ppt for equation of motion

1. Multi Degree of Freedom System
In most physical system, the motion of the significant masses cannot be described by a
single variable, such system must be treated as multiple-degree-of-freedom system
(MDOF).
Idealised Two Storey Shear Frame

2. Assumptions are –
1. The total mass of the structure is concentrated at the levels of the floors.
2. The girders on the floors are infinitely rigid as compared to the columns.
3. The deformation of the structure is independent of the axial forces present
in the columns.
Multi Degree of Freedom System

3. Multi Degree of Freedom System
A two storey frame subjected to external forces 𝑝1 𝑡 and 𝑝2 𝑡
• The structure has been idealized as shear
building
• The structure has two degrees of freedom

4. Multi Degree of Freedom System
A two storey frame subjected to external forces 𝑝1 𝑡 and 𝑝2 𝑡
• The structure has been idealized as shear
building
• The structure has two degrees of freedom

5. Multi Degree of Freedom System
Equation of motion
• The external force pj(t)
• The elastic or inelastic forces fsj
• The damping force fdj
• The inertial force fij

6. Multi Degree of Freedom System
Forces acting on mass 1
Equation of motion

7. Multi Degree of Freedom System
Forces acting on mass 2
Equation of motion

8. Multi Degree of Freedom System
Then for each mass
𝑝𝑗 − 𝑓𝑠𝑗 − 𝑓𝑑𝑗 = 𝑚𝑗𝑢𝑗
𝑚𝑗𝑢𝑗 + 𝑓𝑠𝑗 + 𝑓𝑑𝑗 = 𝑝𝑗 𝑡 … … … . . (1)
for j=1 and 2
𝑚1𝑢1 + 𝑓𝑠1 + 𝑓𝑑1 = 𝑝1(𝑡)
𝑚2𝑢2 + 𝑓𝑠2 + 𝑓𝑑2 = 𝑝2(𝑡)

9. Multi Degree of Freedom System
Equation contains two equation for j= 1 and 2 and can be
written in matrix form
𝑚1 0
0 𝑚2
𝑢1
𝑢2
+
𝑓𝑠1
𝑓𝑠2
+
𝑓𝑑1
𝑓𝑑2
=
𝑝1(𝑡)
𝑝2(𝑡)
……………(2)
The equation 2 can be compactly written as
𝑚 𝑢 + 𝑓𝑠 + 𝑓𝑑 = 𝑝1(𝑡) …………… (3)
𝑣𝑗=𝑘𝑗∆𝑗………………………(4)

10. Multi Degree of Freedom System
fs1=fas1+fbs1
∆1= 𝑢1
∆2= 𝑢2 − 𝑢1
And the forces fs2 at the 2nd floor
fs2=k2(u2-u1)
Equation of motion

11. Multi Degree of Freedom System
It can be seen that fas1 and fs2 are equal in
magnitude but opposite in direction.
fas1-fs2= -k2(u1-u2)
fbs1=k1u1
fs2=k2(u2-u1)
𝑓𝑠1
𝑓𝑠2
=
𝑘1 + 𝑘2 −𝑘2
−𝑘2 𝑘2
𝑢1
𝑢2
Equation of motion

12. Multi Degree of Freedom System
similarly
𝑓𝑑1
𝑓𝑑2
=
𝑐1 + 𝑐2 −𝑐2
−𝑐2 𝑐2
𝑢1
𝑢2
The equation of motion is
𝑚 𝑢 + 𝑐 𝑢 + 𝑘 𝑢 = 𝑝(𝑡)
Equation of motion

13. Natural Frequency and Mode
A mode shape is the deformation that the component would show when
vibrating at the natural frequency.

14. Natural Frequency and Mode
The free vibration of an undamped system can be described mathematically
by,
𝑢(𝑡) = 𝑞𝑛 𝜑𝑛
Deflected shape 𝜑𝑛 does not vary with time. The time variation of
displacement is described by the harmonic motion.
𝑞𝑛 𝑡 = 𝐴𝑛 cos 𝜔𝑛𝑡 + 𝐵𝑛 sin 𝜔𝑛 𝑡
Where, 𝐴𝑛 𝑎𝑛𝑑 𝐵𝑛 are constants of integration that can be determined from
the initial conditions that initiate the motion. Combining equation is
𝑢 𝑡 = 𝜑𝑛 𝐴𝑛 cos 𝜔𝑛𝑡 + 𝐵𝑛 sin 𝜔𝑛 𝑡
Where, 𝜔𝑛 𝑎𝑛𝑑 𝜑𝑛are unknowns.

