Question 11.3: Dynamic Force Analysis of a Fourbar Linkage. (See Figure 11-...
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:
\begin{array}{cccc}\theta_{2} \text { deg } & \omega_{2} rad / sec & \alpha_{2} rad / sec ^{2} & a_{G_{2}} in / sec ^{2} \\60 & 25 & -40 & 1878.84 \text { @ 273.66 }^{\circ}\end{array}
The coupler (link 3) is 15 in long and weighs 7.7 lb. Its CG is at 9 in @ 45° off the line of centers (LRCS). Its mass moment of inertia about its CG is 1.5 lb-in-sec². Its kinematic data are:
\begin{array}{cccc}\theta_{3} \text { deg } & \omega_{3} \text { rad/sec } & \alpha_{3} \text { rad/sec }^{2} & a_{G_{3}} \text { in/sec }^{2} \\20.92 & -5.87 & 120.9 & 3646.1 \text { @ } 226.5^{\circ}\end{array}
The ground link is 19 in long. The rocker (link 4) is 10 in long and weighs 5.8 lb. Its CG is at 5 in @ 0° on the line of centers (LRCS). Its mass moment of inertia about its CG is 0.8 lb-in-sec². There is an external torque on link 4 of 120 lb-in (GCS). An external force of 80 lb @ 330° acts on link 3 in the GCS, applied at point P at 3 in @ 100° from the CG of link 3 (LRCS). The kinematic data are:
\begin{array}{cccc}\theta_{4} deg & \omega_{4} rad / sec & \alpha_{4} rad / sec ^{2} & a_{G_{4}} in / sec ^{2} \\104.41 & 7.93 & 276.29 & 1416.8 @ 207.2^{\circ}\end{array}
Find: Forces F _{12}, F _{32}, F _{43}, \text { and } F _{14} \text { at the joints and the driving torque } T _{12} needed to maintain motion with the given acceleration for this instantaneous position of the link.
1 Convert the given weight to proper mass units, in this case blobs:
\text { mass }_{\text {link } 2}=\frac{\text { weight }}{g}=\frac{1.5 lb }{386 in / sec ^{2}}=0.004 \text { blob } (a)
\text { mass link } 3=\frac{\text { weight }}{g}=\frac{7.7 lb }{386 in / sec ^{2}}=0.020 \text { blob } (b)
\text { mass }_{\text {link } 4}=\frac{\text { weight }}{g}=\frac{5.8 lb }{386 \text { in } / sec ^{2}}=0.015 \text { blob } (c)
2 Set up an LNCS xy coordinate system at the CG of each link, and draw all applicable vectors acting on that system as shown in the figure. Draw a free-body diagram of each moving link.
3 Calculate the x and y components of the position vectors R _{12}, R _{32}, R _{23}, R _{43}, R _{34}, R _{14}, \text { and } R _{P} in the link’s LNCS. R _{43}, R _{34}, \text { and } R _{14} will have to be calculated from the given link geometry data using the law of cosines and law of sines. Note that the current value of link 3’s position angle ( \theta_{3} ) in the GCS must be added to the angles of all position vectors before creating their x, y components in the LNCS if their angles were originally measured with respect to the link’s embedded, local rotating coordinate system (LRCS).
\begin{array}{llll}R _{12}=3.00 @ \angle 270.00^{\circ} ; & R_{12_{x}}=0.000, & R_{12_{y}}=-3 \\R _{32}=2.83 @ \angle 28.00^{\circ} ; & R_{32_{x}}=2.500, & R_{32_{y}}=1.333 \\R _{23}=9.00 @ \angle 245.92^{\circ} ; & R_{23_{x}}=-3.672, & R_{23_{y}}=-8.217 \\R _{43}=10.72 @ \angle-15.46^{\circ} ; & R_{43_{x}}=10.332, & R_{43_{y}}=-2.858 \\R _{34}=5.00 @ \angle 104.41^{\circ} ; & R_{34_{x}}=-1.244, & R_{34_{y}}=4.843 \\R _{14}=5.00 @ \angle 284.41^{\circ} ; & R_{14_{x}}=1.244, & R_{14_{y}}=-4.843 \\R _{P}=3.00 @ \angle 120.92^{\circ} ; & R_{P_{x}}=-1.542, & R_{P_{y}}=2.574\end{array} (d)
4 Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system (GCS):
\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} (e)
5 Calculate the x and y components of the external force at P in the GCS:
F _{P 3}=80 @ \angle 330^{\circ} ; \quad F_{P 3_{x}}=69.28, \quad F_{P 3_{y}}=-40.00 (f)
