Finite Element Analysis

Unit-4
Dynamic Analysis

Dr. M Krishna
Professor & Dean
CESD, RVCE
 Content

✓ Equations of motion
✓ Mass and stiffness matrices
✓ Distributed and Consistent mass matrices
✓ Eigen values and Eigen vectors
✓ Numerical
 Introduction

✓ Dynamic analysis using the FEM is essential for studying how structures and mechanical
components respond to time-dependent forces, vibrations, and dynamic loads.
Governing Equations of Motion in FEM
The general form of the equation of motion for a system with n DOF is given by:
where:
[M] = Mass matrix
[C]= Damping matrix
[K][= Stiffness matrix
{u} = Displacement vector
ሶ Velocity vector
{𝑈}=
{𝑈}ሷ = Acceleration vector
{F(t)} = Time-dependent external force vector
 Difference between Static and Dynamic Analysis
Aspect Static Analysis Dynamic Analysis
Determines the response of a structure Analyzes the response of a structure under
Definition under steady loads without considering ent or varying loads, considering inertia and
inertia and damping effects. damping.
Governing Equation [K]{u}={F} ሷ
[M]{𝑈}+[C]{ ሶ
𝑈}+[K]{u}={F(t)}
Time-independent (loads are applied Time-dependent (loads vary with time and
Consideration of Time
gradually and remain constant). cause motion).
Inertia Effects Neglected Considered
Damping Effects Not included Included (in most cases)
Constant forces, pressures, thermal Time-varying loads, impact loads, harmonic
Load Types
loads, etc. excitations, seismic forces, etc.
Displacement, stress, and strain are Displacement, velocity, acceleration, and
Response Type
determined at equilibrium. vibration characteristics are analyzed
 Difference Between Static and Dynamic Analysis
Aspect Static Analysis Dynamic Analysis
Modal analysis, time integration methods
Direct stiffness method, iterative
Solution Techniques (Newmark-β, Wilson-θ), mode
solvers.
superposition.
Vibrations of an aircraft wing, seismic
Load-bearing capacity of a beam, stress
Examples analysis of a building, vehicle crash
analysis of a bracket.
simulation.
Lower (solves a single set of Higher (solves time-dependent equations,
Computational Complexity
equilibrium equations). often requiring numerical integration).
 FE Procedure for Static and Dynamic Analysis
Step
FEM Steps Static Analysis Dynamic Analysis
No.
Discretization of the Domain Divide the structure into finite Same as static analysis (generate
1
(Mesh Generation) elements (1D, 2D, or 3D elements). finite element mesh).
Selection of Displacement Choose elements (beam, shell, solid) Same as static analysis but shape
2 model / element / shape and define shape functions for functions may also approximate
function displacement approximation. velocity and acceleration.
Compute mass matrix [M],
Formulation of Element Compute stiffness matrix [Ke] for each
3 damping matrix [C] and stiffness
Matrices element.
matrix [K].
Assemble global mass [M],
Assemble the global stiffness matrix
4 Assembly of Global Matrices damping [C] and stiffness [K]
[K].
matrices.
 FE Procedure for Static and Dynamic Analysis
Step
FEM Steps Static Analysis Dynamic Analysis
No.
Apply initial conditions 𝑢ሶ 0 and
Application of Boundary Apply displacement constraints and
5 and time-dependent boundary
Conditions and Loads external forces.
conditions.
Solve
Solution of Equilibrium ሷ
[M]{𝑈}+[C]{ ሶ
𝑈}+[K]{u}={F(t)}
6 Solve: [K]{u}={F}using direct solvers
Equations using modal analysis, time
integration methods

Evaluate displacements,
velocities, accelerations, stresses
Compute displacements, stresses, and
7 Post-Processing of Results over time. Generate time-history
strains. Generate deformation plots.
plots, frequency response, and
mode shapes.
 Basics of Dynamic Analysis
A dynamic system is characterized by three independent properties:
✓ Mass (m) – Represents the inertia or resistance of a body to changes in motion.
✓ Stiffness (k) – Defines the ability of a system to resist deformation.
✓ Damping (c) – Represents the energy dissipation mechanism, which reduces
oscillations over time.

