1. MULTI DEGREE OF FREEDOM (M-DOF)
ROSLI ASMAWI
Faculty of Mechanical & Manufacturing Engineering
Universiti Tun Hussein Onn Malaysia

2. Outline-Part 2
Using Newton’s Second Law to derive Equations of Motion
2 Influence Coefficients
Eigenvalue problem

3. Using Newton’s 2nd Law to derive Equations of Motion
Step 1: Set up suitable coordinates to describe positions of the various masses in the system.
Step 2: Measure displacements of the masses from their static equilibrium positions
Step 3: Draw free body diagram and indicate forces acting on each mass
Step 4: Apply Newton’s 2nd Law to each mass

4. Example 1
Derive the equations of motion of the spring-mass damper system shown below.

5. Solution Free-body diagram is as shown:
Applying Newton’s 2nd Law gives:
Set i=1 with x0=0 and i=n with xn+1=0:

6. Solution
Notes:
The equations of motion can be expressed in matrix form:
[m] is the mass matrix
[c] is the damping matrix
[k] is the stiffness matrix

7. Influence Coefficients
One set of influence coefficients is associated with each matrix involved in the equations of motion
Equation of Motion
Influence coefficient
Stiffness matrix
Stiffness influence coefficient
Mass matrix
Inertia influence coefficient
Inverse stiffness matrix
Flexibility influence coefficient
Inverse mass matrix
Inverse inertia coefficient

8. Stiffness Influence Coefficient, kij
kij is the force at point i due to unit displacement at point j and zero displacement at all other points.
Total force at point i,
Matrix form:

9. Stiffness Influence Coefficient, kij
Note:
kij = kji
kij for torsional systems is defined as the torque at point i due to unit angular displacement at point j and zero angular displacements at all other points.

10. Example 2
Find the stiffness influence coefficients of the system shown below.

11. Solution
Let x1, x2 and x3 be the displacements of m1, m2 and m3 respectively.
Set x1=1 and x2=x3=0.
Horizontal equilibrium of forces:
Mass m1: k1 = -k2 + k (E1)
Mass m2: k21 = -k (E2)
Mass m3: k31 = (E3)
Solving E1 to E3 gives
k11=k1+k2
k21 = -k2
k31 = 0

12. Solution Next set x2=1 and x1=x3=0. Horizontal equilibrium of forces:
Mass m1: k12 + k2 = 0 (E4)
Mass m2: k22 - k3 = k2 (E5)
Mass m3: k32 = - k (E6)
Solving E4 to E6 gives
k22 = k2+k3
k12 = -k2
k32 = -k3

13. Solution Finally set x3=1 and x1=x2=0.
Horizontal equilibrium of forces:
Mass m1: k13 = (E7)
Mass m2: k23 + k3 = 0 (E8)
Mass m3: k33 = k (E9)
Solving E7 to E9 gives
k33 = k3
k13 = 0
k23 = -k3
Thus the stiffness matrix is:

14. Flexibility Influence Coefficient, aij
Have to solve n sets of linear equations to obtain all the kij’s in an n DOF system
Generating aij’s is simpler.
aij is defined as the deflection at point i due to unit load at point j,
xij = aijFj, where xij is the displacement at point i due to external force Fj
Matrix form:

15. Flexibility Influence Coefficient, aij
Note:
Stiffness and flexibility matrices are the inverse of each other.
aij = aji
aij for torsional systems is defined as the angular deflection of point i due to unit torque at point j.

16. Example 3
Find the flexibility influence coefficients of the system shown below.

17. Solution
Let x1, x2 and x3 be the displacements of m1, m2 and m3 respectively.
Set F1=1 and F2=F3=0.
Horizontal equilibrium of forces:
Mass m1: k1a11 = k2(a21 – a11) + 1 (E1)
Mass m2: k2(a21 – a11) = k3(a31 – a21) (E2)
Mass m3: k3(a31 – a21) = 0 (E3)
Solving E1 to E3 gives a11 = a21 = a31 = 1/k1

18. Solution Next set F2=1 and F1=F3=0. Horizontal equilibrium of forces:
Mass m1: k1a12 = k2(a22 – a12) (E4)
Mass m2: k2(a22 – a12) = k3(a32 – a22) +1 (E5)
Mass m3: k3(a32 – a22) = 0 (E6)
Solving E4 to E6 gives
a12 = 1/k1
a22 = a32 = 1/k1 + 1/k2

19. Solution Next set F3=1 and F1=F2=0. Horizontal equilibrium of forces:
Mass m1: k1a13 = k2(a23 – a13) (E7)
Mass m2: k2(a23 – a13) = k3(a33 – a23) (E8)
Mass m3: k3(a33 – a23) = 1 (E9)
Solving E7 to E9 gives
a13 = 1/k1
a23 = 1/k1 + 1/k2
a33 = 1/k1 + 1/k2 + 1/k3

20. Eigenvalue Problem (Eigenvalue problem)
For a non-trivial solution, determinant Δ of the coefficient matrix must be zero.
i.e. Δ= |kij-ω2mij| = |[k]- ω2[m]| =0 (Characteristic equation)
ω2 is the eigenvalue

21. Solution of the Eigenvalue Problem
Multiplying by [k]-1:
[I] is identity matrix
[D]=[k]-1[m] is dynamical matrix.
For a non-trivial solution of
characteristic determinant Δ=|λ[I]-[D]|=0
Use numerical methods to solve if DOF of system is large

22. Example 4
Find the natural frequencies and mode shapes of the system shown below for k1=k2=k3=k and m1=m2=m3=m

23. Solution Dynamical matrix [D]=[k]-1[m] ≡[a][m] Flexibility matrix
Mass matrix
Thus

24. Solution
Frequency equation:
Δ=|λ[I]-[D]|=
Dividing throughout by λ,

25. Solution
Once the natural freq are known, the eigenvectors can be calculated using

26. Solution 1st mode: Substitute λ1=5.0489 into (E1):
3 unknowns X1(1),X2(1), X3(1) in 3 equations
Can express any 2 unknowns in terms of the remaining one.

27. Solution X2(1) + X3(1) = 4.0489 X1(1) 3.0489 X2(1) – 2X3(1) = X1(1)
Solving the above, we get
X2(1)=1.8019X1(1) and X3(1)=2.2470X1(1)
Thus first mode shape
where X1(1) can be chosen arbitrarily.

28. Solution 2nd mode: Substitute λ2=0.6430 into (E1):
3 unknowns X1(2),X2(2), X3(2) in 3 equations
Can express any 2 unknowns in terms of the remaining one.

29. Solution –X2(2) – X3(2) = 0.3570X1(2) -1.3570X2(2) – 2X3(2) = X1(2)
Solving the above, we get
X2(2)=0.4450X1(2) and X3(2)= X1(2)
Thus 2nd mode shape
where X1(2) can be chosen arbitrarily.

30. Solution 3rd mode: Substitute λ3=0.3078 into (E1):
3 unknowns X1(3),X2(3), X3(3) in 3 equations
Can express any 2 unknowns in terms of the remaining one.

31. Solution -X2(3) - X3(3) = 0.6922X1(3) -1.6922X2(3) – 2X3(3) = X1(3)
Solving the above, we get
X2(3)= X1(3) and X3(3)=0.5544X1(3)
Thus 3rd mode shape
where X1(3) can be chosen arbitrarily.

32. Solution
When X1(1) = X1(2) = X1(3) =1, the mode shapes are as follows: