solution 2: two dimensional analysis for beams and framesdrsekaran.tripod.com/sol2.pdf · ... two...

Università degli studi di Napoli Federico II Facoltà di Ingegneria - a.a. 2007/2008

Corso di TECNICA DELLE COSTRUZIONI II prof. G. Serino (allievi Civili, Civili_SS, STReGA) Lecture by: Dr (Eng) Chandrasekaran

pag. 1/14

Solution 2: Two dimensional analysis for beams and frames

1. What do you understand by degree-of-freedom? Distinguish this from kinematic

indeterminacy.

� By definition, degree-of-freedom is the number of independent coordinates required to

fully describe the displacement of a structural element subjected to any arbitrary loading

with respect to any standard reference coordinate system. Whereas kinematic indeterminacy

refers, in particular, to the number of unrestrained joint displacements (rotations and

translations) of the structure. For a give beam element, knowing the nature of support (roller,

hinged or fixed), the number of unrestrained joint displacements can be identified (for ex,

refer to Table 1 of Lesson 1). Hence they become the unknowns to be solved for a given

external loading. On the other hand, degree-of-freedom is the number of independent

displacements required to specify the geometric position of the structure under action of any

given loading. At this point of time, it should be appreciated that they have only a marginal

difference between them and normally leads to a confusion.

2. Define stiffness coefficient.

� Stiffness coefficient, kij is defined as the force required (generated or developed) in ith

degree-of-freedom by giving unit displacement (may be rotation or translation) in jth degree-

of-freedom by restraining all other degrees-of-freedom (except that of i and j under

consideration) from any type of displacement (neither translational nor rotational). It is to be

appreciated here that the jth shall be always subjected to unit translation (either unit rotation

or unit displacement). Therefore stiffness coefficient can also be understood as force

generated (developed) by giving unit displacement to the structure at appropriate degree-of-

freedom.

3. Derive [K] for a 2d beam element, considering and neglecting the axial deformation.

Before we derive the stiffness coefficients of the beam element, let us understand some basic

equations related to fixed beam. Though these are extended fundamentals of the classical

indeterminate structural analysis classes, for completion sake they are repeated here.

a) Let us consider a beam of length ‘L’ and flexural rigidity ‘EI’, fixed at both ends as

shown in Fig. 2.3. The end B is given a rotation amounting θ.

+

The basic equations of compatibility are

θ=θ+θ rm ( 2.1)

0rm =∂+∂ (2.2)

by substituting in Eq. (2.1), we get

Fig. 2.3 Fixed beam subjected to joint rotation

MA MB

RA RB

θ

MB

RB



pag. 2/14

θ=+−

EI2

LR

EI

LM 2BB (2.3)

L

EI2LRM2 BB

θ=+−∴ (2.4)

Also, 0EI3

LR

EI2

LM 3B

2B =+

− (2.5)

BBBB M2

3LR0LR2M3 =⇒=+−∴ (2.6)

Substituting Eq. (2.6) in Eq. (2.4), we get L

EI4MB

θ−= (-ve sign is due to the fact that moment

is applied in the direction opposite to rotation.

Now LRMM BBA −= (2.7)

L

EI2M

2

3M BB

θ=−= (2.8)

Further by substituting Eq. (2.8) in Eq. (2.6), we get 2B

L

EI6R

θ−=

Finally,

L

EI2 θ

L

EI4 θ

2L

EI6 θ

2L

EI6 θ

b) Let us now consider a beam of length ‘L’ and flexural rigidity ‘EI’, fixed at both

ends as shown in Fig. 2.5 with support B settles by Λ.