15. Substituting this form of 𝑢 𝑡 in equation of motion of undamped vibration
gives,
𝑚 𝑢 + 𝑘 𝑢 = 0
This equation can be satisfied in one of two ways either 𝑞𝑛 𝑡 = 0. Which
implies that 𝑢 𝑡 = 0 and there is no motion of the system (Trivial solution), or
the natural frequencies 𝜔𝑛 and modes 𝜑𝑛 must satisfy the following algebraic
equation.
𝑘 𝜑𝑛 = 𝜔𝑛
2
𝑚 𝜑𝑛
The mass and stiffness matrices 𝑘 𝑎𝑛𝑑 𝑚 are known the problem is to
determine the scalar 𝜔𝑛
2
and vector 𝜑𝑛 .
Natural Frequency and Mode

16. To indicates the formal solution to equation, it is rewritten as -
−𝜔𝑛
2 𝑚 + 𝑘 𝑞𝑛 𝑡 = 0
This set always has the trivial solution 𝜑𝑛 = 0, which is not useful
because it implies no motion. It has nontrivial solution if,
−𝜔𝑛
2 𝑚 + 𝑘 = 0
𝑜𝑟, 𝑚 −1
𝑘 − 𝜔𝑛
2
𝑚 𝑚 −1
= 0 × 𝑚 −1
𝑜𝑟, 𝑘 𝑚 −1
− 𝜔𝑛
2
𝐼 = 0
𝐴 = 𝑘 𝑚 −1, 𝜆 = 𝜔𝑛
2, 𝐼 = 𝐼
𝐴 − 𝜆𝐼 𝑋 = 0
Natural Frequency and Mode

17. These represents the Eigen value problem, then-
𝑑𝑒𝑡 −𝜔𝑛
2 𝑚 + 𝑘 = 0
This characteristics equation has N real and positive roots for 𝜔𝑛
2
because 𝑚 and 𝑘 , the structural mass and stiffness matrices, and
symmetric and positive definite.
The N roots of the frequency equations 𝜔1
2, 𝜔2
2, 𝜔3
2 … … … . . 𝜔𝑁
2
represent the frequencies at which the undamped system can oscillate in
the absence of external forces.
Natural Frequency and Mode

18. Mode Shape
When a natural frequency 𝜔𝑛 is known can be solved for the
corresponding vector 𝜑𝑛 to within a multiplicative constant.
The Eigen value problem does not fix the absolute amplitude of the
vector 𝜑𝑛 . Only the shape of the vector given by the relative values of
the n displacement 𝜑𝑗𝑛 (j=1,2………N). Corresponding to the N
natural vibration frequencies 𝜔𝑛 of an N-DOF system, there are N
independent vectors 𝜑𝑛 which are known as natural modes of vibration
or natural shapes of vibration. These vectors are also known as Eigen
vector, characteristic vectors or normal modes.

19. Mode Shape
A vibrating system will N-DOFs has N natural vibration frequencies
𝜔𝑛(1,2,3…………..N)
Which are arranged in sequence from smallest to largest ( 𝜔1 <
𝜔2 … … … … … . < 𝜔𝑛) corresponding to N modes of vibration occur at a
particular natural frequency and cause the structure to deform with a
particular natural mode shape 𝜑𝑛 .
The mode corresponding to lowest natural frequency is called first mode
(n=1) or the fundamental mode.

20. Mode Shape
Compute the natural frequencies for all the storey and then compute the
mode shape for the structure
2000kN/m
2500kN/m
3000kN/m
10 kN-𝑠𝑒𝑐2/𝑚
12 kN-𝑠𝑒𝑐2
/𝑚
15 kN-𝑠𝑒𝑐2
/𝑚

21. Mode Shape