SOLVED EXAMPLES ON TWO-DIMENSIONAL PLANE SOLIDS

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FEM For Two-Dimensional Solids - SOLVED EXAMPLES ON TWO-DIMENSIONAL PLANE SOLIDS

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Prof. Dr. Ahmed Hasan Alwathaf 23-7

SOLVED EXAMPLES ON TWO-DIMENSIONAL PLANE SOLIDS Example 1 (3-nodded triangular element)

The figure below shows a 2-dimensiola plane stress solid structure subjected to a horizontal force at the free end at the top. Implementing the finite element procedure, determine the displacement at the free ends and the developed stresses in the structures, using 3-nodded triangular element. The structure having the following properties:

Thickness h = 80 mm Modulus of elasticity E = 10932 N/mm2 Poisson ratio v = 0.2

A coarse mesh is needed to obtain accurate results. However, two triangular elements will be used (as shown in the figure) to illustrate the finite element calculation procedure. Using Eq. (7.15) to obtain the area of the elements. Because the triangles are right triangles, it’s easy to obtain the area without using Eq.

(7.15)

Ae1=Ae2= 30000 mm2

Coordinate of the nodes:

Node x y

1 -150 0

2 150 0

3 150 200

4 -150 200

1 2

4 3

2

1

D

5

D6 v

3

D

7

D

8

200 mm 10000 N

300 mm

x y

u

1

v

1

u

2

v

2

D

3

D

4

D

1

D

2

u

3

v

3

u

1

v

1

u

2

v

2

u

3

v

3

2

1

(3)

Arrangement of nodes as the local numbering (according to shape function identification)

Element Node Global DOF

Local numbering

Local DOF

1

1 D1

1 u1

D2 v1

2 D3

2 u2

D4 v2

3 D5

3 u3

D6 v3

2

1 D1

1 u1

D2 v1

3 D5

2 u2

D6 v2

4 D7

3 u3

D8 v3

To determine B matrix for triangular elements ( Eq. 7.38), coefficients b

i

and c

i

should be calculated from Eq. (7.23) and put them in the B matrix as shown below.

Element 1 Be1 = ^-0.00333 ⁰ ^0.003333 ⁰ ⁰ ⁰

0 0 0 -0.005 0 0.005

0 -0.003333 -0.005 0.003333 0.005 0

Element 2 Be2 = 0 0 0.003333 0 -0.0033333 0

0 -0.005 0 0 0 0.005

-0.005 0 0 0.003333 0.005 -0.003333

The elasticity matrix of the material (constant matrix) c f or plane stress, isotropic materials,

(4)

or

11387.5 2277.5 0

c1=c2 = 2277.5 11387.5 0

0 0 4555

The element stiffness matrix k

e

of the element can be obtained from Eq. 7.39

Since the strain matrix B is a constant matrix and the thickness of the element (h) is assumed to be uniform, the integration can be carried out very easily, which leads to

h.Ae = 2400000 mm³

D1 D2 D3 D4 D5 D6

u1 v1 u2 v2 u3 v3

303666.7 0 -303666.7 91100 0 -91100 u1 D1

0 121466.7 182200 -121466.7 -182200 0 v1 D2

ke1 = h.AeB1TCB1= -303667 182200 576966.7 -273300 -273300 91100 u2 D3 91100 -121466.7 -273300 804716.7 182200 -683250 v2 D4

0 -182200 -273300 182200 273300 0 u3 D5

-91100 0 91100 -683250 0 683250 v3 D6

D1 D2 D5 D6 D7 D8

u1 v1 u2 v2 u3 v3

ke2= h.AeB2TCB2= 273300 0 0 -182200 -273300 182200 u1 D1

0 683250 -91100 0 91100 -683250 v1 D2

0 -91100 303666.7 0 -303666.67 91100 u2 D5

-182200 0 0 121466.7 182200 -121466.7 v2 D6

-273300 91100 -303666.7 182200 576966.67 -273300 u3 D7 182200 -683250 91100 -121466.7 -273300 804716.7 v3 D8

Putting the element stiffness matrix in the global stiffness matrix

D1 D2 D3 D4 D5 D6 D7 D8

303666.7 0 -303666.7 91100 0 -91100 D1

0 121466.7 182200 -121466.7 -182200 0 D2

-303667 182200 576966.7 -273300 -273300 91100 D3

K1= 91100 -121466.7 -273300 804716.7 182200 -683250 D4

0 -182200 -273300 182200 273300 0 D5

-91100 0 91100 -683250 0 683250 D6

D7

(5)

