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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 mechanicsdirect 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 ApplicationsExercises 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 by use of Castiglianostheorems. 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 Castiglianos 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

P

E, A

, LE, A, L

45o 45o

B

C

A

P

EI, L

EI,

L

A

B

M0

EI, L

k

P

k EI 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 = 3 in the current application. Determine , 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 EI 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 EI L3= kM EI 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 Hllfasthetslra , Eds. P.-L. Larsson & R.Lundell, KTH, Stockholm, january 2001. The solutions to these problems (in Swedish) are based onCastiglianos 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 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( ) =

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( )= =

M1 M0RAM03L-------= M1

M2 M2

RBM03L-------=

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

Complementary elastic energy: WLM1

2

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

2LM22

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

M02L

6EI----------- W

M0----------

M0L

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

tion forces: gives:

PR MR

MF

Introduce a fictitious bending moment, MFand 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: BW

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+( )=

WMR----------- 0 MR

PL2

-------= =

WP-------- 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:M04

------- MR VL 0=

=> 1 statically indeterminate, e.g. MR

Total complementary elastic energy:

W 4L

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

2MR

M04

-------M04

-------

2

+ + =

WMR----------- 0 = MR

M08

------- ;= WM0----------

M0L

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

properties!

MR

V

Equilibrium, beam:

: V N P+ 0=

=> 1 statically indeterminate,


L3

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

2N

21

3---+

2PN+ =

WN-------- 0 = N

P3 +------------- ;=

PWP-------- PL

3

EI--------- 1

3 +( )-----------------= =

P

N

k EIL

3------=

: 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 (Castiglianos 2nd theorem):

L6EI--------- PL NL( )2 N

2

2EI L3-----------------------+=

Equilibrium gives

Complementary

WM0---------- 3

2---

M0L

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

M0 M0 M0MR

k6EI

L3

---------=

M0

RN

NM0 MR

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

Note! one staticallyindeterminate exists!

elastic enerrgy: W L6EI--------- MR

2MRM0 M0

2+ +( ) L

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

2( ) N2

2k------+ +

L4EI--------- MR

23M0

2+( )= =

Condition to determine the unknownW

MR----------- L

4EI---------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 M0L

-------

2 2P0M0L

-----------------+ L

3

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

2 M0L

-------

2 2P0M0L

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

Complementary elastic energy:

(given condition) M03 2+3 +

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

k EIL

3------=

NM0

R

P0P0M2M1N

Equilibrium:

: M0 P0L NL+ 0=

N P0 M0 L=

P0WP0--------- 0= =

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

WP--------

M 0=

=PL

3

EI---------=

Case (b): statically indeterminate problem,

WM-------- 0 M PL

23 2+----------------= = b

WP-------- L

3

EI------ P

23---M

L-----

PL3

EI---------9 2+

9 6+----------------= = =

According to the given conditions: b12---a

92---= =

where M is an internal indeterminate quantity, thus

Equilib:


PWP-------- 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 (Castiglianos 2nd theorem):

M0 PP

P P2P

M0 M0 PP(statically indeterminate probl.)

1.7 (10)


1.10

1.11

1.12

Equilibrium:

WMA----------- 0 MA

QL2

-------- ;= =

MA

RAQ RA RB Q+ 0=

RBMA 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+ +( )=

BWQ--------

Q 0=

718------PL

3

EI---------= =Displacement at B (Castiglianos 2nd theorem):

L L L

Complementary elastic energy in the beam: WL

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

240RAP 8P

2+( )=

CWP-------- 52

57------PL

3

EI---------= =Displacement in point C (Castiglianos 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: WRA--------- 0 RA

519------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 mechanicsdirect 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

45ok1k2

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 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 , 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 .

