mechanics of solids week 9 lectures

8/12/2019 mechanics of solids week 9 lectures

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Lecture Week 9 MECH3361

1

Bar Elements in 2-D and 3-D Space

Step 1: Element analysis (given in general)

Step 2: Transformation of nodal Displacements and Forces

i

iiiiii

v

umlvuvuu sincos90coscos'

i

iiii

v

ulmvuv cossin'

where direction cosines: L

YY

mL

XX

l

ijij

sin90cos,cos

In a matrix form: iii

i

i

i orv

u

lm

ml

v

uuTu

~''

'

(transform global to local)

where transformation matrix:

lm

mlT~

is orthogonal, which means TTT ~~ 1

For the two nodes of the bar element, we have

j

j

i

i

j

j

i

i

v

u

v

u

lm

ml

lm

ml

v

u

v

u

00

00

00

00

'

'

'

'

Or Tuu' with

T

TT ~

0

0~

The nodal forces are transformed in the same way,

yj

xj

yi

xi

yj

xj

yi

xi

f

f

f

f

lm

ml

lm

ml

f

f

f

f

00

00

00

00

'

'

'

'

Or Tff' (transform global f to local f )Step 3 Elemental stiffness matrices:Since lateral displacement vi does not contribute to the stretch of the bar, within the linear

theory, in the local coordinate system (x-y), we have

'

'

'

'

2

1

11

11

11

11

xj

xi

j

i

f

f

u

u

L

EA

u

u

kk

kk

Augmenting this equation, we re-write:

0

0

0000

0101

0000

0101

'

'

'

'

'

'

xj

xi

j

j

i

i

f

f

v

u

v

u

L

EA

vi


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2

Or ''' fuk Using transformation derived, we obtain

TfTuk ' Multiplying both sides by TTand noticing that TTT = I (orthogonal), we obtain

fTukT 'T

fuTkT 'T fku

Thus, the element stiffness matrix k in the global coordinate system is

TkTk 'T

In the explicit form,

22

22

22

22

mlmmlm

lmllml

mlmmlm

lmllml

L

EAk

The structural (global) stiffness matrix is assembled by using the elemental stiffness matrices

in the similar way to that in the 1-D bar cases.Step 4 Calculate elemental strain and stress:

'

'

j

i

u

uEE B

LL

NNdx

dji

11)()( B

Thus

j

j

i

i

j

j

i

i

j

i

v

u

v

u

mlmlL

E

v

u

v

u

ml

ml

LLE

u

uE

00

0011'

'

B

Example 8.4A simple plane truss is made of two identical bars (with E, A,

andL), and loaded as shown in the figure. Find

1) displacements of node 2; 2) stress in each bar.

Solution:

Step 1: Local elemental stiffness matrix:This simple structure is used here to demonstrate the assembly

and solution process using the bar element in 2-D space.

In local coordinate systems, we have

''kk 21

11

11

L

EA

Step 2: Global stiffness matrix:These two matrices cannot be assembled together di rectly,

because they are in different coordinate systems. We need to convert them to global

coordinate systemXOY.

Element 1: 2245cos,2245cos,45 ml . The global stiffness matrix is:

1111

1111

1111

1111

2

22

22

22

22

11T11

L

EA

mlmmlm

lmllml

mlmmlm

lmllml

L

EATkTk

' ,

0

000000

000000

001111

101111

001111

001111

21

L

EAk


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3

Element 2:2

245cos,

2

2135cos,135 ml . The global stiffness matrix is:

11111111

1111

1111

222

22

22

22

22T22

L

EA

mlmmlmlmllml

mlmmlm

lmllml

L

EATkTk

' ,

111100

111100

111100

111100

000000

000000

22 L

EA

k

Step 3: Assemble the global stiffness matrix and FE equation

Step 4: Apply boundary condition (BC) and loads

22123311 ,,0 PFPFvuvu Yx Step 5: Condense (deleting all zero-displacement rows/columns) FE equation

2

1

2

2

20

02

2 P

P

v

u

L

EA

Step 6: solve for the equation and we obtain the displacement of node 2:

2

1

2

2

P

P

EA

L

v

u

Step 7: Elemental stress in the two bars:

21

2

11

2

20

0

11112

2PP

A

P

PEA

L

L

E

v

u

v

u

mlmlL

EE

j

j

i

i

Bu

212

1

22

2

0

01111

2

2PP

A

P

P

EA

L

L

E

8.3 Beam Element

Simple Plane Beam ElementL length

I 2nd

moment of inertia of the cross-sectional area A dAyI 2

E elastic modulus

v = v(x) deflection (lateral displacement) of the neutral axis

dxdv/ slope or rotation about the z-axisF =F(x) shear force

M =M(x) moment about z-axis

As per Beam Theory (Mechanics of Solids I): )(2

2xM

dx

vdEI andI

My


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4

Formal Approach

Step 1: Introduce the shape functions:

3

2

2

4

3

3

2

23

3

2

22

3

3

2

21

11

)(,

23

)(

12)(,

231)(

xLxLxNxLxLxN

xL

xL

xxNxL

xL

xN

Step 2: Calculate the deflection:

j

j

i

i

v

v

xNxNxNxNxv

)()()()()( 4321Nu

which is a cubic function w.r.t. the coordinate ofxand uis nodal displacement vector.

Curvature of the beam is, BuuNNu

2

2

2

2

2

2

dx

d

dx

d

dx

vd

where the strain-displacement matrix B is given by,

232232

''4

''3

''2

''12

2 6212664126)()()()(

L

x

LL

x

LL

x

LL

x

LxNxNxNxN

dx

dNB

Step 3: Strain energystored in the beam element is

uBBuBuBu

dxEIdxEIdxdx

vdEI

dx

vd

MdxEI

MdAdxI

My

EI

MydVU

L

TT

L

T

L

T

L

T

V

T

V

T

2

1

2

1

2

1

2

1

1

2

11

2

1

2

1

2

2

2

2

We conclude that the stiffness matrix for the simple beam element is

dxEIL

TBBk

2

1

After integrating, equilibrium equation is:

Combining the bar element, we obtain the stiffness matrix of ageneral 2-D beam element,

8.4 Distributed LoadEquivalent nodal load

Uniformly distributed axial load q (N/mm, N/m, lb/in) can be converted to two equivalent

nodal forcesof magnitude qL/2. We verify this by considering the work done by the load q,

2/2/

21

22211

2

)()(2

)(2

)()(2

1

2

1

10

1

0

1

0

1

00

qLqLuu

uuqLqL

uudqL

du

uNN

qLdu

qLLdquuqdxW

jij

i

j

i

j

iji

L

q


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5

That is qT

qW fu2

1 where

2/

2/

qL

qLqf

Thus, from U=W principle for element, we should have

loadddistribute

qT

forcenodalother

Tq

T WU fufukuu

2

1

2

1

2

1

which yields qffku . Thus the new nodal force vector is

2/

2/

2/

2/

qLf

qLf

qL

qL

f

f

j

i

j

iqff

in an assemble bar system

Multiple bar element systemFor a multiple bar element system, e.g. two elements (as figure next page), one can expand

the force vector in each element and then add the expanded force vectors, as

2/

2/

2/

2/

2/

2/

0

0

2/

2/

3

2

1

)1(2

)2(1

)1(2

)1(1

11

qL

qL

qL

qLf

qLf

qLf

qLf

qLfqLf

qLf

elementqelementq ffff

Equivalent nodal load for transverse force

This can be verified by considering the work done by the distributed load q. For the 2 beam

elements, the equivalent loading can be calculated as follows:

Example 8.5Given: A cantilever beam with distributed lateral load p as shown above.

Find: The deflection and rotation at the right end, the reaction force and moment at the left

end.


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Solution:

Step 1: Calculate the quivalent nodal load: The work-equivalent nodal loads are shown on

right, in which the equivalent nodal loads are: 12/,2/ 2pLmpLf .Step 2: Applying the FE equilibrium equation, we have

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

Step 3: Load and constrains (BCs) are: 0,, 1122 vmMfFY Reduce the FE equation into

Thus:

m

fv

LL

L

L

EI

2

223 46

612

Step 4: Solve for the reduced FE equation

)6/(

)8/(

63

32

6 3

42

2

2

EIpL

EIpL

mLf

LmfL

EI

Lv

These nodal values are the same as the engineering beam theorey solution. Note that the

deflection v(x) (for 0


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7

which is related to nodes #3 and #4 with displacement v3and v4

Beam element 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 element 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 conditionsPFMMvvv 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:

Beam element 1

Beam element 2

spring element


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8

kN

kN

mkN

kN

F

F

M

F

Y

Y

Y

488.3

2.116

78.69

78.69

4

2

1

1

Thus the free-body diagram can be drawn as below:

8.5 Two-Dimensional Problems

General Formula for the Stiffness Matrix

Recall the displacement in terms of shape function in bar element

Nu

j

i

ji u

uNNu ][

We can extend it to 2D, where the displacements (u, v) in a plane element are interpolated

fr om nodal displacements(ui, vi) using shape functionsNias follows,

2211

2211

),(),(),(

),(),(),(

vyxNvyxNyxvv

uyxNuyxNyxuu

In the matrix form: Ndu

2

2

1

1

21

21

0000

v

u

v

u

NNNN

vu

where N is the shape function matr ix, u the displacement vector inside each element and d

the elemental nodal displacement vector. Here we assume that u depends on the nodal values

of u only, and v on nodal values of v only.

From strain-displacement relation, the strain vector can be derived as:

dDNDu Or Bd

where DNB

is the strain-displacement matrix and D is derivative matrix (see below).

Consider the strain energy stored in an element,

V

T

V

T

V

T

V

xyxyyyyyxxxx

V

T

dVdVdV

dVdVU

))2

1

2

1

2

1

)(2

1][

2

1

dd

Since nodal displacement vector dis independent on the elemental coordinate dV.

kdddd T

V

TTdVU

2

1


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9

From this, we obtain the general formula for the element sti f fness matr ix,

Remarks:

Note that unlike 1-D cases, E here is a matrix that is given by the stress-strain relation(generalised Hookes rule) The stiffness matrix k is symmetric since E is symmetric. Also note that given the material property, the behaviour of k depends on the B matrix

only, which in turn on the shape functions. Thus, the quality of finite elements in

representing the behaviour of a structure is entirely determined by the choice of shape

functions.

Constant Strain Triangle (CST or T3)This is the simplest 2-D element, which is also called li near tr iangular element.

For this element, we have three nodes at the vertices of

the triangle, which are numbered around the element inthe counterclockwise direction. Each node has two

degrees of freedom(can move in thex andy directions).

The displacements u and v are assumed to be linear

functions within the element, that is,

ybxbbv

ybxbbu

654

321

where bi(i = 1, 2, ..., 6) are constants. From these, the

strains are found to be,

2321 bybxbbxxu

xx

6654 bybxbbyy

vyy

533216542 bbybxbby

ybxbbxx

v

y

uxy

which are constant throughout the element. Thus, we have the name constant strain

triangle (CST)element.

Displacement functions should satisfy the following six equations,

363543

333213

262542

232212

161541

131211

ybxbbv

ybxbbu

ybxbbv

ybxbbu

ybxbbv

ybxbbu

Solving these six equations, we can find the coefficients b1, b2, ..., b6in terms of nodal

displacements and coordinates. Substituting these coefficients into displacement equations

and rearranging the terms, we obtain:


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10

3

3

2

2

1

1

321

321

000

000

v

u

v

u

v

u

NNN

NNN

v

u (li near distri bution)

where the shape functions (note that they are linear functions ofx andy) are

yxxxyyyxyxA

N

yxxxyyyxyxA

N

yxxxyyyxyxA

N

)()()(2

1

)()()(2

1

)()()(2

1

122112213

311331132

233223321

where

33

22

11

1

1

1

det2

1

yx

yx

yx

A is the area of the triangle

Using the strain-displacement relation, we have,

3

3

2

2

1

1

122131132332

211332

123123

000

000

2

1

vu

v

u

v

u

yxyxyx

xxx

yyy

Axy

yy

xx

Bd

wherexij=xi-xjandyij=yi-yj (i, j = 1, 2, 3). Again, we see constant strains within the

element (because variablesxandydisappeared). From stress-strain relation (Hookes law),

we see that stresses obtained in the CST element should also be constant.

Applying general formulafor stiffness matrix, we obtain it for the CST element,

in which t is the thickness of the element. Notice that k for CST is a 6 by 6symmetric matrix.

The matrix multiplication can be carried out by a computer program.

Applications of the CST Element:

Use in the areas where the strain gradient is small. Use in the mesh transition areas (fine mesh to coarse mesh) to accommodate different

mesh topologies.

Avoid using CST in stress concentration or other crucial areas in the structure, such asedges of holes and corners.

Recommended for quick and preliminary FE analysis of 2-D problems.

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mechanics of solids week 9 lectures