▲

+

-RBL/EI

- MB/EI

Pl note that the

moment is in the

same direction as

that of the rotation

given

Fig. 2.4 Fixed beam subjected to joint rotation with final values

Fig. 2.5 Fixed beam subjected to settlement of support

MB RB

Fig. 2.6 conjugate beam loaded with M/EI diagram

MB MA

RB

RA



pag. 3/14

From the classical structural mechanics, it is known that the rotation at point B in the real

beam is same as that of shear force at B in conjugate beam; and deflection at point B in the

real beam is same as that of moment at the same point in the conjugate beam. Hence the

equations for rotation and deflections at point B from the two conjugate beams loaded with

equivalent M/EI diagrams is given by:

EI2

LMand

EI

LM 2B

MB

M−

=δ−

=θ (2.9)

EI3

LR

3

L2

EI2

LRand

EI2

LRL

EI

LR

2

13

B2

BR

2BB

R =

=δ=

=θ (2.10)

The compatibility conditions are: ∆−=δ+δ=θ+θ RMRM and0 (2.11)

where rotations and displacements due to moments carry suffix M and that due to reactions

carry suffix R.

0RM =θ+θ will yield 0EI2

LR

EI

LM 2BB =+

− (2.12)

L

M2R0LRM2 BBBB =⇒=+− (2.13)

∆−=δ+δ RM will yield ∆−=+−

EI3

LR

EI2

LM 3B

2B

(2.14)

substituting the value from Eq. (2.13), we now get 2B

L

EI6M

∆−= (2.15)

Similarly, for getting MA, we use the compatibility condition, ∆=δ+δ RM yielding

2AL

EI6M

∆+= (2.16)

By substituting Eq. (2.15) in Eq. 2.13), we get

3A3BL

EI12Rand

L

EI12R

∆=

∆−= (2.17)

6EI ▲/L2

▲ 6EI ▲/L2

12EI ▲/L3

12EI ▲/L

3

c) Let us now consider a beam of length ‘L’ and flexural rigidity ‘EI’, fixed at both ends as

shown in Fig. 2.8 with axial deformation taken into account

Fig. 2.7 Fixed beam subjected to support settlement with final values

Fig. 2.8 Fixed beam considered for the analysis

Pl note that the

moment is applied to

counter act the

settlement. i.e. in the

direction to lift the

beam which is settled



pag. 4/14

The degrees-of-freedom are to be marked at both the supports. The common confusion starts here

with a fundamental question whether to consider the un-restrained degree-of-freedom or restrained

degree-of-freedom (or both). In fact, if you consider a fixed beam, you actually do not have any un-

restrained degree-of-freedom and while you think of marking only the un-restrained degree-of-

freedom, as required for stiffness method of analysis, then you cannot really start the derivation.

You should always consider the beam as a general case and consider all possible degrees-of-

freedom I every support. The restrained and un-restrained degrees-of-freedom will be identified

actually when this beam becomes a part of any structure or a frame. The order of numbering

degrees-of-freedom is very important in this method of analysis as this shall enable to write a

computer program very easily and systematically. Let the beam has structural properties as A, E, I,

L which are known before hand. One may get a confusion here that usually structural design, which

actually gives some of these values (for ex. b, d etc) following the analysis and therefore how they

are considered as known values in the analysis. It is important to understand that structural design is

an iterative process wherein dimensions are initially assumed (usually known as preliminary values

in the design) and analysis is done. On the basis of the results obtained from the analysis, the stress

values at critical cross-sections are checked with respect to relevant design code in practice and

revised, if required.

The order of degrees-of-freedom can be remembered as below:

1. First start from the left hand support and number the rotation at this support

as 1 and that at right hand support as 2.

Always follow an unique sign convention. As this generally

causes confusion, the author strictly advise to assume anti-clockwise

rotations and upward reactions at all supports. It is very important to note that these

conventions do not have any sign attached to it (neither +ve nor –ve).

2. Next number the reactions at left and right supports as 3 and 4 respectively. Remember to

mark the directions of these degrees-of-freedom as upwards. (They do not have sign

attached to them)

3. Next number axial degrees-of-freedom in left and right support as 5 and 6 respectively.

Remember to mark them acting from left to right at both the supports. (They do not have

any sign attached to them)

Fig. 2.9 shows the beam considered with all degrees-of-freedom marked as explained above.

