Question 3.CS.3A: Automobile Scissors-Jack Loading Analysis Problem: Determine...

See Figures 3-5 through 3-8 and Table 3-4, parts 1 and 2.1    Figure 3-5 shows a schematic of a simple scissors jack used to raise a car. It consists of six links which are pivoted and/or geared together and a seventh link in the form of a lead screw which is turned to raise the jack. While this is clearly a three-dimensional device, it can be analyzed as a two-dimensional one if we assume that the applied load (from the car) and the jack are exactly vertical (in the z direction). If so, all forces will be in the xy plane. This assumption is valid if the car is jacked from a level surface. If not, then there will be some forces in the yz and xz planes as well. The jack designer needs to consider the more general case, but for our simple example we will initially assume two-dimensional loading. For the overall assembly as shown in Figure 3-5, we can solve for the reaction force F_{g}, given force P, by summing forces: F_{g} = –P.

2    Figure 3-6 shows a set of free-body diagrams for the entire jack. Each element or subassembly of interest has been separated from the others and the forces and moments shown acting on it (except for its weight, which is small compared to the applied forces and is thus neglected for this analysis). The forces and moments can be either internal reactions at interconnections with other elements or external loads from the “outside world.” The centers of gravity of the respective elements are used as the origins of the local, nonrotating coordinate systems in which the points of application of all forces on the elements are located. In this design, stability is achieved by the mating of two pairs of crude (noninvolute) gear segments acting between links 2 and 4 and between links 5 and 7. These interactions are modeled as forces acting along a common normal shared by the two teeth. This common normal is perpendicular to the common tangent at the contact point.

There are 3 second-law equations available for each of the seven elements, allowing 21 unknowns. An additional 10 third-law equations will be needed for a total of 31. This is a cumbersome system to solve for such a simple device, but we can use its symmetry to advantage in order to simplify the problem.

3    Figure 3-7 shows the upper half of the jack assembly. Because of the mirror symmetry between the upper and lower portions, the lower half can be removed to simplify the analysis. The forces calculated for this half will be duplicated in the other. If we wished, we could solve for the reaction forces at A and B using equations 3.3b from this free-body diagram of the half-jack assembly.

\sum F_x=0 \quad \sum F_y=0 \quad \sum M_z=0     (3.3b)

4    Figure platex]3-8a[/latex] shows the free-body diagrams for the upper half of the jack assembly which are essentially the same as those of Figure 3-6. We now have four elements but can consider the subassembly labeled 1 to be the “ground,” leaving three elements on which to apply equations 3.3. Note that all unknown forces and moments are initially assumed positive in the equations.

5    Link 2 has three forces acting on it: F_{42} is the unknown force at the gear tooth contact with link 4; F_{12} and F_{32} are the unknown reaction forces from links 1 and 3 respectively. Force F_{12} is provided by part 1 on part 2 at the pivot pin, and force F_{32} is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and directions of these pin forces and the magnitude of F_{42} are unknown. The direction of F_{42} is along the common normal shown in Figure 3-8b. Write equations 3.3b for this element to sum forces in the x and y directions and sum moments about the CG (with the cross products expanded):*

\begin{array}{l} \sum F_x=F_{12 x}+F_{32 x}+F_{42 x}=0 \\ \sum F_y=F_{12 y}+F_{32 y}+F_{42 y}=0 \\ \sum M_z=R_{12 x} F_{12 y}-R_{12 y} F_{12 x}+R_{32 x} F_{32 y}-R_{32 y} F_{32 x}+R_{42 x} F_{42 y}-R_{42 y} F_{42 x}=0 \end{array}        (a)

6    Link 3 has three forces acting on it: the applied load P, F_{23}, and F_{43}. Only P is known. Writing equations 3.3b for this element gives

\begin{array}{l} \sum F_x=F_{23 x}+F_{43 x}+P_x=0 \\ \sum F_y=F_{23 y}+F_{43 y}+P_y=0 \\ \sum M_z=R_{23 x} F_{23 y}-R_{23 y} F_{23 x}+R_{43 x} F_{43 y}-R_{43 y} F_{43 x}+R_{P x} P_y-R_{P y} P_x=0 \end{array}      (b)

