chapter05_Plane Frame and Grid Eq.doc

CIVL 7117 Finite Elements Methods in Structural Mechanics Page 181

Plane Frame and Grid Equations

Introduction

Many structures, such as buildings and bridges, are composed of framesand/or grids. This chapter develops the equations and methods for solution ofplane frames and grids. First, we will develop the stiffness matrix for a beamelement arbitrarily oriented in a plane. We will then include the axial nodaldisplacement degree of freedom in the local beam element stiffness matrix. Thenwe will combine these results to develop the stiffness matrix, including axialdeformation effects, for an arbitrarily oriented beam element. We will alsoconsider frames with inclined or skewed supports.

Two-Dimensional Arbitrarily Oriented Beam Element

We can derive the stiffness matrix for an arbitrarily oriented beam element,shown in the figure below, in a manner similar to that used for the bar element.The local axes x and y are located along the beam element and transverse toˆ ˆ

the beam element, respectively, and the global axes x and y are located to beconvenient for the total structure.

The transformation from local displacements to global displacements is given inmatrix form as:


sin

cosˆ

ˆ

S

C

d

d

CS

SC

d

d

y

x

y

x

Using the second equation for the beam element, we can relate local nodaldegrees of freedom to global degree of freedom:

1

1 1

1 1

22

22

2

ˆ 0 0 0 0ˆ 0 0 1 0 0 0 ˆ

ˆ 0 0 0 0

0 0 0 0 0 1ˆ

X

y y

y x yXy

y

dd dS C

dSd CddS Cdd

For a beam we will define the following as the transformation matrix :

100000

0000

000100

0000

CS

CS

T

Notice that the rotations are not affected by the orientation of the beam.Substituting the above transformation into the general form of the stiffness matrix

TkTk T ˆ gives:

Let’s know consider the effects of an axial force in the general beamtransformation.

2 2

2 2

2 2

3 2 2

2 2

2 2

12 12 6 12 12 6

12 12 6 12 12 6

6 6 4 6 6 2

12 12 6 12 12 6

12 12 6 12 12 6

6 6 2 6 6 4

SSC LS S SC LS

SC C LC SC C LC

LS LC L LS LC LEIk

L SSC LS S SC LS

SC C LC SC C LC

LS LC L LS LC L


Recall the simple axial deformation, define in the spring element:

x

x

x

x

d

d

L

AE

f

f

2

1

2

1

ˆ

ˆ

1

1

1

1ˆ

ˆ

Combining the axial effects with the shear force and bending moment effects, inlocal coordinates,gives:

where

321 L

EIC

L

AEC

11 1 1

11 2 2 2 22 2

1 2 2 2 2 1

1 1 22

2 2 2 222

2 22 2 2 2

2 2

ˆˆ 0 0 0 0ˆˆ 0 12 6 0 12 6ˆˆ 0 6 4 0 6 2

ˆ ˆ0 0 0 0

0 12 6 0 12 6 ˆˆ0 6 2 0 6 4ˆ ˆ

xx

xy

xx

yy

df C CdfCLC C LC

mLC C L LC C L

C C dfCLC C LC df

LC C L LC C Lm


Therefore:

2

22

2

22

2222

11

2

22

2

22

2222

11

460260

61206120

0000

260460

61206120

0000

ˆ

LCLCLCLC

LCCLCC

CC

LCLCLCLC

LCCLCC

CC

k

The above stiffness matrix include the effects of axial force in the x direction,ˆshear force in the y , and bending moment about the z axis. The local degrees ofˆ ˆ

freedom may be related to the global degrees of freedom by:

2

2

2

1

1

1

2

2

2

1

1

1

100000

0000

0000

000100

0000

0000

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

y

x

x

x

y

x

x

x

d

d

d

d

CS

SC

CS

SC

d

d

d

d

where the transformation matrix, including axial effects is:

100000

0000

0000

000100

0000

0000

CS

SC

CS

SC

T


Substituting the above transformation into the general form of the stiffness matrix TkTk T ˆ gives:

The analysis of a rigid plane frame can be undertaken by applying stiffnessmatrix. A rigid plane frame is defined here as a series of beam elements rigidlyconnected to each other; that is, the original angles made between elements attheir joints remain unchanged after the deformation. Furthermore, moments aretransmitted from one element to another at the joints. Hence, moment continuityexists at the rigid joints. In addition, the element centroids, as well as the appliedloads, lie in a common plane. We observe that the element stiffnesses of a frameare functions of E, A, L, I, and the angle of orientation of the element withrespect to the global-coordinate axes.

Ek

L


Rigid Plane Frame Example

Consider the frame shown in the figure below.

The frame is fixed at nodes 1 and 4 and subjected to a positive horizontal forceof 10,000 lb applied at node 2 and to a positive moment of 5,000 lb-in. applied atnode 3. Let E = 30 x 10 psi and A = 10 in. for all elements, and let I = 200 in.6 2 4

for elements 1 and 3, and I = 100 in. for element 2.4

Element 1: The angle between x and x is 90°ˆ

10 SC

where

32

220.10

120

)200(66167.0

120

)200(1212in

L

Iin

L

I

36

/000,250120

1030inlb

L

E


Therefore, for element 1:

(1)

1 1 1 2 2 2

0.167 0 10 0.167 0 10

0 10 0 0 10 0

10 0 800 10 0 400250,000

0.167 0 10 0.167 0 10

0 10 0 0 10 0

10 0 400 10 0 800

d d d dx y x y

lbk in


01 SC

32

220.5

120

)100(660835.0

120

)100(1212in

L

Iin

L

I


inlbk

yd

xd

yd

xd

4005020050

50835.0050835.00

00100010

2005040050

50835.0050835.00

00100010

000,250)2(

333222


10 SC

32

220.10

120

)200(66167.0

120

)200(1212in

L

Iin

L

I

36

/000,250120

1030inlb

L

E



inlbk

yd

xd

yd

xd

800010400010

01000100

100167.0100167.0

400010800010

01000100

100167.0100167.0

000,250)3(

444333

The boundary conditions for this problem are:

