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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. Elastic Energy and Energy principles

1.1 A simply supported beam of length 3L and withbending stiffness EI is subjected to an externalbending moment M0, see the figure to the right.Determine the rotation θ by use of Castigliano’stheorems. Neglect possible contributions fromshear forces when evaluating the complementaryelastic energy.

1.2 Two beams, each with bending stiffness EI andlength L, are connected at B, see the figure to left.The left beam is clamped to a rigid wall at A and apoint force is applied on the right beam at point C.Determine the rotation of the corner B by use of anenergy method (neglect possible contributionsfrom normal forces and shear forces to the energyexpression).

1.3 Determine the horizontal displacement atpoint B in the beam structure to the right. Allbeams are of length L and have the bendingstiffness EI. Use complementary elastic energyand Castigliano’s theorems in the analysis.

1.4 A plane structure is composed of four equal beamsas shown to the right. The beams are of the length L andhave the bending stiffness EI. A moment M0 is appliedat the centre point of the structure. Use a suitableenergy method based on complementary elastic energyto determine the rotation of the centre point.

1.5 A cantilever beam of length L and with bending stiffness EI isclamped at its left end, and attached to a spring at its right end. Thespring constant is . A point force, P, is applied on thebeam according to the figure to the right. Determine, by use of asuitable energy method, the deflection of the right end of the beam.

1.6 A beam of length 2L and with bending stiffness EI issubjected to a bending moment applied at its right end,see the figure to the right. The left end of the beam isclamped, and the mid point is attached to a vertical springwith spring constant . Evaluate the rotationof the right end of the beam by use of an energy method.

M0

L 2L

θ

P

E, A, L

E, A, L

45o 45o

B

C

A

P

EI, L

EI,

L

A

B

M0

EI, L

k

P

k ηEI L3⁄=

M0EI

kLL

k 6EI L3⁄=

– 1.1 (10) –


1.7 A beam with bending stiffness EI and total length 2L,is simply supported at its mid point. The left end of thebeam is attached to a linear spring with the spring constant

. The beam is subjected to a point force P0and a moment M0. Determine M0 such that the deflectionof 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 withbending stiffness EI and length L, and aspring kN, see Figure (a) to the right, hasproven to be to compliant in an application.The system is therefore made stiffer by useof a torsion spring kM, see Figure (b). Thecomplementary 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 forceP. 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 afactor 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 beamwith bending stiffness EI and length 2L is shown in theright hand figure. Determine the vertical displacement ofthe beam at the point force. Use an energy based method.

1.10 A beam with bending stiffness EI and totallength 3L is subjected to a uniformly distributed loadwith the resultant Q, see the figure to the right.Determine all reaction forces acting on the beam.Use an energy method, based on the complementaryelastic energy, for statically indeterminate quantities.

1.11 A freely supported beam with bending stiff-ness EI and total length 3L is loaded by a pointforce, P, according to the right hand figure. Deter-mine the deflection at point B.

P0

L LM0

kk ηEI L3⁄=

P

EI, L

kN

EI,

L

P

EI, L

kN

EI,

L

kM

(a) (b)

WL

3

EI------ 1

6--- P

2P

ML-----–⎝ ⎠

⎛ ⎞2

+⎝ ⎠⎛ ⎞ 1

2--- P M L⁄–( )2

kNL3

EI⁄----------------------------- 1

2--- M L⁄( )2

kML EI⁄--------------------+ +=

kN αEI L3⁄= kM βEI L⁄=

P

L L

Q

2LL

P

A B

– 1.2 (10) –


1.12 A plane frame is composed of three beams connectedat the stiff joints B and C, see the figure to the right. Thebending stiffness of all beams are EI. The frame is loadedby a point force. Determine the vertical displacement ofpoint 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 onCastigliano’s theorems.

1.13 Determine the horizontal displacement at point B. Thebending stiffness of each beam in the planar frame is EI.

1.14 A planar frame constructed by two beams, eachwith bending stiffness EI, is loaded by a uniformlydistributed load with the resultant P and a point forceP according to the right hand figure. Calculate thevertical displacement at the point force.

1.15 A beam with circular cross section (diameterd) is shaped as a U, see the figure to the right. Thebeam is clamped at point A and loaded by a pointforce P, acting perpendicular to the plane of thebeam, at point D. Calculate the displacement atpoint D in the direction of the point force.

P

A B

C

D

2L

2LL

A

BC

L

L

M

P PL

L L

PDA

L

B CL

d

– 1.3 (10) –


1.16 A planar frame, according to the right hand fig-ure, is clamped at point A and point B. The frame isloaded perpendicular to its plane by a couple of pointforces, each of magnitude P. The beam has a circulartube shaped cross section with mean radius r andthickness t (assume that ). The material has theelastic modulus E and the shear modulus G. Calculatethe out-of-plane displacement at points B and C.Consider the case when L = l.

1.17 A rectangular planar frame with measurementsaccording to the figure is clamped at point A and point D.Two point forces is applied perpendicular to the frame inopposite directions at points B and C, respectively. Thebending stiffness of the frame is everywhere equal to EIand its torsional stiffness is GK, with ,where κ being a non-dimensional constant. Calculate thedisplacements perpendicular to the plane at the points Band C, respectively.

1.18 A planar frame of quadratic shape is freely supported at thecorner points A, B, C and D, such that only reaction forces per-pendicular to the frame may occur. The side of the frame is oflength L. The bending stiffness and the torsional stiffness are EIand GK, respectively. The frame is subjected to a uniformly dis-tributed load with the resultant Q acting perpendicular to theframe on the side AB. Calculate the displacement at the midpoint 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

2LEI GK( )⁄ κ=

Q

B C

A D

L, EI, GK

– 1.4 (10) –


Solutions

1.1

1.2

1.3

Introduce reac-

M0

RB

RA

Equilibrium RB M0 3L( )⁄=

RA RB M 3L( )⁄= =

M1M0RA

M0

3L-------= M1 M2 M2

RB

M0

3L-------=

Equilibrium requires: M1 M0 3⁄= M2 2M– 0 3⁄=

Complementary elastic energy: WLM1

2

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

2LM22

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

M02L

6EI----------- θ⇒

∂W∂M0----------

M0L

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

tion forces: gives:

PRMR

MF

Introduce a fictitious bending moment, MF

and reaction forces, R & MR

Equilibrium:R = PMR = MF

P PMF

M1

PP

M2PP

M1 M2MF

M1 MF PL 2⁄+=

M2 PL 2⁄=

Equilibrium:

WMF

2L

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

MFPL2

2 2EI------------------ P

2L

3

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

Complementary elastic energy:

Rotation at B: θB∂W

∂MF

-----------

MF 0=

PL2

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

M0

M0

M0MR

P

1 statically indeterminate, chose e.g. MR

M0 PL–=Equilibrium:

WL

6EI--------- MR

2MR– PL 2P

2L

2+( )=

∂W∂MR

----------- 0 MR⇒PL2

-------= =

δ ∂W∂P-------- 7

12------PL

3

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

Displacement at point B:

– 1.5 (10) –


1.4

1.5

1.6

Cut and use the symmetry

M0

M0/4M0/4

M0/4M0/4

M0/4MR

V

V

V

V

V

V

Equilibrium:M0

4------- MR– VL– 0=

=> 1 statically indeterminate, e.g. MR

Total complementary elastic energy:

W 4L

6EI--------- MR

2MR

M0

4-------

M0

4-------⎝ ⎠⎛ ⎞

2

+ +⎝ ⎠⎛ ⎞⋅=

∂W∂MR

----------- 0 = MR⇒M0

8------- ;–= θ ∂W

∂M0----------

M0L

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

properties!

MR

V

Equilibrium, beam:

: V– N– P+ 0=

=> 1 statically indeterminate,


L3

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

2N

21

3η---+⎝ ⎠

⎛ ⎞ 2PN–+⎝ ⎠⎛ ⎞=

∂W∂N-------- 0 = N⇒

ηP3 η+------------- ;=

δP∂W∂P-------- PL

3

EI--------- 1

3 η+( )-----------------= =

P

N

k ηEI

L3

------=

: MR PL– NL+ 0=

chose e.g. N

W Wbeam Wspring+L

6EI---------MR

2 N2

2k------+= =

Principle of least work:

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

L6EI--------- PL NL–( )2 N

2

2ηEI L3⁄

-----------------------+=

Equilibrium gives

Complementary

θ ∂W∂M0

---------- 32---

M0L

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

M0 M0 M0MR

k6EI

L3

---------=

M0

RN

NM0 MR–

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

Note! one staticallyindeterminate exists!

elastic enerrgy: WL

6EI--------- MR

2MRM0 M0

2+ +( ) L

6EI--------- 3M0

2( ) N2

2k------+ +

L4EI--------- MR

23M0

2+( )= =

Condition to determine the unknown∂W

∂MR

----------- L4EI---------2MR 0 MR⇒ 0= = =

Rotation at the point where the external moment is applied :

reaction force:

– 1.6 (10) –


1.7

1.8

1.9

Free body diagram:

M1 NL P0L M0–= =

M2 P0L=

Equilibrium:

WM1

2L

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

M22L

6EI----------- N

2

2k------+ +

L3

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

2 M0

L-------⎝ ⎠⎛ ⎞

2 2P0M0

L-----------------–+⎝ ⎠

⎛ ⎞ L3

2ηEI------------- P0

2 M0

L-------⎝ ⎠⎛ ⎞

2 2P0M0

L-----------------–+⎝ ⎠

⎛ ⎞+= =

Complementary elastic energy:

(given condition) M0⇒3 2η+3 η+

---------------- P0L=

k ηEI

L3

------=

NM0

R

P0P0M2M1N

Equilibrium:

: M0 P0L– NL+ 0=

N⇒ P0 M0 L⁄–=

δP0

∂W∂P0--------- 0= =

Case (a): no torsion spring (M = 0), kN = 3EI / L3 δa∂W∂P--------

M 0=

=⇒PL

3

EI---------=

Case (b): statically indeterminate problem,

∂W∂M-------- 0 M⇒ PL

2β3 2β+----------------= = δb

∂W∂P-------- L

3

EI------ P

23---M

L-----–⎝ ⎠

⎛ ⎞ PL3

EI---------9 2β+

9 6β+----------------= = =

According to the given conditions: δb12---δa β⇒

92---= =

where M is an internal indeterminate quantity, thus

Equilib:


δP∂W∂P-------- 2

3---PL

3

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

: M0 PL=

W 2M

2

2EI---------dx

0

L

∫ M x( ) PLxL---=

⎩ ⎭⎨ ⎬⎧ ⎫ P

2L

3

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

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

M0 PP

P P2P

M0 M0 PP(statically indeterminate probl.)

– 1.7 (10) –


1.10

1.11

1.12

Equilibrium:

∂W∂MA

----------- 0 MA⇒QL2

--------– ;= =

MA

RAQ RA RB Q–+ 0=

RB

MA RBL Q2L–+ 0=

Note! one staticallyindeterminate, choosefor instance MA.

M(x)

Qx/L

x

M x( ) QL4

-------- xL---⎝ ⎠⎛ ⎞

2=

Complementary elastic energy: WL

6EI--------- MA

2MAQL QL( )2

+ +( ) M x( )2

2EI---------------dx

0

2L

∫+=

RA 3Q2

------- ;–= RB5Q2

-------=

P QIntroduce a fictitious force Q when the comple-

mentary elastic energy, , is calculated.W

The comp. elastic energy in the beam becomes: WL

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

27PQ 4Q

2+ +( )=

δB∂W∂Q--------

Q 0=

718------PL

3

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

L L L

Complementary elastic energy in the beam: WL

6EI--------- 76RA

240RAP– 8P

2+( )=

δC∂W∂P-------- 52

57------PL

3

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

P

RD

RA

MD

2 equilibrium Eqs. and 3 unknown reac-tion forces (RA, RD and MD). 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∂RA

--------- 0 RA⇒5

19------P= =

1819------PL

1019------PL

1019------PL

1019------PL

Bending moment diagram:

– 1.8 (10) –


1.13

1.9

1.14

1.15

1.16

– 1.9 (10) –


1.171.18

– 1.10 (10) –


2. Matrix formulated structural mechanics—direct method

2.1 A system of five springs are connected asshown in the figure to the right. All the springconstants ki are in the current applicationequal to k. Furthermore, D1 = D4 = 0, F3 = 0and F2 = P. Determine the displacements andreaction forces.

2.2 Three springs are connected accordingto the figure to the right, also showing theapplied external force P. The spring con-stants are: k1 = 5k, k2 = k and k3 = 2k.Determine the displacements at the pointswhere the springs are connected and evalu-ate all the reaction forces.

2.3 Determine the displacement at thepoint force in the spring system shown tothe right.

2.4 The plane structure to the right consists of foursprings with spring constants: k1 = k2 = k4 = 2k andk3 = 4k. The springs are attached to five nodes withcoordinates shown in the right hand figure. The struc-ture is loaded by two point forces acting in node 4and node 5, as shown in the figure. Calculate the dis-placement at each node and the normal force actingin spring element number 4.

k1

k2

k3

k4k5

D1, F1 D4, F4D3, F3D2, F2

k1

k2

k3

rigid beam

3a

4a

P

P

45o 45o

45ok1

k2

k3

k1 = k

k2 = 2k

k3 = 2k

x/L

y/L

k1 k2

k3k4

(−1,1)(1,1)

(1,0)

(1,−1)

P

P

2

5

3

1

4

– 2.1 (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 crosssectional area A and elastic modulus E. The length of each trusselement is evident by the figure. A point force, P, is acting on node4. Calculate the displacements at the nodes and the reaction forcesat nodes 1 and 2, respectively. Show also that global equilibrium issatisfied in the vertical direction.

2.6 The plane frame structure to the right contains twotruss 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 pointforce P according to the figure. The displacement willfor the present structure always be in the direction of theforce P. Determine the relation between δ and P.

2.7 The plane structure in the figure to the right containstwo truss elements and two spring elements. The trusselements 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, k2, such that thedisplacement always is aligned with the force P, i.e. in thedirection of the angle ϕ.

2.8 A mass m0 is attached to three similar springs. Thedivision between the springs is 120o and the spring con-stants are k1 = k2 = k3 = k. The springs are attached to arigid ring of radius R. The coordinate system shown in thefigure is fixed to the ring, where the y-axis is located inthe direction of spring k1. The circumferential position ofthe ring is determined by the angle ϕ. Calculate the dis-placement of m0 in the x- and y-directions for an arbitraryangle. The acceleration of gravity g (see the figure) isassumed 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

EA

,L

EA, L φ

P

α

π2--- φ–

k ηEA L⁄=

P30o

k1 k2

30o

ϕE, A

, LE, A, L

k1 0.75 EA L⁄=

xy

ϕ

gm0

k3

rigidring

k2

k1

– 2.2 (12) –


2.9 An adjustable crane consists of two rods which are connected at node3, see the figure to the right. The elastic modulus of the rods is E. Rod e1has a cross sectional area A and a length l. Rod e2 is composed of twocylindrical 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 crosssectional area of rod e2 is . The relations and are valid here. Determine the displacement of node 3 and the normalforce acting in rod e2 (the force acting on the hydraulic actuator), whenthe crane is loaded by a mass m for the case . The accelerationof 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 pointforces (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 matrixformulated method and determine the reaction forces at all thenodes.

.2.11 The truss structure to the right contains two truss elements andone spring element, with spring constant . The structureis 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 modulusE. Determine the displacements at the nodes where the elements areconnected. 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

ϕl

e1

e2

1

2

3

g

3A θ 2ϕ= L 2l ϕcos=

ϕ 30°=

E, A, L

2P

P

E, A

, L

E 2A 2L,,

45ok

EA

EA, L

PL

45o

90o

k 2EA L⁄=

P

3k

2k

k

x

y

1

2

3

4

3 2⁄– 1 2⁄–

3 2⁄– 1 2⁄

– 2.3 (12) –


2.13 The planar truss structure to the right con-sists of four spring elements, each of length L andwith spring constant k. All springs are orientedwith a 45o angle with respect to the horizontalplane. Determine all displacements and reactionforces at the nodes.

