beam stiffness analysis 2012

UWI/Civil & Environmental Engineering – CVNG 3001 – Structural Engineering (2012)

1

MATRIX STIFFNESS ANALYSIS – BEAMS

LECTURER: Dr. William Wilson

Assumptions:

(i) Beam is prismatic between nodes (i.e. constant cross-section)

(ii) Loads are applied at the nodes (loading along the member is transformed to

equivalent loads acting at the nodes).

(iii) Local and global coordinates are collinear

Idealisation: The beam is sub-divided into discrete elements with nodes at the ends.

Location of nodes:

At supports

Connection points of members

Points of application of external loads

Sudden change in cross-section

Where displacements (deflections, rotations) are to be determined

y

x 1 2 3 4 (1) (2) (3)

1 2

3

4 5

6 7 8 P

Fig. 3.1 Beam discretised into 3 elements

Nodes numbers 1, 2, 3, 4

Members numbers (1) (2) (3)

DOFs 1, 2, 3, 4, 5, 6, 7


2

BEAM ELEMENT STIFFNESS MATRIX

The Beam element s shown in Figure 3.2

Coordinates confirm to right-hand screw rule

Local x' and global x coordinates are collinear. (Transformation not necessary)

diy vertical translation y' direction at node i

diz rotational DOF about z-axis at node i

fiy member force in y' direction at node i

fiz, member force in y' direction at node i

Considering the effects of bending and shear deformation:

DOFs - 2 degrees of freedoms at each node (1 translation and 1 rotation)

Transformation not necessary - Since local and global coordinates are collinear (x', x

are parallel)

1 2

f1z, d1z

f2z, d2z

x', x

y', y

f1y, d1y

f2y, d2y

Fig. 3.2 Beam member – Local and Global coordinate (+ve sign conventions)


3

Considering displacements:

1 2

1z 2y2

6EIf d

Lf1z, d1z

x'

y'

1y 2y3

12EIf d

L

Fig. 3.4 - Imposing +ve displacement d2y at joint 2 (all other possible

displacements restrained)

d2y 2y 2y3

12EIf d

L

2 2y2

6EIf d

Lz

1 2

1z 1y2

6EIf d

Lf1z, d1z

2z 1y2

6EIf d

L

x'

y'

1y 1y3

12EIf d

L

Fig. 3.3 - Imposing +ve displacement d1y at joint 1 (all other possible


d1y

2y 1y3

12EIf d

L


4

Considering Rotations

1 2

1y 1z2

6EIf d

Lf1z, d1z

x'

y'

1 1z

4EIf d

Lz

Fig. 3.5 - Imposing +ve rotation d1z at joint 1 (all other possible displacements

restrained)

d1z

2 1z

2EIf d

Lz

2y 1z2

6EIf d

L

1 2

1y 2z2

6EIf d

Lf1z, d1z

x'

y'

1 2z

2EIf d

Lz

Fig. 3.6 - Imposing +ve rotation d2z at joint 2 (all other possible


d2z

2y 2z2

6EIf d

L

2 2z

4EIf d

Lz


5

By superposition of these results we obtain the force-displacement relations for the beam

element as follow.

3 2 3 2

1y 1y

2 21z 1z

2y 2y3 2 3 2

2z 2z

2 2

12EI 6EI 12EI 6EI-

L L L Lf d

6EI 4EI 6EI 2EI-

f dL L L L=

f d12EI 6EI 12EI 6EI- - -

L L L Lf d6EI 2EI 6EI 4EI

-L L L L

Eqn. 2.1

If E, I and L are constants we get;

2 2

1y 1y

1z 1z

2y 2y2 2

2z 2z

12 6 12 6-

L L L Lf d

6 64 - 2

f dEI L L=

f d12 6 12 6L- - -

L L L Lf d6 6

2 - 4L L

Or, in symbolic form

[f] = [k] [d] Eqn. 2.2

Where,

[k] is the member stiffness matrix.

