mechanics of solids week 13 lectures

8/12/2019 mechanics of solids week 13 lectures

1/8

Week 13 - Review 2013 (Part 2)

1

Chapter 7 Plasticity and Failure

Strain energy density(strain energy per unit volume) (SED)

)(22

1zxzxyzyzxyxyzzzzyyyyxxxxU

(DED)densityenergydistortioncalledstress,sheartoduechangeShape

213

232

221

stresschydrostatibycausedchangevolumetodueSED

2

1

13322123

22

21

)()()(6

1

2

1

)(2)(2

1

EIK

EU

Tresca criterion

Tresca observed that material flow seems to be along the direction of maximum shear stress.

Tresca stress: 312

1 Tresca

Tresca criterion: Y 31

Failure criterion (Ductile materials):

Y 31

Von Mises criterion:

plastic yielding occurs when the distortion energy density equal or exceeds that of the same

material under uniaxial tension.

Von Mises stress: 213232221 )()()(2

1 vm

Von Mises criterion 22132

322

21 2)()()( Y

Von Mises Failure criterion (Ductile materials):

2

213

232

221

2)()()( Y

Brittle materials

There is no yield for brittle material in general.

The maximum normal stress theory

f1

f is the failure normal stress

The maximum normal strain theory

f1

f

is the failure normal strain.

Since from Hookes law: )(1

3211 E

f )( 321


2/8


2

Chapter 8 Finite Element Method1D elements:Spring element, Bar element, Beam element and their combination

Spring element Bar element Beam element

dof=2:

j

i

j

i

f

f

u

u

kk

kk

dof =2:

j

i

j

i

f

f

u

u

L

EA

11

11 dof =4

2D element: CST (T3), LST (T6), Q4, Q8 element

Displacement Shape function Strain

T3,

3-node

constant strain triangle

(CST)

Linear:

ybxbbv

ybxbbu

654

321

(6 d.o.f.)

)(2

1yx

AN iiii

(i=1,2,3)

33

22

11

1

1

1

2

1

yx

yx

yx

A

Constant strain

2bx

uxx

6by

vyy

53

2

bb

x

v

y

uxy

T6

quadratic triangular

elementlinear strain triangle

(LST)

Quadratic:

21211

210

987

265

24

321

ybxybxb

ybxbbv

ybxybxb

ybxbbu

(12 d.o.f.)

)1(4

)1(4

4

1)1(2

)1(

)12(

)12(

6

5

4

3

2

1

N

N

N

N

N

N

Fully-linear

ybxbbxx 542 2

ybxbbyy 12119 2

ybbxbb

bbxy

116103

53

222

2

Linear Quadrilateral

Element (Q4)

Bilinear

xybyb

xbbv

xybyb

xbbu

87

65

43

21

(8 d.o.f.)

)1)(1(4

11 N ,

)1)(1(4

12 N ,

)1)(1(4

13 N ,

)1)(1(4

14 N

Half-linear

xbbybb

y

u

x

v

xbby

v

ybbx

u

xy

yy

xx

4386

87

42

2


3/8


3

Quadratic QuadrilateralElement (Q8)

Quadratic

216

215

21413

212

11109

28

27

265

24

321

xybyxb

ybxybxb

ybxbbv

xybyxb

ybxybxb

ybxbbu

(16 d.o.f.)

)1)(1)(1(41

)1)(1)(1(4

1

)1)(1)(1(4

1

)1)(1)(1(4

1

4

3

2

1

N

N

N

N

)1)(1(2

1

)1)(1(2

1

)1)(1(2

1

)1)(1(2

1

28

27

26

25

N

N

N

N

Half-quadratic

21615

131210

82

7

653

16215

141311

287

542

2

2

2

2

2

2

2

2

ybxyb

ybxbxb

xybxb

ybxbb

xybxb

ybybyb

ybxyb

ybxbxb

xy

yy

xx

3D element: Tet (4-node and 10-node) and Brick (8-node and 20-node)

4-node tet 10-node tet 8 node brick 20 node brick

3D

elements

dof 12 30 24 60

Example 17 (Quiz 2012).In the following barspring structure as shown, use finite element

method to (1) Derive global equilibrium equation; (2) Determine displacement at the free end

point O, u1, mid point u2and reaction force in the right hand side wall (point C). (3) Sketch

the displacement distribution in thex-coordinate. Assume thatEA=1,L=1, ks=6.

