Hand-Operated Crimping-Tool Loading Analysis Problem: Determine the forces on the elements of the crimping tool shown in Figure 3-3 during a crimp operation. Given: The geometry is known and the tool develops a crimp force of 2 000 lb (8 896 N) at closure in the position shown. Assumptions: The

Question

Hand-Operated Crimping-Tool Loading Analysis

Problem: Determine the forces on the elements of the crimping tool shown in Figure 3-3 during a crimp operation.

Given: The geometry is known and the tool develops a crimp force of 2 000 lb (8 896 N) at closure in the position shown.

Assumptions: The accelerations are negligible. All forces are coplanar and two dimensional. A Class 1 load model is appropriate and a static analysis acceptable.

Accepted Answer

See Figures 3-3 and 3-4, and Table 3-3, parts 1 and 21 Figure 3-3 shows the tool in the closed position in the process of crimping a metal connector onto a wire. The user’s hand provides the input forces between links 1 and 2, shown as the reaction pair [latex]F_{h}[/latex]. The user can grip the handle anywhere along its length, but we are assuming a nominal moment arm of [latex]R_{h}[/latex] for the application of the resultant of the user’s grip force (see Figure 3-4). The high mechanical advantage of the tool transforms the grip force to a large force at the crimp.

Figure 3-3 is a free-body diagram of the entire assembly, neglecting the weight of the tool which is small compared to the crimp force. There are four elements, or links, in the assembly, all pinned together. Link 1 can be considered to be the “ground” link, with the other links moving with respect to it as the jaw is closed. The desired magnitude of the crimp force [latex]F_{c}[/latex] is defined and its direction will be normal to the surfaces at the crimp. The third law relates the action-reaction pair acting on links 1 and 4:

[latex] \begin{array}{l} F_{c 1 x}=-F_{c 4 x} \ F_{c 1 y}=-F_{c 4 y} \end{array} [/latex] (a)

2 Figure 3-4 shows the elements of the crimping-tool assembly separated and drawn as free-body diagrams with all forces applied to each element, again neglecting their weights as being insignificant compared to the applied forces. 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.*

3 We will consider link 1 to be the ground plane and analyze the remaining moving links. Note that all unknown forces and moments are initially assumed positive. Link 2 has three forces acting on it: [latex]F_{h}[/latex] is the unknown force from the hand, and [latex]F_{12}[/latex] and [latex]F_{32}[/latex] are the reaction forces from links 1 and 3, respectively. Force [latex]F_{12}[/latex] is provided by part 1 on part 2 at the pivot pin and force [latex]F_{32}[/latex] is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and directions of both these pin forces are unknown. We can write equations [latex]3.3b[/latex] for this element to sum forces in the [latex]x[/latex] and [latex]y[/latex] directions and sum moments about the CG (with cross products expanded).

[latex] \sum F_x=0 \quad \sum F_y=0 \quad \sum M_z=0 [/latex] (3.3b)

[latex] \begin{array}{l} \sum F_x=F_{12 x}+F_{32 x}=0 \ \sum F_y=F_{12 y}+F_{32 y}+F_h=0 \ \sum M_z=F_h R_h+\left(R_{12 x} F_{12 y}-R_{12 y} F_{12 x} ight)+\left(R_{32 x} F_{32 y}-R_{32 y} F_{32 x} ight)=0 \end{array} [/latex] (b)

4 Link 3 has two forces on it, [latex]F_{23}[/latex] and [latex]F_{43}[/latex]. Write equations [latex]3.3b[/latex] for this element:

[latex] \begin{aligned} \sum F_x &=F_{23 x}+F_{43 x}=0 \ \sum F_y &=F_{23 y}+F_{43 y}=0 \ \sum M_z &=\left(R_{23 x} F_{23 y}-R_{23 y} F_{23 x} ight)+\left(R_{43 x} F_{43 y}-R_{43 y} F_{43 x} ight)=0 \end{aligned} [/latex] (c)

