chapter2_2015

1

Chapter 2:

1D Elements and Computational Procedures

2

1D System of Springs

Textbook: 2.1, 2.2, 2.5, 2.7

3

1-D SYSTEM OF SPRINGS

• Bodies move only in horizontal direction

• External forces, F2, F3, and F4, are applied

• No need to discretize the system (it is already discretized!)

• Rigid body (including walls) NODE

• Spring ELEMENT

F2

F3

F41 2

3

4

5

6

1

2

3

4 5

4

SPRING ELEMENT

• Element e

– Consist of Nodes i and j

– Spring constant k(e)

– Force applied to the nodes:

– Displacement ui and uj

– Elongation:

– Force in the spring:

– Relation b/w spring force and nodal forces:

– Equilibrium:

e

i j

( ), ei iu f ( ), e

j ju f

(e)j iu u

e e e ej iP k k u u

e ejf P

e e e ei j i jf f 0 or f f

e ei jf , f

5

SPRING ELEMENT cont.

• Spring Element e

– Relation between nodal forces and displacements

– Matrix notation:

– k: stiffness matrix

– q: vector of DOFs

– f: vector of element forces

e ei i j

e ej i j

f k u u

f k u u

e e (e)i i

(e)e ej j

u fk k

u fk k

(e)i i(e)

(e)j j

(e) (e) (e)

u f

u f

[ ]

k

k q f

k q f

6

SPRING ELEMENT cont.

• Stiffness matrix– It is square as it relates to the same number of forces as the

displacements.

– It is symmetric.

– It is singular, i.e., determinant is equal to zero and it cannot be inverted.

– It is positive semi-definite

• Observation– For given nodal displacements, nodal forces can be calculated by

– For given nodal forces, nodal displacements cannot be determined uniquely

(e) (e) (e)[ ]k q f

7

SYSTEM OF SPRINGS cont.

• Element equation and assembly

F2

F3

F41 2

3

4

5

6

1

2

3

4 5

(1)1 1 1 1

(1)1 1 2 2

k k u f

k k u f

(1)11 1 1

(1)21 1 2

3

4

5

uk k 0 0 0 f

uk k 0 0 0 f

u0 0 0 0 0 0

u0 0 0 0 0 0

u0 0 0 0 0 0

(2)2 2 2 2

(2)2 2 4 4

k k u f

k k u f

(1)11 1 1

(1)21 1 2

3

4

(2)2 2 2

(2)2 2 4

5

uk k 0 0 0 f

uk k 0 0 f

u0 0 0 0 0 0

u0 0 0

k

u0 0 0

k f

k f

0 0 0

k

8


(3)3 3 2 2

(3)3 3 3 3

k k u f

k k u f

(4)14 4 1

(4)34 4 3

uk k f

uk k f

(5)5 5 3 3

(5)5 5 4 4

k k u f

k k u f

(1)11 1 1

(1)21 1 2

(2)2 2 2

(2)2

(3)

3

4

5

3 3 2(3)

4

3 3 3

2

uk k 0 0 0 f

uk k 0 f

u0 0 0

u0 0 0

u0 0 0 0 0 0

k k f

k k

k f

k k f

f

k

(1)11 1 1

(1)21 1 2

3

4

5

(3)3 3 2

(3)

(4)4 4 1

(2)2 2 2

(2)2 2

3

4

3 3(4)

4 4 3

k k f

uk k 0 0 f

uk k 0 f

u0 0

u0 0 0

u0 0

k

k k f

k k k fk

k f

k

0

f

0 0

k f

0

(

(2

(5)5

)2 2 2

(2)2 2

(1)11 1 1

(1)21 1 2

5 3(5)

(3)3

4)4 4 1

(4)

3

4

4 4 3

2

3

4

5

5 5

3

4

( )3 3 3

uk k 0 0 f

uk k 0 f

u0

u0 0

u0 0 0 0

k k f

k k

k k f

k k f

k k

f

k k f

k k f

f

k k f

0 0

9


(6)6 6 4 4

(6)6 6 5 5

k k u f

k k u f

(1)1 1 1

(3)3 3 2

(3)3

(4

(2)2 2 2

(

(5)5 5 3

(5)5 5 4

1(1)

1 1 2 2

(6)6 6 4

(

3

)4 4 1

(4

42)

2 2 4

3

6)6 6 5

)4 4 3

5

3

k k 0 0 u f

k k 0 u f

0 u

0 u

0 0 0 u

k k f

k k

k k f

k k fk

k k f

k k

k k f

k k

fk k f

f

k ff

F2

F3

F41 2

3

4

5

6

1

2

3

4 5

10


• Relation b/w elementforces and external force

• Force equilibrium

• At each node, the summation of element forces is equal to the applied, external force

F2

F3

F41 2

3

4

5

6

1

2

3

4 5

F3

3

(5)3f

(3)3f

(4)3f

e

e

ie

i ie 1

ie

i ie 1

F f 0

F f , i 1,...ND

(1) (4)11 1

(1) (2) (3)22 2 2

(3) (4) (5)33 3 3

(2) (5) (6)44 4 4

(6)55

Rf f

Ff f f

Ff f f

Ff f f

Rf

11


• Assembled System of Matrix Equation:

• [Ks] is square, symmetric, singular and positive semi-definite.

• When displacement is known, force is unknown

R1 and R5 are unknown reaction forces

1 4 1 4 1 1

1 1 2 3 3 2 2 2

4 3 3 5 4 5 3 3

2 5 2 5 6 6 4 4

6 6 5 5

k k k k 0 0 u R

k k k k k k 0 u F

k k k k k k 0 u F

0 k k k k k k u F

0 0 0 k k u R

s s s[ ]{ } { }K Q F

1 5u u 0

12


• Imposing Boundary Conditions

– Ignore the equations for which the RHS forces are unknown and strike

out the corresponding rows in [Ks].

– Eliminate the columns in [Ks] that multiply into zero values of

displacements of the boundary nodes.

