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FEM for Engineering Applications

Exercises with Solutions

Jonas Faleskog

KTH Solid Mechanics

August 2008

1. Elastic Energy and Energy principles

2. Matrix formulated structural mechanics—direct method

3. Strong/weak form and FEM-equations

4. FEM: trusses and beams

5. FEM: planar frames of trusses and beams

6. FEM: 2D/3D solids

7. FEM: heat conduction



FEM for Engineering Applications—Exercises with Solutions / August 2008 / J. Faleskog

– 1.2 (10) –

1.7 A beam with bending stiffness EI and total length 2 L,

is simply supported at its mid point. The left end of the

beam is attached to a linear spring with the spring constant

. The beam is subjected to a point force P0

and a moment M 0. Determine M 0 such that the deflection

of the right end of the beam becomes zero. Carry out theanalysis using an energy method, based on complementary elastic energy.

1.8 A system of two beams, each with

bending stiffness EI and length L, and a

spring k N, see Figure (a) to the right, has

proven to be to compliant in an application.

The system is therefore made stiffer by use

of a torsion spring k M, see Figure (b). The

complementary elastic energy for the sys-

tem is

,

where M is the moment arising in the torsion spring when the system is loaded by a point force

P. The stiffness of the springs can be expressed as and , where α = 3 in the current application. Determine β , such that the stiffness of the system increases by a

factor of two, i.e. such that the displacement at P in (b) becomes half compared to the case in

(a). Hint: problem (b) is statically indeterminate.

1.9 The constraint and the boundary conditions of a beam

with bending stiffness EI and length 2 L is shown in the

right hand figure. Determine the vertical displacement of

the beam at the point force. Use an energy based method.

1.10 A beam with bending stiffness EI and total

length 3 L is subjected to a uniformly distributed load

with the resultant Q, see the figure to the right.

Determine all reaction forces acting on the beam.Use an energy method, based on the complementary

elastic energy, for statically indeterminate quantities.

1.11 A freely supported beam with bending stiff-

ness EI and total length 3 L is loaded by a point

force, P, according to the right hand figure. Deter-

mine the deflection at point B.

P0

L L M 0

k k η EI L3

⁄ =

P

EI, L

k N

E I ,

L

P

EI, L

k N

E I ,

L

k M

(a) (b)

W L

3

EI ------

1

6--- P

2P

M

L-----–

2

+

1

2---

P M L ⁄ –( )2

k N L3

EI ⁄ -----------------------------

1

2---

M L ⁄ ( )2

k M L EI ⁄ --------------------+ +=

k N α EI L3

⁄ = k M β EI L ⁄ =

P

L L

Q

2 L L

P

A B




– 1.3 (10) –

1.12 A plane frame is composed of three beams connected

at the stiff joints B and C, see the figure to the right. The

bending stiffness of all beams are EI . The frame is loaded

by a point force. Determine the vertical displacement of

point C and evaluate the distribution of bending moment inthe frame.

The examples below are taken from “Exempelsamling i Hållfasthetslära” , Eds. P.-L. Larsson & R.

Lundell, KTH, Stockholm, january 2001. The solutions to these problems (in Swedish) are based on

Castigliano’s theorems.

1.13 Determine the horizontal displacement at point B. The

bending stiffness of each beam in the planar frame is EI .

1.14 A planar frame constructed by two beams, each

with bending stiffness EI , is loaded by a uniformly

distributed load with the resultant P and a point force

P according to the right hand figure. Calculate the

vertical displacement at the point force.

1.15 A beam with circular cross section (diameter

d ) is shaped as a U, see the figure to the right. The

beam is clamped at point A and loaded by a point

force P, acting perpendicular to the plane of the

beam, at point D. Calculate the displacement at

point D in the direction of the point force.

P

A B

C

D

2 L

2 L L

A

BC

L

L

M

P P L

L L

P

DA

L

B C L

d




– 1.4 (10) –

1.16 A planar frame, according to the right hand fig-

ure, is clamped at point A and point B. The frame is

loaded perpendicular to its plane by a couple of point

forces, each of magnitude P. The beam has a circulartube shaped cross section with mean radius r and

thickness t (assume that ). The material has the

elastic modulus E and the shear modulus G. Calculate

the out-of-plane displacement at points B and C.

Consider the case when L = l.

1.17 A rectangular planar frame with measurements

according to the figure is clamped at point A and point D.

Two point forces is applied perpendicular to the frame in

opposite directions at points B and C, respectively. The

bending stiffness of the frame is everywhere equal to EI

and its torsional stiffness is GK , with ,

where κ being a non-dimensional constant. Calculate the

displacements perpendicular to the plane at the points B

and C, respectively.

1.18 A planar frame of quadratic shape is freely supported at the

corner points A, B, C and D, such that only reaction forces per-

pendicular to the frame may occur. The side of the frame is of

length L. The bending stiffness and the torsional stiffness are EI

and GK , respectively. The frame is subjected to a uniformly dis-

tributed load with the resultant Q acting perpendicular to the

frame on the side AB. Calculate the displacement at the mid

point between A and B.

L

L

D

C

A

B

P

P

D

L

L

C

A

l

P

P

B

t r «

P P

B C

A D

L

2L EI GK ( ) ⁄ κ =

Q

B C

A D

L, EI, GK




– 1.5 (10) –

Solutions

1.1

1.2

1.3

Introduce reac-

M 0

RB

RA

Equilibrium R B M 0 3 L( ) ⁄ =

R A

R B

M 3 L( ) ⁄ = =

M 1 M 0 R A

M 0

3 L-------=

M 1 M 2 M 2

R B

M 0

3 L-------=

Equilibrium requires: M 1 M 0 3 ⁄ = M 2 2 M – 0 3 ⁄ =

Complementary elastic energy: W LM 1

2

6 EI -----------

2 LM 22

6 EI --------------+

M 02 L

6 EI ----------- θ⇒

∂W

∂ M 0----------

M 0 L

3 EI -----------= = = =

tion forces: gives:

P R M R

M F

Introduce a fictitious bending moment, M F

and reaction forces, R & M R

Equilibrium:

R = P

M R = M F

P P M F

M 1

PP

M 2PP

M 1 M 2 M F

M 1 M F PL 2 ⁄ +=

M 2 PL 2 ⁄ =

Equilibrium:

W M F

2 L

2 EI -----------

M F PL2

2 2 EI ------------------

P2 L

3

6 EI ------------+ +=

Complementary elastic energy:

Rotation at B: θ B∂W

∂ M F

-----------

M F 0=

PL2

2 2 EI ----------------= =

M 0

M 0

M 0 M R

P

1 statically indeterminate, chose e.g. M R

M 0 PL–=Equilibrium:

W L

6 EI --------- M R

2 M R– PL 2P

2 L

2+( )=

∂W

∂ M R----------- 0 M R⇒

PL

2-------= =

δ ∂W

∂P--------

7

12------

PL3

EI ---------= =

Displacement at point B:




– 1.6 (10) –

1.4

1.5

1.6

Cut and use the symmetry

M 0

M 0 /4 M 0 /4

M 0 /4 M 0 /4

M 0 /4 M R

V

V

V

V

V

V

Equilibrium: M 0

4------- M R– VL– 0=

=> 1 statically indeterminate, e.g. M R

Total complementary elastic energy:

W 4 L

6 EI --------- M R

2 M R

M 0

4-------

M 0

4-------

2

+ +

⋅=

∂W

∂ M R----------- 0= M R⇒

M 0

8------- ;–= θ

∂W

∂ M 0----------

M 0 L

16 EI ------------= =Rotation:

properties!

M R

V

Equilibrium, beam:

: V – N – P+ 0=

=> 1 statically indeterminate,


L3

6 EI --------- P

2 N

21

3

η---+

2PN –+

=

∂W

∂ N -------- 0= N ⇒

ηP

3 η+------------- ;=

δP∂W ∂P-------- PL3

EI --------- 1

3 η+( )-----------------= =

P

N

k η EI

L3

------=

: M R PL– NL+ 0=

chose e.g. N

W W beam W spring+ L

6 EI --------- M R

2 N 2

2k ------+= =

Principle of least work:

Deflection at the right end (Castigliano’s 2nd theorem):

L

6 EI --------- PL NL–( )

2 N 2

2η EI L3

⁄ -----------------------+=

Equilibrium gives

Complementary

θ ∂W

∂ M 0----------

3

2--- M 0 L

EI -----------= =

M 0 M 0 M 0 M R

k 6 EI

L3

---------=

M 0

R N

N

M 0 M R–

L---------------------=

Note! one statically

indeterminate exists!

elastic enerrgy: W L

6 EI --------- M R

2 M R M 0 M 0

2+ +( )

L

6 EI --------- 3 M 0

2( )

N 2

2k ------+ +

L

4 EI --------- M R

23 M 0

2+( )= =

Condition to determine the unknown ∂W

∂ M R-----------

L

4 EI ---------2 M R 0 M R⇒ 0= = =

Rotation at the point where the external moment is applied :

reaction force:




– 1.7 (10) –

1.7

1.8

1.9

Free body diagram:

M 1 NL P0 L M 0–= =

M 2 P0 L=

Equilibrium:

W M 1

2 L

6 EI -----------

M 22 L

6 EI -----------

N 2

2k ------+ +

L3

6 EI --------- 2P0

2 M 0

L-------

2 2P0 M 0

L-----------------–+

L3

2η EI ------------- P0

2 M 0

L-------

2 2P0 M 0

L-----------------–+

+= =

Complementary elastic energy:

(given condition) M 0⇒

3 2η+

3 η+---------------- P0 L=

k η EI

L3

------=

N

M 0

R

P0

P0

M 2

M 1 N

Equilibrium:

: M 0 P0 L– NL+ 0=

N ⇒ P0 M 0 L ⁄ –=

δP0

∂W

∂P0--------- 0= =

Case (a): no torsion spring ( M = 0), k N = 3 EI / L3 δa

∂W

∂P--------

M 0=

=⇒

PL3

EI ---------=

Case (b): statically indeterminate problem,

∂W

∂ M

-------- 0 M ⇒ PL2 β

3 2 β +

----------------= = δb

∂W

∂P

-------- L

3

EI

------ P2

3

--- M

L

-----–

PL3

EI

---------9 2 β +

9 6 β +

----------------= = =

According to the given conditions: δb

1

2---δa β ⇒

9

2---= =

where M is an internal indeterminate quantity, thus

Equilib:


δP

∂W

∂P--------

2

3---

PL3

EI ---------= =

: M 0 PL=

W 2 M 2

2 EI ---------d x

0

L

∫

M x( ) PL x L---=

P2

L3

3 EI ------------= = =

Displacement at the point force (Castigliano’s 2nd theorem):

M 0 PP

P P2P

M 0 M 0 PP(statically indeterminate probl.)




– 1.8 (10) –

1.10

1.11

1.12

Equilibrium:

∂W

∂ M A----------- 0 M A⇒

QL

2--------– ;= =

M A

R AQ

R A R B Q–+ 0=

R B

M A R B L Q2 L–+ 0=

Note! one statically

indeterminate, choose

for instance M A.

M(x)

Qx/L

x

M x( )QL

4

-------- x

L

---

2

=

Complementary elastic energy: W L

6 EI --------- M A

2 M AQL QL( )

2+ +( )

M x( )2

2 EI --------------- d x

0

2 L

∫

+=

R A 3Q

2------- ;–= R B

5Q

2-------=

P QIntroduce a fictitious force Q when the comple-

mentary elastic energy, , is calculated.W

The comp. elastic energy in the beam becomes: W L

18 EI ------------ 4P

27PQ 4Q

2+ +( )=

δ B

∂W

∂Q--------

Q 0=

7

18------

PL3

EI ---------= =Displacement at B (Castigliano’s 2nd theorem):

L L L

Complementary elastic energy in the beam: W L

6 EI --------- 76 R A

240 R AP– 8P

2+( )=

δC

∂W

∂P--------

52

57------

PL3

EI ---------= =Displacement in point C (Castigliano’s 2nd theorem):

P

RD

RA

M D

2 equilibrium Eqs. and 3 unknown reac-

tion forces ( RA, RD and M D). Thus, theproblem has one statically indetermi-

nate. Treat RA as known when calculat-

ing the complementary elastic energy.

The unknown RA is given by:∂W

∂ R A

--------- 0 R A⇒

5

19------P= =

18

19------PL

10

19------PL

10

19------PL

10

19------PL

Bending moment diagram:




– 1.9 (10) –

1.13

1.9

1.14

1.15

1.16




– 1.10 (10) –

1.171.18




– 2.1 (12) –

2. Matrix formulated structural mechanics—direct method

2.1 A system of five springs are connected as

shown in the figure to the right. All the spring

constants k i are in the current application

equal to k . Furthermore, D1 = D4 = 0, F 3 = 0

and F 2 = P. Determine the displacements and

reaction forces.

2.2 Three springs are connected according

to the figure to the right, also showing the

applied external force P. The spring con-

stants are: k 1 = 5k , k 2 = k and k 3 = 2k .

Determine the displacements at the points

where the springs are connected and evalu-

ate all the reaction forces.

2.3 Determine the displacement at the

point force in the spring system shown to

the right.

2.4 The plane structure to the right consists of four

springs with spring constants: k 1 = k 2 = k 4 = 2k and

k 3 = 4k . The springs are attached to five nodes with

coordinates shown in the right hand figure. The struc-

ture is loaded by two point forces acting in node 4

and node 5, as shown in the figure. Calculate the dis-

placement at each node and the normal force acting

in spring element number 4.

k 1

k 2

k 3

k 4k 5

D1, F 1 D4, F 4 D3, F 3 D2, F 2

k 1

k 2

k 3

rigid beam

3a

4a

P

P

45o 45o

45ok 1

k 2

k 3

k 1 = k

k 2 = 2k

k 3 = 2k

x/L

y/L

k 1 k 2

k 3k 4

(−1,1)(1,1)

(1,0)

(1,−1)

P

P

2

5

3

1

4




– 2.2 (12) –

2.5 A plane truss structure consists of three truss elements con-

nected to four nodes, as shown to the right. All trusses have cross

sectional area A and elastic modulus E . The length of each truss

element is evident by the figure. A point force, P, is acting on node4. Calculate the displacements at the nodes and the reaction forces

at nodes 1 and 2, respectively. Show also that global equilibrium is

satisfied in the vertical direction.

2.6 The plane frame structure to the right contains two

truss elements and two spring elements. The spring con-

stant for both springs is . The truss ele-

ments are of length L, have cross sectional area A andelastic modulus E . The structure is subjected to a point

force P according to the figure. The displacement will

for the present structure always be in the direction of the

force P. Determine the relation between δ and P.

2.7 The plane structure in the figure to the right contains

two truss elements and two spring elements. The truss

elements have the same length L, cross sectional area Aand elastic modulus E . The stiffness of the left spring is

. A point force P is applied on the struc-

ture, acting at an angle ϕ, as shown in the figure. Deter-

mine the stiffness of the right spring, k 2, such that the

displacement always is aligned with the force P, i.e. in the

direction of the angle ϕ .

2.8 A mass m0 is attached to three similar springs. The

division between the springs is 120o and the spring con-

stants are k 1 = k 2 = k 3 = k . The springs are attached to arigid ring of radius R. The coordinate system shown in the

figure is fixed to the ring, where the y-axis is located in

the direction of spring k 1. The circumferential position of

the ring is determined by the angle ϕ. Calculate the dis-

placement of m0 in the x- and y-directions for an arbitrary

angle. The acceleration of gravity g (see the figure) is

assumed to be known.

Hint: derive the equation system with reference to the given xy-coordinate system.

P

L

L L/ 2

2

1

4

3

k

k

E A , L

EA, L φ

P

α

π 2--- φ –

k η EA L ⁄ =

P

30o

k 1 k 2

30o

ϕ E , A

, LE , A

, L k 1 0.75 EA L ⁄ =

x

y

ϕ

g

m0k 3

rigidring

k 2

k 1




– 2.3 (12) –

2.9 An adjustable crane consists of two rods which are connected at node

3, see the figure to the right. The elastic modulus of the rods is E . Rod e1

has a cross sectional area A and a length l. Rod e2 is composed of two

cylindrical tubes to facilitate adjustment of its length by use of a hydrau-

lic actuator, where its length L is given by the angle ϕ . The effective cross

sectional area of rod e2 is . The relations andare valid here. Determine the displacement of node 3 and the normal

force acting in rod e2 (the force acting on the hydraulic actuator), when

the crane is loaded by a mass m for the case . The acceleration

of gravity is known and denoted g.

Hint: the normal force can be determined by use the reaction forces act-

ing on node 2.

2.10 A structure of three truss elements is loaded by two point

forces (P and 2P), see the figure to the right. The elastic modu-

lus, the cross sectional area and the length of each truss areshown in the figure. Analyse the structure by use of a matrix

formulated method and determine the reaction forces at all the

nodes.

.

2.11 The truss structure to the right contains two truss elements and

one spring element, with spring constant . The structure

is loaded by a point force P according to the figure. The truss ele-ments are of length L, have cross sectional area A and elastic modulus

E . Determine the displacements at the nodes where the elements are

connected. Evaluate also all reaction forces.

The examples below are taken from “Exempelsamling i Hållfasthetslära” , Ed. P.-L. Larsson & R.

Lundell, KTH, Stockholm, january 2001.

2.12 Determine the displacements and the reaction

forces at the nodes.

Node x/L y/L

1

2

3 0 1

4 0 0

m

θ

ϕ l

e1

e2

1

2

3

g

3 A θ 2ϕ = L 2l ϕ cos=

ϕ 30°=

E, A, L

2P

P

E ,

A ,

L E 2 A 2 L,,

45ok

EA

EA, L

P L

45o

90o

k 2 EA L ⁄ =

P

3k

2k

k

x

y

1

2

3

4

3 2 ⁄ – 1 2 ⁄ –

3 2 ⁄ – 1 2 ⁄




– 2.4 (12) –

2.13 The planar truss structure to the right con-

sists of four spring elements, each of length L and

with spring constant k . All springs are oriented

with a 45o angle with respect to the horizontal

plane. Determine all displacements and reaction

forces at the nodes.

2.14 Determine the displacements and the reac-

tion forces at the nodes, and the normal forces in

the springs.

2.15Determine the displacements and the reaction forces

at the nodes.

2.16 Determine the displacements and the reactionforces at the nodes, and the normal forces in the

spring elements. The stiffness and length of each

spring is shown in the figure to the right.

