Hello Veeky Forums

The solution is approx 2.34

I also don't get the aluminium vs steel part. The members were not given a specific diameter or shape. Either would work. This question really seems unclear.

oh diameter is also being asked in this question i forget to write that.

Same for all members or can it be different?
What about compression, how did your prof tell you to deal with that? Unlike tension it's very much dependent on the shape.

I scanned my notes. We always consider the unknown beam forces are tension. so when we solve the equibilium the tension forces becomes + and compression becomes -

All these retarded faggots in this thread.

Obviously this is a catch 22 as anyone worth a damn wouldn't use steel or aluminum, carbon nanotubes are the answer OP.

I'm sorry not everyone is a superior third-year student. Please, do tell us exactly which high schools are giving their students courses in statics.

Yeah but for like breaking how do you find it for compression? You just see if the pressure exceeds the material strength? That's not a very accurate way of doing it but it simplifies the problem. And are the beam allowed to have different diameters?

If they aren't going to help at all in solving the problem they can shut up. Looking at the problem initially some parts were in fact missing, saying "lulz 2 ez" was pretty retarded when the problem wasn't even solvable.

Knowing the structure materials is enough to solve for statically indeterminate systems, given that you put the answer down as a function of the cross-sectional geometries of the trusses.

Yeah sure but I very much doubt that's what either OP or the prof wants.

no they aren't and the book out professor follows (hibbiler's mechanical systems statics) says most of the time the weight of the beams are ignored

But then why give a density?

Fixed your circuit OP.

yeah right the book has no example with beams with weights

Shit.
Well, solving it without weight is trivial. Use the external force to find all the external forces on the truce, then split it apart to find the internal forces, and finally find the diameter needed to support that force, using the material you want I guess.
I still don't get how he wants you to deal with compression failure.

>8832462

>homework get you extra points on your final

wow american schools are a joke, i see now why so many faggots here have 4.0 GPAs

OP I solved this but I will only give it to you for .1 BTC

let me know when you are ready for the address

t. ME PhD Student

Seems legit
>american
baka

100% legit, this really isn't that hard. Sophmore level shit right here.

I scanned the solution on two pages and had to look up a few things but I am fairly sure I have the answer and solution.

Drop a burner if you want to discuss. For future reference sci will not do your homework for free

s1-9 are all sides of (assumedly) isomentric triangles. Therefore all the sides are 3m long.
200N at thehalfway between the two pivots + 350N at the haflway between the centre of s6 = 550N at 1/4 of 3m to the right of the 200n force point.
assuming that the pivots are on a ground that does not cave in under any force, the normal force should be distributed so that these forces cancel. the least amount of force will be applied on the far left pivot so that leaves us knowing that the majority of stress is on the structure over the right pivot.
The majority of the force is acting on the edge to the far right of the structure.

Niw i'm no engineer, but i am a physics student and a bright lad. So if i understand correctly then forces acting on the structure will be distributed accros the points by tension and conpression. Aluminium is more maliable and less flexible than steel but cheaper.

the force acting on both sides of the right-most pivot makes me think that s8 will be under a lot of stress by tension, while s9 and s7 will be under much compressive force.
The force.
s4 will be under compressive forces, s2 and 26 will be under tensile forces.

Try to figure the rest out on your own if you can help it OP. You got this far so you must have an idea of what you are doing.

Except I am not the OP.
And how did you deal with compression?

All the things you mentioned are just the obvious stuff, saying you wont do the rest doesn't change that.

you did physics 12 right?
Imagine that the sides are all both ropes and sticks (shrodingers beam if you will). You simply hae to figure out how forces are being dispersed across each of the beams (which are all 3m long) and determine from that whether aluminium or steel will be able to hold against the forces in any one of the connecting beams.

you are right s8 is under 1.875N tension force and the s9 is under 3.125N compression force

>And how did you deal with compression?
brainlet pls

I know but im not engineer and i only know principle physics. What i do know is that the solution involves the obvious: figure out how the forces of compression and tension are distributed across each beam after determining how the forces at each connecting point interact with them. then determine which material can withstand it.

Only if you exclude the weight of the material. Doing it without the weight of the beam is piss easy.
How do you calculate the compressive strength of the beam in this instance?

Yeah that's right you couldn't.
Well yeah so you know what we do and you don't know what we don't. You don't really get to school him on that.

How do i add weight?

>Yeah that's right you couldn't.
how about you tell me what to do about compression?

Like I said I have this solved and I am waiting for BTC payment until I post or email it to someone.

lel i'm broke

>b8
I don't, and neither do you, because the problem is unclear.

>how
That's kind of the point.

>implying
I do know, I just am not giving hints.

will you be able to make .05 BTC?

this is what he wrote on the board

>I know the answer I just don't want to tell you >:(
Literally 5 grader behavior. I am not even asking for a solution just how you did a part of it, because it's either undoable or you added your own specifications that were not included in the problem. Come up with an answer, otherwise you're just baiting.

