One-Dimensional Problemstaminmn/SME 3033 Finite Element Method/04-1D... · UNIAXIAL BAR ELEMENTS...

UNIAXIAL BAR ELEMENTS

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One-Dimensional Problems


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We wish to use FEM for solving the following problems:

d = 2 x 10-2 mm

10 kN

x

x

Calculate displacement of

bar ABC, take E = 200GPa


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3-1 Objectives

1. To develop a system of linear equations for one-dimensional

problem.

2. To apply FE method for solving general problems involving

bar structures with different support conditions.

3-2 General Loading Condition

Consider a non-uniform bar subjected

to a general loading condition, as

shown.

Note: The bar is constrained by a fix support at

the top and is free at the other end. The positive

x-direction is taken downward.


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Types of Loading

a) Body force, f

Distributed force per unit volume (N/m3)

Example: self-weight due to gravity

b) Traction force, T

Force per unit area (N/m2)

For a 1-D problem,

area ofperimeter area

force

T

Examples: Frictional forces, Viscous drag,

and Surface shear.

c) Point load, Pi

Concentrated load (in Newton) acting at any point i.


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3-3 Finite element Modeling

3-3-1 Element Discretization

The first step is to subdivide the bar into several sections – a

process called discretization.

Note: The bar is discretized

into 4 sections, each has a

uniform cross-sectional area.

The non-uniform bar is

transformed into a stepped

bar.

We will use the stepped bar

as a basis for developing a

finite element model of the

original non-uniform bar.


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3-3-2 Numbering Scheme

To analyze the stepped bar systematically, a global numbering

scheme is assigned as shown. The x-direction is considered as

the global coordinate direction.

Note:

F1, …, F5 represent global

forces acting on the points

connecting all sections of the

stepped bar.

Q1, …, Q5 represent global

displacements of the points

resulting from the forces

acting on these points. TQQQQQQ 54321

TFFFFFF 54321The stepped bar is transformed into

a finite element model using 1-D

(line) elements.


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Connectivity between global and local nodes must be established for

each element, as tabulated in the table shown.

3-3-3 Element Connectivity

Element connectivity table

Consider a single line element. It lies in a local coordinate

system, denoted by x.

Note: Node number in local

coordinate is denoted by a

number with a hat on top.

q1 and q2 are nodal displacements

in the local coordinate direction.

^

^ ^ x̂

1̂ 2̂


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3-4-1 Natural Coordinate

3.4 Natural Coordinate and Shape Functions

We define a natural or intrinsic coordinate system, ,

1

21

12

xxxx

Note: The -coordinate will be used to define shape functions,

required to establish interpolation function for the displacement

field within the element.

Consider a single element. Local node 1 is at distance x1 from a

datum, and node 2 is at x2, measured from the same datum point.


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3-4-2 Shape Functions

We establish a linear interpolation function to represent the linear

displacement field within the element. To implement this, linear shape

functions are defined, given by,

2

1

2

121

ξξN

ξξN

and

The displacement field, u(x), within the element is not known.

For simplicity, it is assumed that the displacement varies

linearly from node 1 to node 2 within the element.


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The linear displacement field, u(x), within the element can now

be expressed in terms of the linear shape functions and the

local nodal displacement q1 and q2 as:

21

^

2211

^

2

1

2

1)(

)(

qqxu

qNqNxu

In matrix form:

qN)x(u^

where 21 NNN

and Tqqq

qq 21

2

1


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Coordinate x of any point on the element (measured from the

same datum point as x1 and x2) can be expressed in terms of

the same shape functions, N1 and N2 as

3-4-3 Isoparametric Formulation

21

2211

2

1

2

1xxx

xNxNx

When the same shape functions N1 and N2 are used to establish

interpolation function for coordinate of a point within an element and

the displacement of that point, the formulation is specifically referred

to as an isoparametric formulation.


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Example 2-1

(a)Evaluate , N1, and N2 at point P.