Node x/L y/L

1 -1 1

2 1 0

3 0 0

Node x/L y/L

1 0 2

2 1 2

3 1 1

4 0 0

P

1 2

3

4

x

y

5

P

k

1

2

3

y x k

stelQ

k

2---

k

2-------

k

1 2

3

4

k

2-------

x

y

stel

Q

3 L, 2k

4 L, k

5 L, 5k

1

23

x

y




– 2.5 (12) –

Solutions

2.1 The equation system: . Solution:

The reaction forces are obtained from Eqs. (1) and (4) as: .

2.2

2.3

2k k – k – 0

k – 3k k – k –

k – k – 3k k –

0 k – k – 2k

0

D2

D3

0

R1

P

0

R4

=

D2

D3

P

8k ------3

1=

R1 R4 P 2 ⁄ –= =

5k

k

2k

D1

D4

D3

D2

D6

D5

k

5---

16 12 16– 12– 0 0

12 19 12– 9– 0 10–

16– 12– 21 12 5– 0

12– 9– 12 9 0 0

0 0 5– 0 5 0

0 10– 0 0 0 10

0

D2

D3

0

0

0

R1

P–

0

R4

R5

R6

=

Equation-

system

Boundary conditions & prescribed forces:

Solution: D2

D3

P

17k --------- 7–

4–= Reaction forces: R4 3P 17 ⁄ = R1 4P 17 ⁄ –=

R5 4P 17 ⁄ = R6 14P 17 ⁄ =

D1 = D4 = D5 = D6 = 0 and F 2 = −P, F 3 = 0

D8

D7

D6

D5

D4

D3 D2

D1

x

ye1

e3

e2 x

x1

2

45o

Element 1:

Element 3:

ke k ir r–

r– r

=Element stiffness matrix:

k 1 k = r1 0

0 0=,

k 2 2k = r1 2 ⁄ 1 2 ⁄

1 2 ⁄

1 2 ⁄

=,

Assembly:

K k1 k2 k3+ + k

1 0 1– 0

0 0 0 0

1 1– 1– 1

1– 1 1 1–

1 1 1– 1–

1 1 1– 1–

1– 0 1– 1 1– 1– 3 0

0 0 1 1– 1– 1– 0 2

= =

D1 D2 0= = D3 D4 0= =

D5 D6 0= = D8 0=

B.C.:

0

0

00

0

0 F

R1

R2

R3

R4

R5

R6

P

R8

=

Reactions- forces

Equation (7) gives

3kD7 P=

D7⇒

P

3k ------=

x1

2

−45oElement 2: k 3 2k = r1 2 ⁄ 1– 2 ⁄

1– 2 ⁄ 1 2 ⁄ =,




– 2.6 (12) –

2.4

Boundary conditions:

e1& e4:

Assembly of global stiffness matrix:

D2

Element stiffness matrices:

Ke k ia a–

a– a

a, c2

sc

sc s2

c φcos=

s φsin=,= =

e2: k 2 2k a, φ 45°={ }1

2--- 1 1

1 1= = = e3: k 3 4k a, φ 90°={ } 0 0

0 1= = =

Eqs. (8) & (9):

D1

D4

D3

D10

D9

D6

D5

D8

D7

e2

e3

e1

e4

F 8 = −P, F 9 = −P

D1 = D2 = D3 = D4 = D5 = D6 = D7 = D10 = 0

k 1 k 4 2k a, φ 45– °={ }1

2--- 1 1–

1– 1= == =

u1

u2

Tde1

2------- 1 1– 0 0

0 0 1 1–

D9

D10

D7

D8

1

2-------– D9

D8

= = =

Normal force in element 4: N k 4δ=

k 4 2k = δ u2 u1–=where

N ⇒

3 2

7----------P=

k 5 1

1 3

D8

D9

P–

P–

D8

D9

⇒ P

14k ---------–

2

4= =

K k

1 1– 1– 1

1– 1 1 1–

1 1 1– 1–

1 1 1– 1–

0 0 0 0

0 4 0 4–

0 0 1 1– 1– 1

0 4– 1– 5 1 1–

1– 1 1– 1– 1– 1 3 1–

1 1– 1– 1– 1 1– 1– 3

=

0

0

0

0

0

0

00

0 0

0

0

and




– 2.7 (12) –

2.5

2.6


e1:

Assembly of stiffness:

D2

Element stiffness matrices: Ke k ia a–

a– a

a, c2

sc

sc s2

c φcos=

s φsin=,= =

e3: k 3 EA

L------- a, φ 45°={ }

1

2--- 1 1

1 1= = =

e2: k 22 EA

L----------- a, φ 90 °={ } 0 0

0 1= = =

K EA

2 L-------

1 1– 1– 1

1– 1 1 1–0 0 0 0

0 4 0 4–

1– 1 0 0 2 0 1– 1–

1 1– 0 4– 0 6 1– 1–

1– 1– 1 1

1– 1– 1 1

=

Eqs. (6) & (8) (reduced system of equations):

D1 D4

D3

D6

D5

8

D7

D1 = D2 = D3 = D4 = D5 = D7 = 0; F 6 = 0; F 8 = −P

k 1 EA

L------- a, φ 45– °={ }

1

2--- 1 1–

1– 1= ==

0

0

0

0

0

0

Reaction forces in node 1 & 2 ( D.O.F.:s 1 - 4):

EA2 L------- 6 1–

1– 1 D6

D8

0P–

D6

D8

⇒ PL5 EA-----------– 2

12= =

e1e2

e3

R2

EA

2 L------- D– 6( )

P

5---= =

R3 0= R4

4 EA

2 L----------- D– 6( )

4P

5-------= =

R1

EA

2 L------- D6

P

5---–= =

Global equilibrium in vertical dir.: R2 R4 F 6 F 8+ + +P

5---

4P

5------- 0 P–+ + 0= = OK!

φ

2

4

3

51π

2--- φ –

1 21

2

1

2

1

2

e1

e2

e4

e3

Boundary conditions: D1x= D1y= D2x= D2y= D3x= D3y= D4x= D4y=0

The element stiffness contribution to node 5:

EA

L------- 1 0

0 0;

η EA

L------- c

2sc

sc s2

c φ cos=

s φ sin=

e1:

e4:

e3:

EA

L------- 0 0

0 1;e2:

π2--- φ –

–

cos φ sin=

π2--- φ –

–

sin φ cos–=

η EA

L------- s

2

s– cs– c c

2⇒

Assembly:K

EA

L------- 1 η c

2s

2+( )+ η sc sc–( )

η sc sc–( ) 1 η c2

s2

+( )+

1 η +( ) EA

L------- 1 0

0 1= =

(only node 5)

Eq. system: 1 η +( ) EA

L------- 1 0

0 1

D5 x

D5 y

P α cos

α sin=

D5 x

D5 y

⇒

PL

1 η +( ) EA------------------------- α cos

α sin=

Thus: α

δ

D5 x

D5 y

δ ⇒

PL

1 η +( ) EA-------------------------=




– 2.8 (12) –

2.7

2.8

Boundary Condistions:

e1:

Assembly Ki and implementation of B.C. gives the reduced Equation system:

D2

Element sttiffness matrix:

Ke k ia a–

a– a

a, c2 sc

sc s2

c φcos=

s φsin=,= =

e4: k 2 k 0 a, φ 120°={ }1

4--- 1 3–

3– 3

= = =

D1

D4

D3

D6

D5

D9

D1 = D2 = D3 = D4 = D5 = D6 = 0; F 7 = Pcosϕ ; F 8 = Psinϕ

k 13

4---k 0 a, φ 0°={ } 1 0

0 0= ==

D9

D10

⇒ D0ϕ cos

ϕ sin=

e2

P

ϕ

e4

e1

e3

e3: k 3 k 0 a, φ 60°={ }1

4--- 1 3

3 3

= = =

e2: k 2 a φ 0°={ } 1 0

0 0= =,

where k 0 EA

L-------=

D10

1.25k 0 k 2+ 0

0 1.5k 0

D9

D10

P ϕ cos

ϕ sin=

But D should aligned

with the external force

Eq. 9: 1.25k 0 k 2+ P D0 ⁄ =

Eq. 10: 1.5k 0 P D0 ⁄ =

1.25k 0 k 2+⇒ 1.5k 0 k 2⇒ 0.25k 0 EA

4 L-------= = =

Boundary Conditions:

Assembly of the stiffness matrix:

D2

Element stiffness matrix: Ki k iai ai–

ai– ai

=

Eq. (7) & (9):

D1

D4

D3

D6

D5

D8 D7

e2 e3

e1

D1 = D2 = D3 = D4 = D5 = D6 = 0

3

2---k 1 0

0 1

D7

D8

mg ϕsin

ϕcos–

D7

D8

⇒

2mg

3k ----------- ϕsin

ϕcos–= =

F 7 mg ϕ F 8 mg ϕcos–=,sin–=

mg

ϕ

a10 0

0 1a2

1

4--- 3 3

3 1

a3

1

4--- 3 3–

3– 1

=,=,= x

y

where

K k

a1 0 0 a1–

0 a2 0 a2–

0 0 a3 a3–

a– 1 a2– a3– a1 a2 a3+ +

=




– 2.9 (12) –

2.9

2.10

Boundary Cond.:

Assembly of stiffness matrix: D2


Ki

ai

ai–

ai– ai

=

D1

D4

D3

D6

D5

D1 = D2 = D3 = D4 = 0,

EA

2l------- 2 3

3 2

D5

D6

mg0

1–

D5

D6

⇒

2mgl

EA------------- 3

2–= =

F 5 0 F 6, mg–==

a1 EA4l------- 3 3

3 1a2 EA

4l------- 1 3

3 3=,=where

K

a2 0 a2–

0 a1 a1–

a2– a1– a1 a2+

EA

4l-------

1 3 0 0 1– 3–

3 3 0 0 3– 3–

0 0 3 3 3– 3–

0 0 3 1 3– 1–

1– 3– 3– 3– 4 2 3

3– 3– 3– 1– 2 3 4

= =

60o

30o

e1

e2

mg

Eq. (5) & (6):

x

y

R1

EA

4l------- 1 D5– 3 D6–( )

3

2-------mg R2

EA

4l------- 3 D5– 3 D6–( )

3

2---mg= =,= =

Reaction forcesat node 1:

N 2

R1

R2

Equilibrium in y-dir. gives: R2 N 2 30cos+ 0 N 2⇒ 3mg–= =

2

31 1 2

1

2

1e1

e2 e3

B.C.: D1 = D2 = D3 = D6 = 0, F 4 = 2P, F 5 = −P

e1:

Assembly of

D6

D5

D4

D3

D2

2Element stiffness matrices: Ke k i

a a–

a– a

a, c2

sc

sc s2

c φcos=

s φsin=,= =

k 1 EA

L------- a, φ 0°={ } 1 0

0 0= = =

e2: k 2 EA

L------- a, φ 90°={ } 0 0

0 1= = = e3: k 2

EA

L------- a, φ 135°={ }

1

2--- 1 1–

1– 1= = =

K EA

2 L-------

2 0 0 0 2– 0

0 2 0 2– 0 0

0 0 1 1– 1– 1

0 2– 1– 3 1 1–

2– 0 1– 1 3 1–

0 0 1 1– 1– 1

=

Eqs. (4) & (5):

EA

2 L------- 3 1

1 3

D4

D5

2P

p–

D4

D5

⇒

PL

4 EA----------- 7

5–= =

Reaction forces:

(1): R1

EA

2 L------- 2 D5–( )

5P

4-------= =

(2): R2

EA

2 L

------- 2 D4–( )7P–

4

----------= =

(3): R3 EA

2 L------- D4– D5–( )

P

4--- (Node 2)–= =

(6): R6

EA

2 L

------- D4– D5–( )P

4

--- (Node 3)–= =

(Node 1)

D1

stiffness:




– 2.10 (12) –

2.11

2.12

Boundary conditions: D2 = D3 = D5 = D6 = 0,

Assembly:

Element stiffness Ke k i a a–a– a

a, c2

sc

sc s2

c φcos=s φsin=

,= =

e1: k 1 EA

L------- a, φ 45– °={ }

1

2--- 1 1–

1– 1= = =

e3: k 32 EA

L----------- a, φ 0°={ } 1 0

0 0= = =

K

EA

2 L-------

2 0 1– 1– 1– 1

0 2 1– 1– 1 1–

1– 1– 5 1 4– 01– 1– 1 1 0 0

1– 1 4– 0 5 1–

1 1– 0 0 1– 1

=

Eqs. (1) & (4):

EA2 L------- 2 1–

1– 1

D1

D4

P0

D1

D4

⇒

2PL EA---------- 1

1= =

Reaction forces: (2): R2 EA

2 L------- D4–( ) P– (Node 1)= =

(3): R3

EA

2 L------- D– 1 D4+( ) 0 (Node 2)= =

(5): R5

EA

2 L------- D1–( ) P (Node 3)–= =

(6): R6

EA

2 L------- D1( ) P (Node 3)= =

2

3

1

1

21 2

1

e1 e2

D6

D5

D4

D3

D2

2 D1

e3

e2: k 2 EA

L------- a, φ 45°={ }

1

2--- 1 1

1 1= = =

matrices:

F 1 = P, F 4 = 0

1

2

3

4 5

6

7

8 Kk

4---

3 3 0 0 0 0 3– 3–

3 1 0 0 0 0 3– 1–

0 0 6 2 3– 0 0 6– 2 3

0 0 2 3– 2 0 0 2 3 2–

0 0 0 0 0 0 0 0

0 0 0 0 0 12 0 12–

3– 3– 6– 2 3 0 0 9 3–

3– 1– 2 3 2– 0 12– 3– 15

=

D1 D2 0= =

D3 D4 0= =

D5 D6 0= =

F 7 P F 8, 0= =


D K1–F

D7

D8

⇒

P

33k ---------

15

3

P

k --- 0.4545

0.0525= = =

F 1

F 2

F 3

F 4

F 5

F 6

P

33------

12–

4 3–

21–

7 3

0

3 3–

P

0.3636–

0.2099–

0.3664–

0.3674

0

0.1575–

= =Reaction-

forces:

Equation-

system:




– 2.11 (12) –

2.13

2.14

1

2

3

4

5

6

7

8

9

10K

k

2---

1 1– 0 0 0 0 0 0 1– 1

1– 1 0 0 0 0 0 0 1 1–

0 0 1 1 0 0 0 0 1– 1–

0 0 1 1 0 0 0 0 1– 1–0 0 0 0 1 1 1– 1– 0 0

0 0 0 0 1 1 1– 1– 0 0

0 0 0 0 1– 1– 2 0 1– 1

0 0 0 0 1– 1– 0 2 1 1–

1– 1 1– 1– 0 0 1– 1 3 1–

1 1– 1– 1– 0 0 1 1– 1– 3

=

D1 D20= =

D3

D4

0= =

D5 D6 0= =

F 7 0 F 9, P F 10,– 0= = =


D8 0=

D K1–F

D7

D9

D10

⇒

P

6k

------2–

5–

1–

= =

F 1

F 2

F 3

F 4F 5

F 6

F 8

P

6---

2

2–

3

31

1

2–

=

Reaction-

forces:

Equation-

system:

1

2

3

4

5

6

Kk

2---

1 1– 0 0 1– 1

1– 1 0 0 1 1–

0 0 2 0 2– 0

0 0 0 0 0 0

1– 1 2– 0 3 1–

1 1– 0 0 1– 1

=

D1 D2 0= =

D3 D4 0= =

D6 0=

F 5 P–=


D K1–F D5

⇒

2P

3k ------- 1–= =

F 1

F 2

F 3

F 4

F 6

P

3---

1

1–

2

0

1

=

Reaction-

forces:

Equation-

system:

Element 1:

The normal force, N , in one element is given by ,

where .

f e kTDe=

N f 2=

k k 1 1–

1– 1 T;

1

2------- 1 1– 0 0

0 0 1 1– De;

D1

D2

D5

D6

f ⇒

k

2-------

D1 D2 D5– D6+–

D– 1 D2 D5 D6–+ + N

⇒

2

3-------P–= = = = =

Element 2:

k k 1 1–

1– 1 T;

1 0 0 0

0 0 1 0 De;

D5

D6

D3

D4

f ⇒ k D5 D3

–

D– 5 D3+ N ⇒

2

3---P= = = = =

1

21 2

e2

e1




– 2.12 (12) –

2.15

2.16

stel

1

2

3

4

7

8

5

6

D1 D2 0= =

D3 D4 0= =

D5

0=

F 6 F 7 0 F 8, Q–= = =


(p.g.a. stel bom)

K

k

2 2----------

1 1– 0 0 0 0 1– 1

1– 1 2+ 0 0 0 2– 1 1–

0 0 0 0 0 0 0 0

0 0 0 2 2 0 0 0 2 2–0 0 0 0 1 1 1– 1–

0 2– 0 0 1 1 2+ 1– 1–

1– 1 0 0 1– 1– 2 0

1 1– 0 2 2– 1– 1– 0 2 2 2+

=

D K1–F

D6

D7

D8

⇒

Q

k ----

6– 4 2+

3– 2 2+

5 4 2–

Q

k ----

0.3431–

0.1716–

0.6569–

= = =

F 1

F 2

F 3

F 4

F 5

Q

2-------

3 2– 4+

6 2 8–

0

8 5 2–

3 2 4–

Q

0.1716–

0.3431

0

0.6569

0.1716

= =Reaction-forces

Equation-

system:

Kk

5---

16 12 16– 12– 0 0

12 19 12– 9– 0 10–

16– 12– 21 12 5– 0

12– 9– 12 9 0 0

0 0 5– 0 5 0

0 10– 0 0 0 10

=

“rigid”1

2

3

4

5

6

D40=

D5 D6 0= =

D1 0=

F 2 Q F 3,– 0= =


(rigid support)

D K1–F

D2

D3

⇒

Q

17k --------- 7–

4–

Q

k ---- 0.4118–

0.2353–= = =

F 1

F 4

F 5

F 6

Q

85------

20–

15

20

70

Q

0.2353–

0.1765

0.2353

0.8235

= =

Reaction-

forces

Equation-

system:

Element 1: k 5k 1 1–

1– 1 T;

1

5--- 4 3 0 0

0 0 4 3 D

e;

D1

D2

D3

D4

f ⇒

k 3 D2 4 D3–

3 D– 2 4 D3+ N

⇒

5

17------Q= = = = =

e1

e2

e3 2

1

21

1

2

Element 2:

Element 3:

k k 1 1–

1– 1 T;

1

5--- 1 0 0 0

0 0 1 0 De;

D5

D6

D3

D4

f ⇒

k D3–

D3

N ⇒

4

17------– Q= = = = =

k 2k 1 1–

1– 1

T;0 1 0 0

0 0 0 1

De;

D1

D2

D5

D6

f ⇒

2k D2

D– 2

N ⇒

14

17------Q= = = = =

The normal force, N , in one element is given by ,

where .

f e kTDe=

N f 2=




– 3.1 (15) –

3. Strong/weak form and FEM-equations

3.1 The solution to a specific one-dimensional problem is governed by the differential equa-

tion (strong form)

for ,

where the primary variable φ depends on x. Also D, q and Q may depend on x. Derive theweak form and identify the essential and natural boundary conditions.

