UNIT IV TWO DIMENSIONAL VECTOR VARIABLE PROBLEMS

1. TO
ByBy
Dr.G.PAULRAJDr.G.PAULRAJ
Professor & HeadProfessor & Head
Department of Mechanical EngineeringDepartment of Mechanical Engineering
Vel Tech (Owned by RS Trust)Vel Tech (Owned by RS Trust)
Avadi, Chennai-600062.Avadi, Chennai-600062.

2. Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
UNIT IV TWO DIMENSIONAL
VECTOR VARIABLE
PROBLEMS

3. UNIT IV TWO DIMENSIONAL
VECTOR VARIABLE PROBLEMS
 Equations of elasticity – Plane stress, plane
strain and axisymmetric problems – Body
forces and temperature effects – Stress
calculations - Plate and shell elements.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

4. STRUCTURAL MECHANICS
PROBLEM (2-DIMENSIONAL)
1.Thin structure:
 σz , τxz, τyz = 0 ( for thin wall)
 σx , σy , τxy ≠ 0
 The stress-strain constitutive relation for linear
Elastic, isotropic material under such plane-stress
Condition is given by
 σx 1 v 0 εx
σy = E/(1-v2
) v 1 0 εy - {ε}0
{ε}0
= Initial strain
τxy 0 0 (1-v)/2 γxy γxy = Shear strain
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
y
x
Thin Disk under Plane Stress

5. 2.Long -length member:
 The stress-strain constitutive relation for a linear Elastic, isotropic
material under such plane-strain condition is given by
σx 1-v v 0 εx
σy =E/(1+v)(1-2v) v 1-v 0 εy - {ε}0
{ε}0
= Initial strain
τxy 0 0 (1-2v)/2 γxy γxy = Shear strain
σz ≠ 0
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
y
T
T
z x Long Prismatic shaft under plane strain

6. Axisymmetric problem:
If the geometry of the structure, material properties,
support conditions and loading are all axisymmetric, then
it is reasonable to assume that the resulting deformation will
also be axisymmetric, i.e., it does not vary along the
circumferential direction. The non-zero stresses & strain
in a perfect axisymmetric Situation are related as follows:
σr 1 v/(1-v) v/(1-v) 0 εx
σθ =E/(1+v)(1-2v) 1 v/(1-v) 0 εy - {ε}0
σz 1 0 εz
τrz (1-2v)/2(1-v) γrz
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
r
z
Thick cylinder under
axisymmetric loads
Symmetric

7. DYNAMIC ANALYSIS
 For static analysis, the loads are applied very slowly and the
displacements are produced gradually and the movement of mass is
negligible and invisible.
 On the other hand, when the load is applied suddenly, the mass of the
body is largely disturbed and it will move in a faster rate from its
initial position.
 The analysis of such type of fast moving mass and the resultant
displacement is termed as dynamic analysis.
 In dynamic problems, the displacement, velocities, strains, stresses and
loads are all time-dependent. That is their magnitudes vary w.r.t time.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

8. VIBRATION
 The oscillatory motion of the body due to inertia force of the mass and
strain energy of the body is called as vibration.
 The vibration can also be defined as any motion that repeats itself after
an interval of time.
 A vibratory system, includes a means for storing potential energy
(spring /elasticity), a means for storing kinetic energy(mass/inertia),
and a means by which energy is gradually lost(damper) as shown in
figure
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

9.  Vibratory system:
 Where, F = Applied Load
m = Mass
k = Stiffness
c = Damping capacity
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
Mass
Spring representing
the elasticity of the bar
Equilibrium position
Inner extreme position
Outer extreme
position
m F
c
k

10. TYPES OF VIBRATION
 The vibratory motion is classified in three ways such as:
 According to actuating forces on the body:
 Free vibration: A system oscillates only under an initial disturbances with
no external forces acting after the initial disturbance. Ex: Simple pendulum.
 Forced vibration: If a system is subjected to an external forces (often, a
repeating type of force) the resulting vibration is known as forced vibration.
 Ex:M/C Tools, electric bells, diesel engines etc.,
 According to direction of motion:
 Longitudinal vibration: The movement of the particles of the body is
parallel to its axis, the resulting vibration is stated as longitudinal vibration.
 Transverse vibration: Due to external force, if the system like a rod vibrates
radially (┴ to its axis), the vibration is called as transverse vibration.
 Torsional vibration: When the rod rotates CW and CCW repeatedly due to
the applied load, that action is called as torsional vibration.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

