant M1 was solved, the response of the original structure can be obtained bysuperposing the responses of the primary structure and the responses of the structure I multiplied by

M1. Thus, the remaining support reactions of the original structure are given by

3

o

2

o

2

AyL8

EI3

4

qL7

L4

EI3

2

qL

L2

1

2

qL3R

;

3

o

2

o

2

ByL8

EI3

4

qL13

L4

EI3

2

qL

L2

1

2

qL7R

The shear force diagram and bending moment diagram of the entire beam is shown below.

A

qL

BC

2q

RAy = 7qL/4

RBy = 13qL/4

BMD

qL

qL2

M1 = qL2/2

SFD

7qL/4

9qL/417qL2/64

qL2/2


23/42



455

The deflection at the point C of the original structure, vC, is equal to the sum of the deflection at the

point C of the primary structure, vCo, and the deflection at the point C of the structure I, vCI,

multiplied by the redundant M1. To compute such displacements, the principle of complementary

virtual work is employed with a special choice of the virtual structure shown below.

The displacement vCo is obtained as follows

2/3v)(3/2v1W oCooCoC

BC

C

AB

CC UUU dxEI

MMdx

EI

MMBCAB L

0

0L

0

0

3EI

qL

3

2LL

EI

qL

2

1

4

3L2L

EI

4qL

3

1

3

2L2L

EI

3qL

2

1 4222

CC UW Cov 4 3qL3EI 2

o Downward

The displacement vCI is obtained as follows

CICIC vv1W

BC

C

AB

CC UUU dxEI

MMdx

EI

MM BCABL

0

IL

0

I

3EI

L

3

L2L

EI

1

2

1 2

CC UW CIv 2L3EI

Downward

Therefore, the vertical displacement at the point 4 of the original structure, v C, is equal to

4 2 2

oC Co CI 1 2

3 3EIqL L qLv v v M

3EI 2 3EI 2 4L

o

4 7qL

2EI 4

o Downward

It is important to emphasize again that the virtual structure chosen for computing the deflection or

rotation of a statically indeterminate beam is not necessary to be of identical geometry to the

original beam but can be its (statically determinate and statically stable) released structure. Use of

this statically determinate structure in the analysis significantly reduces the computational effort.

1/2 3/2

M

L

A

B

C

P = 1

Virtual Structure


24/42



456

Example 10.4 Determine all support reactions and sketch the SFD and BMD of a statically

indeterminate beam subjected to uniformly distributed load q shown below. The flexural rigidity of

the segments AB and BC is given by EI and 2EI, respectively. In the analysis, lets consider only

the bending effect.

Solution First, we determine the degree of static indeterminacy of the beam: DI = (2 + 1 + 1) +

(22) (32) = 8 6 = 2. Thus, the structure is statically indeterminate with degree of staticindeterminacy equal to 2. Next, we choose a moment reaction M1 at point A and the bending

moment M2 at point B as redundants and the corresponding primary structure, denoted by astructure 0, is given in the figure below. Consider also the other two systems used for computing

the flexibility matrix f: one associated with a released structure subjected to a unit moment at a

point A, denoted by a structure I, and the other associated with a released structure subjected to a

pair of unit and opposite moments at a point B, denoted by a structure II.

L L

A

B C

q

A B C

q

qL/2 qL/2qL

qL2/2

qL2/2

M0

1

MI

1

MII

A B C

1/L 01/L

1


A

B

C

1/L 01/L

11

Structure II


25/42



457

Since all three structures (i.e. the structure 0, the structure I and the structure II) are statically

determinate, all support reactions and bending moment diagram can readily be obtained from static

equilibrium with results reported in above figures. To determine the redundants M 1 and M2, we

need to set up the following compatibility equations

2

1

2221

1211

o2

o1

r2

r1

M

M

ff

ff

u

u

u

u(e10.4.1)

Since there is no support rotation at the point A and no rotational discontinuity at point B in the

original structure, then u1r = u2r = 0. The rotation at the point A, u1o, and the relative rotation at the

hinge point B, u2o, of the primary structure (structure 0) can be computed using the principle of

complementary virtual work with the structure I and structure II be chosen as the virtual structure,

respectively. Details of calculations are given below.