1 1 1 4 4 4 0x y x yd d d d

After applying the boundary conditions the global beam equations reduce to:

3

3

3

2

2

2

5

120051020050

50835.10050835.00

100167.100010

200501200510

50835.0050835.100

0010100167.10

105.2

000,5

0

0

0

0

000,10

y

x

y

x

d

d

d

d

Solving the above equations gives:

2

2

2

3

3

3

0.211

0.00148

0.00153

0.209

0.00148

0.00149

x

y

x

y

d in

d in

rad

d in

d in

rad


Element 1: The element force-displacement equations can be obtained using dTkf ˆˆ . Therefore, dT is:

rad

in

in

rad

ind

ind

d

d

dT

y

x

y

x

00153.0

211.0

00148.0

0

0

0

00153.0

00148.0

211.0

0

0

0

100000

001000

010000

000100

000001

000010

2

2

2

1

1

1

Recall the elemental stiffness matrix is:

2

22

2

22

2222

11

2

22

2

22

2222

11

460260

61206120

0000

260460

61206120

0000

ˆ

LCLCLCLC

LCCLCC

CC

LCLCLCLC

LCCLCC

CC

k

Therefore, the local force-displacement equations are:

rad

in

indTkf

00153.0

211.0

00148.0

0

0

0

800100400100

10167.0010167.00

100100010

400100800100

10167.0010167.00

00100010

105.2ˆˆ 5)1(

Simplifying the above equations gives:

ink

lb

lb

ink

lb

lb

m

f

f

m

f

f

y

x

y

x

223

990,4

700,3

376

990,4

700,3

ˆ

ˆ

ˆˆ

ˆ

ˆ

2

2

2

1

1

1


Element 2: The element force-displacement equations are:

rad

in

in

rad

in

in

rad

ind

ind

rad

ind

ind

dT

y

x

y

x

00149.0

00148.0

209.0

00153.0

00148.0

211.0

00149.0

00148.0

209.0

00153.0

00148.0

211.0

100000

010000

001000

000100

000010

000001

3

3

3

2

2

2


rad

in

in

rad

in

in

dTkf

00149.0

00148.0

209.0

00153.0

00148.0

211.0

4005020050

50833.0050833.00

00100010

2005040050

50833.0050833.00

00100010

105.2ˆˆ 5)2(


ink

lb

lb

ink

lb

lb

m

f

f

m

f

f

y

x

y

x

221

700,3

010,5

223

700,3

010,5

ˆ

ˆ

ˆˆ

ˆ

ˆ

3

3

3

2

2

2


0

0

0

00149.0

209.0

00148.0

0

0

0

00149.0

00148.0

209.0

100000

001000

010000

000100

000001

000010

4

4

4

3

3

3

rad

in

in

d

d

rad

ind

ind

dT

y

x

y

x



0

0

0

00149.0

209.0

00148.0

800100400100

10167.0010167.00

100100010

400100800100

10167.0010167.00

00100010

105.2ˆˆ 5)3( rad

in

in

dTkf


ink

lb

lb

ink

lb

lb

m

f

f

m

f

f

y

x

y

x

375

010,5

700,3

226

010,5

700,3

ˆ

ˆ

ˆˆ

ˆ

ˆ

4

4

4

3

3

3



The frame is fixed at nodes 1 and 3 and subjected to a positive distributed load of1,000 lb/ft applied along element 2. Let E = 30 x 10 psi and A = 100 in. for all6 2

elements, and let I = 1,000 in. for all elements.4


First we need to replace the distributed load with a set of equivalent nodalforces and moments acting at nodes 2 and 3. For a beam with both end fixed,subjected to a uniform distributed load, w, the nodal forces and moments are:

2 3

(1,000 / )4020

2 2y y

wL lb ft ftf f k

2 2

2 3

(1,000 / )(40 )133,333 1,600

12 12

wL lb ft ftm m lb ft k in

If we consider only the parts of the stiffness matrix associated with the threedegrees of freedom at node 2, we get:

Element 1: The angle between x and x is 45ºˆ

707.0707.0 SC

where

6

3 222

3

30 10 12 12(1,000)58.93 / 0.0463

509 12 30 2

6 6(1,000)11.78551

12 30 2

E Ikin in

L L

Iin

L


2 2 2

(1)

50.02 49.98 8.33

58.93 49.98 50.02 8.33

8.33 8.33 4000

x yd d

kk in

Simplifying the above equation:

2 2 2

(1)

2,948 2,945 491

2,945 2,948 491

491 491 235,700

x yd d

kk in


Element 2: The angle between x and x is 0°

01 SC

where

6

3 222

3

30 10 12 12(1,000)62.5 / 0.0521

480 12 40

6 6(1,000)12.5

12 40

E Ikin in

L L

Iin

L


2 2 2

(2)

100 0 0

62.50 0 0.052 12.5

0 12.5 4,000

x yd d

kk in

Simplifying the above equation:

2 2 2

(2)

6,250 0 0

0 3.25 781.25

0 781.25 250,000

x yd d

kk in

The global beam equations reduce to:

2

2

2

700,485290491

290951,2945,2

491945,2198,9

600,1

20

0

y

x

d

d

ink

k


rad

in

in

d

d

y

x

0033.0

0097.0

0033.0

2

2

2



rad

in

in

rad

in

indT

0033.0

0092.0

00452.0

0

0

0

0033.0

0097.0

0033.0

0

0

0

100000

0707.0707.0000

0707.0707.0000

000100

0000707.0707.0

0000707.0707.0

Recall the elemental stiffness matrix is a function of values C , C , and L1 2

ink

ink

L

EIC

L

AEC 2273.0

23012

)000,1(1030893,5

23012

1030)100(3

6

32

6

1


(1)