2.14 Determine the displacements and the reac-tion forces at the nodes, and the normal forces inthe springs.

2.15 Determine 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 thespring elements. The stiffness and length of eachspring 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

k1

2

3

yx k

stelQ

k2---

k

2-------

k

1 2

3

4

k

2-------

x

y

stel

Q

3L, 2k

4L, k

5L, 5k

1

23

x

y

– 2.4 (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

P8k------ 3

1=

R1 R4 P 2⁄–= =

5k

k

2k

D1

D4

D3

D2

D6

D5

k5---

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

P17k--------- 7–

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

R5 4P 17⁄= R6 14P 17⁄=

D1 = D4 = D5 = D6 = 0 and F2 = −P, F3 = 0

D8

D7

D6

D5

D4

D3D2

D1

x

ye1

e3

e2 x

x1

2

45o

Element 1:

Element 3:

ke kir r–

r– r=Element stiffness matrix:

k1 k= r 1 0

0 0=,

k2 2k= r 1 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⇒P3k------=

x1

2

−45oElement 2: k3 2k= r 1 2⁄ 1– 2⁄1– 2⁄ 1 2⁄

=,

– 2.5 (12) –


2.4

Boundary conditions:

e1& e4:

Assembly of global stiffness matrix:

D2

Element stiffness matrices:

Ke kia a–

a– aa, c

2sc

sc s2

c φcos=

s φsin=,= =

e2: k2 2k a, φ 45°={ } 12--- 1 1

1 1= = = e3: k3 4k a, φ 90°={ } 0 0

0 1= = =

Eqs. (8) & (9):

D1

D4

D3

D10

D9

D6

D5

D8

D7

e2

e3

e1

e4

F8 = −P, F9 = −P

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

k1 k4 2k a, φ 45– °={ } 12--- 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 k4δ=

k4 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

– 2.6 (12) –


2.5

2.6


e1:

Assembly of stiffness:

D2

Element stiffness matrices: Ke kia a–

a– aa, c

2sc

sc s2

c φcos=

s φsin=,= =

e3: k3EAL

------- a, φ 45°={ } 12--- 1 1

1 1= = =

e2: k22EA

L----------- a, φ 90°={ } 0 0

0 1= = =

K EA2L-------

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):

D1D4

D3

D6

D5

D8

D7

D1 = D2 = D3 = D4 = D5 = D7 = 0; F6 = 0; F8 = −P

k1EAL

------- a, φ 45– °={ } 12--- 1 1–

1– 1= ==

0

0

0

0

0

0

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

EA2L------- 6 1–

1– 1

D6

D8

0

P–

D6

D8

⇒ PL

5EA-----------– 2

12= =

e1 e2

e3

R2EA2L------- D– 6( ) P

5---= =

R3 0= R44EA2L

----------- D– 6( ) 4P5

-------= =

R1EA2L-------D6

P5---–= =

Global equilibrium in vertical dir.: R2 R4 F6 F8+ + +P5--- 4P

5------- 0 P–+ + 0= = OK!

φ

2

4

3

51π2--- 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:

EAL

------- 1 0

0 0;

ηEAL

------- c2

sc

sc s2

c φcos=

s φsin=

e1:

e4:

e3:

EAL

------- 0 0

0 1;e2:

π2--- φ–⎝ ⎠⎛ ⎞–⎝ ⎠

⎛ ⎞cos φsin=

π2--- φ–⎝ ⎠⎛ ⎞–⎝ ⎠

⎛ ⎞sin φcos–=⎭⎪⎪⎬⎪⎪⎫

ηEAL

------- s2

s– c

s– c c2

⇒

Assembly: K EAL

------- 1 η c2

s2

+( )+ 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+( )EAL

------- 1 0

0 1

D5x

D5y

P αcos

αsin=

D5x

D5y

⇒PL

1 η+( )EA------------------------- αcos

αsin=

Thus: α

δ

D5x

D5y

δ⇒PL

1 η+( )EA-------------------------=

– 2.7 (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 kia a–

a– aa, c

2sc

sc s2

c φcos=

s φsin=,= =

e4: k2 k0 a, φ 120°={ } 14--- 1 3–

3– 3= = =

D1

D4

D3

D6

D5

D9

D1 = D2 = D3 = D4 = D5 = D6 = 0; F7 = Pcosϕ; F8 = Psinϕ

k134---k0 a, φ 0°={ } 1 0

0 0= ==

D9

D10

⇒ D0ϕcos

ϕsin=

e2

P

ϕe4

e1

e3

e3: k3 k0 a, φ 60°={ } 14--- 1 3

3 3= = =

e2: k2 a φ 0°={ } 1 0

0 0= =,

where k0EAL

-------=

D10

1.25k0 k2+ 0

0 1.5k0

D9

D10

P ϕcos

ϕsin=

But D should alignedwith the external force

Eq. 9: 1.25k0 k2+ P D0⁄=

Eq. 10: 1.5k0 P D0⁄= ⎭⎬⎫

1.25k0 k2+⇒ 1.5k0 k2⇒ 0.25k0EA4L-------= = =

Boundary Conditions:

Assembly of the stiffness matrix:

D2

Element stiffness matrix: Ki kiai ai–

ai– ai

=

Eq. (7) & (9):

D1

D4

D3

D6

D5

D8D7

e2 e3

e1

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

32---k 1 0

0 1

D7

D8

mg ϕsinϕcos

– D7

D8

⇒2mg3k

----------- ϕsinϕcos

–= =

F7 mg ϕ F8 mg ϕcos–=,sin–=

mg

ϕ

a10 0

0 1a2

14--- 3 3

3 1a3

14--- 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+ +

=

– 2.8 (12) –


2.9

2.10

Boundary Cond.:

Assembly of stiffness matrix:D2


Kiai ai–

ai– ai

=

D1

D4

D3

D6

D5

D1 = D2 = D3 = D4 = 0,

EA2l------- 2 3

3 2

D5

D6

mg 0

1–

D5

D6

⇒2mglEA

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

2–= =

F5 0 F6, mg–==

a1EA4l------- 3 3

3 1a2

EA4l------- 1 3

3 3=,=where

K

a2 0 a2–

0 a1 a1–

a2– a1– a1 a2+

EA4l-------

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

R1EA4l------- 1D5– 3D6–( ) 3

2-------mg R2

EA4l------- 3D5– 3D6–( ) 3

2---mg= =,= =

Reaction forcesat node 1:

N2

R1

R2Equilibrium in y-dir. gives: R2 N2 30cos+ 0 N2⇒ 3mg–= =

2

31 1 2

1

2

1e1

e2 e3

B.C.: D1 = D2 = D3 = D6 = 0, F4 = 2P, F5 = −P

e1:

Assembly of

D6

D5

D4

D3

D2

2Element stiffness matrices: Ke ki

a a–

a– aa, c

2sc

sc s2

c φcos=

s φsin=,= =

k1EAL

------- a, φ 0°={ } 1 0

0 0= = =

e2: k2EAL

------- a, φ 90°={ } 0 0

0 1= = = e3: k2

EAL

------- a, φ 135°={ } 12--- 1 1–

1– 1= = =

K EA2L-------

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):

EA2L------- 3 1

1 3

D4

D5

2P

p–

D4

D5

⇒PL

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

5–= =

Reaction forces:

(1): R1EA2L------- 2D5–( ) 5P

4-------= =

(2): R2EA2L------- 2D4–( ) 7P–

4----------= =

⎭⎪⎬⎪⎫ (3): R3

EA2L------- D4– D5–( ) P

4--- (Node 2)–= =

(6): R6EA2L------- D4– D5–( ) P

4--- (Node 3)–= =

(Node 1)

D1

stiffness:

– 2.9 (12) –


2.11

2.12

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

Assembly:

Element stiffness Ke kia a–

a– aa, c

2sc

sc s2

c φcos=

s φsin=,= =

e1: k1EAL

------- a, φ 45– °={ } 12--- 1 1–

1– 1= = =

e3: k32EA

L----------- a, φ 0°={ } 1 0

0 0= = =

KEA2L-------

2 0 1– 1– 1– 1

0 2 1– 1– 1 1–

1– 1– 5 1 4– 0

1– 1– 1 1 0 0

1– 1 4– 0 5 1–

1 1– 0 0 1– 1

=

Eqs. (1) & (4):

EA2L------- 2 1–

1– 1

D1

D4

P

0

D1

D4

⇒2PLEA

---------- 1

1= =

Reaction forces: (2): R2EA2L------- D4–( ) P– (Node 1)= =

(3): R3EA2L------- D– 1 D4+( ) 0 (Node 2)= =

(5): R5EA2L------- D1–( ) P (Node 3)–= =

(6): R6EA2L------- D1( ) P (Node 3)= =

23

1

1

21 2

1

e1 e2

D6

D5

D4

D3

D2

2 D1

e3

e2: k2EAL

------- a, φ 45°={ } 12--- 1 1

1 1= = =

matrices:

F1 = P, F4 = 0

1

2

3

4 5

6

7

8 Kk4---

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= =

F7 P F8, 0= =


D K1–F

D7

D8

⇒P

33k--------- 15

3

Pk--- 0.4545

0.0525= = =

F1

F2

F3

F4

F5

F6

P33------

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:

– 2.10 (12) –


2.13

2.14

1

2

3

4

5

6

7

8

9

10K

k2---

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 D2 0= =

D3 D4 0= =

D5 D6 0= =

F7 0 F9, P F10,– 0= = =


D8 0=

D K 1– F

D7

D9

D10

⇒P6k------

2–

5–

1–

= =

F1

F2

F3

F4

F5

F6

F8

P6---

2

2–

3

3

1

1

2–

=Reaction-forces:

Equation-system:

1

2

3

4

5

6K

k2---

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=

F5 P–=


D K1–F D5⇒

2P3k------- 1–= =

F1

F2

F3

F4

F6

P3---

1

1–

2

0

1

=

Reaction-forces:

Equation-system:

Element 1:

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

where .

fe kTDe=

N f2=

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⇒

23

-------P–= = = = =

Element 2:

k k 1 1–

1– 1 T; 1 0 0 0

0 0 1 0 De;

D5

D6

D3

D4

f⇒ kD5 D3–

D– 5 D3+N⇒

23---P= = = = =

1

21 2

e2

e1

– 2.11 (12) –


2.15

2.16

stel

1

2

3

4

7

8

5

6

D1 D2 0= =

D3 D4 0= =

D5 0=

F6 F7 0 F8, 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 K 1– F

D6

D7

D8

⇒Qk----

6– 4 2+

3– 2 2+

5 4 2–

Qk----

0.3431–

0.1716–

0.6569–

= = =

F1

F2

F3

F4

F5

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:

K k5---

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

D4 0=

D5 D6 0= =

D1 0=

F2 Q F3,– 0= =


(rigid support)

D K1–F

D2

D3

⇒Q

17k--------- 7–

4–

Qk---- 0.4118–

0.2353–= = =

F1

F4

F5

F6

Q85------

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;

15--- 4 3 0 0

0 0 4 3 De;

D1

D2

D3

D4

f⇒ k3D2 4D3–

3D– 2 4D3+N⇒

517------Q= = = = =

e1

e2

e3 2

1

21

1

2

Element 2:

Element 3:

k k 1 1–

1– 1 T;

15--- 1 0 0 0

0 0 1 0 De;

D5

D6

D3

D4

f⇒ kD3–

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⇒ 2kD2

D– 2

N⇒1417------Q= = = = =

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

where .

fe kTDe=

N f2=

– 2.12 (12) –


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 alinear 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, Kx [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 3L 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!

ddx------ D

dφdx------

⎝ ⎠⎛ ⎞ qφ Q+– 0= x1 x x2≤ ≤

ddx------ D

dφdx------

⎝ ⎠⎛ ⎞ qφ Q+– 0=

dvdx------D

dφdx------dx

x1

x2

∫ vqφdx

x1

x2

∫+ vQdx vDdφdx------

x1

x2

+

x1

x2

∫=

xx = 0 x = L

KxE, A

ddx------ EA

dudx------

⎝ ⎠⎛ ⎞ AKx+ 0=

dvdx------EA

dudx------dx

0

L

∫ vKxAdx v σA( )[ ]0L

+

0

L

∫=

x

KxE, A

x = 0 x = L x = 3L

u x( )

29---QL

EA-------- x

L--- 0 x L≤ ≤

136------ QL

EA-------- 3

xL---– 9

xL---⎝ ⎠⎛ ⎞

23

xL---⎝ ⎠⎛ ⎞

3–+

⎩⎪⎪⎨⎪⎪⎧

=

N x 0=( ) 2Q9

-------, N x 3L=( ) 7Q9

-------–==

Exactsoln.

kede fe=

Kx Q 2AL( )⁄ x L 1–⁄( )=

– 3.1 (15) –


3.4 The right figure shows a uniaxial bar coupled toa set of continues springs with spring constant perunit length kx [(N/m) / m]. The bar has elastic modu-lus E, cross sectional area A and is loaded by a bodyforce Kx [N/m3]. The displacement, u, in the bar isgiven 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, ρg, 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 = Lkx

KxE, A

ddx------ EA

dudx------

⎝ ⎠⎛ ⎞ kxu– AKx+ 0=

dvdx------EA

dudx------dx vkxudx

0

L

∫+

0

L

∫ vKxAdx v σA( )[ ]0L

+

0

L

∫=

kede fe=

u 0=σA Q=kx 3EA 2L

2( )⁄= Kx Q AL( )⁄=

(a) (b)

x = 2L

P

x = 3L

x = L

k

xx

x = L

g

EAρ

ddx------ EA

dudx------

⎝ ⎠⎛ ⎞ Aρg+ 0=

EAdvdx------du

dx------dx

0

L

∫ v σA( )[ ]0L ρgA vdx

0

L

∫+=

kede fe=

– 3.2 (15) –


(c) In an application, a bar (E, A) is connected to a linear spring with spring constant k, seeFigure (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, 2L and 3L. 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 kz [(N/m) / m]. Adistributed load per unit length q [N/m] is applied on thebeam. 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. Theboundary 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 2L 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 kw [N/m] and a torsion spring with stiffness kθ [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 2EA L⁄=

x

x = −L kz

q

x = L

z,w

M M

T

T

d2

dx2

-------- EId

2w

dx2

---------⎝ ⎠⎜ ⎟⎛ ⎞

kzw q–+ 0=

d2v

dx2

--------EId

2w

dx2

---------dx vkzwdx

L–

L

∫+

L–

L

∫ vT[ ]L–

L dvdx------M

L–

L

– vqdx

L–

L

∫+=

T EIw″( )′–= M EIw″–=

kede fe=

M0

z,w

x

x = −L x = L

w′ L( )kz λEI L

4⁄=

q Q 2L( )⁄=

Q

kw

kθ

2L, EI

kw3EI

2L3

---------= kθEIL------=

– 3.3 (15) –


3.7 The figure to the right shows a beam, which is loadedby its dead weight per unit volume ρg, where ρ is the den-sity and g acceleration of gravity. The beam has the elas-tic modulus E, moment of inertia I and cross sectionalarea A. The deflection of the beam w is given by the solu-tion to the differential equation

.

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

,

where v is an arbitrary weight function, T is shear force and M is moment (definedaccording to the figure above). The relations and havebeen utilized at the boundaries.


(c) The figure to the right shows a cantilever beamattached to a linear spring with spring constant

. The beam is loaded by its dead weightand by a point force P. Analyse the beam by use of atwo-node beam element and calculate its deflectionand rotation (slope) at for the special case that

, and EI = constant.

3.8 The figure to the right shows a beam with bending stiff-ness EI subjected to a distributed triangular load acting down-wards with a total resultant force equal to Q. The beam haselasticity modulus E and moment of inertia I. The deflectionof the beam (vertical displacement), w, is given by the solu-tion to the differential equation

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

,

where v is an arbitrary weight function, T is a shear force and M is a moment (see thefigure above). The relations and have been used at theboundaries.