Disturbance and Response

✓ Disturbance: An external force f(t) applied to the system.
✓ Response: The system’s reaction, which includes:
❖ Displacement (u(t)) – Position of the system over time.
ሶ
❖ Velocity (𝑼(t)) – Rate of change of displacement.
ሷ
❖ Acceleration (𝑼(t)) – Rate of change of velocity.
 Basics of Dynamic Analysis…
Basic Definitions in Dynamic Analysis
✓ Time: A measure of the sequence of events occurring in a system.
✓ Mass: A quantitative measure of inertia, representing resistance to motion.
✓ Rigid Body: A body whose deformations are negligible compared to its overall
dimensions.
✓ Force: A vector quantity that describes the interaction between two bodies.
✓ Free Vibration: The periodic motion of a system due to restoring strain energy, in the
absence of external forces.
✓ Frequency: The number of oscillation cycles per unit time.
✓ Amplitude: The maximum displacement from the equilibrium position..
 Hamilton’s Principle
The motion of a system between times t1​ and t2​ is such that the integral:

𝑡2 𝑡2
I = න 𝐿𝑑𝑡 = න 𝑇− 𝑑𝑡
𝑡1 𝑡1
𝐿 =𝑇−𝜋

has a stationary value for the actual path of motion. Where

✓ T represents the kinetic energy of the system
✓ Π represents the potential energy of the system.
 Lagrange’s Equation of Motion
Lagrange's equation provides a general method for deriving the equations of motion for
dynamic systems:

where:
L=T−Π is the Lagrangian function (difference between kinetic and potential energy).
qi​ are the generalized coordinates representing the system's motion.
 Problem- 1
Derivation of the Equations of Motion for a Two-Mass, Two-Spring System
L = T -

T = 1Τ2 𝑚1 𝑥ሶ 12 + 1Τ2 𝑚2 𝑥ሶ 12
 = 1Τ2 𝑘1 𝑥12 + 1Τ2 𝑘2 𝑥2 − 𝑥1 2

L = 1Τ2 𝑚1 𝑥ሶ 12 + 1Τ2 𝑚2 𝑥ሶ 12 - 1Τ2 𝑘1 𝑥12 − 1Τ2 𝑘2 𝑥2 − 𝑥1 2

First Step
𝑑 𝐿 𝐿
- =0
𝑑𝑡 𝑥ሶ 1 𝑥1
𝐿 𝑑 𝐿
= 𝑚 𝑥ሶ then = 𝑚1 𝑥ሷ 1
𝑥ሶ 1 1 1 𝑑𝑡 𝑥ሶ 1
𝐿
= − 𝑘1 𝑥1 + 𝑘2 (𝑥2 − 𝑥1 ) = − 𝑘1 𝑥1 + 𝑘2 𝑥2 − 𝑘2 𝑥1
𝑥1
= - x1(k1+k2) + x2k2 c
 L = 1Τ2 𝑚1 𝑥ሶ 12 + 1Τ2 𝑚2 𝑥ሶ 12 - 1Τ2 𝑘1 𝑥12 − 1Τ2 𝑘2 𝑥2 − 𝑥1 2

Problem- 1
𝑑 𝐿 𝐿
- = 𝑚1 𝑥ሷ 1+ x1(k1+k2) - x2k2 = 0 (1)
𝑑𝑡 𝑥ሶ 1 𝑥1
Second Step
𝑑 𝐿 𝐿
- =0
𝑑𝑡 𝑥ሶ 2 𝑥2

𝐿 𝑑 𝐿
= 𝑚 𝑥ሶ 𝑡ℎ𝑒𝑛 = 𝑚2 𝑥ሷ 2
𝑥ሶ 2 2 2 𝑑𝑡 𝑥ሶ 2
𝐿
= −𝑘2 𝑥2 − 𝑥1 = −𝑘2 𝑥2 + 𝑘2 𝑥1
𝑥2
𝑑 𝐿 𝐿
- = 𝑚2 𝑥ሷ 2 - 𝑘2 𝑥1 + 𝑘2 𝑥2 = 0 (2)
𝑑𝑡 𝑥ሶ 2 𝑥2
From (1) and (2) we get
𝑚1 0 𝑥ሷ 1 𝑘 + 𝑘2 −𝑘2 𝑥1 0
= 1 𝑥2 =
0 𝑚2 𝑥ሷ 2 −𝑘2 𝑘2 0
 Problem- 2
L = T -

k3
T = 1Τ2 𝑚1 𝑥ሶ 12 + 1Τ2 𝑚2 𝑥ሶ 12
 = 1Τ2 𝑘1 𝑥12 + 1Τ2 𝑘2 𝑥2 − 𝑥1 2 + 1Τ2 𝑘3 𝑥22
L = 1Τ2 𝑚1 𝑥ሶ 12 + 1Τ2 𝑚2 𝑥ሶ 12 - 1Τ2 𝑘1 𝑥12 − 1Τ2 𝑘2 𝑥2 − 𝑥1 2 + 1Τ2 𝑘3 𝑥22

