HCBBS Forum (English)
Submit Chemical Projects / Find Solutions
Amplify Your Requirements on a Broader Chemical Platform *Engineering · Technology · Equipment · Solutions*
Submit Request

Regarding the derivation of the thrust formula for fixed supports in thermal pipelines

2020-07-07View Original

Thread Content

I’ve been looking at the formulas in the textbook for a long time; I keep feeling there’s something wrong with them, and I can’t find the derivation process. Dear experts, please take a look; I think there’s an issue with the friction part. Isn’t it the weight of the pipe between the two supports, which should be shared equally between them? Why is it all assigned to one support here?
Reply #22020-07-08
What is shared is the vertical load; the thrust calculation here refers to the axial load
Reply #32020-07-08
qμL1 refers to the total weight of a uniform load over a length of L1
Reply #42020-07-08
Isn’t frictional force caused by vertical loads? The textbook clearly states that the load borne by a fixed support is half of the weight of the pipeline between the two supports
Reply #52020-07-08
But this bracket only accounts for half of the mass of L1; how is the entire weight calculated based on this bracket?
Reply #62020-07-09
During operation, the pipeline heats up and compresses the compensator; the friction generated by the contact between the pipeline on the right side of the compensator and its sliding support pushes it in the right direction toward the H fixed support. The same principle applies to the left side of the compensator; however, according to the formula, there is no sliding bracket between the fixed bracket on the left side and the compensator, as is standard in such setups.
Reply #72020-07-09
You’re spanning continuously; there’s a tube on your left and another on your right.
Reply #82020-07-10
The friction force generated by the weight of the tube on the right side has had a coefficient of 0.7 applied to it; it was reduced
Reply #92020-07-10
My problem with this formula is that I don’t quite understand the concept of friction force being equal to uqL. According to the textbooks, a support bears only half of the weight of a pipe – that is, it carries half of the weight on the left side, corresponding to L1, and half of the weight on the right side, corresponding to L2. But why is the friction force on the left side simply uqL1, and on the right side simply uqL2?
Reply #102020-07-14
It feels like there are top-tier experts everywhere here
Reply #112020-07-16
A forum that is closely linked to reality; more people are welcome to participate in the discussions. Many sailors have been exchanging ideas here for five or eight years now

Submit a Project

**Looking for Chemical Technology, Equipment & Solutions?** No Registration Required Broader Platform Exposure | Global Chemical Service Provider Connections

Submit Request — Free Consultation

Disclaimer

This is an automated machine translation of the original thread. Some technical terms may have inaccuracies; the original text shall prevail. Click "View Original" at the top right to access the source page, which supports IP-based automatic real-time language translation. Please watch out for contact details and sales inducements to prevent fraud. All content and translations are for reference only, representing solely the poster's personal views. For enquiries, email service@hcbbs.com.