D8

D1 D2 D3 D4 D5 D6 D7 D8

273300 0 0 -182200 -273300 182200 D1

0 683250 -91100 0 91100 -683250 D2

D3

D4

K2= 0 -91100 303666.67 0 -303667 91100 D5

-182200 0 0 121466.7 182200 -121467 D6

-273300 91100 -303666.67 182200 576966.7 -273300 D7

182200 -683250 91100 -121466.7 -273300 804716.7 D8

Adding the element participation in the final global stiffness matrix, K=K1+K2

D1 D2 D3 D4 D5 D6 D7 D8

576966.7 0 -303666.7 91100 0 -273300 -273300 182200 D1

0 804716.7 182200 -121466.7 -273300 0 91100 -683250 D2

-303667 182200 576966.7 -273300 -273300 91100 0 0 D3

K= 91100 -121466.7 -273300 804716.7 182200 -683250 0 0 D4

0 -273300 -273300 182200 576966.67 0 -303667 91100 D5

-273300 0 91100 -683250 0 804716.7 182200 -121467 D6

-273300 91100 0 0 -303666.67 182200 576966.7 -273300 D7

182200 -683250 0 0 91100 -121466.7 -273300 804716.7 D8

Applying the boundary restraint at the supports (D1=D2=D3=D4=0); the stiffness matrix of the structure

D5 D6 D7 D8

576966.7 0 -303666.7 91100 D5

K = 0 804716.7 182200 -121466.7 D6

-303667 182200 576966.7 -273300 D7 91100 -121466.7 -273300 804716.7 D8

And the external load vector

F5 0 N

F = F6 0 N

F7 10000 N

F8 0 N

Solving the equation to obtain the unknown displacements

D5 0.014941 mm

D = D6 -0.00577 mm

D7 0.030752 mm

D8 0.007881 mm

Strain and stress

The strain and stress can be determined using Eq. 7.36

(6)

or

Stress vector can be obtained by the expression

Or

Element 1   = B1 de1

Rearrange the displacements as the order of the local numbering to obtain d

e1

u1 D1 0

v1 D2 0

de1 = u2 D3 0

v2 D4 0

u3 D5 0.0149408 v3 D6 -0.005773

Strain and stress vectors

x = -0.06574 N/mm²

y= -0.3287 N/mm²

xy= 0.340277 N/mm²

Element 2 :  = B2de2

Rearrange the displacements as the order of the local numbering to obtain d

e2

u1 D1 0

de2 = v1 D2 0 u2 D5 0.0149408 v2 D6 -0.005773 u3 D7 0.0307519 v3 D8 0.0078812

x = 0.000E+00

y = -2.887E-05

xy = 7.470E-05

 







 







=

3 3 2 2 1 1

3 2

1

3 2

1

0 0

0

0 0

0

b c b c b c

c c

c

b b

b

B

(7)

Strain and Stress vectors

x = -0.51042 N/mm²

y= 0.328704 N/mm²

xy= 0.493057 N/mm²

Example 2 coordinate mapping of iso-parametric element

The Figure below shows a quadrilateral element in global coordinates. Show that the mapping described by Equation 7.70 correctly describes the line connecting nodes 2 and 3 and determine the (x , y) coordinates corresponding to (ξ, η) = (1, 0.5)

Solution

First, we determine the equation of the line passing through nodes 2 and 3 strictly by geometry, using the equation of a two-dimensional straight line y = mx + b. Using the known coordinates of nodes 2 and 3, we have

Node 2: 1 = 3m + b Node 3: 2 = 2.5m + b

Solving simultaneously, the slope is m = −2

and the y intercept is b = 7

Therefore, element edge 2-3 is described by y = −2x + 7

Using the interpolation functions given in Equation 7.51 and substituting nodal x and y coordinates, the geometric mapping of Equation 7.70 becomes

x =1/4[(1 − ξ )(1 − η)(1) +(1 + ξ )(1 − η)(3) +(1 + ξ )(1 + η)(2.5) +(1 − ξ )(1 + η)(1.25)]

y =1/4[(1 − ξ )(1 − η)(1) +(1 + ξ )(1 − η)(1) +(1 + ξ )(1 + η)(2) +(1 − ξ )(1 + η)(1.75)]

Noting that edge 2-3 corresponds to ξ = 1, the last two equations become x =3/2 (1 − η) +2.5/2 (1 + η) =5.5/2 −0.5/2 η

y =1/2 (1 − η) + (1 + η) =3/2 −1/2 η Eliminating η gives

2x + y =14/2 which is the same as

x = -5.270E-05

y = 3.941E-05

xy = 1.082E-04

(8)

y = −2x + 7 as desired.

