The beam should be strong enough to bear compressive as well as tension. It transfers the load to the column. Bar bending schedule of the beam means to calculate the total cutting length and quantity of steel rods required in the beam for a safe and durable structure.

A beam is a horizontal load-bearing structure supported by two members like a column or wall. The load provided on the beam is generally uniformly distributed load.

Beam has a two-zone compressive zone at the upper portion and a tension zone at the lower portion. As we know that concrete is weak in tension, we should make it strong by placing a steel rod in the tension zone.

Before proceeding to the calculation of bar bending schedule of beam I would recommend you to read the article Different Types of Beam- Why cantilever beam are made trapezoidal so that you will be easy for calculation.

**Bar Bending Schedule of Simply supported Beam**

Example:

Given,

Width of beam = 300 mm

Height of Beam = 500 mm

Length of Beam = 4 m = 4000 mm

The Bottom main bar is of diameter 20 mm and top bar is of diameter 16 mm, The diameter of stirrups is 12 mm at both end having C/C 100 mm and 150 mm at the middle span.

Here we have to calculate the cutting length of all steel rods and then we will move toward the weight of all steel rods.

Let us see how to do.

**For Main Rods,**

Now, Cutting Length of one main rod = length of beam – Clear cover at both ends + development length – Bend length

**Where,**

**Length of beam = 4000 mm**

**Clear cover at both end = 25 mm + 25 mm = 50 mm**

**Development Length =**** **(18d x 2) = (18 x 20 x2) = 720 mm

(Assuming development length 18d, where d is the diameter of main rods) It has two development lengths so, we have to multiply with 2.

Bend Length = (2d x 2) = 2 x 20 x 2 = 80 mm

Since,

**Bend at 45 is taken = 1d,****Bend at 90 is taken = 2d,****Bend at 135 is taken = 3d**

Now,

Now, Cutting Length of one main rod (L) = length of beam – Clear cover at both ends + development length – Bend length

Total Length of main rods (L) = 4000 mm – **50**** ****mm**** **+ 720 mm – 80 mm = 4590 mm

But, the beam have two main rod so, we have to multiply with 2.

Total length of main rods = 2 x 4590 = 9180 mm = 9.180 meter

**Total Weight of Main Rods**

= d²/162.25 x length

= (20 x 20)/162.25 x 9.180

= 22.631 kg

**For Top Bars,**

Now, Cutting Length of one Top rod = length of beam – Clear cover at both ends + development length – Bend length

**Where,**

**Length of beam = 4000 mm**

**Clear cover at both end = 25 mm + 25 mm = 50 mm**

**Development Length =**** **(18d x 2) = (18 x 16 x2) = 576 mm

(Assuming development length 50d, where d is diameter of main rods) It have two development length so, we have to multiply with 2.

Bend Length = (2d x 2) = 2 x 16 x 2 = 64 mm

Now,

Total Cutting Length of Top bars (L) = length of beam – Clear cover at both ends + development length – Bend length

Total Length of main rods (L) = 4000 mm – **50**** ****mm**** **+ 576 mm – 64 mm = 4462 mm

But, the beam have two main rods so, we have to multiply with 2.

Total length of main rods = 2 x 4462 = 8924 mm = 8.924 meter

**Total Weight of Main Rods**

= d²/162.25 x length

= (16 x 16)/162.25 x 8.924

**= 14 kg**

**For Stirrups,**

**Stirrups:**

Length has divided into two parts 1^{st} is near support and 2^{nd} is at mid span of the beam.

**For end supports: (spacing 100 mm)**

So, Length near the end support = (Length/ 3)

= 4000/3

= 1333.3 m

Therefore, No. of stirrups

= Length / spacing + 1

= 1333.3/100 + 1

**= 14.33 Nos.**

**There are two end support So, Total no of stirrups will be,**

= 14.33 x 2

= 28.66 nos. Say 29 nos.

**For mid supports: (spacing 150 mm)**

So, Length near the end support = (Length/ 3)

= 4000/3

= 1333.3 m

Therefore, No. of stirrups

= Length / spacing + 1

= 1333.3/150 + 1

**= 7.88 use 8 Nos.**

**Now, Cutting Length of stirrups:**

Given,

Width of stirrups = Width of column – Clear Cover – Clear Cover

W = 300 – 25 -25 = 250 mm

Height of stirrups = Height of column – Clear cover – Clear cover

**H = 500 – 25 – 25 = 450**

**Cutting length of one stirrups,**

**Cutting length = Perimeter of stirrups + Hook Length – Bend Length **

L = (2 x W) + (2 x H) + (2 x 10D) – (3d x 2) – (2d x3)

**Where**,

10d is hook length, 3d is subtracted for bend 135 and 2d subtracted for bend for 90. Where d is diameter of stirrups.

Stirrups have 3 bend of 90 and two bend of 135.

**Therefore Total Cutting Length of one stirrup:**

= ( 2 x 250) + (2 x 450) + (2 x 10 x 12) – (3 x 12 x 2) – (2 x 12 x 3)

= 500 + 900 + 240 -72-72

= 1496 mm or 1.496 m

We have calculated total no. of stirrups = 38 nos. (Both end span and mid span stirrups)

**So, ****Total Cutting Length of all stirrups**

= 1.496 x 37

= 55.352 meter

**Weight of steel required for stirrups is,**

= d²/162 x length

= 12²/162 x 55.352

**= 49.12 kg**

**Hence Total weight of Steel Bars required in the Beam,**

**Total weight of steel **= Weight of main bars + Weight of top bars + Weight of stirrups

= 22.631 +14 + 49.12

**= 85.75 kg**

I hope this article on “**Bar bending schedule of Beam**” remains helpful for you.

Happy Learning – Civil Concept

**Contributed by,**

**Civil Engineer – Ranjeet Sahani**

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