1 4 1 4 1 1

1 1 2 3 3 2 2 2

4 3 3 5 4 5 3 3

2 5 2 5 6 6 4 4

6 6 5 5

k k k k 0 0 u R

k k k k k k 0 u F

k k k k k k 0 u F

0 k k k k k k u F

0 0 0 k k u R

13


• Global Matrix Equation

• Global Stiffness Matrix [K]– square, symmetric and positive definite and hence non-singular

• Solution

• Once nodal displacements are obtained, spring forces can be calculated from

1 2 3 3 2 2 2

3 3 4 5 5 3 3

2 5 2 5 6 4 4

k k k k k u F

k k k k k u F

k k k k k u F

[ ]{ } { }K Q F

1{ } [ ] { }Q K F

e e e ej iP k k u u

14

Formal Procedure for Applying BC

• Known displacement {Dc} and unknown displacement {Dx}

• Known force {Fc} and unknown reaction {Fx}

• Decomposed matrix equation

• Handle the known force part first

• After calculating unknown displacement {Dx}, calculate unknown reaction from the second part

11 12

21 22

x c

c x

DK K F

DK K F

111 12 11 12[ ]{ } [ ]{ } { } { } [ ] { } [ ]{ }x c c x c c

K D K D F D K F K D

21 22{ } [ ]{ } [ ]{ }x x c F K D K D

15

Alternative Procedure for Applying BC

• Let assume D2 = 2 is a given nodal value

• Move known product Ki22 to RHS

• Replace the second equation with D2 = 2

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

K K K D F

K K K D F

K K K D F

11 13 1 1 12 2

21 23 2 2 22 2

31 33 3 3 32 2

0

0

0

K K D F K

K K D F K

K K D F K

11 13 1 1 12 2

2 2

31 33 3 3 32 2

0

0 1 0

0

K K D F K

D

K K D F K

16

Homework

• Four rigid bodies, 1, 2, 3, and 4, are connected by four springs as shown in the figure. A horizontal force of 1,000 N is applied on Body 1 as shown in the figure. Using finite element analysis, (a) find the displacements of the two bodies (1 and 3), (2) find the element force (tensile/compressive) of spring 1, and (3) the reaction force at the right wall (Body 2). Assume the bodies can undergo only translation in the horizontal direction. The spring constants (N/mm) are k1 = 400, k2 = 500, k3 = 500, and k4 = 300.

F12

3

4

4

3

1

2

1

x

17

Bar Element

Textbook: 2.4, 2.6, 2.10

18

Bar and Beam Elements

• Static analysis: forces are constant in time or change very slowly.

• Linear analysis: deflections are small so that material behavior is elastic. No failure, no gaps that open or close.

• Truss elements (bars, rods): pinned (hinged) at connection points; resist axial forces only. Hence it has axial DOFs only.

• Frame elements (beams): welded (or, connected with multiple fasteners) at connection points; resist axial and transverse forces and bending moments. Has axial, transverse and rotational DOFs.

19

Bars

• Bars are structural members that can only carry axial loads. This is usually the case when the end connections are hinged

1

2 3

4

20

UNIAXIAL BAR

• For general uniaxial bar, we need to divide the bar into a set of elements and nodes

• Elements are connected by sharing a node

• Forces are applied at the nodes (distributed load must be converted to the equivalent nodal forces)

• Assemble all elements in the same way with the system of springs

• Solve the matrix equationfor nodal displacements

• Calculate stress and strainusing nodal displacements

Fp(x)x

Fp(x)

Statically indeterminate

Statically determinate

21

1D BAR ELEMENT

• Two-force member

• Only constantcross-section

• Element force isproportional torelative displ

• First node: isecond code: j

• Force-displacement relation

x

f1 f2

L A

Node i Node j

ui uj

K=EA/L ( )eif

( )ejf

(e)(e)i i j

(e)(e) (e)j i j i

AEf (u u )

L

AEf f (u u )

L

Similar to the spring element

22

1D BAR ELEMENT cont.

• Element equation

– Either force or displacement (not both) must be given at each node.

– Example: ui = 0 and fj = 100 N.

– What happens when fi and fj are given?

• Element forces– After solving nodal displacements, the element force can be calculated

– Element stress

(e)(e)ii

(e)jj

uf 1 1AEuf L 1 1

Node i Node j

ui uj

K=EA/L ( )eif

( )ejf

(e) (e) (e){ } [ ]{ }f k q

(e)(e) (e)

j i j

AEP u u f

L

(e)(e)i

(e)j

u1 1P AEuL 1 1P

(e)

(e)

P

A

23

EXAMPLE

• 3 elements and 4 nodes

• At node 2:

• Equation for each element:

Element 1

x

F1 K1 Element 2

Element 3

K2

K3

Element 1

u1

F1 K1

Element 2

Element 3

K2

K3

N1

N3

N4

N2

u4

u3u2

F3

F4

(1) (2) (3)2 2 2 2F f f f

(1)1 1 11

(1)1 1 22

K K uf

K K uf

(2)22 22

(2)32 23

uK Kf

uK Kf

(3)3 3 22

(3)3 3 44

K K uf

K K uf

24

EXAMPLE cont.

• How can we combine different element equations? (Assembly)– First, prepare global matrix equation:

– Write the equation of element 1 in the corresponding location

(1)11 11

(1)21 12

3

4

uK K 0 0f

uK K 0 0f

u0 0 0 00

u0 0 0 00

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

Displacement vector

Stiffness matrix

Applied force vector

Element 1

u1

F1 K1

Element 2

Element 3

K2

K3

N1

N3

N4

N2

u4

u3u2

F3

F4

25

EXAMPLE cont.

– Write the equation of element 2:

– Combine two equations of elements 1 and 2

1(2)

22 2 2(2)

33 2 2

4

u0 0 0 0 0

uf 0 K K 0

uf 0 K K 0

u0 0 0 0 0

(1)11 11

(1) (2)21 1 2 22 2

(2)32 23

4

uK K 0 0f

uK K K K 0f f

u0 K K 0f

u0 0 0 00

Element 1

u1

F1 K1

Element 2

Element 3

K2

K3

N1

N3

N4

N2

u4

u3 u2

F3

F4

26

EXAMPLE cont.

