Bicycle Brake Arm Loading Analysis Problem: Determine the forces acting in three dimensions on the bicycle brake arm in its actuated position as shown in Figure 3-9. This brake arm has been failing in service and may need to be redesigned. Given: The existing brake arm geometry is known and the arm

Question

Bicycle Brake Arm Loading Analysis

Problem: Determine the forces acting in three dimensions on the bicycle brake arm in its actuated position as shown in Figure 3-9. This brake arm has been failing in service and may need to be redesigned.

Given: The existing brake arm geometry is known and the arm is acted on by a cable force of 1 046 N in the position shown. (See also Case Study 1A.)

Assumptions: The accelerations are negligible. A Class 1 load model is appropriate and a static analysis is acceptable. The coefficient of friction between the brake pad and wheel rim has been measured and is 0.45 at room temperature and 0.40 at 150°F.

Accepted Answer

See Figures 3-9 and 3-10, and Table 3-5.

1 Figure 3-9 shows a center-pull brake arm assembly commonly used on bicycles. It consists of six elements or subassemblies, the frame and its pivot pins (1), the two brake arms (2 and 4), the cable spreader assembly (3), the brake pads (5), and the wheel rim (6). This is clearly a three-dimensional device and must be analyzed as such.

2 The cable is the same one that is attached to the brake lever in Figure 3-1. The 267-N (60-lb) hand force is multiplied by the mechanical advantage of the hand lever and transmitted via this cable to the pair of brake arms as was calculated in Case Study 1A. We will assume no loss of force in the cable guides, thus the full 1 046-N (235-lb) cable force is available at this end.

3 The direction of the normal force between the brake pad and the wheel rim is shown in Figure 3-9 to be at θ = 172° with respect to the positive [latex]x[/latex] axis, and the friction force is directed along the [latex]z[/latex] axis. (See Figures 3-9 and 3-10 for the [latex]xyz[/latex] axis orientations.)

4 Figure 3-10 shows free-body diagrams of the arm, frame, and cable spreader assembly. We are principally interested in the forces acting on the brake arm. However, we first need to analyze the effect of the cable spreader geometry on the force applied to the arm at [latex]A[/latex]. This analysis can be two dimensional if we ignore the small [latex]z[/latex]-offset between the two arms for simplicity. A more accurate analysis would require that the [latex]z[/latex]-directed components of the cable-spreader forces acting on the arms be included. Note that the cable subassembly (3) is a concurrent force system. Writing equations 3.3b in two dimensions for this subassembly, and noting the symmetry about point A, we can write from inspection of the FBD:

[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_{23 x}+F_{43 x}=0 \ \sum F_y=F_{23 y}+F_{43 y}+F_{ ext {cable }}=0 \end{array} [/latex] (a)

This equation set can be easily solved to yield

[latex] \begin{array}{l} F_{23 y}=F_{43 y}=-\frac{F_{ ext {cable }}}{2}=-\frac{1046}{2}=-523 N \ F_{23 x}=\frac{F_{23 y}}{ an \left(56^{\circ} ight)}=\frac{-523}{1.483}=-353 N \ F_{43 x}=-F_{23 x}=353 N \end{array} [/latex] (b)

Newton’s third law relates these forces to their reactions on the brake arm at point [latex]A[/latex]:

[latex] \begin{array}{l} F_{32 x}=-F_{23 x}=353 N \ F_{32 y}=-F_{23 y}=523 N \ F_{32 z}=0 \end{array} [/latex] (c)

5 We can now write equations 3.3a for the arm (link 2).

[latex] \begin{array}{lll} \sum F_x=0 & \sum F_y=0 & \sum F_z=0 \ \sum M_x=0 & \sum M_y=0 & \sum M_z=0 \end{array} [/latex] (3.3a)

For the forces:

[latex] \begin{array}{l} \sum F_x=F_{12 x}+F_{32 x}+F_{52 x}=0 ; \quad F_{12 x}+F_{52 x}=-353\ \sum F_y=F_{12 y}+F_{32 y}+F_{52 y}=0 ; \quad F_{12 y}+F_{52 y}=-523\ \sum F_z=F_{12 z}+F_{32 z}+F_{52 z}=0 ; \quad F_{12 z}+F_{52 z}=0 \end{array} [/latex] (d)

