plane frame and grid equations

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 frames and/or grids. This chapter develops the equations and methods for solution of plane frames and grids. First, we will develop the stiffness matrix for a beam element arbitrarily oriented in a plane. We will then include the axial nodal displacement degree of freedom in the local beam element stiffness matrix. Then we will combine these results to develop the stiffness matrix, including axial deformation effects, for an arbitrarily oriented beam element. We will also consider 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 to the beam element, respectively, and the global axes x and y are located to be convenient for the total structure.

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


θθ

sincos

ˆˆ

==

−

=

SC

dd

CSSC

dd

y

x

y

x

Using the second equation for the beam element, we can relate local nodal degrees 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 00 0 0 0 0 1ˆ

φ φ

φφ

− = = − + −

X

y y

y x yXy

y

dd dS C

d Sd CddS Cdd

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

−

−

=

10000000000001000000

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 beam transformation.

2 2

2 2

2 2

3 2 2

2 2

2 2

12 12 6 12 12 612 12 6 12 12 66 6 4 6 6 212 12 6 12 12 6

12 12 6 12 12 66 6 2 6 6 4

− − − − − − − −

= − −

− − − − − −

S SC LS S SC LSSC C LC SC C LCLS LC L LS LC LEIk

L S SC LS S SC LSSC C LC SC C LCLS LC L LS LC L


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

−−

=

x

x

x

x

dd

LAE

ff

2

1

2

1

ˆˆ

11

11

ˆˆ

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

where

321 LEIC

LAEC ==

11 1 1

11 2 2 2 22 2

1 2 2 2 2 1

1 1 22

2 2 2 222 2 2

2 2 2 22 2

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

ˆ ˆ0 0 0 00 12 6 0 12 6 ˆˆ0 6 2 0 6 4ˆ ˆ

φ

φ

− − − = − − − −

−

xx

xy

xx

yy

df C Cdf C LC C LC

m LC C L LC C LC C df

C LC C LC dfLC C L LC C Lm


Therefore:

−−−−

−−−

−

=

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

The above stiffness matrix include the effects of axial force in the x direction, shear force in the y , and bending moment about the z axis. The local degrees of freedom may be related to the global degrees of freedom by:

−

−

=

2

2

2

1

1

1

2

2

2

1

1

1

1000000000000000010000000000

ˆˆˆˆˆˆ

φ

φ

φ

φ

y

x

x

x

y

x

x

x

dd

dd

CSSC

CSSC

dd

dd

where the transformation matrix, including axial effects is:

−

−

=

1000000000000000010000000000

CSSC

CSSC

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 stiffness matrix. A rigid plane frame is defined here as a series of beam elements rigidly connected to each other; that is, the original angles made between elements at their joints remain unchanged after the deformation. Furthermore, moments are transmitted from one element to another at the joints. Hence, moment continuity exists at the rigid joints. In addition, the element centroids, as well as the applied loads, lie in a common plane. We observe that the element stiffnesses of a frame are functions of E, A, L, I, and the angle of orientation θ of the element with respect to the global-coordinate axes.

EkL

=


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 force of 10,000 lb applied at node 2 and to a positive moment of 5,000 lb-in. applied at node 3. Let E = 30 x 106 psi and A = 10 in.2 for all elements, and let I = 200 in.4 for elements 1 and 3, and I = 100 in.4 for element 2. Element 1: The angle between x and x is 90°

10 == SC where

( )32

22 0.10120

)200(66167.0120

)200(1212 inLIin

LI

====

3

6

/000,250120

1030 inlbLE

=×

=


Therefore, for element 1:

(1)

1 1 1 2 2 2

0.167 0 10 0.167 0 100 10 0 0 10 010 0 800 10 0 400

250,0000.167 0 10 0.167 0 10

0 10 0 0 10 010 0 400 10 0 800

φ φ

− − − − −

= − −

−

d d d dx y x y

lbk in

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

01 == SC

( )32

22 0.5120

)100(660835.0120

)100(1212 inLIin

LI

====


inlbk

yd

xd

yd

xd

−−−

−−

−

=

400502005050835.0050835.00

00100010200504005050835.0050835.0000100010

000,250)2(

333222φφ


10 −== SC

( )32

22 0.10120

)200(66167.0120

)200(1212 inLIin

LI

====

3

6

/000,250120

1030 inlbLE

=×

=



inlbk

yd

xd

yd

xd

−−

−−−−

−−

=

80001040001001000100100167.0100167.0

40001080001001000100

100167.0100167.0

000,250)3(

444333φφ

The boundary conditions for this problem are:

0414411 ====== φφyxyx dddd After applying the boundary conditions the global beam equations reduce to:

−−−−

−−

−−

×=

3

3

3

2

2

2

5

12005102005050835.10050835.00

100167.10001020050120051050835.0050835.1000010100167.10

105.2

000,50000000,10

φ

φ

y

x

y

x

dd

dd

Solving the above equations gives:

