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Author Structural Engineers - Method of Sections
AK
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4th Jun 12 at 14:27   View User's Profile U2U Member Reply With Quote
I'm having a bit of brain fade today for some reason I just can get my head around this problem.

I'm looking to resolve the forces in section GH (so, Fgh)

I worked out the reaction force at support A to be 113.3kN ^

Method of joints to give forces AF as 160.2 kN and AC as 113.3kN (think this is all to shit)

Orig structure



I've split the structure like so




any help resolving Force at GH using sections???
alan-g-w
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4th Jun 12 at 15:55   View User's Profile U2U Member Reply With Quote
3
AK
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4th Jun 12 at 16:34   View User's Profile U2U Member Reply With Quote
I'll take that as an attempt at humour?
alan-g-w
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4th Jun 12 at 17:48   View User's Profile U2U Member Reply With Quote
I'll take the op as an over zealous attempt at showing off how technical your work is? Either that or I'll take it as you being the dobber at work that no-one talks to, so you need to ask a corsa website a structural eng problem?

Aye, mine was an attempt at humour to maybe get the ball rolling and a few answers, but I'll give you 10/10 for consistency on the arsey attitude front yet again.
_Allan_
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4th Jun 12 at 17:56   View User's Profile U2U Member Reply With Quote
Alan let it go.
It's all water under the bridge
Ian
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4th Jun 12 at 17:59   View Garage View User's Profile U2U Member Reply With Quote
Truss you to come up with that
_Allan_
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4th Jun 12 at 18:06   View User's Profile U2U Member Reply With Quote
Thanks for the support Ian
AK
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4th Jun 12 at 18:41   View User's Profile U2U Member Reply With Quote
alan-g-w

What the fuck has got your back up? My post being over zealous showing how technical my work is!?!? This has NOTHING to do with my work! I'm doing an OU degree, and I remembered there were a couple of folk on here doing mech/structural engineering so thought I'd ask for some help.

I have an answer, but I just doesnt feel right. -13.3kN

Arent you a engineer of sorts in fact?


[Edited on 04-06-2012 by AK]
cpcrampton
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4th Jun 12 at 19:05   View User's Profile U2U Member Reply With Quote
Google nodal analysis , it's pretty straight forward.

If you require any additional help I am willing to help you out.
AK
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4th Jun 12 at 19:09   View User's Profile U2U Member Reply With Quote
I know the theory but I think I'm being a spastic

If you can (i.e if its quick and easy for you) can you tell me if the 13.3kN is correct.

I dont think it is.

I'll go have a scan of Nodal Analysis on google, ta

[Edited on 04-06-2012 by AK]
cpcrampton
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4th Jun 12 at 19:20   View User's Profile U2U Member Reply With Quote
Il run through it in 30 min sitting with my little one at the moment.
cpcrampton
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4th Jun 12 at 19:33   View User's Profile U2U Member Reply With Quote
http://www.sqa.org.uk/pastpapers/papers/papers/2011/H_Technological-Studies_all_2011.pdf

OK go to question 9 here ^^^^
And find the working a 9. here ----- :-

http://www.sqa.org.uk/pastpapers/papers/instructions/2011/mi_H_Technological-Studies_all_2011.pdf

I hope this helps. im kind of tied up at the moment otherwise i would solve it for you.
LeeM
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4th Jun 12 at 20:03   View User's Profile U2U Member Reply With Quote
Pretty sure it's not 13.3 but too drunk to work it out lol! Seems relatively simple though
AK
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4th Jun 12 at 20:52   View User's Profile U2U Member Reply With Quote
ok,

so I'll have a bash again...

Taking moments about D

SUM Md = 0

Reaction Force at A = 113.3kN

-(113.3*2m)+(30kN*1m)-(Fgh*1m) = 0
so Fgh = -196.6kN

Na, thats not fucking right
cpcrampton
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4th Jun 12 at 20:56   View User's Profile U2U Member Reply With Quote
Start from the fixed point and work your way to the roller at the left hand side. start your nodal at the 30kn
AK
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4th Jun 12 at 21:02   View User's Profile U2U Member Reply With Quote
I have to do it via sections, not joints (working from fixed point) :S
AK
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4th Jun 12 at 21:06   View User's Profile U2U Member Reply With Quote
so I'll have a bash again... Noticed a cock up in the 1st one

Taking moments about D

SUM Md = 0

Reaction Force at A = 113.3kN

-(113.3*2m)+(30kN*1m)-(Fgh*2m) = 0
so Fgh = -98.3kN

So 98.3kN (compression)



[Edited on 04-06-2012 by AK]
AK
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5th Jun 12 at 08:14   View User's Profile U2U Member Reply With Quote
had a chat with another guy thats doing the course and he did it the same way.

So.... i'll just go with that :S
csweatherston
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5th Jun 12 at 11:15   View User's Profile U2U Member Reply With Quote
i made it 132.2
AK
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5th Jun 12 at 11:22   View User's Profile U2U Member Reply With Quote
whet :S

Can you explain how you got to that?

feckin question
shaned12345
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5th Jun 12 at 11:31   View User's Profile U2U Member Reply With Quote
WTF is that?
csweatherston
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5th Jun 12 at 12:09   View User's Profile U2U Member Reply With Quote
https://courses.cit.cornell.edu/arch264/calculators/example1.6/index.html
AK
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5th Jun 12 at 12:34   View User's Profile U2U Member Reply With Quote
None of the pre-defined models are quite like the structure above. i'll have a fanny about with it.
AK
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5th Jun 12 at 20:35   View User's Profile U2U Member Reply With Quote
up... anyone else?
Bart
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6th Jun 12 at 08:04   View User's Profile U2U Member Reply With Quote
If I could be bothered to get my college work out my loft id have a go.

Did this around 5 years ago, but like anything use it or loose it.

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