Computation of u1o: the structure 0 is treated as the actual structure whereas the structure I is chosen

as the virtual structure

1o1oC uu1W

BC

C

AB

CC UUU dxEI

MMdx

EI

MM BCABL

0

I0L

0

I0

24EI

qL0

4

1L

2EI

qL

3

1

3

1L

2EI

qL

2

1 322

CCU

W

1ou 24EI

qL3

CW

Computation of u2o: the structure 0 is treated as the actual structure whereas the structure II is

chosen as the virtual structure

2o2oC uu1W

BC

C

AB

CC UUU dxEI

MMdx

EI

MM BCABL

0

II0L

0

II0

12EIqL

)2(4

3

L2EI

qL

3

1

)2(3

2

L2EI

qL

2

1 322

CC UW 2ou12EI

qL3 CW

Note that the minus sign of u1o and u2o indicates that the point A of the primary structure rotates in a

clockwise direction and the relative rotation at the point B of the primary structure is in THE

clockwise direction.

The flexibility coefficient fij can also be computed from the principle of complementary

virtual work by properly choosing a pair of actual and virtual structures from the structure I and the

structure II as shown below. By using the symmetric property of the flexibility matrix, only threeflexibility coefficients f11, f12 and f22 needs to be computed.


26/42



458

Computation of f11: the structure I is treated as the actual and virtual structures

1111C ff1W

BC

C

AB

CCU

U

U dxEI

MM

dxEI

MM BCABL

0

IIL

0

II

3EIL

3

2

LEI

1

2

1

CC UW 3EI

Lf11

Computation of f12: the structure I is treated as the actual structure whereas the structure II is chosen


1212C ff1W

BCC

ABCC UUU dxEI

MMdxEIMM

BCAB L

0

IIIL

0

III

6EIL31LEI121

CC UW 2112 f6EI

Lf

Computation of f22: the structure II is treated as the actual and virtual structures

2222C ff1W

BC

C

AB

CC UUU dxEI

MM

dxEI

MMBCAB L

0

IIIIL

0

IIII

3EI2L

)2(3

2

LEI

1

2

1

CC UW 2222 f3EI

2Lf

By substituting u1r, u1r, u1o, u2o, and finto the compatibility equations (e10.4.1), the two redundants

M1 and M2 can be solved as follow:

2

13

M

M

41

12

EI6

L

2

1

EI24

qL

0

0

3

2

28

qL

M

M 2

2

1

Once the redundants M1 and M2 were solved, the responses of the original structure can be obtained

by superposing the responses of the primary structure and the responses of the structure I multipliedby M1 and the responses of the structure II multiplied by M2. Thus, all support reactions of the

original structure are given by

28

qL13

28

3qL

L

1

14

qL

L

1

2

qLR

22

Ay

;

14

qLMM

2

1A

7

qL8

28

3qL

L

2

14

qL

L

1

qLR

22

By

; 28qL11

28

3qL

L

1

14

qL

02

qL

R

22

Cy


27/42



459

The total shear force diagram (SFD) and the total bending moment diagram (BMD) of the original

structure are given below.

Example 10.5 Consider a statically indeterminate rigid frame as shown in the figure below. Giventhat EI for the vertical and horizontal segments are 2EI and EI, respectively. The frame is subjected

to external loads as shown in the figure and, during load applications, the roller support at a point C

settles downward with an amount ofo. Determine all support reactions, sketch the SFD and BMDfor the case that o = 0, and compute the rotation at point C. Consider only deformation due tobending.