First Step
𝑑 𝐿 𝐿
- =0
𝑑𝑡 𝑥ሶ 1 𝑥1
𝐿 𝑑 𝐿
= 𝑚 𝑥ሶ then = 𝑚1 𝑥ሷ 1
𝑥ሶ 1 1 1 𝑑𝑡 𝑥ሶ 1
𝐿
= − 𝑘1 𝑥1 + 𝑘2 (𝑥2 − 𝑥1 ) = − 𝑘1 𝑥1 + 𝑘2 𝑥2 − 𝑘2 𝑥1
𝑥1
= - x1(k1+k2) + x2k2

𝑑 𝐿 𝐿
- = 𝑚1 𝑥ሷ 1 + x1(k1+k2) - x2k2 = 0 (1)
𝑑𝑡 𝑥ሶ 1 𝑥1
 Problem- 2

k3
Second Step
𝑑 𝐿 𝐿
- =0
𝑑𝑡 𝑥ሶ 2 𝑥2
𝐿 𝑑 𝐿
= 𝑚 𝑥ሶ 𝑡ℎ𝑒𝑛 = 𝑚2 𝑥ሷ 2
𝑥ሶ 2 2 2 𝑑𝑡 𝑥ሶ 2
𝐿
= −𝑘2 𝑥2 − 𝑥1 − 𝑘3 𝑥2 = −𝑘2 𝑥2 + 𝑘2 𝑥1 − 𝑘3 𝑥2
𝑥2
𝐿
= −𝑥2 (k2+k3) + 𝑘2 𝑥1
𝑥2
𝑑 𝐿 𝐿
- = 𝑚2 𝑥ሷ 2 - 𝑘2 𝑥1 + 𝑥2 (𝑘2 + 𝑘3 ) = 0 (2)
𝑑𝑡 𝑥ሶ 2 𝑥2
From (1) and (2) we get
𝑚1 0 𝑥ሷ 1 𝑘 + 𝑘2 −𝑘2 𝑥1 0
= 1 =
0 𝑚2 𝑥ሷ 2 −𝑘2 𝑘2 + 𝑘3 𝑥2 0
 Types of Mass Representation in FEA
✓ Consistent Mass Model: Distributes mass over the element volume, improving accuracy for
dynamic problems.

✓ Lumped Mass Model: Approximates mass at discrete points (nodes), simplifying

calculations but reducing accuracy.

✓ Reduced Mass Model: Uses simplifications to balance computational efficiency and

precision.

In FEA a solid body with distributed mass refers to a structure in which the mass is
continuously spread throughout its volume rather than being concentrated at discrete points.
 Difference between Consistence, lumped and Reducing mass
Aspect Consistent Mass Model Lumped Mass Model Reduced Mass Model
Uses simplifications to
Distributes mass over the Approximates mass by
reduce computation while
Definition element volume in a concentrating it at discrete
maintaining acceptable
physically accurate way. nodal points.
accuracy.
Reduced-order
Fully populated (non- Diagonal matrix (simpler
Mass Matrix Type approximation of consistent
diagonal) matrix. form).
mass matrix.
Higher due to full matrix Lower due to diagonal Moderate, as it balances
Computational Cost
inversion. nature of the matrix. accuracy and efficiency.
High accuracy, especially in Lower accuracy, may lead Moderate accuracy,
Accuracy dynamic and vibration to unrealistic dynamic optimized for practical
problems. behavior. applications.
 Difference between Consistence, lumped and Reducing mass
Aspect Consistent Mass Model Lumped Mass Model Reduced Mass Model
Used in modal analysis, Used in practical
Suitable for static problems
wave propagation, and engineering problems
Application Suitability and approximate dynamic
transient dynamics where requiring a balance between
analysis.
accuracy is crucial. speed and precision.
Provides reasonable
Provides more accurate Can cause frequency shifts
approximations with
Eigenvalue Computation natural frequencies and and inaccurate mode
reduced computational
mode shapes. shapes.
effort.
Applied in reduced-order
Recommended for detailed Used in simplified
Use in Structural models where
dynamic analysis (vibration, structural models for quick
Analysis computational resources are
impact, wave motion). simulations.
limited.
Provides an approximate
Properly accounts for Over-simplifies kinetic
Energy Distribution energy distribution with
distributed kinetic energy. energy distribution.
acceptable error margins.
 Solid Body with Distributed Mass
Let  is density (mass/ unit area), u is displacement (m), 𝑢ሶ is velocity (m/s), 𝑢ሷ is acceleration
(m/s2) and f is body force. The kinetic energy (T) is given

