Tied Arch - Roark's Formula

[vmirat]

I'd like someone to double check my calcs on a tied arch. I'm using Roark, Table 9.3.1.h. The arch has the following:

Span = 100'
Radius = 97.7'
Theta = 31 degrees = 0.54 rad
w = 625 plf
The top chord of the arch is 2-L5x3-1/2x1/2 angles, so this would be a thin arch.
I assumed k1 and k2 to be 1.
I assumed Rg = R.

I get LPh = 611 lbs and Ahh = 0.019
Therefore, Ha = 32,158#.

 

[rb1957]

is your w down (ie weight) or radial ?

if down, then the vertical component of the reactions is 31250 lbs, and the lateral componentis tan(0.54)* (yes?) which is 18615 lbs.

another day in paradise, or is paradise one day closer ?

 

[vmirat]

rb1957,
This is snow+dead. Are you using Roark's equation?

 

[rb1957]

ok, we're on the same page.

No, I'm assuming the reaction is inclined, along the tangent.

another day in paradise, or is paradise one day closer ?

 

[BAretired]

I don't have Roark, but if I understand your geometry correctly, the horizontal reaction is not 32,158#. It is more like 56,000#.



BA

 

[vmirat]

BAretired,

That's what I thought too. That's why I want someone to check the Roark formula.

I used (wL^2)/(8f) to check it and got 55,804# (f=14', which is the distance from the bottom chord to the top chord at mid-span). Something doesn't seem right.

 

[BAretired]

As I said, I don't have the Roark formula, so can't check it but your last calculation seems about right to me.

BA

 

[rb1957]

why wouldn't the reaction be tangential ? isn't that the solution to the three-force body ?

take 1/2 the arch, horizontal reaction at the top CL, inclined reaction at the ground.
vertical load on 1/2 arch is 32,500 lbs, horizontal off-set between load and reaction is 25'
if height 14', then horizontal reaction is 32500*25/14 = 58305 ...

dumba$$, took tan of the wrong angle !
i get the 1/2 angle as asin(50/97.7) = 0.537rad (about 30deg, about right ... the span is close to the radius, total angle about 60deg)
so the horizontal reaction is tan(pi/2-0.537)*32500 = 52460

another day in paradise, or is paradise one day closer ?

 

[BAretired]

rb1957,

Your answer is close, but no cigar. There is no reason why the reaction should be tangential because the arch is a circular curve and the moment diagram is parabolic.

BA

 

[vmirat]

rb1957,
I think your reasoning is right. The difference between answers is probably rounding error.

 

[BAretired]

What was the formula in Roark?

BA

 

[IDS]

Using Roark's Formula (7th Edition) I get:

Using Theta = ASin(50/97.7) = 0.537245 radians I get:

Lph = -651.496
Ahh = 0.011295
Ha = -57682

Or for Theta = 31 degrees (0.541052 radians):
Lph = -673.814
Ahh = 0.011691
Ha = -57634

So they are reasonably consistent with the other estimates.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

A linear frame analysis gives 56130, so BA's guess looks pretty good

By the way, it looks like a very slender section for a 100 foot span!

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

if this was a frame (a section supporting bending) then there'd be a single redundancy in the 1/2 arch (a moment in the arch at the CL); maybe a second one at the base.

if this is a ring (a section that doesn't support bending) then the 1/2 arch is a three force body and the reactions are tangential.

another day in paradise, or is paradise one day closer ?

 

[BAretired]

rb1957,

The arch, hinged at both ends and under uniform load, is not in pure axial compression, so it must be capable of resisting bending moment.

The reactions are not tangential to a circular arch. They would be tangential if the arch was parabolic.





BA

 

[KootK]

BA has the right of this. The support reaction is only tangential if the arch is in pure compression. And the arch is only in pure compression if its shape matches the funicular curve, which it doesn't. Interestingly, the dead load curve probably is a circular profile.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[BAretired]

I assume this thread is in reference to thread507-374346. As mentioned there, unbalanced load will produce more severe bending in the top chord (arch). Wind uplift will tend to put the bottom chord (arch tie) in compression.

BA

 

[IDS]

No, the dead load curve for a constant cross section is a catenary.

As Robert Hooke wrote, "as hangs the flexible line, so but inverted stands the rigid arch".

Well actually he wrote "abcccddeeeeefggiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuvx", but that's what he meant.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

I didn't make myself clear above IDS. I meant the curved shape of the applied dead load diagram. It would be cereal bowl-ish.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[SAIL3]

what allowable stress is being used for snap-thru for unbalanced loading?...