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

x

x = −L x = L

z,w

M M

T

T

g

d2

dx2

-------- EId

2w

dx2

---------⎝ ⎠⎜ ⎟⎛ ⎞

ρgA+ 0=

d2v

dx2

--------EId

2w

dx2

---------dx

L–

L

∫ vT[ ]L–

L dvdx------M

L–

L

– ρgA vdx

L–

L

∫–=

T EIw″( )′–= M EIw″–=

kede fe=

P

z,w

x

x = −L x = L

kg

k ηEI L3⁄=

x L=η 1 2⁄= P ρgAL=

x

x = −L x = L

z,w

M M

T

T

Q

d2

dx2

-------- EId

2w

dx2

---------⎝ ⎠⎜ ⎟⎛ ⎞

1xL---+⎝ ⎠

⎛ ⎞ Q2L------+ 0=

d2v

dx2

--------EId

2w

dx2

---------dx

L–

L

∫ vT[ ]L–

L dvdx------M

L–

L

– Q v12--- 1

xL---+

⎝ ⎠⎛ ⎞ dx

L------

L–

L

∫–=

T EIw″( )′–= M EIw″–=

kede fe=

– 3.4 (15) –


(c) In an application, the beam is clamped at x = L andthe 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 3L/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).


,

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


(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 ofthe 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

xx = −L

x = Lz,wQ

2L, EI P

ξ x L⁄=

Heat conductionConvection

x 0= x 3L 2⁄=T x( )

x

ddx------ kA

dTdx------

⎝ ⎠⎛ ⎞ qA hP T T∞–( )–+ 0=

T∞

dvdx------kA

dTdx------dx vhPTdx

0

3L 2⁄

∫+

0

3L 2⁄

∫ v Q–( )[ ]0

3L 2⁄v qA hPT∞+( )dx

0

3L 2⁄

∫+=

Q k– AdT dx⁄=

kede fe=

T 4T∞= T T∞= x 3L 2⁄=hP 12kA L

2⁄=

x 3L 2⁄=

LP A 96⁄=

Heat conductionConvection

T x( )

x

x = x1 x = x2T∞

– 3.5 (15) –


.


,

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

ddx------ kA

dTdx------

⎝ ⎠⎛ ⎞ hP T T∞–( )– 0=

dvdx------kA

dTdx------dx vhPTdx

x1

x2

∫+

x1

x2

∫ v Q–( )[ ]x1

x2 vhPT∞dx

x1

x2

∫+=

Q k– AdT dx⁄=

keTe fe=

hPL 16kA L⁄=

Q hPLT∞=x 0= T T∞= x L=

2L, EI

0 1ξ

−1

d3

d4d2

d1

w ξ( ) N1d1 N2d2 N3d3 N4d4+ + + Nde B, d2N

dx2

---------- 1

L2

-----d2N

dξ2----------= = = =Beam element:

BTBdx

L–

L

∫1

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 3 3L–

3L 2L2

3L– 4L2

= NTNdx

L–

L

∫L

105---------

78 22L 27 13L–

22L 8L2

13L 6L2

–

27 13L 78 22L–

13L– 6L2

– 22L– 8L2

=

N1 2 3ξ– ξ3+( ) 4⁄ N2 L 1 ξ– ξ2

– ξ3+( ) 4⁄=,=

N3 2 3ξ ξ3–+( ) 4 N4 L 1– ξ– ξ2

+ ξ3+( ) 4⁄=,⁄=

Deflection:

φ ξ( ) N1φ1 N2φ2+ N1 N2

φ1

φ2

= = N1 1 ξ–= N2 ξ=

1 2φ1 φ2

L

0 1ξ

1D:

NTNdx

0

L

∫ dx Ldξ={ } L6--- 2 1

1 2= =

N

dNT

dx---------- N

dx------dx

0

L

∫1L--- 1 1–

1– 1=

NT 12--- 1

xL---+

⎝ ⎠⎛ ⎞ dx

L------

L–

L

∫1

30------

9

4L

21

6L–

=

– 3.6 (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

dx------ D

dφdx------

⎝ ⎠⎛ ⎞ qφ Q+–

⎝ ⎠⎛ ⎞ dx

x1

x2

∫ 0=

vddx------ D

dφdx------

⎝ ⎠⎛ ⎞

⎝ ⎠⎛ ⎞ dx

x1

x2

∫ vDdφdx------

x1

x2 dvdx------D

dφdx------dx

x1

x2

∫–=

dvdx------D

dφdx------dx

x1

x2

∫ vqφdx

x1

x2

∫+ vQdx vDdφdx------

x1

x2

+

x1

x2

∫= φ φ0 eller Ddφdx------ D

dφdx------

0

på x xi= = =

Essential B.C. Natural B.C.

Weak form:

φ ξ( ) Nφe= N 1 ξ– ξ= φeφ1

φ2

=

dφdx------ dφ

dξ------dξ

dx------ 1

l---dN

dξ-------φe Bφe= = =⇒

v ξ( ) Nβ βTNT= = βT

β1 β2=

dvdx------ dv

dξ------dξ

dx------ βT1

l---dNT

dξ---------- βTBT

= = =⇒

βT BTDBldξ

0

1

∫ NTqNldξ

0

1

∫+ φe βT NTQldξ NT

Ddφdx------

0

1+

0

1

∫=

kqfQ fs

keφe⇒ fe= fe fQ fs+=därβT is an arbitrary vector

The element matrices becomes:

kD BTDBldξ

0

1

∫1l--- 1–

1D

1l--- 1– 1 ldξ

0

1

∫1l--- 1 1–

1– 1Ddξ

0

1

∫= = =

kq NTqNldξ

0

1

∫1 ξ–

ξq 1 ξ– ξ ldξ

0

1

∫ l 1 ξ–( )21 ξ–( )ξ

1 ξ–( )ξ ξ2qdξ

0

1

∫= = =

kD

ke kD kq+= och

fQ NTQldξ

0

1

∫ l 1 ξ–

ξQdξ

0

1

∫= =

– 3.7 (15) –


3.3 (b) , where and .

3.3 (c)

3.4(a)

3.4(b) Displacement interpolation:

Weight function:

3.4(c)

kede fe= ke B0

L

∫TEAB xd= fe N

0

L

∫TKxA x NT σA( )[ ]

0

L+d=

D3D2D1

EA2L-------

2 2– 0

2– 3 1–

0 1– 1

0

D2

0

R1

Q 3⁄R3 2Q 3⁄+

=Eqn. system:

D229---QL

EA--------=

R129---Q–=

R379---Q–=

⇒

Node/element division:

Weighted residual: vd

dx------ EA

dudx------

⎝ ⎠⎛ ⎞ kxu– AKx+ dx

0

L

∫ 0=

Integration by parts: vddx------ EA

dudx------

⎝ ⎠⎛ ⎞ dx

0

L

∫ v EAdudx------

⎝ ⎠⎛ ⎞

0

L dvdx------EA

dudx------dx

0

L

∫–=

dvdx------EA

dudx------dx

0

L

∫ vkxudx

0

L

∫+ vAKxdx

0

L

∫ v EAdudx------

⎝ ⎠⎛ ⎞

0

L+=

(2) inserted into (1)

(1)

(2)

gives the weak form:

u Nde=dudx------ dN

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

v Nbe beTNT

= =dvdx------ be

TdNT

dx---------- be

TBT

= =

beT BT

EAB x NTkxN xd

Le

∫+dLe

∫ de beT NT

KxA x NT σA( )[ ]0Le

+dLe

∫=

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

ke fe

but beT

is arbitrary kede⇒ fe=

D2D1 D3

Le = L/2Le = L/2N 1 ξ–( ) ξ= B

2L--- 1– 1=

Element-

NT QAL-------A

L2---dξ

0

1

∫Q4---- 1

1=

BTEABL

2---dξ

0

1

∫2EA

L----------- 1 1–

1– 1=

NT3EA

2L2

-----------NL2---dξ

0

1

∫EA8L------- 2 1

1 2=

⎭⎪⎪⎪⎬⎪⎪⎪⎫

keEA8L------- 18 15–

15– 18;=⇒

Assembly: EA8L-------

18 15– 0

15– 36 15–

0 15– 18

D1

D2

D3

Q4----

1

2

1

R

0

Q

+=point force

reaction force

Equation (2) & (3):D2

D3

⇒QL

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

140

QLEA-------- 0.525

0.993= =

matrices:

– 3.8 (15) –


3.5(a)


Weight function:

3.5(c)

Weighted residual: vddx------ EA

dudx------

⎝ ⎠⎛ ⎞ Aρg+ dx

0

L

∫ 0=

Integration by parts: vddx------ EA

dudx------

⎝ ⎠⎛ ⎞ dx

0

L

∫ v EAdudx------

⎝ ⎠⎛ ⎞

0

L dvdx------EA

dudx------dx

0

L

∫–=

with

(1)

(2)

gives the weak form EA dvdx------

dudx------dx

0

L

∫ v σA( )[ ]0

LAρg vdx

0

L

∫+=σ Edudx------=

(2) inserted into (1)

u Nde=dudx------ dN

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

v Nbe beTNT

= =dvdx------ be

T dNT

dx---------- be

T B

T= =

ke febut be

Tis arbitrary kede⇒ fe=

beT

EA BTBdx

0

L

∫ de beT NT σA( )[ ]

0

LAρg NT

dx

0

L

∫+=

⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

Inserted into theweak form gives:

Element matrices:

ke EA BTBLdξ0

1

∫EAL

------- 1 1–

1– 1= = fe Aρg NT

Ldξ0

1

∫AρgL

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

1= =

ke k 1 1–

1– 1= where k 2

EAL

-------=

Truss element:

Spring element:

D1 D2 D3 D4

Element lengths: L1 = L2 = L3 = LL3L2L1

Boundary conditions: D1 = D4 = 0

Diskretization:

Assembly: KEAL

-------

1 1– 0 0

1– 2 1– 0

0 1– 3 2–

0 0 2– 2

= F

R1

P

0

R4

AρgL2

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

1

2

1

0

+=

(reaction forces: R1 & R4)

EAL

------- 2 1–

1– 3

D2

D3

P AρgL+

AρgL 2⁄

D2

D3

⇒PL

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

1

AρgL2

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

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

– 3.9 (15) –


3.6(a)


Weight function:

3.6(c)

Weighted residual: v EIw″( )″ kzw q–+[ ]dx

L–

L

∫ 0=

Integration by parts: v EIw″( )″[ ]dx

L–

L

∫ v EIw″( )′[ ]L–

Lv′ EIw″( )′[ ]dx

L–

L

∫–= =

(2) inserted into (1) with

(1)

(2)v EIw″( )′[ ]L–

Lv′ EIw″( )[ ]

L–

L– v″EIw″dx

L–

L

∫+=

gives the weak form:

v″EIw″dx vkzwdx

L–

L

∫+

L–

L

∫ vT[ ]L–

Lv′M[ ]

L–

L– vqdx

L–

L

∫+=

T EIw″( )′–= and M EIw″–=

w Nde=d

2w

dx2

--------- d2N

dx2

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

v Nbe beTNT

= =d

2v

dx2

-------- beT d

2NT

dx2

------------ beT B

T= =

beT BT

EIB

L–

L

∫ x NT

L–

L

∫ kzN xd+d de beT NT

L–

L

∫ q x NTT[ ]

L–

L dNT

dx----------M

L–

L

–+d=

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

ke fe

but beT

is arbitrary kede⇒ fe=

K BTEIBdx NTλEI

L4

---------Ndx

L–

L

∫+

L–

L

∫EI

2L3

---------

3 3L 3– 3L

3L 4L2

3L– L2

3– 3L– 3 3L–

3L L2

3L– 4L2

λEI

105L3

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

78 22L 27 13L–

22L 8L2

13L 6L2

–

27 13L 78 22L–

13L– 6L2

– 22L– 8L2

+= =

F

R1

R2

R3

M0

=

Reactionforces/moments

Stiffnessmatrix

Loadvector

2EIL

--------- 8λEI105L-------------+

⎝ ⎠⎛ ⎞ d4 M0=

w′ L( ) d4105

210 8λ+( )--------------------------

M0L

EI-----------= =⇒

Eq. (4):

– 3.10 (15) –


3.6(d)

3.7(a)


Weight function:

Element

Kbeam BTEIBdx

L–

L

∫EI

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 3 3L–

3L 2L2

3L– 4L2

= =

Beam (dof 1 to 4) Tensile & torsion springs

The total stiffness matrix is obtained by assemblyK

EI

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 6 3L–

3L 2L2

3L– 6L2

=of the stiffnesses from the beam and the springs:

FbQ2L------ NT

dx

L–

L

∫Q2----

1

L 3⁄1

L– 3⁄

= = Fs

R1

R2

0

0

=

Force vector: F Fb Fs+=

where

Reactionforce & moment

External pointforce & moment

EI

2L3

---------6 3L–

3L– 6L2

w2

θ2

Q2---- 1

L– 3⁄

w2

θ2

⇒QL

3

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

1= =

Displacement boundary conditions: w1 = θ1 = 0. The reduced equation system becomes

stiffnessmatrices

(only dof 3 and 4)

kw3EI

2L3

---------= kθEIL------=

Weighted residual: v EIw″( )″ ρgA+[ ]dxL–

L

∫ 0=

Integration by parts: v EIw″( )″[ ]dxL–

L

∫ v EAw″( )′[ ]L–


L–

L

∫–= =

(1)

(2)v EAw″( )′[ ]L–

Lv′ EAw″( )[ ]

L–

L– v″EIw″dx

L–

L

∫+=

v″EIw″dxL–

L

∫ vT[ ]L–

Lv ′M[ ]

L–

L– ρgA vdx

L–

L

∫–=

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

w Nde=d

2w

dx2

--------- d2N

dx2

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

v Nbe beTN

T= =

d2v

dx2

-------- beT d

2NT

dx2

------------ beT B

T= =

beT BT

EIB

L–

L

∫ xd de beT NT

T[ ]L–

L dNT

dx----------M

L–

L

– ρgA NTxd

L–

L

∫–=

⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

ke fe

but beT

is arbitrary

kede⇒ fe=

– 3.11 (15) –


3.7(c)

3.8 (a)

3.8 (b) Displacement interpolation:

Weight function:

Inserted into the weak form gives

Element

Kbalk BTEIBdx

L–

L

∫EI

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 3 3L–

3L 2L2

3L– 4L2

= = Kfjäder kEI

2L3

---------= =

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

Assembly of total stiffness matrix

K Kbalk Kfjäder+EI

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 4 3L–

3L 2L2

3L– 4L2

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

Fdistributed ρgA NTdx

L–

L

∫– ρgAL

1

L 3⁄1

L– 3⁄

–= = Fpoint

R1

R2

P–

0

=

Force vector: F Fdistributed Fpoint+=

where

Reaktionforce/moment

External pointforce/moment

P ρgAL= inserted gives:

F

R1 ρgAL–

R2 ρgAL2

3⁄–

2ρgAL–

ρgAL2

3⁄

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

EI

2L3

---------4 3L–

3L– 4L2

w2

θ2

ρgAL 2–

L 3⁄

w2

θ2

⇒ ρgAL

3

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

4 3⁄–= =

Displacement boundary conditions: w1 = θ1 = 0

stiffnessmatrices:

Weighted residual: v EIw″( )″ 1xL---+⎝ ⎠

⎛ ⎞ Q2L------+ dx

L–

L

∫ 0=

Integration by parts: v EIw″( )″[ ]dx

L–

L

∫ v EAw″( )′[ ]L–


L–

L

∫–= =

(1)

(2)v EAw″( )′[ ]L–

Lv′ EAw″( )[ ]

L–

L– v″EIw″dx

L–

L

∫+=

v″EIw″dx

L–

L

∫ vT[ ]L–

Lv′M[ ]

L–

L– Q v

12--- 1

xL---+⎝ ⎠

⎛ ⎞ dxL------

L–

L

∫–=

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

w Nde=d

2w

dx2

--------- d2N

dx2

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

v Nbe beTNT

= =d

2v

dx2

-------- beT d

2NT

dx2

------------ beT B

T= =

beT BT

EIB

L–

L

∫ xd de beT NT

T[ ]L–

L dNT

dx----------M

L–

L

– Q NT12--- 1

xL---+

⎝ ⎠⎛ ⎞ dx

L------

L–

L

∫–=

⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

ke fe

but beT

is arbitrary

kede⇒ fe=

– 3.12 (15) –


3.8 (c)

