Bicycle Brake Lever Loading Analysis Problem: Determine the forces on the elements of the bicycle brake lever assembly shown in Figure 3-1 during braking. Given: The geometry of each element is known. The average human’s hand can develop a grip force of about 267 N (60 lb) in the lever position

Question

Bicycle Brake Lever Loading Analysis

Problem: Determine the forces on the elements of the bicycle brake lever assembly shown in Figure 3-1 during braking.

Given: The geometry of each element is known. The average human’s hand can develop a grip force of about 267 N (60 lb) in the lever 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 is acceptable.

Accepted Answer

See Figures 3-1, 3-2, and Table 3-2, parts 1 and 2.1 Figure 3-1 shows the handbrake lever assembly, which consists of three subassemblies: the handlebar (1), the lever (2), and the cable (3). The lever is pivoted to the handlebar and the cable is connected to the lever. The cable runs within a plastic-lined sheath (for low friction) down to the brake caliper assembly at the bicycle’s wheel rim. The sheath provides a compressive force to balance the tension in the cable ([latex]F_{sheath} = –F_{cable}[/latex]). The user’s hand applies equal and opposite forces at some points on the lever and handgrip. These forces are transformed to a larger force in the cable by the lever ratio of part 2.

Figure 3-1 is a free-body diagram of the entire assembly since it shows all the forces and moments potentially acting on it except for its weight, which is small compared to the applied forces and is thus neglected for this analysis. The “broken away” portion of the handlebar can provide [latex]x[/latex] and y force components and a moment if required for equilibrium. These reaction forces and moments are arbitrarily shown as positive in sign. Their actual signs will “come out in the wash” in the calculations. The known applied forces are shown acting in their actual directions and senses.

2 Figure 3-2 shows the three subassembly elements separated and drawn as free body diagrams with all relevant forces and moments applied to each element, again neglecting the weights of the parts. The lever (part 2) has three forces on it, [latex]F_{b2}, F_{32}[/latex], and [latex]F_{12}[/latex]. The two-character subscript notation used here should be read as, force of element 1 on 2 ([latex]F_{12}[/latex]) or force at [latex]B[/latex] on 2 ([latex]F_{b2}[/latex]), etc. This defines the source of the force (first subscript) and the element on which it acts (second subscript).

This notation will be used consistently throughout this text for both forces and position vectors such as [latex]R_{b2}, R_{32}[/latex], and [latex]R_{12}[/latex] in Figure 3-2 which serve to locate the above three forces in a local, nonrotating coordinate system whose origin is at the center of gravity (CG) of the element or subassembly being analyzed.*

On this brake lever, [latex]F_{b2}[/latex] is an applied force whose magnitude and direction are known. [latex]F_{32}[/latex] is the force in the cable. Its direction is known but not its magnitude. Force [latex]F_{12}[/latex] is provided by part 1 on part 2 at the pivot pin. Its magnitude and direction are both unknown. We can write equations 3.3[latex]b[/latex] for this element to sum forces in the [latex]x[/latex] and [latex]y[/latex] directions and sum moments about the CG. Note that all unknown forces and moments are initially assumed positive in the equations. Their true signs will come out in the calculation.† However, all known or given forces must carry their proper signs.

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

[latex] \begin{aligned} \sum F_x &=F_{12 x}+F_{b 2 x}+F_{32 x}=0 \ \sum F_y &=F_{12 y}+F_{b 2 y}+F_{32 y}=0 \ \sum M _z &=\left( R _{12} imes F _{12} ight)+\left( R _{b 2} imes F _{b 2} ight)+\left( R _{32} imes F _{32} ight)=0 \end{aligned} [/latex] (a)

The cross products in the moment equation represent the “turning forces” or moments created by the application of these forces at points remote from the CG of the element. Recall that these cross products can be expanded to