Axis of analysis

It is now important to understand that the axis under consideration in such 2-d analysis of beams.

Imagine a X-Y plane of which is the beam is lying such that the length of the beam is towards the

positive side of X axis. Therefore, depth of the beam shall be normal to the X axis (parallel to XZ

plane) whereas breadth of the beam shall be measured along Y axis (parallel to XY plane) and

length of the beam is measured normal to XZ plane as shown in the Fig. 2.10

Important to

remember

Fig. 2.9 Fixed beam with degrees-of-freedom marked

1 2

3 4

5 6



pag. 5/14

Please note that the moments marked in Fig. 2.9 are about y axis and their values shall be taken as

12

bdI

3

yy = . Coefficients of [K] shall be derived by giving unit displacement in jth degree-of-freedom

and finding the forces in ith degree-of-freedom, by keeping all other degrees-of-freedom as

restrained.

a) by giving unit rotation in 1st degree-of-freedom and obtaining the forces

8Dr. Chandrasekaran24/07/2007

12

11yy

kEI4

=θ

l

1=θ

21yy

kEI2

=θ

l

312

yyk

EI6=

θ

l

412

yyk

EI6=

θ

l

3 4

−

0

0

/EI6

/EI6

/EI2

/EI4

2

2

l

l

l

l

First column of K

could be

b

d

x

y A, E, I and length

Fig. 2.10 Fixed Beam shown with the axis considered for the analysis

Fig. 2.11 Obtaining first column of [K]



pag. 6/14

b) by giving unit rotation in 2nd

degree-of-freedom and obtaining the forces


12

12yy

kEI2

=θ

l

1=θ

22yy

kEI4

=θ

l

322

yyk

EI6=

θ

l

422

yyk

EI6=

θ

l

3 4

−

0

0

/EI6

/EI6

/EI4

/EI2

2

2

l

l

l

l

second column of

K could be

c) by giving unit displacement along 3rd



12

132

yyk

EI6=

∆

l232

yyk

EI6=

∆

l

333

yyk

EI12=

∆

l

433

yyk

EI12=

∆

l

3 4

−

0

0

/EI12

/EI12

/EI6

/EI6

3

3

2

2

l

l

l

l

third column of K

could be

1=∆

Fig. 2.12 Obtaining second column of [K]

Fig. 2.13 Obtaining third column of [K]



pag. 7/14

d) by giving unit displacement along 4th



12

3 4

−

−

−

0

0

/EI12

/EI12

/EI6

/EI6

3

3

2

2

l

l

l

l

fourth column of K

could be

1=∆

142

yyk

EI6=

∆

l

242

yyk

EI6=

∆

l

343

yyk

EI12=

∆

l

443

yyk

EI12=

∆

l

e) by giving unit axial displacement along 5th



12

3 4

− l

l

/AE

/AE

0

0

0

0

fifth column of K

could be

65

5

55kAE

=∆

l

65kAE

=∆

l

6

1=∆

Fig. 2.14 Obtaining fourth column of [K]

Fig. 2.15 Obtaining fifth column of [K]



pag. 8/14

f) by giving unit axial displacement along 6th



12

3 4

−

l

l

/AE

/AE

0

0

0

0

sixth column of K

could be

1=∆

65

5

56kAE

=∆

l

66kAE

=∆

l

6

Thus the complete stiffness matrix of the fixed beam is given as:

The partition lines above show the 4x4 stiffness matrix of the fixed beam if axial deformation

degrees-of-freedom are not considered.

Fig. 2.16 Obtaining sixth column of [K]

−

−

−−−

−

−

−

ll

ll

llll

llll

llll

llll

AEAE0000

AEAE0000

00EI12EI12EI6EI6

00EI12EI12EI6EI6

00EI6EI6EI4EI2

00EI6EI6EI2EI4

3322

3322

22

22



pag. 9/14

4. In your opinion, why a fixed beam should be considered as a general case for derivation.?

Does it have any significant advantage?