7    Link 4 has three forces acting on it: F_{24} is the unknown force from link 2; F_{14} and F_{34} are the unknown reaction forces from links 1 and 3 respectively.

\begin{aligned} \sum F_x &=F_{14 x}+F_{24 x}+F_{34 x}=0 \\ \sum F_y &=F_{14 y}+F_{24 y}+F_{34 y}=0 \\ \sum M_z &=R_{14 x} F_{14 y} -R_{14 y} F_{14 x}+R_{24 x} F_{24 y}-R_{24 y} F_{24 x}+R_{34 x} F_{34 y}-R_{34 y} F_{34 x}=0 \end{aligned}       (c)

8    The nine equations in sets a through c have 16 unknowns in them, F_{12x}, F_{12y}, F_{32x}, F_{32y}, F_{23x}, F_{23y}, F_{43x}, F_{43y}, F_{14x}, F_{14y}, F_{34x}, F_{34y}, F_{24x}, F_{24y}, F_{42x}, and F_{42y} . We can write the third-law relationships between action-reaction pairs at each of the joints to obtain six of the seven additional equations needed:

\begin{array}{ll} F_{32 x}=-F_{23 x} & F_{32 y}=-F_{23 y} \\ F_{34 x}=-F_{43 x} & F_{34 y}=-F_{43 y} \\ F_{42 x}=-F_{24 x} & F_{42 y}=-F_{24 y} \end{array}     (d)

9     The last equation needed comes from the relationship between the x and y components of the force F_{24} (or F_{42}) at the tooth/tooth contact point. Such a contact (or half) joint can transmit force (excepting friction force) only along the common normal^{[4]} which is perpendicular to the joint’s common tangent as shown in Figure 3-8b. The common normal is also called the axis of transmission. The tangent of the angle of this common normal relates the two components of the force at the joint:

F_{24 y}=F_{24 x} \tan \theta        (e)

10    Equations (a) through (e) comprise a set of 16 simultaneous equations that can be solved either by matrix reduction or by iterative root-finding methods. Putting them in standard form for a matrix solution gives:

\begin{aligned} F_{12 x}+F_{32 x}+F_{42 x} &=0 \\ F_{12 y}+F_{32 y}+F_{42 y} &=0 \\ R_{12 x} F_{12 y}-R_{12 y} F_{12 x}+R_{32 x} F_{32 y}-R_{32 y} F_{32 x}+R_{42 x} F_{42 y}-R_{42 y} F_{42 x} &=0 \\ F_{23 x}+F_{43 x} &=-P_x \\ F_{23 y}+F_{43 y} &=-P_y \\ R_{23 x} F_{23 y}-R_{23 y} F_{23 x}+R_{43 x} F_{43 y}-R_{43 y} F_{43 x} &=-R_{P x} P_y+R_{P y} P_x \\ F_{14 x}+F_{24 x}+F_{34 x} &=0 \\ F_{14 y}+F_{24 y}+F_{34 y} &=0 \\ R_{14 x} F_{14 y}-R_{14 y} F_{14 x}+R_{24 x} F_{24 y}-R_{24 y} F_{24 x}+R_{34 x} F_{34 y}-R_{34 y} F_{34 x} &=0 \\ F_{32 x}+F_{23 x} &=0 \\ F_{32 y}+F_{23 y} &=0 \\ F_{34 x}+F_{43 x} &=0 \\ F_{34 y}+F_{43 y} &=0 \\ F_{42 x}+F_{24 x} &=0 \\ F_{42 y}+F_{24 y} &=0 \\ F_{24 y}-F_{24 x} \tan \theta &=0 \end{aligned}      (f)