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 . 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

P

2---

k EA L=

P30o

k1 k2

30o

E, A

, LE, A, L

k1 0.75 EA L=

xy

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 . 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 Hllfasthetslra , 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

l

e1

e2

1

2

3

g

3A 2= L 2l cos=

30=

E, A, L

2P

P

E, A

, L

E 2A 2L,,

45ok

EA

EA, L

PL

45o

90ok 2EA L=

P

3k2k

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

D2D30

R1P

0

R4

=D2D3

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

D2D30

0

0

R1P

0

R4R5R6

=

Equation-system

Boundary conditions & prescribed forces:

Solution: D2D3

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 rr r

=Element stiffness matrix:

k1 k= r1 0

0 0=,

k2 2k= r1 2 1 21 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

R2R3

R4R5

R6P

R8

=

Reactions-forces

Equation (7) gives

3kD7 P=

D7P3k------=

x1

2

45oElement 2: k3 2k= r 1 2 1 21 2 1 2

=,

2.5 (12)


2.4

Boundary conditions:

e1& e4:

Assembly of global stiffness matrix:

D2

Element stiffness matrices:

Ke kia aa a

a, c2

sc

sc s2

c cos=s sin=

,= =

e2: k2 2k a, 45={ }12--- 1 1

1 1= = = e3: k3 4k a, 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, 45 ={ }12--- 1 1

1 1= == =

u1u2

Tde1

2------- 1 1 0 0

0 0 1 1

D9D10D7D8

1

2-------

D9D8

= = =

Normal force in element 4: N k4=k4 2k= u2 u1=where

N3 2

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

k 5 1

1 3

D8D9

P

P

D8D9

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 aa a

a, c2

sc

sc s2

c cos=s sin=

,= =

e3: k3EAL

------- a, 45={ } 12--- 1 1

1 1= = =

e2: k22EA

L----------- a, 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, 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

D6D8

0

P

D6D8

PL

5EA----------- 2

12= =

e1 e2

e3

R2EA2L------- D 6( )

P5---= =

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!

2

4

3

512---

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;

EAL

------- c2

sc

sc s2

c cos=s sin=

e1:

e4:

e3:

EAL

------- 0 0

0 1;e2:

2---

cos sin=

2---

sin cos=

EAL

------- s2

s c

s c c2

Assembly: K EAL

------- 1 c2

s2

+( )+ sc sc( )

sc sc( ) 1 c2 s2+( )+1 +( )EA

L------- 1 0

0 1= =

(only node 5)

Eq. system: 1 +( )EAL

------- 1 0

0 1

D5xD5y

P cossin

=D5xD5y

PL

1 +( )EA------------------------- cos

sin=

Thus:

D5x

D5y

PL1 +( )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 aa a

a, c2

sc

sc s2

c cos=s sin=

,= =

e4: k2 k0 a, 120={ }14--- 1 3

3 3= = =

D1

D4

D3

D6

D5

D9

D1 = D2 = D3 = D4 = D5 = D6 = 0; F7 = Pcos; F8 = Psin

k134---k0 a, 0={ }

1 0

0 0= ==

D9D10

D0cossin

=

e2

P

e4

e1

e3

e3: k3 k0 a, 60={ }14--- 1 3

3 3= = =

e2: k2 a 0={ }1 0

0 0= =,

where k0EAL

-------=

D10

1.25k0 k2+ 0

0 1.5k0

D9D10

P cossin

=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

D7D8

mg sincos

D7D8

2mg3k

----------- sincos

= =

F7 mg F8 mg cos=,sin=

mg

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

Element stiffness matrices:

Kiai ai

ai ai=

D1

D4D3

D6

D5

D1 = D2 = D3 = D4 = 0,

EA2l------- 2 3

3 2

D5D6

mg 0

1

D5D6

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( )

32

-------mg R2EA4l------- 3D5 3D6( )

32---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

D4D3

D2

2Element stiffness matrices: Ke ki

a aa a

a, c2

sc

sc s2

c cos=s sin=

,= =

k1EAL

------- a, 0={ } 1 00 0

= = =

e2: k2EAL

------- a, 90={ } 0 00 1

= = = e3: k2EAL

------- a, 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

PL4EA----------- 7

5= =

Reaction forces:

(1): R1EA2L------- 2D5( )

5P4

-------= =

(2): R2EA2L------- 2D4( )

7P4

----------= = (3): R3

EA2L------- D4 D5( )

P4--- (Node 2)= =

(6): R6EA2L------- D4 D5( )

P4--- (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 aa a

a, c2

sc

sc s2

c cos=s sin=

,= =

e1: k1EAL

------- a, 45 ={ } 12--- 1 1

1 1= = =

e3: k32EA

L----------- a, 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

D6D5

D4

D3

D2

2 D1

e3

e2: k2EAL

------- a, 45={ } 12--- 1 1

1 1= = =

matrices:

F1 = P, F4 = 0

1

2

3

4 5

6

7

8 K k4---

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 K1F

D7D8

P

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

3

Pk--- 0.4545

0.0525= = =

F1

F2F3

F4F5F6

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

D7D9

D10

P6k------

2

5

1

= =

F1

F2F3

F4F5

F6F8

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 K1F D5

2P3k------- 1= =

F1F2F3

F4F6

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 11 1

T; 1

2------- 1 1 0 0

0 0 1 1 De;

D1D2

D5D6

f k

2-------

D1 D2 D5 D6+

D 1 D2 D5 D6+ +N

23

-------P= = = = =

Element 2:

k k 1 11 1

T; 1 0 0 00 0 1 0

De;

D5D6D3D4

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

D6D7

D8

Qk----

6 4 2+

3 2 2+

5 4 2

Qk----

0.3431

0.1716

0.6569

= = =

F1F2

F3F4F5

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

=

rigid1

2

3

4

5

6

D4 0=

D5 D6 0= =

D1 0=

F2 Q F3, 0= =


(rigid support)

D K1F

D2

D3

Q17k--------- 7

4

Qk---- 0.4118

0.2353= = =

F1

F4F5F6

Q85------

20

15

20

70

Q

0.2353

0.1765

0.2353

0.8235

= =

Reaction-forces

Equation-system:

Element 1: k 5k 1 11 1

T;15--- 4 3 0 0

0 0 4 3 De;

D1D2

D3D4

f k3D2 4D3

3D 2 4D3+N

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

e1

e2

e3 2

1

21

1

2

Element 2:

Element 3:

k k 1 11 1

T; 15--- 1 0 0 0

0 0 1 0 De;

D5D6

D3D4

f kD3

D3N

417------ Q= = = = =

k 2k 1 11 1

T; 0 1 0 00 0 0 1

De;

D1

D2D5

D6

f 2kD2D 2

N1417------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 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 Galerkins 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/m

3]. 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 denotes the normal stress and v is an arbitrary weight function.(b) Derive the FEM-equation (use Galerkins 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

ddx------

q Q+ 0= x1 x x2

ddx------ D

ddx------

q Q+ 0=

dvdx------D

ddx------dx

x1

x2

vqdxx1

x2

+ vQdx vDddx------

x1

x2+

x1

x2

=

xx = 0 x = L

KxE, A

ddx------ EA

dudx------

AKx+ 0=

dvdx------EA

dudx------dx

0

L

vKxAdx v A( )[ ]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/m

3]. The displacement, u, in the bar isgiven by the solution to the differential equation

.

(a) Show that the weak form is

,

where denotes the normal stress and v is an arbitrary weight function.(b) Derive the FEM-equation (use Galerkins method) to the weak form above for one ele-


.

(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, g, where 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, and g are constants andshow that the weak form is

,

where is the normal stress in the bar and v an arbitrary weight function.(b) Derive the FEM-equation (use Galerkins method) to the weak form above for one ele-


.

x

x = 0 x = Lkx

KxE, A

ddx------ EA

dudx------

kxu AKx+ 0=

dvdx------EA

dudx------dx vkxudx

0

L

+0

L

vKxAdx v A( )[ ]0L

+

0

L

=

kede fe=

u 0=A Q=kx 3EA 2L

2( )= Kx Q AL( )=

(a) (b)

x = 2L

P

x = 3L

x = L

k

xx

x = L

g

EA

ddx------ EA

dudx------

Ag+ 0=

EAdvdx------du

dx------dx

0

L

v A( )[ ]0L gA 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 Galerkins 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 k [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[ ] LL dv

dx------M

L

L

vqdx

L

L

+=

T EIw( )= M EIw=

kede fe=

M0

z,w

x

x = L x = L

w L( )kz EI L

4=

q Q 2L( )=

Q

kw

k

2L, EI

kw3EI

2L3

---------= kEIL------=

3.3 (15)


3.7 The figure to the right shows a beam, which is loadedby its dead weight per unit volume g, where 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 , 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 Galerkins method).

x

x = L x = L

z,w

M M

T

T

g

d2

dx2

-------- EId

2w

dx2

---------

gA+ 0=

d2v

dx2

--------EId

2w

dx2

---------dx

L

L

vT[ ] LL dv

dx------M

L

L

gA vdxL

L

=

T EIw( )= M EIw=

kede fe=

P

z,w

x

x = L x = L

kg

k EI L3=

x L= 1 2= P gAL=

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[ ] LL dv

dx------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(), where 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 (Fouriers 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

