plane stress vs strain

7/30/2019 Plane Stress vs Strain

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Chapter 5

PLANE STRESS AND PLANE STRAIN

Many problems in theory of elasticity are two dimensional in nature and they can be modeled

as plane stress and plane strain. On the other hand the solution of three-dimensional elaticity

problems are generaly very difficult. If the structure enables some necessary conditions, it can

be analyzed using a two-dimensional mathematical model.

i. Plane Stress : A thin plate subjected to in-plane loading acting in its own plane, the

state of stress and deformation within the plate is called plane stress, Fig. 5.1. In this

case only two dimensions (in the plane of plate) are required for the analysis.

ii. Plane Stress: If a long body is subjected to transverse loading and its cross section and

loading do not vary significantly in the longitudinal direction, a small thickness in the

loaded area can be treated as subjected to plane strain, Fig. 5.2.

Now, strain-displacement and stress-strain relationships will be developed.

5.1 BASIC EQUATIONS

5.1.1 Strain-Displacement Relationships

The displacement vector has two components for two dimensional problems.

{ }

=v

u(5.1)

The strain-displacement relationships are given as follows. For a two dimensional problem,

there will be only three independent strain components (x, y, xy) and the strain-displacementrelations, Eqs. (2.1), reduced to

+

=

0xy

0y

0x

xy

y

x

v

u

xy

y0

0x

(5.2)


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Figure 5.1 A thin plate under inplane loading.

Figure 5.2 A long cylinder under internal pressure.

z

yz

x

y

x

y

z

x

y


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Here the second term at the right hand side is the initial strains vector, such as thermal strains.

The first two types, x and y, are normal strains in the x and y directions, and xy is shearingstrain. As before, they may be written in matrix form as:

{ } [ ]{ } { }= + 0 d (5.3)

where

[ ]

0

0

x

y

y x

=

d (5.4)

and in the case of thermal strains

{ }

=

=0

1

1

T

0xy

0y

0x

0 (5.5)

for plane stress problems and

{ } ( )0

0 0

0

1

1 1

0

x

y

xy

T

= = +

(5.6)

for the plane strain problems.

5.1.2 Stress-Strain Relationships

Assuming an isotropic material, we shall develop relationships between stresses and strains

for both plane stress and plane strain.

(i) Plane Stress

0yzzxz === 0zxyz ==

If the thermal strains are taken into the account, the 3-D strain-stress relations, given by Eq.

(2.8), reduced to


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( )

+

+

=

0xy

0y

0x

xy

y

x

xy

y

x

1200

01

01

E

1(5.7)

In the case of plane stress, the component of strain in the z direction will be nonzero and isgiven by

( ) TE

yxz ++=

(5.8)

Solving for stresses in terms of the strains, we find:

( )

=

0xy

0y

0x

xy

y

x

2

xy

y

x

2

100

01

01

1

E(5.9)

We can write the constitutive relations as the matrix expressions:

{ } [ ]{ } { }0C += { } [ ] { } { }( )0E = (5.10)

where

[ ]( )

+ =

1200

0101

E

1C and [ ]

( )

=

2

100

0101

1

EE

2(5.11)

For a material that is orthotropic in the x and y directions, the case of plane stress gives the

following strain-stress and stress-strain operators:

[ ]

=

xy

yx

xy

y

xy

x

G

100

0E

1

E

0EE

1

C [ ]( )

=

xyyxxy

yyyx

xxyx

G100

0EE

0EE

E (5.12)

Here the symbol xy denotes the strain in the x direction due to strain in the y direction. If weuse for an approximation shear modulus as

y

xy

x

yx

xy E

1

E

1

G

1 ++

+ (5.13)


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we need only three independent constants.

From the reciprocal theorem, we have

yx

xy

x

y

E

E

= (5.14)

(ii) Plane Strain

The case of plane strain is based on the assumptions that:

w = 0 and 0z

w=

at every cross section. Here the independent variables are assumed to be

functions of only the x and y coordinates provided we consider a cross section of the body

away from the ends.

0zxyz == 0yzzxz ===

If the thermal strains are taken into the account, the 3-D strain-stress relations, given by Eq.

(2.3), reduced to

+

+=

0xy

0y

0x

xy

y

x

xy

y

x

200

01

01

E

1(5.15)

Here, in the case of thermal strains

{ } ( )

+=

=0

1

1

T1

0xy

0y

0x

0 (5.16)

As before we can solve for the stresses in terms of the corresponding strains to obtain:

( )( )

+

=

0xy

0y

0x

xy

y

x

xy

y

x

2

2100

0101

211

E(5.17)

The component of stress in the z-direction will be nonzero and is given by

( )z x y = + (5.18)

The strain-stress operator [C] is found as


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[ ]

+

=200

01

01

E

1C (5.19)

In addition, the stress-strain operator [E] is seen from above equation to be:

[ ]( )( )

+=

2

2100

01

01

211

EE (5.20)

On the other hand, the case of plane strain in an orthotropic material has the following strain-

stress and stress-strain operators:

[ ]

0

0

10 0

x y

x y

xy

d b

E E

c a

E E

G

=

C [ ]( )

01

0

0 0

x x

y y

xy

aE bE

cE dE ad bc

ad bc G

=

E (5.21)

where

zxxzzxyzyx

zyxzxyzyyz

1dc

b1a

=+=

+==

From the reciprocal theorem we find

c

b

E

E

x

y =

5.2 STRESS TRANSFORMATION

Also of interest to us are inclined stresses. For this purpose Figure shows stresses in the

directions of axes x and y, which are inclined at the angle with axes x and y. The stressesin the inclined axes can be obtained as follows:


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Figure 5.3 Stress transformations between the inclined axis.

cos 2 sin 22 2

x y x yx xy

+ = + + (5.22)

cos 2 sin 22 2

x y x yy xy

+ = (5.23)

sin 2 cos 22

x yxy xy = + (5.24)

It is possible to find the directions of principal stresses by differentiating this equation with

respect to q and setting the results equal to zero.

0d

d x =

Hence, we obtain

x

x

yy

x

y

y

xxy

yx

xy

yx

xxyyx

y

xxy yx y

xxy

y

yx

x

xy

yx

y

y

yx

x

xy


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yx

xy

P

22tan

= (5.25)

Principle normal stresses and the accompanying shear stress can be obtained by substitution.

0

22

22

xy

min

2

xy

2

yxyx

2P

max

2

xy

2

yxyx

1P

=

=+

+=

=+

+

+=

(5.26)

Thus, the shearing stress is zero when the normal stresses have principal values (maximum

and minimum)

On the other hand, the maximum shearing stress max occurs at an angle that can be found bydifferentiating Eq. (5.24) with respect to and setting the results equal to zero, as follows:

0d

d yx =

Hence,

xy

yxS

22tan

= (5.27)

where S = P /4 is the value of for which the shearing stress is maximum. The followingexpression can be found for maximum shear stress:

2

xy

2

yx

max2

+

= (5.28)

The accompanying normal stresses as:

2

yx

yx

+== (5.29)

5.3 TRIANGULAR ELEMENTS

Finite elements with a triangular shape prove to be quite versatile for the purpose of

discretizing any two dimensional continuum. One of the earliest and best-known elements is

the constant strain triangle described by the Turner et al. In the present section stiffnesses and

equivalent nodal loads will be developed in detail for this triangle. In addition, a brief

description will be given for the linear strain triangle of Fraejis de Veubeke.


