plane (2d) truss and frame elements read ... - tamu...

JN Reddy

The Finite Element Method

Read: Sections 4.6 and 5.4

Plane (2D) Truss and Frame Elements

• Review of bar finite elementin the local coordinates

• Plane truss element• Review of beam finite

element in the localcoordinates

• Plane frame element• Numerical examples

CONTENTS

JN Reddy

FINITE ELEMENT ANALYSIS OF PLANE TRUSSES AND FRAMES

Trusses and Frames: 2

JN Reddy

REVIEW OF THE BAR ELEMENT

Linear bar element in the element coordinate system

K u Fe e e=

Trusses & Frames: 3

1

2

1 1 11 1 12

K F,e

e ee e e ee

e

QE A f hh Q

− = = + −

he

211 1,e eQ f 2 2,e eQ fex

ey

Horizontal

EeAehe

⎡⎢⎣1 0 −1 00 0 0 0−1 0 1 00 0 0 0

⎤⎥⎦⎧⎪⎨⎪⎩ue1ve1ue2ve2

⎫⎪⎬⎪⎭ =

⎧⎪⎨⎪⎩F e10F e20

⎫⎪⎬⎪⎭K Fe e eD =

he 2

1θe

y

xeF1

eF2 ex

eyev1

eu1

2eu

2

ev

Inclined

JN Reddy

Bar Element in Global Coordinates

EeAehe

⎡⎢⎣1 0 −1 00 0 0 0−1 0 1 00 0 0 0


⎫⎪⎬⎪⎭ =

⎧⎪⎨⎪⎩F e10F e20

⎫⎪⎬⎪⎭ he 2

1

θe

y

xeF1

eF2 ex

eyev1

eu1

2eu

2

ev

Bar element in the element coordinates

K Fe e eD =

xe = x cos θ + y sin θ

ye = −x sin θ + y cos θxeye

=cos θ sin θ− sin θ cos θ

xy

2

1

θe

y

x

ex

ey( , )e ex y

( , )e ex y P

Transformation relations between the two coordinate systems

Trusses & Frames: 4

●

JN Reddy

Transformation relations between the displacements of the two coordinate systems

he 2

1

θe

y

x

,F e2 ex

eyev1

eu2

ev2

exF1

eu1

eyF1

ev1

eyF2

ev2

exF2

eu, 2

eF1

eu1

1 1

1 1

2 2

2 2

cos sin,

sin cos

cos sinsin cos

e e

e e

e e

e e

u uv v

u uv v

θ θθ θ

θ θθ θ

= − = −

0 0

{∆e} = [T e]{∆e}

⎧⎪⎨⎪⎩ue1ve1ue2ve2

⎫⎪⎬⎪⎭ =

⎡⎢⎣cos θe sin θe 0 0− sin θe cos θe 0 00 0 cos θe sin θe0 0 − sin θe cos θe


⎫⎪⎬⎪⎭

Bar Element in Global Coordinates

Trusses & Frames: 5

{ } { }[ ]e e eF T F=

JN Reddy

Bar Element in Global Coordinates: Truss

[Ke][T e]{∆e} = [T e]{F e}, [T e]−1 = [T e]T

[T e]T[Ke][T e]{∆e} = {F e} or [Ke]{∆e} = {F e}[Ke] = [T e]T[Ke][T e], {F e} = [T e]T{F e}

K Fe e eD =

Trusses & Frames: 6

[Ke] =EA

h

⎡⎢⎢⎣cos2 θ 1

2 sin 2θ − cos2 θ −12 sin 2θ12 sin 2θ sin2 θ −12 sin 2θ − sin2 θ− cos2 θ −12 sin 2θ cos2 θ 1

2 sin 2θ

−12 sin 2θ − sin2 θ 12 sin 2θ sin2 θ

⎤⎥⎥⎦

{F e} =

⎧⎪⎨⎪⎩F e1F e2F e3F e4

⎫⎪⎬⎪⎭ =

⎧⎪⎨⎪⎩P e1 cos θeP e1 sin θeP e2 cos θeP e2 sin θe

⎫⎪⎬⎪⎭+⎧⎪⎨⎪⎩fe1 cos θefe1 sin θefe2 cos θefe2 sin θe

⎫⎪⎬⎪⎭

JN Reddy

EXAMPLE 1

Given truss

P = 100 kNy

xA B

C2P = 200 kN

L

L

P = 100 kN

x

2P = 200 kN

1

yF2yF1

xF2

xy

x

y

y

xF1

)2,1(1)4,3(21 90q =

2 45q =

)6,5(3

2

Finite element discretization

Trusses & Frames: 7

Element Global Geom. Mater. Orient.

number nodes prop. prop.

