# How to reduce a distributed load to a resultant force

• benanderson08
In summary: Let's say the pipe has an inside diameter of d1. Let's say the slurry inside the pipe has a depth of h at the pipe centerline, and the slurry depth is constant along the pipe length. Therefore, the pipe inside radius is r = 0.5*d1. Looking down the center of the pipe, the cross-sectional area of the slurry is,A1 = (h - r)[h*(2*r - h)]^0.5 + (r^2){0.5*pi + asin[(h - r)/r]},where pi = 3.1416. Therefore, the total weight of the slurry in the pipe is
benanderson08
Hi everyone,

So I am trying to study how different types of slurry will deflect a PVC pipe. The first step in my understanding is to reduce distributed load into a resultant force. I'm assuming that the resultant force will be placed in the center of the pipe (assuming a homogeneous slurry), but the magnitude eludes me.

It's a 31in pipe, 4in diameter and the slurry (simple for the sake of calculation) is water @ 75F. The PVC is schedule 40, although that would matter for the reduction of the distributed load.

Would anyone be able to point me in the right direction?

You take the integral of the load along the pipe as usual.
This means you need to know how the load is distributed over the pipe.

Although the integral method works for all loadings, it cannot be used if you look at the distributed load along the length of the pipe, because it is not uniformly distributed along the "b" distance of the pipe. So would I have to start off by looking down the center of the pipe i.e. a circle and then apply the integral method that way? so (integral) (b)(r^2-x^2)^0.5 would be the proper equation maybe?

benanderson08: Let's say the pipe has an inside diameter of d1. Let's say the slurry inside the pipe has a depth of h at the pipe centerline, and the slurry depth is constant along the pipe length. Therefore, the pipe inside radius is r = 0.5*d1. Looking down the center of the pipe, the cross-sectional area of the slurry is,

A1 = (h - r)[h*(2*r - h)]^0.5 + (r^2){0.5*pi + asin[(h - r)/r]},​

where pi = 3.1416. Therefore, the total weight of the slurry in the pipe is,

W1 = (rho1*g)*A1*L,​

where rho1*g = 9810 N/m^3 = 9.81e-6 N/mm^3, and L = pipe length. Therefore, to reduce the slurry uniformly-distributed load to a pipe midspan point load, place a force W1 at the pipe midspan.

Let's try an example. In post 1 you have, d1 = 101.6 mm, and L = 787.4 mm. Therefore, r = 50.8 mm. And let's say the slurry depth in the pipe is, h = 70 mm. Therefore,

A1 = (70 - 50.8)[70(2*50.8 - 70)]^0.5 + (50.8^2){0.5*3.1416 + asin[(70 - 50.8)/50.8]}
= 19.20(47.032) + 2580.6{1.5708 + asin(0.3780)}
= 19.20(47.032) + 2580.6{1.5708 + [(22.210 deg)/(57.2958 deg/rad)]}
= 19.20(47.032) + 2580.6(1.5708 + 0.3876)
= 903.01 + 5053.8
= 5957 mm^2.​

Therefore, the slurry total weight is, W1 = (9.81e-6 N/mm^3)(5957 mm^2)(787.4 mm) = 46.01 Newtons. Therefore, the slurry can be represented as a pipe midspan force of W1 = 46.01 N.

Did you want to include the pipe wall (self) weight? If so, compute the pipe self weight, W2, and add it to W1.

Last edited:
... and of course there is the possibility of doing a surface integral and/or exploiting what you know of the symmetry.

... and there you have it. Enjoy.

benanderson08 said:
Hi everyone,

So I am trying to study how different types of slurry will deflect a PVC pipe. The first step in my understanding is to reduce distributed load into a resultant force. I'm assuming that the resultant force will be placed in the center of the pipe (assuming a homogeneous slurry), but the magnitude eludes me.

It's a 31in pipe, 4in diameter and the slurry (simple for the sake of calculation) is water @ 75F. The PVC is schedule 40, although that would matter for the reduction of the distributed load.

Would anyone be able to point me in the right direction?
You can't analyze the deflection of the pipe in this way. The bending moment variation is different for a distributed load than for a concentrated load. You need to take the specific load distribution into account in determining the deflection. This has to be analyzed like a beam under a distributed load.

Chet

## 1. How do I calculate the resultant force from a distributed load?

The resultant force from a distributed load can be calculated by integrating the load function over the given area. This will give you the total force acting on the system. Alternatively, you can break the distributed load into smaller sections and calculate the resultant force for each section, then sum them together to get the total resultant force.

## 2. What is the difference between a distributed load and a concentrated load?

A distributed load is a force that is spread out over a given area, while a concentrated load is a single point load. Distributed loads are typically easier to work with as they can be broken down into smaller sections, whereas concentrated loads require more precise calculations.

## 3. How can I reduce a distributed load to a single resultant force?

To reduce a distributed load to a single resultant force, you can use the concept of centroids. The centroid is the point where the distributed load can be considered to act. By finding the centroid and multiplying it by the total load, you can calculate the resultant force.

## 4. Can a distributed load have a direction?

Yes, a distributed load can have a direction. This is often seen in structural engineering, where the distributed load may be acting in a specific direction, such as wind or water pressure on a building. In these cases, the resultant force will also have a direction.

## 5. How can I use the concept of distributed loads in real-world applications?

Distributed loads are commonly used in structural engineering, such as in the design of bridges, buildings, and other structures. They are also used in the analysis of forces on objects, such as in the study of fluid mechanics. In everyday life, distributed loads can be seen in actions like carrying a heavy load on your shoulders, where the weight is distributed over a larger area to reduce the pressure on a single point.

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