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( )[ ]03L 2

v qA hPT+( )dx0

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 Galerkins 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( )[ ]x1x2 vhPTdx

x1

x2

+=

Q k AdT dx=

keTe fe=

hPL 16kA L=

Q hPLT=x 0= T T= x L=

2L, EI

0 1

1

d3

d4d2

d1

w ( ) N1d1 N2d2 N3d3 N4d4+ + + Nde B,d

2N

dx2

---------- 1

L2

-----d2N

d2----------= = = =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 3 3

+( ) 4 N2 L 1 2

3+( ) 4=,=

N3 2 3 3

+( ) 4 N4 L 1 2

+ 3+( ) 4=,=

Deflection:

( ) N11 N22+ N1 N212

= = N1 1 = N2 =1 2

1 2L

0 1

1D:

NTNdx

0

L

dx Ld={ }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. (Galerkins method): where (arbitrary)

Inserted into the weak form gives

3.3 (a) See solution to 3.1 and 3.2

vd

dx------ D

ddx------

q Q+

dx

x1

x2

0=

vddx------ D

ddx------

dx

x1

x2

vDddx------

x1

x2 dvdx------D

ddx------dx

x1

x2

=

dvdx------D

ddx------dx

x1

x2

vqdxx1

x2

+ vQdx vDddx------

x1

x2+

x1

x2

= 0 eller Dddx------ D

ddx------

0

p x xi= = =

Essential B.C. Natural B.C.

Weak form:

( ) Ne= N 1 = e12

=

ddx------ d

d------d

dx------ 1

l---dN

d-------e Be= = =

v ( ) N TNT= = T 1 2=

dvdx------ dv

d------d

dx------ T1

l---dN

T

d---------- TBT= = =

T BTDBld0

1

NTqNld

0

1

+ e T NTQld NTDd

dx------

0

1+

0

1

=

kq fQfs

kee fe= fe fQ fs+=drT is an arbitrary vector

The element matrices becomes:

kD BTDBld

0

1

1l--- 1

1D

1l--- 1 1 ld

0

1

1l--- 1 1

1 1Dd

0

1

= = =

kq NTqNld

0

1

1

q 1 ld

0

1

l1 ( )2 1 ( )

1 ( ) 2qd

0

1

= = =

kDke kD kq+= och

fQ NTQld

0

1

l1

Qd

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 N

T A( )[ ]0

L+d=

D3D2D1

EA2L-------

2 2 0

2 3 1

0 1 1

0

D20

R1Q 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

vkxudx0

L

+ vAKxdx0

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= = dv

dx------ be

TdNT

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

TBT

= =

beT BTEAB x NTkxN xd

Le

+dLe

de beT NTKxA x N

T A( )[ ]0Le

+dLe

=

ke febut be

Tis arbitrary kede fe=

D2D1 D3

Le = L/2Le = L/2N 1 ( ) = B

2L--- 1 1=

Element-

NT QAL-------A

L2---d

0

1

Q4---- 1

1=

BTEABL2---d

0

1

2EA

L----------- 1 1

1 1=

NT3EA

2L2

-----------NL2---d

0

1

EA8L------- 2 1

1 2=

keEA8L------- 18 15

15 18;=

Assembly: EA8L-------

18 15 0

15 36 15

0 15 18

D1D2D3

Q4----

1

2

1

R

0

Q

+=point force

reaction force

Equation (2) & (3):D2D3

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------

Ag+ 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 A( )[ ]0L

Ag vdx0

L

+= 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


beT

EA BTBdx

0

L

de beT NT A( )[ ]

0

LAg NTdx

0

L

+=

Inserted into theweak form gives:

Element matrices:

ke EA BTBLd

0

1

EAL

------- 1 1

1 1= = fe Ag N

TLd

0

1

AgL

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

1= =

ke k1 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

R1P

0

R4

AgL2

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

1

2

1

0

+=

(reaction forces: R1 & R4)

EAL

------- 2 1

1 3

D2D3

P AgL+AgL 2

D2D3

PL

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

1

AgL2

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

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

3.9 (15)


3.6(a)