For the moments:

[latex] \begin{aligned} \sum M_x=M_{12 x}+\left(R_{12 y} F_{12 z}-R_{12 z} F_{12 y} ight) &+\left(R_{32 y} F_{32 z}-R_{32 z} F_{32 y} ight) \ &+\left(R_{52 y} F_{52 z}-R_{52 z} F_{52 y} ight)=0 \end{aligned} \ \begin{aligned} \sum M_y=M_{12 y}+\left(R_{12 z} F_{12 x}-R_{12 x} F_{12 z} ight) &+\left(R_{32 z} F_{32 x}-R_{32 x} F_{32 z} ight) \ &+\left(R_{52 z} F_{52 x}-R_{52 x} F_{52 z} ight)=0 \end{aligned} \ \begin{aligned} \sum M_z=\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) \ &+\left(R_{52 x} F_{52 y}-R_{52 y} F_{52 x} ight)=0 \end{aligned} [/latex] (e)

Note that all unknown forces and moments are initially assumed positive in the equations, regardless of their apparent directions on the FBDs. The moments [latex]M_{12x}[/latex] and [latex]M_{12y}[/latex] are due to the fact that there is a moment joint between the arm (2) and the pivot pin (1) about the [latex]x[/latex] and [latex]y[/latex] axes. We assume negligible friction about the [latex]z[/latex] axis, thus allowing [latex]M_{12z}[/latex] to be zero.

6 The joint between the brake pad (5) and the wheel rim (6) transmits a force normal to the plane of contact. The friction force magnitude, [latex]F_{f}[/latex], in the contact plane is related to the normal force by the Coulomb friction equation,

[latex] F_f=\mu N [/latex] (f)

where μ is the coefficient of friction and [latex]N[/latex] is the normal force. The velocity of the point on the rim below the center of the brake pad is in the [latex]z[/latex] direction. The force components [latex]F_{52x}[/latex] and [latex]F_{52y}[/latex] are due entirely to the normal force being transmitted through the pad to the arm and are therefore related by Newton’s third law.

[latex] \begin{array}{l} F_{52 x}=-N_x=-N \cos heta=-N \cos 172^{\circ}=0.990 N \ F_{52 y}=-N_y=-N \sin heta=-N \sin 172^{\circ}=-0.139 N \end{array} [/latex] (g)

The direction of the friction force [latex]F_{f}[/latex] must always oppose motion and thus it acts in the negative [latex]z[/latex] direction on the wheel rim. Its reaction force on the arm has the opposite sense.

[latex] F_{52 z}=-F_f [/latex] (h)

7 We now have 10 equations (in the sets labeled [latex]d, e, f, g[/latex], and [latex]h[/latex]) containing 10 unknowns:

[latex] F_{12 x}, F_{12 y}, F_{12 z}, F_{52 x}, F_{52 y}, F_{52 z}, M_{12 x}, M_{12 y}, N ext {, and } F_f ext {. Forces } F_{32 x}, F_{32 y} ext {, and } F_{32 z} [/latex] are known from equations [latex]c[/latex]. These 10 equations can be solved simultaneously either by matrix reduction or by iterative root-finding methods. First arrange the equations with all unknowns on the left and all known or assumed values on the right.

[latex] \begin{aligned} F_{12 x}+F_{52 x} &=-353 \ F_{12 y}+F_{52 y} &=-523 \ F_{12 z}+F_{52 z} &=0 \ M_{12 x}+R_{12 y} F_{12 z}-R_{12 z} F_{12 y}+R_{52 y} F_{52 z}-R_{52 z} F_{52 y} &=R_{32 z} F_{32 y}-R_{32 y} F_{32 z} \ M_{12 y}+R_{12 z} F_{12 x}-R_{12 x} F_{12 z}+R_{52 z} F_{52 x}-R_{52 x} F_{52 z} &=R_{32 x} F_{32 z}-R_{32 z} F_{32 x} \ R_{12 x} F_{12 y}-R_{12 y} F_{12 x}+R_{52 x} F_{52 y}-R_{52 y} F_{52 x} &=R_{32 y} F_{32 x}-R_{32 x} F_{32 y} \ F_f-\mu N &=0 \ F_{52 x}+N \cos heta &=0 \ F_{52 y}+N \sin heta &=0 \ F_{52 z}+F_f &=0 \end{aligned} [/latex] (i)

