Determine the maximum safe tensile load that can be supported by a 1 m section of a double-riveted butt joint with 15-mm-thick main plates and two 8-mm-thick cover plates. There are six rivets in each of the outer rows and seven rivets in each of the inner rows.The rivets are all 20 mm in diameter

Question

Determine the maximum safe tensile load that can be supported by a 1 m section of a double-riveted butt joint with 15-mm-thick main plates and two 8-mm-thick cover plates. There are six rivets in each of the outer rows and seven rivets in each of the inner rows. The rivets are all 20 mm in diameter. 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

In analyzing a double-riveted joint, it is only necessary to analyze one side due to symmetry.

The safe tensile load based on double shearing of the rivets is equal to the number of rivets times the number of shearing planes per rivet times the cross-sectional area of the rivet times the allowable shearing stress.

[latex]L=n imes 2 imes A_{r} au_{d}[/latex]

n = 6 + 7 = 13

[latex]L=13 imes 2 imes \pi \frac{0.02^{2}}{4} imes 60 imes 10^{6}=490.1 kN[/latex]

The safe tensile load based on bearing stress is given by [latex]L=n A_{b} \sigma_{c}[/latex]

[latex]L=13 imes 0.02 imes 0.015 imes 131 imes 10^{6}=510.9 kN[/latex]

The cover plates have a combined thickness greater than that of the main plates, so detailed analysis need not be considered here.

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

[latex]A_{p}=0.015(1-6(0.02+0.0015))=0.01307 m ^{2}[/latex]

[latex]L=0.01307 imes 75 imes 10^{6}=980.3 kN[/latex]

To complete the analysis, it is necessary to consider the sum of the load that would cause tearing between rivets in the inner section plus the load carried by the rivets in the outer section. The sum of the loads is used because if the joint is to fail, it must fail at both sections simultaneously.

[latex]L=n A_{b} \sigma_{c}=6 imes 0.02 imes 0.015 imes 131 imes 10^{6}=235.8 kN[/latex]

The safe tensile load based on the tensile strength of the main plate between the holes in the inner section is

[latex]L=A_{p} \sigma_{t}=0.015(1-7(0.02+0.0015)) imes 75 imes 10^{6}=955.7 kN[/latex]

The total safe tensile load based on the sum of the above two values is 1.191 MN.

The safe tensile load for the riveted joints would be the least of the values calculated, i.e. 490.1 kN.

The efficiency of the riveted joint is given by

[latex]\eta=\frac{490.1 imes 10^{3}}{0.015 imes 1 imes 75 imes 10^{6}}=0.436=43.6 \%[/latex]

Question 16.5: Determine the maximum safe tensile load that can be supporte...

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