5 Link 4 has three forces acting on it: [latex]F_{c4}[/latex] is the known (desired) force at the crimp, and [latex]F_{14}[/latex] and [latex]F_{34}[/latex] are the reaction forces from links 1 and 3, respectively. The magnitudes and directions of both these pin forces are unknown. Write equations [latex]3.3b[/latex] for this element:

[latex] \begin{aligned} \sum F_x &=F_{14 x}+F_{34 x}+F_{c 4 x}=0 \ \sum F_y &=F_{14 y}+F_{34 y}+F_{c 4 y}=0 \ \sum M_z &=\left(R_{14 x} F_{14 y}-R_{14 y} F_{14 x} ight)+\left(R_{34 x} F_{34 y}-R_{34 y} F_{34 x} ight) \ & +\left(R_{c 4 x} F_{c 4 y}-R_{c 4 y} F_{c 4 x} ight)=0 \end{aligned} [/latex] (d)

6 The 9 equations in sets b through d have 13 unknowns: [latex]F_{12x}, F_{12y}, F_{32x}, F_{32y}, F_{23x}, F_{23y}, F_{43x}, F_{43y}, F_{14x}, F_{14y}, F_{34x}, F_{34y}[/latex], and [latex]F_{h}[/latex]. We can write the third-law relationships between action-reaction pairs at each of the joints to obtain the four additional equations needed:

[latex] F_{32 x}=-F_{23 x} ; \quad F_{34 x}=-F_{43 x} ; \quad F_{32 y}=-F_{23 y} ; \quad F_{34 y}=-F_{43 y} [/latex] (e)

7 The 13 equations [latex]b–e[/latex] can be solved simultaneously by matrix reduction or by iteration with a root-finding algorithm. For a matrix solution, the unknown terms are placed on the left and known terms on the right of the equal signs.

[latex] \begin{aligned} F_{12 x}+F_{32 x} &=0 \ F_{12 y}+F_{32 y}+F_h &=0 \R_h F_h+R_{12 x} F_{12 y}-R_{12 y} F_{12 x}+R_{32 x} F_{32 y}-R_{32 y} F_{32 x} &=0 \ F_{23 x}+F_{43 x} &=0 \ F_{23 y}+F_{43 y} &=0 \ R_{23 x} F_{23 y}-R_{23 y} F_{23 x}+R_{43 x} F_{43 y}-R_{43 y} F_{43 x} &=0 \ F_{14 x}+F_{34 x} &=-F_{c 4 x} \ F_{14 y}+F_{34 y} &=-F_{c 4 y} \ R_{14 x} F_{14 y}-R_{14 y} F_{14 x}+R_{34 x} F_{34 y}-R_{34 y} F_{34 x} &=-R_{c 4 x} F_{c 4 y}+R_{c 4 y} F_{c 4 x} \ F_{32 x}+F_{23 x} &=0 \ F_{34 x}+F_{43 x} &=0 \ F_{32 y}+F_{23 y} &=0 \ F_{34 y}+F_{43 y} &=0 \end{aligned} [/latex] (f)

8 Substitute the given data from Table 3-3 part 1.

[latex] \begin{aligned} F_{12 x}+F_{32 x} &=0 \ F_{12 y}+F_{32 y}+F_h &=0 \ -4.4 F_h+1.4 F_{12 y}-0.05 F_{12 x}+2.2 F_{32 y}-0.08 F_{32 x} &=0 \ F_{23 x}+F_{43 x} &=0 \ F_{23 y}+F_{43 y} &=0 \ -0.6 F_{23 y}-0.13 F_{23 x}+0.6 F_{43 y}+0.13 F_{43 x} &=0 \ F_{14 x}+F_{34 x} &=1956.3 \ F_{14 y}+F_{34 y} &=-415.82 \ -0.16 F_{14 y}+0.76 F_{14 x}+0.16 F_{34 y}-0.76 F_{34 x} &=-0.45(415.82)-0.34(1956.3) \ &=-852.26 \ F_{32 x}+F_{23 x} &=0 \ F_{34 x}+F_{43 x} &=0 \ F_{32 y}+F_{23 y} &=0 \ F_{34 y}+F_{43 y} &=0 \end{aligned} [/latex] (g)