3.2 The weak form to is

,

where v( x) is an arbitrary weight function. Derive the FEM-equation for one element. Use a

linear interpolation for the primary variable and use Galerkin’s method, regarding the choiceof the weight function.

3.3 The figure to the right shows a rod with elasticmodulus E and cross sectional area A. The rod isloaded by a body force, K

x

[N/m3]. The displace-ment, u, in the rod is given by the solution to the dif-ferential equation

.

(a) Show that the weak form is ,

where σ denotes the normal stress and v is an arbitrary weight function.

(b) Derive the FEM-equation (use Galerkin’s method) to the weak form above for one ele-ment, i.e. identify the quantities in the equation

.

(c) The rod shown to the right is of length 3 L andloaded by , whereQ corresponds to the total axial force actingon the rod. Both ends of the rod are clamped.Divide the rod into two elements of lengths Land 2L respectively and determine the nodedisplacements and the reaction forces. Com-pare with the exact solution. Redo the analy-

sis with more elements!

dd x------ D

dφd x------

qφ Q+– 0= x1 x x2≤ ≤

d

d x------ D

dφd x------

qφ Q+– 0=

dvd x------ Ddφd x------d x

x1

x2

∫

vqφd x

x1

x2

∫

+ vQd x vDdφd x------ x1

x2

+

x1

x2

∫

=

x

x = 0 x = L

K x E, A

d

d x------ EA

du

d x------

AK x+ 0=

dv

d x------ EA

du

d x------d x

0

L

∫

vK x Ad x v σ A( )[ ]0 L

+

0

L

∫

=

x

K x E, A

x = 0 x = L x = 3 L

u x( )

2

9---

QL

EA--------

x

L--- 0 x L≤ ≤

1

36------

QL

EA-------- 3

x

L---– 9

x

L---

2

3 x

L---

3

–+

=

N x 0=( )2Q

9-------, N x 3 L=( )

7Q

9-------–==

Exact

soln.

kede f e=

K x Q 2 AL( ) ⁄ x L 1– ⁄ ( )=




– 3.2 (15) –

3.4 The right figure shows a uniaxial bar coupled to

a set of continues springs with spring constant perunit length k x [(N/m) / m]. The bar has elastic modu-lus E , cross sectional area A and is loaded by a bodyforce K x [N/m3]. The displacement, u, in the bar is

given by the solution to the differential equation

.

(a) Show that the weak form is

,

where σ denotes the normal stress and v is an arbitrary weight function.


.

(c) Divide the bar into two linear elements of the same length and analyse the problem.Evaluate the node displacements. Apply the boundary conditions: for x = 0 and

for x = L. Assume that E and A are constants and that the spring constant and the body force .

3.5 Figure (a) to the right shows a uniform bar loadedby its dead weight, ρ g, where ρ is the density of the barand g is the acceleration of gravity. The bar has elasticmodulus E and cross sectional area A. The displace-ment, u, is given by the solution to the differentialequation

.

(a) Assume that E, A, ρ and g are constants andshow that the weak form is

,

where σ is the normal stress in the bar and v an arbitrary weight function.


.

x

x = 0 x = Lk x

K x E, A

d

d x------ EA

du

d x------

k xu– AK x+ 0=

dv

d x------ EA

du

d x------d x vk xud x

0

L

∫

+

0

L

∫

vK x Ad x v σ A( )[ ]0 L

+

0

L

∫

=

kede f e=

u 0=σ A Q=k x 3 EA 2 L

2( ) ⁄ = K x Q AL( ) ⁄ =

(a) (b)

x = 2 L

P

x = 3 L

x = L

k

x x

x = L

g

EA ρ

d

d x------ EA

du

d x------

A ρ g+ 0=

EAdv

d x------

du

d x------d x

0

L

∫

v σ A( )[ ]0

L ρ gA vd x

0

L

∫

+=

kede f e=




– 3.3 (15) –

(c) In an application, a bar ( E, A) is connected to a linear spring with spring constant k , see

Figure (b) above. The bar is loaded by a point force applied at the x = L and by its deadweight. Divide the bar/spring structure in three elements with nodes placed in thepoints: x = 0, L, 2 L and 3 L. Thus, the bar should be divided into two equal elements.Let , where EA is constant, and calculate the displacements at the nodes.

3.6 The figure to the right shows a beam with bendingstiffness EI attached to an elastic foundation character-ized by a spring constant per unit length k z [(N/m) / m]. Adistributed load per unit length q [N/m] is applied on the

beam. The deflection of the beam w is given by the solu-tion to the differential equation

.

(a) Show that the weak form is

,

where v is an arbitrary weight function, T is a shear force and M is a moment. The rela-tions and have been utilized at the boundaries.

(b) Derive the FEM-equation (use Galerkin’s method) to the weak form above for one ele-ment, i.e. identify the quantities in the equation .

(c) Divide the beam into a two-node beam element (2D.O.F. per node) and determine . Assumethat EI is constant, and q = 0. The

boundary conditions are shown in the figure to theright.

(d) The beam shown below is subjected to a uniformly distributed load ,

where Q is the resultant of the total distributed load acting on the beam. The totallength of the beam is 2 L and its bending stiffness is EI . The left end of the beam isclamped, whereas the right end support is flexible, here modelled by a combination of atension spring with stiffness k w [N/m] and a torsion spring with stiffness k θ [Nm].Model the beam with one beam element (2 node element with 2 D.O.F. per node) andcalculate the deflection and rotation of the right end of the beam. Use the values of thespring constants shown in the figure.

k 2 EA L ⁄ =

x

x = − L k z

q

x = L

z,w

M M

T

T

d2

d x2

-------- EI d

2w

d x2

---------

k zw q–+ 0=

d2v

d x2

-------- EI d

2w

d x2

---------d x vk zwd x

L–

L

∫

+

L–

L

∫

vT [ ] L–

L dv

d x------ M

L–

L

– vqd x

L–

L

∫

+=

T EIw″ ( )′–= M EIw″ –=

kede f e=

M 0

z,w

x

x = − L x = L

w′ L( )k z λ EI L

4 ⁄ =

q Q 2 L( ) ⁄ =

Q

k w

k θ

2 L, EI

k w3 EI

2 L3

---------= k θ EI

L------=




– 3.5 (15) –

(c) In an application, the beam is clamped at x = L and

the rotation is prevented at x = − L, see the figureto the right. The beam is subjected to the triangu-lar load and a point force according to the figure.Analyse the beam by use of one 2-node beam ele-

ment and calculate the deflection w(ξ ), where ξ is natural coordinate defined as.

3.9 A one dimensional model of a cooling fin is shownto the right. The cooling fin has a cross sectional area A

[m2], length 3 L /2 and coefficient of thermal conductiv-ity k [W/m/ oC]. The convection coefficient is h [W/m2 / oC] and the perimeter of the fin is P [m]. The tempera-ture distribution in the fin T [oC] at steady state condi-tions is given by the solution to the differential equation

.

Here, q [W/m3] is a continues distributed heat source and is the ambient temperature (thelast term represents convection to the surrounding medium).

(a) Show that the weak form to the differential equation is

,

where v is an arbitrary weight function and Q is heat flow, where(Fourier’s law) has been used at the boundaries.

(b) Derive the FEM-equation (use Galerkin’s method) to the weak form above for one ele-ment, i.e. identify the quantities in the equation .

(c) Divide the cooling fin into three linear elements (two nodes per element and one tem-perature d.o.f. per nod) and determine the temperature at the nodes. The boundary con-ditions are described by at x = 0 and at . Assume that k,

A and P are constants, q = 0 and that .

(d) Change the boundary conditions in from prescribed temperature to convec-tion. Assume that the relation between the surface of the perimeter and the surface of

the end of the fin is .

3.10 The figure to the right shows a model for heat con-

duction in a one-dimensional rod, where heat exchangeby convection between the surface of the rod and thesurrounding medium is taken into consideration. Theambient temperature of the surrounding medium is .The rod has cross sectional area A [m2], perimeter P

[m], thermal conductivity k [W/m/ oC] and convectionheat transfer coefficient h [W/m2 / oC]. The temperature T [oC] in the rod as a function of posi-tion at steady state conditions is given by the solution to the differential equation

x x = − L

x = L z,w

Q

2 L, EI

P

ξ x L ⁄ =

Heat conduction

Convection

x 0= x 3 L 2 ⁄ =T x( )

x

dd x------ kAdT d x------

qA hP T T ∞–( )–+ 0=

T ∞

dv

d x------kA

dT

d x------d x vhPT d x

0

3 L 2 ⁄

∫

+

0

3 L 2 ⁄

∫

v Q–( )[ ]0

3 L 2 ⁄ v qA hPT ∞+( )d x

0

3 L 2 ⁄

∫

+=

Q k – AdT d x ⁄ =

kede f e=

T 4T ∞= T T ∞= x 3 L 2 ⁄ =

hP 12kA L2

⁄ =

x 3 L 2 ⁄ =

LP A 96 ⁄ =

Heat conduction

Convection

T x( )

x

x = x1 x = x2T ∞




– 3.6 (15) –

.

(a) Show that the weak form to the differential equation is

,

where v is an arbitrary function of x and Q is heat flow, where (Fou-riers law) has been used at the left and right boundary of the rod.

(b) Derive the FEM equation of the weak form above for one element, i.e. identify thequantities in the equation (use Galerkin’s method).

(c) Assume that the total length of the rod is L and that k, h, A and P are constants, relatedas . Divide the rod into two equal linear elements (two nodes per ele-ment with one temperature d.o.f. per node) and calculate the temperature at the nodes.The boundary conditions are described by a prescribed heat flow at

and a prescribed temperature at .

FORMULAS

d

d x------ kA

dT

d x------

hP T T ∞–( )– 0=

dv

d x------kA

dT


x1

x2

∫

+

x1

x2

∫

v Q–( )[ ] x1

x2vhPT ∞d x

x1

x2

∫

+=

Q k – AdT d x ⁄ =

keTe f e=

hPL 16kA L ⁄ =

Q hPLT ∞=

x 0= T T ∞= x L=

2 L, EI

0 1ξ

−1

d 3

d 4d 2

d 1

w ξ( ) N 1d 1 N 2d 2 N 3d 3 N 4d 4+ + + Nde B,d

2N

d x2

----------1

L2

-----d

2N

dξ2

----------= = = =Beam element:

BT Bd x

L–

L

∫

1

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– 2 L2

3– 3 L– 3 3 L–

3 L 2 L2

3 L– 4 L2

= NT

Nd x

L–

L

∫

L

105---------

78 22 L 27 13 L–

22 L 8 L2

13 L 6 L2

–

27 13 L 78 22 L–

13 L– 6 L2

– 22 L– 8 L2

=

N 1 2 3ξ– ξ3

+( ) 4 ⁄ N 2 L 1 ξ– ξ2

– ξ3

+( ) 4 ⁄ =,=

N 3 2 3ξ ξ3

–+( ) 4 N 4 L 1– ξ– ξ2

+ ξ3

+( ) 4 ⁄ =, ⁄ =

Deflection:

φ ξ( ) N 1φ1 N 2φ2+ N 1 N 2

φ1

φ2

= = N 1 1 ξ–= N 2 ξ=

1 2φ 1 φ 2

L

0 1

ξ

1D:

NT Nd x

0

L

∫

d x Ldξ={ } L6--- 2 1

1 2= =

N

dN

T

d x---------- Nd x------d x

0

L

∫

1 L--- 1 1–1– 1

=

NT 1

2--- 1

x

L---+

dx

L------

L–

L

∫

1

30------

9

4 L

21

6 L–

=




– 3.7 (15) –

Solutions

3.1 (i) Weighted residual: , where v is an arb. weight fcn.

(ii) Integration by parts gives

3.2 Approximation function: , ,

, where l is the element length

Weight fcn. (Galerkin’s method): where (arbitrary)

Inserted into the weak form gives

3.3 (a) See solution to 3.1 and 3.2

vd

d x------ D

dφd x------

qφ Q+–

d x

x1

x2

∫

0=

vd

d x------ D

dφd x------

d x

x1

x2

∫

vDdφd x------

x1

x2 dv

d x------ D

dφd x------d x

x1

x2

∫

–=

dv

d x------ D

dφd x------d x

x1

x2

∫

vqφd x

x1

x2

∫

+ vQd x vDdφd x------

x1

x2

+

x1

x2

∫

= φ φ0 eller Ddφd x------ D

dφd x------

0

på x xi= = =

Essential B.C.Natural B.C.

Weak form:

φ ξ( ) Nφe= N 1 ξ– ξ= φe

φ1

φ2

=

dφd x------

dφdξ------

dξd x------

1

l---

dN

dξ-------φe Bφe= = =⇒

v ξ( ) Nβ βT N

T = = β

T β1 β2

=

dv

d x------

dv

dξ------

dξd x------ β

T 1

l---

dNT

dξ---------- β

T B

T = = =⇒

βT

BT DBldξ

0

1

∫

NT qNldξ

0

1

∫

+ φe βT

NT Qldξ N

T D

dφd x------

0

1

+

0

1

∫

=

kq f Q f s

keφe⇒ f e= f e f Q f s+=därβT is an arbitrary vector

The element matrices becomes:

k D BT DBldξ

0

1

∫

1l--- 1–

1 D1

l--- 1– 1 ldξ

0

1

∫

1l--- 1 1–

1– 1 Ddξ

0

1

∫

= = =

kq NT qNldξ

0

1

∫

1 ξ–

ξq 1 ξ– ξ ldξ

0

1

∫

l1 ξ–( )

21 ξ–( )ξ

1 ξ–( )ξ ξ2

qdξ

0

1

∫

= = =

k D

ke k D kq+= och

f Q NT Qldξ

0

1

∫

l1 ξ–

ξQdξ

0

1

∫

= =




– 3.8 (15) –

3.3 (b) , where and .

3.3 (c)

3.4(a)

3.4(b) Displacement interpolation:

Weight function:

3.4(c)

kede f e= ke B0

L

∫

T

EAB xd= f e N0

L

∫

T

K x A x NT

σ A( )[ ]0

L+d=

D3 D2 D1

EA2 L-------

2 2– 0

2– 3 1–

0 1– 1

0

D2

0

R1

Q 3 ⁄

R3 2Q 3 ⁄ +

=Eqn. system:

D2

2

9---

QL

EA--------=

R1 29---Q–=

R37

9---Q–=

⇒

Node/element division:

Weighted residual: vd

d x------ EA

du

d x------

k xu– AK x+ d x

0

L

∫

0=

Integration by parts: vd

d x------ EA

du

d x------

d x

0

L

∫

v EAdu

d x------

0

L dv

d x------ EA

du

d x------d x

0

L

∫

–=

dv

d x------ EA

du

d x------d x

0

L

∫

vk xud x

0

L

∫

+ vAK xd x

0

L

∫

v EAdu

d x------

0

L

+=

(2) inserted into (1)

(1)

(2)

gives the weak form:

u Nde=du

d x------

dN

d x-------de Bde= =

v Nbe be

T N

T = =

dv

d x------ be

T dNT

d x---------- be

T B

T = =

be

T

B

T

EAB x N

T

k xN xd Le

∫ +d Le

∫ de be

T

N

T

K x A x N

T

σ A( )[ ]0

Le

+d Le

∫ =

ke f e

but be

T is arbitrary kede⇒ f e=

D2 D1 D3

Le = L /2 Le = L /2N 1 ξ–( ) ξ= B

2

L--- 1– 1=

Element-

NT Q

AL------- A

L

2---dξ

0

1

∫

Q

4---- 1

1=

BT EAB

L

2

---dξ

0

1

∫

2 EA

L

----------- 1 1–

1– 1

=

NT 3 EA

2 L2

-----------N L

2---dξ

0

1

∫

EA

8 L------- 2 1

1 2=

ke

EA

8 L------- 18 15–

15– 18;=⇒

Assembly: EA

8 L-------

18 15– 0

15– 36 15–

0 15– 18

D1

D2

D3

Q

4----

1

2

1

R

0

Q

+=point force

reaction force

Equation (2) & (3):

D2

D3

⇒

QL

141 EA-----------------74

140

QL

EA--------0.525

0.993= =

matrices:




– 3.9 (15) –

3.5(a)


Weight function:

3.5(c)


d x------ EA

du

d x------

Aρg+ d x

0

L

∫

0=


d x------ EAdu

d x------

d x0

L

∫ v EAdu

d x------

0

L dv

d x------ EAdu

d x------d x0

L

∫ –=

with

(1)

(2)

gives the weak form EA dv

d x------

du

d x------d x

0

L

∫

v σ A( )[ ]0

L Aρg vd x

0

L

∫

+=σ E du

d x------=

(2) inserted into (1)

u Nde=du

d x------

dN

d x-------de Bde= =

v Nbe be

T N

T = =

dv

d x------ be

T dN

T

d x---------- be

T B

T = =

ke f ebut be


be

T EA B

T Bd x

0

L

∫

de be

T N

T σ A( )[ ]

0

L Aρg N

T d x

0

L

∫

+=

Inserted into theweak form gives:

Element matrices:

ke EA BT B Ldξ

0

1

∫

EA

L------- 1 1–

1– 1= = f e Aρg N

T Ldξ

0

1

∫

AρgL

2-------------- 1

1= =

ke k 1 1–

1– 1= where k 2

EA

L-------=

Truss element:

Spring element:

D1 D2 D3 D4

Element lengths: L1 = L2 = L3 = L L3 L2 L1

Boundary conditions: D1 = D4 = 0

Diskretization:

Assembly: K

EA

L-------

1 1– 0 0

1– 2 1– 0

0 1– 3 2–

0 0 2– 2

= F

R1

P

0

R4

AρgL

2--------------

1

2

1

0

+=

(reaction forces: R1 & R4)

EA

L------- 2 1–

1– 3

D2

D3

P AρgL+

AρgL 2 ⁄

D2

D3

⇒

PL

5 EA----------- 3

1

AρgL2

10 E ---------------- 7

4+= =Eqs. (2) & (3):