– Write the equation of element 3

– Combine with other two elements

1(3)

3 32 2

3(3)

3 34 4

0 0 0 00 u

0 K 0 Kf u

0 0 0 00 u

0 K 0 Kf u

(1)1 11 11

(1) (2) (3)1 1 2 3 2 32 22 2 2

(2)2 23 33

(3)3 34 44

K K 0 0F uf

K (K K K ) K KF uf f f

0 K K 0F uf

0 K 0 KF uf

Structural Stiffness Matrix

Element 1

u1

F1 K1

Element 2

Element 3

K2

K3

N1

N3

N4

N2

u4

u3 u2

F3

F4

27

EXAMPLE cont.

• Substitute boundary conditions and solve for the unknown displacements.– Let K1 = 50 N/cm, K2 = 30 N/cm, K3 = 70 N/cm and f1 = 40 N.

– Knowns: F1, F2, u3, and u4

– Unknowns: F3, F4, u1, and u2

1 1

2 2

3 3

4 4

F u50 50 0 0

F u50 (50 30 70) 30 70

F u0 30 30 0

F u0 70 0 70

1

2

3

4

40 50 50 0 0 u

0 50 (50 30 70) 30 70 u

F 0 30 30 0 0

F 0 70 0 70 0

28

EXAMPLE cont.

– Remove zero-displacement columns: u3 and u4.

– Remove unknown force rows: F3 and F4.

– Now, the matrix should not be singular. Solve for u1 and u2.

– Using u1 and u2, Solve for F3 and F4.

1

3 2

4

40 50 50

0 u50 150

F u0 30

F 0 70

1

2

u40 50 50

u0 50 150

1

2

u 1.2 cm

u 0.4 cm

3 1 2

4 1 2

F 0u 30u 12 N

F 0u 70u 28 N

29

EXAMPLE cont.

• Recover element data

1.2 cm

F1 = 40 N K1

K2

K3

0.4 cm

-12 N

-28 N

(1)1 1 11

(1)1 1 22

K K u 50 50 1.2 40f

K K u 50 50 0.4 40f

Element force

(2)22 22

(2)32 23

uK K 30 30 0.4 12f

uK K 30 30 0.0 12f

(3)3 3 22

(3)3 3 44

K K u 70 70 0.4 28f

K K u 70 70 0.0 28f

30

EXAMPLE

• Statically indeterminate bars

• E = 100 GPa

• F = 10,000 N

• A1 = 10−4 m2, A2 = 2×10−4 m2

• Element stiffness matrices:

• Assembly

0.25 m

A B C

0.4 m

F RRRL

11 41(1) 7

2

u1 1 4 410 10[ ] 10

u0.25 1 1 4 4k

11 42(2) 7

3

u1 1 5 510 2 10[ ] 10

u0.4 1 1 5 5k

1 17

2

3 3

4 4 0 u F

10 4 9 5 u 10,000

0 5 5 u F

31

EXAMPLE cont.

• Applying BC

• Element forces or Element stresses

• Reaction forces

7 42 210 9 u 10,000 u 1.11 10 m

j i

AEP u u

L

2 1

(1) 7

(2) 73 2

P 4 10 u u 4,444N

P 5 10 u u 5,556N

(1)L

(2)R

R P 4,444N

R P 5,556N

(1)(1) 7

(1)

(2)(2) 7

(2)

P4.444 10 Pa 44.44MPa

A

P5.556 10 Pa 55.56MPa

A

32

What if elements are not horizontal?

• Transform each element to global coordinate and then assemble

• Assembly must be performed in the forces and displacements in the global coordinate system

• Assume a element coordinate system x'y' (parallel to element direction)

• Transform the displacements and forces from the element coordinate system to the global coordinate system

1

2 3

4

k1

k2

k3

k4

33

Bar with Arbitrary Orientation

1 1 1

2 2 2

cos sin

cos sin

u u v

u u v

Displacements

fxi, fyi :forces in the directions of ui and vi

1

1 1

2 2

2

cos 0

sin 0

0 cos

0 sin

x

y

x

y

f

f f

f f

f

1 1

1 1

2 2

1 2

cos

sin

cos

sin

x

y

x

y

f f

f f

f f

f f

1

1 1

2 2

2

cos sin 0 0

0 0 cos sin

u

u v

u u

v d Td

Tr T r

1

2

k=AE/L

x

yu1

v1

u2

v2

x'

34

Coordinate Transformation

In element coordinate system x'y'

1 1

2 2

u fk k

u fk k

Substituting the element coordinate displacements and forces in terms of global coordinate displacement and

forces k d r k d r

k Td r

T TT k Td T r r

Transforming from local to global coordinate system is accomplished by pre-multiplying

displacement/force vector by the transverse of the transformation matrix TT

TT k T d r

Tk T k T

1

2

k=AE/L

x

yu1

v1

u2

v2

x'

Element stiffness in the global coordinate

35

Stiffness of 2D Truss

2 2

2 2

2 2

2 2

coswhere

sin

c cs c cs

cs s cs sAE

L c cs c cs

cs s cs s

c

s

k1

23

4

k1

k2

k3

k4

x

y u1

v1

u2

v2

u3

v3

u4

v4

For element 1: cos = 0, sin = 1

For element 2: cos = 1, sin = 0

For element 3: cos = 0, sin = -1

36

Coordinate Transformation in 3D

• The transformation equation in 3-D are same as before

• The stiffness matrix now becomes a 6x6 matrix and has displacement DOF u, v and w

Tk T k T

1 1 1

1 1 1

0 0 0

0 0 0

l m n

l m nT

x y z

x' l1 m1 n2

y' l2 m2 n2

z' l3 m3 n3

Direction cosines between axes

37

50 N

12 cm

Element 1

Element 2

N1

N2

N3

8 cm

EXAMPLE

• Two-bar truss– Diameter = 0.25 cm

– E = 30106 N/cm2

• Element 1– Connectivity: N1 N2

– In local coordinate

N1

N21

x

y

K

1 = 33.7o

E = 30 x 106 N/cm2

A = r2 = 0.049 cm2

L = 14.4 cm

1v1u

2v2u

1xf

2xf

1x 1

1y 1

2x 2

2y 2

f 1 0 1 0 uf 0 0 0 0 vEA

Lf 1 0 1 0 u

f 0 0 0 0 v

(1) (1) (1){ } [ ]{ }f k q

38

EXAMPLE cont.