11.  According to damping property:
 If no energy is lost or dissipated in friction or other resistance during
oscillation, the vibration is known as un-damped vibration whereas any
energy is lost during vibratory motion, is called as damped vibration.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

12. EQUATIONS OF MOTION USING
LAGRANGE’S APPROACH
 Consider a single d.o.f dynamic system with single mass m, single
spring of stiffness k and single damper having damping coefficient c as
shown in figure.
 The Lagrange equation is,
(d/dt)(∂L/∂q) - ∂L/∂q + ∂R/∂q =Fq --(1)
 and L = T -  (Here L = Lagrange function)
 Where T = Kinetic energy
 = Strain energy/Potential energy
q = Displacement coordinate
q = Velocity in the direction of q-coordinates
Fq= Applied force in q-coordinate
R = Damping energy
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
m F(t)
c
k

13.  If displacement occurs in X-direction then q can be specified as u.
 Now, the Kinetic energy, T = ½ m u2
where u = du/dt
 And the Potential energy,  = ½ k u2
 Damping energy, R = ½ c u2
 The equation (1) can be expanded as
 (d/dt)(∂T/∂q) - (d/dt)(∂/∂q) - ∂T/∂q + ∂/∂q +∂R/∂q =Fq
 By changing q into u, we get
 (d/dt)(∂T/∂u) - (d/dt)(∂/∂u) - ∂T/∂u + ∂/∂u +∂R/∂u =Fu -----(2)
 Now ∂T/∂u = ½ x 2mu = mu
 (d/dt)(∂T/∂u) = (d/dt) (mu) = mu
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

14.  (d/dt)(∂/∂u) = 0; ∂T/∂u = 0
 Similarly
 ∂/∂u = ½ x 2 ku = ku and
 ∂R/∂u = ½ x 2cu = cu
 Substituting the above, in the equation (2), we get
M u + ku + c u = F
Which is the dynamic equation of motion
For undamped and free vibration, the above equation will become,
M u + ku = 0
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

15. EQUATIONS OF MOTION
BASED ON WEAK FORM
(a)Axial Vibration of a Rod:
 The governing equation for free axial vibration of a rod
(considering the dynamic equilibrium of a differential element)
is given by
AE (∂2
u / ∂x2
) = ρA (∂2
u / ∂t2
) ---- (1)
 Using the technique of separation of variables and assuming
harmonic vibration,
u(x, t) = U (x)e-iωt
---- (2)
 Subtituting from equ. 2 in equ. 1,
AE (d2
U / dx2
) + ρA ω2
U = 0 ---- (3)
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

16.  The Weighted- Residual (WR) statement can be written as,
 ∫AE (d2
U / dx2
) + ρA ω2
U = 0 ---- (3)
 The weighted Residual (WR) statement can be written as,
L
W(x) AE (d2
U / dx2
) + ρA ω2
U dx = 0 ------(4)
0
 Integrating by parts, the weak form of the WR statement can be
written as
L L L
W(x) AE (dU / dx) - AE(dU/dx)(dW/dx)dx+ W(x) ρA ω2
U(x)dx=0
0 0 0 -------- (5)
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

17.  Observe that the first two terms are identical to the general
Weighted Residual weak form statement and an additional term
involving the mass density (inertia) effects.
 For a FE mesh, the integrals are evaluated over each element and
then summation is done over all elements.
 For a typical bar element, have two nodes and axial deformation
d.o.f at each nodes. The interpolation function is given by,
 U(x)= 1- x/l U1 + x/l U2
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
X=0
X=l
1
→F1
2
X
→F2
→u2
→u1

18.  Following Galerkin formulation, the weighted functions are the
same as the shape fuction.
 W1(x)= 1- x/l , W2 (x)= x/l
 Writing the weak form equations w.r.t W1 and W2
 We obtain from equation (5),
1 -1 U1 -P0 2 1 U1
 AE/l = + (ρA ω2
) /6 ------ (6)
-1 1 U2 Pl 1 2 U2
 Where P stands for AE (dU/dx)
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

