Question 17.3: A friction-drive stainless steel metal belt runs over two 4-...

(a) From step 1, \phi = θ_{d} = π, therefore exp(0.35π) = 3.00. From step 2,

(S_{f} )_{10^{6}} = 14.17(10^{6})(10^{6})^{−0.407} = 51  210  psi

From steps 3, 4, 5, and 6,

F_{1a} =\left[ 51  210 −\frac{28(10^{6})0.003}{(1 − 0.285^{2})4}\right]0.003b =85.1b  lbf                     (1)

ΔF = 2T/D = 2(30)/4 = 15  lbf
F_{2} = F_{1a} − ΔF = 85.1b − 15  lbf                     (2)
F_{i} =\frac{F_{1a} + F_{2}}{2} =\frac{85.1b + 15}{2}  lbf               (3)

From step 7,

b_{min} =\frac{ΔF}{a} \frac{exp( f  \phi)}{exp( f  \phi) − 1} =\frac{15}{85.1} \frac{3.00}{3.00 − 1} = 0.264  in

Select an available 0.75-in-wide belt 0.003 in thick.

Eq. (1):                   F_{1} = 85.1(0.75) = 63.8 lbf
Eq. (2):                  F_{2} = 85.1(0.75) − 15 = 48.8 lbf
Eq. (3):                F_{i} = (63.8 + 48.8)/2 = 56.3 lbf

f^{′} =\frac{1}{\phi} ln \frac{F_{1}}{F_{2}} =\frac{1}{π} ln \frac{63.8}{48.8} = 0.0853

Note f^{′} < f , that is, 0.0853 < 0.35.

The selection of a metal flat belt can consist of the following steps:

1 Find exp( f \phi) from geometry and friction
2 Find endurance strength

S_{f} = 14.17(10^{6})N^{−0.407}_{p}                      301, 302 stainless
S_{f} = S_{y}/3                           others

3 Allowable tension

F_{1a} =\left[ S_{f} −\frac{Et}{(1 − ν^{2})D}\right] tb = ab

4   ΔF = 2T/D
5    F_{2} = F_{1a} − F = ab − ΔF
6     F_{i} =\frac{F_{1a} + F_{2}}{2} =\frac{ab + ab − ΔF}{2} = ab − \frac{ΔF}{2}
7      b_{min} =\frac{ΔF}{a} \frac{exp( f \phi)}{exp( f \phi) − 1}
8      Choose b > b_{min}, F_{1} = ab, F_{2} = ab − ΔF, F_{i} = ab − F/2, T = FD/2