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Q1) Please answer the following questions: Part a) consider the system shown in Figure Q1a

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Q1) Please answer the following questions: Part a) consider the system shown in Figure Q1a. Assume that the rigid bar with length ?, is massless and is attached to the wall with a pivot joint as shown in the figure. A mass ? is attached at the tip and is supported by spring with stiffness ?. Assuming small angle approximation, i.e. sin ? ≅ ? and cos ? ≅ 1, derive the static and dynamic equations of motion. Determine the tip static deflection in mm and natural frequency of the system in rad/s for ? = 106 N/m, ? = 100 kg, and ? = 1 m. Assume gravitational acceleration ? = 10 m/s2. [5 marks] Figure Q1a Part b) Now add a dashpot to the system of part (a) as shown in Figure Q1b. The dashpot is added at distance ? from the pivot point as shown in the figure and has a damping coefficient of ?. Assuming small angle approximation, i.e. sin ? ≅ ? and cos ? ≅ 1, and considering the mass moment of inertia of the rigid bar about pivot point is ?, derive the dynamic equation of motion. Now assume ? = 106 N/m, ? = 100 kg, ? = 1.6 × 105 N.s/m, ? = 300 kg.m2 and ? = 1 m. Determine the distance ? (in m) for the system to be critically damped. [5 marks] Figure Q1b 2 ? Part c) Consider the system of part (b) but add a harmonic force ?(?) = ?0 cos(?? + 4 ) at the tip as shown in Figure Q1c. Derive the dynamic equation of motion. Now assume ? = 106 N/m, ? = 100 kg, ? = 104 N.s/m, ? = 300 kg.m2 , ? = 0.5 m, ? = 1 m, ? = 49 rad/s and ?0 = 1000 N. Determine the steady state tip displacement response of the system at ? = 10 s (in mm). [7 marks] Figure Q1c Part d) Consider the system shown in Figure Q1c, but now replace the periodic force ?(?) by ?0 (1 − cos(??)) ? 0≤? ≤2s ?(?) = { 0 ? ? >2s and remove the dashpot. The new system is shown in Figure Q1d. If ?0 = 1000 N, derive an expression for the tip displacement response for ? > 2 s and determine the tip displacement response of the system at ? = 4 s (in mm). [8 marks] Figure Q1d Note: formula for integration (for ? ≠ ?) 3 ? ∫ (1 − cos(??))sin(?(? − ?))?? ?0 1 1 ? ? −( + 2 cos(??0 )) cos(?(? − ?0 )) + 2 cos(??) 2 ? ? ? −? ? − ?2 ? + 2 sin(??0 ) sin(?(? − ?0 )) ? − ?2 = Part e) Now replace the rigid bar with a flexible beam with total mass ?? (uniformly distributed). Also remove the dashpot ? and spring ?. Support the flexible beam by fixing it to the wall, as shown in Figure Q1e (a). Using the kinetic energy of the beam and assuming the displacement of the beam as ?? 2 3? ( − ?) 6?? 2 ?(?) = ?? 2 ? (3? − ) {24?? 2 0≤ ?≤ ? 2 ? ≤ ?≤? 2 where ? is the load applied at the mid-point of the beam, ?? is flexural rigidity, and ? is the coordinate measured from the root as shown in Figure Q1e (a), determine the effective mass of the flexible beam at the mid-point of the beam (?1 ). Figure Q1e (a) Considering the effective mass of the beam at the mid-point of the flexible beam, the system may be considered as a 2 degrees of freedom system shown in Figure Q1e (b). Determine the first two natural frequencies in Hz assuming ?? = 106 N.m2, ? = 100 kg, ???? = 50 kg and ? = 1 m. Figure Q1e (b) [15 marks] 4 DATA SHEET Combinations of springs Parallel combination: ??? = ?1 + ?2 + ? + ?? Series combination ??? = 1 1 1 1 + + ?+ ?1 ?2 ?? Equivalent tip lateral stiffness and mass of cantilever beam 3?? 33? ??? = ?3 , ??? = 140 Equivalent tip Axial stiffness and mass of rod ??? = EA ? ? 3 , ??? = The undamped, damped natural frequency of free SOF ??? + c?? + k? = 0 Undamped natural frequency 2? ? = 2?? = √ ? ? Critical damping ?? = ??? = 2√?? = 2??? Free SDOF system Undamped case (? = 0) ?? + ??2 ? = 0 ?