2

2

2

3

3

3

0.2110.00148

0.001530.209

0.001480.00149

φ

φ

−

= −

−

x

y

x

y

d ind in

radd ind in

rad


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

−−

=

−===

===

−

−

=

radinin

radind

ind

dd

dT

y

x

y

x

00153.0211.0

00148.0000

00153.000148.0

211.0000

100000001000010000000100000001000010

2

2

2

1

1

1

φ

φ

Recall the elemental stiffness matrix is:

−−−−

−−−

−

=

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

Therefore, the local force-displacement equations are:

−−

−−−−

−−

−−

×==

radinin

dTkf

00153.0211.0

00148.0000

80010040010010167.0010167.00

10010001040010080010010167.0010167.0000100010

105.2ˆˆ 5)1(

Simplifying the above equations gives:

⋅−

⋅

−

=

inklblbinklblb

mffmff

y

x

y

x

223990,4700,3

376990,4700,3

ˆ

ˆˆˆ

ˆˆ

2

2

2

1

1

1


Element 2: The element force-displacement equations are:

−−

−

−

=

−=−==

−===

=

radinin

radinin

radind

indradind

ind

dT

y

x

y

x

00149.000148.0

209.000153.0

00148.0211.0

00149.000148.0209.0

00153.000148.0

211.0

100000010000001000000100000010000001

3

3

3

2

2

2

φ

φ


−−

−

−−−−

−−

−−

×==

radinin

radinin

dTkf

00149.000148.0

209.000153.0

00148.0211.0

400502005050833.0050833.00

00100010200504005050833.0050833.0000100010

105.2ˆˆ 5)2(


⋅−

−⋅−

−

=

inklblbinklblb

mffmff

y

x

y

x

221700,3010,5

223700,3010,5

ˆ

ˆˆˆ

ˆˆ

3

3

3

2

2

2


−=

===

−=−==

−

−

=

000

00149.0209.0

00148.0

000

00149.000148.0209.0

100000001000010000000100000001000010

4

4

4

3

3

3

radin

in

dd

radind

ind

dT

y

x

y

x

φ

φ



−

−−−−

−−

−−

×==

000

00149.0209.0

00148.0

80010040010010167.0010167.00

10010001040010080010010167.0010167.0000100010

105.2ˆˆ 5)3( radin

in

dTkf


⋅−−

⋅=

inklblbinklblb

mffmff

y

x

y

x

375010,5700,3

226010,5700,3

ˆ

ˆˆˆ

ˆˆ

4

4

4

3

3

3



The frame is fixed at nodes 1 and 3 and subjected to a positive distributed load of 1,000 lb/ft applied along element 2. Let E = 30 x 106 psi and A = 100 in.2 for all elements, and let I = 1,000 in.4 for all elements.


First we need to replace the distributed load with a set of equivalent nodal forces 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:

kwLff yy 202

40)000,1(232 −=−=−==

inkftlbwLmm ⋅=⋅−=−=−=−= 600,133,13312

40)000,1(12

22

32

If we consider only the parts of the stiffness matrix associated with the three degrees of freedom at node 2, we get: Element 1: The angle between x and x is 45º

707.0707.0 == SC where

( )3

222

36

78551.1123012)000,1(66

0463.023012)000,1(1212/93.58

5091030

inLI

inL

IinlbLE

=×

=

=×

==×

=


−−=400033.833.8

33.802.5098.4933.898.4902.50

93.58)1(

222

k

dd yx φ

Simplifying the above equation:

2 2 2

(1)

2,948 2,945 4912,945 2,948 491491 491 235,700

φ

= − −

x yd d

k



01 == SC where

( )3

222

36

5.124012

)000,1(66

0521.04012

)000,1(1212/5.62480

1030

inLI

inL

IinlbLE

=×

=

=×

==×

=


inlbk

dd yx

=

000,45.1205.12052.00

0010050.62)2(

222 φ

Simplifying the above equation:

inlbk

dd yx

=

000,25025.781025.78125.30

00250,6)2(

222 φ

The global beam equations reduce to:

=

⋅−−

2

2

2

700,485290491290951,2945,2491945,2198,9

600,1200

φy

x

dd

inkk


−−=

radin

indd

y

x

0033.00097.0

0033.0

2

2

2

φ



−−−

=

−−

−

−

=

radinin

radin

indT

0033.00092.0

00452.0000

0033.00097.0

0033.0000

1000000707.0707.00000707.0707.00000001000000707.0707.00000707.0707.0

Recall the elemental stiffness matrix is a function of values C1, C2, and L

( ) ink

ink

LEIC

LAEC 2273.0

23012

)000,1(1030893,5230121030)100(

3

6

32

6

1 =×

×===

××

==


(1)

5,893 0 10 5,893 0 0 00 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 0.00920 694.8 117,000 0 694.8 235,800 0.0033

