truss (1 Viewer)

[richz]

hi ppl,

could anyone plz help me with a past hsc question,

i cant seem to get it here but heres the website

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2002exams/pdf_doc/engineer_studies_02.pdf

and the questions is 12a) thnx, its sum truss one..

 

[A2RAYA]

off the top of my head( i havent done any engineering work since my hsc ) i got around 3.46 Kn...using the method of sections...i'm not 100% sure of that though

 

[richz]

off the top of my head( i havent done any engineering work since my hsc ) i got around 3.46 Kn...using the method of sections...i'm not 100% sure of that though

ok can u teach me the steps, on how to do it?

 

[A2RAYA]

actually mate i did something wrong in the calculations so that answer is incorrect...but basically u split the truss by drawing a line through the members, anywhere along the bridge that runs through the member which u want to find the size of (call it member A)...next you find the point where the two other members (not member A) intersect and take sum of moments about that point...BUT u have to take into account the distance of A from the point you're taking moments about...you do this through simple trigonometry......to help you get started i took sum of moments about the second 6kN force and took the vertical (or Y) direction as positive....anyone wanna add anything to this..or make any corrections? because this is purely from memory.

hope it made sense

 

[richz]

ok i'll give it a go thnx

 

[mr_guy99493]

i got 56kN, can anyone verify this.
uhm, i did it by looking at all the forces in each joint.
would be interesting to know if i got this right, as i just started this a few hours ago, and came here looking to test my skillz

 

[mr_guy99493]

seems i have also learnt new skills after trying this other method.

what happens if you have them complex trusses with like, two of these trusses on top of each other
then the line goes through more than 2 other members....

 

[Xayma]

If:

A line can not be drawn that cuts through three members (not necessairly the one you want, since you could use method of sections, then method of joints) you must only use method of joints.

 

[richz]

so do u have to solve the reactions at each pt and then you use the method of sections

 

[A2RAYA]

yeah but if it's obvious what the reactions are going to be ie if there are for example 20 kN acting on the bridge and the bridge is an even 10m or something then you know it's going to be 20kN opposing it so then you can say "by inspection reaction A and reaction B = "10" kN each"

 

[richz]

ya i guess so just wandering if u had to solve them -_-

 

[SDuke]

i got 56kN, can anyone verify this.
uhm, i did it by looking at all the forces in each joint.
would be interesting to know if i got this right, as i just started this a few hours ago, and came here looking to test my skillz

My magnitude is 55.4 kN, pretty much the same as yours, u probably rounded off some figures. I'm not sure about the nature of A, i think it's in tension.

I summed the moments on the RHS of the sectioned line, this is my final line of working out:

Sum the moments where the two cut members intersect, where clockwise is positive = 3m x (-16kN) + (1.5tan30)A

A = 55.4kN (2 dp)

 

[fakingtheday]

Yeah my bad. Fark sill mistake. That won't get me 95% in HSC now will it?

 

[SDuke]

I summed moments around the same piont as you did.

In my calculation, i have a 16kN acting upwards, this being the fixed pin on the RHS of the rail system.

However, i found an error in my calculation, i did not include the 2kn acting downwards.

This is my final solution.

0 = A(1.5tan30) + 2x1.5kn - 3x16kn
A = 52.0 kn (2dp)

 

[mattmcg]

i get 51.96kN (2dp.)

in compression