3.9(a)

3.9 (b) Temperature interpolation:

Weight function:

Inserted into the weak form gives:

Element stiffness matrix: K BTEIBdx

L–

L

∫EI

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 3 3L–

3L 2L2

3L– 4L2

= =

Fb Q NT12--- 1

xL---+

⎝ ⎠⎛ ⎞ dx

L------

L–

L

∫–Q30------

9

4L

21

6L–

–= = Fpoint

P–

R2

R3

R4

=

Nodal force vector: F Fb Fpoint+=

where Reactionforce/moment

External point force

w ξ( ) N1 ξ( )d123---PL

3

EI--------- 1

5---QL

3

EI----------+⎝ ⎠

⎛ ⎞ 2 3ξ– ξ3+

4---------------------------⎝ ⎠⎛ ⎞ where ξ,–

xL---= = =

Reduced equation system, Eq. (1): EI

2L3

---------3d1 P–3Q10-------– d1⇒

23---PL

3

EI---------–

15---QL

3

EI----------–= =

Displacement boundary conditions: d2 = d3 = d4 = 0 => Reaction forces

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

FEM, discretization: 2L, EI

0 1ξ

−1

d3

d4d2

d1

(one element)


dx------ kA

dTdx------

⎝ ⎠⎛ ⎞ qA hP T T∞–( )–+ dx

0

3L 2⁄

∫ 0=

Integration by parts: vd

dx------ kA

dTdx------

⎝ ⎠⎛ ⎞ dx

0

3L 2⁄

∫ v kAdTdx------

⎝ ⎠⎛ ⎞

0

3L 2⁄ dvdx------ kA

dTdx------

⎝ ⎠⎛ ⎞ dx

0

3L 2⁄

∫–=


(1)

dvdx------kA

dTdx------dx vhPTdx

0

3L 2⁄

∫+

0

3L 2⁄

∫ v Q–( )[ ]0

3L 2⁄v qA hpT∞+( )dx

0

3L 2⁄

∫+=

Q– kAdTdx------= gives the weak form:

(2)

T NTe=dTdx------ dN

dx-------Te BTe= =

v Nbe beTNT

= =dvdx------ be

T dNT

dx---------- be

T B

T= =

beT BT

kABx1

x2

∫ x NT

x1

x2

∫ hPN xd+d Te beT NT

x1

x2

∫ qA hpT∞+( ) x NTQ–( )[ ]

x1

x2+d=

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

ke febut be

Tis arbitrary keTe⇒ fe=

– 3.13 (15) –


3.9 (c)

3.9 (d)

3.10 (a)

K kAL

------

4 1– 0 0

1– 8 1– 0

0 1– 8 1–

0 0 1– 4

=

For one element applies:

ke1

L 2⁄---------- 1–

1kA

1L 2⁄---------- 1– 1

L2---dξ 1 ξ–

ξhP 1 ξ– ξ

L2---dξ

0

1

∫+

0

1

∫ hPL 12kAL

------=⎩ ⎭⎨ ⎬⎧ ⎫ kA

L------ 4 1–

1– 4= = =

Assembly of system matrix gives

e3e2e1 4321

T4T3T2T1

FEM - model:

fe1 ξ–

ξqA hPT∞+( )L

2---dξ

0

1

∫ q 0 hPL 12kAL

------=;=⎩ ⎭⎨ ⎬⎧ ⎫

3kAL

------T∞1

1= = =

F 3kAL

------T∞

1

2

2

1

QR1

0

0

QR4–

+=and the r.h.s

Equation system:

kAL

------

4 1– 0 0

1– 8 1– 0

0 1– 8 1–

0 0 1– 4

4T∞

T2

T3

T∞

3kAL

------T∞

1

2

2

1

QR1

0

0

QR4–

+=

4 1– 0 0

1– 8 1– 0

0 1– 8 1–

0 0 1– 4

4

T2 T∞⁄

T3 T∞⁄

1

⇔

3

6

6

3

QR1L kAT∞( )⁄

0

0

QR4L– kAT∞( )⁄

+=

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

8 1–

1– 8

T2 T∞⁄

T3 T∞⁄6 1 4 0 1⋅+⋅–{ }–

6 0 4 1 1⋅–⋅{ }–

T2

T3

T∞21------ 29

22=⇒ T∞

1.3809

1.0476= =

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

Q–( )[ ]ξ1

ξ2c

0

Q2–

Q1–

0–=

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

NT Q–( )[ ]ξ1

ξ2 kA8L------ 0 0

0 1

T1

T2

kA8L------ 0

T∞+=⇒

Q2 hA T2 T∞–( )=

where hA h PL 96⁄( ) kA 8L( )⁄ ·= =

The equation system is modified according to:

4 1– 0 0

1– 8 1– 0

0 1– 8 1–

0 0 1– 418---+

4

T2 T∞⁄

T3 T∞⁄

T4 T∞⁄

3

6

6

318---+

QR1L kAT∞( )⁄

0

0

0

T2

T3

T4

⇒+ 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


dx------ kA

dTdx------

⎝ ⎠⎛ ⎞ hP T T∞–( )– dx

x1

x2

∫ 0=

Integration by parts: vddx------ kA

dTdx------

⎝ ⎠⎛ ⎞ dx

x1

x2

∫ v kAdTdx------

⎝ ⎠⎛ ⎞

x1

x2 dvdx------ kA

dTdx------

⎝ ⎠⎛ ⎞ dx

x1

x2

∫–=


(1)

(2)

dvdx------kA

dTdx------dx vhPTdx

x1

x2

∫+

x1

x2

∫ v Q–( )[ ]x1

x2 vhpT∞dx

x1

x2

∫+=

Q– kAdTdx------= gives the weak form:

– 3.14 (15) –


3.10 (b) Temperature interpolation:

Weight function:

Inserted into the weak form gives:

3.10 (c)

T NTe=dTdx------ dN

dx-------Te BTe= =

v Nbe beTNT

= =dvdx------ be

T dNT

dx---------- be

T B

T= =

beT BT

kABx1

x2

∫ x NT

x1

x2

∫ hPN xd+d Te beT NT

x1

x2

∫ hpT∞ x NTQ–( )[ ]

x1

x2+d=

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

ke fe

but beT

is arbitrary keTe⇒ fe=

K23---kA

L------

7 1– 0

1– 14 1–

0 1– 7

=

For one element applies:

ke1

L 2⁄---------- 1–

1kA

1L 2⁄---------- 1– 1

L2---dξ 1 ξ–

ξhP 1 ξ– ξ

L2---dξ

0

1

∫+

0

1

∫ hPL 16kAL

------=⎩ ⎭⎨ ⎬⎧ ⎫ 2kA

3L---------- 7 1–

1– 7= = =

Assembly of the stiffness matrix gives:

e2e1 321

T3T2T1

FEM - model:

fe1 ξ–

ξhPT∞

L2---dξ

0

1

∫hPLT∞

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

1= =

Equation system:

23---kA

L------

7 1– 0

1– 14 1–

0 1– 7

T1

T2

T3 T∞=

kAL

------

20

8

44QR

hPLT∞-----------------–

T∞=7 1– 0

1– 14 1–

0 1– 7

T1

T2

T∞

30

12

66QR

hPLT∞-----------------–

T∞=⇔

Eq. (1) & (2) gives:

System matrix:

7 1–

1– 14

T1

T2

30 0( )–

12 1–( )–T∞

T1

T2

⇒197------ 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

1QR

hPLT∞-----------------–

kAL

------ hPLkA L⁄--------------⎝ ⎠⎛ ⎞ T∞

4------

⎩ ⎭⎨ ⎬⎧ ⎫ kA

L------

20

8

44QR

hPLT∞-----------------–

T∞= = =

= 16

– 3.15 (15) –


4. FEM: trusses and beams

4.1 A uniaxial bar is modelled by a linear truss element.For a certain applied load, the node displacementshown in the figure results. (a) Show that the straindeveloping in the element is and (b) show

that if ε0 = 0, the element is subjected to a rigid bodymotion equal to δ0.

4.2 Derive the four shape functions forthe uniaxial “cubic” element shown tothe right. Express the shape functionsusing the natural coordinate ξ.

4.3 The figure to the right shows a uniaxial isoparamet-ric element where a 2nd degree polynomial is used forthe interpolation of the displacement. The coordinates ofthe 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} areknown 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 anuniformly distributed axial load Kx = q0 and a point forceP. Analyse the bar by use of the finite element methodwith (a) one linear element and (b) two linear element.Compare the solutions with the exact solution givenbelow the figure.

x = 0 x = Lx

u1 δ0= u2 δ0 ε0L+=ε x( ) ε0=

Nod 1 Nod 3 Nod 4 Nod 2

ξ = −1 ξ = 1ξ = −1/3 ξ = 1/3ξ

xL−L λL

Nod 1 Nod 3 Nod 2

1 λ 1< <–

N1 1 ξ–( )ξ– 2⁄=1 23

Coordinates:

x1 x2x3

x

1 23

−1 10ξx ξ( ) Nkxk

k 1=

3

∑=

Primary variable:

φ ξ( ) Nkφkk 1=

3

∑=N2 1 ξ+( )ξ 2⁄=

N3 1 ξ2–=

x = 0 x = LE, A

Kx q0=P

Exact solution:

u x( ) PxEA-------

q0L2

E----------- x

L--- 1

2--- x

L---⎝ ⎠⎛ ⎞

2–

⎝ ⎠⎛ ⎞+=

σ x( ) PA--- q0L 1 x L⁄–( )+=

– 4.1 (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 displacementinterpolation 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 offreedom at each node, this means thatthe deflection of the beam must be ableto take the form

,

where δ and θ are parameters describing an arbitrary rigid body motion as illustrated in thefigure to the right. Show that the displacement interpolation of the element can satisfy a rigidbody motion as described above.

4.7 The figure to the leftshows an initially straightbeam element that is sub-jected to a deformation statethat results in a constant cur-vature , where R0is 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 δ0 at the midpoint of the element (x = 0).

4.8 A cantilever beam is loaded by a point force P, amoment M and a uniformly distributed force per unitlength (Q is the total resultantforce) according to the figure to the right. The bend-ing stiffness of the beam is EI and its length 2L. Ana-lyse the beam with FEM and use a 2-node beamelement (3rd degree polynomial for the interpolationof the deflection). Carry out the analysis using (a)one element and (b) two elements. Compare theresults with the exact solution shown below the fig-ure.

x = 0 x = 2Lx32---L=

E, A

Kx q0 N

m3

-------=1 2L

ξ0 1

N1 1 ξ N2,– ξ= = dx Ldξ=

u1 u2

(a) (b)

θ2

w2

w1

θ1

θ

δ

x

− L 0 L

θ1 θ2 θ= =

w1 δ θL–=

w2 δ θL+=

w x( ) δ θx+=

θ2

w2

w1

θ1

x

− L 0 L

R01κ0-----=

w1 a b– c–=

θ1 b L 2c L⁄+⁄=

w2 a b c–+=

θ2 b L 2c L⁄–⁄=

δ0θ0

κ0 1 R0⁄=

w″ κ–=κ0

P

M

q x( )

x x = 2L

[N/m]

w x( ) PL3

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

xL---⎝ ⎠⎛ ⎞

2 xL---⎝ ⎠⎛ ⎞

3–

⎝ ⎠⎛ ⎞ ML

2

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

L---⎝ ⎠⎛ ⎞

2+=

QL

3

EI---------- 1

2--- x

L---⎝ ⎠⎛ ⎞

2 16--- x

L---⎝ ⎠⎛ ⎞

3–

148------ x

L---⎝ ⎠⎛ ⎞

4+

⎝ ⎠⎛ ⎞+

Exact solution:

q x( ) Q 2L( )⁄=

– 4.2 (6) –


4.9 The figure below shows a cantilever beam, which is subjected to a distributed force, amoment 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 inthe FEM-model are: . Determine the force vector, where also the reactionforces should be indicated.

FORMULAS

x 0 2L 3L,,{ }=

x = 0x = 2L

P

x

z

x = 3L

q x( )Q2L------ x

L---

Nm----⋅=

M

2L, EI

0 1ξ

−1

d3

d4d2

d1

w ξ( ) N1d1 N2d2 N3d3 N4d4+ + + Nde B, d2N

dx2

---------- 1

L2

-----d2N

dξ2----------= = = =Balkelement:

BTBdx

L–

L

∫1

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 3 3L–

3L 2L2

3L– 4L2

= NTNdx

L–

L

∫L

105---------

78 22L 27 13L–

22L 8L2

13L 6L2

–

27 13L 78 22L–

13L– 6L2

– 22L– 8L2

=

N1 2 3ξ– ξ3+( ) 4⁄ N2 L 1 ξ– ξ2

– ξ3+( ) 4⁄=,=

N3 2 3ξ ξ3–+( ) 4 N4 L 1– ξ– ξ2

+ ξ3+( ) 4⁄=,⁄=

Deflection:

φ ξ( ) N1φ1 N2φ2+ N1 N2

φ1

φ2

= = N1 1 ξ–= N2 ξ=

1 2φ1 φ2

L

0 1ξ

1D:

NTNdx

0

L

∫ dx Ldξ={ } L6--- 2 1

1 2= =

N

dNT

dx---------- N

dx------dx

0

L

∫1L--- 1 1–

1– 1=

NTdξ

1–

1

∫

1

L 3⁄1

L 3⁄–

= NTξdξ1–

1

∫1

15------

6–

L–

6

L–

=

– 4.3 (6) –


Solutions

4.1

4.2 Use Lagrange interpolation:

4.3

4.4

u N1u1 N2u2+ δ0 ε0ξL+ x ξL={ } δ0 ε0x+= = = =

ε x( ) dudx------ 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=

N1 1 ξ–( ) 1 3ξ–( ) 1 3ξ+( ) C1⁄= N1 1–( ) 1= C1⇒ 16–=

N2 1 ξ+( ) 1 3ξ–( ) 1 3ξ+( ) C2⁄= N2 1( ) 1= C2⇒ 16–=

N3 1 ξ+( ) 1 ξ–( ) 1 3ξ–( ) C3⁄= N3 1– 3⁄( ) 1= C3⇒ 16 9⁄=

N4 1 ξ+( ) 1 ξ–( ) 1 3ξ+( ) C4⁄= N4 1 3⁄( ) 1= C4⇒ 16 9⁄=

ε dudx------ du

dξ------dξ

dx------= =

dx∂Nk

∂ξ---------xkdξ

k 1=

3

∑ L 1 2λξ–( )dξ= =⎭⎪⎪⎬⎪⎪⎫

ε⇒u2 u1–

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

u1 u2 2u3–+( )L

-------------------------------------ξ+1

1 2λξ–( )-----------------------=

Strain:

Note! singular for 2λξ 1=

(a) One element solution, discretization:

keEAL

------- 1 1–

1– 1= fb NT

KxALdξ0

1

∫ ALq01 ξ–

ξdξ

0

1

∫ALq0

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

1= = =

Uniform load contribution to the nodal force vector:

Eq. EAL

------- 1 1–

1– 1

D1 0=

D2

R

P

ALq0

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

1+=

Element stiffness matrix:D2D1

D3D2D1

e1 e2

D2PLEA-------

q0L2

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

R P– ALq0–=

Eq. 2 gives:system:

Eq. 1 then gives:

(b) Two element solution, discretization:

Element stiffness matrix:

k1 k2EAL

------- 1 1–

1– 1= = fb1 fb2 NT

KxAL2---dξ

0

1

∫AL2

-------q01 ξ–

ξdξ

0

1

∫ALq0

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

1= = = =

Uniform load contribution to the nodal force vector:

Eq.2EA

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

1 1– 01– 2 1–

0 1– 1

D1 0=

D2

D3

R

0

P

ALq0

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

12

1

+=D2

D3

PL2EA----------- 1

2

q0L2

8E----------- 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.