Weight function:

3.6(c)

Weighted residual: v EIw( ) kzw q+[ ]dxL

L

0=

Integration by parts: v EIw( )[ ]dxL

L

v EIw( )[ ] LL

v EIw( )[ ]dxL

L

= =

(2) inserted into (1) with

(1)

(2)v EIw( )[ ]L

Lv EIw( )[ ]

L

L vEIwdx

L

L

+=

gives the weak form:

vEIwdx vkzwdxL

L

+L

L

vT[ ] LL

vM[ ]L

L vqdx

L

L

+=

T EIw( )= and M EIw=

w Nde=d

2w

dx2

--------- d2N

dx2

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

v Nbe beTNT= =

d2v

dx2

-------- beT d

2NT

dx2

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

T= =

beT BTEIB

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 febut be


K BTEIBdx NTEI

L4

---------Ndx

L

L

+L

L

EI

2L3

---------

3 3L 3 3L

3L 4L2

3L L2

3 3L 3 3L

3L L2

3L 4L2

EI

105L3

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

78 22L 27 13L

22L 8L2

13L 6L2

27 13L 78 22L

13L 6L2

22L 8L2

+= =

F

R1R2

R3M0

=

Reactionforces/moments

Stiffnessmatrix

Loadvector

2EIL

--------- 8EI105L-------------+

d4 M0=

w L( ) d4105

210 8+( )--------------------------

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------ NTdx

L

L

Q2----

1

L 31

L 3

= = Fs

R1R20

0

=

Force vector: F Fb Fs+=

where

Reactionforce & moment

External pointforce & moment

EI

2L3

---------6 3L

3L 6L2

w22

Q2---- 1

L 3

w22

QL

3

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

1= =

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

stiffnessmatrices

(only dof 3 and 4)

kw3EI

2L3

---------= kEIL------=

Weighted residual: v EIw( ) gA+[ ]dxL

L

0=


L

v EAw( )[ ] LL

v EIw( )[ ]dxL

L

= =

(1)

(2)v EAw( )[ ]L

Lv EAw( )[ ]

L

L vEIwdx

L

L

+=

vEIwdxL

L

vT[ ] LL

v M[ ]L

L gA 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 BTEIB

L

L

xd de beT NTT[ ]

L

L dNT

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

L

L

gA NT xdL

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 BT

EIBdx

L

L

EI

2L3

---------

3 3L 3 3L

3L 4L2

3L 2L2

3 3L 3 3L

3L 2L2

3L 4L2

= = Kfjder kEI

2L3

---------= =

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

Assembly of total stiffness matrix

K Kbalk Kfjder+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 gA NTdx

L

L

gAL1

L 31

L 3

= = Fpoint

R1R2P

0

=

Force vector: F Fdistributed Fpoint+=

where

Reaktionforce/moment

External pointforce/moment

P gAL= inserted gives:

F

R1 gAL

R2 gAL2

3

2gAL

gAL2 3

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

EI

2L3

---------4 3L

3L 4L2

w22

gAL 2L 3

w22

gAL3

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

4 3= =

Displacement boundary conditions: w1 = 1 = 0

stiffnessmatrices:

Weighted residual: v EIw( ) 1xL---+

Q2L------+ dx

L

L

0=


L

v EAw( )[ ] LL

v EIw( )[ ]dxL

L

= =

(1)

(2)v EAw( )[ ]L

Lv EAw( )[ ]

L

L vEIwdx

L

L

+=

vEIwdxL

L

vT[ ] LL

vM[ ]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= =

d2v

dx2

-------- beT d

2NT

dx2

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

T= =

beT BTEIB

L

L

xd de beT NTT[ ]

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 BTEIBdxL

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

R2R3R4

=

Nodal force vector: F Fb Fpoint+=

where Reactionforce/moment

External point force

w ( ) N1 ( )d123---PL

3

EI--------- 1

5---QL

3

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

2 3 3

+4

--------------------------- where , x

L---= = =

Reduced equation system, Eq. (1): EI

2L3

---------3d1 P3Q10------- 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 1

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( )[ ]03L 2

v qA hpT+( )dx0

3L 2

+=

Q kAdTdx------= gives the weak form:

(2)