5,893 0 10 5,893 0 0 0

0 2.730 694.8 0 2.730 694.8 0

10 694.8 117,900 0 694.8 117,000 0ˆ ˆ5,893 0 0 5,983 0 0 0.00452

0 2.730 694.8 0 2.730 694.8 0.0092

0 694.8 117,000 0 694.8 235,800 0.0033

fkTdin

in

rad


1

1

1

2

2

2

ˆ 26.64ˆ 2.268ˆ 389.1

ˆ 26.64

2.268ˆ778.2ˆ

x

y

x

y

f k

f k

m kin

kfkf

kinm



0

0

0

0033.0

0097.0

0033.0

0

0

0

0033.0

0097.0

0033.0

100000

010000

001000

000100

000010

000001

rad

in

in

rad

in

in

dT

Recall the elemental stiffness matrix is a function of values C , C , and L1 2

ink

ink

L

EIC

L

AEC 2713.0

4012

)000,1(1030250,6

4012

1030)100(3

6

32

6

1


(2)

6,250 0 0 6,250 0 0 0.0033

0 3.25 781.1 0 3.25 781.1 0.0097

0 781.1 250,000 0 781.1 125,000 0.0033ˆ ˆ6,250 0 0 6,250 0 0 0

0 3.25 781.1 0 3.25 781.1 0

0 781.1 125,000 0 781.1 250,00 0

in

in

radfkTd


ink

k

k

ink

k

k

dk

50.412

58.2

63.20

57.832

58.2

63.20

ˆˆ


To obtain the actual element local forces, we must subtract the equivalent nodalforces.

2

2

2

3

3

3

ˆ 20.63 0 20.63ˆ 2.58 20 17.42ˆ 832.57 1600 767.4

ˆ 20.63 0 20.63

2.58 20 22.58ˆ412.50 1600 2,013ˆ

x

y

x

y

f k k

f k k k

m kin kin kin

k kfk k kfkin kin kinm


Consider the frame shown in the figure below. In this example will illustrate theequivalent joint force replacement method for a frame subjected to a load actingon an element instead of at one of the joints of the structure. Since no distributedloads are present, the point of application of the concentrated load could betreated as an extra joint in the analysis.

This approach has the disadvantage of increasing the total number of joints,as well as the size of the total structure stiffness matrix K . For small structuressolved by computer, this does not pose a problem. However, for very largestructures, this might reduce the maximum size of the structure that could beanalyzed.

The frame is fixed at nodes 1, 2, and 3 and subjected to a concentrated load of15 k applied at mid-length of element 1. Let E = 30 x 10 psi, A = 8 in , and let I =6 2

800 in for all elements.4


Solution Procedure

1. Express the applied load in the element 1 local coordinate system (herex is directed from node 1 to node 4).ˆ

2. Next, determine the equivalent joint forces at each end of element 1,using the table in Appendix D (see figure below).


3. Then transform the equivalent joint forces from the local coordinatesystem forces into the global coordinate system forces, using theequation fTf T ˆ . These global joint forces are shown below.

4. Then we analyze the structure, using the equivalent joint forces (plusactual joint forces, if any) in the usual manner.


5. The final internal forces developed at the ends of each element may beobtained by subtracting Step 2 joint forces from Step 4 joint forces.

Element 1: The angle between x and x is 63.43°ˆ

895.0447.0 SC

where

2 3

22

12 12(800) 6 6(800)0.0334 8.95

44.7 1244.7 12 I I

in inL L

6330 10

55.9 /44.7 12

Ekin

L


(1)

4 4 4

90.0 178 448

178 359 244

448 244 179,000

kin

d dx y

k

Element 2: The angle between x and x is 116.57°ˆ

895.0447.0 SC

where

2 3

22

12 12(800) 6 6(800)0.0334 8.95

44.7 1244.7 12 I I

in inL L

6330 10

55.9 /44.7 12

Ekin

L


(2)

4 4 4

90.0 178 448

178 359 244

448 244 179,000

kin

d dx y

k


Element 3: The angle between x and x is 0° (The author of your textbookˆdirected the element from node 4 to 3. In general, as we have discussed in class,we usually number the element numerically or from 3 to 4. In this case the anglebetween x and x is 180°)ˆ

6330 10

1 0 50 /50 12

EC S k in

L

2 3

22

12 12(800) 6 6(800)0.0267 8.0

50 1250 12 I I

in inL L


(2)

4 4 4

400 0 0

0 1.334 400

0 400 160,000

kin

d dx y

k

The global beam equations reduce to:

4

4

4

7.5 582 0 896

0 0 719 400

900 896 400 518,000

x

y

k d

d

kin


4

4

4

0.0103

0.000956

0.00172

x

y

din

din

rad


Element 1: The element force-displacement equations can be obtainedu s i n g f k T d . T h e r e f o r e , d T ˆ ˆ is:

895.0447.0

100000

0000

0000

000100

0000

0000

SC

CS

SC

CS

SC

T

0.447 0.895 0 0 0 0 0 0

0.895 0.447 0 0 0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 0.447 0.895 0 0.0103 0.00374

0 0 0 0.895 0.447 0 0.000956 0.00963

0 0 0 0 0 1 0.00172 0.00172

Tdin in

in in

rad rad

Recall the elemental stiffness matrix is:

222

222

2222

11

222

222

2222

11

460260

61206120

0000

260460

61206120

0000

ˆ

LCLCLCLC

LCCLCC

CC

LCLCLCLC

LCCLCC

CC

k

ink

ink

L

EIC

L

AEC 155.0

72.4412

)800(10302.447

72.4412

1030)8(3

6

32

6

1



rad

in

indTkf

00172.0

00963.0

00374.0

0

0

0

000,1795.5000490,895.5000

5.500868.105.500868.10

0044700447

490,895.5000000,1795.5000

5.500868.105.500868.10

0044700447

ˆˆ)1(


(1)

1.67

0.88

158ˆ ˆ ˆ1.67

0.88

311

k

k

kinfkd

k

k

kin

To obtain the actual element local forces, we must subtract the equivalent nodalforces.