• Element 1 cont.– Element equation in the global coordinates

• Element 2• Connectivity: N2 N3

(1) (1) (1){ } [ ]{ }f k q

(1)1x 1(1)1y 1

(1)22x

(1)22y

f 0.692 0.462 0.692 0.462 uf 0.462 0.308 0.462 0.308 v

1021500.692 0.462 0.692 0.462 uf

0.462 0.308 0.462 0.308 vf

N3

N2

2

x

y

K

2 = –90o

E = 30 x 106 N/cm2

A = r2 = 0.049 cm2

L = 8 cm

2u

2v

3u

3v

3xf

2xf

(2)2x 2(2)2y 2

(2)33x

(2)33y

f 0 0 0 0 uf 0 1 0 1 v

1841250 0 0 0 uf

0 1 0 1 vf

39

EXAMPLE cont.

• Assembly– After transforming to the global coordinates

• Boundary Conditions– Nodes 1 and 3 are fixed.

– Node 2 has known applied forces: F2x = 50 N, F2y = 0 N

1x 1

1y 1

2x 2

2y 2

33x

33y

F 70687 47193 70687 47193 0 0 uF 47193 31462 47193 31462 0 0 vF 70687 47193 70687 47193 0 0 uF 47193 31462 47193 215587 0 184125 v

0 0 0 0 0 0 uF

0 0 0 184125 0 184125 vF

Element 1

Element 2

40

EXAMPLE cont.

• Boundary conditions (striking-the-columns)

– Striking-the-rows

• Solve the global matrix equation

1x

1y

2

2

3x

3y

F 70687 47193 70687 47193 0 0 0F 47193 31462 47193 31462 0 0 0

50 70687 47193 70687 47193 0 0 u

0 47193 31462 47193 215587 0 184125 v

F 0 0 0 0 0 0 0

F 0 0 0 184125 0 184125 0

2

2

u50 70687 47193

v0 47193 215587

42

42

u 8.28 10 cm

v 1.81 10 cm

41

EXAMPLE cont.

• Support reactions

– The reaction force is parallel to the element length (two-force member)

• Element force and stress (Element 1)– Need to transform to the element local coordinates

1x

41y

43x

3y

F 70687 47193 50F 47193 31462 33.398.28 10

NF 0 0 01.81 10

F 0 184125 33.39

1

14

2 24

2 2

u 0.832 .555 0 0 0

v 0.555 .832 0 0 0

u u0 0 .832 .555 5.89 10

v v0 0 .555 .832 6.11 10

42

EXAMPLE cont.

• Element force and stress (Element 1) cont.– Element force can only be calculated using local element equation

– There is no force components in the local y-direction

– In x-direction, two forces are equal and opposite

– The force in the second node is equal to the element force

– Normal stress = 60.2 / 0.049 = 1228 N/cm2.

1x

1y

42x

42y

f 1 0 1 0 0 60.2f 0 0 0 0 0 0EA

NLf 1 0 1 0 5.89 10 60.2

f 0 0 0 0 6.11 10 0

60.2 N–60.2 N

1 2

43

EXAMPLE

• Directly assembling global matrix equation (applying BC in the element level)

• Element property & direction cosine table

• Since u3 and v3 will be deleted after assembly, it is not necessary to keep them

1

2

34

1

2

3

F

45

Elem AE/L i -> j l = cos m = sin1 206×105 1 -> 3 -30 0.866 0.52 206×105 1 -> 2 90 0 13 206×105 1 -> 4 210 0.866 0.5

1 1 3 3

2 21

(1) 2 2(1) 1

2 23

2 23

u v u v

ul lm l lm

vlm m lm mEA

L ul lm l lm

vlm m lm m

k

1 1

(1) 2(1) 1

21

u v

ul lmEA

L vlm mk

44

EXAMPLE

• Space truss

4

1

2

3

x

y

z

E1

E2

E3

10,000N

Node x y z1 0 0 02 0 –1 13 0 1 14 1 0 1

Elem EA/L i -> j l m n

1 1 -> 4 0

2 2 -> 4 0

3 3 -> 4 0

535 2 10 1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

535 2 10

535 2 10

2 2 i

2 2i

2 2i

2j

2

j2

j

ul lm l lmvm mn lm m mnwn mn nEA

[ ]L l lm ln u

sym m mn vn w

ln ln

lnk

45

Thermal Load

• Temperature change causes thermal strain

• Constraints cause thermal stresses

• Thermo-elastic stress-strain relationship

(a) at T = Tref (b) at T = Tref + T

No stress, no strain No stress, thermal strain

Thermal stress, no strain

L L L

= E T

Thermal expansion coefficient

= + T

E

46

THERMAL STRESSES cont.

• Force-displacement relation

• Finite element equation (1D)

• For plane truss, transform to the global coord.

L LP = AE T AE AE T

L L

(e) (e) (e) (e)T{ } [ ]{ } { }f k q f

i(e)T

j

u1{ } AE T

u1f

Thermal force vector

T{ } [ ]{ } { }f k q f

i

i

Tj

j

ulvm

{ } AE Tul

m v

f T[ ]{ } { } { }k q f f

s s s Ts[ ]{ } { } { }K Q F F

47

Homework

• Use FEM to determine the axial force P in each portion, AB and BC, of the uniaxial bar. What are the support reactions? Assume: E = 100 GPa, area of cross sections of the two portions AB and BC are, respectively, 10−4 m2 and 2×10−4 m2

and F = 10,000 N. The force F is applied at the cross section at B.

• Use FEM to solve the plane truss shown below. Assume AE = 106 N, L = 1 m. Determine the nodal displacements, forces in each element and the support reactions.