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x

y

node i

Q2i-1

Q2i

Figure 5.4 Plane problem and finite element model.

5.3.1 Constant Strain Triangles

Figure shows a constant strain triangle of thickness t, having the following generic

displacements (translations) in the x-y plane:

{ }

Tu v=

(5.30)

x

y


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Its three corners, points 1, 2, and 3, serve as nodes that are numbered in a counterclockwise

sequence. At each node there are two nodal translations in the directions of x and y, with the

former preceding the latter. We denote the element displacement vector as,

{ } T654321 qqqqqqq = (5.31)

A triangular element of area A appears in the Figure and any point P(x,y) on the triangle may

be located by dividing it into sub triangles having areas A1, A2, and A3. Dimensionless area

coordinates for the triangle are defined as

A

A

A

A

A

A 221 === (5.32)

By inspection, we see that:

AAAA 321 =++

Thus,

1=++ (5.33)

which shows that , , and are interdependent. Figure indicates that = 1 at point 1 and =0 along edge 2-3. Also indicated is a linear variation of from point 1 to the opposite edge,and similarly for and .

The displacements inside the element are now written using the shape functions and the nodalvalues of the unknown displacement field.

1 1 2 3 3 5

1 2 2 4 3 6

u N q N q N q

v N q N q N q

= + +

= + +(5.34)

Figure 5.5 Constant strain triangles.

(x3, y3)q5

q6

3

(x1, y1)q1

q2

1

(x2, y2)q3

q4

2

x

(x, y)e

u

v

te

Ae


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N3 = 1- = 1

1

N3

2

3

x

Figure 5.6 Natural coordinates.

Figure 5.7 Shape functions of CST.

3

= 01

2

x

(x, y)

= 1

= 0

= 1A1

A2

A3

N1 = = 1

1

N1

2

3

x

N2 = = 1

1

N2

2

3

x


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The independent shape functions are conveniently represented by the pair, , as follows

=== 1NNN 321 (5.35)

The relations can be expressed in a matrix format

{ } [ ]{ }= N q (5.36)

where [N] is the shape function matrix, given by

[ ] 1 2 31 2 3

0 0 0

0 0 0

N N N

N N N

=

N (5.37)

If we substitute Eq. (5.36) into the strain-displacement relations, we obtain

{ } [ ][ ] [ ][ ]{ } [ ]{ }

0

0

x

u

vy

y x

= = = =

N q B q (5.38)

The strain-displacement matrix [B] can be written as

{ } [ ]{ } 1 2 31 2 3

31 2

31 2

3 31 1 2 2

0

0 0 00

0 0 0

0 0 0

0 0 0

x

N N N

N N Ny

y x

NN N

x x x

NN N

y y y

N NN N N N

y x y x y x

= =

=

B N

(5.39)

A finite element is said to be isoparametric if the same interpolation formulas define both the

geometric and the displacement shape functions. Such elements satisfy both geometric and

displacement compatibility conditions. If the geometric interpolation functions are of lower

than the displacement shape functions, the element is called subparametric. If the reverse istrue, the element is referred to as superparametric.


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Because isoparametric elements are usually curved, they tend to be more suitable than

subparametric elements for modeling geometric boundary conditions. However, strain-

displacement relationships are complicated by the fact that generic displacements are

expressed in terms of local coordinates, whereas differentiations with respect to the global

coordinates are required. Also, it becomes necessary to employ numerical integration

whenever explicit integrations are impossible.

For the triangular element, the coordinates x, y can also be represented in terms of nodal

coordinates using the same shape functions, as follows

1 1 2 2 3 3

1 1 2 2 3 3

x N x N x N x

y N y N y N y

= + +

= + +(5.40)

In matrix form we may write

=

3

3

2

2

1

1

321

321

y

x

y

x

yx

N0N0N0

0N0N0N

y

x(5.41)

Using the notation, xij = xi xj and yij = yi yj we can write Eq. (5.40) as

13 23 3

13 23 3

x x x x

y y y y

= + += + +

(5.42)

This equation relates x and y coordinates to the and coordinates.

Using the chain rule for partial derivatives of shape functions, we have

i i i

i i i

N N Nx y

x y

N N N y

x y

= +

= +

i = 1,2,3 (5.43)

which can be written in matrix notation as

i i

ii

N x y N

x

NN x y

y

=

i = 1,2,3 (5.44)

where the (2 x 2) square matrix is denoted as the Jacobian of the transformation, [ J]:


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[ ] 13 1323 23

x y

y

x yx y

= =

J (5.45)

Also, from Eq. (5.44),

11 12

21 22

ii

i i

NN

x

N N

y

=

i = 1,2,3 (5.46)

where [ ] is the inverse of the Jacobian [J], given by

[ ] [ ][ ]

1 23 13

23 13

1

det

y y

x x

= = J

J(5.47)

det[J] = x13y23 x23y13

From the knowledge of the area of the triangle, it can be seen that the magnitude of det [J] is

twice the area of the triangle. If the points 1, 2, and 3 are ordered in a counterclockwise

manner, det [J] is positive sign. We have

[ ]1 det2A = J (5.48)

where represents the magnitude. Most computer codes use a counterclockwise order for

the nodes and use det [J] for evaluating the area.

Let us write Eq. (5.46) in following expanded form,

[ ] [ ]

31 231 2

11 12

31 2 21 22 31 2

G L

NN NNN N

x x x

NN N NN N

y y y

= D D

144424443 144424443

(5.49)

where [DG] and [DL] are the matrices of global and local derivatives of shape functions,

respectively. Substitution of Eqs.(5.35) and (5.47) into Eq. (5.49) yields

[ ] 23 13 23 13 13 2323 13 23 13 23 13

1 0 11 1

0 1 1G

y y y y y y

x x x x x

= =

DJ J

Considering xij = xji and xi,j xkj = xik, we obtain the global derivatives as follows


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[ ] 23 31 1232 13 21

1G

y y y

x x

==

D

J

The matrix [B] given by Eq. (5.39) can be written in terms of the global derivatives.

[ ]11 12 13

21 22 23

21 11 22 12 23 13

0 0 0

0 0 0

G G G

G G G

G G G G G G

D D D

D D D

D D D D D D

=

B (5.50)

After the multiplication we obtain element strain-displacement matrix,

[ ]23 31 12

32 13 2132 23 13 31 21 12

0 0 01

0 0 0

y y y

x x

y x y x y

=

BJ

(5.51)

Here [B] is a (3 x 6) matrix relating the three strains to the six nodal displacements. It may benoted that all the elements of the [B] matrix are constants expressed in terms of the nodal

coordinates. The element stiffness matrix given by

[ ] [ ] [ ][ ] [ ] [ ][ ]e e

T T

V V

dV dV = = k B E B B E B

Considering eeV

AtdVe

= we obtain the stiffness matrix as

[ ] [ ] [ ][ ]Te et A=k B E B (5.52)

Equivalent Nodal Forces

Now, the equivalent nodal loads due to the body forces and surface tractions will becalculated using the finite element formulations.