(b) If q1 = 0.003 in and q2 = -0.005 in, determine the value of

displacement u at point P.

Solution

(a) The coordinate of point P is given by

1

2 1

21

224 20 1

36 20

0.5

P

P

x xx x


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1

2

1 1 0.50.75

2 2

1 1 0.50.25

2 2

N

N

(b) Displacement of point P

1 1 2 2

0.75 0.003 0.25 0.005

0.001 in

P

P

u N q N q

u

The shape functions are:


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3-5 Strain-Displacement Relation

Normal strain is related to displacement by

du

dx

Using the chain rule of differentiation

du d

d dx

1

2 1 2 1

2 21

dx x

x x dx x x

1 2

1 2

1 1

2 2 2

q qduu q q

d

The two terms of the above relation are obtained as follows


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which can be written in matrix form as

2

1

q

qB

where [B] is a row matrix called the strain-displacement matrix,

given by

2 1

1 11 1 1 1

e

Bx x l

since x2 – x1 = element length = le.

Thus the normal strain relation can be written as

1 2

2 1

1q q

x x


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3-6 Stress-Strain Relation

1

2

qE B

q

Normal stress is related to the normal strain by a Hooke’s

law, E

where E is modulus of elasticity.

Substitute for the normal strain ,

we get,

Robert Hooke (1635-1703);

(Experimental Philosopher)


Theory of Minimum Potential Energy

17


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3-7 Element Stiffness Matrix

We will use the potential energy approach to derive the

element stiffness matrix [k] for the 1-D element.

p U

For the non-uniform bar, its total potential energy is given by

1

2

T T T

p i iL L L

i

A dx u fA dx u T dx Q P

U = internal strain energy;

= potential energy of external forces.

Total potential energy of a body subjected to loads is given by,

Since the bar has been discretized into finite elements

1

2

T T T

p i ie e e

e e e i

A dx u fA dx u T dx Q P


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We will derive the element stiffness matrix of the 1-D element

using the internal strain energy term, U as follows,

1

2

T

ee

U A dx

E B q

Recall, the stress and strain are given by

and B q

Substitute these into the expression for Ue,

1

2

1

2

1

2

T

ee

T T

e

T T

ee

U E B q B q A dx

q B E B q A dx

U q B E B A dx q


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2

2

e

e

lddx d

dx l

Substitute and simplifying the expression yields,

1

1

2

1

2 2

12

2 2

11 1 11 1

12

11 11 1

12

1 11

1 12

T T ee e e

T T ee e

T

e e e

e e

T

e e e

e

T e ee

e

lU q B E B A d q

lq B E B A q

q A l E ql l

q A l E ql

A EU q q

l

Recall again,


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The internal strain energy for the 1-D element can now be

written in the form,

1

2

T e

eU q k q

where [k]e represents the element stiffness matrix for the 1-D

element, i.e.

1 1

1 1

e e e

e

E Ak

l

Note: Ee = elastic modulus;

Ae = cross-sectional area;

le = element length.


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3-8 Element Force Vector

a) Due to body force, fb

The potential energy due to body force fb in a single element is

given by the second term, i.e.

1 1 2 2

1 1 2 2

T

f be

T

be

T

f e be

u f A dx

N q N q f A dx

A f N q N q dx

The forces acting on 1-D structures can be of body force, fb,

traction force, T, and concentrated force, P. They may act

individually in various combination.

1

2

T T T

p b i ie e e

e e e i

A dx u f A dx u T dx Q P

The total potential energy of the structure,


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1

2

e bT e

f

e be

A f N dxq

A f N dx

1

11

1

21

1

2 2 2

1

2 2 2

e e

e

e e

e

l lN dx d

l lN dx d

Recall that,

2

eldx d

12

12

2

e b e

T T e e bf

e b e

A f l

A l fq q

A f l

Rewrite,

Substitute and simplifying, yields

Also, - Show details of

this integration.


