Problem 117 Find the smallest diameter bolt that can be used in the clevis shown in Fig. 1-11b if P = 400 kN. The shearing strength of the bolt is 300 MPa.

Solution 117

$400(1000) = 300[ \, 2(\frac{1}{4} \pi d^2) \, ]$

$d = 29.13 \, \text{mm}$ answer

How did it became 400*1000?

the 400KN force was converted to Newton (N), ie (400 * 1000)N = 400,000N

why use 2 @ area calculation

The bolt is subjected to double shear.

ok

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## Comments

## How did it became 400*1000?

How did it became 400*1000?

## the 400KN force was

the 400KN force was converted to Newton (N), ie (400 * 1000)N =

400,000N## why use 2 @ area calculation

why use 2 @ area calculation

## The bolt is subjected to

The bolt is subjected to double shear.

## Why have you converted kN to

ok