𝐿 General from of Kinetic and potential energy

𝑇 = 1Τ2 ‫׬‬0 𝑈ሶ 𝑇 𝑈𝜌
ሶ 𝑑𝑣 where 𝑢ሶ = 𝑈ሶ 𝑉ሶ 𝑊ሶ 𝑇
T = 1Τ2 𝑄ሶ 𝑇 𝑀 𝑄ሶ
U=Nq  = 1Τ2 𝑄𝑇 𝐾𝑄 − 𝑄𝑇 𝐹
𝑈ሶ = 𝑁 𝑞ሶ
𝐿
𝑇 = 1Τ2 ‫׬‬0 𝑞ሶ 𝑇 𝑁 𝑇 𝑁 𝑞
ሶ 𝑑𝑣 L = T-  = 1Τ2 𝑄ሶ 𝑇 𝑀 𝑄ሶ -1Τ2 𝑄𝑇 𝐾𝑄 + 𝑄𝑇 𝐹
It is in the form of T = 1Τ2 𝑞ሶ 𝑇 𝑚𝑒 𝑞ሶ compare both
𝐿
me = ‫׬‬0 𝑁 𝑇 𝑁 𝑑𝑣
General form of Mass Matrix equation.
d L L
− =0 MQ + KQ = F = 0
dt Q Q
Solid Body with Distributed Mass
d L L
− =0 MQ + KQ = F = 0
dt Q Q

− M 2U + KU = 0

KU = −M 2U
KU = −MU
Where =  2
 Derivation of the Consistent Mass Matrix for a 1D Bar
A consistent mass matrix is derived by distributing the mass over the element using shape functions,
ensuring accurate representation of inertia effects in dynamic analysis. Consider a 1D bar element of
length L, with density (), Cross sectional area(A) and total mass m= AL.

𝐿 𝐴𝑙
𝑀𝑒 = ‫׬‬0 𝑁 𝑇 𝑁  𝑑𝑣 where N = [N1 N2] dv = A dx = 𝑒 d
2
1 𝑁 𝐴𝑙𝑒
Me = ‫׬‬−1 1 𝑁1 𝑁2 d 1 1 1−
2
1 1+2 −2 2 1
𝑁2 2 2
‫׬‬−1 𝑁1 d = ‫׬‬−1 2 d = ‫׬‬−1 𝑑 = = ‫׬‬−1 𝑁22 d
4 3

1 1 1− 1+ 1 1−2 1

𝐴𝑙𝑒 1 𝑁12 𝑁1 𝑁2 ‫׬‬−1 𝑁1 𝑁2 d = ‫׬‬−1 d = ‫׬‬−1 𝑑 =
Me = ‫׬‬−1 d 2 2 4 3
2 𝑁1 𝑁2 𝑁22

AE  1 − 1
2 1
𝐴𝑙𝑒 𝐴𝑙𝑒 2
Stiffness matrix K e =
3 3 1
le − 1 1 
Me = 1 2 =
2 6 1 2
3 3
 Derivation of the Lumped Mass Matrix for a 1D Bar
The lumped mass approximation places half the total mass at each node.
𝑚 𝐴𝐿
M1 = M2 = =
2 2

Since mass is lumped at the nodes, the mass matrix is diagonal:

𝐴𝐿
𝑀1 0 0 𝐴𝐿 1 0
2
M= = 𝐴𝐿 =
0 𝑀2 0 2 0 1
2

𝐴𝐿 1 0
M =
2 0 1
AE  1 − 1
Stiffness matrix : K e =
le − 1 1 
 Derivation of the consistence Mass Matrix for a 1D truss element
We know that k=LTk′Lwhere k represents the global stiffness matrix, k′ denotes the local
stiffness matrix, and Lis the direction cosine matrix of the truss element. Similarly,
m=LTm′L where m is the global mass matrix, m′ is the local mass matrix, and L is the
direction cosine matrix of the truss element.