GLOBAL (STRUCTURE) STIFFNESS MATRIX

The Global stiffness matrix [K] is obtained by assembly of the individual member

stiffness matrices [k] for n elements.

n

i

1

K k

The structure force-displacement relationship is, as before

[F] = [K][D]

Where,

[K] is the structure stiffness matrix.


6

INTERMEDIATE LOADS

Matrix structural analysis formulated on the basis of all external loads are applied at

the joints

All loads occurring along the member/element must be transferred to nodes.

This is achieved by considering the beam loaded by equivalent fixed-end loads and

support reactions and the joints.

By superposition the equivalent loaded beam can be achieved, as follows:

Member Forces

The shear and moment at the ends of each beam element can be determined from:

[f] = [k'] [d] + [fo] Eqn. 2.3

= +

w

L

L

2

w

L

2

w

2L

12

w2L

12

w

2L

12

w

2L

12

w

L

2

w L

2

w

w

Actual Loading Fixed-end loading and

reaction on joints

Actual loading and reactions

on fixed-supported element

w = -fo + w + fo


7

BEAM EXAMPLE 1

Determine the reaction at the support of beam given in Figure B1. Assume EI is constant.

(2)

5 kN

Fig B1 (a) Node, members and joint displacement numbers

(1)

1 2

3

1

2

x

y

6 4

5 3

2m 2m

5 kN

Fig B1 - Given beam (EI constant)


8

Member Stiffness Matrices

Member 1: L = 2

6 4 5 3

1

1.5 1.5 1.5 1.5

1.5 2 1.5 1k EI

1.5 1.5 1.5 1.5

1.5 1 1.5 2

Member 2

5 3 2 1

2

1.5 1.5 1.5 1.5

1.5 2 1.5 1k EI

1.5 1.5 1.5 1.5

1.5 1 1.5 2

Structure Stiffness Matrix

1 2K = k + k

1 2 3 4 5 6

2 1.5 1 0 1.5 0

1.5 1.5 1.5 0 1.5 0

1 1.5 4 1 0 1.5K =EI

0 0 1 2 1.5 1.5

1.5 1.5 0 1.5 3 1.5

0 0 1.5 1.5 1.5 1.5

Solving for Displacements using

F = K D

and partitioning in the form:

k 11 12 u

u 21 22 k

F K K D=

F K K D


9

1

2

3

4

5

6

0 2 1.5 1 0 1.5 0 D

5 1.5 1.5 1.5 0 1.5 0 D

0 1 1.5 4 1 0 1.5 DEI

0 0 0 1 2 1.5 1.5 D

1.5 1.5 0 1.5 3 1.5 0

0 0 1.5 1.5 1.5 1.5 0

F

F

Hence the unknown displacements [Du] are found from

k 11 uF K D

1

u k 11D F K

1

1

2

3

4

D 0 2 1.5 1 0

D 5 1.5 1.5 1.5 0 1

D 0 1 1.5 4 1 EI

D 0 0 0 1 2

Giving

1

2

3

4

D 16.67

D 26.67 1=

D 6.67 EI

D 3.33

Solving for Reactions

Using u 21 uF K D

5

6

16.67

F 1.5 1.5 0 1.5 26.67 1EI

F 0 0 1.5 1.5 6.67 EI

3.33

Giving 5

6

F 10

F 5


10

BEAM EXAMPLE 2

Determine the moment at support A of the beam shown in Figure B2. Use E = 210 x 106

kN/mm2 and I = 24300 cm

4.