− − −

= = − − − − − −

− −

f kTdin

inrad


1

1

1

2

2

2

ˆ 26.64ˆ 2.268ˆ 389.1ˆ 26.64

2.268ˆ778.2ˆ

− − ⋅ = −

− ⋅

x

y

x

y

f kf km k in

kfkf

k inm



−−

=

−−

=

000

0033.00097.0

0033.0

000

0033.00097.0

0033.0

100000010000001000000100000010000001

radin

in

radin

in

dT

Recall the elemental stiffness matrix is a function of values C1, C2, and L

( ) ink

ink

LEIC

LAEC 2713.0

4012)000,1(1030250,6

40121030)100(

3

6

32

6

1 =×

×===

××

==


(2)

6,250 0 0 6,250 0 0 0.00330 3.25 781.1 0 3.25 781.1 0.00970 781.1 250,000 0 781.1 125,000 0.0033ˆ ˆ

6,250 0 0 6,250 0 0 00 3.25 781.1 0 3.25 781.1 00 781.1 125,000 0 781.1 250,00 0

− − − − − −

= = − − − −

−

inin

radf kTd


⋅−

−⋅−

−

=

inkkkinkkk

dk

50.41258.263.20

57.83258.263.20

ˆˆ


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

2

2

2

3

3

3

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

2.58 20 22.58ˆ412.50 1600 2,013ˆ

− − − ⋅ − ⋅ ⋅ = − = − − −

− ⋅ ⋅ − ⋅

x

y

x

y

f k kf k k km k in k in k in

k kfk k kfk in k in k inm


Consider the frame shown in the figure below. In this example will illustrate the equivalent joint force replacement method for a frame subjected to a load acting on an element instead of at one of the joints of the structure. Since no distributed loads are present, the point of application of the concentrated load could be treated 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 structures solved by computer, this does not pose a problem. However, for very large structures, this might reduce the maximum size of the structure that could be analyzed.


The frame is fixed at nodes 1, 2, and 3 and subjected to a concentrated load of 15 k applied at mid-length of element 1. Let E = 30 x 106 psi, A = 8 in2, and let I = 800 in4 for all elements. Solution Procedure

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


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 coordinate

system forces into the global coordinate system forces, using the equation fTf T ˆ= . These global joint forces are shown below.


4. Then we analyze the structure, using the equivalent joint forces (plus

actual joint forces, if any) in the usual manner. 5. The final internal forces developed at the ends of each element may be

obtained by subtracting Step 2 joint forces from Step 4 joint forces.

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

895.0447.0 == SC where

( )2 3

2212 12(800) 6 6(800)0.0334 8.95

44.7 1244.7 12= = = =

××

I Iin inL L

3

6

/9.55127.44

1030 inlbLE

=×

×=


(1)

4 4 4

90.0 178 448178 359 244448 244 179,000

φ

= − −

d dx y

k


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

895.0447.0 =−= SC where

( )2 3

2212 12(800) 6 6(800)0.0334 8.95

44.7 1244.7 12= = = =

××

I Iin inL L

3

6

/9.55127.44

1030 inlbLE

=×

×=


(2)

4 4 4

90.0 178 448178 359 244448 244 179,000

φ

− = −

d dx y

k

Element 3: The angle between x and x is 0° (The author of your textbook directed 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 angle between x and x is 180°)

6330 101 0 50 /

50 12×

= = = =×

EC S lb inL

( )2 3

2212 12(800) 6 6(800)0.0267 8.0

50 1250 12= = = =

××

I Iin inL L


(2)

4 4 4

400 0 00 1.334 4000 400 160,000

φ

=

d dx y

k


The global beam equations reduce to:

4

4

4

7.5 582 0 8960 0 719 400

900 896 400 518,000 φ

− = − ⋅

x

y

k dd

k in


4

4

4

0.01030.0009560.00172φ

− = −

x

y

d ind in

rad


895.0447.0

1000000000000000010000000000

==

−

−

= SC

CSSC

CSSC

T

0.447 0.895 0 0 0 0 0 00.895 0.447 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 0.447 0.895 0 0.0103 0.003740 0 0 0.895 0.447 0 0.000956 0.009630 0 0 0 0 1 0.00172 0.00172

−

= = − − − − −

Tdin inin in

rad rad


Recall the elemental stiffness matrix is:

−−−−

−−−

−

=

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

( ) ink

ink

LEIC

LAEC 155.0

72.4412)800(10302.447

72.44121030)8(

3

6

32

6

1 =××

===×

×==


−

−

−−−−

−−−

−

==

radinin

dTkf

00172.000963.000374.0

000

000,1795.5000490,895.50005.500868.105.500868.10

0044700447490,895.5000000,1795.5000

5.500868.105.500868.100044700447

ˆˆ)1(


(1)