19.  So, the element stiffness matrix as,
1 -1
 [k]e
=AE/l -------- (7)
-1 1
 The element mass matrix as,
1 -1
 [m]e
=ρAl/6 -------- (8)
-1 1
 Equation (6) can be rewritten as,
-P0
 [k]e
[δ]e
= + [m]e
ω2
[δ]e
----- (9)
Pl
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

20.  Where [δ]e
contains the nodal d.o.f.
 So, the typical free vibration equation
 [K]{U} = ω2
[M][U]
 Where [K] & [M] = Global, assembled stiffness and mass
matrices.
{U} = Nodal d.o.f.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

21. (b)Transverse Vibration of a beam:
 The Governing equation for free transverse vibration of a beam
based on the Euler- Bernoulli theory is given by
 EI (∂4
v / ∂x4
) + ρA (∂2
v / ∂t2
) = 0 ---- (1)
 Using the technique of separation of variables and assuming
harmonic vibration,
V(x, t) = V (x)e-iωt
---- (2)
 Subtituting in equ. 1,
EI (d4
V / dx4
) - ρA ω2
V = 0 ---- (3)
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

22.  The Weighted- Residual (WR) statement can be written as,
L
W(x) EI (d4
V / dx4
) - ρA ω2
V dx = 0 ------(4)
0
 Integrating by parts, the weak form of the WR statement can be
written as
L L L
W(x) EI (d4
V / dx4
) - EI(d3
V/dx3
)(dW/dx)dx- ρA ω2
W(x)V(x)dx= 0
0 0 0 -------- (5)
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

23.  Once again performing integration by parts,
L L L
W(x) EI (d3
V/dx3
) - (dW/dx)EI(d2
V/dx2
) + EI(d2
V/dx2
)(d2
W/dx2
) dx
0 0 0
L
- ρA ω2 W(x)V(x)dx= 0 -------- (6)
0
For a typical Euler- Bernoulli beam element,
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
x=Lx=0
V2
V1
θ1 θ2

24.  Where the shape function N are given by,
 N1 = 1 – 3x2
/L2
+ 2X3
/L3
, N2 = x – 2x2
/L + x3
/L2
 N3 = 3x2
/L2
-2x3
/L3
, N4 = -x2
/L + x3
/L2
 To solve the equation (6), we have the mass matrix as,
156 22L 54 -13L
22L 4L2
13L -3L2
 [m]e
= ρAL/420 54 13L 156 -22L ------ (7)
-13L -3L2
-22L 4L2
 We observe that the coefficients of the mass matrix
corresponding to translational d.o.f (viz. ρAL/420
(156+54+54+156)) sum up to ρAL, the mass of the element.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

25. CONSISTENT AND LUMPED
MASS MATRICES
 Equation for the mass matrix is
[m] = ρ ∫[N]T
[N] dv
Where ρ = Density and N = Shape function or interpolation function.
 It is called consistent because the same displacement model that is used
for deriving the element stiffness matrix is used for the derivation of
mass matrix.
 Consistent mass matrix equation for bar element:
2 1
 [m] = ρAL/6
1 2
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

26.  Consistent mass matrix equation for beam element:
156 22L 54 -13L
22L 4L2
13L -3L2
 [m] = ρAL/420 54 13L 156 -22L
-13L -3L2
-22L 4L2
 Consistent mass matrix equation for Truss element:
2 0 1 0
 [m] = ρAL/6 0 2 0 1
1 0 2 0
0 1 0 2
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

27.  Consistent mass matrix equation for CST element:
2 0 1 0 1 0
0 2 0 1 0 1
[m] = ρAL/12 1 0 2 0 1 0
0 1 0 2 0 1
1 0 1 0 2 0
0 1 0 1 0 2
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
u2
1
x1,y1
vi
3
x3,y3
2
x2, y2
u1
u3
v2
v3
X
Y
CST element
O

28. SOLUTION OF EIGENVALUE
PROBLEMS
The eigenvalue- eigenvector evaluation procedure fall into the following
basic categories:
 1. Characteristics polynomial methods
 2. Transformation based methods
 3. vector iteration based methods
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

29.  1. Characteristics polynomial methods:
 The generalized formula for free vibration is,
 [K]{U} = ω2
[M][U] [λ=ω2
]
 Where [K] & [M] = Global, assembled stiffness and mass matrices.
{U} = Nodal d.o.f. & λ= Eigen value
 (K – λM)U = 0 ----- (1)
 If the eigenvector is to be nontrivial, the required condition is,
 det(K – λM) = 0
 This represents the characteristic polynomial in λ
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

