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.