– 3.10 (15) –

3.6(a)


Weight function:

3.6(c)

Weighted residual: v EIw″ ( )″ k zw q–+[ ]d x

L–

L

∫

0=

Integration by parts: v EIw″ ( )″ [ ]d x

L–

L

∫

v EIw″ ( )′[ ] L–

L v′ EIw″ ( )′[ ]d x

L–

L

∫

–= =

(2) inserted into (1) with

(1)

(2)v EIw″ ( )′[ ] L–

Lv′ EIw″ ( )[ ]

L–

L– v″ EIw″ d x

L–

L

∫

+=

gives the weak form:

v″ EIw″ d x vk zwd x

L–

L

∫

+

L–

L

∫

vT [ ] L–

Lv′ M [ ]

L–

L– vqd x

L–

L

∫

+=

T EIw″ ( )′–= and M EIw″ –=

w Nde=d2w

d x2

---------d2N

d x2

----------de Bde= =

v Nbe be

T N

T = =

d2v

d x2

-------- be

T d

2N

T

d x2

------------ be

T B

T = =

be

T B

T EI B

L–

L

∫

x NT

L–

L

∫

k zN xd+d de be

T N

T

L–

L

∫

q x NT T [ ]

L–

L dNT

d x---------- M

L–

L

–+d=

ke f ebut be


K BT EI Bd x N

T λ EI

L4

---------Nd x

L–

L

∫

+

L–

L

∫

EI

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– L2

3– 3 L– 3 3 L–

3 L L2

3 L– 4 L2

λ EI

105 L3

---------------

78 22 L 27 13 L–

22 L 8 L2

13 L 6 L2

–

27 13 L 78 22 L–

13 L– 6 L2

– 22 L– 8 L2

+= =

F

R1

R2

R3

M 0

=

Reaction

forces/moments

Stiffness

matrix

Load

vector

2 EI

L---------

8λ EI

105 L-------------

+

d 4 M 0=

w′ L( ) d 4105

210 8λ +( )--------------------------

M 0 L

EI -----------= =⇒

Eq. (4):




– 3.11 (15) –

3.6(d)

3.7(a)


Weight function:

Element

Kbeam BT EI Bd x

L–

L

∫

EI

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– 2 L2

3– 3 L– 3 3 L–

3 L 2 L2 3 L– 4 L2

= =

Beam (dof 1 to 4) Tensile & torsion springs

The total stiffness matrix is obtained by assembly

K EI

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– 2 L2

3– 3 L– 6 3 L–

3 L 2 L2

3 L– 6 L2

=of the stiffnesses from the beam and the springs:

FbQ

2 L------ N

T d x

L–

L

∫

Q

2----

1

L 3 ⁄

1

L– 3 ⁄

= = Fs

R1

R2

00

=

Force vector: F Fb Fs+=

where

Reaction

force & moment

External pointforce & moment

EI

2 L3

---------6 3 L–

3 L– 6 L2

w2

θ2

Q

2---- 1

L– 3 ⁄

w2

θ2

⇒

QL3

27 EI ------------ 5 L

1= =

Displacement boundary conditions: w1 = θ 1 = 0. The reduced equation system becomes

stiffness

matrices

(only dof 3 and 4)

k w3 EI

2 L3

---------= k θ EI

L

------=

Weighted residual: v EIw″ ( )″ ρgA+[ ]d x L–

L

∫

0=


L–

L

∫

v EAw″ ( )′[ ] L–

Lv′ EIw″ ( )′[ ]d x

L–

L

∫

–= =

(1)

(2)v EAw″ ( )′[ ] L–

Lv′ EAw″ ( )[ ]

L–


L–

L

∫

+=

v″ EIw″ d x

L–

L

∫

vT [ ] L–

Lv′ M [ ]

L–

L– ρgA vd x

L–

L

∫

–=

(2) inserted into (1) with gives the weak form:T EIw″ ( )′–= and M EIw″ –=

w Nde=d

2w

d x2

---------d

2N

d x2

----------de Bde= =

v Nbe be

T N

T = =

d2v

d x2

-------- be

T d

2N

T

d x2

------------ be

T B

T = =

be

T B

T EI B

L–

L

∫

xd de be

T N

T T [ ]

L–

L dNT

d x---------- M

L–

L

– ρgA NT

xd

L–

L

∫

–=

ke f e

but be

T

is arbitrary

kede⇒ f e=




– 3.12 (15) –

3.7(c)

3.8 (a)

3.8 (b) Displacement interpolation:

Weight function:

Inserted into the weak form gives

Element

Kbalk BT EI Bd x

L–

L

∫

EI

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– 2 L2

3– 3 L– 3 3 L–

3 L 2 L2 3 L– 4 L2

= = Kfjäder k EI

2 L3

---------= =

Beam (d.o.f 1 to 4) Spring (only d.o.f. 3)

Assembly of total stiffness matrix

K Kbalk Kfjäder+ EI

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– 2 L2

3– 3 L– 4 3 L–

3 L 2 L2

3 L– 4 L2

= =(the spring only contributes to d.o.f. 3):

Fdistributed ρgA NT d x

L–

L

∫

– ρgAL

1

L 3 ⁄

1

L– 3 ⁄

–= = Fpoint

R1

R2

P–

0

=

Force vector: F Fdistributed Fpoint+=

where

Reaktion

force/moment

External point

force/momentP ρgAL= inserted gives:

F

R1 ρgAL–

R2 ρgAL2

3 ⁄ –

2ρgAL–

ρgAL2

3 ⁄

=Reduced equation system (Eq. (3) & (4)):

EI

2 L3

---------4 3 L–

3 L– 4 L2

w2

θ2

ρgAL2–

L 3 ⁄

w2

θ2

⇒ ρgAL

3

EI ---------------- 2

4 3 ⁄ –= =

Displacement boundary conditions: w1 = θ 1 = 0

stiffness

matrices:

Weighted residual: v EIw″ ( )″ 1 x L---+

Q2 L------+ d x

L–

L

∫

0=


L–

L

∫

v EAw″ ( )′[ ] L–

Lv′ EIw″ ( )′[ ]d x

L–

L

∫

–= =

(1)

(2)v EAw″ ( )′[ ] L–

Lv′ EAw″ ( )[ ]

L–


L–

L

∫

+=

v″ EIw″ d x

L–

L

∫

vT [ ] L–

Lv′ M [ ]

L–

L– Q v

1

2--- 1

x

L---+

d x

L------

L–

L

∫

–=

(2) inserted into (1) with gives the weak form:T EIw″ ( )′–= and M EIw″ –=

w Nde=d

2w

d x2

---------d

2N

d x2

----------de Bde= =

v Nbe be

T N

T = =

d2v

d x2

-------- be

T d

2N

T

d x2

------------ be

T B

T = =

be

T B

T EI B

L–

L

∫

xd de be

T N

T T [ ]

L–

L dNT

d x---------- M

L–

L

– Q NT 1

2--- 1

x

L---+

d x

L------

L–

L

∫

–=

ke f e

but be

T

is arbitrary

kede⇒ f e=




– 3.13 (15) –

3.8 (c)

3.9(a)

3.9 (b) Temperature interpolation:

Weight function:

Inserted into the weak form gives:

Element stiffness matrix: K BT EI Bd x

L–

L

∫

EI

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– 2 L2

3– 3 L– 3 3 L–

3 L 2 L2

3 L– 4 L2

= =

Fb Q NT 1

2--- 1

x

L---+

d x

L------

L–

L

∫

–Q

30------

9

4 L

21

6 L–

–= = Fpoint

P–

R2

R3

R4

=

Nodal force vector: F Fb Fpoint+=

where Reaction

force/moment

External point force

w ξ ( ) N 1 ξ ( )d 12

3---

PL3

EI ---------

1

5---

QL3

EI ----------+

2 3ξ– ξ3

+

4---------------------------

where ξ,– x

L---= = =

Reduced equation system, Eq. (1): EI

2 L3

---------3d 1 P–3Q

10-------– d 1⇒

2

3---

PL3

EI ---------–

1

5---

QL3

EI ----------–= =

Displacement boundary conditions: d 2 = d 3 = d 4 = 0 => Reaction forces

The deflection of the beam is obtained by the displacement interpolation (approx.) as:

FEM, discretization:2 L, EI

0 1

ξ

−1

d 3

d 4d 2

d 1

(one element)


d x------ kA

dT

d x------

qA hP T T ∞–( )–+ d x

0

3 L 2 ⁄

∫

0=


d x------ kA

dT

d x------

d x

0

3 L 2 ⁄

∫

v kAdT

d x------

0

3 L 2 ⁄ dv

d x------ kA

dT

d x------

d x

0

3 L 2 ⁄

∫

–=


(1)

dv

d x------kA

dT


0

3 L 2 ⁄

∫

+

0

3 L 2 ⁄

∫

v Q–( )[ ]0

3 L 2 ⁄ v qA hpT ∞+( )d x

0

3 L 2 ⁄

∫

+=

Q– kAdT

d x------= gives the weak form:

(2)

T NTe=dT

d x------

dN

d x-------Te BTe= =

v Nbe be

T N

T = =

dv

d x------ be

T dN

T

d x---------- be

T B

T = =

be

T B

T kAB

x1

x2

∫

x NT

x1

x2

∫

hPN xd+d Te be

T N

T

x1

x2

∫

qA hpT ∞+( ) x NT

Q–( )[ ] x1

x2

+d=

ke f ebut be

T is arbitrary keTe⇒ f e=




– 3.14 (15) –

3.9 (c)

3.9 (d)

3.10 (a)

KkA

L------

4 1– 0 0

1– 8 1– 0

0 1– 8 1–

0 0 1– 4

=

For one element applies:

ke

1

L 2 ⁄ ---------- 1–

1kA

1

L 2 ⁄ ---------- 1– 1

L

2---dξ 1 ξ–

ξhP 1 ξ– ξ

L

2---dξ

0

1

∫

+

0

1

∫

hPL 12kA

L------=

kA

L------ 4 1–

1– 4= = =

Assembly of system matrix gives

e3e2e1 4321

T 4T 3T 2T 1

FEM - model:

f e1 ξ–

ξqA hPT ∞+( )

L

2---dξ

0

1

∫

q 0 hPL 12kA

L------=;=

3kA

L------T ∞

1

1= = =

F 3kA

L------T ∞

1

2

2

1

Q R1

0

0

Q R4–

+=and the r.h.s

Equation system:

kA

L------

4 1– 0 0

1– 8 1– 0

0 1– 8 1–

0 0 1– 4

4T ∞

T 2

T 3

T ∞

3kA

L------T

∞

1

2

2

1

Q R1

0

0

Q R4–

+=

4 1– 0 0

1– 8 1– 0

0 1– 8 1–

0 0 1– 4

4

T 2 T ∞ ⁄

T 3 T ∞ ⁄

1

⇔

3

6

6

3

Q R1 L kAT ∞( ) ⁄

0

0

Q R4 L– kAT ∞( ) ⁄

+=

Reduced Eq. system (2) & (3): “Reaction flux”

8 1–

1– 8

T 2 T ∞ ⁄

T 3 T ∞ ⁄

6 1 4 0 1⋅+⋅–{ }–

6 0 4 1 1⋅–⋅{ }–

T 2

T 3

T ∞

21------ 29

22=⇒ T ∞

1.3809

1.0476= =

Convection at x=3 L /2 gives for element 3: NT

Q–( )[ ]ξ1

ξ2

c0

Q2–

Q1–

0–=

Consider only the contribution from element node 2, since the contribution from

NT

Q–( )[ ]ξ1

ξ2 kA

8 L------ 0 0

0 1

T 1

T 2

kA

8 L------

0

T ∞

+=⇒

Q2 hA T 2 T ∞–( )=

where hA h PL 96 ⁄ ( ) kA 8 L( ) ⁄ ·

= =

The equation system is modified according to:

4 1– 0 0

1– 8 1– 0

0 1– 8 1–

0 0 1– 41

8---+

4

T 2 T ∞ ⁄

T 3 T ∞ ⁄

T 4 T ∞ ⁄

3

6

6

31

8---+

Q R1 L kAT ∞( ) ⁄

0

0

0

T 2

T 3

T 4

⇒ + T ∞

1.3811

1.0491

1.0119

= =

the element node 1 is cancelled by the contribution from element node 2 in element 2,

furthermore use that


d x------ kA

dT

d x------

hP T T ∞–( )– d x

x1

x2

∫

0=


d x------ kA

dT

d x------

d x

x1

x2

∫

v kAdT

d x------

x1

x2 dv

d x------ kA

dT

d x------

d x

x1

x2

∫

–=


(1)

(2)

dv

d x------kA

dT

d x------d x vhPT d x x1

x2

∫ + x1

x2

∫ v Q–( )[ ] x1

x2

vhpT ∞d x x1

x2

∫ +=

Q– kAdT

d x------= gives the weak form:




– 3.15 (15) –

3.10 (b) Temperature interpolation:

Weight function:

Inserted into the weak form gives:

3.10 (c)

T NTe=dT

d x------

dN

d x-------Te BTe= =

v Nbe be

T N

T = =

dv

d x------ be

T dN

T

d x---------- be

T B

T = =

be

T B

T kAB

x1

x2

∫

x NT

x1

x2

∫

hPN xd+d Te be

T N

T

x1

x2

∫

hpT ∞ x NT

Q–( )[ ] x1

x2

+d=

ke f e

but be

T is arbitrary keTe⇒ f e=

K2

3---

kA

L------

7 1– 0

1– 14 1–

0 1– 7

=

For one element applies:

ke

1

L 2 ⁄ ---------- 1–

1kA

1

L 2 ⁄ ---------- 1– 1

L

2---dξ 1 ξ–

ξhP 1 ξ– ξ

L

2---dξ

0

1

∫

+

0

1

∫

hPL 16kA

L------=

2kA

3 L---------- 7 1–

1– 7= = =

Assembly of the stiffness matrix gives:

e2e1 321

T 3T 2T 1FEM - model:

f e1 ξ–

ξhPT ∞

L

2---dξ

0

1

∫

hPLT ∞

4----------------- 1

1= =

Equation system:

2

3---

kA

L------

7 1– 0

1– 14 1–

0 1– 7

T 1

T 2

T 3 T ∞=

kA

L------

20

8

44Q R

hPLT ∞-----------------–

T ∞=7 1– 0

1– 14 1–

0 1– 7

T 1

T 2

T ∞

30

12

66Q R

hPLT ∞-----------------–

T ∞=⇔

Eq. (1) & (2) gives:

System matrix:

7 1–

1– 14

T 1

T 2

30 0( )–

12 1–( )–T ∞

T 1

T 2

⇒

1

97------ 433

121T ∞

4.46

1.25T ∞≈= =

and r.h.s. FhPLT ∞

4-----------------

1

2

1

hPLT ∞

1

0

Q– R

hPLT ∞-----------------

+= =

F⇒

hPLT ∞

4-----------------

5

2

1Q R

hPLT ∞-----------------–

kA

L------

hPL

kA L ⁄ --------------

T ∞

4------

kA

L------

20

8

44Q R

hPLT ∞-----------------–

T ∞= = =

= 16




– 4.1 (6) –

4. FEM: trusses and beams

4.1 A uniaxial bar is modelled by a linear truss element.

For a certain applied load, the node displacement

shown in the figure results. (a) Show that the strain

developing in the element is and (b) show

that if ε0 = 0, the element is subjected to a rigid body

motion equal to δ0.

4.2 Derive the four shape functions for

the uniaxial “cubic” element shown to

the right. Express the shape functions

using the natural coordinate ξ .

4.3 The figure to the right shows a uniaxial isoparamet-

ric element where a 2nd degree polynomial is used for

the interpolation of the displacement. The coordinates of

the nodes can be seen in the figure, where λ is a non-

dimensional parameter in the interval: .

Assume that the node displacements {u1, u2, u3} are

known and calculate the strain in the element.

Hint: express the strain as a function of the natural coordinate ξ , see below.

4.4 The bar in the figure to the right is subjected to an

uniformly distributed axial load K x = q0 and a point force

P. Analyse the bar by use of the finite element method

with (a) one linear element and (b) two linear element.

Compare the solutions with the exact solution given

below the figure.

x = 0 x = L x

u1 δ0= u2 δ0 ε0 L+=ε x( ) ε0=

Nod 1 Nod 3 Nod 4 Nod 2

ξ = −1 ξ = 1ξ = −1/3 ξ = 1/3ξ

x L− L λ L

Nod 1 Nod 3 Nod 2

1 λ 1< <–

N 1 1 ξ–( )ξ– 2 ⁄ =1 23

Coordinates:

x 1 x 2 x 3

x

1 23

−1 10ξ x ξ( ) N k xk

k 1=

3

∑

=

Primary variable:

φ ξ( ) N k φ

k

k 1=

3

∑

= N 2 1 ξ+( )ξ 2 ⁄ =

N 3 1 ξ2

–=

x = 0 x = L E, A

K x q0=P

Exact solution:

u x( )Px

EA-------

q0 L2

E -----------

x

L---

1

2---

x

L---

2

–

+=

σ x( )P

A--- q0 L 1 x L ⁄ –( )+=




– 4.2 (6) –

4.5 Carry out a finite element analysis of the uniaxial bar problem shown in figure (a) below.

Divide the bar in two linear elements of the same length. A linear element with shape func-

tions is shown in figure (b) below.

4.6 One requirement the displacement

interpolation in an element must satisfy,

is that it should be able to model an arbi-

trary rigid body motion. For a plane 2-

node beam element with 2 degrees of freedom at each node, this means that

the deflection of the beam must be able

to take the form

,

where δ and θ are parameters describing an arbitrary rigid body motion as illustrated in the

figure to the right. Show that the displacement interpolation of the element can satisfy a rigid

body motion as described above.

4.7 The figure to the left

shows an initially straightbeam element that is sub-

jected to a deformation state

that results in a constant cur-

vature , where R0

is the radius of curvature.

Curvature is here defined as

(small deformations is assumed). The displacements of the two nodes of the ele-

ment are shown in the figure. (a) Calculate the curvature and (b) find the slope (angle θ0)

and the displacement δ 0 at the midpoint of the element ( x = 0).

