DEAR MEMBERS,

DURING ONE ANALYSIS, I GOT NOZZLE LOAD HIGH. NEAR TO NOZZLE ONE SPRING IS THERE.I INCREASED THE OPERATING LOAD OF SPRING, SO THAT I REDUCED THE NOZZLE LOADS. WHAT CAESAR IS DOING TO ACHEIVE THE RESULT? COULD SOMEBODY EXPLAIN ME.

I will try to explain with an example. Let us say from nozzle to the spring the direction vector is in -Z and you have a vertical riser above the spring.Also let us consider that you were getting
a high -Mx.By increasing the load on the spring you were trying to counter the negative moment with a counterclokwise (+) moment and thereby reduce the resultant moment on the nozzle.

Try to visualize the free body diagram and the rationale of the solution will immediately come out.

Anindya Bhattacharya

I have assume u'r nozzle is in x dir'n and after a bend u have verical rise and then u have spring . for dead load based on stiffness u'r dead load is distributed to the nozzle and spring ( in this case it will treat spring as a resting support). Since in operating condition u'r simple rest does not share any load ( line going upward) So u have to put such a support which will work in operating case for that u have to put a spring that will pull u'r system in sustain case to share the operating load. In operating condition based on the downward or upward movement and the stiffness of u'r system the operating load is distributed in the system. If the movement is such that u'r spring is not able to either pull u'r system or release the system it will not properly serve u'r sytem.

To make u'r system free from vertical load u select such a spring that will help u in making the system change in displacement in the Y direction zero, for the straight run before vertical rise.

Depending upon u'r system behaviour u have to pull or release u'r system to make the nozzle free from vertical load.

For understanding the role of spring selection in making u'r system free read CAESAR II manuals.