CHAPTER 4:PRINCIPLES OF STIFFNESS METHOD FOR BEAMS AND PLANE

FRAMES4.1 INTRODUCTION:

In chapter 3 the analysis of trusses using stiffness method was discussed. In this chapter application of stiffness method will be extended to beams and plane frames. The procedure for application of this method is the same as that of the trusses but the difference is only in member stiffness matrix and deformation transformation matrix, which will be developed in the subsequent section.

4.2 MEMBER OR ELEMENT STIFFNESS MATRIX (FLEXURAL ELEMENT):

As a frame element is subjected not only to axial forces but also to shear forces and bending moments, therefore three degrees of freedom per joint of a frame element are present. A degree of freedom is an independent deformation of a joint or a node. These are:i Axial deformation.ii) End rotations.iii) Normal translations.

Out of these three, axial deformation is normally neglected, so element or member stiffness matrix for an element subjected to shear force and bending moment will only be developed at this stage.

Consider a member/element shown in figure 4.1. The x-y coordinate system shown is local coordinate system. Origin is always at the near end. There are two forces (shear force w3 and a moment w1) acting at near end of the joint and correspondingly there are two deformations (vertical translations 3, and rotation 1).Similarly there are two forces (shear force w4 and a moment w2) acting at the far end of the joint and correspondingly two deformations (vertical translation 4 and rotation 2).

4.2.1 SIGN CONVENTION Moments (w1, w2) and rotations (1 and 2) are positive when

clockwise and negative when counter clockwise. Translation 3, 4 are positive when upward and negative when

downward. The local x-axis runs along the member from the first joint to the second joint

4.2.2 DERIVATIONThe load-stiffness-deformation relationship for this element is the same as

that for a truss element as expressed in equation 3.17.[w]m = [k]m []m---------------------------------- (3.17)

As there are four forces and four corresponding deformations then the equation 3.17 can be expanded in the following form:

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

kkkk

kkkk

kkkk

kkkk

w

w

w

w

---------------------------------- (4.1)

Where each element of the stiffness matrix is called stiffness coefficient as discussed in chapter 2. It represents the place occupied by it with respect to row and columns. Any stiffness coefficient may be represented by kij; where i and j are number of rows and columns. The above mentioned element stiffness matrix [k]m is formed by applying a unit value of each end deformation in turn and the corresponding column of the matrix of equation 4.1 gives the various end forces developed at the member ends while other deformations are restrained. This procedure is as follows:

Apply unit positive deformation (clockwise rotation) 1 = 1 and equating all other deformations to zero (2 = 3 = 4 = 0). The element would be deformed as shown in figure 4.2.a. From the definition of stiffness, the forces induced at both ends due to unit clockwise rotation of near end are as under.

w1 = k11 = Moment produced at ‘1’ due to unit clockwise rotation at 1.w2 = k21 = Moment produced at ‘2’ due to unit clockwise rotation at 1.w3 = k31 = Vertical reaction produced at ‘3’ due to unit clockwise rotation at

1.w4 = k41 = Vertical reaction produced at ‘4’ due to unit clockwise rotation at

1.The values of k11, k21, k31 and k41 can be obtained by using the

moment area theorems.

k

(+)

(a)

(b)

k

k

k11 21

31

41

1

A B

k11

(-)

L

Figure 4.2

(c)

(d)

k k

k k

2111

4131

k 21

A B

As according to moment area theorem no.1 change in slope between two points on an elastic curve is equal to area of the M/EI diagram between these two points. Looking at figure 4.2’a’ change in slope between two ends is equal to unity. Adding the areas of figure 4.2 (b) and figure 4.2 (c).

------------------------------- (.4.2)0.12

.

2

. 2111 EI

Lk

EI

Lk

03

2

2

.

32

. 2111

L

EI

LkL

EI

Lk

According to theorem no.2 of moment area method tangential deviation of a certain point with respect to the tangent at another point is equal to the moment of M/EI diagram between the two points calculated about the point where the deviation is to be determined. From the above definition the moment of M/EI diagram (figures 4.2 (b) and 4.2 (c) ) about the left of the member is equal to zero.

----------------------- (4.3)

Following values of k11 and k21 are obtained by solving equations 4.2 and 4.3.

L

EIk

L

EIk

2

,4

21

11

---------------------------------- (4.4)

Reaction k41 and k31 can be obtained using equation of equilibrium.

