problem on bar

8/13/2019 Problem on Bar

1/26

27th to 31st Jan 20091

Dr. T. JAGADISHProfessor, Department of Mechanical Engineering,

Bangalore Institute of Technology

FACULTY DEVELOPMENT PROGRAMMEFACULTY DEVELOPMENT PROGRAMME

ONON

FINITE ELEMENT METHOD FOR ENGINEERING ANALYSISFINITE ELEMENT METHOD FOR ENGINEERING ANALYSIS

PROBLEMSPROBLEMS

Dr.Dr. T. JAGADISHT. JAGADISH

Professor, Department of Mechanical Engineering,Professor, Department of Mechanical Engineering,

BANGALORE INSTITUTE OF TECHNOLOGY,BANGALORE INSTITUTE OF TECHNOLOGY,

K.R.Road, V. V. Puram, BangaloreK.R.Road, V. V. Puram, Bangalore 560 004560 004


2/26





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1-D BAR ELEMENT


3/26





ONON


For the simple bar shown in figure determine the displacement, strain, stress and

reaction using two element discretization. Given length of the bar is 1000mm,

Cross Section of the bar is 10mmx50mm = 500mm2 , P=1000N and Youngs

Modulus 2x105 N/mm2

PA, L, E

Analytical Solution:

Displacement in the bar at any section ux = xP

AEx

1000

500x2x105=

Displacement in the bar at midpoint x=500mm is = = 5x10-31000x500500x2x105

= 0.005mm

Displacement in the bar at the free end x=1000 is =1000x1000

500x2x105= 1x10-2 = 0.01mm

Axial strain in the bar at any section x = = 1x10-5(U)x=lL 1x10-2

1000= (Tensile)

or Ex = orPAAxial Stress in the bar at any section x =1000

500

Reaction Force at the fixed end Rx = - 1000 N/mm2

(Tensile)

1x10-5x2x105 = 2N/mm2


4/26





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Finite Element Analysis Solution:

Discretize the given structure with two 1-D bar element of

equal length le= 500mm

Q1 Q2

1 2 3

Q21 2

Elemental Stiffness Matrix for a 1-D Bar Element of uniformcross section is given by

EAle

[ke] = +1 -1-1 +1

Then the Elemental Stiffness Matrix Element 1 is given by[k1] =2x105x500

500

Then the Elemental Stiffness Matrix Element 2 is given by[k2] =

[k1] =105

+2 -2

-2 +2Then we have[k

2

] =105+2 -2

-2 +2and

Then the Overall Stiffness Matrix 2 is given by

2x105x500

500

+1 -1

-1 +1

2

3

2 3

+1 -1

-1 +1

1

2

1 2

[Ko] = 105

2 -2 0

-2 2+2 -2

0 -2 2

1

2

3

1 2 3

=1052 -2 0

-2 4 -2

0 -2 2

1

2

3

1 2 3


5/26





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Since in the element 1 there is no force acting at either of thenode the elemental Nodal force vector for the element 1 will be

Elemental Nodal force vector for aconcentrated force Px acting at x = ais given by

[fc]=[N]T= a [Px] = N1N2

Px

=a=

(1-)/2(1+)/2Px =a

(1-a)

(1+a)=

Px2

{fc1}= 00

1

2

In the element 2 there is a force Px =1000 N is acting at the

second node the elemental Nodal force vector for the element 2

will be

{fc2}=0

1

2

31000

Overall Nodal force vector will be {Fo}=

0

0

1

1

2

3

1000

Characteristic equation for the over all problem is given by [Ko] {Q} = {Fo}

Apply the boundary conditions. Since at node 1 the

bar is fixed hence Q1=0, Then by elimination

approach eliminating the first row and column in the

characteristic equation we have

1052 -2 0

-2 4 -2

0 -2 2

1

2

3

1 2 3

Q1Q2Q3

0

0

1

= 1000


6/26





ONON


For the Element 1 q1 = Q1 = 0 and q2 = Q2 = 0.005mm

=

1052 -2 0

-2 4 -2

0 -2 2

Q1Q2Q3

0

0

1

= 1000105

4 -2

-2 2Q2Q3

0

1= 1000Then we

Solving we get

Q2 = 0.005mm and Q3 = 0.01mm

Thus the solution for the displacements are Q1= 0, Q2 = 0.005mm and Q3 = 0.01mm

DETERMINATION OF OTHER UNKNOWNS

= -1 =+1=01 2

q2q1

Displacement in an element is given by {U} = [N]{q}

In which [N] = [ N1 N2 ] is the Shape function matrix

{q} = q1 is unknown nodal displacement vector

q2

in which N1=(1-)