T NTe=dTdx------ dN

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

v Nbe beTNT= = dv

dx------ be

T dNT

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

T B

T= =

beT BTkAB

x1

x2

x NT

x1

x2

hPN xd+d Te beT NT

x1

x2

qA hpT+( ) x NT

Q( )[ ]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---d 1

hP 1

L2---d

0

1

+0

1

hPL 12kAL

------= kA

L------ 4 1

1 4= = =

Assembly of system matrix gives

e3e2e1 4321

T4T3T2T1

FEM - model:

fe1

qA hPT+( )

L2---d

0

1

q 0 hPL 12kAL

------=;=

3kAL

------T1

1= = =

F 3kAL

------T

1

2

2

1

QR10

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

4TT2

T3T

3kAL

------T

1

2

2

1

QR10

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 T6 1 4 0 1+{ }6 0 4 1 1{ }

T2T3

T21------ 29

22= T

1.3809

1.0476= =

Convection at x=3L/2 gives for element 3: NT Q( )[ ]12

c0

Q2

Q1

0=

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

NT Q( )[ ]12 kA

8L------ 0 0

0 1

T1T2

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

T3T4

+ 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( )[ ]x1x2 vhpTdx

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= = dv

dx------ be

T dNT

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

T B

T= =

beT BTkAB

x1

x2

x NT

x1

x2

hPN xd+d Te beT NT

x1

x2

hpT x NT

Q( )[ ]x1

x2+d=

ke febut be

Tis 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---d 1

hP 1

L2---d

0

1

+0

1

hPL 16kAL

------= 2kA

3L---------- 7 1

1 7= = =

Assembly of the stiffness matrix gives:

e2e1 321

T3T2T1

FEM - model:

fe1

hPT

L2---d

0

1

hPLT

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

1= =

Equation system:

23---kA

L------

7 1 0

1 14 1

0 1 7

T1T2

T3 T=

kAL

------

20

8

44QR

hPLT-----------------

T=7 1 0

1 14 1

0 1 7

T1T2T

30

12

66QR

hPLT-----------------

T=

Eq. (1) & (2) gives:

System matrix:

7 1

1 14

T1T2

30 0( )12 1( )

TT1T2

197------ 433

121T

4.46

1.25T= =

and r.h.s. FhPLT

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

1

2

1

hPLT

1

0

Q RhPLT-----------------

+= =

FhPLT

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 0 = 0, the element is subjected to a rigid bodymotion equal to 0.

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

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 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 , 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 0= u2 0 0L+= x( ) 0=

Nod 1 Nod 3 Nod 4 Nod 2

= 1 = 1 = 1/3 = 1/3

xLL L

Nod 1 Nod 3 Nod 2

1 1<


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 and 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 0)and the displacement 0 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

0 1

N1 1 N2, = = dx Ld=

u1 u2

(a) (b)

2

w2w1

1

x

L 0 L

1 2 = =

w1 L=

w2 L+=

w x( ) x+=

2w2

w11

x

L 0 L

R010-----=

w1 a b c=

1 b L 2c L+=

w2 a b c+=

2 b L 2c L=

00

0 1 R0=

w =0

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 1

1

d3

d4d2

d1

w ( ) N1d1 N2d2 N3d3 N4d4+ + + Nde B,d

2N

dx2

---------- 1

L2

-----d2N

d2----------= = = =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 3 3

+( ) 4 N2 L 1 2

3+( ) 4=,=

N3 2 3 3

+( ) 4 N4 L 1 2

+ 3+( ) 4=,=

Deflection:

( ) N11 N22+ N1 N212

= = N1 1 = N2 =1 2

1 2L

0 1

1D:

NTNdx

0

L

dx Ld={ }L6--- 2 1

1 2= =

N

dNT

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

dx------dx

0

L

1L--- 1 1

1 1=

NTd1

1

1

L 31

L 3

= NTd1

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+ 0 0L+ x L={ } 0 0x+= = = =

x( ) dudx------ 0= =

Displacement in the element:

(a) Strain in the element:

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

N1 1 ( ) 1 3( ) 1 3+( ) C1= N1 1( ) 1= C1 16=

N2 1 +( ) 1 3( ) 1 3+( ) C2= N2 1( ) 1= C2 16=

N3 1 +( ) 1 ( ) 1 3( ) C3= N3 1 3( ) 1= C3 16 9=

N4 1 +( ) 1 ( ) 1 3+( ) C4= N4 1 3( ) 1= C4 16 9=

dudx------ du

d------d

dx------= =

dxNk

---------xkdk 1=

3

L 1 2( )d= =

u2 u1

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

u1 u2 2u3+( )L

-------------------------------------+ 11 2( )