Solution First, we determine the degree of static indeterminacy of the given rigid frame: DI = (3 +

1) + (23) (33) = 10 9 = 1. Thus, the structure is statically indeterminate with degree of static

indeterminacy equal to 1. Next, we choose the support reaction of the roller support at the point C,denoted by R1, as a redundant and the corresponding primary structure, denoted by a structure 0,

BMDqL

2/14

SFD

RAy = 13qL/28

MA = qL2/14

RBy = 8qL/7 RCy = 11qL/28

13qL/28

15qL/28

17qL/28

11qL/28

3qL2/28

57qL2/1568

121qL2/1568

L L

A B C

q

L/2

L

qL

q

L/2

A

B

C


28/42



460

is given in the figure below. Consider also a released structure (used for computing the flexibility

matrix f) subjected only to a unit load at the point C, denoted by a structure I, as shown in thefigure below.

Since both structure 0 and structure I are statically determinate, all support reactions and thebending moment diagrams can readily be obtained from static equilibrium with results reported in

above figures. To determine the redundant R1, we need to set up the following compatibility

equation

111o1r1

Rfuu (e10.5.1)

Since there is downward settlement at the roller support equal to o, then u1r = o. The verticaldisplacement at the point C, u1o, of the primary structure (structure 0) can readily be computedusing the principle of complementary virtual with the structure I be chosen as the virtual structure as

follows:

1o1oC u0u1W

BC

C

AB

CC UUU dxEI

MMdx

EI

MMBCAB L

0

I0L

0

I0

16EI5qL

6L5

2L

2EIqL

21LL

4EIqL

31LL

4EIqL

21

4222

qL

qL2/2

qL2/2

qL2

qL2qL

A

BC

L

L

B C

qL

q

A

B

C

1

L0

1

A

B

C

A


M

0

M

I


29/42



461

CC UW 41o 5qLu16EI

Downward

The flexibility coefficient f11 can also be computed from the principle of complementary virtual

work by choosing the structure I as both the actual and virtual structures as shown below.

1111C ff1W

BC

C

AB

CC UUU dxEI

MMdx

EI

MMBCAB L

0

IIL

0

II

6EI5L

3

2LL

EI

L

2

1LL

2EI

L 3

CC UW 11f 35L6EI

UpwardBy substituting u1r, u1o, and f11 into the compatibility equation (e10.5.1), the redundant R1 can be

solved as follows:

134

o R6EI

5L

16EI

5qL

3

o1

L5

EI6

8

3qLR

After the redundant R1 was solved, the response of the original structure can be obtained by

superposing the response of the primary structure and the response of the structure I multiplied byR1. Thus, all support reactions of the original structure are given by

oAx 36EI3qL

R qL 0 qL8 5L

Leftward

o oAy 3 36EI 6EI3qL 5qLR qL 1 8 5L 8 5L

Upward

2

o

2

3

o2

AL5

EI6

8

5qL

L5

EI6

8

3qLLqLM

CCW

oCy 1 3

6EI3qLR R

8 5L

The total shear force diagram (SFD) and the total bending moment diagram (BMD) of the original

structure for the case that o = 0 are given below.

C

3qL/8

qL

qL2

qL

qL

q

A

B

C

3qL2/16

qL2/8

qL2/8

A

B

5qL2

/8

qL

5qL/8

3qL/8

A

B

C

SFD BMD


30/42



462

Similarly, the rotation at the point C of the original structure, C, is equal to the sum of the rotationat the point C of the primary structure, Co, and the rotation at the point C of the structure I, CI,multiplied by the redundant R1. To proceed, we again employ the principle of complementary

virtual work with a virtual structure shown below.