[latex] \begin{aligned} \sum M_z=\left(R_{12 x} F_{12 y}-R_{12 y} F_{12 x} ight)+(&\left.R_{b 2 x} F_{b 2 y}-R_{b 2 y} F_{b 2 x} ight) \ &+\left(R_{32 x} F_{32 y}-R_{32 y} F_{32 x} ight)=0 \end{aligned} [/latex] (b)

We have three equations and four unknowns ([latex]F_{12x}, F_{12y}, F_{32x}, F_{32y}[/latex]) at this point, so we need another equation. It is available from the fact that the direction of [latex]F_{32}[/latex] is known. (The cable can pull only along its axis.) We can express one component of the cable force [latex]F_{32}[/latex] in terms of its other component and the known angle, [latex] heta[/latex] of the cable.

[latex] F_{32 y}=F_{32 x} an heta [/latex] (c)

We could now solve the four unknowns for this element, but will wait to do so until the equations for the other two links are defined.

Part 3 in Figure 3-2 is the cable which passes through a hole in part 1. This hole is lined with a low-friction material which allows us to assume no friction at the joint between parts 1 and 3. We will further assume that the three forces [latex]F_{13}, F_{23}[/latex], and [latex]F_{cable}[/latex] form a concurrent system of forces acting through the CG and thus create no moment. With this assumption only a summation of forces is necessary for this element.

[latex] \begin{array}{l} \sum F_x=F_{ ext {cable }_x}+F_{13 x}+F_{23 x}=0 \ \sum F_y=F_{ ext {cable }_y}+F_{13 y}+F_{23 y}=0 \end{array} [/latex] (d)

4 The assembly of elements labeled part 1 in Figure 3-2 can have both forces and moments on it (i.e., it is not a concurrent system), so the three equations 3.3b are needed.

[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_{21 x}+F_{b 1 x}+F_{31 x}+P_x+F_{ ext {sheath }_x}=0 \ \sum F_y=F_{21 y}+F_{b 1 y}+F_{31 y}+P_y=0 \ \sum M _z= M _h+\left( R _{21} imes F _{21} ight)+\left( R _{b 1} imes F _{b 1} ight)+\left( R _{31} imes F _{31} ight)+\left( R _p imes P ight)+\left( R _d imes F _{ ext {sheath }} ight)=0 \end{array} [/latex] (e)

Expanding cross products in the moment equation gives the moment magnitude as

[latex] \begin{aligned} \sum M_z=M_h &+\left(R_{21 x} F_{21 y}-R_{21 y} F_{21 x} ight)+\left(R_{b 1 x} F_{b 1 y}-R_{b 1 y} F_{b 1 x} ight)+\left(R_{31 x} F_{31 y}-R_{31 y} F_{31 x} ight)\ &+\left(R_{P x} P_y-R_{P y} P_x ight)+\left(R_{d x} F_{ ext {sheath }_y}-R_{d y} F_{ ext {sheath }_x} ight)=0 \end{aligned} [/latex] (f)

5 The total of unknowns at this point (including those listed in step 2 above) is 21: [latex] F_{b1x}, F_{b 1 y}, F_{12 x}, F_{12 y}, F_{21 x}, F_{21 y}, F_{32 x}, F_{32 y}, F_{23 x}, F_{23 y}, F_{13 x}, F_{13 y}, F_{31 x}, F_{31 y}, F_{cablex}, F_{cabley}, F_{sheathx}, F_{sheathy}, P_{x}, P_{y} [/latex] , and [latex]M_{h}[/latex]. We have only nine equations so far, three in equation set ([latex]a[/latex]), one in set ([latex]c[/latex]), two in set ([latex]d[/latex]) and three in set ([latex]e[/latex]). We need twelve more equations to solve this system. We can get seven of them from the Newton’s third-law relationships between contacting elements:

[latex] \begin{array}{l} F_{23 x}=-F_{32 x} \quad F_{23 y}=-F_{32 y}\ F_{21 x}=-F_{12 x} \quad F_{21 y}=-F_{12 y}\ F_{31 x}=-F_{13 x} \quad F_{31 y}=-F_{13 y}\ \quad F_{ ext {sheath }_x}=-F_{ ext {cable }_x} \end{array} [/latex] (g)