(?) = [? sin(?? ?) + ? cos(?? ?)] = ? sin(?? ? + ?) ? = ?0 /?? , ? = ?0 ?= 1 √(?0 )2 + (?0 ?? )2 ?? ?0 ?? ) ?0 Underdamped case (? < 1) ?? + 2??? ?? + ??2 ? = 0 ? = tan−1 ( ?(?) = ? −???? [? sin(?? ?) + ? cos(?? ?)] = ?? −???? sin(?? ? + ?) ? = (?0 + ??? ?0 )/?? , ? = ?0 1 √(?0 + ??? ?0 )2 + (?0 ?? )2 ?= ?? ?0 ?? ) ? = tan−1 ( ?0 + ??? ?0 Critically damped case (? = 1): ?(?) = [? + ?t]? −??? Damping ratio ?= ? ??? ? = ?0 and ? = ?0 + ?0 ?? Damped natural frequency ?? = ?? √1 − ?2 Overdamped case (? > 1): ?(?) = ? −???? [?? (−?? √? 2 −1)? + ?? (?? √? 2 −1)? ] The Logarithmic Decrement −?0 + (−? + √? 2 − 1)?0 ?? ?(t) 2?? ?= ? = ln ( )= 2?? √? 2 − 1 x(t + T) √1 − ? 2 ?0 + (−? + √? 2 − 1)?0 ?? ? ? = ?= 2?? √? 2 − 1 √4? 2 + ? 2 DATA SHEET CONTINUES ON THE FOLLOWING PAGE TURN OVER DATA SHEET CONTINUED Energy Method ? (? + ?) = 0 ?? where ? is kinetic energy and U is the potential energy. 1 ? = 2 ??? 2 for translational motion Undamped case (? = 0, ? = ?? ) ??? + k? = ?0 cos(??) ?0 ?0 ?(?) = sin ?? + ?0 cos ?? + ? sin ?? ? 2? Harmonic excitation (? = ?/?? ) 1 ? = 2 ??? 2 for rotational motion 1 Underdamped case (? < 1) 2 ? = 2 ?? potential energy in spring ? = ??? potential energy due to gravity Harmonic excitation (? = ?/?? ) Undamped case (? = 0, ? ≠ ?? ) ??? + k? = ?0 cos(??) ?0 ?(?) = sin(?? ?) ?? ?0 ) cos(?? ?) 2 ?? − ? 2 ?0 ( 2 ) cos(??) ?? − ? 2 + (?0 − + where ?0 = ?0 /? ??? + c?? + k? = ?0 cos(??) ?(?) = ? −???? [? cos(?? ?) + ? sin(?? ?)] + ? cos(?? − ?) ? and ? are determined from initial conditions. Note : If ? = ?0 cos(?? + ?) where ? is a given angle, then steady state response will be ? cos(?? + ? − ?). Steady state amplitude in the above equation. ?= ?0 1 ? √(1 − ? 2 )2 + (2??)2 Note : If ? = ?0 cos(?? + ?) where ? is a given angle, then steady state response Steady state phase angle ?0 will be (?2 −? and 2?? 2 ) cos(?? + ?) ? ) ? = tan−1 ( coefficients of free response 1 − ?2 (? and ? in ? cos(?? ?) + ? sin(?? ?)) will Complex Numbers: be determined from initial conditions. ? ?? = cos(?) + ? sin(?) For zero initial conditions, i.e. ?0 = ?0 = 0, and ?(?) = ?0 cos(??), or/and beating Small Angle Approximations (in any case), the response is ?(?) sin(?) ≈ ?, cos(?) ≈ 1 2?0 ?? − ? ?? + ? ) sin ( =( 2 ?) sin ( ?) ?? − ? 2 2 2 DATA SHEET CONTINUES ON THE FOLLOWING PAGE TURN OVER DATA SHEET CONTINUED Response to impact ?? Base excitation (? = ?/?? ) ??? + c?? + k? = c?? + k? ? = ? cos(?? + ?) ?(?) = ? −??? ? [? cos(?? ?) + ? sin(?? ?)] + ? cos(?? + ? − ?) Steady state Response is ? cos(?? + ? − ?), where: ? 1 + (2??)2 =√ ? (1 − ? 2 )2 + (2??)2 ??? + ??? + ?? = ?(?) = ?? ?(?) ?+? ?? = ∫ ?(?)?? = ?(?)?? ?−? Undamped case ?? ?(?) = sin(?? ?) ??? Underdamped case ?? ? −???? sin(?? ?) ??? Response to an Arbitrary Input ?(?) = ? 2?? 3 ? = tan−1 ( ) (1 − ? 2 ) + (2??)2 ?(?) = ∫ ?(?)?(? − ?)?? 0 ? = ∫ ?(? − ?)?(?)?? Transmitted force: 0 ?? = ??? 2 (√ 1 + (2??)2 ) (1 − ? 2 )2 + (2??)2 Relative motion (? = ? − ?) ? ?2 = ? √(1 − ? 2 )2 + (2??)2 Rotating imbalance (? = ?/?? ) Steady state amplitude ??? + c?? + k? = ?0 ??2 sin(??) ?(?) = ? −???? [? cos(?? ?) + ? sin(?? ?)] + ? sin(?? − ?) Response is ? sin(?? + ? − ?), when force is ?0 ??2 sin(?? + ?). ?0 ? ?2 ?= ? √(1 − ? 2 )2 + (2??)2 Steady state phase angle 2?? ) ? = tan−1 ( 1 − ?2 where undamped case: 1 ?(?) = sin(?? ?) ??? Underdamped case ? −???? sin(?? ?) ??? Multiple degrees of freedom ?(?) = ??? + ?? = ? To obtain the natural frequencies ? solve: det(−?2 ? + ?) = 0 To obtain the mode shapes ?1 and ?2 solve: (−?12 ? + ?)?1 = 0 (−?22 ? + ?)?2 = 0 System with proportional damping ? = ?? + ?? ???? ? ?? = + 2??? 2