1.670.88

158ˆ ˆ ˆ1.670.88

311

− − ⋅

= = − − ⋅

kk

k inf kd

kk

k in

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


1

1

1

4

4

4

ˆ 1.67 3.36 5.03ˆ 0.88 6.71 7.59ˆ 158 900 1,058ˆ 1.67 3.36 1.68

0.88 6.71 5.83ˆ311 900 589ˆ

− − − − ⋅ ⋅ − ⋅ = − = − − −

− ⋅ − ⋅ ⋅

x

y

x

y

f k k kf k k km k in k in k in

k k kfk k kf

k in k in k inm


895.0447.0

1000000000000000010000000000

=−=

−

−

= SC

CSSC

CSSC

T

−

=

−

−

−−−

−−−

=

radinin

radinin

dT

00172.000879.000546.0

000

00172.0000956.0

0103.0000

1000000447.0895.00000895.0447.00000001000000447.0895.00000895.0447.0


( ) ink

ink

LEIC

LAEC 155.0

72.4412)800(10302.447

72.44121030)8(

3

6

32

6

1 =××

===×

×==


−

−−−−

−−−

−

==

radinin

dTkf

00172.000879.000546.0

000

000,1795.5000490,895.50005.500868.105.500868.10

0044700447490,895.5000000,1795.5000

5.500868.105.500868.100044700447

ˆˆ)2(


(2)

2.440.877

158ˆ ˆ ˆ2.440.877

312

− − ⋅

= = − ⋅

kk

k inf kd

kk

k in

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


−

−

=

−

−

=

000

00172.0000956.0

0103.0

000

00172.0000956.0

0103.0

100000010000001000000100000010000001

radinin

radinin

dT



( ) ink

ink

LEIC

LAEC 111.0

5012)800(1030400

50121030)8(

3

6

32

6

1 =×

×===

××

==

−

−

−−−−

−−−

−

==

000

00172.0000956.0

0103.0

000,1604000000,804000400335.10400335.100040000400000,804000000,1604000

400335.10400335.100040000400

ˆˆ)3(

radinin

dTkf


(3)

4.120.687

275ˆ ˆ ˆ4.120.687

137

− − − ⋅

= = − ⋅

kk

k inf kd

kk

k in

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




The frame is fixed at nodes 2 and 3 and subjected to a concentrated load of 500 kN applied at node 1. For the bar, A = 1 x 10-3 m2, for the beam, A = 2 x 10-3 m2, I = 5 x 10-5 m4, and L = 3 m. Let E = 210 GPa for both elements.

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

01 == SC where

345

252

5

2 103

)105(661067.6)3(

)105(1212 mLIm

LI −

−−

−

=×

=×=×

=

36

6

/10703

10210 mkNLE

×=×

=


mkNk

yx dd

×=

20.010.0010.0067.00002

1070 3)1(

111 φ

500 kN


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

707.0707.0 == SC

where

( )m

kNm

mkNmk

yx dd

×=

−

5.05.05.05.0

24.4/1021010 2623

)2(

11

mkNk

yx dd

×=

354.0354.0354.0354.0

1070 3)2(

11

Assembling the elemental stiffness matrices we obtain the global stiffness matrix

mkNK

×=

20.010.0010.0421.0354.00354.0354.2

1070 3

The global equations are:

×=

−

1

1

13

20.010.0010.0421.0354.00354.0354.2

10700

5000

φy

x

dd

mkNkN


−=

radmm

dd

y

x

0113.00225.0

00388.0

1

1

1

φ

Bar Element: The bar element force-displacement equations can be obtained using dTkf ˆˆ = .

−

−=

y

x

y

x

x

x

dddd

SCSC

LAE

ff

3

3

1

1

3

1

0000

1111

ˆˆ


Therefore, the forces in the bar element are:

( ) kNSdCdL

AEf yxx 670ˆ111 −=+=

( ) kNSdCdL

AEf yxx 670ˆ113 =+−=

Beam Element: The beam element force-displacement equations can be obtained using dkf ˆˆˆ = . Since the local axis coincides with the global coordinate system, and the displacements at node 2 are zero. Therefore, the local force-displacement equations are:

32

1

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LEIC

LAEC

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

=

=

−−−−

−−−

−

=

⋅−

−−−−

−−−

−

×==

000

0113.00225.0

00388.0

20.010.0010.010.0010.0067.0010.0067.00

00200210.010.0020.010.0010.0067.0010.0067.00002002

1070ˆˆˆ 3)1(

mkNmm

dkf

Substituting numerical values into the above equations gives:

⋅−

−

−

=

mkNkNkN

kNkN

mffmff

y

x

y

x

3.785.26

4730.05.26

473

ˆ

ˆˆˆ

ˆˆ

2

2

2

1

1

1




The frame is fixed at nodes 1 and 3 and subjected to a moment of 20 kN-m applied at node 2. Assume A = 2 x 10-2 m2, I = 2 x 10-4 m4, and E = 210 GPa for all elements.