1

1

1

4

4

4

ˆ 1.67 3.36 5.03ˆ 0.88 6.71 7.59ˆ 158 900 1,058

ˆ 1.67 3.36 1.68

0.88 6.71 5.83ˆ311 900 589ˆ

x

y

x

y

f k k k

f k k k

m kin kin kin

k k kfk k kf

kin kin kinm



895.0447.0

100000

0000

0000

000100

0000

0000

SC

CS

SC

CS

SC

T

rad

in

in

rad

in

indT

00172.0

00879.0

00546.0

0

0

0

00172.0

000956.0

0103.0

0

0

0

100000

0447.0895.0000

0895.0447.0000

000100

0000447.0895.0

0000895.0447.0


ink

ink

L

EIC

L

AEC 155.0

72.4412

)800(10302.447

72.4412

1030)8(3

6

32

6

1

rad

in

indTkf

00172.0

00879.0

00546.0

0

0

0

000,1795.5000490,895.5000

5.500868.105.500868.10

0044700447

490,895.5000000,1795.5000

5.500868.105.500868.10

0044700447

ˆˆ)2(



(2)

2.44

0.877

158ˆ ˆ ˆ2.44

0.877

312

k

k

kinfkd

k

k

kin

Since there are no applied loads on element 2, there are no equivalent nodalforces to account for. Therefore, the above equations are the final local nodalforces

Element 3: The element force-displacement equations can be obtainedu s i n g f k T d . T h e r e f o r e , d T ˆ ˆ is:

0

0

0

00172.0

000956.0

0103.0

0

0

0

00172.0

000956.0

0103.0

100000

010000

001000

000100

000010

000001

rad

in

in

rad

in

in

dT



ink

ink

L

EIC

L

AEC 111.0

5012

)800(1030400

5012

1030)8(3

6

32

6

1

0

0

0

00172.0

000956.0

0103.0

000,1604000000,804000

400335.10400335.10

0040000400

000,804000000,1604000

400335.10400335.10

0040000400

ˆˆ)3(

rad

in

in

dTkf


(3)

4.12

0.687

275ˆ ˆ ˆ4.12

0.687

137

k

k

kinfkd

k

k

kin

Since there are no applied loads on element 3, there are no equivalent nodalforces to account for. Therefore, the above equations are the final local nodalforces. The free-body diagrams are shown below.



The frame shown on the right is fixed at nodes 2 and3 and subjected to a concentrated load of 500 kNapplied at node 1. For the bar, A = 1 x 10 -3 m , for the2

beam, A = 2 x 10 -3 m , I = 5 x 102 -5 m , and L = 3 m.4

Let E = 210 GPa for both elements.


Beam Element 1: The angle between x and x is 0°ˆ

01 SC

where

345

25

2

5

210

3

)105(661067.6

)3(

)105(1212m

L

Im

L

I

366

/10703

10210mkN

L

E


mkNk

yx dd

20.010.00

10.0067.00

002

1070 3)1(

111

Bar Element 2: The angle between x and x is 45°ˆ

707.0707.0 S C

where

m

kNm

mkNmk

yx dd

5.05.0

5.05.0

24.4

/1021010 2623)2(

11

mkNk

yx dd

354.0354.0

354.0354.01070 3)2(

11

Assembling the elemental stiffness matrices we obtain the global stiffness matrix

mkNK

20.010.00

10.0421.0354.0

0354.0354.2

1070 3


The global equations are:

1

1

1

3

20.010.00

10.0421.0354.0

0354.0354.2

1070

0

500

0

y

x

d

d

mkNkN


rad

m

m

d

d

y

x

0113.0

0225.0

00388.0

1

1

1

Bar Element: The bar element force-displacement equations can be obtainedu s i n g f k T d . ˆ ˆ

Therefore, the forces in the bar element are:

kNSdCdL

AEf yxx 670ˆ

111

kNSdCdL

AEf yxx 670ˆ

113

Beam Element: The beam element force-displacement equations can beo b t a i n e d u s i n g f k d . S i n c e t h e l o c a l a x i s c o i n c i d e s w i t h t h e g l o b a l c o o r d i n a t e

ˆ ˆ ˆ

system, and the displacements at node 2 are zero. Therefore, the local force-displacement equations are:

y

x

y

x

x

x

d

d

d

d

SC

SC

L

AE

f

f

3

3

1

1

3

1

00

00

11

11ˆ

ˆ


32

1

2

22

2

22

2222

11

2

22

2

22

2222

11

460260

61206120

0000

260460

61206120

0000

ˆ

L

EIC

L

AEC

LCLCLCLC

LCCLCC

CC

LCLCLCLC

LCCLCC

CC

k

0

0

0

0113.0

0225.0

00388.0

20.010.0010.010.00

10.0067.0010.0067.00

002002

10.010.0020.010.00

10.0067.0010.0067.00

002002

1070ˆˆˆ 3)1(

mkN

m

m

dkf

Substituting numerical values into the above equations gives:

mkN

kN

kN

kN

kN

m

f

f

m

f

f

y

x

y

x

3.78

5.26

473

0.0

5.26

473

ˆ

ˆ

ˆˆ

ˆ

ˆ

2

2

2

1

1

1




The frame is fixed at nodes 1 and 3 and subjected to a moment of 20 kN-mapplied at node 2. Assume A = 2 x 10 -2 m , I = 2 x 102 -4 m , and E = 210 GPa for4

all elements.