The vector of body forces is

{ } [ ] { }dVbNf TV

be

=

where

{ }

=y

x

b

bb and dV = tedA


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Figure 5.8 Body forces and equivalent nodal forces on a CST.

{ } dA

bN

bN

bN

bN

bN

bN

tdAb

b

N0

0N

N0

0N

N0

0N

tfee A

y3

x3

y2

x2

y1

x1

e

y

x

T

A

3

3

2

2

1

1

eb

=

= (5.53)

The force components, bx and by, can be expressed in terms of and using the isoparametricformulation.

Integrals of polynomial terms in the area coordinates and can be obtained as follows [2]:

( ) ec

A

ba A2!2cba

!c!b!adA

e+++

=

If c = 0, it becomes

( ) eA

ba A2!2ba

!b!adA

e++

=

If the body force is constant, we need the integrations of dAN

eA

i . For N1 = , we calculate the

integral as follows:

( ) eeA

1 A3

1A2

!201

!0!1dA

e

=++

=

Similarly, eA

3

A

2 A3

1dANdAN

ee

== . Hence, the body force vector is given as,

1

2

3

fb1

fb2

fb3

fb4

fb5

fb6

b(x,y)

x


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{ } Tyxyxyxeeb bbbbbb3

tAf = (5.54)

A traction force is distributed load acting on the surface of the body. We can integrate the

traction force to obtain a distributed load on the midline of the surface.

et = (5.55)

Consider an edge 2 3 acted on by traction ,x y , shown in Fig. 5.7. We have

{ } [ ] { }2 3

T

s

l

f N dl

= (5.56)

where

{ } xy

= (5.57)

Substituting Eqs. (5.37) and (5.57) into Eq. (5.56) we obtain

Figure 5.9 Surface tractions and equivalent nodal forces on a CST.

1

2

3

fs1

fs2

fs3

fs4

fs5

fs6

x,

x

y

x

y ( ),x y

x

y


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{ } 2 3 2 3

11

11

22

22

3 3

3 3

0

0

0

0

0

0

Tx

y

x x

s y yl l

x

y

NN

NN

NNf dl dl

NN

N N

N N

= =

(5.58)

If the traction force is constant, we need the integrations of ll

j-i

dN i . We note here that N1 = 0

along the edge 2 - 3 and 0l

3-l

=2

dN1

On the other hand, for N2 = and li-j = l2-3 we calculate the integral as follows:

( ) 3-23-2l

lll

3-2

2

1

!1010

!0!1!0d010 =

+++=

Similarly,( ) 3-23-2

ll

llll

3-23-2

2

1

!1100

!1!0!0ddN3 =+++

= . Hence, the vector of surface forces is given

as,

{ } 2 3 0 02T

x y x yl = sf (5.59)

We may designate the element temperature load as

{ } [ ] [ ]{ }e

T

V

dV= T 0f B E

where {0} is initial strain vector due to the temperature change, T. From the theory ofelasticity, {0} can be represented by

{ }0

0

0

1

1

0

x

y

xy

T

= =

0 (5.60)

for plane stress and

{ } ( )

0

0

0

1

1 10

x

y

xy

T

= = + 0

(5.61)


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for plane strain.

If the temperature change is constant on the element the integration may be performed as

{ } [ ] [ ]{ } [ ] [ ]{ }e

T Te e

V

dV A t = =T 0 0f B E B E (5.62)

The stresses in an element are then obtained by using following equation:

{ } [ ] { } { }( ) [ ] [ ]{ } { }( )= = 0 0 E E B q (5.63)

Finally, we may use the constant strain triangles in order to model a thin plate subjected to

inplane loads, given in Fig. 5.4. We need to arrange two tables for nodal coordinates andelement connectivity.

Table 5.1 Element connectivity for the FE model given in Fig. 5.4.

Elements Nodes

1 11 12 13

2 14 11 13

3 13 4 14

4 12 7 13

5 11 1 12

6 2 11 14

7 5 2 14

8 6 5 14

9 4 6 14

10 8 4 13

11 7 8 13

12 9 7 12

13 10 9 12

14 1 10 12

15 3 1 11

16 2 3 11

Table 5.2 Nodal coordinates for the FE model given in Fig. 5.4.

NODE X Y Z

1 0.15 -0.4 0

2 -0.3 -0.2 0

3 -9.77E-02 -0.351 0

4 5.00E-02 0.4 0

5 -0.39993 9.50E-02 0

6 -0.25379 0.36834 0

7 0.4 0.1 0

8 0.25992 0.2914 0

9 0.44994 -0.11375 0

10 0.34652 -0.3001 0

11 -1.91E-04 -0.15168 0

12 0.2518 -0.13182 0

13 0.14752 0.10795 014 -0.12607 0.10327 0


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500 N/mm

40 kN

200 mm

300 mm

The nodal coordinates are stored in a two-dimensional array represented by the total number

of nodes and the two coordinates per node. The connectivity may be clearly seen by isolatinga typical element, as shown in Fig. 5.5. For the three nodes designated locally as 1,2, and 3,

the corresponding global node numbers are defined in Fig. 5.4. This element connectivityinformation becomes an array of the size of number of elements and three nodes per element.

A typical connectivity representation is shown in Table 5.2. Most standard finite elementcodes use the convention of going around the element in a counterclockwise direction toavoid calculating a negative area.

Example 5.1 For the two-dimensional loaded plate shown in Figure E5.1, determine the

displacements of points A and B and the element stresses using plane stress conditions. Body

force may be neglected in comparison to the external forces. The thickness of plate is 5 mm,the Young modulus is 70 GPa and, the Poisson ratio is 0.3.

A

Figure E5.1-a

Solution:

Finite element model is depicted in Fig. E5.1-b.

Figure E5.1-b

Q6

1

Q7

3

2

1Q8

Q1

Q2

Q3

Q4

Q52

4

y

x

B

A


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3

1

1

2

4

2

3

1

2

3

For the plane stress conditions, the material property matrix is given by

[ ] ( )

=

=69.200069.731.2

031.269.7

10

2

100

01

01

1

E

E4

2

Figure E5.1-c

In Fig. E5.1-c, the bold numbers are the global ones and the others are the local numbers.

Then, we establish the connectivity as follows

Element Number Nodes

1 2 3

1 1 2 4

2 3 4 2

The table of nodal coordinates for the coordinate system depicted in Fig. E5.1-b, is given in

the following table

Node

Number

x

(mm)

y

(mm)

1 300 0

2 300 200

3 0 200

4 0 0

We found first Jacobian matrices for the elements:

[ ]

=

=

200300

0300

yx

yxJ

2323

13131 and [ ] 241 mm10x6200300

0300Jdet ==

[ ]

=

=

200300

0300

yx

yxJ

2323

13132 and [ ] 242 mm10x6200300

0300Jdet =

=

The strain-nodal displacement matrices [Be] can be determined as

Local numbers

Global numbers

2

1


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[ ]

=

200323

003030

020002

10x6

1B

2

1

[ ]

=

200323

003030

020002

10x6

1B

2

2

On performing the matrix multiplication [E][B]e , we get

[ ] 1 427.69 2.31 0 2 0 0 0 2 0

110 2.31 7.69 0 0 3 0 3 0 0

6 100 0 2.69 3 2 3 0 0 2

256 115.5 0 115.5 256 0

77 384.5 0 384.5 77 0

134.5 89.7 134.5 0 0 89.7

E Bx

=

=

[ ][ ]

