When designing and detailing the reinforcement in a concrete structure, the dimensions referred to for bars and wires are based on “**nominal sizes**“.

The word nominal size is used in place of diameter, referring in this way to diameter of a circle with an area equal to the effective cross-sectional area of bar or wire.

For example, for a 16 bar, because of surface deformations, there are no cross dimensions measuring 16 mm. Usually, the deformed bar can be contained in a circumscribing circle 10% (with peak values of 13 or 14%) larger than the nominal size of bar.

Preferred sizes of high yield reinforcing bars are 8, 10, 12, 16, 20, 25, 32 and 40 mm. Additional sizes, not preferred, are 6 and 50. The reason for these last to being not preferred regards the availability of the material by fabricators.

The following list below provides a comparison between the nominal size and the maximum size (in red are shown the non-preferred sizes):

- 6 mm (nominal size) => 9 mm (maximum size);
- 8 => 11;
- 10 => 13;
- 12 => 14;
- 16 => 19;
- 20 => 23;
- 25 => 29;
- 32 => 37;
- 40 => 46;
- 50 => 57.

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

*Eng. Onorio Francesco Salvatore*

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*

Pipe racks are structures used in the petrochemical, chemical and power plants industries. They are meant to support pipes, power cables and instrument cable trays. Sometimes, pipe racks support mechanical equipment, vessels and valve access platforms. Basically, they are a connection between equipment and storage or utility areas.

The main structural components of the pipe rack superstructure are:

– **transverse beams** (supports for piping, etc.);

– **columns**;

– **longitudinal struts**;

– **vertical bracing**.

When designing pipe racks, the Structural Engineer shall take into account the maintenance, allowing a clear access under the pipe rack. For this reason, moment resisting frames (MRF) are used or the bracing is limited only to few bays.

There are primarily two types of pipe racks:

– Strutted;

– Unstrutted.

The first type, **strutted**, has longitudinal struts between the transverse frames. This addition provides stiffness in the longitudinal direction. In addition, diagonal bracing can be added in order to act together with the longitudinal struts to resist the horizontal loads. This arrangement is very common.

The **unstrutted** pipe rack doesn’t have the longitudinal struts, hence the columns act as cantilevers to resist lateral loads longitudinal to the pipe rack.

Usually, if using strutted pipe racks, the columns shall normally be designed with pinned or fixed base, depending on the lateral drifting requirements.

For unstrutted pipe racks, columns shall be considered pinned in the transverse direction and fixed in the longitudinal direction. The major axis of columns should be perpendicular to the longitudinal direction of the pipe rack.

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

*Eng. Onorio Francesco Salvatore*

In this post we are going to discuss about the loads defined in the Process Industry Practices (PIP). The PIP is a self-funded consortium of process industry companies that publishes common industry practices for projects and maintenance work. PIP develops “Practices” that are a compilation of Engineering standards from different Engineering disciplines.

According to PIP, but the definition is common in every standard, dead loads are the actual weight of materials forming the building, structure, foundation and all permanently attached appurtenances.

Fixed process equipment machinery, piping, valves, electrical cable tray, including their contents, are considered as dead loads.

In PIP STC01015 – Structural Design Criteria, PIP adopts a specific nomenclature for the various types of loads: Ds, Df, De, Do and Dt.

They are defined as:

**– Ds = Structure dead load**, that is the weight of the materials forming the structure, excluding process equipment, vessels, tanks, piping and cable trays. Are included in this category also the foundation, soil above the foundation resisting uplift and, in general, all permanently attached appurtenance such as lightning, instrumentation, HVAC, sprinkler and other similar systems.

**– Df = Erection dead load**, that is the fabricated weight of process equipment or vessels. It is normally taken from the certified equipment or vessel drawing.

**– De = Empty dead load of the empty process equipment**. The difference with the erection dead load is that in this case all the attachments, trays, internals, insulation, fireproofing, agitators, piping, ladders, platforms, etc. are included.

**– Do = Operating dead load**. It is defined as the empty weight of the process equipment, vessels, tanks, piping and cable trays plus the maximum weight of the contents (fluid load) during normal operation.

**– Dt = Test dead load**. It is defined as the empty weight of process equipment, vessels, tanks and/or piping plus the weight of the test medium contained in the system. The test medium is usually specified in the contract documents or by the owner. Unless otherwise specified, PIP suggests to use a minimum specific gravity of 1.0 as test medium (water). Equipment and pipes that may be simultaneously tested shall be included. Cleaning load shall be used for test dead load if the cleaning fluid is heavier than the test medium.

