Determine the safe tensile, shear, and bearing loads and the efficiency for a 300 mm section of single-riveted lap joint made from ¼ in. plates using six 16-mm-diameter rivets. Assume that the drilled holes are 1.5 mm larger in diameter than the rivets. The values for the design limits for tensile

Question

Determine the safe tensile, shear, and bearing loads and the efficiency for a 300 mm section of single-riveted lap joint made from ¼ in. plates using six 16-mm-diameter rivets. Assume that the drilled holes are 1.5 mm larger in diameter than the rivets. The values for the design limits for tensile, shear, and bearing stress can be taken as 75, 60, and 131 MPa, respectively.

Accepted Answer

The safe tensile load, L, based of shear of the rivets is given by

[latex]L=n A_{r} au_{d}=6 \pi \frac{0.016^{2}}{4} imes 75 imes 10^{6}=90.48 kN[/latex]

The safe tensile load based on bearing or compressive stress is given by

[latex]L=6 imes 0.016 imes 6.35 imes 10^{-3} imes 131 imes 10^{6}=79.86 kN[/latex]

The safe load based on tensile load is given by [latex]L=A_{p} \sigma_{t}.[/latex]

The area of the plate between the rivet holes, [latex]A_p[/latex], is given by

[latex]A_{p}=0.00635\left(0.3-6\left(16 imes 10^{-3}+1.5 imes 10^{-3} ight) ight)=1.238 imes 10^{-3} m ^{2}[/latex]

[latex]L=A_{p} \sigma_{t}=1.238 imes 10^{-3} imes 131 imes 10^{6}=162.2 kN[/latex]

The safe tensile load would be the least of the three values determined, i.e. L = 79.86 kN.

[latex]\eta=\frac{79.86 imes 10^{3}}{0.00635 imes 0.3 imes 75 imes 10^{6}}=0.56[/latex]

The efficiency is 56%.

Question 16.4: Determine the safe tensile, shear, and bearing loads and the...

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