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 5–17. 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 SAE 1020 cold-drawn steel. “Design Properties of Carbon and Alloy Steels” lists s_y = 51 ksi and s_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: The Goodman criterion, expressed in Equation (5–32) (\frac{K_{t} \sigma_{a}^{\prime}}{s_{n}^{\prime}}+\frac{\sigma_{m}^{\prime}}{s_{u}}=\frac{1}{N}) in Section 5–7, 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, \tau= F/2A .
Now we will use the basic forms of Equations (5–1) (\sigma_{m}=\left(\sigma_{\max }+\sigma_{\min }\right) / 2 and (5–2) (\sigma_{a}=\left(\sigma_{\max }-\sigma_{\min }\right) / 2,  \sigma_{a}=\left(\sigma_{\max }-\sigma_{m}\right)) to compute the values for the mean and alternating forces on the pin

The stresses will be found from \tau_{m}=\frac{F_{m}}{2 A} and \tau_{a}=\frac{F_{a}}{2 A}.
From stress transformation, the three mean principal stresses are as follows:
\sigma_{m 1}=\frac{F_{m}}{2 A}, \sigma_{m 2}=0, \sigma_{m 3}=-\frac{F_{m}}{2 A}
Similarly, the three alternating principal stresses are as follows:
\sigma_{a 1}=\frac{F_{a}}{2 A}, \sigma_{a 2}=0, \sigma_{a 3}=-\frac{F_{a}}{2 A}
We can now apply the Goodman criterion [Equation (5-32)]
\begin{gathered}\frac{K_{t} \sigma_{a}^{\prime}}{s_{n}^{\prime}}+\frac{\sigma_{m}^{\prime}}{s_{u}}=\frac{1}{N} \\\frac{K_{t}\left(\sigma_{a 1}-\sigma_{a 3}\right)}{s_{n}^{\prime}}+\frac{\sigma_{m 1}-\sigma_{m 3}}{s_{u}}=\frac{1}{N}\end{gathered}
Therefore,
\frac{K_{t} F_{a}}{A s_{n}^{\prime}}+\frac{F_{m}}{A s_{u}}=\frac{1}{N}
The material strength values needed are s_{u}=61000 \mathrm{psi} and s_{n}^{\prime}. We must find the value of s_{n}^{\prime} using the method from Section 5-6. We find from Figure 5-11 that s_{n}=23 \mathrm{ksi} for the machined pin having a value of s_{u}=61 \mathrm{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_{s t}=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 use C_R = 0.75 to produce a reliability of 0.999 (see Table 5–3).

Then

Because the pins will be of uniform diameter, K_t = 1.0. Let’s use N = 4 because mild shock is expected. Therefore,

Results Based on the above analysis, we can solve for the required area, A = 0.03919 in^2

Then the required diameter is D = 0.223 in.

Final Design Decisions and Comments:
The computed value for the minimum required diameter for the pin, 0.223 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.