NOTE: Unconstrained freedoms are D1 and D2 (rotations at Joints B and C)

D3 = 0 since joint A is fixed

(2)

(b) Beam showing Coordinate System, Members, joints and freedoms

(1) 1

2 3

1

6

x

y

3

4 5

2

24 kN/m

8 m

E = 210 x 106 kN/mm

2

I = 24300 cm4

Fig. B2 (a) Given Beam

48 kN

A B C

3 m


11

Member (1):

224 8

FEM = 128 kNm12

Member (2)

248 3

FEM = 54 kNm8

(2)

24 kN

(c) Beam subjected to actual loads and fixed-supported reactions

(1)

B

2 C

3

1

x

y

A

3

96kN 96kN 24kN

128 kNm 54 kNm

24 kN/m 48kN

(2)

24 kN

(b) Beam showing loading analysed by stiffness method

(1) 1

2 3

x

y

96 kN 24kN

128kNm -54kNm = 74 kNm 54 kNm

96kN

128 kNm


12


Member 1: L = 8 m

3 2 5 4

33

12 210 x 10 N/m 24.3 x 10 m12EI1.196

L 8 m

3 5

22

6 210 x 10 24.3 x 106EI4.784

L 8

3 54 210 x 10 24.3 x 104EI25.515

L 8

3 52 210 x 10 24.3 x 102EI12.758

L 8

4 3 5 2

1

1.196 4.784 1.196 4.784

4.784 25.515 4.784 12.758k

1.196 4.784 1.196 4.784

4.784 12.758 4.784 25.515

Member 2: L = 3 m

3 5

33

12 210 x 10 24.3 x 1012EI22.68

L 3

3 5

22

6 210x10 24.3x106EI34.02

L 3

3 54 210x10 24.3x104EI68.04

L 3

3 52 210x10 24.3x102EI34.02

L 3


13

Hence, assembling the member stiffness matrix, we get

5 2 6 1

2

22.68 34.02 22.68 34.02

34.02 68.04 34.02 34.02k

22.68 34.02 22.68 34.02

34.02 34.02 34.02 68.04

The Global stiffness relationship is given as;

[F] = [K] [D]

Assembling the Global Stiffness Matrix and setting the force matrix introducing the

known applied external, we get;

1 2 3 4 5 6

3

4

5

6

54 68.04 34.02 0 0 34.02 34.02

74 34.02 93.555 12.758 4.784 29.236 34.02

F 0 12.758 25.515 4.784 4.784 0

F 0 4.784 4.784 1.196 1.196 0

F 34.02 29.236 4.784 1.196 23.876 22.68

F 34.02 34.02 0 0 22.68 22.68

1

2

D

D

0

0

0

0

Solving for Displacements

1

2

D54 68.04 34.02

D74 34.02 93.555

1

1

2

D 54 68.04 34.02

D 74 34.02 93.555

Giving 1 3

2

D 6191.64x10

D 8766.15 rads

The support reactions are determined from:

3

4

5

6

F 0 12.758

F 0 4.784 6191.64

F 34.02 29.236 8766.15

F 34.02 34.02


14

Giving

3 ABF 0 12.758 8766.15 111838.5kNmm + F.E.M 111.8 kNm 128 kNm

3 ABF =M =111.8 kNm + 128 kNm = 239.8 kNm

The internal moments and shears for members can be determined from:

[f] = [k] [d] + [fo]

Member 1 - Nodes 1, 2

4

3 3

5

2

f 1.196 4.784 1.196 4.784 0 96

f 4.784 25.515 4.784 12.758 0 128x10

f 1.196 4.784 23.876 29.236 0 96

f 4.784 12.758 29.236 93.555 8766.15 128


15

EXAMPLE 3 CONTINUOUS BEAM

Analyse the continuous beam shown in Figure B.3 for flexural deformations only;

ignore shear deformations and axial deformations.

Use E = 210 x 106

kN/m2.

Displacements are the rotations at joints (no translation) members axially rigid. Shear

deformations are ignored.

DOF = 4

20 kN/m

4m

E = 210 x 106 kN/m

2

Fig. B3 - Given Beam

25 kN/m

A B C

5m 6m

30 kN/m

D


16

40

130 152.5 62.5

(1) (2) (3)

-26.67 -63.33 -37.92 52.08

Restraining Forces at joints

x

40 40 90 90 62.5 62.5

26.67 -26.67 90 -90 52.08 -52.08

Fixed-end forces

1 2 3 4 (1) (2) (3)

D1 D2 D3 D4

Joint Freedoms (DOFs)

x

y


17

The member properties are given in the following table.

