Chapter 3

LOAD DETERMINATION

81

 

 

 

 

R

y

32

F 3 2

F b 2

 

2

F

C

x

F

 

32

 

b 2

F

B

 

12

 

A

3

Vector diagram for brake lever

F12

R

R

12

b 2

 

F

13

F cable

3

F 2 3

1

 

 

F31

R31

 

F sheath

 

y A

F21

 

D

 

R d

R

P

21

x

x

 

M h

R p

1

P y

 

R

F

b 1

b 1

F I G U R E 3 - 2

 

 

Bicycle Brake Lever Free-Body Diagrams

 

 

 

 

F =

+

+

+

+

=

x F 21 x Fb 1 x F 31 x Px Fsheath 0

x

 

 

F =

+

+

+

=

y F 21 y Fb 1 y F 31 y Py 0

( e)

M = +

)

)

)

)

z M h (R21 F21 + (R b 1 F b 1 + (R31 F31 + (R p P) + (R d F sheath = 0

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

+

)

)

)

z

Mh ( R 21 x F 21 y R 21 y F 21 x + ( Rb 1 x Fb 1 y Rb 1 y Fb 1 x + ( R 31 x F 31 y R 31 y F 31 x ( f )

+( R

)

)

Px Py

RPy Px + ( Rdx Fsheath R

= 0

y

dy Fsheathx

 

 

5 The total of unknowns at this point (including those listed in step 2 above) is 21: Fb 1 x, Fb 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, Fcable , F

, F

, F

, P

x

cabley

sheathx

sheathy

x, Py, and Mh. We have only nine equations so far, three in equation set ( a), one in set ( c), two in set ( d) and three in set ( e). 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:

 

F

=

=

23 x F 32 x

F 23 y F 32 y

F

=

=

21 x F 12 x

F 21 y F 12 y

( g)

F

=

=

31 x F 13 x

F 31 y F 13 y

F

=

sheath

F

x

cablex

82

MACHINE DESIGN -

An Integrated Approach

 

 

 

Two more equations come from the assumption (shown in Figure 3-1) that the two Table 3–2 - part 1 repeated

forces provided by the hand on the brake lever and handgrip are equal and opposite:*

Case Study 1A

Given Data

F

=

b 1 x Fb 2 x

( h)

Variable Value

Unit

F

=

b 1 y Fb 2 y

3

F 13 x

0.0

N

The remaining three equations come from the given geometry and the assumptions F

made about the system. The direction of the forces F

b 2 x

0.0

N

cable and F sheath are known to

be in the same direction as that end of the cable. In the figure it is seen to be hori-Fb 2 y

–267.0

N

zontal, so we can set

184.0

deg

 

 

180.0

deg

F

=

=

cable

0;

F

0

y

 

sheathy

 

( i)

Rb 2 x

39.39

mm

Because of our no-friction assumption, the force F 31 can be assumed to be normal to Rb 2 y

2.07

mm

the surface of contact between the cable and the hole in part 1. This surface is hori-R 32 x

–50.91

mm

zontal in this example, so F 31 is vertical and

R 32 y

4.66

mm

F

=

31 x 0

( j)

R 12 x

–47.91

mm

R 12 y

–7.34

mm

6 This completes the set of 21 equations (equation sets a, c, d, e, g, h, i, and j), and they R 21 x

7.0

mm

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.

R 21 y

19.0

mm

However, the problem can be simplified by manually substituting equations c, g, h, i, Rb 1 x

47.5

mm

and j into the others to reduce them to a set of eight equations in eight unknowns. The Rb 1 y

–14.0

mm

known or given data are as shown in Table 3-2, part 1.

R 31 x

–27.0

mm

7 As a first step, for link 2, substitute equations b and c in equation a to get: R 31 y

30.0

mm

R

F

+

+

=

px

–27.0

mm

12 x Fb 2 x F 32 x 0

R

F

+

+

py

0.0

mm

12 y Fb 2 y F 32 x tan = 0

( k)

Rdx

–41.0

mm

( R

)

)

)

12 x F 12 y R 12 y F 12 x + ( Rb 2 x Fb 2 y Rb 2 y Fb 2 x + ( R 32 x F 32 x tan − R 32 y F 32 x = 0

Rdy

27.0

mm

8 Next, take equations d for link 3 and substitute equation c and also – F 32 x for F 23 x, and

 

F 32 y for F 23 y from equation g to eliminate those variables.

F

+

=

cable

F

F

0

x

13 x

32 x

 

 

( l)

F

+

cable

F

F

y

13 y

32 x tan = 0

9 For link 1, substitute equation f in e and replace F 21 x with – F 12 x, F 21 y with – F 12 y, F 31 x with – F 13 x , F 31 y with – F 13 y, and Fsheath with – F

from equation g,

x

cablex

F

+

+

=

12 x Fb 1 x

F 13 x Px Fcable

0

x

 

F

+

=

12 y Fb 1 y F 32 x tan + Py 0

( m)

M +

+

)

)

h (− R 21 x F 12 y R 21 y F 12 x + ( Rb 1 x Fb 1 y Rb 1 y Fb 1 x

+(− R

+

)

)

=

31 x F 13 y R 31 y F 13 x + ( RPx Py RPy Px + Rdy Fcable 0

x

 

 

 

10 Finally, substitute equations h, i, and j into equations k, l, and m to yield the following set of eight simultaneous equations in the eight remaining unknowns: F 12 x, F 12 y, F

, P

 

32 x , F 13 y, Fcablex

x, Py, and Mh. Put them in the standard form which has all un-

* But not necessarily colinear.

known terms on the left and all known terms to the right of the equal signs.