The fixed beam is the most probable case of a structural element which is a subset of the structural

frame. In the analysis, if beams with other types of supports are encountered, only the degrees-of-

freedom has to be changed from restrained to un-restrained and so on. This makes the analysis very

simple and systematic. It has a major advantage because we do not have to derive the stiffness

matrix for each beam in relationship with its type of support and just add or delete the respective

rows and columns from the original matrix (appending) to generate a new stiffness matrix suitable

for the specific beam under consideration. This also saves time (during hand calculation) and

optimizes the memory allocation in computer program.

5. What are the important characteristics of [K] of a beam element

The important characteristics of [K] can be listed as:

� It is a square matrix, the size of being the degrees-of-freedom considered

� The matrix will be always a symmetrical matrix

� [K] is written for a member and NOT FOR A JOINT

� all leading diagonal elements of this matrix are positive

� It is diagonally dominant

� The elements of [K] are functions of E, I, l and ▲and not a function of applied force and

the member response.

IF ANY OF THE ABOVE CHARACTERISTICS ARE NOT SEEN IN THE DERIVED

[K], THEN IT IS DERIVED FOR A SPECIAL MEMBER.

6. Derive [K] for the following

The degrees-of-freedom are to be labelled for the given problem, which is the most important step

in the analysis. For the continuous beam shown in Fig. 2.17, starting from left support A, let us

mark the un-restrained degrees-of-freedom in each support. These labels are shown in grey shade

and please notice that they are continuously numbered. Then start labelling restrained degrees-of-

freedom in each support (again starting from left support A). These labels are shown without shade

as 4,5 and 6. Now this gives us an idea that [K] for the beam ABC will of size 6x6 (as there are 6

degrees-of-freedom in total. The stiffness matrix derived for beam AB and beam BC are given

below:

6m, EI 4m, EI

6

1 2 3

4 5

A B C

Fig. 2.17 Continuous beam ABC – Problem statement

Axial

deformation is

not considered



pag. 10/14

- 0. 375 - 0. 375 - 0. 1875

0. 375 0. 375 0. 1875

0. 5 1 0. 375

1 0. 5 0. 375

1 2

1

5

4

5

2

- 0. 375

- 0. 375

- 0. 1875

0. 1875

KAB = EI

- 0. 125 - 0. 375 - 0. 375

0. 125 0. 375 0. 375

0. 375 1. 5 0. 75

1. 5 0. 75 0. 375

2 3

2

5 6

5

6

3

- 0. 375

- 0. 375

- 0. 125

0. 125

KBC = EI

0. 1875

2

1

-0. 375

3 -0. 375

0.0 4 0. 1875

5 - 0. 125

KABC = EI

6

0.0

6 0. 125

5

-0. 375

0.0

0. 375

0. 3125

- 0. 125

4

0. 375

0. 375

0.0

- 0. 1875

0.0

3

0. 0

0. 75

1. 50

0.0

0. 375

- 0. 375

2

0. 5

2. 5

0.75

0. 375

0.0

0. 375

1

1.0

0. 5

0.0

0. 375

0. 375

0.0



pag. 11/14

The above matrix can also be re-written in a general form as below:

=

RRRU

URUUABC

SS

SSEIK

where SUU, SUR etc are elements of the complete stiffness matrix of the beam ABC. “U” stands for

un-restrained and “R” stands for restrained degree-of-freedom respectively. The reader can easily

identify that size of all the four sub matrices shall be 3x3 (as this problem has 3 un-restrained and 3

restrained degree-of-freedom. The advantage of partitioning the complete structure stiffness matrix

into sub-matrices as above shall be discussed later.

The above matrix KABC has some special characteristics, which are interesting to note. The left

square area of 3x3 show matrix of unrestrained degrees-of-freedom and right square area show 3x3

of restrained degrees-of-freedom. It is arranged in this way as they are labelled in a specific pattern.