Member L I x 106 A x 10

6

1 4 23.48 3230

2 6 55.44 4740

3 5 34.04 3210

Because of coordinate system and beam orientation, for all elements,

1.0l 0m


Member 1

1y

1z 1 13

12y

2z 2 2

f =0 00.93 1.85 0.93 1.85

f =F D4.93 1.85 2.47k 10

f =0 00.93 1.85

f =F D4.93

sym

From which,

1 13

2 2

F D4.93 2.4710

F D2.47 4.93

Element 2

1y

1z 2 23

2y

2z 3 3

f 00.65 1.94 0.65 1.94

f =F D7.76 1.94 3.8810

f 00.65 1.94

f =F D7.76

sym

From which,

2 23

3 3

F D7.76 3.8810

F D3.88 7.76


18

Element 3

1y

1z 3 33

2y

2z 4 4

f 00.69 1.72 0.69 1.72

f =F D5.72 1.72 2.8610

f 00.69 1.72

f =F D5.72

sym

Assemble the Structure Stiffness Matrix

3

4.93 2.47 0 0

4.93 7.76 3.88 0K 10

7.76 5.72 2.86

5.72

3

4.93 2.47 0 0

12.69 3.88 010

. 13.48 2.86

5.72

sym

Determining the joint loads iF =ΣFi by calculating the sum of the fixed-end forces at the

joints in the direction of Di due to all the elements at the joint;

1 1

2 2

3 3

4 4

F =ΣF = - -26.67 = 26.67

F =ΣF = -(26.67-90) = 63.33

F =ΣF = -(90-52.08) = -37.92

F =ΣF = - 52.08 = -52.08

The stiffness relation for the structure becomes

F = K D

And the structure displacements are calculated from

-1

D = K F

1

1

2 3

3

4

D 4.93 2.47 0 0 26.67

D 12.69 3.88 0 63.3310

D 13.48 2.86 37.92

D 5.72 52.08


19

Giving

1

2 3

3

4

D 2.8

D 5.310 radians

D 2.7

D 7.8

Calculate the bending moments at the ends of the members caused by the displacement

from the formulae:

1 1 2 1x 2x 1y 2y

2 1 2 1x 2x 1y 2y

EI 6M = 4 2 d -d d -d

L L

EI 6M = 2 4 d -d d -d

L L

m l

m l

Example for element (1) 1x 1y 2x 2y 1 1 2 2d =d =d =d =0, θ =D and θ =D

Substituting the values for EI, L we get

M1 = 26.67 M2 = 32.83

Tabulating the forces in elements due to displacements;

Element M1 M2

1 26.67 32.83

2 30.50 -0.37

3 -37.55 -52.08

Calculating the Total Forces in the Members

Total Forces = Fixed end forces + Forces due to displacements

Element M1 M2

1 -26.67 + 26.67 = 0 26.67 + 32.38 = 59.50

2 -90.0 + 30.5 = -59.5 90.0 + 0.37 = 89.63

3 -52.08 + -37.55 = -89.63 52.08 – 52.08 = 0


20

Final B.M. Diagram

10.26

59.5

0

60.44

89.63

33.31

Final B. M. Diagram


21

PROBLEM SHEET 2 – Beams

30 kN 2 kN/m

25 kN/m 4 kN/m 4 kN/m 2 kN/m

6m 2m

10m 6m 10m 6m

2m 6m

EI constant

(1) Determine reactions at supports

EI constant

(2) Determine reactions at supports

(4) Determine moments at A, B and C

1.5I I

(3) Determine reactions and moment at support B

EI constant

A B

C

beam stiffness analysis 2012

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