=

7.89005.1347.895.134

0775.38405.38477

02565.11505.115256

BE2

The stiffness matrices can be calculated as

[ ]1 1 11 1 2

3

2 0 3

0 3 2256 115.5 0 115.5 256 0

0 0 330000 577 384.5 0 384.5 77 0

0 3 06 10134.5 89.7 134.5 0 0 89.7

2 0 0

0 0 2

229 125 101 57.8 128.2 67.2

125 333 67.2 288.4 57.8 4

10

T xk A t B E B

x

= =

=

4.8

101 67.2 101 0 0 67.2

57.8 288.4 0 288.4 57.8 0

128.2 57.8 0 57.8 128.2 0

67.2 44.8 67.2 0 0 44.8


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89

[ ] [ ] [ ][ ]

==

8.44002.678.442.67

02.1288.5708.572.128

08.574.28804.2888.57

2.67001012.67101

8.448.574.2882.67333125

2.672.1288.57101125229

10BEBtAk32T2

22

2

In the above matrices, the global dof association is shown on top. In the problem under

consideration, Q2, Q5, Q6, Q7, and Q8 are all zero. Using the elimination approach, it is now

sufficient to consider the stiffness associated with the degrees of freedom Q1, Q3, Q4. Theforce vector due to the distributed load on the 3-1 edge of element 2 can be calculated as:

{ } 3 1300

0 0 0 500 0 0 0 5002 2

T Tx y x y

l = = 2sf

The forces are neglected. A concentrated load of 40 kN is applied in the direction of Q1.

The set of equations is given by the matrix representation

=

75

0

40

10

Q

Q

Q

33308.57

0229101

8.57101229

10 3

4

3

1

3

Solving for Q1, Q

3and Q

4, we get

Q1 = 0.30326 mm, Q3 = 0.13361 mm, Q4 = -0.27749 mm

For element 1, the element nodal displacement vector is given by

{ } T1 0027749.013361.00030326.0q =

The element stresses 1 are calculated from [E][B] 1{q}1 as

{ } MPa839.22398.83742.45

T1 =

Similarly

{ } T2 27749.013361.00000q =

The element stresses 2 are calculated from [E][B] 2{q}2 as

{ } MPa903.24277.10258.34 T2 =


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90

5.3.2 Linear Strain Triangles

The linear strain triangles in the Figure 5.10 have a constant thickness t and the followinggeneric displacements.

{ } Tvu=

In addition to the three corner nodes (numbered 1, 2, and 3), there are also three mid-edgenodes (numbered 4, 5, and 6). The linear strain triangle is shown in Fig. 5.12(a) with its

natural coordinates, and its isoparametric counterpart element appears in Fig. 5.12(b). We willapproach this isoparametric element by first examining its straight-sided parent. Nodal

displacements for either element are:

Figure 5.10 Element T6. (a) Triangular parent (LST) (b) Isoparametric counterpart.

= 03

vi

i

1

q22

x

y

(x, y)

uv

ui

q14

5

6

= = 1

= 0

= 1

= 0 = 1

=

=

= 03

vi

i

1

q22

x

y

(x, y)

u

v

ui

q14

5

6

=

= 1

= 0

= 1

= 0 = 1

= =

(a)

(b)


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91

{ } T611T

1221 v,...,vuq,...,qq == (5.64)

Geometric interpolation functions for the parent element are linear, as is the case for any

triangle with straight sides. Quadratic displacement shape functions may be written in naturalcoordinates as:

==

==6

1i

ii

6

1i

ii vNvuNu (5.65)

where

4N)12(N

4N)12(N

4N)12(N

63

52

41

======

(5.66)

The strain-displacement matrix can be written as follows

{ } [ ]{ } 1 2 61 2 6

61 2

61 2

6 61 1 2 2

0

0 0 00

0 0 0

0 0 0

0 0 0

x

N N N

N N Ny

y x

NN N

x x x

NN N

y y y

N NN N N N

y x y x y x

= =

=

...B d N

...

...

...

...

(5.67)

Eq. (5.67) can be written in the format

{ } [ ] [ ] [ ]{ }{ }

{ }

1

21 2 6

6

=

q

qB B ... B

q

(5.68)

and hence it can be concisely expressed as:

{ } [ ] { }iii qB= (i = 1,2,3,4) (5.69)

where


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92

[ ] [ ][ ]

=

==

x

N

y

N

y

N0

0x

N

N0

0N

xy

y0

0x

NB

ii

i

i

i

i

ii (5.70)

(i) Subparametric Triangles: The derivatives in the strain-displacement relationships are

found by using the linear interpolations for the geometry given by Eqs. (5.40) through (5.42).The global derivatives can be obtained the same relation given with Eq. (5.46) on condition

with the upper index 3 changed to 6.

11 12

21 22

ii

i i

NN

x

N Ny

=

i = 1,2,,6

where the shape functions are the ones given in Eq. (5.66).

(ii) Isoparametric Triangles: For the isoparametric triangle T6 in Fig 3.12(b), we take thegeometric interpolation functions to be:

==

==6

1i

ii

6

1i

ii yNyxNx (5.71)

where Ni are given by Eqs. (5.5-7). Thus, the edges of the element become quadratic curves,as indicated in the figure. Because the natural coordinates are curvilinear, the Jacobian matrix

is required. Thus,

[ ]

6 6

1 1

6 6

1 1

i ii i

i i

i ii i

i i

N Nx yx y

x y N Nx y

= =

= =

= =

J (5.72)

For the T6 element, the Jacobian matrix is

[ ]

[ ]

[ ]

1 1

2 23 5 61 2 4

3 3

4 43 5 61 2 4

5 5

6 6L

N

x y

x yN N NN N N

y

x yN N NN N N

x y

x y

=

D

C

J

14444444244444443

14243


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93

Here, [DL] matrix can be evaluated using the shape functions given by Eq. (5.66).

[ ] ( )( )4 1 0 4 4 3 4 4 4 1 3

0 4 1 4 4 3 4 4 1 3 4L

+ = +

D

The inverse of the Jacobian matrix, [], has already given by Eq. (5.47), and is repeated here.

[ ] [ ][ ]

==

1121

12221

JJ

JJ

Jdet

1J

The matrix of global derivatives is given as follows

[ ] [ ][ ]G L=D D

The terms of this matrix can be determined as

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

[ ] [ ]

( ) ( )

( )

11 22 21 21

12 12 22 11

13 22 12 23 11 21

14 22 12 24 21 11

15 22 12 25 21 11

16 22

1 14 1 4 1

1 14 1 4 1

1 14 4 3 4 4 3

4 4

4 41 3 1 3

41 3

G G

G G

G G

G G

G G

G

D J D J

D J D J

D J J D J J

D J J D J J

D J J D J J

D J

= =

= =

= + = +

= = +

= = +

=

J J

J J

J J

J J

J J

J( )12 26 21 11

41 3GJ D J J + = J

(5.73)

Referring to Eqs. (5.73), we see that the submatrix [B]i may also be written as follows:

[ ]1

2

2 1

0

0

0 0

i

G ii

G ii

G i G ii i

N

x DN

Dy

D DN N

y x

= =

B i = 1,2,3,4 (5.74)

The matrix [B] is a (3 x 12) element strain-displacement matrix relating the three strains to the

twelve nodal displacements.