For (ξ, η) = (1, 0.5), we obtain x =5.5/2 −0.5/2 (0.5) = 2.625 y =3/2 +1/2 (0.5) = 1.75

Example 3 (4-nodded iso-parametric rectangular element)

Resolve the previous example using one plane stress 4-nodded iso-parametric rectangular element. The structure has the same previous properties.

Coordinate of the nodes

Node x y

1 -150 0

2 150 0

3 150 200

4 -150 200

Arrangement of nodes as the local numbering (according to the shape functions identification)

Element Node Global DOF

Local numbering

Local DOF

1

1 D1

1 u1

D2 v1

2 D3

2 u2

D4 v2

3 D5 3 u3

10000 N

300 mm

200 mm

1 2

4 3

D

5

D6 v

3

D

7

D

8

x y

u

2

v

2

D

3

D

4

D

1

D

2

u

3

v

3

u

1

v

1

u

4

v

4

ξ

η

(9)

D6 v3

4 D7

4 u4

D8 v4

For iso-parametric element, the Gauss numerical integration will be used to determine the stiffness matrix for the element as shown in Eqs. (7.80) and (7.59). Four Gaussian points will be used to carry out the integration. The location of these points is shown in Fig. 7.11’ (a). The point location and the weight coefficients are shown also in Table 7.1. Therefore, four

B matrices will be calculated at each Gaussian

point.

The calculations are shown in the following table.

Gauss point 1 Gauss point 2 Gauss point 3 Gauss point 4

ξi -0.57735 -0.57735 0.57735 0.57735

ηj -0.57735 0.57735 -0.57735 0.57735

wi 1 1 1 1

wj 1 1 1 1

N1 0.622008 0.166667 0.166667 0.044658

N2 0.166667 0.044658 0.622008 0.166667

N3 0.044658 0.166667 0.166667 0.622008

N4 0.666667 2.488034 0.178633 0.666667

dN1/dξ -0.394338 -0.105662 -0.394338 -0.105662

dN2/dξ 0.394338 0.105662 0.394338 0.105662

dN3/dξ 0.105662 0.394338 0.105662 0.394338

dN4/dξ -0.105662 -0.394338 -0.105662 -0.394338

dN1/dη -0.394338 -0.394338 -0.105662 -0.105662

dN2/dη -0.105662 -0.105662 -0.394338 -0.394338

dN3/dη 0.105662 0.105662 0.394338 0.394338

dN4/dη 0.394338 0.394338 0.105662 0.105662

[J] ¹⁵⁰ ⁰ ¹⁵⁰ ⁰ ¹⁵⁰ ⁰ ¹⁵⁰ ⁰

0 100 0 100 0 100 0 100

(10)

IJI 15000 15000 15000 15000

[J]^-1 ^0.006667 ⁰ ^0.006667 ⁰ ^0.006667 ⁰ ^0.006667 ⁰

0 0.01 0 0.01 0 0.01 0 0.01

wiwj IJI 15000 15000 15000 15000

hwiwj IJI 1200000 1200000 1200000 1200000

c

11387.5 2277.5 0 11387.5 2277.5 0 11387.5 2277.5 0 11387.5 2277.5 0 2277.5 11387.5 0 2277.5 11387.5 0 2277.5 11387.5 0 2277.5 11387.5 0

0 0 4555 0 0 4555 0 0 4555 0 0 4555

It can be seen from the table that, [J] matrices are the same for all Gauss points because of the regularity of the geometry of the rectangular element. As a result the

IJI

and

[J]^-1

are also have the same values.

However, if the element have irregular shape, the mentioned values ([J], IJI and

[J]^-1

) will take different values at the Gauss Points. Also, all the material constant matrices ( c ) at all Gauss Points are the same for loading stages which is the feature of the linear material. However, for nonlinear material, this will not be valid and the material constant matrices ( c ) at the Gauss Points will vary according the loading level.