– 4.4 (6) –


4.5

4.6

4.7

4.8

D3D2D1

e1 e2 Element stiff-ness matrices:

k1 k2EAL

------- 1 1–

1– 1= =

Element 2: fQ NTKxALdξ

0

1

∫ ALq01 ξ–

ξdξ

1 2⁄

1

∫ALq0

8------------- 1

3= = =

Assembly: EAL

-------1 1– 0

1– 2 1–

0 1– 1

D1 0=

D2

D3 0=

ALq0

8-------------

0

1

3

R1

0

R3

+=

B.C. Reaction forces

D2⇒q0L

2

16E-----------=

Reaction forces: R1EAL

------- D2–( ) ALq0

16-------------–= = R3

EAL

------- D2–( )3ALq0

8----------------–

7ALq0

16----------------–= =

Note! R1 R3+( )– KxAdx

3L 2⁄

2L

∫ALq0

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

w x( ) w ξL( ) Nidi

i 1=

4

∑ N1 δ Lθ–( ) N2θ N3 δ Lθ+( ) N4θ+ + += = =

δ N1 N3+( ) θ N2 N4 L N3 N1–( )+ +( )+= δ θξL+ δ θx+= =

κ0 w″–d

2Ni

dx2

-----------di

i 1=

4

∑–N1

″

L2

---------w1

N2″

L2

---------θ1

N3″

L2

---------w2

N4″

L2

---------θ2+ + +⎝ ⎠⎜ ⎟⎛ ⎞

–2c

L2

------= = = =

δ0 w x 0=( ) Ni ξ 0=( )di

i 1=

4

∑ N1w1 N2θ1 N3w2 N4θ2+ + +( )ξ 0=

a= = = =

θ0 w′ x 0=( )dNi ξ 0=( )

dx---------------------------di

i 1=

4

∑N1

′L

--------w1

N2′

L--------θ1

N3′

L--------w2

N4′

L--------θ2+ + +⎝ ⎠

⎛ ⎞

ξ 0=

bL---= = = =

(a) Discretization, one element:D1 D3

D2 D4

ke BTEIBdx

L–

L

∫EI

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 3 3L–

3L 2L2

3L– 4L2

= =


fb NT Q2L------Ldξ

1–

1

∫Q6----

3

L

3

L–

= =

Distributed load contributionto the nodal force vector:

– 4.5 (6) –


4.8 cont.

4.9

D5

D6

Eq.

EI

2L3

---------

3 3L 3– 3L

3L 4L2

3L– 2L2

3– 3L– 3 3L–

3L 2L2

3L– 4L2

D1 0=

D2 0=

D3

D4

R

MR

P

M

Q6----

3

L

3

L–

+=

R P– Q–=

Eq. 3 & 4 give:

system:

Eq. 1 & 2 then give the reaction forces: MR 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

EIBdx

L– 2⁄

L 2⁄

∫4EI

L3

---------

3 3L 2⁄ 3– 3L

3L 2⁄ L2

3L 2⁄– L2

2⁄3– 3L 2⁄– 3 3L 2⁄–

3L 2⁄ L2

2⁄ 3L 2⁄– L2

= = = fb1 fb2 NT Q2L------L

2---dξ

1–

1

∫Q24------

6

L

6

L–

= = =

D3

D4

D5

D6

⇒PL

3

6EI---------

5

9 L⁄16

12 L⁄

ML2

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

1

2 L⁄4

4 L⁄

QL3

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

17

28 L⁄48

32 L⁄

+ +=

4EI

L3

---------

3 3L 2⁄ 3– 3L 0 0

3L 2⁄ L2

3L 2⁄– L2

2⁄ 0 0

3– 3L 2⁄– 6 0 3– 3L 2⁄

3L 2⁄ L2

2⁄ 0 2L2

3L 2⁄– L2

2⁄0 0 3– 3L 2⁄– 3 3L 2⁄–

0 0 3L 2⁄ L2

2⁄ 3L 2⁄– L2

D1 0=

D2 0=

D3

D4

D5

D6

R

MR

0

0

P

M

Q24------

6

L

12

0

6

L–

+=

D3

D4

PL3

EI--------- 8 3⁄

2 L⁄ML

2

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

2 L⁄QL

3

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

1 L⁄+ +=

Eq. 3 - 6 give: Eq. 1 & 2 then give the reactionR P– Q–=

MR 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 forcesacting at the nodes are in global equilibrium, i.e. external loads are in balancewith 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+( ) dx⇒ Ldξ ,= = fb NTqLdξ where q

1–

1

∫Q2L------ 1 ξ+( )= =

fb12 3ξ– ξ3

+( )4

-------------------------------- Q2L------ 1 ξ+( )Ldξ

1–

1

∫3Q10-------= = fb2

2QL15

-----------= fb37Q10-------= fb4

QL5

--------–=

FTR1

3Q10-------+ R2

2QL15

-----------+ R37Q10-------+ M

QL5

--------– P– 0=⇒

– 4.6 (6) –


5. FEM: planar frames of trusses and beams

5.1 The figure to the right shows a structure with three linearelastic truss elements with elastic modulus E. Element 1 and 3have cross sectional area A and length L. Element 2 has crosssectional area and length . The structure is sub-jected to a point force Q and a force per unit volume (bodyforce) according to the figure. Calculate thedisplacement at node 2, all reaction forces and the normalstress in each one of the truss elements.

5.2 A truss structure containing three trusses, all with elasticmodulus E and cross sectional area A, is shown to the right. Thestructure is loaded by one point force Q/2 and a body force oftotal magnitude Q acting on the vertical truss member down-wards. Model the structure by use of three linear elements andcalculate the displacements and possible reaction forces at thenodes. Note, the displacements at the nodes will in the currentcase agree with the exact solution. Will the numerical solutiondeviate 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 thefigure. Note that with the load applied in the present case, theFEM 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 beamwith elastic modulus E and a cross sectional area Aand a moment of inertia I. The left end of the beamis 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, wherethe beam is modelled by one two-node element, and evaluate the deflection of thebeam. 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 exactsolution. The deflection between the nodes for will deviate somewhat from theexact solution due to distributed load q(x).

x/LKx

Q

3 2

1

y/L

(0,1) (1,1)

(1,0)

2A 2L

Kx Q AL( )⁄=

Q 2⁄

Q2L

L

L

45o

2Ph

h

h L⁄ 1 10⁄=

q(x)

x

y

kx = −2L

x=2L

L

P

k ηEI L3⁄=

η 3 2⁄= q x( ) Q 2L( ) x L⁄( )⁄–=

0 x 2L≤ ≤

– 5.1 (6) –


5.5 Analyse the linear elastic planar frame work shown to theright by use of FEM. Use a 2-node combined truss/beam ele-ment allowing for both axial and bending deformation. Thecross section of the frame is displayed in the figure and the elas-tic modulus is E. Will the FEM solution agree with the exactsolution, 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 machinemember. 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 problemis fully utilized, only a quarter of the ring needs to be modelled. The problem can for instancebe analysed by the Matlab based FEM program “frame2D”, available at the home page of thecourse. If the displacement, δ, primarily is due to bending deformation (a good approximationif ), 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 FEMsolution to come close to this result, the FEM model requires that .

hh

P

L

L

k P δ⁄=

R h»

k4π

π28–( )

-------------------EI

R3

------=

R h»

P

R

x

y

R

δ 2⁄

δ 2⁄

P

R

ηh

P 2⁄δ 2⁄

“symmetric quarter”

– 5.2 (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

φk

x

y

a c2

sc

sc s2

=

Ke k a a–

a– a=

al122

l12m12

l12m12 m122

=

c φcos=

s φsin=

l12 φxcos x2 x1–( ) L⁄= =

m12 φycos y2 y1–( ) L⁄= =

L x2 x1–( )2y2 y1–( )2

+=

where

alternativly

Global forulation for truss & spring elements

spring constant for a truss element kEAL

-------=

θ1y, M1y

w1, f1z

w2, f2z

θ2y, M2y

u2, f2x

xz

u1, f1xke

EA2L------- 0 0

EA2L-------– 0 0

03EI

2L3

--------- 3EI

2L2

--------- 03EI

2L3

---------–3EI

2L2

---------–

03EI

2L2

--------- 2EIL

--------- 03EI

2L2

---------–EIL------

EA2L-------– 0 0

EA2L------- 0 0

03EI

2L3

---------–3EI

2L2

---------– 03EI

2L3

--------- 3EI

2L2

---------–

03EI

2L2

---------–EIL------ 0

3EI

2L2

---------–2EI

L---------

=

2L

Local formulation

Global formulation

(local coordinate system)

(global coordinate system)

D1, F1

xz

D2, F2 D4, F4D6, F6

D5, F5

D3, F3

X

Z

{X1, Z1}

{X2, Z2}

de TDe=

fe kede= ⎭⎬⎫

fe⇒ keTDe=

Fe TTfe=⎭⎪⎪⎬⎪⎪⎫

Fe⇒ TTkeT De Ke De= =

deT

u1 w1 θ1y u2 w2 θ2y=

feT

f1x f1z M1y f2x f2z M2y=

FeT

F1 F2 F3 F4 F5 F6=

DeT

D1 D2 D3 D4 D5 D6=

Transformation scheme:

TT2 0

0 T2

där T2

lx mx 0

lz mz 0

0 0 1

= =

Transformation matrix:

lz φzXcosZ2 Z1–

2a-----------------– ϕsin–= = =

lx φxXcosX2 X1–

2a------------------ ϕcos= = =

mz φzZcosX2 X1–

2a------------------ ϕcos= = =

mx φxZcosZ2 Z1–

2a----------------- ϕsin= = =

“Directioncosines”

lx2

lz2

+ 1=

mx2

mz2

+ 1=⎩⎪⎨⎪⎧

⇒

Fe TT

fe= Tl12 m12 0 0

0 0 l12 m12

=där

Local/global transformation of nodal force vector:

– 5.3 (6) –


Solutions

5.1

5.2

D8

D7


k2EA2L-------

1 1 1– 1–

1 1 1– 1–

1– 1– 1 1

1– 1– 1 1

=k1EAL

-------

1 0 1– 0

0 0 0 0

1– 0 1 0

0 0 0 0

= k3EAL

-------

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

fb1 NKxALdx

0

L

∫Q2---- 1

1–= =

Fb1 TTfb1

Q2----

1

0

1

0

–= = T 1 0 0 0

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

EA2L-------

2 0 2– 0

0 0 0 0

1 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

QL8EA----------- 1–

5–=

Eq. 1 - 6 then give: R15Q8

------- R2, 0 R3, 3Q8

-------= = =

R43Q8

------- R5, 0 R6, 5Q8

-------= = =

The normal stress in one element: σ = EBde = EBTDe

σ1 E 1L---–

1L---

1 0 0 0

0 0 1 0

D7

D8

D1

D2

Q8A-------= =

σ2 E 1

2L----------–

1

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

1

2------- 1

2------- 0 0

0 01

2------- 1

2-------

D7

D8

D3

D4

3Q8A-------= =

σ3 E 1L---–

1L---

0 1 0 0

0 0 0 1

D7

D8

D5

D6

5Q8A-------= =

with

EAL

-------

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

⇒QL2EA----------- 3 2–

1 2–

QLEA-------- 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.

– 5.4 (6) –


5.3

5.4

5.5

P

PEquivalentE, I, A, 2L

Ih

4

12------ A, h

2= =

xy

D3

D2

D1 D6

D5

D4problem:

Discretization:

EI

2L3

---------

λ 0 0 λ– 0 0

0 3 3L 0 3– 3L

0 3L 4L2

0 3– L 2L2

λ– 0 0 λ 0 0

0 3– 3– L 0 3 3– L

0 3L 2L2

0 3– L 4L2

D1 0=

D2 0=

D3 0=

D4

D5

D6

Rx

Rz

My

P

P

0

=

Eq.system:

λ AL2

I--------- 12

Lh---⎝ ⎠⎛ ⎞

2= =Here

Eq. 4 - 6 give: D4

D5

D6

PL3

EI---------

2 λ⁄8 3⁄–

8 4L( )⁄–

2PL EA( )⁄

P– 2L( )33⁄

P– 2L( )22⁄

= =

Eq. 1 - 3 then give the reaction forces:Rx P Ry,– P My, P2L= = =

x

yq x( ) Q

2L------

xL---–=

D3

D2

D1D6

D5

D4

D7

D8

One element solution:

k3EI

2L3

---------=

EI, EA

K

EA2a------- 0 0

EA2a-------– 0 0 0 0

03EI

2a3

--------- 3EI

2a2

--------- 03EI

2a3

---------–3EI

2a2

--------- 0 0

03EI

2a2

--------- 2EIa

--------- 03EI

2a2

---------–2EI

a--------- 0 0

EA2a-------– 0 0

EA2a------- 0 0 0 0

03EI

2a3

---------–3EI

2a2

---------– 03EI

2a3

--------- k+3EI

2a2

---------– 0 k–

03EI

2a2

--------- 2EIa

--------- 03EI

2a2

--------- 2EIa

--------- 0 0

0 0 0 0 0 0 0 0

0 0 0 0 k– 0 0 k

0

0

0

D4

D5

D6

0

0

R1

R2Q10------–

R37QL60

-----------–

P

9Q10-------–

23QL60

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

0

R8

= =

Eqn.system:

where a = 2L och

D4

D5

D6

QL2

EI----------

0

225– L45

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

15---

P4LEA----------

1

0

0

+=⇒

xy

D3

D2

D1 D6

D5

D4

Discreti-

D9

D8

D7

x

y

el. 1el. 2 ke1 ke2

EI

2L3

---------

λ 0 0 λ– 0 0

0 3 3L 0 3– 3L

0 3L 4L2

0 3– L 2L2

λ– 0 0 λ 0 0

0 3– 3– L 0 3 3– L

0 3L 2L2

0 3– L 4L2

A C

CT

B= = =

Element stiffnessmatrices (local):

λ AL2

I--------- 12

Lh---⎝ ⎠⎛ ⎞

2= =here

Element stiffness

zation:

X

Z

Ke1 ke1= Ke2 TTke1TT2

T0

0 T2T

A C

CT

B

T2 0

0 T2

T2TAT2 T2

TCT2

T2TC

TT2 T2

TBT2

= = =

matrix (globally):

T2

0 1– 0

1 0 0

0 0 1

=Transformation matrix:

– 5.5 (6) –


5.5 cont.

5.6

Considering the displacement boundary conditions (D1=D2=D3=0)the reduced equation system becomes:

EI

2L3

---------

3 λ+ 0 3L 3– 0 3L

0 3 λ+ 3L– 0 λ– 0

3L 3L– 8L2

3L– 0 2L2

3– 0 3L– 3 0 3L–

0 λ– 0 0 λ 0

3L 0 2L2

3L– 0 4L2

D4

D5

D6

D7

D8

D9

0

0

0

P

0

0

D4

D5

D6

D7

D8

D9

⇒4PL

3

EI-------------

h L⁄( )224⁄

1

1 L⁄

8 3 h L⁄( )224⁄+⁄

1

3 2L( )⁄

= =

The reaction forces are given by Eqs. 1 - 3 as R1 = −P, R2 = 0, R3 = −2PL

The FEM solution coincides with the exact solution, since the frame only is subjectedto a point force.

0 4 8 12 161.0

1.1

1.2

1.3

x

y

Ne = number of elements

1 10 1000.001

0.01

0.1

1

2

1

Ne = 1

Ne = 2

Ne = 4

Number of elements

k FEM

k anal

ytic

–

k anal

ytic

--------

--------

--------

--------

-----

k FEM

k anal

ytic

⁄

The FEM analysis is carried out by use of a 2 node combined truss/beam element. To facilitate a comparison with the analytical solution, the

geometry must be chosen such that .

Since, , the cross section of the ring has

been choose such that .

h R⁄( )21«

h2

I A⁄∼

R2

I A⁄»

Number of elements

– 5.6 (6) –


6. FEM: 2D/3D solids

6.1 Derive the element stiffness matrixfor the CST element shown in the figureto the right. Assume that the material isisotropic, linear elastic with elastic mod-ulus E and Poisson’s number ν = 0.