10 SC

where

344

24

2

4

2103

4

)102(66105.1

)4(

)102(1212m

L

Im

L

I

376

/1025.54

10210mkN

L

E

Therefore, the stiffness matrix for element 1, considering only the partsassociated with node 2, is:

mkNk

yx dd

08.0003.0

020

03.00015.0

1025.5 5)1(

222



01 SC

where

344

25

2

4

2104.2

5

)102(66106.9

)5(

)102(1212m

L

Im

L

I

376

/102.45

10210mkN

L

E

Therefore, the stiffness matrix for element 2, considering only the partsassociated with node 2, is:

mkNk

yx dd

08.0024.00

024.00096.00

002

102.4 5)2(

222

Assembling the elemental stiffness matrices we obtain the global stiffness matrix:

mkNK

0756.00101.00158.0

0101.00500.10

0158.008480.0

106


2

2

2

6

0756.00101.00158.0

0101.00500.10

0158.008480.0

10

20

0

0

y

x

d

d

mkN



62

62

42

4.95 10

2.56 10

2.66 10

x

y

d m

d m

rad

Element 1: The beam element force-displacement equations can be obtainedu s i n g f k T d . ˆ ˆ

6 6

6 6

4 4

0 1 0 0 0 0 0 0

1 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 0 1 0 4.95 10 2.56 10

0 0 0 1 0 0 2.56 10 4.95 10

0 0 0 0 0 1 2.66 10 2.66 10

Tdm m

m m

rad rad


2

22

2

22

2222

11

2

22

2

22

2222

11

460260

61206120

0000

260460

61206120

0000

ˆ

LCLCLCLC

LCCLCC

CC

LCLCLCLC

LCCLCC

CC

k

2 66

1

6 4

2 33

(2 10 )210 101.05 10

4

210 10 (2 10 )656.25

4

kNm

kNm

AEC

L

EIC

L


rad

m

mdTkf

4

6

6

3

)1(

1066.2

1095.4

1056.2

0

0

0

830430

35.1035.10

0020000200

430830

35.1035.10

0020000200

1025.5ˆˆ

Solving for the forces and moments gives:

mkN

kN

kN

mkN

kN

kN

m

f

f

m

f

f

y

x

y

x

17.11

2.4

69.2

59.5

2.4

69.2

ˆ

ˆ

ˆˆ

ˆ

ˆ

2

2

2

1

1

1

Element 2: The beam element force-displacement equations can be obtainedu s i n g f k T d . ˆ ˆ

rad

m

m

rad

m

mdT

4

6

6

4

6

6

1066.2

1056.2

1095.4

0

0

0

1066.2

1056.2

1095.4

0

0

0

100000

010000

001000

000100

000010

000001


2 66

1

6 4

2 33

(2 10 )210 100.84 10

5

210 10 (2 10 )336

5

kNm

kNm

AEC

L

EIC

L


0

0

0

1066.2

1056.2

1095.4

840.20440.20

40.296.0040.296.00

0020000200

440.20840.20

40.296.0040.296.00

0020000200

102.4ˆˆ4

6

6

3

)2(

rad

m

m

dTkf


2

2

2

3

3

3

ˆ 4.16ˆ 2.69ˆ 8.92

ˆ 4.16

2.69ˆ4.47ˆ

x

y

x

y

f kN

f kN

m kN m

kNfkNfkN mm

Inclined or Skewed Supports

If a support is inclined, or skewed, at some angle for the global x axis, asshown below, the boundary conditions on the displacements are not in the globalx-y directions but in the x’-y’ directions.


We must transform the local boundary condition of d’3y = 0 (in local coordinates)into the global x-y system. Therefore, the relationship between of thecomponents of the displacement in the local and the global coordinate systemsat node 3 is:

3

3

3

3

3

3

100

0cossin

0sincos

'

'

'

y

x

y

x

d

d

d

d

We can rewrite the above expression as:

100

0cossin

0sincos

][' 3333

tdtd

We can apply this sort of transformation to the entire displacement vector as:

'][or][' dTddTd Tii

where the matrix [T ] is:i

][

]0[

]0[

]0[

][

]0[

]0[

]0[

][

][

3t

I

I

Ti

Both the identity matrix [I] and the matrix [t ] are 3 x 3 matrices.3

The force vector can be transformed by using the same transformation.

fTf i ]['

In global coordinates, the force-displacement equations are:

dKf ][


Applying the skewed support transformation to both sides of the force-displacement equation gives:

dKTfT ii ]][[][

By using the relationship between the local and the global displacements, theforce-displacement equations become:

']][][['']][][[][ dTKTfdTKTfT Tii

Tiii

Therefore the global equations become:

1

3

3

2

2

2

1

1

1

3

3

3

2

2

2

1

1

1

'

'

]][][[

'

'

y

x

y

x

y

x

T

ii

y

x

y

x

y

x

d

d

d

d

d

d

TKT

M

F

F

M

F

F

M

F

F

Grid Equations

A grid is a structure on which the loads are applied perpendicular to the planeof the structure, as opposed to a plane frame where loads are applied in theplane of the structure. Both torsional and bending moment continuity aremaintained at each node in a grid element. Examples of a grid structure arefloors and bridge deck systems. A typical grid structure is shown in the figurebelow.


A representation of the grid element is shown below:

The degrees of freedom for a grid element are: a vertical displacement diyˆ

(normal to the grid), a torsional rotation ix about the x axis, and a bendingˆ

rotation iz about the z axis. The nodal forces are: a transverse force fˆiy a

torsional moment m ixˆ about the x axis, and a bending moment m about the zˆ

izˆ ˆ

axis.

Let’s derive the torsional rotation components of the element stiffness matrix.Consider the sign convention for nodal torque and angle of twist shown the figurebelow.

A linear displacement function is assumed.

xaa ˆ21

Applying the boundary conditions and solving for the unknown coefficients gives:

xxx x

L 112 ˆˆˆˆ


Or in matrix form:

x

xNN2

121 ˆ

ˆˆ

where N and N are the interpolation functions gives as:1 2

L

xN

L

xN 21

ˆ1

To obtain the relationship between the shear strain and the angle of twist consider the torsional deformation of the bar as shown below.