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

Strain:

Note! singular for 2 1=

(a) One element solution, discretization:

keEAL

------- 1 1

1 1= fb N

TKxALd

0

1

ALq01

d

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

ALq02

------------- 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 N

TKxA

L2---d

0

1

AL2

-------q01

d

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=

D2D3

R

0

P

ALq04

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

1

+=D2D3

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 NTKxALd

0

1

ALq01

d

1 2

1

ALq0

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

3= = =

Assembly: EAL

-------1 1 0

1 2 1

0 1 1

D1 0=

D2D3 0=

ALq08

-------------0

1

3

R10

R3

+=

B.C. Reaction forces

D2q0L

2

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

Reaction forces: R1EAL

------- D2( ) ALq0

16-------------= = R3

EAL

------- D2( )3ALq0

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

7ALq016

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

Note! R1 R3+( ) KxAdx3L 2

2L

ALq0

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

w x( ) w L( ) Nidii 1=

4

N1 L( ) N2 N3 L+( ) N4+ + += = =

N1 N3+( ) N2 N4 L N3 N1( )+ +( )+= L+ x+= =

0 wd

2Ni

dx2

-----------dii 1=

4

N1

L2

---------w1N2

L2

---------1N3

L2

---------w2N4

L2

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

2c

L2

------= = = =

0 w x 0=( ) Ni 0=( )dii 1=

4

N1w1 N21 N3w2 N42+ + +( ) 0= a= = = =

0 w x 0=( )dNi 0=( )

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

i 1=

4

N1L

--------w1N2L

--------1N3L

--------w2N4L

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

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

= =

Element stiffness matrix:

fb NT Q2L------Ld

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=

D3D4

R

MRP

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 D4Element stiffness matrices: Distributed load contribution

to 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 23 3L 2 3 3L 2

3L 2 L2 2 3L 2 L2

= = = fb1 fb2 NT Q

2L------L

2---d

1

1

Q24------

6

L

6

L

= = =

D3D4D5D6

PL

3

6EI---------

5

9 L16

12 L

ML2

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

1

2 L4

4 L

QL3

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

17

28 L48

32 L

+ +=

4EI

L3

---------

3 3L 2 3 3L 0 0

3L 2 L2 3L 2 L2 2 0 03 3L 2 6 0 3 3L 2

3L 2 L2 2 0 2L2 3L 2 L2 20 0 3 3L 2 3 3L 2

0 0 3L 2 L2 2 3L 2 L2

D1 0=

D2 0=

D3D4

D5D6

R

MR

0

0

P

M

Q24------

6

L

12

0

6

L

+=

D3D4

PL3

EI--------- 8 3

2 LML

2

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

2 LQL

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

D6e2e1

B.C. & kinematical constraint: D1 = D2 = D3 = 0(give reaction forces/moments: R1, R2 & R3)

Element 1: x L 1 +( ) dx Ld ,= = fb NTqLd where q

1

1

Q2L------ 1 +( )= =

fb12 3 3+( )

4-------------------------------- Q

2L------ 1 +( )Ld

1

1

3Q10-------= = fb2

2QL15

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

QL5

--------=

FT R13Q10-------+ 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 EI L3=

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, , 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 =

R h

k4

2 8( )-------------------EI

R3

------=

R h

P

R

x

y

R

2

2

P

R

h

P 2 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

kx

y

a c2

sc

sc s2

=

Ke ka aa a

=

al122

l12m12

l12m12 m122

=

c cos=s sin=

l12 xcos x2 x1( ) L= =

m12 ycos y2 y1( ) L= =

L x2 x1( )2

y2 y1( )2

+=

where

alternativly

Global forulation for truss & spring elements

spring constant for a truss element kEAL

-------=

1y, M1y

w1, f1z

w2, f2z

2y, M2y

u2, f2x

xz

u1, f1xke

EA2L------- 0 0

EA2L------- 0 0

03EI

2L3

--------- 3EI

2L2

--------- 03EI

2L3

---------3EI

2L2

---------

03EI

2L2

--------- 2EIL

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