Department of Architecture & Civil Engineering ( ) 2 2a. L = 65 2 ρπa4 L. + asinα = 3aθ 2. ( ) = a 1 cos( θ ρπa4 L.

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1 MODE ANSWER age: 1 QUESTION Mass of tube = ρπ 3a ( ) ( a) Moment of inertia of tube = ρπ 3 Mass of bar = ρπa Moment of inertia of bar = = 5ρπa ( 3a) 4 a ( ) 4 ρπ ( a )4 = 1 3 ρπa4 Horizontal displacement of centre of mass of tube = 3aθ = 65 ρπa4 Horizontal displacement of centre of mass of bar = 3aθ + asinα = 3aθ + asin( θ 1 θ ) Vertical displacement of centre of mass of bar = a 1 cosα Thus, kinetic energy, ( ) + 1 T = 1 ( 5ρπa ) 3aθ + 1 ρπa { } 65 ρπa4 θ { ( )( θ 1 θ )} ( ) 3aθ + acos θ 1 θ ( ) θ 1 θ + asin θ 1 θ θ θ ( ) ρπa4 θ 1 + ρπa4 9θ + 6 cos θ1 θ ( ) ( ) = a 1 cos( θ 1 θ ) ( ) θ 1θ θ The gravitational potential energy, G = ( ρπa g)a( 1 cos( θ 1 θ )) Thus the derivatives, d T θ 1 = d 3θ ρπa 4 ( ) θ 1+ 6cos( θ 1 θ )θ + θ 1 θ ( ) θ 1 4θ 6sin( θ 1 θ ) θ 1 θ + 6cos( θ 1 θ )θ ( ) + θ 1 θ

2 MODE ANSWER age: d T θ = d ρπa 4 4θ ( ) θ +6cos θ 1 θ T = ρπa4 6sin θ 1 θ 155θ + 18θ + 6 cos θ 1 θ ( ) ( 4 1 θ θ ) ( )( 1 4θ θ ) ( ) θ 1 4θ θ 6sin( θ 1 θ ) θ 1 θ ( 1 4θ ) ( ) θ 1θ θ T 1sin θ 1 θ θ ( ) θ 1θ θ G = ( ρπa 3 g)sin( θ 1 θ ) and G = ( ρπa 3 g)sin( θ 1 θ ) θ Substituting into d T θ 1 T + G and d T θ T + G gives agrange s θ θ equations of motion

3 MODE ANSWER age: 3 QUESTION 3 We will need the orthogonality conditions Δ s T MΔ r Δ s T KΔ r Δ r T MΔ r = m r Δ r T KΔ r = s r r s in which m r and s r are the mass and stiffness of the r th mode The orthogonality conditions are obtained from Δ s T Δ r T Mδ + Kδ ( K ω r M)Δ r ( K ω s M)Δ s N Substituting δ = Δ r f r ( t) into Mδ + Dδ N + Kδ = p gives ( M!! f r + D!f r + Kf r )Δ r = p and r N T hence Δ s ( M!! f + D!f + Kf )Δ r r r r = Δ T s p r Thus using the orthogonality conditions and if we ignore coupling of modes through damping, Δ s T MΔ s!! fs + Δ s T DΔ s! fs + Δ s T KΔ s f s = Δ s T p or m s!! fs + λ s! fs + s s f s = p s where p s is the load exciting the s th mode The damping λ s is estimated from the nondimensional damping coefficient, s λ s s m s hysically each mode lives a separate life as a single degree of freedom system with its own mass, damping, stiffness and load The single degree of freedom equations can either be solved by numerical integration (eg Verlet) or by performing the Fourier transform of p s r

4 MODE ANSWER age: 4 QUESTION 4 The tension in the spring is sy 1 and taking moments about the left hand end, y = sy 1 Thus M = d y dx = (y y) sy 1 ( x) = (y y) y when x ( x) M = d y dx = (y y) when x Using Macaulay brackets we can write both these as d y dx + y = y y x = x y y ( ) y Note the Macaulay bracket { } is ignored when the contents are negative The solution to the differential equation is y = Acos x + Bsin x + x y y + y sin The term y sin The boundary conditions are y when x, is included to give continuity of dy dx at x = y = y 1 = y s when x = y = y when x = Thus

5 MODE ANSWER age: 5 0 = A y s = Acos + Bsin + y y = Acos + Bsin + y y + y sin so that 0 = Bsin 0 = Bsin and = B cos y B = Finally, sin + y 1 s + y + y 1 s sin sin = cos tan = 1 s which has to be solved numerically THIS IS THE END OF THE ANSWER, REQUIRED TO GET 100% If s is infinite we have tan =

6 MODE ANSWER age: 6 π which is satisfied by = 1166 Thus = π 1166 effective length of 69 If s is not infinite, the effective length will be longer = π which corresponds to an ( 69)