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If **Development length **values are not given in the structural drawing or If you’re designing the structure drawing for any construction work, then what must be the development length values you should use? In this article, you’ll learn How to calculate **Development length** using the Development length formula. Also, know how the data are extracted from Limit State Method for Deformed bars in tension and compression. And finally, get the tables of development length values L_{d}

**23 April 2022**

**IS Code Specification For Development Lenght****Formula For Development Length****Calculating Development Length for Fe 250 & M20****Calculating Development Length for Fe 415 & M20****Development Length For all Bar and All grade****Video For it**

A **Development length** can be defined as the amount of reinforcement (bar) length needed to be embedded or projected into the column to establish the desired bond strength between the concrete and steel (or any other two types of material)

Watch this Video till the end to get the full details about development length-

When you open a structural Drawing- You Find L_{d}

Sometimes It may be-

L_{d}=30d

L_{d} =40d

L_{d} =50d

But the question is where it is coming from or where it is Derive From and Depends on which factors.

By the Thumb rule, these are taken in the site work as anything but that's not the correct way to take it, maybe failure occurs.

Because it depends upon both the Grade of Concrete and The type of Steel you use in the construction members. so we can't take the value of development length by guessing work.

All we know from this article.

Now to find the development length which is actually used in the construction site, IS 456:2000 will help you to find the Required Data-

For Formula to Calculate Development Length of Bar as per IS 456:2000, Go to clause No 26.2.1

σ_{s}- Is actually the Stress in bars which depends upon the grade of steels like Fe250, Fe415, Fe500

**For Design bond stress, go to IS 456:2000 Clause No -** 26.2.1, you’ll find this Table** **

These are all data for Plain bars but we Generally use steels in the market or for House Construction are Deformed Bars.

These are all data for Plain bars but we Generally use steels in the market or for House Construction are Deformed Bars.

**For Design bond stress, go to IS 456:2000 Clause No -** 26.2.1, you’ll find this Table** **

For Plain Bar (clause 26.2.1.1)-No Ribs in the Bars.

When you Purchase Bars from the Market, You get Bars with Ribs- Which are called Deformed Bars and we calculate Values for Those Bars.

But Before calculation, knowing all the important factors to be considered.

But Deformed bar calculation is not given in the IS Code but they have outlined two specifications.

You’ll Find This In clause No. 26.2.1.1 Here are the important factors to be considered.

In short-

- For Deformed Bar = Tension increased 60 % = Tension * 1.60
- For Compression = Tension increased 25 % = Tension * 1.25

Design Bond Stress (τ_{bd}) in Limit State Method for Plain Bars in Tension shall be as below: | ||||||

Grade Of Concrete | M20 | M25 | M30 | M35 | M40 & Above | |

For Plain bars in Tension (τ_{bd}) | 1.2 | 1.4 | 1.5 | 1.7 | 1.9 | IS Code |

For Deformed bars in Tension (τ_{bd}) | 1.92 | 2.24 | 2.40 | 2.72 | 3.04 | τ_{bd}*1.6 |

The value of Bond stress can be increased by 60% when high yield strength deformed (HYSD) bars are used.

It can be further increased by 25% when we calculate the development length for bars in compression.

Design Bond stress **(τ _{bd})** in Limit state Method for Plain Bars in Compression shall be as Below-

For Plain bars in compression for M20= t_{bd }(of plain bar) x 1.25 = 1.2*1.25= 1.5 For Deformed bars in compression for M20= t_{bd }(of plain bar) x 1.25 = 1.92*1.25= 2.4

Similarly, you can calculate the Values for all grades of Concrete.

Design Bond Stress (τ_{bd}) in Limit State Method for Plain Bars in Compression shall be as below: | ||||||

Grade of Concrete | M20 | M25 | M30 | M35 | M40 & Above | IS Code |

For Plain bars in compression (τ_{bd}) | 1.5 | 1.75 | 1.875 | 2.125 | 2.375 | τ_{bd}*1.25 |

For Deformed bars in Compression (τ_{bd}) | 2.4 | 2.8 | 3.0 | 3.4 | 3.8 | τ_{bd}*1.25*1.60 |

Here you’ll calculate the development length for the plain bar in tension, by the limit state method.

Grade of concrete = M20

Reinforcement bar = Fe250 (fy = 250)

Stress in Bar = 0.87 * fy= 0.87 X 250

For M20 Concrete Grade, Design bond stress for the plain bar in Tension

τ_{bd} =1.2 N/mm^{2}

Therefore, by using the Formula for Development Length, Put the above data in Following Formula-

For Fe-250 | M20 | M25 | M30 | M35 | M40 & Above |

Tension (Plain Bar) | 45.3Ф | 38.8Ф | 36.3Ф | 32.0Ф | 28.6Ф |

Compression (Plain Bar) | 36.3Ф | 31.1Ф | 29.0Ф | 25.36Ф | 22.9Ф |

**Data Given:-**

Grade of concrete = M25

Reinforcement bar = Fe415 (Fy_{ }= 415)

Stress in Bar = 0.87 * 415= 0.87 X 415

For M20 Concrete Grade, Design bond stress for the Deformed bar in Tension

τ_{bd} =1.2 N/mm^{2} * 1.6 = 1.9 N/mm^{2}

Therefore, by using the Formula for Development Length Here

**Similarly checking with all the Diameters of Bars, You’ll find the Following Results**

For Fe415 | M20 | M25 | M30 | M35 | M40 & above |

Tension (Deformed Bar) | 47.0Ф | 40.3Ф | 37.6Ф | 33.2Ф | 29.7Ф |

Compression (Deformed Bar) | 30.1Ф | 25.8Ф | 24.1Ф | 21.2Ф | 19.0Ф |

That’s How You can find the Development length values for Different grades of concrete and Steel.

As per the General Thumb rule, the development length requirement is expressed *as **’**40Ø**’* to *’**50 Ø**’. But as a Civil Engineer, * you must use the correct calculation.

Let's say You suggest clients use a 12mm diameter deformed Bar for Beam construction. So you must use M25 (1:1:2-ration of Cement, sand, and aggregate), and The Required Development length (Tension Deformed Bar) for the Bar will be 41*12=492 mm.

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