Question 5.DE.2: A part of a conveyor system for a production operation is sh...

Objective      Design the pin for attaching the fixture to the conveyor system.

Given       The general arrangement is .shown in Figure 5_18. The fixture places a shearing load that is alternately 85 lb and 310 lb (85 + 225) on the pin many thousands of times in the expected life of the system.

Basic Design Decisions        It is proposed to make the pin from AISI 1020 cold-drawn steel. Appendix 3 listss_{y} = 51ksi ands_{u}= 61 ksi. The steel is ductile with 15°⁄° elongation. This material is inexpensive,and it is not necessary to achieve a particularly small size for the pin.The connection of the fixture to the bar is basically a clevis joint with two tabs at the top of the fixture, one on each side of the bar. There will be a close fit between the tabs and the bar to minimize bending action on the pin. Also, the pin will be a fairly close fit in the holes while still allowing rotation of the fixture relative to the bar.

Analysis       Case H from Section 5_9 applies for completing the design analysis because fluctuating shearing stresses are experienced by the pin. Therefore, we will have to determine relationships for the mean and alternating stresses (\tau _{m}and \tau _{a}) in terms of the applied loads and the cross-sectional area of the bar. Note that the pin is in double shear, so two cross sections resist the applied shearing force. In general, τ= F/2A. Now we will use the basic forms of Equations (5_1) and (5_2)\sigma_{m}=\left(\sigma_{\max }+\sigma_{\min }\right)/2

\sigma_{a}=\left(\sigma_{\max }-\sigma_{\min }\right)/2 to compute the values for the mean and alternating forces on the pin:F_{m}=\left(F_{\mathrm{ma}}+F_{\mathrm{min}}\right) / 2=(310+85) / 2=198 \mathrm{lb}

F_{a}=\left(F_{\max }-F_{\min }\right) / 2=(310-85) / 2=113 \mathrm{lb} The stresses will be found from\tau _{m}=F_{m}/2A and\tau _{a}=F_{a}/2A . The material strength values needed in Equation (5_28) \frac{K_{t} \tau_{\mathrm{a}}}{s_{s n}^{\prime}}+\frac{\tau_{\mathrm{m}}}{s_{\mathrm{su}}}=\frac{1}{N} for Case H are s_{\mathrm{su}}=0.75 s_{u}=0.75(51 \mathrm{ksi})=38.3 \mathrm{ksi}=38300 \mathrm{psi}

s^{\prime }_{sn}=0.577s^{\prime }_{n} We must find the value ofs^{\prime }_{n}using the method from Section 5_4. We find from Figure 5_8 that s_{n}= 21 ksi for the machined pin having a value ofs_{n}= 61 ksi. It is expected that the pin will be fairly small, so we will use C_{s}= 1.0. The material is wrought steel rod, so C_{m}= 1.0. Let’s use C_{sI}= 1.0 to be conservative because there is little information about such factors for direct shearing stress. A high reliability is desired for this application, so let’s useC_{R} = 0.75 to produce a reliability of 0.999 (see Table 5_1).

Thens_{n}^{\prime}=C_{R}\left(s_{n}\right)=(0.75)(21 \mathrm{ksi})=15.75 \mathrm{ksi}=15750 \mathrm{psi} Finally, s_{s n}^{\prime}=0.577 s_{n}^{\prime}=0.577(15750 \mathrm{psi})=9088 \mathrm{psi} We can now apply Equation (5_28) \frac{K_{t} \tau_{\mathrm{a}}}{s_{s n}^{\prime}}+\frac{\tau_{\mathrm{m}}}{s_{\mathrm{su}}}=\frac{1}{N} from Case H: \frac{1}{N}=\frac{\tau_{m}}{s_{s n}}+\frac{K_{t} \tau_{a}}{s_{s n}^{\prime}} Because the pins will be of uniform diameter. K_{1} =1.0. Substituting \tau _{m}=F_{m}/2A and\tau _{a}=F_{a}/2Afound earlier gives \frac{1}{N}=\frac{F_{m}}{2 A s_{s u}}+\frac{F_{a}}{2 A s_{s n}^{\prime}} Let’s use N = 4 because mild shock can be expected. Note that we now know all factors in this equation except the cross-sectional area of the pin. A, We can solve for the required area:A=\frac{N}{2}\left[\frac{F_{m}}{s_{s u}}+\frac{F_{a}}{s_{sn}^{\prime}}\right] Finally, we can compute the minimum allowable pin diameter, D, from A =\pi D^{2}/4 and D=\sqrt{4A/\pi } .

Results         The required area is A=\frac{4}{2}\left[\frac{198 \mathrm{lb}}{38300 \mathrm{lb} / \mathrm{in}^{2}}+\frac{113 \mathrm{lb}}{9088 \mathrm{lb} / \mathrm{in}^{2}}\right]=0.0352 \mathrm{in}^{2} Now the required diameter is D=\sqrt{4 A / \pi}=\sqrt{4\left(0.0352 \mathrm{in}^{2} / \pi\right.}=0.212 \mathrm{in}

Final Design Decisions and Comments                 The computed value for the minimum required diameter for the pin, 0.212 in, is quite small.Other considerations such as bearing stress and wear at the surfaces that contact the tabs of the fixture and the bar indicate that a larger diameter would be preferred. Let’s specify D = 0.50 in for the pin at this location. The pin will be of uniform diameter within the area of the bar and the tabs. It should extend beyond the tabs, and it could be secured with cotter pins or retaining rings.This completes the design of the pin. But the next design example deals with the horizontal bar for this same system. There are pins at the conveyor hangers to support the bar.They would also have to be designed. However, note that each of these pins carries only half the load of the pin in the fixture connection. These pins would experience less relative motion as well, so wear should not be so severe. Thus, let’s use pins with D = 3/8 in = 0.375 in at the ends of the horizontal bar.