√ ◦1 2 3 A, h1 = L E θ1 = 90

◦

2 1 3 A, h2 = 2L E θ2 = 45

JN Reddy

EXAMPLE 1 (continued)

P = 100 kN

x

2P = 200 kN

1

yF2yF1

xF2

xy

x

y

y

xF1

)2,1(1)4,3(21 90q =

)6,5(3

2

[K3] =EA

L

⎡⎢⎣0.3536 0.3536 −0.3536 −0.35360.3536 0.3536 −0.3536 −0.3536−0.3536 −0.3536 0.3536 0.3536−0.3536 −0.3536 0.3536 0.3536

⎤⎥⎦

The element stiffness matrices are [1/(2√2) = 0.3536]

Trusses & Frames: 8

2[ ]K =

2 45q = [K2] =EA

L

⎡⎢⎣0 0 0 00 1 0 −10 0 0 00 −1 0 1

⎤⎥⎦1[ ]K =

JN Reddy


Assembled stiffness coefficients

Trusses & Frames: 9

Connectivity array for 2 DoF per node truss

3 4 5 61 2 5 6é ùê úê úë û

1 2 3 4

[B] =(1)(2)

(2) (2)11 11 12 12 13

(2) (2)14 15 13 16 14

(2)22 22 23 24

(2) (2)25 23 26 24

, , 0,0, , ,

, 0, 0,, ,

K K K K KK K K K KK K K KK K K K

= = =

= = =

= = =

= =

(1) (1) (1)33 11 34 12 34 12

(1) (1) (1)35 13 36 14 34 12

(1) (1) (1)35 13 36 14 44 22

(1) (1) (1) (2)45 23 46 24 55 33 33

(1) (2) (1) (2)56 34 34 66 44 44

, , ,, , ,, , ,, , ,

, .

K K K K K KK K K K K KK K K K K KK K K K K K KK K K K K K

= = =

= = =

= = =

= = = +

= + = +

P = 100 kN

x

2P = 200 kN

1

yF2yF1

xF2

xy

x

y

y

xF1

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

1 90q =

3 5 6( , )

2

)2,1(1

2 3 4( , )

)2,1(1

2 3 4( , )

2 45q =

JN Reddy

Assembled system of equations of the truss


The displacement continuity conditions are

u11 = u31 = U1, v

11 = v

31 = V1

u12 = u21 = U2, v

12 = v

21 = V2

u22 = u32 = U3, v

22 = v

32 = V3

Trusses & Frames: 10

EA

L

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1.3536 0.3536 −1.0 0.0 |− 0.3536 −0.35360.3536 0.0 0.0 |− 0.3536 −0.3536

1.0 0.0 | 0.0 0.01.0 | 0.0 −1.0

symm. −−− −−− −−− |−−−− −−−−| 0.3536 0.3536| 1.3536

⎤

⎦

11

11

22

22

33

33

x

y

x

y

x

y

FUFVFUFV

FUFV

ì üì ü ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ïï ï ï ï=í ý í ýï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ïî þ ï ïî þ

0 0.

0 0.

0.