30. 2. Transformation based methods:
 The basic approach here is to transform the matrices to a simpler
form and then determine the eigenvalues and eigenvectors.
 The major methods in this category are the generalized Jacobi
method and the QR method.
 In the QR method, the matrices are first reduced to tridiagonal
form using Householder matrices.
 The Jacobi method uses the transformation to simultaneously
diagonalize the stiffness and mass matrices.This method needs
the full matrix locations and is quite efficient for calculating all
eigenvalues and eigenvectors for small problems.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

31. 3. vector iteration based methods:
 Various vector iteration methods use the properties of the
Rayleigh quotient.
 For the generalized eigenvalue problems given by the above
equation, define the Rayleigh quotient
Q(v) = (vT
Kv)/(vT
Mv), where v is arbitrary vector.
 A fundamental property of the Rayleigh quotient is that it lies
between the smallest and largest eigenvalue:
λ ≤ Q(v) ≤ λn
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

32. 4 -20 -10
 P1: Find the eigen values and eigen vectors of -2 10 4
6 -30 -13
Solution:
First find the characteristic equation:
4 -20 -10
 Let matrix A = -2 10 4
6 -30 -13
The characteristic equation is
λ3
- a1 λ2
+ a2 λ – a3 = 0 ------(1)
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

33.  Where a1 = Sum of leading diagonal elements
a1 = 4 + 10 – 13 = 1
 a2 = Sum of minor of the leading diagonal elements
10 4 4 -10 4 -20
= + +
-30 -13 6 -13 -2 10
 a2 = -130 + 120 -52 + 60 + 40 -40 = -2
4 -20 -10
 a3 = A = -2 10 4 = 4[-130+120]+20[26-24]-10[60-60]=0
6 -30 -13
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

34.  Substitute the a1, a2 and a3 values in equation (1),
 λ3
- λ2
- 2λ = 0
 Second find the eigen values:
 λ3
- λ2
- 2λ = 0
 λ( λ2
–λ-2) = 0
 λ2
- λ -2= 0
 When λ = 0,
 λ = (1± √1+8 )/2 = 2 or -1
 Eigen values are λ = 0, -1, 2
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

35.  Third find the eigen vectors:
 The eigen vectors are given by the equation (A-λI)X=0
 Where λ is the eigen value.
 X is the eigen vector.
4 -20 -10 x1 0
 i.e 2 10 4 x2 = 0
6 -30 -13 x3 0
 (4- λ)x1 – 20x2 – 10x3= 0
 -2x1 – (10- λ)x2 +4x3 = 0 -------- (2)
 6x1 – 30x2 –(13+ λ)x3 = 0
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

36.  From the above three equations, we can determine the
components x1, x2, x3 of the eigen vectors X for different values of
λ.
 Case-1:
 To find the eigen vector X1 = (x1, x2, x3 ) corresponding to λ = 0.
 When λ = 0 the above equations (2) becomes,
 4x1 – 20x2 – 10x3= 0
 -2x1 – 10x2 +4x3 = 0
 6x1 – 30x2 –13x3 = 0
 To find x1, x2, x3 , take any two of the above equations, say first
two equations. By cross rule method,
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

37.  x1/(-80+100) = x2/(20-16) = x3/(40-40)
 x1 /20 = x2 /4 = x3 /0 = k
 x1 =20k, x2 =4k, x3 =0
20k
 Hence the general eigen vector corresponding to λ = 0 is 4k
0
And by taking k = ¼. 5
We get the simplest eigen vector, X1 = 1
0
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

38.  Case-2:
 To find the eigen vector X2 = (x1, x2, x3 )
 When λ = -1 the above equations (2) becomes,
 5x1 – 20x2 – 10x3= 0
 -2x1 +11x2 +4x3 = 0
 6x1 – 30x2 –12x3 = 0
 To find x1, x2, x3 , take any two of the above equations, say first
two equations. By cross rule method,
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