If we assume that all radial lines, such as OA , remain straight during twisting or

torsional deformation, then the arc length AB is:

ˆˆmax RdxdAB

Therefore;

xd

Rdˆ

ˆmax

At any radial position, r, we have, from similar triangles OAB and OCD :


xxL

r

xd

dr 12

ˆˆˆ

ˆ

The relationship between shear stress and shear strain is:

G

where G is the shear modulus of the material. From elementary mechanics ofmaterials, we get:

ˆ x

Jm

Rwhere J is the polar moment of inertia for a circular cross section or thetorsional constant for non-circular cross sections. Rewriting the above equationwe get:

2 1ˆ ˆˆ xxx

GJm

L

The nodal torque sign convention gives:

xxxx mmmm ˆˆˆˆ21

Therefore;

1 1 2 2 2 1ˆ ˆ ˆ ˆˆ ˆ xxx x x x

GJ GJm m

L L

In matrix form the above equations are:

x

x

x

x

L

GJ

m

m

2

1

2

1

ˆ

ˆ

11

11

ˆ

ˆ


Combining the torsional effects with shear and bending effects, we obtain thelocal stiffness matrix equations for a grid element.

z

x

y

z

x

y

LEI

LEI

LEI

LEI

LGJ

LGJ

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LGJ

LGJ

LEI

LEI

LEI

LEI

z

x

y

z

x

y

d

d

m

m

f

m

m

f

2

2

2

1

1

1

4626

612612

2646

612612

2

2

2

1

1

1

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

00

0000

00

00

0000

00

ˆ

ˆ

ˆˆ

ˆ

ˆ

22

2323

22

2323

The transformation matrix relating local to global degrees of freedom for a gridis:

CS

SC

CS

SC

TG

0000

0000

001000

0000

0000

000001

where is now positive taken counterclockwise from x to x in the x-z plane:ˆtherefore;

cos sinj ijix xzzC S

L L

The global stiffness matrix for a grid element arbitrary oriented in the x-z plane isgiven by:

GG

T

GG TkTk ˆ


Grid Example


The frame is fixed at nodes 2, 3, and 4, and is subjected to a load of 100 kipsapplied at node 1. Assume I = 400 in. , J = 110 in. , G = 12 x 10 ksi, and E = 304 4 3

x 10 ksi for all elements.3

To facilitate a timely solution, the boundary conditions at nodes 2, 3, and 4 areapplied to the local stiffness matrices at the beginning of the solution.

0

0

0

444

333

222

zxy

zxy

zxy

d

d

d

Beam Element 1:

447.036.22

1020sin894.0

36.22

200cos

)1(

12

)1(

12 L

zzS

L

xxC

where

3 3

3 3 2 2

12 12(30 10 )(400) 6 6(30 10 )(400)7.45 1,000

(22.36 12) (22.36 12)

EI EIk kinL L


3 34 4(30 10 )(400) (12 10 )(110)179,000 4,920

(22.36 12) (22.36 12)

EI GJkin kin

L L

The global stiffness matrix for element 1, considering only the parts associatedwith node 1, and the following relationship:

GG

T

GG TkTk ˆ

894.0447.00

447.0894.00

001

894.0447.00

447.0894.00

001T

GG TT

inkk

zxyd

000,1790000,1

0920,40

000,1045.7ˆ )1(

111

Therefore, the global stiffness matrix is

inkk

zxyd

000,144600,69894

600,69700,39447

89444745.7)1(

111

Beam Element 2:

447.036.22

100sin894.0

36.22

200cos

)2(

13

)2(

13 L

zzS

L

xxC

where

3 3

3 3 2 2

12 12(30 10 )(400) 6 6(30 10 )(400)7.45 1,000

(22.36 12) (22.36 12)

EI EIk kinL L


3 34 4(30 10 )(400) (12 10 )(110)179,000 4,920

(22.36 12) (22.36 12)

EI GJkin kin

L L


GG

T

GG TkTk ˆ

894.0447.00

447.0894.00

001

000,1790000,1

0920,40

000,1045.7

894.0447.00

447.0894.00

001)2(k


inkk

zxyd

000,144600,69894

600,69700,39447

89444745.7)2(

111

Beam Element 3:

110

100sin0

10

2020cos

)3(

14

)3(

14 L

zzS

L

xxC

where

kL

EIink

L

EI000,5

)1210(

)400)(1030(66/3.83

)1210(

)400)(1030(12122

3

23

3

3

inkL

GJink

L

EI

000,11

)1210(

)110)(1012(000,400

)1210(

)400)(1030(44 33


GG

T

GG TkTk ˆ


010

100

001

000,4000000,5

0000,110

000,503.83

010

100

001)3(k


000,1100

0000,400000,5

0000,53.83

)3(

111

k

zxyd

Superimposing the three elemental stiffness matrices gives:

1 1 1

98.2 5,000 1,790

5,000 479,000 0

1,790 0 299,000

y x zd

K


1 1

1 1

1 1

100 98.2 5,000 1,790

0 5,000 479,000 0

0 1,790 0 299,000

y y

x x

z z

F k d

M

M


rad

rad

ind

z

x

y

0169.0

0295.0

83.2

1

1

1


Element 1: The grid element force-displacement equations can be obtainedusing dTkf GG

ˆˆ .

0

0

0

00192.0

0339.0

83.2

0

0

0

0169.0

0295.0

83.2

894.0447.00000

447.0894.00000

001000

000894.0447.00

000447.0894.00

000001

rad

rad

in

rad

rad

in

dTG


0

0

0

00192.0

0339.0

83.2

000,1790000,1500,890000,1

0920,400920,40

000,1045.7000,1045.7

500,890000,1000,1790000,1

0920,400920,40

000,1045.7000,1045.7

ˆˆ)1(

rad

rad

in

dTkf


1

1

1

2

2

2

ˆ 19.2ˆ 167ˆ 2,480

ˆ 19.2

167ˆ

2,260ˆ

y

x

z

y

x

z

f k

m kin

m kin

kfkinmkinm


Element 2: The grid element force-displacement equations can be obtainedusing ˆ ˆ

G GfkT d .