Soln: Step 1:Elemental stiffness matrices

11

11

11

11

11

111

L

EA

kk

kkK ,

66

662

ss

ss

kk

kkK

,

66

663

ss

ss

kk

kkK

Step 2:Expanded elemental stiffness matrices

660

660

000

,

660

660

000

,

000

011

011

321 KKK

Step 3 (sub-question (1)):Global FE equation; FuK :

3

2

1

3

2

1

12120

12131

011

F

F

F

u

u

u

Step 4: Apply boundary conditions, 03u

3

2

1

2

1

012120

12131

011

F

F

F

u

u

P

P

F

F

u

u

4131

11

2

1

2

1

x

y

x

y

12

3

4

5

6

7

8

1m 1m

EA=1 ks

xP4P

A BC

Oks


4/8


4

P

P

u

u

4131

11

2

1

Puu

Puu

413 21

21

4/

4/3

2

1

Pu

Pu

Displacement vector:

0

4/

4/3

3

2

1

P

P

u

u

u

Step 5 Plot displacement functions:

jijjiink kk uuuNuNuNxu )1()()()()( 1

, in which Lx /

Element 1 (AB): local Cartesian coordinate

)1(x is the same as global Cartesian, i.e.

xx )1(

PxPPxPxuLxuLxxu BAAB 4/3)4/()4/3(1)/()/1()( )1()1(

Element 2 (BC): local Cartesian )2(x differs from global Cartesian, i.e. 1)2( xLxx 2/4/)0(1)4/()1(1)/()/1()( )2()2( PPxxPxuLxuLxxu CBBC

Step 6: Reaction forceat the wall (use the dropped first equation)

PP

P

FRC 3

0

4/

4/3

121203

(4) If remove the load at node B and consider a stopper

on the right hand-side as shown. Determine the

displacements and reaction from the stopper and C.

3

2

1

012120

12131

011

F

R

Pu

2

1

131

11

R

Pu

21

1

13 Ru

Pu

PR

Pu

122

1

Displacement vector:

12

0

121203

P

FRC

03

2

1

P

u

u

u

Example 18 (Quiz 2012): Two 3-node triangular elements are used to model a plane stress

problem. The stress distributions are shown below. Which are the possible correct plots and why?

a) should be the answer. Note that these two 3-node triangular elements are the constant strain

triangular (CST) elements (need mention this, otherwise lose 3 marks), which have constant

strain and stress.

x

y

1

2

3

yy

4

x

y

1

2

3

yy

4

y

1

2

3

yy

4

(b)(a) (c)

x

3P/4

-P/4

u

x

3P/4

-P/4

u

1m 1m

EA=1 ks

xP

A BC

Oks

Stopper


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5

Example 19 (Quiz 2013)A uniform beam

(EI = 1) is fully clamped at the left end (point

A, node 1), and supported at both the mid-

span (point B, node 2) and the right end

(point C, node 3) by rollers. A bending

moment (M = 14 Nm) is applied at the mid-span as shown.

a. Express the global equilibrium equation using a finite element formulation.b. Determine the slopes at the mid-span and at the right end.c. Find all reaction forces and moments.d. In order to make 3=1, what moment loadingM3is needed?

Soln

Step 1: Element stiffness matrices

4626612612

2646

612612

4626612612

2646

612612

22

22

3

21

LLLLLL

LLLL

LL

L

EIKK

Step 2: Install theGlobal FE equilibrium equation

3

3

2

2

1

1

3

3

2

2

1

1

462600

61261200

268026

612024612

002646

00612612

M

F

M

F

M

F

v

v

v

uK

Step 3: Apply Boundary condition and external loadings

0

14

0

0

0

0

462600

61261200

268026

612024612

002646

00612612

3

2

1

1

3

2

F

F

M

F

0

14

42

28

3

2

Step 4: Solve for the unknowns:

1

2

3

2

Step 5: Reaction forces and moments

0

14

0

0

0

0

462600

61261200

268026

612024612

002646

00612612

3

2

1

1

3

2

F

F

M

F

;

0

6

14

6

4

12

1

2

42

66

28

60

02

06

42

66

28

60

02

06

3

2

3

3

2

2

1

1

M

F

M

F

M

F

Step 6: Add boundary condition of 3=1 into the global FE equilibrium equation

0

14

0

0

0

0

462600

61261200

268026

612024612

002646

00612612

3

2

1

1

3

2

F

F

M

F

A B

L=1

EI=1x

L=1

CEI=1

1 2 3M


6/8


6

3

3

2

1

1

2 14

1

0

0

0

0

462600

61261200

268026

612024612

002646

00612612

M

F

F

M

F

1428 2 5.18/122 . From the last equation: 745.1242 23 M

Example 20 (Quiz 2011): If one Gaussian point was used for the 2D Q8 element, determine

the displacements at this Gaussian point if the nodal displacements

are given as

01011012

12101101

v

u

Soln: The displacement functions as per Q8s shape functions Ni

and nodal displacements ui and vi can be expressed as:

ii

iii

i vNvuNu

4

1

4

1

,

25.2)0,0(

)1()01)(01(5.0)2()01)(01(5.0)1()01)(01(5.0)0()01)(01(5.0

)1()001)(01)(01(25.0)1()001)(01)(01(25.00)001)(01)(01(25.01)001)(01)(01(25.0

)0,0(),(

2222

8877665544332211

8

1

u

uNuNuNuNuNuNuNuNuNuu ii

i

5.0)0,0(

)0()01)(01(5.0)1()01)(01(5.0)0()01)(01(5.0)1()01)(01(5.0

)1()001)(01)(01(25.0)0()001)(01)(01(25.0

1)001)(01)(01(25.02)001)(01)(01(25.0

)0,0(),(

2222

8877665544332211

8

1

v

vNvNvNvNvNvNvNvNvNvv ii

i

Example 21 (Quiz 2012):For the Q4 element as shown in the

Cartesian coordinate system, the nodal displacement of the

element that was generated from the FEA is T1,1,2,2,0,0,0,0 u .