Then the stiffness matrix for the element is obtained from:


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94

[ ] [ ] [ ][ ]e

T

e

A

t dA= k B E B

Hence, the element stiffness matrix given by

[ ] [ ] [ ] [ ] [ ] [ ] [ ]12 3 3 3 3 12 12 3 3 3 3 12

ee

T T

ex x x x x x

AV

dV t dA= = k B E B B E B (5.75)

Integration in the Natural Coordinates:

In order to obtain any term of the stiffness matrix, the type of integration to be performed as:

( )dxdy,fI = (5.76)

However, the integral in Eq. (5.76) is more easily evaluated if it is first transformed to the

natural coordinates and . In addition the limits of each integration must be changed tobecome 1 to 1; and the infinitesimal area dA = dxdy must be replaced by an appropriate

expression in terms of d and d. For this purpose Fig. 5.11 shows an infinitesimal area dA inthe natural coordinates. Vector r locates a generic point in the Cartesian coordinates x and y as

follows:

jiyxr yx +=+= (5.77)

The rate of change of r with respect to is:

Figure 5.11 Infinitesimal area in naturel coordinates.

y

y

r

dA

dr

r

+

dy

dx

dy

dx

d

r

dr

i

k

dr

r

+


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95

jir

+

= yx

(5.78)

Also, the rate of change of r with respect to is:

jir

+

= yx

(5.79)

When multiplied by d and d, the derivatives in Eqs. (5.78) and (5.79) form two adjacentsides of the infinitesimal parallelogram of area dA in the figure. This area may be determined

from the following vector triple product:

krr

=

dxddA (5.80)

Substitution of Eqs. (5.78) and (5.79) into Eq. (5.80) gives:

ddyxyx

dA

= (5.81)

The expression in the parentheses of Eq. (5.81) may be written as a determinant. That is,

x y

dA d d d d x y

= =

J (5.82)

in which [J] is the Jacobian matrix and det[J] is its determinant. Thus, the new form of the

integral in Eq. (5.77) becomes:

( ) ddJ,fI1

1

1

1

= (5.83)

Gaussian quadrature procedure will be applied for the numerical integration. For triangles in

natural coordinates the numerical integration formula is [4]:

( )=

=n

1j

jj,fWAI (5.84)

in which Wj is the weighting factor for the jth sampling point. Integration points for n = 1, 3,

and 4 appear in Fig 5.12, and their locations and weighting factors are given in Table 5.3.


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96

Figure 5.12 Integration points for n = 1,3, and 4.

Table 5.3 The locations of integration points and weighting factors.

N Order Points Wj1 Linear a

31

31 1

3 Quadratic

a2

1 2

1 3

1

b 0 21

31

c21 0 3

1

4 Cubic

a31

31

4827

b 0.6 0.2 4825

c 0.2 0.6 4825

d 0.2 0.2 4825

5.4 QUADRILATERAL ELEMENTS

5.4.1 Bilinear Displacement Rectangle

In two-dimensional problems the isoparametric quadrilaterals serve a useful purpose. This

element is the isoparametric parent of the bilinear displacement rectangle, developed by

Melosh [6]. The elements depicted in Fig. 5.13. The generic displacements indicated in the

figure are:

{ } Tv,u=

An x and a y translation are shown at each node. Thus, the nodal displacement vector is:

{ } T411T

821 v,...,v,u,q,...,q,qq == (5.85)

Now let us assume the following displacement functions

=

=

=+++=

=+++=

6

1iii44332211

6

1i

ii44332211

vNuNuNuNuNv

uNuNuNuNuNu

(5.86)

1

2

3

1

2

3

1

2

3

a

a

a

b

b

cc

d

Linear Quadratic Cubic


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97

in which N1, N2, N3, and N4 are the interpolation functions and they are given as follows:

( )( ) ( )( )( )( ) ( )( )

+=++=

+==

11N11N

11N11N

41

441

3

41

241

1(5.87)

We may write in the matrix form:

{ }

=

=

8

2

1

4321

4321

q

q

q

N0N0N0N0

0N0N0N0N

v

u

M (5.88)

Figure 5.13 Bilinear displacement rectangle and its isoparametric quadrilateral.

x

y

=-1

q7

q2

q1

q4

q3

q5

q6

v u

q8

= - =

= -

=

=0

= 0 = -1

= 1

= 1

1

2

3

4

q1q2

q4q3

q5

q6q

8

q7

= y/b

= y/bu

v

2a

2b


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98

Eq. (5.88) can be arranged as follows

{ } [ ]{ } [ ] [ ] [ ] [ ]

{ }{ }{ }{ }

1

2

1 2 3 43

4

= =

q

qN q N N N N

q

q

(5.89)

or it may be written in a concise form as

{ } [ ] { }iii qN= (i = 1,2,3,4) (5.90)

For example, for node 1 we may write following equations:

{ } [ ] { }111 qN= (5.91)

1 1 11

1 2 21

0 1 0

0 0 1

N q quN

N q qv

= =

(5.92)

The generic displacements {}i in Eq. (5.91) represent the translations at any point due to thedisplacements {q}i at node i. As a further efficiency of notation, we can write the functions N i

as

( )( )00i 114

1N ++= (5.93)

where

0 = 0 = (i = 1,2,3,4) (5.94)

[B] matrix can be expressed as:

{ } [ ] [ ] [ ] [ ]{ }{ }{ }{ }

1

2

1 2 3 43

4

=

qq

B B B Bq

q

(5.95)

This relation can be concisely expressed as:

{ } [ ] { }i i i

= B q (i = 1,2,3,4) (5.96)

where


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99

[ ] [ ][ ]

00

00 0

0

i

i i

i ii

i i

N

xx

N N

Ny y

N N

y x y x

= = =

B N (5.97)

In the isoparametric formulation, we use the same shape functions N i to also express the

coordinates of a point within the element in terms of nodal coordinates. Thus,

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

x N x N x N x N x

y N y N y N y N y

= + + += + + +

(5.98)

In matrix form we may write

=

4

4

3

3

2

2

1

1

4321

4321

y

x

y

x

y

x

y

x

N0N0N0N0

0N0N0N0N

y

x(5.99)

The Jacobian matrix, [J], is given as follows:,

[ ]

=

=

==

==4

1i

ii

4

1i

ii

4

1i

ii

4

1i

ii

yN

xN

yN

xN

yx

yx

J

(5.100)

From Eqs. 6.2.1 and 6.2.2, for the special case of four-node plane element

[ ] [ ]{ }L N=J D C (5.101)

The matrix [DL] given by this expression consists of derivatives of N i with respect to local

coordinates. That is,


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100

[ ] ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

31 2 4

31 2 4

1 1 1 11

1 1 1 14L

NN N N

NN N N

+ +

= = + +

D (5.102)

[CN] is the matrix of nodal coordinates as follows

[ ]

1 1

2 2

3 3

4 4

N

x y

y

x y

y

=

C

The matrix [DG] given by this expression consists of derivatives of Ni with respect to globalcoordinates. It can be written in terms of local derivatives of shape functions using the inverse

of the Jacobian matrix, [ ].