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The potential energy due to the body force can now be

expressed in the form,

T e

f q f

where the force vector due to body force fb is,

1

12

e e e bA l ff

Quiz: Can you give the physical interpretation of {f}e?


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b) Due to traction force, T

The potential energy due to traction force T is given by,

1 1 2 2

TT

Te eu T dx N q N q T dx

Rearranging and simplifying,

1

2

2

2

e

T Te

T

ee

lT N dx

q qlT N dx

1

11

1

21

1

2 2 2

1

2 2 2

e e

e

e e

e

l lN dx d

l lN dx d

2

eldx d

Recall,


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The last equation is in the form,

1

12

T eT

lq T

i.e.

T e

T q T

Thus, element traction force vector due to traction T,

1

12

e eTlT

Quiz: Can you give the physical interpretation of this?


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Summary

1 1

1 1

e e e

e

A Ek

l

3. Element force vector

due to body force, fb

1

12

e e e bA l ff

4. Element force vector

due to traction force, T

1

12

e eTlT

1

2

1 1e

qE

ql

We have established, for 1-D problems,

1. Stress-strain relation

2. Element stiffness matrix


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Example 3-2

A thin steel plate has a uniform thickness t = 1 in., as shown. Its elastic modulus, E = 30 x 106 psi, and weight density, r = 0.2836 lb/in3.

The plate is subjected to a point load P = 100 lb at its midpoint and a traction force T = 36 lb/ft.

Determine:

a) Displacements at the mid-point and at the free end,

b) Normal stresses in the plate, and

c) Reaction force at the support.


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Solution

1. Transform the given plate into 2 sections, each having

uniform cross-sectional area.

Area at midpoint is

Amid = 4.5 in2.

Average area of section 1 is

A1 = (6 + 4.5)/2 = 5.25 in2.

Average area of section 2 is

A2 = (4.5 + 3)/2 = 3.75 in2.

2. Model each section using 1-D

(line) element.

Note:


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element 2: 6

(2) 1 13.75 30 10

1 112k

4. Assemble global stiffness matrix,

6

5.25 5.25 030 10

5.25 9.00 3.7512

0 3.75 3.75

K

3. Write the element stiffness matrix for each element

element 1: 6

(1) 1 15.25 30 10

1 112k

Note: The main diagonal must contain positive numbers only!


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5. Write the element force vector for each element

a) Due to body force, fb = 0.2836 lb/in3

element 1 (1) 15.25 12 0.2836

12bf

element 2 (2) 13.75 12 0.2836

12bf

Assemble global force vector due to body force,

5.25 8.912 0.2836

9.00 15.32

3.75 6.4

bF


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b) Due to traction force, T = 36 lb/ft

element 1 (1)

3612

1 11218

1 12T

element 2 (2)

3612

1 11218

1 12T

Assemble global force vector due to traction force,

1 18

18 2 36

1 18

TF


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c) Due to concentrated load, P = 100 lb at node 2

0

100

0

PF

6. Assemble all element force vectors to form the global force

vector for the entire structure.

8.9 18 0 26.9

15.3 36 100 151.3

6.4 18 0 24.4

F lb


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7. Write system of linear equations (SLEs) for entire model

K Q F

16

2

3

5.25 5.25 0 26.930 10

5.25 9.00 3.75 151.312

0 3.75 3.75 24.4

Q

Q

Q

The SLEs can be written in condensed matrix form as

Expanding all terms and substituting values, we get

Note:

1. The global force term includes the unknown reaction force R1 at

the support. But it is ignored for now.

2. The SLEs have no solutions since the determinant of [K] = 0;

Physically, the structure moves around as a rigid body.


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8. Impose boundary conditions (BCs) on the global SLEs

There are 2 types of BCs:

a) Homogeneous = specified zero displacement;

b) Non-homogeneous = specified non-zero displacement.

In this example, homogeneous BC exists at node 1. How to

impose this BC on the global SLEs?

DELETE ROW AND COLUMN #1 OF THE SLEs!

