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 106 belt passes with smooth torque (Ks = 1), (a) select the belt if the torque is to be 30 lbf · in, and (b) find the initial tension Fi .

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]\left(K_{s}=1\right)[/latex], (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= heta_{d}=\pi[/latex], therefore exp(0.35π) = 3.00. From step 2,

[latex]\left(S_{f} ight)_{10^{6}}=14.17\left(10^{6} ight)\left(10^{6} ight)^{-0.407}=51210 psi[/latex]

From steps 3, 4, 5, and 6,

[latex]F_{1 a}=\left[51210-\frac{28\left(10^{6} ight) 0.003}{\left(1-0.285^{2} ight) 4} ight] 0.003 b=85.1 b lbf[/latex] (1)

[latex]\Delta F=2 T / D=2(30) / 4=15 lbf[/latex]

[latex]F_{2}=F_{1 a}-\Delta F=85.1 b-15 lbf[/latex] (2)

[latex]F_{i}=\frac{F_{1 a}+F_{2}}{2}=\frac{85.1 b+15}{2} lbf[/latex] (3)

From step 7,

[latex]b_{\min }=\frac{\Delta F}{a} \frac{\exp (f \phi)}{\exp (f \phi)-1}=\frac{15}{85.1} \frac{3.00}{3.00-1}=0.264 ext { 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 lbf[/latex]

Eq. (2):

[latex]F_{2}=85.1(0.75)-15=48.8 lbf[/latex]

Eq. (3):

[latex]F_{i}=(63.8+48.8) / 2=56.3 lbf[/latex]

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

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

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

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