2

N2=(1+)

2

are the shape functions of 1-D Bar Element.and

Thus U(-1) = 0 at node 1 and U(+1) = 0.005mm = 5x10-3

mm at node 2

Thus {U} = N1q1 + N2q2 =(1-)

2

(1+)2

(0) + (0.005) 2.5x10-3(1+) U()=2.5x10-3(1+)Thus the displacement any where with in the element is given by U()=2.5x10-3(1+)


7/26





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Strain {} = [B] {q} But [B] = 1le

[ -1 1]

{} = 1500

[ -1 1]

Thus {} = 1le

[ -1 1]q1q2

0

0.005

Stress {} = [D] {} = E = (2x105)x(1x10-5) = 2 N/mm2 (Tensile)= 1x10-5 (Tensile)

For the Element 2 q1 = Q2 = 0.005mm and q2 = Q3 = 0.01 mm

=

= -1 =+1=01 2q

2q1


In which [N] = [ N1 N2 ] is the Shape function matrix{q} = q1 is unknown nodal displacement vector

q2

in which N1=(1-)

2N2=

(1+)2


Thus U(-1) = 0.005mm = 5x10-3 mm at node 1 and U(+1) = 0.01mm at node 2

Thus {U} = N1q1 + N2q2 =(1-)

2(1+)

2(0.005) + (0.01) 2.5x10-3(3+)

Thus the displacement any where with in the element is given by U()=2.5x10-3(3+)


8/26





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[ -1 1]

{} = 1500

[ -1 1]

Thus {} = 1le

[ -1 1] q1q2

0.005

0.01

Stress {} = [D] {} = E = (2x105)x(1x10-5) = 2 N/mm2 (Tensile)

= 1x10-5 (Tensile)

Reaction forces are determined from the eliminated row of the characteristic equation

of the system i.e.

Thus 105[2 -2 0]Q1Q2Q3

= R1 Thus R1=105[2 -2 0]

0

0.005

0.01

Therefore R1 = -1000 N

COMPARISON OF RESULTS

Analytical FEM

Displacements At Fixed End 0 0

Displacements Mid Length of Bar 0.005 mm 0.005 mmDisplacements Free End of the Bar 0.01 mm 0.01 mm

Strain 1x10-5 1x10-5

Stress 2 N/mm2 2N/mm2

Reactions -1000 N -1000 N


9/26





ONON


For the taper bar shown in figure determine the displacement, strain, stress and

reaction using two element discretization. Given length of the bar is 600mm, Cross

Section Area at the fixed end A1 = 1200mm2 and at the free end A2 = 600 mm

2, Load

P=1000N and Youngs Modulus 2x105 N/mm2

PA1 L, E A2For the taper bar in the given length of 700mm,Cross Section Area at the mid section is found to be

Am = 900 mm2


Discretize the given structure with two 1-D bar element ofequal length le= 300 mm

Q1 Q2

1 2 3

Q21 2

Elemental Stiffness Matrix for a 1-D Bar Element

of uniform varying cross section is given byE(A1+A2)

le 2[ke] =

+1 -1

-1 +1

Then the Elemental Stiffness Matrix Element 1 is [k1] =2x105x(1200+900)

300 2

+1 -1

-1 +1

1

2

1 2

[k1] = 105 +7 -7

-7 +7

1

2

1 2

Similarly [k2] =105

+5 -5

-5 +5

2

3

2 3


10/26

27th to 31st Jan 200910




ONON


Thus the overall stiffness matrix is [Ko] =

+7 -7 0

-7 +12 -5

0 -5 5

1

2

3

1 2 3

105

Elemental Nodal force vector for aconcentrated force Px acting at x = ais given by

[fc]=[N]T= a [Px] = N1N2

Px

=a(1-a)(1+a)

= Px

2

Since in the element 1 there is no force acting at either of the

node the elemental Nodal force vector for the element 1 will be{fc

1}=

0

0

1

2

In the element 2 there is a force Px =1000 N is acting at the

second node the elemental Nodal force vector for the element 2

will be

{fc2}=0

1

2

31000


0

0+0

1

1

2

3

1000 {Fo}=

0

0

1

1

2

3

1000


11/26

27th to 31st Jan 200911




ONON





approach eliminating the first row and column in

the characteristic equation we have

1057 -7 0

-7 12 -5

0 -5 5

Q1Q2Q

3

0

0

1

= 1000

1057 -7 0

-7 12 -5

0 -5 5

Q1Q2Q3

0

0

1

= 1000 10512 -5

-5 5Q2Q3

0

1= 1000Then we Solving we get

Q2 = 1.43x10-3

mm and Q3 = 3.43x10-3

mmThus the solution is Q1 = 0, Q2 = 1.43x10

-3 mm and Q3 = 3.43x10-3 mm

For the Element 1 q1 = Q1 = 0 and q2 = Q2 = 1.43x10-3 mm


= -1 =+1=01 2

q2q1

Displacement in an element is given by {U} = [N]{q}In which [N] = [ N1 N2 ] is the Shape function matrix

