Determine Bolt dia from Shear stress and force

Question

The Diagram below shows a lap joint where two plates are held together by a bolt. The applied force (F) is $42~\text{kN}$ and the shear stress is $593~\frac{\text{N}}{\text{m}^2}$ Assuming that the joint will be held together by a standard metric bolt, determine the minimum diameter permissible and identify the standard bolt diameter that will need to be used.

This is my answer is this correct?

$$\sum f_x=0$$

$$V = \frac{p}{1} = \frac{42}{1} = 42kN$$

$$\text{shear Stress} = 593N/m^2$$

$$593^2 = 351.649 Kn$$

$$ \frac {42000}{351.649} = 119$$

$$ a = \frac{pi}{4} D^2$$

$$D = \sqrt\frac{4A}{pi}$$

$$D = \sqrt\frac{(4) (119)}{pi}$$

$$ = 12.309$$

Is this the correct answer, and the correct way to complete this question?

Answer 1

Your units don't match, and you are calculating a required area of 17.7m², not a diameter of 17.7mm. I guess, the shear stress should have been an allowed stress in MPa, not Pa.

Why do you divide the force by 4? Are there four bolts even though your diagram only shows one?

When converting the required area to a bolt diameter, you also need to consider whether the shear cut is placed in the shank of the bolt or in the thread.

And finally you need to check the contact stress between bolt and plate (the bearing resistance), and adjust for the eccentricity of the load in a single lap joint with just the one bolt.