Archinect
anchor

how would you crack this problem?

RTVSkaarchitecture

it's a boom at a right angle with a weight pulling down

I've never had a statics class, I'm just wondering.

 
Mar 5, 17 10:07 am
threadkilla

No idea what 'the problem' is, but the solution involves trigonometry. And the steel bible. I recommend taking a statics class, it's not rocket science.

Mar 5, 17 5:31 pm  · 
 · 
Wood Guy

I don't see a specific question, but in general, it's a torque problem, acting relative to the pivot point--the lower connection at the wall. (Torque is force times distance; units typically ft•lbs or Newtons.) In the counterclockwise direction you have the force pulling down times the distance to the pivot point. Acting in the other direction is the tension load in the cable; that distance is measured from the pivot point to the point on the cable perpendicular to the pivot point. The two torques have to cancel each other out, so force times distance of one equals force times distance in the other.

Once you know the tension load in the cable, you can resolve it to its vertical and horizontal components, which happens to relate proportionately to the dimensions of the triangular shape you are analyzing.

Mar 5, 17 6:30 pm  · 
 · 
starrchitect

I would hire a structural engineer and let him take the liability. 

Mar 5, 17 9:19 pm  · 
 · 
chigurh

that has nothing to do with torque.  wood guy has no idea what he is talking about.

Mar 5, 17 10:13 pm  · 
 · 
randomised

"how would you crack this problem?"

Ask archinect of course! Can you transfer liability to a website though?

Mar 6, 17 5:18 am  · 
 · 
randomised

Maybe archinect could set up a PayPal system linked to the user accounts?

Mar 6, 17 11:49 am  · 
 · 
Non Sequitur

Just use the side of the pan like you would with an egg.

Mar 6, 17 9:29 am  · 
 · 
Wood Guy

Chigurgh, you're funny. I have a BS in engineering (90% of a civil/structural degree as part of a pre-arch program) and know exactly what I'm talking about. Sure, you can just apply trigonometry and get an answer, but the way I learned to visualize what was happening in a problem like that is to resolve the forces using torque equations. The question was "how would you crack this problem." If architects need to call an engineer to understand the forces in a simple diagram like that, that is sad. 

Mar 6, 17 11:07 am  · 
 · 
s=r*(theta)

I would start by mirroring the text so I could read it more legibly

Mar 6, 17 11:21 am  · 
 · 
Wood Guy

RVTS, here is a decent video solving your problem using classic statics techniques: https://www.youtube.com/watch?v=dbjNX91EQh0&t=18s.

Here's one using torque to actually visualize what is happening: https://www.youtube.com/watch?v=V_lxELlEOFg.

Unless you're a geometry wiz, there's nothing much intuitive about the first approach, though it should be simple enough once you learn the principles. I'm a visual learner and the concept of torque (which, again, is simply force times distance) makes more sense to me. 

Mar 6, 17 11:22 am  · 
 · 
Bloopox

First I'd go find the textbook, so I could see the problem in a not-blurry and not-backwards manner.  Then I'd read the book, and do my homework.

Mar 6, 17 11:24 am  · 
 · 
Wood Guy

I'm wondering if the OP was specific with his title wording just to see what kind of responses he'd get...

Mar 6, 17 11:33 am  · 
 · 
JLC-1

I would call my structural engineer.

Mar 6, 17 11:34 am  · 
 · 
Non Sequitur

 To the regulars here, let's not forget the OP's already stated in previous posts that he's a structural wizard and will force the entire shut-down of US construction via strike among many other things.

Mar 6, 17 11:50 am  · 
 · 
archietechie

Wasn't that koz? lol

Mar 6, 17 12:13 pm  · 
 · 
Non Sequitur

Look up his history. Not Koz level inanity, but not far.

Mar 6, 17 12:29 pm  · 
 · 
tduds

You weren't kidding

http://archinect.com/forum/thread/149937076/this-nitwit-b-arch-stuff

Mar 6, 17 1:24 pm  · 
 · 
x-jla

I would solve the problem by sending it to my engineer.  

Mar 6, 17 12:42 pm  · 
 · 
tduds

Cop out.

Mar 6, 17 2:18 pm  · 
 · 
x-jla

It's not a design problem. It's an engineering problem. Intuitively I know enough to know what can and cannot be done. Let the engineers size the members and prove me right. Checks and balances.

Mar 6, 17 3:02 pm  · 
 · 
x-jla

That said, there is a turning moment at the bottom member where it connects to wall, that member is under compression and will buckle if not sized correctly, the top connection has a pulling force, and the member is in tension. The angle of that force is complicated because it's wanting to rotate...essentially like a lever where the distance from the fulcrum point will increase the turning moment force. The points of most concern are the top connection point and the bottom members compressive strength/section.

Mar 6, 17 3:07 pm  · 
 · 
tduds

Sure, I just don't like the trend in this thread (and elsewhere in this forum) of people responding this way whenever non-design problems come up. Especially because it's mostly from students in courses we all took and that we all know have value.

IMO it sends a message of being proud to not know something, which is one of the most offensive things I can think of.

Mar 6, 17 3:08 pm  · 
 · 

Do your homework.

Mar 6, 17 2:04 pm  · 
 · 
mightyaa

Wood guy is close; it's a moment arm type problem.  So the horizontal is pushing against the wall (where the moment radius starts), the top arm is trying to pull out.  So you use 'geometry' and vector forces to figure out the loads and size everything correctly.

It's highschool level physics & geometry.  So if you can't solve this, you really need an engineer.  Simple overkill if in a rush: if it's 1000# hanging on a 3' horizontal, figure 3000# pullout at the top (3x1000), and 3000# push on the wall at the horizontal.  Also don't forget that wall you are connecting to is also going to want to rotate now, so it needs to be designed for it.

Mar 6, 17 2:37 pm  · 
 · 
Wood Guy

.

Mar 27, 17 9:40 am  · 
 · 
your conscience

how much harder than statics is rocket science ?

Mar 6, 17 11:35 pm  · 
 · 
Bloopox

^ Plagiarized from "Cracking A Problem:  Another Use for a Fashionable, New Materials", The Economist, 2011.

Mar 7, 17 10:12 am  · 
 · 
randomised

Sharp! It's just to establish an on-line persona before the spamming of obscure websites commences.

Mar 7, 17 10:32 am  · 
 · 
MrVSNET

1. YouTube

2. Google

Mar 26, 17 11:26 pm  · 
 · 
Volunteer

The force is dependent of the weight on the end of the beam and the angle the smaller beam, or cable, makes with the horizontal beam. Oddly enough, the length of the horizontal beam is not a factor.

Mar 27, 17 9:01 am  · 
 · 
Wood Guy

It's a proportional relationship. If the beam had no mass, a 10' beam with a support 5' above would have the same reaction forces as a 100' beam with a support 50' above. I know that's what you are saying as well, but to say the beam length is not a factor isn't really correct.

Mar 27, 17 9:34 am  · 
 · 

Block this user


Are you sure you want to block this user and hide all related comments throughout the site?

Archinect


This is your first comment on Archinect. Your comment will be visible once approved.

  • ×Search in: