Shear Centre of asymmetrical section.

[truckdesigner]

Good afternoon folks,

I am currently attempting to calculate the shear centre and shear flows in a "built-up" asymmetrical section we are using in a particular fabrication. I have not done this type of thing for quite some time and am finding it quite difficult to find any worked examples using asymmetrical sections - every text I pick up has the same old easy C channel section.

Can anyone assist me?

Some help would be much appreciated.

Regards.

 

[Lion06]

Check Salmon and Johnson, if there's not an example there try Timoshenko's Mechanics of Materials.

Draw the cross section with a shear force diagram (of the cross section, not the typical shear diagram of the beam). Using VQ/I, get the shears at the corners and the max along the vertical legs (the vert legs will be parabolic). Now draw a vertical shear outside the section and use this force to determine the "eccentricity" of the shear forces acting on the section. Now determine the value of that eccenticity such that the shear forces acting on the section have no net moment. That is your shear center.

 

[ToadJones]

try this for starters....

 

[rb1957]

maybe a pic of the section would help us understand the issues ?

 

[BAretired]

This link may shed a little light on the subject:

http://theconstructor.org/2009/10/the-shear-center-concept-2/

BA

 

[Lion06]

what kind of cross section are you looking at? The Mechanics of Materials text I have by Timoshenko has a number of cross section examples for the shear center.

 

[truckdesigner]

I have a copy of Gere & Timoshenko Mechanics of Materials 4th Edition and it only has the basic sections in it.

Attached is a drawing of the section. For whatever reason it has come back from the fabricators as this so we need to run with it.

Once again - any help greatly appreciated.

 

[BAretired]

The shear center, by inspection, is going to be at mid height and a little to the right of the web. Using the principles outlined by SEIT, you should be able to pinpoint the exact location. It isn't very difficult.

Just out of curiosity, why do you want to know this?

BA

 

[truckdesigner]

Probably barking up the wrong tree anyhow, but originally it was requested of me as we were to do a "flange check" as we were to hang an underslung trolley and hoist off it. This has now changed however to a top running crab (running on both sides of a shipping container). We are having another section the same fabricated as I speak.

Now simply curiosity has got to me, and I would like to know a simple way of doing this in case of similar events in the future.

By the way - what is SEIT? If you have a solution handy could you please post it?

Regards.

 

[BAretired]

Sorry about that. I don't like acronyms either. SEIT refers to StructuralEIT who has posted a couple of comments on this thread. It is nearly midnight here and I do not have a solution handy, but if nobody else has provided you the answer by the time I wake up in the morning (don't forget, I am retired, so I can sleep in as long as I want to), I will endeavor to do my best. Good night.

BA

 

[Lion06]

BA has it right on. It's at mid-height of the section and to the right of the web. My Timoshenko text has an example of this. I'll post when I get to work, but the procedure is as I outlined above.

 

[prex]

Using the classical method, outlined by StructuralEIT (but not so simple as it appears from his words), I find (hopefully checked by someone else):
e=3(b²-c²)/[6(b+c)+h]
with
h=section depth (at mid thickness) (200)
b=flange width on larger side (100)
c=flange width on shorter side (50)
thickness assumed constant everywhere
e=shear center distance from web (located towards the shorter flange)=20.5
The formula correctly gives zero for a doubly symmetrical section and reduces to the known (Roark) formula for a 'C' section (c=0).

prex

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[rb1957]

"flange check" ... i'd be looking at flange bending ...

 

[ToadJones]

I get the shear center as 20.45 mm to the right of the center line of the web and in the center of the height.

 

[truckdesigner]

The flange check was a bending check.

SEIT could you tell me what edition of Timoshenko you have?

Prex that formula is great! Could you show me how it is derived?

Regards.

 

[Lion06]

I'll have to check in the morning.

 

[BAretired]

Proof of formula (see attached)

I = h³/12 + (b + c)(h/2)²*2

Q_b = (b+c)*h/2

v_b = V*Q_b/I = 6V(b+c)/(h(h+6(b+c))

H_t&b = (b²-c²)v_b/(2(b+c))

For no torsion, H_t&b*h = V*e

so e = 3(b² - c²)/(h(h + 6(b+c))

BA