0.25 m

A B C

0.4 m

F RRRL

1

2

3

4

1

2

310,000 N

L

L

L

x

y

48

Beam Element

Textbook: 2.3

49

BEAM THEORY

• Euler-Bernoulli Beam Theory– can carry the transverse load

– slope can change along the span (x-axis)

– Cross-section is symmetric w.r.t. xy-plane

– The y-axis passes through the centroid

– Loads are applied in xy-plane (plane of loading)

LF

x

y

F

Plane of loading

y

z

Neutral axis

A

50

BEAM THEORY cont.

• Euler-Bernoulli Beam Theory cont.– Plane sections normal to the beam axis remain plane and normal to

the axis after deformation (no shear stress)

– Transverse deflection (deflection curve) is function of x only: v(x)

– Displacement in x-dir is function of x and y: u(x, y)

y

y(dv/dx)

= dv/dx

v(x)L

F

x

y

Neutral axis

0

dvu(x,y) u (x) y

dx

20

xx 2

duu d vy

x dx dx

dv

dx

51

BEAM THEORY cont.

• Euler-Bernoulli Beam Theory cont.– Strain along the beam axis:

– Strain xx varies linearly w.r.t. y; Strain yy = 0

– Curvature:

– Can assume plane stress in z-dir basically uniaxial status

• Axial force resultant and bending moment

20

xx 2

duu d vy

x dx dx

0 0du / dx

2 2d v / dx

2

xx xx 0 2

d vE E Ey

dx

2

xx 0 2A A A

22

xx 0 2A A A

d vP dA E dA E ydA

dx

d vM y dA E ydA E y dA

dx

Moment of inertia I(x)

0

2

2

P EA

d vM EI

dx

EA: axial rigidity

EI: flexural rigidity

52

BEAM THEORY cont.

• Beam constitutive relation– We assume P = 0 (We will consider non-zero P in the frame element)

– Moment-curvature relation:

• Sign convention

– Positive directions for applied loads

2

2

d vM EI

dx Moment and curvature is linearly dependent

+P +P

+M+M+Vy

+Vy

yx

p(x)

F1 F2 F3

C1 C2 C3

y

x

53

GOVERNING EQUATIONS

• Beam equilibrium equations

– Combining three equations together:

– Fourth-order differential equation

yy

dVV dx

dx

dMM dx

dx

yVM

dx

p

yy y y

dVf 0 p(x)dx V dx V 0

dx

ydV

p(x)dx

y

dM dxM M dx pdx V dx 0

dx 2

4

4

d vEI p(x)

dx

y

dMV

dx

54

STRESS AND STRAIN

• Bending stress

– This is only non-zero stress component for Euler-Bernoulli beam

• Transverse shear strain

– Euler beam predicts zero shear strain (approximation)

– Traditional beam theory says the transverse shear stress is

– However, this shear stress is in general small compared to the bending stress

2

xx 2

d vEy

dx

2

2

d vM EI

dx

xx

M(x)y(x,y)

I

xy

u v v v0

y x x x

0

dvu(x,y) u (x) y

dx

Bending stress

xy

VQ

Ib

55

POTENTIAL ENERGY

• Potential energy

• Strain energy– Strain energy density

– Strain energy per unit length

– Strain energy

U V

2 22 22 2

0 xx xx xx 2 2

1 1 1 d v 1 d vU E( ) E y Ey

2 2 2 dx 2 dx

2 22 22 2

L 0 2 2A A A

1 d v 1 d vU (x) U (x,y,z)dA Ey dA E y dA

2 dx 2 dx

Moment of inertia

22

L 2

1 d vU (x) EI

2 dx

22L L

L 20 0

1 d vU U (x)dx EI dx

2 dx

56

POTENTIAL ENERGY cont.

• Potential energy of applied loads

• Potential energy

– Potential energy is a function of v(x) and slope

– The beam is in equilibrium when has its minimum value

CF NNL

ii i i0

i 1 i 1

dv(x )V p(x)v(x)dx Fv(x ) C

dx

CF2 NN2L L

ii i i20 0

i 1 i 1

dv(x )1 d vU V EI dx p(x)v(x)dx Fv(x ) C

2 dx dx

vv*

0v

57

Plane Beam Element

• Resists transverse shear force and in-plane bending only

• The corresponding displacements are transverse displacement, v=v(x), and rotation, =(x).

• The element has two degrees of freedom (DOF) at each node: (v1, 1) and (v2, 2)

F1 F2

C2C1 x

v1 v2

21

Lx

Applied loads Nodal DOFs

d

d

v

x

58

FINITE ELEMENT INTERPOLATION

• Beam element– Divide the beam using a set of elements

– Elements are connected to other elements at nodes

– Concentrated forces and couples can only be applied at nodes

– Consider two-node bean element

– Positive directions for forces and couples

– Constant or linearlydistributed load

F1 F2

C2C1

p(x)

x

59

FINITE ELEMENT INTERPOLATION cont.

• Nodal DOF of beam element– Each node has deflection v and slope – Positive directions of DOFs

– Vector of nodal DOFs

• Scaling parameter s– Length L of the beam is scaled to 1 using scaling parameter s

• Will write deflection curve v(s) in terms of s

v1 v2

q2q1

Lx1

s = 0x2

s = 1

x

T1 1 2 2{ } {v v } q

1x x 1s , ds dx,

L Lds 1

dx Lds,dx L

60


• Deflection interpolation– Interpolate the deflection v(s) in terms of four nodal DOFs

– Use cubic function:

– Relation to the slope:

– Apply four conditions:

– Express four coefficients in terms of nodal DOFs

2 30 1 2 3v(s) a a s a s a s

21 2 3

dv dv ds 1(a 2a s 3a s )

dx ds dx L

1 1 2 2

dv(0) dv(1)v(0) v v(1) v

dx dx

1 0

1 1

2 0 1 2 3

2 1 2 3

v v(0) a

dv 1(0) a

dx Lv v(1) a a a a

dv 1(1) (a 2a 3a )

dx L

0 1

1 1

2 1 1 2 2

3 1 1 2 2

a v

a L

a 3v 2L 3v L

a 2v L 2v L

61


• Deflection interpolation cont.