Nonlinear finite element analysis is not straightforward analysis and needs to apply an incremental loading procedure. Every increment will be analyzed as linear analysis.

B matrix

Gauss point 1

-0.002629 0 0.002629 0 0.000704 0 -0.000704 0

B1 0 -0.00394 0 -0.001057 0 0.00105662 0 0.003943

-0.003943 -0.00263 -0.001057 0.002629 0.001057 0.00070442 0.003943 -0.000704

Gauss point 2

-0.000704 0 0.000704 0 0.002629 0 -0.002629 0

B2 0 -0.00394 0 -0.001057 0 0.00105662 0 0.003943

-0.003943 -0.0007 -0.001057 0.000704 0.001057 0.00262892 0.003943 -0.002629

Gauss point 3

-0.002629 0 0.002629 0 0.000704 0 -0.000704 0

B3 0 -0.00106 0 -0.003943 0 0.00394338 0 0.001057

-0.001057 -0.00263 -0.003943 0.002629 0.003943 0.00070442 0.001057 -0.000704

Gauss point 4

-0.000704 0 0.000704 0 0.002629 0 -0.002629 0

B4 0 -0.00106 0 -0.003943 0 0.00394338 0 0.001057

-0.001057 -0.0007 -0.003943 0.000704 0.003943 0.00262892 0.001057 -0.002629

(11)

Stiffness matrix Gauss point 1

179439.1 84997.5 -71666.6 -49073.3 -48080.6 -22775.0 -59691.9 -13149.2 84997.5 250270.3 -13149.2 19160.9 -22775.0 -67059.7 -49073.3 -202371.4 -71666.6 -13149.2 100544.2 -22775.0 19203.0 3523.3 -48080.6 32400.8

hwiwjB1TcB1 IJI

-49073.3 19160.9 -22775.0 53033.0 13149.2 -5134.1 58699.2 -67059.7 -48080.6 -22775.0 19203.0 13149.2 12883.1 6102.5 15994.4 3523.3 -22775.0 -67059.7 3523.3 -5134.1 6102.5 17968.6 13149.2 54225.3 -59691.9 -49073.3 -48080.6 58699.2 15994.4 13149.2 91778.1 -22775.0 -13149.2 -202371.4 32400.8 -67059.7 3523.3 54225.3 -22775.0 215205.9 Gauss point 2

91778.1 22775.0 15994.4 -13149.2 -48080.6 -58699.2 -59691.9 49073.3 22775.0 215205.9 -3523.3 54225.3 -32400.8 -67059.7 13149.2 -202371.4 15994.4 -3523.3 12883.1 -6102.5 19203.0 -13149.2 -48080.6 22775.0

hwiwjB2TcB2 IJI

-13149.2 54225.3 -6102.5 17968.6 -3523.3 -5134.1 22775.0 -67059.7 -48080.6 -32400.8 19203.0 -3523.3 100544.2 22775.0 -71666.6 13149.2 -58699.2 -67059.7 -13149.2 -5134.1 22775.0 53033.0 49073.3 19160.9 -59691.9 13149.2 -48080.6 22775.0 -71666.6 49073.3 179439.1 -84997.5 49073.3 -202371.4 22775.0 -67059.7 13149.2 19160.9 -84997.5 250270.3

Gauss point 3

100544.2 22775.0 -71666.6 13149.2 -48080.6 -32400.8 19203.0 -3523.3 22775.0 53033.0 49073.3 19160.9 -58699.2 -67059.7 -13149.2 -5134.1 -71666.6 49073.3 179439.1 -84997.5 -59691.9 13149.2 -48080.6 22775.0

hwiwjB3TcB3 IJI

13149.2 19160.9 -84997.5 250270.3 49073.3 -202371.4 22775.0 -67059.7 -48080.6 -58699.2 -59691.9 49073.3 91778.1 22775.0 15994.4 -13149.2 -32400.8 -67059.7 13149.2 -202371.4 22775.0 215205.9 -3523.3 54225.3 19203.0 -13149.2 -48080.6 22775.0 15994.4 -3523.3 12883.1 -6102.5 -3523.3 -5134.1 22775.0 -67059.7 -13149.2 54225.3 -6102.5 17968.6