[ ] [ ][ ]

31 2 4

31 2 4

G L

NN N N

x x xx

NN N N

y y y y

= =

D D (5.103)

Evaulating terms in [DG] yields:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

11 22 12 21 21 11

12 22 12 22 21 11

13 22 12 23 21 11

14 22 12 24 21 11

1 11 1 1 1

4 4

1 11 1 1 1

4 4

1 11 1 1 1

4 4

1 11 1 1 14 4

G G

G G

G G

G G

D J J D J J

D J J D J J

D J J D J J

D J J D J J

= + =

= + + = +

= + + = + + +

= + = + +

J J

J J

J J

J J

(5.104)

By this approach we can find all the terms in [DG] numerically.

Because of the appearance of the determinant of [J] in denominator positions, we usually

cannot integrate explicitly to obtain stiffnesses and equivalent nodal loads. Instead, it becomes

necessary to use numerical integration.

Then, we can obtain the submatrix [B]i using Eqs. (5.104) as follows


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101

[ ]1

2

2 1

0

0

G i

G ii

G i G i

D

D

D D

=

B i = 1,2,3,4 (5.105)

After determining the [B] matrix, we express the stiffness matrix for Q4 element (withconstant thickness t) in Cartesian coordinates. Thus,

[ ] ( ) [ ] ( ) ( )1 1

3 38 3 3 81 1

, , ,T

e xx xt d d

= k B E B J (5.106)

Similarly, equivalent nodal loads due to body forces may be written in natural coordinates as:

[ ] ( ) ( ) ( )1 1

8 2 2 11 1

, , ,T

b e x xt d d

= f N b J (5.107)

In addition, equivalent nodal loads due to initial strains in natural coordinates are:

[ ] ( ) [ ] ( ) ( )1 1

0 3 38 3 3 11 1

, , ,T

e xx xt d d

= 0f B E J (5.108)

Except in special cases, the integrals in Eqs. (5.107) and (5.108) must be performed by

numerical integration. However, if the element is rectangular, direct explicit integration may

be used. Also, line loadings with or constant may be handled by explicit line integrations.Of course, if the body forces consist of point loads, no integration is required at all.

Furthermore, note that the determinant of the Jacobian matrix appears in the denominators of

all the terms in matrix [B]. Therefore, the determinant J in Eq. (5.108) cancels throughout;so the equivalent loads for initial strains may be integrated either explicitly or numerically. If

a temperature change T varies bilinearly, it is defined as:

=

=4

1i

ii TNT

Numerical Integration

For quadrilaterals in natural coordinates, the type of integration to be performed as

( ) ddJ,fI1

1

1

1

=

Two successive applications of Gaussian quadrature produce:

( ) ( )kjkjn

1k

n

1j

kj ,J,fRRI = = =

(5.109)

where Rj and Rkare weighting factors for the point (j, k). Integration points for n = 1,2,3each way on a quadrilateral are shown in Fig. 5.5.


36/53

102

x

y

(7,1)

1

2

3

4

(20,5)

(14,14)

(5,10)

G1

G2

G3

G4

Figure 5.14 Integration points for n = 1,3, and 4.

Table 5.4 The locations of integration points and weighting factors for quadrilaterals.

n k Rj Rk1 0 0 2 2

2 3

1 3

1 1 1

3

0.7745966692 0.7745966692 0.5555555556 0.5555555556

0.0 0.0 0.8888888889 0.8888888889

Example:

Derive numerically the stiffness term k12

for the isoparametric Q4 element in the

figure, using Gaussian integration with n = 2

each way. Assume that thickness t is

constant .The coordinates of nodes are given

in the figure.

Solution:

The formula for numerical integration of

terms in [k] is:

[ ] ( ) [ ] ( ) ( )3 38 3 3 81 1

, , ,n n T

e j k j k j k j k xx xk j

t R R = =

= k B E B J

Matrices [B] and [J] in this expression are functions of the coordinates j and kfor theintegration points (or Gauss points) G1, G2, G3, and G4 shown in the figure. In particular, for

k12 with n = 2, Rj = Rk= 1 so that

[ ] [ ] [ ]12 1,1 3 3 1,2

1 1

n nT

e xk j

k t = == B E B J

1

2

3

4

1

2

3

4

1

2

3

4

n = 1 n = 2 n = 3


37/53

103

In this formula the symbol [B]1,1 denotes the first column of submatrix [B]1 and [B]1,2 is the

second column. Substituting these columns in above equation, we obtain:

( )12 12 33 11 211 1

n n

e G G

k j

k t E E D D= =

= + J

in which the terms E12 and E33 are left in unspecified form.

To evaluate this equation, we first calculate Jacobian matrix as:

[ ] [ ][ ] ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

7 1

1 1 1 1 20 5 11 2 41 1

1 1 1 1 14 14 4 2 94 2

5 10

L N

+ + = = = + +

J D C

Then the determinant of J is:

( )1 115 18 84

= +J

Terms required from DG are:

( ) ( )( )

( ) ( )( )

( )

11 22 12

21 21 11

1 5 9 41 1

4 2 115 18 8

3 5 2 31

1 14 2 115 18 8

G

G

D J J

D J J

+ = + = +

+ +

= = +

J

J

Evaluating the product DG11DG21| J | at each of the four integration points and summing the

results can be obtained as:

( )3312e12 EEt1578.0k +=

which can be finalized using numerical values of t, E12, and E33.

5.4.2 Isoparametric Quadrilateral

Figure 5.15(a) shows the rectangular parent of the isoparametric quadrilateral (Q8) element,

which appears in Fig. 5.15(b). In order to understand this higher-order isoparametric element,

it is helpful to study first its rectangular parent. Nodal displacements for either element consist

of x and y translations at each node. Thus,

{ } TT vvuqqqq 8111621 ,...,,,...,, == (5.110)

Now let us assume the following displacement functions

==

==8

1

8

1 i

ii

i

ii vNvuNu (5.111)


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104

Figure 5.15 (a) Parent rectangular element (b) Isoparmetric quadrilateral (Q8) element.

in which N1, N2, , N8 are the interpolation functions and they are given as follows:

( )( )( )

( )( )

( )( ) )8,6(112

1

)7,5(112

1

)4,3,2,1(1114

1

2

0

0

2

0000

=+=

=+=

=+++=

iN

iN

iN

i

i

i

(5.112)

where

ii == 00

q1q2

q4q3

q5q6q8

q7

= y/b

= y/bu

v

2a

2b

q9 q10

q12

q11

q14q13

q16

q15

x

q7

q2

q1

q4

q3

q5

q6

v u

q8

1

2

3

4

5

6

7

8

q10

q9

q11

q12q13

q14

q16

q15

(b)

(a)


39/53

105

This rectangle is called a Serendipity element [1] because nodes appear only on its edges.

If we take the geometric interpolation functions to be the same as the displacement shape

functions in Eqs. (5.112), we formulate the isoparametric (Q8) element. Physically, this

means that the natural coordinates and are curvilinear, and all sides of the element becomequadratic curves [6]. Thus,

==

==8

1

8

1 i

ii

i

ii yNyxNx

The formulation of stiffnesses and equivalent nodal loads for element Q8 is very similar to

that for Q4 element given earlier. Numerical integration also follows the same pattern as

before, even though the local coordinates are curved.