4.8 A cantilever beam is loaded by a point force P, a

moment M and a uniformly distributed force per unit

length (Q is the total resultant

force) according to the figure to the right. The bend-

ing stiffness of the beam is EI and its length 2 L. Ana-

lyse the beam with FEM and use a 2-node beam

element (3rd degree polynomial for the interpolation

of the deflection). Carry out the analysis using (a)

one element and (b) two elements. Compare the

results with the exact solution shown below the fig-ure.

x = 0 x = 2 L x3

2--- L=

E, A

K x q0 N

m

3-------=

1 2 L

ξ 0 1

N 1 1 ξ N 2,– ξ= = d x Ldξ=

u1 u2

(a) (b)

θ 2

w2

w1θ1

θ

δ

x

− L 0 L

θ 1 θ 2 θ = =

w1 δ θ L–=

w2 δ θ L+=

w x( ) δ θ x+=

θ 2

w2

w1

θ1

x

− L 0 L

R0

1

κ 0-----=

w1 a b– c–=

θ1 b L 2c L ⁄ + ⁄ =

w2 a b c–+=

θ2 b L 2c L ⁄ – ⁄ =

δ 0θ 0

κ 0 1 R0 ⁄ =

w″ κ –=

κ 0

P

M

q x( )

x x = 2 L

[N/m]

w x( )PL

3

6 EI --------- 6

x

L---

2 x

L---

3

–

ML2

2 EI -----------

x

L---

2

+=

QL3

EI ----------

1

2---

x

L---

2 1

6---

x

L---

3

–

1

48------

x

L---

4

+

+

Exact solution:

q x( ) Q 2 L( ) ⁄ =




– 4.3 (6) –

4.9 The figure below shows a cantilever beam, which is subjected to a distributed force, a

moment and a point force. The bending stiffness of the beam is EI . The beam is during a FEM-

analysis modelled by two 2-nodes beam element. The coordinates of the three nodes used in

the FEM-model are: . Determine the force vector, where also the reaction

forces should be indicated.

FORMULAS

x 0 2 L 3 L,,{ }=

x = 0 x = 2 L

P

x

z

x = 3 L

q x( )Q

2 L------

x

L---

N

m----⋅=

M

2 L, EI

0 1ξ

−1

d 3

d 4d 2

d 1

w ξ( ) N 1d 1 N 2d 2 N 3d 3 N 4d 4+ + + Nde B,d

2N

d x2

----------1

L2

-----d

2N

dξ2

----------= = = =Balkelement:

BT

Bd x

L–

L

∫

1

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– 2 L2

3– 3 L– 3 3 L–

3 L 2 L2 3 L– 4 L2

= NT Nd x

L–

L

∫

L

105---------

78 22 L 27 13 L–

22 L 8 L2

13 L 6 L2

–

27 13 L 78 22 L–

13 L– 6 L2– 22 L– 8 L2

=

N 1 2 3ξ– ξ3

+( ) 4 ⁄ N 2 L 1 ξ– ξ2

– ξ3

+( ) 4 ⁄ =,=

N 3 2 3ξ ξ3

–+( ) 4 N 4 L 1– ξ– ξ2

+ ξ3

+( ) 4 ⁄ =, ⁄ =

Deflection:

φ ξ( ) N 1φ1 N 2φ2+ N 1 N 2

φ1

φ2

= = N 1 1 ξ–= N 2 ξ=

1 2φ 1 φ 2

L

0 1ξ

1D:

NT Nd x

0

L

∫

d x Ldξ={ } L

6--- 2 1

1 2= =

N

dNT

d x----------

N

d x------d x

0

L

∫

1

L--- 1 1–

1– 1=

NT dξ

1–

1

∫

1

L 3 ⁄

1

L 3 ⁄ –

= NT ξdξ

1–

1

∫

1

15------

6–

L–

6

L–

=




– 4.4 (6) –

Solutions

4.1

4.2 Use Lagrange interpolation:

4.3

4.4

u N 1u1 N 2u2+ δ0 ε0ξ L+ x ξ L={ } δ0 ε0 x+= = = =

ε x( )

du

d x------ ε0= =

Displacement in the element:

(a) Strain in the element:

(b) The case ε0 = 0 results in the rigid body motion since u x( ) δ0=

N 1 1 ξ–( ) 1 3ξ–( ) 1 3ξ+( ) C 1 ⁄ = N 1 1–( ) 1= C 1⇒ 16–=

N 2 1 ξ+( ) 1 3ξ–( ) 1 3ξ+( ) C 2 ⁄ = N 2 1( ) 1= C 2⇒ 16–=

N 3 1 ξ+( ) 1 ξ–( ) 1 3ξ–( ) C 3 ⁄ = N 3 1– 3 ⁄ ( ) 1= C 3⇒ 16 9 ⁄ =

N 4 1 ξ+( ) 1 ξ–( ) 1 3ξ+( ) C 4 ⁄ = N 4 1 3 ⁄ ( ) 1= C 4⇒ 16 9 ⁄ =

εdu

d x------

du

dξ------

dξd x------= =

dx∂ N k

∂ξ--------- xk dξ

k 1=

3

∑

L 1 2λξ–( )dξ= =

ε⇒

u2 u1–

2 L----------------

u1 u2 2u3–+( )

L-------------------------------------ξ+

1

1 2λξ–( )-----------------------=

Strain:

Note! singular for 2λξ 1=

(a) One element solution, discretization:

ke

EA

L------- 1 1–

1– 1= f b N

T K x ALdξ

0

1

∫

ALq01 ξ–

ξdξ

0

1

∫

ALq0

2------------- 1

1= = =

Uniform load contribution to the nodal force vector:

Eq. EA

L------- 1 1–

1– 1

D1 0=

D2

R

P

ALq0

2------------- 1

1+=

Element stiffness matrix:

D2 D1

D3

D2

D1

e1 e2

D2

PL

EA-------

q0 L2

2 E -----------+=

R P– ALq0–=

Eq. 2 gives:system:

Eq. 1 then gives:

(b) Two element solution, discretization:


k1 k2

EA

L------- 1 1–

1– 1= = f b1 f b2 N

T K x A

L

2---dξ

0

1

∫

AL

2-------q0

1 ξ–

ξdξ

0

1

∫

ALq0

4------------- 1

1= = = =

Uniform load contribution to the nodal force vector:

Eq.2 EA

L-----------

1 1– 0

1– 2 1–

0 1– 1

D1 0=

D2

D3

R

0

P

ALq0

4-------------

1

2

1

+= D2

D3

PL

2 EA----------- 1

2

q0 L2

8 E ----------- 3

4+=

R P– ALq0–=

Eq. 2 & 3:syst.:

Eq. 1 then gives:

Reaction force

Note! The point force solution is exact and independent of the number of element used,

whereas the distributed load solution is approximate. The forces acting at the nodes

Boundary conditions

are in global equilibrium, i.e. external loads are in balance with internal (reaction) forces.




– 4.6 (6) –

4.8 cont.

4.9

D5

D6

Eq.

EI

2 L3

---------

3 3 L 3– 3 L

3 L 4 L2

3 L– 2 L2

3– 3 L– 3 3 L–

3 L 2 L2

3 L– 4 L2

D1 0=

D2 0=

D3

D4

R

M R

P

M

Q

6----

3

L

3

L–

+=

R P– Q–=

Eq. 3 & 4 give:

system:

Eq. 1 & 2 then give the reaction forces: M R 2PL– M – QL–=

(b) Discretization, two elements:

D1 D3

D2 D4

Element stiffness matrices: Distributed load contributionto the nodal force vector:

Eq.system:

k1 k2 BT EI Bd x

L– 2 ⁄

L 2 ⁄

∫

4 EI

L3

---------

3 3 L 2 ⁄ 3– 3 L

3 L 2 ⁄ L2

3 L 2 ⁄ – L2

2 ⁄

3– 3 L 2 ⁄ – 3 3 L 2 ⁄ –

3 L 2 ⁄ L2

2 ⁄ 3 L 2 ⁄ – L2

= = =f b1 f b2 N

T Q

2 L------

L

2---dξ

1–

1

∫

Q

24------

6

L

6

L–

= = =

D3

D4

D5

D6

⇒

PL3

6 EI ---------

5

9 L ⁄

16

12 L ⁄

ML2

2 EI -----------

1

2 L ⁄

4

4 L ⁄

QL3

48 EI ------------

17

28 L ⁄

48

32 L ⁄

+ +=

4 EI

L3

---------

3 3 L 2 ⁄ 3– 3 L 0 0

3 L 2 ⁄ L2

3 L 2 ⁄ – L2

2 ⁄ 0 0

3– 3 L 2 ⁄ – 6 0 3– 3 L 2 ⁄

3 L 2 ⁄ L2

2 ⁄ 0 2 L2

3 L 2 ⁄ – L2

2 ⁄

0 0 3– 3 L 2 ⁄ – 3 3 L 2 ⁄ –

0 0 3 L 2 ⁄ L2

2 ⁄ 3 L 2 ⁄ – L2

D1 0=

D2 0=

D3

D4

D5

D6

R

M R

0

0

P

M

Q

24------

6

L

12

0

6

L–

+=

D3

D4

PL3

EI --------- 8 3 ⁄

2 L ⁄

ML2

EI ----------- 2

2 L ⁄

QL3

3 EI ---------- 3

1 L ⁄ + +=

Eq. 3 - 6 give: Eq. 1 & 2 then give the reaction

R P– Q–=

M R 2PL– M – QL–=

forces:

Note! The solutions for P and M are exact independent of the number of beam elementsused, whereas the distributed load solution is approximate. Also note that the forces

acting at the nodes are in global equilibrium, i.e. external loads are in balance

with internal (reaction) forces.

D1 D3

D2 D4

D5

D6

e2e1B.C. & kinematical constraint: D1 = D2 = D3 = 0

(give reaction forces/moments: R1, R2 & R3)

Element 1: x L 1 ξ+( ) d x⇒ Ldξ ,= = f b NT qLdξ where q

1–

1

∫

Q

2 L------ 1 ξ+( )= =

f b1

2 3ξ– ξ3

+( )4

--------------------------------Q

2 L------ 1 ξ+( ) Ldξ

1–

1

∫

3Q

10-------= = f b2

2QL

15-----------= f b3

7Q

10-------= f b4

QL

5--------–=

FT

R1

3Q

10

-------+ R2

2QL

15

-----------+ R3

7Q

10

-------+ M QL

5

--------– P– 0=⇒




– 5.1 (6) –

5. FEM: planar frames of trusses and beams

5.1 The figure to the right shows a structure with three linear

elastic truss elements with elastic modulus E . Element 1 and 3

have cross sectional area A and length L. Element 2 has cross

sectional area and length . The structure is sub-

jected to a point force Q and a force per unit volume (body

force) according to the figure. Calculate the

displacement at node 2, all reaction forces and the normal

stress in each one of the truss elements.

5.2 A truss structure containing three trusses, all with elastic

modulus E and cross sectional area A, is shown to the right. The

structure is loaded by one point force Q /2 and a body force of total magnitude Q acting on the vertical truss member down-

wards. Model the structure by use of three linear elements and

calculate the displacements and possible reaction forces at the

nodes. Note, the displacements at the nodes will in the current

case agree with the exact solution. Will the numerical solution

deviate from the exact one? If so, how?

5.3 Analyse the cantilever beam to the right by use of FEM.

Use a 2-node element, which allows for both axial deforma-

tion and development of curvature (bending). The elastic mod-ulus of the beam is E and the cross section is shown in the

figure. Note that with the load applied in the present case, the

FEM solution will agree with the exact solution. Especially,

evaluate the solution for the case .

5.4 A force per unit length q( x) is applied on a beam

with elastic modulus E and a cross sectional area A

and a moment of inertia I . The left end of the beam

is clamped and the right end rests on an elastic sup-port, here modelled by a vertical spring with spring

constant .

(a) Carry out a finite element analysis, where

the beam is modelled by one two-node element, and evaluate the deflection of the

beam. Here: and .

(b) Divide the beam into two element of equal to length and redo the analysis.

Note that the deflection at the nodes will in the current case always coincide with the exact

solution. The deflection between the nodes for will deviate somewhat from the

exact solution due to distributed load q(x).

x/LK x

Q

3 2

1

y/L

(0,1) (1,1)

(1,0)

2 A 2 L

K x Q AL( ) ⁄ =

Q 2 ⁄

Q2 L

L

L

45o

2P

h

h

h L ⁄ 1 10 ⁄ =

q(x)

x

y

k x = − 2 L

x=2 L

L

P

k η EI L3

⁄ =

η 3 2 ⁄ = q x( ) Q 2 L( ) x L ⁄ ( ) ⁄ –=

0 x 2 L≤ ≤




– 5.2 (6) –

5.5 Analyse the linear elastic planar frame work shown to the

right by use of FEM. Use a 2-node combined truss/beam ele-

ment allowing for both axial and bending deformation. The

cross section of the frame is displayed in the figure and the elas-

tic modulus is E . Will the FEM solution agree with the exact

solution, i.e. with the Euler-Bernoulli beam theory, in the presentcase?

5.6 The figure below shows a circular ring, which is an integral part of a flexible machine

member. The ring is subjected to diametrically opposed forces according to the figure. Deter-

mine the spring constant defined as by use of FEM. If the symmetry of the problem

is fully utilized, only a quarter of the ring needs to be modelled. The problem can for instance

be analysed by the Matlab based FEM program “frame2D”, available at the home page of the

course. If the displacement, δ, primarily is due to bending deformation (a good approximation

if ), the spring constant of the ring can analytically be expressed as

,

where E is the elastic modulus and I area moment of inertia. Note that in order for the FEM

solution to come close to this result, the FEM model requires that .

h

h

P

L

L

k P δ ⁄ =

R h»

k 4π

π2

8–( )-------------------

EI

R3

------=

R h»

P

R

x

y

R

δ 2 ⁄

δ 2 ⁄

P

R

η h

P 2 ⁄ δ 2 ⁄

“symmetric quarter”




– 5.3 (6) –

FORMULAS

Frames of truss/spring members (based on 2-node elements):

Frames of beam members (based on 2-node elements):

1

2

D4

D3

D2

D1

φ k

x

y

ac

2sc

sc s2

=

Ke k a a–

a– a

=

al12

2l12m12

l12m12 m12

2=

c φ cos=

s φ sin=

l12 φ xcos x2 x1–( ) L ⁄ = =

m12 φ ycos y2 y1–( ) L ⁄ = =

L x2 x1–( )2

y2 y1–( )2

+=

where

alternativly

Global forulation for truss & spring elements

spring constant for a truss element k EA

L-------=

θ 1y, M 1y

w1, f 1 z

w2, f 2 z

θ 2y, M 2y

u2, f 2 x

x z

u1, f 1 xke

EA

2 L------- 0 0

EA

2 L-------– 0 0

03 EI

2 L3

---------3 EI

2 L2

--------- 03 EI

2 L3

---------–3 EI

2 L2

---------–

03 EI

2 L2

---------2 EI

L--------- 0

3 EI

2 L2

---------– EI

L------

EA

2 L

-------– 0 0 EA

2 L

------- 0 0

03 EI

2 L3

---------–3 EI

2 L2

---------– 03 EI

2 L3

---------3 EI

2 L2

---------–

03 EI

2 L2

---------– EI

L------ 0

3 EI

2 L2

---------–2 EI

L---------

=

2 L

Local formulation

Global formulation

(local coordinate system)

(global coordinate system)

D1, F 1

x z

D2, F 2 D4, F 4 D6, F 6

D5, F 5

D3, F

3 X

Z

{ X 1, Z 1}

{ X 2, Z 2}

de TDe=

f e kede=

f e⇒ keTDe=

Fe TT f e=

Fe⇒ TT keT De Ke De= =

de

T u1 w1 θ 1 y u2 w2 θ 2 y

=

f e

T f 1 x f 1 z M 1 y f 2 x f 2 z M 2 y=

Fe

T F 1 F 2 F 3 F 4 F 5 F 6=

De

T D1 D2 D3 D4 D5 D6

=

Transformation scheme:

TT2 0

0 T2

där T2

l x m x 0

l z

m z

0

0 0 1

= =

Transformation matrix:

l z φ zX cos Z 2 Z 1–

2a-----------------– ϕ sin–= = =

l x φ xX cos X 2 X 1–

2a------------------ ϕ cos= = =

m z φ zZ cos X 2 X 1–

2a------------------ ϕ cos= = =

m x φ xZ cos Z 2 Z 1–

2a----------------- ϕ sin= = =

“Direction

cosines”

l x2

l z2

+ 1=

m x

2m z

2+ 1=

⇒

Fe TT f e= T

l12 m12 0 0

0 0 l12 m12

=där

Local/global transformation of nodal force vector:




– 5.4 (6) –

Solutions

5.1

5.2

D8

D7


k2

EA

2 L-------

1 1 1– 1–

1 1 1– 1–

1– 1– 1 1

1– 1– 1 1

=k1

EA

L-------

1 0 1– 0

0 0 0 0

1– 0 1 0

0 0 0 0

= k3

EA

L-------

0 0 0 0

0 1 0 1–

0 0 0 0

0 1– 0 1

=

Distributed load contribution to nodal force vector (el. 1):

3 2

1

D4

D3

D6

D5

D2

D1

f b1 NK x ALd x

0

L

∫

Q

2---- 1

1–= =

Fb1 TT f b1

Q

2----

1

0

1

0

–= = T1 0 0 0

0 0 1 0=Transformation to global coordinate system gives:

EA

2 L-------

2 0 2– 0

0 0 0 01 1 1– 1–

1 1 1– 1–

0 0 0 0

0 2 0 2–

2– 0 1– 1– 0 0 3 1

0 0 1– 1– 0 2– 1 3

D1

D2

D3

D4

D5

D6

D7

D8

R1 Q 2 ⁄ –

R2

R3

R4

R5

R6

Q 2 ⁄ –

Q–

=

0

00

0

00

Eq.

system:Bound. cond.: D1= D2= D3= D4= D5= D6=0

Give rise to the reaction forces R1 t.o.m. R6

Eq. 7 and 8 give: D7

D8

QL

8 EA----------- 1–

5–=

Eq. 1 - 6 then give: R1

5Q

8------- R2, 0 R3,

3Q

8-------= = =

R4

3Q

8------- R5, 0 R6,

5Q

8-------= = =

The normal stress in one element: σ = E Bde = E BTDe

σ1 E

1

L---–

1

L---

1 0 0 0

0 0 1 0

D7

D8

D1

D2

Q

8 A-------= =

σ2 E 1

2 L----------–

1

2 L----------

1

2-------

1

2------- 0 0

0 01

2-------

1

2-------

D7

D8

D3

D4

3Q

8 A-------= =

σ3 E 1

L

---–1

L

---0 1 0 0

0 0 0 1

D7

D8

D5

D6

5Q

8 A

-------= =

with

EA

L-------

11

2 2----------+

1

2 2----------– 1– 0

1

2 2----------–

1

2 2----------

1

2 2----------–

1

2 2---------- 0 0

1

2 2----------

1

2 2----------–

1– 0 1 0 0 0

0 0 0 1 0 1–

1

2 2----------–

1

2 2---------- 0 0

1

2 2----------

1

2 2----------–

1

2 2----------

1

2 2----------– 0 1–

1

2 2----------– 1

1

2 2----------+

D1

0

0

0

0

D6

Q

R2

Q 2 R3+ ⁄

R4

R5

0

=

D1

D4

D3

D2

D6

D5

Displacement B.C.: D2 = D3 = D4 = D5 = 0

Eq.system:

D1

D6

⇒

QL

2 EA----------- 3 2–

1 2–

QL

EA-------- 0.793

-0.207= = R2 R3 R4 R5

⇒ Q -0.207 -1.293 0.207 -0.207=

The numerical solution results in a linear displacement variation in vertical element,

whereas the exact solution yields a quadratic displacement variation, due to thedistributed load.