0

0

0

0283.0

0188.0

83.2

0

0

0

0169.0

0295.0

83.2

894.0447.00000

447.0894.00000

001000

000894.0447.00

000447.0894.00

000001

rad

rad

in

rad

rad

in

dTG


0

0

0

0283.0

0188.0

83.2

000,1790000,1500,890000,1

0920,400920,40

000,1045.7000,1045.7

500,890000,1000,1790000,1

0920,400920,40

000,1045.7000,1045.7

ˆˆ)2(

rad

rad

in

dTkf


1

1

1

3

3

3

ˆ 7.23ˆ 92.5ˆ 2,240

ˆ 7.23

92.5ˆ

295ˆ

y

x

z

y

x

z

f k

m kin

m kin

kfkinm

kinm



G GfkT d .

0

0

0

0295.0

0169.0

83.2

0

0

0

0169.0

0295.0

83.2

010000

100000

001000

000010

000100

000001

rad

rad

in

rad

rad

in

dTG


0

0

0

0295.0

0169.0

83.2

000,4000000,5000,2000000,5

0000,1100000,110

000,503.83000,503.83

000,2000000,5000,4000000,5

0000,1100000,110

000,503.83000,503.83

ˆˆ)3(

rad

rad

in

dTkf


1

1

1

4

4

4

ˆ 88.1ˆ 186ˆ 2,340

ˆ 88.1

186ˆ

8,240ˆ

y

x

z

y

x

z

f k

m kin

m kin

kfkinmkinm


To check the equilibrium of node 1 the local forces and moments for eachelement need to be transformed to global coordinates. Recall, that:

1ˆˆ TTfTfTff TT

Since we are only checking the forces and moments at node 1, we need only theupper-left-hand portion of the transformation matrix T .G

Therefore; for Element 1:

1

1

1

1 0 0 19.2 19.2

0 0.894 0.447 167 1,260

0 0.447 0.894 2,480 2,150

y

x

z

fkk

mkin k in

mkin k in



1

1

1

1 0 0 7.23 7.23

0 0.894 0.447 92.5 1,080

0 0.447 0.894 2,240 1,960

y

x

z

fkk

mkin k in

mkin k in


1

1

1

1 0 0 88.1 88.1

0 0 1 2,340 2,340

0 1 0 186 186

y

x

z

fkk

mkin k in

mkin k in

The forces and moments that are applied to node 1 by each element are equal inmagnitude and opposite direction. Therefore the sum of the forces and momentsacting on node 1 are:

The forces and moments accurately satisfy equilibrium considering theamount of truncation error inherent in results of the calculationspresented in this example.

kF y 07.01.882.1923.71001

inkM x 0.0340,2080,1260,11

inkM z 0.4186060,1150,21


Grid Example


The frame is fixed at nodes 1 and 3, and is subjected to a load of 22 kN appliedat node 2. Assume I = 16.6 x 10 -5 m , J = 4.6 x 104 -5 m , G = 84 GPa, and E = 2104

GPa for all elements.

To facilitate a timely solution, the boundary conditions at nodes 1 and 3 areapplied to the local stiffness matrices at the beginning of the solution.

0

0

333

111

zxy

zxy

d

d

Beam Element 1: the local x axis coincides with the global x axis

03

0sin1

3

3cos

)1(

12

)1(

12 L

zzS

L

xxC

where

mkN /L

EI1055.1

)3(

)106.16)(10210(1212 4

3

56

3

kNL

EI 4

2

56

21032.2

)3(

)106.16)(10210(66

mkN·L

EI1065.4

3

)106.16)(10210(44 456

mkN·L

GJ10128.0

3

)106.4)(1084( 456


The global stiffness matrix for element 1, considering only the parts associatedwith node 2, may be obtained from the following relationship:

GG

T

GG TkTk ˆ

mkNk

100

010

001

65.4032.2

0128.00

32.2055.1

100

010

001

104)1(


mkNk

zxyd

65.4032.2

0128.00

32.2055.1

104

)1(

222

Beam Element 2: the local x axis is located from node 2 to node 3ˆ

3 2 3 2(2) (1)

0 3cos 0 sin 1

3 3

x x z zC S

L L

The global stiffness matrix for element 2, considering only the parts associatedwith node 2, may be obtained using:

GG

T

GG TkTk ˆ

mkNk

010

100

001

65.4032.2

0128.00

32.2055.1

010

100

001

104)2(


mkNk

zxyd

128.000

065.432.2

032.255.1

104)2(

222


Superimposing the two elemental stiffness matrices gives:

mkNK

zxyd

78.4032.2

078.432.2

32.232.210.3

10 4

222


x

x

y

z

x

y d

M

M

kNF

2

2

2

4

2

2

2

78.4032.2

078.432.2

32.232.210.3

10

0

0

22


2

2

2

0.00259

0.00126

0.00126

y

x

z

d m

rad

rad


G GfkT d .

rad

rad

m

rad

rad

mdTG

00126.0

00126.0

00259.0

0

0

0

00126.0

00126.0

00259.0

0

0

0

100000

010000

001000

000100

000010

000001



rad

rad

mdTkf

00126.0

00126.0

00259.0

0

0

0

65.4032.233.2032.2

0128.000128.00

32.2055.132.2055.1

33.2032.265.4032.2

0128.000128.00

32.2055.132.2055.1

10ˆˆ 4

)1(


1

1

1

2

2

2

ˆ 11.0ˆ 1.50ˆ 31.0

ˆ 11.0

1.50ˆ1.50ˆ

y

x

z

y

x

z

f kN

m kN m

m kN m

kNfkN mmkN mm


G GfkT d .