10 == SC where

344

242

4

2 1034

)102(66105.1)4(

)102(1212 mLIm

LI −

−−

−

×=×

=×=×

=

37

6

/1025.54

10210 mkNLE

×=×

=

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

mkNk

yx dd

×=

08.0003.002003.00015.0

1025.5 5)1(

222 φ



01 == SC where

344

252

4

2 104.25

)102(66106.9)5(

)102(1212 mLIm

LI −

−−

−

×=×

=×=×

=

37

6

/102.45

10210 mkNLE

×=×

=

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

mkNk

yx dd

×=

08.0024.00024.00096.00002

102.4 5)2(

222 φ

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

mkNK

=

0756.00101.00158.00101.00500.100158.008480.0

106


=

⋅ 2

2

26

0756.00101.00158.00101.00500.100158.008480.0

1020

00

φy

x

dd

mkN


62

62

42

4.95 102.56 102.66 10φ

−

−

−

− × = − ×

×

x

y

d md m

rad


Element 1: The beam element force-displacement equations can be obtained using dTkf ˆˆ = .

6 6

6 6

4 4

0 1 0 0 0 0 0 01 0 0 0 0 0 0 0

0 0 1 0 0 0 0 00 0 0 0 1 0 4.95 10 2.56 100 0 0 1 0 0 2.56 10 4.95 100 0 0 0 0 1 2.66 10 2.66 10

− −

− −

− −

−

= = − × − × − − × ×

× ×

Tdm mm m

rad rad


−−−−

−−−

−

=

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

( ) mkN

mkN

LEIC

LAEC

25.6564

)102(10210

1005.14

10210)102(

3

46

32

662

1

=××

==

×=××

==

−

−

×××−

−−−−

−−−

−

×==

−

−

−

radmm

dTkf

4

6

63

)1(

1066.21095.41056.2

000

83043035.1035.10

002000020043083035.1035.100020000200

1025.5ˆˆ


Solving for the forces and moments gives:

⋅−

−⋅

=

mkNkNkNmkN

kNkN

mffmff

y

x

y

x

17.112.4

69.259.5

2.469.2

ˆ

ˆˆˆ

ˆˆ

2

2

2

1

1

1

Element 2: The beam element force-displacement equations can be obtained using dTkf ˆˆ = .

××−×−

=

××−×−

=

−

−

−

−

−

−

radmm

radmm

dT

4

6

6

4

6

6

1066.21056.21095.4

000

1066.21056.21095.4

000

100000010000001000000100000010000001


( ) mkN

mkN

LEIC

LAEC

3365

)102(10210

1084.05

10210)102(

3

46

32

662

1

=××

==

×=××

==

−

−

××−×−

−−−−

−−−

−

×==−

−

−

0001066.2

1056.21095.4

840.20440.2040.296.0040.296.00

0020000200440.20840.2040.296.0040.296.000020000200

102.4ˆˆ4

6

6

3)2(

radmm

dTkf



2

2

2

3

3

3

ˆ 4.16ˆ 2.69ˆ 8.92ˆ 4.16

2.69ˆ4.47ˆ

− ⋅ = −

⋅

x

y

x

y

f kNf kNm kN m

kNfkNfkN mm

Inclined or Skewed Supports

If a support is inclined, or skewed, at some angle α for the global x axis, as shown below, the boundary conditions on the displacements are not in the global x-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 the components of the displacement in the local and the global coordinate systems at node 3 is:


−=

3

3

3

3

3

3

1000cossin0sincos

'''

φαααα

φy

x

y

x

dd

dd

We can rewrite the above expression as:

{ } { } [ ]

−==

1000cossin0sincos

][' 3333 αααα

tdtd

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

{ } { } { } { }'][or][' dTddTd Tii ==

where the matrix [Ti] is:

=

][]0[]0[

]0[][]0[

]0[]0[][

][

3tI

ITi

Both the identity matrix [I] and the matrix [t3] are 3 x 3 matrices.

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, the force-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

''

]][][[

''

φ

φ

φ

y

x

y

x

y

x

Tii

y

x

y

x

y

x

dd

dd

dd

TKT

MFFMFFMFF

Grid Equations

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


A representation of the grid element is shown below:

The degrees of freedom for a grid element are: a vertical displacement iyd (normal to the grid), a torsional rotation ixφ about the x axis, and a bending rotation izφ about the z axis. The nodal forces are: a transverse force iyf a torsional moment ixm about the x axis, and a bending moment izm about the z 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 figure below.

