Question 11.2: Dynamic Force Analysis of a Threebar Crank-Slide Linkage wit...
Dynamic Force Analysis of a Threebar Crank-Slide Linkage with Half Joint. (See Figure 11‑2.)
Given: The 5-in long crank (link 2) shown weighs 2 lb. Its CG is at 3 in and 30° from the line of centers. Its mass moment of inertia about its CG is 0.05 lb-in-sec². Its acceleration is defined in its LNCS, x,y. Its kinematic data are:
\begin{array}{cccc}\theta_{2} deg & \omega_{2} rad / sec & \alpha_{2} rad / sec ^{2} & a_{G_{2}} in / sec ^{2} \\60 & 30 & -10 & 2700.17 @-89.4^{\circ}\end{array}
The coupler (link 3) is 15 in long and weighs 4 lb. Its CG is at 9 in and 45° from the line of centers. Its mass moment of inertia about its CG is 0.10 lb-in-sec². Its acceleration is defined in its LNCS, x,y. Its kinematic data are:
\begin{array}{cccc}\theta_{3} deg & \omega_{3} rad / sec & \alpha_{3} rad / sec ^{2} & a_{G_{3}} in / sec ^{2} \\99.59 & -8.78 & -136.16 & 3453.35 @ 254.4^{\circ}\end{array}
The sliding joint on link 3 has a velocity of 96.95 in/sec in the +Y direction.
There is an external force of 50 lb at – 45°, applied at point P which is located at 2.7 in and 101° from the CG of link 3, measured in the link’s embedded, rotating coordinate system or LRCS x’, y’ (origin at A and x axis from A to B). The coefficient of friction µ is 0.2.
Find: The forces F _{12}, F _{32}, F _{13} \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 weights to proper mass units, in this case blobs:
\text { mass }_{\text {link } 2}=\frac{\text { weight }}{g}=\frac{2 lb }{386 in / sec ^{2}}=0.0052 \text { blob } (a)
\operatorname{mass}_{\operatorname{link} 3}=\frac{\text { weight }}{g}=\frac{4 lb }{386 in / sec ^{2}}=0.0104 \text { blob } (b)
2 Set up a local, nonrotating xy coordinate system (LNCS) at the CG of each link, and draw all applicable position and force vectors acting within or on that system as shown in Figure 11-2. Draw a free-body diagram of each moving link as shown.
3 Calculate the x and y components of the position vectors R _{12}, R _{32}, R _{23}, R _{13}, \text { and } R _{P} in the LNCS coordinate system:
\begin{array}{llll}R _{12}=3.00 @ \angle 270.0^{\circ} ; & R_{12_{x}}=0.000, & R_{12_{y}}=-3.0 \\R _{32}=2.83 @{@} \angle 28.0^{\circ} ; & R_{32_{x}}=2.500, & R_{32_{y}}=1.333 \\R _{23}=9.00 \text { @ } \angle 324.5^{\circ} ; & R_{23_{x}}=7.329, & R_{23_{y}}=-5.224 \\R _{13}=10.72 \text { @ } \angle 63.14^{\circ} ; & R_{13_{x}}=4.843, & R_{13_{y}}=9.563 \\R _{P}=2.70 \text { @ } \angle 201.0^{\circ} ; & R_{P_{x}}=-2.521, & R_{P_{y}}=-0.968\end{array} (c)
These position vector angles are measured with respect to the LNCS which is always parallel to the global coordinate system (GCS), making the angles the same in both systems.
4 Calculate the x and y components of the acceleration of the CGs of all moving links in the global coordinate system:
\begin{array}{llll}a _{G_{2}} & =2700.17 @ \angle-89.4^{\circ} ; & a_{G_{2 x}} & =28.28, \quad a_{G_{2 y}}=-2700 \\a _{G_{3}} & =3453.35 @ \angle 254.4^{\circ} ; & a_{G_{3 x}} & =-930.82, \quad a_{G_{3 y}}=-3325.54\end{array} (d)
5 Calculate the x and y components of the external force at P in the global coordinate system:
F _{P}=50 @ \angle-45^{\circ} ; \quad F_{P_{x}}=35.36, \quad F_{P_{y}}=-35.36 (e)
6 Substitute these given and calculated values into the matrix equation 11.7.
\left[\begin{array}{cccccc}1 & 0 & 1 & 0 & 0 & 0 \\0 & 1 & 0 & 1 & 0 & 0 \\-R_{12_{y}} & R_{12_{x}} & -R_{32_{y}} & R_{32_{x}} & 0 & 1 \\0 & 0 & -1 & 0 & 1 & 0 \\0 & 0 & 0 & -1 & -\mu S G N\left(V_{31}\right) & 0 \\0 & 0 & R_{23_{y}} & -R_{23_{x}} & \left(\mu R_{13_{x}}-R_{13_{y}}\right) & 0\end{array}\right] \times\left[\begin{array}{c}F_{12_{x}} \\F_{12_{y}} \\F_{32_{x}} \\F_{32_{y}} \\F_{13_{x}} \\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}}\end{array}\right] (11.7)
\begin{aligned}&\left[\begin{array}{cccccc}1 & 0 & 1 & 0 & 0 & 0 \\0 & 1 & 0 & 1 & 0 & 0 \\3 & 0 & -1.333 & 2.5 & 0 & 1 \\0 & 0 & -1 & 0 & 1 & 0 \\0 & 0 & 0 & -1 & -0.2 & 0 \\0 & 0 & -5.224 & -7.329 & {[(0.2) 4.843-(9.563)]} & 0\end{array}\right] \times\left[\begin{array}{c}F_{12_{x}} \\F_{12_{y}} \\F_{32_{x}} \\F_{32_{y}} \\F_{13_{x}} \\T_{12}\end{array}\right]=\\&\left[\begin{array}{c}(0.005)(28.28) \\(0.005)(-2700) \\(0.05)(-10) \\(0.01)(-930.82)-35.36 \\(0.01)(-3325.54)-(-35.36) \\(0.1)(-136.16)-(-2.521)(-35.36)+(-0.968)(35.36)\end{array}\right]=\left[\begin{array}{r}0.141 \\-13.500 \\-0.500 \\-44.668 \\2.105 \\-136.987\end{array}\right]\end{aligned} (f)
7 Solve this system either by inverting matrix A and premultiplying that inverse times matrix C using a pocket calculator with matrix capability; using Mathcad or Matlab; 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}{l}F_{12_{x}} \\F_{12_{y}} \\F_{32_{x}} \\F_{32_{y}} \\F_{13_{x}} \\T_{12}\end{array}\right]=\left[\begin{array}{r}-39.232 \\-10.336 \\39.373 \\-3.164 \\-5.295 \\177.590\end{array}\right] (g)
Converting the forces to polar coordinates:
\begin{aligned}& F _{12}=40.57 lb @ \angle 194.76^{\circ} \\& F _{32}=39.50 lb @ \angle-4.60^{\circ} \\& F _{13}=5.40 lb @ \angle 191.31^{\circ}\end{aligned} (h)