6.2 One way to satisfy compatibility across elementboundaries between regions of high to low order elementsis to use transition elements. A plane triangular transitionelement is shown in Figure 4. The shape functions for thevertex 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 shapefunctions.

6.3 A plate containing a circular hole with radius R is mod-elled by use of plane 8-noded bi-quadratic isoparametricelements. 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 onequarter of the circumference of the hole, i.e. the straight ele-

ment sides: 2-6-3 and 1-8-4, respectively, form 45o angleswith respect to the y-axis. The nodes 1, 5 and 2 are placed atthe 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:

, ,

, ,

, , , .

N1 1 ξ– η–=

N2 ξc= N3 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

N1 1 ξ 3 2 ξ η+( )–( )– η–=

N2 ξ 2 ξ η+( ) 1–( )=

N3 η=

1 2

345 6

7

8

x

y

x0 y0,{ }

ξ ξ0 1 2⁄ η, η0 1–= = = =

x02

y02

+( )

N114--- 1 ξ–( ) 1 η–( ) 1 ξ η+ +( )–= N2

14--- 1 ξ+( ) 1 η–( ) 1 ξ– η+( )–=

N314--- 1 ξ+( ) 1 η+( ) 1 ξ– η–( )–= N4

14--- 1 ξ–( ) 1 η+( ) 1 ξ η–+( )–=

N512--- 1 ξ2

–( ) 1 η–( )= N612--- 1 ξ+( ) 1 η2

–( )= N712--- 1 ξ2

–( ) 1 η+( )= N812--- 1 ξ–( ) 1 η2

–( )=

– 6.1 (24) –


6.4 The Figure to the right shows a plane bi-linear isoparametric4-node element. The order of node numbering is opposite theone 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 elementwith the erroneous node numbering to the right.

6.5 The figure to the right shows a three dimensional20 node element of serendipity type. Assume that theshape functions associated with the nodes 1 to 19 (N1,..., N19) are known functions of the natural coordi-nates: ξ, η, ζ. The element is shaped as a cube in thenatural coordinate system, defined in the interval −1 to1 for each coordinate. Determine the shape functionassociated with node 20, i.e. N20(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 a4-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 unitvolume (body force) . Let and determinethe contributions to the nodal force vector from

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

L

x

y

L

1

2

3

4−L

−L

1

2

3

45

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)

ξ1 2

34η

1

1

−1−1

N1 1 ξ–( ) 1 η–( ) 4⁄=

N2 1 ξ+( ) 1 η–( ) 4⁄=

N3 1 ξ+( ) 1 η+( ) 4⁄=

N4 1 ξ–( ) 1 η+( ) 4⁄=

J ∂x ∂ξ⁄ ∂y ∂ξ⁄∂x ∂η⁄ ∂y ∂η⁄

=

B B1 B2 B3 B4=

λ λ

λλ

x/L

y/L

1

1

−1

−1

4 3

21

ρ

p0 g

x ξ η,( ) y ξ η,( )J

Ky ρg–= λ 1 2⁄=

– 6.2 (24) –


6.9 Calculate the contribution from the force per unit volume Ky to the nodal force vector inthe 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 2L.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 theplate is modelled by

(a) one isoparametric 4-node quadrilateral element,

(b) one isoparametric 8-node quadrilateral element, where the mid nodes are placed in themiddle between their corresponding corner nodes.

Assume that the traction vectors along AB are composed of a constant normal stress σ0 and aconstant shear stress τ0, such that

.

Use the results in (a) and (b) to evaluate the contribution to the total nodal force vector alongAB 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 isloaded by its dead weight (gives rise to abody 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×

t12--- 1

sL---–⎝ ⎠

⎛ ⎞ tA12--- 1

sL---+⎝ ⎠

⎛ ⎞ tB+=

tA tBtx

ty

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

12

34

5

6

7

8

12

34

1

2

3

4

56

7

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

x

yA

B

12

34

x

y

a

b3

1 2

4

thickness hBody force

Kx = 0

Ky = −ρ gg

– 6.3 (24) –


6.12 A triangular 2D domain is modelled by one plane triangular CST element according tothe figure below. The bottom side of the triangle is rigidly supported and the left side (x = 0) issubjected 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 elementand the element stiffness matrix are given in the figure below.

6.13 A CST element for 2D linear elastic analysis is shownto the right. The material is isotropic, linear elastic withelastic 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 outFEM 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 nodedisplacements and stresses in the elements. Forsimplicity, 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.

tTp y( ) 0,[ ]=

ν 1 3⁄=

x/L

y/L

3

21

(0,1)

(1,0)

p y( ) =

p0 1yL---–

⎝ ⎠⎛ ⎞

N1 1 ξ– η N2 ξ N3 η=,=,–=

Ke3Eh16

----------

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⁄=


deT

d1x d1y d2x d2y d3x d3y=

Node displacement vector:

1

2

3

x/L

y/L

1/2

1

deT

0 014--- 1–

108--------- 0

1–6

------ Lε0=

N1 1 ξ– η N2;– 2ξ N3; η ξ–= = =ξ x L⁄= η y L⁄=

de

x/l

y/l

e2

e1

1 2

34

σ0 σ0τ0

τ0

ρg

(a) (b) (c)

l l×

ν 0=

– 6.4 (24) –


6.15 A rectangular sheet metal of thickness h is subjected to an uniaxial load corresponding toa normal stress σ0. 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 isapplied 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 bemarked. Use the coordinate s (see the figure) when calculating the consistent nodalforces. 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 σ0. An exactanalysis 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 theelement 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 materialis 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 f1x = R1x 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 thetraction t agrees with the answer in (b) above. The element stiffness matrixKe and the resulting node displacement vector De are given below.

ν 1 3⁄=

s 0= s 2L=x L s 2⁄–= y s 2⁄=

εx νσ– 0 E⁄= εy σ0 E⁄= γxy 0=

x

y

3

21

N1 1xL---–

yL--- N2

xL--- N3

yL---=,=,–=

Ke3Eh16

----------

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:


deT


Node displacement vector:

L

L

σ0

σ0

t1

2------- 0

σ0

=

s

x L⁄ 1±= y L⁄ 1±=

ν 1 3⁄=

DeT

d1x d1y … d4y=

– 6.5 (24) –


6.17 Figure (a) to the right shows a planerectangular plate of thickness h, subjectedto a pure bending moment. Utilizing thesymmetry in the problem, only half of theplate needs to be modelled. A crude FEMmodel 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 = 2L, asshown in the figure. The material is iso-tropic, linear elastic with elastic modulus Eand Poisson’s ratio ν. Here plane stress con-ditions are assumed to prevail.

(a) Calculate/evaluate the nodal forcevector F, where also the reactionforces should be indicated.

(b) Calculate the normal stress in the x-direction as a function of position.Make use of the node displacementsolution is given in figure (b).

6.18 A quadratic plate with edges of length and of thickness h is loaded by its deadweight 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 materialis isotropic, linear elastic with elastic modulus E and Poisson’s ratio ν. Assume further thatplane stress conditions prevail and that . A rather coarse finite element model that tosome extent utilizes the symmetry of the problem is shown in figure (b) below. The modelconsists of only one triangular element with a linear interpolation for the displacements (CST

2L

σ0

2L

σ0

y/L

x/L

1 2

34

t σ01

0=

KeEh16-------

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

σ0L

E---------

0

2 3⁄2

2 3⁄2

0

0

0

=

N1 1 ξ–( ) 1 η–( ) 4⁄=

N2 1 ξ+( ) 1 η–( ) 4⁄=

N3 1 ξ+( ) 1 η+( ) 4⁄=

N4 1 ξ–( ) 1 η+( ) 4⁄=1 2

34

ξ

ηShape functions:

MM

2L 2L

2L

y/L

x/L

(2,−1)

(2, 1)

1 2

34

t σ0y L⁄

0=

DTD1x D1y D2x D2y D3x D3y D4x D4y

=

43---

σ0

E------L 0 0 1– 1– 1 1– 0 0=

(a)

(b)

2L

ν 0=

– 6.6 (24) –


element). The inertia forces due to the angular velocity ω and the dead weight can beaddressed by introducing the body forces and into the FEM analysis.The shape functions Ni and the element stiffness matrix ke are given in the figure below.

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

(b) Calculate all the displacements at the nodes de and the reaction forces.

(c) Calculate the stresses in the element.

6.19 A rectangular plate of thickness h rotates with a constant angular velocity ω, see Figure(a) below. The material of the plate is isotropic linear elastic with elasticity modulus E, Pois-son’s ratio ν and has density ρ. Plane stress is assumed to prevail in the plate and that ν = 0. Ifthe symmetry of the problem is considered and utilized, the plate can be modelled by only twobi-linear isoparametric elements according to Figure (b) below. The node coordinates is evi-dent from the figure. The inertia forces due to the angular velocity ω can be addressed byintroducing the body forces and into the FEM analysis.

(a) Determine the contribution from the body force to the node force vector in element 2. Hint: the coordinate transformation , is useful in element 2.

(b) The resultant node displacements, DT, corresponding to the current load is given in thefigure below. Calculate the stresses at the three points: {x = 0, y = 0}, {x = L, y = 0}, {x= 2L, y = 0}. The exact solution for the normal stress in the x-direction can beexpressed as . How much does the FEM solutiondeviates from the exact solution at the three points? At which point do the solutionsdeviate the least?Hint: the Jacobi matrix of the coordinate transformation is , where I is a unit

Kx ρω2x= Ky ρ– g=

g

ω

2L

2L

y/L

x/L

(0, −1)

3

2

1

(0, 1)

(1, 0)

(a) (b)

N112--- 1

xL---–

yL---–⎝ ⎠

⎛ ⎞ N2xL---=,=

N312--- 1

xL---–

yL---+

⎝ ⎠⎛ ⎞= ke

Eh8

-------

3 1 4– 2– 1 1

1 3 0 2– 1– 1–

4– 0 8 0 4– 0

2– 2– 0 4 2 2–

1 1– 4– 2 3 1–

1 1– 0 2– 1– 3

=

Shape functions: Element stiffness matrix:

deT


Nodal displacement vector:

Kx ρω2x= Ky 0=

x 3 ξ+( )L= y ηL=

σxx x( ) ρω2L

22⁄( ) 16 x L⁄( )2

–( )=

J L I=

– 6.7 (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 ofthe boundary as shown in the figure to the right. Thematerial 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 isindicated in the figure, where also the displacementvector D and the nodal force vector F are shown.

(a) Define the displacement boundary conditions.

(b) Calculate/evaluate the nodal force vector. Markthe presence of possible reaction forces as F1x= R1x and so on.

(c) Calculate the stresses in element e1. Assumeplane stress prevail and that the displacementvector 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 arrangedaccording 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 alinear elastic, isotropic material with elastic modulus E and Poisson’s ratio ν = 0. Plane stressconditions is assumed to be valid. The left end of the beam is welded on a wall, which in thepresent model is assumed to be rigid. The beam is subjected to a shearing load acting on itsright end as illustrated by the figure. The components of the displacement vector and the forcevector 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 contributionsfrom the traction vector.

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

4L

4L

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)

DTD1x D1y D2x … D6x D6y

=

43---ρω2

L3

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 = 2L

y = 0

y = L

y = 2L

e7

e6e5

e4e3

e2e1

p x( ) p0xL---=

DTD1x D1y D2x D2y D8x D8y

=

FT

F1x F1y F2x F2y F8x F8y=

– 6.8 (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

τ0

τ0

DT

D1x D1y D2x D2y D30x D30y= F

TF1x F1y F2x F2y F30x F30y

=

2a

2b

– 6.9 (24) –


FORMULAS

d3y

d3x

d1y

d1x

d2y

d2x

x

y

N11

2Ae--------- y2 y3–( ) x x2–( ) x3 x2–( ) y y2–( )+[ ]=

N21

2Ae--------- y3 y1–( ) x x3–( ) x1 x3–( ) y y3–( )+[ ]=

N31

2Ae--------- y1 y2–( ) x x1–( ) x2 x1–( ) y y1–( )+[ ]=

Ae

u x y,( )v x y,( )

N1 0 N2 0 N3 0

0 N1 0 N2 0 N3

de Nde= = de

d1x

d1y

d2x

d2y

d3x

d3y

=Displace-ments:

B B1 B2 B3= Bi

∂Ni ∂x⁄ 0

0 ∂Ni ∂y⁄

∂Ni ∂y⁄ ∂Ni ∂x⁄

=

εxx

εyy

γxy

B de=

Strains:

σxx

σyy

σxy

E

1 ν2–( )

-------------------1 ν 0

ν 1 0

0 0 1 ν–( ) 2⁄

εxx

εyy

γxy

=

Stresses:(Plane stress)

x

y

d1y

d1x

d2y

d2x

d3y

d3x

d4y

d4x

N1 1 ξ–( ) 1 η–( ) 4⁄= N2 1 ξ+( ) 1 η–( ) 4⁄=,

N3 1 ξ+( ) 1 η+( ) 4⁄= N4 1 ξ–( ) 1 η+( ) 4⁄=,

∂Ni ∂x⁄

∂Ni ∂y⁄J 1– ∂Ni ∂ξ⁄

∂Ni ∂η⁄= J ∂x ∂ξ⁄ ∂y ∂ξ⁄

∂x ∂η⁄ ∂y ∂η⁄=

u ξ η,( )v ξ η,( )

N1 0 N2 0 N3 0 N4 0

0 N1 0 N2 0 N3 0 N4

de Nde= =

1 2

34

ξ

η

B B1 B2 B3 B4= Bi

∂Ni ∂x⁄ 0

0 ∂Ni ∂y⁄


=

x Nixi

i

4

∑=

y Niyi

i

4

∑=

Displace-ments:

σxx

σyy

σxy

E

1 ν2–( )

-------------------1 ν 0

ν 1 0

0 0 1 ν–( ) 2⁄

εxx

εyy

γxy

=

Stresses:(Plane stress)

Deformation:

where

Plane (2D) triangular linear element:

Plane (2D) quadrilateral bi-linear element:

– 6.10 (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

N3 = N4 = N6 = N7 = N8 = 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):

B 1L---

1– 0 1 0 0 0

0 1– 0 0 0 1

1– 1– 0 1 1 0

= C E1 0 0

0 1 0

0 0 1 2⁄

=

ke BTCBhdA

Ve

∫hL

2

2---------B

TCB

Eh4

-------

3 1 2– 1– 1– 0

3 0 1– 1– 2–

2 0 0 0

1 1 0

1 0

2

= = =

Sym.