Summation of moments about the right end is equal to zero (see figure 4.2 (d).

kk k

L3111 21

MB=0

k31 L - k11 - k22 = 0---------------------------------- (4.5)

- --------------------- (4.6)

kEI L EI L

L314 2

/ /

k31 = 2

6

L

EI----------------------------------

(4.7)

Applying force equation of equilibrium to figure 4.2 (d)Fy = 0k41 - k31 = 0

2

6

L

EI

k41 = k31 ------------------------------ (4.8)

k41 = ----------------------------- (4.9)

L

EI4

These are the forces and moments as shown in figure 4.2 a, b, c, d. On comparison with figure 4.1 the correct signs are obtained and these are defined by the following equations.

k11 = k21 =

L

2EI

k31 = 2

6

L

EI k41 =

2L

2EI ------------

(4.10)

This gives the first column of the element stiffness matrix. As this matrix is symmetric so it also provides the first row. To obtain 2nd column of stiffness matrix deformation (rotation) 2 = 1 is imposed on the far end equating all other deformations to zero 1 = 3 = 4 = 0. The element would be deformed as shown in figure 4.3 (a). From the definition of stiffness as mentioned in chapter 2, the forces induced at both ends due to unit rotation at far end can be defined as

w1 = k12 = Moment produced at ‘1’ due to unit clockwise rotation at 2.

w2 = k22 = Moment produced at ‘2’ due to unit clockwise rotation at 2.

w3 = k32 = Vertical reaction produced at ‘3’ due to unit clockwise rotation at 2.

w4 = k42 = Vertical reaction produced at ‘4’ due to unit clockwise rotation at 2.

The values of k12, k22, k32 and k42 can be obtained by using the moment area theorems.

(a)

(b)

(+)

k

k

k

k

k

22

4232

12

12

L

2

AB

(-)

L

k k

k k

k

A B

Figure 4.3

22

4232

12 22

(c)

(d)

Applying moment area theorem no. 1 and using bending moment diagram of figure 4.3 (b,c). For this case change in slope between both ends is equal to unity so

k L

EI

k L

EI22 12

2 21

------------------------------ (4.11)

However according to moment area theorem no.2 the moment of M/EI diagram about the right end of the member is equal to zero.

03

2

2

.

32

. 1222

L

EI

LkL

EI

Lk------------------------ (4.12)

Solving equations 4.11 and 4.12

L

EIk

422

L

EIk

212

Reaction k42 and k32 can be obtained using equation of equilibrium. Summation

of moments about the left end is equal to zero (see figure 4.3 (d)).

--------------------------- (4.13)،

(a)

BENDING MOMENT DIAGRAMS FOR THE ELEMENT

k33 Figure 4.4

k13

A

k43

k23(d)

k23(c)

(b)

k43

B

k13

k33 k23

L

k13

Applying equation of force equilibrium to figure 4.5(d)Fy = 0

L

k34

k14

k24

k44

k24

k14

(a)

(d)

k14 k24

k44k34

L

BENDING MOMENT DIAGRAM FOR THE ELEMENT

Figure 4.5

A

B

(b)

(c)

kEI

L34 3

12

Correct signs can be obtained by comparing these values with figure 4.1. These are defined as

kEI

L14 2

6 k

EI

L24 2

6 k

EI

L34 3

12

kEI

L44 3

12

On combining the calculation given in equation 4.10, 4.17, 4.27 and 4.35, following element stiffness matrix is obtained.

--------- (4.35)

--------------------------- (4.34)

3322

3322

22

22

121266

121266

6642

6624

L

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EI

k m

4

3

2

1

121266

121266

6642

6624

4

3

2

1

3322

3322

22

22

L

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EI

w

w

w

w

And the force, stiffness and deformation relationship is as under:-

--------------------------- (4.36)

--------------------------- (4.37)

4.3 MEMBER OR ELEMENT STIFFNESS MATRIX FOR A BEAM / FRAME ELEMENT SUBJECTED TO MOMENTS ONLY:

The element stiffness matrix for members whose ends cannot translate but can rotate is obtained by removing third and fourth row corresponding to 3, 4 and third and fourth columns corresponding to w3 and w4 from the above mentioned element stiffness matrix of equation 4.36 as shown below:-

4

3

2

1

121266

121266

6642

6624

4 3 2 1

3322

3322

22

22

L

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EI

k m

The resulting stiffness matrix for a beam element is as follows:

42

24

L

EImk ----------------------------(4.38)

And the force stiffness deformation relationship for a beam element is as fallows:-

w

w

EI

L1

2

1

2

4 2

2 4

--------------------------- - (4.39)

4.4 MEMBER OR ELEMENT STIFFNESS MATRIX OF A FRAME ELEMENT SUBJECTED TO AXIAL LOADING IN ADDITION TO SHEARING FORCES AND BENDING MOMENTS.

The final element that is capable of axial loading in addition to shear forces and bending moments is now considered.