𝑙 0
𝐴𝑙𝑒 2 1 𝑙 𝑚 0 0
M = Ltm’L = 𝑚 0
0 𝑙 6 1 2 0 0 𝑙 𝑚,
0 𝑚

2𝑙2 2𝑙𝑚 𝑙2 𝑙𝑚
𝐴𝑙𝑒 2𝑙𝑚 2𝑚2 𝑚𝑙 𝑚2
M=
6 𝑙2 𝑙𝑚 2𝑙2 2𝑙𝑚
𝑙𝑚 𝑚2 2𝑙𝑚 2𝑚2
Derive Lumped Mass Matrix for a 1D truss element

1 0 0 0
𝐴𝐿 0 1 0 0
M=
2 0 0 1 0
0 0 0 1
 Derive Consistence Mass Matrix for a 1D Beam element
𝐿 𝑙𝑒 𝑙𝑒 𝐴𝑙𝑒
𝑀𝑒 = ‫׬‬0 𝐻 𝑇 𝐻  𝑑𝑣 where H = 𝐻1 𝐻2 𝐻3 𝐻4 and dv = A dx = d
2 2 2

1 1 1
Where H1 = 2 − 3 + 3 𝐻2 = 1 −  − 2 + 3 𝐻3 = 2 + 3 − 3 𝐻4 =
4 4 4
1
−1 −  +  + 3 2
4

𝐻1
𝑙
1 𝐻2 𝑒 𝑙𝑒 𝑙𝑒 𝐴𝑙𝑒
2
Me = ‫׬‬−1 𝐻1 𝐻2 𝐻3 𝐻4 𝑑 =
𝐻3 2 2 2
𝑙𝑒
𝐻4
2
Derive Lumped Mass Matrix for a 1D Beam element
 Difference between Consistent Mass & Lumped Mass Matrix
Consistent Mass Matrix Lumped Mass Matrix
✓ Obtained using the same shape functions used ✓ Usually obtained by inspection, without using
for deriving the stiffness matrix. shape functions.
✓ Total element mass is distributed equally to
✓ Total element mass is distributed unequally to
the nodes and is associated only with
the nodes based on shape functions.
translational degrees of freedom.
✓ Provides more accurate results, especially for ✓ Provides less accurate results compared to the
flexural elements (e.g., beams). consistent mass matrix.
✓ More complex and difficult to handle due to ✓ Easier to handle as it results in a diagonal
coupling terms. matrix.
✓ Natural frequencies obtained are closer to ✓ Natural frequencies obtained are lower than
exact values. exact values.
 Eigen Values and Eigen Vectors

✓ Eigenvalues (λ): These represent the natural frequencies squared ( = 2) of the system.
Each eigenvalue corresponds to a natural mode of vibration.

✓ Eigenvectors (U): These represent the mode shapes, i.e., the relative displacements of the
system when vibrating at a particular natural frequency.

✓ A natural mode of vibration (or simply, mode shape) refers to a specific pattern in which a
structure or mechanical system tends to vibrate naturally when disturbed and allowed to
vibrate freely (number of natural modes = No. of DoF)
 Significance of Natural Modes / eigenvalues

Aspect Significance
Resonance Avoidance Prevents failure due to excessive vibrations
Design Optimization Improves stiffness, safety, and performance
Simplifies transient and forced vibration
Dynamic Response Analysis
problems
Buckling Analysis Determines critical load and buckling shape
Enhances acoustic comfort and reduces vibration
Noise Control
noise
Experimental Validation Matches simulated mode shapes with test data
 Different Natural Modes of Vibration
No. Mode Type Deformation Type Common Example