• Shape functions

– Hermite polynomials

– Interpolation property

2 3 2 3 2 3 2 31 1 2 2v(s) (1 3s 2s )v L(s 2s s ) (3s 2s )v L( s s )

2 31

2 32

2 33

2 34

N (s) 1 3s 2s

N (s) L(s 2s s )

N (s) 3s 2s

N (s) L( s s )

1

11 2 3 4

2

2

v

v(s) [N (s) N (s) N (s) N (s)]v

v(s) { } N q

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0

N1 N3

N2/L

N4/L

62


• Properties of interpolation– Deflection is a cubic polynomial (discuss accuracy and limitation)

– Interpolation is valid within an element, not outside of the element

– Adjacent elements have continuous deflection and slope

• Approximation of curvature– Curvature is second derivative and related to strain and stress

– B is linear function of s and, thus, the strain and stress

– Alternative expression:

– If the given problem is linearly varying curvature, the approximation is accurate; if higher-order variation of curvature, then it is approximate

1

2 21

2 2 2 22

2

v

d v 1 d v 1[ 6 12s, L( 4 6s), 6 12s, L( 2 6s)]

vdx L ds L

2

2 24 11 4

d v 1{ }

dx L

B q B: strain-displacement vector

2T T

2 24 11 4

d v 1{ }

dx L

Bq

63


• Approximation of bending moment and shear force

– Stress is proportional to M(s); M(s) is linear; stress is linear, too

– Maximum stress always occurs at the node

– Bending moment and shear force are not continuous between adjacent elements

2

2 2

d v EIM(s) EI { }

dx L B q

3

y 3 3

dM d v EIV EI [ 12 6L 12 6L]{ }

dx dx L q

Linear

Constant

64

FINITE ELEMENT EQUATION FOR BEAM

• Finite element equation using PMPE– A beam is divided by NEL elements with constant sections

• Strain energy– Sum of each element’s strain energy

– Strain energy of element (e)

p(x)

F1 F2 F3

C1 C2 C3

y

x

F4 F5

C4 C5

1 2 3 45

( )11x

( ) ( )1 22 1x x= ( ) ( )2 3

2 1x x= ( ) ( )3 42 1x x= ( )4

2x

eT 2

e1

NEL NELL x eL L0 x

e 1 e 1

U U (x)dx U (x)dx U

e2

e1

2 22 2x 1e

2 3 2x 0

1 d v EI 1 d vU EI dx ds

2 dx L 2 ds

65

FE EQUATION FOR BEAM cont.

• Strain energy cont.– Approximate curvature in terms of nodal DOFs

– Approximate element strain energy in terms of nodal DOFs

• Stiffness matrix of a beam element

22 2 2 T Te e

2 2 21 4 4 14 1 1 4

d v d v d v{ } { }

ds ds ds

q B B q

e

1 Te e e e e(e) T T3 0

1 EI 1U { } ds { } { } [ ]{ }

2 L 2 q B B q q k q

1e

3 0

6 12s

L( 4 6s)EI[ ] 6 12s L( 4 6s) 6 12s L( 2 6s) ds

L 6 12s

L( 2 6s)

k

66


• Stiffness matrix of a beam element

• Strain energy cont.

– Assembly

2 2

e

3

2 2

12 6L 12 6L

6L 4L 6L 2LEI[ ]

L 12 6L 12 6L

6L 2L 6L 4L

k

Symmetric, positive semi-definite

Proportional to EI

Inversely proportional to L

NEL NEL

e e e(e) T

e 1 e 1

1U U { } [ ]{ }

2

q k q

Ts s s

1U { } [ ]{ }

2 Q K Q

67

PRINCIPLE OF MINIMUM POTENTIAL ENERGY

• Potential energy (quadratic form)

• PMPE– Potential energy has its minimum when

• Applying BC– The same procedure with truss elements (striking-the-rows and

striking-he-columns)

• Solve for unknown nodal DOFs {Q}

T Ts s s s s

1U V { } [ ]{ } { } { }

2 Q K Q Q F

s s s[ ]{ } { }K Q F

[ ]{ } { }K Q F

[Ks] is symmetric & PSD

[K] is symmetric & PD

68

Plane Frame Element

• Beam– Vertical deflection and slope. No axial deformation

• Frame structure– Can carry axial force, transverse shear force, and bending moment

(Beam + Truss)

• Assumption– Axial and bending effects

are uncoupled

– Reasonable when deformation is small

• 3 DOFs per node

• Need coordinate transfor-mation like plane truss

F

p

1

2 3

4

1u2u

1v 2v

1q

2q

1u

2u

1v

2v

1q

2q

1u

2u

1v

2v

1q

2q 1

2 3

i i i{u , v , }

69

Plane Frame Element

• Element matrix equation (local coord.)

• Element matrix equation (global coord.)

• Same procedure for assembly and applying BC

x11 1 1

y12 2 2 2 12 2

2 2 2 2 1 1

1 1 2 x2

2 2 2 2 2 y22 2

2 2 2 2 2 2

fa 0 0 a 0 0 uf0 12a 6La 0 12a 6La v

0 6La 4L a 0 6La 2L a c

a 0 0 a 0 0 u f

0 12a 6La 0 12a 6La v f

0 6La 2L a 0 6La 4L a c

1

2 3

EAa

LEI

aL

[ ]{ } { }k q f

[ ][ ]{ } [ ]{ }k T q T f T[ ] [ ][ ]{ } { }T k T q f [ ]{ } { }k q f

T[ ] [ ] [ ][ ]k T k T

70

Plane Frame Element

• Calculation of element forces– Element forces can only be calculated in the local coordinate

– Extract element DOFs {q} from the global DOFs {Qs}

– Transform the element DOFs to the local coordinate

– Then, use 1D bar and beam formulas for element forces

– Axial force

• Bending moment and shear force:

{ } [ ]{ }q T q

2 1

AEP u u

L

y1 12 2

1 13

y2 22 2

22

V v12 6L 12 6L

M 6L 4L 6L 2LEIV vL 12 6L 12 6L

6L 2L 6L 4LM

Bending Moment

Shear Force

71

Homework

• Solve Problem 2.3-7

72

Sparsity and Symmetry

Textbook: 2.8, 2.9, 2.11

73

Sparsity

• Sparsity is a term used to quantify the number of zeros in a stiffness matrix.