Two more equations come from the assumption (shown in Figure 3-1) that the two forces provided by the hand on the brake lever and handgrip are equal and opposite:*

[latex] \begin{array}{l} F_{b 1 x}=-F_{b 2 x} \ F_{b 1 y}=-F_{b 2 y} \end{array} [/latex] (h)

The remaining three equations come from the given geometry and the assumptions made about the system. The direction of the forces [latex]F_{cable}[/latex] and [latex]F_{sheath}[/latex] are known to be in the same direction as that end of the cable. In the figure it is seen to be horizontal, so we can set

[latex] F_{ ext {cable }_y}=0 ; \quad F_{ ext {sheath }_y}=0 [/latex] (i)

Because of our no-friction assumption, the force [latex]F_{31}[/latex] can be assumed to be normal to the surface of contact between the cable and the hole in part 1. This surface is horizontal in this example, so [latex]F_{31}[/latex] is vertical and

[latex] F_{31 x}=0 [/latex] (j)

6 This completes the set of 21 equations (equation sets [latex]a, c, d, e, g, h, i[/latex], and [latex]j[/latex]), and they can be solved for the 21 unknowns simultaneously “as is,” that is, all 21 equations could be put into matrix form and solved with a matrix-reduction computer program. However, the problem can be simplified by manually substituting equations [latex]c, g, h, i,[/latex] and [latex]j[/latex] into the others to reduce them to a set of eight equations in eight unknowns. The known or given data are as shown in Table 3-2, part 1.

7 As a first step, for link 2, substitute equations [latex]b[/latex] and [latex]c[/latex] in equation [latex]a[/latex] to get:

[latex] \begin{aligned} F_{12 x}+F_{b 2 x}+F_{32 x} &=0 \ F_{12 y}+F_{b 2 y}+F_{32 x} an heta &=0 \ \left(R_{12 x} F_{12 y}-R_{12 y} F_{12 x} ight)+\left(R_{b 2 x} F_{b 2 y}-R_{b 2 y} F_{b 2 x} ight)+\left(R_{32 x} F_{32 x} an heta-R_{32 y} F_{32 x} ight) &=0 \end{aligned} [/latex] (k)

8 Next, take equations d for link 3 and substitute equation c and also [latex]–F_{32x}[/latex] for [latex]F_{23x}[/latex], and [latex]–F_{32y}[/latex] for [latex]F_{23y}[/latex] from equation g to eliminate those variables.

[latex] \begin{aligned} F_{ ext {cable }_x}+F_{13 x}-F_{32 x} &=0 \ F_{ ext {cable }_y}+F_{13 y}-F_{32 x} an heta &=0 \end{aligned} [/latex] (l)

9 For link 1, substitute equation [latex]f[/latex] in [latex]e[/latex] and replace [latex]F_{21x}[/latex] with [latex]–F_{12x}, F_{21y}[/latex] with [latex]–F_{12y}, F_{31x}[/latex] with [latex]–F_{13x} , F_{31y}[/latex] with [latex]–F_{13y} [/latex], and [latex]F_{sheathx}[/latex] with [latex]–F_{cablex}[/latex] from equation [latex]g[/latex],

[latex] \begin{array}{r} -F_{12 x}+F_{b 1 x}-F_{13 x}+P_x-F_{ ext {cable }_x}=0 \ -F_{12 y}+F_{b 1 y}-F_{32 x} an heta+P_y=0 \ M_h+\left(-R_{21 x} F_{12 y}+R_{21 y} F_{12 x} ight)+\left(R_{b 1 x} F_{b 1 y}-R_{b 1 y} F_{b 1 x} ight) \ +\left(-R_{31 x} F_{13 y}+R_{31 y} F_{13 x} ight)+\left(R_{P x} P_y-R_{P y} P_x ight)+R_{d y} F_{ ext {cable }_x}=0 \end{array} [/latex] (m)

