Question 3.CS.5A: Fourbar Linkage Loading Analysis Problem: Determine the theo...
Fourbar Linkage Loading Analysis
Problem: Determine the theoretical rigid-body forces acting in two dimensions on the fourbar linkage shown in Figure 3-11.
Given: The linkage geometry, masses, and mass moments of inertia are known and the linkage is driven at up to 120 rpm by a speed-controlled electric motor.
Assumptions: The accelerations are significant. A Class 4 load model is appropriate and a dynamic analysis is required. There are no external loads on the system; all loads are due to the accelerations of the links. The weight forces are insignificant compared to the inertial forces and will be neglected. The links are assumed to be ideal rigid bodies. Friction and the effects of clearances in the pin joints also will be ignored.
See Figures 3-11 through 3-13 and Table 3-6.
1 Figures 3-11 and 3-12 show the fourbar linkage demonstrator model. It consists of three moving elements (links 2, 3, and 4) plus the frame or ground link (1). The motor drives link 2 through a gearbox. The two fixed pivots are instrumented with piezoelectric force transducers to measure the dynamic forces acting in x and y directions on the ground plane. A pair of accelerometers is mounted to a point on the floating coupler (link 3) to measure its accelerations.
2 Figure 3-12 shows a schematic of the linkage. The links are designed with lightening holes to reduce their masses and mass moments of inertia. The input to link 2 can be an angular acceleration, a constant angular velocity, or an applied torque. Link 2 rotates fully about its fixed pivot at O_{2}. Even though link 2 may have a zero angular acceleration \alpha _{2}, if run at constant angular velocity \omega _{2}, there will still be time-varying angular accelerations on links 3 and 4 since they oscillate back and forth. In any case, the CGs of the links will experience time-varying linear accelerations as the linkage moves. These angular and linear accelerations will generate inertia forces and torques as defined by Newton’s second law. Thus, even with no external forces or torques applied to the links, the inertial forces will create reaction forces at the pins. It is these forces that we wish to calculate.
3 Figure 3-13 shows the free-body diagrams of the individual links. The local, nonrotating, coordinate system for each link is set up at its CG. The kinematic equations of motion must be solved to determine the linear accelerations of the CG of each link and the link’s angular acceleration for every position of interest during the cycle. (See reference 1 for an explanation of this procedure.) These accelerations, A_{Gn} and \alpha _{n}, are shown acting on each of the n links. The forces at each pin connection are shown as xy pairs, numbered as before, and are initially assumed to be positive.
4 Equations 3.1 can be written for each moving link in the system. The masses and the mass moments of inertia of each link about its CG must be calculated for use in these equations. In this case study, a solid modeling CAD system was used to design the links’ geometries and to calculate their mass properties.
\sum F =m a \quad \sum M _G=\dot{ H }_G (3.1a)
\sum F_x=m a_x \quad \sum F_y=m a_y \quad \sum F_z=m a_z (3.1b)
H _G=I_x \omega_x \hat{ i }+I_y \omega_y \hat{ j }+I_z \omega_z \hat{ k } (3.1c)
\begin{array}{l} \sum M_x=I_x \alpha_x-\left(I_y-I_z\right) \omega_y \omega_z \\ \sum M_y=I_y \alpha_y-\left(I_z-I_x\right) \omega_z \omega_x \\ \sum M_z=I_z \alpha_z-\left(I_x-I_y\right) \omega_x \omega_y \end{array} (3.1d)
5 For link 2:
\begin{array}{l} \sum F_x=F_{12 x}+F_{32 x}=m_2 A_{G 2 x} \\ \sum F_y=F_{12 y}+F_{32 y}=m_2 A_{G 2 y} \\ \sum M_z=T_2+\left(R_{12 x} F_{12 y}-R_{12 y} F_{12 x}\right)+\left(R_{32 x} F_{32 y}-R_{32 y} F_{32 x}\right)=I_{G 2} \alpha_2 \end{array} (a)
6 For link 3:
\begin{array}{l} \sum F_x=F_{12 x}+F_{32 x}=m_2 A_{G 2 x} \\ \sum F_y=F_{12 y}+F_{32 y}=m_2 A_{G 2 y} \\ \sum M_z=T_2+\left(R_{12 x} F_{12 y}-R_{12 y} F_{12 x}\right)+\left(R_{32 x} F_{32 y}-R_{32 y} F_{32 x}\right)=I_{G 2} \alpha_2 \end{array} (b)
7 For link 4:
\begin{aligned} \sum F_x &=F_{14 x}+F_{34 x}=m_4 A_{G 4} \\ \sum F_y &=F_{14 y}+F_{34 y}=m_4 A_{G 4} \\ \sum M_z &=\left(R_{14 x} F_{14 y}-R_{14 y} F_{14 x}\right)+\left(R_{34 x} F_{34 y}-R_{34 y} F_{34 x}\right)=I_{G 4} \alpha_4 \end{aligned} (c)
8 There are 13 unknowns in these nine equations: F_{12 x}, F_{12 y}, F_{32 x}, F_{32 y}, F_{23 x}, F_{23 y}, F_{43 x}, F_{43 y}, F_{14 x}, F_{14 y}, F_{34 x}, F_{34 y}, \text { and } T_2 . Four third-law equations can be written to equate the action-reaction pairs at the joints.