6 Substitute these given and calculated values into the matrix equation 11.9.
\left[\begin{array}{ccccccccc}1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\-R_{12 y} & R_{12_{x}} & -R_{32_{y}} & R_{32_{x}} & 0 & 0 & 0 & 0 & 1 \\0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\0 & 0 & R_{23_{y}} & -R_{23_{x}} & -R_{43_{y}} & R_{43_{x}} & 0 & 0 & 0 \\0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 \\0 & 0 & 0 & 0 & R_{34_{y}} & -R_{34_{x}} & -R_{14_{y}} & R_{14_{x}} & 0\end{array}\right] \times\left[\begin{array}{c}F_{12_{x}} \\F_{12_{y}} \\F_{32_{x}} \\F_{32_{y}} \\F_{43_{x}} \\F_{43_{y}} \\F_{14_{x}} \\F_{14_{y}} \\T_{12}\end{array}\right]=\left[\begin{array}{c}m_{2} a_{G_{2 x}} \\m_{2} a_{G_{2 y}} \\I_{G_{2}} \alpha_{2} \\m_{3} a_{G_{3 x}}-F_{P_{x}} \\m_{3} a_{G_{3 y}}-F_{P_{y}} \\I_{G_{3}} \alpha_{3}-R_{P_{x}} F_{P_{y}}+R_{P_{y}} F_{P_{x}} \\m_{4} a_{G_{4 x}} \\m_{4} a_{G_{4 y}} \\I_{G_{4}} \alpha_{4}-T_{4}\end{array}\right] (11.9)
\left[\begin{array}{ccccccccc}1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\3 & 0 & -1.330 & 2.5 & 0 & 0 & 0 & 0 & 1 \\0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\0 & 0 & -8.217 & 3.673 & 2.861 & 10.339 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 \\0 & 0 & 0 & 0 & 4.843 & 1.244 & 4.843 & 1.244 & 0\end{array}\right] \times\left[\begin{array}{c}F_{12_{x}} \\F_{12_{y}} \\F_{32_{x}} \\F_{32_{y}} \\F_{43_{x}} \\F_{43_{y}} \\F_{14_{x}} \\F_{14_{y}} \\T_{12}\end{array}\right]
\left[\begin{array}{c}(0.004)(119.94) \\(0.004)(-1875.01) \\(0.4)(-40) \\(0.02)(-2509.35)-(69.28) \\(0.02)(-2645.23)-(-40) \\(1.5)(120.9)-[(-1.542)(-40)-(2.574)(69.28)] \\(0.015)(-1259.67) \\(0.015)(-648.50) \\(0.8)(276.29)-(120)\end{array}\right]=\left[\begin{array}{r}0.480 \\-7.500 \\-16.000 \\-119.465 \\-12.908 \\298.003 \\-18.896 \\-9.727 \\101.031\end{array}\right] (g)
7 Solve this system either by inverting matrix A and premultiplying that inverse times matrix C using a pocket calculator with matrix capability, or by inputting the values for matrices A and C to program Matrix downloadable with this text, which gives the following solution:
\left[\begin{array}{c}F_{12_{x}} \\F_{12_{y}} \\F_{32_{x}} \\F_{32_{y}} \\F_{43_{x}} \\F_{43_{y}} \\F_{14_{x}} \\F_{14_{y}} \\T_{12}\end{array}\right]=\left[\begin{array}{r}-117.65 \\-107.84 \\118.13 \\100.34 \\-1.34 \\87.43 \\-20.23 \\77.71 \\243.23\end{array}\right] (h)
Converting the forces to polar coordinates:
\begin{array}{lrlr}F _{12}= & 159.60 lb @ \angle 222.52^{\circ} \\F _{32}= & 154.99 lb @ \angle 40.35^{\circ} \\F _{43}= & 87.44 lb @ \angle 90.88^{\circ} \\F _{14}= & 80.30 lb @ \angle 104.59^{\circ}\end{array} (i)
8 The pin-force magnitudes in (i) are needed to size the pivot pins and links against failure and to select pivot bearings that will last for the required life of the assembly. The driving torque T_{12} defined in (h) is needed to select a motor or other device capable of supplying the power to drive the system. See Section 2.19 for a brief discussion of motor selection. Issues of stress calculation and failure prevention are beyond the scope of this text, but note that those calculations cannot be done until a good estimate of the dynamic forces and torques on the system has been made by methods such as those shown in this example.