Gauss point 4

12883.1 6102.5 15994.4 3523.3 -48080.6 -22775.0 19203.0 13149.2 6102.5 17968.6 13149.2 54225.3 -22775.0 -67059.7 3523.3 -5134.1 15994.4 13149.2 91778.1 -22775.0 -59691.9 -49073.3 -48080.6 58699.2 hwiwjB4TcB4 IJI 3523.3 54225.3 -22775.0 215205.9 -13149.2 -202371.4 32400.8 -67059.7 -48080.6 -22775.0 -59691.9 -13149.2 179439.1 84997.5 -71666.6 -49073.3 -22775.0 -67059.7 -49073.3 -202371.4 84997.5 250270.3 -13149.2 19160.9 19203.0 3523.3 -48080.6 32400.8 -71666.6 -13149.2 100544.2 -22775.0 13149.2 -5134.1 58699.2 -67059.7 -49073.3 19160.9 -22775.0 53033.0

K

e

=∑ hw

i

w

j

B

^T

cB IJI = (hw

i

w

j

B

^T

cB IJI)

GP1

+ (hw

i

w

j

B

^T

cB IJI)

GP2

+ (hw

i

w

j

B

^T

cB IJI)

GP3+

(hw

i

w

j

B

^T

cB IJI)

GP4

u1 v1 u2 v2 u3 v3 u4 v4

384644.4 136650 -111344 -45550 -192322 -136650 -80977.8 45550 u1 136650 536477.8 45550 146772.2 -136650 -268239 -45550 -415011 v1

(12)

-111344 45550 384644.4 -136650 -80977.8 -45550 -192322 136650 u2 Ke= -45550 146772.2 -136650 536477.8 45550 -415011 136650 -268239 v2 -192322 -136650 -80977.8 45550 384644.4 136650 -111344 -45550 u3 -136650 -268239 -45550 -415011 136650 536477.8 45550 146772.2 v3 -80977.8 -45550 -192322 136650 -111344 45550 384644.4 -136650 u4 45550 -415011 136650 -268239 -45550 146772.2 -136650 536477.8 v4

No need to perform assembly, because there is only one element. The finite element equation becomes K=

Ke. Applying the boundary condition at the support; we can simply remove the first four equations because

D

1

= D

2

= D

3

= D

4

=0. The stiffness matrix of the structure

D5 D6 D7 D8

384644.4 136650.0 -111344.4 -45550.0 D5 K= ^136650.0 ^536477.8 ^45550.0 ^146772.2 ^D6 -111344.4 45550.0 384644.4 -136650.0 D7 -45550.0 146772.2 -136650.0 536477.8 D8

The external load vector

F5 0 N

F = F6 0 N

F7 10000 N

F8 0 N

Solving the equation to obtain the unknown displacements

D5 0.016245 mm

D = D6 -0.01105 mm

D7 0.036913 mm

D8 0.013805 mm

Strain and stress

As before, the strain can be determined using

And stress can be obtained by the expression

In this case the strain and stress will be obtained at each Gauss point using B matrix at that point (B1, B2, B3 and B4).

Arrange the displacements as the order of the local numbering to obtain d

e

. The displacement vector d

e

will be used for all Gauss points.

u1 D1 0

v1 D2 0

de = u2 D3 0

v2 D4 0

(13)

u3 D5 0.016245 v3 D6 -0.01105 u4 D7 0.036913 v4 D8 0.013805

Strains and stresses can be tabulated as follows

Example 4 (8-nodded iso-parametric rectangular element)

Resolve the previous example using one plane stress 8-nodded iso-parametric rectangular element. The structure has the same previous properties.

Gauss point 1 Gauss point2 Gauss point 3 Gauss point4

x = -1.4559E-05 -5.433E-05 -1.45588E-05 -5.43E-05 y = 4.27622E-05 4.276E-05 -2.89836E-05 -2.9E-05

xy = 0.000145221 9.739E-05 8.55583E-05 3.77E-05

x (N/mm²)= -0.06839745 -0.5213396 -0.231798469 -0.684741

y (N/mm²) = 0.453796765 0.3632083 -0.363208333 -0.453797

xy (N/mm²) = 0.661483327 0.4436153 0.389718033 0.17185

1 (N/mm²) = 0.903847948 0.5473534 0.097714619 -0.362227

2 (N/mm²) = -0.51844863 -0.7054847 -0.692721421 -0.77631

u

1

v

1

1 2

3 4 6 5

7

8

1 2

3 4 6 5

7

8

10000 N

300 mm

200 mm

D

5

D6 v

3

D

7

D

8

x y

u

2

v

2

D

3

D

4

D

1

D

2

u

3

v

3

u

4

v

4

1 2

3 4 6 5

7

8

D

9

D

10

D

11

D

12

D

13

D

14

D

15

D

16

u

8

v

8

u

7

v

7

u

6

v

6

u

5

v

5

ξ

η

(14)

