Question 11.4: Analysis of a Fourbar Linkage by the Method of Virtual Work....
Analysis of a Fourbar Linkage by the Method of Virtual Work. (See Figure 11‑3.)
Given: The 5-in-long crank (link 2) shown weighs 1.5 lb. Its CG is at 3 in at +30° from the line of centers. Its mass moment of inertia about its CG is 0.4 lb-in-sec². Its kinematic data are:
\begin{array}{cccl}\theta_{2} deg & \omega_{2} rad / sec & \alpha_{2} rad / sec ^{2} & V_{G_{2}} in / sec \\60 & 25 & -40 & 75 \text { @ } 180^{\circ}\end{array}
The coupler (link 3) is 15 in long and weighs 7.7 lb. Its CG is at 9 in at 45° off the line of centers. Its mass moment of inertia about its CG is 1.5 lb-in-sec². Its kinematic data are:
\begin{array}{cccc}\theta_{3} deg & \omega_{3} rad / sec & \alpha_{3} rad / sec ^{2} & V_{G_{3}} in / sec \\20.92 & -5.87 & 120.9 & 72.66 @ 145.7^{\circ}\end{array}
There is an external force on link 3 of 80 lb at 330°, applied at point P which is located 3 in @ 100° from the CG of link 3. The linear velocity of that point is 67.2 in/sec at 131.94°.
The rocker (link 4) is 10-in long and weighs 5.8 lb. Its CG is at 5 in at 0° off the line of centers. Its mass moment of inertia about its CG is 0.8 lb-in-sec². Its data are:
\begin{array}{cccc}\theta_{4} deg & \omega_{4} rad / sec & \alpha_{4} rad / sec ^{2} & V_{G_{4}} in / sec \\104.41 & 7.93 & 276.29 & 39.66 \text { @ 194.41 }^{\circ}\end{array}
There is an external torque on link 4 of 120 lb-in. The ground link is 19-in long.
Find: The driving torque T _{12} needed to maintain motion with the given acceleration for this instantaneous position of the link.
1 The torque, angular velocity, and angular acceleration vectors in this two-dimensional problem are all directed along the Z axis, so their dot products each have only one term. Note that in this particular example there is no force F _{P_{4}} \text { and no torque } T _{3} .
2 The cartesian coordinates of the acceleration data were calculated in Example 11-3.
\begin{array}{lll}a _{G_{2}}=1878.84 @ \angle 273.66^{\circ} ; & a_{G_{2_{x}}}=119.94, & a_{G_{2} y}=-1875.01 \\a _{G_{3}}=3646.10 @ \angle 226.51^{\circ} ; & a_{G_{3_{x}}}=-2509.35, & a_{G_{3 y}}=-2645.23 \\a _{G_{4}}=1416.80 @ \angle 207.24^{\circ} ; & a_{G_{4 x}}=-1259.67, & a_{G_{4 y}}=-648.50\end{array} (a)
3 The x and y components of the external force at P in the global coordinate system were also calculated in Example 11-3:
F _{P_{3}}=80 @ \angle 330^{\circ} ; \quad F_{P_{3 x}}=69.28, \quad F_{P_{3 y}}=-40.00 (b)
4 Converting the velocity data for this example to cartesian coordinates:
\begin{array}{llcl}V _{G_{2}}=75.00 & @ \angle 180.00^{\circ} ; & V_{G_{2 x}}=-75.00, & V_{G_{2} y}=1 \\V _{G_{3}}=72.66 & @ \angle 145.70^{\circ} ; & V_{G_{3 x}}=-60.02, & V_{G_{3}}=40.95 \\V _{G_{4}}=39.66 & @ \angle 194.41^{\circ} ; & V_{G_{4 x}}=-38.41, & V_{G_{4 y}}=-9.87 \\V _{P_{3}}=67.20 & @ \angle 131.94^{\circ} ; & V_{P_{3 x}}=-44.91, & V_{P_{3 y}}=49.99\end{array} (c)
5 Substituting the example data into equation 11.16c:
\begin{aligned}\left(F_{P_{3 x}} V_{P_{3 x}}\right.&\left.+F_{P_{3 y}} V_{P_{3 y}}\right)+\left(F_{P_{4 x}} V_{P_{4 x}}+F_{P_{4 y}} V_{P_{4 y}}\right)+\left(T_{12} \omega_{2}+T_{3} \omega_{3}+T_{4} \omega_{4}\right) \\=m_{2}(&\left.a_{G_{2 x}} V_{G_{2 x}}+a_{G_{2 y}} V_{G_{2 y}}\right)+m_{3}\left(a_{G_{3 x}} V_{G_{3 x}}+a_{G_{3 y}} V_{G_{3 y}}\right) \\&+m_{4}\left(a_{G_{4 x}} V_{G_{4 x}}+a_{G_{4 y}} V_{G_{4 y}}\right)+\left(I_{G_{2}} \alpha_{2} \omega_{2}+I_{G_{3}} \alpha_{3} \omega_{3}+I_{G_{4}} \alpha_{4} \omega_{4}\right)\end{aligned} (11.6c)
\begin{gathered}{[(69.28)(-44.91)+(-40)(49.99)]+[0]+\left[25 T_{12}+(0)+(120)(7.93)\right]} \\=\frac{1.5}{386}[(119.94)(-75)+(-1875.01)(0)] \\+\frac{7.7}{386}[(-2509.35)(-60.02)+(-2645.23)(40.95)] \\+\frac{5.8}{386}[(-1259.67)(-38.41)+(-648.50)(-9.87)] \\+[(0.4)(-40)(25)+(1.5)(120.9)(-5.87)+(0.8)(276.29)(7.93)]\end{gathered} (d)
6 The only unknown in this equation is the input torque T _{12} which calculates to:
T _{12}=243.2 \hat{ k } (e)
which is the same as the answer obtained in Example 11-3.