39.  x1/(-80+110) = x2/(20-20) = x3/(55-40)
 x1 /30 = x2 /0 = x3 /15 = k
 x1 =30k, x2 =0k, x3 =15k
30k
 Hence the general eigen vector corresponding to λ = -1 is 0k
15k
And by taking k = 1/15 . 2
We get the simplest eigen vector, X2 = 0
1
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

40.  Case-3:
 To find the eigen vector X3 = (x1, x2, x3 )
 When λ = 2 the above equations (2) becomes,
 2x1 – 20x2 – 10x3= 0
 -2x1 +8x2 +4x3 = 0
 6x1 – 30x2 –15x3 = 0
 To find x1, x2, x3 , take any two of the above equations, say first
two equations. By cross rule method,
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

41.  x1/(-80+80) = x2/(20-8) = x3/(16-40)
 x1 /0 = x2 /12 = x3 /-24 = k
 x1 =0k, x2 =12k, x3 =-24k
0k
 Hence the general eigen vector corresponding to λ = 2 is 12k
-24k
And by taking k = 1/12 . 0
We get the simplest eigen vector, X3 = 1
-2
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

42.  P2: Determine the eigen values and natural frequencies of a
system whose stiffness and mass matrices are given below.
 [k] = 2AE/L 3 -1 , [m] = ρAL/12 6 1
-1 1 1 2
 Solution:
 General F.E equation for longitudinal free vibration of rod is
given by,
 {[k] –ω2
[m] } {u} = 0 ----------- (1)
 Substituting [k] and [m] matrix in equation (1),
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

43.  2AE/L 3 -1 -ρAL ω2
/12 6 1 u1 = 0
-1 1 1 2 u2
 3 -1 u1 = ρL ω2
/24E 6 1 u1
-1 1 u2 1 2 u2
Let λ = ρL2
ω2
/24E
Characteristic equation is given by |[k] – λ[m]| = 0
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

44.  3 -1 - λ 6 1 = 0
-1 1 1 2
(3-6 λ) (-1- λ)
(-1- λ) (1- 2λ) = 0
11 λ2
- 14 λ+2 = 0 ------- (2)
By solving quadratic equation (2)
λ1 = 1.1, λ2 = 0.168
The eigen values are λ1 = 1.1, λ2 = 0.168
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

45.  To find the natural frequencies,
 λ1 = ρL2
ω1
2
/24E
 ω1
2
= 1.1 24E/ (ρL2
)
 ω1 = (5.13/L) √E/ρ rad/sec
 λ2 = ρL2
ω2
2
/24E
 ω2
2
= 0.168 24E/ (ρL2
)
 ω2 = (2/L) √E/ρ rad/sec
 Natural frequencies are ω1 = (5.13/L) √E/ρ rad/sec,
ω2 = (2/L) √E/ρ rad/secDr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

46. SOLUTION OF EIGENVALUE
PROBLEMS
 The general form of G.E for un-damped free vibration of the structure is
given by
 [k]nxn {ui}nx1 = ω2
i [m]nxn{ui}nx1, i= 1, 2, 3.., n ---(1)
 Where, [k] = stiffness matrices
[m] = mass matrices
ωi = Natural frequencies
{ui} = Mode shape
 Then equ (1) can be rewrite as
 [m]-1
[k] {ui} = ω2
i {ui}
 [A] {ui} = λi {ui}
 Where [A] = [m] λ-1
[k] and λi = ω2
i
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

47.  [A] {u} = λ {u} is known as Standard form of eigen value problem.
 [k] {u} = λ [m]{u} is known as non-standard form of eigen value
problem.
 The determinant | k – λm | = 0 is called as the characteristic equation.
 {k – λm } {u} = 0 is called Eigen vector.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

48.  P1: Find the natural frequencies of longitudinal vibration of the
unconstrained stepped bar shown in figure.
 Solution:
 The dynamic equation of motion
for the un-damped free vibration
of whose system is given by
 [ [K] – [M] ω2
] (u) = 0 ------(1)
Where
[K] = Global stiffness matrix
[M] = Global mass matrix
ω = Natural frequency
{u} = Displacement vector
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
u3
A(1) = 2A
A(1) = A
Element 1 Element 2
u1 u2
x
1
2
3
l(1) = L/2 l(2) = L/2
Stepped bar with axial degree of freedom