The rotation Co is obtained as follows:

CoCoC 1W

BC

C

AB

CC UUU dx

EI

MMdx

EI

MMBCAB L

0

0L

0

0

3EI

qL1

2

L

2EI

qL

2

11L

4EI

qL

3

11L

4EI

qL

2

1 3222

CC UW Co 3EI

qL3

The rotation CI is obtained as follows:

CICIC 1W

BC

C

AB

CC UUU dxEI

MMdxEI

MM BCAB L

0

IL

0

I

EIL1L

EI

L

2

11L2EI

L 2

CC UW CI EI

L2 Therefore, the rotation at the point C of the original structure, C, is equal to

5L

6

24EI

qL

5L

6EI

8

3qL

EI

L

3EI

qLR 0

3

3

0

23

1CIC0C

The rotation at the point C is in the counter clockwise direction if C > 0; otherwise, it is in theclockwise direction.

0

1

0

1

A

BC

1

1B

C

A

Virtual Structure M


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463

Example 10.6 Consider a statically indeterminate rigid frame as shown in the figure below. Given

that E, G, A, I, are throughout. The frame is subjected to external loads as shown in the figurebelow and, during load applications, the roller support at a point C settles downward with an

amount of o. Choose a proper set of redundants and then set up the corresponding set ofcompatibility equations to solve for those redundants. In the analysis, lets consider all possible

effects.

Solution First, we determine the degree of static indeterminacy of the given frame as follow: DI =(3 + 1 + 1) + (33) (43) = 14 12 = 2. Thus, the structure is statically indeterminate withdegree of static indeterminacy equal to 2. Next, we choose a reaction R1 at the point C and areaction R2 at the point D as redundants and the corresponding primary structure, denoted by a

structure 0, is given in the figure below. Consider also the other two systems used for computingthe flexibility matrix f: one associated with a released structure subjected to a unit vertical load atthe point C, denoted by a structure I, and the other associated with a released structure subjected

to a unit vertical load at the point D, denoted by a structure II.Since all three structures (i.e. the structure 0, the structure I and the structure II) are

statically determinate, all support reactions, the axial force diagram, the shear force diagram and thebending moment diagrams can readily be obtained from static equilibrium with results reported inthe figures below. To determine the redundants R1 and R2, we need to set up the following

compatibility equations

L

qL

A

B

CL

L

D

q

qL

A

B C D

qL

qL

3qL2/2V

0M

03qL2

/2

qL

V0

M0

qL

qL

qL

qL2/2

qL2/2

qL

Structure 0


32/42



464

2

1

2221

1211

o2

o1

r2

r1

R

R

ff

ff

u

u

u

u(e10.6.1)

Since the roller support at the point C settles downward with an amount of o and there is nosupport settlement of the roller support at the point D in the original structure, then u1r= o and u2r= 0.

The vertical displacement at the point C, u1o, and the vertical displacement at the point D,

u2o, of the primary structure (structure 0) can be computed using the principle of complementary

virtual work with the structure I and structure II be chosen as the virtual structure, respectively.Details of calculations are given below.

VI

MI

0

VI

MI

1

L

L

1

A

B

C

D

1

0

L

1

00

Structure I

VIIM

II

0

VII

MII

1

2L

2L

1

A

B C

D

1

0

2L

1

00

Structure II


33/42



465

Computation of u1o: the structure 0 is treated as the actual structure whereas the structure I is chosen


1o1oC uu1W

CsCaCbC UUUU

CD

Cb

BC

Cb

AB

CbCb UUUU dxEI

MMdx

EI

MMdx

EI

MMCDBCAB L

0

I0L

0

I0L

0

I0

8EI

9qL0

4

L3L

2EI

qL

3

1LL

EI

2qL

2

1 422

CD

Ca

BC

Ca

AB

CaCa UUUU dxEA

FFdx

EA

FFdx

EA

FFCDBCAB L

0

I0L

0

I0L

0

I0

EA

qL001L

EA

qL 2

CD

Cs

BC

Cs

AB

CsCs UUUU dxGA

VVdx

GA

VVdx

GA

VVCDBCAB L

0

I0L

0

I0L

0

I0

2GA

qL01L

GA

qL

2

10

2

CC

UW 1o

u2GA

qL

EA

qL

8EI

9qL 224

Computation of u2o: the structure 0 is treated as the actual structure whereas the structure II is