N1 N2 N3 N4+ + + 1 N4⇒ 1 N1 N2 N3+ +( )– 4 1 ξ– η–( )ξ= = =

N4 ξ η,( )

N4 0 0,( ) N4 1 0,( ) N4 0 1,( ) 0= = =

N4 1 2⁄ 0,( ) 1=

ξ0 1 2⁄ η0, 1–= = N12 1–4

---------------- N22 1+4

---------------- N5,=,–12---= =

x1 y1,( ) 1– 1,( ) R

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

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

x0 x ξ0 η0,( ) Nixi∑R2---= = =

y0 y ξ0 η0,( ) Niyi∑1 2+( )2 2

---------------------R= = =

x02

y02

+ R 5 2 2+8

------------------- 0.989R≈=

y ξ η,( ) Niyi

i 1=

4

∑ L N1 N3–( ) L2--- ξ– η–( )= = =

Partial derivatives: ∂x∂ξ------ L

2---

∂x∂η------;

L2---

∂y∂ξ------;–

L2---

∂y∂η------;–

L2---–= = = =

JL2--- 1 1–

1– 1–J⇒

L2

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

x ξ η,( ) Nixi

i 1=

4

∑ L N2 N4–( ) L2--- ξ η–( )= = =

– 6.11 (24) –


6.5 Use the property , Note that N1, ..., N19 are known!

Alternatively, derive the shape functions by use of the properties:

N20 = 0 in nodes 1 to 19 and that N20 = 1

6.6 (a) ;

;

(b)

6.7(a)

(b) and

(c)

Ni

i 1=

20

∑ 1= N20⇒ 1 Ni

i 1=

19

∑–=

N20⇒14--- 1 ξ–( ) 1 η–( ) 1 ζ2

–( )=

∂x∂ξ------ 1

4--- 1 η–( ) x2 x1–( ) 1 η+( ) x3 x4–( )+[ ] 3

2--- l= =

∂y∂ξ------ 0=

∂x∂η------ 0=

∂y∂η------ 1

4--- 1 ξ–( ) y4 y1–( ) 1 ξ+( ) y3 y2–( )+[ ] 1

2--- l= = J⇒

l2--- 3 0

0 1=

∂N1 ∂x⁄∂N1 ∂y⁄

J 1– ∂N1 ∂ξ⁄∂N1 ∂η⁄

2l--- 1 3⁄ 0

0 1

1 η–( ) 4⁄–

1 ξ–( ) 4⁄– 12l----- 1 η–( ) 3⁄

1 ξ–( )–= = =

B1⇒ ""12l-----

1 η–( ) 3⁄ 0

0 1 ξ–( )1 ξ–( ) 1 η–( ) 3⁄

–=

x Nixi∑ L 1 λη+( )ξ y Niyi∑ Lη= =,= =

J ∂x ∂ξ⁄ ∂y ∂ξ⁄∂x ∂η⁄ ∂y ∂η⁄

L 1 λη+ 0

λξ 1= = J L

21 λη+( )=

B B1 B2 B3 B4= Bi

∂Ni ∂x⁄ 0

0 ∂Ni ∂y⁄


=∂Ni ∂x⁄

∂Ni ∂y⁄J 1– ∂Ni ∂ξ⁄

∂Ni ∂η⁄=where with

∂N1 ∂x⁄

∂N1 ∂y⁄1

4L------

1 η–( )–1 λη+

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

1 ξ–( )–1 η–( )λξ1 λη+

------------------------+

=∂N2 ∂x⁄

∂N2 ∂y⁄1

4L------

1 η–1 λη+----------------

1 ξ+( )–1 η–( )λξ1 λη+

------------------------–

=

∂N3 ∂x⁄

∂N3 ∂y⁄1

4L------

1 η+1 λη+----------------

1 ξ+( ) 1 η+( )λξ1 λη+

------------------------–

=∂N4 ∂x⁄

∂N4 ∂y⁄1

4L------

1 η+( )–1 λη+

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

1 ξ–( ) 1 η+( )λξ1 λη+

------------------------+

=

Strain: ε = Bde, where the B-matrix is given by

Inverse of the Jacobi matrix becomes J 1– 1L--- 1 1 λη+( )⁄ 0

λ– ξ 1 λη+( )⁄ 1=

where

– 6.12 (24) –


14 2L=

NTtdS

Se

∫=

f1x f4x= =

V h J dξ=

J L2(=

2ρg fb3,

0 5 0 5 0

(d)6.8

6.9 Numerical integration with Gauss-Legendre quadrature

where

ke BTCBhdA

Ae

∫ BT

CBh J dξdη

1–

1

∫1–

1

∫

B1T

CB1 B1TCB2 B1

TCB3 B1

TCB4

B2TCB1 B2

TCB2 B2TCB3 B2

TCB4

B3TCB1 B3

TCB2 B3TCB3 B3

TCB4

B4T

CB1 B4TCB2 B4

TCB3 B4

TCB4

hL2

1 λη+( )dξdη

1–

1

∫1–

1

∫= = =1 λ2

+ ϕcos, 1

1 λ2+

------------------- ϕsin, λ

1 λ2+

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

dS hl14

2------dη=

ξ 1–=⎩ ⎭⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

NTξ 1–=

thL 1 λ2+ dη

1–

1

∫= =

p0hL f1y f4y λp0hL= =,

dη} NTfVh J dξdη1–

1

∫1–

1

∫=

λ 12---=1 λη+ ) one obtains with that

y fb4y 76---h– L

2ρg= =

Note! fbiy∑ 4hL2ρg– Veρg–= =7 0 7

fbIy fI ξ η,( )dξdη1–

1

∫1–

1

∫ fI ξi ηj,( )wiwj

j

nη

∑i

nξ

∑= = fI NI 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=

fb1y⇒ N1 0 0,( ) ρg–( )hL2

2 2⋅( ) h– L2ρg= =

fb2y fb3y fb4y 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 η11–

3-------= = ξ2 η2

1

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

fb1y⇒ ρghL2

N1 ξ1 η1,( ) 1 λη1+( ) 1 1⋅( ) N1 ξ1 η2,( ) 1 λη2+( ) 1 1⋅( )

N1 ξ2 η1,( ) 1 λη1+( ) 1 1⋅( ) N1 ξ2 η2,( ) 1 λη2+( ) 1 1⋅( )

+

+ +

[

]

–

56---ρghL

2–

=

=

fb2y56---ρghL

2–= fb3y fb4y

76---h– L

2ρg= =

– 6.13 (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 , , .

fs NTtdS

Se

∫= dS hds h dx2

dy2

+= =

dx∂x∂ξ------dξ ∂x

∂η------dη+= dy

∂y∂ξ------dξ ∂y

∂η------dη+= ξ 1=

dξ 0= dS h ∂x ∂η⁄( )2 ∂y ∂η⁄( )2+ dη=

N2 ξ 1=1 η–( ) 2⁄= N3 ξ 1=

1 η+( ) 2⁄=

x N2x2 N3x3+( )ξ 1=

∂x ∂η⁄⇒ x3 x2–( ) 2⁄= =

y N2y2 N3y3+( )ξ 1=

∂y ∂η⁄⇒ y3 y2–( ) 2⁄= =

dS hx3 x2–

2----------------⎝ ⎠⎛ ⎞

2 y3 y2–

2----------------⎝ ⎠⎛ ⎞

2

+ dη hLdη= =

f2x N2txhLdη1–

1

∫ N21 η–

2------------ tAx

1 η–2

------------ tBx+⎝ ⎠⎛ ⎞ hLdη

1–

1

∫hL3

------ 2tAx tBx+( )= = =

f2yhL3

------ 2tAy tBy+( )= f3xhL3

------ tAx 2tBx+( )= f3yhL3

------ tAy 2tBy+( )=

(b) 8-node element: , , N2 ξ 1=

1 η–( ) η–( )2

-----------------------------= N3 ξ 1=

1 η+( )η2

---------------------= N6 ξ 1=1 η2

–=

x N2x2 N3x3 N6x6+ +( )ξ 1=

x6

x2 x3+

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

⎩ ⎭⎨ ⎬⎧ ⎫

N2

N6

2------+⎝ ⎠

⎛ ⎞ x2 N3

N6

2------+⎝ ⎠

⎛ ⎞ x3+⎝ ⎠⎛ ⎞

ξ 1=

= = =

y N2y2 N3y3 N6y6+ +( )ξ 1=

y6

y2 y3+

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

⎩ ⎭⎨ ⎬⎧ ⎫

N2

N6

2------+⎝ ⎠

⎛ ⎞ y2 N3

N6

2------+⎝ ⎠

⎛ ⎞ y3+⎝ ⎠⎛ ⎞

ξ 1=

= = =

.

In the same way we obtain , , ,

and .

∂x∂η------

x3 x2–

2---------------- ∂y

∂η------

y3 y2–

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

⎩ ⎭⎨ ⎬⎧ ⎫

dS hx3 x2–

2----------------⎝ ⎠⎛ ⎞

2 y3 y2–

2----------------⎝ ⎠⎛ ⎞

2

+ dη hLdη= =⇒ ⇒

f2x N2txhLdη1–

1

∫ N21 η–

2------------ tAx

1 η–2

------------ tBx+⎝ ⎠⎛ ⎞ hLdη

1–

1

∫hL3

------ tAx= = =

f2yhL3

------tAy= f3xhL3

------tBx= f3yhL3

------tBy=

f6yhL3

------ 2tAx 2tBx+( )= f6yhL3

------ 2tAy 2tBy+( )=

– 6.14 (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:

tAx tBx tx σ0 θ τ0 θsin–cos= = =

tAy tBy ty σ0 θ τ0 θcos+sin= = =

f2x f3x hLetx= = f2y f3y hLety= =

Le L 3⁄=

FsT

hLe tx ty 2tx 2ty 2tx 2ty tx ty …=

f2x f3x

hLe

3--------tx= = f2y f3y

hLe

3--------ty= = f6x

hLe

3--------4tx=

f6y

hLe

3--------4ty= Le

L3---=

FsT hLe

3-------- tx ty 4tx 4ty 2tx 2ty 4tx 4ty 2tx 2ty 4tx 4ty tx ty …=

fb NTfvh

ab4

------dξdη1–

1

∫1–

1

∫= NN1 0 N2 0 N3 0 N4 0

0 N1 0 N2 0 N3 0 N4

= fv0

ρg–=

NTfv

N1 0

0 N1 0

ρg–

0

f1y= = f1yρghab

4---------------- 1 ξ–( ) 1 η–( )

4----------------------------------dξdη

1–

1

∫1–

1

∫–ρghab

4----------------–= =

fbT

ρghab

4---------------- 0 1 0 1 0 1 0 1–=⇒

– 6.15 (24) –


6.12

6.13(a)

t p01 η–

0=

fs NTξ 0=

thLdη

0

1

∫= f2x 0 f3x 0≠,≠⇒ others equal to zero!

f1x1 η–( )

2-----------------ηhLdη

1–

1

∫σ0hL

3-------------–= = f3x

1 η+( )2

-----------------ηhLdη1–

1

∫σ0hL

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

The consistent nodal force vector becomes (use subst. ξ = x/L, η = y/L):

x/L

y/L

1 2

3

Displacement boundary conditions: d1x = d1y = d2x = d2y = 0

give rise to the reaction forces: R1x, R1y, R2x, R2y

The external load acts on side x = 0, between nodes 1 and 3,

there, the shape functions reduces to

N1 1 η– N3 η N2 0=,=,=

Load vector: F

R1x p0hL 3⁄+

R1y

R2x

R2y

p0hL 6⁄

0

=

Reduced Eq. syst. (5) & (6):

3Eh16

---------- 1 0

0 3

d3x

d3y

p0hL

6------------ 1

0=

d3x

d3y

⇒89---

p0

E-----L 1

0=

Reaction forces:

Eq. (1): R1x3Eh16

---------- d3x–( )p0hL

3------------–

p0hL

2------------–= = Eq. (2): R1y

3Eh16

---------- d3x–( )p0hL

6------------–= =

Eq. (3): R2x 0= Eq. (4): R2y3Eh16

---------- d3x( )p0hL

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

With AeL

2

4-----=

x1 0=

y1 0=

x2 L 2⁄=

y2 L 2⁄=

x3 0=

y3 L=the shape functions

N21

2Ae

--------- y3 y1–( ) x x3–( ) x1 x3–( ) y y3–( )+[ ] 2ξ= =

N11

2Ae

--------- y2 y3–( ) x x2–( ) x3 x2–( ) y y2–( )+[ ] 1 ξ– η–= =

N31

2Ae

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

are obtained as

– 6.16 (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.

ν 13---=

⎩ ⎭⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

3E8

-------3 1 0

1 3 0

0 0 1

= =

Thus

∂N1 ∂x⁄ 1 L ∂N1 ∂y⁄;⁄– 1– L⁄= =

∂N2 ∂x⁄ 2 L ∂N2 ∂y⁄;⁄ 0= =

∂N3 ∂x⁄ 1– L⁄ ∂N3 ∂y⁄; 1 L⁄= = ⎭⎪⎬⎪⎫

B B1 B2 B3=⇒ B1

1L---

1– 0

0 1–

1– 1–

= B21L---

2 0

0 0

0 2

=

B31L---

1– 0

0 1

1 1–

=

σσxx

σyy

σxy

C B1 B2 B3

d1

d2

d3

C B1d1 B2d2 B3d3+ +[ ] Eε0

1 2⁄0

1 18⁄

= = = =

d21 4⁄1 108⁄–

=

KEh4

-------

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 0

0 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): D1x = D1y = D2y = D4x = 0

Force vector: FTF1x F1y F2x F2y F3x F3y F4x F4y R1x R1y

σ0hl

2----------- R2y

σ0hl

2----------- 0 R4x 0= =

DT

⇒σ0

E------l 0 0 1 0 1 0 0 0= el. 1 & el. 2: σT σ0 1 0 0=

(b) Displ. B.C. (remove rigid body motion): D1x = D1y = D2y = 0

Force vector: FT

R1x

τ0hl

2----------–

⎝ ⎠⎛ ⎞ R1y

τ0hl

2----------–

⎝ ⎠⎛ ⎞ τ0hl

2----------– R2y

τ0hl

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

⎝ ⎠⎛ ⎞ τ0hl

2----------

τ0hl

2----------

τ0hl

2----------

τ0hl

2----------–

⎝ ⎠⎛ ⎞=

DT⇒

τ0

E 2⁄----------l 0 0 0 0 1 0 1 0= el. 1 & el. 2: σT τ0 0 0 1=

(c) Displ. B.C. (remove rigid body motion): D1x = D1y = D2y = 0

Force vector: FT

R1x R1yρghl

2

6--------------–

⎝ ⎠⎛ ⎞ 0 R2y

ρghl2

3--------------–

⎝ ⎠⎛ ⎞ 0

ρghl2

6--------------– 0

ρghl2

3--------------–=

DT⇒ ""

ρgl2

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

24-------- 1 15– 1– el. 2: σT ρgl

24-------- 1– 9– 1==

exact!

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

– 6.17 (24) –


6.15(a)

6.15(b)

6.15(c)

6.16(a) Displacement boundary conditions: d1x=d3y=d4x=d4y=0

Consistent nodal force vector:

f2y N2tyh sd

0

2L

∫ x Ls

2-------–=

⎩ ⎭⎨ ⎬⎧ ⎫ 1

L--- L

s

2-------–

⎝ ⎠⎛ ⎞ σ0

2-------h sd

0

2L

∫σ0Lh

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

fs NT2 3–

th s f2y 0 f3y 0≠,≠⇒d

0

2L

∫=

FTR1x R1y 0 R2y

σ0Lh

2-------------+⎝ ⎠

⎛ ⎞ R3x

σ0Lh

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

f3y N3tyh sd

0

2L

∫ ys

2-------=

⎩ ⎭⎨ ⎬⎧ ⎫ 1

L--- s

2-------

σ0

2-------h sd

0

2L

∫σ0Lh

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

Zero displacement B.C.: d1x=d1y=d2y=d3x=0 => reaction forces: R1x, R1y, R2y, R3x

Node force vector:

3Eh16

---------- 3 1

1 3

d2x

d3y

σ0Lh

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

1

d2x

d3y

⇒ Lσ0

E------ 1 3⁄–

1= =

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

Reaction forces: Eq. (1): R1x3Eh16

---------- 3d2x– d3y–( ) 0= =

Eq. (2): R1y3Eh16

---------- d2x– 3d3y–( ) σ0Lh

2-------------–= =

Eq. (4): R2y

σ0Lh

2-------------+ 0 R2y⇒

σ0Lh

2-------------–= =

Eq. (6): R3x 0=

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

N1 1 x L⁄– y L⁄ –= ∂N1 ∂x⁄⇒ 1 L ∂N1 ∂y⁄;⁄– 1– L⁄= =

N2 x L⁄= ∂N2 ∂x⁄⇒ 1 L ∂N2 ∂y⁄;⁄ 0= =

N3 y L⁄= ∂N3 ∂x⁄⇒ 0 ∂N3 ∂y⁄; 1 L⁄= = ⎭⎪⎬⎪⎫ B B1 B2 B3

= =

1L---

1– 0 1 0 0 0

0 1– 0 0 0 1

1– 1– 0 1 1 0

=

⇒

εεx

εy

γxy

1L---

1– 0 1 0 0 0

0 1– 0 0 0 1

1– 1– 0 1 1 0

Lσ0

E------

00

1 3⁄–

0

0

1

σ0

E------

1 3⁄–

1

0

= = =Agrees with theexact solutionwhen ν = 1/3

– 6.18 (24) –


6.16(b)