The stiffness matrix for this element can be formed by superposition of the element stiffness matrix of truss element and frame element formed as given in the equations 2.16 and 4.20. Figure 4.6 defines the positive forces (axial forces, shear forces and bending moments) and deformations (axial deformations, vertical translations and rotations) for the element.

w ,w ,

w ,w ,

w ,w ,

Figure 4.6

EI,A

L

y

x

11

55

22

44

33

66

By superposition of equation 2.16 and 4.36.The following element stiffness matrix for six element forces and deformations is obtained.

k

EI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

LAE

L

AE

LAE

L

AE

L

m

4 2 6 60 0

2 4 6 60 0

6 6 12 120 0

6 6 12 120 0

0 0 0 0

0 0 0 0

2 2

2 2

2 2 3 3

2 2 3 3 --------------- (4.40)

And the force, stiffness and deformation relationship is as under:

6

5

4

3

2

1

3322

3322

22

22

6

5

4

3

2

1

0000

0000

00121266

00121266

006642

006624

L

AE

L

AE

L

AE

L

AE

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

L

EI

w

w

w

w

w

w

-------- (4.41)

4.5 DEFORMATION TRANSFORMATION MATRIX:As discussed in the case of trusses, deformation transformation matrix is used to transform the element deformations from local coordinates to structure deformations in global co-ordinates. Using this deformation transformation matrix structure stiffness matrix is obtained as given in the following equation:[K]m = [T]T

m [k]m [T]m --------------------- (3.3)

4.5.1 DEFORMATION TRANSFORMATION MATRIX OF A FRAME ELEMENT SUBJECTED TO AXIAL FORCE, SHEAR FORCE AND BENDING MOMENT.

A deformation transformation matrix will also be developed to carry out the transformation from element to global coordinates.

Consider the frame member shown in figure 4.7. Member axis, x-axis of member coordinate system makes an angle x with x-axis of the structure coordinate system as shown in figure 4.7 (b), similarly member axis, x-axis of member coordinate system makes angle y with y-axis of the structure coordinate system. The cosines of these angles are given below

l= Cos xm = Cos y

w ,w ,

w ,w ,

w ,w ,

EI,A

L

y

x

11

55

22

44

33

66

Element forces and deformations

W ,

W ,W ,

W ,

W ,

W ,

Figure 4.7

x

y

4 4

2 2

66

5

3 3

5

1 1 Structure forces and deformations

Again consider the frame member shown in figure 4.7, 1 and 2 are the

element deformations (rotation) whereas 1 and 2 are the structure

deformations (rotations). As Z-axes for both element and structure coincides, therefore both element and structure rotations will be the same when structure deformation 1 = 1, then the element deformation

1

2

3

4

5

6

1

0

0

0

0

0

-------------------------------- (4.43)

and when structure deformation 2 = 1, then the element deformations

1

2

3

4

5

6

0

1

0

0

0

0

-------------------------------- (4.44)

Relationship between structure and element deformations can also be obtained in the similar manner for a unit vertical translation of the left end i.e. 3 = 1 (see figure 4.8)

3 = 3.Cos x, 5 = 3.Cos y

3 = 1. Cos x, 5 = 1. Cos y

3=1. = , 5=1.m=m

3

X-axis

Y-axis

x

y

3=

cos3

x

x

5=

cos3

y

FIGURE 4.85=3Cosy=1.

m=m3=3Cosx=1.

l=l1=2=4=6=0

mm

1

0

.1

0

1.1

0

0

6

5

4

3

2

1

y

x

4

4

X-axis

Y-axis

Figure 4.94=4Cosx=1.l=l6=4Cosy=1.m=m

For a unit vertical translation of the right end i.e. 4 = 1 (Fig.4.9) element

deformations are as under:

mmyCos

llxCos

.1

.1

0

0

0

0

46

5

44

3

2

1

-------------------------------- (4.46)

where 3 and 4 are vertical translations of the element.

for a unit horizontal translation (along x-axis) of the left end.i-e; 5=1

(Fig.4.10)element deformations are as under:-

1

2

3

4

5

6

0

0

0

0

m

l

-------------------------------- (4.47)

For a unit horizontal translation the right end I-e;6 = 1, element

deformations are as under:-

1

2

3

4

5

6

0

0

0

0

m

l

-------------------------------- (4.48)

Figure 4.10

cosx=sinx=-m

Figure 4.11

sinx -m

cosx =1

X-axis

Y-axis

5

5

3x

y

x

y

6

X-axis

Y-axis

6

Writing the above six equations in matrix form.

