As seen in previous posts, a pipe rack is composed by transverse beam required to support the longitudinal piping. The piping arrangement can be very different from one project to another and specific Engineering judgment is required. The main problem discussed in this post is how to model the loads applied to the transverse beam.

Let’s discuss some different cases that a Structural Engineer could need to deal with:

– **Case 1: rack full of same size pipes equally spaced**;

– **Case 2: rack full of different size pipes**;

– **Case 3: rack full of small pipes except one or two large pipes**;

– **Case 4: rack almost empty, very few pipes**.

As appear obvious, a standard approach is not feasible for all the four cases listed above. Let’s see then how to manage these loading conditions.

**Case 1: rack full of same size pipes equally spaced**

Two different approaches are suggested for this case 1.

– **Case 1a: pipe diameter ≤ 12″**;

– **Case 1b: pipe diameter > 12**“.

In the case 1a, we have small pipes with all the same size and equally spaced. The pipes are modeled as an uniform distributed load multiplying the weight of the single pipe for the number of pipes and then dividing it for the length B.

In the case 1B, we have big pipes with all the same size and equally spaced. The pipes are modeled as concentrated loads.

**Case 2: rack full of different size pipes**

Again, two different approaches are suggested for this case 2 too.

– **Case 2a: pipe diameter ≤ 12″**;

– **Case 2b: pipe diameter > 12**“.

In the case 2a, we have small pipes with all different sizes. In this case, that is similar to 1a, we can make sum all the different weights and then divide the result for the length B to obtain an uniform distributed load.

In the case 2B, we have big pipes with all different sizes. The approach is the same of 1a: we can consider concentrated loads.

**Case 3: rack full of small pipes except one or two large pipes**

As per figure above, let’s assume that D1 and D2 larger than 12″ and all the others smaller. In this case, the approach is a mix of the previous: D1 and D2 will be concentrated loads, all the others can be summed and then divided for the length B of the transverse beam to obtain the uniform distributed load.

**Case 4: rack almost empty, very few pipes**

Again, two different approaches are suggested for this case 4.

– **Case 4a: pipe diameter ≤ 12″**;

– **Case 4b: pipe diameter > 12**“.

In the case 4a, we have few small pipes. A uniform distributed load can be applied. The specific weight value can be suggested by codes or projects specifications.

In the case 4B, we have few big pipes. The approach is the same of 1a: we can consider concentrated loads.

For any issues or questions, you can contact the author at:

*Eng. Onorio Francesco Salvatore*

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