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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
8/13/2019 Problem on Bar
2/26
27th to 31st Jan 20092
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
1-D BAR ELEMENT
8/13/2019 Problem on Bar
3/26
27th to 31st Jan 20093
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
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
8/13/2019 Problem on Bar
4/26
27th to 31st Jan 20094
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
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
8/13/2019 Problem on Bar
5/26
27th to 31st Jan 20095
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
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
8/13/2019 Problem on Bar
6/26
27th to 31st Jan 20096
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
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+)
8/13/2019 Problem on Bar
7/26
27th to 31st Jan 20097
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
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
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-)
2N2=
(1+)2
are the shape functions of 1-D Bar Element.and
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+)
8/13/2019 Problem on Bar
8/26
27th to 31st Jan 20098
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
Strain {} = [B] {q} But [B] = 1le
[ -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
8/13/2019 Problem on Bar
9/26
27th to 31st Jan 20099
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
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
Finite Element Analysis Solution:
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
8/13/2019 Problem on Bar
10/26
27th to 31st Jan 200910
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
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
Overall Nodal force vector will be {Fo}=
0
0+0
1
1
2
3
1000 {Fo}=
0
0
1
1
2
3
1000
8/13/2019 Problem on Bar
11/26
27th to 31st Jan 200911
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
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
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
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
8/13/2019 Problem on Bar
12/26
27th to 31st Jan 200912
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
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+)
Strain {} = [B] {q} But [B] = 1le
[ -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
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-)
2N
2
=(1+)
2are the shape functions of 1-D Bar Element.and
8/13/2019 Problem on Bar
13/26
27th to 31st Jan 200913
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
in which N1=(1-)
2N2=
(1+)2
are the shape functions of 1-D Bar Element.and
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)
Reaction forces are determined from the eliminated row of the characteristic equation
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
8/13/2019 Problem on Bar
14/26
27th to 31st Jan 200914
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
Finite Element Analysis Solution:
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
8/13/2019 Problem on Bar
15/26
27th to 31st Jan 200915
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
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
Overall Nodal force vector will be {Fo}=
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
8/13/2019 Problem on Bar
16/26
27th to 31st Jan 200916
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
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
Characteristic equation for the over all problem is given by [Ko] {Q} = {Fo}
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
8/13/2019 Problem on Bar
17/26
27th to 31st Jan 200917
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
For the Element 1 q1 = Q1 = 0 and q2 = Q2 = 0.068 mm
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 vectorq2
in which N1=(1-)
2N2=
(1+)2
are the shape functions of 1-D Bar Element.and
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)
FACULTY DEVELOPMENT PROGRAMMEFACULTY DEVELOPMENT PROGRAMME
8/13/2019 Problem on Bar
18/26
27th to 31st Jan 200918
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
For the Element 2 q1 = Q2 = 0.068 mm and q2 = Q3 = 0
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 vectorq2
in which N1=(1-)
2N2=
(1+)2
are the shape functions of 1-D Bar Element.and
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)
FACULTY DEVELOPMENT PROGRAMMEFACULTY DEVELOPMENT PROGRAMME
8/13/2019 Problem on Bar
19/26
27th to 31st Jan 200919
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
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
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Dr. T. JAGADISHProfessor, Department of Mechanical Engineering,
Bangalore Institute of Technology
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FINITE ELEMENT METHOD FOR ENGINEERING ANALYSISFINITE ELEMENT METHOD FOR ENGINEERING ANALYSIS
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
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FINITE ELEMENT METHOD FOR ENGINEERING ANALYSISFINITE ELEMENT METHOD FOR ENGINEERING ANALYSIS
Finite Element Analysis Solution:
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
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FINITE ELEMENT METHOD FOR ENGINEERING ANALYSISFINITE ELEMENT METHOD FOR ENGINEERING ANALYSIS
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
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
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FINITE ELEMENT METHOD FOR ENGINEERING ANALYSISFINITE ELEMENT METHOD FOR ENGINEERING ANALYSIS
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
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) = 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+)
=
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FINITE ELEMENT METHOD FOR ENGINEERING ANALYSISFINITE ELEMENT METHOD FOR ENGINEERING ANALYSIS
Strain {} = [B] {q} But [B] = 1le
[ -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
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-)
2N2=
(1+)2
are the shape functions of 1-D Bar Element.and
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)
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FINITE ELEMENT METHOD FOR ENGINEERING ANALYSISFINITE ELEMENT METHOD FOR ENGINEERING ANALYSIS
Strain {} = [B] {q} But [B] = 1le
[ -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)
Reaction forces are determined from the eliminated row of the characteristic equation
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
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FINITE ELEMENT METHOD FOR ENGINEERING ANALYSISFINITE ELEMENT METHOD FOR ENGINEERING ANALYSIS