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By
Sourav Dutta
Manager-Civil
We generally designate stirrups by its external dimensions. For example, a 250mm x 300mm stirrup has the 250mm and 300mm dimensions in perpendicular directions as shown in Fig 1.
Now a stirrup has 2 bends, types of which are shown in Fig 1 and Fig 4. Guideline of minimum bend length is as per Cl.3.8 of BIS2502 (shown in Fig 3).
Fig 1: 250mm x 300mm stirrup
To calculate cutting length of stirrup, we need to do so along centre-line. The calculation is done as per Table VIII of BIS2502 (Fig 4) for most common sizes of stirrups as shown.
As per IS:2502 – 1963 (reaffirmed 2004) Indian standard code for Bending and fixings of bars for concrete reinforcements
Section 3.8 Binders, Stirrups, Links and the Like – In the case of binders, stirrups, links etc, the straight portion beyond end of the curve at the end shall not be less than eight times the nominal size of the bar.
Fig 3: Guideline of minimum bend length as per BIS2502
Fig 4: Cutting length calculation of stirrup as per BIS2502 [“d”=nominal dia of stirrup]
It is to be noticed in Table VIII that the dimensions “A”, “E” or “C” as shown are all internal dimensions. But since the size of any stirrup is denoted by its external dimensions, therefore to calculate cutting length, the internal dimensions should be converted to external dimensions before using the formula of cutting length.
For example, for the stirrup shown in Ref. No. A whose internal dimensions are “A” and “E”, let the corresponding external dimensions be “X” and “Y”. Thus, A=X-2d, E=Y-2d.
Therefore, cutting length = 2 (A+E) + 24d = 2 (X-2d+Y-2d) + 24d = 2 (X+Y) + 16d.
We at engineeringcivil.com are thankful to Sir Sourav Dutta for submitting this very useful information to us.
If you have a query, you can ask a question here.
sir please tell about cutting length of bars in roof slab.
In this expression, bending deduction is taken or we have to deduct it.
How 24d comes out in this formula
For bending of both hooks of 135degree.