6.16(c)

6.17(a) Nodal force vector:

Consistent nodal force vector:

f2x N2txh yd

L–

L

∫1 η–

2------------σ0hL ηd

1–

1

∫ σ0hL= = =

fs NT2 3–

th s f2x 0 f3x 0≠,≠⇒d

L–

L

∫=

FeT

R1x 0 σ0Lh 0 σ0Lh R3y R4x R4y=

Displacement boundary conditions give the reaction forces: R1x, R3y, R4x, R4y

Node force vector:

f3x N3txh 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=

F1x⇒ R1x σ0hL F2x;– σ0hL F3x; σ0hL F4x; R4x σ0hL–= = = = = =

F1y 0 F2y 0 F3y;=; R3y 0 F4y; R4y 0= = = = =

t σ0y L⁄

0=

x

y

1 2

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

the shape functions take the values:

N1 N4 0 och N21 η–( )

2----------------- N3

1 η+( )2

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

fe NTξ 1=

thLdη

1–

1

∫= f2x 0 f3x 0≠,≠⇒ others zero!

f2x1 η–( )

2-----------------ηhLdη

1–

1

∫σ0hL

3-------------–= = f3x

1 η+( )2

-----------------ηhLdη1–

1

∫σ0hL

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

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

Inclusive of reaction forces, theFT

R1x 0σ0hL

3-------------– 0

σ0hL

3------------- 0 R4x R4y

=nodal force vector becommes:

– 6.19 (24) –


6.17(b) Stresses are calculated as: where

B B1 B2 B3 B4=

B21

4L------

1 η– 0

0 1 ξ+( )–

1 ξ+( )– 1 η–

B31

4L------

1 η+ 0

0 1 ξ+

1 ξ+ 1 η+

=,=

deT

d1T d2

T d3T d4

T where d1 d4 0 and d243---

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:

εεx

εy

γxy

B2d2 B3d3+

14L------4

3---

σ0

E------L 1 η–( )– 1 η+( )+[ ]

εy

γxy

23---

σ0

E------η

εy

γxy

= = = =

Stresses:

σσx

σy

τxy

CεE 0 0

0 E 0

0 0 E 2⁄

23---

σ0

E------η

εy

γxy

23---σ0η

σy

τxy

= = = =

plane stress (ν = 0.3)

Note!The solution based on one FEM element deviates from the exact solution:σx σ0y L⁄=

σy τxy 0= =

σ Cε= ε Bde=

– 6.20 (24) –


6.18 (a)

6.18 (b)

Fb NTKdV

Ve

∫ NTKhdy

L x–( )–

L x–( )

∫⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞

dx

0

L

∫= =The contribution from the volume force is:

Here, both fbix 0≠ fbix 0≠ i 1 2 3, ,=and

fb1x 12--- 1

xL---–

yL---–

⎝ ⎠⎛ ⎞ρω2

xhdy

L x–( )–

L x–( )

∫⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞

dx

0

L

∫ρω2

hL

------------- L x–( )2xdx

0

L

∫ρω2

hL3

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

fb2x xL---ρω2

xhdy

L x–( )–

L x–( )

∫⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞

dx

0

L

∫ρω2

hL

------------- 2x2

L x–( )dx

0

L

∫ρω2

hL3

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

fb3x 12--- 1

xL---–

yL---+⎝ ⎠

⎛ ⎞ρω2xhdy

L x–( )–

L x–( )

∫⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞

dx

0

L

∫ρω2

hL

------------- L x–( )2xdx

0

L

∫ρω2

hL3

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

The total nodal force vector becomes: f ρω2hL

3

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

1

0

2

0

1

0

ρghL2

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

0

1

0

1

0

1

–

R1x

R1y

0

0

R3x

0

+=

fb1y 12--- 1

xL---–

yL---–

⎝ ⎠⎛ ⎞ ρg–( )hdy

L x–( )–

L x–( )

∫⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞

dx

0

L

∫ ρgh

L----------– L x–( )2

dx

0

L

∫ ρghL

2

3----------------–= = =

fb2y xL--- ρg–( )hdy

L x–( )–

L x–( )

∫⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞

dx

0

L

∫ 2ρgh

L-------------– x L x–( )dx

0

L

∫ ρghL

2

3----------------–= = =

fb3y 12--- 1

xL---–

yL---+⎝ ⎠

⎛ ⎞ ρg–( )hdy

L x–( )–

L x–( )

∫⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞

dx

0

L

∫ ρgh

L----------– L x–( )2

dx

0

L

∫ ρghL

2

3----------------–= = =

The displacement B.C:s : d1x = d1y = d3x = 0 give rise to reaction forces!

Reduced equation system withrespect to the boundary:conditions, Eq. (3,4,6):

Displacement boundary conditions: d1x = d1y = d3x = 0

Eh8

-------8 0 0

0 4 2–

0 2– 3

d2x

d2y

d3y

ρω2hL

3

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

2

0

0

ρghL2

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

0

1

1

–=

d2x

d2y

d3y

⇒ρω2

L3

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

1

0

0

ρgL

2

3E------------–

0

5

6

–=

– 6.21 (24) –


6.18 (c) The stresses in the element is given by ,where

6.19(a)

σ Cε=ε Bde B1d1 B2d2 B3d3+ += =

Here we have

C ν 0={ } E1 0 0

0 1 0

0 0 1 2⁄

d1 0 d2;ρω2

L3

6E---------------- 1

0

ρgL2

3E------------ 0

5 d3

ρgL2

3E------------ 0

6–=;–= =;= =

σ⇒

σx

σy

τxy

ρω2L

2

6----------------

1

0

0

ρgL3

----------0

3

1

–= =

B2

∂N2 ∂x⁄ 0

0 ∂N2 ∂y⁄∂N2 ∂y⁄ ∂N2 ∂x⁄

1L---

1 0

0 0

0 1

B3

∂N3 ∂x⁄ 0

0 ∂N3 ∂y⁄∂N3 ∂y⁄ ∂N3 ∂x⁄

12L------

1– 0

0 1

0 1–

= =;= =

σσx

σy

τxy

E

1 0 0

0 1 0

0 012---

0

0

0

1L---

1 0

0 0

0 1

ρω2L

3

6E---------------- 1

0

ρgL2

3E------------ 0

5–

⎝ ⎠⎜ ⎟⎛ ⎞ 1

2L------

1– 0

0 1

0 1–

ρgL2

3E------------ 0

6–⎝ ⎠⎜ ⎟⎛ ⎞

+ +

⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞

= =

Thus

Note that the stress solution is approximate and rather poor in this case!To obtain a better solution, many more elements are needed!

fb2T

⇒ρω2

hL3

3------------------- 8 0 10 0 10 0 8 0=

f1x1 ξ–( ) 1 η–( )

4----------------------------------ρω2

L 3 ξ+( )hL2dξdη

1–

1

∫1–

1

∫=⇒ f4x f1x=

fb2 NTKdVVe

∫ NTK h J dξdη1–

1

∫1–

1

∫= = K ρω2x

0

ρω2L 3 ξ+( )

0= =

J L2

=

Body force vector in element 2:

where

f2x1 ξ+( ) 1 η–( )

4-----------------------------------ρω2

L 3 ξ+( )hL2dξdη

1–

1

∫1–

1

∫= f3x f2x=

– 6.22 (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= =

σxx

σyy

τxy

⇒ E1 0 0

0 1 0

0 0 1 2⁄

14L------

δx δx+

0

1 ξ+( )δx– 1 ξ+( )δx+

Eδx 2L( )⁄0

0

223------ρω2

L2

1

0

0

= = =

Bi1L---

Ni ξ, 0

0 Ni η,

Ni η, Ni ξ,

B21

4L------

1 0

0 1 ξ+( )–

1 ξ+( )– 1

B31

4L------

1 0

0 1 ξ+( )1 ξ+( ) 1

=;=⇒=

σxxFEM σxx

Exact–

σxxExact

---------------------------------- 100%×

C E1 0 0

0 1 0

0 0 1 2⁄

= dTd1

T d2T d3

T d4T

e1=Here,

d2 d3δx

0= =

δx443

------ρω2L

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 = Lx = 0 x = 2L

D1x D1y D2x D3x D3y D4y D7y 0= = = = = = =

3 2

1

t0

p0xL---–

=

e4

N1 2yL--- N2,–

xL--- y

L--- 2 N3,–+ 1

xL---–= = =Shape functions:

fe NTy 2L=

thdx

0

L

∫= f2y 0 f3y 0≠,≠⇒ others zero!

f2yxL--- p0

xL---–⎝ ⎠

⎛ ⎞ hdx

0

L

∫hLp0

3------------–= = f3y 1

xL---–⎝ ⎠

⎛ ⎞ p0xL---–⎝ ⎠

⎛ ⎞ hdx

0

L

∫hLp0

6------------–= =

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

F1x R1x F1y; R1y F2x; R2x F3x; R3x F4y; R4y F7y; R7y= = = = = =

F3y hLp0 6⁄ F6y;– hLp0 3⁄–= =

– 6.23 (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.);= d2D4x

0; d3

0

D2y

==C P.S.{ } E

1 ν2–

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

1 ν 0

ν 1 0

0 01 ν–( )

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

= =

σσxx

σyy

σxy

C B1d1 B2d2 B3d3+ +[ ] CD4x L⁄

D2y L⁄E

1 ν2–

--------------D4x νD2y+( ) L⁄

D2y νD4x+( ) L⁄= = = =Thus

N1 1 x L⁄– y L⁄ –= ∂N1 ∂x⁄⇒ 1 L ∂N1 ∂y⁄;⁄– 1– L⁄= =

N2 x L⁄= ∂N2 ∂x⁄⇒ 1 L ∂N2 ∂y⁄;⁄ 0= =

N3 y L⁄= ∂N3 ∂x⁄⇒ 0 ∂N3 ∂y⁄; 1 L⁄= = ⎭⎪⎬⎪⎫ B B1 B2 B3

= =

1L---

1– 0 1 0 0 0

0 1– 0 0 0 1

1– 1– 0 1 1 0

=

⇒

Shape functions:

D1x D1y D2x D2y D3x D3y D4x D4y D5x D5y 0= = = = = = = = = =

2

34

1t

0

τ0–= fs NT

ξ 1=thbdη

1–

1

∫= f2y 0 f3y 0≠,≠⇒ others are equalto zero!

f2y N2 ξ 1=thbdη

1–

1

∫ τ0hb1 η–( )

2-----------------dη

1–

1

∫– τ0hb–= = = f3y N3 ξ 1=thbdη

1–

1

∫ τ0hb1 η+( )

2-----------------dη

1–

1

∫– τ0hb–= = =

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

F26y F30y τ0hb och F27y– F28y F29y 2τ0hb–= = = = =according to:

σ CBde=

de

d1

d2

d3

d4

= d1 d4 0 (B.C.);= = d2D7x

D7y

; d3D6x

D6y

==where2

34

1D7x

D7y

D6x

D6y

CP.S.

ν 0=⎩ ⎭⎨ ⎬⎧ ⎫

E1 0 0

0 1 0

0 0 1 2⁄

;= = B B1 B2 B3 B4=

∂Ni

∂x-------- 1

a---

∂Ni

∂ξ--------=

∂Ni

∂y-------- 1

b---

∂Ni

∂η--------=

⎩⎪⎪⎨⎪⎪⎧

where

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

B214---

1 a⁄ 0

0 1 b⁄–

1 b⁄– 1 a⁄

; B314---

1 a⁄ 0

0 1 b⁄1 b⁄ 1 a⁄

==

σσxx

σyy

σxy

CB2d2 CB3d3+E4---

D6x D7x+( ) a⁄

D6y D7y–( ) b⁄

D6x D7x–( ) 2b( ) D6y D7y+( ) 2a( )⁄+⁄

= = =

Thus

– 6.24 (24) –


7. FEM: Heat conduction

7.1 The Figure to the right shows a one dimensionalmodel of a cooling fin. At the left boundary thetemperature is constant at 80o C. Along the remain-ing boundary, heat is lost to the surrounding air byconvection. 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 tothermal expansion, where it can be assumed thatthe fin is undeformed at 20o C. Analyse the prob-lem by FEM and use two linear elements. Carry outthe 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 kand 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 theright hand figure. The temperature at the right side of the wall iskept constant at 20o C. At the left side, a heat flux arises due toconvection, where the ambient temperature is equal to −5o C. Thethermal convection coefficient is h and the thermal conductivity ofthe two materials is k1 and k2, respectively. Determine the tempera-ture distribution through the wall by FEM-analysis. In the presentcase it suffice to model each layer by one linear element.

Data: k1 = 0.2 W/cm/oC, k2 = 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 rigidwalls. At point B, the bar is influenced by a point forceP0 = 2kN and a heat source keeping the temperatureconstant at 100o C. The right boundary (point C) isinsulated 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 surroundingmedium 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.

ab

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 2L

P0

AB

C

bb

T∞ 20°C=

Konvektion

α 1.8 105–⋅=

– 7.1 (3) –


Solutions

7.1

0 2 4 6 8x / cm

10

20

30

40

50

60

70

80

Tem

pera

ture

/ o C

Exact solution2 element4 element8 element

0 2 4 6 8x / cm

0

5

10

15

20

Dis

plac

emen

t / μ

m

Exact solution2 element4 element8 element

T1 T3T2

FEM-analysis: 2 linear elements

0.6733 0.1133– 0

0.1133– 1.3467 0.1133–

0 0.1133– 0.7133

T1 80°C=

T2

T3

11.20 QR+

22.40

12.00

=Equationsystem


x = L: kA∂T∂x------– hA T T∞–( )=

unit [W/oC] unit [W]

“Reactionheat flow”

T2

T3

⇒ 25.12

20.81°C= QR⇒ 39.82 W=

(a) Temperature distribution

(b) Displacement

D1 D3D2

FEM-analysis: 2 linear elements Boundary conditions:

x = L: σA = 0

80 106⋅

1 1– 01– 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 (3) –


7.2

7.3

T1 T3T2


l2l1Boundary conditions, x = 0:

x = L: T = -5 oC

kA∂T∂x------– hA T T∞–( )=

A

k1

l1

----- h+k1

l1

-----– 0

k1

l1

-----–k1

l1

-----k2

l2

-----+ k2

l2

-----–

0k2

l2

-----–k2

l2

-----

T1

T2

T3 20°C=

A

hT∞

0

QR A⁄

T1

T2

2.58–

0.161–°C=

QR A⁄ 0.242 W cm2⁄[ ]=

⎩⎪⎪⎨⎪⎪⎧

⇒=

Equationsystem

D1 D3D2


D5D4

T5T4T3T2T1

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 T3 is prescribed.

T1 T3T2 T3 T5T4

0.8467 0.7467– 0

0.7467– 1.6933 0.7467–

0 0.7467– 0.8467

T1 20°C=

T2

T3 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

T3 100°C=

T4

T5

4 Q+ R3

8

4

=

(b) Displacement and stress calculations

Valid for each element: keEAli

------- 1 1–

1– 1 fT; EAαΔTi

1–

1= = = average temperature

change in element iΔTi

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

D5 0=

R1

0

P0

0

R5

EAα

ΔT4–

ΔT1 ΔT2–

ΔT2 ΔT3–

ΔT3 ΔT4–

ΔT4

+

3.968–

9.00–

2.532

6.877

5.559

103

R1

0

0

0

R5

+= =

e1 e2 e3 e4

R1

R5

⇒7487

9487–N=

D2

D3

D4

⇒

14.08–

7.84

31.43

106–m=

Stress in the elements:

σ1 ED2 D1–

l1------------------- αΔT1– 74.9 MPa, σ2 E

D3 D2–

l2------------------- αΔT2– 74.9 MPa–= =–= =

σ3 ED4 D3–

l3------------------- αΔT3– 94.9 MPa, σ4 E

D5 D4–

l4------------------- αΔT4– 94.9 MPa–= =–= =

T2 55.27°C=⇒T4

T5

⇒50.54

38.87°C=

– 7.3 (3) –

FEM for Engineering Applications Exercises with Solutions Jonas …/Menu/... · FEM for Engineering...

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