The bell-crank mechanism shown in Figure P1.33 is in equilibrium for an applied load of P = 7 kN applied at A. Assume a = 200 mm, b = 150 mm, and θ = 65°. Determine the minimum diameter d required for pin B for each of the following conditions:
(a) The average shear stress in the pin may not exceed 40 MPa.
(b) The bearing stress in the bell crank may not exceed 100 MPa.
(c) The bearing stress in the support bracket may not exceed 165 MPa.

Question

Accepted Answer

Equilibrium: Using the FBD shown, calculate the reaction forces that act on the bell crank.

[latex]\begin{gathered}\Sigma M_{B}=(7,000 N ) \sin \left(65^{\circ} ight)(200 mm ) \-F_{2}(150 mm )=0 \ herefore F_{2}=8,458.873 N \\Sigma F_{x}=B_{x}+(7,000 N ) \cos \left(65^{\circ} ight) \\quad+8,458.873 N =0 \ herefore B_{x}=-11,417.201 N \\Sigma F_{y}=B_{y}-(7,000 N ) \sin \left(65^{\circ} ight)=0 \ herefore B_{y}=6,344.155 N\end{gathered}[/latex]

The resultant force at B is

[latex]|B|=\sqrt{(-11,417.201 N )^{2}+(6,344.155 N )^{2}}=13,061.423 N[/latex]

(a) The average shear stress in the pin may not exceed 40 MPa. The shear area required for the pin at B is

[latex]A_{V} \geq \frac{13,061.423 N }{40 N / mm ^{2}}=326.536 mm ^{2}[/latex]

Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin is

[latex]A_{ pin }=\frac{A_{V}}{2}=163.268 mm ^{2}[/latex]

and therefore, the pin must have a diameter of

[latex]d \geq \sqrt{\frac{4}{\pi}\left(163.268 mm ^{2} ight)}=14.42 mm[/latex]

(b) The bearing stress in the bell crank may not exceed 100 MPa. The projected area of pin B on the bell crank must equal or exceed

[latex]A_{b} \geq \frac{13,061.423 N }{100 N / mm ^{2}}=130.614 mm ^{2}[/latex]

The bell crank thickness is 8 mm; therefore, the projected area of the pin is [latex]A_{b}[/latex] = (8 mm)d. Calculate the required pin diameter d:

[latex]d \geq \frac{130.614 mm ^{2}}{8 mm }=16.33 mm[/latex]

(c) The bearing stress in the support bracket may not exceed 165 MPa. The pin at B bears on two 6- mm-thick support brackets. Thus, the minimum pin diameter required to satisfy the bearing stress limit on the support bracket is

[latex]\begin{aligned}&A_{b} \geq \frac{13,061.423 N }{165 N / mm ^{2}}=79.160 mm ^{2} \&d \geq \frac{79.160 mm ^{2}}{2(6 mm )}=6.60 mm\end{aligned}[/latex]

Question 1.33: The bell-crank mechanism shown in Figure P1.33 is in equilib...

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