10 Finally, substitute equations [latex]h, i[/latex], and [latex]j[/latex] into equations [latex]k, l[/latex], and [latex]m[/latex] to yield the following set of eight simultaneous equations in the eight remaining unknowns: [latex]F_{12x}, F_{12y}, F_{32x} , F_{13y}, F_{cablex} , P_{x}, P_{y}[/latex], and [latex]M_{h}[/latex]. Put them in the standard form which has all unknown terms on the left and all known terms to the right of the equal signs.

[latex] \begin{array}{c} F_{12 x}+F_{32 x}=-F_{b 2 x} \ F_{12 y}+F_{32 x} an heta=-F_{b 2 y} \ F_{c a b l e_x}-F_{32 x}=0 \ F_{13 y}-F_{32 x} an heta=0 \ -F_{12 x}+P_x-F_{c a b l e_x}=F_{b 2 x} \ -F_{12 y}-F_{13 y}+P_y=F_{b 2 y} \ R_{12 x} F_{12 y}-R_{12 y} F_{12 x}+\left(R_{32 x} an heta-R_{32 y} ight) F_{32 x}=-R_{b 2 x} F_{b 2 y}+R_{b 2 y} F_{b 2 x} \ M_h-R_{21 x} F_{12 y}+R_{21 y} F_{12 x}-R_{31 x} F_{13 y}+R_{P x} P_y-R_{P y} P_x+R_{d y} F_{c a b l e_x}=R_{b 1 x} F_{b 2 y}-R_{b 1 y} F_{b 2 x} \end{array} [/latex] (n)

11 Form the matrices from equation [latex]n[/latex].

[latex] \left[\begin{array}{cccccccc} 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & an heta & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & - an heta & 1 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \ 0 & -1 & 0 & -1 & 0 & 0 & 1 & 0 \ -R_{12 y} & R_{12 x} & R_{32 x} an heta R_{32 y} & 0 & 0 & 0 & 0 & 0 \ R_{21 y} & -R_{21 x} & 0 & -R_{31 x} & R_{d y} & -R_{P y} & R_{P x} & 1 \end{array} ight] imes \quad \left[\begin{array}{c} F_{12 x} \ F_{12 y} \ F_{32 x} \ F_{13 y} \ F_{ ext {cable }_x} \ P_x \ P_y \ M_h \end{array} ight]= \ \left[\begin{array}{c} -F_{b 2 x} \ -F_{b 2 y} \ 0 \ 0 \ F_{b 2 x} \ F_{b 2 y} \ -R_{b 2 x} F_{b 2 y}+R_{b 2 y} F_{b 2 x} \ R_{b 1 x} F_{b 2 y}-R_{b 1 y} F_{b 2 x} \end{array} ight] [/latex] (o)

12 Substitute the known data as shown in Table 3-2 part 1 (repeated opposite).

[latex] \left[\begin{array}{cccccccc} 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0.070 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & -0.070 & 1 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & -1 & 1 & 0 & 0 \ 0 & -1 & 0 & -1 & 0 & 0 & 1 & 0 \ 7.34 & -47.91 & -0.324 & 0 & 0 & 0 & 0 & 0 \ 19 & -7 & 0 & 27 & 27 & 0 & -27 & 1 \end{array} ight] imes \left[\begin{array}{c} F_{12 x} \ F_{12 y} \ F_{32 x} \ F_{13 y} \ F_{c a b l e_x} \ P_x \ P_y \ M_h \end{array} ight]=\left[\begin{array}{c} 0 \ 267 \ 0 \ 0 \ 0 \ -267 \ 93.08 \ -112.25 \end{array} ight] [/latex] (p)

13 The solution is shown in Table 3-2 part 2. This matrix equation can be solved with any one of a number of commercially available matrix solvers such as Mathcad, MATLAB, Maple, or Mathematica, as well as with many engineering pocket calculators. A custom-written program called MATRIX is provided on the CD-ROM in this book that can be used to solve a linear system of up to 16 equations. Equation [latex]p[/latex] was solved with program MATRIX to find the 8 unknowns listed in step 10. Those results were then substituted into the other equations to solve for the previously eliminated variables.