– 6.1 (24) –

6. FEM: 2 D /3 D solids

6.1 Derive the element stiffness matrix

for the CST element shown in the figure

to the right. Assume that the material is

isotropic, linear elastic with elastic mod-

ulus E and Poisson’s number ν = 0.

6.2 One way to satisfy compatibility across element

boundaries between regions of high to low order elements

is to use transition elements. A plane triangular transition

element is shown in Figure 4. The shape functions for the

vertex nodes of the element are displayed below the ele-

ment. Determine the shape function associated with node 4and show that it fulfil standard requirements put on shape

functions.

6.3 A plate containing a circular hole with radius R is mod-

elled by use of plane 8-noded bi-quadratic isoparametric

elements. The figure to the right shows one such elementlocated at the hole of the plate. The element is symmetri-

cally located with respect to the y-axis and extends one

quarter of the circumference of the hole, i.e. the straight ele-

ment sides: 2-6-3 and 1-8-4, respectively, form 45o angles

with respect to the y-axis. The nodes 1, 5 and 2 are placed at

the border of the hole. Determine the distance from the cen-

tre of the hole to the point , defined by the natural

coordinates , i.e. calcu-

late . How much does this point deviate from the geometric boundary (the radius)

of the circular hole?

The shape functions of the element are:

, ,

, ,

, , , .

N 1 1 ξ– η–=

N 2

ξc= N 3

η=

ξ x L ⁄ η y L ⁄ =,= x/L

y/L

1 2

3

1

1

1.0

1.0

ξ

η

2

3

41

N 1 1 ξ 3 2 ξ η+( )–( )– η–=

N 2 ξ 2 ξ η+( ) 1–( )=

N 3 η=

1 2

34

5 6

7

8

x

y

x0 y0,{ }

ξ ξ0 1 2 ⁄ η, η0 1–= = = =

x02 y0

2+( )

N 11

4--- 1 ξ–( ) 1 η–( ) 1 ξ η+ +( )–= N 2

1

4--- 1 ξ+( ) 1 η–( ) 1 ξ– η+( )–=

N 31

4--- 1 ξ+( ) 1 η+( ) 1 ξ– η–( )–= N 4

1

4--- 1 ξ–( ) 1 η+( ) 1 ξ η–+( )–=

N 51

2--- 1 ξ

2–( ) 1 η–( )= N 6

1

2--- 1 ξ+( ) 1 η

2–( )= N 7

1

2--- 1 ξ

2–( ) 1 η+( )= N 8

1

2--- 1 ξ–( ) 1 η

2–( )=




– 6.2 (24) –

6.4 The Figure to the right shows a plane bi-linear isoparametric

4-node element. The order of node numbering is opposite the

one used in the natural coordinate system (ξ, η). As a conse-

quence, the determinant of the Jacobi matrix becomes negative.

The element stiffness matrix, which is calculated by area inte-

gration in the natural coordinate system, will then also becomenegative. A FEM-analysis with such an element will “crash”.

Calculate the determinant of the Jacobi matrix for the element

with the erroneous node numbering to the right.

6.5 The figure to the right shows a three dimensional

20 node element of serendipity type. Assume that the

shape functions associated with the nodes 1 to 19 ( N 1,

..., N 19) are known functions of the natural coordi-

nates: ξ, η, ζ . The element is shaped as a cube in the

natural coordinate system, defined in the interval −1 to

1 for each coordinate. Determine the shape functionassociated with node 20, i.e. N 20(ξ,η,ζ ).

6.6 Figure (a) to the right shows a plane

(2D) isoparametric element with its shape

functions given in figure (b) to the right.

(a) Determine the Jacobi matrix of the

element, i.e

(b) Determine the sub-matrix B1 in the

B-matrix of the element

.

6.7 Model the plate (thickness h) shown to the right by use of a

4-noded bi-linear isoparametric element. Determine

(a) the coordinate transformation and ,

(b) the Jacobi matrix J and its determinant ,

(c) the strain vector ε (assume that the displacement vector de

is known),

(d) an expression for the element stiffness matrix ke.

6.8 The plate in the above problem is loaded by a pressure p0

(uniform traction) acting on the side 1-4 and by its dead weight

(density ρ). The dead weight can be modelled as a force per unit

volume (body force) . Let and determine

the contributions to the nodal force vector from

(a) the pressure p0 and (b) the force per unit volume K y.

L

x

y

L

1

2

3

4

− L

− L

1

2

3

4 5

67

8

9

10

11

1213

1415

16

17

18

19

20

ξ

ζ

η

x

y

1 2

3

a

4

a + 3l

b + l

b

(a)

(b)

ξ

1 2

34η

1

1

−1

−1

N 1 1 ξ–( ) 1 η–( ) 4 ⁄ =

N 2 1 ξ+( ) 1 η–( ) 4 ⁄ =

N 3 1 ξ+( ) 1 η+( ) 4 ⁄ =

N 4 1 ξ–( ) 1 η+( ) 4 ⁄ =

J

∂ x ∂ξ ⁄ ∂ y ∂ξ ⁄

∂ x ∂η ⁄ ∂ y ∂η ⁄ =

B B1 B2 B3 B4=

λ λ

λ λ

x/L

y/L

1

1

−1

−1

4 3

21

ρ

p0 g

x ξ η,( ) y ξ η,( )

J

K y ρg–= λ 1 2 ⁄ =




– 6.3 (24) –

6.9 Calculate the contribution from the force per unit volume K y to the nodal force vector in

the above problem by use of numerical integration based on Gauss-Legendre quadrature. Use:

(a) and (b) point integration scheme in the element.

6.10 A traction vector t (force per unit surface) is acting between points A and B located onthe edge of a plate of thickness h. The segment between A and B is straight and of length 2 L.

Consider a linear variation of the traction vector according to

,

where s is a natural coordinate, tA and tB are the traction vectors at the points A and B, respec-

tively, see the figure below. Determine the contribution to the total nodal force vector if the

plate is modelled by

(a) one isoparametric 4-node quadrilateral element,

(b) one isoparametric 8-node quadrilateral element, where the mid nodes are placed in the

middle between their corresponding corner nodes.

Assume that the traction vectors along AB are composed of a constant normal stress σ0 and a

constant shear stress τ0, such that

.

Use the results in (a) and (b) to evaluate the contribution to the total nodal force vector along

AB if the boundary is modelled by(c) three equal isoparametric 4-node quadrilateral elements and

(d) three equal isoparametric 8-node quadrilateral elements, see the figure below.

6.11 A bi-linear rectangular element is

loaded by its dead weight (gives rise to a

body force), see the figure to the right.

Determine the nodal force vector.

Assume that the acceleration of gravity,g, is known.

1 1× 2 2×

t1

2--- 1

s

L---–

tA

1

2--- 1

s

L---+

tB+=

tA tB

t x

t y

σ0 θ τ0 θsin–cos

σ0 θ τ0 θcos+sin= = =

ta

tb

A

B

θ

s

L

− L

0

A

B

A

B

A

B

1

2

34

5

6

7

8

1

2

34

1

2

3

4

5

67

(a) (b) (c) (d)

x

yA

B

123

4

x

y

a

b3

1 2

4

thickness hBody force

K x = 0

K y = −ρ g

g




– 6.4 (24) –

6.12 A triangular 2D domain is modelled by one plane triangular CST element according to

the figure below. The bottom side of the triangle is rigidly supported and the left side ( x = 0) is

subjected to a linearly varying pressure, p( y), described by the traction vector .

The material is isotropic, linear elastic with elastic modulus E and Poisson’s ratio .

Determine the node displacements and the reaction forces. The shape functions of the element

and the element stiffness matrix are given in the figure below.

6.13 A CST element for 2D linear elastic analysis is shown

to the right. The material is isotropic, linear elastic with

elastic modulus E and Poisson’s ratio ν = 1/3.

(a) Show that the shape functions of the element is:, where

and .

(b) Calculate the stresses in the element. The displace-

ment vector of the element, , is given in the figure,

where ε0 is a reference strain.

6.14 Consider a thin quadratic sheet metal of size

and thickness h of a linear elastic material ( E ,

ν). Model the sheet metal by use of two linear tri-

angular elements (CST-element) and carry out

FEM analyses for the three different load cases

(a), (b) and (c). Introduce appropriate displace-

ment boundary conditions, where symmetry con-

ditions can be utilized, and determine node

displacements and stresses in the elements. For

simplicity, let Poisson’s ratio be .

Load cases: (a) uniaxial tension,

(b) pure shear,(c) dead weight, where ρ is the density and g acceleration of gravity.

tT

p y( ) 0,[ ]=

ν 1 3 ⁄ =

x/L

y/L

3

21

(0,1)

(1,0)

p y( ) =

p0 1 y

L---–

N 1 1 ξ– η N 2 ξ N 3 η=,=,–=

K e3 Eh

16----------

4 2 3– 1– 1– 1–

2 4 1– 1– 1– 3–

3– 1– 3 0 0 1

1– 1– 0 1 1 0

1– 1– 0 1 1 0

1– 3– 1 0 0 3

=

Shape functions:

där ξ x L ⁄ = η y L ⁄ =


de

T d 1 x d 1 y d 2 x d 2 y d 3 x d 3 y=

Node displacement vector:

12

3

x/L

y/L

1/2

1

de

T 0 0

1

4---

1–

108--------- 0

1–

6------ Lε0=

N 1 1 ξ– η N 2;– 2ξ N 3; η ξ–= = =

ξ x L ⁄ = η y L ⁄ =

de

x/l

y/l

e2

e1

12

34

σ 0 σ 0τ 0

τ 0

ρ g

(a) (b) (c)

l l×

ν 0=




– 6.5 (24) –

6.15 A rectangular sheet metal of thickness h is subjected to an uniaxial load corresponding to

a normal stress σ 0. An exact analysis can for this case be carried out using a plane CST ele-

ment according to the FEM model shown in the figure below. Here, the uniaxial load is

applied as a traction vector, t, acting on the element side (edge) between node 2 and node 3.

The material is isotropic, linearly elastic with elastic modulus E and Poisson’s ratio .

The shape functions of the element and the element stiffness matrix Ke (plane stress) are givenin the figure.

(a) Calculate/evaluate the nodal force vector F, where also the reaction forces should be

marked. Use the coordinate s (see the figure) when calculating the consistent nodal

forces. Note that at node 2 and at node 3, which give the relations

and .

(b) Calculate the node displacements, de, and the reaction forces.

(c) Calculate the strains in the element and show that they agree with the exact solution, i.e.

, and .

6.16 A rectangular sheet metal of thickness h is subjected to an uniaxial stress σ 0. An exact

analysis of the problem can for instance be carried out by use of only one plane bi-linear 4-

node quadrilateral element as in the FEM model shown in the figure below. The position of the

element nodes are and . The uniaxial load is introduced in the model asa traction vector applied on the element side (edge) between node 2 and node 3. The material

is isotropic, linearly elastic with elastic modulus E and Poisson’s ratio .

(a) Introduce the displacement vector and define the dis-

placement boundary conditions.

(b) Calculate/evaluate the nodal force vector Fe. Mark the reaction forces accord-

ing to f 1 x = R1 x etc. (the shape functions of the element, isoparametric for-

mulation, are given in the figure below).

(c) Calculate all the reaction forces and also check that the nodal forces due to the

traction t agrees with the answer in (b) above. The element stiffness matrix

Ke and the resulting node displacement vector De are given below.

ν 1 3 ⁄ =

s 0= s 2 L=

x L s 2 ⁄ –= y s 2 ⁄ =

ε x νσ– 0 E ⁄ = ε y σ0 E ⁄ = γ xy 0=

x

y

3

21

N 1 1 x

L---–

y

L--- N 2

x

L--- N 3

y

L---=,=,–=

Ke

3 Eh

16----------

4 2 3– 1– 1– 1–

2 4 1– 1– 1– 3–

3– 1– 3 0 0 1

1– 1– 0 1 1 0

1– 1– 0 1 1 0

1– 3– 1 0 0 3

=

Shape functions:


de

T d 1 x d 1 y d 2 x d 2 y d 3 x d 3 y=

Node displacement vector:

L

L

σ 0

σ 0

t1

2-------

0

σ0

=

s

x L ⁄ 1±= y L ⁄ 1±=

ν 1 3 ⁄ =

De

T d 1 x d 1 y … d 4 y=




– 6.6 (24) –

6.17 Figure (a) to the right shows a plane

rectangular plate of thickness h, subjected

to a pure bending moment. Utilizing the

symmetry in the problem, only half of the

plate needs to be modelled. A crude FEM

model consisting only of one 4-node (bi-lin-

ear) quadrilateral element is shown in figure

(b). The coordinates of the nodes are evi-

dent from the figure. The moment isreplaced by an equivalent traction vector, t,

prescribed on the boundary x = 2 L, as

shown in the figure. The material is iso-

tropic, linear elastic with elastic modulus E

and Poisson’s ratio ν. Here plane stress con-

ditions are assumed to prevail.

(a) Calculate/evaluate the nodal force

vector F, where also the reaction

forces should be indicated.

(b) Calculate the normal stress in the x-

direction as a function of position.

Make use of the node displacement

solution is given in figure (b).

6.18 A quadratic plate with edges of length and of thickness h is loaded by its dead

weight and rotates around its diagonal with a constant angular velocity ω , see figure (a) below.

The lower corner of the plate is mounted on a bearing. The plate has density ρ and the material

is isotropic, linear elastic with elastic modulus E and Poisson’s ratio ν. Assume further that

plane stress conditions prevail and that . A rather coarse finite element model that to

some extent utilizes the symmetry of the problem is shown in figure (b) below. The model

consists of only one triangular element with a linear interpolation for the displacements (CST

2 L

σ 0

2 L

σ 0

y/L

x/L

1 2

34

t σ01

0=

Ke

Eh

16-------

8 3 5– 0 4– 3– 1 0

3 8 0 1 3– 4– 0 5–

5– 0 8 3– 1 0 4– 3

0 1 3– 8 0 5– 3 4–

4– 3– 1 0 8 3 5– 0

3– 4– 0 5– 3 8 0 1

1 0 4– 3 5– 0 8 3–

0 5– 3 4– 0 1 3– 8

= De

σ0 L

E ---------

0

2 3 ⁄

2

2 3 ⁄

2

0

0

0

=

N 1 1 ξ–( ) 1 η–( ) 4 ⁄ =

N 2 1 ξ+( ) 1 η–( ) 4 ⁄ =

N 3 1 ξ+( ) 1 η+( ) 4 ⁄ =

N 4 1 ξ–( ) 1 η+( ) 4 ⁄ =

1 2

34

ξ

η Shape functions:

M M

2 L 2 L

2 L

y/L

x/L

(2,−1)

(2, 1)

1 2

34

t σ0 y L ⁄

0=

D

T

D1 x D1 y D2 x D2 y D3 x D3 y D4 x D4 y=

4

3---

σ0

E ------ L 0 0 1– 1– 1 1– 0 0=

(a)

(b)

2 L

ν 0=




– 6.8 (24) –

matrix of dimension 2.

6.20 A plane solid of thickness h is subjected to a lin-

early varying pressure, p(x), acting on a segment of

the boundary as shown in the figure to the right. The

material is isotropic, linear elastic with elastic modu-

lus E and Poisson’s ratio ν. The plane solid is mod-

elled by triangular CST elements. The mesh is

indicated in the figure, where also the displacement

vector D and the nodal force vector F are shown.

(a) Define the displacement boundary conditions.

(b) Calculate/evaluate the nodal force vector. Mark

the presence of possible reaction forces as F 1 x= R1 x and so on.

(c) Calculate the stresses in element e1. Assume

plane stress prevail and that the displacement

vector D is known, where the zero displace-

ment boundary conditions may be enforced.

6.21 A cantilever beam is modelled by 20 plane 4-node quadrilateral elements arranged

according to the figure below, where the global numbering of nodes and elements are indi-

cated. All elements are rectangular, of equal size and of thickness h. The beam is made of a

linear elastic, isotropic material with elastic modulus E and Poisson’s ratio ν = 0. Plane stress

conditions is assumed to be valid. The left end of the beam is welded on a wall, which in the

present model is assumed to be rigid. The beam is subjected to a shearing load acting on its

right end as illustrated by the figure. The components of the displacement vector and the force

vector is also indicated in the figure.

(a) Define the displacement boundary conditions.

(b) Calculate the external load contributions to the node force vector, i.e. the contributions

from the traction vector.