Should the labelling pattern changed, the unrestrained and restrained degrees-of-freedom will not

get automatically grouped as obtained now. This grouping has an advantage in inverse of the

matrix, which is required to obtain the complete solution. The elements of KABC are obtained by

collecting the respective terms from KAB and KBC and adding them. For ex, to obtain the element 1-

1, we shall collect elements of 1-1 from KAB and KBC which shall be

(1.0 + 0 as there is no 1-1 in KBC ).

Problem 2:

For the frame, the unrestrained and restrained degrees-of-freedom are identified in the same manner

as that of the previous problem. It can be easily seen that since there are 4 unrestrained degree-of-

freedom and 5 restrained degree-of-freedom, the size of [K] shall be 9x9. In this case, there will be

three stiffness matrices, one for each element of the frame, namely KAB, KBC and KDC respectively.

Let us first derive KBC which is a simple beam element and is given as below:

Fig. 2.18 Single storey- single bay frame- Problem statement

Note that supports B and C do

not have separate lateral dof .

Axial deformations are

neglected

4m, EI

4m, EI C

D

B

3m, EI

A

1 2

3

4

5

6

7

8

9



pag. 12/14

The column elements AB and DC can be simply viewed as beam elements, by doing the following:

For column element AB, hold the support A and rotate the column clockwise by 90° and for

member DC, hold support D and rotate the member clockwise similarly. As a result, we get

horizontal members AB and DC as shown in the Fig.

- 0. 375 - 0. 375 - 0. 1875

0. 375 0. 375 0. 1875

0. 5 1 0. 375

1 0. 5 0. 375

1 2

1

6 8

6

8

2

- 0. 375

- 0. 375

- 0. 1875

0. 1875

KBC = EI

- 0. 375 - 0. 375 - 0. 1875

0. 375 0. 375 0. 1875

0. 5 1 0. 375

1 0. 5 0. 375

5 1

5

7 3

7

3

1

- 0. 375

- 0. 375

- 0. 1875

0. 1875

KAB = EI

A

5 1

7 3

B 4m, EI

D

4 2

9 3

C 3m, EI

Fig. 2.19 Beam elements derived from the frame



pag. 13/14

KDC = EI

1.3333 0.6667 0.6667 -0.6667

0.6667 1.3333 0.6667 -0.6667

0.6667 0.6667 0.4445 -0.4445

-0.6667 -0.6667 -0.4445 0.4445

4 2 9 3

3

9

2

4

SUU = EI

2. 00 0. 50 - 0. 375 0. 00

0.50 2. 3333 - 0. 6667 0. 6667

- 0. 375 - 0. 6667 0. 632 -0. 6667

0. 0 0. 6667 - 0. 6667 1. 3333

4 2 9 3

3

9

2

4



pag. 14/14

1

2.0 0.5 -0.375 0.0 0.5 0.375 0.375 -0.375 0.0

0.5 2.3333 -0.6667 0.6667 0.0 0.375 0.0 -0.375 0.6667

-0.375 -0.6667 0.632 -0.6667 -0.375 0.0 -0.1875 0.0 -0.4445

0.0 0.6667 -0.6667 1.3333 0.0 0.0 0.0 0.0 0.6667

0.5 0.0 -0.375 0.0 1.0 0.0 0.375 0.0 0.0

0.375 0.375 0.0 0.0 0.0 0.1875 0.0 -0.1875 0.0

0.375 0.0 -0.1875 0.0 0.375 0.0 0.1875 0.0 0.0

-0.375 -0.375 0.0 0.0 0.0 -0.1875 0.0 0.1875 0.0

0.0 0.6667 -0.4445 0.6667 0.0 0.0 0.0 0.0 0.4445

2 3 4

1

2

3

4

5 6 7 8 9

5

6

7

8

9

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