The figure shown above represents the global and local degree of freedoms. The same procedure will be followed here as the previous example. The shape functions and their derivatives will be obtained from equations 7.59’ and 7.59’’. Nine Gaussian points will be used to carry out the integration to determine the stiffness matrix of the solid. Therefore, nine

B matrices will be calculated at each Gaussian point. The

location of these points is shown in Fig. 7.11’ (a).

The point location and the weight coefficients are shown in the following table.

Gauss point 1 Gauss point 2 Gauss point 3

ξi -0.7746 -0.7746 -0.7746

ηj -0.7746 0 0.774597

wi 0.555556 0.555556 0.555556

wj 0.555556 0.888889 0.555556

Gauss point 4 Gauss point5 Gauss point 6

ξi 0 0 0

ηj -0.7746 0 0.774597

wi 0.888889 0.888889 0.888889

wj 0.555556 0.888889 0.555556

Gauss point 7 Gauss point 8 Gauss point 9

ξi 0.774597 0.774597 0.774597

ηj -0.7746 0 0.774597

wi 0.555556 0.555556 0.555556

wj 0.555556 0.888889 0.555556

Stiffness matrix

Ke=∑ hwiwjB^TcB IJI = (hw1w1B^TcB IJI)GP1+ (hw1w2B^TcB IJI)GP2+ (hw1w3B^TcB IJI)GP3+ (hw2w1B^TcB IJI)GP4+(hw2w2B^TcB IJI)GP5+ (hw2w3B^TcB IJI)GP6+ (hw3w1B^TcB IJI)GP7+ (hw3w2B^TcB IJI)GP8+(hw3w3B^TcB IJI)GP9

Applying the boundary condition at the support; we can simply see D

1

= D

2

= D

3

= D

4

=D

5

= D

6

=0. The

stiffness matrix of the structure

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D7 D8 D9 D10 D11 D12 D13 D14 D15 D16

1295635 0 -445379 -182201 0 242931 -283421 60733 161965 0 D7

0 2558826 -60730 -1198435 242931 0 60733 -623511 0 1085066 D8

-445379 -60730 666707 258114 -503396 -182201 292189 15184 -283421 -60733 D9 -182201 -1198435 258114 929871 -60730 -124843 -15184 333688 -60733 -623511 D10

0 242931 -503396 -60730 1371192 0 -503396 60730 0 -242931 D11

K 242931 0 -182201 -124843 0 1160660 182201 -124843 -242931 0 D12

-283421 60733 292189 -15184 -503396 182201 666707 -258114 -445379 60730 D13 60733 -623511 15184 333688 60730 -124843 -258114 929871 182201 -1198435 D14

161965 0 -283421 -60733 0 -242931 -445379 182201 1295635 0 D15

0 1085066 -60733 -623511 -242931 0 60730 -1198435 0 2558826 D16

The external load vector

F7 0 N

F8 0 N

F9 0 N

F10 0 N

F= F11 0 N

F12 0 N

F13 10000 N

F14 0 N

F15 0 N

F16 0 N

Solving the equation to obtain the unknown displacements

D7 0.0145914 mm

D8 -0.0061448 mm

D9 0.0140342 mm

D10 -0.0046319 mm

D= _D11 _0.0296501 _mm

D12 -0.0041518 mm

D13 0.0597711 mm

D14 0.0285088 mm

D15 0.0167878 mm

D16 0.0165587 mm

Strain and stress

As before, the strain can be determined using

And stress can be obtained by the expression

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Strain and stress will be obtained at each Gauss point using B matrix at that point (B1, B2, B3, B4, B5, B6, B7, B8, and B9).

