Question 3.21: A rectangular bar is used as a hanger, as shown in Figure 3–...
A rectangular bar is used as a hanger, as shown in Figure 3–18. Compute the allowable load on the basis of bearing stress at the pin connection if the bar and the clevis members are made from 6061-T4 aluminum. The pin is to be made from a stronger material.
Objective Compute the allowable load on the hanger.
Given Loading in Figure 3–18. Pin diameter = d = 18 mm.
Thickness of the hanger = t_{1} = 25 mm; width = w = 50 mm.
Thickness of each part of clevis = t_{2} = 12 mm.
Hanger and clevis material: aluminum 6061-T4 (s_{y} = 145 MPa).
Pin is stronger than hanger or clevis.
Analysis For cylindrical pins in close-fitting holes, the bearing stress is based on the projected area in bearing, found from the diameter of the pin times the length over which the load is distributed.
\sigma_{b} = \frac{F}{A_{b}} = \frac{F}{dL}
Let \sigma_{b} = \sigma_{bd} = 0.65 s_{y} for aluminum 6061-T4
Bearing area for hanger: A_{b1} = t_{1}d = (25 mm)(18 mm) = 450 mm².
This area carries the full applied load, W.
For each side of clevis, A_{b2} = t_{2}d = (12 mm)(18 mm) = 216 mm².
This area carries 1/2 of the applied load, W/2.
Because A_{b2} is less than 1/2 of A_{b1} , bearing on the clevis governs.
Results \sigma_{bd} = 0.65=y_{s} (145 MPa) = 94.3 MPa = 94.3 N/mm²
\sigma_{b} = \sigma_{bd} = (W/2)/A_{b2}
Then, W = 2(A_{b2})(\sigma_{bd}) =2(216 mm²)(94.3 N/mm²) = 40 740 N = 40.74 kN
Comment This is a very large force, and other failure modes for the hanger would have to be analyzed. Failure could occur by shear of the pin or tensile failure of the hanger bar or the clevis.