It still doesn't help us you have yet to explain how your prof want's you to find compressive failure.

from what i have learned: trionometry.
We know the triangles have an isometric geometry, so we know that all the angles in the triangles are the same: 60 degrees.
knowing this, we know that if a force is applied at any point, it is evenly distributend down two arms compressively and those two arms are held together by one arm under a tensile strength equal to the sum of 2T un the image i have left.
I hope this helps because it's been a year since i last did this, and im also not an engineer.

teacher has the same idea as me it seems:
although i feel stupid for not considering that s1 would be under tension.

Either way i know you are capable of doing it OP. You are a competent engineer (or can be one) if you study and work hard. Think about the question, it is not as hard as it may seem.

If you need help, walk us through your thought process as you go about solving it and we will correct you if you make a mistake.

As I said that's obvious, that's not what I am asking for. I am asking for how do you calculate the maximum compresive load a beam can sustain before it breaks. Normally you'd look up in a table of different cross sectional shapes, find something suitable, like a specific I beam, and use that shape in your calculations. From what I am getting, this is NOT how his prof wants him to solve this.

I do have the answer and I will only divulge it and my methods upon payment.

If you are interested give me an email address or express the interest to drop the money to my BTC address

btw the compression shit you are talking about isn't even that complex in the context of this problem

CE tutor here. Try doing this

This OP.

this truss problem doesn't work like that

don't you guys know how to choose which trusses to solve for first? literally first thing you learn when learning to solve truss problems

Assume all the forces are in tension and if you get a negative number, you assumed wrong.

But that's insane.
You know I can give you a proof or the colatz conjecture if you give me three fiddy. for legit

sorry I'm not capable of doing that yet but i am researching pdf's but i still can't find how am i going to use weight. I understand the last part
we start from which has less members.

How is that insane? That is how it is taught

no not wrong that means that is compression

Is each member of the truss weighted?

Well maybe but it's still insane. It's not even a simplification it's outright wrong. Not sure why they teach that.

So it doesn't.
Can you lead the way Einstein? I'm new to this engineering stuff.

>we start from which has less members.

there are some exceptions

starting at the wrong truss can give you redundant equations in later steps. it's all about knowing where to base the system of equations

yes they are weighted

I am the one who has the solution for .1 BTC

It looks like OP doesn't want to pay so I won't post full/complete solution nor will I give significant help

However ask a question or present a solution and I can tell you if things are right I s'pose

I'll give you a tip. The sum of forces acting at any point of the truss is zero.

Unless it's a dynamic truss :^)

pls lets not go there

skyciv.com/free-truss-calculator/

you're welcome

but there are wheels on one of teh pivots.
we all know what that means (even me, the not-engineer): dynamic trusses!!!!! (i assume)

You're a fuckimg retard, no wonder your whole class failed. If you assume something is in tension, and get a negative answer, what does that mean? What is the opposite of tension?

nah those are just rollers they are used all the time in statics, that just means the bar cannot be moved in the upwards direction so there is a force in the upwards direction to keep it in statics equilirium.

compression i'm OP he is someone else. I'm saying the same thing for 2 hours

>that just means the bar cannot be moved in the upwards direction so there is a force in the upwards direction to keep it in statics equilirium.
please elaborate.

>I am in need of brilliant civil, mechanical or mechatronics engineers.
You realize this is a simple first year problem, right? Drop out of engineering and switch into arts you little babby.

You say you assume all beams are in tension. You are the retard. Of course with those rules a negative answer would be compression. And I am not OP.

If you have a roller, you can rotate the beam and move it in the x direction no problem, but there is a force acting on it in the y direction so it cannot move.

A wheeled pivot carries no moment or vertical reaction force.
There is literally nothing wrong with assuming all members are in tension or compression. This is how these problems have been solved for decades. Educate yourself.

threads like this make me feel all warm and fuzzy with job security.

keep struggling you glorious pleb. have fun being a """"cost analyst"""" or """project manager""" when you graduate. lel.

>A wheeled pivot carries no moment or HORIZONTAL reaction force.
I am tired.

user i'm a mechatronics student i won't use this for rest of my life.

>mechatronics student
>can't solve a simple force balance problem
Sounds about right. Have fun being a jack of all trades, master of none.

>3.488kN
>-.438kN

figure out the rest yourself

>user i'm a mechatronics student i won't use this for rest of my life.

yeah, because you are going to end up in sales or customer service or some other bullshit where you are just moving money around in excel all day. you sure as shit aren't going to be doing engineering.

OP, you won't make it past first year.

Cost analysis seems like a pretty cozy job, actually. It's just a small fraction of the work an engineer does, but with a large fraction of the pay.

>but with a large fraction of the pay.
lol