9 Form the matrices for solution.

[latex] \left[\begin{array}{ccccccccccccc} 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ -0.05 & 1.4 & -0.08 & 2.2 & -4.4 & 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 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & -0.13 & -0.6 & 0.13 & 0.6 & 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 & 1 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.76 & -0.16 & -0.76 & 0.16 \ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array} ight] \left[\begin{array}{c} F_{12 x} \ F_{12 y} \ F_{32 x} \ F_{32 y} \ F_h \ F_{23 x} \ F_{23 y} \ F_{43 x} \ F_{43 y} \ F_{14 x} \ F_{14 y} \ F_{34 x} \ F_{34 y} \end{array} ight] = \left[\begin{array}{c} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1956.30 \ -415.82 \ -852.26 \ 0 \ 0 \ 0 \ 0 \end{array} ight] [/latex] (h)

10 Table 3-3 part 2 shows the solution to this problem for the data given in Table 3-3 part 1, assuming a 2000-lb (8 896-N) force applied at the crimp, normal to the crimp surface. Program MATRIX (on the CD-ROM) was used. The force generated in link 3 is 1 547 lb (6 888 N), the reaction force against link 1 by link 2 ([latex]F_{21}[/latex]) is 1 561 lb (6 943 N) at 166°, the reaction force against link 1 by link 4 ([latex]F_{41}[/latex]) is 451 lb (2 008 N) at 169°, and a –233.5 lb-in (–26.6-N-m) moment must be applied to the handles to generate the specified crimp force. This moment can be obtained with a 53.1-lb (236-N) force applied at mid-handle. This force is within the physiological grip-force capacity of the average human.

* Again, in a static analysis it is not necessary to take the CG as the coordinate system origin (any point can be used), but we do so to be consistent with the dynamic analysis approach in which it is quite useful to do so.

Table 3–3 - part 1 Case Study 2A Given Data
Variable	Value	Unit
[latex]F_{c4x}[/latex]	–1 956.30	lb
[latex]F_{c4y}[/latex]	415.82	lb
[latex]R_{c4x}[/latex]	0.45	in
[latex]R_{c4y}[/latex]	0.34	in
[latex]R_{12x}[/latex]	1.40	in
[latex]R_{12y}[/latex]	0.05	in
[latex]R_{32x}[/latex]	2.20	in
[latex]R_{32y}[/latex]	0.08	in
[latex]R_{h}[/latex]	–4.40	in
[latex]R_{23x}[/latex]	–0.60	in
[latex]R_{23y}[/latex]	0.13	in
[latex]R_{43x}[/latex]	0.60	in
[latex]R_{43y}[/latex]	–0.13	in
[latex]R_{14x}[/latex]	–0.16	in
[latex]R_{14y}[/latex]	–0.76	in
[latex]R_{34x}[/latex]	0.16	in
[latex]R_{34y}[/latex]	0.76	in

Table 3–3 - part 2 Case Study 2A Calculated Data
Variable	Value	Unit
[latex]F_{h}[/latex]	53.1	lb
[latex]F_{12x}[/latex]	1 513.6	lb
[latex]F_{12y}[/latex]	–381.0	lb
[latex]F_{32x}[/latex]	–1 513.6	lb
[latex]F_{32y}[/latex]	327.9	lb
[latex]F_{43x}[/latex]	–1 513.6	lb
[latex]F_{43y}[/latex]	327.9	lb
[latex]F_{23x}[/latex]	1 513.6	lb
[latex]F_{23y}[/latex]	–327.9	lb
[latex]F_{34x}[/latex]	1 513.6	lb
[latex]F_{34y}[/latex]	–327.9	lb
[latex]F_{14x}[/latex]	442.7	lb
[latex]F_{14y}[/latex]	–87.9	lb
- - - - - - - - - -
[latex]F_{21x}[/latex]	–1 513.6	lb
[latex]F_{21y}[/latex]	381.0	lb
[latex]F_{41x}[/latex]	–442.7	lb
[latex]F_{41y}[/latex]	87.9	lb

Question 3.CS.2A: Hand-Operated Crimping-Tool Loading Analysis Problem: Determ...

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