When higher-order elements are available, it is not always clear whether to use them or torefine network of lower-order elements to achieve the same accuracy. Figure 5.16 depicts a

two-dimensional continuum discretized into Q4 elements and a smaller number of Q8

elements. In this case it seems clear that using Q8 elements instead of tour times as many Q4

elements has advantages. First, we see that the number of nodes is smaller when Q8 elements

are used. Second, this type of element is able to model the curved boundary with superior

accuracy. Third, we can expect greater numerical accuracy in the results because of quadratic

displacement functions in place of linear functions. Of course, the validity of these predictions

can only be verified by numerical experimentation with practical applications.

Figure 5.16 Discretization with Q4 and Q8 elements.


40/53

106

PROBLEMS

Problem 1. The element shown in the figure is taken from a finite element model of a thin

aluminum plate with 5 mm thickness. The calculated nodal displacements are given below.

Determine (a) the natural coordinates (,) of point P, (b) the nodal displacement at the pointP, and (c) element stresses.

u1 = 0.05 mm

v1 = 0.02 mm

u2 = 0.04 mm

v2 = 0.0 mm

u3 = 0.03 mm

v3 = 0.0 mm

Problem 2. A cantilevered beam is made of an aluminum alloy and is subjected a

concentrated load of 10 kN on the free end. The dimensions of the beam are shown in the

Figure.

(a) Model the portion of the beam, shown in the Figure, with two CST elements.

(b) Apply the loads due to the bending and shear. Use simple calculations of beam theory

to obtain the loads.

(c) Apply the appropriate boundary conditions.

(d) Obtain the nodal displacements and element stresses.

(e)

Model the entire beam using FE software. Obtain converged solutions and comparethe results with yours. (E = 70 GPa, = 0.3)

Problem 3. The aluminum disc shown in the figure has a radius of 50 mm and a

thickness of 5 mm. A linearly varying pressure load is applied to right and left hand sides

of the disc and its maximum value is 800 MPa.

a. model the disc with two constant strain triangles (use the symmetry of geometry and

loads)

b. determine the nodal displacements and draw the deformed and undeformed shapesc. calculate the element stresses.

u1

v1

u2

v2

u3

v3

(5,6)

(10,5)

(9,8)

1

2

3

P(8,6)

Coordinates are in mm.

Thickness = 10100

1200 mm100

200

10 kN


41/53

107

2

34 =

=

uv

2a

2b

TT

2

34 =

=

uv

2a

2b

T

d. model the disc using FE software. Obtain converged solutions and compare the results

with yours. (E = 70 GPa, = 0.3)

Problem 4. Suppose that a bilinear displacement rectangle is subjected to a temperature

change that varies linearly from edge 1-4 to edge 2-3, as indicated in the figure. Develop

equivalent nodal loads for node 2 due to such temperature variation, assuming isotropic

material in plane stress.

Problem 5. For the bilinear displacement rectangle is subjected to a temperature change that

varies linearly from edge 1-4 to edge 2-3, as indicated in the figure. Develop equivalent nodal

loads for node 2 due to such temperature variation, assuming isotropic material in plane

stress.

(Weaver & Johnston, prob. 2.3-6, answer:( ) ( )3 4

2,2 1 3 1

t tE T E Tf tb f ta

= =

)

x

y

Pmax =800 MPa45

50 mm


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Problem 6. Derive numerically the stiffness term k66 for the isoparametric Q4 element in the

figure using Gaussian numerical integration with n=2 each way. Assume that the thickness t is

constant and the material is isotropic. The coordinates for 1,2,3, and 4 are (2,1), (8,1), (7,6)

and (1,5), respectively. Give the numerical results to four significant figures (Weaver &

Johnston, prob. 3.4-5, answer: k55 = (0.5312E22 + 0.1929E33)t




and (1,5), respectively. Give the numerical results to four significant figures.




and (1,5), respectively. Give the numerical results to four significant figures


figure using Gaussian numerical integration with n = 2 each way. Assume that the thickness t

is constant and the material is isotropic. The coordinates for 1,2,3, and 4 are (2,1), (8,1), (7,6)

and (1,5), respectively. four significant figures.

Problem 10. The aluminum plate is subjected the distributed and concentrated loads. The

dimensions of plate are given in the figure. E = 70 GPa, = 0.3 and thickness of the plate 5mm.

(a)Model the plate with two finite elements

(b)Solve the nodal displacements

(c)Obtain the element stresses

(d)Calculate the support reactions.

(e)Optional: Solve the problem using a FE software and compare the results.

You may use MATLAB software to solve the linear algebraic equations and to perform other

matrix operations.

3

x

y


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Problem 11. A linearly varying load (force per unit length) is aplied on the 2-3 edge of Q4

element shown in the Figure.. Obtain the equivalent nodal loads due to the distributed load.For this purpose, express bx in terms of. The length of edge is l23 and = 1 on this edge. Thecoordinates for 1,2,3, and 4 are (2,1), (8,1), (7,6) and (1,5), respectively.

Problem 12. The cross-section of a long steel axle shown in the figure has a radius of 50 mm.

A linearly varying pressure load is applied to right and left sides of the axleand its maximum

value is 800 MPa.

a) model the disc with two constant strain triangles (use the symmetry of geometry and

loads)

b) determine the nodal displacements and draw the deformed and undeformed shapes

c) calculate the element stresses.

600 mm

800 mm

40 kN

90

50 kN/m

500 mm

500 mm

y

4

3bx3

bx221

x

y

Pmax =800 MPa45

50 mm


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1

2

34

= y/b

= x/a

uv

2a

2b

by0by

Problem 13. For the bilinear displacement rectangle, derive the equivalent nodal loads due to

the parabolically distributed force by = by0(1-x2) per unit length, apllied in the h direction

along edge 1-2 in the figure.



is constant and the material is isotropic. The coordinates for 1,2,3, and 4 are (4,2), (10,2),

(8,8) and (2,6), respectively. Give the numerical results to four significant figures.

Problem 15. The geometry and loading of a plate which is made from A-36 steel are shown

in the figure. The thickness of plate is 10 mm. (a) model the plate with two constant strain

triangles (b) determine the nodal displacements and draw the deformed and undeformed

shapes (c) calculate the element stresses.

x

y

4

12

3


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111

1

2

34

= y/b

= x/a

u

v

2a

2b

T

-T

x

4

1

2

3

Problem 16. Suppose that a bilinear displacement rectangle is subjected to a temperaturechange that varies linearly from edge 1-2 to edge 3-4, as indicated in the figure. Develop

equivalent nodal loads at node 4 due to such a temperature variation, assuming isotropic

material in plane stress.



is constant and the material is isotropic. The coordinates for 1,2,3, and 4 are (4,4), (10,6),

(10,12) and (2,8), respectively. Give the numerical results to four significant figures.

20 N/mm

200 mm

400 mm

400 mm

40 N/mm

y


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Problem 18. For the Q4 element shown in the figure, the element displacement vector {q} is

given as { } 0 0 0.20 0 0.15 0.10 0 0.05 T= q (a)Find the x, y coordinates of a point whose location in the natural coordinates is given

by = 0.5, = 0.5.

(b)Find the u, v displacements of the point P.

Problem 19.