11    Substituting the given data from Table 3-4 part 1 gives:

\begin{aligned} F_{12 x}+F_{32 x}+F_{42 x} &=0 \\ F_{12 y}+F_{32 y}+F_{42 y} &=0 \\ -3.12 F_{12 y}+1.80 F_{12 x}+2.08 F_{32 y}-1.20 F_{32 x}+2.71 F_{42 y}-0.99 F_{42 x} &=0 \\ F_{23 x}+F_{43 x} &=0.0 \\ F_{23 y}+F_{43 y} &=1000 \\ -0.78 F_{23 y}+0.78 F_{23 x}+0.78 F_{43 y}+0.78 F_{43 x} &=-500 \\ F_{14 x}+F_{24 x}+F_{34 x} &=0 \\ F_{14 y}+F_{24 y}+F_{34 y} &=0 \\ 3.12 F_{14 y}+1.80 F_{14 x}-2.58 F_{24 y}-1.04 F_{24 x}-2.08 F_{34 y}-1.20 F_{34 x} &=0 \\ F_{32 x}+F_{23 x} &=0 \\ F_{32 y}+F_{23 y} &=0 \\ F_{34 x}+F_{43 x} &=0 \\ F_{34 y}+F_{43 y} &=0 \\ F_{42 x}+F_{24 x} &=0 \\ F_{42 y}+F_{24 y} &=0 \\ F_{24 y}+1.0 F_{24 x} &=0 \end{aligned}      (g)

12    Put these equations into matrix form.

\begin{bmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1.80 & -3.12 & -1.20 & 2.08 & -1.00 & 2.71 & 0 & 0 & 0& 0& 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0& 0& 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0.78 & -0.78 & 0.78 & 0.78 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1.80 & 3.12 & -1.04 & -2.58 & -1.20 & -2.08 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \end{bmatrix}  \times \left[\begin{array}{l} F_{12 x} \\ F_{12 y} \\ F_{32 x} \\ F_{32 y} \\ F_{42 x} \\ F_{42 y} \\ F_{23 x} \\ F_{23 y} \\ F_{43 x} \\ F_{43 y} \\ F_{14 x} \\ F_{14 y} \\ F_{24 x} \\ F_{24 y} \\ F_{34 x} \\ F_{34 y} \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 1000 \\ -500 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]    (h)

13    Table 3-4 part 2 shows the solution to this problem from program MATRIX for the given data in Table 3-4 part 1, which assumes a vertical 1 000-lb (4 448-N) applied force P.

14    The forces on link 1 can also be found from Newton’s third law.

\begin{array}{l} F_{A x}=-F_{21 x}=F_{12 x} \\ F_{A y}=-F_{21 y}=F_{12 y} \\ F_{B x}=-F_{41 x}=F_{14 x} \\ F_{B y}=-F_{41 y}=F_{14 y} \end{array}       (i)

* Note the similarity to equations (b) in Case Study 2A. Only the subscript for the reaction moment is different because a different link is providing it. The consistent notation of this force analysis method makes it easy to write the equations for any system.

Table 3–4 – part 1 Case Study 3A Given Data
Variable	Value	Unit
P_x	0.00	lb
P_y	–1 000.00	lb
R_{p x}	–0.50	in
R_{p y}	0.87	in
\theta	–45.00	deg
R_{12 x}	–3.12	in
R_{12 y}	–1.80	in
R_{32 x}	2.08	in
R_{32 y}	1.20	in
R_{42 x}	2.71	in
R_{42 y}	1.00	in
R_{23 x}	–0.78	in
R_{23 y}	–0.78	in
R_{43 x}	0.78	in
R_{43 y}	–0.78	in
R_{14 x}	3.12	in
R_{14 y}	–1.80	in
R_{24 x}	–2.58	in
R_{24 y}	1.04	in
R_{34 x}	–2.08	in
R_{34 y}	1.2	in

Table 3–4 – part 2 Case Study 3A Calculated Data
Variable	Value	Value
F_{12 x}	877.8	lb
F_{12 y}	530.4	lb
F_{32 x}	–587.7	lb
F_{32 y}	–820.5	lb
F_{42 x}	–290.1	lb
F_{42 y}	290.1	lb
F_{23 x}	587.7	lb
F_{23 y}	820.5	lb
F_{43 x}	–587.7	lb
F_{43 y}	179.5	lb
F_{14 x}	–877.8	lb
F_{14 y}	469.6	lb
F_{24 x}	290.1	lb
F_{24 y}	–290.1	lb
F_{34 x}	587.7	lb
F_{34 y}	–179.5	lb