A linear displacement function φ is assumed.

xaa ˆ21 +=φ Applying the boundary conditions and solving for the unknown coefficients gives:

xxx x

L 112 ˆˆˆˆ

φφφφ +

−=


Or in matrix form:

[ ]

==x

xNN2

121 ˆ

ˆˆφφ

φ

where N1 and N2 are the interpolation functions gives as:

LxN

LxN =−= 21

ˆ1

To obtain the relationship between the shear strain γ and the angle of twist φ 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:

φγ ˆˆmax RdxdAB == Therefore;

xdRd

ˆˆ

max

φγ =

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


( )xxLr

xddr 12

ˆˆˆˆ

φφφγ −==

The relationship between shear stress and shear strain is:

τ γ=G where G is the shear modulus of the material. From elementary mechanics of materials, we get:

ˆ τ=x

JmR

where J is the polar moment of inertia for a circular cross section or the torsional constant for non-circular cross sections. Rewriting the above equation we get:

( )2 1ˆ ˆˆ φ φ= −x x x

GJmL

The nodal torque sign convention gives:

xxxx mmmm ˆˆˆˆ 21 =−= Therefore;

( ) ( )1 1 2 2 2 1ˆ ˆ ˆ ˆˆ ˆφ φ φ φ= − = −x x x x x x

GJ GJm mL L

In matrix form the above equations are:

−

−=

x

x

x

x

LGJ

mm

2

1

2

1

ˆˆ

1111

ˆˆ

φφ


Combining the torsional effects with shear and bending effects, we obtain the local 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

mmf

mmf

2

2

2

1

1

1

4626

612612

2646

612612

2

2

2

1

1

1

ˆˆˆˆˆˆ

000000

0000

000000

ˆˆ

ˆˆˆ

ˆ

22

2323

22

2323

φφ

φφ

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

−

−=

CSSC

CSSC

TG

00000000

00100000000000000001

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

Lzz

qSL

xxC ijij −

==−

== sincosθ

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

GGT

GG TkTk ˆ=


Grid Example


The frame is fixed at nodes 2, 3, and 4, and is subjected to a load of 100 kips applied at node 1. Assume I = 400 in.4, J = 110 in.4, G = 12 x 10 3 ksi, and E = 30 x 10 3 ksi for all elements.

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

000

444

333

222

=========

zxy

zxy

zxy

ddd

φφφφφφ

Beam Element 1:

447.036.221020sin894.0

36.22200cos )1(

12)1(

12 =−

=−

==−=−

=−

==L

zzSL

xxC θθ

where

3 3

3 3 2 212 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 GJk in k inL L

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

GGT

GG TkTk ˆ=

−−−=

−−−=

894.0447.00447.0894.00001

894.0447.00447.0894.00001

TGG TT

inkk

zxyd

=

000,1790000,10920,40

000,1045.7ˆ )1(

111 φφ

Therefore, the global stiffness matrix is

inkk

zxyd

−−

−−=

000,144600,69894600,69700,39447

89444745.7)1(

111 φφ

Beam Element 2:

447.036.22100sin894.0

36.22200cos )2(

13)2(

13 −=−

=−

==−=−

=−

==L

zzSL

xxC θθ

where

3 3

3 3 2 212 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 GJk in k inL L


GGT

GG TkTk ˆ=

−−−

−−−=

894.0447.00447.0894.00001

000,1790000,10920,40

000,1045.7

894.0447.00447.0894.00001

)2(k


inkk

zxyd

−−−−

=000,144600,69894600,69700,39447

89444745.7)2(

111 φφ

Beam Element 3:

110

100sin010

2020cos )3(14

)3(14 −=

−=

−===

−=

−==

LzzS

LxxC θθ

where

kLEIink

LEI 000,5

)1210()400)(1030(66/3.83

)1210()400)(1030(1212

2

3

23

3

3=

××

==×

×=

inkL

GJinkLEI

⋅=×

×=⋅=

××

= 000,11)1210(

)110)(1012(000,400)1210(

)400)(1030(44 33


GGT

GG TkTk ˆ=


−

−=

010100

001

000,4000000,50000,110000,503.83

010100001

)3(k


=

000,11000000,400000,50000,53.83

)3(

111

k

zxyd φφ

Superimposing the three elemental stiffness matrices gives:

1 1 1

98.2 5,000 1,7905,000 479,000 01,790 0 299,000

φ φ

− = −

y x zd

K


1 1

1 1

1 1

100 98.2 5,000 1,7900 5,000 479,000 00 1,790 0 299,000

φφ

= − − = = = −

y y

x x

z z

F k dMM


−

−=

radrad

ind

z

x

y

0169.00295.0

83.2

1

1

1

φφ


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

ˆˆ = .