(a) Sketch a four-node plane element for which J is a function of but not of.(b) Use Gauss quadrature to evaluate the integral

1 1 2

21 1

3

2I d d

+=

+

Use (i) one point, (ii) four points. Determine the percentage of error of each result.

Problem 20. Natural coordinates other than the , system may be chosen. One alternativeis the r,s system shown. Write shape functions of the bilinear element in terms of r and s.

Problem 21. The figure shows a linearly varying force bx (force per unit length) applied in

the x direction along side 2-3 of the Q4 element. By explicit integration derive the equivalent

nodal loads due to such an edge loading. For this purpose, let L23 be the length of edge 2-3

(where =1), and express the intensity of bx in terms of before integration.(a) Determine the displacement components,

(b) Obtain the strain components

(c) Figure 2b shows a four-node quadrilateral. The (x,y) coordinates of each node are given

in the figure. Find the x,y coordinates of a point P whose location is given by = 0.5, =0.5.

4

1 2

y3

x

bx3

4

1 2

y3

x

r

s

r = 1

s = 1


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Problem 23. Figure 2a shows that a four-node element shown was initially a square. Nodal

displacement components, each of magnitude c in both x and y directions, create the

displaced shape indicated by dashed lines. For the point x = y = a/2,

(a) Determine the displacement components,

(b) Obtain the strain components

(c) Figure 2b shows a four-node quadrilateral. The (x,y) coordinates of each node aregiven in the figure. Find the x,y coordinates of a point P whose location is given by =0.5, = 0.5.

Figure 2a Figure 2b

Problem 24. The Q8 element in the figure has a parabolicaly distributed force bx (per unit

length) applied in the x direction on edge 2-6-3. Find the eqivalent loads at node 6 due to this

influence.

Problem 25. For the aluminum square plate shown in the figure, (i) model by two CST

elements (ii) model by one bilinear displacement rectangle. Take Poissons ratio of 0. The

dimension a = 100 mm and the thickness of plate is 1 mm. The distributed load b y is 100

N/mm.

(a)Obtain the nodal displacements

(b) Calculate the element stresses. Compare and comment on the results of two solutions.

(c)Find the strain energy of rectangular element.

=x/a

=x/a

2a

2a

4

1

(1,4)

(1,1) 2(5,1)

(8,8)

y

3

x

x

1

2

3

4

5

6

7

8

bx0


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Problem 26. The Q8 element in the figure has a linearly distributed force by (per unit length)

applied in the y direction on edge 1-5-2. Find the eqivalent loads at node 5 due to this

influence.

Problem 27.

(a) Write shape functions of the plane stress element in terms of and .(b) Explain the isoparametric approach in the element formulations.(c) Obtain the J11 term of the Jacobian matrix.

a

a

b

x

1

2

3

4

5

6

7

8

by1

by2


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Problem 28. For the bilinear displacement rectangle, derive the equivalent nodal loads

due to parabolically distributed force bx = bx0(1-2)per unit length, applied in

the direction along edge 2-3 in the figure.

Problem 29. Figure shows that a four-node element shown was initially a square. Nodal

displacement components, each of magnitude c in both x and y directions, create the displaced

shape indicated by dashed lines. Write the displacement functions using the bilinear shape

functions for the Q4 element.

(a) Determine the displacement components for the point x = y = a/2,

(b) Obtain the strain components at this point.

Problem 30. Figure shows a four-node quadrilateral. The (x,y) coordinates of each node are

given in the figure. Assume that the thickness t is constant and the material is isotropic.

(a) Derive the stiffness term k46 for the isoparametric Q4 element until the application of

numerical integration.(b) Explain the application of Gaussian numerical integration with n=2 each way.

=x/a

=x/a

2a

2a

4

1

2

3

= x/a

= y/b

bx0

2a

2b

4

1 2

y3

x

6

5


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Problem 30. Write the shape functions of the triangular plane-stress element given in the

figure. The node 4 is at the centroid of triangle. Use the non-dimensional area coordinates and .

Problem 31. The nodal coordinates and displacements of a quadrilateral plane-stress element aregiven in the figure.

(a)Calculate the displacement components at the point P(12.0, 13.4). The corresponding

natural coordinates are = 0.25 and = 0.60 at the point P.(b)Obtain the Jacobian matrix of the element.

(c)Calculate the normal strains and normal stresses at the centroid of the element.

E = 200 GPa, = 0.3, element thickness t = 4 mm.

1

2

3

4

x, u

y, v

4

1

(8,16)

(8,8)

2 (16,8)

(14,14)y

3

x

Coordinates in mm.

P

4

1

(1,4)

(1,1) 2(5,1)

(8,8)

y

3

x


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6 5

4a a

b

b

=x/a

=y/2b

1

2

3

Problem 32. A special plane stress element in a shape of right triangle is shown in the figure.

The element has three corner nodes and hence it is a constant strain triangle.

(a)Write the linear shape functions of element. Notice that the shape function must be

zero at the edge across the related node.

(b)Obtain the [B] matrix. Why this element has the constant strains? Which type of

triangular element provides the linear strain variation within the element? Pleaseexplain.

(c)Calculate the k12 term of stiffness matrix.

Problem 33. The linear strain triangle (LST) element shown in the figure is a plane-stress

element.

(a) Determine shape functions N3 and N4 in terms of non-dimensional coordinates and.

(b) Sketch their variations on the element.

(c) Let u3, v3, u4, and v4 be the only nonzero d.o.f. In terms of these d.o.f., , , a, and b,what is the element strain field?

Problem 34. The figure shows a linearly varying load bx (force per unit length) applied in the

x direction along side 2-3 of the Q4 element. By explicit integration derive the equivalent

nodal loads due to such an edge loading. For this purpose, let L23 be the length of edge 2-3(where = 1), and express the intensity of bx in terms of before integrating.

=x/a

=y/b

a

b

1 2

3


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Problem 35. A CST element with three nodes is taken from a FE mesh. The nodalcoordinates are shown in the figure and the nodal displacements under the applied loads

and 60C temperature increase are given in the table. The material of element is steel withE = 70 GPa, = 0.32, and = 23(10-6)/ C. Recover the element stresses.

Coordinates are in mm.

Problem 36. (a) Write the shape functions in terms of non-dimensional coordinates for

the rectangular element shown in figure.

(b) If the nodal displacements of a rectangular element a = b = 2 are u1 = 0.215, u2 =

0.180, u3 = u4 = 0.125, compute u andu

x

at the point x = y = 1.5.

Node No. u (mm) v (mm)

13 0.020 -0.015

17 0.025 0.010

19 0.015 0.0

1

2

3

4

bx2

bx3

x

y

x, = x/a

y, = y/b

a

b

1 2

34

(20,10)

(60,20)

(40,40)

17

13

19

x

y


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Problem 37. A distributed loading is normal to side 2-3 of the Q4 element. Derive explicit

expression for the equivalent nodal load at node 2.

Problem 38. Natural coordinates other than the , system may be chosen. One alternativeis the r,s system shown. Write the shape function N1 in terms of r and s. Plot the shape

function.

4

1 2

y3

x

r

s

r = 1

5

6

8

s = 1

l

(2,1)(9,1)

(6,6)

1 2

4

x

y (1,5)3

4 N/mm

2 N/mm

plane stress vs strain

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