• In very large models only a few nodes are connected to each other. This creates a lot of zeros.

• To economize storage (computer memory) commercial programs often use different ways to avoid storing the zero locations.

• Node numbering in a FE model affects the topology of the stiffness matrix– For a linear assembly of bar or beam elements you obtain a banded

matrix.

– However for most 3-D complex structures the bandwidth increases and can often result in a “dense” matrix.

• We need to find a topology that favors storage and efficient solving of the equation

74

Skyline of a Matrix

• Due to symmetry only terms on the diagonal and above need to be stored.

• Skyline of a matrix encloses the uppermost nonzero coefficients in each column

• The coefficients are then stored in a one-dimensional array.

• The array that describes the profile of the matrix is needed to develop a 1-D storage for the stiffness matrix

• Zeros under the skyline need to be stored as they will become filled in the solution process

1

23

4

k1

k2

k3

k4

x

y u1

v1

u2

v2

u3

v3

u4

v4

1

23

4

k1

k2

k3

k4

x

y u1

v1

u1

v1

u2

v2

u2

v2

u3

v3

u3

v3

u4

v4

u4

v4

111 12 15 16 1

121 22 24 25 26 1

233 35 2

242 44 2

351 52 53 55 56 3

61 62 65 66 68 3

4

68 88 4

0 0 0 0

0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0

0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0

x

y

x

y

x

y

fK K K K u

fK K K K K v

fK K u

fK K v

fK K K K K u

fK K K K K v

u

K K v

3

4

4

x

y

f

f

75

Profile of a Matrix

• The auxiliary array that describes the skyline is

• The sum of the entries in the profile description is the number of coefficients that must be stored and is referred to as the profile of the matrix

• The profile for the above case is 21

30653121

111 12 15 16 1

121 22 24 25 26 1

233 35 2

242 44 2

351 52 53 55 56 3

61 62 65 66 68 3

4

68 88 4

0 0 0 0

0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0

0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0

x

y

x

y

x

y

fK K K K u

fK K K K K v

fK K u

fK K v

fK K K K K u

fK K K K K v

u

K K v

3

4

4

x

y

f

f

76

Bandwidth of a Matrix

• Semi-bandwidth bi of any row i is equal to the number of columns from the diagonal to the rightmost non-zero term

• A matrix of large profile also has a large bandwidth

• The root mean square of the semi-bandwidths of the rows is used as a measure of the bandwidth of the matrix

• For our example the bandwidth is 3.28

• The solution procedure has to create fewer fills in the matrix when the coefficients are tightly clustered around the diagonal

111 12 15 16 1

121 22 24 25 26 1

233 35 2

42 44 2

51 52 53 55 56 3

61 62 65 66 68 3

4

68 88 4

0 0 0 06

0 0 05

0 0 0 0 0 03

0 0 0 0 0 01

0 0 02

0 0 03

0 0 0 0 0 0 0 01

0 0 0 0 0 01

x

y

x

fK K K K u

fK K K K K v

fK K u

K K v

K K K K K u

K K K K K v

u

K K v

2

3

3

4

4

y

x

y

x

y

f

f

f

f

f

77

Solution of Matrix Equation

• We have developed FE structural equations of the form

• and indicated that this can be solved (when the system is non-singular) as

• This must be simply interpreted as solution of the equation set.

• Seldom do we need to obtain the inverse matrix and multiply them as shown above.

• Solution procedure are of two types:– Direct solvers (e.g. Gauss elimination, LU factorization)

– Iterative solvers (e.g. Gauss-Siedel iteration)

K D R

1D K R

78

Direct Solvers

• Direct solvers use methods to transform the equations into an upper or lower triangle matrix that facilitates the solution (Gauss elimination), or decomposes the matrix into a product of upper and lower triangle (LU decomposition)

• The effort of solving a system of equations using the direct solvers is a function of its sparsity and profile (or bandwidth)

• Typically the number of operation required to solve a nxnmatrix system of equations is nb2 when the matrix bandwidth is b. (For dense matrix this is n3/3 )

• Review basics of Gauss elimination

• For very large structures, the matrix equations can be solved even before the entire equation can be assembled. These methods are called frontal solvers.

79

Iterative Solvers

• These methods start with a guess and iterate till it converges to a solution– (e.g Newton-Raphson method for solving algebraic equations)

– For matrices review Gauss-Siedel Iteration

• The effort required is difficult to predict in iterative solvers

• The convergence depends on the “condition number” of the stiffness matrix

• Condition number is the ratio of the largest and smallest eigenvalues of the stiffness matrix

• In structures the eigenvalues relate to the natural frequencies of the structure

• Pre-conditioned iterative solvers transform the system of equations to improve its conditioning before solving it.

80

APPLIED LOADS

• Potential energy of applied loads– Concentrated forces and couples

– Distributed load (Work-equivalent nodal forces)

ND

i i i ii 1

V Fv C

1

1T

1 1 2 ND s s2

ND

F

C

V v v ...... { } { }F

C

Q F

e2

e1

NEL NELx (e)

xe 1 e 1

V p(x)v(x)dx V

e2

e1

1x(e) (e)

x0

V p(x)v(x)dx L p(s)v(s)ds

1

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

0

1 1 1 1(e) (e) (e) (e)

1 1 1 2 2 3 2 4

0 0 0 0

(e) (e) (e) (e)1 1 1 1 2 2 2 2

V L p(s) v N N v N N ds

v L p(s)N ds L p(s)N ds v L p(s)N ds L p(s)N ds

v F C v F C

81

EXAMPLE – WORK-EQUIVALENT NODAL FORCES

• Uniformly distributed load

pL/2 pL/2

pL2/12 pL2/12

p

Equivalent

1 1 2 31 10 0

pLF pL N (s)ds pL (1 3s 2s )ds

2

21 12 2 3

1 20 0

pLC pL N (s)ds pL (s 2s s )ds

12

1 1 2 32 30 0

pLF pL N (s)ds pL (3s 2s )ds

2

21 12 2 32 40 0

pLC pL N (s)ds pL ( s s )ds

12

2 2T pL pL pL pL

{ }2 12 2 12

F

82


• Finite element equation for beam

– One beam element has four variables

– When there is no distributed load, p = 0

– Applying boundary conditions is identical to truss element

– At each DOF, either displacement (v or ) or force (F or C) must be

known, not both

– Use standard procedure for assembly, BC, and solution

1 12 2 2

1 13

2 22 2 2

2 2

v F12 6L 12 6L pL / 2

C6L 4L 6L 2L pL / 12EIv FL 12 6L 12 6L pL / 2

C6L 2L 6L 4L pL / 12

83

BENDING MOMENT & SHEAR FORCE

• Bending moment

– Linearly varying along the beam span

• Shear force

– Constant

– When true moment is not linear and true shear is not constant, many elements should be used to approximate it

