# How to Calculate Reinforcement in Beams A customary individual can evaluate reinforcement as a singular amount. We see numerous individuals gauge reinforcement as a level of solid volume, for example, 2% or 1.5% of solid volume. In any case, as a development proficient you shouldn't gauge rebar amount as a percent of solid volume. You should appraise that as appeared in the basic drawing.

Evaluating rebar amount is simple. All you need to get the cutting length for each kind of bar in the beam. In this post, we will talk about how to appraise beam reinforcement from the drawing. We'll utilize the standard ideals as an auxiliary drawing of the beam.

Instructions to Estimate Beam Reinforcement

For assessing beam reinforcement, you will first need to realize which kinds of bars are accessible in the beam. From the ideal plan we have discovered the accompanying information:

• Top longitudinal bar, T – 3D20
• Base longitudinal bar, B – 3D20
• The additional top bar on support1, Et1 – 2D20
• The additional top bar on support2, Et2 – 2D20
• Additional base Bar, Eb – 2D20
• Stirrup, S1 – D10 @ 4″ c/c

You will likewise require the accompanying information to appraise the beam reinforcement:

• Clear cover for reinforcement
• Lap length
• Bend length

You can get this information from the overall notes sheet of auxiliary drawing. So, now we have enough information to gauge the beam reinforcement. How about we start the estimate?

Top Longitudinal Bar (T)
Apply the accompanying formula to get the cutting length of the top longitudinal bar.
Cutting Length of top longitudinal bar is,
= Length of beam – 2 × clear cover + 2 × bend length – 2 × bend deduction length + lap length

Now, Length of beam: 20″ + (22′-8″) + 20″ + (8′-4″) + 20″ = 36′ (from above)
Clear cover: 1½"
Bend length: 12D = 12 × 20 = 240 mm, state 9½"
Bend deduction length: Bend deduction length for 90° bend is equivalent to multiple times of bar dia = 2 × 20 mm = 40mm = 1½"

Thus, the cutting length of the top longitudinal bar is,
= 36′ – 2 × 1½" + 2 × 9½ – 2 × 1½" + 0
= 37′-1″.

There are three quantities of the top longitudinal bar in the above picture. So, the complete length of the top longitudinal bar is,

= 3 × (37′-1″)
= 111′-3″.

We know the full length of a strengthening bar is about 40′. The length of a top longitudinal bar, we are assessing, doesn't surpass the length of a stocked bar. So, you don't need to include lap length.

Base Longitudinal Bar (B)
The recipe for computing the cutting length of the base longitudinal bar is equivalent to the equation for the top longitudinal bar. So, the length is additionally equivalent to T1. That is 37′-1″.
Therefore, the absolute length of the B1 bar is,
= number of B1 bar × cutting length of a B1 bar
= 3 × (37′-1″)
=111′-3″.

Additional Top Bar on Support1 (Et1)
Cutting length of Et1 = expanded length from support + width of support – clear cover + bend length – bend deduction length
Expanded length from support: 7′-5″
Width of support: 20″
Clear cover: 1½"
Bend length: 9½"
Bend deduction length: 1½"

Therefore, the cutting length of Et1 is,
= (7′-5″) + 20″ – 1½" + 9½" – 1½"
= 9′-7½"

There are two quantities of Et1 bar appeared in the drawing. So the absolute length of Et1 is,
=2 × (9′-7½")
19′-3″

Additional Top Bar on Support2 (Et2)
Expanded length from support: 7′-5″
Width of support: width of support for Et2 ought to be the width of support2 + separation somewhere in the range of support2 and support3 + width of support3 = 20″ + (8′-4″) + 20″ = 11′-8″
Clear cover: 1½"
Bend length: 9½"
Bend deduction length: 1½"

In this way, cutting length of Et2 = (7′-5″) + (11′-8″) – 1½" + 9½" – 1½"
= 19′-7½"

All out length of Et2 = 2 × (19′-7½")
= 39′-3″

Cutting length of the additional base bar is,
=Distance between support – 2 × separation between additional base bar and closest support
= (22′-8″) – 2 × (2′-10″)
= 17′

The absolute length of the additional base bar is,
= 2 × 17′ (there are 2 bars for Eb appeared in the picture above)
= 34′

Stirrup (S1)
For evaluating stirrup, you need to figure the necessary number of stirrups for the beam and cutting length of the bar for a stirrup.
Number of stirrups: Required number of stirrups for first range,
= (22′-8″)/4″ +1 = 69 + 1 = 70 nos.
Required number of stirrups for second range,
= (8′-4″)/4″ + 1 = 25 + 1 = 26 nos.
All out number of stirrups = 70 + 26 = 96 nos.

Equation for computing cutting length of a stirrup's bar is,
= 2 × (a+b) + 24D (for 135° snare)

Where,
a = length of stirrup's long side
b = length of stirrup's short side
D = dia of stirrup bar

Therefore, cutting length of the bar is,
= 2 × (21″+9″) + 24 × 10
= 69½" (240 mm = 9½")
= 5′-9½"

All out length of bars for all stirrups,
= 96 × (5′-9½")
= 556′

Rundown: The total fortifying bar we have evaluated for the beam up until this point –

• 20 mm ø bar
• T = 111′-3″
• B = 111′-3″
• Et1 = 19′-3″
• Et2 = 19′-7½"
• Eb = 34′

So absolute 20 mm ø bar is,
= 111′-3″ + 111′-3″ + 19′-3″ + 19′-7½" + 34′ = 296′.
10 mm ø bar,
S1 = 556′.

Be that as it may, steel fortifying bars are estimated in kilograms in the market. So, you need to change over the bar length to kilograms. Unit weight of 10 mm ø bar is 0.188 kg/ft and 20 mm ø bar is 0.75 kg/ft. So, the reinforcement required for our model beam,

• 20 mm ø bar = 296′ × 0.75 = 222 kilograms.
• 10 mm ø bar = 556 × 0.188 = 105 kilograms.

Note: When you need to buy strengthening bar for the beam you should include 5% more with your evaluated amount as wastage.

This is it! You’ve got the reinforcement amount in your beam.