Question 3.21: The lug for a clevis joint of the type shown in Figure 3–33 ...

Objective: Specify w, h, and t and the nominal clearance.

Given: \quad d=12.0 \mathrm{~mm} ; d / w=0.40 . F=8.75 \mathrm{kN}
Analysis: Lug thickness, t: Recommended t / d \leq 0.50. Using the upper limit gives: t=0.50 d=(0.50)(12.0 \mathrm{~mm})=6.0 \mathrm{~mm}
Lug width, w : Using d / w=0.40, w=d / 0.40=12 \mathrm{~mm} / 0.40=30.0 \mathrm{~mm}
End distance, h : Recommended ratio h / w=1.0. Then,
h=w=30.0 \mathrm{~mm}
Close-fitting pin: d_{\text {hole }} \cong d_{\text {pin }}(1.002)=12.0 \mathrm{~mm}(1.002)=12.024 \mathrm{~mm}
Maximum stress:

\begin{aligned}&\sigma_{\max }=K_{t} \times \sigma_{\text {nom }} \\&\sigma_{\text {nom }}=F /(w-d) t=\left[8.75 \mathrm{kN}/(30.0-12.0)(6.0) \mathrm{mm}^{2}\right](1000 \mathrm{~N} / \mathrm{kN}) \\&\sigma_{\text {nom }}=81.8\mathrm{~N}/ \mathrm{mm}^{2}=81.8 \mathrm{MPa}\end{aligned}
To find K_{t}: At d / w=0.40 in Figure 3-33, K_{t} \cong 3.0. Then
\sigma_{\max }=K_{t} \times \sigma_{\text {nom }}=(3.0)(81.8 \mathrm{MPa})=243 \mathrm{MPa}

Results: The design details for the lug of the clevis joint in Figure 3–33 are:

d=12.0 \mathrm{~mm} ; w=30.0 \mathrm{~mm} ; t=6.0 \mathrm{~mm} ; h=30.0 \mathrm{~mm}
Nominal clearance between pin and hole: \delta=0.02 \mathrm{~mm}
Maximum stress in lug at hole: \sigma_{\max }=243 \mathrm{~mm}