49.  For two element stepped bar, the equation of motion(1) can be derived
as follows:
 For element (1): (Between nodes 1&2)
1 -1
 Element stiffness matrix [k1] = (A1E1)/l1
-1 1
1 -1 1 -1
= (2A E)/(L/2) = (4A E)/(L) ------(2)
-1 1 -1 1
2 1
Element stiffness matrix [m1] = (ρ1 A1 l1)/6
1 2
(Assumingconsistent mass matrix)Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

50. 2 1 2 1
= (ρ.2A. L/2)/6 = (ρAL)/6 ---(3)
1 2 1 2
For element (2): (Between nodes 2&3)
1 -1
 Element stiffness matrix [k2] = (A2E2)/l2
-1 1
1 -1 1 -1
= (A E)/(L/2) = (2A E)/(L) ------(4)
-1 1 -1 1
2 1
Element stiffness matrix [m2] = (ρ2A2 l2)/6
1 2
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

51. 2 1 2 1
= (ρ.A. L/2)/6 =(ρAL)/12 ------(5)
1 2 1 2
By combining the equas.(2) & (4), we get global stiffness matrix as
2 -2 0
[K] = (2AE)/L -2 3 -1
0 -1 1
Similarly by combining the equas.(3) & (5), we get global mass matrix as
4 2 0 u1
[M]= (ρAL)/12 2 6 1 and {u} = u2
0 1 2 u3
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

52.  Now , by substituting the values of above matrices in equation (1) we get
2 -2 0 4 2 0 u1
(2AE)/L -2 3 -1 - (ρAL)/12 2 6 1 ω2
u2 = 0 --(6)
0 -1 1 0 1 2 u3
 We know that, for getting the non-zero solution of circular frequency ω
for {u}, the determinant of the coefficient matrix [ [K] – [M] ω2
] should be
zero.
 i.e., | [K] – [M] ω2
| = 0
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

53. 2 -2 0 4 2 0
i.e., (2AE)/L -2 3 -1 - (ρAL)/12 2 6 1 ω2
= 0
0 -1 1 0 1 2
(or) 2 -2 0 4 2 0
-2 3 -1 -(ρL2
ω2
)/24E 2 6 1 = 0 ------(7)
0 -1 1 0 1 2
Let λ = (ρL2
ω2
)/24E
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

54.  Then, the equation (7) becomes,
(2-4λ) (-2-2λ) 0
(-2-2λ) (3 -6λ) (-1- λ ) = 0 ----(8)
0 (-1-λ) (1-2 λ)
i.e., (2-4λ) [(3 -6λ) (1-2 λ) - (-1-λ)(-1-λ)] - (-2-2λ) [(-2-2λ) (1-2 λ)-0] +0 = 0
i.e., -36 λ3
+ 90λ2
+ -36 λ = 0
i.e., 18λ(1-2λ) (λ-2) = 0
(or) λ(1-2λ) (λ-2) = 0
In the above equation either λ=0 or (1-2λ)=0 or (λ-2) = 0
When λ = 0 we get (ρL2
ω2
)/24E= 0 => ω2
= 0 (or) ω=0
i.e., the first natural frequency ω1 = 0
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

55.  When (1-2λ)=0, we get 1- ( 2 (ρL2
ω2
)/24E)= 0
 i.e., ω2
=12E/ ρL2
=> ω = 3.46( E/(ρL2
))1/2
 The second natural frequency ω2 = 3.46( E/(ρL2
))1/2
 When (λ-2)=0, we get (ρL2
ω2
)/24E) - 2= 0
 i.e., ω2
=48E/ ρL2
=> ω = 6.93( E/(ρL2
))1/2
 The third natural frequency ω3 = 6.93( E/(ρL2
))1/2
 That is the natural frequencies of unconstrained stepped bar are,
 ω1 = 0
 ω2 = 3.46( E/(ρL2
))1/2
rad/sec
 ω3= 6.93( E/(ρL2
))1/2
rad/sec
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

56.  P2: Find the mode shapes (i.e., eigenvectors) for the natural
frequencies of longitudinal vibration of the unconstrained stepped
bar shown in figure.
 Solution:
 The natural frequencies of the
unconstrained stepped bar as
described in the above problem are :
 ω1 = 0
 ω2 = 3.46( E/(ρL2
))1/2
rad/sec
 ω3= 6.93( E/(ρL2
))1/2
rad/sec
 The natural frequencies follow some well defined deformation patterns
called mode shapes. It is observed that the first frequency ω1 = 0
corresponds to rigid body mode, whereas the second and third
frequencies (ω2 , ω3) correspond to elastic deformation modes.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
u3
A(1) = 2A
A(1) = A
Element 1 Element 2
u1 u2
x
1
2
3
l(1) = L/2 l(2) = L/2
Stepped bar with axial degree of freedom