Arrange the displacements as the order of the local numbering to obtain d

e

. The displacement vector d

e

will be used for all Gauss points.

u1 D1 0

v1 D2 0

u2 D3 0

v2 D4 0

u3 D5 0

v3 D6 0

u4 D7 0.0145914

de= v4 D8 -0.0061448

u5 D9 0.0140342

v5 D10 -0.0046319

u6 D11 0.0296501

v6 D12 -0.0041518

u7 D13 0.0597711

v7 D14 0.0285088

u8 D15 0.0167878

v8 D16 0.0165587

Strains and stresses can be tabulated as follows

GP 1 GP 2 GP 3 GP 4 GP 5 GP 6

x = 1.9389E-06 -4.477E-05 -0.0001742 1.04E-05 -7.32E-06 -0.0001077 y = 1.1668E-04 9.169E-05 6.6699E-05 -3.26E-05 -2.08E-05 -0.0000089

xy = 5.2404E-05 9.981E-05 0.00017175 8.48E-05 7.257E-05 0.0000848

x (N/mm²)= 2.8780E-01 -0.3010247 -1.8314379 0.044034 -0.130646 -1.2469096

y (N/mm²) = 1.3330E+00 0.9421039 0.36284491 -0.347219 -0.25306 -0.3472188

xy (N/mm²) = 2.3870E-01 0.4546316 0.78230678 0.38644 0.3305676 0.3864402

1 (N/mm²) = 1.3850E+00 1.0906254 0.61319122 0.281543 0.1443333 -0.2040238

2 (N/mm²) = 2.3587E-01 -0.4495461 -2.0817842 -0.584727 -0.528039 -1.3901046

GP 7 GP 8 GP 9

x = 1.8822E-05 3.0131E-05 -4.1249E-05 y = -8.5273E-05 -3.6665E-05 1.1943E-05

xy = 1.6079E-04 8.8851E-05 4.1446E-05

x (N/mm²)= 2.0131E-02 2.5960E-01 -4.4251E-01

y (N/mm²) = -9.2815E-01 -3.4889E-01 4.2052E-02

xy (N/mm²) = 7.3239E-01 4.0472E-01 1.8879E-01

1 (N/mm²) = 4.1846E-01 4.6167E-01 1.0692E-01

2 (N/mm²) = -1.3265E+00 -5.5096E-01 -5.0738E-01

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Problems

1. Prove that, the thickness that can be used to obtain the stiffness matrix of a triangular element whose thickness varies linearly is

h=( h

1

+h

2

+h

3

)/3

where h

1

, h

2

and h

3

are the nodal thickness.

Hint:

The area coordinate can be used in the solution as shown in the figure in which:

2. Derive the shape functions and their derivatives with respect to ξ and η

for 8-nodded iso- parametric rectangular element shown in the figure below.

3. Make a comparison between the finite element analysis results that used in the solved examples. The comparison should include the following elements:

1. 2D triangular element,

3 1

2 7

6 5

8

4 η

ξ

Y

X

u v

(0,-1) (-1,-1)

(1,-1) (1,0) (1, 1) (0, 1)

(-1, 1)

(-1,0)

h

1

h

2

h

3

A

1

A

2

A

3

P(x,y)

Fig. Q1

Fig. Q2

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2. 2D 4-nodded iso-parametric element 3. 2D 8-nodded iso-parametric element

The comparison should cover the displacement (deformed shape) , stresses values and stresses distribution.

The properties are as follows:

Thickness h = 80 mm Modulus of elasticity E = 10932 N/mm2 Poisson ratio v = 0.2

4. Analyze the previous structure identified in Q3 using finite rectangular plane-stress element implemented by StaadPro Package. Finer mesh should be used to obtain accurate results. A rectangular element 10 mm x10 mm is suggested.

5. Determine the deflected shape and stresses contours (σ1 and σ2) for the cantilever beam shown in Fig Q5. The beam is subjected to a concentrated moment at the free end. The analysis will be conducted using 2D plane stress element available in StaadPro Package. One of possible mesh that can be used is suggested in the figure below. The properties of the beam are as follows:

Thickness h = 0.25 m Modulus of elasticity E = 20000 N/mm

²

Poisson ratio v = 0.2

Fig. Q3 200 mm

10000 N

300 mm

x y

M = 45.0 kN.m

3.0 m 1.5 m

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Fig. Q5

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