\begin{array}{l} F_{32 x}=-F_{23 x} \\ F_{32 y}=-F_{23 y} \\ F_{34 x}=-F_{43 x} \\ F_{34 y}=-F_{43 y} \end{array} (d)
9 The set of thirteen equations in a through d can be solved simultaneously to determine the forces and driving torque either by matrix reduction or by iterative root-finding methods. This case study was solved by both techniques and files for both are on the CD. Note that the masses and mass moments of inertia of the links are constant with time and position, but the accelerations are time-varying. Thus, a complete analysis requires that equations a–d be solved for all positions or time steps of interest. The models use lists or arrays to store the calculated values from equations a–d for 13 values of the input angle \theta _{2} of the driving link ( 0 to 360° by 30° increments). The model also calculates the kinematic accelerations of the links and their CGs which are needed for the force calculations. The largest and smallest forces present on each link during the cycle can then be determined for use in later stress and deflection analyses. The given data and results of this force analysis for one crank position (\theta _{2} = 30°) are shown in Table 3-6, parts 1 and 2. Plots of the forces at the fixed pivots for one complete revolution of the crank are shown in Figure 3-14.
| Table 3–6 – part 1 Case Study 5A Given and Assumed Data |
||
| Variable | Value | Unit |
| \theta _{2} | 30.00 | deg |
| \omega _{2} | 120.00 | rpm |
| mass_{2} | 0.525 | kg |
| mass_{3} | 1.050 | kg |
| mass_{4} | 1.050 | kg |
| Icg_{2} | 0.057 | kg-m² |
| Icg_{3} | 0.011 | kg-m² |
| Icg_{4} | 0.455 | kg-m² |
| R_{12x} | –46.9 | mm |
| R_{12y} | –71.3 | mm |
| R_{32x} | 85.1 | mm |
| R_{32y} | 4.9 | mm |
| R_{23x} | –150.7 | mm |
| R_{23y} | –177.6 | mm |
| R_{43x} | 185.5 | mm |
| R_{43y} | 50.8 | mm |
| R_{14x} | –21.5 | mm |
| R_{14y} | –100.6 | mm |
| R_{34x} | –10.6 | mm |
| R_{34y} | 204.0 | mm |
| Table 3–6 – part 2 Case Study 5A Calculated Data |
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| Variable | Value | Unit |
| F_{12x} | –255.8 | N |
| F_{12y} | –178.1 | N |
| F_{32x} | 252.0 | N |
| F_{32y} | 172.2 | N |
| F_{34x} | –215.6 | N |
| F_{34y} | –163.9 | N |
| F_{14x} | 201.0 | N |
| F_{14y} | 167.0 | N |
| F_{43x} | 215.6 | N |
| F_{43y} | 163.9 | N |
| F_{23x} | –252.0 | N |
| F_{23y} | –172.2 | N |
| T_{12} | –3.55 | N-m |
| \alpha _{3} | 56.7 | rad/sec² |
| \alpha _{4} | 138.0 | rad/sec² |
| A_{cg2x} | –7.4 | rad/sec² |
| A_{cg2y} | –11.3 | rad/sec² |
| A_{cg3x} | –34.6 | rad/sec² |
| A_{cg3y} | –7.9 | rad/sec² |
| A_{cg4x} | –13.9 | rad/sec² |
| A_{cg4y} | 2.9 | rad/sec² |