(c) Calculate the stresses at the centroid of element e1. Assume that the displacement vec-

4 L

4 L

ρ , E , ν = 0

5 4 3

6 21

y/L

x/L

(0, -1)

(0, 1)

(4, 1)

Elem.1 Elem. 2

ω (a) (b)(2, 1)

DT

D1 x D1 y D2 x … D6 x D6 y=

4

3--- ρω

2 L

3

E ---------------- 11 0 16 0 16 0 11 0 0 0 0 0=

y

x1

2

4

5

7

8

3 6

x = 0 x = L x = 2 L

y = 0

y = L

y = 2 L

e7

e6

e5

e4e3

e2

e1

p x( ) p0

x

L---=

DT D1 x D1 y D2 x D2 y D8 x D8 y=

FT

F 1 x F 1 y F 2 x F 2 y F 8 x F 8 y=




– 6.9 (24) –

tor D is known.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

e1

e2

e3

e4

e5

e6

e7

e8

e9

e10

e11

e12

e13

e14

e15

e17

e16

e18

e19

e20

x

y

τ 0

τ 0

DT

D1 x D1 y D2 x D2 y D30 x D30 y= F

T F 1 x F 1 y F 2 x F 2 y F 30 x F 30 y

=

2a

2b




– 6.10 (24) –

FORMULAS

d 3 y

d 3 x

d 1 y

d 1 x

d 2 y

d 2 x

x

y

N 11

2 Ae

--------- y2 y3–( ) x x2–( ) x3 x2–( ) y y2–( )+[ ]=

N 2

1

2 Ae

--------- y3

y1

–( ) x x3

–( ) x1

x3

–( ) y y3

–( )+[ ]=

N 31

2 Ae

--------- y1 y2–( ) x x1–( ) x2 x1–( ) y y1–( )+[ ]=

Ae

u x y,( )

v x y,( )

N 1 0 N 2 0 N 3 0

0 N 1 0 N 2 0 N 3

de Nd e= = de

d 1 x

d 1 y

d 2 x

d 2 y

d 3 x

d 3 y

=Displace-

ments:

B B1 B2 B3= Bi

∂ N i ∂ x ⁄ 0

0 ∂ N i ∂ y ⁄

∂ N i ∂ y ⁄ ∂ N i ∂ x ⁄

=

ε xx

ε yy

γ xy

B de=

Strains:

σ xx

σ yy

σ xy

E

1 ν2

–( )-------------------

1 ν 0

ν 1 0

0 0 1 ν–( ) 2 ⁄

ε xx

ε yy

γ xy

=

Stresses:

(Plane stress)

x

y

d 1 y

d 1 x

d 2 y

d 2 x

d 3 y

d 3 x

d 4 y

d 4 x

N 1 1 ξ–( ) 1 η–( ) 4 ⁄ = N 2 1 ξ+( ) 1 η–( ) 4 ⁄ =,

N 3 1 ξ+( ) 1 η+( ) 4 ⁄ = N 4 1 ξ–( ) 1 η+( ) 4 ⁄ =,

∂ N i ∂ x ⁄

∂ N i ∂ y ⁄ J

1– ∂ N i ∂ξ ⁄

∂ N i ∂η ⁄ = J

∂ x ∂ξ ⁄ ∂ y ∂ξ ⁄

∂ x ∂η ⁄ ∂ y ∂η ⁄ =

u ξ η,( )

v ξ η,( )

N 1 0 N 2 0 N 3 0 N 4 0

0 N 1 0 N 2 0 N 3 0 N 4

de Nde= =

1 2

34

ξ

η

B B1 B2 B3 B4= B i

∂ N i ∂ x ⁄ 0

0 ∂ N i ∂ y ⁄

∂ N i ∂ y ⁄ ∂ N i ∂ x ⁄

=

x N i xi

i

4

∑

=

y N i yi

i

4

∑

=

Displace-ments:

σ xx

σ yy

σ xy

E

1 ν2

–( )-------------------

1 ν 0

ν 1 0

0 0 1 ν–( ) 2 ⁄

ε xx

ε yy

γ xy

=

Stresses:

(Plane stress)

Deformation:

where

Plane (2D) triangular linear element:

Plane (2D) quadrilateral bi-linear element:




– 6.11 (24) –

Solutions

6.1 ; plane stress with ν = 0 gives


6.2 Use that

must satisfy the condition:

(i) zero in nodes 1, 2 and 3, i.e.

(ii) unity in node 4, i.e.

Both conditions are satisfied!

6.3

For we obtain , and

N 3 = N 4 = N 6 = N 7 = N 8 = 0.

Coordinates at the nodes: , ,

Coord. in xy-plane: ,

Distance from origin: , i.e. the deviation of the FE-mesh

from the boundary of the hole is 1%.

6.4

The coordinates in the isoparametric element is given by the interpolation (coord.-transform):

B1

L---

1– 0 1 0 0 0

0 1– 0 0 0 1

1– 1– 0 1 1 0

= C E

1 0 0

0 1 0

0 0 1 2 ⁄

=

ke BT CBhd A

V e

∫

hL2

2---------B

T CB

Eh

4-------

3 1 2– 1– 1– 0

3 0 1– 1– 2–

2 0 0 0

1 1 0

1 0

2

= = =

Sym.

N 1 N 2 N 3 N 4+ + + 1 N 4⇒ 1 N 1 N 2 N 3+ +( )– 4 1 ξ – η –( )ξ = = =

N 4 ξ η ,( )

N 4 0 0,( ) N 4 1 0,( ) N 4 0 1,( ) 0= = =

N 4 1 2 ⁄ 0,( ) 1=

ξ0 1 2 ⁄ η0, 1–= = N 12 1–

4---------------- N 2

2 1+

4---------------- N 5,=,–

1

2---= =

x1 y1,( ) 1– 1,( )

R

2-------= x2 y2,( ) 1 1,( )

R

2-------= x5 y5,( ) 0 1,( ) R=

x0 x ξ0 η0,( ) N i xi∑

R

2---= = =

y0 y ξ0 η0,( ) N i yi∑

1 2+( )

2 2--------------------- R= = =

x0

2 y0

2+ R

5 2 2+

8------------------- 0.989 R≈=

y ξ η ,( ) N i yi

i 1=

4

∑

L N 1 N 3–( ) L

2--- ξ– η–( )= = =

Partial derivatives: ∂ x

∂ξ ------

L

2---

∂ x∂η ------;

L

2---

∂ y∂ξ ------;–

L

2---

∂ y∂η ------;–

L

2---–= = = =

J

L

2---

1 1–

1– 1– J⇒

L2

2-----–= =The Jacobi matrix and its determinant becomes:

x ξ η ,( ) N i xi

i 1=

4

∑

L N 2 N 4–( ) L

2--- ξ η–( )= = =




– 6.13 (24) –

(d)

6.8

6.9 Numerical integration with Gauss-Legendre quadrature

where

k e BT CBhd A

Ae

∫

BT

CBh J dξdη

1–

1

∫

1–

1

∫

B1

T CB1 B1

T CB2 B1

T CB 3 B1

T CB 4

B2

T CB1 B2

T CB2 B2

T CB 3 B2

T CB 4

B3

T CB1 B3

T CB2 B3

T CB 3 B3

T CB 4

B4

T CB1 B4

T CB2 B4

T CB 3 B4

T CB 4

hL2

1 λη+( )dξdη

1–

1

∫

1–

1

∫

= = =

4 2 L 1 λ 2

+ ϕcos,1

1 λ 2

+

------------------- ϕsin, λ

1 λ 2

+

-------------------= = =

NT tdS

S e

∫

dS hl14

2------dη=

ξ 1–=

NT

ξ 1–=thL 1 λ

2+ dη

1–

1

∫

= =

1 x f 4 x p0hL f 1 y f 4 y λ p0hL= =,= =

h J dξdη= } NT f V h J dξdη

1–

1

∫

1–

1

∫

=

λ 1

2---=J L

21 λη+( )= one obtains with that

2ρg f b3 y f b4 y

76---h– L

2ρg= =,

Note! f biy∑

4hL2ρg– V eρg–= =5 0 5 0 7 0 7

bIy f I ξ η,( )dξdη

1–

1

∫

1–

1

∫

f I ξi η j,( )wiw j

j

nη

∑

i

nξ

∑

= = I N I ξ η,( ) ρg–( )hL2

1 λη+( )=

(a) 1 x 1 scheme: , and .

in the same way we obtain

nξ nη 1= = ξ1 η1 0= = w1 2=

f b1 y⇒ N 1 0 0,( ) ρg–( )hL2

2 2⋅( ) h– L2ρg= =

b2 y f b3 y f b4 y h– L2ρg= = =

(b) 2 x 2 scheme: , , and .

in the same way we obtain and ,

i.e. the numerical integration are exact!

nξ nη 2= = ξ1 η1

1–

3-------= = ξ2 η2

1

3-------= = w1 w2 1= =

f b1 y⇒ ρghL2

N 1 ξ1 η1,( ) 1 λη1+( ) 1 1⋅( ) N 1 ξ1 η2,( ) 1 λη2+( ) 1 1⋅( )

N 1 ξ2 η1,( ) 1 λη1+( ) 1 1⋅( ) N 1 ξ2 η2,( ) 1 λη2+( ) 1 1⋅( )

+

+ +

[

]

–

5

6---ρghL

2–

=

=

b2 y

5

6---ρghL

2–= f b3 y f b4 y

7

6---h– L

2ρg= =




– 6.14 (24) –

6.10

Contributions to the nodal force vector: , with .

Here and , but along 2-3 and thus

, which give .

(a) 4-node element: and

,

,

which give .

,

and , , .

f s NT tdS

S e

∫

= dS hds h d x2

dy2

+= =

d x ∂ x

∂ξ------d

ξ

∂ x

∂η------d

η+= d y

∂ y

∂ξ------d

ξ

∂ y

∂η------d

η+=

ξ1=

dξ 0= dS h ∂ x ∂η ⁄ ( )2

∂ y ∂η ⁄ ( )2

+ dη=

N 2ξ 1=

1 η–( ) 2 ⁄ = N 3ξ 1=

1 η+( ) 2 ⁄ =

x N 2 x2 N 3 x3+( )ξ 1=

∂ x ∂η ⁄ ⇒ x3 x2–( ) 2 ⁄ = =

y N 2 y2 N 3 y3+( )ξ 1=

∂ y ∂η ⁄ ⇒ y3 y2–( ) 2 ⁄ = =

dS h

x3 x2–

2----------------

2 y3 y2–

2----------------

2

+ dη hLdη= =

2 x N 2t xhLdη

1–

1

∫

N 21 η–

2------------t A x

1 η–

2------------t B x+

hLdη

1–

1

∫

hL

3------ 2t A x t B x+( )= = =

2 y

hL

3------ 2t A y t B y+( )= 3 x

hL

3------ t A x 2t B x+( )= 3 y

hL

3------ t A y 2t B y+( )=

(b) 8-node element: , , N 2ξ 1=

1 η–( ) η–( )2

-----------------------------= N 3ξ 1=

1 η+( )η2

---------------------= N 6ξ 1=

1 η2

–=

x N 2 x2 N 3 x3 N 6 x6+ +( )ξ 1=

x6 x2 x3+

2----------------=

N 2 N 62

------+

x2 N 3 N 62

------+

x3+

ξ 1=

= = =

y N 2 y2 N 3 y3 N 6 y6+ +( )ξ 1=

y6

y2 y3+

2----------------=

N 2 N 6

2------+

y2 N 3 N 6

2------+

y3+

ξ 1=

= = =

.

In the same way we obtain , , ,

and .

∂ x∂η------

x3 x2–

2----------------

∂ y∂η------

y3 y2–

2----------------=,=

dS h x3 x2–

2----------------

2 y3 y2–

2----------------

2

+ dη hLdη= =⇒ ⇒

2 x N 2t xhLdη

1–

1

∫

N 2 1 η–2

------------t A x1 η–

2------------t B x+

hLdη

1–

1

∫

hL3

------ t A x= = =

2 y

hL

3------t A y= 3 x

hL

3------ t B x= 3 y

hL

3------t B y=

6 y

hL

3------ 2t A x 2t B x+( )= 6 y

hL

3------ 2t A y 2t B y+( )=




– 6.15 (24) –

6.10 cont.

6.11 where and .

;

In (c) and (d) we have that and

(c) For a 4-node element this give and , where

. The nodal force vector becomes (nodes 1 to 4, global node numbering)

after assembly:

(d) For a 8-node element this give , ,

and , where . The nodal force vector becomes (nodes 1 to 7,

global node numbering) after assembly:

t A x t B x t x σ0 θ τ0 θsin–cos= = =

t A y t B y t y σ0 θ τ0 θcos+sin= = =

2 x f 3 x hLet x= = 2 y f 3 y hLet y= =

Le L 3 ⁄ =

Fs

T hLe t x t y 2t x 2t y 2t x 2t y t x t y …=

2 x f 3 xhLe

3--------t x= = 2 y f 3 y

hLe

3--------t y= = 6 x

hLe

3--------4t x=

6 y

hLe

3--------4t y= Le

L

3---=

FsT

hLe

3-------- t x t y 4t x 4t y 2t x 2t y 4t x 4t y 2t x 2t y 4t x 4t y t x t y …=

f b NT f vh

ab

4------dξdη

1–

1

∫

1–

1

∫

= N N 1 0 N 2 0 N 3 0 N 4 0

0 N 1 0 N 2 0 N 3 0 N 4

= f v0

ρ g–=

NT f v

N 1 0

0 N 1 0

ρ g–

0

f 1 y= = 1 y

ρ ghab

4----------------

1 ξ–( ) 1 η–( )4

----------------------------------dξdη

1–

1

∫

1–

1

∫

– ρ ghab

4----------------–= =

f bT

ρ ghab

4---------------- 0 1 0 1 0 1 0 1–=⇒




– 6.17 (24) –

6.13(b) The stresses in the element are given by

6.14

σ CBde=

de

d1

d2

d3

= d10

0= d3

0

1 6 ⁄ –=whereC

P.S.

ν1

3---=

3 E

8-------

3 1 0

1 3 0

0 0 1

= =

Thus

∂ N 1 ∂ x ⁄ 1 L ∂ N 1 ∂ y ⁄ ; ⁄ – 1– L ⁄ = =

∂ N 2 ∂ x ⁄ 2 L ∂ N 2 ∂ y ⁄ ; ⁄ 0= =

∂ N 3

∂ x ⁄ 1– L ⁄ ∂ N 3

∂ y ⁄ ; 1 L ⁄ = =

B B1 B2 B3=⇒ B1

1

L---

1– 0

0 1–

1– 1–

= B2

1

L---

2 0

0 0

0 2

=

B3

1

L---

1– 0

0 1

1 1–

=

σ

σ xx

σ yy

σ xy

C B1 B2 B3

d1

d2

d3

C B1d1 B2d2 B3d3+ +[ ] E ε01 2 ⁄

0

1 18 ⁄

= = = =

d21 4 ⁄

1 108 ⁄ –=

K Eh

4-------

3 1 2– 1– 0 0 1– 0

1 3 0 1– 0 0 1– 2–

2– 0 3 0 1– 1– 0 1

1– 1– 0 3 0 2– 1 0

0 0 1– 0 3 1 2– 1–

0 0 1– 2– 1 3 0 1–

1– 1– 0 1 2– 0 3 00 2– 1 0 1– 1– 0 3

=

Stiffnessmatrix:

D K1–F=

σ CBD=

Displacements are given by:

Stresses are given by:

(a) Displ. B.C. (symmetry & remove rigid body motion): D1 x = D1 y = D2 y = D4 x = 0

Force vector: FT

F 1 x F 1 y F 2 x F 2 y F 3 x F 3 y F 4 x F 4 y R1 x R1 y

σ0hl

2----------- R2 y

σ0hl

2----------- 0 R4 x 0= =

DT

⇒

σ0

E ------ l 0 0 1 0 1 0 0 0= el. 1 & el. 2: σ

T σ0 1 0 0=

(b) Displ. B.C. (remove rigid body motion): D1 x = D1 y = D2 y = 0

Force vector: F

T

R1 x

τ0

hl

2----------–

R1 y

τ0

hl

2----------–

τ0

hl

2----------– R2 y

τ0

hl

2----------+

τ0

hl

2----------τ

0

hl

2----------τ

0

hl

2----------τ

0

hl

2----------–

=

DT

⇒

τ0

E 2 ⁄ ---------- l 0 0 0 0 1 0 1 0= el. 1 & el. 2: σ

T τ0 0 0 1=

(c) Displ. B.C. (remove rigid body motion): D1 x = D1 y = D2 y = 0

Force vector: FT

R1 x R1 y

ρgh l2

6--------------–

0 R2 y

ρgh l2

3--------------–

0 ρgh l

2

6--------------– 0

ρgh l2

3--------------–=

DT

⇒

""ρgl

2

24 E -----------– 0 0 1– 0 3 9 2 15= el. 1: σ

T ρgl

24-------- 1 15– 1– el. 2: σ

T ρgl

24-------- 1– 9– 1==

exact!

exact!

Note! E /2 = G (shear modulus) in this case!




– 6.18 (24) –

6.15(a)

6.15(b)

6.15(c)

6.16(a) Displacement boundary conditions: d 1 x=d 3 y=d 4 x=d 4 y=0

Consistent nodal force vector:

f 2 y N 2t yh sd0

2 L

∫ x Ls

2-------–=

1

L--- Ls

2-------–

σ0

2-------h sd0

2 L

∫

σ0 Lh

2-------------= = = =

f s NT

2 3–th s f 2 y 0 f 3 y 0≠,≠⇒ d

0

2 L

∫

=

FT

R1 x R1 y 0 R2 y

σ0 Lh

2-------------+

R3 x

σ0 Lh

2-------------=

f 3 y N 3t yh sd

0

2 L

∫

ys

2-------=

1

L---

s

2-------

σ0

2-------h sd

0

2 L

∫

σ0 Lh

2-------------= = = =

Zero displacement B.C.: d 1 x=d 1 y=d 2 y=d 3 x=0 => reaction forces: R1 x , R1 y , R2 y , R3 x

Node force vector:

3 Eh

16---------- 3 1

1 3

d 2 x

d 3 y

σ0 Lh

2------------- 0

1

d 2 x

d 3 y

⇒ L

σ0

E ------ 1 3 ⁄ –

1= =

Eq. (3) and (6) gives the reduced equation system

Reaction forces: Eq. (1): R1 x

3 Eh

16---------- 3d 2 x– d 3 y–( ) 0= =

Eq. (2): R1 y

3 Eh

16---------- d 2 x– 3d 3 y–( )

σ0 Lh

2-------------–= =

Eq. (4): R2 y

σ0 Lh

2

-------------+ 0 R2 y⇒ σ0 Lh

2

-------------–= =

Eq. (6): R3 x 0=

The strains in the element are given by: ε Bde=

N 1 1 x L ⁄ – y L ⁄ –= ∂ N 1 ∂ x ⁄ ⇒ 1 L ∂ N 1 ∂ y ⁄ ; ⁄ – 1– L ⁄ = =

N 2 x L ⁄ = ∂ N 2 ∂ x ⁄ ⇒ 1 L ∂ N 2 ∂ y ⁄ ; ⁄ 0= =

N 3 y L ⁄ = ∂ N 3 ∂ x ⁄ ⇒ 0 ∂ N 3 ∂ y ⁄ ; 1 L ⁄ = =

B B1 B2 B3= =

1

L

---1– 0 1 0 0 0

0 1– 0 0 0 1

1– 1– 0 1 1 0

=

⇒

ε

ε x

ε y

γ xy

1

L---

1– 0 1 0 0 0

0 1– 0 0 0 1

1– 1– 0 1 1 0

Lσ0

E ------

0

0

1 3 ⁄ –

0

0

1

σ0

E ------

1 3 ⁄ –

1

0

= = =Agrees with the

exact solution

when ν = 1/3




– 6.19 (24) –

6.16(b)

6.16(c)

6.17(a) Nodal force vector:

Consistent nodal force vector:

f 2 x N 2t xh yd

L–

L

∫

1 η–2

------------σ0hL ηd

1–

1

∫

σ0hL= = =

f s NT

2 3–th s f 2 x 0 f 3 x 0≠,≠⇒ d

L–

L

∫

=

Fe

T R1 x 0 σ0 Lh 0 σ0 Lh R3 y R4 x R4 y

=

Displacement boundary conditions give the reaction forces: R1 x , R3 y , R4 x , R4 y

Node force vector:

f 3 x N 3t xh yd

L–

L

∫

1 η+

2-------------σ0hL ηd

1–

1

∫

σ0hL= = =

others are

equal to

zero!

