Restraints summary in ceasar report

HI,

I am woking on a system in which my line is oriented at an angle from the global axis. Now for getting the guide loads, i have to calculate the resultant from loads in two global axis. is there any shortcut in C-II to find out loads on these restraints in local coordinates.. C-II only gives forces on local elemnts but doesn't have any idea about restraints...

Forces on local elements is the same with the restraint loads in terms of magnitude the only thing to remember here is to flip the
signs on all of the forces and moments. Precisely because in local element forces report the force is acting on the element (Free-body Diagram) while the local restraint forces, forces are acting on the restraints.
You can refer also to Technical Reference (Local coordinate)or use the Global to local transformation utility in COADE website.

Thanks for your response Nald, But when I am viewing the local element forces in output report, for a same node no where restraint is given,which occurs in two consecutive element force value is different. So which value should i use?


Node No. fx(N) fy(N) fz(N) mx(N.m) my(N.m) mz(N.m)

40 27872 182 -11775 11746 -38921 2847
50 -27872 -182 800 -11746 44705 -2680

50 232846 -38 -800 11792 -1593 -42
60 -232846 38 -55 -11792 1620 39

add the two nodes (50) then flip the sign of the results and that will be the forces acting on your restraint. Read Technical Reference manual.
If that is skewed line, you can check the results using the Global-to-Local transformation utility. Download that utility from COADE website, generate restraint report, get the restraint load for node 50, input it together with the distance bet. two nodes (DX,Dy&DZ) including sign.
Try and compare the results..!

Hi I cannot find on the website the utility Nald mentions.
Can someone give me any advice?
Regards

http://www.coade.com/AllProductDownloads/PD#CAESARII

Regards,