14 Table 3-2 part 2 shows the solution data for the data given in Figure 3-2 and Table 3-2 part 1. This assumes a 267-N (60-lb) force is applied by the person’s hand normal to the brake lever. The force generated in the cable (Fcable) is then 1 909 N (429 lb) and the reaction force against the handlebar ([latex]F_{21}[/latex]) is 1 951 N (439 lb) at –168°.

* Actually, for a simple static analysis such as the one in this example, any point (on or off the element) can be taken as the origin of the local coordinate system. However, in a dynamic force analysis it simplifies the analysis if the coordinate system is placed at the CG. So, for the sake of consistency, and to prepare for the more complicated dynamic analysis problems ahead, we will use the CG as the origin even in the static cases here.

† You may not have done this in your statics class but this approach makes the problem more amenable to a computer solution. Note that regardless of the direction shown for any unknown force on the FBD, we will assume its components to be positive in the equations. The angles of the known (given) forces (or the signs of their components) do have to be correctly input to the equations, however.

Table 3–2 - part 1 Variable Unit Value Case Study 1A Given Data
Variable	Unit	V
[latex]F_{13x}[/latex]	0.0	N
[latex]F_{b2x}[/latex]	0.0	N
[latex]F_{b2y}[/latex]	–267.0	N
[latex] heta[/latex]	184.0	deg
[latex]\phi[/latex]	180.0	deg
[latex]R_{b2x}[/latex]	39.39	mm
[latex]R_{b2y}[/latex]	2.07	mm
[latex]R_{32x}[/latex]	–50.91	mm
[latex]R_{32y}[/latex]	4.66	mm
[latex]R_{12x}[/latex]	–47.91	mm
[latex]R_{12y}[/latex]	–7.34	mm
[latex]R_{21x}[/latex]	7.0	mm
[latex]R_{21y}[/latex]	19.0	mm
[latex]R_{b1x}[/latex]	47.5	mm
[latex]R_{b1y}[/latex]	–14.0	mm
[latex]R_{31x}[/latex]	–27.0	mm
[latex]R_{31y}[/latex]	30.0	mm
[latex]R_{px}[/latex]	–27.0	mm
[latex]R_{py}[/latex]	0.0	mm
[latex]R_{dx}[/latex]	–41.0	mm
[latex]R_{dy}[/latex]	27.0	mm

Table 3–2 - part 2 Case Study 1A Calculated Data
Variable	Unit	V
[latex]F_{32x}[/latex]	–1 909	N
[latex]F_{32y}[/latex]	–133	N
[latex]F_{12x}[/latex]	1 909	N
[latex]F_{12y}[/latex]	400	N
[latex]F_{23x}[/latex]	1 909	N
[latex]F_{23y}[/latex]	133	N
[latex]F_{13y}[/latex]	–133	N
[latex]F_{cablex}[/latex]	–1 909	N
[latex]F_{cabley}[/latex]	0	N
[latex]F_{b1x}[/latex]	0	N
[latex]F_{b1y}[/latex]	267	N
[latex]F_{31xy}[/latex]	0	N
[latex]F_{31y}[/latex]	133	N
[latex]F_{21x}[/latex]	–1 909	N
[latex]F_{21y}[/latex]	–400	N
[latex]P_{x}[/latex]	0	N
[latex]P_{y}[/latex]	0	N
[latex]M_{h}[/latex]	9	N - m
[latex]F_{sheath_x}[/latex]	1 909	N

Question 3.CS.1A: Bicycle Brake Lever Loading Analysis Problem: Determine the ...

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