BULETINUL INSTITUTULUI POLITEHNIC DIN IAŞI

40 Radu Ibănescu and Cătălin Ungureanu
4. Numerical Example
The following values are considered for a numerical example: R=0.05 m,
l=0.2 m, e=0.02 m, a=0.04 m, Φ=0.01 m, d=0.01 m, r=0.005 m, k e =500 N/m,
G=30 N, μ=0.02, μ 1 =0.01, F=200 N and φ=π/2 (for the most inconvenient
situation). In this case, all the unknowns will be functions of the distance h. The
following expressions for the unknowns N A , N B and NC
are obtained by
using these numerical values
⎡
0.184h
+ 0.5759568
⎤
⎢
⎥
⎢
0.039984h
− 0.0000761568
⎥
⎢ − 0.57561606144h
+ 0.001096365477888
⎥
Insolved( M,v)
→ ⎢⎢
.
(0.039984h−0.0000761568)( − 0.9996h+
0.00190392) ⎥
⎥
⎢
9.2h
+ 0.00552
⎥
⎢
⎣
0.039984h
− 0.0000761568
⎥
⎦
(15)
The normal force N C is then
9.2h
+ 0.00552
NC( h) =
.
0.039984h
− 0.0000761568
(16)
The normal force N C is infinity for h=0.00190468 m.
The unknowns H, V and M m are the solution of the very simple system of
equations (8), (9) and (10). The driving torque M m is given by the following
function of the distance h:
2 2 2
Mn( h) = μr μ NC( h) + ( G+ NC( h)) + NC( h)[ ecos φ+μ( R+esin φ)] + Gecosφ.
(17)
30
20
10
Mm( h)
0
− 10
− 20
− 30
110 × − 3 210 × − 3 310 × − 3 410 ×
− 3
h
Fig. 4 – The function M m (h).