A friction-drive stainless steel metal belt runs over two 4-in metal pulleys ( f = 0.35).
The belt thickness is to be 0.003 in. For a life exceeding $10^{6}$ belt passes with smooth torque ( $K_{s}$ = 1), (a) select the belt if the torque is to be 30 lbf · in, and (b) find the initial tension $F_{i}$ .

Question

A friction-drive stainless steel metal belt runs over two 4-in metal pulleys ( f = 0.35).
The belt thickness is to be 0.003 in. For a life exceeding [latex]10^{6}[/latex] belt passes with smooth torque ([latex]K_{s}[/latex ]= 1), (a) select the belt if the torque is to be 30 lbf · in, and (b) find the initial tension [latex]F_{i}[/latex] .

Accepted Answer

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

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

From steps 3, 4, 5, and 6,

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

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

From step 7,

[latex]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[/latex]

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

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

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

Note [latex]f^{′} < f[/latex] , that is, 0.0853 < 0.35.

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

1 Find exp( f [latex]\phi[/latex]) from geometry and friction
2 Find endurance strength

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

3 Allowable tension

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

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

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

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