{q} = q1 is unknown nodal displacement vector

q2

in which N1=

(1-)2 N2=

(1+)2 are the shape functions of 1-D Bar Element.and


12/26

27th to 31st Jan 200912




ONON


Thus U(-1) = 0 at node 1 and U(+1) = 0.00143mm =1.43x10-3 mm at node 2

Thus {U} = N1q1 + N2q2 =(1-)

2(1+)

2(0) + (1.43x10-3)= 7.15x10-4(1+)

U()=7.15x10-4(1+)Thus the displacement any where with in the element is given by U()=7.15x10-4(1+)


[ -1 1]

{} = 1300

[ -1 1]

Thus {} = 1le

[ -1 1]q1q20

1.43x10-3

Stress {} = [D] {} = E = (2x105)x(4.77x10-6) = 0.9534 N/mm2 (Tensile)= 4.77x10-6 (Tensile)

For the Element 2 q1 = Q2 = 1.43x10-3mm and q2 = Q3 = 3.43x10

-3 mm

= -1 =+1=01 2

q2q1



q2

in which N1=(1-)

2N

2

=(1+)

2are the shape functions of 1-D Bar Element.and


13/26

27th to 31st Jan 200913




ONON


in which N1=(1-)

2N2=

(1+)2


Thus U(-1) = 1.43x10-3 mm at node 1 and U(+1) = 3.43x10-3 mm at node 2

Thus {U} = N1q1 + N2q2 =(1-)

2

(1+)2

(1.43x10-3) + (3.43x10-3) = 10-3(2.43+)Thus the displacement any where with in the element is given by U()=10-3(2.43+)Strain {} = [B] {q} But [B] = 1

le[ -1 1]

{} =1

300[ -1 1]

Thus {} = 1le

[ -1 1]q1q2

1.43x10-3

3.43x10-3

Stress {} = [D] {} = E = (2x105)x(6.67x10-6) = 1.334 N/mm2 (Tensile)= 6.67x10-6 (Tensile)


of the system i.e.Thus 105[7 -7 0]

Q1Q2Q3

= R1 Thus R1 = 105[7 -7 0]

0

1.43x10-3

3.42X10-3

Therefore R1 = -1000 N

{x} = 6.67x10-6 (Tensile)

= -1000 N


14/26

27th to 31st Jan 200914




ONON



Discretize the given structure with two 1-D bar element Q1 Q2

1 2 3

Q21 2

Elemental Stiffness Matrix for a 1-D Bar Element of

uniform varying cross section is given by

Aluminum Steel10 kN

For the Stepped bar shown in figure an axial load of10 kN is applied at 20oC as shown. When the

temperature is raised to 80oC determine the

displacement, strain, stress and reaction using two

element discretization. Rise in Temperature is 60o

C

Aluminum Steel

L 500mm 250 mm

A 1000 mm

2

500 mm

2

E 0.7x105 N/mm2 2x105 N/mm2

A 23x10-6 /0C 11.7x10-6 /oC

EA

le[ke] =

+1 -1

-1 +1For the Element 1 [k1] =

0.7x105x1000

500

+1 -1

-1 +1

1

2

1 2

1051.4 -1.4

-1.4 1.4

1

2

1 2

=

Similarly [k2] =105 4 -4

-4 4

2

3

2 3

105[Ko] =

1.4 -1.4 0

-1.4 5.4 -4

0 -4 4

1 2 31

2

3

Overall stiffness matrix

Elemental Nodal force vector for aconcentrated force Px acting at x = a

is given by

[fc]=[N]T= a [Px] = N1N

2

Px

=a

(1-a)

(1+a)=

Px2


15/26

27th to 31st Jan 200915




ONON


Since in the element 2 there is no force acting at either of the

node the elemental Nodal force vector for the element 2 will be{fc2}=

0

0

1

2

In the element 1 there is a force Px =10 kN acting at the secondnode the Elemental Nodal force vector for the element 2 will be {f

c1} =

0

1

1

2104


0

1+0

0

1

2

3

104 {Fco}=

0

1

0

1

2

3

104

The Elemental nodal force vector due to temperature change T is {fint} =-1

1

{Fo} = {Fco} +{F

into}

EATElemental Nodal force vector due to temperature in the element 1 and 2 will be