Force vector (external forces + reaction forces) can be calculated as Fe KeDe=

F 1 x⇒ R1 x σ0hL F 2 x;– σ0hL F 3 x; σ0hL F 4 x; R4 x σ0hL–= = = = = =

F 1 y 0 F 2 y 0 F 3 y;=; R3 y 0 F 4 y; R4 y 0= = = = =

t σ0 y L ⁄

0=

x

y

1 2

34 On the boundary x = 2 L, between nodes 2 and 3,

the shape functions take the values:

N 1 N 4 0 och N 21 η–( )

2----------------- N 3

1 η+( )2

-----------------=,== =

f e NT

ξ 1=thLdη

1–

1

∫

= f 2 x 0 f 3 x 0≠,≠⇒ others zero!

f 2 x

1 η–( )

2-----------------ηhLdη

1–

1

∫

σ0hL

3-------------–= = f

3 x

1 η+( )

2-----------------ηhL dη

1–

1

∫

σ0hL

3-------------= =

Consistent nodal force vector becomes (subst. η = y/L):

Inclusive of reaction forces, theF

T

R1 x 0σ0hL

3-------------– 0

σ0hL

3------------- 0 R4 x R4 y

=nodal force vector becommes:




– 6.20 (24) –

6.17(b) Stresses are calculated as: where

B B1 B2 B3 B4=

B2

1

4 L------

1 η – 0

0 1 ξ +( )–

1 ξ +( )– 1 η –

B3

1

4 L------

1 η + 0

0 1 ξ +

1 ξ + 1 η +

=,=

de

T

d1

T d2

T d3

T d4

T where d1 d4 0 and d2

4

3---

σ0

E ------ L 1–

1– d3,

4

3---

σ0

E ------ L 1

1–= = = = =

Nodal force vector (given):

B-matrix:

J ∂ x ∂ξ ⁄ ∂ y ∂ξ ⁄

∂ x ∂η ⁄ ∂ y ∂η ⁄

L 0

0 L= =with Jacobi matrix

Strains:

ε

ε x

ε yγ xy

B2d2 B3d3+

1

4 L------

4

3---

σ0

E ------ L 1 η –( )– 1 η +( )+[ ]

ε y

γ xy

2

3---

σ0

E ------η

ε y

γ xy

= = = =

Stresses:

σ

σ x

σ y

τ xy

Cε E 0 0

0 E 0

0 0 E 2 ⁄

2

3---

σ0

E ------η

ε y

γ xy

2

3---σ0η

σ y

τ xy

= = = =

plane stress (ν = 0.3)

Note!

The solution based on one

FEM element deviates from the

exact solution:σ x σ0 y L ⁄ =

σ y τ xy 0= =

σ Cε= ε Bd e=




– 6.21 (24) –

6.18 (a)

6.18 (b)

Fb NT KdV

V e

∫

NT Khd y

L x–( )–

L x–( )

∫

d x

0

L

∫

= =The contribution from the volume force is:

Here, both f bix 0≠ f bix 0≠ i 1 2 3, ,=and

f b1 x 1

2--- 1

x

L---–

y

L---–

ρω 2 xhd y

L x–( )–

L x–( )

∫

d x

0

L

∫

ρω 2h

L------------- L x–( )

2 xd x

0

L

∫

ρω 2hL

3

12-------------------= = =

f b2 x x

L---ρω

2 xhd y

L x–( )–

L x–( )

∫

d x

0

L

∫

ρω 2h

L------------- 2 x

2 L x–( )d x

0

L

∫

ρω 2hL

3

6-------------------= = =

f b3 x 1

2--- 1

x

L---–

y

L---+

ρω 2 xhd y

L x–( )–

L x–( )

∫

d x

0

L

∫

ρω 2h

L------------- L x–( )

2 xd x

0

L

∫

ρω 2hL

3

12-------------------= = =

The total nodal force vector becomes: f ρω

2hL

3

12-------------------

1

0

2

0

1

0

ρghL2

3----------------

0

1

0

1

0

1

–

R1 x

R1 y

0

0

R3 x

0

+=

f b1 y 1

2--- 1

x

L---–

y

L---–

ρ g–( )hd y

L x–( )–

L x–( )

∫

d x

0

L

∫

ρ gh

L----------– L x–( )

2d x

0

L

∫

ρ ghL

2

3----------------–= = =

f b2 y x

L--- ρ g–( )hd y

L x–( )–

L x–( )

∫

d x

0

L

∫

2 ρ gh

L-------------– x L x–( )d x

0

L

∫

ρ ghL

2

3----------------–= = =

f b3 y 1

2

--- 1 x

L

---– y

L

---+

ρ g–( )hd y

L x–( )–

L x–( )

∫

d x

0

L

∫

ρ gh

L

----------– L x–( )2d x

0

L

∫

ρ ghL

2

3

----------------–= = =

The displacement B.C:s : d 1 x = d 1 y = d 3 x = 0 give rise to reaction forces!

Reduced equation system with

respect to the boundary:

conditions, Eq. (3,4,6):

Displacement boundary conditions: d 1 x = d 1 y = d 3 x = 0

Eh

8-------

8 0 0

0 4 2–

0 2– 3

d 2 x

d 2 y

d 3 y

ρω 2hL

3

12-------------------

2

0

0

ρghL2

3----------------

0

1

1

–=

d 2 x

d 2 y

d 3 y

⇒

ρω 2 L

3

6 E ----------------

1

0

0

ρ gL

2

3 E ------------–

0

5

6

–=




– 6.23 (24) –

6.19(b) Stresses in the element is given by

6.20(a) Boundary conditions:

6.20(b) For element e4 yields:

σ Cε CBde= =

σ xx

σ yy

τ xy

⇒ E

1 0 0

0 1 0

0 0 1 2 ⁄

1

4 L------

δ x δ x+

0

1 ξ+( )δ x– 1 ξ+( )δ x+

E δ x 2 L( ) ⁄

0

0

22

3------ ρω 2

L2

1

0

0

= = =

Bi

1

L---

N i ξ, 0

0 N i η,

N i η, N i ξ,

B2

1

4 L------

1 0

0 1 ξ+( )–

1 ξ+( )– 1

B3

1

4 L------

1 0

0 1 ξ+( )

1 ξ+( ) 1

=;=⇒ =

σ xx

FE Mσ xx

Ex ac t –

σ xx

Ex ac t ---------------------------------- 100%×

C E

1 0 0

0 1 0

0 0 1 2 ⁄

= dT

d1

T d2

T d3

T d4

T

e1=Here,

d2 d3

δ x

0= =

δ x

44

3------

ρω 2 L

3

E ----------------=where

d1 d4 0= =

σ C B1d1 B2d2 B3d3 B4d4+ + +( )=⇒ C B2d2 B3d3+( )=

Comparison with the exact solution:

Relative eroor = 8.3 %– 2.2 %– 22.2 %

x = L x = 0 x = 2 L

D1 x D1 y D2 x D3 x D3 y D4 y D7 y 0= = = = = = =

3 2

1

t0

p0

x

L---–

=

e4

N 1 2 y

L--- N 2,–

x

L---

y

L--- 2 N 3,–+ 1

x

L---–= = =Shape functions:

f e

NT

y 2 L=

thd x

0

L

∫

= f 2 y

0 f 3 y

0≠,≠⇒ others zero!

f 2 y x

L--- p0

x

L---–

hd x

0

L

∫

hLp0

3------------–= = f 3 y 1

x

L---–

p0

x

L---–

hd x

0

L

∫

hLp0

6------------–= =

Assembly of global nodal force vector including the reaction forces gives:

F 1 x R1 x F 1 y; R1 y F 2 x; R2 x F 3 x; R3 x F 4 y; R4 y F 7 y; R7 y= = = = = =

F 3 y hLp0 6 ⁄ F 6 y;– hLp0 3 ⁄ –= =




– 6.24 (24) –

6.20(c) Stress in element 1 is given by

6.21(a) Boundary conditions:

6.21(b) For one element yields:

6.21(c) Stresses in element 1 is given by

σ CBde=

de

d1

d2

d3

= d1 0 (R.V.);= d2

D4 x

0; d3

0

D2 y

==C P.S.{ } E

1 ν2

–--------------

1 ν 0

ν 1 0

0 01 ν–( )

2----------------

= =

σ

σ xx

σ yy

σ xy

C B1d1 B2d2 B3d3+ +[ ] C D4 x L ⁄

D2 y L ⁄

E

1 ν2

–--------------

D4 x ν D2 y+( ) L ⁄

D2 y ν D4 x+( ) L ⁄ = = = =Thus

N 1 1 x L ⁄ – y L ⁄ –= ∂ N 1 ∂ x ⁄ ⇒ 1 L ∂ N 1 ∂ y ⁄ ; ⁄ – 1– L ⁄ = =

N 2

x L ⁄ = ∂ N 2

∂ x ⁄ ⇒ 1 L ∂ N 2

∂ y ⁄ ; ⁄ 0= =

N 3 y L ⁄ = ∂ N 3 ∂ x ⁄ ⇒ 0 ∂ N 3 ∂ y ⁄ ; 1 L ⁄ = =

B B1 B2 B3= =

1

L---

1– 0 1 0 0 0

0 1– 0 0 0 1

1– 1– 0 1 1 0

=

⇒

Shape functions:

D1 x D1 y D2 x D2 y D3 x D3 y D4 x D4 y D5 x D5 y 0= = = = = = = = = =

2

34

1t

0

τ 0–= f s N

T

ξ 1=thbdη

1–

1

∫

= f 2 y 0 f 3 y 0≠,≠⇒

others are equal

to zero!

f 2 y N 2ξ 1=

thbdη

1–

1

∫

τ 0hb1 η –( )

2-----------------dη

1–

1

∫

– τ 0hb–= = = f 3 y N 3ξ 1=

thb dη

1–

1

∫

τ 0hb1 η +( )

2----------------- dη

1–

1

∫

– τ 0hb–= = =

Assembly of consistent nodal forces from elements gives non-zero components

F 26 y F 30 y τ 0hb och F 27 y– F 28 y F 29 y 2τ 0hb–= = = = =according to:

σ CB de=

de

d1

d2

d3

d4

= d1 d4 0 (B.C.);= = d2

D7 x

D7 y

; d3

D6 x

D6 y

==where2

34

1 D7 x

D7 y

D6 x

D6 y

CP.S.

ν 0=

E

1 0 0

0 1 0

0 0 1 2 ⁄

;= = B B1 B2 B3 B4=

∂ N i

∂ x

--------1

a---

∂ N i

∂ξ

--------=

∂ N i

∂ y--------

1

b---

∂ N i

∂η --------=

where

B2 and B3 evaluated in the centroid of element 1 (ξ = η = 0) becomes:

B2

1

4---

1 a ⁄ 0

0 1 b ⁄ –

1 b ⁄ – 1 a ⁄

; B3

1

4---

1 a ⁄ 0

0 1 b ⁄

1 b ⁄ 1 a ⁄

==

σ

σ xx

σ yy

σ xy

CB2d2 CB3d3+ E

4---

D6 x D7 x+( ) a ⁄

D6 y D7 y–( ) b ⁄

D6 x D7 x–( ) 2b( ) D6 y D7 y+( ) 2a( ) ⁄ + ⁄

= = =

Thus




– 7.1 (3) –

7. FEM: Heat conduction

7.1 The Figure to the right shows a one dimensional

model of a cooling fin. At the left boundary the

temperature is constant at 80o C. Along the remain-

ing boundary, heat is lost to the surrounding air by

convection. Determine the cooling effect of the fin,

i.e. calculate the heat flow across its left boundary.

Also determine the displacement in the fin due to

thermal expansion, where it can be assumed that

the fin is undeformed at 20o C. Analyse the prob-

lem by FEM and use two linear elements. Carry out

the analysis in two steps: (a) calculate the tempera-

ture distribution in the fin and the heat flow at x = 0,

and (b) calculate the displacement in the fin.

The material has elasticity modulus E , thermal expansion coefficient α, thermal conductivity k and convection coefficient h. The material data and the geometry is shown in the figure.

7.2 A wall made of two layers of different material is shown in the

right hand figure. The temperature at the right side of the wall is

kept constant at 20o C. At the left side, a heat flux arises due to

convection, where the ambient temperature is equal to −5o C. The

thermal convection coefficient is h and the thermal conductivity of

the two materials is k 1 and k 2, respectively. Determine the tempera-

ture distribution through the wall by FEM-analysis. In the present

case it suffice to model each layer by one linear element.

Data: k 1 = 0.2 W/cm/ oC, k 2 = 0.06 W/cm/ oC, h = 0.1 W/cm2 / oC, L1 = 2 cm and L2 = 5 cm.

7.3 A bar of copper is clamped between two rigid

walls. At point B, the bar is influenced by a point force

P0 = 2kN and a heat source keeping the temperature

constant at 100o C. The right boundary (point C) is

insulated and at the left boundary (point A) the tem-

perature is kept constant at 20o C. Between the endpoints of the bar, a heat flux occurs by convection,

where the ambient temperature of the surrounding

medium is equal to 20o C. Conduct a FEM-analysis to determine the distribution of tempera-

ture, displacement and normal stress in the bar. Use two linear elements in each of the inter-

vals A-B and B-C, respectively. Carry out the analysis in the two consecutive steps: (a)

calculate the temperature and (b) calculate the displacement and normal stress.

Data: L = 10 cm, b = 1 cm, k = 3.9 W/cm/ oC, h = 0.01 W/cm2 / oC, E = 125 GPa and

.

a

b

xT ∞ 20°C=

T 80°C=

Convection

x = L

Data:

a = 1 cm, b = 0.4 cm, L = 8 cm,

E = 80 GPa,

k = 3 W/cm/ oC, h = 0.1 W/cm2 / oC

α 1.4 105–

⋅=

L1 L2

1 2

L 2 L

P0

AB

C

b

b

T ∞ 20°C=

Konvektion

α 1.8 105–

⋅=




– 7.2 (3) –

Solutions

7.1

0 2 4 6 8

x / cm

10

20

30

40

50

60

70

80

T e m p e r a t u r e

/ o C

Exact solution

2 element

4 element

8 element

0 2 4 6 8

x / cm

0

5

10

15

20

D i s p l a c e m e n t / µ m

Exact solution

2 element

4 element

8 element

T 1 T 3T 2

FEM-analysis: 2 linear elements

0.6733 0.1133– 0

0.1133– 1.3467 0.1133–

0 0.1133– 0.7133

T 1 80°C=

T 2

T 3

11.20 Q R+

22.40

12.00

=Equationsystem


x = L: kA∂T

∂ x------– hA T T ∞–( )=

unit [W/ oC] unit [W]

“Reaction

heat flow”

T 2

T 3

⇒

25.12

20.81°C= Q

R

⇒ 39.82 W=

(a) Temperature distribution

(b) Displacement

D1 D3 D2

FEM-analysis: 2 linear elements Boundary conditions:

x = L: σA = 0

80 106

⋅1 1– 0

1– 2 1–

0 1– 1

D1 0=

D2

D3

1.459– 103

⋅ R+

1.326 103

⋅

0.133 103⋅

D2

D3

0.01823

0.0198910

3– m=

R 0=

⇒ =

unit [N/m]unit [N]

Comparison between the exact solution and FEM-solutions based on 2, 4 and 8 elements

x = 0: T = 80 oC

x = 0: u = 0




7.2

7.3

T 1 T 3T 2


l2l1Boundary conditions, x = 0:

x = L: T = -5o

C

kA∂T

∂ x------– hA T T ∞–( )=

A

k 1

l1

----- h+k 1

l1

-----– 0

k 1

l1

-----–k 1

l1

-----k 2

l2

-----+k 2

l2

-----–

0k 2

l2

-----–k 2

l2

-----

T 1

T 2

T 3 20°C=

A

hT ∞

0

Q R A ⁄

T 1

T 2

2.58–

0.161–°C=

Q R A ⁄ 0.242 W cm2

⁄ [ ]=

⇒ =

Equationsystem

D1 D3 D2


D5 D4

T 5T 4T 3T 2T 1

Prescribed values:

x = xA: T = 20o C

x = xB: T = 100o C

x = xC: Q kA∂T

∂ x------– 0= =

(a) Temperature distribution, divide into two separate analysis, since T 3 is prescribed.

T 1 T 3T 2 T 3 T 5T 4

0.8467 0.7467– 0

0.7467– 1.6933 0.7467–

0 0.7467– 0.8467

T 1 20°C=

T 2

T 3 100°C=

2 Q+ R1

4

2 Q+ R3

=

0.5233 0.3233– 0

0.3233– 1.0467 0.3233–

0 0.3233– 0.5233

T 3 100°C=

T 4

T 5

4 Q+ R3

8

4

=

(b) Displacement and stress calculations

Valid for each element: ke

EA

li

------- 1 1–

1– 1 f T ; EAα∆ T i

1–

1= = = average temperature

change in element i

∆T i

125 106

⋅

2 2– 0 0 0

2– 4 2– 0 0

0 2– 3 1– 0

0 0 1– 2 1–

0 0 0 1 1

D1

0=

D2

D3

D4

R1

0

P0

0

EA α

∆T 4–

∆T 1 ∆T 2–

∆T 2 ∆T 3–

∆T 3 ∆T 4–

+

3.968–

9.00–

2.532

6.877

5 559

103

R1

0

0

0

+= =

e1 e2 e3 e4

T 2 55.27°C=⇒

T 4

T 5

⇒

50.54

38.87°C=

fem problem booklet

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