Design of Machinery

68 SOLVED PROBLEMS

Question: 16.4

Improvement to the Design of Example 16-3 Given: The design of Example 16-3 proved to be dynamically poor. In that example the required position and velocity constraints were applied to the slider motion and the resulting function was transformed by equations relating slider motion to crank ...

1 The functions chosen to provide the motions will...
Question: 16.3

Slider-Crank Linkage Driven by a Servomotor to Perform a Forming Operation Given: A slider-crank linkage mechanism (Figure 16-11) has been designed to provide the geometry for a suitable motion of a die in a forming press. The required motion begins from a dwell and moves the slider through a 2-in ...

1 The functions chosen to provide the motions defi...
Question: 16.2

Cam-Driven Fourbar Slider With Motion Functions Applied to the Output Link Given: A cam-driven fourbar crank-slider with the geometry shown in Figure 16-3 is driven by a cam with a constant velocity motion program similar to that developed in Example 8-12 and Figure 8-42. The slider is the end ...

1 The linkage geometry is the same as shown in Fig...
Question: 16.1

Cam-Driven Fourbar Slider With Motion Functions Applied to the Input Link Given: A cam-driven fourbar crank-slider with the geometry shown in Figure 16-3 is driven by a cam with a constant velocity motion program similar to that developed in Example 8-12 and Figure 8-42. The slider is the end ...

1 The equations for the position of the crank-slid...
Question: 11.5

Determining the Energy Variation in a Torque-Time Function. Given: An input torque-time function which varies over its cycle. Figure 11-11 shows the input torque curve from Figure 11-8. The torque is varying during the 360° cycle about its average value. Find: The total energy variation over one ...

1 Calculate the average value of the torque-time f...
Question: 11.3

Dynamic Force Analysis of a Fourbar Linkage. (See Figure 11-3) Given: The 5-in-long crank (link 2) shown weighs 1.5 lb. Its CG is at 3 in @ +30° from the line of centers (LRCS). Its mass moment of inertia about its CG is 0.4 lb-in-sec². Its kinematic data are: θ2 deg ω2 rad/sec α2 rad/sec² aG2 ...

1 Convert the given weight to proper mass units, i...
Question: 9.8

Determining the Efficiency of an Epicyclic Gear Train.* Find the overall efficiency of the epicyclic train shown in Figure 9-43. The basic efficiency E0 is 0.9928 and the gear tooth numbers are: NA= 82t, NB= 84t, NC = 86t, ND = 82t, NE = 82t, and NF = 84t. Gear A (shaft 2) is fixed to the frame, ...

1 Find the basic ratio ρ for the gear train using ...
Question: 9.7

Analyzing Ferguson’s Paradox by the Formula Method. Consider the same Ferguson paradox train as in Example 9-6 which has the following tooth numbers and initial conditions (see Figure 9-37): Sun gear #2 N2 = 100-tooth external gear Sun gear #3 N3 = 99-tooth external gear Sun gear #4 N4 = 101-tooth ...

1 We will have to apply equation 9.14 twice, once ...
Question: 9.1