1 Rigid Body Mode No deformation Free-floating object, aircraft in space

Mass-spring system, lateral vibration of

2 Translational Mode Linear displacement
slabs

3 Rotational Mode Angular displacement Free-spinning disc, rotor motion

Cantilever beam deflection, bridge

4 Bending (Flexural) Mode Bending deformation
vibration

5 Torsional Mode Twisting about axis Shaft twisting, propeller blade twist

6 Local Mode Localized structural motion Vibrating panel corner, bracket vibration

7 Acoustic Mode Pressure wave in fluid Air in pipe, speaker cavity resonance
 Properties of Eigen Values and Eigen Vectors
1. Real and Positive Eigenvalues: In undamped and conservative systems with symmetric
stiffness [K] and mass [M] matrices, all eigenvalues (λ=ω2) are real and non-negative.
Zero eigenvalues correspond to rigid body modes, while positive eigenvalues represent
the natural frequencies of vibration.
2. Ordered Set: The eigenvalues can be arranged in ascending order as λ1< λ2 <λ3 < ⋯ < λn.
Each eigenvalue corresponds to an increasing mode of natural frequency, and the
associated eigenvectors are denoted as U1,U2,U3,…,Un representing the respective mode
shapes.
3. Orthogonality of Eigenvectors: Eigenvectors (mode shapes) are orthogonal with
respect to both:
the mass matrix 𝑈𝑖 𝑇 𝑀 𝑈𝑗 = 0 𝑖 𝑗
the stiffness matrix 𝑈𝑖 𝑇 𝐾 𝑈𝑗 = 0 𝑖𝑗
 Properties of Eigen Values and Eigen Vectors
4. Normalization of Eigenvectors
the mass matrix 𝑈𝑖 𝑇 𝑀 𝑈𝑖 = 1
the stiffness matrix 𝑈𝑖 𝑇 𝐾 𝑈𝑗 =
 Problem-1
1. Consider the axial vibration of the steel bar shown in the
Figure. Develop the global stiffness matrix, mass matrix
and eigen values. E = 2 x 105 MPa and density = 7840
kg/m3
Stiffness Matrices
A1 E1  1 − 1 5 8 − 8
k1 = =
l1 − 1 1  − 8 8 
10
 
 8 −8 0 
A2 E2  1 − 1 5 6 − 6 k = k 1 + k 2 = 10 5 − 8 14 − 6
k =
2
= 10 
l2 − 1 1   − 6 6 
  0 − 6 6 
 Problem-1
Mass Matrices Equilibrium equation (KQ=MQ)
 8 − 8 0   q1  0.9408 0.4704 0   q1 
A11l1 2 1 0.9408 0.4704 
m = =     
10 − 8 14 − 6 q2  =  0.4704 1.6464 0.3528  q2 
1
6 1 2 0.4704 0.9408 
5

 0 − 6 6   q3   0 0.3528 0.7056   q3 

A2 2l2 2 1 0.7056 0.3528  Elimination condition

m =
2
=
6 1 2 0.3528 0.7056  8 −8 0   q1  0.9408 0.4704 0   q1 
10 5 − 8 14 − 6 q2  =  0.4704 1.6464 0.3528  q2 
 0 −6 6   q3   0 0.3528 0.7056   q3 
0.9408 0.4704 0 
m = m1 + m 2 = 0.4704 1.6464 0.3528 
 0 0.3528 0.7056   14 − 6 q2  1.6464 0.3528  q2 
10 5    =  0.3528 0.7056  q 
− 6 6   q3    2 
 Problem-1
Required condition for evaluation of Eigen
values
 14 − 6 1.6464 0.3528 
10 5   −  0.3528 0.7056  = 0
 − 6 6   

 14e5 − 1.6464  − 6e5 − 0.3528  

− 6e5 − 0.3528  =0
 6e5 − 0.7056  

4.8e11-239940 +1.0372322 =0

1 =442486.1 2=4183364

n1 =665.1963 n2=2045.327

 Problem-2 A2=500 mm3
A1=1000 mm3 E1=0.7e5 Mpa
2. Consider the axial vibration of the stepped bar shown in the E1=2e5 Mpa 1 = 2100 kg/m3
Figure. Develop the global stiffness matrix, mass matrix and 1 = 7840 kg/m3
eigen values
1000 mm 1500 mm
Stiffness Matrices

