Eccentric Loading of Bolted Joint

[Ski8839]

It may be that I came across this issue at hour 8 of studying today, but either I am just really missing something (perhaps very simple) or there is an error in the solution.

This is from the 6 Minute Practice Problems, Depth Section, Problem # 47

There a four equally sized fasteners, at the corners of a rectangle, 10" wide x 12" tall.

What I don't have issues with understanding/agreeing to in the solution:

1) The moment at the centroid. It is a Clockwise Moment.

2) The polar moment of the bolt group

3) The secondary (moment) shear stress for each bolt

What I am having troubles with:

1) Resolving the secondary shear stress into it's component portions (x & y / horizontal and vertical) to evaluate the critical fasteners (the right two of the group).

So my geometry / trig shows that the ratio of stresses follow that of a triangle with a horizontal leg of length 5 (10/2) and a vertical leg of length 6 (12/2), with a hypotenuse of length 7.81.

So, in calculating the vertical component of the magnitude of the secondary shear stress, I multiply (6/7.81) x Secondary Shear Stress = Vertical component.

The solution manual shows that what I think is the vertical component is the horizontal.

The solution manual shows that what I think is the horizontal component is the vertical.

I can post a drawing if this helps.

What am I doing wrong?? I have drawn, re-drawn and thought about this for way too long today.

Hopefully someone can illuminate what is happening here.

Thanks!

 

[Ramnares P.E.]

Can you post a drawing and your solution so far?

 

[monty01]

Hey Ski, remember that the direction of that torsional shear is 90 degrees from the line you are using for your length ratios (5 & 6).  Since it is 90 deg from that line, the actual leg ratios of the stress are now reversed from that of the "r" that you used.  In other words, the rise becomes the run and vice versa for a line normal to the "reference" line.

In order to check this, draw a 3/4/5 triangle on engineering paper, then draw a line normal to that hypotenuse.  Compare the slope of your new line to the original hypotenuse.  

To summarize, the actual stress vector for torsional shear has a rise of 5 and a run of 6 (opposite from the line from center of bolt group to bolt):

Fx = F(5/7.81)

Fy = F(6/7.81)  

 

[Ski8839]

Hi Monty,

Thanks for explaining that. Frustrated that I missed that detail, but sure enough, after sketching it out (to scale!) per your suggestion, simple inspection shows that to be the case.

Not to beat a dead horse here, but if you feel like it, mathematically why is that? I can see by inspection and logic that it is correct, but it now has me wondering what property have I forgotten that explains why a line 90 degrees off of another has an inverted slope.

Math nerd stuff aside, thanks again.

Thanks Ramnares, Monty saved me from scanning and posting. Next time I certainly will.

 

[monty01]

Ski, im not sure if this property has an official name.  Maybe "reciprocal slope" or something.  But if you have a line defined by y=2x and you need to find the line perpindicular to it, it woud be the recipricol and of course negative slope, y=-x/2.

Now for the easy way to remember or think about it.  What about the x and y axes?  Y has an infinite rise, X has an infinite run.

 

[JHW 3d]

If you know the angle (α) of a line from +X, the slope is

mα = tan(α) = sin(α)/cos(α).

For a perpendicular line (increase angle by 90°), you then have
mα+90 = tan(α + 90) = sin(α + 90)/cos(α + 90).

[Apply angle addition identities:
sin(a + b)=sin(a)cos(b) + sin(b)cos(a)
cos(a + b)=cos(a)cos(b) - sin(a)sin(b)]

sin(α + 90)= cos(α)
cos(α + 90)= -sin(α)

Therefore
mα+90 = cos(α)/(-sin(α)) = -1/ mα

same result applies for a line at an angle -90° from original line, although in that case the negative sign follows the cosine term.

 

[Ski8839]

Monty,

Thank you.

What did you mean however by the axes having infinite rise/run? I believe that they do, however, how does this help you remember in this case?

Johnny,

Thanks! That's the dense math I new had to exist out there somewhere. Now, just for me to stare at it long enough to sink back into my brain!

 

[monty01]

Hey Ski, yeah...it's just an easy way to remember that perpendicular slopes are inverses.  Formally, the negative inverse.  But in our case, we don't care about signage, as that is taken into account with the component equations of the vector. 

 

[Ski8839]

Thanks Monty!

 

[volzyl]

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