MODE ANSWER age: 1 4. The students are given approximately 4 hours of lectures devoted to this topic. Thus the emphasis in the answer must be in demonstrating an understanding of the physical principals involved and how mathematics can be used to predict the behaviour of a real structure. 5 [ ] = 5ρπa. Mass of tube = ρπ ( 3a) ( a) Moment of inertia of tube = ρπ 3a 3 [ ( ) 4 ( a ) 4 ] = 65 ρπa4. Mass of bar = ρπa. Moment of inertia of bar = ρπ ( a )4 = 1 3 ρπa4. Horizontal displacement of centre of mass of tube = 3aθ. 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) = a 1 cos θ 1 θ Thus, inetic energy, ( ( )).
MODE ANSWER age: ( ) 3aθ T = 1 5ρπa + 1 ρπa + 1 65 ρπa4 θ ( ) 3aθ + acos θ 1 θ + asin θ 1 θ + 1 1 = ρπa4 + ρπa4 ( ) θ 1 θ ρπa4 θ 1 155θ + ρπa4 θ 1 ( ) θ 1 θ 9θ + 6θ cos θ 1 θ ( ) θ 1 θ + θ 1 θ = ρπa4 3 θ 1 1 cos( θ 1 θ )θ 1θ + { 178 1cos ( θ 1 θ )}θ The gravitational potential energy, G = ( ρπa g)a( 1 cos( θ 1 θ )). Thus the derivatives, d T dt θ = d 1 dt = ρπa4 ρπa 4 3θ 1 1 cos( θ 1 θ )θ 3θ 1 1cos( θ 1 θ )θ +1sin θ 1 θ ( ) θ 1 θ θ
MODE ANSWER age: 3 d T dt θ = d dt = ρπa4 ρπa 4 +sin θ 1 θ T θ 1 = 1 cos( θ 1 θ )θ 1 + { 178 4 cos( θ 1 θ )}θ 1 cos( θ 1 θ )θ 1 + { 178 4 cos( θ 1 θ )}θ ( ) θ 1 θ ρπa4 T = ρπa4 θ 1θ 1 4θ θ 1θ +θ 1sin θ 1 θ ( ) θ 1θ + θ 1sin ( θ 1 θ ) G = ρπa 3 g θ 1 ( )sin( θ 1 θ ) and G = ρπa 3 g θ ( )sin( θ 1 θ ). Substituting into d T dt θ T + G = 0 and d T 1 θ 1 θ 1 dt θ T + G = 0 gives agrange s θ θ equations of motion. 6 Stiffness matrix = EI 3 6 6 3 6 6 3 3 3 (note the factor of ).
MODE ANSWER age: 4 m 156 54 13 54 156 1 0 0 +1m 0 0 0 40 13 4 0 0 0 Mass matrix = (again note the factor of ). 156 + 5040 54 13 = m 54 156 40 13 4 The natural frequencies are obtained by setting the determinant 6 6 3 5196 54 13 3EI 3 6 6 3 ω m 54 156 = 0 Writing λ = ω m 4, 40 EI 3 3 13 4 adding the second row to the first and dividing by the common factor of, 0 0 0 550 10 35 λ 6 6 3 54 156 = 0. 3 3 13440 13 4 Dividing first row by 35, 0 0 0 150 6 1 λ 6 6 3 54 156 = 0. 3 3 13440 13 4 Now add 6 times the final column to the middle column and 150 times the final column to the first column. This will give a cubic in λ which has one root equal to zero leaving a quadratic. 7.
MODE ANSWER age: 5 B C θ s A D The strain energy in the springs is U = 4 1 β β 0 ( ) + 1 s cos π 4 β = 4 1 β β 0 cos π 4 β 0 ( ) ( ) + s cosβ + sinβ ( cosβ 0 + sinβ 0 ) The potential energy of the loads is G = cos π 4 β + sin π 4 β = cosβ For equilibrium, 0 = U β + G β ( ) + s cosβ + sinβ ( cosβ 0 + sinβ 0 ) ( ) = 4 β β 0 sinβ Hence = ( β β 0 ) + sinβ β 0 = 0, = β sinβ + ( ) sinβ + cosβ s ( ( ))( cosβ sinβ ) and when sinβ cosβ + sinβ cosβ 0 + sinβ 0 s ( cosβ + sinβ 1) ( cosβ sinβ ). sinβ
MODE ANSWER age: 6 β β The diagram on the left shows the contribution of the moment springs to the load and the diagram on the right is that from the linear spring. The moment springs mae a stable post bucling contribution but the contribution from the linear spring is unstable.