A1 E1  1 − 1 5 2 − 2 Mass Matrices
k1 = =
l1 − 1 1  − 2 2 
10
  A11l1 2 1 2.61 1.31
m1 = =
6 1 2 1.31 2.61
A2 E2  1 − 1 5  0.23 − 0.23
k2 = = 2 1 0.53 0.26 
l2 − 1 1  − 0.23 0.23  A2 2l2
10
  m2 = =
6 1 2 0.26 0.53
2 −2 0 
2.61 1.31 0 
k = k 1 + k 2 = 10 5 − 2 2.23 − 0.23 m = m1 + m 2 = 1.31 3.14 0.26 
 0 − 0.23 0.23   0 0.26 0.53
 Problem-2
Equilibrium equation (KQ=MQ) Required condition for evaluation of Eigen
values
2 −2 0   q1  2.61 1.31 0   q1 
 2.23 − 0.23 3.14 0.26 
5    
10 − 2 2.23 − 0.23 q2  =  1.31 3.14 0.26  q2  10 5   −   =0
 0 − 0.23 0.23   q3   0 0.26 0.53  q3  − 0.23 0.23  0.26 0.53
 2.23e5 − 3.14 − 0.23e5 − 0.26 
Elimination condition
− 0.23e5 − 0.26 =0
 0.23e5 − 0.53 
 2 −2 0   q1  2.61 1.31 0   q1 
10 − 2
5
2.23 − 0.23 q2  =  1.31 3.14
   0.26  q2  4.67e9-202778 +1.5787192 =0
 0 − 0.23 0.23   q3   0 0.26 0.53  q3 

1 =60105.27 2=196720.4
 2.23 − 0.23 q2  3.14 0.26  q2 
10 5    =  0.26 0.53 q  n1 =245.16 n2=443.5
− 0.23 0.23   q3    2 
 Problem-3
A2=600 mm3
3. Determine the eigenvalues and eigenvector for the stepped E1=0.7e5 Mpa
A1=1200 mm3
bar as shown in figure. 1 = 2100 kg/m3
E1=1e5 Mpa
Stiffness Matrices 1 = 8400 kg/m3

A1 E1  1 − 1 5  0.8 − 0.8 1500 mm

k1 = = 2000 mm
l1 − 1 1  − 0.8 0.8 
10
  Mass Matrices
A2 E2  1 − 1 5  0.21 − 0.21 2 1 5.04
A11l1 2.52
k =
2
= 10  m1 = =
6 1 2 2.52
l2 − 1 1   5.04 
− 0.21 0.21 
 0.8 − 0.8 0  A  l 2 1 0.84 0.42 
m2 = 2 2 2  =
k = k 1 + k 2 = 10 5 − 0.8 1.01 − 0.21 6 1 2 0.42 0.84 
 0 − 0.21 0.21  5.04 2.52 0 
m = m1 + m 2 = 2.52 5.88 0.42 
 0 0.42 0.84 
 Problem-3
Equilibrium equation (KQ=MQ) Required condition for evaluation of Eigen
values
 0.8 − 0.8 0   q1  5.64 2.52 0   q1 
10 5 − 0.8 1.01 − 0.21 q2  =  2.52 5.88 0.42  q2  1.01e5- 5.88 =0
 0 − 0.21 0.21   q3   0 0.42 0.84   q3  =17176.87
Elimination condition
n =131
 0.8 − 0.8 0  q1  5.64 2.52 0   q1  As per III properties of eigenvector and eigenvalues
0 5 − 0.8   
1.01 − 0.21 q2  =  2.52 5.88 0.42  q2 
 0 − 0.21 0.21   q3   0 0.42 0.84   q3  U iT MUi = 1 Modal Shape
q2 5.88 q2 = 1
1.01e5q2 =5.88q2

q2 = 0.412393 mm
 Problem-3
Equilibrium equation (KQ=MQ) Required condition for evaluation of Eigen
values
 0.8 − 0.8 0   q1  5.64 2.52 0   q1 
10 5 − 0.8 1.01 − 0.21 q2  =  2.52 5.88 0.42  q2  1.01e5- 5.88 =0
 0 − 0.21 0.21   q3   0 0.42 0.84   q3  =17176.87
Elimination condition
n =131
 0.8 − 0.8 0  q1  5.64 2.52 0   q1  As per III properties of eigenvector and eigenvalues
0 5 − 0.8   
1.01 − 0.21 q2  =  2.52 5.88 0.42  q2 
 0 − 0.21 0.21   q3   0 0.42 0.84   q3  U iT MUi = 1 Modal Shape
q2 5.88 q2 = 1
1.01e5q2 =5.88q2

q2 = 0.412393 mm
Thank You