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

*Eng. Onorio Francesco Salvatore*

Similarly to other international standards, the British Standard 8110-1 defines different exposure conditions based on the design environment.

The environment can be:

– Mild;

– Moderate;

– Severe;

– Very severe;

– Most severe;

– Abrasive.

The **mild environment** is defined as:

“*Concrete surfaces protected against weather or aggressive conditions*”

The **moderate environment** is defined as:

“*Exposed concrete surfaces but sheltered from severe rain or freezing whilst wet*

* Concrete surfaces continuously under non-aggressive water*

* Concrete in contact with non-aggressive soil (see sulfate class 1 of Table 7a in*

* BS 5328-1:1997)*

* Concrete subject to condensation*”

The **severe environment** is defined as:

“*Concrete surfaces exposed to severe rain, alternate wetting and drying or occasional*

* freezing or severe condensation*”

The **very severe environmen**t is defined as:

“*Concrete surfaces occasionally exposed to sea water spray or de-icing salts (directly or*

* indirectly)*

* Concrete surfaces exposed to corrosive fumes or severe freezing conditions whilst wet*”

The **most severe environment** is defined as:

“*Concrete surfaces frequently exposed to sea water spray or de-icing salts (directly or *

*indirectly)*

*Concrete in sea water tidal zone down to 1 m below lowest low water*”

The **abrasive environment** is defined as:

“*Concrete surfaces exposed to abrasive action, e.g. machinery, metal tyred vehicles or *

*water carrying solids*”

The correct design is based on the correct combination of **concrete class** and **concrete cover**.

The following table coming from the BS 8110-1 has to be used for the concrete design:

So, if we are in a **severe environment** the minimum choice is to use a **C40** concrete class with a nominal cover of **40 mm**. If, instead of a C40, we use a C50, then the concrete cover can be reduced to 25 mm.

In case of **most severe environment** the concrete class to be used is raised to C50 and the minimum cover is 50 mm.

Special analysis on the abrasion are required for the **abrasive environment**.

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

*Eng. Onorio Francesco Salvatore*

Another addition for the Downloads section. In this post we’ll discuss about an Excel spreadsheet for the evaluation of the crack width in concrete structures.

The inputs required are:

INPUT | |||

Characteristic compressive cylinder strength | f_{ck} |
30 | N/mm² |

Characteristic tensile strength of reinforcement | f_{yk} |
500 | N/mm² |

Base | b | 300 | mm |

Height | h | 500 | mm |

QP moment | M | 100.0 | kNm |

Age at cracking | 14 | days | |

Cement type | N | [-] | |

Creep factor | j | 2.0 | [-] |

Area of tension steel | A_{s} |
600 | mm² |

Area of compression steel | A_{s2} |
314 | mm² |

Bar diameter | Ø_{eq} |
12 | mm |

Maxmum tension bar spacing | S | 200 | mm |

Short term or Long term? | Long | [-] | |

Cover to A_{s} |
c | 25 | mm |

Depth for tension bars | d | 469 | mm |

Depth for compression bars | d_{2} |
31 | mm |

The first output is related to the evaluation of the cracking moment. The software checks firstly if the section is cracked:

CRACKING MOMENT | |||

Modulus of elasticity of concrete | E_{cm} |
32.84 | GPa |

Modulus of elasticity of steel | E_{s} |
200.0 | GPa |

Modular ratio | a_{e} |
18.27 | GPa |

Reinforcement ratio for compressive steel | r’ | 0.0022 | [-] |

Reinforcement ratio for tension steel | r | 0.0043 | [-] |

Mean concrete strength at cracking | f_{cm,t} |
34.26 | MPa |

Mean concrete tensile strength | f_{ct,eff} |
2.61 | MPa |

Uncracked neutral axis depth | x_{u} |
256.53 | mm |

Uncracked 2^{nd} moment of area |
I_{u} |
3875 | mm^{4} 10^{6} |

Cracking moment | M_{cr} |
41.56 | kNm |

Moment comparison | The section is cracked |

The second output is related to the stresses in the section, for both concrete and steel:

STRESSES | |||

Fully cracked | x_{c} |
140.71 | mm |

Stress limit coefficient for concrete | k_{1} |
0.60 | [-] |

Concrete compression stress limit | s_{c,lim} |
18.00 | MPa |

Concrete stress | s_{c} |
9.29 | MPa |

Stress limit coefficient for steel | k_{3} |
0.80 | [-] |

Steel tension stress limit | s_{s,lim} |
400.00 | MPa |

Steel stress | s_{s} |
396.18 | MPa |

The last output is related to the evaluation of the crack width:

CRACK WIDTH | |||

Effective tension area | A_{c,eff} |
23250 | mm² |

A_{s} / A_{c,eff} |
r_{p,eff} |
0.0258 | [-] |

Equation (7.11) factor | k_{3} |
3.4 | [-] |

Equation (7.11) factor | k_{4} |
0.425 | [-] |

Maximum final distance between cracks | s_{r,max} |
164.1 | mm |

Average strain for crack width calculation | e_{sm}-e_{cm} |
1683.1 | [-] |

Calculated crack width | w_{k} |
0.276 | mm |

Some screenshots below.

You can download the Excel spread sheet from here:

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

*Eng. Onorio Francesco Salvatore*

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In this post an Excel spreadsheet based on the AS/NZS 1170 is available. Scope of the tool is to define the Annual Probability of Exceedance, P (and, hence, Average Recurrence Level, R), for Ultimate State Limit and Serviceability Limit States. The values are referred to **New Zealand**, as per Section 3, Table 3.3.

The **Annual Probability of Exceedance** is the probability that a given event over a given duration will be exceeded in any one year; it is the inverse of the **Return Period**, also know as **Recurrence Interval**.

The Return Period/Recurrence Interval is an estimate of the likelihood of an event, such as an earthquake. It is a statistical measurement based on historic data denoting the average recurrence interval over an extended period of time. Any times a Structural Engineer refers to return periods for the evaluation of an action, he is basically doing a risk analysis. Hence, it is used to design structures to withstand an event with a certain return period.

These parameter are based on the following inputs, according to AS/NZS 1170:

– **Design working life** (temporary, less than 6 months, 5 years or less, 25 years, 50 years, 100 years or more);

– **Type of structure** (minor, normal, major, post-disaster, exceptional);

– **Type of analysis** (ULS, SLS 1, SLS 2).

Output:

You can download the Excel spread sheet from here:

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

*Eng. Onorio Francesco Salvatore*

In this post, an Excel spreadsheet for the evaluation of the torsional section properties of concrete sections according to the ACI 318 is available. The sections are:

– rectangular near the edge;

– rectangular in interior position;

– L-shaped;

– inverted tee.

The design for torsion as per ACI 318 is based on a thin-walled tube, space truss analogy. A beam subjected to torsion is idealized as a thin-walled tube with the core concrete cross section in a solid beam neglected as shown below:

Once a reinforced concrete beam has cracked in torsion, its torsional resistance is provided primarily by closed stirrups and longitudinal bars located near the surface of the member. In the thin-walled tube analogy, the resistance is assumed to be provided by the outer skin of the cross section roughly centered on the closed stirrups. Both hollow and solid sections are idealized as thin-walled tubes both before and after cracking.

According to the ACI the fundamental properties are:

– Area enclosed by outside perimeter of concrete cross section;

– Outside perimeter of concrete cross section;

– Area enclosed by centerline of the outermost closed transverse torsional reinforcement;

– Perimeter of centerline of outermost closed transverse torsional reinforcement.

These properties are calculated for the types of sections discussed at the begin of the post, that are:

**RECTANGULAR NEAR THE EDGE**

**RECTANGULAR IN INTERIOR POSITION**

**L-SHAPED**

**INVERTED TEE**

You can download the Excel spread sheet from here:

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

*Eng. Onorio Francesco Salvatore*

Following the Excel spreadsheet for the rectangular section, in the following you can download the spreadsheet for the evaluation of the Moment of Inertia of the cracked section transformed to concrete, Icr, for T-sections. The reference is the **ACI 314R-11** “*Guide to Simplified Design for Reinforced Concrete Buildings*“.

Basically, starting from the gross section:

We can obtain the cracked transformed section:

The relation used is:

For a doubly reinforced section:

The cracked transformed section is:

And the equation to use in this case is:

You can download the Excel spread sheet from here:

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

*Eng. Onorio Francesco Salvatore*

New addition for the Download section: the following Excel spread sheet calculates the Moment of Inertia of the cracked section transformed to concrete, Icr, for rectangular sections. The reference is the **ACI 314R-11** “*Guide to Simplified Design for Reinforced Concrete Buildings*“.

Basically, starting from the gross section:

We can obtain the cracked transformed section:

The relation used is:

For a doubly reinforced section:

The cracked transformed section is:

And the equation to use in this case is:

You can download the Excel spread sheet from here:

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

*Eng. Onorio Francesco Salvatore*