chosen as the virtual structure

2o2oC uu1W

CsCaCbC UUUU

CD

Cb

BC

Cb

AB

CbCb UUUU dx

EI

MMdx

EI

MMdx

EI

MMCDBCAB L

0

II0L

0

II0L

0

II0

24EI

55qL0

4

L7L

2EI

qL

3

1L2L

EI

2qL

2

1 422

CD

Ca

BC

Ca

AB

CaCa UUUU dxEA

FFdx

EA

FFdx

EA

FFCDBCAB L

0

I0L

0

I0L

0

I0

EA

qL001L

EA

qL 2

CD

Cs

BC

Cs

AB

CsCs UUUU dxGA

VVdxGA

VVdxGA

VVCDBCAB L

0

I0L

0

I0L

0

I0


34/42



466

2GA

qL01L

GA

qL

2

10

2

CC UW 2ou2GA

qL

EA

qL

24EI

55qL224

The flexibility coefficient fij can also be computed from the principle of complementary virtual

work by properly choosing a pair of actual and virtual structures from the structure I and thestructure II as shown below. By using the symmetric property of the flexibility matrix, only three

flexibility coefficients f11, f12 and f22 needs to be computed.

Computation of f11: the structure I is treated as the actual and virtual structures

1111C ff1W

CsCaCbC UUUU

CD

Cb

BC

Cb

AB

CbCb UUUU dxEI

MMdx

EI

MMdx

EI

MMCDBCAB L

0

IIL

0

IIL

0

II

3EI

4L0

3

L2L

EI

L

2

1LL

EI

L 3

CD

Ca

BC

Ca

AB

CaCa UUUU dxEA

FFdx

EA

FFdx

EA

FFCDBCAB L

0

IIL

0

IIL

0

II

EAL001L

EA1

CD

Cs

BC

Cs

AB

CsCs UUUU dxGA

VVdx

GA

VVdx

GA

VVCDBCAB L

0

IIL

0

IIL

0

II

GA

L01L

GA0

CC UW GA

L

EA

L

3EI

4Lf

3

11

Computation of f12: the structure I is treated as the actual structure whereas the structure II is chosenas the virtual structure

1212C ff1W

CsCaCbC UUUU

CD

Cb

BC

Cb

AB

CbCb UUUU dxEI

MMdx

EI

MMdx

EI

MMCDBCAB L

0

IIIL

0

IIIL

0

III

6EI

17L03L5L

EIL

21L2L

EIL 3


35/42



467

CD

Ca

BC

Ca

AB

CaCa UUUU dxEA

FFdx

EA

FFdx

EA

FFCDBCAB L

0

IIIL

0

IIIL

0

III

EA

L001L

EA

1

CD

Cs

BC

Cs

AB

CsCs UUUU dxGA

VVdx

GA

VVdx

GA

VVCDBCAB L

0

IIIL

0

IIIL

0

III

GA

L01L

GA0

CC UW 21312 fGA

L

EA

L

6EI

17Lf

Computation of f22: the structure II is treated as the actual and virtual structures

2222C ff1W

CsCaCbC UUUU

CD

Cb

BC

Cb

AB

CbCb UUUU dxEI

MMdx

EI

MMdx

EI

MMCDBCAB L

0

IIIIL

0

IIIIL

0

IIII

3EI

20L

3

L42L

EI

2L

2

1L2L

EI

2L 3

CD

Ca

BC

Ca

AB

CaCa UUUU dxEA

FFdx

EA

FFdx

EA

FFCDBCAB L

0

IIIIL

0

IIIIL

0

IIII

EA

L001L

EA

1

CD

Cs

BC

Cs

AB

CsCs UUUU dxGA

VVdx

GA

VVdx

GA

VVCDBCAB L

0

IIIIL

0

IIIIL

0

IIII

GA

L2

012LGA0

CC UW GA

L2

EA

L

3EI

20Lf

3

22

By substituting u1r, u1r, u1o, u2o, and finto (e10.6.1), we obtain a set of two compatibility equations

governing the two unknown redundants R1 and R2:

2

1

33

33

224o

R

R

GAL2

EAL

3EI20L

GAL2

EAL

3EI20L

GA

L

EA

L

6EI

17L

GA

L

EA

L

3EI

4L

1

1

2GA

qL

1

1

EA

qL

55

27

24EI

qL

0


36/42



468

Example 10.7 Consider a statically indeterminate rigid frame as shown in the figure below. The

flexural rigidity EI is assumed to be constant throughout and the axial member AD has constant

axial rigidity EA. The frame is subjected to a uniformly distributed load q on the top segment BC.

Determine all support reactions, the member force in the axial member, the shear force diagram and

the bending moment diagram for the entire frame, and then compute the displacement at a point D.

Consider only deformation due to bending effect for the segments AB, BC, and CD and assume thatI = AL2/3.

Solution First, we determine the degree of static indeterminacy of the given structure as follow: DI

= (2 + 1) + (33+1) (34) = 13 12 = 1. Thus, the structure is statically indeterminate withdegree of static indeterminacy equal to 1. Next, we choose the axial force in member AD, denoted

by F1, as a redundant and the corresponding primary structure, denoted by a structure 0, is given

in the figure below. Consider also a released structure (used for computing the flexibility matrix f)

subjected only to a pair of unit and opposite forces at the point of release, denoted by a structure

I, as shown in the figure below.

L

L

A

B C

D

q

A

B C

q

qL/2

0

qL/2

DA

B C

0

0

0

D

1

1

A

B C

DA

B C

D0 0

qL2/8

0 0

1 1

LL

L


M0 MI


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469

Since both the structure 0 and the structure I are statically determinate, all support reactions and the

bending moment diagram can readily be obtained from static equilibrium with results given in

above figures. To determine the redundant F1, we need to set up the following compatibility

equation

111o1r1 Ffuu (e10.7.1)

Since there is no internal discontinuity in the member AD, then u1r = 0. The overlapping generated

in the member AD of the primary structure (structure 0) , denoted by u1o, can readily be computed

using the principle of complementary virtual work with the structure I being chosen as the virtual

structure as follows:

1o1oC u0u1W

AD

Ca

CD

Cb

BC

Cb

AB

CbC UUUUU

EA

LFFdx

EI

MMdx

EI

MMdx

EI

MM ADI0L

0

I0L

0

I0L

0

I0 CDBCAB

2 42 qL qL

0 L L 0 03 8EI 12EI

CC UW 12EI

qLu

4

1o GapThe flexibility coefficient f11 can also be computed from the principle of complementary virtual

work by choosing the structure I as the actual and virtual structures as shown below.

1111C

ff1W

AD

Ca

CD

Cb

BC

Cb

AB

CbC UUUUU EA

LFFdx

EI

MMdx

EI

MMdx

EI

MM ADIIL

0

IIL

0

IIL

0

II CDBCAB

EA

L

3EI

5L

EA

LLL

EI

L2

3

L2L

EI

L

2

1 3

CC UW 11fEI

2L

3EI

L

3EI

5L

EA

L

3EI

5L 3333 Overlapping

By substituting u1r, u1o, and f11 into the compatibility equation (e10.7.1), the redundant R1 can besolved as follow:

134

FEI

2L

12EI

qL0

24

qLF1

Once the redundant F1 was solved, the response of the original structure can be obtained by

superposing the responses of the primary structure and the responses of the structure I multiplied by

R1. Thus, all support reactions of the original structure are given by

024qL00RAx

;

2qL

24qL0

2qLRAy ;

2qL

24qL0

2qLRDy


38/42



470

The shear force and bending moment diagrams of the original structure are given below.