6

5

4

3

2

1

6

5

4

3

2

1

0000

0000

0000

0000

000010

000001

lm

lm

ml

ml

-------------------------------- (4.49)

= T ---------------------------------- (3.2)Comparing equation (4.49) with equation (3.2) deformation transformation matrix (T) is obtained which is as follows:

lm

lm

ml

mlT

0000

0000

0000

0000

000010

000001

-------------------- (4.50)

This is the deformation transformation matrix of a frame element subjected to shear force, bending moment and axial forces.To obtain the deformation transformation matrix of a beam/frame element subjected only to shear force and bending moment, the axial deformations are ignored. Therefore by deleting last two rows and last two columns of the matrix in equation (4.50) following deformation transformation matrix is obtained:

Tl

l

1 0 0 0

0 1 0 0

0 0 0

0 0 0 -------------------------------- (4.51)

Similarly to obtain the deformation transformation matrix of a beam/frame element subjected only to bending moment, the axial and shear deformations are ignored Therefore by deleting last two rows and last two columns of the matrix in equation (4.51) following deformation transformation matrix is obtained:

T

1 0

0 1------------------------------- (4.52)

4.6 STRUCTURE STIFFNESS MATRICES4.6.1 For a beam/frame element subjected to bending moment only:Using the element stiffness matrix and deformation transformation matrix

structure stiffness matrix is formed. Following equation is used for this purpose:

[K]m = [T]Tm [k]m [T]m --------------------------- (4.53)

Where,[T]m = Deformation Transformation matrix[k]m = Element stiffness matrix

In this case:

T m

1 0

0 1-------------------------------- (4.52)

And k EI

Lm

4 2

2 4-------------------------------- (4.38)

and T m

T

1 0

0 1

Substituting these values in equation (4.53) and solving, following structure stiffness matrix is obtained:

K EI

Lm

4 2

2 4-------------------------------- (4.54)

4.6.2 Beam/Frame element subjected to Shear Force and Bending Moment only:In this case:

Tl

l

1 0 0 0

0 1 0 0

0 0 0

0 0 0

-------------------------------- (4.51)

k

EI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

L

m

4 2 6 6

2 4 6 6

6 6 12 12

6 6 12 12

2 2

2 2

2 2 3 3

2 2 3 3

-------------------------------- (4.36)

Tl

l

mT

1 0 0 0

0 1 0 0

0 0 0

0 0 0

Substituting these values in equation (4.53) the following structure stiffness matrix is obtained:

K

EI

L

EI

L

EI

Ll

EI

Ll

EI

L

EI

L

EI

Ll

EI

Ll

EI

Ll

EI

Ll

EI

Ll

EI

Ll

EI

Ll

EI

Ll

EI

Ll

EI

Ll

m

4 2 6 6

2 4 6 6

6 6 12 12

6 6 12 12

2 2

2 2

2 2 32

32

2 2 32

32

. .

. .

. . . .

. . . .

4.6.3 Frame element subjected to Shear Force/ Bending Moment/Axial Forces:

Here

Tl m

l m

m l

m l

m

1 0 0 0 0 0

0 1 0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

-------------------------------- (4.50)

k

EI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

LEI

L

EI

L

EI

L

EI

LAE

L

AE

LAE

L

AE

L

m

4 2 6 60 0

2 4 6 60 0

6 6 12 120 0

6 6 12 120 0

0 0 0 0

0 0 0 0

2 2

2 2

2 2 3 3

2 2 3 3 -------------------------------- (4.40)

and

T l m

l m

m l

m l

m

T

1 0 0 0 0 0

0 1 0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

Substituting these values in equation (4.53) and multiplying, structure stiffness matrix is given on the next page, is obtained.

223

12223

12

3

12

3

12

2

6

2

6

223

12223

12

3

12

3

12

2

6

2

6

3

12

3

12223

12223

12

2

6

2

6

3

12

3

12223

12223

12

2

6

2

6

2

6

2

6

2

6

2

642

2

6

2

6

2

6

2

624

lL

AEm

L

EIl

L

AEm

L

EIlm

L

AE

L

EIlm

L

AE

L

EIm

L

EIm

L

EI

lL

AEm

L

EIl

L

AEm

L

EIlm

L

AE

L

EIlm

L

AE

L

EIm

L

EIm

L

EI

lmL

AE

L

EIlm

L

AE

L

EIm

L

AEl

L

EIm

L

AEl

L

EIl

L

EIl

L

EI

lmL

AE

L

EIlm

L

AE

L

EIm

L

AEl

L

EIm

L

AEl

L

EIl

L

EIl

L

EI

mL

EIm

L

EIl

L

EIl

L

EI

L

EI

L

EI

mL

EIm

L

EIl

L

EIl

L

EI

L

EI

L

EI

mK