JN Reddy


F 11 + F31 = F

1x , F

12 + F

32 = F

1y

F 13 + F21 = F

2x , F

14 + F

22 = F

2y

F 23 + F33 = F

3x , F

24 + F

34 = F

3y

{∆} =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

U1V1U2V2U3V3

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭, {F} =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

F 11 + F31

F 12 + F32

F 13 + F21

F 14 + F22

F 23 + F33

F 24 + F34

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

F 1xF 1yF 2xF 2yF 3xF 3y

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭Boundary conditions

U1 = V1 = U2 = V2 = 0, F 3x = P, F3y = −2P

where the global forces and displacements are

100 kN

x

− 200 kN

1

yF2yF1

xF2

xy

x

y

y

xF1

)2,1(1)4,3(21 90q =

2 45q =

)6,5(3

2

1U

1V2V

2U

3U

3V


JN Reddy


EA

L

0.3536 0.35360.3536 1.3536

U3V3

=P−2P⎧⎪⎨⎪⎩

F 1xF 1yF 2xF 2y

⎫⎪⎬⎪⎭ =EA

L

⎡⎢⎣−0.3536 −0.3536−0.3536 −0.35360.0 0.00.0 −1.0

⎤⎥⎦⎧⎨⎩U3V3

⎫⎬⎭U3 = (3 + 2

√2)PL

EA= 5.828

PL

EA, V3 = −3PL

EA

F 1x = −P, F 1y = −P, F 2x = 0.0, F 2y = 3P

Solution100 kN

x

− 200 kN

1

yF2yF1

xF2

xy

x

y

y

xF1

)2,1(1)4,3(21 90q =

2 45q =

)6,5(3

2

1U

1V2U2V

3U

3U


(P = 105 N, EA = 108 N)

(m)

(N)1 1 2 21 2 1 23 2( ) ( ) ( ) ( );F F P F F P=- = =- =-

JN Reddy

σe = − Pe1

Ae=P e2Ae

P e1P e2

=AeEehe

1 −1−1 1

ue1ue2⎧⎪⎨⎪⎩

ue1ve1ue2ve2

⎫⎪⎬⎪⎭ =

⎡⎢⎣cos θe sin θe 0 0− sin θe cos θe 0 00 0 cos θe sin θe0 0 − sin θe cos θe


⎫⎪⎬⎪⎭

Post-computation of member displacements and stresses


100 kN

x

− 200 kN

1

yF2yF1

xF2

xy

x

y

y

xF1

)2,1(1)4,3(21 90q =

2 45q =

)6,5(3

2

1U

1V2U2V

3U

3U


1 1 2 2 1 2 1 21 1 1 1 2 2 3 2 2 3

30 3 2 2; ( ) ,PL PLu v u v u u U v v VAE AE

= = = = = = = + = = =-

We have

JN Reddy

Post-computation of member displacements and stresses



( )

12 3 1 3 1 3

22 3 2 3 2 3 3

1 1 2 21 2 1 2

1 2

3

12

3 2

3 2( ) ( )

cos sin

cos sin

,

,

PLu U V VA

u U V U V

P P P P P P

P PA A

q q

q q

s s

= + = =-

= + = +

=- = =- =-

=- =

JN Reddy

PLANE FRAME STRUCTURES

he1 2

eee fQF 111 +=

eee qQF 233 +=

eee qQF 122 +=

eee qQF 466 +=eee fQF 244 +=

eee qQF 355 +=

1 2

eS1 eS2

ew1ew2

eu1eu2

he

Displ. degrees of freedom in the element coordinates

Force degrees of freedom in the element coordinates

2

1 α

z

xeu1

eS1

ew1

ew2 eu2 exeS2

ez

Displ. degrees of freedom in the local coordinates

eu1

eS1

ew1

ew2

eu2

eS2 2

1α

z

x

Displ. degrees of freedom in the global coordinates


JN Reddy

Frame Element in Global Coordinates

[Ke][T e]{∆e} = [T e]{F e}, [T e]−1 = [T e]T

[T e]T[Ke][T e]{∆e} = {F e} or [Ke]{∆e} = {F e}[Ke] = [T e]T[Ke][T e], {F e} = [T e]T{F e}

Te e eD D=

1 1 2 2

1 1 2 2

1 1 2 2

0 00 0

0 0 1 0 0 1

cos sin cos sinsin cos , sin cos

e e e e

e e e e

e e e e

u u u uw w w w

α α α αα α α α

θ θ θ θ

= − = −

1 1

1 1

1 1

2 2

2 2

2 2

00

0 0 100

0 0 1

cos sinsin cos

cos sinsin cos

e e

e e

e e

e e

e e

e e

u uw w

u uw w

α αα α

θ θα αα α

θ θ

− =

−

0

0


JN Reddy Trusses & Frames: 17

Frame Element in Global Coordinates (the Euler-Bernoulli beam frame element)

