corsasport.co.uk
 

Corsa Sport » Message Board » Off Day » Structural Engineers - Method of Sections


New Topic

New Poll
  <<  1    2  >> Subscribe | Add to Favourites

You are not logged in and may not post or reply to messages. Please log in or create a new account or mail us about fixing an existing one - register@corsasport.co.uk

There are also many more features available when you are logged in such as private messages, buddy list, location services, post search and more.


Author Structural Engineers - Method of Sections
cpcrampton
Member

Registered: 16th Feb 07
Location: Stirling
User status: Offline
6th Jun 12 at 09:01   View User's Profile U2U Member Reply With Quote

Hello AK

I worked this out for you.

GH = 131kn and it is in compression.

reaction at A = 113kn


you can confirm by using this link.
http://www.jhu.edu/virtlab/bridge/bridge.htm

hope this helps you in the future.
AK
Member

Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
6th Jun 12 at 09:30   View User's Profile U2U Member Reply With Quote

cheers, that app is quite handy...

It agrees with my answers for reactions and AF/AC.

But still not fully following how to get 132 for GH.

I've worked it out on the prev page... which part of the working is to shit?
cpcrampton
Member

Registered: 16th Feb 07
Location: Stirling
User status: Offline
6th Jun 12 at 09:32   View User's Profile U2U Member Reply With Quote

One other thing. if you are just getting into this i recommend getting a book.

This will take you from very simple beam elements through to full on bridge design (resonance, vibration etc.)

The book :-

"McGraw-Hill Series In Structural Engineering & Mechanics"

If you require any input for anything engineering related feel free to ask.
AK
Member

Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
6th Jun 12 at 09:36   View User's Profile U2U Member Reply With Quote

using that java app shows the load GH to be 197kN

2x113.3 - 30 = 197
cpcrampton
Member

Registered: 16th Feb 07
Location: Stirling
User status: Offline
6th Jun 12 at 09:46   View User's Profile U2U Member Reply With Quote

SUM Md = 0

Reaction Force at A = 113.3kN

(113.3*2m)-(30kN*1m)-(Fgh*1m) = 0
so 226.6-30-?=0
so ? = 226.6-30 = 196.6KN

You were screwing up by mutiplying fGH by 2. why were you doing that? its one element from D not 2. ignore my first answer of 132 i was looking at someone elses reply when i was writing it.
cpcrampton
Member

Registered: 16th Feb 07
Location: Stirling
User status: Offline
6th Jun 12 at 09:47   View User's Profile U2U Member Reply With Quote

quote:
Originally posted by AK
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


you had it correct here ^^^
AK
Member

Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
6th Jun 12 at 10:02   View User's Profile U2U Member Reply With Quote

yup.... that was my whole issue....

I'd had it right, but still think its too easy for the marks!!!
AK
Member

Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
6th Jun 12 at 10:07   View User's Profile U2U Member Reply With Quote

oh, and I was popping 2 in there as it was 2m from A... but thats just daft
LeeM
Member

Registered: 26th Sep 05
Location: Liverpool
User status: Offline
6th Jun 12 at 10:09   View User's Profile U2U Member Reply With Quote

I think you need to do it all from one point, aren't angular forces calculated from the components of a vector? I did this ages ago so without digging some paperwork out I can't remember exactly lol
cpcrampton
Member

Registered: 16th Feb 07
Location: Stirling
User status: Offline
6th Jun 12 at 10:16   View User's Profile U2U Member Reply With Quote

Lee you are correct, you can also do it point by point , using nodal analysis where you split the force up across each member using like 30kn/sin45 give the force down one member. so you would pick a point and draw a free body diagram where the sum of the forces = 0 then you move to the next point and do it one by one.
LeeM
Member

Registered: 26th Sep 05
Location: Liverpool
User status: Offline
6th Jun 12 at 10:20   View User's Profile U2U Member Reply With Quote

There we go, that's more what I was getting at
AK
Member

Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
6th Jun 12 at 10:23   View User's Profile U2U Member Reply With Quote

yup - nodal (method of joints)... that was the method I had to use to get the forces for AF and AC. Prevoius part of the Q

AK
Member

Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
6th Jun 12 at 13:55   View User's Profile U2U Member Reply With Quote

oh, thanks by the way
cpcrampton
Member

Registered: 16th Feb 07
Location: Stirling
User status: Offline
6th Jun 12 at 14:32   View User's Profile U2U Member Reply With Quote

no worries. I will introduce myself next time i see you at knockhill. Time attack !
AK
Member

Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
6th Jun 12 at 18:18   View User's Profile U2U Member Reply With Quote

I wont be there Working in Norway.

Fee will be there though...
cpcrampton
Member

Registered: 16th Feb 07
Location: Stirling
User status: Offline
6th Jun 12 at 21:46   View User's Profile U2U Member Reply With Quote

aww well, unlucky. il be in stavanger very soon, got ops meetings with conocophillips for a development well with some prototype turbodrills.

[Edited on 06-06-2012 by cpcrampton]
AK
Member

Registered: 5th Jul 00
Location: Aberdeen City
User status: Offline
7th Jun 12 at 08:39   View User's Profile U2U Member Reply With Quote

I'll be in Stavanger on the 18/19th June

  <<  1    2  >>
New Topic

New Poll

Corsa Sport » Message Board » Off Day » Structural Engineers - Method of Sections 22 database queries in 0.0133438 seconds