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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.

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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?

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  • $\begingroup$ If the red portion is a slot and is to scale I'd expect the bolt to slide out before it fails. But the force of the bolt will put friction on the joint and reduce shear of the bolt. Bolted plates in shear are typically held together at least as much by friction as by bolt shear strength. $\endgroup$ – Tiger Guy Feb 12 at 18:33
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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.

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  • $\begingroup$ I am dividing by 4 as i have seen it done like that. They had 2 bolts and still divided by 4. $\endgroup$ – Kyle Anderson Feb 12 at 11:01
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    $\begingroup$ @KyleAnderson, was that for the same layout? Maybe it was a double lap joint with two shear cuts per bolt? I am just guessing, but that would distribute the force on a total of four shear cuts. $\endgroup$ – ingenørd Feb 12 at 11:07
  • $\begingroup$ Yes it was actually, just checked. So i should just divide by 1? $\endgroup$ – Kyle Anderson Feb 12 at 11:21
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    $\begingroup$ @KyleAnderson Yes, that will do it. $\endgroup$ – ingenørd Feb 12 at 11:29
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    $\begingroup$ I gather you had a 4 that isn't there now. shear Stress=593N/m2. Is this number correct? That is a very small number for a square meter. Also I think you are rooting too many times. Just solve for A as area then determine the diameter. A = 42k/593 $\endgroup$ – Tiger Guy Feb 12 at 14:29

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