Author Topic: Math Formula Needed (Read 5398 times)

CAB · « **on:** November 04, 2008, 04:38:12 PM »

This is a engineering problem but thought it would get more views here.

I'm trying to find the formula to solve for the Rod Load in this example.
Anyone have an equation for this?

jnieman · « **Reply #1 on:** November 04, 2008, 04:48:11 PM »

What's the rod spacing and length of canopy?

Strucmad · « **Reply #2 on:** November 04, 2008, 04:49:05 PM »

P = 500/sin(tan^-1(2'11.5 / 4'6.5))

This will find you the load, not the stress (ultimate or allowable)..

The 500# is the tributary load right ? (imperial ??)

Atook · « **Reply #3 on:** November 04, 2008, 04:56:23 PM »

The way I calc it:

T=1.83218*W

Where T=rod tension, W=Tributary Load

@W=500, T=916.1

Strucmad · « **Reply #4 on:** November 04, 2008, 05:05:24 PM »

Quote from: Atook on November 04, 2008, 04:56:23 PM

The way I calc it:

T=1.83218*W

Where T=rod tension, W=Tributary Load

@W=500, T=916.1

Yours is simpler...but we get the same answer :kewl:......

mjfarrell · « **Reply #5 on:** November 04, 2008, 05:07:21 PM »

did he say rod load?

CAB · « **Reply #6 on:** November 04, 2008, 05:10:07 PM »

Thanks so much for the replies.

Imperial, Yes
500 pounds tributary load, Yes

I would think you needed to consider the angle of the rod.
Thanks PS_port

Atook is the 1.83218 constant for a specific angle?

CAB · « **Reply #7 on:** November 04, 2008, 05:11:39 PM »

PS_Port is there a difference in this situation for the Load & Stress?

Maverick® · « **Reply #8 on:** November 04, 2008, 05:13:49 PM »

I saw that CAB.

Greg B · « **Reply #9 on:** November 04, 2008, 05:14:25 PM »

Quote from: Maverick© on November 04, 2008, 05:13:49 PM

I saw that CAB.

Ditto

Strucmad · « **Reply #10 on:** November 04, 2008, 05:14:38 PM »

Yes CAB that is for that angle...

1/sin(tan^-1(2'11.5 / 4'6.5)) = 1.83218

(tan^-1(2'11.5 / 4'6.5)) = your angle

CAB · « **Reply #11 on:** November 04, 2008, 05:25:07 PM »

Quote from: Greg B on November 04, 2008, 05:14:25 PM

Quote from: Maverick© on November 04, 2008, 05:13:49 PM
I saw that CAB.

Ditto

You're all too quick.

CAB · « **Reply #12 on:** November 04, 2008, 05:26:33 PM »

Quote from: PS_Port on November 04, 2008, 05:14:38 PM

Yes CAB that is for that angle...

1/sin(tan^-1(2'11.5 / 4'6.5)) = 1.83218

(tan^-1(2'11.5 / 4'6.5)) = your angle

Thanks for the clarification.

Atook · « **Reply #13 on:** November 04, 2008, 05:26:45 PM »

Stress will be the tension in lbs divided by the cross sectional area of the rod.

And Port is right, my solution was for the drawing shown. Port's solution can be used for other triangles, just substitute your H & W values.

My solution stayed away from trig (sort of) and is:

T=W*L/H

Where T=tension in lbs, W= load, L=length of hypotenuse, and H=height.

Strucmad · « **Reply #14 on:** November 04, 2008, 05:28:19 PM »

CAB,
Load is just the load being applied to the rod, the diameter of the rod can vary,

The load will remain the same but the stress developed in the rod will be dependant on the diameter,

stress = load / area of steel.

This is where imperial start to loose me. (here Newtons / millimetres ² = MPa)
I believe you guys work in KIPs ' (Pounds per square inch) so as long as you keep your units the same you should be ok.

Might have to wait for Bob G. to check..?

ronjonp · « **Reply #15 on:** November 04, 2008, 05:28:26 PM »

Quote from: Atook on November 04, 2008, 05:26:45 PM

Stress will be the tension in lbs divided by the cross sectional area of the rod.

And Port is right, my solution was for the drawing shown. Port's solution can be used for other triangles, just substitute your H & W values.

My solution stayed away from trig (sort of) and is:

T=W*L/H

Where T=tension in lbs, W= load, L=length of hypotenuse, and H=height.

Nerd

CAB · « **Reply #16 on:** November 04, 2008, 05:40:18 PM »

I understand now.

I love this stuff

CAB · « **Reply #17 on:** November 04, 2008, 05:49:53 PM »

I was trying to get the formula in a LISP form but don't know what to do with the -1 power.

Code: [Select]

(/ load (sin(tan (/ H W))))Hight & Width

jnieman · « **Reply #18 on:** November 04, 2008, 05:56:36 PM »

I'm sticking with my reply

Strucmad · « **Reply #19 on:** November 04, 2008, 05:56:49 PM »

tan^-1 is the inverse tan or (arctan)

found this it might help. (note radians)

http://www.afralisp.net/Tips/code20.htm

Strucmad · « **Reply #20 on:** November 04, 2008, 06:00:22 PM »

CAB
sorry mate a man of your experience would know the (radians) thing....
I didn't need to point that one out..

Spike Wilbury · « **Reply #21 on:** November 04, 2008, 06:06:52 PM »

Quote from: CAB on November 04, 2008, 05:49:53 PM

I was trying to get the formula in a LISP form but don't know what to do with the -1 power.
Code: [Select]
(/ load (sin(tan (/ H W))))Hight & Width

just in case - load is a lisp function - .....

