Flexural strength calculation question

[jsu0512]

Could someone help me with the calculation of flexural strength of the clip when it is subjected to the 2kN as shown in the attached drawing?

I would appreciate it if someone tell me what to check when this clip is subjected to the load as shown on the attached drawing.

Thank you

 

[FEA way]

Of course you could solve this with FEA. But if you want to do it analytically then treat this clip as a frame:

 

[BAretired]

It depends on what kind of precision you need. You need to know the height of the vertical legs if you need a precise elastic solution.

If you just need an approximate solution, I would assume that failure occurs when a plastic hinge occurs at the ends and middle of the horizontal beam where the load is applied.

L = 70mm
b = 50mm
t = 1/4" = 6.35mm
Plastic Modulus, Z = bt^2/4 = 50(6.35)^2/4 = 504mm^3
Plastic Moment = Z*Fy = 504*260 = 131,000N-mm or 131kN-mm

Mp = PL/8

P = 8*131/70 = 15kN (well in excess of 2kN)

Should really check the 1/2" bolt for shear stress, but would need the height of vertical legs for that.

But even in a worst case scenario, where the beam acts as a simple span, the moment cannot be more than PL/4 which would mean P at failure would be about 7.5kN. Still well in excess of 2kN.

BA

 

[jsu0512]

Hi BAretired,

The height is 40mm. How does the the bolt shear is calculated with the height of vertical leg? I appreciate for your response.

 

[dik]

Your FBD looks like...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[BAretired]

Hi BAretired,

The height is 40mm. How does the the bolt shear is calculated with the height of vertical leg? I appreciate for your response.

The structure is indeterminate, which means that to solve the problem elastically, one must either have a 2D frame program or tables. The curves at the bend points would also need to be taken into account for a thorough analysis.

The bolts add several degrees of indeterminacy to the problem. If, instead of bolts, you assume the clip is supported on a flat surface with a pin under one wall and a roller under the other, the problem becomes determinate. In that case, the load will cause the supports to spread apart which they can do freely. In that case, the top beam cannot develop moment at the ends, so it spans as a simple span with concentrated load of 2kN at midspan. The failure load would be P = 7.5kN as shown in my earlier post. To be consistent with normal practice, a phi factor should be included to account for unknown properties of the material. In Canadian practice, a phi factor of 0.9 is normally used, giving a calculated failure load of 6.75kN instead of 7.5kN.

In my judgment, the problem, including bolts each side, tightened down to prevent rotation, is closer to a rigid frame fixed at the bottom of each vertical wall; closer, but not exactly correct. So I tend to believe that the failure load is likely closer to 15kN or, with phi factor included, 13.5kN.

Like many engineering problems, more precise calculations can be made. The question is, does it really matter in this case? If the expected maximum load is 2kN, then either solution would seem to provide an adequate solution without going into further refinement.

BA

 

[BAretired]

FEWway,

Fixed end at point 1 is pretty close to reality, but there should be a roller support under point 2 as it does not appear to be free to move vertically. There should be a curved section at points 2 and 3 and the uniform load of 0.026kN/mm is a complete mystery to me.

dik,

I believe the intention was that W would act at Joint 4 (or S2). A vertical roller is shown at S2 and I think the idea was that only half the frame would need to be solved.

In my opinion, any attempt to solve this problem would be a waste of time and effort. If it were to be done, probably it should be done using plastic design rather than elastic design, since the original question asks for the strength of the clip. If, on the other hand, the question is asking for stresses in the clip under the 2kN load, then an elastic analysis would be required. So, the OP's question is not entirely clear.

BA

 

[dik]

BART... the approach I would use, if I had to analyse it... plastic design not such a good idea... moment is max throughout most a lot of the structure... redistribution is limited... like a cantilever or a column.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[FEA way]

Since the force is applied to the whole top surface of the bracket, I converted it to line load (UDL). The value may not be correct, it’s there just to give an idea on how the frame should like like.

When it comes to curved beams, they are not available in the app used for this scheme and for this reason (but also to make it simpler) there are only straight lines.

 

[BAretired]

dik,

If you are not going to use plastic design, what do you suggest? Unless there is good reason to do otherwise, a conservative approach should be taken.

FEA way,

I interpreted the 2kN force as a concentrated load at midspan, not as a uniform load applied over the top surface of the bracket. Since the concentrated load is more critical, conservatively, it makes sense to use it.

BA

 

[BAretired]

BART... the approach I would use, if I had to analyse it... plastic design not such a good idea... moment is max throughout most a lot of the structure... redistribution is limited... like a cantilever or a column.

Your message is not complete, dik. What approach would you use to analyze it, if not plastic design?

BA

 

[Enable]

Your message is not complete, dik. What approach would you use to analyze it, if not plastic design?

The FBD Dik provided is determinate and I assume he would resolve for forces in that system. He appears to be placing a pin at an assumed moment inflection point on the right most side of the top plate. I think what he's trying to get across is that if he had to analyze this thing that is how he would do it, probably juice the load with an ample safety factor, and call it a day. If someone wanted to be more precise than that it would just be modeled.

 

[BAretired]

Another approximation: make it a 3 hinged arch with hinges at points A, C and E.
Then H = 0.4375P and Mb = Md = 40H = 17.5P
Since plastic moment Mp = 131kN-mm, P(failure) = about 7.5kN (without phi factor)
And H = about 3.3kN (shear in bolt each side).

This is a very conservative estimate since there will not be zero moment at pin locations, but it may be adequate for the purpose at hand. A more realistic estimate could be made by assuming fixity at points A and E and a rigid connection at point C.

BA

 

[BAretired]

Enable,

Okay, but looking at the original diagram, wouldn't there be a vertical reaction at point 2 and its mirror image?

EDIT: Is it determinate? Looks like a 2 hinged arch.

BA

 

[Enable]

Okay, but looking at the original diagram, wouldn't there be a vertical reaction at point 2 and its mirror image?

I haven't done too many of these things by hand in nearly a decade so I am apt to be a bit rusty on the details but I believe dik's roller is in the wrong direction. Here is a look at a similar problem from Structural Analysis by Hibbler