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

Outline-Part 2 1 Using Newton’s Second Law to derive Equations of Motion 2 Influence Coefficients 3 Eigenvalue problem

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

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

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:

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

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

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:

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.

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

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 + k11 (E1) Mass m2: k21 = -k2 (E2) Mass m3: k31 = 0 (E3) Solving E1 to E3 gives k11=k1+k2 k21 = -k2 k31 = 0

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 = - k3 (E6) Solving E4 to E6 gives k22 = k2+k3 k12 = -k2 k32 = -k3

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

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:

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.

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

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

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

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

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

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

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

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

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

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

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.

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.

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.

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)=-0.8020X1(2) Thus 2nd mode shape where X1(2) can be chosen arbitrarily.

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.

Solution -X2(3) - X3(3) = 0.6922X1(3) -1.6922X2(3) – 2X3(3) = X1(3) Solving the above, we get X2(3)=-1.2468X1(3) and X3(3)=0.5544X1(3) Thus 3rd mode shape where X1(3) can be chosen arbitrarily.

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