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.

143801 3-18
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Objective               Compute the allowable load on the hanger.

Given                     Loading in Figure 3–18. Pin diameter = d = 18 mm.

Thickness of the hanger =t1 t_{1} = 25 mm; width = w = 50 mm.

                               Thickness of each part of clevis = t2 t_{2} = 12 mm.

Hanger and clevis material: aluminum 6061-T4 (sys_{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.

 σb=FAb=FdL \sigma_{b} = \frac{F}{A_{b}} = \frac{F}{dL}

Let σb=σbd=0.65sy \sigma_{b} = \sigma_{bd} = 0.65 s_{y} for aluminum 6061-T4

Bearing area for hanger: Ab1=t1d A_{b1} = t_{1}d = (25 mm)(18 mm) = 450 mm².

This area carries the full applied load, W.

For each side of clevis,  Ab2=t2d  A_{b2} = t_{2}d = (12 mm)(18 mm) = 216 mm².

This area carries 1/2 of the applied load, W/2.

Because Ab2 A_{b2}  is less than 1/2 of Ab1 A_{b1} , bearing on the clevis governs.

Results                 σbd=0.65=ys \sigma_{bd} = 0.65=y_{s} (145 MPa) = 94.3 MPa = 94.3 N/mm²

 σb=σbd=(W/2)/Ab2 \sigma_{b} = \sigma_{bd} = (W/2)/A_{b2}

Then, W = 2(Ab2)(σbd) 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.

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