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 $b > b_{min}, F_{1} = ab, F_{2} = ab − ΔF, F_{i} = ab − F/2, T = FD/2$