−−

=

−

−

−−−

−−−

=

000

00192.00339.0

83.2

000

0169.00295.0

83.2

894.0447.00000447.0894.00000001000000894.0447.00000447.0894.00000001

radradin

radradin

dTG


−−

−−

−−−−

−−

==

000

00192.00339.0

83.2

000,1790000,1500,890000,10920,400920,40000,1045.7000,1045.7500,890000,1000,1790000,10920,400920,40

000,1045.7000,1045.7

ˆˆ)1(

radradin

dTkf


1

1

1

2

2

2

ˆ 19.2ˆ 167ˆ 2,480ˆ 19.2

167ˆ2,260ˆ

− − ⋅ − ⋅ = ⋅

− ⋅

y

x

z

y

x

z

f km k inm k in

kfk inmk inm



ˆˆ = .

−−

=

−

−

−−−

−−−

=

000

0283.00188.0

83.2

000

0169.00295.0

83.2

894.0447.00000447.0894.00000001000000894.0447.00000447.0894.00000001

radradin

radradin

dTG


−−

−−

−−−−

−−

==

000

0283.00188.0

83.2

000,1790000,1500,890000,10920,400920,40000,1045.7000,1045.7500,890000,1000,1790000,10920,400920,40

000,1045.7000,1045.7

ˆˆ)2(

radradin

dTkf


1

1

1

3

3

3

ˆ 7.23ˆ 92.5ˆ 2,240ˆ 7.23

92.5ˆ295ˆ

− ⋅ − ⋅ =

− ⋅

− ⋅

y

x

z

y

x

z

f km k inm k in

kfk inm

k inm



ˆˆ = .

−

=

−

−

−

−

=

000

0295.00169.0

83.2

000

0169.00295.0

83.2

010000100000

001000000010000100000001

radradin

radradin

dTG


−

−−

−−−−

−−

==

000

0295.00169.0

83.2

000,4000000,5000,2000000,50000,1100000,110000,503.83000,503.83000,2000000,5000,4000000,5

0000,1100000,110000,503.83000,503.83

ˆˆ)3(

radradin

dTkf


1

1

1

4

4

4

ˆ 88.1ˆ 186ˆ 2,340ˆ 88.1

186ˆ8,240ˆ

− ⋅ − ⋅ = − ⋅

− ⋅

y

x

z

y

x

z

f km k inm k in

kfk inmk inm


To check the equilibrium of node 1 the local forces and moments for each element 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 the upper-left-hand portion of the transformation matrix TG. Therefore; for Element 1:

1

1

1

1 0 0 19.2 19.20 0.894 0.447 167 1,2600 0.447 0.894 2,480 2,150

− − = − − − ⋅ = ⋅ − − ⋅ ⋅

y

x

z

f k km k in k inm k in k in


Therefore; for Element 2:

1

1

1

1 0 0 7.23 7.230 0.894 0.447 92.5 1,0800 0.447 0.894 2,240 1,960

− = − − ⋅ = ⋅ − − − ⋅ − ⋅

y

x

z


Therefore; for Element 3:

1

1

1

1 0 0 88.1 88.10 0 1 2,340 2,3400 1 0 186 186

− − = − ⋅ = − ⋅ − − ⋅ − ⋅

y

x

z


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

The forces and moments accurately satisfy equilibrium considering the amount of truncation error inherent in results of the calculations presented 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 applied at node 2. Assume I = 16.6 x 10-5 m4, J = 4.6 x 10-5 m4, G = 84 GPa, and E = 210 GPa for all elements.

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

00

333

111

======

zxy

zxy

dd

φφφφ

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

030sin1

33cos )1(

12)1(

12 ==−

====−

==L

zzSL

xxC θθ

where

mkNLEI /1055.1

)3()106.16)(10210(1212 4

3

56

3 ×=××

=−

kNLEI 4

2

56

2 1032.2)3(

)106.16)(10210(66×=

××=

−

mkNLEI ·1065.4

3)106.16)(10210(44 4

56

×=××

=−

mkNL

GJ ·10128.03

)106.4)(1084( 456

×=××

=−


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

GGT

GG TkTk ˆ=

mkNk

−

−

=

100010001

65.4032.20128.0032.2055.1

100010001

104)1(


mkNk

zxyd

−

−=

65.4032.20128.0032.2055.1

104)1(

222 φφ

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

133sin0

30cos )1(

23)2(

23 −=−

=−

====−

==L

zzqSL

xxC θ

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

GGT

GG TkTk ˆ=

mkNk

−

−

−

−=

010100

001

65.4032.20128.0032.2055.1

010100001

104)2(


mkNk

zxyd

=

128.000065.432.2032.255.1

104)2(

222 φφ


Superimposing the two elemental stiffness matrices gives:

mkNK

zxyd

−

−=

78.4032.2078.432.2

32.232.210.310 4

222 φφ


−

−=

==

−=

x

x

y

z

x

y d

MM

kNF

2

2

24

2

2

2

78.4032.2078.432.232.232.210.3

1000

22

φφ


2

2

2

0.002590.001260.00126

φφ

− = −

y

x

z

d mradrad


ˆˆ = .

