Chapter 3

LOAD DETERMINATION

79

 

C A S E S T U D Y 1 A

 

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.

3

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.

Solution:

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 ( Fsheath =

Fcable). 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 x 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 neglect-F b 2

 

 

F

 

F cable

3

sheath

 

2

brake lever

 

 

cable

pivot

 

 

P x

 

M

1

 

h

 

 

P y

 

handlebar

handgrip

 

F b 1

F I G U R E 3 - 1

Bicycle Brake Lever Assembly

80

MACHINE DESIGN -

An Integrated Approach

 

 

ing the weights of the parts. The lever (part 2) has three forces on it, F b 2, F32, and Table 3–2 - part 1

F12. The two-character subscript notation used here should be read as, force of ele-Case Study 1A

ment 1 on 2 (F12) or force at B on 2 (F b 2), etc. This defines the source of the force Given Data

(first subscript) and the element on which it acts (second subscript).

 

Variable Value

Unit

This notation will be used consistently throughout this text for both forces and posi-3

F

tion vectors such as R b 2, R32, and R12 in Figure 3-2 which serve to locate the above 13 x

0.0

N

three forces in a local, nonrotating coordinate system whose origin is at the center of F

 

b 2 x

0.0

N

gravity (CG) of the element or subassembly being analyzed.*

b

F 2 y

–267.0

N

 

On this brake lever, F b 2 is an applied force whose magnitude and direction are known.

184.0

deg

 

F32 is the force in the cable. Its direction is known but not its magnitude. Force F12

 

180.0

deg

is provided by part 1 on part 2 at the pivot pin. Its magnitude and direction are both

 

Rb 2 x

39.39

mm

unknown. We can write equations 3.3 b for this element to sum forces in the x and y

 

 

R

directions and sum moments about the CG. Note that all unknown forces and moments b 2 y

2.07

mm

 

are initially assumed positive in the equations. Their true signs will come out in the R 32 x

–50.91

mm

 

calculation.† However, all known or given forces must carry their proper signs.

32

R y

4.66

mm

R 12 x

–47.91

mm

F =

+

+

=

x F 12 x Fb 2 x F 32 x 0

R 12 y

–7.34

mm

R

F =

+

+

=

y F 12 y Fb 2 y F 32 y 0

( a)

21 x

7.0

mm

R 21 y

19.0

mm

M =

)

)

)

z (R12 F12 + (R b 2 F b 2 + (R32 F32 = 0

Rb 1 x

47.5

mm

The cross products in the moment equation represent the “turning forces” or moments Rb 1 y

–14.0

mm

created by the application of these forces at points remote from the CG of the element.

R 31 x

–27.0

mm

Recall that these cross products can be expanded to

R 31 y

30.0

mm

R

M =

)

)

z ( R 12 x F 12 y R 12 y F 12 x + ( Rb 2 x Fb 2 y Rb 2 y Fb 2 x px

–27.0

mm

R

+

py

0.0

mm

( R F

( b)

R

F

= 0

)

32 x 32 y

32 y 32 x

 

Rdx

–41.0

mm

 

We have three equations and four unknowns ( F

R

12 x, F 12 y, F 32 x, F 32 y) at this point, so

 

dy

27.0

mm

we need another equation. It is available from the fact that the direction of F32 is

 

 

 

 

known. (The cable can pull only along its axis.) We can express one component of

 

* Actually, for a simple static

the cable force F

analysis such as the one in this

32 in terms of its other component and the known angle,  of the cable.

example, any point (on or off the

element) can be taken as the origin

F

=

32 y F 32 x tan

( c)

of the local coordinate system.

However, in a dynamic force

analysis it simplifies the analysis if

We could now solve the four unknowns for this element, but will wait to do so until the coordinate system is placed at

the equations for the other two links are defined.

the CG. So, for the sake of

consistency, and to prepare for the

more complicated dynamic analysis

3

Part 3 in Figure 3-2 is the cable which passes through a hole in part 1. This hole is problems ahead, we will use the CG

lined with a low-friction material which allows us to assume no friction at the joint as the origin even in the static cases

between parts 1 and 3. We will further assume that the three forces F13, F23, and F cable here.

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.

You may not have done this in

your statics class but this approach

makes the problem more amenable

F =

+

+

=

x Fcable

F 13 x F 23 x 0

to a computer solution. Note that

x

( d)

regardless of the direction shown

for any unknown force on the FBD,

F = F

+ F

+

=

13 y F 23 y 0

we will assume its components to

y

cabley

be positive in the equations. The

angles of the known (given) forces

4 The assembly of elements labeled part 1 in Figure 3-2 can have both forces and mo-

(or the signs of their components)

ments on it (i.e., it is not a concurrent system), so the three equations 3.3 b are needed.

do have to be correctly input to the

equations, however.

 

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