Similarly, the displacement at the point D of the original structure, denoted by uD, is equal to thesum of the displacement at the point D of the primary structure, denoted by u Do, and the

displacement at the point D of the structure I, denoted by uDI, multiplied by the redundant F1. To

proceed, we again employ the principle of complementary virtual work with a virtual structure

shown below.

The displacement uDo is obtained as follows

DoDoC uu1W

AD

Ca

CD

Cb

BC

Cb

AB

CbC UUUUU EA

LFFdx

EI

MMdx

EI

MMdx

EI

MM AD0L

0

0L

0

0L

0

0 CDBCAB

2 42 qL qL

0 L L 0 03 8EI 12EI

CC UW 12EIqLu4

Do Rightward

D A

BC

D A

B C

D

qL/24 qL/24

qL/2

qL/2

qL2/24qL2/24 qL2/24

qL2/12

qL2/24

A

B C

q

qL/2

0

qL/2

(qL/24)

SFD BMD

A

B C

D0 0

L

L L

A

B C

0

1

0

D1

Virtual Structure

M


39/42



471

The displacement uDI is obtained as follows

DIDIC uu1W

AD

Ca

CD

Cb

BC

Cb

AB

CbC UUUUU EA

LFFdx

EI

MMdx

EI

MMdx

EI

MM ADIL

0

IL

0

IL

0

I CDBCAB

3EI

5LLL

EI

L2

3

L2L

EI

L

2

1 3

CC UW 3EI

5Lu

3

DI LeftwardTherefore, the displacement at the point D of the original structure (u D) is equal to

4 3

D Do DI 1

qL 5L qLu u u F

12EI 3EI 24

4qL

72EI Rightward

1. Analyze following statically indeterminate trusses for support reactions and member forces by

the method of consistent deformation. The axial rigidity of each member, the support settlement

(if exists), the temperature change, and error from fabrication are shown in the figure.

Exercises

P

2P

o

EA

L

L

EA

2EA

2EA 2EA P

2P

P

EA EA

EA

2EA

2EA 2EA

2EA

EA EA

L L

L

L

P

EA is constant

, T

eo

L L

L

L L

L

L

EA is constant

, T, T

eo


40/42



472

2. Analyze following statically indeterminate beams for all support reactions, SFD and BMD, thedeflection at a point a and the rotation at a point b by the method of consistent deformation. In

the analysis, lets consider only bending effect. The flexural rigidity and length of each member

and the support settlement are clearly indicated in the figure.

L

P

L

o

P

L LL L

L

L

2EA2EA

EA

EA is constant

eo, T

2EI, 3L

qL2q

EI, L a, b

q

EI, L EI, Lb

P

2EI, LEI, L

a

o

P

EI, L1 EI, L2a, b

o

M

2EI, L EI, La b

qqL

EI, L EI, L EI, 2La, b

M

EI, L b

q

EI, LEI, L


41/42



473

3. Analyze following statically indeterminate frames for all support reactions, AFD, SFD, BMD,

the deflection at a point a and the rotation at a point b by the method of consistent deformation.

In the analysis, lets consider only the bending effect except for axial members. The flexural

rigidity and length of each segment and the support settlement are clearly indicated in the figure.

q

2EI, L

EI, L

a

b

2qL

q EI, L

2EI, L 2EI, La

b

P

o

EI, L EI, L

EI, 2L

EI, L EI, L

a

b

P

q

EI, L/2

EI, L/2

EI, 2L

2EI, L

EI, L EI, L

2EI, 2L

EA, L

, T

q

EA, L

EI, 2L

P

2PL

o

a, b

a, b

a, b

q

EI, L

EI, L

EI, L

2EI, L

EA, L

a, b

a, b


42/42

FUNDAMENTAL STRUCTURAL ANALYSISJaroon Rungamornrat Method of Consistent Deformation474

P

2P

EI, L 2EI, LEA, L EA, L

EA, LEA, 2 LEA, 2 L

a

b

2P 2P2PP P

3P

2P

EIEI

EA constant

L LL L

L

3L

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