We will introduce an example of calculating the support spacing for piping support.

In order to properly secure the piping without bending, it is essential to support it with supports.

However, when it comes to how much support to create, sometimes it is surprisingly difficult to construct it properly.

If it sways easily when you touch it, you can fix it with supports, but knowing the support spacing will be helpful in terms of adding support.

In terms of piping design, you will likely be using tables found in handbooks, but I would like to introduce an example strength calculation.

It’s for people who want to know more deeply.

## calculation model

The calculation model used is as follows.

- 50A
- SGP
- Inner diameter 52.9mm
- At normal temperature

Since we are assuming a situation in which the piping is supported by supports, we will use the mechanical mechanics formula for uniformly distributed loads for beams supported at both ends .

The following two calculation formulas are used.

bending stress

$$ σ =MZ=\frac{wl^2}{8}Z $$

Deflection

$$ δ =\frac{5wl^4}{384EI} $$

We will treat this expression here as if it had fallen from heaven.

In calculations, the key is to derive the moment of inertia I and section modulus Z.

## parameters

Let’s organize the parameters one by one.

### Longitudinal elastic modulus E

Since it is SGP piping at room temperature, we use **2×10 ^{11} N/m ^{2}(200GPa).**

This number is also a number that fell from the sky.

There is some variation depending on the temperature, but since it is within the error range, it is rounded to make calculations easier.

### weight w

Weight w is the piping weight per unit length.

50A SGP piping has an inner diameter of 52.9mm and a thickness of 3.8mm, so be sure to calculate carefully.

First, calculate the cross-sectional area.

π*(52.9/1000)*(3.8/1000)+π/4*(3.8/1000) ^{2} =(631+45)*10 ^{-6} =677*10 ^{-6} m ^{2}

Since the specific gravity of iron is 7850kg/m3 ^{,} the weight w is

677*10 ^{-6} ×7850= **5.31kg/m**

### Moment of inertia I

For the moment of inertia of area, use the formula for a hollow circular section.

π/32*{(60.5/1000) ^{4} -(52.9/1000) ^{4} }= **2.73×10- ^{7} m ^{4}** (27.3cm

^{4})

Here, the SGP outer diameter of 50A uses the value 52.9+3.8*2=60.5.

I don’t want to make calculation mistakes or wrong units, so I prefer to match in m units.

### Section modulus Z

The section modulus Z is the moment of inertia I divided by the outer radius.

2.73×10^{-7} /(60.5/2/1000)= **9.03×10 ^{-6} m^{3}** (9.03cm

^{3})

## Bending stress σ

Let’s calculate the bending stress σ.

First, calculate the bending stress when l=1m. Let’s set σ to 0.

Since l=1, the bending stress σ_{0} is

5.31/8/9.03×10-6 ^{×} 9.8=0.72× 10^{6} Pa(0.72MPa)

It becomes. Note that the gravitational acceleration is multiplied by 9.8m/s^{2}.

Generally, the support spacing is determined so that the bending stress does not exceed 10MPa.

All you have to do is change l, but all you have to do is multiply the calculation result of σ_{0} by l^{2}.

So,

0.72×l^{2} < 10

You will have to calculate this.

l < √(10/0.72) = 3.72m

From the calculation result, from the viewpoint of bending stress, the support spacing is required to be **3.72m or less.**

## Deflection δ

Calculate the deflection in the same way.

If the deflection when l=1 is δ_{0} , then

(5*5.31)/(384*2×10^{11} *2.73×10^{-7} )=26.55/20.9664=1.26×10^{-6} m(1.23×10^{-3} mm)

Deflection must be kept to 2 to 3 mm.

2 < 1.23×10^{-3} *l^{4} < 3

Let’s do the calculation.

2/1.23×10^{3} < l^{4} < 3/1.23×10^{3} → 1.63×10^{3} < l^{4} < 2.44×10^{3} → 6.35 < l < 7.03

As a result, let’s consider that from the viewpoint of deflection, the support spacing is required to be **6.35m or less.**

## summary

Let’s summarize the calculation results.

- Bending stress 3.72m or less
- Deflection 6.35m or less

The restrictions due to bending stress are more severe, and in the case of 50A, the idea would be to support at intervals of about 3m to 4m.

The same calculation can be done for different calibers.

The calculation itself is not complicated, but mistakes can occur, such as when converting units.

## reference

## related article

related information

## lastly

We introduced an example of calculating support spacing for piping support design.

This is a calculation of both ends support and uniformly distributed load of a beam in mechanical mechanics.

If you use a handbook, you can get results quickly, but if you can understand the calculation formula, I think it will have a wider range of applications.

Sometimes I want to do calculations by hand.

Please feel free to post any concerns, questions, or concerns you may have regarding the design, maintenance, and operation of chemical plants in the comments section. (The comment section is at the bottom of this article.)

*We will read all comments received and respond seriously.