Atook · « **Reply #22 on:** November 04, 2008, 06:24:47 PM »

CAB, my equation is the same as Port's, just without any trig functions (you could say they've been evaluated already).

SohCahToa right?

Strucmad · « **Reply #23 on:** November 04, 2008, 06:39:13 PM »

Atook, your right there.

CAB just divide the hypotenuse by the height and multiply by load

Code: [Select]

(* load (/ (sqrt (+ (* h h) (*w w))) h))
I'm not good at lisp, someone might need to check the syntax

CAB · « **Reply #24 on:** November 04, 2008, 06:55:14 PM »

Thanks Luis I just threw the load in there at the posting for readability.
PS_Port, Arctan I understand but did not recognize it as tan^-1
This works just fine.

Code: [Select]

_$ (setq l 500 h 35.5 w 54.5)(* l (/  (sqrt (+ (* h h) (* w w))) h))
916.089

Strucmad · « **Reply #25 on:** November 04, 2008, 06:57:44 PM »

Cool bananas

I get thrown with arctan

CAB · « **Reply #26 on:** November 04, 2008, 07:04:51 PM »

This works as well.

Code: [Select]

 (/ 500(sin(atan (/ 35.5 54.5))))
916.089

So here is the final test.

Code: [Select]

Where T=tension in lbs, W= load, L=length of hypotenuse, and H=height.

(setq W 500 H 35.5 L 54.5)

(setq T (/ W (sin (atan (/ L H)))))
916.089

(setq T (* W (/  (sqrt (+ (* H H) (* L L))) H)))
916.089

Bob Garner · « **Reply #27 on:** November 04, 2008, 08:26:48 PM »

My company wouldn't let us use the net today (!?). I'm finally home from a really long day and not presently capable of doing any math at all so I'm worthless. But looks like everyone's covered it well. All day I had that song on my mind, "Whiskey for my horses, champaigne for my men". After reading all your posts, I'm singing, "Whiskey for my horses, Pay raises for the Swampers".
Kips are in units of 1000 pounds or one kip = one thousand pounds. Anybody know who the hell Kip is?

Happy evening all. May your votes work out for you.

Bob

CAB · « **Reply #28 on:** November 04, 2008, 10:10:04 PM »

Bob,
Hope tomorrow is a better one.

Notsober · « **Reply #29 on:** November 04, 2008, 10:32:53 PM »

Quote from: Bob Garner on November 04, 2008, 08:26:48 PM

Anybody know who the hell Kip is?

yes. Kip is that burned-out engineer who needs help with his "engineering" from a guy named Robert that just so happens to be gifted with pointing out all of his mistakes.

Dr. George: "Good catch, Robert, wow! **makes sounds of a crowd screaming**

I need to make a youtube video of the engineer here... y'all would laugh your (_Y_)'s off!

Maverick® · « **Reply #30 on:** November 05, 2008, 01:16:27 AM »

Quote from: Bob Garner on November 04, 2008, 08:26:48 PM

Anybody know who the hell Kip is?

Like you people don't know.

She's only seventeennnnnn

craigr · « **Reply #31 on:** November 05, 2008, 07:26:51 AM »

I am VERY envious of you folks that can figure this out.

It makes me wish I would have gone further in Math & Engineering.

I bow to you....

craigr

Bob Garner · « **Reply #32 on:** November 05, 2008, 09:56:22 AM »

If y'all can find the book, "Simplified Mechanics and Strength of Materials", by Parker, it's a gem for figuring this stuff out. I told a welder about this book recently and he was able to find it. I think the author is now Parker and Ambrose.

Bob

Bob Garner · « **Reply #33 on:** November 06, 2008, 10:51:07 AM »

I usually figgur these brackets graphically since I have to draw them anyway. Attached is such a graphical drawing.
I hope this helps.

Bob G

Kate M · « **Reply #34 on:** November 11, 2008, 10:56:58 PM »

Quote from: Bob Garner on November 04, 2008, 08:26:48 PM

Kips are in units of 1000 pounds or one kip = one thousand pounds. Anybody know who the hell Kip is?

Short for kilopound?

Strucmad · « **Reply #35 on:** November 11, 2008, 11:27:01 PM »

Quote from: Kate M on November 11, 2008, 10:56:58 PM

Quote from: Bob Garner on November 04, 2008, 08:26:48 PM
Kips are in units of 1000 pounds or one kip = one thousand pounds. Anybody know who the hell Kip is?
Short for kilopound?

That would be so confusing if it were true.....
S.I. units, 'kilo' surely you guys wouldn't work in anything so logical as a 10³ system..

It would be more like a `kappa' which would be 4 & 3/4 of a drizzle which if you times by 67 is a quarter of a delta mile..

Kerry · « **Reply #36 on:** November 12, 2008, 12:17:01 AM »

confusing , but true.

jnieman · « **Reply #37 on:** November 12, 2008, 08:17:20 AM »

I like to think of it as "killa-pound" as in; a killer-pound. One bad motha, SHUTCHO MOUF pound... just talkin' bout kips...

Ya know.. wouldn't want to think the metrics are infesting our number system, now would I?

News:

Author Topic: Math Formula Needed (Read 5398 times)

jnieman

Strucmad

Strucmad

Strucmad

Strucmad

jnieman

Strucmad

Strucmad

Spike Wilbury

Strucmad

Strucmad

Bob Garner

Notsober

craigr

Bob Garner

Bob Garner

Kate M

Strucmad

jnieman