{fint1} =-1

1

1

20.7x105x1000x23x10-6x60 =

-9.66

9.66

1

2104

{fint1} =-11

1

22x105x500x11.7x10-6x60 = -7.02

7.02

2

3104

Thus {Finto} =-9.66

2.64

7.02

1

2

3

104

Thus {Fint

o} =

-9.66

9.66 7.02

7.02

1

2

3104

Overall Nodal force vector due to

combined temperature and

concentrated load will be


16/26

27th to 31st Jan 200916




ONON


Thus the overall Nodal force vector

due to combined temperature and

concentrated load will be

{Fo} =0

1

0

1

2

3

104 +

-9.66

2.64

7.02

1

2

3

104 Fo =

-9.66

3.64

7.02

1

2

3

104


Apply the boundary conditions. Since at node 1 and 3 the bar is fixed hence Q1= Q3=0,Then by elimination approach eliminating the first and the third row and column in

the characteristic equation we have

Then we 5.4x105 Q2 =3.64x104 Solving we get

Thus the solution is Q1 = 0, Q2 = 0.068 mm and Q3 = 0

Q1Q2Q3

1051.4 -1.4 0

-1.4 5.4 -4

0 -4 4

1 2 31

2

3

-9.66

3.64

7.02

1

2

3

= 104

Q1Q

2Q3

1051.4 -1.4 0

-1.4 5.4 -4

0 -4 4

1 2 31

2

3

-

9.66

3.647.02

1

2

3

= 104

Q2 = 0.068 mm Q2 = 0.068 mm

FACULTY DEVELOPMENT PROGRAMME


17/26

27th to 31st Jan 200917




ONON


For the Element 1 q1 = Q1 = 0 and q2 = Q2 = 0.068 mm


= -1 =+1=01 2

q2q1



{q} = q1 is unknown nodal displacement vectorq2

in which N1=(1-)

2N2=

(1+)2


Thus U(-1) = 0 at node 1 and U(+1) = 0.068mm at node 2

Thus {U} = N1q

1+ N

2q

2 =

(1-)2

(1+)2(0) +

(0.068)= 0.034(1+) U()=0.034(1+)

Thus the displacement any where with in the element is given by U()=0.034(1+)

Strain { -int}=[B] {q} -T 1500

[ -1 1]0

0.068

Stress {} = [D] {t} = E t = (0.7x105)x(-1.244x10-3) = -87.08 N/mm2 (Compressive)

=1

le

[ -1 1]q1

q2= T - 23x10-6x60

(Compressive)Strain t ={ -int} = -1.244x10-3

Stress {} = - 87.08 N/mm2 (Compressive)



18/26

27th to 31st Jan 200918




ONON


For the Element 2 q1 = Q2 = 0.068 mm and q2 = Q3 = 0


= -1 =+1=01 2

q2q1



{q} = q1 is unknown nodal displacement vectorq2

in which N1=(1-)

2N2=

(1+)2


Thus U(-1) = 0.068 mm at node 1 and U(+1) = 0 at node 2

Thus {U} = N1q

1+ N

2q

2 =

(1-)2

(1+)2(0.068) + (0) = 0.034(1-) U()=0.034(1-)

Thus the displacement any where with in the element is given by U()=0.034(1-)

Strain { -int}=[B] {q} -T 1250

[ -1 1]0.068

0

Stress {} = [D] {t} = E t = (2x105)x(-9.74x10-4) = -194.8 N/mm2 (Compressive)

=1

le[ -1 1]

q1

q2= T - 11.7x10-6x60

(Compressive)Strain t ={ -int} = - 9.74x10-4

Stress {} = - 194.8 N/mm2 (Compressive)



19/26

27th to 31st Jan 200919




ONON


Reaction forces are determined from the eliminated rows of the characteristic equation

of the system i.e.

Therefore R1 = -9520 N R2 = -2720 N

Thus

Q1Q2

Q3

1051.4 -1.4 0

0 -4 4

1 2 3

1

3=

R1

R3

Then we have 1051.4 -1.4 0

0 -4 4

1 2 3

1

3=

R1R

3

0

0.068

0

= 105R1R3

1.4x0 1.4x0.068 + 0x0

0x0 4x0.068 + 4x0=

-9520

-2720



20/26

27th to 31st Jan 200920




ONON


For the simple bar shown in figure determine the displacement,

strain, stress caused due to self weight using two element

discretization. Given length of the bar is 0.5m, Cross Section Area of

the bar is 0.1m2, =7848 kg/m3 and Youngs Modulus 2x1011 N/m2 .A,

L,

EAnalytical Solution:

Displacement in the bar at any section ux = xgl2E x7848x9.81x0.52x2x1011=

Displacement in the bar at the free end x=0.5 is = = 4.812x10-8

7848x9.81x0.52

2x2x1011



21/26

27th to 31st Jan 200921




ONON



Discretize the given structure with two 1-D bar element of

equal length le= 0.25 m

Q1

Q2

1

2

3

Q2

1

2

Elemental Stiffness Matrix for a 1-D Bar Element of uniformcross section is given by EA

le[ke] =

+1 -1

-1 +1

Then the Elemental Stiffness Matrix Element 1 is given by[k1] =2x1011x0.1

0.25

Then the Elemental Stiffness Matrix Element 2 is given by[k2] =

[k1

] = 1010+8 -8

-8 +8Then we have [k2] = 10

10 +8 -8

-8 +8and

Then the Overall Stiffness Matrix 2 is given by

2x1011x0.1

0.25

+1 -1

-1 +1

2

3

2 3

+1 -1

-1 +1

1

2

1 2

[Ko] = 1010

8 -8 0

-8 8+8 -8

0 -8 8

1

2

3

1 2 3

=1058 -8 0

-8 16 -8

0 -8 8

1

2

3

1 2 3



22/26

27th to 31st Jan 200922




ONON





approach eliminating the first row and column in the

characteristic equation we have

10108 -8 0

-8 16 -8

0 -8 8

1

2

3

1 2 3

Q1Q2Q3

1

2

1

962.36

The elemental load vector due to body force is given by gle

2[fb] =

1

1

Then the Elemental load vector in the element 1 is given by

7848x9.81x0.1x0.252

[fb1] = 11

1

2= 962.36 1

1

Similarly Elemental load vector in the element 2 is given by 2

3= 962.36

1

1

Then the Overall Nodal Force Vector due to self weight will be {Fo}=1

2

1

962.36

1

2

3



23/26

27th to 31st Jan 200923




ONON


For the Element 1 q1 = Q1 = 0 and q2 = Q2 = 3.61x10-8 mm

10108 -8 0

-8 16 -8

0 -8 8

Q1Q2Q3

1

2

1

962.361010

16 -8

-8 8Q2Q3

2

1962.36Then we

Solving we get Q2 = 3.61x10-8mm and Q3 = 4.81x10

-8mm

Solution for the displacements are Q1= 0, Q2 = 3.61x10-8mm and Q3 = 4.81x10-8mm


= -1 =+1=01 2

q2q1



q2

in which N1=(1-)

2

N2=(1+)

2


Thus U(-1) = 0 at node 1 and U(+1) = 3.61x10-8

mm at node 2

Thus {U} = N1q1 + N2q2 =(1-)

2

(1+)2

(0) + (3.61x10-8)

2.5x10-3(1+)

U()=1.805x10-8(1+)Displacement any where with in the element is given by U()=1.805x10-8(1+)

=



24/26

27th to 31st Jan 200924




ONON



[ -1 1]

{} = 10.25

[ -1 1]

Thus {} = 1le

[ -1 1] q1q2

0

3.61x10-8

Stress {} = [D] {} = E = (2x1011)x(1.444x10-7) = 2.888x104 N/mm2 (Tensile)= 1.444x10-7 (Tensile)

For the Element 2 q1 = Q2 = 3.61x10-8mm and q2 = Q3 = 4.61x10

-8 mm

=

= -1 =+1=01 2q

2q1



q2

in which N1=(1-)

2N2=

(1+)2


Thus U(-1) = 3.61x10-8

mm at node 1 and U(+1) = 4.81x10-8

mm at node 2

Thus {U} = N1q1 + N2q2 = (1-)2 (1+)2(3.61x10-6) + (4.81x10-8) 10-8(4.21+0.6)Thus the displacement any where with in the element is given by U()=10-8(4.21+0.6)



25/26

27th to 31st Jan 200925




ONON



[ -1 1]

{} = 10.25

[ -1 1]

Thus {} = 1le

[ -1 1] q1q23.61x10-8

4.81x10-8

Stress {} = [D] {} = E = (2x1011)x(4.8x10-8) = 9600 N/mm2 (Tensile)

= 4.8x10-8 (Tensile)


of the system i.e.

Thus 1011[8 -8 0]Q1Q2

Q3

= R1 Thus R1=1011[8 -8 0]

0

3.61x10-8

4.81x10-8

R1 = -28.88x103 N



26/26

27th to 31st Jan 200926



ONON


problem on bar

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