−

−=

−

−

=

radrad

m

radrad

mdTG

00126.000126.0

00259.0000

00126.000126.0

00259.0000

100000010000001000000100000010000001



−

−

−−

−−−−

−−

==

radrad

mdTkf

00126.000126.0

00259.0000

65.4032.233.2032.20128.000128.0032.2055.132.2055.1

33.2032.265.4032.20128.000128.0032.2055.132.2055.1

10ˆˆ 4)1(


1

1

1

2

2

2

ˆ 11.0ˆ 1.50ˆ 31.0ˆ 11.0

1.50ˆ1.50ˆ

− ⋅ ⋅ =

− ⋅

⋅

y

x

z

y

x

z

f kNm kN mm kN m

kNfkN mmkN mm


ˆˆ = .

−

=

−

−

−

−

=

000

00126.000126.0

00259.0

000

00126.000126.0

00259.0

010000100000

001000000010000100000001

radrad

m

radrad

m

dTG



−

−−

−−−−

−−

==

000

00126.000126.0

00259.0

65.4032.233.2032.20128.000128.0032.2055.132.2055.1

33.2032.265.4032.20128.000128.0032.2055.132.2055.1

10ˆˆ 4)2(

radrad

m

dTkf


2

2

2

3

3

3

ˆ 11.0ˆ 1.50ˆ 1.50ˆ 11.0

1.50ˆ31.0ˆ

− ⋅ − ⋅ = − ⋅

− ⋅

y

x

z

y

x

z

f kNm kN mm kN m

kNfkN mm

kN mm


Beam Element Arbitrarily Oriented in Space

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

The axis y is the principle axis for which the moment of inertia is minimum, Iy. The right-hand rule is used to establish the z axis and the maximum moment of inertia, Iz. Bending in the zx ˆˆ − plane: The bending in the zx − plane is defined by ym . 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

4626

612612

2646

612612

ˆ

LLLL

LL

LL

LLLL

LL

LL

LEI

k yY

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


Bending in the yx ˆˆ − plane: The bending in the yx − plane is defined by zm . The stiffness matrix for bending the in the yx − plane is:

−−

−−

−−=

3

2

3

2

2

2

3

2

3

2

2

2

4

4626

612612

2646

612612

ˆ

LLLL

LL

LL

LLLL

LL

LL

LEIk z

z

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

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

zyxzyxzyxzyx dddddd 222222111111ˆˆˆˆˆˆˆˆˆˆˆˆ φφφφφφ


The global stiffness matrix may be obtained using:

TkTk T ˆ= where

=

33

33

33

33

x

x

x

x

T

λλ

λλ

where

=

zzzyzx

yzyyyx

xzxyxx

x

CCCCCCCCC

ˆˆˆ

ˆˆˆ

ˆˆˆ

33λ

where the direction cosines,

jiC ˆ , are defined as shown below

The direction cosines of the x axis are:

kjix xzxyxx ˆˆˆ coscoscosˆ θθθ ++= where

nL

zzmL

yylL

xxxzxyxx =

−==

−==

−= 12

ˆ12

ˆ12

ˆ coscoscos θθθ


The y axis is selected to be perpendicular to the x and the z axes is such a way that the cross product of global z with x results in the y axis as shown in the figure below.

jDli

Dm

nml

kji

Dyxz +−===× 1001ˆˆ

where 22 mlD +=

The z axis is determined by the condition that yxz ˆˆˆ ×=

kDjD

mniDnl

lmnmlkji

Dyxz +−−=

−=×=

0

1ˆˆˆ

Therefore, the transformation matrix

=

zzzyzx

yzyyyx

xzxyxx

x

CCCCCCCCC

ˆˆˆ

ˆˆˆ

ˆˆˆ

33λ

becomes

yxz ˆˆ =×


−

−=

DD

mnDnl

Dl

Dm

nml

x 033λ

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

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

−=

001000100

λ

−=

001000100

λ


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


In this case the symbol φ are:

2 2

12 12y zy z

s s

EI EIGA L GA L

φ φ= =

where As is the effective beam cross-section in shear. Recall the shear modulus of elasticity or the modulus of rigidity, G, is related to the modulus of elasticity and the Poisson’s ratio, ν as:

( )2 1

EGν

=+

If φy and φz are set to zero, the stiffness matrix reduces the that shown previously on page 235. This is the form of the stiffness matrix used by SAP2000 for its frame element.


Example Frame Application

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

Concept of Substructure Analysis

Sometimes structures are too large to be analyzed as a single system or treated as a whole; that is, the final stiffness matrix and equations for solution exceed the memory capacity of the computer. A procedure to overcome this problem is to separate the whole structure into smaller units called substructures. For example, the space frame of an airplane, as shown below, may require thousands of nodes and elements to completely model and describe the 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, then we 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 your textbook “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 your textbook “A First Course in the Finite Element Method” by D. Logan. You may use the SAP 2000 to do frame analysis.

Documents

plane frame and grid equations