1. Dynamics of Machinery Problems and Solutions

2. Q1) In the gear train shown, the driver gear is gear-2
Q1) In the gear train shown, the driver gear is gear-2. Metric module of the gears are 4 mm and pressure angles are 20. Gear-2 rotates at 500 rpm and transmits 20 kW to the gear set. Find constraint force vectors F23, F13, and output torque T14.

3. S1

4. Q2) In the slider crank mechanism below, crank 2 is balanced and rotates with constant angular velocity. In this system, equivalent mass approach will be used. (System is working on horizontal plane, friction is neglected)
a) Find constraint force F34 due to gas force P at this instant that the piston is at 80% expansion position. (Bore diameter of the cylinder is 100 mm).
b) Find crank-shaft moment M12 due to inertia effect of the mass of the connecting rod reduced to bearing B.
(AO2=7.6 cm; AB=30 cm; AG3=12 cm ; m3=1.6 kg), θ=ωt=30o , ω=10r/s (CCW) ,

5. S2
𝑃= 𝑃 𝑟 𝜋 =0.69( 10 6 ) 𝜋 =5419 N a)
𝑡𝑎𝑛∅= 7.6 ( 10 −2 ) 30 ( 10 −2 ) 𝑠𝑖𝑛 ( 10 −2 ) ( 10 −2 ) 2 𝑠𝑖𝑛 2 30
𝑡𝑎𝑛∅=0.1276
∅= 𝑜
𝐹 34 𝑐𝑜𝑠∅=P, 𝐹 34 = 𝑃 → 𝐹 34 =5463.2N

6. 𝑥=𝑙− 𝑟 2 4𝑙 +𝑟 𝑐𝑜𝑠𝜔𝑡+ 𝑟 4𝑙 𝑐𝑜𝑠2𝜔𝑡 𝑥 =−𝑟𝜔 𝑠𝑖𝑛𝜔𝑡+ 𝑟 2𝑙 𝑠𝑖𝑛2𝜔𝑡 b)
𝑚 3𝐵 + 𝑚 3𝐴 = 𝑚 3 =1.6 𝑘𝑔
𝐴 𝐺 3 𝑚 3𝐴 = 𝐵 𝐺 3 𝑚 3𝐵
12 𝑚 3𝐴 =18 𝑚 3𝐵
𝑚 3𝐵 𝑚 3𝐵 =1.6
3 2 𝑚 3𝐵 =1.6
𝑚 3𝐵 =0.64 kg
𝑚 3𝐴 =0.96𝑘𝑔
𝑥=𝑙− 𝑟 2 4𝑙 +𝑟 𝑐𝑜𝑠𝜔𝑡+ 𝑟 4𝑙 𝑐𝑜𝑠2𝜔𝑡
𝑥 =−𝑟𝜔 𝑠𝑖𝑛𝜔𝑡+ 𝑟 2𝑙 𝑠𝑖𝑛2𝜔𝑡
𝑥 =−𝑟 𝜔 2 𝑐𝑜𝑠𝜔𝑡+ 𝑟 𝑙 𝑐𝑜𝑠2𝜔𝑡
𝑥 = − 𝑐𝑜𝑠 𝑐𝑜𝑠60
𝑥 = -7.6(0.993) = -7.54m/s2
𝐹 3𝐵 𝑖 =− 𝑚 3𝐵 𝑎 𝐵 =−0.64 −7.54 𝑖 =4.82𝑁

7. 𝑅cos∅= 𝐹 3𝐵 𝑖
𝐹 23 =𝑅= 4.82 𝑐𝑜𝑠7.27 =4.86𝑁
𝑀 𝑂 2 =0
- 𝐹 32 𝑐𝑜𝑠∅ 𝐴 𝑂 2 𝑠𝑖𝑛𝜔𝑡− 𝐹 32 𝑠𝑖𝑛∅ 𝐴 𝑂 2 𝑐𝑜𝑠𝜔𝑡+ 𝑀 12 =0
𝑀 12 =4.86𝑐𝑜𝑠 −2 𝑠𝑖𝑛 𝑠𝑖𝑛 −2 𝑐𝑜𝑠30
𝑀 12 = N.m

8. Q3) In the four bar mechanism shown, crank 2 is balanced and rotates with constant angular velocity. Gravity force and all frictions are neglected.
Find all the pin (joint) forces and the external torque that must be applied to link 2 of the mechanism by using the analytical method.
Find the external torque that must be applied to link 2 of the mechanism by using the Virtual Work method.
m/s2
O2A=20 cm
AB=60 cm
O4B=30 cm
AG3=30 cm
O4G4=15 cm
m3=5 kg
m4=3 kg
F=100N
m/s2
α3=115.2 r/s2 (CCW)
α4==162.8 r/s2 (CCW)
ω3=1.53 r/s (CW)
ω4 =8.85 r/s (CW)
ω2 =10 r/s (Constant, CCW)
I=mL2/12
m/s
m/s

9. S4

10. a)

12. b)

13. S4) In the cam mechanism shown in the figure, cam rotates by 1500 and gives a harmonic motion 30 mm rise to its follower to the right. Then, follower dwells for 300, returns with parabolic motion for 1500 and finally dwells for 300 cam rotations. Follower’s mass is 20 kg, spring constant is 5000 N/m.
a) Before any computation, draw graphs of the follower’s displacement, acceleration and force F23 for the 3600 cam rotation. Show on the drawing the first cam angle that jump may occur.
b) If the first critical cam speed 20r/s that jump may occur, determine the amount of minimum pre- compression deflection of the spring 2 3 1

14. 𝐹 23 =𝑚 𝑦 +𝑘𝑦+𝑘𝛿>0
𝑦= 𝐿 2 1−𝑐𝑜𝑠 𝜋𝜃 𝛽 = 30 (10 −3 ) 2 1−𝑐𝑜𝑠𝜋 =30 (10 −3 )𝑚
𝑦 = 𝜋 2 𝐿 2 𝛽 2 𝜔 2 𝑐𝑜𝑠 𝜋𝜃 𝛽 = 𝜋 2 30 (10 −3 ) 𝜋 𝜋𝑐𝑜𝑠𝜋=− 8.64𝑚 𝑠 2
𝛿>− 𝑚 𝑦 +𝑘𝑦 𝑘 →𝛿=− 20 − − =4.56𝑚𝑚 b)