A single riveted lap joint of two similar plates as shown in the figure below has the following geometrical and material details.

width of the plate $w = 200 m m,$ thickness of the plate $t = 5 m m$ , number of rivets $n = 3$ , diameter of the rivet $d_{r} = 10 m m$ , diameter of the rivet hole $d_{n} = 11 m m$ , allowable tensile stress of the plate $σ_{p} = 200 M P a$ , allowable shear stress of the rivet $σ_{s} = 100 M P a$ and allowable bearing stress of the rivet $σ_{c} = 150 M P a$ .
If the rivets are to be designed to avoid crushing failure, the maximum permissible load $P$ in $k N$ is

22.50

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30.00

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7.50

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15.00

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Solution

The correct option is A $22.50$
$P_{crushing} = 3 (d \times t \times σ_{c}) = 3 (10 \times 5 \times 150) = 22.50 k N$

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The correct option is A 22.50Pcrushing=3(d×t×σc)=3(10×5×150)=22.50 kN

The correct option is A $22.50$
$P_{crushing} = 3 (d \times t \times σ_{c}) = 3 (10 \times 5 \times 150) = 22.50 k N$