Steel calculation for a triangular slab/ How to calculate the quantity of reinforcement in a triangular slab?

 Let us now calculate the cutting length of the reinforcements in a triangular slab as shown below.

Given data:

Type of slab = right angle triangular slab.

 Length of the slab = 4m.

The breadth of the slab = 3m.

Main bar =T12 @ 150 c/c

Distribution bar =T10 @ 150 c/c

Clear cover on all sides = 20mm.=0.02m.

First, we will calculate the cutting length of the main bars.

As you can observe in the below drawing, I have redrawn the slab with the main bars. I have shown the red dotted triangle, joining both the end of all the bars. This red triangle is drawn, after deducting the clear cover from all the sides.

The lengths of all the bars are different & let us name them as L1, L2, L3, etc. as shown below.

Length of bar L1 

= [slab breadth - ( clear cover on both sides.)]

= [3m - ( 2nos. × 0.02m.)]

= 2.96m.

To calculate the length of other bars, we have to find out the angle of the red triangle as shown below.

Tanθ1

 = [opposite side ÷ adj. side.]

=[ 3.96m ÷ 2.96m.] 

= 1.3378

θ1 = 1.3378tan-1

θ1 = 53.22°

Now, from the below drawing, you can observe that the length of the bar L2 = L1 -a

Consider the small triangle formed between the steel bars L1 & L2 as shown above.

Here,

  tanθ1  = [opposite side ÷ adj. side.]

  tan53.22° = [ 150mm.÷ a.] 

1.3378m =  [ 0.15m.÷ a.] 

By cross multiplication,

a = [1.3378÷ 0.15]

 = 0.112m.

    L2 =[ L1 - a]

         = [2.96m. -0.112m]

        = 2.848m.

As all the steel bars are at equidistance,

L3 = [L2 - a] or [L1 - 2a]

      = [ 2.848m -0.112m.]

      = 2.736m.

Similarly,

L4 = [L3 - a] or [L1 - 3a]

     = [ 2.736m -0.112m.]

     = 2.624m.

 L5  = [ 2.624m -0.112m.] = 2.512m

 L6  = [ 2.512m -0.112m.] = 2.4m  

 L7  = [ 2.40m -0.112m.]   = 2.288m 

L8  = [ 2.288m -0.112m.]  = 2.176m

L9  = [ 2.176m -0.112m.]  = 2.064m

L10  = [ 2.064m -0.112m.] = 1.952m

L11  = [ 1.952m -0.112m.] = 1.84m

L12  = [ 1.84m -0.112m.]   = 1.728m

L13  = [ 1.728m -0.112m.] = 1.616m

L14  = [ 1.616m -0.112m.] = 1.504m

L15  = [ 1.504m -0.112m.] = 1.392m

L16  = [ 1.392m -0.112m.] = 1.28m

L17  = [ 1.28m -0.112m.]   = 1.168m

L18  = [ 1.168m -0.112m.] = 1.056m

L19  = [ 1.056m -0.112m.] = 0.944m

L20 = [ 0.944m -0.112m.]  = 0.832m

L21  = [ 0.832m -0.112m.] = 0.72m

L22  = [ 0.72m -0.112m.]   = 0.608m

L23  = [ 0.608m -0.112m.] = 0.496m

L24  = [ 0.496m -0.112m.] = 0.384m

L25  = [ 0.384m -0.112m.] = 0.272m

L26  = [ 0.272m -0.112m.]  = 0.16m

L27 = [ 0.16m -0.112m.]    = 0.048m

                                             ------------

The total length of bars       = 40.608m.

The total weight of the bars     = 40.608m ×0.89kg/m

                                               = 36.14kg

Add 5% wastage                     =   1.81kg.

                                                --------------

The total wt. of the main bars = 37.95 kg.

NOTE: 

1. I have made this elaborate calculation so that you will understand the concept clearly. Practically, we should make these calculations on an excel sheet.

2. By following a similar procedure, you can find out the weight of distribution bars. Here, you have to calculate θ2 instead of θ1.

Thank you for going through this article❤. Have a good day😄.

If you have any quarries, you can ask me in the comment box👇.

           

     

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