16

2

3

5.25 5.25 0 26.930 10

5.25 9.00 3.75 151.312

0 3.75 3.75 24.4

Q

Q

Q


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62

3

9.00 3.75 151.330 10

3.75 3.75 24.412

Q

Q

Solve using Gaussian elimination method, yields

52

53

1.339 10

1.599 10

Qin

Q

9. Solve the reduced SLEs for the unknown nodal

displacements

The reduced SLEs are,

Quiz: Does the answers make sense? Explain…


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10. Estimate stresses in each elements

1( )

2

11 1e

e

qE B q E

ql

element 1

1 6

5

0130 10 1 1 33.48 psi

1.339 1012

element 2

5

2 6

5

1.339 10130 10 1 1 6.5 psi

12 1.599 10

Recall,


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11. Compute the reaction force R1 at node 1

165

5

5.25 5.25 0 0 26.930 10

5.25 9.00 3.75 1.339 10 151.312

0 3.75 3.75 1.599 10 24.4

R

6

5

1

5

030 10

5.25 5.25 0 1.339 10 26.933412

1.599 10

R

1 202.68 lbR

We now include the reaction force term in the global SLEs.

From the 1st. equation we get,

We have,


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Example 3-3

A concentrated load P = 60 kN is applied at the midpoint of a uniform bar as shown.

Initially, a gap of 1.2 mm exists between the right end of the bar and the support there.

If the elastic modulus E = 20 x 103 N/mm2, determine the:

a) displacements field,

b) stresses in the bar, and

c) reaction force at the support.

P

150 mm 150 mm

1.2 mm

x

250 mm2


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Solution

1. Write the element stiffness matrices and assemble the

global stiffness matrix.

3

1 1 020 10 250

1 2 1150

0 1 1

K

2. Write the element force vectors and assemble the global force

vector.

30, 60 10 , 0T

F

3. Write the global system of linear equations.

133

2

3

500 500 0 010

500 1000 500 10 6015

0 500 500 0

Q

Q

Q


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4. Impose the boundary conditions.

We have; Q1 = 0; Q3 = 1.2 mm. Using Gaussian elimination

method:

a) Delete 1st row and column.

b) Delete 3rd row and column and modify the force term.

133

2

500 500 0 010

500 1000 500 10 6015

0 500 500 1.2 0

Q

Q

The reduced SLE becomes,

3

3

2

500 1.2101000 10 60

15 15Q

Modification to

force term


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7. Solve the reduced SLE, we get

8. Compute stresses in the bar,

2 1.5 mmQ

3

1

1

0120 10 1 1

1.5150

200 MPa

3

2

2

1.5120 10 1 1

1.2150

40 MPa

9. Compute reaction forces at supports

Using the 1st and 3rd equations, we obtain,

R1 = -50 x 103 N; R3 = -10 x 103 N.


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Exercise 2-1

A composite bar ABC is subjected to axial forces as shown. Given, the elastic moduli, E1 = 200 GPa and E2 = 70 GPa. Estimate:

a) Displacement of end C; [Answer: dC = 6.62x10-2 mm]

b) Stress in section 2, and

c) Reaction force at support A.

Verify your results with analytical solution.


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Exercise 2-2

Reconsider Exercise 2-1. Suppose a gap of d = 2 x 10-2 mm exists between end C and a fixed support there. Estimate:

a) Displacement of point B;

b) Stress in section 1, and

c) Reaction forces at both supports.

2 x 10-2 mm

10 kN


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Assignment 2-1

Find a journal paper on the application of finite element method to model and simulate real world problems, from various journals on the internet. (e.g. : www.sciencedirect.com).

Download the paper (in PDF format), and print it.

Read the paper and make one (1) page summary on the content of the paper - typewritten.

Submit the summary and copy of the paper to me. Use cover page.

Due in: 7 days time.

One-Dimensional Problemstaminmn/SME 3033 Finite Element Method/04-1D... · UNIAXIAL BAR ELEMENTS...

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