Dynamic Force Analysis

PSG COLLEGE OF TECHNOLOGY, COIMBATORE-4 DEPARTMENT OF MECHANICAL ENGINEERING 08M404 Dynamics of Machinery Assignment: Dynamic force Analysis

1. Find the external torque which must be applied to link 2 of the four-bar linkage shown to drive it at the given velocity. RAO2 =75 mm , RO4O2 =175 mm, RBA=200 mm, RBO4 =150 mm , RG3A =100 mm , RG4O4 =75 mm ,m3 =0.32 kg, m4 =0.32 kg,IG2 =2.9154 X 10-3 kgm2 , IG3 =5.808 X 10-3 kgm2, IG4 =60.003 X 10-3 kgm2 ,ω2 = 180k rad/s , α2 =0 , α3 = 4950k rad/s2 , α4 =-8900k rad/s2 ,AG3 = 1928i+229j m/s2 , AG4 =1610i+229j m/s2 . 2. Crank 2 of the four-bar linkage shown in the figure is balanced. For the given angular velocity of link 2, find the forces acting at each joint and the external torque which must be applied to link 2. RAO2 = 50 mm, RO4O2 = 325 mm, RBA=425 mm, RBO4 = 200 mm, RG3A = 210 mm, RG4O4 = 100 mm, m3 = 1.2 kg, m4 = 3.04 kg,IG2 = 2.700 X 10-3 kgm2 , IG3 = 6.84 X 10-3 kgm2, IG4 = 60.003 X 10-3 kgm2,ω2 = 200k rad/s , α2 =0 , α3 =-6500k rad/s2 , α4 =-240k rad/s2 ,AG3 = -964i+80j m/s2 , AG4 =-244i643.5j m/s2 . 3. The angular velocity of crank 2 given in the figure, find the reactions at each joint and the external torque to be applied to the crank.RAO2 =75 mm, RBA= 300 mm, RBO4 = 200 mm, RG3A = 113 mm, m3 = 1.54 kg, m4 = 1.3 kg, IG2 =39.77 X 10-3 kgm2, IG3 =12.204 X 10-3 kgm2, ω2 = 210k rad/s, α2 =0, α3 =7670k rad/s2, AG3 = -2385i-1487j m/s2 ,AG4 = -2394i m/s2 . 4. The figure shows a slider-crank mechanism with an external force FB applied to the piston. For the given crank velocity, find all the reaction forces in the joints and the crank torque. RAO2 =75 mm, RBA=300 mm, RG2O2 = 32 mm, RG3A = 88 mm, m2 = 0.43 kg, m3 = 1.59 kg, m4 = 0.283 kg, IG2 = 4.39 X 10-4 kgm2, IG3 = 0.0124 X 10-3 kgm2, ω2 = 160k rad/s, α2 =0 , α3 =-3090k rad/s2, AG2 = 805 m/s2 @ 1500, AG3 = 1870 m/s2 @158.30 , AG4 = 1916 m/s2 @1800,FB = 363 N @ 1800.

5. Analyze the dynamics of the offset slider-crank mechanism shown in the figure using the following data:

A=0.06 m, RAO2 =0.1 m, RBA=0.38 m, RCA=0.4 m, θC = 32°, RG3A = 0.26 m, α = 22°, m2=2.5 kg, m3=7.4 kg, m4=2.5 kg, IG2=0.0005 kgm2, IG3=0.0136 kgm2, θ2= 120° and ω2 =18 rad/sec with α2 =0, FB = -2100i N, FC = -1000i N. Assume a balanced crank and no friction forces.

6. The following data apply to the four –bar linkage shown in the figure : RA02 = 0.3 m, RO4O2 =0.9 m, RBA=1.5 m, RBO4 =0.8 m, RCA=0.85 m, θC = 330 , RDO4 =0.4 m, θD = 530, RG2O2 =0, RG3A =0.65 m, α= 160 , RG4O4 =0.45 m, β=170, m2 = 5.2 kg, m3 = 65.8 kg, m4 = 21.8 kg, IG2 = 2.3kg.m2, IG3 = 4.2kg.m2, IG4 = 0.51kg.m2 . A kinematic analysis at θ2 = 530 and ω2 = 12 rad/sec ccw with α2 = 0, θ3 = 0.70 , θ4 = 550 , α3 = 85.6 rad/sec2 cw, α4 = 172 rad/sec2 cw, AG3 = 96.4 m/s2 @2590and AG4 = 97.8 m/s2 2700. Find all the pin reactions and the torque to be applied to link 2.