8 Substitute the known and assumed values from Table 3-5 part 1:

[latex] \begin{aligned} F_{12 x}+F_{52 x} &=-353 \ F_{12 y}+F_{52 y} &=-523 \ F_{12 z}+F_{52 z} &=0 \ M_{12 x}-27.2 F_{12 z}-23.1 F_{12 y}-69.7 F_{52 z}-0 F_{52 y} &=0(523)-38.7(0)=0 \ M_{12 y}+23.1 F_{12 x}-5.2 F_{12 z}+0 F_{52 x}+13 F_{52 z} &=-75.4(0)-0(353)=0 \ 5.2 F_{12 y}+27.2 F_{12 x}-13 F_{52 y}+69.7 F_{52 x} &=38.7(353)+75.4(523)=53095 \ F_f-0.4 N &=0 \ F_{52 x}-0.990 N &=0 \ F_{52 y}+0.139 N &=0 \ F_{52 z}+F_f &=0 \end{aligned} [/latex] (j)

9 Form the matrices for solution.

[latex] \left[\begin{array}{cccccccccc} 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & -23.1 & -27.2 & 0 & 0 & -69.7 & 1 & 0 & 0 & 0 \ 23.1 & 0 & -5.2 & 0 & 0 & 13 & 0 & 1 & 0 & 0 \ 27.2 & 5.2 & 0 & 69.7 & -13 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -0.4 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -0.990 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0.139 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \end{array} ight] imes \left[\begin{array}{c} F_{12 x} \ F_{12 y} \ F_{12 z} \ F_{52 x} \ F_{52 y} \ F_{52 z} \ M_{12 x} \ M_{12 y} \ F_f \ N \end{array} ight]=\left[\begin{array}{c} -353 \ -523 \ 0 \ 0 \ 0 \ 53095 \ 0 \ 0 \ 0 \ 0 \end{array} ight] [/latex] (k)

10 Table 3-5 part 2 shows the solution to this problem from program MATRIX for the given data in Table 3-5 part 1. This problem can be solved using any one of several commercial equation-solver programs such as Mathcad, MATLAB, Maple, Mathematica or with the program MATRIX provided with the text.

Table 3–5 - part 1 Case Study 4A Given and Assumed Data
Variable	Value	Unit
[latex] \mu [/latex]	0.4	none
[latex] heta [/latex]	172.0	deg
[latex] R_{12x} [/latex]	5.2	mm
[latex] R_{12y} [/latex]	-27.2	mm
[latex] R_{12z} [/latex]	23.1	mm
[latex] R_{32x} [/latex]	–75.4	mm
[latex] R_{32y} [/latex]	38.7	mm
[latex] R_{32z} [/latex]	0.0	mm
[latex] R_{52x} [/latex]	-13.0	mm
[latex] R_{52y} [/latex]	–69.7	mm
[latex] R_{52z} [/latex]	0.0	mm
[latex] F_{32x} [/latex]	353.0	N
[latex] F_{32y} [/latex]	523.0	N
[latex] F_{32z} [/latex]	0.0	N
[latex] M_{12z} [/latex]	0.0	N–m

Table 3–5 - part 2 Case Study 4A Calculated Data
Variable	Value	Unit
[latex] F_{12x} [/latex]	–1 805	N
[latex] F_{12y} [/latex]	–319	N
[latex] F_{12z} [/latex]	587.0	N
[latex] F_{52x} [/latex]	1 452	N
[latex] F_{52y} [/latex]	–204	N
[latex] F_{52z} [/latex]	-587	N
[latex] M_{12x} [/latex]	32 304	N–mm
[latex] M_{12y} [/latex]	52 370	N–mm
[latex] N [/latex]	1 467	N
[latex] F_{f} [/latex]	587	N

Question 3.CS.4A: Bicycle Brake Arm Loading Analysis Problem: Determine the fo...

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