• Bending stress

• Shear stress for rectangular section

2 2

2 2 2 2

d v EI d v EIM(s) EI { }

dx L ds L B q

1

3 31

y 3 3 3 32

2

v

dM d v EI d v EIV (s) EI [ 12 6L 12 6L]

vdx dx L ds L

x

My

I

2y

xy 2

1.5V 4y(y) 1

bh h

84

EXAMPLE – CLAMPED-CLAMPED BEAM

• Determine deflection & slope at x = 0.5, 1.0, 1.5 m

• Element stiffness matrices

F2 = 240 N

y

x

1 2

1 m

3

1 m1 1 2 2

1

1(1)

2

2

v v

v12 6 12 6

6 4 6 2[ ] 1000

v12 6 12 6

6 2 6 4

k2 2 3 3

2

2(2)

3

3

v v

v12 6 12 6

6 4 6 2[ ] 1000

v12 6 12 6

6 2 6 4

k

11

11

2

2

33

33

F12 6 12 6 0 0 v

C6 4 6 2 0 0

24012 6 24 0 12 6 v1000

06 2 0 8 6 2

F0 0 12 6 12 6 v

C0 0 6 2 6 4

85

EXAMPLE – CLAMPED-CLAMPED BEAM cont.

• Applying BC

• At x = 0.5 s = 0.5 and use element 1

• At x = 1.0 either s = 1 (element 1) or s = 0 (element 2)

2

2

v24 0 2401000

0 8 0

2

2

v 0.01

0.0

12

1 1 1 1 1 11 1 1 2 2 3 2 4 32 2 2 2 2 2

3122 (1)

s

v( ) v N ( ) N ( ) v N ( ) N ( ) 0.01 N ( ) 0.005m

dN1( ) v 0.015rad

L ds

2 3 3

32(1)

s 1

v(1) v N (1) 0.01 N (1) 0.01m

dN1(1) v 0.0rad

L ds

2 1 1

12(2)

s 0

v(0) v N (0) 0.01 N (0) 0.01m

dN1(0) v 0.0rad

L ds

Will this solution be accurate or approximate?

86

EXAMPLE – CANTILEVERED BEAM

• One beam element

• No assembly required

• Element stiffness

• Work-equivalent nodal forces

C = –50 N-m

p0 = 120 N/m

EI = 1000 N-m2

1

1s

2

2

v12 6 12 6

6 4 6 2[ ] 1000

v12 6 12 6

6 2 6 4

K

2 31e

2 311e

0 02 302e

2 32e

F 1/ 2 601 3s 2s

C L / 12 10(s 2s s )Lp L ds p L

F 1/ 2 603s 2s

C L / 12 10( s s )L

L = 1m

87

EXAMPLE – CANTILEVERED BEAM cont.

• FE matrix equation

• Applying BC

• Deflection curve:

• Exact solution:

1 1

1 1

2

2

v12 6 12 6 F 60

6 4 6 2 C 101000

v12 6 12 6 60

6 2 6 4 10 50

2

2

v12 6 601000

6 4 60

2

2

v 0.01

0.03

m

rad

33 4v(s) 0.01N (s) 0.03N (s) 0.01s

4 3 2v(x) 0.005(x 4x x )

88


• Support reaction (From assembled matrix equation)

• Bending moment

• Shear force

2 2 1

2 2 1

1000 12v 6 F 60

1000 6v 2 C 10

1

1

F 120N

C 10N m

2

1 1 2 22

EIM(s) { }

LEI

( 6 12s)v L( 4 6s) (6 12s)v L( 2 6s)L1000[ 0.01(6 12s) 0.03( 2 6s)]

60s N m

B q

y 1 1 2 23

EIV 12v 6L 12v 6L

L1000[ 12 ( 0.01) 6( 0.03)]

60N

89


• Comparisons

-0.010

-0.008

-0.006

-0.004

-0.002

0.000

0 0.2 0.4 0.6 0.8 1x

v

FEM

Exact

-0.030

-0.025

-0.020

-0.015

-0.010

-0.005

0.000

0 0.2 0.4 0.6 0.8 1x

FEM

Exact

-60

-50

-40

-30

-20

-10

0

10

0 0.2 0.4 0.6 0.8 1x

M

FEM

Exact

-120

-100

-80

-60

-40

-20

0

0 0.2 0.4 0.6 0.8 1x

Vy

FEM

Exact

Deflection Slope

Bending moment Shear force

90

Using Symmetry

• Can reduce model size and save computation time

• Can provide necessary boundary conditions

pp

Symmetry plane

Modeled portion

pp

91

Using Symmetry

• Load must be halved if it is on the symmetric plane

• Beam

px

y

(a) One symmetric plane

px

y

(b) Two symmetric planes

P

L

P/2

L/2= 0

92

Anti-symmetry

• Symmetry– Translational motion has no component normal to a plane of symmetry

– Rotation vectors have no components in a plane of symmetry

• Anti-symmetry– Translational motion has no component in a plane of anti-symmetry

– Rotation vectors have no components normal to a plane of anti-symmetry

93

94

Homeworks

• Solve Problem 2.8-7

• A linearly varying distributed load is applied to the beam finite element of length L. The maximum value of the load at the right side is q0. Calculate “work equivalent” nodal forces and couples.

xy

q0

F1 F2

M1 M2

L

chapter2_2015

Documents

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