57.  To find the mode shape corresponding to natural frequencies ωi, we
must solve the equation[ [K] – [M] ω2
] (u) = 0.
 The equation (6) of the above problem is given by
2 -2 0 4 2 0 u1
(2AE)/L -2 3 -1 -(ρAL)/12 2 6 1 ω2
u2 = 0 ------ I
0 -1 1 0 1 2 u3
 By selecting λ = (ρL2
ω2
)/24E the above equation can be simplified as
(2-4λ) (-2-2λ) 0 u1
(-2-2λ) (3 -6λ) (-1- λ ) u2 = 0 ---- II
0 (-1-λ) (1-2 λ) u3
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

58.  By equating the determinant of coefficient matrix for {u} to zero, we get
 λ(1-2λ) (λ-2) = 0 which implies λ = 0 (or) λ = 0.5 (or) λ = 2.
 i.e., the eigen values are
 λ1 0
λ2 = 0.5 writing in the increasing order.
λ3 2
 To find 1st
mode shape:
 Let λ = 0
 Now the equation II implies
2u1 – 2u2 = 0 ---(i)
-2u1 + 3u2 – u3 = 0 ---(ii)
-u2 + u3 = 0 ---(iii)
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

59.  Equa. (i) => u1 = u2 and Equa (iii) u2 = u3
 Hence u1 = u2 = u3
 For ω1 = 0 (or) λ = 0, corresponding to rigid body
 u1 = u2 = u3 = 1
 That is , 1st
mode shape is
u1 1
Um1 = u2 = 1 u1
u3 1
 To find 2nd
mode shape:
 Let λ = 0.5
 Now the equation II implies
– 3u2 = 0 ---(iv)
-3u1–1.3u3 = 0 ---(v)
-1.5u2 = 0 ---(vi)Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

60.  Equa(iv) => u2 = 0 and Equa(v) => 3u1 = -1.5u3 (or) 2u1 = -u3
 That is , 2nd
mode shape is
u1 1 1
Um2 = u2 = 0 = 0 u1
u3 -2u1 -2
 To find 3rd
mode shape:
 Let λ = 2
 The equation II implies
-6u1– 6u2 = 0 ---(vii)
-6u1-9u2–3u3 = 0 ---(viii)
- 3u2 -3u3= 0 ---(ix)
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

61.  Equa(vii) => u1= -u2 and Equa(ix) => u2 = -u3
 That is , 3rd
mode shape is
u1 u1 1
Um3 = u2 = -u1 = -1 u1
u3 u1 1
The three mode shapes corresponding to λ = 0, λ = 0.5 and λ = 2
(i.e.,corresponding to ω1 , ω2 and ω3) are shown in figure.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

62. Types of mode-shapes
(Eigenvectors)
1st
mode shape
2nd
mode shape
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.
u3u1
u21
2
3
L/2 L/2
x
1 1 1
u1 u2 u3
0 L/2 L
x
u1
u2
u3
0
L0
1
-2
x
u1
u2
u3
0 0
1 1
-1
3rd
mode shape

63. TRANSIENT VIBRATION
ANALYSIS
Modelling of Damping:
 The diminishing of vibration is called as damping. The damping
had a little effect in most cases, on the natural frequencies.
 In most practical situations, time dependent excitation forces are
always present and it is of interest so find the actual dynamic
response of the structure to such an excitation.
 An equivalent viscous damping model is commonly used –
viscous friction force, being linearly proportional to relative
velocity, offers significant modelling advantage.
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.

64.  Generalized governing equation for damping is given by,
 [M] { x } + [C]{x} + [K]{x} = F{t} Where C = Damping matrix
 There are two categories of methods of solving the above G.E.
They are:
 1) Mode superposition methods
 2) Direct Integration methods
Dr.G.PAULRAJ,
Professor&Head(Mech.), VTRS,
AVADI, CHENNAI.