0

0

0

00126.0

00126.0

00259.0

0

0

0

00126.0

00126.0

00259.0

010000

100000

001000

000010

000100

000001

rad

rad

m

rad

rad

m

dTG



0

0

0

00126.0

00126.0

00259.0

65.4032.233.2032.2

0128.000128.00

32.2055.132.2055.1

33.2032.265.4032.2

0128.000128.00

32.2055.132.2055.1

10ˆˆ 4)2(

rad

rad

m

dTkf


2

2

2

3

3

3

ˆ 11.0ˆ 1.50ˆ 1.50

ˆ 11.0

1.50ˆ

31.0ˆ

y

x

z

y

x

z

f kN

m kN m

m kN m

kNfkN mm

kN mm

The resulting free-body diagrams:


Beam Element Arbitrarily Oriented in Space

In this section,we will develop a beam element that is arbitrarily oriented inthree-dimensions. This element can be used to analyze three-dimensionalframes. Let consider bending about axes, as shown below.

The y axis is the principle axis for which the moment of inertia is minimum, I .ˆ y

The right-hand rule is used to establish the z axis and the maximum moment ofˆinertia, I .z

Bending in the zx ˆˆ plane: The bending in the zx plane is defined by m .yˆ

The stiffness matrix for bending the in the x-z plane is:

3

2

3

2

2

2

3

2

3

2

2

2

4

4

6

2

6

6

12

6

12

2

6

4

6

6

12

6

12

ˆ

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

EIk y

Y

where I is the moment of inertia about the y axis (the weak axis).y ˆ

Bending in the yx ˆˆ plane: The bending in the yx plane is defined by m .ˆz

The stiffness matrix for bending the in the yx plane is:

3

2

3

2

2

2

3

2

3

2

2

2

4

4

6

2

6

6

12

6

12

2

6

4

6

6

12

6

12

ˆ

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

EIk z

z

where I is the moment of inertia about the z axis (the strong axis).z ˆ


Direct superposition of the bending stiffness matrices with the effects of axialforces and torsional rotation give:

The global stiffness matrix may be obtained using:

TkTk T ˆ

where

33

33

33

33

x

x

x

x

T

where

zyxzyxzyxzyx dddddd 222222111111ˆˆˆˆˆˆˆˆˆˆˆˆ


zzzyzx

yzyyyx

xzxyxx

x

CCC

CCC

CCC

ˆˆˆ

ˆˆˆ

ˆˆˆ

33

where the direction cosines, C , are defined as shown belowji ˆ

The direction cosines of the x axis are:ˆ

kjix xzxyxx

GGGˆˆˆ coscoscosˆ

where

nL

zzm

L

yyl

L

xxxzxyxx

12ˆ

12ˆ

12ˆ coscoscos

The y axis is selected to be perpendicular to the x and the z axes is such a wayˆ ˆthat the cross product of global z with x results in the y axis as shown in theˆ ˆ

figure below.

yxz ˆˆ


3 3 0x

lmn

m lD Dln mn

DD D

jD

li

D

m

nml

kji

Dyxz

GG

GGG

1001

ˆˆ

where 22 mlD

The z axis is determined by the condition thatˆ yxz ˆˆˆ

kDjD

mni

D

nl

lm

nml

kji

Dyxz

GGG

GGG

0

1ˆˆˆ

Therefore, the transformation matrix becomes:

zzzyzx

yzyyyx

xzxyxx

x

CCC

CCC

CCC

ˆˆˆ

ˆˆˆ

ˆˆˆ

33

There are two exceptions that arise when using the above expressions formapping the local coordinates to the global system: (1) when the positive xcoincides with z; and (2) when the positive x is in the opposite direction as z. Forˆthe first case, it is assumed that y is y.ˆ

0 0 1

0 0 0

1 0 0

In case two, it is assumed that y is y.ˆ

0 0 1

0 0 0

1 0 0


If the effects of axial force, both shear forces, twisting moment, and both bendingmoments are considered, the stiffness matrix for a frame element is:


In this case the symbol are:

2 2

12 12y zy z

s s

EI EI

GA L GA L

where A is the effective beam cross-section in shear. Recall the shear moduluss

of elasticity or the modulus of rigidity, G, is related to the modulus of elasticityand the Poisson’s ratio, as:

2 1

EG

If y and z are set to zero, the stiffness matrix reduces to that shown previouslyon page 235. This is the form of the stiffness matrix used by SAP2000 for itsframe element.

zyxzyxzyxzyx dddddd 222222111111ˆˆˆˆˆˆˆˆˆˆˆˆ


Example Frame Application

A bus subjected to a static roof-crush analysis. In this model 599 frameelements and 357 nodes are used.

Concept of Substructure Analysis

Sometimes structures are too large to be analyzed as a single system ortreated as a whole; that is, the final stiffness matrix and equations for solutionexceed the memory capacity of the computer. A procedure to overcome thisproblem is to separate the whole structure into smaller units calledsubstructures . For example, the space frame of an airplane, as shown below,may require thousands of nodes and elements to completely model and describethe response of the whole structure. If we separate the aircraft into substructures,


such as parts of the fuselage or body, wing sections, etc., as shown below, thenwe can solve the problem more readily and on computers with limited memory.


Problems

14. Do problems 5.3, 5.8, 5.13, 5.28, 5.41, and 5.43 on pages 240 - 263 in yourtextbook “A First Course in the Finite Element Method” by D. Logan.

15. Do problems 5.23, 5.25, 5.35, 5.39 , and 5.55 on pages 240 - 263 in yourtextbook “A First Course in the Finite Element Method” by D. Logan. Youmay use the SAP2000 to do frame analysis.

Documents

chapter05_Plane Frame and Grid Eq.doc