Use shape function to calculate the displacement (u, v) at the

mid point E on side BE.

Soln: TT 1,2,0,0,1,2,0,0 vu

ii

iii

i vNvuNu

4

1

4

1

,

4/)1)(1(1 N , 4/)1)(1(2 N , 4/)1)(1(3 N , 4/)1)(1(4 N

A B

L=1

EI=1x

L=1

CEI=1

1 2 3M2

M3

3

3

2

1

1

2 14

1

0

0

0

0

462600

61261200

268026

612024612

002646

00612612

M

F

F

M

F

A(0,0) B(6,0)

C(4,6)

D(0,3)

x

y

A(0,0) B(4,0)

C(4,6)

D(0,3)

x

y

E(4,3)


7/8


7

At the mid point E: 0,1

1)1)(01)(11(

4

1)2)(01)(11(

4

1)0)(01)(11(

4

1)0)(01)(11(

4

1

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

14321

4

1

uuuuuNu ii

i

1)1)(01)(11(4

1)2)(01)(11(

4

1)0)(01)(11(

4

1)0)(01)(11(

4

1

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

14321

4

1

vvvvvNv ii

i

Displacement: (1.0, 1.0)

Example 22 (Quiz 2011)if four Gaussian points are used in a Q4

element, determine the coordinates of Gaussian point G1 in

Cartesian coordinate system of iso-parametric element Q4 as

shown. Compute Jacobian matrix [J] and its determinant |J|.Soln

Step 1:Node numbering: Node (1,2,3,4) = Node (A,B,C,D)

Step 2:Coordinate in terms of Shape functions

)1)(1(4

11 N , )1)(1(

4

12 N , )1)(1(

4

13 N , )1)(1(

4

14 N

Cartersian coordinate (x,y) in terms of natural coordinate (, )

22),(

0)1)(1(4

14)1)(1(

4

14)1)(1(

4

10)1)(1(

4

1

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

1

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

14321

4

1

x

xxxx

xxxxxNx

DCBA

ii

i

22),(

4)1)(1(4

14)1)(1(

4

10)1)(1(

4

10)1)(1(

4

1

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

1

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

14321

4

1

y

yyyy

yyyyyNy

DCBA

ii

i

Thus G1 coordinate:

8452.0)5774.0(2222)5774.0,5774.0( x

8452.0)5774.0(2222)5774.0,5774.0( y

Step 3:Jacobian matrix

20

02

2222

2222

yx

yx

J

Jacobian determinant: 422det JJ

A(0,0) B(4,0)

C(4,4)D(0,4)

x

y

* *

**

G3G4

G1 G2


8/8


8

Example 23 (Exam 2009) Given: P = 50 kN, k = 200 kN/m, L

= 3 m, E = 210 GPa, I = 210-4m4.

Find:Deflections, rotations and reaction forces.

Soln: This is a combined problem of beam and spring

elements

Step 1: The system has 4 nodesas well as 2 beamelementsand 1 springelement.

Step 2: Spring elementhas stiffness matrix is

j

i

j

i

f

f

u

u

kk

kk which is related to nodes

#3 and #4 with displacement v3and v4

Beam 1:

2

2

1

1

2

2

1

1

22

22

3

4626

612612

2646

612612

m

f

m

f

v

v

LLLL

LL

LLLL

LL

L

EI

Y

Y

Beam 2:

3

3

2

2

3

3

2

2

22

22

3

4626

612612

2646

612612

m

f

m

f

v

v

LLLL

LL

LLLL

LL

L

EI

Y

Y

Step 3: The global FE equations can be assembled as (where )/(' 3

EIkLk )

Step 4: Apply the boundary conditions

PFMMvvv Y 3324211 ,0,0

The reduced FE equation becomes:

0

0

462

6'126

268

3

3

2

22

22

3

Pv

LLL

LkL

LLL

L

EI

Step 5 Solve for the reduced FE equation:

rad

m

rad

LkEI

PLv

007475.0

01744.0

002492.0

9

7

3

)'712(

2

3

3

2

Step 6: Reaction force and moment can be found from the eliminated equations:

kN

kN

mkN

kN

F

F

M

F

Y

Y

Y

488.3

2.116

78.69

78.69

4

2

1

1

Free-body diagram can be drawn as

Beam element 1

Beam element 2

spring element

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mechanics of solids week 13 lectures