JN Reddy

AN EXAMPLE OF A FRAME STRUCTURE

Given structure

a

b

F

A B

C P

Finite element discretization

1 2 3 4 5 64 5 6 7 8 9é ùê úê úë û

1 2 3 4 5 6 (1)

(2) [B] =

Connectivity array for 3 DoF per node

ab

F

A B

C

2

1

1 2

3

1 2

2

1(1,2,3) (4,5,6)

(7,8,9)P

• •

•

(1,2,3)1

(4,5,6)1

(4,5,6)2

(1,2,3)2

x

z

0 0 01 20 , 90 or 270a a= = −

JN Reddy 2-D Problems: 19

(1) (1) (1)11 11 12 12 16 16, , ... ,K K K K K K= = =

(1) (1) (1)22 22 23 23 26 26, , ... ,K K K K K K= = =

(1) (2) (1) (2)44 44 11 45 45 12

(1) (2) (1) (2)46 46 13 55 55 22

(1) (2) (1) (2)56 56 23 66 66 33

(2) (2) (2)47 14 48 15 49 16

(2) (2) (2)57 24 58 25 59 26

(2) (2)67 24 68 35 69

, ,, ,, ,

, , ,, , ,, ,

K K K K K KK K K K K KK K K K K KK K K K K KK K K K K KK K K K K

= + = +

= + = +

= + = +

= = =

= = =

= = = (2)26

(2) (2) (2)77 44 78 45 79 46

(2) (2) (2)88 55 89 56 99 66

,, , ,, , .

KK K K K K KK K K K K K

= = =

= = =

Global stiffness coefficients in terms of element stiffness coefficients

(1) (1) (1) (1)33 33 34 34 35 35 36 36, , ,K K K K K K K K= = = =

ab

F

2

1

1 2

2

1(1,2,3) (4,5,6)

(7,8,9)P

• •

•

(1,2,3)1

(4,5,6)1

(4,5,6)2

(1,2,3)2

ASSEMBLY OF ELEMENT MATRICES

1 2 3 4 5 64 5 6 7 8 9é ùê úê úë û

1 2 3 4 5 6 (1)

(2) [B] =

JN Reddy 20

EXAMPLE 2 (from the textbook)

The y-axis is into the plane of the paper, and is measured counterclockwise from x-axis to x-axis

A = 10 in.2 , I = 10 in.4 , E = 30×106 psi

1°

°

11

12 qQ +

2

11

11 fQ +

12

14 fQ +

13

15 qQ +

12

13 qQ +

14

16 qQ +

x

z1

1

22Q 2

21Q

24Q

25Q

23Q

26Q

2P4P

xz2

x

z

( )1 32 4tana −=

4P

2P

144 in.

72 in.

11

2

3

B

43

lb/in.72P

1 90 ,a =

2a

in.144

in.1082

1a

a

44 45 46 4 4

54 55 56 5 5

64 65 66 6 6

K K K U FK K K U FK K K U F

=

● (1,2,3)2(4,5,6)1

(4,5,6)

(1) (2) (1) (2)44 44 11 45 45 12

(1) (2) (1) (2)46 46 13 55 55 22

(1) (2) (1) (2)56 56 23 66 66 33

(1) (2)4 4 1

(1) (2)5 5 2

(1) (2)6 6 3

, ,, ,, ,

( 2 ) ( 2 ) 4P24 72 48

K K K K K KK K K K K KK K K K K KF F F PF F F P PF F F P P P

= + = +

= + = +

= + = +

= + =

= + = - + - =-

= + = - =-

The rest of the calculations canbe found on pp.278-281 of the Book.

JN Reddy

SUMMARY

In this lecture we have covered the following topics:

• Plane truss element in the local and global coordinates• Transformation of element equations from element

coordinates to global coordinates• Examples, illustrating assembly, application of

boundary conditions, and calculation of elementforces and stresses

• Plane frame element in the local and globalcoordinates

• Transformation of element equations from elementcoordinates to global coordinates


plane (2d) truss and frame elements read ... - tamu...

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